SAMPLE PAPER CHEMISTRY THEORY_CLASS_XII (Lesson: - Solution) MM:70 Time: 3 hours
General Instructions: Read the following instructions carefully.
a) There are 35 questions in this question paper with internal choice.
b) SECTION A consists of 18 multiple-choice questions carrying 1 mark each.
c) SECTION B consists of 7 very short answer questions carrying 2 marks each.
d) SECTION C consists of 5 short answer questions carrying 3 marks each.
e) SECTION D consists of 2 case- based questions carrying 4 marks each.
f) SECTION E consists of 3 long answer questions carrying 5 marks each.
g) All questions are compulsory.
h) Use of log tables and calculators is not allowed
SECTION A
The following questions are multiple-choice questions with one correct answer. Each question
carries 1 mark. There is no internal choice in this section.
MULTIPLE CHOICE QUESTIOS
1 The value of Henry’s constant KH is:
(a) greater for gases with higher solubility
(b) greater for gases with lower solubility.
(c) constant for all gases.
(d) not related to the solubility of gases.
2 Increasing the temperature of an aqueous solution will cause:
(a) decrease in molality
(b) decrease in molarity
(c) decrease in mole fraction
(d) decrease in % (w/w)
3 Colligative properties depend on:
(a) the nature of the solute
(b) the number of solute particles in solution
(c) the physical properties of solute
(d) the nature of the solvent
4 The unit of ebullioscopic constant is:
(a) K kg mol-1
(b) K-1 kg mol
(c) K kg-1 mol-1
(d) K kg-1 mol
5 The most suitable colligative property to determine molecular weight of biomolecules is:
(a) Lowering of vapour pressure
(b) Osmotic pressure
(c) Elevation of boiling point
(d) Depression of freezing point
6 The number of moles of NaCl in 3 litres of 3M solution is:
(a) 1
(b) 3
(c) 9
(d) 7
7 Which has highest freezing point:
(a) 1 M Glucose
(b) 1 M NaCl
(c) 1 M CaCl2
(d) 1 M AlF3
8
9
10
11
12
13
14
15
16
Which of the following condition is not satisfied by an ideal solution?
(a) ΔHmixing = 0
(b) ΔVmixing = 0
(c) Raoult’s Law is obeyed
(d) Formation of an azeotropic mixture
Considering the formation, breaking and strength of hydrogen bond, predict which of the
following mixtures will show a negative deviation from Raoult’s law?
(a) Methanol and acetone.
(b) Chloroform and acetone.
(c) Phenol and aniline.
(d) Cyclohexane and ethanol
The boiling point of an azeotropic mixture of water and ethanol is less than that of water and
ethanol. The mixture shows:
(a) no deviation from Raoult’s Law.
(b) positive deviation from Raoult’s Law.
(c) negative deviation from Raoult’s Law.
(d) that the solution is unsaturated.
If 2 gm of NaOH is present is 200 ml of its solution, its molarity will be
(a) 0.25
(b) 0.5
(c) 5
(d) 10
The atmospheric pollution is generally measured in the units of
(a) mass percentage
(b) volume percentage
(c) volume fraction
(d) ppm
A 5% solution of cane-sugar (molecular weight = 342) is isotonic with 1% solution of substance
A. The molecular weight of X is
(a) 342
(b) 171.2
(c) 68.4
(d) 136.8
234.2 gm of sugar syrup contains 34.2 gm of sugar. What is the molal concentration of the
solution.
(a) 0.1
(b) 0.5
(c) 5.5
(d) 55
Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A): When NaCl is added to water a depression in freezing point is observed.
Reason (R): The lowering of vapour pressure of a solution causes depression in the freezing
point.
Select the most appropriate answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A): Azeotropic mixtures are formed only by non-ideal solutions and they may have
boiling points either greater than both the components or less than both the components.
Reason (R): The composition of the vapour phase is same as that of the liquid phase of an
azeotropic mixture.
Select the most appropriate answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
17 Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A): Molarity of a solution in liquid state changes with temperature.
Reason (R): The volume of a solution changes with change in temperature.
Select the most appropriate answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
18 Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A): Molecular mass of benzoic acid when determined by colligative properties is
found high.
Reason (R): Dimerization of benzoic acid.
Select the most appropriate answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
SECTION B
19
20
21
22
23
24
This section contains 7 questions with internal choice in two questions. The following questions
are very short answer type and carry 2 marks each.
Equimolal solutions of NaCl and BaCl2 are prepared in water. Freezing point of NaCl is found to
be – 20C, What freezing point do you expect for BaCl2 solution?
Write down four important points of differences between an ideal and a non-ideal solution.
OR
Calculate the mole fraction of benzene in a solution containing 30% by mass
of it in CCl4.
What do you mean by Raoult’slaw ? What are the limitations of Raoult’s law ?
OR
How many types of azeotropes are there? Define them along with one example of each.
Plot a graph between vapour pressure and mole fraction of a solution obeying Raoult’s Law at
constant temperature?
A mixture of chlorobenzene and bromobenzene is a nearly an ideal solution but a mixture of
chloroform and acetone is not Explain?
What is the molality of 1.0 M solution of sodium nitrate (NaNO3) if its density is 1.25 g cm-3?
25 The vapour pressure of water is 12.3 kPa at 300K; calculate the
vapour pressure of 1 molar solution of a solute in it.
SECTION C
This section contains 5 questions with internal choice in two questions. The following
questions are short answer type and carry 3 marks each.
26 Miscible liquid pairs often show positive and negative deviation from Raoult’s law. Explain the
reason for such deviation? Give one example of each of liquid pairs.
27 A solution containing 18 g of non-volatile solute in 200g of water freezes at 272.07 K. calculate
the molecular mass of solute (given Kf = 1.86 K/m)
OR
Calculate the osmotic pressure at 270C of a solution formed by mixing equal volumes of two
solutions, one containing 0.05 mole of glucose in 250 ml of solution and the other containing
3.42 g of C12H22O11 in 250 ml of solution. [R = 0.082 L atm mol-1K-1]
28 An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Determine (i) Molality
of solution. (ii) Boiling point of solution (iii) Lowering of vapour pressure of water at 298 K.
29
The element A and B formed purely covalent compounds having molecular formula AB2 and
AB4. When dissolved in 20 gram of benzene 1 gram of AB2 lowers the freezing point by 2.3 K and
of AB4 by 1.3 K. Calculate atomic mass of A and B. The molar depression constant for freezing is
5.1 KKg mol-1
30
An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal
boiling point of the solvent. What is the molar mass of the solute? (Vapour pressure of pure
water at the boiling point (P0) = 1 atm = 1.013 bar)
OR
Explain the following colligative properties in brief under the following head: (Definition,
Graphical representation and related mathematical expression)
SECTION D
The following questions are case-based questions. Each question has an internal choice and
carries 4 (1+1+2) marks each. Read the passage carefully and answer the questions that follow.
31. Osmotic pressure results from a reduction in the chemical potential of a solvent in the
presence of a solute. The tendency of a system to have equal cemical potentials over its entire
volume and to reach a state of lowest free energy gives rise to the osmotic diffusion of matter. In
ideal and dilute solutions, the osmotic pressure is independent of the nature of the solvent and
solutes. At constant temperature it is determined only by the number of kinetically active partic
les—ions, molecules, associated species, and colloidal particles in a unit volume of the solution.
For very dilute solutions of nondissociating compounds, osmotic pressure is described with
sufficient accuracy by the equation πV = nRT, where n is the number of moles of solute, V is the
volume of the solution, R is the universal gas constant, and T is the absolute temperature.
The following questions are multiple choice questions. Choose the most appropriate answer
(i) Name and define the process/method which is used for purification of water?
(ii) Which colligative property is most suitable to measure molecular mass of proteins and why?
(iii) 200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure
of such a solution at 300 K is found to be 2.57 × 10-3 bar. Calculate the molar mass of the protein.
OR
A solution contains 0.8960g of K2SO4 in 500ml solution. Its osmotic pressure is found to be
0.690atm at 270C. Calculate the value of Van’t Hoff factor. (K=39.0, S=32, O=16, R=0.082atm
mol-1K-1)
32.
Read
the
passage
carefully
and
answer
the
questions
that
follow.
(i) What type of liquids form the ideal solution?
(ii) Give one example of an ideal solution.
(iii) (a) Write two characteristics of a non-ideal solution.
(b) Which type of deviation will be shown by the solution if yAB < yAA.
OR
Plot a graph between vapour pressure and mole fraction of a non-ideal solution showing positive
and negative deviations from an ideal solution.
SECTION E
The following questions are long answer type and carry 5 marks each. Two questions have an
internal choice.
33. (i) Two liquids A and B on mixing form an ideal solution. At 300C vapour pressure of solution
containing 3 mol of A and 1 mol of B is 550 mmHg. But when 4 mol of A and 1 mol of B are mixed.
The vapour pressure of solution thus formed is 560 mm Hg. What would be the V.P of pure A and
B?
(ii) Explain the fact that Raoult’s Law is a special case of Henry’s Law.
(iii) According to Raoult’s law, the vapour pressure of a volatile component in a given solution
is given by pi = xi pi0.
OR
(a) 0.6 mL of acetic acid (CH3COOH), having density 1.06 g mL–1, is dissolved in 1 litre of water.
The depression in freezing point observed for this strength of acid was 0.0205°C. Calculate the
van’t Hoff factor and the dissociation constant of acid.
(b) Define Van’t Hoff factor. Give the value of Van’t Hoff factor in case of the following:
Association, Dissociation and neither association nor dissociation of solute when dissolved in a
solvent.
34. (a) What is meant by abnormal molar mass of solute? Discuss the factors which bring
abnormality in the experimentally determined molecular masses of solutes using colligative
properties.
(b) If N2 gas is bubbled through water at 293 K, how many millimoles of N 2 gas would dissolve in 1
litre of water. Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henry’s law
constant for N2 at 293 K is 76.48 kbar.
l Molar Mass/ van’t Hoff factor = 120/4 = 30 g/mol
35. (a) Why is glycol and water mixture used in car radiators in cold countries?
(b) Give reason When 30 ml of ethyl alcohol and 30ml of water are mixed, the volume of
resulting solution is more than 60ml.
(c) Define cryoscopic constant?
(d) State (i) Azeotropes and (ii) Henry’s Law constant.
SAMPLE PAPER CHEMISTRY THEORY_CLASS_XII (Lesson: - Solution) MM:70 Time: 3 hours
General Instructions: Read the following instructions carefully.
a) There are 35 questions in this question paper with internal choice.
b) SECTION A consists of 18 multiple-choice questions carrying 1 mark each.
c) SECTION B consists of 7 very short answer questions carrying 2 marks each.
d) SECTION C consists of 5 short answer questions carrying 3 marks each.
e) SECTION D consists of 2 case- based questions carrying 4 marks each.
f) SECTION E consists of 3 long answer questions carrying 5 marks each.
g) All questions are compulsory.
h) Use of log tables and calculators is not allowed
SECTION A
The following questions are multiple-choice questions with one correct answer. Each question
carries 1 mark. There is no internal choice in this section.
1
2
3
4
5
6
7
MULTIPLE CHOICE QUESTIOS
The value of Henry’s constant KH is:
(a) greater for gases with higher solubility
(b) greater for gases with lower solubility.
(c) constant for all gases.
(d) not related to the solubility of gases.
Increasing the temperature of an aqueous solution will cause:
(a) decrease in molality
(b) decrease in molarity
(c) decrease in mole fraction
(d) decrease in % (w/w)
Colligative properties depend on:
(a) the nature of the solute
(b) the number of solute particles in solution
(c) the physical properties of solute
(d) the nature of the solvent
The unit of ebullioscopic constant is:
(a) K kg mol-1
(b) K-1 kg mol
(c) K kg-1 mol-1
(d) K kg-1 mol
The most suitable colligative property to determine molecular weight of biomolecules is:
(a) Lowering of vapour pressure
(b) Osmotic pressure
(c) Elevation of boiling point
(d) Depression of freezing point
The number of moles of NaCl in 3 litres of 3M solution is:
(a) 1
(b) 3
(c) 9
(d) 7
Which has highest freezing point:
(a) 1 M Glucose
(b) 1 M NaCl
(c) 1 M CaCl2
8
9
10
11
12
13
14
15
(d) 1 M AlF3
Which of the following condition is not satisfied by an ideal solution?
(a) ΔHmixing = 0
(b) ΔVmixing = 0
(c) Raoult’s Law is obeyed
(d) Formation of an azeotropic mixture
Considering the formation, breaking and strength of hydrogen bond, predict which of the
following mixtures will show a negative deviation from Raoult’s law?
(a) Methanol and acetone.
(b) Chloroform and acetone.
(c) Phenol and aniline.
(d) Cyclohexane and ethanol
The boiling point of an azeotropic mixture of water and ethanol is less than that of water and
ethanol. The mixture shows:
(a) no deviation from Raoult’s Law.
(b) positive deviation from Raoult’s Law.
(c) negative deviation from Raoult’s Law.
(d) that the solution is unsaturated.
If 2 gm of NaOH is present is 200 ml of its solution, its molarity will be
(a) 0.25
(b) 0.5
(c) 5
(d) 10
The atmospheric pollution is generally measured in the units of
(a) mass percentage
(b) volume percentage
(c) volume fraction
(d) ppm
A 5% solution of cane-sugar (molecular weight = 342) is isotonic with 1% solution of substance
A. The molecular weight of X is
(a) 342
(b) 171.2
(c) 68.4
(d) 136.8
234.2 gm of sugar syrup contains 34.2 gm of sugar. What is the molal concentration of the
solution.
(a) 0.1
(b) 0.5
(c) 5.5
(d) 55
Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A): When NaCl is added to water a depression in freezing point is observed.
Reason (R): The lowering of vapour pressure of a solution causes depression in the freezing
point.
Select the most appropriate answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
16 Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A): Azeotropic mixtures are formed only by non-ideal solutions and they may have
boiling points either greater than both the components or less than both the components.
Reason (R): The composition of the vapour phase is same as that of the liquid phase of an
azeotropic mixture.
Select the most appropriate answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
17 Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A): Molarity of a solution in liquid state changes with temperature.
Reason (R): The volume of a solution changes with change in temperature.
Select the most appropriate answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
18 Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A): Molecular mass of benzoic acid when determined by colligative properties is
found high.
Reason (R): Dimerization of benzoic acid.
Select the most appropriate answer from the options given below:
(a) Both A and R are true and R is the correct explanation of A
(b) Both A and R are true but R is not the correct explanation of A
(c) A is true but R is false
(d) A is false but R is true
[Ans: 1 (b), 2 (b), 3 (b), 4 (a), 5 (b), 6 (c), 7 (a), 8 (d), 9 (b), 10 (b), 11 (b), 12 (d), 13 (c), 14 (b), 15
(a), 16 (b), 17 (a), 18 (a)]
SECTION B
This section contains 7 questions with internal choice in two questions. The following questions
are very short answer type and carry 2 marks each.
19 Equimolal solutions of NaCl and BaCl2 are prepared in water. Freezing point of NaCl is found to
be – 20C, What freezing point do you expect for BaCl2 solution?
Ans: i for NaCl = 2 and i for BaCl2 = 3 , (ΔTf)NaCl / (ΔTf)BaCl2 = 2/3
(ΔTf)BaCl2 = 3 X 2 /2 = 30C So that Tf for BaCl2 = - 30C
20 Write down four important points of differences between an ideal and a non ideal solution.
Ans:
Ideal Solution
Non-Ideal Solution
Follows Raoult’s law at all temperature
Does not follow Raoult’s law at all temperature
and concentrations. P = PA + PB
and pressure.
P ≠ PA + PB
Intermolecular forces in resulting solution Intermolecular forces in resulting solution are
are same as in pure components.
different from the inter molecular force of
A–B=A–A=B–B
pure components. A – B ≠ A – A, B – B
No change in volume while mixing
Change in volume while mixing components.
components. Δ V mix = 0
Δ V mix ≠ 0
No heat change take place while mixing
Heat changes take place while mixing the
the components.
components.
Δ H mix = 0
Δ H mix ≠ 0
Eg: n – hexane + n – heptane
Eg: Acetone + Water
& benzene + toluene
& Acetone + CHCl3
OR
Calculate the mole fraction of benzene in a solution containing 30% by mass
of it in CCl4.
Ans: Mass of benzene =30g
Mass of CCl4 =70g
Molar mass of benzene= 78 g
Molar mass of CCl4= 154g mol-1
nC6H6=30g/78g=0.385mol
nCCl4=70g/154g=0.454mol
X C6H6= nC6H6/nC6H6 + nCCl4 =0.385 mol /0.385mol + 0.454mol = 0.459
21 What do you mean by Raoult’slaw ? What are the limitations of Raoult’s law ?
Ans. Raoult’s law – partial vapour pressure of any volatile component of a
solution at any temperature is equal to the product of vapour pressure of the
pure component and mole fraction of the component in solution.
pA = pA0xA and pB = pB0 xB.
Limitations: (i)Applicable only to dilute sol.
(ii)Applicable to solutions of only non- electrolyte.
OR
How many types of azeotropes are there? Define them along with one example of each.
Ans:
Azeotropes: (Constant Boiling Mixture) :-Solution in which components are present in a fixed
proportion, boils at a constant temperature irrespective of boiling point of pure components.
Types: - Two types
๏ Minimum Boiling Azeotropes: Boils at a temperature lower than b.ps. of pure components.
[95% Alcohol by volume]
๏ Maximum Boiling Azeotropes: Boils at a temperature higher than b.p.s. of pure components.
[68% HNO3 by mass]
22 Plot a graph between vapour pressure and mole fraction of a solution obeying Raoult’s Law at
constant temperature?
Ans.
23
A mixture of chlorobenzene and bromobenzene is a nearly an ideal solution but a mixture of
chloroform
and
acetone
is
not
Explain?
Ans. Chlorobezene & bromobenzene both have similar structure and polarity. Therefore the
various interactions (solute – solute, solvent – solvent & solute – solvent) are same whereas in
chloroform and acetone initially there is no hydrogen bonding but after mixing solute solvent
interactions (H –bond ) become stronger and solution deviates from ideal behaviour.
24 What is the molality of 1.0 M solution of sodium nitrate (NaNO3) if its density is 1.25 g cm-3?
Ans:
25 The vapour pressure of water is 12.3 kPa at 300K; calculate the
vapour pressure of 1 molar solution of a solute in it.
Ans: 1 molar solution = 1mol/1000 g of H2O (assuming solution to
diluted)Mole fraction of solute = 1/1+55.5 = 0.0177
P0-Ps / P0 = X2
so 12.3- Ps/12.3 = 0.0177
Therefore Ps = 12.08 kPa
SECTION C
This section contains 5 questions with internal choice in two questions. The following
questions are short answer type and carry 3 marks each.
26 Miscible liquid pairs often show positive and negative deviation from Raoult’s law. Explain the
reason for such deviation? Give one example of each of liquid pairs.
Ans:
Miscible liquid pairs often show positive and negative deviation from Raoult’s law are
called non ideal solutions. Explanation of the reason of such deviations is as follows:
Negative deviation
(i) The vapour pressure of solution is lower
than that of an ideal solution of the same
composition.
(ii) A negative deviation is executed by liquid
pairs for which A-B molecular interaction
forces are stronger than the A-A and B-B
molecular interaction forces. This means that
molecules of A and B will find it difficult to
escape from a solution than from the pure
components. This will result in decrease in
vapour pressure which is known as –ve
deviation.
Eg. Acetone + Chloroform
Intermolecular force of interaction between:
Acetone – Acetone, Diople dipole interaction
Chloroform – Chloroform, Diople dipole
interaction
Acetone – Chloroform, H- Bonding (Stronger
than above two)
Positive deviation
(i) The vapour pressure of solution is higher
than that of an ideal solution of the same
composition.
(ii) A negative deviation is executed by liquid
pairs for which A-B molecular interaction
forces are weaker than the A-A and B-B
molecular interaction forces. This means
that molecules of A and B will find it easier
to escape from a solution than from the
pure components. This will result in increase
in vapour pressure which is known as +ve
deviation.
Eg. Ethanol + Cyclohexane
Intermolecular force of interaction
between:
Ethanol – Ethanol, H- Bonding
Cyclohexane – Cyclohexane, London
dispersions
Ethanol – Cyclohexane (Weaker than above
two)
27 A solution containing 18 g of non-volatile solute in 200g of water freezes at 272.07 K. calculate
the molecular mass of solute (given Kf = 1.86 K/m)
Ans. W2 = 18 g W1= 200g, kf = 1.86k/m Tf= 273K – 272.07K = 0.93K
Tf =
1000×kf ×W2
M2 ×W1
1000×1.86×18
 M2 =
0.93×200
 M2 =
1000×kf ×W2
∆Tf ×W1
= 180 amu
OR
270C
Calculate the osmotic pressure at
of a solution formed by mixing equal volumes of two
solutions, one containing 0.05 mole of glucose in 250 ml of solution and the other containing
3.42 g of C12H22O11 in 250 ml of solution. [R = 0.082 L atm mol-1K-1]
Ans:
0.05
3.42 1000
 = CRT = (C1 + C2) RT = [
 1000 +

]  .082  300 = 0.59112 atm (Ans)
250
342 250
28 An aqueous solution freezes at 272.4 K while pure water freezes at 273 K. Determine (i) Molality
of solution. (ii) Boiling point of solution (iii) Lowering of vapour pressure of water at 298 K.
Ans:
Given: Kf = 1.86 K Kg mol -1. Kb = 0.512 K Kg mol -1,
V.P.of pure water at 298 K = 23.756 mm Hg.
Tf
0.6
Solution: (i) ΔTf = Kf  m Therefore m =
=
= 0.32mol / Kg
Kf
1.86
(ii) ΔTb = Kb  m = 0.512 x 0.32 = 0.164 Therefore Tb = 373 + 0.164 = 373.16 K
(iii) ∆P = P0.m.MA= 23.756 x 0.32 x 0.018 = 0.137 mm Hg.
29
The element A and B formed purely covalent compounds having molecular formula AB2 and
AB4. When dissolved in 20 gram of benzene 1 gram of AB2 lowers the freezing point by 2.3 K and
of AB4 by 1.3 K. Calculate atomic mass of A and B. The molar depression constant for freezing is
5.1 KKg mol-1
Ans:
K  WB  1000
Solution: M B = f
Tf  WA
5.1  1  1000
5.1  1  1000
Case 1: M AB2 =
= 110.87 g/mol Case 2: M AB4 =
= 196 g/mol
2.3  20
1.3  20
Now 2B = AB4 – AB2 = 196 – 110.87 = 85.13 g/mol, Therefore B = 42.56 g/ mol
Now A = AB2 – 85.13 = 25.74 g/ mol.
30 An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal
boiling point of the solvent. What is the molar mass of the solute? (Vapour pressure of pure
water at the boiling point (P0) = 1 atm = 1.013 bar)
Ans:
Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar
Vapour pressure of pure water at normal boiling
Mass of solute, (w2) = 2 g
Mass of solvent (w1) = 98 g mol-1
Molar mass of solvent (water),(M1)=18 g According
to Raoult’s law,
= 41.35 g mol−1
OR
Explain the following colligative properties in brief under the following head: (Definition,
Graphical representation and related mathematical expression)
1. Elevation of boiling point 2. Depression of freezing point
Elevation of Boiling Point:
ΔTb m [molality]
Difference between boiling
ΔTb = Kb m
of solution containing non
Kb=Molal Elevation Constant
volatile solute and B.P. of
(Ebullioscopic Constant).
pure solvent is called
The unit of Kb is K kg mol-1
elevation of B.P.
ΔTb = Tb – Tob
ΔTf  m [molality]
ΔTf = Kfm
Kf =Molal Depression Constant
(Cryoscopic Constant).
The unit of Kf is K kg mol-1
Depression of Freezing
Point:
Difference in freezing point
of pure solvent and freezing
point of solution is called
‘Depression in Freezing
Point’.
ΔTf = T0f – Tf
Ans:
SECTION D
The following questions are case-based questions. Each question has an internal choice and
carries 4 (1+1+2) marks each. Read the passage carefully and answer the questions that follow.
31. Osmotic pressure results from a reduction in the chemical potential of a solvent in the
presence of a solute. The tendency of a system to have equal cemical potentials over its entire
volume and to reach a state of lowest free energy gives rise to the osmotic diffusion of matter. In
ideal and dilute solutions, the osmotic pressure is independent of the nature of the solvent and
solutes. At constant temperature it is determined only by the number of kinetically active partic
les—ions, molecules, associated species, and colloidal particles in a unit volume of the solution.
For very dilute solutions of nondissociating compounds, osmotic pressure is described with
sufficient accuracy by the equation πV = nRT, where n is the number of moles of solute, V is the
volume of the solution, R is the universal gas constant, and T is the absolute temperature.
The following questions are multiple choice questions. Choose the most appropriate answer
(i) Name and define the process/method which is used for purification of water?
Ans: Reverse Osmosis which is defined as: When an excess pressure greater than the osmotic
pressure is applied on solution, the solvent flows out of the solution. This phenomenon is known as
reverse osmosis.
(ii) Which colligative property is most suitable to measure molecular mass of proteins and why?
Ans: Osmotic pressure. This is because all measurements in osmotic pressure method are done at
room temperature and biomolecules are stable only at room temperature.
(iii) 200 cm3 of an aqueous solution of a protein contains 1.26 g of the protein. The osmotic pressure
of such a solution at 300 K is found to be 2.57 × 10-3 bar. Calculate the molar mass of the protein.
Ans. The various quantities known to us are as follows: Π = 2.57 × 10–3 bar,
V = 200 cm3 = 0.200 litre T = 300 K R = 0.083 L bar mol-1 K-1
Π = CRT = n x RT/V (n = W2/M2), Π = W2/M2 X RT/V
Therefore M2 = W2/ Π X RT/V
Substituting values:
M2 = 1.26 g × 0.083 L bar K−1 mol−1 × 300 K / 2.57×10−3 bar × 0.200 L
= 61,022 g mol-1
OR
A solution contains 0.8960g of K2SO4 in 500ml solution. Its osmotic pressure is found to be
0.690atm at 270C. Calculate the value of Van’t Hoff factor. (K=39.0, S=32, O=16, R=0.082atm
mol-1K-1)
Ans: Observed molar mass,
MB =
WB  R  t
, WB=0.8960g, V=500ml=0.5L
 V
R=0.082L atm mol-1 K-1, x=0.690atm, T=300K
MB =
0.896  0.082  300
= 63.9
0.690  0.5
Normal molar mass = 2  39 + 32 + 4  32 =174
Van’t Hoff Factor, i =
32.
