JEE Main 10 April 2023 Shift 1 Memory-Based Questions

PHYSICS

1. A conducting rod of length 1 m is moved across a magnetic field of 0.15 T, with a constant speed of 4 m/s. Find the force (in N) on the rod.
2. A monoatomic gas, initially at pressure P and volume V, is compressed to 1/8th of its volume adiabatically. What will be the final pressure of the gas?
3. A particle of mass m moving with a velocity v collides with a particle of mass 2m at rest and sticks to it. Find out the velocity of the combined mass.
4. A particle, when projected at 15° with the horizontal, has a range of 50 m. Find the range when it is projected at 45° with the horizontal.
5. A point-sized object is placed 4 cm from the double convex lens of a focal length of 8 cm. Find the change in the position of the image when the object is moved 2 cm towards the lens.📷
6. A radioactive nuclei x decays simultaneously into two nuclei: y and z. If the time taken for x decaying to given y is t1/2 = 12 minutes and that of x decaying into z is t'1/2

= 3 minutes, then find the time in which nuclei x decays 50%.

1. A solenoid having 60 turns and a length of 15 cm produces a magnetic field of 2.4 x 10-3 T, find the current in the solenoid.
2. An object is placed in front of a plane mirror 12 cm away from it. If the object is kept fixed while the plane mirror is shifted towards the object by a distance of 4 cm, then what will be the length of the shift in the position of the image in cm?
3. Decay constant for a radioactive nuclide is given to be 2 x 103. If the molar mass of the sample is 60 gm, then find the activity of the 0.3µg sample in disintegration/seconds.
4. Find the equivalent capacitance across points A and B in the given electrical circuit. (diagram given)
5. For a given circuit (triangular in shape with 4 Ω resistances on its 2 sides and all 3 medians; the remaining side has a 16 Ω resistance), find the equivalent resistance across its ends. (diagram given)
6. For a particle performing linear SHM, its position (x) as a function of time (t) is given by x = A sin(ωt + δ). Given that, at t = 0, the particle is at at + A/2 and is moving towards x = +A. Find δ.
7. For an object radiating heat at 300K, the wavelength corresponding to the maximum intensity is λ. If the temperature of the body is further increased by 300K, what will be the new wavelength corresponding to the maximum intensity?
   
8. If 10 resistors, each of 10 resistance, when connected together in all possible combinations give minimum equivalent resistance R1 and maximum equivalent resistance R2, then find R2/R1.
   
9. If an object weighs 200 N at the surface of the earth, what will be its weight at a depth of 2 where R is the radius of the earth?
   
10. If the earth shrinks to 1/64 times its initial volume, the time period of the rotation of the earth is found to be 24/x hrs. Find the value of x.
    
11. In an AM wave, the amplitude of modulating wave = 3 units and the amplitude of the carrier wave = 15 units. Find the ratio of maximum to minimum intensity i.e., Imax/Imin.
    
12. State, whether the following statement(s) is/, are correct/incorrect.
    

Statement I: An LCR circuit connected to an AC source has the maximum average power at resonance.
Statement II: A resistor-only circuit with zero phase difference has the maximum average power.📷

1. The angular momentum of an e- in the first Bohr's orbit is L. What will be the change in angular momentum if it jumps into the second orbit?
2. The equation of progressive wave is y = 5 sin(6t + 0.03x). Find the speed of the wave.
3. The equivalent resistance for a given circuit is x/7. Find x. (diagram given).
4. Three concentric shells A, B and C have surface charge density σ, - σ and a respectively. The radius of A and B is 2 cm and 3 cm respectively. The electric potential at surface A is VA and at C is Vc. If VA = VC, then find the radius of C in cm.
5. Two blocks of mass 2 kg and 1.14 kg are hung by steel and brass wires respectively as shown in the figure (diagram given). Find the change in the length of the steel wire if Ysteel = 2 x 1011 N/m2 and Ybrass = 1 x 1010 N/m2.
6. What is the maximum percentage error in the measurement of quantity l, if it is given by l = (a2b3)/(c√d)? Consider the percentage error in the calculation of a, b, c and d as 1%, 2%, 3% and 4% respectively.

JEE Main 10 April 2023 Shift 1 Memory-Based Questions

CHEMISTRY

1. 2CO (g) + O2 (g) → 2CO2 (g) ; ΔH = -x kJ/mol

C (graphite) + O2 (g) → CO2 (g) ; ΔH = -y kJ/mol Find ΔH for C (graphite) + 1/2O2 (g) → CO (g).

1. A radioactive element C decays into A at t1/2 = 30 minutes and into B at t1/2 = 60 minutes. Calculate the overall half-life of C in minutes.
2. At STP conditions, find out the number of molecules and moles in 2.8375 litres of O2.
3. Enthalpy of adsorption and enthalpy of formation of micelle are respectively...
   1. Positive, Positive
   2. Positive, Negative
   3. Negative, Positive📷
   4. Negative, Negative
4. FeO42- → (at E10 = +2.20 V) → Fe3+ → (at E20 = +0.77 V) → Fe2+ → (at E30 = -0.44 V)

→ Fe
Calculate E40 if it corresponds to the transition between FeO42- and Fe.

1. Find out I T1 - T2 l for a solution of 0.1 molal weak acid HA, if Kf of water = 1.86 K kg mol-1

T1 = Freezing point of solution assuming no dissociation of acid
T2 = Freezing point of solution assuming the degree of dissociation (α) = 0.3.

1. How many compounds can be easily prepared by Gabriel phthalimide synthesis, which on reaction with Hinsberg reagent produces a compound which is soluble in KOH?
2. Match the following.

Column 1:
A. Cotton mills, B. Paper mills, C. Fertilizer, D. Thermal power plant Column 2:

1. Biodegradable waste, ii. Gypsum, iii. Non-biodegradable waste, iv. Fly ash
2. Match the following.

Column 1:
A. Dacron, B. Urea formaldehyde resin, C. Nylon-2, Nylon-6, D. Nylon-6,6 Column 2:

1. Thermosetting, ii. Biodegradable, iii. Polyester, iv. Used for making bristles of brushes
2. Mixture of A and B is added to a column containing an adsorbent for separation using a solvent. A is eluted first and B is eluted last. Then B has:
   
   1. High R, less adsorption
   2. Low R, strongly adsorbed
   3. High R, strong adsorption
   4. Low R, weakly adsorbed
3. State whether the following statement(s) is/are correct/incorrect. Statement I: Potassium dichromate is used in volumetric analysis. Statement II: K2Cr2O, is more soluble in water than Na2Cr2O.
   
4. State whether the following statement(s) is/are correct/incorrect. Statement I: Reduction potential M3+/M2+ is more for Fe than Mn. Statement II: V2+ has a magnetic moment between 4.4 - 5.2 BM.
   
5. The degree of dissociation of monobasic acid is 0.3. By what per cent will the observed depression in the freezing point be greater than the calculated depression in the freezing point?
   
6. The pair of compounds from the following pairs having both the compounds with net zero dipole moment is:
   
   1. CH2Cl2: CHCI,
   2. 1,4-dichlorobenzene; 1,3,5-trichlorobenzene📷
   3. Benzene; P-Anisidine
   4. Cis-dichloroethene; Trans-dichloroethene
7. The pressure value of a given gas is 930.2 mm Hg. Its volume is then reduced to 40% of its initial value at a constant temperature. Find out its final pressure (in mm Hg).
   
8. What will be the sum of number of lone pairs in the central atom in IF5 and IF7.
   
9. Which of the following compounds do not exist?
   
   1. BeCl2, ii. NaO2, iii. PbEt4, iv. (NH4)2B
10. Which of the following compounds is diamagnetic with low spin?
    
    1. [Co(NH3)6]3+, ii. [CoCl6]3-, iii. [CoF6]3-, iv. [Fe(H2O)6]3+
11. Which of the following does not stabilize the secondary and tertiary proteins?
    
    1. H-H linkage
    2. S-S linkage
    3. Vander Waal’s forces
    4. Hydrogen bonding
12. Which of the following molecules are bent in shape?
    
    1. SO2, ii. O3, iii. I3-, iv. N3-
13. Which stabilizer is used for concentrating sulphuric ore?
    
14. Why is the prolongated heating of ferrous ammonium sulphate avoided?
    

JEE Main 10 April 2023 Shift 1 Memory-Based Questions

MATHEMATICS

1. 96 cos(π/33) cos(2π/33) cos(4π/33) ... cos(16π/33) = ?
2. An AP series is given as 3, 8, 13, ..., 373. What will be the sum of terms which are not divisible by 3?
3. Find the number of points of non-differentiability for:

𝑥𝑥; −2 < 𝑥 ≤ 0
𝑓(𝑥) = { 𝑥 − 3 + 𝑥 + 1 − 2𝑥 − 2; 0 < 𝑥 ≤ 2
𝑥(𝑥2 − 𝑥); 2 < 𝑥 ≤ 3

1. Find the shortest distance between the lines:

25(x + 1)/7 = (y + 1)/-6 = (z + 1)/1 and (x - 3)/1 = (y - 1)/-2 = (z - 7)/1

1. Find the coefficient of x7 in (1 - 2x + x3)10.📷
2. Find the number of integral values of x which satisfy the inequality x2 - 10x + 19 < 6.
3. From a square of side 30cm, the squares of side x cm are cut off to make a cuboid of maximum volume. What will be the surface area of the cuboid with an open top?
4. Identify the negation of the statement (p V q) A ~ r.
5. If a2 + (ar)2 + (ar2)2 = 33033, (a,r E N), then find the value of a + ar + ar2.
6. If the coefficient of x7 in the expansion of (ax - 1/bx2)13 is equal to the coefficient of x- 5 in the expansion of (ax + 1/bx2)13, then find a4b4.
7. If the number of ways in which a mixed double badminton can be played such that no couples played into a same game is 840. Then find the number of players.
8. If the order of matrix A is 3 x 3 and A = 2, then 3 adj ( 3A A2) = ?
9. If the slope of the tangent to a curve at a variable point is (x2 + y2)/2xy and y(2) = 0, then find y(8).
10. Let f be a differentiable function such that x2 f(x) - x = 4 ∫0x t f(t) dt. If f(1) = 2/3 then find 18 f(3).
11. The mean of the following data is 26. Find the variance of the data.
12. Two dice are rolled and the sum of the numbers of two dice is N then the probability that 2N < N! is m/n, where m and n are coprime, then find 11m – 3n.
13. Using the numbers 1, 2, 3, 4, 6, 6 and 7, find the total numbers of 7-digit numbers which do not contain a string of 154 or 2367. Repetition is not allowed.

https://docs.google.com/document/d/1q3fojxt_oG7HJNBigiFXM4ZXnFv3LBIi/edit?usp=share_link&ouid=109474854956598892099&rtpof=true&sd=true
Organic Compounds Containing Oxygen
General methods of preparation, properties, reactions and uses. ALCOHOLS, PHENOLS AND ETHERS Alcohols : Identification of primary, secondary and tertiary alcohols; mechanism of dehydration. Phenols : Acidic nature, electrophilic substitution reactions : halogenation, nitration and sulphonation, Reimer - Tiemann reaction. Ethers : Structure. A ldehyde and Ketones : Nature of carbonyl group; Nucleophilic addition to >C=O group, relative reactivities of aldehydes and ketones; Important reactions such as - Nucleophilic addition reactions (addition of HCN, NH3 and its derivatives), Grignard reagent; oxidation; reduction (Wolff Kishner and Clemmensen); acidity of α - hydrogen, aldol condensation, Cannizzaro reaction, Haloform reaction; Chemical tests
to distinguish between aldehydes and Ketones. CARBOXYLIC ACIDS Acidic
strength and factors affecting it.

ALCOHOLS

Molecules containing –OH group are termed as alcohols. Classification of alcohols they are classified as primary, secondary or tertiary alcohol according to the carbon that is bonded with –OH.
Again when any molecule contain 1, 2 or 3 –OH groups then it is called mono, di or tri hydric alcohols respectively. (as in case of alkyl halides)
C H A P T E R
THIS CHAPTER INCLUDES

• Alcohols
• Phenols

OH OH OH
OH OH

• Ethers

CH3 – CH2OH
Ethyl alcohol (monohydric)
CH2 – CH2
Ethylene glycol (dihydric)
CH2 – CH – CH2
Glycerol (trihydric)

• Carbonyl

compounds

• Preparation

GENERAL METHODS OF PREPARATION

1. From Alkenes :
   1. By direct hydrolysis :

OH
CH – CH = CH + H O H2SO4 CH – CH – CH

1. Oxymercuration demercuration :

CH – CH = CH + H O Hg (OAc)2

• Properties
• Carboxylic acid
• Derivatives of carboxylic acid
• Acid Halide

3 2 2
THF
OH
CH – CH – CH
NaBH OH
CH – CH – CH

• Acid Anhydride
• Ester

2 2 OH– 3 3

1. Hydroboration oxidation :

Hg(OAc)

• Acid Amide

6CH – CH = CH B2H6
2(CH – CH – CH –) B H2O2/OH
OH
6CH3 – CH2 – CH2 + 2H3BO3
Overall result of above reaction is anti Markwonikoff addition of water and with no rearrangement.

1. Oxo process followed by hydrogenation :

O
CH3 – CH = CH2 + CO + H2
[Co(CO)4]2
high temperature and high pressure
CH3 – CH2 – CH2 – C – H
H2/Pd
CH3 – CH2 – CH2 – CH2 – OH
Product has one more carbon.

1. From Alkyl Halides :

When alkyl halides are treated with aq. KOH or aq. NaOH or moist Ag2O, alcohols are formed.
−
R − X + OH
⎯⎯→R − OH + X–

1. Reduction of Carbonyl Compounds, Carboxylic Acids and their Derivatives :

O
R – C – H red. agent R – CH OH
O
R – C – R′ red. agent
O
R – C – OR′ red. agent
O
OH
📷📷
R – CH2OH + R′ – OH
R – C – X O
red. agent
O📷
R – CH2OH
R – C – O – C – R
red. agent
2R – CH2OH

Table : Reducing nature of different reagents

Group
Product
LiAlH4/H2O
NaBH4/C2H5OH
B2H6/THF
H2/Pt
O
– C – H
– CH2 – OH
Yes
Yes
Yes
Yes
> C = O
> CH – OH
Yes
Yes
Yes
Yes
– COOH
– CH2OH
Yes
No
Yes
Yes
O
– C – Cl
– CH2OH
Yes
Yes
No
Yes
O
(R – C – O)2O
R – CH2OH
Yes
No
Yes
Yes
O
– C – OR
​
– CH2OH + R – OH
​
Yes
​
No
​
Yes
​
Yes
> C = C <
> CH – CH <
No
No
Yes
Yes
(LiAIH4 reduces)
C = C only
when it is
conjugated with
Phenylic system
Meerwein-Ponndorf-Verley Reduction (MPV Reduction) : Its a name reaction of reduction of
O → alcohol📷
Ketones can be reduced to secondary alcohols with aluminium isopropoxide in 2-propanol solution.
OH
R
C = O + CH3 – CH – CH3
[(CH3)2CHO]3Al R
R′
CH – OH + (CH3)2C = O

1. Using Grignard Reagent :
   1. From aldehydes or ketones :

OMgX OH📷📷
H O/H+ OH
– C – + R – MgX – C – R 2
– C – R + Mg
X
In this reaction
Formaldehyde gives 1°-alcohol Other aldehyde gives 2°-alcohol Ketones give 3°-alcohol

1. From carboxylic acid and their derivatives :

O Z📷
R – C – R′ + Mg
X
CH3 – CH — CH + R – MgX📷
OMgX CH3 – CH – CH2
R
H O/H+
OH OH
CH3 – CH – CH2 + Mg
R X

1. Hydrolysis of Ether :

Ether undergo acid hydrolysis with dilute H2SO4 under pressure to give corresponding alcohols.
R − O − R′ + H2O ⎯⎯dil.
2S⎯O⎯4 →R − OH + R′ − OH

Physical Properties

1. Physical state : At ordinary temperature, lower members are colourless liquids with burning taste and a pleasant smell.
2. Boiling Point :

The boiling point of alcohols rise regularly with the rise in the molecular mass. Amongst isomeric alcohols, the boiling point decrease in the order.
1° > 2° > 3°

1. Solubility : The extent of solubility of any alcohol in water depends upon the capability of its molecules to form hydrogen bonds with water molecule.
2. Alcohols are lighter than water however, the density increases with the increase in molecular mass.

Chemical Properties

1. Reactions involving cleavage of O – H Bond

Alcohols are acidic in nature but they are less acidic than water hence they do not give H+ in aqueous solution. They do not change the colour of litmus paper.
Their acidic strength increases by increasing–I strength of the groups attached and decreases by increasing
+I strength of the groups.

1. Alcohols do not react with aqueous alkali, as it does not give H+ in aqueous solution.
2. Action of active metal : When alcohols are treated with active metal they form alkoxides with the liberation of H2 gas.

− +
2ROH + 2Na ⎯⎯→2R − ONa + H2 ↑

1. Esterification : When carboxylic acid is treated with alcohols in the presence of acid as catalyst, esters are formed.

O
R – C – OH + H – O – R′ H
O
R – C – OR′ + HOH

1. Reaction with Grignard Reagent : When Grignard reagents are treated with alcohol (or any proton donor) they form alkanes.📷

R – MgX + H – OR′ R – H + R′ – OMgX
Other proton donors can be carboxylic acids, phenols, alkynes, H2O, Amines, NH3 etc.

1. Reactions Involving Cleavage of C– O Bond
   1. Reaction with HX : Most alcohols undergo SN1.

(a) R − OH + HCl(g) ⎯⎯⎯⎯⎯⎯2 →R − Cl + H2O
(b)
R − OH + HBr
H+ / H2SO4
conc.
R − Br + H2O
(c)
R − OH + HI
H+ / H2SO4
conc.
R − I + H2O
Reactivity order of HX is HI > HBr > HCl

1. Dehydration : Alkyl chlorides can also be prepared by following methods : R – OH + PCl5 R – Cl + POCl3 + HCl

3R – OH + PCl3 3R – Cl + H3PO3
R – OH + SOCl2 R – Cl + SO2 + HCl (Darzen's process) Darzen’s process is the best method as the other products are gases.

1. Reduction :

Alcohols are reduced to alkanes when they are treated with Zn-dust or red P + HI.
R − OH ⎯⎯Zn d⎯u⎯st →R − H + ZnO

1. Oxidation :

O O
CH3OH
[O]
H – C – H
[O]
[O]
H – C – OH
CO2
3°-alcohols can’t be oxidised.

1. Strong oxidising agent like KMnO4 or K2Cr2O7 cause maximum oxidation as above.
2. If 1°-alcohol has to be converted into aldehyde PCC + CH2Cl2 or CrO3 should be used among which PCC + CH2Cl2 is the best.
3. 2°-alcohol can converted to ketone best by PCC + CH2Cl2 or CrO3 or H2CrO4 in aq. acetone (Jones reagent).
4. MnO2 selectively oxidises the –OH group of allylic and benzylic 1° and 2° alcohols to aldehydes and ketones respectively.
5. Action of Heated Copper :

(i)
(ii)
CH3 – CH2 – OH
OH
CH3 – CH – CH3
Cu 573K
Cu 573K
O
CH3 – C – H + H2 (Dehydrogenation)
O
CH3 – C – CH3 + H2 (Dehydrogenation)
(iii) Tertiary alcohols undergo dehydration to give alkene under similar condition.
CH3
CH – C – OH Cu
CH2
CH – C + H O
3
CH3
573 K
3 2
CH3

Distinction Between 1°, 2° and 3° Alcohols📷

1. Lucas Test :

Any alcohol is treated with Lucas reagent (HCl + an hyd. ZnCl2) at room temperature if

1. Solution becomes cloudy immediately, alcohol is 3°.
2. Solution becomes cloudy after 5-min, alcohol is 2°.
3. In solution cloud does not form at room temperature, alcohol is 1°.
4. Victor Meyer’s Method :

R – CH2OH
(1°-alcohol)
P + I2 R – CH I
AgNO2 R – CH – NO + HNO
–H2O
R – C – NO2
N
OH
(Nitrolic acid)
NaOH
blood red colour
R2CH – OH
(2°-alcohol)
P + I2 R – CH – I
AgNO2 R CH – NO + HNO
–H2O
R2C – NO2 NO
R – C – OH P + I2 R C – I
(3°-alcohol)
Pseudonitrol
NaOH
blue colour
AgNO2 R C – NO HNO2 No reaction NaOH Colourless

PHENOLS

OH📷
and their derivatives are called phenols. In phenol R– of alcohol is replaced by aryl ring.

Comparison of bond Angles in Phenols, Alcohols and Ethers :📷

H H–C
H
..
O📷
H
108.5°
109° H
H H–C
H
..
H
C
111.7° H H
Bond angle increases with the increase in hindrance.

Method of Preparation

1. From Aryl Sulphonic Acids :
When aryl sulphonic acids are fused with NaOH at 570 – 620 K followed by hydrolysis phenols are formed.
SO3H📷📷
SO3Na
Δ 📷
ONa OH
H+/H O📷📷
Cl📷
(aq.) 623 K
320 atm
+
ONa OH📷📷

1. From Benzene Diazonium Salts :

NH2📷📷
+ N2 Cl
OH
+ N2 + H–Cl📷📷📷
📷📷

1. Cumene Process :

H
H PO
CH3
CH3
C H📷
light
CH3
CH3
C–O–O–H OH O📷📷
+CH – C = CH 3 4
+ O2
+ CH —C—CH
3 2
Propene
Benzene
3 3
Cumene Cumene hydroperoxide

1. Grignard's Synthesis :

1
H2O / H+ OH
C6H5MgBr + 2 O2
⎯⎯→ C6H5OMgBr ⎯⎯⎯⎯→ C6H5OH + Mg
Br

1. From Salicylic Acid :

OH OH📷📷
COOH
+ NaOH

Chemical Properties

1. Acidic Nature :

COONa
+ NaOH
OH
CaO, Δ 📷
–Na2CO3

1. Phenol behave as a weak acid forming phenoxide ion with strong alkalies.

C6H5OH
Phenol
+ NaOH →
C6H5ONa
Sodiumphenoxide
+ H2O

1. It also reacts with sodium metal to form sodium phenoxide and hydrogen is evolved

C H OH
+ Na →
+ + 1 H
6 5 C6H5 ONa 2 2📷
Phenol

1. Effect of substituents on the acidity of phenols : It should be noted that the presence of electron withdrawing groups like –NO2, – CN, –CHO,–X, –COOH, etc. increases the acidic strength (because of the greater polarity of O–H bond the greater stability of the phenoxide ion by the dispersal of –ve charge, by –R effect). On the other hand, electron-releasing groups like –CH3, –NH2, –OH, etc., tend to destabilize the phenoxide ion by intensifying its –ve charge by +R effect and hence decreases the acidic strength.

o - chlorophenol > m - chlorophenol > p - chlorophenol
Ka = 7.7 × 10–9 1.6 × 10–9 6.3 × 10–10
In case of haloarenes –I effect of halogens dominates over it's +M effect. (except for fluorine) p-nitrophenol > o-nitrophenol > m - nitrophenol > phenol

1. Steric Effect : 3, 5 -dimethyl-4-nitrophenol is weaker acid than the isomeric 2, 6-dimethyl - 4 - nitrophenol.
2. Alkylation or Etherification : When sodium phenoxide is treated with alkyl halides (but not with aryl halides as they are inert) form phenolic ethers.

C6H5OH ⎯⎯NaO⎯H →C6H5ONa ⎯⎯CH⎯3I →
C6H5OCH3

• NaI

Phenol
−H2O
Sodium phenoxide
Methyl phenylether (Anisole)
C6H5OH ⎯⎯NaO⎯H →C6H5ONa ⎯⎯C2H⎯5B⎯r →C6H5OC2H5 + NaI
Phenol
Sodium phenoxide
Ethyl phenyl ether (Phenetole)

1. Claisen rearrangement :

C6H5ONa + BrCH2 – CH = CH2 → C6H5 − O − CH2CH = CH2 + NaBr
Allyl phenyl ether
When aryl allyl ether is heated to 475 K, the allyl group of the ether migrates from ethereal oxygen to the ring carbon at ortho position.
O–C* H2–CH = CH2📷
⎯⎯475⎯K →
OH
CH –CH = C* H📷
o-Allyl phenol

1. Acylation and benzoylation :

O O
C6H5

• OH

+ CH3 – C – Cl ⎯⎯Pyr⎯idin⎯e →
Acetyl chloride
C6H5 – O – C – CH3
Phenol
Phenyl acetate

1. Fries Rearrangement : When heated with anhydrous aluminium chloride, phenyl esters undergo Fries rearrangement forming a mixture of o- and p-hydroxy ketones.

O
O – C – CH3📷
⎯⎯He⎯at →
AlCl3
OH O OH
C – CH3 +📷📷
Phenyl acetate📷
o-hydroxyacetophenone
O = C – CH3
p-Hydroxyacetophenone
The para isomer is formed predominantly at low temperature while at higher temperatures o - isomer is predominant.

1. Reactions due to C–O Bond :
   1. Reaction with PCl5 :

C6H5OH + PCl5 → C6H5Cl + POCl3 + HCl
3C6H5OH + PCl3 → P(OC6H5 )3 + 3HCl
Triphenyl phosphate
The yields of C6H5Cl is very poor due to the formation of triaryl phosphate.

1. Reaction with Ammonia :

C6H5OH+ NH3 ⎯⎯ZnC⎯l2 →C6H5NH2 + H2O
Phenol
573 K
Aniline

1. Reaction with Zinc Dust :

C6H5OH+ Zn ⎯⎯Δ → C6H6 + ZnO
Phenol Benzene

1. Reaction with Neutral FeCl3: ( Test for phenol)

3C6H5OH + FeCl3
⎯⎯→
(C6H5O)3 Fe + 3HCl
Ferric phenoxide (Violet)

1. Electrophilic Substitution Reaction on the Benzene Ring :

From the contributing structure of phenol, it is clear that ortho- and para-position on it become rich in
electron density. Thus the electrophilic attack at these positions is facilitated. Again benzene ring is the very powerful ring activator towards electrophilic aromatic substitution.📷
present on the

1. Bromination :

OH📷
+ 3Br2
⎯⎯H2⎯O →
OH
Br Br📷
+ 3HBr
Phenol
Br
2,4, 6-Tribromophenol (yellow ppt.)
OH
📷
SO3H
p-Phenolsulphonic acid
+ 3Br2 (aq.) →
OH
Br Br📷
Br
2,4, 6-Tribromophenol (yellow ppt.)
+ 3HBr + H2SO4

1. Nitration :

OH📷
(a)
Phenol
+ dil HNO3
⎯⎯293⎯K →
OH
OH📷📷
NO2 +
NO
o-Nitrophenol (40% yield)📷
2
p-Nitrophenol (13% yield)
(b) With concentrated nitric acid and sulphuric acid, it forms 2, 4, 6-trinitrophenol (Picric acid).
OH📷
+ 3HNO3 ⎯⎯H2S⎯O4⎯co⎯n⎯c. →
(conc.)
O2N
OH
NO2📷
NO2
2,4, 6-trinitrophenol (Picric acid)

1. Sulphonation : When heated with conc. sulphuric acid, phenol forms hydroxy benzene sulphonic acid.

OH📷📷📷
H
H2SO4, 298 K 3📷
–H2O
OH📷📷
–H2O
SO3H
p-Hydroxy benzene sulphonic acid

1. Friedel-Crafts Alkylation and Acylation : Phenol undergo both these reaction to form mainly p-isomer.

OH📷
+ CH3Cl

• A⎯nh⎯yd. A⎯lC⎯l 3 →

OH OH
+📷📷
CH3
Phenol
CH3
(Major product) p–Cresol
o-Cresol
OH📷
+ CH3COCl

• A⎯nh⎯yd. A⎯lC⎯l 3 →

OH
COCH3 +📷
OH
COCH3📷
o- p-
Hydroxy acetophenone

1. Kolbe’s reaction📷📷

📷📷

1. Riemer Tiemann Reaction :
2. On heating with chloroform and alkali phenols are converted to phenolic aldehydes📷

OH
+ CHCl3
+ 3NaOH
⎯⎯333⎯−34⎯3 →
(aq.) H+
OH
CHO
+ 3NaCl + 2H2O
In this reaction dichloro carbene is formed as intermediate which attack on benzene ring as electrophile.

1. If instead of chloroform, carbon tetrachloride is used, salicylic acid is formed. Some para isomers is also formed.📷

OH📷
+ CCl4
⎯N⎯aO⎯H →
340 K
ONa
CCl3
⎯3⎯Na⎯OH(⎯a⎯q.)→
–3NaCl
ONa
COONa
⎯⎯Dil. ⎯H 2S⎯O 4 →
OH
COOH📷
In this reaction ⊕ CCl3 is formed as intermediate which attack on benzene ring as electrophile.📷

1. Coupling with Diazonium Salts :

C H N Cl + C H OH
⎯⎯0−5⎯°⎯C →
N = N OH
6 5 2
6 5 pH 9−10
p-Hydroxy azobenzene (An orange dye)

1. Test for phenol
   1. Neutral FeCl3 test → Aqueous solution of phenol gives a violet colouration with FeCl3.
   2. Br2 water test → Aqueous solution of phenol gives a yellow precipitate of 2, 4, 6-tribromophenol with bromine water.
   3. Phenol gives Liebermann's nitroso reaction.

Phenol
⎯⎯NaN⎯O⎯2
→ Red colour ⎯⎯NaO⎯H →blue colour
(in conc. H SO ) excess of water
excess
2 4

ETHERS📷

Ethers are those organic compounds which contain two alkyl groups attached to an oxygen atom, i.e., R–O–R. They are regarded as dialkyl derivatives of water or anhydrides of alcohols.
H – O – H ⎯⎯–2⎯H → R – O – R ←⎯⎯⎯ R – OH + HO – R
Water

• 2R

Ether
−H2O
Alcohol (2 moles)
Ethers may be of two types : (i) Symmetrical or simple ether are those in which both the alkyl groups are identical and (ii) unsymmetrical or mixed ethers are those in which the two alkyl groups are different.
CH3–O–CH3; C6H5–O–C6H5 CH3–O–C2H5; CH3–O–C6H5
Symmetrical (simple) ethers Unsymmetrical (mixed)ethers
Like water, ether has two unshared pair of electrons on oxygen atom, yet its angle is greater than normal tetrahedral (109°28´) and different from that in water (105°). This is because of the fact that in ethers the repulsion between lone pairs of electrons is overcome by the repulsion between the bulky alkyl groups.

Preparation of Ethers :

1. By dehydrating excess of alcohols : Simple ethers can be prepared by heating an excess of primary alcohols with conc. H2SO4 at 413K. Alcohol should be taken in excess so as to avoid its dehydration to alkenes.

C2H5 – OH + HO – C2H5 ⎯⎯Con⎯c. ⎯H2S⎯O⎯4 → C2H5 – O – C2H5 + H2O
Ethanol (2 molecules)
413 K
Diethyl ether
Dehydration may also be done by passing alcohol vapours over heated catalyst like alumina under high pressure and temperature of 200 – 250°C.

1. By heating alkyl halide with dry silver oxide (only for simple ethers) :

C2H5I + Ag2O + IC2H5 → C2H5OC2H5 + 2AgI
Dry
Remember that reaction of alkyl halides with moist silver oxide (Ag2O + 2H2O = 2AgOH) gives alcohols C2H5I + Ag2O (moist) → C2H5OH + AgI

1. By heating alkyl halide with sod. or pot. alkoxides (Williamson synthesis): C2H5ONa + ICH3 → C2H5OCH3 + NaI

ONa + BrCH3 OCH3 + NaBr📷📷
Sod. phenoxide Methoxybenzene (Anisole)
However
CH3
+
CH3
CH3–Cl + NaO–C–CH3
CH3
CH3–O–C–CH3 + NaCl
CH3
If alkyl halide is other than methyl halide and it is treated with tertiary alkoxide ion, Hoffmann elimination takes place instead of Williamson's ether synthesis.
📷
CH3–CH2–CH=CH2 + HCl
(Major)

1. Methyl ethers can be prepared by treating primary or secondary alcohol or phenol with diazomethane in presence of BF3.

C2H5OH + CH2N2 ⎯⎯B⎯F3 →

Chemical Properties :

C2H5OCH3
Ethylmethyl ether

• N2

1. Properties due to Alkyl Groups :
   1. Halogenation : When ethers are treated with chlorine or bromine in the dark, substitution occurs at the α-carbon atom. The extent of substitution depends upon the reaction conditions.

β α α´ β´
dark
CH3 – CH2 – O – CH2 – CH3 + Cl2
⎯⎯⎯→ CH3 .CHCl – O – CH2.CH3
α– Chlorodiethyl ether
↓ Cl2
CH2Cl.CHCl – O – CH2 .CH3 + CH3CHCl – O – CHCl.CH3
α, β–Dichlorodiethyl ether α, α´–Dichlorodiethyl ether
CH3CH2 – O – CH2 .CH3 + 10Cl2
⎯⎯lig⎯ht → CCl3 .CCl2 – O – CCl2 .CCl3
Perchlorodiethyl ether

1. Combustion :

C2H5.O.C2H5 + 6O2 4CO2 + 5H2O📷

1. Properties due to Ethereal Oxygen :
   1. Chemical inertness : Since ethers do not have an active group, in their molecules, these do not react with active metals like Na, strong bases like NaOH, reducing or oxidising agents.
   2. Formation of peroxide (Autoxidation) : On standing in contact with air and light ethers are converted

into unstable peroxides (R2O O) which are highly explosive even in low concentrations.📷

1. Basic nature : Owing to the presence of unshared electron pairs on oxygen, ether behave as Lewis bases. Hence they dissolve in strong acids (e.g. conc. HCl, conc. H2SO4) at low temperature to form oxonium salts.

(C2H5 )2 O + H2SO4 →
Diethyl ether
[(C2H5 )2 OH]+HSO–
Diethyloxonium hydrogen sulphate
On account of this property, ether is removed from ethyl bromide by shaking with conc. H2SO4.
Being Lewis bases, ethers also form coordination complexes with Lewis acids like BF3, AlCl3, RMgX, etc.
R2O + BF3 R2O BF3
2R2O + RMgX📷
R2O R
Mg📷📷
R2O X
It is for this reason that ethers are used as solvent for Grignard reagents.

1. Properties due to carbon-oxygen bond :📷
2. Hydrolysis :

C2H5 – O – C2H5 + H2O ⎯⎯H2S⎯O⎯4 → 2C2H5OH
Ethyl alcohol
The hydrolysis may also be effected by boiling the ether with water or by treating it with steam.

1. Action of conc. sulphuric acid :

C2H5 – O – C2H5 + H2SO4 (conc.) → C2H5OH +
Ethyl alcohol
C2H5HSO4
Ethyl hydrogen sulphate
C2H5OH + H2SO4 (conc.) → C2H5HSO4 + H2O

1. Action of hydroiodic or hydrobromic acid :

In cold, ether react with HI or HBr to give the corresponding alkyl halide and alcohol. In case of mixed ethers, the halogen atom attaches itself to the smaller alkyl group.
C2H5 – O – C2H5 + HI →
Ethyl ether
C2H5I
Ethyl iodide

• C2H5OH

Ethyl alcohol
The order of reactivity of halogen acids is : HI > HBr > HCl
If one of the the group around oxygen is aryl group then I– will always attack on the group other than aryl group.
O–CH3 + HI OH + CH I📷📷
CH3 CH
3📷
O–C–CH3 + HI OH + I–C–CH📷
CH3

1. Properties Due to Benzene Nucleus :

3
CH3
Alkoxy group, being o-, p- directing, anisole undergoes substitution in o- and p- positions. However,
–OR group is less activating than the phenolic group.

1. Nitration :

OCH3 OCH3📷📷📷📷
conc. HNO3 conc. H2SO4
NO2
+
OCH3
Methylphenyl ether (Anisole)📷📷📷

1. Bromination :

Methyl 2-nitrophenyl ether or o-Nitroanisole📷📷
NO2
Methyl 4-nitrophenyl ether or p-Nitroanisole
OCH3
Br
Br2/Fe
OCH3
Br
Br
Anisole 2, 4, 6, Tribromoanisole📷
OCH3📷
+ Br2
CS2
OCH3
Br📷
+
OCH3
Br
Anisole 2-Bromoanisole 4-Bromoanisole

1. Sulphonation :

OCH3 OCH3 OCH3
H2SO4 SO3📷📷
SO3H
SO3H
Anisole p-Methoxybenzene sulphonic acid
o-Methoxybenzene sulphonic acid

CARBONYL COMPOUNDS (ALDEHYDES AND KETONES)

Aldehydes and ketones are the compounds containing carbonyl group (>C=O).
O
– C –
Carbonyl group
O
R – C – H
An aldehyde
O
R – C – R
Ketone
Structure of the carbonyl group: Like the carbon-carbon double bond of alkenes, the carbon-oxygen double bond of the carbonyl group is composed of one σ and one π bond.
In the carbonyl group, carbon atom is in state of sp2 hybridisation. The C–O σ bond is produced by overlap of an sp2 orbital of carbon with a p-orbital of oxygen. On the other hand, the C–O π bond is formed by the sideways overlap of p orbitals of carbon and that of p orbital of oxygen. The remaining two sp2 orbitals of carbon form σ bonds with the s orbital of hydrogen or sp3 orbital of carbon of the alkyl group.
p p
X X📷📷📷📷
O O
Y Y
(a)
(b)
The polar nature of the carbonyl group causes intermolecular attraction (dipole-dipole attraction) in aldehydes and ketones and hence accounts their higher boiling points than that of hydrocarbons and ethers of comparable mol. wt. However, the high values of dipole moments (2.3 - 2.8 D) of aldehydes and ketones can’t be accounted for, only by inductive effect; this can be accounted for if carbonyl group is a resonance hybrid of the following two structures.📷
+ −

• C = O ↔ > C — O

I know the defn's of arrhenius, bronsted lowry, lewis acids and bases, but i still don't find i can answer these questions confidently even when using all those criteria and drawing lewis dot diagrams. I basically just asked myself 6 questions for each compound (or at least checking if one of the base definitions or acid definitions was applicable) 1. does this fit the description of an arrhenius acid (AA) 2. does this fit the description of an arrhenius base (AB) 3. does this fit the description of a bronsted lowry acid (BLA) 4. does this fit the description of a bronsted lowry base (BLB) 5. does this fit the description of a lewis acid (LA) 6. does this fit the description of a lewis base (LB)
HNO3 - could this be a BLA? well the protonated O could lose it's H, so yes - could this be a BLB? the -1 charge on the single bonded O with no H seems like it would accept an H+, so yes but it's not amphoteric, it's an acid... so the proposition of it being a BLB must be wrong. how? why could it not become H2NO3+ ? is it just because that's unlikely to occur? that hardly seems like a reason to conclude it can never become that. after all the defn of a BLB is that it can operate as a H+ acceptor
H2O is amphoteric i just know this is amphoteric from memory
HCO3- - could this be a BLA? the protonated O could lose it's H, so yes - could this be a BLB? the -1 charge on the single bonded H with no H seems like it would accept an H+, so yes it's amphoteric
CH3NH2 - could this be a BLA? well why couldn't it donate an H+ from N or C? apparently I'm meant to say no here because i know the answer says this molecule is a base. but why?
i don't understand what procedural method i can go through to answer these questions that results in a correct answer
https://lens.google.com/search?p=AfVzNa85uCNhn1U6V0fa-1An3SXylloMLTcDaaZiSaAGj-W6vtIBWd2buX1TWq9fvSikM1rH0-Nrcq_MaOcDxXBCg4nVnNtR0jB0eIv5e580Uv66Vu8bjlxqnk_Bu0Rde5kWPF_dkkPQaLqgm3c5shH6eqfUbOzf3EqbUSVUqE_LlJAfweX64WeLS2MCQdtPzxqHm-v3iorQ0mhm9CwiFtOlkEhWIyQYG7AGNFlX_TaM6BkTdb_goVZnCA%3D%3D#lns=W251bGwsbnVsbCxudWxsLG51bGwsbnVsbCxudWxsLG51bGwsIkVrY0tKRE5sWTJJd01tTTJMVGRpTkRrdE5HWXdZeTFpTldZMkxUTmxaak13TmpabVlUaGlZeElmUVROT1kyNWFRVXR2V2xGVldVeHVaVFo2U1ROWFNFOUdZa2s1VUZsQ1p3PT0iXQ==
Here is the list of compounds I was asked about

https://docs.google.com/document/d/1ts2KjX27bln5tXUumLsIuSnDQU-P0bfN/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true
preparation, properties and reactions. Alkanes - Conformations : Sawhorse and Newman projections (of ethane); Mechanism of halogenation of alkanes. Alkenes - Geometrical isomerism; Mechanism of electrophilic addition: addition of hydrogen, halogens, water, hydrogen halides (Markownikoff’ s and peroxide effect); O zonolysis, oxidation, a nd polymerization. Alkynes - Acidic character; Addition of hydrogen, ha logens, water a nd hydrogen halides; Polymerization. Aromatic hydrocarbons - Nomenclature, benzene - structure and aromaticity; Mechanism of electrophilic substitution: halogenation, nitration, Friedel – Craft’s alkylation and acylation, directive influence of functional group in
mono-substituted benzene.

HYDROCARBONS AND THEIR CLASSIFICATIONS

Compounds of only Carbon and Hydrogen are called hydrocarbons. They are parent organic compounds and all other organic compounds have been derived by replacing one or more H atoms from hydrocarbon.
Hydrocarbons📷📷
CHAPTER INCLUDES

• Alkane
• Alkene

Acyclic
(open chain hydrocarbon)
📷📷
Alicyclic📷📷
Cyclic
Aromatic
CH3

• Alkyne
• Aromatic

hydrocarbons

Saturated hydrocarbon/
Alkanes/Paraffins
Unsaturated
hydrocarbons (having= and ≡)
(cycloalkanes)
i.e., 📷 ,
etc. i.e.,
,
Toluene

ALKANES

Methods of Preparation :

1. Reactions where number of carbon atoms are increased
   1. Wurtz Reaction

2R − X + 2Na ⎯⎯dry⎯eth⎯er →R − R + NaX
Here other metals in the finely divided state may also be used such as Cu, Ag etc.

1. Methane cannot be prepared by this method.
2. Only symmetrical alkane can be prepared by this method in good yield.

TEACHING CARE Online Live Classes https://www.teachingcare.com/ +91-9811000616

1. Corey-House Synthesis:

R - X + Li ⎯⎯→RLi + LiX
RLi + CuI ⎯⎯→R2LiCu
(a)
(b)
R2LiCu + 2R − X ⎯⎯→ 2R − R + LiX + CuX R2LiCu + 2R′ − X ⎯⎯→ 2R − R′ + LiX + CuX
etc.
It can be used for preparing both symmetrical and unsymmetrical alkanes.

1. Kolbe’s Electrolytic Decarboxylation

2RCOONa (aq) ⎯⎯ele⎯ct⎯rol⎯yti⎯c →R - R + 2NaOH + 2CO2 + H2

Mechanism:

RCOONa RCOO– + Na+ 2H O 2H+ + 2OH–

Anodic Reaction

2RCOO– 2RCOO• + 2e–
2RCOO• R - R + 2CO

Cathodic Reaction📷

2H+ + 2e– H

1. Methane cannot be prepared by this method
2. Unsymmetrical hydrocarbon (alkane) cannot be prepared
3. Reactions where number of carbon atoms are retained
   1. Sabatier - Sanderen’s Reduction

H H
C = C📷📷
+ H2
Raney Ni
200 - 300°C
📷C – C
📷
H H
– C ≡ C – + 2H
Raney Ni
200 - 300°C

— C — C —

H H

1. Reduction of Alkyl Halides
2. R − X + Zn ⎯⎯H⎯Cl →R − H+ HX

Zn – Cu and C2H5OH or Na and alcohol can also be used

1. 4R–X + LiAlH4 ⎯⎯→ 4RH + LiX+AlX

This is a nucleophilic substitution reaction with the nucleophilic H– coming from LiAlH .

1. R – X + (n – C4 H9)3SnH R – H + (n –C4H9)3SnX.📷
2. Reduction of Alcohols, Aldehydes, Ketones and Carboxylic Acids with HI/Red P.
   
3. ROH + 2HI
   

R
⎯⎯Re d⎯⎯P →R − H + I2 + H2O
150°C
(b)
H
R1
C = O + 4HI
2H 2H
Red P 150°C
R–CH3 + H2O + 2I2

C = O

R2
+ 4HI
⎯⎯Re d⎯P →
150°C
R1 – CH2

• R2

+ H2O + 2I2
2H

1. R — C

2H O
+ 6HI ⎯⎯Red⎯P →R − CH3 + 2H2O + 3I2
150°C
OH H

1. Clemmensen’s Reduction

CH 3 − C − H + 4H ⎯ ⎯Zn−⎯Hg⎯/ H⎯Cl → CH 3 − CH 3 + H 2 O
Δ
O
R1 − C− R2 + 4H ⎯⎯Zn−⎯Hg⎯/ H⎯Cl →H2O + R1 − CH2 − R2

O
Clemmensen’s reduction should not be used when the carbonyl compound has a basic end in it.

1. Wolf-Kishner’s Reduction:

R1 – C– H + NH2

• NH2

→ R – C = N − NH2

• H2O

KOH+Glycol 453 k − 473 k
R – CH3 + N2
O
CH3 – C– C2H5 + NH2 – NH2

O
H
CH3 C H
Hydrazone
C=N–NH2 +H📷📷
O ⎯⎯KO⎯H⎯+ ⎯Gl⎯yc⎯o⎯l → CH3CH2C2H5 + N2
2 453 K –473 K
2 5 Hydrazone
This reaction should not be used when the carbonyl compound has an acidic end in it.

1. (a) Using Grignard’s Reagent

R - X + Mg
⎯⎯eth⎯er → RMgX
RMgX + H – OH R – H + MgX(OH)📷
+ H – NH2 R – H + MgX(NH2)📷
+ H – OR1 R – H + MgX(OR1)📷
+H–O–C–R′

O
R–H + MgX (O–C–R′)

O

(b) Using Alkyl lithium compound

R1 NH + R–Li (R1NH) Li + R – H.📷

1. Reaction where number of carbon atom are decreased Sodalime Decarboxylation

RCOONa + NaOH
CaO
630 k
R – H + Na2CO3
CH3📷
CaO
CH—CH —COONa + NaOH
CH3
CH—CH + Na CO📷
2
CH3
630 k
CH3
3 2 3

1. Some other methods of preparation
2. Preparation of Methane from carbides

(a) Al4C3 + 12H2O 4Al(OH)3 + 3CH4

1. Be2C + 4H2O 2Be(OH)2 + CH4
2. Methane from carbon monoxide

CO + 3H2
⎯⎯Ni +⎯C →
250°C
CH4
+ H2O
This is also called Sabatier Sanderen’s reduction

Physical Properties Boiling Point (B.P.)

B.P. increases with the increase of molecular mass. Among the isomers, straight chain alkane have higher
b.p. than branched chain alkane.

Melting Point (M.P.)

The melting points do not show regular variation with increase in molecular size. The even number members have higher m.p. as compared to next alkanes with odd number of carbon atoms (ALTERATION EFFECT).

Solubility

They are soluble in non polar solvents but insoluble in polar solvents such as water.

Chemical Properties📷

Alkanes are generally inert towards acids, bases, oxidising and reducing agents but they give following reactions:

1. Halogenation. Alkanes undergoes substitution reaction with halogen. Cl2 and Br2 only in presence of ultra violet light or high temperature (573 – 773K). But in presence of direct sunlight reaction is as

CH + 2Cl ⎯⎯⎯⎯⎯⎯→ C + 4HCl
direct sunlight
CH4 + Cl2 ⎯⎯h⎯ν → CH3Cl + HCl CH3Cl + Cl2 ⎯⎯h⎯ν → CH2Cl2 + HCl CH2Cl2 + Cl2 ⎯⎯h⎯ν → CHCl3 + HCl CHCl3 + Cl2 ⎯⎯h⎯ν → CCl4 + HCl
Decreasing order of reactivity of halogens towards alkanes. F2 > Cl2 > Br2 > I2
CH – CH – CH + Cl
298 K
CH CH – CH + CH CH CH –Cl
3 2 3
CH3
2 Light 3
Cl
3 3 2 2
CH – C – H + Cl
298 K
(CH ) C – Cl + (CH ) CHCH Cl
3
CH3
2 Light
3 3 3 2 2
Similarly
CH3
CH – C – H + Br
298 K
(CH ) C – Br + (CH ) CH–CH Br
3
CH3
2 Light
3 3 3 2 2
99% 1%

1. General mechanism : Halogenation of alkanes takes place in 3 steps :📷📷
2. Chain initiation steps : X + X

sunlight

1. Propagation :
2. Termination

R − H + X∙ ⎯⎯→R∙ + HX R∙ + X − X ⎯⎯→RX + X∙
X∙ + X∙ ⎯⎯→ X − X
R∙ + R∙ ⎯⎯→R − R📷
R∙ + X∙ ⎯⎯→R − X
Among these three steps, propagation step is rate determing step.
The relative rates of abstraction of various types of hydrogen follow the order : 3ºC : 2ºC : 1ºC
Chlorination → 5 : 3.8 : 1
Bromination → 1600 : 82 : 1
Liquid phase For higher alkanes – Fuming HNO3, at 413 K

1. Nitration :

e.g., :
C6H13 – H + HONO2
C6H13NO2 + H2O
Vapour phase For lower alkanes – conc. HNO3, at 670-750 K
e.g., :
CH3 −
NO2 ⎯⎯→CH3NO2 + H2O

1. Oxidation : Alkanes undergo oxidation under special conditions to yield a variety of products.

(a)
(b)
2CH + O ⎯⎯⎯C⎯u ⎯⎯→ 2CH OH
573 / 1100 atm
CH + O ⎯⎯⎯⎯→HCHO + H O
Mo2O3

1. 2R – CH + 3O

Manganese acetate
O
2R – C – OH + 2H O
3 2 373-430 K 2
or Ag2O

1. Isomerization : In presence of Anhy. AlCl3 + HCl or AlBr3 + HBr, straight chain alkanes, get converted in branched alkane.

CH – CH

• CH
• CH
  AlCl3 + HCl
  

CH3
CH – CH – CH
3 3 3 3
200ºC, 35 atm 3 3
CH (CH ) – CH AlCl3 + HCl
CH3 – CH – CH2 – CH2 – CH3 + CH3
CH3 – CH2 – CH– CH2 – CH3 CH3
2-methyl pentane 3-methyl pentane

1. Aromatization : Alkanes having minimum 6 or more carbons when heated at 773 K under high pressure (10-20 atm) in presence of Cr2O3, V2O3, Mo2O3 supported on Alumina gets converted into aromatic hydrocarbon.

CH ( CH ) CH Cr2O3
+ 4H ↑
3 2 4 3
773K, 10-20 atm
2
Benzene
n-heptane Cr2O3 773K, 10-20 atm

1. Reaction with steam :

CH4 + H2O ⎯⎯Ni → CO + 3H2📷

Conformations of Alkanes

CH3
Toluene
+ 4H2↑
Conformations isomers/conformers are compounds which arises due to rotation around C–C. In fact C–C rotation is hindered by an energy barrier of 1 to 20 kJ × mol–1. There are infinite number of conformers possible. Out of infinite number of conformers extremes can be discussed as

Conformers of ethane :

H H H H H📷📷
H H📷📷
H
H H H H H H
H H H H
Eclipsed
Staggered
Eclipsed Staggered
Sawhorse projection Newmann projection

1. It may be noted that one extreme conformation of ethane can be converted into other extreme conformer by rotation of 60º about C–C bond.
2. Conformers lying between two extreme are called skew conformations.

📷📷
0 60º 120º 180º
dihedral angle
Staggered> Eclipsed Decreasingorder of stability📷

Conformers of propane :

CH3📷📷📷
H H
CH3
H📷📷
H
H H H H H H
Eclipsed Staggered

Conformers of Butane :

CH3
CH3
CH3
CH CH3
H H H📷📷
CH3 H 3
60º
60º
H
60º
H
H H
CH3
(I)
H CH3
H
(II)
H H
H
(III)
H H H
(IV)
Staggered📷📷
Partially eclipsed
Gauche
(V)📷
Gauche
CH3
H
60º
Fully eclipsed
CH3 H📷📷
CH3
H H H
(VI)
Partially eclipsed
Stability order : I > III ≈ V > II ≈ VI > IV

ALKENES

Alkene have the structural unit
〉C = C〈 . The carbon atoms carrying the unsaturation are sp2 hybridized
with the p orbital laterally overlaping to form π-bonds. They have the general formula CnH2n. They are isomeric with cycloalkanes e.g. – C4H8
CH3 – CH2 – CH = CH2
Butene

Nomenclature :

CH2 CH2
CH2 CH2
Cyclobutane
The IUPAC rules for naming alkenes are similar in many respects to those for naming alkanes. Hence few solved examples are taken.
1 2 3 4 5
(i)
CH3 – CH = CH– CH2 – CH3
2 - pentene
CH3
C2H5

1. CH3– C = CH– CH2 – CH – CH2–CH2–CH3 1 2 3 4 5 6 7 8

5-Ethyl–2–methyl–2–octene

1. Alkene – H = Alkenyl

e.g., CH2 = CH– ethenyl (Vinyl group); CH2 = CH – CH2 – propenyl (Allyl group)
e.g., CH2 = CH – Cl - Vinyl chloride; CH2 = CH–CH2–OH Allyl alcohol

Preparations of Alkenes

From Alcohols

By heating alcohol with H2SO4 or H3PO4 at 170°C.
CH – CH

• OH

95% H2SO4 CH
= CH
3 2

Mechanism :

170°C 2 2
1st step : H2SO4
⎯⎯→
H+ + HSO –
2nd step : CH – CH – OH + H+ +
3 2
3rd step : CH – CH ⊕ H📷📷
CH3 – CH2 – OH2
⊕
CH3 – CH2 + H2O
4th step :
+ CH2 – CH2
H📷
–
CH2 = CH2 +H2SO4
HSO4
Here the 3rd step i.e. formation of carbocation is rate determining step.
The ease of dehydration of alcohols depends on the stability of carbocations formed. Hence the order of reactivity of alcohols is ter–> sec–> pri–> CH3 – OH because the incipient carbonium ion stability is ter–> sec–> pri–> CH +.

Hydride shift or

Example (Hydride shift)
CH3 – CH – CH – CH3 CH3 OH

Mechanism :

H2SO4
170°c📷
+
CH3 – C = CH2 – CH3 +
CH3
(Major)
CH2 = C – CH2 – CH3
CH3
(Minor)
CH3 – CH – CH – CH3 + H CH3 O.. H
H
CH3– C – CH – CH3📷
CH3 – CH – CH – CH2 CH3 OH2
+ +
CH3 – C – CH – CH3 CH3 – C – CH2 – CH3
CH3 CH3
sec-carbocation ter-carbocation
(Less stable) (More stable)
H (a) CH – C = CH – CH
(A)
H – CH2 – C – CH – CH3 CH3
CH3
CH2 = C – CH2 – CH3 CH3
(B)

1. Dehydration of alcohols follow Saytzeff’s rule.

Hence (B) product is maximum.
Saytzeff’s Rule: It states that “During dehydration of alcohols and dehydrohalogenation of alkyl halides the product formed is preferentially the one in which maximum number of alkyl groups are attached to the doubly bonded carbon atoms.

1. Dehydration by passing over alumina (Lewis Acid)

CH – CH

• OH Al2O3 CH = CH + H O

3 2
CH3📷
350°C 2 2 2
Al O
CH3
2 3
250°C
CH3 – CH = CH2
(CH ) C – OH Al2O3 CH – C = CH📷
3 3 250°C 3 2
The experimental conditions change with the structure of alcohols
pri - alcohols
⎯⎯con⎯c. H⎯2S⎯O⎯4 → alkene
180 C
sec - alcohols
⎯⎯85%⎯H⎯3PO⎯3 →
170°C
alkene
OH ⎯8⎯5%⎯H3⎯P⎯O3→📷📷
170°C
ter - alcohols
CH2 – OH📷
⎯2⎯0%⎯H2⎯S⎯O4→
85°C
H+ / heat
alkene

1. Dehydrohalogenation of alkyl halides

β α

1. CH3 – CH2 – CH2 – CH2 – Br
   alc KOH
   

Δ
CH3
– CH2
– CH = CH2
β α β
CH – CH – CH – CH
alc KOH
CH – CH = CH – CH + CH – CH – CH = CH

1. 3 2

3 Δ 3 3
3 2 2
Br (Major) (Minor)
CH3 CH3
CH2

CH – CH – C – CH
alc KOH
CH3 – C – CH – CH3 + (CH3)2 C = C(CH3)2

1. 3

3 Δ
Br
(Minor)
CH3 (Major)

1. The base used may be strongly basic anions like OH–, RO–, C2H5O– (CH3)3CO– etc.)
2. The group leaving (i.e., halogen) is a good leaving group if it is the conjugate base of a strong acid. (i.e., weakly basic halide ion).
3. One may also use sulphonates.
4. Dehydrohalogenation of 2° and 3° alkyl halides follow Saytzeff’s Rule.
5. Exception to Saytzeff’s Rule (Hofmann Rule)

When dehydrohalogenation is carried out with potassium, tertiary butoxide there is formation of less substituted alkene.
CH3
(CH ) COH
CH3
CH – CH – C = CH
(CH ) CO– + CH CH C – Br 3 3
CH – CH = C
3 2 2
3 3 3 2
CH3
75°C
3
(Minor)
CH3
CH3 (Major)

1. Dehalogenation Reactions

alc
– C – C –

Br Br
+ Zn
KOH
– C = C – + ZnBr2
VICINAL
2 R – CH
Br alc
+ Zn KOH📷
Br
R – CH = CH – R + 2ZnBr2
GEMINAL

1. By controlled hydrogenation of alkynes

(i)
CH3 – C ≡ C – CH3 + H2
Pd/BaSO4 LINDLAR’S CATALYST
CH3 C = C CH3 Cis - 2 - Butene H H
Here we may use small amounts of sulphur or quinoline also.
CH – C ≡ C – CH
Na/LiqNH3
CH3
H
C = C
3 3 Δ H
CH3
trans-2-Butene

1. By heating Quaternary Ammonium hydroxide

(C2H5)NOH Δ
CH2 = CH2 + (C2H5)3N + H2O

1. Kolbe’s Electrolysis

CH2COONa
CH2COONa
+ 2H2O
electrolysis
CH2 = CH2 + 2CO2 + 2NaOH + H2
Sodium succinate
Mechanism is similar to preparation of alkanes.

1. Pyrolysis

CH CH CH CH ⎯⎯Δ → CH –CH=CH + CH
3 2 2 3 770 K 3 2 4
CH CH CH CH CH CH ⎯⎯Δ → CH CH=CH
+ CH CH CH
3 2 2 2 2 3 770 K 3
2 3 2 3

1. By heating saturated hydrocarbon with SiO2

C H – CH

• CH – C H ⎯⎯SiO⎯2 → C H –CH=CH–C H + H

6 5 2
2 6 5 Δ 6 5
6 5 2

Physical Properties

Solubility

They are insoluble in water but soluble in organic solvents.

Boiling point

The boiling point of cis–alkenes is usually higher than corresponding trans–alkenes (More polarity).

Melting point

The melting point of trans-alkenes is usually greater than cis-alkene. (trans form is more symmetrical).📷

Chemical Properties

In alkenes C = C bond is made of stable σ-bond and reactive π-bond. As π-bond can easily be broken, alkenes undergo addition reactions.📷📷📷📷
C = C
+ XY
C –
C
X
Y
Being electron rich species they react with electrophiles in three ways.

1. Ionic Mechanism

C = C E – Nu C – C📷📷📷📷
+
–
+ Nu
E

1. Free Radical Mechanism

📷 C = C📷 E – E C – C + E📷📷
E

1. Transition State

📷 C = C📷+ E – Nu C – C📷📷
E Nu

Reactions :

1. Hydrogenation :

C = C + H2📷📷
Pd or Pd or Ni Δ
C – C
H H📷📷📷
The relative rates of hydrogenation is CH2 = CH2 > RCH = CH2 > R2 C = CH2 > RCH = CHR > R2 C
= CR2. This is due to the fact that as number of alkyl groups increase the steric hindrance increases and there by rate decreases.

1. Halogenation :

X
CH = CH + X ⎯⎯⎯C⎯C⎯l4 ⎯⎯→
2 2 2 Or
inert solvent
CH2 – CH2
X
CH = CH

• Cl

Cl
⎯⎯⎯C⎯C⎯l4 ⎯⎯→ 📷
2 2 2
Or inert solvent
CH2 – CH2
Cl
The addition always leads to the formation of trans addition product.

Mechanism :

δ– X
δ+ X
📷 C = C📷 + X2 C …— C📷📷
Transition State (T.S.)
δ–
X
📷 C …– C📷 📷 C – C + X📷
X📷📷
📷 📷 C — C📷
X
X
e.g., CH3 – CH = CH – CH3 + Cl2
– 9°C
CH3– CH – CH – CH3

Cl Cl

1. Addition of Halogen Acid

C = C + HX📷📷
C — C
H X📷📷
The reactivity order of halogen acid is HI > HBr > HCl

Mechanism :

R–CH=CH2 + HX
b R–CH–CH2
T.S.
(a)
– R–CH–CH3 + X
–📷
(b) R– CH2–CH2 + X
Here the transition state cleaves to from the most stable carbocation hence (a) cleavage takes place
and hence, + – are formed
R – CH2 – CH3+ X
i.e., R – CH – CH3+ X
R – CH – CH3
X📷
Markownikoff’s Rule may therefore be applied. It states that, “During the addition of unsymmetrical reagents to unsymmetrical alkenes, the negative part of the addendum goes to carbon of double bond with least number of atoms”.

Example :

δ—
X
H b
CH3–CH=CH2 + HX
(a)
CH3– CH – CH2
CH3–CH–CH3 + X A
CH3–CH2–CH2 + X
B
Since A is more stable than B. Hence A is formed and we get
+
CH3 – CH – CH3 X
CH3 – CH – CH3
X
CH3 – CH = CH2 + HBr
CH3– CH– CH3
Br

Kharasch - Mayo Effect

If the above reaction is carried out in the presence of some peroxide then addition takes place contrary to Markownikoff’s Rule
CH3 – CH = CH2+ HBr
⎯P⎯ero⎯x⎯ide→ CH3– CH2 – CH2 – Br

Explanation

This can be explained on the basis of free radical formation
•
Step 1 : R – O – O – R 2R – O
• •
Step 2 : R – O + HBr ROH + Br
CH – CH – CH • (A)

• 2📷

Step 3 : CH3 – CH = CH2 + Br
Br 1° Carbon free radical
•
CH3 – CH – CH2 – Br (B)
(B) is more stable than (A). 2° Carbon free radical
📷

1. Addition of Hypohalous Acid

📷 C = C📷+ HOX
CH2 = CH2 + HOCl
CH2 – CH = CH2 + HOBr

1. Addition of Water

C — C OH X
CH2 — CH2📷📷
OH Cl
CH2 – CH – CH2
OH Br
This reaction also takes place via carbocation mechanism (Rearrangement possible).
CH3
(CH3)2 C = CH2 + H2O
H+ 10% H2SO4
CH3 – C – OH
CH3

1. Addition of cold and conc. H2SO4

Carbocation Mechanism Followed, (Rearrangement Possible)
R – CH = CH2 + H2SO4
R – CH – CH3
H2O
Heat
R – CH – CH3
HSO4 OH

1. Oxy-Mercuration - Demercuration

Synthesis of alcohols from alkenes is in accordance with Markownikoff’s Rule (No carbocation formed)
R – CH = CH

1. Hg(OAc)2/THF,H2O

R – CH – CH
2 ii) NaBH /OH– 3
OH

1. Addition of Oxygen:

Ag CH2 CH2
CH2 = CH2 + ½ O2 570 K
O

1. Ozonolysis:

C = C + O3📷📷
O
CCl4 C📷📷
Zn/H2O
–H2O2📷
2 C = O
O— O📷

Example :

CH3CH = CH2 + O3
CCl4
O
CH — CH CH Zn/H2O 📷
–H2O2
CH3 – CHO + HCHO
O — O
CH3
+
O
📷

1. O3/CCl4
2. Zn/H2O (–H2O2)

CH3
Ozonalysis helps to locate the positions of double bonds in alkene.

1. Hydroboration Oxidation

R – CH = CH2 + B2H6
0°C
ether
(R – CH2 – CH2)3 – B
H2O2 OH–
R – CH2 – CH2 – OH + H3BO3
(Product is Antimarkownikov product.)

1. Oxidation Reactions
   1. Reaction with Baeyer’s Reagent (Cold dilute Alkaline KMnO4, Hydroxylation)

C = C + alk KMnO C – C📷📷📷📷

OH OH
The addition is a syn addition to form vicinal dihydroxy compounds.

1. With hot KMnO4 or acidic KMnO4

CH2 = CH2
⎯⎯[O⎯] →
Hot KMnO 4
2HCOOH ⎯⎯→ 2CO2 + 2H2O
CH3 − CH = CH2 ⎯⎯KM⎯nO⎯4 →CH3COOH + CO2 + H2O
H
(CH ) C = C(CH ) ⎯⎯Ho⎯t → 2 CH3
3 2 3 2 KMnO4
CH3
C = O

1. Substitution Reaction

CH3–CH = CH2 + Cl2 ⎯⎯500⎯−60⎯0⎯°C →
Cl–CH2–CH=CH2 + HCl–
This type of reaction takes place at a carbon atom attached to double bond carbon. This is called allylic substitution.

Mechanism

Step - 1 : Cl – Cl
⎯5⎯00 ⎯−60⎯0⎯°C→ 2Cl•
Step - 2 : CH =CH–CH
+Cl•
HCl + CH
= CH–
CH•
2 3 ⎯⎯→ 2 2

Step - 3 : CH2

= CH -
CH• +Cl2
⎯⎯→
CH2
= CH – CH2

• Cl + Cl•

1. Wohl Zeigler Reaction

CH3 – CH = CH2 + Br2
⎯L⎯ow ⎯con⎯c. o⎯f B⎯r2→
CH2 = CH–CH2–Br
The low concentration of Br2 is obtained from NBS📷
O
CH3 – CH = CH2 +CH2 — C
CH2 — C
O
N – Br
CH2 = CH – CH2Br +
O

CH2 — C
CH2 — C

O
N – H

1. Addition of Carbenes

(i)
CH2 – N ≡ N
hν ..
CH2 + N2
(ii) CH3 – CH = CH2 +

1. Isomerization :

C&H2 ⎯⎯h⎯ν → CH3 – CH – CH2
CH2
CH – CH – CH – CH = CH
700 - 970 atm
CH – CH – CH = CH – CH
2 2 2

1. Polymerization :

2 or Al2 (SO4)3; 470 K
3 2 3
CH = CH Pressure
O2
( CH2 – CH2)n
CF2 = CF2
Trans O2
( F2C – CF2)n

ALKYNES

Compounds containing the structural unit – C≡C– are called Alkynes. Like the double bond it is unsaturated and highly reactive towards the reagent that double bonds react with and also towards others. The simplest member of the alkyne family is acetylene, C2H2.Each of the carbon atoms carrying the triple bond are sp hybridized.
1s 2s 3p📷
C📷
sp
C📷
sp
Alkynes the compounds having general formula CnH2n – 2 where n ≥ 2 it can be categorize by two ways.

1. Terminal alkynes : Alkynes having triple bond at one end of the carbon attached to H

e.g. CH3–C ≡ C–H, CH3 – CH2–C≡CH. Terminal hydrogen is acidic in nature.

1. Non-terminal alkynes : Alkynes in which both triple bonded carbons are attached to alkyl group.

Preparation of Alkynes

1. From Dehydrohalogenation of vicinal or geminal dihalides

CH2
— CH2
alc. KOH
Δ
CH ≡ CH + 2H2O + KBr
Br Br
R– CH – CH alc. KOH
R – C ≡ CH + 2H O + KBr
2 Δ 2
Br Br

1. Dehalogenation Reaction📷

(a)
X X

– C – C –

X X
Zn dust
Δ
– C ≡ C + 2ZnX2
X📷

1. 2R – C X

X
Zn dust
Δ
R – C ≡ C – R + 3ZnX2

1. Kolbe’s Electrolytic Decarboxylation

R COOK
C
C
R COOK
+ 2H O Electrolysis
R – C ≡ C – R + 2KOH + CO2 + H2
cis or trans

1. Formation of Higher alkyne
   
   1. CH ≡ CH NaNH2

CH ≡ C Na
📷📷
CH ≡ C Na + CH3Cl CH ≡ C – CH3 + NaCl

1. R–C ≡ CH + NaNH2 RC ≡ CNa + NH3

R–C ≡ C Na + R′ X R–C ≡ C–R′ + NaX

Chemical Properties

Alkynes undergo electrophilic addition generally but in the presence of salt of heavy metals which forms complexes with multiple bonds it undergo nucleophilic addition reaction.

1. Addition Reaction
   1. Electrophilic addition reaction :
      1. Addition of halogen

CH
+ Cl2
CH
CHCl = CHCl Cl2
acetylene dichloride
CHCl2
CHCl2
acetylene tetrachloride (westron)
CH3 – C ≡ CH
Br2/CCl4

CH3 – C = C

Br
Br Br2/CCl4

H

Br Br CH3 – C – C – H

Br Br

Trans-1, 2-dibromopropene
1, 1, 2, 2-tetra bromo propane

1. Addition of halogen acids

HC ≡ CH HCl CH = CHCl📷
HCl CH CHCl
CH3COOH
2
Vinyl chloride
CH3COOH
2 2
Ethylidene chloride
CH ≡ CH
Cl2/H2O
or
CH = CHOH
OH
HOCl CHCl – CH – OH
–H2O
HCCl2 – CHO
HOCl Cl
Dichloro acetaldehyde

1. Nucleophilic addition reaction :

Because of greater electronegativity of sp hybridized C as compared to sp2 hybridized carbons, Alkynes are more susceptible to nucleophilic addition reactions than alkenes. It is due to formation of some📷📷
sort of complex of heavy metal ion with π electrons like
Hg2+
and this results decrease in
electron density around triply bonded carbon atoms and this can be attacked by nucleophiles.

1. Addition of H2O or hydration of alkyne or Kucherov reaction

+ 2+
H , Hg
H OH O
tautomerise
+ H2O
CH
333 K
H – C = C – H
Vinyl alcohol
CH3 – C – H
Acetaldehyde
R – C ≡ C – H + H2O

1. Addition of HCN

+ 2+
H , Hg
OH
R – C = CH2
tautomerise
O
R – C – CH3
Ketone
CH + HCN Ba(CN)2
CH
CH2
CHCN
Vinyl cyanide
Similarly alkynes adds acids in presence of lewis acid catalyst or Hg2+ give vinyl ester.

O

HC≡C – H + H3CCOOH
Hg2+/353 H

CH3 – C – O – CH = CH2

Vinyl ethanoate

O O R

R – C
≡ CH + CH – C – OH BF3 CH – C – O – C = CH
a-alkyl vinyl ethanoate

1. Reaction of Acidic H Atom
2. Alkynes having acidic H atom reacts with metals like Na, K, evolves H2 gas.

CH
+ Na
CH (1 mol)
C – Na
C – H
+ 1/2 H2↑
CNa C – H
+ Na
Monosodium acetylide
C – Na
+ 1/2 H2↑
C – Na
Disodium acetylide
HC ≡ C − H + NaNH2 → HC ≡ C − Na + NH3 ↑ ⎯⎯NaN⎯H⎯2 →NaC ≡ C − Na + NH3 ↑📷

1. Reaction with Tollens reagent : When alkyne reacts with tollens reagent (Ammonical AgNO3 solution) at forms white precipitate of silver acetylide.

C – H
C – H
+ 2AgNO3 + 2NH4OH
(Tollens reagent)
C – Ag C – Ag
¯ + 2NH4Cl + 2H2O
(White precipitate)
These acetylide are not decomposed by H2O like acetylide of Na but by mineral acids like dil HNO3.
C – Ag C – Ag
+ 2HNO3
CH
2AgNO3 +
CH
R − C ≡ CH + Ag+ → R − C ≡ C − Ag + H⊕

1. Reaction with Ammonical Cuprous Chloride :

HC
+ Cu2Cl2 + 2NH4OH
HC
CCu
CCu
(red ppt)
↓+ 2NH4Cl + 2H2O
copper acetylide
These reactions are used to distinguish terminal alkynes from other alkynes.

1. Polymerization Reaction
2. When acetylene is passed in red hot cutube or retube. It converted into benzene.

CH 773K
3
CH
C6H6
CH3📷
Similarly CH3 – C ≡ CH
red hot Cu or Fe tube
H3C
CH3
(Mesitylene)
CH

1. 2

CuCl/NH4Cl
CH = CH – C ≡ CH
HCl
Cl
CH = CH – C = CH
CH Cu2Cl2/NH4Cl
2
Vinyl acetylene
2 2
Chloroprene
Chloroprene on polymerization gives polymer called neoprene; used as artificial rubber.

1. Under high pressure and in presence Ni(CN)2 acetylene tetramerises.

4 CH ≡ CH
Ni(CN)2
Cycloocta tetraene

1. Reaction with S8, N2, NH3 and HCN

Acetylene reacts with S8, N2, NH3 and HCN to form different heterocyclic compounds.

1. CH

CH
CH
+
+ CH
300°C
S

1. CH

CH
+ N2
Electric spark
2HCN
1 S
(Thiophene)
8 8

1. CH CH Δ

+
CH + CH N

1. (iv)

2 CH + HCN red hot tube
CH📷
N
(Pyridine)
NH3
H
(Pyrole)

AROMATIC HYDROCARBONS

Hydrocarbons which follow Huckel rule are termed as Aromatic hydrocarbons.