Read
the
normal molar mass 174
=
=2.72
observed molar mass 63.9
passage
carefully
and
answer
the
questions
(i) What type of liquids form the ideal solution?
Ans: Liquids that have similar structures and polarities from ideal solutions.
(ii) Give one example of an ideal solution.
Ans: Benzene + Toluene.
that
follow.
(iii) (a) Write two characteristics of a non-ideal solution.
(b) Which type of deviation will be shown by the solution if yAB < yAA.
Ans: (a) Enthalpy change on mixing is not equal to zero and volume change on mixing is also not
equal to zero.
(b) Solution will show positive deviation.
OR
Plot a graph between vapour pressure and mole fraction of a non-ideal solution showing positive
and negative deviations from an ideal solution.
Ans:
SECTION E
The following questions are long answer type and carry 5 marks each. Two questions have an
internal choice.
33. (i) Two liquids A and B on mixing form an ideal solution. At 300C vapour pressure of solution
containing 3 mol of A and 1 mol of B is 550 mmHg. But when 4 mol of A and 1 mol of B are mixed.
The vapour pressure of solution thus formed is 560 mm Hg. What would be the V.P of pure A and
B?
(ii) Explain the fact that Raoult’s Law is a special case of Henry’s Law
Ans (i)
Formula: Ps = PA0 X
A =
 A + PB0 X  B
B = 1
0
3 PA
P
+ B
 550 =
4
4
4
0
0
0
3 PA
P
Case I:
or (550 =
+ B ) X 4,
4
4
2200 = 3 PA0 + PB0 eq 1
0
0
0
0
4 PA
4 PA
PB
PB
1
4


Case II: A = and B =  560 =
+
or (560 =
+
) X 5,
5
5
5
5
5
5
2800 = 4 PA0 + PB0 eq 2
0
0
Subtracting equation 1 from 2: 2800 = 4 PA + PB
2200 = 3 PA0 + PB0
PA0 = 600 mm Hg
0
Substituting PA in equation 1 or 2
PB0 = 400 mm Hg.
(ii) According to Raoult’s law, the vapour pressure of a volatile component in a given solution
is given by pi = xi pi0.
3
and
4
Ans (ii) In the solution of a gas in a liquid, one of the components is so volatile that it exists
as a gas. Its solubility is given by Henry’s law. p = KH x.
If we compare the equations for Raoult’s law and Henry’s law, it can be seen that the partial
pressure of the volatile component or gas is directly proportional to its mole fraction in
solution. Only the proportionality constant KH differs from p10. Thus, Raoult’s law becomes
a special case of Henry’s law in which KH becomes equal to p10.
OR
(a) 0.6 mL of acetic acid (CH3COOH), having density 1.06 g mL–1, is dissolved in 1 litre of water.
The depression in freezing point observed for this strength of acid was 0.0205°C. Calculate the
van’t Hoff factor and the dissociation constant of acid.
(b) Define Van’t Hoff factor. Give the value of Van’t Hoff factor in case of the following:
Association, Dissociation and neither association nor dissociation of solute when dissolved in a
solvent.
Ans: (a)
Acetic acid is a weak electrolyte and will dissociate into two ions: acetate and hydrogen
Ans (b) Van’t Hoff factor is defined as the ratio of normal molar mass to the abnormal molar mass
of solute when dissolved in water. Its value for Association is < 1, for dissociation is >1 and for
neither association nor dissociation = 1
34. (a) What is meant by abnormal molar mass of solute? Discuss the factors which bring
abnormality in the experimentally determined molecular masses of solutes using colligative
properties.
(b) If N2 gas is bubbled through water at 293 K, how many millimoles of N2 gas would dissolve in 1
litre of water. Assume that N2 exerts a partial pressure of 0.987 bar. Given that Henry’s law constant
for N2 at 293 K is 76.48 kbar.
Ans: (a) The molecular mass obtained with the help of colligative property sometimes is different
from normal molecular mass, it is called abnormal molecular mass. The factors which bring
abnormality are:
Association: When the solute molecules undergo association, the number of particles becomes less
and molecular mass determined with the help of colligative property will be more.
Dissociation: It leads to increase in number of particles, therefore, increase in colligative property,
therefore, decrease in molecular weight because colligative property is inversely proportional to
molecular weight.
Ans: (b)The mole fraction of the gas in the solution is calculated by applying Henry’s law.
Thus: x(Nitrogen)=p(nitrogen)/KH=0.987bar/76,480bar=1.29×10-5
As 1 litre of water contains 55.5 mol of it, therefore if n represents no. of moles of N 2in solution
x(Nitrogen) = n mol/ n mol + 55.5 mol =n/55.5= 1.29 x 10-5
Thus n = 1.29 × 10–5 × 55.5 mol = 7.16 × 10–4 mol
=7.16×10−4 mol × 1000 millimoles /1 mol = 0.716 millimoles
OR
(a) What will happen when Red Blood Cell are placed in 0.5% NaCl Solution?
(b) The vapour pressure of solvent gets lowered, when a non- volatile solute is added to it. Why?
(c) What is de-icing agent? How does it work?
(d) What do you understand by the term that Kf for water is 1.86 K Kg mlo-1?
(e)The molecular mass of a solute is 120 g/mol and van’t Hoff factor is 4. What is its abnormal
molecular mass?
Ans (a) RBC’s will swell due to endosmosis.
Ans (b) When a non-volatile solute is added to a solvent, the surface area for escape of solvent
molecules decreases and vapour pressure gets lowered.
Ans (c) Common salt is called de-icing agent as it lowers the freezing point of water to such an
extent that it does not freeze to form ice. Thus, it is used to clear snow from roads.
Ans (d) It means that the freezing point of water is lowered by 1.86 K when 1 mole of a nonvolatile solute is dissolved in 1000 g of water.
Ans (e)Abnormal molecular mass = Normal Molar Mass/ van’t Hoff factor = 120/4 = 30 g/mol
35. (a) Why is glycol and water mixture used in car radiators in cold countries?
(b) Give reason When 30 ml of ethyl alcohol and 30ml of water are mixed, the volume of resulting
solution is more than 60ml.
(c) Define cryoscopic constant?
(d) State (i) Azeotropes and (ii) Henry’s Law constant.
Ans (a): Ethylene Glycol reduces the freezing point of water. Due to this, the coolant in radiators
will not freeze.
Ans (b): Solution of water and ethyl alcohol shows positive deviation and hence 𝛥𝑉 > 0
Ans (c): It is the depression in freezing point when 1 mole of a non-volatile solute is dissolved in
1000 g of solvent.
Ans (d) (i) Azeotropes are constant boiling mixture of two or more liquids having same composition
of the components in liquid as well as in vapour phase.
(ii) Henry’s Law states that at a constant temperature the solubility of a gas in a liquid is directly
proportional to the pressure of the gas.
SAMPLE QUESTION PAPER (2022-23)
CHEMISTRY THEORY (043)
ELECTROCHEMISTRY
M.M:70
Time:3 Hours
General instructions:
Read the following instructions carefully.
A)
B)
C)
D)
E)
F)
G)
H)
There are 35 questions in this question paper with internal choice.
SECTION A consists of 18 multiple choice questions carrying 1 mark each.
SECTION B consists of 7 very short answer questions carrying 2 marks each.
SECTION C consists of 5 short answer questions carrying 3 marks each.
SECTION D consists of 2 case – based questions carrying 4 marks each.
SECTION E consists of 3 long answer questions carrying 5 marks each.
All questions are compulsory.
Use of log tables and calculators is not allowed.
SECTION A
The following questions are multiple choice questions with one correct answer. Each question carry 1 mark.
There is no internal choice in this section.
1. The function of the salt bridge is to
(a) Allow ions to move from anode to cathode
(b) Allow solutions from one half cell to the other half cell
(c) Allow the current to flow through the cell and keep the solutions electrically neutral
(d) Keep the level of solution same.
2. Zinc is used to protect iron from corrosion because
(a) Eoxi of zinc is less than Eoxi of iron
(b) Ered of zinc is less than Ered of iron
(c) Zinc is cheaper than iron
(d) Zinc is abundantly available
3. The charge required for reduction of 1 mol of Cr2O72- ions to Cr3+ is(a) 96500 C
(b) 2 x 96500 C
(c) 6 x 96500 C
(d) 4 x 96500 C
4. Other things being equal , the life of a Daniel cell may be increased by
(a) Keeping low temperature
(b) Using large copper electrodes
(c) Decreasing concentration of copper ions
(d) Using large zinc electrodes
5. The ionic conductivity of the cations and anion of the univalent salt is 140 and 80 respectively. The
molar conductivity of the salt is
(a) 160 Ω-1 cm2 mol-1
(b) 280 Ω-1 cm2 mol-1
(c) 60 moles
(d) 220 Ω-1 cm2 mol-1
6. Farday’s laws of electrolysis are related to
(a) Atomic number of the cation
(b) Atomic number of the anion
(c) Equivalent mass of the electrolytes
(d) Speed of the cation
7. The molar conductivity of an electrolyte increases as
(a) Dilution increases
(b) Temperature decreases
(c) Dilution decreases
(d) None of the above
8. When aqueous solution of NaCl is electrolyzed the product obtained at the cathode is
(a) Hydrogen
(b) Sodium metal
(c) Oxygen
(d) Chlorine
9. A dilute aqueous solution of Na2SO4 is electrolyzed using Pt electrodes. The products at the anode
and cathode are
(a) O2, H2
(b) S2O82-, Na
(c) O2, Na6
(d) S2O82-,H2
10. The ionic conductance of Ba2+ and Cl- ions are respectively 127 and 76 Ω-1 cm2 at infinite dilution.
The equivalent conductance of BaCl2 at infinite dilution is
(a) 203 Ω-1 cm2
(b) 279 Ω-1 cm2
(c) 101.5 Ω-1 cm2
(d) 139.5 Ω-1 cm2
11. The best way to prevent rusting of iron is by
(a) Putting it in to saline water
(b) Cathodic protection
(c) Coating tin on it
(d) Putting it in tap water
12. A lead storage battery is recharged
(a) Lead dioxide dissolves
(b) Sulphuric acid is regenerated
(c) Lead electrodes become coated with lead sulphide
(d) The concentration of sulphuric acid decreases
13. Saturated solution of KNO3 is used to make salt bridge because
(a) Velocity of K+ ion is greater than that of NO3(b) Velocity of K+ ion is lesser than that of NO3(c) Velocity of K+ and NO3- ions are equal
(d) KNO3 is highly soluble in water
14. For the redox reaction Zn + Cu2+(0.1M) ------→ Zn2+(1M) +Cu taking place in a cell E°Cell =1.10 V. Ecell
for the galvanic cell will be
(a) 2.14 V
(b) 1.80 V
(c) 1.07 V
(d) 0.82 V
15. Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A) – The reduction potential of the electrode can be increased by increasing the
concentration of metal cations.
Reason (R ) – E is directly proportional to [Mn+]
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true
16. Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A) – Zinc protect the iron better than tin even after cracks.
Reason (R ) – Oxidation potential of Zn > Fe but oxidation potential of Sn < Fe
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true
17. Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A) – Identification of anode and cathode is done by use of thermometer.
Reason (R ) – Higher the value of reduction potential greater would be its oxidizing power.
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true
18. Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion (A) – E° is an intensive property.
Reason (R ) – E° is constant for a particular electrode at a given temperature
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true
SECTION B
This section contains 7 questions with internal choice in two questions. The following questions are very
short answer type and carry 2 marks of each.
19. The conductivity of 0.20 M solution of KCl at 298 K is 0.0248 S cm-1. Calculate its molar conductivity.
20. How much charge is required for the following reduction of
(i)
(ii)
1 mol of Al3+ to Al
1 mol of MnO4- to Mn2+
OR
How much electricity in terms of Faraday is required to produce?
(i)
20.0 g of Ca from molten CaCl2
(ii)
40.0 g of Al from molten Al2O3
21. What are Fuel cells? Write cell reaction.
OR
What advantages do fuel cell have over conventional methods of generating electrical energy?
22. State Kohlrausch’s law and mention it’s applications.
23. What is corrosion? Give the mechanism of rusting on the basis of electrochemical theory.
24. Calculate the potential of hydrogen electrode in contact with a solution whose pH is 10.
25. Give chemical reaction taking place during discharging of lead storage battery.
SECTION C
This section contains 5 questions with internal choice in two questions. The following questions
are short answer type and carry 3 marks.
26. Calculate the maximum work done that can be obtained from the Daniell cell Zn/Zn 2+//Cu2+/Cu.
Given that E°Zn2+/Zn and E°Cu2+/Cu are -0.76 and +0.34 V respectively.
27. Define molar conductivity. How conductivity and molar conductivity does varies with dilution for
both weak and strong electrolyte?
28. Given the following cell Al/Al3+(0.01M)//Fe2+(0.02M)/Fe. Calculate the value of Ecell at 298K. Given
E°Al3+/Al and E°Fe2+/Fe are -1.66 V and -0.44 V respectively. [ log 2=0.3010]
29. A current of 1.50A was passed through an electrolytic cell containing AgNO3 solution with inert
electrodes .The weight of silver deposited was 1.50 g. How long did the current flow ?
( Ag=108 u , F=96500 C/mol)
OR
The conductivity of a 0.01M solution of acetic acid at 298K is 1.65 x 10-4 Scm-1 . Calculate its molar
conductivity.
30. Represent the cell in which the following reaction takes place. The value of E° for the cell is 1.260V.
What is the value of Ecell? 2Al +3Cd2+(0.1M) ----→ 3Cd + 2Al3+(0.01M)
OR
Calculate the emf of the following cell at 25°C. Al/Al3+(0.001M)//Ni2+(0.1M)/Ni
Given E°Al3+/Al and E°Ni2+/Ni are -1.66 V and -0.25 V respectively. [log2=0.3010,log3=0.4771]
SECTION D
The following questions are case-based questions. Each question has an internal choice and
carries 4 (1+1+2) marks each. Read the passage carefully and answer the questions that follow.
31. Molar conductivity for weak electrolytes can be obtained from molar conductivities of strong
electrolytes at infinite dilution by doing algebraic addition. For example ,molar conductivity of weak
electrolyte like CH3COOH can be obtained from molar conductivities at infinite dilution of strong
electrolytes like CH3COONa , HCl and NaCl according to Kohlrauch’s law
Λ°m CH3COOH = [Λ°mCH3COO- + Λ°mNa+]+[ Λ°mH++ Λ°mCl-]-[ Λ°m Na+ + Λ°mCl-]
(i)
What is the expression of Λ°m for an electrolyte AmBn ?
(ii)
Define limiting molar conductivity.
(iii)
Calculate Λ°m for AgCl if
Scm2mol-1
Λ°m (AgNO3) = 133.4, Λ°m(KCl)=149.9, Λ°m(KNO3)=144.9
OR
Calculate Λ°m for HAc if Λ°m (HCl) = 425.9, Λ°m(NaCl)=126.4, Λ°m(NaAc)=91.0 Scm2mol-1
32. The standard electrode potentials are very important and we can extract a lot of useful information
from them. If the standard electrode potential of an electrode is greater than Zero then it’s reduced
form is more stable compared to hydrogen gas. Similarly, if the standard electrode potential is
negative than hydrogen gas is more stable than the reduced form of the species. It can be seen that
the standard electrode potential for Fluorine is the highest indicating that F 2 gas has the maximum
tendency to get reduced to Fluoride ion F- and therefore Fluorine gas is the strongest oxidizing agent
and Fluoride ion is the weakest reducing agent.
(i)
Which element can act as the strongest reducing agent?
(ii)
How cell potential can be measured?
(iii)
Which electrode can work as anode in galvanic cell?
(iv)
Find out E°Cell for Zn/Zn2+(0.1M//Cu2+(0.10M)/Cu
[given E°Zn2+/Zn and E°Cu2+/Cu are -0.76 V and +0.34 V respectively]
OR
Write cell reaction for the above galvanic cell and mention cathode and anode.
SECTION E
The following questions are long answer type and carry 5 marks each. Two questions have an internal
choice.
33. a. Why does the cell voltage of a mercury cell remain constant during its life time ?
b. Write the reaction occurring at anode and cathode and the products of
electrolysis of aq KCl.
c. What is the pH of HCl solution when the hydrogen gas electrode shows a potential of -0.59 V at
standard temperature and pressure?
OR
a. Molar conductivity of substance “A” is 5.9×103 S/m and “B” is 1 x 10-16 S/m. Which of the two is most
likely to be copper metal and why?
b. What is the quantity of electricity in Coulombs required to produce 4.8 g of Mg from molten MgCl 2? How
much Ca will be produced if the same amount of electricity was passed through molten CaCl 2?
(Atomic mass of Mg = 24 u, atomic mass of Ca = 40 u).
c. What is the standard free energy change for the following reaction at room temperature? Is the reaction
spontaneous?
Sn(s) + 2Cu2+ (aq) ----→ Sn2+ (aq) + 2Cu+ (s)
Given E°Sn2+/Sn and E°Cu2+/Cu are -0.14 V and +0.34 V respectively
34. (i) State Faraday’s Laws of electrolysis?
(ii)Three electrolytic cells A, B, C containing solutions of ZnSO4 , AgNO3 and CuSO4 respectively are
connected in series. A steady current of 1.5 amperes was passed through them until 1.45 g of silver
deposited at the cathode of cell B. How long did the current flow? What mass of copper and zinc were
deposited?
35. The molar conductivity of 0.025 mol L-1 methanoic acid is 46.1 S cm2mol- . Calculate its degree of
dissociation and dissociation constant. Given
= 349.6 S cm2mol-1
and
OR
(a) Depict the galvanic cell in which the reaction takes place.
Zn(s)
+
2Ag+(aq)
→Zn2+(aq)
+
2Ag
Further
show:
(i)
Which
of
the
electrode
is
negatively
charged?
(ii)
The
carriers
of
the
current
in
the
cell.
(iii) Individual reaction at each electrode.
(b) A solution of Ni (NO3)2 is electrolysed between platinum electrodes using a current of 5 amperes
for 20 minutes. What mass of Ni is deposited at the cathode?
SAMPLE QUESTION PAPER (2022-23)
CHEMISTRY THEORY (043)
ELECTROCHEMISTRY
MARKING SCHEME
SECTION –A
1. C
2. B
3. C
4. A
5. D
6. A
7. A
8. A
9. A
10. B
11. B
12. C
13. C
14. C
15. A
16. A
17. D
18. B
19.
SECTION-B
K= 0.0248 Scm-1
c = 0.20 M
Therefore, Molar conductivity,
20. (i) Al3+ +3e -----------→ Al
Quantity of charge required for reduction of 1 mol of Al3+ = 3F = 3x 96500 C=289500C
(ii) MnO4- + 5 e -------------→ Mn2+
Quantity of charge required for reduction of 1 mol of MnO4- = 5F = 5x 96500 C=482500C
OR
(i)
Ca2+
+ 2 e ----→ Ca (40g)
Electricity required for production of 40g of Ca = 2F
Electricity required for production of 20g of Ca = 1F or 96500C
(ii)
Al3+ +3e -----------→ Al(27g)
Electricity required for production of 27g of Al = 3F
Electricity required for production of 40g of Al= 3F x40 /27 = 4.4For 4288888.9C
21. A device in which the chemical energy produced as a result of the oxidation of a fuel is
converted in to electricity is called a fuel cell. 2H2+O2---→2H2O
OR
(i)
They do not cause any pollution. (ii) They are highly efficient 60 to 70%.
22. Kohlrausch’s law – at infinite dilution the molar conductivity of an electrolyte is sum of
the molar conductivities of individual ions . (i). To find the limiting molar conductivity of
weak electrolyte (ii).to find degree of dissociation (iii).to find ionization constant
23. In the process of corrosion, due to the presence of air and moisture, oxidation takes
place at a particular spot of an object made of iron. That spot behaves as the anode. The
reaction at the anode is given by,
The reaction corresponding at the cathode is given by,
The overall reaction is:
24.
, it is given that pH = 10
Therefore,
= 10 – 10M
Now, using Nernst equation:
= – 0.0591 log
= – 0.591 V
25. When the battery is in use, the following cell reactions take place:
At anode:
At cathode:
The overall cell reaction is given by,
SECTION –C
26.
Zn+Cu2+------→
Zn2+
+ Cu
E°Cell = E°cathode - E°anode
=0.34-(-.76)
=1.10V
= – 2 × 96487 × 1.10
= – 21227 J mol – 1
= – 21.227kJ mol – 1
27.
Molar conductivity of a solution at a given concentration is the conductance of
volume V of a solution containing 1 mole of the electrolyte kept between two
electrodes with the area of cross-section A and distance of unit length.
Conductivity always decreases with a decrease in concentration, both for weak and
strong electrolytes. This is because the number of ions per unit volume that carry the
current in a solution decreases with a decrease in concentration.
Molar conductivity increases with a decrease in concentration. This is because the total
volume V of the solution containing one mole of the electrolyte increases on dilution.
28.
2Al +3Fe3+ ----------→ 2 Al3+ + 3 Fe
E°Cell = E°cathode - E°anode
=-0.44-(-1.66)
=1.22V
Applying Nernst equation we will get 1.209 V
29. Given I=1.50A ,
W= 1.50g ,
We know = W= Z x I x t
t= W/Z x I
= 1.50 x 96500 /1080X 1.50 = 893.51s
M = 108 ,
F=96500C
t= ?
K= 1.65 x 10-4 Scm-1
c = 0.01 M
Therefore, Molar conductivity,
30.
1.65 x 10-4 x 1000/0.01
= 16.5 Scm2/mol
SECTION D
31. (i) mΛ°An+ + nΛ°Bm(ii)
Limiting molar conductivity is the molar conductivity at infinite dilution or at zero
concentration
(iii)
Λ°mAgCl = (Λ°mAgNO3 + Λ°mKCl) - Λ°m KNO3
(133.4+149.9)-144.9
= 138.45 Scm2 mol-1
OR
Λ°mHAc = (Λ°mNaAc + Λ°mHCl) - Λ°m NaCl
(91.0+425.9)-126.4
= 390.5 Scm2 mol-1
32. (i) Lithium
(ii)By making a cell with Standard Hydrogen electrode
(iii)Zinc is anode and Cu act as Cathode
(iv)E°Cell = E°C - E°A
= 0.34- (-0.76) = 1.10 V
OR
Zn+Cu2+------→ Zn2+ + Cu
Zn anode , Cu Cathode
SECTION E
33. The cell potential remains constant during its life as the overall reaction does not
involve any ion in solution whose concentration can change during its life time.(1)
b.
KCl (aq) -------→ K+ (aq) + Cl- (aq)
cathode: H2O(l) + e- ---→½ H2 (g) + OH- (aq)
(1/2)
anode: Cl- (aq) -----→ ½ Cl2 (aq) + enet reaction:
KCl (aq) + H2O (l) ------→K+ (aq) +OH- (aq) + ½ H2 (g) + ½ Cl2 (g)
c.
(1/2)
(1)
Given, potential of hydrogen gas electrode = −0.59 V
Electrode reaction: H+ + e– → 0.5 H2
Applying Nernst equation,
E (H+/H2) = Eo (H+/H2) – 0.059/n log [H2]1/2 / /[ H+ ]
(1)
Eo (H+/H2) = 0 V
E (H+/H2) = -0.59 V
n=1
[H2] =1 bar
−0.59 = 0 - 0.059 ( - log [H+] )
−0.59 = −0.059pH
∴ pH = 10
(1/2)
(1/2)
OR
a. “A” is copper, metals are conductors thus have high value of conductivity. (1)
b.
Mg2+ + 2e- ----→ Mg
1 mole of magnesium ions gains two moles of electrons or 2F to form 1 mole of Mg
24 g Mg requires 2 F electricity
4.8 g Mg requires 2 x4.8/24 = 0.4 F = 0.4 x96500 = 38600C
(1)
Ca2+ + 2e-----→Ca
2 F electricity is required to produce 1 mole =40 g Ca
0.4 F electricity will produce 8 g Ca
(1)
c.
F = 96500C, n=2,
Sn2+ (aq) + 2e– → Sn(s)
Cu2+(aq) + e- → Cu+ (aq)
–0.14V
0.34 V
Eocell = Eocathode – Eo anode
= 0.34 – (-0.14) = 0.48V
(1)
ΔGo = -nFEocell
= -2 x96500x 0.29 = 92640J/mol (spontaneous)
(1)
34. (i) Faraday’s Laws of electrolysis
First Law: The amount of chemical reaction which occurs at any electrode during
electrolysis by a current is proportional to the quantity of electricity passed through the
electrolyte.
Second Law: The amount of different substances liberated by the same quantity of
electricity passing through the electrolytic solution is proportional to their chemical
equivalent weights.
.(ii)
According to the reaction:
i.e., 108 g of Ag is deposited by 96487 C.
Therefore, 1.45 g of Ag is deposited by =
= 1295.43 C
Given,
Current = 1.5 A
Therefore, Time
= 863.6 s
= 864 s
= 14.40 min
Again,
i.e.,
C of charge deposit = 63.5 g of Cu
Therefore, 1295.43 C of charge will deposit =
= 0.426 g of Cu
i.e.,
C of charge deposit = 65.4 g of Zn
Therefore, 1295.43 C of charge will deposit =
= 0.439 g of
35.
C = 0.025mol/L
= 349.6 + 54.6 =
Now, degree of dissociation:
= 0.114 (approximately)
Thus, dissociation constant:
OR
a. The galvanic cell in which the given reaction takes place is depicted as:
(i) Zn electrode (anode) is negatively charged.
(ii) Ions are carriers of current in the cell and in the external circuit, current will flow
from silver to zinc.
(iii) The reaction taking place at the anode is given by,
The reaction taking place at the cathode is given by,
b. Given,
Current = 5A
Time =
= 1200 s
Therefore,
= 6000 C
According to the reaction,
Nickel deposited by
= 58.71 g
Therefore, nickel deposited by 6000 C
= 1.825 g
SAMPLE PAPER (2022-23)
CHEMISTRY THEORY
(043)
MM:70
Time: 3 hours
General Instructions:
Read the following instructions carefully.
a)
b)
c)
d)
e)
f)
g)
h)
There are 35 questions in this question paper with internal choice.