Huckel's rule :

A planar molecule having complete delocalisation of (4n + 2) π electrons is termed as aromatic hydrocarbon (where n is any integer)
📷
e.g., 4n + 2 = 6
n = 1
⊕
4n + 2 = 2
n = 0
📷
4n + 2 = 6
n = 1
CH3
4n + 2 = 6📷
n = 1
4n + 2 = 6
n = 1📷📷

Homologues of Benzene

They are all aromatic hydrocarbons. Aromaticity is present due to benzene ring.
CH3 C2H5📷📷
etc.📷

1. Electrophilic Aromatic substitution reactions in Benzene : (EAS)

Benzene and its homologues readily undergo EAS. As a consequence of complete delocalization of π
electrons in benzene, it has π electron cloud over benzene ring which makes electrophile attack over it.

General Mechanism :

+ E+ 📷📷
H
+📷📷
E E + H+

1. Halogenation

+ Cl📷
(resonance stabilised intermediate)
Cl📷
⎯⎯FeC⎯l3 →
+ HCl
Benzene
2 dark
Reaction with I2 is reversible.
I📷📷
+ I2
+ HI
Hence it is carried out in the presence of conc. nitric acid to oxidise the Hydrogen Iodide formed.

1. Nitration📷📷📷

+ HNO3📷
⎯⎯H2S⎯O4⎯co⎯n⎯c. →
NO2

• H2O

⎯⎯H2S⎯O4⎯co⎯n⎯c. →
+HNO3 363−373K
NO2
NO2

1. Sulphonation

Nitro benzene
SO3H📷📷
SO3H📷
+ H2SO4 + SO3📷
⎯⎯He⎯at →
+ H2O or + ClSO3H
⎯⎯He⎯at →

• ΗCl

1. Friedel Crafts Reaction

Chloro Sulphuric Acid Benzene sulphonic acid
Alkylation : Reactive intermediate is carbocation which can undergo rearrangement.
CH3📷📷
+ CH3 – Cl
⎯⎯Anh⎯ydr⎯ous⎯Al⎯C⎯l3 →
+ HCl
+ CH3CH2CH2Cl📷
⎯⎯Anh⎯ydr⎯ous⎯Al⎯C⎯l3 →
+ HCl
⎛ O:📷📷
⎜
Acylation : Reactive intermediate is acylium ion ⎝R–C+
COCH3📷📷
which cannot undergo rearrangement.
+ CH3 – COCl
⎯⎯Anh⎯ydr⎯ous⎯Al⎯C⎯l3 →
+ HCl
Acetyl chloride Acetophenone
Ortho and para substitution : Electron releasing groups like - R (alkyl) — O.. H, — O.. R, — NHR,
— NHCOR are activating groups i.e., they increase electron density at ortho and para position,
therefore, are ortho and para directing towards electrophilic substitution reactions.
Meta substitution : Electron withdrawing groups such as – NO2, – CHO, – COOH, – COCH3, – CN,
– SO3H, – COOR are called deactivating groups. They decrease electron density at ortho and para- position, therefore, electrophillic substitution takes place at meta-position.

1. Halogenation of side chain :

CH3📷
Toluene
or Methyl Benzene
+ Cl
⎯⎯su⎯n →
2 light
CH2Cl
Benzyl chloride📷
or chlorophenyl methane
+ Cl2
⎯⎯hv →
CHCl2
Benzal chloride📷
or dichlorophenyl methane
⎯⎯C⎯l2 →
CCl3📷
+ HCl
Benzotrichloride

1. Oxidation :

or trichlorophenyl methane
O O
📷
Benzene
+ 5O2
+ H2O
2-Butene-1,4-dioic acid
❑ ❑ ❑

https://docs.google.com/document/d/1pSOMGU_vHBOXeDrlD8MvMNsEO8Kyqye5/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true
Work Book (Phase - VIII)
Subjective:
Board Type Questions

1. What will be the sequence of bases on mRNA molecule synthesized on the following strand of DNA?
   
2. Aspartame, an artificial sweetener, is a peptide and has the following structure:
   

📷
(i) Identify the four functional groups.
(ii) Write the Zwitterionic structure.
(iii) Write the structure of amino acids obtained from the hydrolysis of aspartame.
(iv) Which of the two amino acids is more hydrophobic?

1. Could a copolymer be formed in both condensation or not? Explain with example.
   
2. What are biodegradable polymers write structure of PHBV?
   
3. What is Nylon? Write an equation for the chemistry involved when a drop of hydrochlorine acid makes a hole in a nylon stocking.
   
4. (i) Give the Fischer projection of L-glucose.
   

(ii) Give the products of reaction of L-glucose with Tollen’s reagent.

1. Differentiate between

(a) Nucleotide and Nucleoside
(b) Vitamins and Hormones
(c) DNA and RNA

1. Give the IUAPC names of the following compounds:

(a) 📷
(b) Cr(CO)6
(c) K3[Al(C2O4)3]

1. Combination of Pt(IV), NH3, Cl− and K+ results in the formation of seven complexes and one such complex is 📷. Write the formulae of the other six members of the series.
   
2. What is the number of ionizable chlorine atoms in the complex 📷?
   
3. Complete and balance:
   

📷

1. Compete the following

📷

1. Explain why:

Ferrous salt turns brown in air.

1. Which one of Fe2+ and Fe3+ ions is more paramagnetic and why?
   
2. Which of the following is formed when K2Cr2O7, CaCl2 and conc. H2SO4 is heated?
   

📷

1. Though copper, silver and gold have completely filled sets of d-orbitals yet they are considered as transition metals. Why?
   
2. Why does Mn(II) show maximum paramagnetic character amongst the bivalent ions of the first transition series?
   
3. Why Zn2+ salts are white while Ni2+ salts are blue?
   
4. Why Zn2+ salts are white while Cu2+ salts are blue?
   
5. [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless. Explain.
   
6. A co-ordination compound has the formula CoCl3.4NH3. It does not liberate ammonia but precipitates chloride ions as silver chloride. Give the IUPAC name of the complex and write its structural formula.
   
7. Arrange the following complexes in order of increasing electrical conductivity:
   

[CoCl3(NH3)3], [CoCl(NH3)5]Cl2, [Co(NH3)6]Cl3, [CoCl2(NH3)4]Cl

1. Which of the two compounds is more stable and why?

[K4[Fe(CN)6], K3[Fe(CN)6]

1. [NiCl4]2− is paramagnetic while [Ni(CO)4] is diamagnetic though both are tetrahedral. Why?
   
2. Name one polymer formed by step growth polymerization. Give names of its monomers.
   
3. Arrange the following polymers in increasing order of their intermolecular forces. Also classify them as addition and condensation polymers:
   

Nylon-66, Buna-S, Nylon – 66

1. Given reason of the following

(i) Aniline does not undergo Friedel-Craft reaction.
(ii) Diazonium salts of aromatic amines are more stable than those of aliphatic amines

1. Given account of the following

(i) Gabriel phthalamide synthesis is preferred for synthesing primary amines.
(ii) Ethylamine is soluble in water whereas aniline is not.

1. Given reason of the following

(i) Methylamine in water reacts with FeCl3 to precipitate hydrated ferric oxide.
(ii) pKb of aniline is more than that of methylamine.

1. Discuss the nature of bonding in the following co-ordinate compound on the basis of valence bond theory.

(i) 📷 (ii) 📷 (iii) 📷 (iv) 📷

1. (i) Draw figure to show splitting of degenerate d-orbital in an octahedral crystal field?

(ii) What is the significance of Δ0 (stabilization energy).

1. (i) Explain inner and outer orbital complexes with suitable example.

(ii) Draw the figure to show splitting of degenerate d-orbitals in an octahedral crystal field?

1. What are class-b-acceptors? What which type of ligands they form stable complexes?
   
2. (i) Classify the following into monosaccharides and disachharides.
   

Ribose, 2-deoxyribose, maltose, galactose, fructose and lactose.
(ii) What do you understand by the term glycosidic linkage?
(iii) What is glycogen? How is it different from starch?

1. (i) What is the basic structural difference between starch and cellulose?

(ii) What happens when D-glucose is treated with the following reagents?
(a) HI (b) Br2 – water (c) HNO3

1. (i) Write the monomers used for getting the following polymers

(a) Polyvinyl chloride (b) Teflon (c) Bakelite
(ii) How can you differentiate between addition and condensation polymerization.

1. (i) How is Dacron obtained from ethylene glycol and terephthalic acid?

(ii) What is a bidegreadable polymer? Give an example of a biodegradable aliphatic polyester.

1. (a) What are the monomeric repeating units of Nylon-6 and Nylon -66?

(b) Write the names and structure of the monomers of the following polymers.
(i) Buna-S (ii) Buna-N (iii) Dacron (iv) Neoprene

1. Indicate the steps in the preparation of

(i) K2Cr2O7 from chromite ore.
(ii) KMnO4 from pyrolusite ore.

1. Describing the oxidizing action of K2Cr2O7 and write the ionic equation for its reaction with

(i) Iodine (ii) Iron (II) solution (iii) H2S
IIT Level Questions

1. How will you carry out the following conversions?

(i) Methylamine to ethylamine
(ii) Ethylamine to methylamine
(iii) Aniline to benzoic acid
(iv) Nitrobenzene to 2, 4, 6-tribromoaniline
(v) Nitrobenzene to benzamide

1. For M2+/M and M3+/M2+ systems the E0 values for same metals are as follows:

📷
📷
📷
📷
Use the above data to comment upon
(i) the stability of Fe3+ in acid solution as compound to that of Cr3+ or Mn2+ and
(ii) the case with which iron can be oxidized as compared to a similar process for either chromium or manganese metal.

1. Arrange the following:

(i) C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3................increasing basic strength in gaseous.
(ii) C2H5NH2, (C2H5)2NH, (C2H5)3N and NH3..................increasing basic strength in water.

1. How will you convert?

(i) Aniline to p-bromoaniline
(ii) Benzyl chloride to 2-phenyl ethanamine
(iii) Aniline to 2, 4, 6-tribromoflurobenzene

1. Complete the following reaction:

(i) 📷
(ii) 📷
(iii) 📷

1. Identification of A, B and C in the following reactions:

(i) 📷
(ii) 📷
(iii) 📷

1. Write the correct formulae for the following co-ordination compounds

(i) CrCl3.6H2O (violet with 3Cl− ions /unit formula)
(ii) CrCl3.6H2O (light green colour, with 2Cl− ions/unit formula)
(iii) CrCl3.6H2O (dark green colour with Cl− ion/unit formula)

1. (i) Briefly outline the drawback of V.B.T.

(ii) Explain the hybridization and magnetic behaviour.
(a) 📷 (b) 📷

1. (a) Explain with example cis-trans geometrical isomerism.

(b) Explain the facial and meridional isomerism.’

1. (i) Define the following as related to proteins.
   (a) Peptide linage (b) Primary structure (iii) Denaturation
   

(ii) What is the difference between a nucleoside and a nucleotide.
Objective:
Multiple choice questions with single correct options

1. Correct order of basicities of the following compounds is

📷
(A) 2 > 1 > 3 > 4 (B) 1 > 3 > 2 > 4
(C) 3 > 1 > 2 > 4 (D) 1 > 2 > 3 > 4

1. The compound that will react most readily with NaOH to form methanol is

(A) 📷 (B) CH3OCH3
(C) 📷 (D) 📷

1. Which of the following is the strongest base?

(A) C6H5NH2 (B) 📷
(C) 📷 (D) 📷

1. In 📷 the order of proton accepting tendency will be

(A) 📷 (B) 📷
(C) 📷 (D) 📷

1. Ions of the two most common isotopes of the transition metal nickel are shown below:

📷
Which one of the following statements is true?
(A) The electron arrangement of both these Ni2+ ions is 1s2, 2s2, 2p6, 3s2, 3p6, 3d6, 4s2.
(B) The 📷 ion will have more protons in its nucleus than the 📷 ion.
(C) In the same strength magnetic field, the 📷 ion will be deflected more than the 📷 ion.
(D) Both 📷ions have the same number of electrons but a different number of neutrons.

1. Which equation does not involve the reduction of a transition metal compound?

(A) 📷 (B) 📷
(C) 📷 (D) 📷

1. Ag+ forms many complexes, some of these are [Ag(NH3)2]+, [Ag(CN)2]–, [Ag(S2O3)2]3–.

Which of the following statements is true?
(A) In these complexes, Ag+ is a Lewis base.
(B) The hybridization of Ag+ is sp2.
(C) The Ag+ complexes are good reducing agents.
(D) These complexes are all linear.

1. The magnetic moment of an iron compound is 5.918 BM, then the oxidation state of iron in this compound will be

(A) 0 (B) 1
(C) 2 (D) 3

1. Which pair of elements can form an alloy?

(A) Zn + Pb (B) Fe + Hg
(C) Fe + C (D) C + Pt

1. Identify the product P in the following reaction.

📷
(A)
📷
(B)
📷
(C)
📷
(D)
📷

1. The product(s) of the Hofmann exhaustive methylation of the following compound is

📷
(A)
📷
(B)
📷
(C)
📷
(D)
All of the above

1. When aniline is heated with benzene diazonium chloride at low temperature in weakly acidic medium, the final product obtained is

(A)
📷
(B)
📷
(C)
📷
(D)
📷
13.
The major product formed in the elimination reaction of
📷
(A)
📷
(B)
📷
(C)
📷
(D)
📷
14.
📷
Hydrocarbon ‘B’ is,
(A) CH4 (B) 📷
(C) 📷 (D) 📷

1. In which of the following compounds chromium shows maximum ionic radius?

(A) K2Cr2O7 (B) CrO2Cl2
(C) Cr2(SO4)3 (D) CrCl2

1. Zinc tarnishes in moist air due to

(A) ZnCO3.3Zn(OH)2 (B) Zn(OH)2
(C) ZnCO3.ZnCl2 (D) None of the above

1. Which compound does not dissolve in hot dilute 📷

(A) HgS (B) PbS
(C) CuS (D) CdS

1. A compound of mercury which is a strong poison and its antidote being the white of an egg, which eliminates it from the system in the form of a coagulated mass, is

(A) Hg2Cl2 (B) HgCl2
(C) HgI2.HgO (D) K2HgI4

1. The product of reaction of an aqueous solution of Bi3+ salt with sodium thiosulphate gives

(A) Bi2S3 (B) Na3[Bi(S2O3)3]
(C) Na[Bi(S2O3)2] (D) [Bi2(S2O3)2]Cl2

1. An aqueous solution of FeSO4.Al2(SO4)3 and chrome alum is heated with excess of X and filtered, a yellow filtrate and a brown residue is obtained. X is

(A) Na2O2 (B) Na2ZnO2
(C) NaCl (D) KCl

1. When H2O2 in ether reacts with acidified K2Cr2O7 solution blue coloured perchromic anhydride is obtained in etheral solution. It is due to:

(A) CrO3 (B) H2CrO4
(C) CrO5 (D) H2CrO7

1. The atomic radius of Cu is greater than that of Cr but ionic radius of Cr2+ is greater than Cu2+:

(A) because Cr+2 has no stable oxidation state of Cr
(B) because (d – d) transition is not easily available in Cr.
(C) because in Cu, (d – d) electron repulsion is larger due to pairing of electron
(D) because of Cu has larger no of protons in nucleus w.r.t. Cr

1. Dicyclopentadienyliron [(C5H5)-]2 Fe2+ is stable molecule because of

(A) resonance (B) not stable
(C) five member ring (D) none

1. Which one of the following would not react with aqueous silver nitrate to produce a precipitate that is soluble in concentrated aqueous ammonia?

(A) CaBr2 (B) [CoCl4]2–
(C) (CH3)4N+I– (D) CH3COCl

1. From the stability constant which one is strongest ligand?

(A) 📷
(B) 📷
(C) 📷
(D) 📷

1. The crystal field splitting energy for Cr3+ ion in an octahedral field increases for the ligands I–, H2O, NH3, CN– and the order is

(A) CN– < I– < H2O < NH3 (B) I– < H2O < NH3 < CN–
(C) CN– < NH3 < H2O < I– (D) NH3 < H2O < I– < CN–

1. Which of the following complex ion is paramagnetic?

(A) [Fe(CN)6]4– (B) [Fe(CN)6]3–
(C) [Fe(CN)5NO]2– (D) [Co(NO2)6]3–

1. IUPAC nomenclature of the complex compound Na3[Cr(O)2(O2)(O2)2(NH3)] where half of the coordination number are occupied by neutral ligands with magnetic moment 📷 must be

(A) Sodium amminedioxodioxygensuperoxochromate (III)
(B) Sodium amminedioxygendioxodisuperoxochromate (III)
(C) Sodium amminebisdioxygendioxoperoxochromate (III)
(D) Sodium amminebisdioxygendioxoperochromium (III)

1. Polymer used in bullet proof glass is

(A) PMMA (B) Lexame
(C) Nomex (D) Kelvar

1. On hydrolysis which of the following carbohydrates gives only glucose?

(A) Sucrose (B) Lactose
(C) Maltose (D) Galactose

1. Grape sugar is

(A) Glucose (B) Fructose
(C) Maltose (D) Lactose

1. Which of the following is fruit sugar?

(A) Sucrose (B) Fructose
(C) Glucose (D) All of these

1. The urine sample of diabetic patients contains

(A) Sucrose (B) fructose
(C) Cane sugar (D) Starch

1. The fibre obtained by the condensation of hexamethylene diamine and adipic acid is

(A) Dacron (B) Nylon 66
(C) Rayon (D) Teflon

1. Natural silk is

(A) Polyester (B) Polyamide
(C) Epoxide (D) Polyurethane

1. Natural rubber is obtained from Latex which is a

(A) mixture of wood, plants and gums
(B) Colloidal dispersion of rubber in water
(C) mixture of chloroprene and carbohydrate
(D) none of these

1. A raw material used in making Nylon is

(A) adipic acid (B) 1, 3-butadiene
(C) ethyne (D) cyclohexanone

1. Which of the following is not a natural polymer?

(A) Wool (B) Silk
(C) Cotton (D) Teflon

1. In the dichromate dianion

(A) 4Cr – O bonds are equivalent
(B) 6-Cr – O bonds are equivalent
(C) All Cr – O bonds are equivalent
(D) All Cr – O bonds are non – equivalent

1. In the standardization of Na2S2O3 using K2Cr2O7 by iodomaety, the equivalent weight of K2Cr2O7 is

(A) Molecular weight/2 (B) Molecular weight/6
(C) Molecular weight/3 (D) Same as molecular weight

1. The chemical composition of ‘slag’ formed during smelting process in the extraction of copper in

(A) Cu2O + FeS (B) FeSiO3
(C) CuFeS2 (D) Cu2S + FeO

1. Amongst the following, identify the species with an atom in +6 oxidation state

(A) MnO4− (B) Cr(CN)6−3
(C) NiF6−2 (D) CrO2Cl2

1. Ammonium dichromate is used in same fire works. The green coloured powder blown in the air is

(A) CrO3 (B) Cr2O3
(C) Cr (D) CrO(O2)

1. Isoprene on polymerization produces

(A) Synthetic rubber (B) Neoprene
(C) Gutta-Percha (D) Cis-poly(2-Methyl-1, 3-butadiene)

1. Example of thermosetting plastic is/are

(A) bakelite (B) PVC
(C) Polyurethane (D) Nylon

1. The repeating units of PTFE are

(A) CH ≡ CH (B) CF3 – CF3
(C) CH2 = CHCN (D) CF2 = CF2

1. Terylene is a condensation polymer of ethylene glycol and

(A) Benzoic acid (B) acetic acid
(C) terephthalic acid (D) salicylic acid

1. Which of the following is used as rocket fuel?

(A) cyanogens + O3 (B) Cyanogen + O2
(C) water gas + O3 (D) Nitrolium + O3

1. The expected spin only magnetic moment for 📷respectively are

(A) 1.73 and 4.82 (B) 5.92 and 4.82 (C) 1.73 and 3.87 (D) 0.00 and 1.73
Multiple choice questions with more than one option correct

1. Aniline and diethylamine can be distinguished by

(A) coupling reaction (B) carbylamine reaction
(C) reaction with HONO (D) Hoffmann bromamide reaction

1. Cyclobutyl amine is treated with nitrous acid. The products may be

(A)
📷
(B)
📷
(C)
📷
(D)
📷

1. Consider the reaction

📷
Which of the following statements is/are correct?
(A) Product B is an orange coloured compound
(B) Product A contain azo linkage
(C) Product B contain azo linkage
(D) Product A is used as dye

1. The low spin complexes are

(A) K3[Fe(CN)6] (B) [Ni(CO)4]
(C) K3[CoF6] (D) Na2[Ni(CN)4]

1. Which of the following statement(s) is/are correct?

(A) Oxidation number of Fe in Na2[Fe(CN)5(NO)] is +2.
(B) [Ag(NH3)2]+ is linear in shape.
(C) In [Fe(H2O)6]3+, Fe is d2sp3 hybridized.
(D) In [Ni(CO)4], oxidation number of Ni is zero.

1. When MnO2 is fused with KOH, a coloured compound is formed then the compound will be

(A) K2MnO4 (B) Mn3O4
(C) purple in colour (D) brown in colour

1. Which of the following statements is/are correct when a mixture of NaCl and K2Cr2O7 is gently warmed with conc. H2SO4?

(A) A deep red vapour is evolved.
(B) The vapour when passed into NaOH solution gives a yellow solution of Na2CrO4.
(C) Chlorine gas is evolved.
(D) Chlomyl chloride is formed.

1. Which of the following gives an optically active compound when reacted with sodium borohydride, NaBH4?

(A)
📷
(B)
📷
(C)
📷
(D)
📷
59.
📷
For the above reaction,
(A)
📷
(B)
📷
(C)
📷
(D)
📷

1. Consider the following three bases of DNA and RNA.

📷
Order of acidic strength on different sites will be
(A) (2) > (1) (B) (3) > (4)
(C) (7) > (6) (D) (6) > (5)

1. Which of the following statement(s) about D(+) glucose is/are true?

(A) Naturally occurring glucose is dextrorotatory
(B) It reduces ammoniacal AgNO3 solution
(C) On polymerisation it forms galactose
(D) It forms cyanohydrin on reaction with HCN
COMPREHENSION - I
Read the following paragraph and answer the questions given below:
The variety of colours among transition metal complexes has always fascinated the observers. For example, aqueous solutions of octahedral [Co(H2O)6]2+ are pink in colour but those of tetrahedral [CoCl4]2– are blue. The green colour of aqueous [Ni(H2O)6]2+ turns blue when ammonia is added to the solution to give [Ni(NH3)6]2+. The reduction of violet [Cr(H2O)6]3+ gives bright blue [Cr(H2O)6]2+. As with all colours, these arise from electronic transitions between levels whose spacing correspond to the wavelength available in visible light. The magnitude of spacing depends upon the factors such as the geometry of the complex, the nature of the ligands present, and the oxidation state of central atom.

1. Identify the complexes which is/are expected to be coloured?

(A) [Ti(NO3)4] (B) [Cu(NCCH3)4]+📷
(C) [Cr(NH3)6]3+3Cl– (D) [Zn(H2O)6]2+

1. Which of the following is π-acid ligand?

(A) NH3 (B) CO
(C) F– (D) Ethylene diamine

1. The complex which has no ‘d’ electrons in the central metal atom is

(A) [MnO4]– (B) [Fe(CN)6]3–
(C) [Co(NH3)6]3+ (D) [Cr(H2O)6]3+

1. Among 📷(atomic numbers of Ti = 22, Co = 27, Cu = 29, Ni = 28), the colourless species are

(A) 📷and 📷 (B) 📷
(C) Cu2Cl2 and 📷 (D) 📷

1. The ferric ion is detected by the formation of a prussian blue precipitate on addition of potassium ferrocyanide solution. The formula of the prussian blue precipitate is

(A) Fe4[Fe(CN)6]3 (B) Fe3[Fe(CN)6]4
(C) KFe[Fe(CN)6] (D) None of the above
COMPREHENSION - II
Read the following paragraph and answer the questions given below:
Molecules of the amino acids that comprise our proteins have the property of being non-super imposable on their mirror images because, they are said to be chiral or possess ‘handedness’. Most of the amino acids are of left handed form, but with no reasons available. Even more interesting, recent experiments have shown that a 7–9% excess of four amino acids is present in Murchison meteorite discovered in 1970. This analysis shows life could arise outside the confines of earth. The origin of unequal distribution is probably because of electromagnetic radiations emitted in a corkscrew fashion from the poles of spinning neutron stars could lead to a bias of one mirror image isomer over another when molecules form in interstellar space.
A stereogenic carbon can be defined as a carbon atom bearing groups of such nature that an interchange of any two groups will produce a stereoisomer. The device that is used for measuring the effect of optically active compounds on plane polarized light is a polarimeter. The analyser in the polarimeter decides the activity of compound present. This is being decided on the basis of angle by which axis of analyser has to be rotated for complete brightness. Left, right or no adjustment decides 📷activity or optical inactivity respectively.

1. For existence of life, amino acids as ingredient of protein exist mainly in

(A) 📷form (B) d–form
(C) Both (D) None of the above

1. EMR emitted in corkscrew fashion from poles of spinning neutrons are biased to

(A) 📷amino acids (B) d–amino acids
(C) Both (D) None of the above

1. Stereogenic carbons can bring

(A) optical activity (B) optical inactivity
(C) Both (D) None of the above

1. Optically inactive isomer possessing chiral centres are categorized as

(A) optical isomers (B) non-optical isomers
(C) mesomers (D) diastereoisomers

1. In one of the experiment performed by two students “Abraham” and “Fienkelstein” for measurement of optical activity of a substance. Result given by “Abraham” was: It is d–form with +30o rotation and by “Fienkelstein”, it was: It is 📷form with –150o rotation. One of them can be correct only as solution can either be d form or, 📷form.

Which of the following is correct measure to come out of this dispute?
(A) Result with acute angle is correct.
(B) Result with obtuse angle is correct.
(C) Students should perform another experiment by varying the concentrations or length of polarimeter tube to which angle of rotation is directly proportional.
(D) Students should perform the experiment again with precaution. They will come on same conclusion as one of them must be committing observational mistake.
COMPREHENSION - III
Formation of amine salts can be used to isolate and characterize amines. Most amines containing more than six carbon atoms are relatively insoluble in water. In dilute aqueous acids, these amines form their corresponding ammonium salts and they dissolve. Formation of soluble salt is one of the characteristic functional group test for amines.
The formation of amine salts is also used to separate amines from less basic compounds. When the solution is made alkaline (by
addition of NaOH) the free amine is regenerated. The purified free amine either separates out of the aqueous solution or is extracted into an organic solvent. Many drugs and other biologically important amines are commonly stored and used as their salts. Amine salts are less prone to decomposition by oxidation and other reactions and they have virtually no fishy odor. The salts are soluble in water and they are easily converted to solution for syrup and injections.

1. Amine form salt with which of the following reagents

(A) NaOH (B) KOH
(C) HBr (D) All of these

1. In the presence of NaOH amines separate out because

(A) amines react with NaOH (B) amines are less basic than NaOH
(C) amines form slat with NaOH (D) None of these

1. Amines are soluble in solution of many transition metal salts because

(A) Amines act as base with transition metal salt
(B) Amines act as ligand with transition metal salts
(C) Amines do not react with transition metal salt
(D) None of these

1. The hybridisation of N-atom in amine salts is

(A) sp2 (B) dsp2
(C) sp (D) sp3

1. Which of the following reaction involves amine salt?

(A) Hoffmann bromamide reaction (B) Hoffmann elimination
(C) Hoffmann rearrangement (D) None of these
COMPREHENSION - IV
Read the following paragraph and answer the questions given below:
Triglycerides are the oils of plants and the fats of animal origin. They include such common substances as peanut oil, soya bean oil, corn oil, butter etc. Triglycerides that are liquid at room temperature are generally called oils; those that are solids are called fats. In simple triglycerides all three acyl groups are same and in mixed triglycerides acyl groups are different. Hydrolysis of oils and fats produces a mixture of fatty acids.
Most natural fatty acids have unbranched chains and because they are synthesized from two carbon units, they have an even number of carbon atoms. In most unsaturated fatty acids double bonds are in cis configuration. Many naturally occurring fatty acids contain two or three double bonds. The fats or oils that these come from are called polyunsaturated fats or oils.
The carbon chain of saturated fatty acids can adopt many conformations but tend to be fully extended because this minimizes steric repulsions between neighbouring methylene groups. Saturated fatty acids pack efficiently into crystals because of larger van der Waals’ forces these have high melting points. The cis configuration of the double bond of an unsaturated fatty acid puts a rigid bend in the carbon chain that interferes with crystal packing causing reduced van der Waals’ attraction between molecules.

1. The hydrolysis of fats or oils with NaOH is called

(A) esterification (B) trans esterification
(C) glycolysis (D) saponification

1. Consider the reaction,

📷
The product A on oxidation with HIO4 gives
(A) only formaldehyde (B) only formic acid
(C) both formaldehyde and formic acid (D) None of the above

1. Unsaturated fatty acids having trans configuration have

(A) higher boiling points than cis isomer
(B) higher melting points than cis isomer
(C) higher molecular mass than cis isomer
(D) All of the above

1. Aliphatic unsaturated fatty acids can be reduced to corresponding unsaturated alcohol by

(A) NaBH4 (B) LiAlH4
(C) H2/Pt (D) All of the above

1. The boiling points of branched chain fatty acids are less than straight chain isomers due to

(A) less steric hindrance (B) less van der Waals’ forces
(C) more van der Waals’ forces (D) None of the above
Match The Following

1. List – 1 (complex ion) List – 2 (number of unpaired electrons)

(A) [CrF6]-4 (1) one
(B) [MnF6]-4 (2) two
(C) [Cr(CN)6]-4 (3) three
(D) [Mn(CN)6]-4 (4) four
(5) five

1. Match List − I with List − II

List − I
List − II
(A)
Amino acid
(p)
Secondary structure
(B)
α-amino acid
(q)
Neutral
(C)
Albumin
(r)
Simple protein
(D)
α-helix structure
(s)
Ninhydrin

1. Match List − I with List − II

List − I
List − II
(A)
Epimers
(p)
Reserve food for animals
(B)
Sorbitol
(q)
Hetrogenous polysaccharides
(C)
Glycogen
(r)
By the reduction of glucose
(D)
Starch
(s)
Glucose and mannose

1. Match the compounds in List-I with their properties in List-II:

List-I List-II
(A) K2MnO4 (P) Transition element in +6 state
(B) KMnO4 (Q) Oxidising agent in acid medium
(C) K2Cr2O7 (R) Manufactured from pyrolusite ore
(D) K2CrO4 (S) Manufactured from chromite ore

1. Match the complexes in List-I with their information in List-II:

List-I List-II
(A) [Cu(NH3)2]SO4 (P) dsp2
(B) [Pt(NH3)2Cl2] (Q) Octahedral
(C) K4[Fe(CN)6] (R) sp3d2
(D) [Fe(H2O)6]Cl3 (S) Square planar
ANSWERS TO WORK BOOK
Objective:
Single Correct Questions

1. B 2. A 3. D 4. C
   
2. D 6. B 7. D 8. D
   
3. C 10. A 11. B 12. C
   
4. D 14. B 15. D 16. A
   
5. A 18. B 19. A 20. A
   
6. C 22. C 23. A 24. C
   
7. B 26. B 27. B 28. B
   
8. C 30. B 31. C 32. A
   
9. B 34. C 35. B 36. B
   
10. B 38. A 39. D 40. B
    
11. B 42. B 43. D 44. B
    
12. A 46. A 47. D 48. C
    
13. A 50. C
    

Multiple Correct Questions

1. A, B, C 52. A, B, C 53. A, C 54. A, B, D
   
2. A, B, D 56. A, C 57. A, B, D 58. B, C, D
   
3. C, D 60. A, B, D 61. A, B, D
   

Comprehension

1. C 63. B 64. A 65. D
   
2. C 67. A 68. A 69. C
   
3. A 71. C 72. C 73. B
   
4. B 75. D 76. B 77. D
   
5. C 79. B 80. B 81. B
   

Match The Following

1. (A) – (4), (B) – (5), (C) – (2), (D) – (1)
   
2. (A) –(q), (B) – (s), (C) – (r), (D) – (p)
   
3. (A) –(s), (B) – (r), (C) – (p), (D) – (q)
   
4. (A-P, R) (B-Q,R) (C-P, Q,S) (D-P,S)
   
5. (A-P,S) (B-P,S) (C-Q) (D-Q,R)
   

https://docs.google.com/document/d/1IEwSIKqd51VapPCH8DekF0I25ZWeml_4/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true
Work Book (Phase - VII)
Subjective:
Board Type Questions

1. A sample of lead weighing 1.05 g was dissolved in a small quantity of nitric acid to produce aqueous solution of Pb2+ and Ag+ (which is present as impurity). The volume of the solution was increased to 300 ml by adding water, a pure silver electrode was immersed in the solution and the potential difference between this electrode and a standard hydrogen electrode was found to be 0.503 V at 25°C. What was the % of Ag in the lead metal? Given 📷=0.799 V. Neglect amount of Ag+ converted to Ag.
   
2. Specific conductance of a decinormal solution of KCl is 0.0224 📷. The resistance of a cell containing the solution was found to be 64. What is the cell constant?
   
3. Calculate the electrode potential of a copper electrode dipped in a 0.1 M solution of copper sulphate at 📷 the standard electrode potential of 📷 system is 0.34 volts at 298 K.
   
4. Given:
   

I. 📷
II. 📷
Find E0 for the half reaction:
📷

1. NaOH prepared by the electrolysis of NaCl(aq). If current efficiency is 80% then determine the quantity of electricity required to convert 20g NaCl in NaOH
   
2. Iron is corroded by atmospheric oxygen under acidic condition to product Fe2+ (aq) ions initially. The standard reduction potential
   

E0Fe2+/Fe = −0.44 V and for the reaction H2O (📷) ⎯→ 2H+ (g) + 📷
E0 = −1.23 V
Find whether the formation of Fe2+ (aq) is thermodynamically favorable or not.

1. The emf of the cell reaction,

📷
Calculate the entropy change. Given that enthalpy of the reaction is − 216.7 KJ mol−1
And 📷 and 📷 = + 0.34V.

1. Arrange HCl, LiCl, NaCl, KCl and RbCl in order of increasing 📷value in aqueous solution.
   
2. (a) AlCl3 is covalent but AlCl3⋅6H2O is ionic. Why?
   

(b) Explain why BaSO4 is insoluble in water through it is basic.

1. What happens when aluminium forms [AlF6]3– but boron does not form [BF6]3–.
   
2. What happens when?
   

Methyl chloride is passed over silicon in the presence of copper catalyst at 275 − 375°C.

1. Why ‘Al’ extraction from Bauxite is not possible by carbon reduction method?
   
2. Identify (A) based on following facts:
   

(i) (A) reduces HgCl2 solution to white ppt. changing to grey.
(ii) (A) turns FeCl3 yellow colour solution to green.
(iii) (A) give white ppt. with NaOH soluble in excess of NH4OH
(iv) (A) gives yellow dirty ppt. on passing H2S gas soluble in yellow ammonium sulphide.
(v) A gives chromyl chloride test.

1. Give reason why the hydroxides of aluminium and iron are insoluble in water. However, NaOH is used to separate one from the other?
   
2. In the reaction H2+I2 ⎯→ 2HI the rate of disappearance of I2 is found to be 10-6 mole per litre per second. What would be the corresponding rate of appearance of HI?
   
3. Rate of reaction A + B ⎯→ product is given as a function of different initial concentration of A and B Determine the order of reaction with respect to A and with respect to B. What is the value of rate constant?
   

A
(mole L-1)
(B)
(mole L-1)
Initial rate
(mole L-1 min-1)
0.01
0.02
0.01
0.01
0.01
0.02
0.005
0.010
0.005

1. For the formation of phosgene from CO(g) and CO(g) + Cl2(g) ⎯→ COCl2(g),

The experimentally determined rate law is, 📷 K′[CO] [Cl2]3/2, is the following mechanism consistent with the equation?
(i) Cl2 📷 2Cl (fast)
(ii) Cl + CO 📷 COCl (fast)
(iii) COCl + Cl2 📷 COCl2 +Cl(slow)

1. A first order reaction have rate constant of 1.5×10-4 sec-1. How long will 5.0 gm of this reactant take to reduce to 3.0 gm?
   
2. In the Arrhenius’ equation for a certain reaction, the value of A and Ea are 4×1013 sec-1 and 98.6 KJ mol-1 respectively. If the reaction is of first order at what temperature will its half life period be ten minutes?
   
3. The gas phase decomposition 2N2O5 ⎯→ 4NO2+O2 follows the first orde rate law. At given temperature rate constant of the reaction is 7.5×10-3 sec-1. The initial pressure of N2O5 is 0.1mol-1. Calculate the time of decomposition of N2O5 so that the total pressure becomes 0.15 atm
   
4. Calculate the number of α and β particles emitted when 92U238 changes into radioactive 82Pb206
   
5. A radioisotope has t1/2 = 5 years. After a given amount of decays for 15 years, what fraction of the original isotope remains?
   