SECTION A consists of 18 multiple-choice questions carrying 1 mark each.
SECTION B consists of 7 very short answer questions carrying 2 marks each.
SECTION C consists of 5 short answer questions carrying 3 marks each.
SECTION D consists of 2 case- based questions carrying 4 marks each.
SECTION E consists of 3 long answer questions carrying 5 marks each.
All questions are compulsory.
Use of log tables and calculators is not allowed.
Section A
The following questions are multiple-choice questions with one correct answer. Each question carries
1 mark. There is no internal choice in this section.
1. What kind of order of reaction decomposition of Ammonia on platinum surface is:
(a) Zero order reaction
(b) First order reaction
(c) Second order reaction
(d) Fractional order reaction
2.The rate of a chemical reaction tells us about
(a) the reactants taking part in the reaction
(b) the products formed in the reaction
(c) how slow or fast the reaction is taking place
(d) none of the above
3.In the rate equation, when the concentration of reactants is unity then the rate is equal to:
(a) .specific rate constant
(b) average rate constant
(c) instantaneous rate constant
(d) None of the above
4.The rate constant of zero-order reactions has the unit
(a) s-1
(b) mol L-1 s-1
(c) L2 mol-2 s-1
(d) L mol-1 s-1
5.In the reaction 2A + B → A2B, if the concentration of A is doubled and that of B is halved, then the
rate of the reaction will
(a) increase 2 times
(b) increase 4 times
(c) decrease 2 times
(d) remain the same
6.In the Haber process for the manufacture of ammonia the following catalyst is used
(a) Platinized asbestos
(b) Iron with molybdenum as a promoter
(c) Copper oxide
(d) Alumina
7.A catalyst alters, which of the following in a chemical reaction?
(a) Entropy
(b) Enthalpy
(c) Internal energy
(d) Activation energy
8.In chemical equation H2 (g) + I2 (g) ⇌ 2HI (g) the equilibrium constant Kp depends on
(a) total pressure
(b) catalyst used
(c) amount of H2 and I2
(d) temperature
9.If the rate of a reaction is expressed by, rate = A [A]² [B], the order of reaction will be
(a) 2
(b) 3
(c) 1
(d) 0
10. Which of the following, is an example of a fractional order reaction?
(a) NH4NO2 → N2 + 2H2O
(b) NO + O3 → NO2 + O2
(c) 2NO + Br2 → 2NOBr
(d) CH3CHO → CH4 + CO
11. The rate constant of a reaction depends upon
(a) temperature of the reaction
(b) extent of the reaction
(c) initial concentration of the reactants
(d) the time of completion of reaction
12.The order of reaction is decided by
(a) temperature
(b) mechanism of reaction as well as relative concentration of reactants
(c) molecularity
(d) pressure
13. In pseudo unimolecular reactions
(a) both the reactants are present in low concentration
(b) both the reactants are present in same concentration
(c) one of the reactant is present in excess
(d) one of the reactant is non-reactive
14.Radioactive disintegration is an example of
(a) zero order reaction
(b) first order reaction
(c) second order reaction
(d) third order reaction
15.Read the statements given as assertion & reason both and choose the correct option as per the
following instructions.
(A) if both assertion & reason are correct statements and reason is the correct explanation of
assertion.
(B) if both assertion & reason are correct statements and reason is not the correct explanation of
assertion.
(C) if the assertion is the correct statement & the reason is an incorrect statement.
(D) if the assertion is incorrect statement and reason is the correct statement.
Assertion: The order of reaction can be zero or fractional.
Reason: The order of a reaction cannot be determined from a balanced chemical reaction.
16. Assertion: The order and molecularity of a reaction are always the same.
Reason: Order is determined experimentally whereas molecularity by a balanced elementary reaction.
17. Assertion: Rate constant of a zero-order reaction has the same unit as the rate of a reaction.
Reason: Rate constant of a zero-order reaction does not depend upon the concentration of the
reactant.
18. Assertion: In a first-order reaction, the concentration of the reactant is doubled, its half-life is also
doubled.
Reason: The half-life of a reaction does not depend upon the initial concentration of the reactant in a
first-order reaction.
Section B
Answer the following questions:
19. Derive an expression to calculate the time required for completion of a zero order reaction.
20. Why does the rate of a reaction increase with a rise in temperature?
Or
The order of a reaction if the units of its rate constant are : (i) L-1 mol s-1 (ii) L mol-1 s-1
What do you understand by the rate law and rate constant of a reaction? Identify.
21. Differentiate between rate of reaction and rate constant of reaction.
22. Identify the reaction order for each of the following rate constant.
K=3.1 x 10^-4 sec^-1
K=4.2x10^-5 L mol^-1 sec^-1
23. Define half life of a reaction. Show that for a 1st order reaction half life is independent of initial
concentration.
24.Explain the pseudo order reaction with example.
25. Define order of reaction with example.
Section C
26. Define the following :
(i) Order of a reaction
(ii) Activation energy of a reaction
27. The thermal decomposition of HCO2H is a first order reaction with a rate constant of 2.4 × 10-3 s-1 at
a certain temperature. Calculate how long will it take for three-fourths of initial quantity of HCO2 H to
decompose. (log 0.25 = -0.6021) .
28. A first order gas phase reaction: A2B2(g) → 2A(g) + 2B(g) at the temperature 400°C has the rate
constant k = 2.0 × 10-4 sec-1. What percentage of A2B2 is decomposed on heating for 900 seconds?
(Antilog 0.0781 = 1.197)
29. Derive integrated rate equation for rate constant of a first order reaction.
Or
A first order reaction takes 10 minutes for 25% decomposition. Calculate t1/2 for the reaction. (Given:
log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021).
30. (a) With the help of a labelled diagram explain the role of activated complex in a
reaction.
(b) A first order reaction is 15% completed in 20 minutes. How long will it take to complete 60% of the
reaction?
Section-D
The following questions are case-based questions. Each question carries 1 marks each. Read the
passage carefully and answer the questions that follow
31. Read the passage given below and answer the following questions:
The half-life of a reaction is the time required for the concentration of reactant to decrease by half, i.e.,
[A]t = [A]/2
For first order reaction,
t1/2 = 0.693/k
this means t1/2 is independent of initial concentration. Figure shows that typical variation of
concentration of reactant exhibiting first order kinetics. It may be noted that though the major portion
of the first order kinetics may be over in a finite time, but the reaction will never cease as the
concentration of reactant will be zero only at infinite time.
(i) A first order reaction has a rate constant k=3.01 x 10-3 /s. How long it will take to decompose half of
the reactant?
(ii) The rate constant for a first order reaction is 7.0 x 10-4 s-1. If initial concentration of reactant is 0.080
M, what is the half-life of reaction?
(iii) The rate of a first order reaction is 0.04 mol L-1 s-1 at 10 minutes and 0.03 mol L-1 s-1 at 20 minutes
after initiation. The half life of the reaction?
(iv)The plot of t1/2 vs initial concentration [A]0 for a first order reaction .
32. Read the passage given below and answer the following questions :
A first-order reaction can be defined as a chemical reaction for which the reaction rate is entirely
dependent on the concentration of only one reactant. In such reactions, if the concentration of the firstorder reactant is doubled, then the reaction rate is also doubled.
The following reaction, A(g)⟶ΔP(g)+Q(g)+R(g)A(g)⟶ΔP(g)+Q(g)+R(g) follows first order kinetics. The
half-life period of this reaction is 69.3 s at 500°C. The gas A is enclosed in a container at 500°C and at a
pressure of 0.4 atm.
Answer the following question: (i) The rate constant for the reaction.
(ii) The total pressure of the system after 230 s will be.
(iii) Give an example of first order reaction.
(iv) The plot of ln[A] vs t graph is.
Section-E
33. (a) The rate of the chemical reaction doubles for an increase of 10 K in absolute
temperature
from 298 K. Calculate .
(b) Explain collision frequency.
Or
During nuclear explosion, one of the products is
with half-life of 28.1 years. If
of
was
absorbed in the bones of a newly born baby instead of calcium, how much of it will remain after 10 years
and 60 years if it is not lost metabolically.
34. The activation energy for the
reaction is
at
581K.
Calculate the fraction of molecules of reactants having energy equal to or greater than activation
energy?
Or
In a reaction between A and B, the initial rate of reaction
concentrations of A and B as given below:
was measured for different initial
What is the order of the reaction with respect to A and B?
35. The decomposition of
production of
and
if
on platinum surface is zero order reaction. What are the rates of
?
Or
The following results have been obtained during the kinetic studies of the reaction:
Experiment
Initial rate of formation of
I
0.1
0.1
II
0.3
0.2
III
0.3
0.4
IV
0.4
0.1
Determine the rate law and the rate constant for the reaction.
Marking Scheme
1. A
2. C
3. A
4. B
5. A
6. B
7. D
8. B
9. B
10. D
11. D
12. B
13. C
14. B
15. B
16. D
17. A
18. D
19.
For zero order reaction [R]=[R]0−kt
For completion of the reaction [R] = 0
∴t = [R]0k
20. An increase in temperature typically increases the rate of reaction. An increase in
temperature will raise the average kinetic energy of the reactant molecules. Therefore, a
greater proportion of molecules will have the minimum energy necessary for an
effective collision
Or
21.
Rate of reaction
It is the speed average the reaction
are converted into the product at any
moment of time.
Rate constant of reaction
It is constant of proportionality in the
rate law expression.
It generally decreases with progress
of reaction.
It is constant and doesn’t depend on
the progress of the reaction.
22. a) 1st order
b) 2nd order
23. According to integrated law of rate:-
k=(2.303log10[A]/[A]o)/t
where [A]o is at t=0
the concentration of reactant falls to [Ao]/2 at t1/2
∴t=t1/2
[A]=Ao/2
∴ Equation for first order can be written as:k=(2.303log10 2[A]o/[A]o)/t
t1/2=(0.693)/k
24. The reaction which appears to be second order behaves as first order reaction is called
pseud order reaction.
Ex : C12H22O11 + H2O → C6H12O6(Glucose) + C6H12O6(Fructose)
25. The sum of powers of the concentration of the reactants in the rate law expression is
called the order of that chemical reaction.
26.
(i) Order of a reaction :
•
It is the sum of powers of molar concentrations of reacting species in the rate
equation of the reaction.
• It may be a whole number, zero, fractional, positive or negative.
• It is experimentally determined.
• It is meant for the reaction and not for its individual steps.
(ii) Activation energy of a reaction: The minimum extra amount of energy absorbed by the
reactant molecules to form the activated complex is called activation energy.
27.
28. Since the reaction is of the first order
29. In a first order reaction, the rate of reaction, is directly proportional to the
concentration of the reactant.
Let us consider the reaction,
A → Products
The instantaneous reaction rate can be expressed as:
If t = 0 and [A] = [A]0, where [A]0 is the initial concentration of the reactant.
Then equation (ii) becomes
-ln[A]0 = I ……………. (iii)
Substitute the value of I in equation (ii)
-ln[A] = Kt – ln[A]
ln[A]0 – ln[A] = Kt
This is called integrated rate equation for the first order reaction.
Or
30.
(a) In order that the reactants may change into products, they have to cross an energy
barrier as shown in the diagram
This diagram is obained by plotting potential energy vs. reaction coordinate. It is believed
that when the reactant molecules absorb energy, their bonds are loosened and new bonds
are formed between them. The intermediate complex thus formed is called activated
complex. It is unstable and immediately dissociates to form the stable products.
(b) For the first order reaction
31.
(i) 230.3 s
•
•
(ii) 990 s
(iii) 24.086 min
(iv)
32. (i) t1/2 = 69.3 s
For first order reaction,
k=0.693/t1/2=0.693/69.3=0.01 s−1
(ii) For the given reaction,
Initial pressure Final
pressure A(g)⟶ΔP(g)0.40.4−0.36+Q(g)+R(g)00.3600.3600.36A(g)⟶ΔP(g)+Q(g)+
R(g) Initial pressure 0.4000 Final pressure 0.4−0.360.360.360.36
Total pressure = (0.4 - 0.36) + (3 x 0.36) = 1.12 atm
(iv) slope = -k.
33. It is given that T1 = 298 K
Therefore,
= 308 K
We also know that the rate of the reaction doubles when temperature is
increased by
.
Therefore, let us take the value of
and that of
Also,
Now, substituting these values in the equation:
We get:
.
(b) The collision frequency is defined as the average rate at which two reactants
collide in the given system. The number of collisions per second per unit volume
of the reaction mixture is known as collision frequency.
Or
Here,
It is known that,
Therefore,
of
will remain after 10 years.
Again,
Therefore,
of
34. In the given case:
will remain after 60 years.
T = 581 K
Now, the fraction of molecules of reactants having energy equal to or greater
than activation energy is given as:
In
=18.8323
Now, x = Anti log (18.8323)
= Anti log
Or
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore,
]
……….(i)
……….(ii)
…………(iii)
Dividing equation (i) by (ii), we obtain
y=0
Dividing equation (iii) by (ii), we obtain
= 1.496
= 1.5 (approximately)
Hence, the order of the reaction with respect to A is 1.5 and with respect to B is
zero.
35. The decomposition of
following equation.
on platinum surface is represented by the
Therefore,
However, it is given that the reaction is of zero order.
Therefore,
Therefore, the rate of production of
And, the rate of production of
is
is
.
Or
Let the order of the reaction with respect to A be x and with respect to B be y.
Therefore, rate of the reaction is given by,
According to the question,
………(i)
……….(ii)
……….(iii)
……….(iv)
Dividing equation (iv) by (i), we obtain
x=1
Dividing equation (iii) by (ii), we obtain
y=2
Therefore, the rate law is
From experiment I, we obtain
From experiment II, we obtain,
From experiment III, we obtain
From experiment IV, we obtain
Therefore, rate constant, k =
SAMPLE PAPER (2022-23)
CHEMISTRY THEORY (d and f Block elements)
MM:70 Time: 3 hours
General Instructions:
Read the following instructions carefully.
a) There are 35 questions in this question paper with internal choice.
b) SECTION A consists of 18 multiple-choice questions carrying 1 mark each.
c) SECTION B consists of 7 very short answer questions carrying 2 marks each.
d) SECTION C consists of 5 short answer questions carrying 3 marks each.
e) SECTION D consists of 2 case- based questions carrying 4 marks each.
f) SECTION E consists of 3 long answer questions carrying 5 marks each.
g) All questions are compulsory.
h) Use of log tables and calculators is not allowed
SECTION A
The following questions are multiple-choice questions with one correct answer. Each question
carries 1 mark. There is no internal choice in this section.
Q1 Which one of the following is diamagnetic ion?
(a) Co2+
(b) Ni2+
(c) Cu2+
(d) Zn2+
Q2. The number of unpaired electrons in gaseous species of Mn3+, Cl3+ and V3+ respectively are
and the most stable species is
(a) 4, 3 and 2; V3+
(b) 3, 3 and 2; Cr3+
(c) 4, 3 and 2; Cr3+
(d) 3, 3 and 3; Mn3+
Q3. Fe3+ ion is more stable than Fe2+ ion because
(a) more the charge on the atom, more is its stability
(b) configuration of Fe2+ is 3d6 while Fe3+ is 3d5
(c) Fe2+ has a larger size than Fe3+
(d) Fe3+ ions are coloured hence more stable
Q4. Colour of transition metal ions are due to absorption of some wavelength. This results in
(a) p-p transition
(b) f-f transition
(c) s-p transition
(d) d-d transition
Q5. What happens when potassium iodide reacts with acidic solution of potassium dichromate?
(a) It liberates iodine
(b) Potassium sulphate is formed
(c) Chromium sulphate is formed
(d) All the above products are formed
Q6. The melting point of copper is higher than that of zinc because
(a) copper has a bcc structure
(b) the atomic volume of copper is higher
(c) the electrons of copper are involved in metallic bonding
(d) the ‘s’ as well as ‘d’ electrons of copper are involved in metallic bonding
Q7 .Which of the following pairs of ions have the same electronic configuration?
(a) Cu2+, Cr2+
(b) Fe3+, Mn2+
(c) Co3+, Ni3+
(d) Sc3+, Cr3+
Q .8Which of the following element is used in treatment of cancer?
(a) Th
(b) U
(c) Pu
(d) Np
Q 9 Among the following pairs of ions, the lower oxidation state in aqueous solution is more
stable than other, in
(a) Tl+,Tl3+
(b) Cu+,Cu2+
(c) Cr2+,Cr3+
(d) V2+, VO2+
Q10 Which of the following statements is not correct?
(a) La(OH)3 is less basic than Lu(OH)3
(b) La is actually an element of transition series rather than Lanthanoids
(c) Atomic radius of Zr and Hf is same
(d) In Lanthanoid series, the ionic radius of Lu3+ is smallest
Q11 Gadolinium belongs to 4f series. Its atomic number is 64. Which of the following is the
correct electronic configuration of gadolinium?
(a) [Xe] 4f 75d16s2
(b) [Xe] 4f 65d26s2
(c) [Xe] 4f 86d2
(d) [Xe] 4f 95s1
Q12. Among the following actinoide pairs, the maximum oxidation states is shown by
(a) U and Np
b) Np and Pu
c)Pu and Th
(d)U and Pa
Q13 Cerium (Z=58) is an important member of lanthanoids. Which of the following statements
about cerium is incorrect
(a) The common oxidation state of cerium are +3 and +4.
(b)The +3 oxidation state of cerium is more stable than +4 oxidation state.
(c)The +4 oxidation state of cerium is not known in solutions.
(d) Cerium(IV) acts as an oxidizing agent.
Q14. Although Zirconium belongs to 4 d transition series and Hafnium to 5 d transition series
even then they show similar physical and chemical properties because .
(a) both belong to d-block.
(b) both have same number of electrons
( c) both have similar atomic radius.
(d) both belong to the same group of the periodic table.
Q15Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of
metals. Which of the following is not the characteristic property of interstitial compounds?
(a)They have high melting points in comparison to pure metals.
(b)They are very hard
(c)They retain metallic conductivity.
(d)They are chemically very reactive
Q16. Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion ( A) : Reactivity of transition elements decreases almost regularly from Sc to Cu.
Reason (B) : There is regular increase in I. E. across the series.
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
Q17. Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion(A) : KMnO4 acts as an oxidising agent in acidic, basic or neutral medium.
Reason(R) : It oxidises ferrous sulphate to ferric sulphate.
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
Q18 Given below are two statements labelled as Assertion (A) and Reason (R)
4. Assertion(A) : Cu+ ion is not stable in aqueous solution.
Reason(R) : Large value of I.E. of Cu is compensated by much more negative hydration energy of
Cu2+ ( aq).
Select the most appropriate answer from the options given below: a. Both A and R are true
and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
SECTION B
This section contains 7 questions with internal choice in two questions. The following
questions are very short answer type and carry 2 marks each.
19. Explain the following:
(a) The enthalpies of atomization of transition metals are quite high.
(b) The transition metals and many of their compounds act as good catalysts
20. How would you account for the following?
(i) The highest oxidation state of a transition metal is usually exhibited in its oxide.
(ii) The oxidising power of the following three oxoions in the series follows the order:
VO+2<Cr2O7−2< MnO−4
OR
Describe the general trends in the following properties of the first series (3d) of the transition
elements:
(i) Number of oxidation states exhibited
(ii) Formation of oxometal ions
21.Describe the oxidising action of potassium dichromate and write the ionic equations for its
reaction with (i) iodine (ii) H2S.
OR
What is Mischmetal and write its composition. Also write its uses.
22. Explain the following observations:
(a) Silver atom has completely filled d-orbitals (4d10) in its ground state, yet it is regarded as a
transition element.
(b) E0value for Mn3+ /Mn2+ couple is much more positive than Cr3+/Cr2+.
23. a) Which metal in the first transition series (3d series) exhibits +1 oxidation state most
frequency and why?
(b) Which of the following cations are coloured in aqueous solutions and why? SC 3+, V3+, Ti4+,
Mn2+.
(At. nos. Sc = 21, V = 23, Ti = 22, Mn = 25)
24. Write one similarity and one difference between the chemistry of lanthanoids and that of
actinoids.
25. What is meant by ‘disproportionation’? Give an example of a disproportionation reaction in
aqueous solution
SECTION C
This section contains 5 questions with internal choice in one questions. The following
questions are short answer type and carry 3 marks each.
26. How would you account for the following?
(i) Many of the transition elements are known to form interstitial compounds.
(ii) The metallic radii of the third (5d) series of transition metals are virtually the same as those
of the corresponding group members of the second (4d) series.
(iii) Lanthanoids form primarily +3 ions, while the actinoids usually have higher oxidation states
in their compounds, +4 or even +6 being typical.
27. How would you account for the following?
(i) With the same d-orbital configuration (d4) Cr2+ is a reducing agent while Mn3+ is an oxidizing
agent.
(ii) The actinoids exhibit a larger number of oxidation states than the corresponding members in
the lanthanoid series.
(iii) Most of the transition metal ions exhibit characteristic in colours in aqueous solutions.
28. What is meant by the term lanthanoid contraction? What is it due to and what
consequences.
29. Give reasons for the following:
(i) Transition metals exhibit a wide range of oxidation states.
(ii) Cobalt(II) is very stable in aqueous solutions but gets easily oxidised in the presence of
strong ligands.
(iii) Actinoids exhibit a greater range of oxidation states than lanthanoids
30. In the 3d series (Sc = 21 to Zn = 30) :
(i) Which element shows maximum number of oxidation states?
(ii) Which element shows only +3 oxidation state?
(iii) Which element has the lowest enthalpy of atomization
OR
a) Give reasons for the following:
(i) Compounds of transition elements are generally coloured.
(ii) MnO is basic while Mn2O7 is acidic.
(iii) Calculate the magnetic moment of a divalent ion in aqueous medium if its atomic number is
26.
SECTION D Case Study/Passage Based Questions
Q31. The transition elements have incompletely filled d-subshells in their ground state or in
any of their oxidation states. The transition elements occupy positions in-between s- and pblocks in groups 3-12 of the Periodic table. Starting from the fourth period, transition
elements consist of four complete series: Sc to Zn, Y to Cd and La, Hf to Hg and Ac, Rf to Cn.
In general, the electronic configuration of outer orbitals of these elements is (n – 1)d1–10 ns0–
2. The electronic configurations of outer orbitals of Zn, Cd, Hg, and Cn are represented by the
general formula (n – 1)d10 ns2. All the transition elements have typical metallic properties
such as high tensile strength, ductility, malleability. Except for mercury, which is liquid at
room temperature, other transition elements have typical metallic structures. The transition
metals and their compounds also exhibit catalytic property and paramagnetic behavior.
Transition metal also forms alloys. An alloy is a blend of metals prepared by mixing the
components. Alloys may be homogeneous solid solutions in which the atoms of one metal are
distributed randomly among the atoms of the other.
i)Which of the following characteristics of transition metals is associated with higher
catalytic activity?
(a) High enthalpy of atomization
(b) Variable oxidation states
(c) Paramagnetic behavior
(d) Colour of hydrated ions
ii) Transition elements form alloys easily because they have
(a) same atomic number
(b) same electronic configuration
(c) nearly same atomic size
(d) same oxidation
iii)The electronic configuration of tantalum (Ta) is
(a) [Xe]4f 05d16s2
(b) [Xe]4f 14 5d2 6s2
(c) [Xe]4f14 5d36s2
(d) [Xe]4f14 5d4 6s2
iv)Which one of the following outer orbital configurations may exhibit the largest number
of oxidation states?
(a) 3d54s1
(b) 3d54s2
(c) 3d24s2
(d) 3d34s2
OR
iv) The correct statement(s) among the following is/are
(i) all d- and f-block elements are metals
(ii) all d- and f-block elements form coloured ions
(iii) all d- and f-block elements are paramagnetic.
(a) (i) only
(b) (i) and (ii) only
(c) (ii) and (iii) only
(d) (i), (ii) and (iii)
Q32 Transition metal oxides are compounds formed by the reaction of metals with oxygen at
high temperatures. The highest oxidation number in the oxides coincides with the group
number. In vanadium, there is a gradual change from the basic V 2O3 to less basic V2O4 and to
amphoteric V2O5⋅ V2O4 dissolves in acids to give VO 2+ salts. Transition metal oxides are
commonly utilized for their catalytic activity and semiconductive properties. Transition metal
oxides are also frequently used as pigments in paints and plastic. Most notably titanium
dioxide. One of the earliest applications of transition metal oxides to the chemical industry
involved the use of vanadium oxide for catalytic oxidation of sulfur dioxide to sulphuric acid.
Since then, many other applications have emerged, which include benzene oxidation to
maleic anhydride on vandium oxides; cyclohexane oxidation to adipic acid on cobalt oxides.
An important property of the catalyst material used in these processes is the ability of
transition metals to change their oxidation state under a given chemical potential of
reductants and oxidants.
i) Which oxide of vanadium is most likely to be basic and ionic?
(a) VO
(b) V2O3
(c) VO2
(d) V
ii) Vanadium ion is
(a) VO2+
(b) VO2+
(c) V2O+
d) V
iii) Which of the following statements is false?
(a) With fluorine vanadium can form VF 5.
(b) With chlorine vanadium can form VCl 5.
(c) Vanadium exhibits the highest oxidation state in oxohalides VOCl 3, VOBr3 and fluoride VF5.
(d) With iodine vanadium cannot form VI 5 due to oxidising power of V 5+ and reducing nature
of I
iv) The oxidation state of vanadium in V2O5 is
(a) +5/2
(b) +7
(c) +5
(d) +6
OR
iv) Identify the oxidizing agent in the following reaction.
(a) V2O5
(b) Ca
(c) V
(d) None of thes
SECTION E
The following questions are long answer type and carry 5 marks each. Two questions have an
internal choice.
33. The elements of 3d transition series are given as:
Sc, Ti, V, Cr, Mn, Fe, Co, Ni, Cu, Zn
Answer the following:
(i) Write the element which shows maximum number of oxidation states. Give reason. (ii) Which
element has the highest m.p.?
(iii) Which element shows only +3 oxidation state?
(iv) Which element is a strong oxidizing agent in +3 oxidation state and why?
OR
i) Name the element of 3d transition series which shows maximum number of oxidation states.
Why does it show so?
(ii) Which transition metal of 3d series has positive E0(M2+/M) value and why?
(iii) Out of Cr3+ and Mn3+, which is a stronger oxidizing agent and why?
(iv) Name a member of the lanthanoid series which is well known to exhibit +2 oxidation state.
(v) Complete the following equation: MnO4–+ 8H++ 5e–
34. (a) Describe the preparation of potassium dichromate from chromite ore. What is the effect
of change of pH on dichromate ion?
(b) How is the variability in oxidation states of transition elements different from that of nontransition elements? Illustrate with examples.
OR
(a Describe the preparation of potassium permanganate
(b)Complete the following reactions
(i) MnO4 (aq) + S2O32-(aq) + H2O (l) →
ii) MnO4 – + SO32– + 6H+→ 2Mn2+ + 5SO42– + 3H2O
Q35. a) Complete the following chemical reaction equations:
i) Cr2O72- (aq) + I– (aq) + H+ (aq) →
ii) Fe2+ (aq) + MnO4- (aq) + H+ (aq) →
(b) Explain the following observations:
(i) In general, the atomic radii of transition elements decrease with atomic number in
a given series.
(ii) The E°M2+/M, for copper is positive (+ 0.34 V). It is the only metal in the first
series of transition elements showing this type of behaviour.
(iii) The E° value for Mn3+ | Mn2+ couple is much more positive than for Cr3+ | Cr2+ or
Fe3+ | Fe2+ couple.
OR
Assign reasons for the following:
a) The enthalpies of atomization of transition elements are high.
b) The transition metals and many of their good catalysts. Compounds act as
c) E°M2+/M values are not regular for first-row transition metals (3d series).
d) Although ‘F’ is more electronegative than ‘O’, the highest Mn fluoride is MnF4, whereas the
highest oxide is Mn207
e) Sc3+ is colorless in an aqueous solution whereas Ti3+ is colored.
.
MARKING SCHEME
d and f block Elements
SUBJECT-CHEMISTRY CLASS-XII
S.
NO.
MARKS
1.
(d) Zn2+
1
2.
(c) 4, 3 and 2; Cr3+
1
3.
(b) configuration of Fe2+ is 3d6 while Fe3+ is 3d5
1
4.
(d) d-d transition
1
5.
(d) All the above products are formed
1
6.
(c) the electrons of copper are involved in metallic bonding
1
7.
(b)Fe3+,Mn2+
1
8.
(a) Th
1
9.
(b) Cu+,Cu2
1
10.
(a) La(OH)3 is less basic than Lu(OH)3
1
11.
(a) [Xe] 4f 75d16s2
1
12.
(b)Np and Pu
1
13.
(c). The +4 oxidation state of cerium is not known in solutions
1
14.
(c) both have similar atomic radius
1
15.
(d)They are chemically very reactive
1
16.
a
1
17.
b
1
18.
a
1
19.
Ans: (a) In transition elements, there are large number of unpaired electrons in their
atoms, thus they have a stronger inter atomic interaction and thereby stronger
bonding between the atoms. Due to this they have high enthalpies of atomization.
(b) Because of the availability of d-orbitals, they can easily form intermediate products
1+1=2
which are activated. The sizes of transition metal atoms and ions are also favourable for
transition complex formation with the reactants
20.
Ans: (i) The highest oxidation state of a metal is exhibited in its oxide or fluoride due to
its high electronegativity, low ionisation energy and small size.
(ii) As oxidation number of metal atom in metal oxide increases, oxidising power
increases
OR
Ans: (i) The number of oxidation states increases upto middle of series i.e. unto +7
and then decreases.
(ii) Oxometal ions are polyatomic ions with oxygen. Example : VO2+, VO+2, TiO2+
21.
Cr2O7 2– + 14H+ + 6I– → 2Cr3+ + 3I2+ 7H2O
2
Cr2O7 2– + 8H+ + 3H2S → 2Cr3+ + 3S+ 7H2O
OR
Mischmetal is an alloy of lanthanum. Composition: Lanthanoid metal (95%), Iron (5%) and
traces of C,S, Ca and Al.
It is used in Mg based alloy to produce bullets, shell and lighter flint.
22.
a) Because silver has incomplete d-orbital (4d9) in its +2 oxidation state, hence it is a
transition element.
(b) The large positive E° value for Mn3+/Mn2+ shows that Mn2+ is much more stable than
Mn+3 due to stable half-filled configuration (3d5). Therefore the 3rd ionisation energy of
Mn will be very high and Mn3+ is unstable and can be easily reduced to Mn2+. E° value for
Cr3+ | Cr2+ is positive but small i.e. Cr3+ can also be reduced to Cr2+ but less easily. Thus
Cr3+ is more stable than Mn3+.
1+1
OR
(a) In transition elements, there are large number of unpaired electrons in their atoms,
thus they have a stronger inter atomic interaction and thereby stronger bonding
between the atoms. Due to this they have high enthalpies of atomization.
(b) Because of the availability of d-orbitals, they can easily form intermediate products
which are activated. The sizes of transition metal atoms and ions are also favourable for
transition complex formation with the reactants.
23.
(a) Copper exhibits + 1 oxidation state more frequently i.e., Cu+1 because of its
electronic configuration 3d104s1. It can easily lose 4s1electron to give stable 3d10
configuration.
(b) SC3+ = 4S03d3+ = no unpaired electron
1+1
V3+ = 3d24s0= 2 unpaired electron
Ti4+ = 3d04s0= no unpaired electron
Mn2+ = 3d54s0= 5 unpaired electron
Thus V3+ and Mn2+ are coloured in their aqueous solution due to presence of unpaired
electron.
24.
25.
Ans: Similarity : Both lanthanoids and actinoids show contraction in size and
irregularity in their electronic configuration.
Difference: Actinoids show wide range of oxidation states but lanthanoids do not
Disproportionation: In a disproportionation reaction an element undergoes self oxidation
as well as self-reduction forming two different compounds.
2
2
Any one example
26
27.
(i) Due to lanthanoid contraction in second series after lanthanum, the atomic radii
of elements of second and third series become almost same and hence show
similarities in properties.
(ii) The electronic configuration of Mn2+ ion is more symmetrical as compared to
that of Cr2+ ion. So 3rd ionisation potential of Mn2+ is much higher. As a result E°
value of Mn3+/ Mn2+ couple is much more positive than for Cr3+/Cr2+ couple.
(iii) The highest oxidation state of a metal is exhibited in its oxide or fluoride due to its
high electronegativity
(i) Cr2+ has the configuration 3d4 which easily changes to d3due to stable half filled t2g
orbitals. Therefore Cr2+ is reducing agent. While Mn2+ has stable half filled
d5configuration. Hence Mn3+ easily changes to Mn2+ and acts as oxidising agent.
1+1+1
1+1+1
(ii) Due to comparable energies of 5f 6d and 7s orbitals of actinoids, these show
largernumber of oxidation states than corresponding members of lanthanoids
(iii) Due to the presence of unpaired electrons in d-orbital, transition metal exhibits
colours in aqueous solution or due to d-d transition
28.
. Lanthanoid contraction: Steady decrease in thesize of the lanthanoids with increase in
the atomic number
across the period. The electrons of 4f orbitals offer imperfect / poor shielding effect in the
same
subshell.
Consequence:
i) Due to this 5d series elements have nearly same radii as that of 4d series.
ii) Decrease in the basic strength from La(OH)3 to Lu(OH)3.
iii) Due to similar atomic size there is difficulty in separation of lanthanides.
1+2
29.
: (i) The variability of oxidation state of transition elements is due to incompletely filled
d orbitals as ns, and (n – 1) d electrons have very less energy difference.
(ii) Co2+ ion is easily oxidised to Co3+ ion in presence of a strong ligand because of its
higher crystal field energy which causes pairing of electrons to give inner orbital
complexes (d2sp3).
1+1+1
(iii) Actinoids because of very small energy gap between 5f, 6d and 7s subshells all their
electrons can take part in bonding and shows variable oxidation states.
30
(i) Mn
(ii) Sc
(iii) Zn
1+1+1
OR
(i) The catalytic properties of the transition elements are due to the presence of unpaired
electrons in their incomplete d- orbitals and variable oxidation states. The colour of
transition metal ions is due to d-d transition. When electrons jump from one orbital to
another in their partially filled d-orbitals, another light is emitted due to which the
compounds of transition elements seem to be coloured
(ii) MnO is basic while Mn2O7 is acidic because the basic nature decreases as the oxidation
state or number of oxygen atoms increases i.e. MnO (+4) and Mn2O7 (+7) (b) Divalent ion
with atomic number 26 is Fe2+
(iii)5.66BM
31.
31.1 b
1+1+1+1
31,2 c
31.3
31.4 b or a
32
32.1 a
32.2.a
32.3 .b
32.4 .c or a
1+1+1+1
33
: (i) Mn shows, maximum number of oxidation states upto +7. It has the
maximum number of unpaired electrons.
(ii) Cr has the highest melting point.
(iii) Sc shows only +3 oxidation state
5
iv) Mn is a strong oxidizing agent in +3 oxidation state because after reduction it attains
+2 oxidation state in which it has the most stable half-filled (d5) configuration
OR
: (i) Mn has the maximum number of unpaired electrons present in the d-subshell (5
electrons). Hence, Mn exhibits the largest number of oxidation states, ranging from +2
to +7. (ii) Copper has positive E0(Cu2+/Cu) value because of its high enthalpy of
atomization and low enthalpy of hydration. The high energy required to oxidise Cu to
Cu2+ is not balanced by its hydration energy.
(iii) Cr2+ has the configuration 3d4 which easily changes to d3due to stable half filled
t2g orbitals. Therefore Cr2+ is reducing agent, it gets oxidized to Cr3+. While Mn2+ has
stable half filled d5configuration. Hence Mn3+ easily changes to Mn2+ and acts as
oxidising agent. (iv) Eutropium is well known to exhibit +2 oxidation state.
(v) MnO4–+ 8H++ 5e– → Mn2+ + 4H2O
34.
a) Potassium dichromate is prepared from chromate by reacting chromite ore with
Na2CO3
5
4 FeCr2O4 + 8Na2CO3 + 7O2 → 8Na2CrO4 + 2Fe2O3 + 8CO2 .
The yellow solution of sodium chromate is filtered off and acidified with H 2SO4 to give
orange sodium dichromate
2Na2CrO4 + 2H+ → Na2Cr2O7 + H2O + 2Na
Sodium dichromate is then treated with KCl to give potassium dichromate as orange
(3+2)
crystals. Na2Cr2O7 + 2KCl → K2Cr2O7 + 2NaCl
The chromates and dichromates are interconvertible in aqueous solution depending
upon pH of the solution.
2CrO4-2+ 2H+ → Cr2O7-2 + H2O
Cr2O7-2+ 2OH–→ 2CrO4-2+ H2O
(b) In transition elements, the successive oxidation state differs by unity, e.g, Mn shows
all the oxidation states from +2 to +7. On the other hand, non-transition metals exhibit
variable oxidation states which differ by two units, e.g. Pb(II), Pb(IV), Sn(II), Sn(IV) etc
OR
Ans: Potassium Permangante (KMnO4) is prepared from pyrolusite ore (MnO2). The
ore (MnO2) is fused with an alkali metal hydroxide like KOH in the presence of air or
an oxidising agent like KNO3 to give dark green potassium manganate (K2MnO4).
K2‐ From pyrolusite ore
I. Conversion of pyrolusite ore into potassium manganate
II. Conversion of potassium manganate to potassium permanganate Following reactions
take place:‐
2MnO2 + 4KOH + O2 → 2 K2MnO4 + 2H2O
3MnO4-+ 4H+→ 2MnO4-+ MnO2 + 2H2O
(i) 8MnO4 (aq) +3 S2O32-(aq) + H2O (l) → 8MnO2+6SO42-+2OHii) 2MnO4 – + 5SO3 2– + 6H+→ 2Mn2+ + 5SO4 2– + 3H2O
35
(a) 5Fe2+ (aq) + MnO4- (aq) + 8H+ (aq) → Mn2+ (aq) + 5Fe3+ (aq) + 4H2O(l)
(b) Cr2O72- (aq) + 6I– (aq) + 14H+ (aq) → 2Cr3+(aq) + 3I2 (s) + 7H2O(l)
2+3
b) (i) Because of increase in effective nuclear charge and weak shielding
effect of d electrons, the atomic radii decreases.
(ii) The E°M2+/M for any metal is related to the sum of the enthalpy changes taking
place in the following steps :
M(g) + ΔaH → M(g) (ΔaH = enthalpy of atomization)
M(g) + ΔiH → M2+(g) (ΔiH = ionization enthalpy)
M2+(g) + aq → M2+(aq) + ΔhydH (ΔhydH = hydration enthalpy)
Copper has high enthalpy of atomization (i.e. energy absorbed and low enthalpy of
hydration (i.e. energy released). Hence E°M2+/M for copper is positive. The high energy
required to transform Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy.
(iii) The large positive E° value for Mn3+ | Mn2+ shows that Mn2+ is much more stable than
Mn3+ due to stable half-filled configuration (3d5). Therefore the 3rd ionization energy of
Mn will be very high and Mn3+ is unstable and can be easily reduced to Mn2+. E° value for
Fe3+ | Fe2+ is positive but small i.e. Fe3+ can also be reduced to Fe2+ but less easily. Thus
Fe3+ is more stable than Mn3+
OR
(a) This is because transition elements have strong metallic bonds as they have large
number of unpaired electrons, therefore they have greater interatomic overlap
(b)The catalytic activity of transition metals is attributed to the following reasonsi)
Because of their variable oxidation state, transition metals form unstable intermediate
compounds and provide a new path with lower activation energy for the reaction.
ii) In some cases, the transitions metal provides a suitable large surface area with free
valencies on which reactants are adsorbed
1X5
(c) E°(M2+/M) values are not regular in the first transition series metals because of
irregular variation of ionization enthalpies (IE1 + IE2) and the sublimation energies.
(d) ) Among transition elements, the bonds formed in +2 and +3 oxidation states are
mostly ionic. The compounds formed in higher oxidation states are generally formed by
sharing of d-electrons. Therefore, Mn can form MnO4- which has multiple bonds also,
while fluorine cannot form multiple bonds.
(e) The absence of unpaired d-electron in Sc3+ whereas in Ti3+ there is one unpaired
electron or Ti3+ shows the d-d transition.
KENDRIYA VIDYALAYA (Ahmedabad Region)
CLASS: XII
SUBJECT: CHEMISTRY
SAMPLE PAPER FOR COORDINATION COMPOUNDS
TIME: 3 HOURS
General Instructions:
MM:70
Read the following instructions carefully.
a) There are 35 questions in this question paper with internal choice.
b) SECTION A consists of 18 multiple-choice questions carrying 1 mark
each.
c) SECTION B consists of 7 very short answer questions carrying 2 marks
each.
d) SECTION C consists of 5 short answer questions carrying 3 marks
each.
e) SECTION D consists of 2 case- based questions carrying 4 marks each.
f) SECTION E consists of 3 long answer questions carrying 5 marks each.
g) All questions are compulsory.
h) Use of log tables and calculators is not allowed
SECTION A ( MCQs)
1.
Which of the following compounds has tetrahedral geometry?
(a) [Ni(CN)4]2(b) [Pd(CN)4]2(c [PdCl4]2(d) [NiCl4]2-
1
2.
K2[Fe(CN)6] is a/an
(a) double salt
(b) complex salt
(c) acid
(d) base
1
3.
The number of possible isomers for the complex
|Co(C2O4)2 (NH3)2|
(a) 1
(b) 2
(c) 3
(d) 4
1
4.
According to Werner’s theory of coordination compounds
(a) Primary valency is ionisable
(b) Secondary valency is ionisable
(c) Primary and secondary valencies are lonisable
(d) Neither primary nor secondary valency is ionisable
1
5.
The ligand N(CH2CH2NH2)3 is
(a) bidentate
(b) tridentate
(c) tetradentate
(d) pentadentate
1
6.
Among the following which are ambidentate ligands?
(i) SCN–
(ii) NO−3
(iii) NO−2
(iv) C2O2−4
(a) (i) and (iii)
(b) (i) and (iv)
(c) (ii) and (iii)
(d) (ii) and (iv)
1
7.
Among the following ions which one has the highest paramagnetism ?
1
(a)[Cr(H2O)6] 3+
(b) [Fe(H2O)6] 2+
(c) [Cu(H2O)6] 2+
(d) [Zn(H2O)6] 2+
8.
The complex ion which has no d-electrons in the central metal atom is
(a) [MnO4]–
(b) [Co(NH3)6]3(c) [Fe(CN)6]3(d) [Cr(H2O)6]3+
1
9.
The correct IUPAC name of the coordination compound K3|Fe(CN)5NO| is
(a) Potassium pentacyanonitrosylferrate (II)
(b) Potassium pentacyanonitroferrate (II)
(c) Potassium nitritopentacyanoferrate (IV)
(d) Potassium nitritepentacynanoiron (II)
1
10.
Correct formulae of tetraaminechloronitroplatinum (IV) sulphate can be written
as
(a) [Pt(NH3)4 (ONO) Cl]SO4
(b) [Pt(NH3)4Cl2NO2]2
(c) [Pt(NH3)4 (NO2) Cl]SO4
(d) [PtCl(ONO)NH3(SO4)]
1
11.
|Pt(NH3)4| |CuCl4| and |Cu(NH3)4||PtCl4| are known as
(a) hybridization isomers.
(b) coordination isomers
(c) linkage isomers
(d) polymerization isomers
1
12.
The name of the linkage isomer of |C0(NH3)5NO2|Cl2 will be
(a) pentaammonotrocobalt (II) chloride
(b) pentaaminenitrochloridecobaltate (III)
(c) pentaamminenitritocobalt (III) chloride
(d) pentanitrosoamminechlorocobaltate(III)
1
13.
Which of the following complex species is not expected to exhibit optical
isomerism?
(a) [Co(en) (NH3)2Cl2]+
(b) [Co(en)3]3+
(c)[Co(en)2Cl2]
(d) [Co(NH3)3Cl3]
1
14.
EDTA is used for the estimation of
(a) Na+ and K+ ions
(b) Cl– and Br– ions
(c) Cu2+ and Cs+ ions
(d) Ca2+ and Mg2+ ions
1
(A) Both Assertion and Reason are correct and Reason is the correct explanation
for Assertion.
(B) Both Assertion and Reason are correct but Reason is not the correct
explanation for Assertion
1
(C) Assertion is correct but Reason is incorrect
(D) Assertion is incorrect but Reason is correct
15
Assertion : [Ti(H2O)6]3+ is coloured while [Sc(H2O)6]3+ is colourless.
Reason : d-d transition is not possible in [Sc(H2O)6]3 as it has no unpaired electron.
1
16
Assertion : [Fe(CN)6]3– is weakly paramagnetic while [Fe(CN)6]4– is diamagnetic.
Reason : [Fe(CN)6]3– has +3 oxidation state while [Fe(CN)6]4– has +2 oxidation state.
1
17
Assertion: Chelates are less stable than ordinary coordination compounds.
Reason: Chelates can be formed by bidentate ligands.
1
181. Assertion : Linkage isomerism arises in coordination compounds containing
ambidentate ligand.
Reason : Ambidentate ligand has two different donor atoms donating
simultaneously.
1
SECTION : B
19
On the basis of crystal field theory,write the electronic configuration for d4 ion if Δo
>P.Explain
2
20
When a coordination compound CrCl3.6H2O is mixed with AgNO3 ,2 moles of AgCl are
precipitated per mole of the compound. Write structural formula & IUPAC name of
the complex.
2
OR
Discuss the nature of Bonding in metal carbonyls.
21
What is meant by stability of a coordination compound in solution? State the factors
which govern stability of complexes.
2
Or
Giving a suitable example for each, explain the following :
(i) Linkage isomerism
(ii) Ambidentate ligand
22.
1+1
The formula Co(NH3) 5CO3Cl could represent a carbonate or chloride. Write structures 2
and names of possible isomers.
23.
[Co (NH3)6] +3 is an inner orbital complex whereas [Ni(NH3)6] +2 is an outer
complex.
2
24.
What is crystal field splitting energy? What are the various factors affecting CFSE.
2
25.
What is spectrochemical series? Explain the difference between a weak field ligand
and a strong field ligand.
2
SECTION :C
26
a.Which complex ion is formed when undecomposed AgBr is washed with hypo
solution in photography?Give reaction.
1+2
b. Why only transition metals are known to form pi complexes?
27
A complex is prepared by mixing CoCl2 and NH3 In the molar ratio of 1:4, 0.1M
solution of this complex was found to freeze at -0.372°c. what is the formula of the
complex ? Given that molal depression constant of water (kf)=1.86c/m
3
28.
a.Calculate the over all complex dissociation equilibrium constant for the[
cu(NH3)4]2+ ion , given that β for this complex is 2.1 x(10)13?
2+1
b. Predict the number of unpaired electrons in the square planar [Pt(CN)4] 2- ion.
29.
a. CuSO4.5H2O is blue in colour while CuSO4 is colouress. Why ?
2+1
b. Magnetic moment of [MnCl4] 2- is 5.92BM. Explain giving reason .
30
a.Explain Coordination isomerism with example.
1+2
b. Write down the applications of coordination compound
Or
1.5+1.5
a) Describe the shape and magnetic behaviour of following complexes :
(i) [CO(NH3)6]3+
(ii) [Ni(CN)4]2(At. No. Co = 27, Ni = 28)
SECTION :D
31
Read the given passage and answer the questions that follows. Negative ion or
Neutral molecule which bound to the metal ion by secondary valency is called as
1+1+2
ligand. In complex , metal ion acts as Lewis acid and ligand act as Lewis base. Ligands
are classified according to number of electron pair in them. The ligand which can
donates one e pair to the metal atom is called Unidentate ligand . The ligand which
can donate two electron pair to the Metal ion is called didentate ligand. The ligand in
which two or more coordination sites are there is called polydentate ligand. 75
Polydentate ligand forms cyclic structure with metal ion and form Chelate.
i)
ii)
iii)
32
Give an example of unidentate neutral ligand.
What are lewis acids and lewis bases.
How didentate and Ambidentate ligands are different ? Give example.
Read the passage given below and answer the following questions:
1+1+1+
1
Coordination compounds are formulated and named according to the IUPAC system.
Few rules for naming coordination compounds are:
(I) In ionic complex, the cation is named first and then the anion.
(II) In the coordination entity, the ligands are named first and then the central metal
ion.
(III) When more than one type of ligands are present, they are named in alphabetical
order of preference with any consideration of charge.
ANSWER THE FOLLOWING QUESTIONS: (i) The IUPAC name of [Ni(CO)4] is
(ii) The IUPAC name of the complex [Pt(NH3)3Br(NO2)Cl]Cl is
(iii) As per IUPAC nomenclature, the name of the complex [Co(H2O)4(NH3)2]Cl3 is
(iv) Which of the following represents correct formula of dichloridobis(ethane-1,2diamine)cobalt(III) ion?
SECTION :E
33
(a) Account for the different magnetic behavior of hexacyanoferrate (III) and
hexafluoroferrate (III).
2+3
(b) (i) What type of isomerism is shown by the complex [Cr (H2O)6]Cl3 ?
(ii) On the basis of CFT , write the electronic configuration for d4 ion if Δ0 > P .
(iii) Write the hybridization and shape of [CoF6] 3- .
Or
2+3
a.(i) Write down the IUPAC name of the following complex :
[Cr(NH3)2CI3(en)]Cl (en = ethylenediamine)
(ii) Write the formula for the following complex : Pentaamminenitrito-o-Cobalt (III)
b.Define the following terms;
(a) Homoleptic complex
( b) Hetroleptic complex
( c) Coordination sphere
34
a. Which of the following is more stable complex and why?
[Co(NH3)6]3+ and [Co(en)3]3+
1+2+2
b. Why are low spin tetrahedral complexes not formed?
c. Give evidence that [Co(NH3)5Cl]SO4 and [Co(NH3)5SO4]Cl are ionization isomers?
OR
a. Compare the following complexes with respect to their shape, magnetic behaviour
and the hybrid orbitals involved :
(i) [CoF4]2(ii) [Cr(H2O)2(C2O4)2]–
(iii) [Ni(CO)4] (Atomic number : Co = 27, Cr = 24, Ni = 28)
35
For the complex [Fe(en)2Cl2], Cl, (en = ethylene diamine), identify
(i) the oxidation number of iron,
(ii) the hybrid orbitals and the shape of the complex,
(iii) the magnetic behaviour of the complex,
(iv) the number of geometrical isomers,
(v) whether there is an optical isomer also, show and
(vi) name of the complex. (At. no. of Fe = 26)
1.5+2+
1.5
½+1+
1/2+1+
1+1
OR
3+2
a. With proper diagram explain crystal field splitting for octahedral complexes.
b. What are t2g and eg orbitals ? Explain.
KENDRIYA VIDYALAYA (Ahmedabad Region)
CLASS: XII
SUBJECT: CHEMISTRY
MARKING SCHEME SAMPLE PAPER FOR COORDINATION COMPOUNDS
TIME: 3 HOURS
MM:70
SECTION A ( MCQs)
1.
(d) [NiCl4]2-
1
2.
(b) complex salt
1
3.
(b) 2
1
4.
(a) Primary valency is ionisable
1
5.
(c) tetradentate
1
6.
(a) (i) and (iii)
1
7.
(b)
1
8.
(a) [MnO4]–
1
9.
(a) Potassium pentacyanonitrosylferrate (II)
1
10.
(c) [Pt(NH3)4 (NO2) Cl]SO4
1
11.
(b) coordination isomers
1
12.
(c) pentaamminenitritocobalt (III) chloride
1
13.
(d) [Co(NH3)3Cl3]
1
14.
(d) Ca2+ and Mg2+ ions
1
(A) Both Assertion and Reason are correct and Reason is the correct explanation for
Assertion.
(B) Both Assertion and Reason are correct but Reason is not the correct explanation for
Assertion
(C) Assertion is correct but Reason is incorrect
(D) Assertion is incorrect but Reason is correct
15
(B) Both Assertion and Reason are correct but Reason is not the correct explanation for
Assertion
1
16
(A) Both Assertion and Reason are correct and Reason is the correct explanation for
Assertion.