6. The half life period of 📷 is 140 days. In how many days 1 gm of this isotope is reduced to 0.25 gm?
   
7. From the following data for the reaction between A and B
   

Expt. (A) (B) Initial rate at
(mol L-1) (mol L-1) 300 K
(i) 2.5×10-4 3.0×10-5 5.0×10-4
(ii) 5.0×10-4 6.0×10-5 4.0×10-3
(iii) 1.0×10-3 6.0×10-3 1.6×10-2
(i) Calculate the order of the reaction with respect to A and B
(ii) Rate constant of the reaction at 300 K

1. A substance reacts according to I order Kinetics and rate constant for the reaction is 1×10-2 sec-1. If its initial concentration is 1 M

(a) What is initial rate?
(b) What is rate after 1 minute?

1. The beta activity of 1 gm of carbon made from green wood is 15.3 counts per minutes. If the activity of 1 gm of carbon derived form the wood of an Egyptian mummy ease in 9.4 counts per minutes under the same conditions, how old is the wood of the mummy case (t1/2 of C14 = 5770 gm)
   
2. Differentiate between:
   

(a) Artificial transmutation and nuclear fission.
(b) Nuclear fission and nuclear fusion.
(c) Chemical reaction and nuclear reaction.

1. Define rate of a reaction.
   
2. For the hypothetical reaction:
   

📷write the rate equation in term of the disappearance of B2 and formation of AB2.

1. In the reaction

📷
Express the rate of disappearance of Br− in terms of formation of Br2.

1. Define specific rate constant.
   
2. For the reaction 📷, the rate constant is 1.26 × 10−3 L mol−1s−1. What is the order of the reaction?
   
3. The gas – phase decomposition of acetaldehyde.
   

📷 follows the rate law 📷
If pressure is in atmosphere and the time is in minutes. Find the units for (i) rate of reaction (ii) rate constant.

1. The rate law for the reaction

📷
📷
Which is the order of a reaction in the absence of HBr?

1. Give any one example of

(i) zero order reaction
(ii) first order reaction

1. Write the integrated rate equation for a zero order reaction.
   
2. For the decomposition of NH3 on a finely divided Pt surface is represented as 📷 with the rate equation
   

📷
Write the rate equation and order of a reaction when
(i) [NH3] is very slow.
(ii) [NH3] concentration is very high.

1. The following graph is a plot of the rate of reaction versus concentration of the reactant. What is the order of the reaction?

📷

1. Define half-life of a reaction.
   
2. The following graph is a plot of t1/2 and concentration. What is the order of the reaction?
   

📷
IIT Level Questions

1. A hydrogenation reaction is carried at 500 K. If the same reaction is carried in presence of catalyst at the same rate the temperature required is 400 K. Calculate the activation energy of the reaction if the catalyst lowers the activation energy by 20 KJ mol-1
   
2. A radioisotope = Am(t1/2=10days) decays to give Bm-12 (stable atom) along with α - particles. If m gm of A are taken and kept in a sealed tube how much ‘He’ will accumulate in 20 days at STP?
   
3. A small amount of solution containing Na24 radionuclide with activity A = 2×103 dps was administered into blood of a patient in hospital, after five hour sample of blood drown from the patient showed activity of 16 dpm per cc. Find the volume of the blood in the patient . (t1/2Na24), 15 hr
   
4. The EMF of the cell Pt H2 (1 atm), HA (0.1 M, 30 ml) Ag+ (0.8 M) Ag is 0.9V. Calculate the EMF when 0.05 M NaOH (40 ml) is added.(📷,antilog of −2.2 = 6.3×10-3, log 2 = 0.3010, log 2.5 = 0.3979, log 1.56 = 0.194) [6]
   
5. Find the solubility product of a saturated solution of Ag2CrO4 in water at 298K if the emf of the cell: Ag Ag+ (sat Ag2CrO4 solution) Ag+ (0.1 M) Ag is 0.164 at 298K
   
6. An inorganic compound (A) shows the following reaction.
   

(i) It is white solid and exists as dimer; gives fumes of (B) with wet air.
(ii) It sublimes in 180°C and form monomer if heated to 400°C.
(iii) Its aqueous solution turns blue litmus to red.
(iv) Addition of NH4OH and NaOH separately to a solution of (A) gives white ppt. which is however soluble in excess of NaOH.

1. Why we prefer instantaneous rate of reaction over average rate of reaction?
   
2. Differentiate between the rate of reaction and the rate constant.
   
3. Define order of a reaction. Can it be a fractional value? If yes then give an example of a fractional order reaction.
   
4. A first order reaction is 60 percent complete in 100 minutes. Find the time when 90 percent of the reaction will be completed.
   

Objective:
Multiple choice questions with single correct options

1. If in the fermentation of sugar in an enzymatic solution that is 0.12 M, the concentration of the sugar is reduced to 0.06 M in 10 h and to 0.03 M in 20 h, what is the order of the reaction

(A) 1 (B) 2
(C) 3 (D) 0

1. For gaseous reactions, the rate is often expressed in terms of dP/dt instead of dC/dt or dn/dt (where C is concentration and n the number of mol). What is relation among these three expressions?

(A) 📷 (B) 📷
(C) 📷 (D) None is correct

1. Milk turns sour at 40oC three times as faster at as 0oC. Hence Eα in cal (activation energy) of turning of milk sour is:

(A) 📷 (B) 📷
(C) 📷 (D) 📷

1. For the reaction

📷
net rate is
📷
📷
ratio of rate constants of the forward and backward reactions is
(A) 0.66 × 10−9 mol L−1 (B) 1.5 × 10−9 mol L−1
(C) 2.0 × 10−4 s−1 (D) 3.0 × 105 L mol−1 s−1

1. A G.M. counter is used to study the radioactive process. In the absence of radioactive substance A, it counts 3 disintegration per second (dps). At the start in the presence of A, it records 23 dps; and after 10 minutes 13 dps,

(1) What does it count after 20 minutes?
(2) What is half-life of A?
(a) 8 dps, 10 min (B) 5 dps, 10 min
(C) 5 dps, 20 min (D) 5 dps, 5 min

1. Rate of formation of SO3 in the following reaction

📷
is 100 kg min−1. Hence rate of disappearance of SO2 will be
(A) 100 kg min−1 (B) 80 kg min−1
(C) 64 kg min−1 (D) 32 kg min−1

1. Following is the graph between (a − x)−1 and time t for second order reaction θ = tan−1(0.5) OA = 2 L mol−1

📷
hence rate at the start of the reaction is
(A) 1.25 L mol−1min−1 (B) 0.5 mol L−1 min−1
(C) 0.125 L mol−1 min−1 (D) 1.25 L mol−1 min−1

1. 📷

📷
Temperature T/K at which (kA = kC) is
(A) 1000 K (B) 2000 K
(C) 📷 (D) 📷
9.
📷
Reaction kinetics can be studied by
(A) measurement of pH (B) titration with hypo after adding KI
(C) both correct (D) none is correct

1. The rate of a chemical reaction generally increases rapidly even for small temperature increase because of a rapid increase in the

(A) collision frequency
(B) fraction of molecules with energies in excess of the activation energy
(C) activation energy
(D) average kinetic energy of molecules

1. During the electrolysis of aqueous zinc nitrate

(A) Zinc plates out at the cathode
(B) Zinc plates out at the anode
(C) Hydrogen gas, H2 is evolved at the anode
(D) Oxygen gas, O2 is evolved at the anode

1. If 📷 is x2 then 📷will be

(A) 3x2 – 2x1 (B) x2 – x1
(C) x2 + x1 (D) 2x1 + 3x2

1. 📷 Reaction quotient is

📷 Variation of Ecell with log Q is of the type with OA = 1.10 V Ecell will be 1.1591 V when
📷
(A) 📷 (B) 📷
(C) 📷 (D) 📷

1. For the half cell

📷
At pH = 2, electrode potential is
(A) 1.36 V (B) 1.30 V
(C) 1.42 V (D) 1.20 V

1. Given 📷

Which of the following statements is/are correct?
(A) AgNO3 can be stored in copper vessel
(B) Cu(NO3)2 can be stored in magnesium vessel
(C) CuCl2 can be stored in silver steel
(D) HgCl2 can be stored in copper vessel

1. The cell EMF is independent of concentration of the species of the cell in

(A)
📷
(B) Pt⎪(H2)⎪HCl⎪Pt(Cl2)
(C) Zn⎪Zn2+⎪⎪Cu2+⎪Cu
(D) Hg, Hg2Cl2⎪KCl⎪⎪AgNO3⎪Ag

1. Number of electrons lost during electrolysis of 0.355 g of Cl− is

(A) 0.01 (B) 0.01No
(C) 0.02 No (D) 📷

1. During electrolysis of acidified water, O2 gas is formed at the anode. To produce O2 gas at the anode at the rate of 0.224 c.c. per second at STP, current passed is

(A) 0.224 A (B) 2.24 A
(C) 9.65 A (D) 3.86 A

1. 📷0.275 g of metal M is deposited at the cathode due to passage of 1 A of current for 965 s. Hence atomic weight of the metal M is

(A) 27.5 (B) 55.0
(C) 110.0 (D) 13.75

1. Same quantity of current is passed through molten NaCl and molten cryolite containing Al2O3. If 4.6 g of sodium were liberated in one cell, the mass of aluminium liberated in other cell was

(A) 0.9 g (B) 2.7 g
(C) 1.8 g (D) 3.6 g

1. 100 mL of a buffer of 1 M NH3(aq) and 1 M 📷(aq) are placed in two voltaic cells separately. A current of 1.5 A is passed through both cells for 20 minutes. If electrolysis of water only takes place

📷
📷then
pH of the
(A) LHS half-cell will increase (B) RHS half-cell will increase
(C) both half-cells will increase (D) both half-cells will decrease

1. If a steady current of 15.0 is passed through an aqueous solution of CuSO4, how many minutes will it take to deposit 0.250 mol of Cu at the cathode, assuming 100% efficiency?

(A) 3.217 × 103 (B) 1.613 × 103
(C) 53.62 (D) 0.893

1. Cost of electricity for the production of x LH2 at NTP at cathode is Rs. x, then cost of electricity for the production of x LO2 gas at NTP at anode will be (assume 1 mol of electrons as one unit of electricity)

(A) 2x (B) 4x
(C) 16x (D) 32x

1. Which has maximum conductivity?

(A) [Cr(NH3)3Cl3] (B) [Cr(NH3)4Cl2]Cl
(C) [Cr(NH3)5Cl]Cl2 (D) [Cr(NH3)6]Cl3

1. One litre of 1 M CuSO4 solution is electrolysed. After passing 2F of electricity, molarity of CuSO4 will be

(A) 📷 (B) 📷
(C) M (D) 0

1. Borax crystal contains

(A) Two triangular units (B) Two tetrahedral units
(C) (A) and (B) both (D) None of these

1. Borax is converted into B by steps

Borax 📷H3BO3 📷B2O3 📷B
I and II reagents are
(A) acid, Al (B) acid, C
(C) acid, Fe (D) acid, Mg

1. On strong heating lead nitrate gives

(A) PbO, NO,📷 (B) PbO, NO, 📷
(C) 📷PbO,📷 (D) 📷

1. The element with the highest first ionisation potential is

(A) Boron (B) Carbon
(C) Nitrogen (D) Oxygen

1. Among LiCl, BeCl2, MgCl2 and RbCl, the compound with greatest and least ionic character, respectively are

(A) LiCl and RbCl (B) RbCl and BeCl2
(C) RbCl and MgCl2 (D) MgCl2 and BeCl2

1. Which of the following statements is wrong?

(A) Anhydrous AlCl3 exists as Al2Cl6
(B) Anhydrous AlCl3 is a trigonal planar molecule
(C) Anhydrous AlCl3 fumes in air
(D) Anhydrous AlCl3 is ionic

1. The basis for Al as a reducing agent in thermite process is

(A) High exothermic hydration of Al3+
(B) High Eox. Pot. of Al
(C) High exothermic enthalpy of formation of Al2O3
(D) High I.E. of Al

1. The structures of quartz, mica and asbestos have the same common units of

(A) (SiO4)−4 (B) (SiO3)−2
(C) 📷 (D) 📷

1. Choose the wrong statement:

(A) Ge, Sn and Pb form +2 as well as +4 O.S.
(B) Ge2+ and Sn2+ compounds act as reducing agents
(C) Pb2+ compounds act as oxidizing agents
(D) Stability decreases from Ge2+ to Pb2+

1. Identify the correct statement wiht respect to carbon monoxide.

(A) It combines with water to form carbonic acid
(B) It reacts with haemoglobin in red blood cells
(C) It is a powerful oxidising agent
(D) It is used to prepare aerated drinks

1. The ratio of the rate constant of a reaction at any temperature T to the rate constant T → ∞ is equal to

(A) Energy of activation of the reaction
(B) Fraction of molecules in the activated state
(C) Average life of the reaction
(D) Pre-exponential factor in the Arrhenius equation

1. The rate constant of a reaction: A ⎯→ B + C at 27°C is 3.0 × 10–5 s–1 at 27°C and at this temperature 1.5 × 10–4 percent of the reactant molecules are able to cross-over the P.E. barrier. The maximum rate constant of the reaction is

(A) 4.5 × 10–9s–1 (B) 4.5 × 10–11 s–1
(C) 0.2 s–1 (D) 20 s–1

1. For a first order reaction: A 📷 B, whose concentration vs. time curve is as shown in the figure. The rate constant is equal to

(A) 41.58 h–1 (B) 4.158 s–1
(C) 1.155 × 10–3 s–1 (D) 6.93 min–1
📷

1. If ΔH of a reaction be positive and k1 andk2 be the rate constants of forward reaction and backward reaction, respectively, at temperature t°C k′1 and k′2 be the respective rate constants at (t + 10)°C then

(A) 📷 (B0 📷
(C) 📷 (D) None of these

1. Among ΔH, equilibrium constant (K), rate constant and Ea the one or more than one that is (are) not affected by catalyst is (one)

(A) Rate constant and ΔH (B) K and Ea
(C) ΔH and K (D) Only K

1. For the gaseous reaction 📷 is found to be first order with respect to A. If at the starting the total pressure was 100 mm Hg and after 20 minutes it is found to be 400 mm Hg. The rate constant of the reaction is:

(A) 📷 (B) 📷
(C) 📷 (D) 📷

1. The decomposition of ozone is believed to occur by the mechanism:

📷, 📷
When the concentration of 📷 is increased, then the rate
(A) increases (B) decreases
(C) remains same (D) cannot be predicted

1. In the sequence of the reaction, 📷 given that 📷 then the rate determining step of the reaction is:

(A) 📷 (B) 📷
(C) 📷 (D) 📷

1. What will be the fraction of molecule having energy equal to or greater than activation energy Ea?

(A) K (B) A
(C) 📷 (D) 📷

1. The ratio of the atoms of 📷 and 📷 in a sample of rock is found to be 4:1. The 📷 of 📷 is 📷 years. How long ago the solidification of the rock could have occurred? (log 2 = 0.30)

(A) 📷 (B) 📷
(C) 📷 (D) 📷

1. For the reaction, 📷

The kinetic data are as given below:
📷 📷 📷
0.2 0.1 📷
0.4 0.1 📷
0.4 0.2 📷
If 📷 for the above reaction is 📷, then the specific reaction rate 📷for the replacement of −OH group of methanol by Cl atom is:
(A) 📷 (B) 📷
(C) 📷 (D) 10

1. 📷and ΔH values of reactions 📷carried out at the same temperature are as given below:

📷
📷
📷
📷
At a given temperature and assuming that the backward reactions of all these reactions have the same frequency factor, the rates of 📷 in their respective backward reactions are in the increasing order of
(A) 📷 (B) 📷
(C) 📷 (D) 📷

1. For the radioactive sequential chain reaction, 📷

for the steady state of B, assuming that the no. of atoms of A is half of the no. of atoms of B then if the half life of B is 2 hours then the half of life of A (in hours) will be
(A) 4 (B) 2
(C) 1 (D) 3

1. For the given reaction

📷
where K is zero order rate constant and b is a first order rate constant concentration of B at any time can be given by the expression.
(A) 📷 (B) 📷
(C) 📷 (D) can’t be expressed

1. If 10–5% reactant molecule is crossing over the barrier in transition state at 298K, then find out the activation barrier needed to cross the transition state.

(A) 39.94 kJ (B) 49.94kJ
(C) 79.94kJ (D) 97.97 kJ
Multiple choice questions with more than one option correct

1. Which of the following statements are correct about half-life period?

(A) it is proportional to initial concentration for zeroth order
(B) average life = 144 half-life for first order reaction
(C) time of 75% reaction is thrice of half-life period in second order reaction
(D) 99.9% reaction takes place in 100 minutes for the case when rate constant is 0.0693 min−1

1. 📷

Half-life period is independent of concentration of zinc at constant pH. For the constant concentration of Zn, rate becomes 100 times when pH is decreased from 3 to 2. Hence
(A) 📷
(B) 📷
(C) rate is not affected if concentration of zinc is made four times and that of H+ ion is halved
(D) rate becomes four times if concentration of H+ ion is doubled at constant Zn concentration
53.
📷
Which of the following statements are correct?
(A) it is unimolecular nucleophilic substitution reaction 📷if I or II is formed.
(B) it is bimolecular nucleophilic substitution reaction 📷if I or II is formed.
(C) it is 📷 if I and its enantiomer are formed so that mixture is racemic.
(D) it is 📷if II is formed.

1. Rate constant k varies with temperature by equation

📷. We can conclude
(A) pre-exponential factor A is 5 (B) Eα is 2000 kcal
(C) pre-exponential factor A is 105 (D) Eα is 9.212 kcal

1. A reaction is catalysed by H+ ion. In presence of HA, rate constant is 2 × 10−3 min−1 and in presence of HB rate constant is 1 × 10−3 min−1, HA and HB both being strong acids, we may conclude

(A) equilibrium constant is 2
(B) HA is stronger than HB
(C) relative strength of HA and HB is 2
(D) HA is weaker than HB and relative strength is 0.5

1. Electrode potentials of the given half cells

Pt(H2)⎪H+(C1); Pt(Cl2)⎪Cl−(C2); Ag⎪Ag+(C3)
p1 (I) p2 (II) (III)
(A) will increase on increasing C1, C2 and C3
(B) will decrease on increasing C1, C2 and C3
(C) will decrease on increasing C1 and C3 and increase on increasing C2
(D) will remain constant if C1 or C2 is doubled and p1 or p2 is made four times

1. There is blue colour formation if

(A) Cu electrode is placed inside AgNO3 solution
(B) Cu electrode is placed inside ZnSO4 solution
(C) Cu electrode is placed inside dil HNO3
(D) Cu electrode is placed inside dil H2SO4

1. The formation of rust on the surface of iron occurs through the reactions

(A) 📷at anode
(B) 📷at cathode
(C) 📷
(D) 📷

1. In which case 📷is zero

(A) Cu⎪Cu2+(0.001 M)⎪⎪Ag+(0.1 M)⎪Ag
(B) Pt(H2)⎪pH = 1⎪⎪Zn2+(0.01 M)⎪Zn
(C) Pt(H2)⎪pH = 1⎪⎪Zn2+(1 M)⎪Zn
(D) Pt(H2)⎪H+(0.01 M)⎪⎪Zn2+(0.01 M)⎪Zn

1. Reactions taking place in a fuel cell are

(A) 📷at the cathode
(B) reaction in (a) at the anode
(C) 📷at the anode
(D) reaction in (c) at the cathode

1. It is not advisable to

(A) stir sugar solution with a steel spoon
(B) stir copper sulphate solution with a silver spoon
(C) stir copper sulphate solution with zinc spoon
(D) stir silver nitrate solution with a copper spoon

1. By passage of 1 F of electricity

(A) 1 mol of Cu is deposited (B) 0.5 mol of Mg is deposited
(C) 9 g of Al is deposited (D) 5.6 L of O2 gas evolved at anode

1. In the following electrochemical cell

📷
📷. This will be when
(A) [Zn2+] = [H+] = 1 M and pH2 = 1 atm
(B) [Zn2+] = 0.01 M, [H+] = 0.1 M and pH2 = 0.01 atm
(C) [Zn2+] = 1 M, [H+] = 0.1 M and pH2 = 0.01 atm
(D) [Zn2+] = [H+] = 0.1 M and pH2 = 0.1 atm

1. Boron readily dissolves in

(A) conc. HCl
(B) fused NaOH at 673 K
(C) fused Na2CO3 + NaNO3 at 1173 K
(D) a mixture of conc. HNO3 and conc. H2SO4 (1 : 2)

1. The allotropic forms of carbon, good conductor of electricity are

(A) diamond (B) graphite
(C) fullerenes (D) gas carbon
Comprehension-I
Three students performed three different experiments on electrolysis. Student ‘A’ electrolysied one litre of 1M aqueous solution of KMnO4 till after reduction. The final solution is 0.1 M K2MnO4, student ‘B’ electrolyzed 0.5 L of 2M Ni(NO3)2 solution by passing a current of 9.65 A for 3 hours between nickel electrodes. Student ‘C’ electrolysed NiSO4 solution by passing 12 ampere current but the current efficiency was only 60%.
Read the above paragraph carefully and answer the questions given below it:

1. The quantity of electricity used by student A is

(A) 0.1 F (B) 1F
(C) 10F (D) 100F

1. The molality of the solution left after electrolysis used by student ‘B’ is

(A) 0.46 M (B) 0.625 M
(C) 0.92M (D) 1.25M

1. The nickel deposited on the cathode per hour in the solution used by student ‘C’ will be

(A) 7.883 g (B) 3.941g
(C) 5.91 g (D) 2.645 g
Comprehension-II
Order is an experimental fact for a reaction and molecularity is defined for elementary reactions and depends upon stoichiometric coefficient of a balanced reaction. Order can be fractional but molecularity is always a whole number. For elementary reaction order and molecularity can’t be differentiated.
Read the above paragraph carefully and answer the questions given below it:

1. For the multistep reaction

📷 and 📷
Molecularity of reaction
(A) Is n1 + n2 (B) Is 📷
(C) Is n1 + 2n2 (D) Can’t be defined

1. Order of reaction in question number 1

(A) Is n1 + n2 (B) Is 📷
(C) Is n1 + 2n2 (D) Can’t be predicted

1. For the elementary reaction, 📷

Molecularity is
(A) 1.5 (B) 3
(C) 0.5 (D) 1

1. For the reaction is question number 3, order of reaction is

(A) 3 (B) 1.5
(C) 0.5 (D) 1

1. For chlorination of CH4 in which rate depends upon light only, order and molecularity are …………….. respectively.

(A) Two and zero (B) Zero and two
(C) Zero and one (D) One and zero
Comprehension-III
An organic Lewis acid (A) which gives fumes in moist air and intensity of fumes is increased when a rod dipped in NH4OH is brought near to it. An acidic solution of (A) on addition of NH4Cl and NH4OH gives a precipitate (B) which dissolves in NaOH solution. An acidic solution of A does not gives precipitate with H2S.
Read the above paragraph carefully and give the answer of following questions:

1. What will be structure of (A)?

(A) Planar (B) Trigonal planar
(C) Pyramidal (D) Octahedral

1. What will be (A) in above sequence?

(A) BF3 (B) AlCl3
(C) GaCl2 (D) ZnCl2

1. The intensity of fumes increase due to the formation of–

(A) NH4OH (B) HCl
(C) NH4Cl (D) NaAlO2

1. The soluble product obtained when the precipitate (B) dissolves in NaOH, is:

(A) Al2O3 (B) NaAlO2
(C) Na2AlO2 (D) None

1. Chloride of compound (A) does not show 📷 bonding because:

(A) Absence of vacant orbital (B) Larger size of compound (A)
(C) Smaller size of compound (A) (D) None
Comprehension-IV
A metal is regarded as an assembly of metal ions and free electrons. When the metal is in contact with water, some metal ions enter into the liquid due to a tendency in the metal, called by Nernst as “Electrolytic solution tension”. An alloy of 1.45 gm Ag and Si was dissolved in desired amount of HNO3 and volume made upto 500 ml. An Ag electrode was dipped in solution and Ecell of the cell is 📷
Read the above paragraph carefully and answer the questions given below:

1. Observed concentration of Ag+ in the cell

(A) 2.25 × 10–9 M (B) 9.25 × 10–6 M
(C) 9.25 × 10–7 M (D) 9.25 × 10–5 M

1. % weight of Pb metal in above alloy sample is

(A) 99.25% (B) 89.95%
(C) 99.96% (D) 82.35%

1. The emf of the cell involving the reaction, 📷 is 0.80 V. The standard oxidation potential of Ag electrode is

(A) 0.80 V (B) –0.80 V
(C) 0.40 V (D) 0.20 V

1. In the above cell, reaction cathodic cell reaction is

(A) 📷 (B) 📷
(C) Both (A) and (B) (D) None of the above

1. Which of the following statement is correct?

(A) Oxidation number of oxygen in KO2 is +1.
(B) The specific conductance of an electrolytic solution decreases with increase in dilution.
(C) Sn2+ oxidizes Fe3+.
(D) Zn/ZnSO4 is a reference electrode.
Match The Following

1. Match List-I with List-II

List-I List-II
(A) XeF2 (P) Square pyramidal
(B) PCl5 (Q) Linear shape
(C) XeOF4 (R) sp3d hybridization
(D) XeF4 (S) sp3d2

1. Match List-I with List-II:

List-I List-II
(A) He (P) London dispersion force
(B) Xe (Q) Lowest boiling noble gas
(C) XeF6 (R) sp3d3
(D) IF7 (S) Distorted octahedral

1. Match List–I and List–II.

LIST−I
LIST−II
(a)
Rate constant has the same unit as the rate of reaction.
(i) One
(b)
Reactions having apparent molecularity more than three.
(ii) Zero order reaction
(c)
Reactions having molecularity two but order of reaction is one.
(iii) Complex reaction
(d) The syum
For a reaction, A → B, the rate of reaction doubles as the concentration of A is doubled.
(iv) Pseudo unimolecular reaction
a b c d
(A) ii iv iii i
(B) ii iii iv i
(C) iii ii iv i
(D) ii iv i iii

1. Match list – I with list – II and choose the correct answer from the code given below

List – I List – II
(a) Half life period of first order reaction (i) 📷
(b) Molecular concentration (ii) 📷
(c) Temperature coefficient (iii) active mass
(d) Half life period of zero order reaction (iv) 📷
Code is
a b c d
(A) (ii) (iii) (i) (iv)
(B) (i) (ii) (iii) (iv)
(C) (ii) (iii) (iv) (i)
(D) (iii) (iv) (i) (ii)

1. Match List-I with List-II and List-III:

List-I List-II List-III
(Quantity) (Symbol) (Unit)
(a) Conductivity (p) 📷 (u) mho cm-1
(b) Cell constant (q) 📷 (v) cm-1
(c) Molar conductance (r) 📷 (w) ohm-1 cm2 mol-1
(d) Equivalent conductance (s) l/A (x) ohm-1 cm2 eq-1

1. Match the salts in Column-I with their use in Column-II:

Column-I Column-II
(a) Hg2Cl2 (p) Salt bridge
(b) Agar-agar (q) Calomel electrode
(c) 0.1 N KCl (r) Used in ice cream
(d) Quinhydron (s) Redox electrode
ANSWERS TO WORK BOOK
Subjective:
Board Type Questions

1. Percentage of Ag = 0.0296%
   
2. 1.4336 cm-1
   
3. 0.31045 volt
   
4. 1.70 V
   
5. 41239.4 C
   
6. The reactions are
   

(i) Fe (s) ⎯→ Fe+2 (aq) + 2e–, 📷
(ii) 2H+ + 📷(g) + 2e− ⎯→ H2O (l), E0 = +1.23V
Adding equation (i) and (ii), we get
Fe (s) + 2H+ + 📷(g) ⎯→ Fe+ (aq) + H2O (l), 📷 = 1.67V 📷
∴ 📷 = positive = 1.67V
∴ ΔG0 is negative. So, the reaction is thermodynamically favourable or spontaneous.

1. − 14.76 J K−1 mol−1.
   
2. 📷12. 25 mL of a solution of HCl (0.1M) is being titrated potentiometrically against 0.1 M NaOH solution using a hydrogen electrode as the indicator electrode and saturated calomel electrode (SCE) as the reference electrode. What would be the EMF of the cell initially and after the addition of 20 mL of alkali at 25°C? Given Reduction potential of SCE = 0.2422V. [log 9 = 0.95].
   
3. (b) It is because of the presence of polar water molecules which stabilizes Al3+ ion and increases the ionic character of Al–Cl bonds.
   

(c) As the hydration energy of BaSO4 is less than its lattice energy, it is insoluble in water.

1. Maximum coordination of boron is four as it does not have d−orbitals while the maximum coordination numbers of aluminium is 6. Thus, Al forms [AlF6]3– ion while boron does not form [BrF6]3– ion.
   
2. 2CH3Cl + Si 📷 (CH3)2SiCl2
   
3. Because Al is highly active metal and its affinity towards oxygen is very high. At high temperature, reacting with carbon it converts into Al4C3 instead of free metal.
   

📷

1. A gives dirty yellow ppt. soluble in yellow ammonium sulphide

∴ (A) has Sn2+
A gives chromyl chloride test of Cl–
∴ (A) is SnCl2
Reaction
📷
📷
📷
📷
📷
📷
📷

1. Al(OH)3 is soluble in NaOH forming NaAlO2, Fe(OH)3 is insoluble.
   
2. 📷
   
3. K = 📷
   
4. Multiplying equation (ii) by 2 and adding (i) we get
   

Cl2 + 2CO 📷 2COCl ∴K1 = 📷
[COCl] = 📷 …(1)
∴ Slowest step is rate determining step hence
Rate = K1(COCl] [Cl2] …(2)
From equation (1) & (2) we get : Rate = K1📷
∴Rate = K′[CO] [Cl2]3/2

1. 📷
   
2. T = 311.2 K
   
3. 📷
   
4. No of α-particles = 8
   

No of β particles =6

1. t1/2 = 5 year

t = 15 year ∴ n = 📷
Let original amount be No and amount after time t be N
∴ N = 📷
Hence (⅛)th of the original amount left.

1. 280 days.
   
2. K = 2.66×108 lit mol-2 sec-1 at 300 K
   
3. (a) Initial rate (R1) = K [ ]1=1×10-2×(1) = 1×10-2
   

∴ Initial rate = 1×10-2 ml Lt -1 sec-1
(b) K = 📷
⇒ 1×10-2 = 📷 (t = 1 minute= 60 sec)
∴ (1-x) = 0.549
Rate after 1 minute (R2) = K× [ ]1 =10-2×0.549
= 5.49×10-3 mol Lt -1 Sec-1

1. t = 3920 years.
   
2. Rate of a reaction may be defined as the change in any one of the reactants or products per unit time.
   
3. 📷
   

📷

1. Rate of disappearance of Br−

= 📷
= 📷

1. It is defined as the rate of a chemical reaction when the concentration of each reactant appearing in the rate equation is taken as unity.
   
2. The units of rate constant is L mol−1s−1 or (mol L−1)−1s−1+. Equate this with general expression of (mol L−1)1−ns−1.
   

📷
∴ −1 = 1 − n or n = 2
The order of reaction = 2

1. (i) Rate of a reaction = atm (min)−1

(ii) Rate constant = atm−1/2 min−1

1. In the absence of Hbr, the rate equation is reduced to

📷
and the overall order is 📷

1. (i) 📷

r = k[NH3]°
(ii) 📷
r = k[NH4NO2]

1. 📷

where k = rate constant
[R]0 = initial concentration of the reactant
[R] = concentration of the reactant at time, t.

1. (i) At very low concentration of NH3, 1 + k2[NH3] ≈ 1.

∴ Rate = k1 [NH3] and order of reaction = 1
(ii) At very high concentration of NH3,
1 + k2[NH3] ≈ k2[NH3]
∴ 📷order of reaction = 0

1. As rate is independent of the concentration the order of the reaction is zero.
   
2. The half-life of a reaction is defined as the time required for the reaction to be 50% complete.
   

It is also defined as the time during which the concentration of reactant is reduced to one half of its initial concentration.

1. The order of the reaction is one, as t1/2 is independent of the concentration of the reactant.

IIT Level Questions

1. Ea = 100 KJ
   
2. Moles of He formed at STP= 3×3 = 9/4
   

Volume of He at STP = 9/4×22.4 =50.4 lt

1. V = 5.95×103 ml = 5.95 L
   
2. 0.99 V
   
3. Ksp = 2.287 × 10–12 mol3 liltre–3
   
4. (i) (A) is characteristic dimerised compound which sublimes in 180°C and forms monomer if heated to 400°C and thus (A) is (AlCl3)2 or Al2Cl6.
   

Al2Cl6(s) 📷Al2Cl4(s) 📷2AlCl3
(ii) It fumes with wet air
Al2Cl4 + 6H2O 📷2Al(OH)3↓ + 6HCl↑
(iii) 2AlCl3 + 6H2O 📷 2Al(OH)3↓ + 6HCl(s)
(iv) (A) gives white ppt. with NH4OH and NaOH, soluble in excess of NaOH
Al2Cl6 + 6NH4OH ⎯→ 2Al(OH)3 + 6NH4Cl
Al2Cl6 + 6NaOH ⎯→ 2Al(OH)3 + 6NaCl 📷2NaAlO2 + 2H2O

1. The rate of reaction decreases continuously with time except for a zero order reaction. Therefore, average rate of reaction has no significance for the reaction. But instantaneous rate of reaction for a given instant of time does not change with time.

48.
Rate of reaction
Rate constant
​
(i)
It is defined as the change in concentration of the reactant or product with time, each divided by its stoichimetric coefficient.
(i)
It is defined as the rate of chemical reaction when the concentration of each reactant appearing in the rate equation is taken as unity.
​
(ii)
It always has a unit of (conc)/time.
(ii)
Its units depends on the order of a reaction. For nth order reaction the units of rate constant = (conc)1−n time−1
​
(iii)
It depends on the initial concentration of the reactants.
(iii)
It is independent of the initial concentration of the reactants.

1. The order of a reaction is defined as the sum of the exponents to which each concentration terms is raised in the experimently derived rate equation. For example in the reaction

A + B + C ⎯→ Products and the rate law expression is
📷
Then, overall order of reaction = x + y + z.
Order of a reaction can be fractional value. An example of a fractional reaction is gas-phase decomposition of CH3CHO.
📷 and 📷

1. 📷

Objective:
Single Correct Questions

1. A 2. A 3. A 4. A
   
2. A 6. B 7. C 8. D
   
3. B 10. B 11. D 12. A
   
4. B 14. C 15. C 16. A
   
5. B 18. D 19. B 20. C
   
6. B 22. C 23. A 24. D
   
7. D 26. C 27. D 28. D
   
8. C 30. B 31. D 32. C
   
9. A 34. D 35. B 36. B
   
10. D 38. C 39. C 40. C
    
11. C 42. B 43. A 44. D
    
12. B 46. B 47. D 48. C
    
13. C 50. A
    

Multiple Correct Questions

1. A,B,C,D 52. B,C,D 53. C,D 54. C,D
   
2. B,C 56. C,D 57. A,C
   
3. A,B,C,D 59. A,B 60. A,C 61. C,D
   
4. B,C,D 63. A,B,C 64. C,D 65. B,D
   

Comprehension

1. A 67. C 68. A 69. D
   
2. C 71. B 72. A 73. B
   
3. B 75. B 76. C 77. B
   
4. B 79. B 80. C 81. A
   
5. C 83. B
   

Match The Following

1. (A-Q,R) (B-R) (C-P,S) (D-S) 85. (A-P,Q) (B-P) (C-R,S) (D-R)
   
2. B 87. A
   
3. (a−q−u), (b−s−v), (c−p−w ), (d−r−x)
   
4. (a − q), (b − p, r), (c − p), (d − s)
   

https://docs.google.com/document/d/140748SYaKhQzLVxSjONDIhBS6nfDdMjC/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true
Work Book (Phase - III)
Subjective:
Board Type Questions

1. How is anhydrous magnesium chloride prepared from magnesium chloride hexahydrate?
   
2. Explain why halides of Be dissolve in organic solvents while those of Ba do not.
   
3. Why potassium can not be obtained by the electrolysis of fused potassium chloride?
   
4. Explain why halides of beryllium fume in moist air but other alkaline earth metals halides do not.
   
5. A piece of burning magnesium ribbon continuous to burn in sulphur dioxide. Explain.
   
6. Explain
   

(a) Be does not react with water or steam.
(b) Mg does not react readily with water.

1. Blue colour of the solution of alkali metals in liquid ammonia fades on standing why?
   
2. Explain
   

(a) The hydroxides of alkaline earth metals are weaker bases than their corresponding alkali metals.
(b) Ca, Br, Sr are stored in paraffin but Be & Mg are not?

1. Account for the following

(a) 📷is insoluble but 📷 is fairly soluble in water.
(b) 📷is amphoteric while 📷 is basic.

1. Give reasons for following

(i) concentrated HNO3 turns yellow in sun light
(ii) A bottle of liquor ammonia should be cooled before opening
(iii) Nitrous oxide supports combustion more vigorously than air
(iv) NF3 is not hydrolysed but NCl3 is readily hydrolysed
(v) Red phosphorus is densor and chemically less reactive than yellow phosphorus
(vi) N2 is diatomic while phosphrus molecule is tetraatomic
(vii) PCl5 is ionic in solid state and conducts current in fused state
(viii)Orthophosphorus acid is diabsic while orthophosphoric acid is tribasic

1. What happens when?

(a) Phosphene is reacted with acidified cupric sulphate solution.
(b) Silver nitrate solution is treated with phosphene.
(c) Mercuric chloride solution is treated with phosphorous acid.