1
17
(D) Assertion is incorrect but Reason is correct
1
18
(C) Assertion is correct but Reason is incorrect
1
SECTION : B
19
On the basis of crystal field theory,write the electronic configuration for d4 ion if Δo >P
2
T2g4 eg0
20
[CrCl(H2O)5]Cl2⋅H2O Pentaaquachloridochromium(III) chloride
2
OR
In a metal carbonyl, the metal-carbon bond possesses both the σ and π character. The 2
bond between the carbonyl molecule and the metal is further strengthened by the
synergic effect produced by the metal-ligand bond. The two types of bonding that exist
in metal carbonyls are explained below:
Structure of Metal Carbonyls:
21

Due to the donation of electrons by the carbonyl molecules to the vacant orbitals
of the metal, a metal-carbon σ bond is formed.

Due to the donation of a pair of electrons from a filled d orbital metal into the
vacant anti-bonding π* orbital of carbonyl ligand, a metal-carbon π bond is
formed.
Let us consider a general reaction:
2
M+4L ⇔ ML4
The stability constant may be written as:
Stability depends upon: charge on central metal and strength of ligands
Or
Giving a suitable example for each, explain the following :
(i) Linkage isomerism: When more than one atom in an ambidentate ligand is linked
with central metal ion to form two types of complexes, then the formed isomers are
called linkage isomers and the phenomenon is called linkage isomerism.
[Cr(H2O)5(NCS)]2+ :Pentaaquathiocyanate chromium (III) ion
[Cr(H2O)5(NCS)]2+ :Pentaaquaisothiocyanate chromium (III) ion
(ii) Ambidentate ligand: The monodentate ligands with more than one coordinating
atoms is known as ambidentate ligand. Monodentate ligands have only one atom
capable of binding to a central metal atom or ion. For example, the nitrate ion NO 2– can
bind to the central metal atom/ion at either the nitrogen atom or one of the oxygen
atoms.
Example : — SCN thiocyanate, — NCS isothiocyanate
22.
1
1
The formula[ Co(NH3) 5CO3]Cl Pentaamminecarbanatocobalt(III)chloride
1
[ Co(NH3) 5Cl]CO3 Pentaamminechloridecobalt(III)carbonate
1
23.
In [Co(NH3)6]3+ Co is in +3 state and has configuration 3d6. In the presence of NH3, 3d
electrons pair up leaving two d-orbitals empty. Hence, the hybridization is d2sp3
forming and inner orbital complex. In [Ni(NH3)6]2+, Ni is in +2 state and has
configuration 3d8. In presence of NH3, the 3d electrons do not pair up. The
hybridization is sp3d2 forming an outer orbital complex.
2
24.
In a free transition metal ions, all the five d-orbitals are degenerate but when it is
involved in a complex formation , the degeneracy is split, This is called CFSE.
1
There are the following factors that affect the crystal field splitting. These are the
nature of ligands, coordination number, arrangement of ligand, size of a metal atom,
charge on the metal atom.
1
25.
Spectrochemical series gives the arrangement of ligands in the increasing order of crystal field
1
splitting.
Weak field ligands cause less crystal field splitting. They form high spin complexes. Examples
1
include chloride ions, fluoride ions etc.
Strong field ligands cause greater crystal field splitting. They form low spin complexes.
Examples includes cyanide ion and CO.
SECTION :C
26
a. AgBr+2Na2SO3→ Na3[Ag(S2O3)2] + NaBr
1+2
b. Transition metals/ions have empty d orbitals into which the electron-pairs can be
donated by ligands containing pi electrons ,ie ;electrons in their pi molecular orbitals ,
eg, CH2=CH2, C5H5 ,C6H6,etc .
27
Theoretical change in temperature =kf x m =1.86 x 0.1= 0.186
3
Observed change in temperature(freezing)=0.372
As observed value is double of the theoretical value This show that each molecule of
the complex dissociates to form two ions . This can be so only if the formula of the
complex is [Co(NH3)4Cl2]C
28.
a. Overall stability constant (β) =2.1x (10)13 Overall dissociation constant is the
reciprocal of the overall stability constant .Hence overall dissociation constant= 1\ β =
1\2.1X(10)13=4.7X(10)-14 .
b. number of unpaired electrons in the square planar [Pt(CN)4] 2- ion is zero.The Pt (II)
ion has 5d8 electronic configuration. For square planar geometry dsp2 hybridization is
involved. For this one empty d orbital is needed for hybridization. Therefor , pairing of
electrons takes place in the remaining d orbitals . Hence , there are no unpaired in
[Pt(CN)4] 2- ion and it is diamagnetic.
29.
a. In CuSO4.5H2O; 4H2O molecules are present as ligand, Crystal field splitting occurs
and hence d-d trasition occurs which gives blue colour, In CuSO4 ,there are no H2o
molecules present as ligand .No CFS occurs and hence it has no colour .
2
1
2
b. Magnetic moment is 5.92 BM means n=5 ie 5 unpaired electron .Mn2+=3d5 4s0 4p0 .
To form [Mncl4] 2- , h4ybridization will be Sp3 . Hence the structure will be tetrahedral. 1
30
a. It arises from interchange of ligands between cationic and anionic entities of
different metal ions present in a complex. Eg- [Co(NH3)6][Cr(CN)6] and
[Cr(NH3)6][Co(CN)6]
1
2
b. The applications of coordination compound
(a) In analytical chemistry (b) In metallurgy (c) In purification of metal (d) In Industry (e)
In medical field
Or
i) [Co (NH3)6]3+, Co is in +3 oxidation state with the configuration 3d6. In the presence
of NH3 a strong ligand, the 3d electrons pair up leaving two d-orbitals empty. Hence,
the hybridization is d2sp3 forming an inner orbital octahedral com
ii) In [Ni(CN)4]2-, there is Ni2+ ion for which the electronic configuration in the valence
shell is 3d8 4s0.
* In presence of strong field CN- ions, all the electrons are paired up. The empty 3d,
3s and two 4p orbitals undergo dsp2 hybridization to make bonds with CN- ligands in
square planar geometry. Thus [Ni(CN)4]2- is diamagnetic. It is said to be a low spin inner
orbital complex.
1
2
SECTION :D
31
i)
ii)
iii)
32
Ammine NH3
Lewis acids are electron pair acceptor whereas Lewis bases are electron pair
donor.
Didentate ligand- those which bind to the metal ion through two donor
atoms. Eg Ethane 1-2 diamine [H2NCH2CH2NH2]. Ambidentate ligands –
those which bind to the metal ion through two different atom but at a time
bind with any one atom. Eg- NO2 -
1
1
2
(i) Tetracarbonyl Nickel(0)
1
(ii) Triamminebromidochloridonitrito-N Platinum(II)Chloride
1
iii) diamminetetraaquacobalt(III) chloride
1
iv) [CoCl2(en)2]Cl
1
SECTION :E
33
a). In [Fe(CN)6] 3- , CNis a strong ligand and therefore electrons pair up in d subshell
leaving one unpaired electron. The complex is inner orbital complex. On the other
hand, in [FeF6] 3- , F is a weak field ligand and therefore , electrons in 3d subshell do
not get pair up. It is outer orbital complex and has 5 unpaired electrons. Thus [FeF6] 3has greater magnetic moment than[Fe(CN)6] 3- .
2
1
b. (I) Hydrate isomerism.
1
(ii) Δ0 > P so t2g 4 eg 0
1
(iii) sp3 d 2 hybridization and Octahedral shape
Or
1
a.
b.
34
(i) Diammine dichlorido ethylenediamine chromium (III) chloride.
(ii) [Co(NH3)5(ONO)]2+
i) Homoleptic complex-; complexes in which the metal atom or ion is linked
to only one type of ligands are called homoleptic complexes, eg
,[co(NH3)6]3+
ii) Hetroleptic complexes;- complexes in which the metal atom or ion is
linked to more than one kind of ligands are called hetroleptic complexes .
eg ;[Co(NH3)4Cl2] +
iii) Coordination sphere ;--The central atom and the ligands which are
directly attached to it are enclosed in square brackets and are collectively
termed as the coordination sphere.
1
1
1
1
a. [Co(en)3]3+ because it forms chelate ring.
1
b. Low spin tetrahedral complexes are not formed because for tetrahedral complexes,
the crystal field stabilization energy is lower than pairing energy.
2
c. When they are dissolved in water ,they will give different ions in the solution which
can be tested by adding AgNo3 solution and BaSO4 solution . when cl ions are counter
ions a white ppt will be obtained AgNo3 solution , If SO4 2- ions are counter ions a
white ppt will be obtained with BaCl2 solution .
OR
2
(i) [COF4]2- : Tetrafluorido cobalt (III) ion
Coordination number = 4 Shape = Tetrahedral Hybridisation = sp 3
1.5
2
1.5
35
a.
(i) [Fe(en)2Cl2] Cl or x + 0 + 2 (-1) + (-1) = 0
x + (- 3) = 0 or x = + 3
∴ Oxidation number of iron, x = + 3
(ii) The complex has two bidentate ligands and two monodentate ligands. Therefore,
the coordination number is 6 and hybridization will be d 2sp3 and shape will be
octahedral.
(iii) In the complex 26Fe3+ = 3d5 4s0 4p0
Due to presence of one unpaired electrons in d orbitals the complex is paramagnetic.
The number of geometrical isomers are two.
(v) In coordination complex of [Fe(en)2Cl2] Cl, only cis-isomer shows optical isomerism.
(vi) Name of complex: Dichloridobis (ethane-1, 2- diamine) Iron (III) chloride.
OR
½+1+
1/2+1+
1+1
(i) Crystal field splitting: It is the splitting of the degenerate energy levels due to the
presence of ligands. When ligand approaches a transition metal ion, the degenerate dorbitals split into two sets, one with lower energy and the other with higher energy.
This is known as crystal field splitting and the difference between the lower energy set
and higher energy set is known as crystal field splitting energy (CFSE)
Example : 3d5 of Mn2+
(ii) b. In a free transition metal ion , the five d orbitals are degenerate .When it forms a
complex , the degeneracy is split In an octahedral field three d orbital s having lower
energy are called t2g orbitals and remaining two orbital is called eg orbitals .
3
2
CLASS XII CHEMISTRY CHAPTER WISE QUESTION PAPER
UNIT : HALOALKANES AND HALOARENES
MAX MARKS:70
TIME:03 hours
General Instructions:
(a) There are 35 questions in this question paper.
(b) SECTION A consists of 18 multiple choice questions carrying 1 mark each.
(c) SECTION B consists of 07 very short answer questions carrying 2 marks each.
(d) SECTION C consists of 05 short answer questions carrying 3 marks each.
(e) SECTION D consists of 02 Case based question carrying 4 marks each.
(f) SECTION E consists of 03 long answer questions carrying 5 marks each.
(g) All questions are compulsory.
(h) Use of log tables and calculators not allowed.
Section A
Q.1
In which of the following molecules carbon atom marked with asterisk (*) is
asymmetric?
1
(A) (a), (b), (c), (d)
(B) (a), (b), (c)
(C) (b), (c), (d)
(D) (a), (c), (d)
Q.2
Chlorobenzene is formed by reaction of chlorine with benzene in the presence of
1
Q.3
AlCl3. Which of the following species attacks the benzene ring in this reaction ?
(A) Cl–
(B) Cl+
(C) AlCl3
(D) [AlCl4]–
Identify following
reaction:
A) Wurtz Reaction
B) Etard Reaction
C) Reimer Tiemann
Reaction
D) Kolbe Reaction
1
Q.4
The reaction of toluene with chlorine in the presence of iron and in the absence of
1
light yields ____________.
A) Benzyl chloride
B) o-Chlorobenzene
C) p-Choloro benzene
D) Mixture of B and C
1
Q.5
What is Q in the following reaction?
Q.6
Which of the following is the correct order of decreasing SN2 reactivity?
A) RCH2X > R2CHX > R3CX
B) R3CX > R2CHX >RCH2X
C) R2CHX >R3CX > RCH2X
D) RCH2X >R3CX >R2CHX
1
Q.7
Which of the following is correct for the reaction
1
Q8
a) A is major product and B is minor product
b) B is major product and A is minor product
c) Only A will be obtained as a product
d) Only B will be obtained as a product
Tertiary alkyl halides are practically inert to substitution by SN 2 mechanism
because of
(A)
steric hindrance
(B)
inductive effect
(C)
instability
(D)
insolubility
1
Q.9
Match the reactions given in Column I with the names given in Column II.
1
A) i-a,
B) i-b,
C) i-d,
D) i-c,
ii-b
ii-a
ii-b
ii-a
,iii-c, iv-d
, iii-d, iv-c
,iii-c, iv-a
,iii-b, iv-d
2
Q.10
Alkyl fluorides are synthesised by heating an alkyl chloride/bromide in presence of_
1
(A) Ca F2
(B) PF3
(C) Hg2F2
(D) NaF
Q.11
Haloalkanes contain halogen atom (s) attached to the sp3 hybridised carbon atom
1
of an alkyl group. Identify haloalkane from the following compounds.
(A) 2-Bromopentane
(B) Vinyl chloride (chloroethene)
(C) 2-chloroacetophenone
(D) chlorobenzene
Q.12
Alkyl halides are prepared from alcohols by treating with______
1
(A) HCl + ZnCl2
(B) Red P + Br2
(C) PCl5
(D) All the above
Q.13
Match the structures given in Column I with the names in Column II.
1
A) i-a, ii-b ,iii-c, iv-d
B) i-b, ii-a, iii-d, iv-c
C) i-d, ii-b ,iii-c, iv-a
D) i-a, ii-d, iii-c, iv-b
Q.14
Which is the correct increasing order of boiling points of the following compounds?
1-Iodobutane, 1-Bromobutane, 1-Chlorobutane, Butane
(A) Butane < 1-Chlorobutane < 1-Bromobutane < 1-Iodobutane
(B) 1-Iodobutane < 1-Bromobutane < 1-Chlorobutane < Butane
(C) Butane < 1-Iodobutane < 1-Bromobutane < 1-Chlorobutane
(D) Butane < 1-Chlorobutane < 1-Iodobutane < 1-Bromobutane
3
1
Given Given below question no 15-18 consist of an "Assertion" (A) and "Reason" (R) Type 1
questions. Use the following Key to choose the appropriate answer.
A.If both (A) and (R) are true, and (R) is the correct explanation of (A).
B.If both (A) and (R) are true but (R) is not the correct explanation of (A).
C.If (A) is true but (R) is false.
D.If (A) is false but (R) is true.
Assertion : Presence of a nitro group at ortho or para position increases the
Q.15
reactivity of haloarenes towards nucleophilic substitution.
Reason : Nitro group, being an electron withdrawing group decreases the electron
density over the benzene ring.
Q.16
Assertion : It is difficult to replace chlorine by –OH in chlorobenzene in comparison
to that in chloroethane.
Reason : Chlorine-carbon (C—Cl) bond in chlorobenzene has a partial double bond
character due to resonance.
Q.17
Assertion: Chlorobenzene is less reactive than benzene towards the electrophilic
substitution reaction.
Reason: Resonance destabilises the carbo cation.
Assertion:The C–Cl bond length in chlorobenzene is shorter than that in CH3–Cl.
Reason: In haloarenes Cl is attached to sp2 hyridised carbon which is more
electronegative than sp3 hybridised carbon.
Section B
Q.18
1
Q.19
Hydrolysis
of optically
butan-2-ol. Why?
optically
inactive 2
Q.20
a)Which one out of CH3CH(Cl)CH2CH3 and CH3CH2CH2CH2Cl is more easily
2
active
2-bromobutane
forms
hydrolysed by KOH(aq).
b) Arrange the following compounds according to reactivity towards nucleophillic
substitution reaction with CH3ONa
4- nitro chloro benzene, 2,4 di nitro chloro bemzene, 2,4,6, trinitrochlorobenzene
Q.21
Vinyl chloride is less reactive than allyl chloride. Why?
2
Q.22
How is chlorobenzene prepared by (a) direct chlorination (b) diazotization method?
2
Q.23
Why do haloalkenes under go nucleophillic substitution whereas haloarenes
under go electophillic substitution ?
2
Q.24
What happens when
a) Thionyl chloride acts upon
b) Ethanol reacts with PBr3 .
2
Q.25
What are the IUPAC names of the insecticide DDT and benzenehexachloride? Why is
their use banned in India and other countries?
Q.26
propan-1-ol.
Section C
Compound ‘A’ with molecular formula C4H9Br is treated with aq. KOH solution. The
rate of this reaction depends upon the concentration of the compound ‘A’ only.
When another isomer ‘B’ of this compound was treated with aq. KOH solution, the
rate of reaction was found to be dependent on concentration of compound and KOH
both. (i) Write down the structural formula of both compounds ‘A’ and ‘B’. (ii) Out
of these two compounds, which one will be optically active?
4
2
3
Q.27
What is the difference between
one example.
Q.28
Some alkylhalides undergo substitution whereas some undergo elimination reaction
on treatment with bases. Discuss the structural features of alkyl halides with the
help of examples which are responsible for this difference?
3
Q.29
Give the IUPAC names of the following:(a) o-Br-C6H4CH(CH3)CH2CH3
(b) CH3C(Cl)(C2H5)CH2CH3
(c) Cl-CH2 C ≡ C-CH2-Br
Complete the following reactions :
i) CH3CH=C(CH3)2 + HBr -------
ii) CH3CH2CH2OH + SOCl2 -----------
iii) CH3CH2Br + Mg ---------
3
Q.30
Q.31
Q.32
enantiomers and diastereomers. Illustrate with
3
3
Section D
Chloroflouro carbon (CFC) compounds of methane and ethane are collectively
known as freons. They are non-inflammable , extremely stable, non-toxic, noncorrosive and low boiling liquids. CFC and gas emitted from the exhaust system of
supersonics aeroplanes might be slowly depleting the concentration of the ozone
layer in the upper atmosphere. Answer the following questions on the basis oy
your knowledge .in this topic
1. Write the formula of freon.
2. Give two uses of chloroflourocarbons.
3. How does freon-12 deplete ozone layer?
4. Do you think the use of CFCs should be banned? Give reason.
Methyl chloride, methyl bromide, ethyl chloride and some chlorofluoromethanes are
gases at room temperature. Higher members are liquids or solids. As we have
already learnt, molecules of organic halogen compounds are generally polar. Due to
greater polarity as well as higher molecular mass as compared to the parent
hydrocarbon, the intermolecular forces of attraction (dipole-dipole and van der
Waals) are stronger in the halogen derivatives. That is why the boiling points of
chlorides, bromides and iodides are considerably higher than those of the
hydrocarbons of comparable molecular mass. The attractions get stronger as the
molecules get bigger in size and have more electrons.
4
4
i)Draw all the possible isomers structure of bromobutane and arrange them in
increasing order of boiling points. (2 marks)
ii)Even though haloalkanes are polar compounds these compounds are least
soluble in water. Why?(1 mark)
iii) How will you distinguish between chloroethane and bromoethane?(1 mark)
Q.33
i)Identify A, B , C, D and E in the following:+SOCl2
AlcKOH
HBr/Peroxide
CH3CH2CH2OH ──────→ A ──────→ B ───────→
5
AgCN
C ───────→ D
KCN
E
5
Section E
Q.34
i) Arrange in increasing order of boiling points.
(a) Bromomethane, Bromoform, chloromethane, Dibromo-methane
(b)1-chloropropane, Isopropyle chloride, 1-Chlorobutane.
(c) 1-chloropropane, , 1-bromopropane, 1-iodopropane
5
ii) Which compound will react faster in SN2 reaction with OH---?
(a) CH3Br and CH3I (SN2)
(b) (CH3)3C-Cl or CH3Cl (SN2)
Q.35
How the following conversions can be carried out?
i)
ii)
iii)
iv)
v)
But-1-ene to n-butyliodide
2-Chloropropane to 1-propanol
Isopropyl alcohol to iodoform (iv)
Chlorobenzene to p-nitrophenol
2-Bromopropane to 1-bromopropane
6
5
CLASS XII CHEMISTRY CHAPTER WISE QUESTION PAPER
UNIT : HALOALKANES AND HALOARENES - ANSWER KEY
MAX MARKS:70
TIME:03 hours
1B
2B
3C
4D
5C
6B
7A
8A
9B
10 C
11 A
12 D
13 A
14 A
15 A
16 C
17 A
18 B
Q.19 Due to the formation of planar carbocation as intermediate in SN1 mechanism, OH- 2
can attack carbocation equally from both side which result in racemic mixture.
Q.20 a) CH3CH(Cl)CH2CH3
2
Q.21 Vinyl chloride is less reactive than allyl chloride, it is due to resonance in vinyl
2
Q.22
2
b) 2,4,6, trinitrochlorobenzene > 2,4 dinitrochlorobemzene > 4- nitrochlorobenzene
chloride the C- Cl bond gets double bond character and becomes stable.
a)
b)
Q.23 Due to more electro negative nature of halide atomC-X bond is polar in haloalkanes
2
Q.24
2
Q.25
and carbon atom becomes slightly positive and is easily attacked by nucleophillic
reagents. While in haloarenes due to resonance, carbon atom becomes slightly
negative and attacked by electrophillic reagents.
(a) When thionyl chloride acts upon 1- propanol chloropropane is formed .
(b) When ethanol reacts with PBr3, bromoethane is formed
Correct name.
Q.26 A. 1-Bromo butane. B) 2-Bromo butane. B is optically active.
Q.27
Q.28
Enantiomers are those optical isomers of optically active compound which are
non-superimposable images of each other.
Diasteriomers are those optical isomers optically active compound which are nonsuperimposable but not mirror images of each other.
Any one example.
Primary alkyl halides undergo substitution reaction where as tertiary alkyl halides
undergo elimination reactions.
Q.29 Correct Name
Q.30 i)CH3CH2CH(Br)CH3
2
3
3
3
3
ii) CH3CH2CH2Cl +SO2+HCl
7
iii)CH3CH3
3
Q.31 1 CCl2F2
4
Q.32
4
Q.33
2 It is used in air conditioning and in domestic refrigeration for cooling purposes.
3 Freons can diffuse into the stratosphere and in stratosphere, freon is able to
initiate radical chain reactions by using UV rays that can upset the natural ozone
balance.
4 CFCs should be banned and its alternative should be used which would be safer
and has no environmental impact.
Hydrochloroflourocarbons and Hydroflourocarbons are some halogen containing
compounds are useful in daily life. Some compounds of this class are responsible
for exposure of flora and fauna to more and more of UV light which causes
destruction to a great extent. Name the class of these halocompounds.
i) All the isomers and correct order
ii) When haloalkanes dissolved in water they neither form hydrogen bond nor they
can release sufficient energy to break the hydrogen bond in water.
iii)They can be distinguished by silver nitrate test ( treating with aq.NaOH followed
by AgNO3 ). Chloroethane gives white ppt of AgCl (soluble in ammonium hydroxide)
but bromo ethane gives pale yellow ppt of AgBr (partially soluble in ammonium
hydroxide)
A) CH3CH2CH2Cl ,
B) CH3CH=CH2 ,
C)CH3CH2CH2 OH ,
D) CH3CH2CH2NC
E) CH3CH2CH2 CN
5
Q.34 i) (a) chloromethane < Brommethane < Dibromo-methane < Bromoform
(b) Isopropyle chloride <1-chloropropane <1-Chlorobutane
(As Branching increases , boiling point decreases)
5
(c)1-Chloropropane< 1-Bromopropane< 1-iodopropane
ii) (a) CH3I will react faster than CH3Br
(b) CH3Cl will react faster than 30 halide
Q.35 (i)
5
(ii)
8
(iii)
(iv)
(v)
9
CLASS XII CHEMISTRY CHAPTER WISE QUESTION PAPER
ALCOHOLS ,PHENOLS AND ETHERS
MAX MARKS:70
TIME:03 hours
General Instructions:
(a) There are 35 questions in this question paper with internal choice.
(b) SECTION A consists of 18 multiple choice questions carrying 1 mark each.
(c) SECTION B consists of 07 very short answer questions carrying 2 marks each.
(d) SECTION C consists of 05 short answer questions carrying 3 marks each.
(e) SECTION D consists of 02 Case based question carrying 4 marks each.
(f) SECTION E consists of 03 long answer questions carrying 5 marks each.
(g) All questions are compulsory.
(h) Use of log tables and calculators not allowed.
Q.1
Acid-catalysed hydration of alkenes except ethene leads to the formation of :
(a)primary alcohol
(b)secondary or tertiary alcohol
(c)mixture of primary and secondary alcohol
(d)mixture of secondary and tertiary alcohol
1
Q.2
From amongst the following alcohols, the one that would react faster with conc HCl
and anhydrous ZnCl2 ,is
(a)2-methypropanol
(b)1-butanol
(c)2-butanol
(d)2-methylpropan-2-ol
Identify following
reaction:
(a) Wurtz Reaction
(b) Etard Reaction
(c) Reimer Tiemann
Reaction
(d) Kolbe Reaction
1
Ortho-nitrophenol is less soluble in water than p- and m-nitrophenols because
(a)o-nitrophenol shows intramolecular H-bonding
(b)o-nitrophenol shows intermolecular H-bonding
(c)melting point of o-nitrophenol is lower than those of m- and p-nitrophenols
(d)o-nitrophenols is more volatile in steam than those of m- and p-isomers
An organic compound X on treatment with pyridinium chlorochromate in
dichloromethane gives compound Y. Compound Y reacts with I2 and alkali to form
triiodo methane . The compound ‘X’ is
(a)CH2CH2OH
(b)CH3CHO
(c)CH3COCH3
(d)CH3COOH
1
Compound ‘A’ of molecular formula C4H10O on treatment with Lucas reagent at room
temperature gives a compound ‘B’.When compound‘B’ is heated with alcoholic KOH,
it gives isobutene. Compound ‘A’ and ‘B’ are respectively
(a) 2-methyl-2propanol and 2-chloro-2-mehtyl-propane
1
Q.3
Q.4
Q.5
Q.6
1
1
(b) 2-methyl-1-propanol and 1-chloro-2methyl-propane
(c) 2-methyl-1-propanol and 2-chloro-2methyl-propane
(d) butan-2-ol and 2-chlorobutane
Gfhgh
Q.7 Which of the following reagent may be used to distinguish between phenol and 1
benzoic acid?