1. What happens when?

(a) Phosphorus acid is treated with an aqueous solution of silver nitrate.
(b) Carbon dioxide is passed through a concentrated aqueous solution of sodium chloride saturated with ammonia.
(c) Methyl chloride is passed over silicon in the presence of copper catalyst at 275− 375°C.

1. On the basis of the following reactions, identify A, B, C and D.

📷
📷
📷

1. Explain the following

(i) Iodine dissolves more in KI solution, than impure water
(ii) KHF2 is well known, where as KHCl2 or KHBr2 does not exist
(iii) A mixture of He and O2 is used for respiration for deep see divers

1. What happens when?

(i) Sulphur is boiled with caustic soda solution.
(ii) Sulphur dioxide is bubbled through aqueous solution of copper sulphate in presence of potassium thiocyanate.

1. 20 ml of a solution containing 0.2 g of impure sample of 📷 reacts with 0.316 g of 📷 (acidic). Calculate

(a) Purity of 📷.
(b) Volume of dry 📷 evolved at 📷C and 750 mm P.

1. What happens when?

(i) Chromium hydroxide is treated with hydrogen peroxide in the presence of NaOH.
(ii) Hydrazinc reacts with hydrogen peroxide.
(iii) Hydrogen peroxide reacts with 📷.
(iv) Sodium hypochlorite reacts with 📷.

1. In the reaction H2+I2 ⎯→ 2HI the rate of disappearance of I2 is found to be 10-6 mole per litre per second. What would be the corresponding rate of appearance of HI?
   
2. Rate of reaction A + B ⎯→ product is given as a function of different initial concentration of A and B Determine the order of reaction with respect to A and with respect to B. What is the value of rate constant?
   

A
(mole L-1)
(B)
(mole L-1)
Initial rate
(mole L-1 min-1)
0.01
0.02
0.01
0.01
0.01
0.02
0.005
0.010
0.005

1. Following equilibrium is setup when SCN− ion is added to Fe3+ in aqueous solution

📷
When silver nitrate is added to the solution, AgCN gets precipitated. What will happen to the equilibrium?

1. Which of the following reactions will get affected by increase in pressure? Also mention whether the change will cause the reaction to go to the right or left direction.

(i) 📷 (ii) 📷
(iii) 📷 (iv) 📷

1. Given these equilibrium constants at 25°C

S + S2– 📷 S22– K1 = 1.7
S + S22– 📷 S32– K2 = 3.1
Calculate K for 2S + S2– 📷 S32– at 25°C

1. At 540 K, 0.10 mole of PCl5are heated in a 8.0 L flask. The pressure of the equilibrium mixture is found to be 1.0 atm. Calculate Kp and KC for the reaction.
   
2. In the reaction C(s) + CO2(g) 📷CO2(g) the equilibrium pressure is 12 atm. If 50% of CO2 reacts calculate Kp.
   
3. Ammonium carbamate dissociates as: NH2COONH4 (s) 📷2NH3 (g) + CO2 (g)
   

In a closed vessel containing ammonium carbamate in equilibrium with its vapour ammonia is added such that the partial pressure of ammonia now equals to the original total pressure. Calculate the ratio of the total pressure now to the total pressure of the original mixture.

1. Calculate the equilibrium constant for reaction📷, if equilibrium constant is 100 atm-1 for equilibrium📷, at constant temperature
   
2. For the reaction📷. If 20.0 g of 📷were kept in a 10 lit container and heated up to📷, what percentage of 📷would remain unreacted at equilibrium.
   
3. N2O4 is 20% dissociated at 27°C and 1 atm pressure. Calculate (i) Kp and (ii) the percentage dissociation at 0.2 atm and 27°C
   
4. When 0.10 mole of NH3 is dissolved in water to make 1.0 L solution, the solution is found to have📷. Calculate Kb for NH3.
   
5. An acid - base indicator, HIn(KIn= 10-6) displays the colour of its acid form only when it is ionised to a maximum of 9.09%, while it displays its basic colour when it is ionised to the extent of at least 95.24%. What is the pH - range of the indicator?
   
6. When a solution containing Ca2+ and Mg2+ ions is at first saturated with NH4Cl(s) and then treated with (NH4)2CO3, only Ca2+ is precipitated as carbonate. Explain.
   
7. A weak acid type indicator was found to be 40% dissociated at pH = 9.1249. What be % dissociation at pH = 9?
   
8. Calculate pH of 10−8 M HCl.
   
9. A buffer of pH 9.26 is made by dissolving x mole of ammonium sulphate and 0.1 mole of ammonia into 100 mL solution. If pKb of ammonia is 4.74, calculate value of x.
   
10. Calculate [H+] in
    

(a) 0.01 M C6H5COOH solution and
(b) 0.01 M C6H5COOH in 0.01 M C6H5COONa; 📷

1. Predict the direction of the following in aqueous solution:

📷
📷

1. Calculate the pH at the equivalence point when a solution of 0.01 M CH3COOH is titrated with a solution of 0.01 M NaOH, pKa of CH3COOH is 4.74.
   
2. It is found that 0.1 M solution of three sodium salts NaX, NaY and NaZ have pH 7.0, 9.0 and 11.0 respectively. Arrange the acids HX, HY and HZ in order of increasing strength. Where possible, calculate ionization constant of acids. 32. Concentration of HCN and NaCN in a solution is 0.01 M each. Calculate📷 and [OH−] 📷
   
3. Concentration of HCN and NaCN in a solution is 0.01 M each. Calculate📷 and [OH−] 📷
   
4. Determine degree of dissociation of 0.05 M NH3 at 25°C in solution of pH = 11.
   

IIT Level Questions

1. Explain the following

(a) Alkali metals are obtained by the electrolysis of the molten salts and not by the electrolysis of their aqueous solutions.
(b) On exposure to air, sodium hydroxide becomes liquid and after sometimes it changes to white powder.
(c) An aqueous solution of iodine becomes colourless on adding excess of sodium hydroxide solution.

1. To a 25 ml H2O2 solution excess of acidify solution of potassium iodide was added. The iodine liberated required 20 ml of 0.3 sodium thiosulphate solution. Calculate the volume strength of H2O2 solution.
   
2. In a mixture of N2 and H2 initially in a mole ratio of 1:3 at 30 atm and 300oC, the percentage of ammonia by volume under equilibrium is 17.8. Calculate the equilibrium constant (KP) of the mixture, for the reaction, N2(g) + 3H2(g) 📷 2NH3(g)
   
3. An equilibrium mixture at 300 K contains N2O4 and NO2 at 0.28 and 1.1 atm respectively. If the volume of container is doubled, calculate the new equilibrium pressures of the two gases.
   
4. The value of Kc for the reaction 📷is 0.50 at 400°C. Find the value of Kp at 400°C when concentrations are expressed in 📷 and pressure in atmosphere.
   
5. Given 📷
   

📷
📷
The equilibrium constant for 📷 will be.

1. 0.10 mol sample of AgNO3 is dissolved in 1.00 L of 1.00 M NH3. If 0.010 mol NaCl is added to this solution, will AgCl(s) precipitate?

📷

1. 30 ml solution of a weak monobasic acid requires 60 ml 0.25 M NaOH for the end point. During titration the pH of the acid solution upon addition of 20 ml alkali (0.25M) was 4.6. Calculate pH of the acid solution at the following two stages

(a) Before addition of any drop of alkali
(b) At the equivalence point, the contents of the conical flask having been diluted to 150 ml by the addition of distilled water (log 2=0.30) and pKw = 14)

1. A solution of volume V contains ‘a’ mole of MCl and ‘b’ mole of NCl where MOH and NOH are two weak bases having dissociation constants K1, K2 respectively. Show that the pH of the solution can be expressed as pH = 📷
   
2. 0.025 moles of a hydrochloride salt of α– amino acid (NH2 – CH2 – COOH) was dissolved in 250 ml of water and the pH of the solution was 2.7. Same moles of sodium salt of this acid is added to 250 ml water and the pH of solution now is 11, Calculate the pH of α amino acid at which isoelectric point will occur.
   

Objective:
Multiple choice questions with single correct options

1. When a substance (A) reacts with water, it produces a combustible gas (B) and a solution of substance (C) in water. When another substance (D) reacts with the solution of (C), it produces the same gas (B) on warming, but (D) can produce (B) on reaction with dilute H2SO4 at room temperature. (A) imparts a golden yellow colour to brown flame. (A), (B), (C) and (D) could be?

(A) K, H2, KOH, Al (B) Na, H2, NaOH, Zn
(C) CaC2, C2H2, Ca(OH)2, Fe (D) Ca, H2, Ca(OH)2, Sn

1. If NaOH is added to an aqueous solution on Zn2+ ions, a white ppt. appears and on adding excess NaOH, the ppt. dissolves. In this solution zinc exist in the

(A) cationic part (B) anionic part
(C) both in cationic and anionic parts (D) there is no zinc left in the solution

1. Acidified solution of sodium thiosulphate is reactive because in thiosulphate

(A) the sulphur atoms are at unstable oxidation state of +2.
(B) the two sulphur atoms are at different oxidation states of +6 and – 2.
(C) the S – S bonds are less stable bonds.
(D) sulphur is in zero oxidation state.

1. When hydrated MgCl2.6H2O is strongly heated?

(A) MgO is formed (B) Mg(OH)2 is formed
(C) Mg(OH)Cl is formed (D) Anhydrous MgCl2 is formed

1. A metal (X) on heating in nitrogen gas gives (Y). (Y) on treatment with H2O gives a colourless gas which when passed through CuSO4 solution gives a blue colour .(Y) is

(A) Mg(NO3)2 (B) Mg3N2
(C) NH3 (D) MgO

1. Phosphorus when dissolved in alkali forms

(A) PH3 + NaH2PO2 (B) PH3 + Na3PO3
(C) P2H4 + NaH2PO2 (D) Both (A) and (C)

1. Match list – I with list – II and select the correct answer using the codes given below

​
List – I
List – II
(a)
Liquid Na metal
1.
Breathing apparatus in submarine
(b)
Potassium stearate
2.
Explosive
(c)
KNO3
3.
Coolant in nuclear reactors
(d)
KO2
4.
Soft soap
Codes
a b c d
(A) 3 4 1 2
(B) 3 1 4 2
(C) 3 4 2 1
(D) 4 3 2 1

1. Which of the following is correct about the reaction,📷?

(A) It is a disproportionation.
(B) Oxidation number of Cl decreases as well as increases in this reaction.
(C) This reaction is used for the manufacture of halates.
(D) All of the above

1. A greenish yellow gas reacts with an alkali metal hydroxide to form a halate, which can be used in fire works and safety matches. The gas and halate respectively, are

(A) Br2, KBrO3 (B) Cl2, KCIO3
(C) I2, NaIO3 (D) Cl2, NaCIO3

1. Sodium nitrate decomposes above ~800°C to give

(A) N2 (B) O2
(C) Na2O (D) All of the above

1. Which reagent would enable you to remove SO42− ion from solution containing both SO42− and Cl− ions?

(A) NaOH (B) Pb2+
(C) Ba(OH)2 (D) BaSO4

1. The main factor responsible for weak acidic nature of B – F bonds in BF3 is

(A) large electronegativity of fluorine
(B) three centred two electron bonds in BF3
(C) pπ-dπ back bonding
(D) pπ- pπ back bonding

1. Borax crystal contains

(A) Two triangular units (B) Two tetrahedral units
(C) (A) and (B) both (D) None of these

1. Borax is converted into B by steps

Borax 📷H3BO3 📷B2O3 📷B
I and II reagents are
(A) acid, Al (B) acid, C
(C) acid, Fe (D) acid, Mg

1. On strong heating lead nitrate gives

(A) PbO, NO,📷 (B) PbO, NO, 📷
(C) 📷PbO,📷 (D) 📷

1. Hypo is used in photography because of its

(A) Reducing nature (B) Oxidising nature
(C) Complexing nature (D) Sensitivity towards light

1. Disodium hydrogen phosphate in presence of NH4Cl and NH4OH gives a white ppt. with a solution of Mg2+ ions. The precipitate is

(A) Mg(H2PO4)2 (B) Mg3(PO4)2
(C) MgNH4PO4 (D) MgHPO4

1. Commercial 10 volume H2O2 is a solution with a strength of approximately

(A) 30% (B) 3%
(C) 1% (D) 10%

1. Hydrogen peroxide when added to a solution of KMnO4 acidified with H2SO4

(A) forms water only (B) acts as an oxidizing agent
(C) acts as a reducing agent (D) reduces H2SO4

1. In basic medium, H2O2 acts as an oxidizing agent in its reaction with

(A) Cr2(SO4)3 (B) Ag2O
(C) K3[Fe(CN)6] (D) K2Cr2O7

1. The reaction📷, shows

(A) acidic nature of H2O2 (B) alkaline nature of H2O2
(C) reducing action of H2O2 (D) oxidizing action of H2O2

1. For the reaction: CO2(g) + H2(g) 📷 CO(g) + H2O(g)

The Kp for the reaction is 0.11. If the reaction was started with 0.45 moles of CO2 and 0.45 moles of H2 at 700K, the concentration of CO when 0.34 mole of CO2 and 0.34 mole of H2 are added when the first equilibrium is attained, is
(A) 0.52M (B) 0.17M
(C) 0.34M (D) 0.26M

1. For the given reaction:

2A(s) + B(g) 📷 C(g) + 2D(s)+ E(s)
the degree of dissociation of B was found to be 20% at 300K and 24% at 500K the rate of backward reaction
(A) increases with increase in pressure and temperature
(B) increases with increase in pressure and decrease in temperature
(C) depends on temperature only and decreases with increase in temperature
(D) increases with increasing the concentration of B and increasing the temperature

1. For the given equilibrium

L(g) 📷 M(g)
The Kf = 5 × 10–4 mole/litre/seconds
and Kb = 3 × 10–2 litre/mole/seconds
the equilibrium concentration of M is
(A) 0.13M (B) 0.3M
(C) 0.8M (D) Can’t be calculated

1. For the following reaction

📷ΔH < 0. On increasing the temperature at constant pressure, which of the following effect will be observed? (K = equilibrium constant)
(A) K will decrease
(B) K will increase
(C) the degree of dissociation (α) of A2(g) & B2 (g) will increase.
(D) none of these

1. 📷dissolves as

📷
When the initial pressure of 📷 is 500 mm Hg, the total pressure is 700 mm Hg. Calculate equilibrium constant Kp for the reaction, assuming that the volume of the system remains unchanged.
(A) 100 mmHg (B) 133.3 mm Hg
(C) 200 mm Hg (D) 214.6 mm Hg

1. For this reaction at equilibrium, which changes will increase the quantity of Fe(s)?

Fe3O4(s) + 4 H2(g) 📷 3 Fe(s) + 4 H2O(g) ΔH > 0

1. increasing temperature
2. decreasing temperature
3. adding Fe(s)

(A) 1 only (B) 1 and 2 only (C) 2 and 3 only (D) 1,2, and 3

1. One mole each of 📷 and 📷 are heated in presence of little conc. 📷so as to establish the following equilibrium

📷
The moles of 📷 formed at equilibrium is
(A) 1 mol (B) 2 mol
(C) 📷 (D) 📷

1. 24 mL of HI are produced from the reaction of 15 mL of 📷and 17.1 mL of 📷 vapour at 4📷. The equilibrium constant for the reaction: 📷

(A) 27.5 (B) 37.647
(C) 73.647 (D) 57.25

1. The vessel has nitrogen gas and water vapours at a total pressure of 1 atm. The partial pressure of water vapours is 0.3 atm. The content of this vessel are transferred to another vessel having one third of original volume, completely at the same temperature, the total pressure of the system in the new vessel is

(A) 3.0 atm (B) 1 atm
(C) 3.33 atm (D) 2.4 atm

1. A gaseous mixture of 2 moles of A, 3 moles of B, 5 moles of C and 10 moles of D is contained in a vessel. Assuming that gases are ideal and the partial pressure of C is 1.5 atm the total pressure is

(A) 3 atm (B) 6 atm
(C) 9 atm (D) 15 atm

1. The rate of a reaction does not involve gases is not dependent on

(A) pressure (B) temperature
(C) concentration (D) catalyst

1. A reversible reaction is one which

(A) proceeds in one direction (B) proceeds in both directions
(C) proceeds spontaneously (D) all the statements are wrong

1. The law of mass action was proposed by

(A) Goldberg and Waage (B) Le chatelier and Braun
(C) Kossel and Lewis (D) Van’t Hoff

1. The rate at which a substance reacts, depend on its

(A) active mass (B) molecular mass
(C) equivalent mass (D) total volume

1. The state of equilibrium refers to

(A) state of rest (B) dynamic state
(C) stationary state (D) state inertness

1. A buffer solution contains 100 mL of 0.01 M CH3COOH and 200 mL of 0.02 M CH3COONa. 700 ml water is added. pH before and after dilution are: (pKa = 4.74)

(A) 5.04, 5.04 (B) 5.04, 0.504
(C) 5.04, 1.54 (D) 5.34, 5.34

1. 📷

📷
Hence pH of 0.01 M NaH2PO4 is
(A) 9.35 (B) 4.675
(C) 2.675 (D) 7.350
​
39.
📷
​
(Aspirin) is a pain reliever with pKa = 2.
Two tablets each containing 0.09 g of asprin are dissolved in 100 mL solution. pH will be
(A) 0.5 (B) 1.0
(C) 0.0 (D) 2.0

1. Molarity of the liquid HCl if density of the solution is 1.176 g/cc is

(A) 36.5 (B) 18.25
(C) 32.05 (D) 42.10

1. A monoprotic acid in 1.00 M solution is 0.001% ionized. The dissociation constant of formic acid is

(A) 1.0 × 10−3 (B) 1.0 × 10−6
(C) 1.0 × 10−8 (D) 1.0 × 10−10

1. Formic acid is 4.5% dissociated in 0.1 N solution at 20°C. The ionization constant of formic acid is

(A) 21 × 10−4 (B) 21
(C) 0.21 × 10−4 (D) 2.1 × 10−4

1. What is the correct representation of solubility product of Ag2CrO4?

(A) 📷 (B) 📷
(C) 📷 (D) 📷

1. The precipitate of CaF2 (Ksp = 1.7 × 10−10) is obtained when equal volumes of which of the following are mixed?

(A) 10−4 M Ca2+ + 10−4 M F− (B) 10−2 M Ca2+ + 10−3 M F−
(C) 10−2 M Ca2+ + 10−5 M F− (D) 10−3 M Ca2+ + 10−5 M F−

1. Buffering action of a mixture of CH3COOH and CH3COONa is maximum when the ratio of salt to acid is equal to

(A) 1.0 (B) 100.0
(C) 10.0 (D) 0.1
Multiple choice questions with more than one option correct

1. Which of the following illustrate/s the anomalous properties of Li?

(A) The m. pt. and b. pt of Li are comparatively high.
(B) Li forms a nitride Li3N unlike group metals.
(C) Li is much softer than the other group I metals.
(D) Li+ ion and its compounds are more heavily hydrated than those of the rest of the group.

1. For two ionic solids CaO and KI, identify the correct statement/s among the following:

(A) Lattice energy of CaO is much higher than that of KI.
(B) KI is soluble in benzene.
(C) CaO has high m. pt.
(D) KI has high m. pt.

1. Boron readily dissolves in

(A) conc. HCl
(B) fused NaOH at 673 K
(C) fused Na2CO3 + NaNO3 at 1173 K
(D) a mixture of conc. HNO3 and conc. H2SO4 (1 : 2)

1. Which metal can be obtained from electrolysis?

(A) Ca (B) Mg
(C) Cr (D) Al

1. A reaction is catalysed by H+ ion. In presence of HA, rate constant is 2 × 10−3 min−1 and in presence of HB rate constant is 1 × 10−3 min−1, HA and HB both being strong acids, we may conclude

(A) equilibrium constant is 2
(B) HA is stronger than HB
(C) relative strength of HA and HB is 2
(D) HA is weaker than HB and relative strength is 0.5

1. Which of the following statement(s) is(are) wrong?

(A) at equilibrium concentration of reactant and product become constant because the reaction stops
(B) addition of catalyst speeds up the forward reaction more than the backward reaction
(C) equilibrium constant of an exothermic reaction decreases with increase of temperature
(D) Kp is always greater than Kc

1. Which of the following will not affect the value of equilibrium constant of a reaction?

(A) change in concentration of the reactants (B) change in temperature
(C) change in pressure (D) addition of catalyst

1. The equilibrium constant of the reaction 📷and📷are K1 and K2 respectively. The relationship between K1 and K2 is

(A) K1 = K2 (B) 📷
(C) 📷 (D) 📷

1. For which of the following reactions is the equilibrium constant called an acidity constant?

(A) 📷
(B) 📷
(C) 📷
(D) 📷

1. For the reaction 📷which is correct representation?

(A) 📷 (B) 📷
(C) 📷 (D) none of these

1. 📷

📷
Correct relation (s) between K1, K2, K3 and K4 is/are
(A) K1 × K3 = 1 (B) 📷
(C) 📷 (D) none of these

1. For the reaction, 📷the forward reaction at constant temperature is favoured by

(A) introducing an inert gas at constant volume
(B) introducing chlorine gas at constant volume
(C) introducing an inert gas at constant pressure
(D) introducing the volume of the container
(E) introducing PCl5 at constant volume

1. Which of the following will favour the formation of NH3 by Haber’s process?

(A) increase temperature (B) increase of pressure
(C) addition of catalyst (D) addition of promoter

1. Which of the following will not affect the volume of equilibrium constant of a reaction?

(A) change in the concentration of the reactants
(B) change in temperature
(C) change in pressure
(D) addition of catalyst

1. For the gas phase reaction,

📷
Carried out in a reaction vessel, the equilibrium concentration of C2H4 can be increased by
(A) increasing the temperature (B) decreasing the pressure
(C) removing some H2 (D) adding some C2H6

1. When NaNO3 is heated in a closed vessel oxygen is liberated and NaNO2 is left behind. At equilibrium which are not correct?

(A) Addition of NaNO2 favours reverse reaction
(B) Addition of NaNO3 favours forward reaction
(C) Increasing temperature favours forward reaction
(D) Increasing pressure favours reverse reaction

1. 📷

Half life period is independent of concentration of Zn at constant pH. For the constant concentration of Zn, rate becomes 100 times when pH is decreased from 3 to 2. Hence
(A) 📷
(B) 📷
(C) rate is not affected if concentration of Zn is made four times and that of [H+] ion is halved
(D) rate becomes four times if concentration of [H+] ion is doubled at constant Zn concentration

1. All of the following are acid-base conjugate pairs.

(A) 📷 (B) H3O+, OH−
(C) 📷 (D) HS−, S2−

1. In the following reaction

📷
(A) A is an acid and B the base
(B) A is a base and B the acid
(C) C is the conjugate acid of A, and D is the conjugate base of B
(D) C is conjugate base of A, and D is the conjugate acid of B

1. Which does not react with NaOH or which is not acid salt?

(A) NaH2PO2 (B) Na2HPO3
(C) Na2HPO4 (D) NaHCO3

1. An acid-base indicator has a Ka of 3.0 × 10−5. The acid form of the indicator is red and the basic form is blue. Then

(A) pH is 4.05 when indicator is 75% red
(B) pH is 5.00 when indicator is 75% blue
(C) pH is 5.00 when indicator is 75% red
(D) pH is 4.05 when indicator is 75% blue

1. pH of the following solution is not affected by dilution

(A) 0.01 M CH3COONa
(B) 0.01 M NaHCO3
(C) buffer of 0.01 M CH3COONa and 0.01 M CH3COOH
(D) 0.01 M CH3COONH4

1. The term infinite dilution refers for

(A) 📷for weak electrolytes
(B) an electrolyte is 100 % dissociated
(C) all interionic effects disappear
(D) when equivalent conductance of an electrolyte becomes constant
Comprehension - I
Read the paragraph carefully and answer the following questions:
Hydrogen bromide gas is one of the strongest acids available in gas form. In aqueous solution, HBr has pKa value of approximately −12. A pKa value of −12 is indicative of a very strong acid. By comparison, HCl has a pKa of approximately −9. Hydrobromic acid (HBr) can be produced by reaction 1.
H2(g) + Br2(l) 📷2HBr(g) …Reaction 1
A student decides to study the equilibrium distribution for Reaction 1. Under ambient conditions, Reaction 1 Reaction reaches equilibrium at some temperature T and pressure P. Once at equilibrium, measurements are taken for the partial pressure of each gas and the mass of bromine liquid. The values are listed in Table
Br2(l)
10.0 grams
H2(g)
1.36 atm
HBr(g)
2.72 atm
The student takes notice that Reaction 1 shifts with changes in the conditions of the system, such as the addition of bromine liquid, the removal of hydrobromic acid, and increases in volume. The student irreversibly increases the volume of the system by opening a valve on the reaction flask that is connected to an evacuated column. Not all changes shift the reaction. Changes in the equilibrium concentrations are not accurately recorded.

1. Evaluate the value of equilibrium constant K at ambient temperature & pressure as per data shown in table for Reaction – 1
   
2. Addition of which of the either reactant or products to the equilibrium mixture will NOT affect the partial pressure of H2(g)?
   
3. If we put the HBr(g), H2(g) over bromine liquid having partial pressure 3 atm., 1.5 atm. respectively in a closed container the gas whose pressure increases after attaining the equilibrium is ………….
   
4. If there is an increase in mole percent of HBr(g) after increasing the temperature of the above reaction is ……………… in nature.
   

https://docs.google.com/document/d/1Fdgg9mKttdfsHL3z6IhXxT0RuvF-tHtedit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true
TRANSITION ELEMENTS
Transition elements (only the first row, 3d series): Definition, Werner’s approach to coordination compounds, general characteristic properties [viz., variable oxidation states, colour (excluding the details of electronic transition) calculation of spin-only magnetic moment, formation of complexes (stereochemistry excluded), nomenclature of simple coordination compounds, valence bond approach to define geometries of coordination compounds of linear, tetrahedral, octahedral and square planar geometries.
Transition Elements - Introduction

Definition and Electronic Configurations of Atoms

The elements lying between s and p-block elements of the periodic table are collectively known as transition or transitional elements (T.E.’s): These elements either in their atomic state or in any of their common oxidation state have partly filled (n-1)d orbitals of (n-1)th main shell. In these elements the differentiating electron enters (n-1)d orbitals of (n-1)th main shell and as such these are called d-block elements.
The valence shell configurations of these elements can be represented by (n-1)d1–10.ns0, 1, 2. The configurations clearly show that strictly, according to the definition of d-block elements, Cu, Ag and Au should be excluded from d-block elements, since these elements, both in their atomic state [with configuration (n-1)d10ns1] and in their +1 oxidation state [with configuration (n-1)d10], do not have partly filled (n-l)d-orbitals. Similarly Zn, Cd and Hg which both in their atomic state [(n-1)d10ns2] and in +2 oxidation [(n-1)d10] do not contian partly filled (n-1)d orbitals, should also be excluded from d-block elements. Similar is the case with Pd atom with configuration 4d105s0. Yet, in order to maintain a rational classification of elements, these elements (viz Cu, Ag, Au, Zn, Cd, Hg and Pd) are also generally studied with d-block elements.
All the d-block elements are classified into four series viz 3d, 4d, 5d and 6d series corresponding to the filling of 3d, 4d, 5d and 6d orbitals of (n-1)th main shell. Each of 3d, 4d and 5d series has ten elements while 6d series has at present only one element viz Ac80 whose valence shell configuration is 6d1 7s2.
Irregularities in Configurations
The irregularities in the observed configurations of Cr (3d54s1), Cu (3d104s1), Mo (4d55s1), Pd (4d105s0), Ag (4d10 5s1) and Au (5d106s1) are explained on the basis of the concept that half-filled and completely filled d-orbitals are realtively more stable than other d-orbitals.
On the basis of the above concept it is, however, not easy to explain the irregularities found in the observed electronic configurations of the atoms of other elements, since one has to consider the net effect of so many other factors such as (i) nuclear electronic attraction (ii) shielding of one electron by several other electrons (iii) inter-electronic repulsion (iv) the exchange-energy forces etc. All these factors play an important part together in determining the final stability of an electronic configuration of an atom. It is not easy to explain why W unlike Cr(3d54s1) and Mo(4d5 5s1) should have the idealised electronic configuration (4f145d46s2).
The properties of transition elements of any given period are not so much different from one another as those of the same period of non-transition elements. The reason of this fact lies in the electronic configuration of transition elements. We know that electronic configurations of transition elements is invariably (n-1)d1–10 ns0 or 1 or 2 which indicates that (i) the electronic configurations of transition elements differ from one another only in the number of electrons in d orbitals in the (n-1)th shell and (ii) the number of electrons in the outermost shell, ns, is invariably 1 or 2.
Physicochemical Properties
Metallic Character
All the transition elements are metals, since the number of electrons in the outermost shell is very small, being equal to 2. They are hard, malleable and ductile. They exhibit all the three types of structures: face centred cubic (fcc), hexagonal close packed (hcp) and body centred cubic (bcc). Metals of VIII and IB Groups are softer and more ductile than other metals.
It appears that covalent and metallic bonding both exist in the atoms of transition metals. The presence of unfilled d-orbitals favours covalent bonding. These metals are good conductors of heat and electricity.
Melting and Boiling Points
The transition elements have very high melting and boiling points as compared to those of s and p block elements. Zn, Cd and Hg have relatively low values. The reason for these low values is that these metals have completely filled d-orbitals with no unpaired electron that may be available for covalent bonding amongst the atoms of these metals. The formation of covalent bonding occurs in the rest of the d-block elements on account of the presence of unfilled d-orbitals.
Although melting and boiling points show no definite trends in the three transition series, the metals having the highest melting and boiling points are towards the middle of each transition series.
Atomic (Covalent) and Ionic Radii
The atomic and ionic radii (M2+ ions) for the elements of 3d-series are given in. it will be seen that these values decrease generally, on moving from left to right in the period. This is due to the fact that an increase in the nuclear charge tends to attract the electron cloud inwards. The atomic radii for the elements from Cr to Cu are, however, very close to one another. This is due to the fact that simultaneous addition of electron to 3d-level exercise the reverse effect by screening the outer 4s-electrons from the inward pull of the nucleus. As a result of these two opposing effects, the atomic radii do not alter much on moving from Cr to Cu.
The ionic radii of M2+ and M3+ ions follow the same trends as their atomic radii. The radii of M2+ ions, although somewhat smaller than that of Ca2+ ion (=0.99Å) are comparable with it. Thus Mo oxides the transition element should be similar to CaO in many ways, although somewhat less basic and less soluble in water. Similarly the hydration energies of M2+ ions should be similar to but somewhat greater than that of Ca2+ ion. This is borne out by facts, since the hydration energy of Ca2+ ion is 395 kcal and the observed values of hydration energies for the elements Ti2+ … Cu2+ are between 446 kcal and 597 kcal.
The radii of M3+ ions are slightly greater than that of Ga3+ ion (=0.62Å). Thus M2O3 oxides of transitional elements should be similar to but slightly less acidic (more basic) than Ga2O3 and the hydration energies of M3+ ions should be less than 1124 kcal which is the hydration energy of Ga3+ ion. The observed values of hydration energies for the series Sc3+ … Fe3+ are between 947 kcal and 1072 kcal.
Ionisation Potentials
The first ionisation potentials of transitional elements lie between the values of those of s- and p-block elements. The first ionisation potentials of all the transition elements lie between 5 to 10 electron volts. In case of transition elements the addition of the extra electron in the (n-1) d level provides a screening effect which shields the outer ns electrons from the inward pull of positive nucleus on the outer ns electrons. Thus the effects of the increasing nuclear charge and the shielding effect created due to the expansion of (n-1)d orbital oppose each other. On account of these counter effects, the ionisation potentials increase rather slowly on moving in a period of the first transition series.
First ionisation Potentials
It is evident that the values for the first four 3d block elements (Sc, Ti, V and Cr) differ only slightly from one another. Similarly the values for Fe, Co, Ni and Cu also are fairly close to one another. The value of I1 for Zn is considerably higher. This is due to the extra-stability of 3d10 level which is completely filled in Zn-atom.
Second ionisation Potentials
The second ionisation potentials are seen to increase more or less regularly with the increase of atomic number. The value of I2 for Cr and Cu are higher than those of their neighbours. This is due to the fact that the electronic configurations of Cr+ and Cu+ ions have extra stable 3d5 and 3d10 levels.
There is a sudden fall in the values of ionisation potentials in going from II B (Zn-group elements) to IIIA sub-group. This sudden fall is expalined on the basis that in case of IIIA group elements the electron to be removed is from a 4p-orbital which is incompletely filled, while in case of the II B group elements, the electron to be removed is from 4s-orbital which is completely filled. Thus more energy will be required to remove an electron from a filled 4s-orbital on comparison to that used to remove an electron from a 4p-orbital which is incompletely filled.
Electropositive character of transitional elements as compared to that of alkali metals and alkaline earth metals
The values of first ionisation potentials of transitional elements in most cases lie between those of s-and p-block elements. Thus the transitional elements are less electropositive than the elements of I A and II A groups. Thus, although the transitional elements do form ionic compounds, yet they do not form ionic compounds so readily as the alkali and alkaline earth metals do. Also, unlike the alkali and alkaline earth metals, the transitional elements also have a tendency to form the covalent compounds under certain conditions. Generally the compounds in which the transitional elements show a smaller valency are ionic, while those in which a higher valency is exhibited are covalent in character.
Oxidation States
One of the most important property that distinguishes transition elements from s-and p-block elements is that they show variable oxidation states. s-and p-block elements have oxidiation states either equal to their group number, G or equal to (8-G). The transition elements on the other hand exhibit variable oxidation states.
This unique property is due to the fact that the energy levels of 3d, 4d and 5d orbitals are fairly close to those of 4s, 5s and 6s orbitals respectively and, therefore, in addition to ns electrons and variable number of (n-1) d electrons are also lost in getting various oxidation states.
i) Minimum oxidation state: All the transition elements with the exception of Cr, Cu, Ag, Au and Hg which have a minimum oxidation state of +1 exhibit a minimum oxidation state of +2. In most cases this +2 oxidation state arises due to the loss of two s-electrons.
ii) Maximum oxidation state: Each of the elements in groups III B to VII B can show the maximum oxidation state equal to its group number. Thus, Cr in group VIB shows a maximum oxidation state of +6 in Cr2O72– ion. Most of the elements in VIII group show a maximum oxidation state equal to + 6. However, Ru and Os have a maximum oxidation state equal to +8 which is the highest oxidation state shown by any element.
iii) Relative stability of various oxidation states: The relative stabilities of various oxidation states of 3d-series elements can be correlated with the extra stability of 3d0, 3d5 and 3d10 configurations to some extent. Thus Ti4+ (3d0) is more stable than Ti3+ (3d1) and similarly Mn2+ (3d5) is more stable than Mn4+ (3d4). It may, however, be pointed out that such a generalisation for the relative stability of various oxidation states of 4d and 5d series elements is often rather difficult to realise.
The higher oxidation state of 4d and 5d series elements are generally more stable than those of the elements of 3d series, e.g., Mo, Te (4d series elements) and W, Re (5d-series elements) form the oxyanions: MoVIO42–, TcVII O4–, WVI, O42–, ReVIIO4– which are stable and in which the transition elements concerned show their maximum oxidation states. The corresponding oxyanions of Cr and Mn namely CrVI O42– and MnVII O4– are strong oxidising agents.
Furthermore, the highest oxidation states of second and third row elements are encountered in compounds containing the more electronegative elements viz. F, O and Cl. Thus for the compounds RuVIII O4, OsVIII O4, WVI Cl6 and PtVI F6 there are no analogs being formed by the first row elements. The lower oxidation states particularly +2 and +3 are important in the chemistry of aquated and complex ions of the 3d-series (i.e. first row) elements but these ions are not very important in the chemistry of second (i.e. 4d series) and third (5d-series) row elements. In short it may be said that in going down a sub-group the stability of the higher oxidation states increases while that of lower oxidation states decreases.
v) Formation of ionic and covalent compounds: Transition elements cannot form ionic compounds in higher oxidation states because the loss of more than three electrons is prevented by the higher attractive force exerted (on the electrons) by the nucleus. Higher oxidation states of these metals are not formed by the actual loss of electrons but due to the formation of new hybrid orbitals involving (n-1)d, ns and np orbitals.
The transition elements form ionic bonds in the lower oxidation states and the ionic character of the bond decreases with the increase in the oxidation state. With this decrease in the ionic character of the bond the basic character of the oxides decreases, e.g. MnO (oxidatioin states of Mn = +2) is basic, MnO2 (Mn + +4) is amphoteric and MnO3 (Mn = +6) is acidic.
Complex Formation
The transition elements have an unparallel tendency to form coordination compounds with Lewis Base, i.e., with groups which are able to donate an electron pair. These groups are called ligands. A ligand may be a neutral molecule such as NH3 as or ion such as Cl– or CN– etc.
Co3+ + 6(NH3) ⎯→ [Co(NH3)6]3+
Fe2+ + 6CN- ⎯→ [Fe(CN)6]4–
The reason for transition elements are good in forming complexes are:
i) Small size and high effective nuclear charge
ii) Availability of low lying vacant d–orbitals which can accept lone pair of electrons donated by a ligand.
Catalytic Properties
Transition metals and their compounds are known to act as good catalyst due to the following reasons:
i) Due to the variable oxidation states, they form unstable intermediate compounds and provide a new path with lower activation energy for the reaction (Intermediate compound formation theory)
ii) In some cases the finely divided metals or their compounds provide a large surface area for adsortion and the adsorbed reactants react faster due to the closer contact(Adsorption theory)

1. TiCl3 Used as Ziegler – Nalta catalyst
   
2. V2O5 Converts SO2 to SO3 in the contact process for making H2SO4
   
3. MnO2 Used as a catlyst to decompose KClO3 to give O2
   
4. Fe Used in Haber – Bosch process for making NH3
   
5. FeCl3 Production of CCl4 from CS2 and Cl2
   
6. FeSO4 & H2O2 Fenton’s reagent
   
7. PdCl2 Wacker process for converting
   

C2H2 + H2O + PdCl2 to → CH3CHO + 2HCl + Pd

1. Pd For hydrogenation (Phenol → Cyclohexanone)
   
2. Pt/Pto Adam catalyst used for reduction
   
3. Pt SO2 → SO3 contact process
   
4. Pt Cleaning car exhaust fumes
   
5. Cu In manufacture of (CH3)2SiCl2
   
6. Cu/V Oxidation of cyclohexanol
   
7. CuCl2 Deacon process or making Cl2 from HCl
   
8. Ni Raneynickel
   

Magnetic Properties
When a substance is placed in a magnetic field of strength H, the intensity of the magnetic field in the substance may be greater than or less than H.
If the field in the substance is greater than H, the substance is paramagnetic. Thus paramagnetic materials attract lines of force and if it is free to move, a paramagnetic material will move from a weaker to a stronger part of the field. Paramagnetism arises as a result of unpaired electron in the atom. If the field in the substance is less than H, the substance is diamagnetic. They tend to repel lines of force and move from a strong to weaker part of a magnetic field. In diamagnetic substance, electrons are paired up:
It should be noted that Fe,Co and Ni are ferromagnetic. Ferromagnetic materials may be regarded as special case of paramagnetism in which the moments of individuals atoms become aligned and all point in the same direction . It is also possible to get antiferromagnetism by pairing the moments of adjacent atoms which point in opposite directions. It occurs in salts of Fe3+, Mn2+ etc.
Paramagnetism is expressed in terms of magnetic moment, which is related to the number of unpaired electrons are follows
μ = 📷 B.M.
n = number of unpaired electrons
B.M. = Bohr Magneton, unit of magnetic moment
More the magnetic moment more is the paramagnetic behaviour
Exercise 1: Calculate the magnetic moment of V3+.
Illustration 1: Compare the magnetic moments of Fe2+ and Fe3+
Solution: Following the same procedure as above try to calculate the magnetic moments of the two given species
n for Fe2+=4 μ = 📷 B.M.=📷= 📷 B.M
n for Fe3+ = 5 μ = 📷 B.M = 📷= 📷 B.M
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Formation of Alloys
As the transition elements have similar atomic sizes hence in the crystal lattice, one metal can be readily replaced by another metal giving solid solution and smooth alloys. The alloys so formed are hard and have often high melting point.
Interstitial Compound
Transition metals form no. of interstitial compounds, in which they take up atoms of small size e.g. H, C and N in the vacant spaces in the their lattices. The presence of these atoms results in decrease in malleability and ductility of the metals but increases their tensile strength.
Compounds of Iron
Ferrous sulphate (Green vitriol), FeSO4.7H2O
It occurs in nature as copperas and commonly known as hara Kasis.
Preparation i) By dissolving scrap Fe in dil. H2SO4
ii) From Kipp’s waste which contains ferrous sulphate with some free H2SO4; the latter is neutralised with scrap iron forming FeSO4 and hydrogen.
iii) By the action of air and water on iron pyrites. The solution is treated with scrap iron to remove H2SO4 and to reduce Fe3+ sulphate Fe2+ sulphate.
Properties i) Hydrated and anhydrous FeSO4 are green and white in colour respectively. It is isomorphous with epsom salt, MgSO4.7H2O and ZnSO4.7H2O. It effervesces on exposure to air.
ii) Light green crystals of FeSO4 lose water and turn brown on exposure to air, due to oxidation.
iii) On heating at 300°C it gives anhydrous FeSO4 which on further heating gives Fe2O3 and SO2.
iv) Like other ferrous salts, it takes up HNO3 forming brown coloured double compound, Fe(NO)SO4, nitroso ferrous sulphate (Ring test for nitrates).
v) It decolourises acidified potassium permanganate and turns acidified dichromate green (reducing character).
vi) It forms double salts with sulphates of alkali metals with general formula R2SO4.FeSO4.6H2O. With ammonium sulphate, it forms a double salt known as ferrous ammonium sulphate or Mohr’s salt, FeSO4.(NH4)2SO4.6H2O. It does not effervesce. It ionises in solution to gives Fe2+, NH4+ and SO42– ions.
Ferric oxide, Fe2O3
i) It occurs in nature as haematite.
ii) Fe2O3 is a red powder, insoluble in H2O and not acted upon by air or H2O
iii) It is amphoteric in nature and reacts with acids and alkalies.
iv) It is reduced to iron by H2,C and CO.
v) It is used as a catalyst in the oxidation of CO to CO2 in the Bosch process.
Ferric chloride, FeCl3
Preparation i) Hydrated ferric chloride (FeCl3.6H2O) can be prepared by dissolving iron, Fe(OH)3 or ferric oxide in dil. HCl.
ii) Reaction of Fe with dry Cl2 gives anhydrous FeCl3,
Properties i) Anhydrous salt is yellow, deliquescent compound and highly soluble in H2O.
ii) Its aqueous solution is acidic due to hydrolysis.
iii) On heating it gives FeCl2 and Cl2.
iv) It oxidizes H2S to S, SO2 to H2SO4, SnCl2 to SnCl4 and Na2S2O3 to Na2S4O6.
Copper, Silver and gold
a) These metals are commonly called as coinage or currency metals. Their general electronic configuration is (n – 1) d10 ns1. These show variable valencies +1, +2 and +3.
b) Gradation in properteis
i) The nobility increases from copper to gold.
ii) The affinity of oxygen also decreases from Cu to Au.
iii) Copper forms a large number of salts followed by silver followed by gold.
iv) The ease with which the salts of these elements are reduced increases from Cu to Au.
Compounds of Copper
i) Copper sulphate, cupric sulphate or blue vitriol, CuSO4.5H2O.
Preparation i) By treating copper scrap or turnings, cuprous oxide, cupric oxide or malachite with H2SO4.
ii) BY roasting copper pyrites, CuFeS2 in air.

Properties i) It has 5 molecules of H2O of crystallisation; all of which can be removed on heating, to form colourless CuSO4 (again coloured with H2O).

ii) At high temperature it forms cupric oxide.
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iii) It forms double salts with alkali sulphates, e.g. K2SO4.CuSO4.6H2O
iv) When treated with NH4OH, it first forms precipitate of cupric hydroxide copper (II) sulphate (Schweitzer’s reagent), used for dissolving cellulose in the manufacture of aritifical silk.
v) It reacts with KCN forming a complex compound K3[Cu(CN)4].
vi) It liberates iodine from soluble iodides.
Uses in the preparation of Bordeaux mixture (CuSO4 solution + lime) which is used to kill moulds and fungi on wines.
Compounds Silver
Silver Nitrate (Lunar caustic) AgNO3
Preparation i) By dissolving Ag in warm dil. HNO3.
Properties i) It is very soluble in H2O and when comes in contact with organic substances (e.g. skin, clothes, etc.) it produces burning sensation and reduced to metallic silver which is white like the moon Luna hence its name Lunar caustic.
ii) On heating above its m.p. it decomposes to silver nitrite.
iii) When treated with soluble halides, it forms the corresponding silver halide.
iv) When treated with alkali, it forms silver oxide which in case of NH4OH dissolves to form complex ion.
v) It reacts with iodine and gives AgIO3 and AgI (when AgNO3 is in excess) or HIO3 and AgI (when I2 is in excess).
vi) With a very dilute solution of Na2S2O3, it gives white ppt. Which quickly changes to yellow, brown and finally black due to the formation of silver sulphide. With conc. solution of sodium thiosulphate, it does not give any ppt.
Uses: In volumetric analysis, photography and in silvering of mirrors.
Silvering of mirrors The process of based on the reduction of an ammonical solution of silver nitrate by some reducing agent like glucose, formaldehyde, tartarate, etc.
Silver Bromide, AgBr
Preparation By adding AgNO3 solution to soluble bromide solution
Properties i) It is insoluble in water and conc. acid but soluble in excess of strong solution of ammonia (cf. AgCl is soluble in dilute solution of NH4OH, AgI is insoluble in NH4OH solution).
ii) Silver halides, are also soluble in KCN and hypo solutions
iii) On heating, it melts to red liquid
iv) It is used as the light sensitive materail in photographic films. It is the most sensitive AgX to photo-reduction.
Zinc, Cadmium and Mercury
a) These are the elements of group 12 having electronic configuration (n – 1) d10 ns2 and +2 oxidation state. In these elements the d-subshell is full, hence these are regarded as non-transition elements which is evident from the following characteristics.
i) They do not show variable valency except mercury
ii) Many of their compounds are white.
iii) Their melting and boiling points are very low.
b) Unique structure of mercurous ion – Unlike Zn and Cd, Hg exhibits +1 well as +2 oxidation state. Thus mercurous ion exists are Hg22+ and not as Hg+.
c) Structure of mercurous ion – It consists of two atoms linked by a covalent and (–Hg – Hg –)2+ and explains the diamagnetic character of mercurous ion were Hg+ (presence of an unpaired electron in 6s orbital) and mercurous salt should have been paramagnetic.
d) Anomalous behaviour of mercury,
i) It is liquid at ordinary temperature while Zn and Cd are solids.
ii) It is less electropositive than hydrogen and therefore does not displace hydrogen from acids while Zn and Cd does.
iii) It does not form hydroxide or peroxide, while Zn and Cd do so
iv) Mercuric oxide, on heating, gives metallic mercury and oxygen while oxides of Zn and Cd are stable towards heat.
v) HgCl2 is covalent while zinc and Cd chlorides are ionic. With NH3, HgCl2 gives a white ppt. of Hg(NH2)Cl, while Zn and Cd salts form complex ions, [M(NH3)4]2+.

Inner Transition Elements

The f-block elements are known as inner transition elements because they involve the filling for inner sub-shells (4f or 5f)
Lanthanides: It consists of elements that follows lanthanum and involve the filling of 4f subshell
Some Important Properties of Lanthanides
1. Electronic Configuration : [Xe] 4fn+1 5d° 6s2 or [Xe] 4fn 5d1 6s2
2. Oxidation State: Lathanides exhibit the oxidation state of +3. Some of them also exhibit the oxidation state of +2 and +4.
3. Colouration: Many of the lanthanides ions are coloured in solid state as well as in solutions. The colour is due to the f-f transition since they have partly filled f-orbitals.
4. Lanthanide Contraction: The steady decrease in the size of lanthanide ions (M3+) with the increase in atomic no. is called lanthanide contraction.
Causes: As we move down the group from left to right in a lanthanide series, the atomic no. increases and for every proton in the nucleus the extra electron goes to 4f orbital. The 4f orbital is too diffused to shield the nucleus effectively, thus there is a gradual increase in the effective nuclear charge experienced by the outer electrons. Consequently , the attraction of the nucleus for the electrons in the outermost shell increases with the increase of atomic number, thus size decreases.

Consequence of Lanthanide Contraction

1. Separation of Lanthanides: Due to the similar sizes of the lanthanides, it is difficult to separate them but due to lanthanide contraction their properties slightly vary (such as ability to form complexes). The variation in the properties is utilized to separate them.
2. Basic Strength of Hydroxide: Due to the lanthanide contraction, size of M3+ ions decreases and there is increase in covalent character in M–OH and hence basic character decreases.
3. Similarity of second and third transition series: The atomic radii of second row transition elements are almost similar to those of the third row transition elements because the increase in size on moving down the group from second to third transition elements is cancelled by the decrease in size due to the lanthanide contraction.
Actinides
It consists of elements that follow Actinium and involve the filling of 5f subshell.
Co-ordination Chemistry
Introduction

• Co-ordination chemistry is a branch of inorganic chemistry which deals with the study of co-ordination compounds.
• Addition or Molecular compounds : When two or more stable compounds are allowed to combine together in stoichiometric proportions, these result in the formation of crystalline substances.
• Addition compounds are classified into two types:

(i) Double salt (ii) Complex compounds

• Double Salt: Addition compound which are quite stable in solid state but dissociate into ions in their aqueous solutions i.e. they give the test of all its constituent ions.

e.g. K2SO4, Al2 (SO4)3, 24H2O.

• Complex Compound : Addition compound which are stable in solid state as well as in dissolved state. They do not give the test of all its constituent ions.

e.g. K4[Fe(CN)6]

• Complex Ion: An electrically charged species in which central atom or ion is surrounded by a suitable no. of ligands.
• Central atom: Metal atom which accepts the electron pair from ligands
• Ligands: Species (neutral, negative, positive) which donate the electron pair to the central metal atom.

Depending upon the no. of donor sites, the ligands may be classified as:
monodentate, bidentate, tridentate etc.
Eg of monodentate : Cl–, Br– (negative ligands), H2O, NH3 (neutral ligand)
Eg of bidentate: (–CH2—NH2 —)2 and C2O42–

• Ionisation Sphere: Portion present outside the square bracket

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Werner’s Co-ordination Theory
Werner’s Co-ordination theory in 1893 was the first attempt to explain the bonding in coordination complexes. Werner was able to explain the nature of bonding in complexes, and he concluded that in complexes the metal shows two different sorts of valency;

1. Primary Valencies: The complex commonly exists as a positive ion. The primary valency is the number of charges on the complex ion. The complex [Co(NH3)6]Cl3 actually exists as [Co(NH3)6]3+ and 3Cl–. Thus the primary valency is 3, as there are three ionic bonds.
2. Secondary Valencies: These are directional. In modern terms the number of secondary valencies equals the number of ligand atoms coordinated to the metal. This is now called the coordination number. Secondary valencies are directional, and so a complex ion has a particular shape, e.g. the complex ion [Co(NH3)6]3+ is octahedral. Whenever the complex is formed the secondary valency should always be satisfied. This he proved with the following example.

CoCl3.6NH3 + AgNO3 ⎯→ 3AgCl
CoCl3.5NH3 + AgNO3 ⎯→ 2AgCl
CoCl3.4NH3 + AgNO3 ⎯→ 1AgCl
Werner deduced that in CoCl3. 6NH3 the three chlorines acted as primary valencies, and the six ammonias as secondary valencies. In modern terms the complex is written [Co(NH3)6]Cl3. The three Cl– are ionic and hence are precipitated as AgCl by AgNO3. In the second complex two moles of AgCl is precipitated indicating the presence of two chlorine atoms in the outer sphere i.e., in order to fulfill the secondary valency one chlorine from the outer sphere drifts into the inner sphere. Similar is the case with the third complex. So actual structure becomes
[Co(NH3)6]Cl3, [Co(NH3)5 Cl]Cl2, [Co(NH3)4 Cl2]Cl,
IUPAC Nomenclature for Co-Ordination Compounds

1. Positive part of a co-ordination compound is named first and is followed by the negative part.
   
2. In naming the complex ion, the name of the ligands are given in alphabetical order regardless of their charge followed by metal.
   
3. When there are several ligands of the same kind, we normally use the prefixes di, tri, tetra, penta and hexa to show the number of ligands of that type. An exception occurs when the name of the ligand includes a number (di, tri etc). To avoid confusion in such cases bis, tris and tetrakis are used instead of di, tri and tetra and the name of the ligand is placed in brackets.
   
4. i) Negative ligands end in 'O', for example:
   

F− − Fluoro H− − hydrido OH− − hydroxo
Cl− Chloro I− − iodo NO2− − nitro.
ii) Neutral groups have no special endings, Examples include NH3 ammine, H2O aqua, CO carbonyl and NO nitrosyl. The ligands N2 and O2 are called dinitrogen and dioxygen. Organic ligands are usually given their common names for example phenyl, methyl, ethylenediamine, pyridine, triphenylphosphine.
iii) Positive ligands end in - ium e.g. NH2−NH3+ hydrazinium.
The spellings of ammine with two m’s distinguishes it from organic amines.

1. The oxidation state of the central metal is shown by Roman numeral in brackets immediately following its name (i.e. no space, e.g. titanium(IV) ).
   
2. Sometimes a ligand may be attached through different atoms. Thus M−NO2 is called nitro and M−ONO is called nitrito. Similarly SCN group may bond M−SCN thiocyanato or M−NCS isothiocyanato. These may be named systematically thiocyanato−S or thiocyanato−N to indicate which atom is bonded to the metal. This convention may be extended to other cases where the mode of linkage is ambiguous.
   
3. When writing the formula of complexes, the complex ion should be enclosed by square brackets. The metal is named first, then the co-ordinated groups are listed in the order; negative ligands, neutral ligands; positive ligands (and alphabetically according to the first symbol with in each group).
   
4. If a complex contains two or more metal atoms, it is termed polynuclear. The bridging ligands which link the two metal atoms together are indicated by the prefix μ. If there are two or more bridging groups of the same kind, this is indicated by di-μ, tri-μ etc. Bridging groups are listed alphabetically with the other groups unless the symmetry of the molecule allows a simpler name. If a bridging group bridges more than two metal atoms it is shown as μ3, μ4, μ5 or μ6 to indicate how many atoms it is bonded to.
   
5. The complete metal name consists of the name of the metal, followed by-ate if the complex is an anion, which in turn is followed by the oxidation number of the metal, indicated by roman numerals in parenthesis. (An oxidation state of zero is indicated by 0 in parenthesis). When there is a latin name for the metal, it is used to name the anion (except for mercury). These names are given in the following table.
   

English name

Latin name
Anion name
Copper
Cuprum
Cuprate
Gold
Aurum
Aurate
Iron
Ferrum
Ferrate
Lead
Plumbum
Plumbate
Silver
Argentum
Argentate
Tin
Stannum
Stannate
Exercise 2: Write down the IUPAC name of the complex (1) K4[Fe(CN)6].
Illustration 2: IUPAC name of K2[Fe(NC)3Cl2(NH3)2]
Solutions: The positive part is named first followed by the negative part. In the negative part the names are written in alphabetical order followed by metal. So the name is Potassiumdiamminedichlorotricyano-N- ferrate (III)
Illustration 3: IUPAC name of [Co(NH3)4(NO2)2]Cl
Solution: In the earlier two examples the negative part was the complex part while in this case the positive part is the complex. So it is named first with ligands in alphabetical order followed by metal (but not ending in –ate as the metal belong to the positive part of the complex). This is followed by the negative part.
So the name is Tetraamminedinitrocobalt (III) chloride.
Illustration 4: Now let take complex where there is an organic ligand like ethylene diamine. [Co(NH3)2Cl(en)2]Cl2
Solution: The approach is same as the earlier one with the exception that in case of -en which is actually ethylene diammine the term bis – comes to indicate two – en groups instead of – bi
The name is Diamminechlorobis (ethylene diamine) cobalt(III) chloride
Illustration 5: Write IUPAC name of [Pt(Py)4][PtCl4].
Solution: Here both the positive and negative part has the same metal. Procedure is same earlier are the IUPAC name. Tetrapyridineplatinum(II) tetrachloroplatinate(II).
Illustration 6: Write the IUPAC name of [Fe(NH3)4O2C2O4]Cl
Solution: In this charge on the complex part is +1. The ligand oxalato has a charge of –2, so iron should be in +3 state meaning O2 to be neutral. Now had O2 been superoxo (O2–) or peroxo (O2– – ) the negative charge of the ligands should have been –3 and –4 respectively. In that case Iron has to be +4 and +5 which is not possible. So O2 will behave as a neutral ligand and IUPAC name is Tetraammineoxalatodioxygeniron (III) chloride.
Bonding in Co-Ordination Compounds
Pauling proposed a simple valence bond theory to explain bonding in Co-ordination Compounds According to this theory:

1. Central metals loses a requisite no. of electrons to form the ion.
   
2. The cation orbitals hybridize to form a new set of equivalent hybridized orbitals with definite directional properties.
   
3. Each ligand contains a lone - pair of electrons. A covalent bond is formed by the overlap of a vacant hybridized metal orbital and a filled orbital of the ligand.
   
4. If the complex have unpaired electron then it will be paramagnetic if not then it will be diamagnetic.
   

To understand the valence bond concept let us take some examples of co-ordination compounds.
1. [Cr(NH3)6]3+, the central metal is in +3 oxidation state
Orbitals of Cr3+ion
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d2sp3 hybridized orbitals of Cr3+
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d2sp3hybrid orbitals of [Cr(NH3)6]3+
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shape is octahedral
2. [Ni(CN4)]2-, Ni is in +2 oxidation state
Atomic orbitals of Ni
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dsp2 hybridized orbitals of Ni+2
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Hybridized orbitals of [Ni(CN)4]2-
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shape is square planar
Illustration 7: [FeF6]4– is an outer orbital complex while [Fe(CN)6]4– is an inner orbital complex
Solution: The oxidation state of Fe in both complexes is +2. i.e, the configuration is 4s°3d6.
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Now F– is a very weak ligand compared to CN–. Now a strong field ligand is one which can pair up the electrons while a weak field ligand is one which cannot pair up. So in presence of CN– the electrons in the d – orbitals of Fe2+ will be paired up. This will leave two vacant 3d orbitals which will then undergo hybridisation to form d2sp3. As the 3d orbital has participated with 4s and 4p orbitals for hybridisation it is called inner orbital complex. While had the outer d orbital ie 4 d participated with 4s and 4p, then the complex should have been called outer orbital complex. This happens in [FeF6]4–. In presence of F– which is a weak field ligand, electrons are not paired up and therefore there is no vacant 3d orbitals and so it is the outer 4d that participates.
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Moreover in [Fe(CN)6]4– there is no unpaired electrons and hence diamagnetic while in [FeF6]4– there are unpaired electron, hence paramagnetic.
State of Hybridisation and Magnetic Behaviour of Some Co-ordination Complex
Metal complex

Metal ion

Configuration of metal ion
Hybridiation of metal ion orbitals for ligand bonds
Number of unpaired electtrons
Magnetic behaviour
[V(H2O)6]3+
V3+
d2
sp3d2
2
Paramagnetic
[Cr(NH3)6]3+
Cr3+
d3
d2sp3
3
Paramagnetic
[MnF6]3–
Mn3+
d4
sp3d2
4
Paramagnetic
[Mn(CN)6]3–
Mn3+
d4
d2sp3
2
Paramagnetic
[Fe(CN)6]3–
Fe3+
d5
d2sp3
1
Paramagnetic
[FeF6]3–
Fe3+
d5
sp3d2
5
Paramagnetic
[FeCl4]2–
Fe2+
d6
sp3
4
Paramagnetic
[CoF6]3–
Co3+
d6
sp3d2
4
Paramagnetic
[Co(CN)6]3+
Co3+
d6
d2sp3
0
Diagmagnetic
[Ni(Cl)4]2–
Ni2+
d8
sp3
2
Paramagnetic
[Ni(CN)4]2–
Ni2+
d8
dsp2
0
Dimagnetic
[CuCl4]2–
Cu2+
d9
sp3
1
Paramagnetic
[ZnNH3)4]2+
Zn2+
d10
sp3
0
Diamagnetic
aramagnetic

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ORES & METALLURGY

Commercially important ores of iron, copper, lead, magnesium, aluminium, tin and silver. Carbon reduction process (iron and tin), Self reduction process (copper and lead), Electrolytic reduction process (magnesium and aluminium), Cyanide process (silver and gold).
Introduction
The earth’s crust is the main source of metals. The occurrence of metal in native or in combined state in the earth’s crust along with a number of rocky and other impurities depends upon the chemical nature of metals. Metals having less electropositive character have less affinity for oxygen, moisture and occur in free or metallic or native state i.e., in uncombined state e.g. Au, Pt , Ag etc. On the other hand metals with higher electropositive character occurs in combined state i.e., as compounds.
The compound of a metal found in nature is called a mineral. A mineral may be a single compound or a complex mixture. Those minerals from which metal can be economically extracted are called ores. Thus all ores are minerals but all minerals are not ores. For e.g. copper occurs in nature in the form of several minerals like Cu2O, Cu2S, CuFeS2, but copper pyrites is considered as the most economical mineral for the extraction of the metal. Hence copper pyrites is the chief ore of copper.
Ores may be divided into four groups:
i) Native Ores: These ores contain the metal in free state eg. Silver gold etc. These are usually formed in the company of rock or alluvial impurities like clay, sand etc.
ii) Oxidised Ores: These ores consist of oxides or oxysalts (eg. carbonates, phosphates, nitrates, sulphates, silicates etc.) of metal. The important oxide ores include, Fe2O3, Al2O3.2H2O etc. and important cabonate ores are limestone (CaCO3), Calamine (ZnCO3) etc.
iii) Sulphurised Ores: These ores consist of sulfides of metals like iron, lead, mercury etc. Examples are iron pyrites (FeS2). galena (PbS), Cinnabar (HgS)
iv) Halide ores: Examples include horn silver (AgCl), fluorspar (CaF2) etc.
Metallurgy
It is the process of extracting a metal from its ores. The following operations are carried out for obtaining the metal in the pure form.
i) Crushing the ore
ii) Dressing or concentration of the ore.
iii) Working up of the concentrated ore
v) Purification or refining of the metal.

Various Metallurgical Process

Crushing and Grinding: The big stones of ore are first crushed into small pieces in gyratory crushers and then ground to powder form. The process of grinder crushed ore into fine powder is known as pulverization.
Concentration / Dressing of Ore: Presence of earthy matter, sand limestone etc in the ores are called gangue or matrix. The removal of these impurities from ores is known as ore-dressing or ore-concentration. Process used to concentrate an ore is known as beneficiation process. Concentration can be carried out in one of the following ways depending on the nature of metal.
a) Hand-picking: The impurities which are distinguishable to naked eye can be easily separated by hand-picking.
b) Levigation or Gravity separation: This method of concentrating the ore is based on the difference in densities of the ore particles and impurities. Generally oxide and carbonate ores are concentrated by this method.
c) Magnetic concentration: This method is adopted when either the ore or the impurities are magnetic in nature. The two can be separated from each other by means of magnetic separators. By this method chromite ( an ore of chromium) being magnetic can be separated from the siliceous impurity. Similarly SnO2( non magnetic) and (TiO2) magnetic can be separated.
d) Froth Floatation Process: This process is mainly suitable for sulfide ores. The process is based on the different wetting characteristics of the ore and gangue particles with water and oil. The former is preferentially wetted by oil and the later by water. In certain cases other chemical compounds are also added during the process. These reagent act either as collectors, activators or depressants e.g. of collectors is ethyl xanthate, that of activators is copper sulfate while that of depressant is sodium cyanide.
e) Electrostatic concentration: This method is based upon the fact that the particles which are good conductors of electricity becomes electrically charged under the influence of an electrostatic field and are thus repelled by the electrode carrying the like charge. This is used to separate lead sulfide and zinc sulfide which are found together in nature.
f) Chemical Method: This process is done in cases where the ore is required in a very pure form Eg. in the concentration of bauxite containing ferric oxides, TiO2 and silica as impurities.
Calcination and Roasting
Calcination is a process in which the ore is heated, generally in absence of air to expel water from a hydrated oxide carbondioxide from a carbonate at a temperature below their melting points. For eg.
Al2O3.2H2O ⎯→ Al2O3 + 2H2O
CaCO3 ⎯→ CaO + CO2
Roasting is a term used to denote the process in which the ore (usually sulfide) alone or mixed with other materials is heated usually in presence of air, at temperature below their melting points
2ZnS + 3O2⎯→ 2ZnO +2SO2
CuS + 2O2 ⎯→ CuSO4
Calcination proceeds only with expulsion of some small molecules like water, CO2, SO2 etc. without any chemical change while during roasting definite chemical changes like oxidation, chlorination takes place. Calcination and roasting are generally carried out in reverberatory furnace. In this type of furnace the charge is placed on the hearth and heated by the flames deflected from its concave roof.
Leaching
This involves treatment of the ore with suitable reagent (acids bases or other reagents) which can selectively dissolve the ore but not the impurities. For eg. In the purification of white bauxite ore NaOH is used as the reagent to dissolved Al2O3 leaving behind Fe2O3 (impurity)
Al2O3+2NaOH→2NaAlO2+H2O
NaAlO2+2H2O→ Al(OH)3NaOH
2Al(OH)3 📷Al2O3+3H2O
Pure alumina
Reduction to free metal
The calcined or roasted ore is then reduced to the metallic state in either of the following ways.
a) Smelting: The oxides of less electropositive metals like Pb, Zn, Fe, Cu are reduced by strongly heating them with coke or coal. Reduction of oxide with carbon at high temperature is known as smelting. Blast furnace is used in reduction of iron oxides while reverberatory furnace in used in the reduction of tin oxide.
b) Electrolytic Reduction: Highly electropositive metals like Na, K, Al are extracted by electrolysis of their oxides, hydroxides or chlorides in fused state and the metal is liberated at cathode Eg. aluminum is obtained by electrolysis of alumina mixed with cryolite which acts as a electrolyte. Oxides of highly electropositive metals can’t be reduced with carbon because such metals react with carbon to form carbide at high temperature.
c) Reduction by Aluminum ( Alumino – Thermit process): Certain oxides like Fe2O3, Cr2O3 are not satisfactory reduced by carbon. In such cases Al is used as a reducing agent. An example is the extraction of Cr from Cr2O3.
Cr2O3 + 3Al ⎯→ Al2O3 + 2Cr + heat
Fe2O3 + 2Al ⎯→ Al2O3 + 2Fe+ heat
Apart from this reduction can also take place by precipitation and air. The former is known as hydrometallurgy and the latter is auto reduction.
All these processes give metals which are not totally pure. But in some cases we need metals of nearly cent percent purity. That may be achieved in either of the following ways depending on the metal.
a) Liquation process
b) Distillation process
c) Oxidation process (Bessemerisation, cupellation, poling)
d) Electro refining
e) Van Arkel process
f) Zone refining
Flux
The ores even after concentration contain some earthy matter called gangue which is heated combine with this earthy matter to form an easily fusible material called as slag. Such a substance is known as flux.
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Fluxes are of the following two types:
i) Acidic fluxes like silica, borax etc. These are used when the gangue is basic such as lime or other metallic oxides like MnO, FeO, etc. The chemical reaction which takes place in removing a basic gangue is follows:
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ii) basic fluxes like CaO, lime stone (CaCO3), magnesite (MgCO3), haematite (Fe2O3) etc. These fluxes are used when the gangue is acidic like silica, P4O10 etc. For example
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Different types of Furnaces
a) Blast Furnace: It is used for smelting iron, copper and lead ores. It has a tall structure made of steel plates rivetted together lined inside with fire bricks. The furnace is provided with an arrangement for blowing air near the base, a slag hole, a tapping hole for removing the molten metal and an exist for waste gases near the top. The charge is introduced at the top. The mouth of the furnace is closed by a special arrangement through which the charge is introduced.
b) Reverberatory Furnace: It is used for calcination, roasting or for smelting. The charge in the powdered form along with a flux is placed on the hearth of the furnace. The fuel is placed on grate and the flames are deflected from its low sloping roof. This furnace is used in case of copper, tin lead, and wrought iron.
c) Muffle Furnace: A furnace in which the charge contianer is heated from all sides is called muffle furnace. The burnt materials including the vapours of the metal are allowed to escape through the same opening. The metal is obtained by condensing the vapours. In Belgian process for extracting zinc, the retorts act as muffles and prolongs help in completing the condensation.
d) Bessemer Converter: It is a pear-shaped furnace made of steel plates lined inside with lime or magnesium oxide. It is fixed on pivots and can be tilted in any direction. It is provided with pipes through which hot air can be blown for heating purposes.
Illustration 1: Explain the following processes as applied to metallurgy
i) Amalgamation
ii) Hydrometallurgy
iii) Solvent Extraction
Solution: i) Amalgamation: this process is used for the extractioin of noble metals like gold and silver from the native ores. The finely crushed ore is brought into contact with mercury which combines with the metal forming its amalgam. The metal is then recovered by distilling the amalgam.
ii) Hydrometallurgy or precipitation by a more electropositive metal: Hydrometallurgy is the process of bringing the metal into solution by the action of suitable chemical reagents (e..g., sodium cyanide solution or chlorine in present of water etc.) followed by recovery of the metal by the use of a proper precipitating agent which is a more electropositive metal.
For example, poor ores of silver are suspended in a dilute solution of sodium cyanide and air is blown through it when the silver present goes into solution as the argentocyanide complex. From this solution the metal is precipitated by adding zinc turnigs.
4Ag + 8NACN + O2 (air) 2H2O ⎯→ 4Na [Ag(CN)2] + 4NaOH
or AgCl + 2NaCN ⎯→ Na[Ag(CN)2] + NaCl
2Na [Ag(CN)2] + Zn ⎯→ Na2[Zn(CN)4] + 2Ag↓
iii) Solvent Extraction: Solvent extraction is the latest separation technique and has become popular because of its elegance, simplicity and speed. The method is based on preferential solubility principles.
Solvent or liquid-liquid extraction is based on the principle that a solute can distribute itslf in a certain ratio between two immiscible solvents, one of which is usually water and the other an organic solvent such as benzene, carbon tetrachloride or chloroform. In certain cases, the solute can be more or less completely transferred into the organic phase. The technique can be used for purpose of preparation, purification, enrichment, separation and analysis.
Illustration 2: Explain the following process as applied to metallurgy
i) Zone refining (Fractional crystallization)
ii) Electro-refining
iii) Van-Arkel Method
Solution: i) Zone refining (Fractional crystallization): This method is employed for preparing extremely pure metals. This method is based upon the principle that when a molten solution of the impure metal is allowed to cool, the pure metal crystilises cut while the impurities remain in the melt.
ii) Electro-refining: In this method, the impure metal is converted into a block which forms the anode while cathode is a rod or plate of pure metal. These electrodes are suspended in an electrolyte which is the solution of a soluble salt of the metal usually a double salt of the metal. When electric current is passed, metal ions from the electrolyte are deposited at the cathode in the form of pure metal while an equivalent amount of metal dissolves from the anode and goes into the electrolyte solution as metal ion. The soluble impurities present in the crude metal anode go into the solution while the insoluble impurities settle down below the anode as anode mud.
iii) Van-Arkel Method: In this method, the metal is converted into it volatile unstable compound such as iodide leaving behind the impurities. The unstable compound thus formed is decomposed to get the pure metal.
Ti(s) + 2F2(s) 📷TiF4(g) 📷 + 2F2(g)
Exercise 1: i) Which furnace is meant for roasting and calcination?
ii) If FeO is present in the ore as impurity then what type of flux is added to remove it?
Extraction of Sodium
Ores of Sodium
Sodium chloride or common salt (NaCl)
Sodium Nitrate or Chile Salt peter (NaNO3)
Sodium Carbonate (Na2CO3)
Sodium Sulphate (Na2SO4)
Two processes are in practice for extracting Na

1. Castner’s process
   
2. Down’s process
   

Castner process
Principle : Fused caustic soda is electrolysed
Cathode : Steel
Anode : Nickel
Cathode reaction
OH– – e ⎯→ OH
4OH ⎯→ 2H2O + O2↑
The water formed at cathode then dissociates as H2O 📷 H+ + OH– and H+ then gets discharged at cathode
H+ + e ⎯→ H; 2H ⎯→ H2;
at anode OH– – e– ⎯→ OH; 2OH ⎯→ H2O + 📷O2↑
The bath temperature is kept higher than the melting point of Na (318°) to dissolve Na in NaOH. Metallic sodium is taken from the cathode chamber from time to time with spoon.
Advantages

1. Since the temperature is 327°C, not much Na is vaporised
   
2. Valuable H2, O2 are produced
   

Disadvantages

1. Costly raw material caustic soda is used
   
2. 50% of the electrical energy is utilised for the production of Na and 50% wastage for the hydrolysis of water.
   

Down’s Process
Principle
Raw materials used : Anhydrous CaCl2 (66.8%), NaCl (32.2%)
Temperature : 600°C
Anode : Graphite
Cathode : Iron
NaCl ⎯→ Na+ + Cl–
Cathode reaction : Na+ + e ⎯→ Na (Reduction)
Anode reaction : Cl– – e ⎯→ Cl (oxidation)
2Cl ⎯→ Cl2
The metal produced at cathode is collected form the surface of the molten electrolyte and then collected kept under kerosene.
Comparison between Castner’s process and Down’s Process
Castner’s Process
Down’s process

1. Molten NaOH is used
   
2. Mixture of NaCl (32%) and anhydrous CaCl2 (66.8%) are used
   
3. Temperature maintained 320–330°C
   
4. Temperature of the both 600°C
   
5. Iron vessel is the electrolyte cell
   
6. Iron vessel lined inside in the acid resistant refractory.
   
7. 50% electrical energy is utilised for the preparation of Na.
   