(a) Neutral FeCl3
(b) Aqueous NaOH
(c) Tollen’s reagent
(d) Molisch reagent
Ujhjhs
Q8
Reaction of phenol with carbon tetra chloride in presence of dilute sodium hydroxide 1
finally introduces which one of the following functional group?
Dssss( (a) – COOH
(b) – CHCl2
(c) – CHO
(d) – CH2Cl
YHGY
Q.9 Heating of phenyl methyl ether with HI produces
Uhu (a)iodobenzene
(b) phenol (c) benzene (d) ethyl chloride
1
FDDD
Q.10 Lucas reagent is:
1
(a) anhydrousCaCl2 and conc. HCI
(b) anhydrous ZnCl2 and conc. HCI
(c) anhydrous AlCl3 and conc. HCI
(d) anhydrous PdCl2 and conc. HCl
Q.11 Vapours of an alcohol X when passed over hot reduced copper, produce an alkene, 1
the alcohol is
(a) primary alcohol (b) secondary alcohol (c) tertiary alcohol
(d) dihydric alcohol
Iod
Q.12
Iodoform test is not given by
(a) Ethanol
(b) Ethanal
1
(c) Pentan-2-one
(d) Pentan-3-one
Q.13 Phenol when treated with neutral FeCl3 , it gives a complex of
(a) violet colour (b) red colour (c) green colour
(d) yellow colour
Q.14 A tertiary alcohol is obtained by the reaction of Grignard reagent with:
(a) Butanone
(b) Propanone
(c) Acetone
(d) All of the above
1
1
Given
Q.15 Given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the 1
following Key to choose the appropriate answer.
a.
If both (A) and (R) are true, and (R) is the correct explanation of (A).
b.
If both (A) and (R) are true but (R) is not the correct explanation of (A).
c.
If (A) is true but (R) is false.
d.
If (A) is false but (R) is true.
an " Assertion: An ether is more volatile than an alcohol of comparable molecular mass.
d Reason : Ethers are polar in nature.
Q.16 Given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the 1
following Key to choose the appropriate answer.
a.
If both (A) and (R) are true, and (R) is the correct explanation of (A).
b.
If both (A) and (R) are true but (R) is not the correct explanation of (A).
c.
If (A) is true but (R) is false.
d.
If (A) is false but (R) is true.
Asse Assertion: Phenols are more acidic than aliphatic alcohols.
Reason: The phenoxide ion is more resonance stabilised than alkoxide ion.
Q.17 Given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the 1
following Key to choose the appropriate answer.
2
a.
If both (A) and (R) are true, and (R) is the correct explanation of (A).
b.
If both (A) and (R) are true but (R) is not the correct explanation of (A).
c.
If (A) is true but (R) is false.
d.
If (A) is false but (R) is true.
Assertion: Phenol is more reactive than benzene towards electrophilic substitution
reaction.
Reason: In case of phenol, the intermediate carbocation is more stabilized by
resonance.
Q.18 Given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the 1
following Key to choose the appropriate answer.
a.
If both (A) and (R) are true, and (R) is the correct explanation of (A).
b.
If both (A) and (R) are true but (R) is not the correct explanation of (A).
c.
If (A) is true but (R) is false.
d.
If (A) is false but (R) is true.
Assertion: Tertiary alcohols gets converted into an alkene instead of a carbonyl
compounds in the presence of heated metallic copper.
Reason: Tertiary alcohols prefer to undergo dehydrogenation instead of
dehydration in the presence of heated copper.
Ans:
Q.19 Arrange the following compounds in increasing order of their boiling point.
a.Propan-1-ol, Butan-1-ol, Butan-2-ol and Pentan-1-ol.
b.Pentanal, n-Butane, Ethoxyethane and Pentan-1-ol.
2
Q.20 How will you bring about following conversions:
a.Phenol to picric acid b.Phenol to Aspirin
2
Q.21 Write IUPAC names of the following compounds:
2
Q.22 Write structures of the compounds whose IUPAC names are as follows:
a. 2-Methylbutan-2-ol
b. 2,3 – Diethylphenol
Q.23 Arrange the following sets of compounds in order of their increasing boiling points:
(a) Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.
(b) Pentan-1-ol, n-butane, pentanal, ethoxyethane.
2
Q.24 Give mechanism for the Hydration of alkenes
Q.25 Give test to differentiate between following pairs:
a.Phenol and Benzoic acid
b. Phenol and Aniline
2
2
Q.26 Give mechanism for the acidic dehydration of alcohols to give ethers.
3
𝑐𝑛𝑜𝑐 𝐻2𝑆𝑂4 ,413 𝐾
ANS: C2H5-OH + HO- C2H5 →
C2H5 -O- C2H5 + H2O
3
2
Q.27 How will you bring about following conversions:
a. Chlorobenzene to phenol
b.Cumene to phenol
c.Phenol to benzoquinone
OR
a.Methyl magnesium bromide to 2-Methylpropan-2-ol
b.Bromo methane to propan-2-ol
c.Chloro ethane to propan-1-ol
3
Q.28 Give test to differentiate between following pairs:
a.Methanol and Ethanol
b.Propanol and Propan-2-ol
c.2-Methyl Propan-2-ol and Propanol
3
Q.29 Give the major products that are formed by heating each of the following ethers with
HI.
3
Q.30 Give Reasons:
a.In ethers R─ O ─ R bond is slightly larger than normal tetrahedral bond angle?
b. Ethers are fairly soluble in water?
c.Anisole can not be prepared by reaction of bromobenzene or iodo benzene with
sodium salt of corresponding alcohols i.e sodium ethoxide?
3
Q.31 Alcohols and phenols behave as weak acids due to the presence of polar -OH group in
them. Phenols are, however, stronger acids than alcohols because the phenoxide ion
is stabilized by resonance. Presence of electron withdrawing groups on the ring
further increases the acidic strength of phenol.
(a) Out of Phenol & Benzyl alcohol, which has lower value of pKa and why?
(b) Out of o-Cresol and o-Nitrophenol, which has higher value of pKa?
(c) Which compound has shorter bond length- Phenol or Cyclohexanol? Give reason.
(d) Which species out of ethoxide ion and phenoxide ion, is stronger base
4
Q.32 A compound (X) containing C, H and O is unreactive towards sodium. It also does not
4
react with Schiff’s reagent. On refluxing with an excess of hydroiodic acid, (X) yields
only one organic product (Y). On hydrolysis, (Y) yields a new compound (Z) which can
be converted into (Y) by reaction with red phosphorous and iodine. The compound (Z)
on oxidation with potassium permanganate gives a carboxylic acid. The equivalent
weight of this acid is 60.
The following questions are multiple choice questions. Choose the most appropriate
answer:
A. The compound (X) is an
(a) acid
(b) aldehyde (c) alcohol
(d) ether
4
B. The IUPAC name of the acid formed is
(a) methanoic acid (b) ethanoic acid (c) propanoic acid
C. Compound (Y) is
(a) ethyl iodide (b) methyl iodide (c) propyl iodide
D.Compound (Z) is
(a) methanol
(b) ethanol
(c) propanol
(d) butanoic acid
(d) mixture of (a) and (b)
(d) butanol
Q.33 Write the equations involved in the following reactions:
5
a.Kolbe Reaction
b.Williamson’s synthesis
c. Esterification
d. Hydroboration–oxidation
e. Reimer-Tiemann reaction:
Q.34 Give Reasons:
5
a. Phenol is more acidic than alcohols(Ethanol).
b. Phenol has small dipole moment than methanol.
c. Ortho – nitrophenol is more acidic than Ortho – methoxyphenol.
d. The presence of electron withdrawing groups such as nitro group enhances the
acidic strength of phenol.
e. The presence of electron releasing groups such as alkyl group decreases the acidic
strength of phenol.
Q.35 Give Reasons:
5
a.The boiling point of alcohols and phenols are higher in comparison to hydrocarbons,
ethers, and halo alkanes of comparable molecular mass.?
b.Alcohols are comparatively more soluble in water than the corresponding
Hydrocarbons
c. Lower alcohols are soluble in water, higher alcohols are not.
d. The boiling point of alcohols and phenols increase with increase in number of
carbon atoms.
e. In alcohols the boiling point decreases with increase in branching.
****
5
CLASS XII CHEMISTRY CHAPTER WISE QUESTION PAPER
ALCOHOLS ,PHENOLS AND ETHERS
MAX MARKS:70
TIME:03 hours
ANSWER KEY
Q.1
Q.2
Acid-catalysed hydration of alkenes except ethene leads to the formation of :
(a)primary alcohol
(b)secondary or tertiary alcohol
(c)mixture of primary and secondary alcohol
(d)mixture of secondary and tertiary alcohol
ANS: (b)secondary or tertiary alcohol
From amongst the following alcohols, the one that would react faster with conc HCl
and anhydrous ZnCl2 ,is
(a)2-methypropanol
(b)1-butanol
(c)2-butanol
(d)2-methylpropan-2-ol
1
1
ANS: (d)2-methylpropan-2-ol
Q.3
Identify following
reaction:
(e) Wurtz Reaction
(f) Etard Reaction
(g) Reimer
Tiemann
Reaction
(h) Kolbe Reaction
1
ANS: (c) Reimer Tiemann Reaction
Q.4
Ortho-nitrophenol is less soluble in water than p- and m-nitrophenols because
(a)o-nitrophenol shows intramolecular H-bonding
(b)o-nitrophenol shows intermolecular H-bonding
(c)melting point of o-nitrophenol is lower than those of m- and p-nitrophenols
(d)o-nitrophenols is more volatile in steam than those of m- and p-isomers
ANS: (a)o-nitrophenol shows intramolecular H-bonding
Q.5
An organic compound X on treatment with pyridinium chlorochromate in
dichloromethane gives compound Y. Compound Y reacts with I2 and alkali to form
triiodo methane . The compound ‘X’ is
(a)CH3CH2OH
(b)CH3CHO
(c)CH3COCH3
(d)CH3COOH
ANS: (a)CH3CH2OH
6
1
Q.6
Compound ‘A’ of molecular formula C4H10O on treatment with Lucas reagent at room
temperature gives a compound ‘B’.When compound‘B’ is heated with alcoholic KOH,
it gives isobutene. Compound ‘A’ and ‘B’ are respectively
(e) 2-methyl-2propanol and 2-chloro-2-mehtyl-propane
(f) 2-methyl-1-propanol and 1-chloro-2methyl-propane
(g) 2-methyl-1-propanol and 2-chloro-2methyl-propane
(h) butan-2-ol and 2-chlorobutane
1
ANS: (a) 2-methyl-2propanol and 2-chloro-2-mehtyl-propane
Gfhgh
Q.7 Which of the following reagent may be used to distinguish between phenol and 1
benzoic acid?
(e) Neutral FeCl3
(f) Aqueous NaOH
(g) Tollen’s reagent
(h) Molisch reagent
ANS: (a) Neutral FeCl3
Ujhjhs
Q8
Reaction of phenol with carbon tetra chloride in presence of dilute sodium hydroxide 1
finally introduces which one of the following functional group?
Dssss( (a) – COOH
((B)uh (b) – CHCl2
FDGFS (c) – CHO
Fghfg (d) – CH2Cl
(JIJIJIJI
ANS: ( ANS: (a) – COOH
YHGY
Q.9 Heating of phenyl methyl ether with HI produces
1
Uhu (a)iodobenzene
Ijij I (b)phenol
(c) benzene
(d) ethyl chloride
ANS: (b)phenol
FDDD
Q.10 Lucas reagent is:
(a) anhydrousCaCl2 and conc. HCI
(b) anhydrous ZnCl2 and conc. HCI
(c) anhydrous AlCl3 and conc. HCI
(d) anhydrous PdCl2 and conc. HCl
ANS : ANS: (b) anhydrous ZnCl2 and conc. HCI
1
Vapours of an alcohol X when passed over hot reduced copper, produce an alkene,
1
the alcohol is (a) primary alcohol
(b) secondary alcohol
(c) tertiary alcohol
(d) dihydric alcohol
ANS: (c) tertiary alcohol
Iod
Q.12 Iodoform test is not given by
1
(a) Ethanol
(b) Ethanal
(c) Pentan-2-one
(d) Pentan-3-one
ANS:
ANS: (d) Pentan-3-one
ANS:
Q.11
7
Pheno
Q.13 When treated with neutral FeCl3 , it gives a complex of
(a) violet colour (b) red colour (c) green colour
ANS:
(a) violet colour
Q.14
1
(d) yellow colour
A tertiary alcohol is obtained by the reaction of Grignard reagent with:
(a) Butanone
(b) Propanone
(c) Acetone
(d) All of the above
1
ANS: (c) Acetone
Given
Q.15 Given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the 1
following Key to choose the appropriate answer.
a.
If both (A) and (R) are true, and (R) is the correct explanation of (A).
b.
If both (A) and (R) are true but (R) is not the correct explanation of (A).
c.
If (A) is true but (R) is false.
d.
If (A) is false but (R) is true.
an " Assertion(A): An ether is more volatile than an alcohol of comparable molecular mass.
d Reason (R): Ethers are polar in nature.
Ans: b
Q.16
Asse
Ans
Given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the 1
following Key to choose the appropriate answer.
a.
If both (A) and (R) are true, and (R) is the correct explanation of (A).
b.
If both (A) and (R) are true but (R) is not the correct explanation of (A).
c.
If (A) is true but (R) is false.
d.
If (A) is false but (R) is true.
Assertion: Phenols are more acidic than aliphatic alcohols.
Reason: The phenoxide ion is more resonance stabilised than alkoxide ion.
Ans: a
Q.17 Given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the 1
following Key to choose the appropriate answer.
a.
If both (A) and (R) are true, and (R) is the correct explanation of (A).
b.
If both (A) and (R) are true but (R) is not the correct explanation of (A).
c.
If (A) is true but (R) is false.
d.
If (A) is false but (R) is true.
Assertion: Phenol is more reactive than benzene towards electrophilic substitution
reaction.
Reason: In case of phenol, the intermediate carbocation is more stabilized by
resonance.
Ans: Ans: a
Q.18 Given below consist of an "Assertion" (A) and "Reason" (R) Type questions. Use the 1
following Key to choose the appropriate answer.
a.
If both (A) and (R) are true, and (R) is the correct explanation of (A).
b.
If both (A) and (R) are true but (R) is not the correct explanation of (A).
c.
If (A) is true but (R) is false.
d.
If (A) is false but (R) is true.
Assertion: Tertiary alcohols gets converted into an alkene instead of a carbonyl
compounds in the presence of heated metallic copper.
8
Ans:
Reason: Tertiary alcohols prefer to undergo dehydrogenation instead of
dehydration in the presence of heated copper.
Ans: c
Q.19 Arrange the following compounds in increasing order of their boiling point.
a.Propan-1-ol, Butan-1-ol, Butan-2-ol and Pentan-1-ol.
b.Pentanal, n-Butane, Ethoxyethane and Pentan-1-ol.
2
ANS: a. Propan-1-ol, Butan-2-ol, Butan-1-ol and Pentan-1-ol.
b. n-Butane, Ethoxyethane, Pentanal, Pentan-1-ol.
Q.20 How will you bring about following conversions:
a.Phenol to picric acid b.Phenol to Aspirin
2
ANS: a.
b.
Q.21 Write IUPAC names of the following compounds:
2
ANS: a. 2,2,4-Trimethylpentan-3-ol b. 5-Ethylheptane-2,4-diol
Q.22 Write structures of the compounds whose IUPAC names are as follows:
2
a. 2-Methylbutan-2-ol
b. 2,3 – Diethylphenol
ANS: a.
9
b.
Q.23 Arrange the following sets of compounds in order of their increasing boiling points:
(a) Pentan-1-ol, butan-1-ol, butan-2-ol, ethanol, propan-1-ol, methanol.
(b) Pentan-1-ol, n-butane, pentanal, ethoxyethane.
2
ANS: (a) Methanol, ethanol, propan-1-ol, butan-2-ol, butan-1-ol, pentan-1-ol.
(b) n-Butane, ethoxyethane, pentanal and pentan-1-ol.
Q.24 Give mechanism for the Hydration of alkenes
ANS:
2
Q.25 Give test to differentiate between following pairs:
a.Phenol and Benzoic acid
b. Phenol and Aniline
2
ANS:a. Phenol reacts with Neutral FeCl3 to form a complex of Violet colour.
3 PhOH (Phenol) + FeCl3 (Neutral)  Violet colouration (PhO)3 Fe + 3 HCl
Ferric Phenate
b. Phenol reacts with Neutral FeCl3 to form a complex of Violet colour.
3 PhOH (Phenol) + FeCl3 (Neutral)  Violet colouration (PhO)3 Fe + 3 HCl
Ferric Phenate
10
Q.26 Give mechanism for the acidic dehydration of alcohols to give ethers.
𝑐𝑛𝑜𝑐 𝐻2𝑆𝑂4 ,413 𝐾
ANS: C2H5-OH + HO- C2H5 →
3
C2H5 -O- C2H5 + H2O
Q.27 How will you bring about following conversions:
a. Chlorobenzene to phenol
b.Cumene to phenol
c.Phenol to benzoquinone
OR
a.Methyl magnesium bromide to 2-Methylpropan-2-ol
b.Bromo methane to propan-2-ol
c.Chloro ethane to propan-1-ol
𝑎𝑙𝑐 𝐾𝐶𝑁
CH3 CH2 Cl →
CH2OH
𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛
CH3 CH2 CN →
3
𝐻𝑁𝑂2
CH3 CH2 CH2NH2 →
ANS: a.
b.
c.
11
CH3 CH2
OR
b.
𝑎𝑙𝑐 𝐾𝐶𝑁
c. CH3 CH2 Cl →
CH2OH
𝑅𝑒𝑑𝑢𝑐𝑡𝑖𝑜𝑛
CH3 CH2 CN →
𝐻𝑁𝑂2
CH3 CH2 CH2NH2 →
CH3 CH2
Q.28 Give test to differentiate between following pairs:
a.Methanol and Ethanol
b.Propanol and Propan-2-ol
c.2-Methyl Propan-2-ol and Propanol
ANS: a.Ethanol when heated with I2 in presence of NaOH ,yellow ppt of iodoform is
formed.
CH3CH2 OH + I2 + NaOH  HCOONa + CHI3 + NaI + H2O
Yellow Ppt
b. Propan-2-ol when heated with I2 in presence of NaOH ,yellow ppt of iodoform is
formed.
CH3CH(OH) CH3 + I2 + NaOH  CH3COONa + CHI3 + NaI + H2O
Yellow Ppt
c.2-Methyl Propan-2-ol being tertiary alcohol, when treated with Lucas reagent form
turbidity immediately.
Propanol being primary alcohol, when treated with Lucas reagent does not form
turbidity at room pemperature.
12
3
Q.29 Give the major products that are formed by heating each of the following ethers with
HI.
3
ANS:
Q.30 Give Reasons:
a.In ethers R─ O ─ R bond is slightly larger than normal tetrahedral bond angle?
ANS:Due to greater repulsion between bulkier alkyl group.
b. Ethers are fairly soluble in water?
ANS. This is due to the fact that just like alcohols, oxygen
of ether can also form hydrogen bonds with water
molecule as shown:
3
c.Anisole can not be prepared by reaction of bromobenzene or iodo benzene with
sodium salt of corresponding alcohols i.e sodium ethoxide?
ANS: During Williamson synthesis bromobenzene or iodo benzene has to go
nucleophilic substitution and aryl halides are much less reactive than alkyl halides
towards nucleophilic substitution reactions.
Q.31 Alcohols and phenols behave as weak acids due to the presence of polar -OH group in
them. Phenols are, however, stronger acids than alcohols because the phenoxide ion
is stabilized by resonance. Presence of electron withdrawing groups on the ring
further increases the acidic strength of phenol.
(a) Out of Phenol & Benzyl alcohol, which has lower value of pKa and why?
(b) Out of o-Cresol and o-Nitrophenol, which has higher value of pKa?
(c) Which compound has shorter bond length- Phenol or Cyclohexanol? Give reason.
(d) Which species out of ethoxide ion and phenoxide ion, is stronger base
Ans: (a) Phenol, because the phenoxide ion is stabilized by resonance.
(b) o-Cresol
(c) Phenol has shorter bond length due to partial double bond character attributed to
resonance. (d) Ethoxide ion.
Q.32 A compound (X) containing C, H and O is unreactive towards sodium. It also does not
react with Schiff’s reagent. On refluxing with an excess of hydroiodic acid, (X) yields
only one organic product (Y). On hydrolysis, (Y) yields a new compound (Z) which can
13
4
4
be converted into (Y) by reaction with red phosphorous and iodine. The compound (Z)
on oxidation with potassium permanganate gives a carboxylic acid. The equivalent
weight of this acid is 60.
The following questions are multiple choice questions. Choose the most appropriate
answer:
A. The compound (X) is an
(a) acid
(b) aldehyde (c) alcohol
(d) ether
B. The IUPAC name of the acid formed is
(a) methanoic acid (b) ethanoic acid (c) propanoic acid
C. Compound (Y) is
(a) ethyl iodide (b) methyl iodide (c) propyl iodide
D.Compound (Z) is
(a) methanol
(b) ethanol
(c) propanol
(d) butanoic acid
(d) mixture of (a) and (b)
(d) butanol
Ans: A. (d) ether
B. (b) ethanoic acid
C. (a) ethyl iodide
D. (b) ethanol
Since the compound 'X' does not react with sodium, so it is neither an acid, nor an
alcohol (-OH group absent).
compound 'X' does not react with Schiff's reagent, so it is not an aldehyde.
Compound 'X' yields only one product with excess of HI, so it may be an ether and last
product of the given sequence of reactions is an acid with molecular weight 60, so
the acid is acetic acid (CH3COOH) and the ether is diethyl ether.
(because acid is a 2C compound so ether contains 4 carbons.)
Q.33 Write the equations involved in the following reactions:
a.Kolbe Reaction
14
5
b.Williamson’s synthesis :In this method, an alkyl halide reacts with sodium alkoxide
to form an ether.
c. Esterification
Alcohols and phenols react with carboxylic acids, acid chlorides and acid anhydrides
to form esters.
d. Hydroboration–oxidation
e.
Reimer-Tiemann reaction: On
treating phenol with chloroform/CCl4 in the presence of sodium hydroxide, a –CHO/ COOH group is introduced at ortho position of benzene ring. This reaction is known as
Reimer - Tiemann reaction.
Q.34 Give Reasons:
a. Phenol is more acidic
than alcohols(Ethanol).
ANS. Phenol is more
acidic than
alcohols(Ethanol).
Because Phenoxide ion is more stablised than alkoxide ion since in phenoxide ion, the
charge is delocalised.,The delocalisation of negative charge makes phenoxide ion
more stable and favours the ionisation of phenol.
b. Phenol has small dipole moment than methanol.
ANS. Phenol has small dipole moment than methanol due to electron-withdrawing
effect of the benzene ring, the C ―O bond in phenol is less polar but in case of
15
5
methanol due to electron-donating effect of CH3 group, C ―O bond is more polar.
c. Ortho – nitrophenol is more acidic than Ortho – methoxyphenol.
ANS:Ortho-nitrophenol is more acidic
because nitro group is electron withdrawing
and will increase +ve Charge on oxygen to
make it more acidic whereas –OCH3 group is
electron releasing and it will increase
electron density on oxygen thus making it
less acidic.
d. The presence of electron withdrawing groups such as nitro group enhances the
acidic strength of phenol.
ANS: The presence of electron withdrawing groups such as nitro group enhances the
acidic strength of phenol.This effect is more pronounced when such a group is
present at ortho and para positions. It is due to the effective delocalisation of
negative charge in phenoxide ion.
e. The presence of electron releasing groups such as alkyl group decreases the acidic
strength of phenol.
A. The presence of electron releasing groups such as alkyl group decreases the acidic
strength of phenol. Because electron releasing groups, such as alkyl groups, in
general, do not favour the formation of phenoxide ion resulting in decrease in acid
strength. Cresols, for example, are less acidic than phenol.
Q.35 Give Reasons:
5
a.The boiling point of alcohols and phenols are higher in comparison to hydrocarbons,
ethers, and halo alkanes of comparable molecular mass.?
ANS: The boiling point of alcohols and phenols are higher in comparison to
hydrocarbons, ethers, haloalkanes .because the –OH group in alcohols and phenols is
involved in intermolecular hydrogen bonding
b.Alcohols are comparatively more soluble in water than the corresponding
Hydrocarbons
ANS: Alcohols are comparatively more soluble in water than the corresponding
hydrocarbons because it can form intermolecular Hydrogen -bonding with water .
c. Lower alcohols are soluble in water, higher alcohols are not.
ANS. Lower alcohols are soluble in water, higher alcohols are not. as lower alcohols
are able to form stronger hydrogen bonds with water.
d. The boiling point of alcohols and phenols increase with increase in number of
carbon atoms.
ANS. The boiling point of alcohols and phenols increase with increase in number of
carbon atoms because of increase in vander Waals forces.
e. In alcohols the boiling point decreases with increase in branching.
ANS: In alcohols the boiling point decreases with increase in branching. because of
decrease in van der Waals forces with decrease in surface area
****
16
17
SAMPLE PAPER (2022-23)
CHEMISTRY THEORY
(043)
CHAPTER-Aldehydes, Ketones and Carboxylic Acids
MM:70
Time: 3 hours
General Instructions:
Read the following instructions carefully.
a) There are 35 questions in this question paper with internal choice.
b) SECTION A consists of 18 multiple-choice questions carrying 1 mark each.
c) SECTION B consists of 7 very short answer questions carrying 2 marks each.
d) SECTION C consists of 5 short answer questions carrying 3 marks each.
e) SECTION D consists of 2 case- based questions carrying 4 marks each.
f) SECTION E consists of 3 long answer questions carrying 5 marks each.
g) All questions are compulsory.
h) Use of log tables and calculators is not allowed
SECTION A
The following questions are multiple-choice questions with one correct answer. Each question
carries 1 mark. There is no internal choice in this section.
1. An aqueous NaOH solution is added to a mixture of benzaldehyde and formaldehyde to
produce:
a.
benzyl alcohol + sodium formate
b.
sodium benzoate + methanol
c.
benzyl alcohol + methanol
d.
sodium benzoate + sodium formate
2. Formic acid and acetic acid are distinguished by
a.
NaHCO3
b.
FeCl3
c.
Victor Mayer test
d.
Tollen's reagent
3. Which of the following compounds do not react with NaHSO3 ?
a.
b.
c.
d.
HCHO
C6H5COCH3
CH3COCH3
CH3CHO
4. Which compound is obtained when acetaldehydes are treated with a dilute solution of
caustic soda?
a.
b.
c.
d.
Sodium acetate
Resinous mass
Aldol
Ethyl acetate
5. Which of the following has the most acidic hydrogen?
a.
b.
c.
d.
hexane-2,4-dione
hexane-2,3-dione
hexane-2,5-dione
hexane-3-one
6. The formation of cyanohydrin from a ketone is an example of
a.
b.
c.
d.
electrophilic addition
nucleophilic addition
nucleophilic substitution
electrophilic substitution
7. Which of the reactions below can result in ketones?
a. Oxidation of primary alcohols
b. Oxidation of secondary alcohols
c. Dehydrogenation of tertiary alcohols
d. Dehydrogenation of primary alcohols
8. Which of the following compounds is formed when benzyl alcohol is oxidised with
KMnO4?
a. CO2 and H2O
b. Benzoic acid
c. Benzaldehyde
d. Benzophenone
9. At 287K, which of the following is a gas?
a. Propanal
b. Acetaldehyde
c. Formaldehyde
d. Acetone
10. The IUPAC name of CH3-CH=CH-CHO is:
a. But-2-enal
b. Ethane
c.
But-2-en
d. Buten-2-al
11. Propanone can be prepared from ethyne by
a. passing a mixture of ethyne and steam over a catalyst, magnesium at 420°C
b. passing a mixture of ethyne and ethanol over a catalyst zinc chromite
c. boiling ethyne with water in the presence of HgSO4 and H2SO4
d. treating ethyne with iodine and NaOH
12. The product of hydrolysis of ozonide of 1-butene are
a. ethanol only
b. ethanal and methanal
c. propanal and methanol
d. methanal only
13. Which of the following compounds will undergo Cannizzaro reaction?
a. CH3CHO
b. CH3COCH3
c. C6H5CHO
d. C6H5CH2CHO
14. Aldehydes that do not undergo aldol condensation are
1. Propanal
2. Trichloroethanal
3. 2-phenylethanal
4. Ethanal
5. Benzaldehyde
a.
b.
c.
d.
3 and 4 only
2 and 5 only
1, 2 and 3 only
2, 3 and 5 only
15. Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion(A) : The boiling points of aldehydes and ketones are higher than
hydrocarbons and ethers of comparable molecular masses.
Reason(R) : There is a relatively stronger molecular association in aldehydes and
ketones.
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
16. Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion(A) : Compounds containing –CHO group are easily oxidised to corresponding
carboxylic acids.
Reason(R) : Carboxylic acids can be reduced to alcohols by treatment with LiAlH4
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
17. Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion(A) : Aromatic aldehydes and formaldehyde undergo Cannizzaro reaction.
Reason(R) : Aromatic aldehydes are almost as reactive as formaldehyde.
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
18. Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion(A) : The α-hydrogen atom in carbonyl compounds is more acidic then
phenol.
Reason (R): The anion formed after the loss of α-hydrogen atom is resonance
stabilised.
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
SECTION B
This section contains 7 questions with internal choice in two questions. The following
questions are very short answer type and carry 2 marks each.
19. Floroacetic acid is a stronger acid than acetic acid.
20. Aldehydes are more reactive than Ketones towards nucleophilic additions.
OR
Carboxylic acids has higher boiling points than alcohols of same no. of carbon atoms.
21. What happens when:
(a) Propanone is treated with methyl magnesium iodide and then hydrolysed, and
(b) Benzene is treated with CH3COCl in presence of anhydrous AlCl3?
OR
What happens when:
(a) Butanone is treated with methylmagnesium bromide and then hydrolysed, and
(b) Sodium benzoate is heated with soda lime?
22. Arrange the following compounds in increasing order of their property as indicated :
(i) F – CH2COOH, O2N – CH2COOH, CH3COOH, HCOOH — acid character
(ii) Acetone, Acetaldehyde, Benzaldehyde, Acetophenone — reactivity towards
addition of HCN
23. Write structures of compounds A and B in each of the following reactions :
24. Give reasons :
(i) Benzoic acid is a stronger acid than acetic acid.
(ii) Methanal is more reactive towards nucleophilic addition reaction than ethanal.
25. Give a simple chemical test to distinguish between propanal and propanone.
SECTION C
This section contains 5 questions with internal choice in two questions. The following
questions are short answer type and carry 3 marks each.
26. Write the main product formed when propanal reacts with the following reagents:
(i) 2 moles of CH3OH in presence of dry HCl
(ii) Dilute NaOH
(iii) H2N – NH2 followed by heating with KOH in ethylene glycol.
27. Perform the following conversions in not more than two steps :
(i) Benzoic acid to benzaldehyde
(ii) Ethyl benzene to Benzoic acid
(iii) Prapanone to Propene
28. (a) Write the chemical equation for the reaction involved in Cannizzaro reaction.
(b) Draw the structure of the semicarbazone of ethanal.
(c) Why pKa of 𝑭 − 𝑪𝑯𝟐 − 𝑪𝑶𝑶𝑯 is lower than that of 𝑪𝒍−𝑪𝑯𝟐 −𝑪𝑶𝑶𝑯?
29. Give any three answer from given a, b, c, d and e.
(a) Write the chemical reaction involved in Wolf-Kishner reduction.
(b) Arrange the following in the increasing order of their reactivity towards nucleophilic
addition reaction: 𝐂𝟔𝐇𝟓𝐂𝐎𝐂𝐇𝟑, 𝐂𝐇𝟑 − 𝐂𝐇𝐎, 𝐂𝐇𝟑𝐂𝐎𝐂𝐇𝟑
(c) Why carboxylic acid does not give reactions of carbonyl group.
(d) Write the product in the following reaction
(e) A and B are two functional isomers of compound C6H6O. On heating with NaOH and
I2, isomer B forms yellow precipitate of iodoform whereas isomer A does not form
any precipitate. Write the formulae of A and B.
30. Write the products of the following reactions:
OR
Give simple tests to distinguish between the following pairs of organic compounds
(i) Pentan-2-one and Pentan-3-one
(ii) Benzaldehyde and Acetophenone
(iii) Phenol and Benzoic acid
SECTION D
The following questions are case-based questions. Each question has an internal choice
and carries 4 (1+1+2) marks each. Read the passage carefully and answer the questions
that follow.
31. Read the passage given below and answer the following questions :
When an aldehyde with no a-hydrogen reacts with concentrated aqueous NaOH,
half the aldehyde is converted to carboxylic acid salt and other half is converted to
an alcohol. In other words, half of the reactant is oxidized and other half is reduced.
This reaction is known as Cannizzaro reaction
The following questions are multiple choice questions. Choose the most appropriate
answer :
(i) A mixture of benzaldehyde and formaldehyde on heating with aqueous NaOH
solution gives
(a) benzyl alcohol and sodium formate
(b) sodium benzoate and methyl alcohol
(c) sodium benzoate and sodium formate (d) benzyl alcohol and methyl alcohol.
(ii) Which compounds will undergo Cannizzaro reaction?
(iii) Trichloroacetaldehyde is subjected to Cannizzaro's reaction by using NaOH. The
mixture of the products contains sodium trichloroacetate ion and another
compound. What will be the other compound?
OR
will Cannizzaro reaction result in the formation of carbon-carbon bonds? explain
32. The addition reaction of enol or enolate to the carbonyl functional group of aldehyde
or ketone is known as aldol addition. The β-hydroxyaldehyde or β-hydroxyketone so
obtained undergo dehydration in the second step to produce a conjugated enone.
The first part of the reaction is an addition reaction and the second part is an
elimination reaction. The Carbonyl compound having α-hydrogen undergoes an aldol
condensation reaction.
(i)The condensation reaction is the reverse of which of the following reaction?
(a) Lock and key hypothesis
(b) Oxidation
(c) Hydrolysis
(d) Glycogen formation
(ii) Which compounds would be the main product of an aldol condensation of
acetaldehyde and acetone?
(iii) Which combination of carbonyl compounds gives phenyl vinyl ketone by an aldol
condensation?
OR
Write the condition for chemical reaction will undergo aldol condensation?
SECTION E
The following questions are long answer type and carry 5 marks each. Two questions
have an internal choice.
33. (a) An organic compound ‘A’ having molecular formula C5H10O gives negative Tollens’
test, forms n-pentane on Clemmensen reduction but doesn’t give iodoform test.
Identify ‘A’ and give all the reactions involved.
(b) Carry out the following conversions :
(i) Propanoic acid to 2-Bromopropanoic acid
(ii) Benzoyl chloride to benzaldehyde
(c) How will you distinguish between benzaldehyde and acetaldehyde ?
OR
(a) Convert benzaldehyde to Cinnamaldehyde?
(b ) What is IUPAC name of-
(c) Write the structural formula of Isoamyl acetate.
34. What happens when 2 moles of acetone are condensed in presence of Ba(OH)2? Write
chemical equation
(b)What happens when acetic acid is heated with P2O5?
(c) Why is alpha () hydrogen of carbonyl compounds acidic in nature ?
(d) What happens when salicylic acid is heated with zinc dust?
(e) Why is p-hydroxy benzoic acid more acidic than p-methoxy benzoic acid?
OR
(a) Complete the following reactions:
(b) Account for the following :
(i) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction.
(ii) pKa value of 4-nitrobenzoic acid is lower than that of benzoic acid.
35. Carry out the following conversions :
(i) P-nitrotoluene to 2-bromobenzoic acid
(ii) Propanoic acid to acetic acid
(b) An alkene with molecular formula C5H10 on ozonolysis gives a mixture of two
compounds, B and C. Compound B gives positive Fehling test and also reacts with
iodine and NaOH solution. Compound C does not give Fehling solution test but forms
iodoform.
Identify the compounds A, B and C.
OR
(a) Carry out the following conversions :
(i) Benzoic acid to aniline
(ii) Ethanal to but-2-enal
(b) An organic compound contains 69.77% carbon, 11.63% hydrogen and rest oxygen.
The molecular mass of the compound is 86. It does not reduce Tollen’s reagent but
forms an addition compound with sodium hydrogen sulphite and gives positive
iodoform test. On vigorous oxidation it gives ethanoic and propanoic acids. Derive the
structure of the compound.
SAMPLE PAPER (2022-23)
CHEMISTRY THEORY (043)
MARKING SCHEME
CHAPTER -ALDEHYDE KETONE AND CARBOXYLIC ACID
SECTION A
Q1 to 18 each correct answer 1 mark
1. (a) benzyl alcohol + sodium formate
Explanation: When a mixture of formaldehyde and another aldehyde without an α-hydrogen
atom is treated with concentrated alkali, it oxidizes formaldehyde to carboxylic acid and reduces
another aldehyde to alcohol.
2. (d) Tollen's reagent
Explanation: Formic acid has an aldehyde group. It reduces Tollen's reagent to a silver mirror-like
other aldehydes. Tollen's test is not given by Carboxylic acids.
3. (b) C6H5COCH3
Explanation: Compared with aliphatic ketones, aromatic ketones are less reactive than
aldehydes. The reaction between acetophenone and NaHSO3 is therefore not initiated.
4. (c) Aldol
Explanation: Aldol is obtained by heating acetaldehyde molecules with dilute NaOH solutions.
5. (a) hexane-2,4-dione
6. (b) nucleophilic addition
7. (b)
Explanation: Ketones are formed when secondary alcohols are oxidised and dehydrogenated.
Aldehydes are produced by the same processes with primary alcohols.
8. ( b) Benzoic acid
9. (c) Formaldehyde
10. (a) But-2-enal.
11. (c) boiling ethyne with water in the presence of HgSO 4 and H2SO4
12. (c) propanal and methanal
13. (c) C6H5CHO
14. (b) 2 and 5 only
15. (a)
16. (b)
17. (c)
18. (d)
SECTION B
19. In fluoroacetic acid, Fluorine being electron withdrawing group stabilizes the conjugate base
through delocalization of the negative charge
Therefore fluoroacetic acid is a stronger acid than acetic acid.
2
20. Carboxylic acids have more extensive association of molecules through intermolecular hydrogen
bonding than alcohols. Moreover their boiling points are higher than alcohols of same carbon
atoms.
2
Ethanoic acid exists as dimer in vapour state in which two molecules remain together by
hydrogen bonding. This increases the effective molecular mass to 120.
21. (a) When Propanone is treated with methylmagnesium iodide and then hydrolysed, tertiary
butyl alcohol is formed
(b) When Benzene is treated with
is formed
in presence of anhydrous
1
, acetophenone
1
OR
(a) Butanone is treated with methylmagnesium bromide and then hydrolysed, and
(b) Sodium benzoate is heated with soda lime ?
22. (i) The order of electron withdrawing ability is-
1
1
1
−NO2>F−>−H>−CH3
Therefore, the order of acidic character isO2N−CH2COOH>F−CH2COOH>HCOOH>CH3COOH
(ii) Addition of HCN is a nucleophilic addition reaction.
The +I effect is more in ketone than in aldehyde. Thus ketone will be least reactive in nucleophilic
addition reactions. The presence of electron withdrawing group increases the reactivity towards the
addition of HCN while the presence of electron donating group decreases the reactivity of compound
towards nucleophilic addition reaction.
Benzaldehyde does not favour nucleophilic addition reaction due to resonance stabilisation.
Hence the increasing order of reactivity towards the addition of HCN will beAcetophenone < Benzaldehyde < Acetone < Acetaldehyde
23.
1
1
1
24.
(i) In benzoic acid, the -COOH group is attached to sp2- carbon (more electronegative)while in acetic
acid -COOH group is attached to sp3- carbon(less electronegative)
1
(ii) It is due to steric and electronic reasons. Presence of methyl group in
ethanal,
hinders the approach of nucleophile to carbon of
also reduces the electrophilicity of the carbonyl gp. due to
effect.
1
25. Propanal and propanone :
and it
2
Propanal gives silver mirror test with Tollen's reagent. Propanone does not gives this test.
Propanal gives red ppt with Fehlings solution but propanone does not give this test.
SECTION C
26. (i) When propanal reacts with excess of methanol in the presence of HCl, it forms 1,1-dimethoxy propane.
1
(ii) Propanal having α-hydrogen atom undergo aldol condensation in presence of dil. NaOH and forms 3-hydroxy2-methyl pentanal.
1
(iii) The carbonyl group of propanal is reduced to CH2 group on treatment with hydrazine followed by heating
with KOH in ethylene glycol and the product formed will be propane. 1
27.
1+1+1
28.
1+1+1
29. a) Wolff kishner reduction is a chemical name reaction in which aldehydes and ketones are
converted into an alkane in presence of hydrazine, base and thermal conditions. When aldehydes or
ketones are heated with hydrazine and potassium hydroxide in ethylene glycol, it results in the
formation of alkanes.
Let us see the chemical reaction:
1
(b) As we know that, aldehydes are more reactive than alcohol because aldehydes are less hindered
than ketones. Aliphatic ketones are more reactive than aromatic ketones because aromatic ketoses
have delocalisation of electrons. Therefore the reactivity order can be written as:
𝐂𝟔𝐇𝟓𝐂𝐎𝐂𝐇𝟑, < 𝐂𝐇𝟑𝐂𝐎𝐂𝐇𝟑, <𝐂𝐇𝟑 – 𝐂𝐇𝐎
1
(c)Carboxylic acids do not give reactions to carbonyl groups because the lone pair of electrons on oxygen
atoms of the −OH−OH group of carboxylic acid is less electrophilic than the carbonyl group of ketones
and aldehydes.
1
(d) Let us write the product of the given reaction:
1
(e) B is a methyl ketone as it forms yellow precipitate of iodoform, that means it undergoes a haloform
reaction. hence, B is acetone CH3COCH3 . The isomer A does not form any precipitate, therefore it will
not have any methyl ketone. Hence isomer A is propanal CH3CH2CHO
1
30.
1+1+1
OR
i. Pentan-2-one and Pentan -3 -one Pentan – 2-one contains CH3CO group and as such it will give iodoform test with NaOI [NaOH + I2] while no such group is there in pentan -3- one and so it will not give
iodoform test.
1
ii. Benzaldehyde and Acetophenone Benzaldehyde is an aromatic aldehyde while acetophenone is a
methyl ketone. These may be distinguished as follows.
1
(a) Iodoform test: Acetophenone, due to the presence of CH3CO group, will give iodoform test with NaOI
(NaOH +I2), while benzaldehyde will not give this test.
(b) Silver mirror test: Benzaldehyde, being an aromatic aldehyde, will reduce tollens’ reagent to silver.
On other hand acetophenone, a ketone will not give this test.
iii. Ethanal and Propanal: As ethanal contains CH3CO group it will give iodoform test with NaOH and I2.
On the other hand propanal will not give this test.
1
SECTION D
31.
Ans (i) (a): It is an example of cross Cannizzaro reaction where aromatic aldehyde gets reduced
to alcohol and aliphatic aldehyde gets oxidised to its sodium salt (both aldehydes must not
contain any α-hydrogen).
1
(ii) (c) C6H5CHO
(iii) (a): The Cannizzaro product of given reaction yields 2, 2, 2-trichloroethanol.
OR
1
2
No , as it is a bimolecular redox reaction.
32.
(i)Answer:(c) Hydrolysis
1
(ii) Answer:(b)
1
(iii) Acetophenone and Formaldehyde
2
OR
Answer- The Carbonyl compound having α-hydrogen undergoes an aldol condensation reaction.
SECTION E
33.
2+2+1
(a) Since, the compound (A) gives a positive iodoform test but negative Tollen's test, so it must be a
methyl ketone. From the given molecular formula C5H10O(A) can be 2-pentanone
(b)
(i)
(ii)
(c) Distinction between acetaldehyde and benzaldehyde : - Acetaldehyde and benzaldehyde can be
distinguish by Fehling solution. Acetaldehyde gives red coloured precipitate with Fehling solution
while benzaldehyde does not.
OR
(a)
(b)
4–hydroxy–3–methoxy-benzaldehyde
(c)
34.
(a)
2+2+1
(b)
(c) In a carbonyl group, the oxygen is extremely electronegative and it attracts the electron cloud
towards itself developing a partial positive charge on the α-carbon. To reduce the positive charge, αcarbon looses its hydrogen readily and makes it acidic in nature.
OR
(d)
(e)
It is because —OH has more —I effect than —OCH3 group.
OR
3+2
(b)
(i) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction as the carboxyl (−COOH) group is
deactivating and Lewis acid catalyst AlCl3 gets bonded to the carboxyl group. The carboxyl (−COOH)
group withdraws electron density from benzene ring through inductive and resonance effects.
(ii) pKa value of 4-nitrobenzoic acid (3.41) is lower than that of benzoic acid (4.19). Lower is the pKa
value, greater is the acid strength. The electron withdrawing nitro (−NO2) group increases the acidity of
benzoic acid.
35.
(i)
(ii)
(b)
(A)
B= Positive Fenling - aldehyde group
= Positive Iodoform test
C - negative testing
= positive Iodoform
2+3
=
OR
(a)
2+3
(i)
Step-1:—Conversion of benzoic acid to benzamde.
Benzoic acid is treated with thionyl chloride to form benzoyl chloride,which on further treatment with
ammonia gives benzamide.
Step-2:—Conversion of benzamide to aniline.
Benzamide is subjected to Hoffmann bromamde degradation by reacting with Br₂ and NaOH to form
aniline.
(ii)
KENDRIYA VIDYALAYA SANGATHAN RO AHMEDABAD
SUBJECT – CHEMISTRY THEORY (043) 2022-23
CHAPTER- AMINES
MM: 70
TIME : 3 HOURS
General instructions:
Read the following instructions carefully.
a)
b)
c)
d)
e)
f)
g)
h)
There are 35 questions in this question paper with internal choice.
Section A consist of 18 multiple-choice questions carrying 1 mark each.
Section B consists of 7 very short answer questions carrying 2 marks each.
Section C consists of 5 short answer questions carrying 3 marks each.
Section D consists of 2 case based questions carrying 4 marks each.
Section E consist of 3 long answer questions carrying 5 marks each.
All questions are compulsory.
Use of log tables and calculators is not allowed.
Section- A
The following questions are multiple choice questions with one correct answer each question
carries one mark there is no internal choice in this section.
1. When a haloalkane is heated with KCN in aqueous ethanoic solution, the product formed isa) Alkyl nitrite
b) nitro alkane
c) alkane nitrile
d) carbylamine
2. Amine that cannot be prepared by Gabriel phthalimide synthesis isa) Aniline
b) benzyl amine
c) tertiary-butylamine
d) isobutylamine
3. Which of the following amine has highest boiling point?
a) Pentaamine
b) 2-methylbutanamine
c) 2,2-dimethylpropamine
d) 3-methylbutamine
4. The amine that reacts with NaNO2 and HCl to give yellow oily liquid isa) Ethylamine.
b) Diethylamine
c) Isopropyl amine
d) Secondary butylamine
5. Sandmeyer reaction of diazonium salt is a replacement of N2 bya) Halogen Cl or Br
b) Hydroxyl group
c) Coupling
d) Hydrogen
6. One of the following amines will not undergo Hoffmann bromamide reactiona) CH3CONHCH3
b) CH3CH2CONH2
c) CH3CONH2
d) C6H5CONH2
7. The correct order of the basic strength of methyl substituted amines in aqueous solution is
a) CH3NH2 > (CH3)2NH > (CH3)3N
b) (CH3)2NH > CH3NH2 > (CH3)3N
c) (CH3)3N >CH3NH2 > (CH3)2NH
d) (CH3)3N > (CH3)2NH > CH3NH2
8. The correct statement regarding the basicity of arylamine is
a) arylamines are generally more basic than alkylamines because of aryl group
b) arylamines are generally more basic than alkylamines because the nitrogen atom in
aryalamines is sp- hybridised
c) arylamines are generally less basic than alkylamines because the nitrogen lone- pair
electrons are delocalised by intraction with the aromatic ring π- electron system
d) arylamines are generally more basic than alkylamines because the nitrogen loan-pair
electrons are not delocalised by interaction with the aromatic ring π-electron system.
9. What reagent is used in Hinsberg test of amines?
a) (CH3CO)2O and pyridine
b) C6H5SO2Cl followed by aq. NaOH
c) NaNO2 in aq. H2SO4
d) CH3I(excess) followed by AgOH
10. Aniline reacts with sodium nitrite and hydrochloric acid at 273 -278 K to give:
a) Benzene
b) Benzene diazonium chloride
c) Chlorobenzene
d) o-chloroaniline
11. Propane amide on reaction with bromine in aqueous NaOH gives:
a) Propanamine
b) ethanamine
c) N-Methyl ethanamine
d) Propanenitrile
12. The gas evolved when methylamine reacts with nitrous acid is:
a) NH3
b) N2
c) H2
d) C2H6
13. Which of the following is not correct regarding aniline?
a) It is less basic than ethyl amine
b) It can be steam distillated
c) It reacts with sodium to give hydrogen
d) It is soluble in water
14. Acetamide and ethylamine can be distinguished by reacting with:
a) Aq. HCl and heat
b) Aq. NaOH and heat
c) Acidified KMnO4
d) Bromine water
15. Given below are two statements labelled assertion (A) and reason(R).
Assertion(A)- Hinsberg's reagent can be used to distinguish primary amines from secondary
amines.
Reason(R)- Benzenesulphonyl chloride is called Hinsberg's reagent.
Select the most appropriate answer from the option given below :
a. Both A and R are true and R is the correct explanation of A.
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
16. Given below are two statements labelled assertion (A) and reason(R).
Assertion(A)- Solubility of amines in water decreases with increase in molar mass.
Reason(R)- Intermolecular H bonds formed by the higher amines are weaker.
Select the most appropriate answer from the option given below :
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
17. Given below are two statements labelled assertion (A) and reason(R).
Assertion(A)- Amines behave as a Lewis base.
Reason(R)- Amines have an unshared pair of electrons on nitrogen atom.
Select the most appropriate answer from the option given below :
a. Both A and R are true and R is the correct explanation of A.
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
18. Given below are two statements labelled assertion (A) and reason(R).
Assertion(A)- Acylation of amines gives a mono substituted product whereas alkylation of
amines gives poly substituted product.
Reason(R)- Acyl group sterically hinders the approach of further acyl groups.
Select the most appropriate answer from the option given below :
a. Both A and R are true and R is the correct explanation of A.
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
Section - B
This section contains 7 questions with internal choice in two questions. The following questions
are very short answer type and carry 2 marks each.
19. How will you distinguish between the following pairs of compounds:
i) Aniline and ethanamine.
ii) Aniline and N-methylaniline
20. Give reasons:
i) Aniline does not undergo friedal Craft reaction.
ii) (CH3)2NH is more basic than (CH3)3N in an aqueous solution.
OR
i) Although -NH2 group is o/p directing in electrophilic substitution reactions, yet aniline, on
nitration gives good yield of m-nitroaniline.
ii) pKb of methylamine is less than that of aniline.
21. Write chemical equations for the following conversion:
i) nitrobenzene to benzenoic acid
ii) aniline to benzyl alcohol
OR
i) Write the reaction involved in the following: Diazotisation
ii) Give reason:
Aromatic diazonium salts are more stable than aliphatic diazonium salts.
22. Illustrate the following with an example of reaction in each casei) Sandmeyer reaction
ii) coupling reaction
23. Complete the following reactions:
i) CH3CH2NH2 + CHCl3+alc.KOH------->
ii) CH3NH2+ C6H5COCl---------->
24. Ammonolysis of alkyl halides is not a good method to prepare pure primary amines. Explain.
25. State and illustrate the following:
Gabriel phthalimide synthesis.
Section- C
Section contains 5 questions with internal choice in two questions the following questions are
sort answer type and carry 3 marks each.