8. Whole electrical energy is utilised.
   

Extraction of Potassium
Ores of Potassium
Sylvine KCl
Kainite KCl. MgSO4. 3H2O
Carnallite KCl.MgCl2. 6H2O
Felspar KAlSiO8
Saltpetre KNO3
Principle of Extraction of Potassium
Potassium can be extracted by reducing molten KCl in the metallic sodium at 850°C.
Na(g) + KCl(l) 📷 NaCl + K(g)
K being more volatile distills off. Reactivities of K is similar to Na. Potassium thus obtained is stored under kerosene.
Extraction of Calcium
Ores of Calcium
CaCO3 (Lime stone)
CaSO4.2H2O (Gypsum or alabaster)
Ca3(PO4) (Phosphorite)
Ca(PO4)2.CaF2 (Fluorapatite)
CaO.Al2O3.2SiO2 (Lime feldspar)
Extraction of Calcium
Flow chart diagram
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Illustration 3: Fluorspar is added to the anhydrous CaCl2 in the extraction of metallic calcium - Explain why?
Solution: This is because of the fact that, the melting point of CaCl2 is 780°C. Consequently much electrical energy is consumed to fuse CaCl2 at this high temperature. As a result the cost of production of calcium is increased. But if a little fluorsper is added to CaCl2, the melting point of CaCl2 is reduced from 780°C to 664°C. As a result the electrolysis of CaCl2 can be carried out a relatively low temperature. Consequently the consumption of electrical energy becomes much low and at the same time the cost of production becomes low. Besides this, the addition of CaF2 lowers the viscosity of the electrolyte and increases its electrical conductivity.
Extraction of Magnesium
Ores of Magnesium
Magnesite : MgCO3
Dolomite : CaCO3.MgCO3
Kieserite : MgSO4.H2O
Kainite : K2SO4.MgSO4.3H2O
Carnalite : KCl. MgCl2.6H2O
Talc : 3MgO. 4SiO2.H2O
Spinel : MgO. Al2O3
Extraction of Magnesium:
Process : Electrolysis
Anode : Graphite rod
Cathode : Iron pot (rectangular)
Electrolyte : MgCl2 (molten,75%),NaCl (25%)
Temperature : 700°C
Dissociation Step : MgCl2 📷 Mg+2 + 2Cl–
Cathode reaction : Mg2+ + 2e ⎯→ Mg
Anode reaction : 2Cl– – 2e ⎯→ Cl2↑
From sea water anhydrous MgCl2 is obtained by the following reactions
MgCl2 + Ca(OH)2 ⎯→ Mg(OH)2↓ + CaCl2
Mg(OH)2 + 2HCl ⎯→ MgCl2 + 2H2O
MgCl2.6H2O crystallises.
MgCl2. 6H2O📷 MgCl2+6H2O
By thermal reduction Mg can be produced
CaC2 + MgCl2 📷 CaCl2 + Mg + 2C
Exercise 2: In extraction of Magnesium by electrolytic reduction process NaCl is added to MgCl2 – Explain why?
Extraction of Zinc
Ores of Zinc
Zincite (ZnO)
Franklinite (ZnO, Fe2O3)
Zinc Blende (ZnS)
Calamine (ZnCO3)
Willemite (ZnSiO3, ZnO)
Electric Calamine (ZnSiO3, ZnO, H2O)
Principle

1. Concentration of Zinc blende (ZnS) by froth floatation process with pine oil (foaming agent), sodium xanthate (collectors) and little acid. The oil forms froth with air and at first galena rises to surface and are removed. Next the ZnS particles are collected from the top.
   
2. Roasting: Temperature: 860 – 900°C
   

2ZnS + 3O2 📷 2ZnO + 2SO2↑
2ZnS + 2O2 📷 ZnSO4
2ZnSO4 📷 2ZnO + 2SO2 + O2

1. Smelting - Carbon reduction
   
2. Belgian process
   

Raw material: Powdered coke (1/5 th weight of roasted ZnO)
Temperature: 1300° – 1400°C
Heating process is carried out and Zn(vapour) is formed and condenses Blue flame from the prolong is observed till the end of the reaction (due to formation of CO).
ZnO + C 📷 Zn + CO
Refining of Zn
Electrical refining is done by using impure Zn as the anode and ZnSO4 electrolyte and pure Zn as the cathode. The following electrolysis reaction will take place.
ZnSO4 📷 Zn+2 + SO4–2
Anode reaction: Zn – 2e 📷 Zn+2
Cathode reaction: Zn+2 + 2e 📷Zn(s)
Extraction of Mercury
Ore of Mercury: Cinnabar (HgS)
Modern Process: Cinnabar is crushed to powder and the crushed ore is then concentrated by forth flotation process. The concentrated ore is then charged in a shaft furnace through a hopper with charcoal. Furnace is fired and the following reaction takes place.
2HgS + 3O2 ⎯→ 2HgO + 2SO2↑
HgO + C ⎯→ Hg + CO↑
Extraction of Silver
Ores of Silver
Silver glance or Argentite (AgS)
Horn silver (AgCl)
Ruby Silver (3Ag2S, Sb2S3)
Silver Copper glance or stromeyerite (Ag2S, Cu2S)
Extraction of Silver

1. Concentration: Crushed powder is concentrated by froth floatation process.
   
2. The concentrated Ag2S is then treated with 0.5% solution of NaCN in presence of air.
   

Ag2S + 2NaCN 📷2AgCN ↓ + Na2S
AgCN + NaCN 📷Na[Ag(CN)2]
4Na2S + 5O2(air) + 2H2O 📷2Na2SO4 + 4NaOH + 2S

1. Precipitation of Silver

As, 📷
Hence Zn will replace Ag from the complex as follows
2Na[Ag(CN)2] + Zn ⎯→ Na2[Zn(CN)4] + 2Ag↓

1. Electro-refining: Metallic silver thus obtained contains impurities like Zn, Cu, Au etc. It is purified by electro-refining process. The electrolytic bath contains AgNO solution with 10% HNO3. Impure silver sheets are used as anode and thin sheets of pure silver are used as cathode. As electric current is passed through the electrolyte , the anode gradually dissolves and pure silver is deposited on the cathode. Impurities like Zn and Cu go into the solution while gold settles down at the anode as anode mud.

Extraction of Tin
Ores of Tin
Cassiterite or tinstone (SnO2)
Extraction

1. Concentration of ore: Crushed ore is washed by a stream of water and mainly FeWO4 is present at this stage.
   
2. Roasting: S, As present in the one ore are volatalised away in the form of their oxides in a revolving furnace. Iron and copper present in the ore are converted into their oxides and sulphates.
   
3. Electromagnetic separation: From the hopper the dry ore is supplied and from the belt the FeWO4 is seperated by magnetic field. It contains 75%. SnO2 (black tin)
   
4. Carbon reduction:
   

SnO2 + 2C ⎯→ Sn + 2CO
Temperature 1200 – 1300°C
Furnace: Reverberatory furnace.
The molten tin is collected as it got 99% pure Sn. The upper layer of the slag contains 10-20% Sn. This is smelted with coke.
SnSiO3 + CaO + C ⎯→ Sn + CaSiO3 + CO
Slag

1. Refining: This is carried out by sweating process. Sn melts leaving behind Fe, CuO, N. Tin thus obtained is 99.5% pure.

Recovery of Sn from Scrap
The scrap tin plates, free from dirts are reacted by Cl2 to form SnCl4. SnCl4 being liquid is voltalised away and is condensed.
SnCl4 + 2H2O ⎯→ SnO2 + 4HCl
SnO2 + 2C ⎯→ Sn + 2CO↑
Extraction of Copper
Copper glance or Chalcocite: Cu2S
Chalcopyrites(copper pyrites): Cu2S⋅Fe2S3
Cuprite Cu2O
Malachite [CuCO3.Cu(OH)2]
Azurite [2CuCO3.Cu(OH)2]
Extraction of Metallic Copper
1. Concentration of the ore by froth floatation process: Copper pyrites contains only (2-3)% of copper. The rest of the ore contains iron or sulphide, silica, silicious materials, sulphur, arsenic etc. as impurities. Froth flotation by Xanthate and pine oil. The froth is collected and dried when concentrated ore is obtained which contains 25-30% of Cu.
2. Roasting
Cu2S.Fe2S3 + O2 ⎯→ Cu2S + 2FeS + SO2↑
Cu2S.Fe2S3 + 4O2 ⎯→ Cu2S + 2FeO + 3SO2↑
2Cu2S + 3O2 ⎯→ 2Cu2O + 2SO2↑
Cu2O + FeS ⎯→ Cu2S + FeO
3. Smelting of the roasted ore in blast furnace: material required

1. Roasted ore
   
2. Lime stone
   
3. Coke (used as fuel)
   
4. Silica (used as flux)
   
5. Lime stone (used to remove excess silica)
   

Reactions occurring given as follows
2FeS + 3O2 ⎯→ 2FeO + 2SO2↑
Cu2O + FeS ⎯→ Cu2S + FeO
FeO + SiO2 ⎯→ FeSiO3 (removed as slag)
CaO + SiO2 ⎯→ CaSiO3 (removal as slag)
4. Self reduction in Bessemer Converter
Reactions involved are,
2FeS + 3O2 ⎯→ 2FeO + 2SO2
FeO + SiO2 ⎯→ FeSiO3 (slag)
2Cu2S + 3O2 ⎯→ 2Cu2O + 2SO2↑
Cu2S + 2O2 ⎯→ Cu2SO4
When 2/3 of the cuprous sulphide is oxiidsed, the balst is stopped. The produced Cu2O and Cu2SO4 are reduced by the rest of cuprous sulphide to produce metallic copper with the evolution of SO2.
Cu2S + 2Cu2O ⎯→ 6Cu + SO2↑
Cu2SO4 + Cu2S ⎯→ 4Cu + 2SO2↑
As the molten copper cools, it gives off the dissolved of SO2. The SO2 gas escaping in the form of bubbles, leaves the surface of the metal with full of cavities which gives the metal a blistered appearnace. This is why the metal thus obtained is by blister copper.
5. Refining
Electrorefining
Anode: Impure copper obtained above
Cathode: Pure copper
Electrolyte: 15% CuSO4 solution + 5% H2SO4
When electric current is passed through the electrolyte, the anodes gradually dissolve and pure copper is deposited on the cathodes which gradually grow in size. The impurities like Fe, Zn, Ni etc., dissolved in the solution as sulphates while gold, silver, platinum settle down below the anode as anode mud.
Reactions coming are as follows
CuSO4 📷 Cu+2 + SO4–2
At anode: Cu – 2e ⎯→ Cu+2
At cathode: Cu+2 + 2e ⎯→ Cu
Exercise 3: Name the by products obtained during the extraction of copper from copper pyrties.
Extraction of Iron
Ores of Iron
Magnetite (Fe3O4)
Red haematite (Fe2O3)
Brown haematic Limonite or (Fe2O3.3H2O)
Spathic iron or sidenite (FeCO3)
Iron pyrites (FeS2), Copper pyrites (CuFeS2)
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Extraction of Cast Iron

1. Dressing of the ores: The iron ores are first broken into small pieces 3-5 cm in size and washed with water to remove clay & sand.
   
2. Roasting or Calcination
   

During roasting S, As, P are oxidised to the respective oxides.
S + O2 ⎯→ SO2↑
4As + 3O2 ⎯→ 2As2O3↑
FeCO3 decomposes as,
FeCO3 ⎯→ FeO + CO2↑
Fe2O3.3H2O loses water
Fe2O3.3H2O ⎯→ Fe2O3 + 3H2O
Fe3O4 is decomposed to ferrous oxide and ferric oxides.
Fe3O4 ⎯→ FeO + Fe2O3
Ferrous oxide reacts with silica to forms ferrous silicate at high temperature.
FeO + SiO2 ⎯→ FeSiO3
But the conversion of FeO into Fe2O3 will prevent the formation of FeSiO3. Due to escape of gases and moisture entire mass of the ore becomes porous causing the increase in the effective surface area.
3. Semlting in the Blast furnance
Blast furnace is a shaft furnace made of steel plate of 20-30 in with 4 – 4.6 diameter.
Bosh: The diameter of the furnace gradually increases from the top downwards. Widest part of the furnace is called Bosh. At above 2 m. tuyeres are there through which hot air blast is blown into the furnace.
Hearth: Below the bosh this region exists. (1) slag notch is at higher height and (2) tap hole for metal passage at lower position from the bottom. At the top of the furnace the hopper is there which is cup and cone arrangement. Through this charge is introduced till the charge bed in the furnance is 4/5th the of the furnace. Hot air at 700°C is forced into the furnace through the tuyeres. The thermal gradient inside exists from 1800°C (hearth) to 400°C – 900°C in the upper region. Near the bosh the temperature varies from 1200°C – 1300°C chemical reactions which take place are.

1. At 1200°C near the tuyeres,

C + O2 ⎯→ CO2; CO2 + C ⎯→ 2CO

1. Above bosh 600° – 900°C, ferric oxide is partially reduced by CO as

Fe2O3 + 3CO 📷 2Fe + 3CO2↑

1. CaCO3 📷 CaO + CO2

2CO 📷CO2 + C
Fe2O3+ 3C ⎯→ 2Fe + 3CO
CaO + SiO2 ⎯→ CaSiO3 (slag)

1. The reaction at 1500°C, MnO2 is reduced to Mn.

MnO2 + 2C ⎯→ Mn + 2CO
Ca3(PO4)2 + 3SiO3 ⎯→ 3CaSiO3 + P2O5
CaO + SiO2 ⎯→ CaSiO3
2P2O5 + 10C ⎯→ P4 + 10CO
5. Collection of Cast Iron
Metal is cast into ingots or in the laddle for further refining like steel making.
There are two types of cast iron
a) White cast iron: This type of cast iron contains carbon mostly in the form of iron carbide (Fe3C) and small amount of graphite.
b) Grey cast iron: This type of cast iron contains carbon mostly in the form of graphite and small amount of iron carbide.
Wrought Iron
Minimum % of carbon, is 0.1 – 0.15% and other impurities like S, P, Mn, Si less than 0.3%.
Manufacturing Porcess
Cast iron takes in puddling furnance and metled by hot blast of air. The chemical reactions which occur are
S + O2 ⎯→ SO2↑ ; 3S + 2Fe2O3 → 4Fe + 3SO2↑
3Si + 2Fe2O3 ⎯→ 4Fe + 3SiO2
Mn + Fe2O3 ⎯→ 2Fe + 3MnO
MnO + SiO2 ⎯→ MnSiO3 (slag)
3C + Fe2O3 ⎯→ 2Fe + 3CO
4P + 5O2 ⎯→ 2P2O5; Fe2O3 + P2O5 ⎯→ 2FePO4 (slag)
The impurities are removed from iron, the melting point of the metal rises and it becomes a semi solid mass. Metal taken out from the furnace in the form of balls with the help of the rabbles. The balls are then beaten under hammer to separate out the slag. The product thus formed is called wrought iron.
Steel

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CONCEPTS OF ACIDS AND BASES
Strength of acids and bases (qualitative treatment), balancing equations of chemical reaction, (including oxidation-reduction using ion-electron and oxidation number methods)
Introduction
There are several so-called theories of acids and bases, but they are not really theories but merely different definitions of what we choose to call an acid or a base. Since it is only a matter of definition, no theory is more right or wrong than any other, and we use the most convenient theory for a particular chemical situation. So before we talk of strength of acids and bases, we need to know several theories.
Various Theories Regarding Acids & Bases
i) Arrhenius Theory: According to Arrhenius, substances producing H+ ions in solution are acids and those producing OH- ions in solution are bases. Therefore, substances like H2O, HCl, H2SO4, CH3 COOH etc. are acids and the ones like NH4OH, NaOH, KOH, H2O etc. are bases.
ii) Bronsted-Lowry Theory: In 1923, Bronsted and Lowry independently defined acids as proton donors, and bases as proton acceptors. For aqueous solutions the definition does not vary much for acids from the Arrhenius theory but it widens the scope of bases. In this, the bases need not contain OH- ions and simply have to accept protons. So ions like Cl-, CH3COO-, Br- etc. which do not contain OH- ions can be considered as bases under this definition.
Levelling Solvents: Whenever an acid is dissolved in water, it acts as an acid only if the solvent acts as a base. That is, if we dissolve acids like HCl, HNO3 etc in water, their acidic strength is almost the same since water acts as a base for both these acids. Infact, it is known that all strong acids show equal acidic strength when dissolved in water. This is because, water acts as a base to all these acids and thus forces them to donate almost the same amount of protons irrespective of their chemical nature. Since water levels the acidic strength of strong acids, it is referred to as a levelling solvent. In order to measure the strength of strong acids, they are dissolved in glacial acetic acid and the amount of protons measured by conductometry. It is found that the strength of acids varies as
HClO4 > HBr> H2SO4 > HCl > HNO3
Amphiprotic species: Many molecules and ions can behave like water and may either gain or lose a proton under the appropriate conditions. Such species are said to be amphiprotic. eg.
Acid1 Base2 Acid2 Base1
HS- + OH- H2O + S2-
HBr + HS- H2S + Br-
📷 + CN- HCN + 📷
H3O+ + 📷 H2CO3 + H2O
The hydroxides of metals near the boundary between metals and non-metals in the periodic table, are amphiprotic and so react either as acids or as bases.
[Al(H2O)3 (OH)3] + OH- H2O + [Al(H2O)2(OH)4]-
H3O+ + [Al(H2O)3(OH)3] [Al(H2O)4(OH)2]+ + H2O.
Exercise 1: Identify conjugate acid base pair in the following reactions
C6H5OH + H2O 📷 H3O+ + C6H5O–
HCl + OH– 📷 H2O + Cl–
iii) Lewis Theory: Lewis developed a definition of acids and bases that did not depend on the presence of protons, nor involve reactions with the solvent. He defined acids as materials which accept electron pairs, and bases as substances which donate electron pairs. Thus a proton is Lewis acid and ammonia is Lewis base since, the lone pair of electrons on the nitrogen atom can be donated to a proton:
H+ : NH3 → [H ← : NH3]+
📷
Conditions to be a Lewis Acid
i) Compounds whose central atoms have an incomplete octet. E.g. BF3, AlCl3, GaCl3 etc.
ii) Compounds in which the central atom has available d-orbital and may acquire more than an octet of valence electrons.
E.g. SiF4 + 2F- → SiF62-
Other examples are : PF3, SF4, SeF4, TeCl4.
iii) All simple cations : Na+, Ag+, Cu2+, Al3+, Fe3+ etc.
Conditions to be a Lewis Base
i) All simple negative ions, Cl- , F- etc.
ii) Molecules with unshared pair of electrons, H2O, NH3 etc.
iii) Multiple bonded compounds which form co-ordination compounds with transition-metals . E.g., CO, NO, Ethylene, Acetylene etc.
Exercise 2: In the following reaction, identify each of the reactant as a Lewis acid or Lewis base
i) Cr+3 + 6H2O ⎯⎯→ 📷
ii) BF3 + (C2H5)2O ⎯⎯→ F3B: O(C2H5)2
Hard and soft acids and bases (Principle of HSAB)
Lewis acids & bases are classified as hard & soft acids & bases. Hardness is defined as the property of retaining valence electrons very strongly.
Thus hard acid is that in which electron-accepting atom is small with a high positive charge and no electron which are easily polarised or removed.
Soft acid is that in which the acceptor atom is large, carries a low positive charge or it has electrons in orbitals which are easily polarised or distorted.
A lewis base which holds its electrons strongly is called hard base eg. OH-, F-, NH3, H2O etc.
A lewis base in which the position of electrons is easily polarised or removed is called a soft base eg. I-, CO, CH3S-, (CH3)3P etc.
A hard acid prefer to bind to hard bases and soft acids prefer to bind to soft bases. The bonding between hard acids & hard bases is ionic & that between soft acids & bases is mainly covalent.
Exericise-3: Identify the hard acid and hard bases among the following
OH–, H2O, Al+3, Cr+3, Sn+4, 📷
Relative Strength of Acids and Bases
i) Predict the relative acidic strength among the following:
CH3 - H, H2N - H, OH - H, F -H
Assume that all these compounds have lost their protons. So we get -CH3 , -NH2, -OH, F-
We shall qualitatively analyse the charge density in each of the species and we shall follow the rule given below:

• Larger the volume over which the charge is spread.
• Smaller is the charge density.
• Smaller is the basic character of the ion to attract a proton.
• Larger is the acidity of the conjugate acid.

In the above case one can see that, the size of the central atom over which the negative charge is present is decreasing from left to right. But it can also be seen that 3/4 th of the volume of C is overlapped by hydrogens in -CH3, 2/3 of the volume of N is overlapped by hydrogens in -NH2, 1/2 the volume of oxygen is overlapped by hydrogens in -OH where as, the entire volume of F is available for the charge in -F SO, actually the space available for the charge is increasing in the order -CH3 < -NH2 < -OH < -F. Therefore, the following conclusion may be arrived at.
CH4, NH3, H2O, HF.

1. Volume available for the negative charge is increasing in the conjugate bases from left to right.
   
2. Charge density of the conjugate bases is decreasing from left to right.
   
3. Basicity of the conjugate bases is decreasing from left to right.
   
4. Acidity of the acids is increasing from left to right.
   

Therefore the increasing acidic character is CH4 < NH3 ii) Predict the relative acidic strength among the following :
HF, HCl, HBr, HI.
Assume that each has lost a proton
We get F- , Cl- , Br- , I-

1. Volume available for the charge is increasing from left to right
   
2. Charge density is decreasing from left to right.
   
3. Basicity of the ions is decreasing from left to right.
   
4. Acidity of the conjugate acids is increasing from left to right.
   

Therefore increasing acidic strength is : HF iii) Predict the relative acidic strength among the following :
HOCl, HOClO, HOClO2 , HOClO3 .
Assume that each has lost a proton. So we get,
-OCl, -OClO, - OClO2, -OClO3
The species which have more than one oxygen atoms show resonance.
That is:
📷
It can be clearly seen that more the no. of oxygen atoms, more the resonance and more is the volume available for the negative charge.
Therefore -OCl, -OClO, - OClO2, -OClO3

1. Volume available for the negative charge is increasing from left to right
   
2. Charge density is decreasing from left to right.
   
3. Basicity of the anion is decreasing from left to right.
   
4. Acidity of the conjugate acid is increasing from left to right.
   

Therefore, the order of acidic strength is :
HOCl < HOClO < HOClO2 < HOClO3
iv) Predict the relative acidic strength among the following :
H2O, H2S, H2Se, H2Te.
Assume that each has lost a proton : So, we get : HO-, HS- , HSe- , HTe-
It can be easily seen that the volume available for the negative charge is increasing from HO- to HTe-, Therefore :

1. Volume available for the negative charge is increasing from left to right.
   
2. Charge density is decreasing from left to right.
   
3. Basiciy is decreasing from left to right.
   
4. Acidity of conjugate acids is increasing from left to right.
   

v) In general, for oxo-acids with different central atoms of same oxidation states, acidity increases with the decrease in size of the central atom.
This is because, smaller the size of the central atom of an oxo- acid of the type HOY, larger is the electro negativity of Y, more is the pull of the electrons of O-Y bond towards Y. This would result in a more positive character on O which would consequently help the H-O electrons to be pulled by O and release H as H+.
Therefore,
(a) HOI < HOBr (c) HPO3 < HNO3 (c) H3 AsO4 < H3 PO4.
For oxo - acids having the same central atom at different oxidation states, the rule to be followed is :
For oxo- acids of the general formula, (HO)m ZOn , where Z is the central atom, m is the no. of oxygen atoms attached to Z and attached to H, while n is the no of oxygen atoms attached to Z but not attached to H. It is these oxygen atoms which help in resonance and which determine the acidic strength.
For n= 0 Acid is very weak HOCl, (HO)3Br
n = 1 Acid is weak HOClO, HONO
n = 2 Acid is strong HOClO2, HONO2
n = 3 Acid is very strong HOClO3, HOIO3
Origin of Acidity and Basicity in Organic Compounds
i) Among hydrocarbons, acidity increases with the % `s' character . This is because, higher the `s' character, closer are the electrons to the nucleus of C and farther the electrons go from H, easily can H be removed as H+.
Therefore CH3 - CH3 < CH2 = CH2 < CH = CH
ii) Alcohols are more acidic than C because oxygen is more electro negative than C and consequently O - H bond easily give H+ than C - H bond
H3C - H < CH3 - O - H
iii) Carboxylic acids are more acidic than alcohols because carboxylate anion is more resonance stabilised than carboxylic acid as compared to alkoxide anion with respect to alcohol.
📷
Therefore CH3O - H < 📷
v) Phenols are more acidic than alcohols because phenoxide ion is more resonance stabilised as compared to phenol than alkoxide as compared to alcohol.
Therefore CH3 - O - H< C6H5 - O - H
vi) Phenols are less acidic than carboxylic acids. This is because in phenoxide ion some of the resonance structures are of higher energy and this makes phenoxide ion not very much stable as compared to phenol.
Therefore C6H5 O- H < 📷
vii) In simple aliphatic acids, more the no. of CH3 added, less is the acidity. This is due to the +I (inductive) effect of CH3. This effect of adding CH3 decreases with increase in distance between CH3 added to the substituent, which releases H+.
CH3 -COOH > CH3 - CH2 - COOH
(viii) Increase in `s' character increases acidity of carboxylic acids.
(CH3)2 - COOH < CH2 = CHCOOH ix) Increase in electron withdrawing substituents into simple aliphatic acids increases acidity.
CH3 - COOH < I - CH2 - COOH
📷
x) Nearer Cl is placed to COOH group, more is acidity.
📷
📷
xi) Benzoic acid is more acidic than carboxylic acids because benzoate anion shows/has more resonance.
CH3 COOH < C6H5 COOH
Exception to this is formic acid.
There is no destabilising factors for formate anion. Therefore, it is highly stabilised compared to formic acid because of resonance. Where as, Benzoic acid itself is resonance stabilised due to benzene ring. Therefore, benzoate ion is not very stable compared to benzoic acid. Thus C6H5 COOH < HCOOH
xii) More the no. of methyl or ethyl groups in NH3, more is its basicity. This is because of +I effect of methyl or ethyl which helps nitrogen to donate its lone pair of electrons.
NH3 < CH3 - NH2 < (CH3)2 NH
NH3 < EtNH2 < (Et)2 NH
An exception to the above trend is the trisubstituted derivative. It is seen that if we introduce an alkyl group in a secondary amine, the basic strength of amines actually decreases. This is due to the fact that, the basic strength of an amine in water is determined not only by electron - availability on the nitrogen atom, but also by the extent to which the cation, formed by uptake of a proton, can undergo solvation, and so become stabilised. The more hydrogen atoms attached to nitrogen in the cation, the greater the possibilities of powerful solvation via hydrogen bonding between these and water which increases stabilisation by solvation.
📷
Thus, on going along the series NH3 → RNH2 → R2NH → R3N, the inductive effect will tend to increase the basicity, but progressively less stabilisation of the cation by hydration will occur, which will tend to decrease the basicity. The net effect of introducing successive alkyl groups thus becomes progressively smaller, and an actual change over takes place, on going from secondary to a tertiary amine. That is why, (CH3)3 N < (CH3)2 NH
If this is the explanation, no such changeover should be observed, if measurements of basicity are made in solvent in which hydrogen- bonding cannot take place; it has, indeed, been found that, in chlorobenzene the order of basicity of the butylamines is
BuH2 < Bu2 NH xiii) Aniline is less basic than NH3. This is because in aniline the nitrogen atom is again bonded to an sp2 hybridised carbon atom but, more significantly, the unshared electron pair on nitrogen can interact with the delocalised orbital of the nucleus:
If aniline is protonated, any such interaction, with resultant stabilisation in the anilinium cation is prohibited, as the electron pair on N is no longer available :
The aniline molecule is thus stabilised with respect to the anilinium cation, and it is therefore energetically unprofitable for aniline to take up a proton ; it thus functions as a base with almost reluctance.
xiv) Introduction of alkyl, e.g. Me groups on the nitrogen atom of aniline results in small increase in pKa due to the +I effect of alkyl groups.
C6H5NH2 < C6H5 NHMe < C6H5NMe2
xv) Introduction of phenyl groups on N lowers basicity because the substituted amine becomes much more stable than the ion.
ph2NH < phNH2
Answer to Exercises
Exercise 1: Conjugate acid conjugate base
C6H5OH C6H5O–
H3O+ H2O
HCl Cl–
H2O OH–
Exercise 2: i) Lewis acid ⎯⎯→ Cr+3
Lewis base ⎯⎯→ H2O
ii) Lewis acid ⎯⎯→ BF3
Lewis base ⎯⎯→ (C2H5)2O
Exercise 3: Hard acids: Al+3, Cr+3, Sn+4
Hard bases: OH–, H2O, 📷
Solved Problems
Subjective
Problem 1: On the basis of H-bonding explain that the second ionization constant K2 for fumaric acid is greater than for maleic acid.
Solution: We know that H-bonding involving acidic H has an acid weakening effect and H-bonding in conjugate base has an acid strengthening effect.
Both dicarboxylic acids have two ionisable hydrogen atoms. Considering second ionization step.
📷
Since the second ionisable H of the Maleate participates in H-bonding more energy is needed to remove this H because the H-bond must be broken. The maleate mono anion is, therefore, the weaker acid.
Problem 2: Arrange o-nitro, p-nitro, m-nitrophenol in decreasing acid strength
Solution: p-nitrophenol > o-nitrophenol > m-nitrophenol
The –NO2 is electron – withdrawing and acid – strengthening. Its resonance effect, which occurs only from para and ortho positions, predominates over its inductive effect, with occurs also from the meta position between ortho and para derivative, para derivative is more acidic because incase of ortho the hydrogen of phenolic group is attached through hydrogen bonding with oxygen of NO2 group and hence not possible to remove easily.
Problem 3: Among N,N dimethyl aniline and N,N,2,6 – tetramethyl aniline which one is a stronger base and why.
Solution:
📷
In N,N-2,6 – tetramethyl aniline the methyl groups on nitrogen and the ortho position are very close to each other resulting in a steric crowding. Now to avoid steric crowding the C—N bond rotates and becomes perpendicular to the benzene ring. In this process the lone pair on nitrogen becomes perpendicular to the p-orbitals of benzene ring thereby inhibiting resonance. But in N,N dimethyl aniline there is no steric hindrance, so the lone pair is in the same plane as the benzene ring and undergoes resonance. Therefore the lone pair on the tetramethyl derivative is more available and hence it is more basic.
Problem 4: Which is the stronger base towards a proton 📷 or 📷 and why?
Solution: Bond energy (N—H > P—H) ionisation suggests that 📷 will be stronger base. This is constituent with the relative strengths of the respective conjugate acids: NH3 < PH3.
Problem5: Predict the nature of change, if any on the acidity of the solution consequent to the addition (i) HgO to an aqueous solution of KI. (ii) CuSO4 to an aq. solution of (NH)2SO4, (iii) Al(OH)3 to aqueous caustic soda solution.
Solution: i) Due to the formation of the stable complex 📷 species, the concentration of OH– increase according to the reaction
HgO + 4KI + H2O ⎯⎯→ HgI4 + 2OH– + 4K+
Hence acidity decreases
ii) Due to the formation of Cu(NH3)4+2 complex ion increases H+ ion concentration. Hence the acidity increases acid to the reaction
Cu2+(aq) + 4NH4+(aq) ⎯⎯→ 📷+ 4H+(aq)
iii) Formation of the complex 📷 anion consumes. OH– ions. Hence acidity increases.
Al(OH)3(s) + OH– (aq) ⎯⎯→ 📷
Objective
Problem 1: The conjugate base of OH– ion is
(A) H2O (B) O22–
(C) H3O+ (D) O–
Solution: 📷
∴ (B)
Problem 2: Which salt undergoes hydrolysis
(A) CH3COONa (B) KNO3
(C) NaCl (D)K2SO4
Solution: Salt of strong acid and strong base does not undergo hydrolysis.
∴ (A)
Problem 3: The compound HCl behaves as …. in the reaction
Cl + HF ⎯→ H2+Cl + F–
(A) Strong acid (B) Strong base
(C) Weak acid (D) Weak base
Solution: HCl is accepting proton in HF medium and acts as weak base.
∴ (D)
Problem 4: The conjugate acid of 📷 is
(A) H3PO4 (B) H2📷
(C) 📷 (D) 📷
Solution: Conjugate acid and base differs by one proton hence the conjugate acid of PO4–3 is HPO3–2
∴ (D)
Problem 5: The anhydride of H3PO4 is
(A) P2O5 (B) P2O3
(C) PO2 (D) None
Solution: P has +5 oxidation number in P2O5 and H3PO4
∴ (A)
Problem 6: The conjugate base of hydrazoic acid is
(A) HN3– (B) N3H
(C) N3– (D) N2–
Solution: N3H 📷N3–
∴ (C)
Problem 7: Which of the following is a strong acid
(A) HClO4 (B) HBrO4
(C) HIO4 (D) HNO3
Solution: The acidic character of oxy acids decreases down the group and increases along the period. Also acidity increases with increase in oxidation number of central atom.
∴ (A)
Problem 8: H3BO3 is ….acid
(A) Monobasic (B) Dibasic
(C) Tribasic (D) None
Solution: H3BO3 is monobasic acid
H3BO3 + H2O ⎯→ B(OH)4–
∴ (A)
Problem 9: Which is not Lewis base
(A) Ag+ (B) Cl–
(C) CN– (D) H2O
Solution: metal cations are Lewis acids
∴ (A)
Problem 10: Tribasic acid furnishes…. Type of anion
(A) 2 (B) 1
(C) 3 (D) 4
Solution: H3PO4 furnish H2PO4–, HPO42– and PO43– anions
∴ (C)
Assignment
Subjective

LEVEL - I

1. The ionization constant of HCN is given as 4 × 10-10. The ionization constant of NH3 is 1.8 × 10-5. Decide which is the stronger acid, HCN or NH4+.

2. Write down the conjugate base of the following

(i) NH4+, (ii) HCOOH (iii) H3O+, (iv) H2NCONH3+

1. Predict the relative acidic strength among the following:

HI, HIO4, ICl.

1. Predict the relative basic strength among the following :

HOCl, HOClO,HOClO2,HOClO3

1. Compare the strength of formic acid and acetic acid
   
2. Phenol is acidic but why ethanol is neutral
   
3. Among CH4, H2S, HI which one is most acidic? Justify your answer
   
4. Which one of the two i.e., CH3SH and CH2OH more strong acid and why?
   

9. Which of the following oxide is most acidic

Ag2O, V2O5, CO, N2O5

1. Identify soft and hard bases among the following

F–, I–, RS–, RSH

LEVEL - II

1. Which way would the pH shift when CuSO4 is added to pure water. Write a net ionic equation to support your answer.
   
2. What is the order of increasing strength of α-chloro butanoic acid, β-chloro butanoic acid, γ-chloro butanoic acid and n – butanoic acid?
   
3. Write the cation, anion and the neutralisation reaction if thionyl chloride reacts with sodium sulphate.
   
4. Among different hydroxy benzoic acids o-hydroxybenzoic acid is the strongest compared to m-and p-isomers considerably. Why?
   
5. Why diphenyl amine is less basic than aniline?
   
6. Arrange the following in increasing order of basicity
   

Phenol, o-nitrophenol, cresol

1. Arrange the following in the increasing order of basicity. Justify answer

R3—N—, —R2NH, RNH2

1. N-ethyl aniline is more basic than N methyl aniline why?
   
2. Arrange the following in the order of their basic strength. Justify your answer
   

NH2NH2,NH3,NH2OH

1. Arrange the following in the order of their decreasing acidic strength

CH4 , LiH, BeH2, H2O, HF, NH3

LEVEL - III

1. Why CHCl3 is more acidic than CHF3?
   
2. Why F–– is most basic among the halogen anion
   
3. Arrange the following in the increasing order of acidic strength.
   

📷

1. Why pyrole is less basic than pyridine
   
2. Arrange the following compounds in the increasing order of their basic strength.
   

CH3NH–Na+, C2H5NH2, (iso-C3H7)3N and CH3CONH2

1. What should be the order of acidic strength in the series H3PO4, H3PO3 and H3PO2?
   
2. Arrange the following in correct order of acidity
   

📷

1. In dilute benzene solutions, equimolar additions of (C4H9)3N and HCl produce a substance with a dipole moment. In the same solvent, equimolar additions of (C4H9)3N and SO3 produce a substance having an almost identical dipole moment. What is the nature of the polar substances formed and what is the unifying feature of HCl and SO3?
   
2. Arrange according to increasing Lewis acid character,
   

B(n–Bu)3, B(t–Bu)3

1. Arrange according to increasing Lewis acid character,

SiF4, SiCl4, SiBr4, Sil4
Objective
LEVEL - I

1. Which one of following is the strongest acid

(A) ClO3(OH) (B) ClO2(OH)
(C) SO(OH)2 (D) SO2(OH)2

1. Amongst the following the most basic compound is

(A) Benzyl amine (B) Aniline
(C) Acetanilide (D) p –nitro aniline

1. Amongst the trihalides of nitrogen, which one is least basic

(A) NF3 (B) NCl3
(C) NBr3 (D) NI3

1. What is the decreasing order of strength of bases

(A) CH3 – CH2− > NH2− > H – C ≡ C− > OH−
(B) H – C ≡ C− > CH3- CH2− > NH2− > OH−
(C) OH− > NH2− > H- C ≡ C− > CH3−CH2−
(D) NH2− > H – C ≡ C− > OH− > CH3 -CH2−

1. H2O is capable to act as an acid with

(A) NH3 (B) H2SO4
(C) C6H6 (D) HCl

1. Which of the following is Lewis acid

(A) PCl3 (B) AlCl3
(C) NCl3 (D) AsCl3

1. The strongest bronsted base in the following anions is

(A) ClO− (B) ClO2−
(C) ClO3− (D) ClO4−

1. With reference to protonic acid which of the following statement is correct ?
2. PH3 is more basic than NH3
3. PH3 is less basic than NH3
4. PH3 is equally basic than NH3
5. PH3 is amphoteric while NH3 is basic
6. Which of the following is a hard base

(A) F– (B) Li+
(C) I– (D) Mg+

1. The conjugate acid of NH2– is

(A) NH3 (B) NH2OH
(C) NH4+ (D) N2H4

1. The conjugate acid of S2O8–2

(A) H2S2O8 (B) H2SO4
(C) HSO4– (D) HS2O8–

1. The conjugate acid of 📷 is

(A) 📷 (B) 📷
(C) H3PO4 (D) H3PO3

1. Which of the following are amphiprotic in nature

(A) OH– (B) 📷
(C) 📷 (D) HF

1. Among the following compounds, the strongest acid is

(A) HC ≡ CH (B) C6H6
(C) C2H6 (D) CH3OH [IIT-JEE ’98]

1. The following acids have been arranged in order of decreasing acid strength. Identify the correct order.

ClOH (I) BrOH (II) IOH (III)
(A) I > II > III (B) II > I > III
(C) III > II > I (D) I > III > II [IIT-JEE ’96]
LEVEL – II

1. The conjugate acid of HF is

(A) F– (B) 📷
(C) 📷 (D) none of these

1. Which of the following species is an acid and also a conjugate base of another acid

(A) 📷 (B) H2SO4
(C) OH– (D) H3O+

1. Which of the following is the strongest acid

(A) H3PO4 (B) H3PO3
(C) H3PO2 (D) All have same strength

1. In the equilibrium CH3COOH + HF 📷 📷 + F–

(A) F– is the conjugate acid of CH3COOH
(B) F– is the conjugate base of HF
(C) CH3COOH is the conjugate acid of 📷
(D) 📷 is the conjugate base of CH3COOH.

1. The conjugate base of 📷 is

(A) 📷 (B) 📷
(C) 📷 (D) 📷

1. Which of the following is a hard acid

(A) CO2 (B) Br2
(C) Fe3+ (D) Cd2+

1. Pyrophosphoric acid is a

(A) mononbasic acid (B) dibasic acid
(C) tribasic acid (D) tetrabasic acid

1. Which among the following is the strongest base

(A) aniline (B) N,N dimethyl amine
(C) para nitro aniline (D) meta nitro aniline

1. Which of the following statements is/are correct

(A) I– is a weaker base than F–
(B) 📷
(C) HONH2 is a weaker base than NH3
(D) F3C:– is a stronger base than Cl3C:–

1. Which of the following orders regarding acid strength is correct?

(A) HCOOH < CH3COOH < PhCOOH
(B) HCOOH > PhCOOH > CH3COOH
(C) HCOOH > CH3COOH > PhCOOH
(D) CH3COOH > HCOOH > PhCOOH

1. Which of the following statements is/are correct?

(A) Maleic acid is a stronger acid than fumaric acid but maleate monoanion is a weaker acid than fumarate monoanion
(B) Maleic acid is a stronger acid than fumaric acid and also maleate monoanion is a stronger acid than fumarate monoanion
(C) Maleic acid is a weaker acid than fumaric acid but maleate monoanion is a stronger acid than fumarate monoanion
(D) Maleic acid is a weaker acid than Fumaric acid and also maleate monoanion is a weaker acid than fumarate monoanion.

1. Of the following orders regarding basicity, which one is correct

(A) CH3CH2NH2 > PhCONH2 > CH3CONH2
(B) CH3CH2NH2 > PhCONH2 > CH3CONH2
(C) CH3CH2NH2 > PhCONH2 < CH3CONH2
(D) CH3CH2NH2 < PhCONH2 < CH3CONH2

1. In the following compounds:

📷
(A) III > IV > I > II (B) I > IV > III > II
(C) II > I > III > IV (D) IV > III > I > II [IIT-JEE ’96]

1. In the following compounds, the correct order of basicity is

📷
(A) IV > I > III > II (B) III > I > IV > II
(C) II > I > III > IV (D) I > III > II > IV [IIT-JEE ’97]

1. Arrange the following ions in increasing order of acidity.

[Na (H2O)n]+, [Al (H2O)6]3+,[Mn (H2O)6]2+, [Ni (H2O)6]2+
(I) (II) (III) (IV)
(A) I < II < III < IV (B) I < III < IV < II
(C) I < III < II < IV (D) II < IV < III < II
Answers
Subjective
LEVEL - I

1. NH4+ is slightly stronger acid (Ka for NH4+ = 5.6 × 10-10).

[Larger the value of ionization constant, stronger the acid will be.]

1. To form a conjugate base means removal of a proton

So answer is
(i) NH3 (ii) HCOO–
(iii) H2O (iv) H2NCONH2

1. IO4- is more stable than I- and I+ is unstable. Hence HIO4 will be strongest acid.

HIO4 > HI> ICI

1. HOCl < HOClO < HOClO2 < HOClO3
   
2. The –CH3 group which is present in acetic acid decrease the acidic character of the –COOH group.
   

3.  Conjugate base of phenol in resonance stabilized white that of alcohol is not. 

4. Acidic nature
   

CH4 < H2S < HI
Size of the central atoms is greatest in HI hence it is most acidic.

1.  The conjugate base of CH3OH – CH3O– is stronger than the conjugate base of CH3SH i.e CH3SH i.e. CH3S–. Since S is large, the negative charge in CH3S– is dispersed over a large. Hence CH3S– is the weaker base. 

2. The oxide with the highest positive oxidation state on the element other than O should be most acidic. Oxidation states of V in V2O5 and N in N2O5 are both +5. But the electronegativity of N is higher, making N2O5 the most acidic oxide.
   
3. Hard bases ⎯⎯→ F–
   

Soft bases ⎯⎯→ I–, RSH, RS–
LEVEL - II

1. The pH will shift down below 7. It is clear from the following equation.

CuSO4 + H2O 📷 Cu(OH)2 + SO42– + 2H+

1. CH3 – CH2 – CHCl –COOH > CH3 – CHCl – CH2 – COOH > CH2Cl – CH2 – CH2 – COOH > CH3 – CH2 – CH2 – COOH
   
2. Reaction SOCl2 + Na2SO3 ⎯→ 2NaCl + 2SO2
   

Cation SO++
Anion SO32–

1. O–hydroxy benzoate anion is stabilized by chelation due to hydrogen bonding which is absent in meta and para isomers.

📷

1. Diphenyl amine is less basic than aniline because lone pair of electron on N-atom is used by two benzene rings in case of diphenyl amine.
   
2. o-nitrophenol > phenol > cresol
   
3. R2NH > RNH2 > R3N
   
4. N-alkylated anilines are stronger bases than aniline because of steric effects which decrease the resonance of the lone pair of electron on nitrogen and hence makes it more available for protonation. Ethyl group is bigger than the methyl group so N-ethyl aniline is stronger base than N-methyl aniline.
   
5. NH3 > NH2 NH2 > NH2OH
   
6. HF > H2O > NH3 > CH4 > BeH2 > LiH
   

LEVEL - III

1. Cl3C–: is less basic than F3C–: because fluroine can disperse charge only by an inductive effect. White Cl (having empty 3d orbitals) disperses charge by inductive effect as well as by pπ - pπ bonding delocalisatioin. Fluorine is a second period element with no 2d orbital.
   
2. Among the halogens, fluorine has the smallest size hence it has availability of electron most.
   
3. I > V > IV > II > III
   
4. Pyrole uses lone pair of electron on nitrogen atom in delocalisation hence less electron are available for protonation, hence less basic than pyridine.
   
5. CH3CONH2 < (iso-C3H7)3N, C2H5NH2, CH3ONH–Na+
   
6. It can be seen that hydrogens in these molecules are not all bonded to oxygens. It is clear from their structures,
   

📷 📷 📷
that the number of terminal oxygen atoms is 1 in all three acids. The electronegativities of P and H are almost the same. Thus no much difference in acidity is expected.

7. IV > III > I > II

1. Both HCl and SO3 are Lewis acids and can react with amine base to form polar substances which could presumably undergo ionic dissociation in a solvent sufficiently more polar than benzene. The reactioins may be represented as follows (R = C4H9):

📷
Sulphur is a third element which can expand its octet due to availability of vacant d-orbitals. Thus sulphur expands its number of valance electron by attaching to the lone pair on the nitrogen. The N – S bond will be polar because of the big difference in electronegativity between N and S.
📷 ⎯⎯→ 📷
Here the proton of the HCl attaches to the lone pair on the N. The connection between N – H and Cl is designated by (----) symbolizing an electron pair on the Cl connected to nitrogen by a hydrogen bond.

1. The highly branched tertiary butyl group involve appreciable back – strain (B-strain) when the boron atom changes to pyramidal environment on adduct formation. This destabilizes the adduct. Hence the order is

B(t–Bu)3 < B(n–Bu)3

1. The order in this case is the reverse of that for BX3. π-conjugation from the halogen p-orbital to the Si-d orbital is not as intense as in the case of BX3 and the order of acidity follows the increase in electron withdrawing power of the halogen from I to F. Hence the order is

SiI4 < SiBr4 < SiCl4 < SiF4
Objective
LEVEL - I

1. A 2. A
   
2. A 4. A
   
3. A 6. B
   
4. A 8. B
   
5. A 10. A
   
6. D 12. A
   
7. A 14. A
   
8. A
   

LEVEL - II

1. C 2. A
   
2. C 4. B
   
3. D 6. A,C
   
4. D 8. B
   
5. A, C, D 10. B
   
6. A 12. A
   
7. D 14. D
   
8. B
   

6

https://docs.google.com/document/d/1ceBwDU3QWnipIOUKssOzni6ClgUgJ0qD/edit?usp=sharing&ouid=109474854956598892099&rtpof=true&sd=true

CHEMICAL BONDING

Electrovalent (ionic bond) and covalent (including coordinate) bonds; polarity in molecules; dipole moments (qualitative aspects only); concept of hybridization involving s and p orbitals only; hydrogen bond.
Introduction
There are several different theories which explain the electronic structure and shapes of known molecules, and attempt to predict the shapes of molecules whose structures are so far unknown. Each theory has its own virtues and shortcomings. None is rigorous. Theories change in the light of new knowledge. If we knew or could prove what a bond was, we would not need theories, which by definition cannot be proved. The value of a theory lies more in its usefulness than in its truth. Being able to predict the shape of a molecule is important. In many cases all the theories give the correct answers.
The Lewis Theory
The octet rule:- The Lewis theory gave the first explanation of a covalent bond in terms of electrons that was generally accepted. If two electrons are shared between two atoms, this constitutes a bond and binds the atoms together. For many light atoms, a stable arrangement is attained when the atom is surrounded by eight electrons.
This octet can be made up from some electrons which are totally owned and some electrons which are ‘shared’. Thus atoms continue to form bonds until they have made up an octet of electrons. This is called the ‘octet rule’. The octet rule explains the observed valencies in a large number of cases. There are exceptions to the octet rule; for example, hydrogen is stable with only two electrons. Other exceptions are discussed later.Today, the conventional Lewis structure representation of a pair of bonded electrons is by means of a ‘dash’ (-) usually called a ‘bond’. Lone pairs or ‘non-bonded’ electrons are represented by ‘dots’. Some structures are represented below:
Such representations of organic molecules are not usually problematic. However, ‘hit-and-trial’ is generally the method (obviously not very efficient) used by most students in figuring out the structures of inorganic molecules.
Type I: A systematic method, applicable to species in which the octet rule is not violated is described :

• Determine the total number of valence electrons in all the atoms present, including the net charge on the species. (n1)

∙ Determine n2 = [2 × (no. of H atoms) + 8 × (no. of other atoms)].
∙ Determine the number of bonding electrons, n3, which equals n2-n1. No. of bonds equals n3/2.
∙ Determine the number of non-bonding electrons, n4, which equals n1- n3. No. of lone pairs equals n4/2.
∙ Knowing the central atom (you'll need to know some chemistry here, math will not help!), arrange and distribute other atoms and n3/2 bonds. Then complete octets using n4/2 lone pairs.
∙ Determine the `formal charge' on each atom.
Formal charge = [ valence electrons in atom)-(no. of bonds) - (no. of unshared electrons)]
∙ Other aspects like resonance etc. can now be incorporated.
To illustrate with a couple of examples ;
Illustration 1: a) CO32 - ; n1 = 4 + (6 × 3) + 2 = 24 [2 added for net charge]
n2 = (2 × 0) + (8 × 4) = 32 (no H atom, 4 other atoms (1’C’ and 3 ‘O’)
n3 = 32 - 24 = 8, hence 8/2 = 4 bonds
n4 = 24 - 8 = 16, hence 8 lone pairs.
Since carbon is the central atom, arrange 3 oxygen atoms around it, thus,
, but total bonds is equal to 4
Hence we get . Now, arrange lone pairs to complete octetsFigure out the formal charge to get 📷
b) CO2 ; n1 = 4+(6 × 2) = 16
n2 = (2 × 0) + (8 × 3) = 24
n3 = 24 -16 = 8, hence 4 bonds
n4 = 16 - 8 = 8, hence 4 lone-pairs.
Since C is the central atom, arrange the two oxygen atoms around it thus the structure would be; O-C-O, but total no. of bonds = 4
Hence we get O=C=O.
Arrange lone pairs to complete octets
Figure out the formal charges and get the final structure;
c) CO; n1= 10; n2= 16; n3=6 ∴ No. of bonds = 3
n4=4 ∴ no. of lone-pairs =2.
Since C is the central atom, arrange the oxygen atom using the number of bonds and complete octets. Thus, we get
Figure out the formal charge and we get; 📷 or 📷
d) CN- ; n1=10, n2=16, n3=6 ∴ No. of bonds = 3
n4=4 ∴ No. of lone pairs = 2.
Thus the structure of CN- using, the no. of bonds and lone-pairs, would be
📷
e) NO2-; n1=18 , n2=24, n3=6 ∴ No. of bonds = 3
n4=12 ∴ No. of lone pairs = 6
since N is the central atom, arrange the two oxygen atoms around N using the number of bonds and lone-pairs. Hence we get,
Figure out the formal charge and we get;
📷
f) NO3-; n1 = 24, n2 = 32, n3 = 8 ∴ No. of bonds = 4
n4 = 16 ∴ No. of lone pairs = 8
Since N is the central atom, arrange the three oxygen atoms around the N atom using the number of bonds and number of lone pairs.
Hence we get,
Figure out the formal charge and we get
g) NH4+; n1 = 8, n2 = 16, n3 = 8, ∴ No. of bonds = 4; n4 = 0
Since N is the central atom, arrange the four H atoms around the N atom using the no. of bonds.
Thus we get,
The method for determining Lewis structures described above does not hold good for species in which the octet rule is violated (this should be obvious from the procedure. Why ?).
It is therefore useful to remember some important categories of exceptions:
∙ Odd-electron species like NO, NO2, O2 etc. Since it is impossible to distribute an odd number of electrons into pairs, these species necessarily violate the octet rule.
The Valence Bond theory handles such species rather clumsily (odd-electron bonds etc.) the molecular orbital theory copes much better.
∙ Species in which the central atom `expands' its octet (so to speak) PCl5, SF6, many transition metal compounds etc.
∙ Electron -deficient species like BCl3, BeCl2, AlCl3, B2H6 etc. in which the central atom has fewer than eight electrons.
VSEPR (Valence Shell Electron Pair Repulsion) Theory
Type II: The geometric arrangement of atoms in molecules and ions may be predicted by means of the valence-shell electron-pair repulsion (VSEPR) theory. This type includes molecules which may or may not obey the octet rule but have only single bonds.
VSEPR theory may be summarized as follows:

1. The shape of the molecule is determined by repulsions between all of the electron pairs present in the valence shell.
   
2. A lone pair of electrons takes up more space around the central atom than a bond-pair, since the lone pair is attracted to one nucleus whilst the bond pair is shared by two nuclei. It follows that repulsion between two lone pairs is greater than repulsion between a lone pair and a bond pair, which in turn is greater than the repulsion between two bond pairs. Thus the presence of lone pairs on the central atom causes slight distortion of the bond angles from the ideal shape. If the angle between a lone pair, the central atom and a bond pair is increased, it follows that the actual bond angle between the atoms must be decreased.
   
3. The magnitude of repulsions between bonding pairs of electrons depends on the electronegativity difference between the central atom and the other atoms.
   
4. Double bonds cause more repulsion than single bonds and triple bonds cause more repulsion than double bonds.
   

A brief summary of molecular shapes resulting from different configurations of electrons pairs is presented below:
With very few exceptions, the predictions based on the VSEPR theory have been shown to be correct.
Some Examples Using the VSEPR Theory:
To find the shape of a molecule follow the steps given below:
i) Identify the central atom and count the number of valence electrons.
ii) Add to this, number of other atoms.
iii) If it is an ion, add negative charges and subtract positive charges. Call it total N
iv) Divide N by 2 and compare the result with chart I and obtain the shape.
Chart I
Total N/2
Shape of molecule or ion
Example
2
Linear
HgCl2/BeCl2
3
Triangular planar
BF3
3
Angular
SnCl2, NO2
4
Tetrahedral
CH4, BF4-
4
Trigonal Pyramidal
NH3, PCl3
4
Angular
H2O
5
Trigonal bipyramidal
PCl5, PF5
5
Irregular tetrahedral
SF4, IF4+
5
T-shaped
CIF3, BrF3
5
Linear
XeF2, I3-
6
Octahedral
SF6, PF6-
6
Square Pyramidal
IF5
6
Square planar
XeF4, ICI4
Illustration 2: a) BF3 : Central atom is B. There are 3 fluorine atoms. The sum (N) is now 6.
Since there is no charge, nothing is to be added or subtracted. Divide 6 by 2.
The value is 3. Looking at chart I, it can be seen that when the total is 3 the shape of BF3 is predicted to be triangular planar. In fact it is so.
b) BF4- : Central atom is B. There are 4 fluorine atoms.
Valence electrons = 3
Negative charge is = -1
Add all these, 3 + 4 + 1 = 8
No. of bonds (N/2) = 8/2 = 4
Chart I tells us that it is a tetrahedron. It is so.
c) Ammonia NH3 : Central atom is N
Valence electrons = 5
There are 3 Hydrogen atoms
Add these, 5 + 3 = 8
No. of bonds (N/2) = 8/2 = 4
∴ Shape is trigonal pyramidal.
d) H2O : Central atom is O
Valence electrons = 6
There are two H - atoms,
So the sum(N) is 8. No. of bonds = 8/2 = 4.
∴ Shape is angular.
📷
e) PCl5 : Central atom is P
Valence electrons = 5
There are 5-Cl atoms, so the sum (N) is 10.
No. of bonds = 10/2 = 5
Bonding electrons = 5. Non-bonding = 0
∴ Shape is trigonal bipyramidal
f) ClF3 : Central atom is Cl.
Valence electrons = 7
There are 3F atoms, so the number of electron pairs shared is = 3
Sum is 10. No. of bonds = 10/2 = 5
∴ It is T-shaped.
Note : Even though 3 possible structures are possible viz.
📷
Only C is possible because LP-LP repulsion is at 120°, and 2 LP-BP are at 120°. This is because in (A) though the LP-LP repulsion is at 180°, LP-BP repulsion is at 90° each. (B) is not possible because LP-LP repulsion is at 90°. Hence in a trigonal bipyramidal structure, the lone pairs will always occupy the equational position.
g) SF4 (Sulphur tetrafluoride)
Central atom is S
Valence electrons = 6
No of Fluorine atoms (4F atoms) = 4
Total N = 10
N/2 = 10/2 = 5
∴ shape is Irregular tetrahedral
📷
h) I3- :
Central atom is I
Valence electrons = 7
No of I atoms = 2
Negative charge = 1
Total (N) = 7 + 2 + 1 = 10
N/2 = 10/2 = 5
∴ shape is linear
📷
i) Sulphur hexaflouride SF6 :
Central atoms is S
Valence electrons = 6
No. of Fluorine atoms = 6
Total (N) = 12
N/2 = 12/2 = 6
It is Octahedral in shape
📷
j) IF7 (iodine hepta fluoride)
Central atoms = I
Valence electrons = 7
No of Fatoms = 7
Total N = 14
N/2 = 7
Shape is Pentagonal bipyramidal
📷
Hybridization
The mixing or merging of dissimilar orbitals of similar energies to form new orbitals is known as hybridization and the new orbitals formed are known as hybrid orbitals.
The following points are to be remembered about hybridization.
∙ Orbitals belonging to the same atom or ion having similar energies get hybridized.
∙ No of hybrid orbitals is equal to the no. of orbitals taking part in hybridization.
∙ The reason hybridization takes place is to produce equivalent orbitals which give maximum symmetry.
∙ It is not known whether actually hybridization takes place or not. It is a concept which explains the known behaviour of molecules.

1. sp3 Hybridization : In this 1 `s' and 3 `p' orbitals of the same principle quantum number mix to form 4 sp3 hybridized orbitals. These orbitals orient themselves towards the corners of a regular tetrahedron. The angle between the orbitals is 109°28'.

For example CH4
📷

1. sp2 hybridization : In this 1 `s' and 2 `p' orbitals of the same principle quantum number mix to form three sp2 hybridized orbitals. These orbitals orient themselves towards the corners of an equilateral triangle E.g. BF3.

📷
In cases like ethene (C2 H4), the two carbon atoms have one p orbital each with an unpaired electron. The p-orbitals overlap in sideways fashion to give rise to a double bond.
When a bond is formed with head-to-head overlap, it is called a sigma (σ) bond. when a bond is formed with sideways overlap, it is called a pi (π) bond.

1. sp hybridization : In this one `s' and one `p' orbital of the same principle quantum number mix to form 2 sp hybridized orbitals. eg. BeCl2.

In cases like ethyne C2H2, two p orbitals with two unpaired electrons are present in each carbon atom. These overlap in sideways fashion and produce a triple bond.
sp3 d hybridization : In this one `s' three `p' and one `d' orbital of the same principle quantum number mix to give 5 sp 3 d hybridized orbitals.
These orbitals orient themselves towards the corners of a trigonal bipyramidal structure.
e.g. PCl5.
📷

1. sp3d2 hybridization : In this one `s' three `p' and two `d' orbitals mix to form 6 sp3d2 hybridized orbitals.These orient themselves towards the corners of a regular octahedron. e.g. SF6 .

Type III: Molecules which have multiple bonds and may or may not obey the Octet rule.

1. Electronic configuration of the outermost orbit of central atom using boxes.
   
2. Find O.S. of central atom. Exite the electrons from orbits of higher energy in such a way that total no. of unpaired electrons = O.S. of central atom,
   
3. Create a pocket.
   

Add electrons to the pocket starting from s orbital in such a way that total no. of unpaired electrons inside the pocket = No. of atoms other than central atom and that gives hybridisation.
N/2 Shape Hybridisation
2 Linear sp
3 Trigonal planar sp2
4 Tetrahedral sp3
5 Trigonal bipyramidal sp3d
6 Octahedral sp3d2
7 Pentagonal bipyramidal sp3d3
Illustration-3: a) ClO3-
Outermost
conf. of Cl
O.S. of Cl. = 5
El. Conf. after
excitation
📷
∴ No. of unpaired electrons = 5 = O.S. of Cl.
∴Shape is Trigonal pyramidal
b) SO2
Outermost El. Configuration of S
O.S. of S = 4
El. Conf. after excitation
add electrons to pocket
∴Shape of the molecule is angular
📷
c) CO2
Electronic configuration of C
O.S. of C = 4
El. cof. after excitation
add electrons to pocket ∴
📷
∴ Shape of the molecule is linear
​
e) SO3
Electronic configuration of S
O.S. of S. = 6
El. conf. after
excitation
∴ Shape of the molecule is Trigonal planar
📷
f) SO42-
Electronic configuration of S
O.S. of S = 6
El. conf. after
Excitation
∴ Shape is Tetrahedral
📷
g) CO32-
Electronic configuration of C
O.S. of C. = 4
El. conf. after excitation
∴Shape is Trigonal planar
📷 📷
h) NO3- :
Electronic configuration of N
O.S. of N. = 5
El. conf. after
excitation
∴ Shape is Trigonal planar
📷
i) ClO-
Electronic configuration of Cl
O.S. of Cl. = 1
El. conf. after excitation
∴ Shape is Linear
j) ClO2-
Electronic configuration of Cl
O.S. of Cl. = 3
El. conf. after excitation.
∴ Shape is Angular
📷
k) ClO4-
Electronic configuration of Cl
O.S. of Cl. = 7
El. conf. after excitation
∴ Shape is Tetrahedral
i) SO32-
Electronic configuration of S
O.S. of S. = 4
El. conf. after excitation
∴Shape is Trigonal pyramidal
m) PO43-
Electronic configuration of P
O.S. of P. = 5
El. conf. after excitation
∴ Shape is Tetrahedral
n) XeOF2
Electronic configuration of Xe
O.S. of Xe = 4
El. conf. after excitation
∴ The molecule is T shaped .
📷
o) XeOF4
Electronic configuration of Xe
O.S. of Xe = 6
El. conf. after excitation
∴Shape is Square paramidal
📷
p) XeOF6
Electronic configuration of Xe
O.S. of Xe = 8
El. conf. after excitation
∴ Shape is Pentagonal bipyramidal
📷
Exercise -1: Predict the hybridisation and shape of the following:
(a) BeF2 (b) BBr3 (c) PBr3Cl2

1. Fajan's Rule: Although atomic bond in a compound like M+X- is considered to be 100% ionic, actually it also has some covalent character. An explanation for the partial covalent character of an ionic bond has been given by Fajan. According to Fajan, if two oppositely charged ions are brought together, the nature of the bond between them depends upon the effect of one ion on the other.

When two oppositely charged ions (say A+ and B- ) approach each other the positive ion attracts electrons on the outermost shell of the anion and repels its positively charged nucleus. This results in the distortion, deformation or polarization of the anion. If the polarization is quite small, an ionic bond is formed, while if the degree of polarization is large, a covalent bond results.
Thus the power of an ion (cation) to distort the other ion is known as its polarization power and the tendency of the ion(anion) to get polarized by the other ion is known as its polarisability. Greater the polarization power or polarisability of an ion, greater will be its tendency to form a covalent bond.
The polarising power, or polarisability and hence formation of covalent bond is favoured by the following factors:
i) Small Positive Ion (Cation): Due to greater concentration of positive charge on a small area, the smaller cation has high polarising power. This explains why LiCl is more covalent than KCl.
ii) Large Negative Ion (Anion): The larger the anion, the greater is its polarisability, i.e. susceptibility to get polarised. It is due to the fact that the outer electrons of a large anion are loosely held and hence can be more easily pulled out by the cation. This explains why iodides, among halides, are most covalent in nature.
iii) Large Charge on Either of the Two Ions: As the charge on the ion increases, the electrostatic attraction of the cation for the outer electrons of the anion also increases, with the result its ability for forming the covalent bond increases. Thus covalency increases in the order : Na+ Cl-, Mg2+ (Cl2) 2-, Al3+ (Cl3)3 -
iv) Electronic Configuration of the Cation : For the two ions of the same size and charge, one with a pseudo noble gas configuration (i.e., 18 electrons in outer-most shell) than a cation with noble gas configuration (i.e. 8 electrons in outermost shell) will be more polarising. Thus copper (I) chloride is more covalent than sodium chloride although Cu+ ion (0.96A°) and Na+ ion (0.95A°) have same size and charge.
The orbital overlapping involved in covalency reduces, the charge on each ion and so weakens the electrovalent forces throughout the solid, as is evident from the melting point of lithium halides.
LiF = 870°C LiCl = 613°C
LiBr = 547°C LiI = 446°C
From the above discussion, we find that greater the possibility of polarisation, lower is the melting point and heat of sublimation and greater is the solubility in non-polar solvents.
Illustration 4: The melting point of KCl is higher than that of AgCl though the crystal radii of Ag+ and K+ ions are almost the same.
Solution : Now whenever any comparison is asked about the melting point of the compounds which are fully ionic from the electron transfer concept it means that the compound having lower melting point has got lesser amount of ionic character than the other one. To analyse such a question first find out the difference between the 2 given compounds. Here in both the compounds the anion is the same. So the deciding factor would be the cation. Now if the anion is different, then the answer should be from the variation of the anion. Now in the above example, the difference of the cation is their electronic configuration. K+ = [Ar]; Ag+ = [Kr] 4d10. This is now a comparison between a noble gas core and pseudo noble gas core, the analysis of which we have already done. So try to finish off this answer.
Exercise -2: SnCl2 is solid but SnCl4 is liquid - Explain

1. Polar & Non-Polar Covalent Bonds: We had seen in the previous section how an ionic bond could change to a covalent one. The reverse of this is also true. A covalent bond could change bond to partially ionic. This happens when the electronegativity values of the two atoms forming the bond are similar but not same. Such bonds are referred to as polar covalent bonds. For example, bond formed between hydrogen atom and halogen atom, bonds in water, ammonia etc.

δ+ δ - δ+ δ - δ+
H ⎯ Cl ; H ⎯ O ⎯ H.
Dipole Moment: Molecules like H- X having two polar ends are known as dipole or polar molecules. The degree of polarity in a polar compound is given by its dipole moment (μ) which is defined as the product of the net positive or negative charge and distance between the two charged ends, i.e., the bond length.
Dipole moment (μ) = electronic charge (e) × Distance (d) Dipole moment is measured in debye unit (D);
1D = 3.33 × 10-30 Cm
Dipole moment is usually indicated by an arrow, having + on the tail (+ →), above the molecule and pointing towards the negative end, e.g.
📷
Note

1. In a bond H-X, hydrogen atom is the positive end of diplole where X is any atom other than C or H.
   
2. In a bond C-X, the carbon atom is the positive end of the dipole where X is any atom other than carbon. However, in the C-H bonds of hydrocarbons the value and direction of the dipole are not constant and depends upon the state of hybridization of the carbon. For example, the direction and value of dipole moments in C-H bonds of methane and ethylene are as below:
   

📷

1. Dipole moment is a vector quantity. So to get the net dipole moment of a molecule the individual dipole moments must be added vectorally.
   
2. Zero dipole moment of a molecule indicates that the molecule is symmetric.
   

Illustration 5: CO2 has got dipole moment of zero why?
Solutions: The structure of CO2 is📷. This is a highly symmetrical structure with a plane of symmetry passing through the carbon. The bond dipole of C–O is directed towards oxygen as it is the negative end. Here two equal dipoles acting in opposite direction cancel each other and therefore the dipole moment is zero.
Exercise -3: Dipole moment of CCl4 is zero while that of CHCl3 is non zero.
Percentage Ionic Character:
Every ionic compound having some percentage of covalent character according to Fajan’s rule. The percentage of ionic character in a compound having some covalent character can be calculated by the following equation.
The percent ionic character = 📷
Illustration 6: Calculate the % of ionic character of a bond having length = 0.92 Å and 1.91 D as it’s observed dipole moment.
Solution: Calculated μ considering 100% ionic bond
= 4.8×10–10× 0.92 ×10–8esu cm
= 4.8 × 0.92 × 10–18 esu cm
= 4.416 D
∴ % ionic character = 📷 = 43.25
The example given above is of a very familiar compound called HF. The % ionic character is nearly 43.25%, so the % covalent character is (100 – 43.25) = 56.75%. But from the octet rule HF should have been a purely covalent compound but actually it has some amount of ionic character in it which is due to the electronegativity difference of H and F. Similarly knowing the bond length and observed dipole moment of HCl, the % ionic character can be known. It was found that HCl has 17% ionic character. Thus it can be clearly seen that although we call HCl and HF as covalent compounds but it has got appreciable amount of ionic character. So from now onwards we should call a compound having more of ionic less of covalent and vice versa rather than fully ionic or covalent.
Exercise 4: Calculate the percentage of covalent character of HI having bond length = 1.62Å and observed dipole moment = 0.39 D.
Hydrogen Bonding
It is an electrostatic attractive force between covalently bonded hydrogen atom of one molecule or a part of a molecule and an electro negative atom (such as F,O, N) of another molecule or another part of the same molecule. It is very weak (strength about 2 - 10 k cal mol-1 or 8.4 - 42 kJ mol-1) as compared to a covalent bond strength 50 – 100 KCal mol -1 or 209-418.4kJ mol-1)
Conditions for Hydrogen Bonding

1. The molecule must possess a highly electronegative atom linked to hydrogen atom.
   
2. The size of the electronegative atom should be small.
   

Types of Hydrogen Bonding

1. Inter-Molecular Hydrogen Bonding: In this, two molecules of the same compound join to form aggregates. For example,

📷
Intermolecular hydrogen bonding increases the boiling point of the compound and also its solubility in water.

1. Intramolecular Hydrogen Bonding: In this hydrogen bonding occurs between atoms of the same molecules present on different sites. This type of hydrogen bonding is generally known as chelation. Some examples are:

📷
Consequence of hydrogen bonding
Intermolecular hydrogen bonding increases boiling point of the compound and also its water solubility on the other hand intramolecular hydrogen bonding decreases the boiling point of the compound and also its water solubility. High boiling points and melting points of NH3,H2O, HF in comparison to hydrides of other elements of group V, VI and VII to which N,O and F belong respectively are due to hydrogen bonding. H2O exists in liquid state whereas H2S in gaseous state because hydrogen bonding exist in water and no H-bonding exists in H2S. Strength of certain acids and bases can be explained on the basis of hydrogen bonding. The organic compounds like alkane, alkenes, alkynes are insoluble in water due to absence of H-bonding whereas alcohols, organic acids, amines are soluble in water due to H-bonding. Density of ice is less than water and water contracts when heated between 0°C and 4°C because ice has open case structure due to hydrogen bonding.
Illustration-7: Ethyl alcohol (C2H5OH) has got a higher boiling point than dimethyl either (CH3-O-CH3) although the molecular weight of both are same.
Solution: Though ethyl alcohol and dimethyl ether have the same molecular weight but in ethyl alcohol the hydrogen of the O-H groups forms intermolecular hydrogen bonding with the OH group in another molecule. But in case of ether the hydrogen is linked to C is not so electronegative to encourage the hydrogen to from hydrogen bonding.
C2H5 C2H5
⎜ ⎜
-----O ⎯ H ----------O ⎯ H ------
Due to intermolecular H-bonding, ethyl alcohol remains in the associated form and therefore boils at a higher temperature compared to dimethyl ether.
Exercise-5: o-hydroxy benzaldehyde is more soluble in water than p-hydroxy benzaldehyde.
Weak Interactions
Van der Waals forces : Although the covalent, ionic, as well as the metallic bond can be used to explain the structural characteristics and the physical states of many substances, a large number of systems do not fit into these categories e.g., inert gases, non polar molecules like H2, N2, Cl2, CH4 etc. (intermolecular interactions of these.). Obviously, weak attractive forces operate between these molecules (else they would not exist in the liquid state, for example). The existence of these forces was first recognised by van der waals as early as 1813, hence these forces.Three obvious sources of van der waals forces are:
∙ Weak dipole-dipole electrostatic attractive forces which would exist in any molecule with permanent dipole eg. HCl, H2O etc.
∙ Weaker dipole-induced dipole interactions resulting from the polarisation of one molecule by the dipole of a neighbouring molecule.
∙ London dispersion forces. For symmetrical molecules (like Ne, Cl2, H2 CH4 etc. ) only London forces operate. Very weak dipoles form even in non-polar molecules due to temporary fluctuations in the electron density distribution.These transient dipoles can now induce dipoles in neighbouring molecules producing a flux of weak temporary interactions.
Valence bond theory (VBT)
A covalent bond is formed by overlapping of valence shell atomic orbitals of the two atoms having unpaired electron. As a result of overlapping, there is maximum electron density between the bonding atoms and large part of bonding force arises due to electrostatic force of attraction between accumulated electron cloud and two nuclei. Greater the overlapping of atomic orbitals higher is the strength of chemical bond. The paired electron of valence shell of an atom can take part in covalent bonding subject to availability of vacant orbitals of slightly higher energy of the same main energy shell and availability of energy for unpairing of paired electron and their shifting to vacant orbitals. This point explains the trivalency of boron, tetravalency of carbon, pentavalency of phosphorous hexavalency of S and hepta valency of Cl, Br, I.
Depending on type of overlapping atomic orbitals covalent bond can be classified into two types