26. a) Write the chemical reaction of methylamine with benzyl chloride and write the
IUPAC name of the product obtained.
b) Arrange the following in the increasing order of their pkb values:
C6H5NH2 , NH3 , C2H5NH2 ,(C2H5)2 NH
27. Write the structures of main products when aniline reacts with the following reagents:
i) Br2 water.
ii) HCl
iii) (CH3CO)2O/Pyridine
28. Write the structures of A, B and C in the following:
KI/∆
i) C6H5CONH2 ---------- ---------→A ------ ---------------→ B ------ ----→ C
NaNO2+ConcHCl/0-5 0C
Br2/aq. KOH
KCN
LiAlH4
CHCl3+alcKOH
II) CH3Cl ---- ----------→ A ---- -----------→ B -------------------- ---→ C
29. Give reasons for any three of the following observations:
a. Aniline is acetylated before nitration reaction.
b. pKb of aniline is smaller than the m-nitro aniline.
c. Primary amine on treatment with benzene sulfonyl chloride forms a product which is
soluble in NaOH, however secondary amine gives product which is insoluble in NaOH.
d. Aniline does not react with methyl chloride in the presence of anhydrous AlCl3 catalyst.
30. Write the structures of main products when benzene diazonium chloride reacts with
the following reagents.
a) CuCN
b) CH3CH2OH
c) KI
Section- D
The following questions are case based questions. Each question has an internal choice and
carries 4(1+1+2) marks each. Read the passage carefully and answer the questions that follow.
31. Benzene ring in aniline is highly activated. This is due to the sharing of lone pair of nitrogen
with the ring which results in increase in the electron density on the ring and hence facilitates
the electrophilic attack. The substitution mainly takes place at Ortho and para positions
because electron density is more at Ortho and para positions. On reaction with aqueous
bromine all the ortho and para position get substituted resulting in the formation of 2,4,6tribromoaniline. To get a monobromo compound, the amino group is acetylated before
bromination. After bromination, the bromoacetanilide is acid hydrolysed to give the desired
halogenated amine.
a) Why does -NH2 group facilitates electrophilic attack in aniline?
b) Explain why: in aniline, the substitution mainly takes place at Ortho and Para positions.
c) Show the conversion of benzene into aniline.
32. In amines the N atoms of the amino group are SP3 hybridised. Amines readily lose its
electron pair and act as a base. In aliphatic amines ,the alkyl group tends to increase the
electron density on the nitrogen atom due to the +I effect. In aromatic amines, the basic
strengths are influenced by the presence of an electron withdrawing group at the O-, m- and pposition in the ring.
a) Lower member of aliphatic amines are soluble in water . Why?
b) Out of (CH3)3N and and (CH3)2NH which one is more basic in aqueous solution?
c) Which isomer of C3H9N which does not react with Hinsberg’s reagent.
Section- E
The following questions are long answer type and carry five marks each two questions have
internal choice.
33. An organic compound A with molecular formula C7 H7NO reacts with Br2/aq. KOH to give
compound 'B', which upon reaction with NaNO2 and HCl at 00 Celsius gives 'C'. Compound 'C'
on heating with CH3CH2OH gives hydrocarbon 'D'. compound 'B' on further reaction with Br2
water gives white precipitate of compound 'E'. Identify the compound A,B,C,D and 'E'.
Also justify your answer by giving relevant chemical equations.
OR
An aromatic compound 'A' on treatment with aqueous ammonia and heating forms compound
'B' which on heating with Br2 and KOH forms a compound 'C' of molecular formula C6H7N. write
the structure and IUPAC names of the compound A, B and C . Justify your answer by giving
relevant chemical equations.
34. i) Write the structure of main products when benzenediazonium chloride (C6H5N2+Cl-) reacts
with the following reagents:
a) KI
b) C2H5OH
ii) write the structures of A B and C in the following reactions:
LiAlH4
HNO2 /Low temp
CH3Cl+ KCN (alc)----------------->A-- --------------------->B------------------------>C
OR
i) Write the reactions involved in the following:
a) Hoffmann’s Bromamide degradation reaction
b) Carbylamine reaction
35. i) Convert the following:
a) Phenol to N-phenylethanamide
b) Chloroethane to methanamine
c) Propanenitrile to ethanal.
ii) What happens whena) N-ethylethaneamine reacts with benzenesulphonyl chloride.
b) Aniline reacts with chloroform in the presence of alcoholic potassium hydroxide.
********************************
KENDRIYA VIDYALAYA SANGATHAN RO AHMEDABAD
SUBJECT – CHEMISTRY THEORY (043)(2022-23)
MM: 70
TIME : 3 HOURS
MARKING SCHEME
Section- A
1. c) alkyl nitrile
2. a) Aniline
3. a) Pentaamine
4. b) Diethylamine
5. a) Halogen Cl or Br
6. a) CH3CONHCH3
7. b) (CH3)2NH > CH3NH2 > (CH3)3N
8. c) arylamines are generally less basic than alkylamines because the nitrogen lone- pair
electrons are delocalised by intraction with the aromatic ring π- electron system
9. b) C6H5SO2Cl followed by NaOH
10. b) Benzene diazonium chloride
11. b) ethanamine
12. b) N2
13. d) It is soluble in water
14. b) Aq. NaOH and heat
15. b ) Both A and R are true but R is not the correct explanation of A.
16. c) A is true but R is false.
17. a) Both A and R are true and R is the correct explanation of A.
18. c) A is true but R is false.
Section - B
This section contains 7 questions with internal choice in two questions. The following questions
are very short answer type and carry 2 marks each.
19. i) aniline being an aromatic primary amine on treatment with nitrous acid at 273 Kelvin
followed by treatment with an alkaline solution of beta naphthol gives an orange coloured azo
dye. Ethylamine does not give these test.
(1)
ii) aniline gives carbylamine test , on treatment with alcoholic KOH and chloroform followed by
heating it gives offensive order of phenylisocyanide but N- methylaniline being secondary
amine, does not show this test.
(1)
20. i) in Friedel crafts reaction, AlCl3 is added as a catalyst which is a Lewis acid. it forms a salt
with aniline due to which the nitrogen of an aniline acquires positive charge this positively
charged nitrogen acts as a strong deactivating group, hence aniline does not undergo Friedel
Crafts reaction.
(1)
ii) In aqueous solution, two degree amine is more basic than three degree amine due to the
combination of inductive effect, solvation effect and steric hinderance.
(1)
OR
i) In the presence of acids the -NH2 group of aniline gets protonated and is converted into NH3 + group. This positively charged group acts as a strong electron withdrawing and meta
directing group.
(1)
ii) in aniline the lone pair of electrons of N- atom are the localized over the benzene ring as a
result electron density on the nitrogen decreases. In contrast in CH3NH2 , + I effect of -CH3
group increase the electron density on the N- atom there for aniline is a weakrbase than methyl
amine and hence, its Pkb value is higher than that of methylamine.
(1)
21. i)
(
ii)
(1+1)
OR
i) Benzenediazonium chloride is prepared by the reaction of aniline with nitrous acid at 273278K. Nitrous acid is produced in the reaction mixture by the reaction of sodium nitrite with
hydrochloric acid. The conversion of primary aromatic amines into diazonium salts is known
as diazotisation.
(1)
ii) Aromatic diazonium salts carry a nitrogen atom with a positive charge which is dispersed
through resonance.
(1)
22. i) The Cl– , Br– and CN– nucleophiles can easily be introduced in the benzene ring in the
presence of Cu(I) ion. This reaction is called Sandmeyer reaction.
(1)
ii) Benzene diazonium chloride reacts with phenol in which the phenol molecule at its para
position is coupled with the diazonium salt to form p-hydroxyazobenzene. This type of reaction
is known as coupling reaction. Similarly the reaction of diazonium salt with aniline yields paminoazobenzene. This is an example of electrophilic substitution reaction.
(1)
23. i)
(1)
ii)
(1)
24. Ammonolysis of alkyl halides with ammonia is a nucleophilic substitution reaction in which
ammonia acts as a nucleophile by donating the electron pair on nitrogen atom to form primary
amine as the initial product. Now, the primary amine can act as a nucleophile and combine with
alkyl halide to give secondary amine and the reaction continuous in the same way to form
tertiary amine and finally quarterly ammonium salt. Thus a mixture of products is formed and it
is not possible to separate individual amines from the mixture.
(2)
25. In this reaction pthalamide is converted into its potassium salt by treating it with alcoholic
potassium hydroxide. Then potassium thalamide is heated with an alkyl halide to yeild an Nalkylpthalimide which is hydrolysed to phthalic acid and primary amine by alkaline hydrolysis.
(1+1)
Section- C
26. i) CH3NH2+ C6H5COCl -----→ CH3NHCOC6H5
( N- methylbenzamide)
ii) stronger the base lower will be its pkb value.
(C2H5)2 NH < C2H5NH2 < NH3 < C6H5NH2
27. i)
(1+1)
(1)
(1+1+1)
ii) C6H5NH2+ HCl -----→ C6H5N+H3Cliii)
28. i)
(1/2+1/2+1/2+1/2+1/2+1/2)
ii)
29. a) Aniline is acetylated, before nitration reaction in order to avoid formation of tarry
oxidation products and protecting the amino group, so that p -nitro derivative can be obtained
as major product.
b) pKb of aniline is lower than the m-nitro aniline.The basic strength of aniline is more that
m-nitroaniline . pkb value is inversely proportional to basic strength. Presence of Electron
withdrawing group decrease basic strength.
C ) Due to the presence of acidic hydrogen in the N-alkylbenzenesulphonamide formed by the
treatment of primary amines.
d) Aniline does not react with methyl chloride in the presence of AlCl3 catalyst , because
aniline is a base and AlCl3 is Lewis acid which lead to formation of salt. (1+1+1)
30. a) C6H5CN
b) C6H6
c) C6H5I
(1+1+1)
Section- D
31.
a) this is due to the sharing of lone pair of nitrogen with the ring.
(1)
b)electron density is more at Ortho and para position then meta position. (1)
c)on nitration Benzene gives nitro benzene and further H2/Pd in ethanol gives Aniline. (2)
32. a)due to intermolecular H bonding.
b)secondary amines due to steric factor.
c)trimethylamine
(1)
(2)
(1)
Section- E
33.
(1+1+1+1+1)
OR
Ans- formula of the compound 'C' indicates it is an amine. since it is obtained by the reaction of
Br2 and KOH with the compound 'B' so compound 'B' can be an amide. As B is obtained from
compound A by reaction with ammonia followed by heating so, compound A could be an
aromatic acid. formula of compound C show that it to be aniline, then B' is benzamide and
compound A is benzoic acid. the sequence of reactions can be written as follows:
34. i) a)
a)
(1)
(1)
ii) A= CH3CN ,
B= CH3CH2NH2
C = CH3CH2OH
(1+1+1)
OR
a) when an amide is treated with bromine in aqueous or ethanolic solution of sodium
hydroxide, a primary amine with one carbon atom less than the origin amide is produced.
RCONH2+Br2+4NaOH------>RNH2+Na2CO3+2NaBr+2H2O
(2)
b) it is used for detection of primary amines.
(2)
RNH2+CHCl3+3KOH----->RNC+3KCl+3H2O
ii)the acetyl group being electron withdrawing attracts the loan pair of electrons of the N-atom
towards carbonyl group as a result the activation effect of NH2 group is reduced.
(1)
35.
Ans:- i) a)
(1+1+1)
b)
c)
ii) a) when N-ethylethanamine reacts with benzenesulphonyl chloride, N,Ndiethylbenzenesulphonamide is formed.
(1)
b) when aniline reacts with chloroform in the presence of alcoholic potassium hydroxide
phenylisocyanide or phenyl isonitrile is formed.
(1)
********************************
KENDRIYA VIDYALAYA SANGATHAN AHMEDABAD REGION
MODEL QUESTION PAPER -2022-23
Time: 3 hrs.
Class: 12
Marks: 70
Subject: CHEMISTRY (BIOMOLECULE)
_________________________________________________________________________
General Instructions:
Read the following instructions carefully.
a. There are 35 questions in this question paper with internal choice.
b. SECTION A consists of 18 multiple-choice questions carrying 1 mark each.
c. SECTION B consists of 7 very short answer questions carrying 2 marks each.
d. SECTION C consists of 5 short answer questions carrying 3 marks each.
e. SECTION D consists of 2 case-based questions carrying 4 marks each.
f. SECTION E consists of 3 long answer questions carrying 5 marks each.
SECTION-A
1
Which of the following carbohydrates does not satisfy the formula Cx(H2O)y?
a) Fructose
b) Glucose
c) Deoxyribose
d) Lactose
1
2
Which of the following is not a polysaccharide?
a) Cellulose
b) Stachyose
c) Starch
d) Glycogen
Sucrose consists of which of the following monosaccharide units?
a) Fructose, galactose
b) Fructose, glucose
c) Galactose, glucose
d) Glucose, glucose
Nucleotides are joined together by
a) peptide linkage b) disulphide linkage c) glycosidic linkage d) phosphodiester linkage
Which of the following statement is not true for glucose?
a) The pentaacetate of glucose does not react with hydroxylamine to give oxime
b) Glucose gives Schiff's test for aldehyde
c) Glucose exists in two crystalline α and β forms
d) Glucose reacts with hydroxylamine to form oxime
Two monosaccharides are joined through a ______ bond to form a disaccharide.
a) ionic
b) peptide
c) glycosidic
d) phosphodiester
Which of the following is incorrect regarding maltose?
a) It consists of two glucopyranose units
b) It is a non-reducing sugar
c) Glycosidic bond between C1 of one unit and C4 of the other unit
d) It is a disaccharide
1
3
4
5
6
7
Page 1 of 5
1
1
1
1
1
8
9
10
11
12
13
14
Proteins are polymers of ______
a) α-amino acids
b) β-amino acids
c) γ-amino acids
d) δ-amino acids
Which of the following amino acids is optically inactive?
a) Glycine
b) Alanine
c) Lysine
d) Valine
Enzymes are regarded as ______
a) biocatalysts
b) messengers
c) inhibitors
d) antibodies
Which of the following vitamin is soluble in water?
a) A
b) C
c) D
d) E
The following compound is a component of which of the following?
a) RNA
b) DNA
c) Adenine
d) Guanine
In nucleic acids, the sequence is:
a) Base - sugar - phosphate
b) Base - phosphate - sugar
c) Phosphate - sugar - base
d) Sugar - base – phosphate
Identify the saccharide from the Haworth projection shown below.
a) Lactose
b) Maltose
c) Sucrose
d) Trehalose
Page 2 of 5
1
1
1
1
1
1
1
15
16
17
18
Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion :- Uracil is present in DNA
Reason :- DNA undergoes replication
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion :- Insulin is globular protein.
Reason :- Globular Proteins are water soluble.
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion :- D (+) – Glucose is dextrorotatory in nature.
Reason :- D’ represents its dextrorotatory nature.
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
Given below are two statements labelled as Assertion (A) and Reason (R)
Assertion :- In presence of enzyme, substrate molecule can be attacked by the reagent
1
1
1
1
effectively
19
20
21
22
23
24
25
26
27
Reason :- Active sites of enzymes hold the substrate molecule in a suitable position.
Select the most appropriate answer from the options given below:
a. Both A and R are true and R is the correct explanation of A
b. Both A and R are true but R is not the correct explanation of A.
c. A is true but R is false.
d. A is false but R is true.
SECTION-B
What happens when D-glucose is treated with. the following reagents?
(i)HI
(ii) Bromine water
What are essential and non-essential amino acids? Give one examples of each type?
How are vitamins classified? Give example of each type?
OR
Name the disease caused due to deficiency of Vitamin E and Ascorbic acid?
What are the hydrolysis products of (i) Maltose and (ii)lactose
OR
What is Zwitter ion? How is it formed?
What is glycogen? How is it different from starch?
Write the important two structural differences between DNA and RNA?
Differentiate between α- helical and β- pleated sheet structure of secondary protein?
SECTION-C
Enumerate the reactions of D-glucose which cannot be explained by its open chain structure?
(a)What is the biological effect of denaturation of proteins?
(b) What are enzymes?
Page 3 of 5
2
2
2
2
2
2
2
3
3
28
Differentiate between following
(i) Globular protein and Fibrous protein
(ii) Nucleotide and Nucleoside
3
29
What information do we get when D-glucose reacts with following reagent, write the
reaction also
(a)Hydroxylamine
(iii)acetic anhydride
(iii) nitric acid
3
30
(a)What are reducing and non-reducing sugar? Give example of each?
3
(b)What are anomers?
OR
(a)Explain the formation of peptide linkage in protein?
(b)What are the two types of protein on the basis of molecular shape?
SECTION-D
Read the text carefully and answer the questions:
4
When a protein in its native form, is subjected to physical changes like change in temperature
or chemical changes like change in pH, the hydrogen bonds are disturbed. Due to this, globules
unfold and helix gets uncoiled and protein loses its biological activity. This is called the
denaturation of protein. The denaturation causes change in secondary and tertiary structures
but primary structures remain intact. Examples of denaturation of protein are coagulation of
egg white on boiling, curdling of milk, formation of cheese when an acid is added to milk.
31
32
33
(a) Phospholipids form a thin layer on the surface of an aqueous medium. Give reason.
(b) Which structure of proteins remains intact during the denaturation process?
(c ) What type of structure is α -helix and β-pleated structures of proteins?
OR
Why do amino acids have high molar masses comparable to ionic solids?
Read the text carefully and answer the questions:
4
The activity of an enzyme can be affected by a change in the conditions which can alter the
tertiary structure of the protein. These include temperature, pH, and change in substrate
concentration or binding of specific chemicals that regulate its activity.Enzymes generally
function in a narrow range of temperature and pH. Each enzyme shows its highest activity at a
particular temperature and pH called the optimum temperature and optimum pH. Activity
declines both below and above the optimum value. Low temperature preserves the enzyme in
a temporarily inactive state whereas high temperature destroys enzymatic activity because
proteins are denatured by heat. Concentration of Substrate With the increase in substrate
concentration, the velocity of the enzymatic reaction rises at first. The reaction ultimately
reaches a maximum velocity (Vmax) which is not exceeded by any further rise in concentration
of the substrate. This is because the enzyme molecules are fewer than the substrate molecules
and after saturation of these molecules, there are no free enzyme molecules to bind with the
additional substrate molecules.
(a) Name a chemical compound or molecule which is responsible for decrease or stop the
enzyme activity by binding to an enzyme?
(b) Give reason – why most of the enzymes get destroyed above optimum temperature?
(c) Explain the relation between substrate concentration and enzymatic activity?
OR
What are co-enzyme and co-factor ?
SECTION E
(a)Write the name and molecular formula of nitrogenous bases present in DNA and RNA?
5
(b) What are different types of RNA found in the cell?
Page 4 of 5
34
35
(c ) write two important functions of nucleic acid?
(a) Write functional difference between DNA and RNA?
(b) What is the basic structural difference between starch and cellulose?
OR
(a) Which form of glucose is obtained by crystallization from hot and saturated aqueous
solution at 371K?
(b) What is chemical formula of glyceraldehyde?
( c) What are oligosaccharides?
(d) The molecular formula of a compound is C2(H2O)2? Is it a carbohydrate, explain your
answer?
(a) Name the various sugars present in RNA & DNA?
(b)Explain Cyclic structure of Fructose?
OR
(a)Explain the cyclic structure of Glucose?
(b)How Nucleotides are joined together in DNA molecule?
Page 5 of 5
5
5
MARKING SCHEME FOR CHAPTER -BIOMOLECULE
( MODEL QUESTION PAPER).
SECTION-A
MARKS
1
c) Deoxyribose
1
2
b) Stachyose
1
3
b) Fructose, glucose
1
4
d) phosphodiester linkage
1
5
b) Glucose gives Schiff's test for aldehyde
1
6
c) glycosidic
1
7
b) It is a non-reducing sugar
1
8
a) α-amino acids
1
9
a) Glycine
1
10 a) biocatalysts
1
11 b) C
1
12 b) DNA
1
13 c) Phosphate - sugar - base
1
14 a) Lactose
1
15 d. A is false but R is true.
1
16 a. Both A and R are true and R is the correct explanation of A
1
17 c. A is true but R is false.
1
18 a. Both A and R are true and R is the correct explanation of A
1
19 (i)Forms n hexane
(ii) oxidized to six carbon carboxylic acid (gluconic acid)
20 Essential amino acids cannot be synthesized in the body. For example: valine and
leucine
Non-essential amino acids can be synthesized in the body. For example: glycine, and
alanine
21 (i) Fat soluble vitamins – which are soluble in fats and oils. e.g., vitamins A, D, E & K.
(ii) Water soluble vitamins – which are soluble in water e.g., vitamins B& C.
OR
Due to vitamin E-Increased fragility of RBCs and muscular
Due to deficiency of Vitamin C(Ascorbic acid)- Scurvy (bleeding gums)
1+1
Page 1 of 6
½*4
1+1
weakness
22 (i)α-D-Glucose and α-D-Glucose
1+1
(ii)β-D-Galactose and β-D-Glucose
OR
Amino acids contain amino and carboxylic acid group, In aqueous solution, the
carboxyl group can lose a proton and amino group can accept a proton, giving rise to a or
dipolar ion known as zwitter ion. This is neutral but contains both positive and negative
charges.
1
1
23 Glycogen is a carbohydrate (polysaccharide). In animals, carbohydrates are stored as
Glycogen.
Starch is a carbohydrate consisting of two components - amylose (15 - 20%) and
amylopectin (80 - 85%).
However, glycogen consists of only one component whose structure is similar to
amylopectin. Also, glycogen is more branched than amylopectin.
24
DNA
sugar moiety in DNA molecules is
β-D-2 deoxyribose
DNA contains thymine (T). It does not
contain uracil (U).
The helical structure of DNA is double stranded.
Any two correct point
α- helical structure
25
Intramolecular hydrogen bonding forms
within the polypeptide chain to create a
spiral structure.
Amino acids exist in the right-handed
coiled rod-like structure.
½
½
½
½
RNA
The sugar moiety in RNA molecules is
β-D-ribose
RNA contains uracil (U). It does not
contain thymine (T).
The helical structure of RNA is
singlestranded
1+1
Β-pleated structure
Beta sheets are formed by linking two or
more beta strands by intermolecular
hydrogen bonds.
Amino acids exist in an almost entirely
extended conformation, i.e. linear or
sheet-like structure
1+1(
any two
points
Aldehydes give 2, 4-DNP test, Schiff's test, and react with NaHSO4 to form the
26 1.
1
hydrogen sulphite addition product. However, glucose does not undergo these
reactions.
2.
The pentaacetate of glucose does not react with hydroxylamine. This indicates 1
that a free -CHO group is absent from glucose.
3.
Glucose exists in two crystalline forms -∝ and β. The α-form (m.p. = 419 K)
crystallises from a concentrated solution of glucose at 303 K and the β form (m.p = 423 1
Page 2 of 6
K) crystallises from a hot and saturated aqueous solution at 371 K. This behavior cannot
be explained by the open chain structure of glucose.
27 (a)On denaturation, protein globules unfold and unhelix gets uncoiled and protein looses its
biological activity
(b) Enzymes are biocatalyst which are very specific for a particular reaction and a
particular substrate. Almost all enzymes are globular proteins.
Fibrous protein
Globular protein
28
It is a fibre-like structure formed by
polypeptide chain in this protein is
the polypeptide chain
folded around itself, giving rise to a
spherical structure
insoluble in water.
usually soluble in water
Fibrous proteins are usually used for
All enzymes are globular proteins
structural purposes.
1.5+1.5
2+1
Nuceotide
Nucleoside
A nucleotide consists of a nitrogenous
A nucleoside consists of a nitrogenous
base, a sugar (ribose or deoxyribose) and base covalently attached to a sugar
one to three phosphate groups
(ribose or deoxyribose) but without the
phosphate group
29 (a)Confirm the presence of carbonyl group
1+1+1
(b) confirms the presence of five –OH groups
(c) On oxidation with nitric acid, glucose as well as gluconic acid both yield a
dicarboxylic acid, saccharic acid.
30 (a)All those carbohydrates which contain a free aldehyde or Ketonic group and reduce 2
Fehling's solution and Tollen's reagent are referred as a reducing Sugar. Examples of
reducing sugar is galactose, glucose, glyceraldehyde, fructose, ribose.
Disaccharides -lactose and maltose are reducing sugar.
Non-reducing sugars: They do not reduce Fehling's solution and Tollens' reagent.exSucrose.
(b) Anomers are diastereoisomers of cyclic forms of sugars or similar molecules differing 1
in the configuration at the anomeric carbon (C-1 atom of an aldose or the C-2 atom of a
2-ketose). The cyclic forms of carbohydrates can exist in two forms, α- and β- based on
the position of the substituent at the anomeric center.
OR
(a) peptide linkage is an amide formed between –COOH group and –NH2 ,group. The
reaction between two molecules of similar or different amino acids, proceeds through
Page 3 of 6
the combination of the amino group of one molecule with the carboxyl group of the
other. This results in the elimination of a water molecule and formation of a peptide 2
bond –CO–NH–.
(b)Globular and fibrous protein
1
31 (a) Phospholipids form a thin layer on the surface of an aqueous medium due to the 2
simultaneous presence of both polar and non-polar groups in the molecule. As a result,
the phospholipid molecules may arrange themselves in a double-layered membrane in
aqueous media.
(b). Primary structure remains intact during the denaturation process.
(c) secondary structure
1
OR
1
As they exists as Zwitter ion and have ionic species at lattice points which have
electrostatic attraction between them.
32
1
(a) Inhibitor
(b) Enzymes are composed of one or several polypeptide chains. Almost all enzymes
are protein. High temperature condition destroys enzymatic activity because 1
proteins are denatured by heat.
(c) Concentration of Substrate With the increase in substrate concentration, the
velocity of the enzymatic reaction rises at first. The reaction ultimately reaches a
maximum velocity (Vmax) which is not exceeded by any further rise in 2
concentration of the substrate. This is because the enzyme molecules are fewer
than the substrate molecules and after saturation of these molecules, there are
no free enzyme molecules to bind with the additional substrate molecules.
OR
Cofactors are non-protein chemical compounds which are termed helper
molecules .Cofactors are of 2 types, one with inorganic and other organic, where
the organic ones are called as coenzymes and they are of 2 types, based on bond,
if bond is permanent, it is prosthetic group, if the bond is temporary, it is 2
substrate.
Coenzymes are small organic molecules that bind to the enzymes, assisting the
function of the enzyme.
33
(a) DNA contains four bases viz. adenine (A), guanine (G), cytosine (C) and thymine 2
(T). RNA also contains four bases, the first three bases are same as in DNA but
the fourth one is uracil (U).
Page 4 of 6
(b) (i) Messenger RNA (m-RNA)
(ii) Ribosomal RNA (r-RNA)
(iii) Transfer RNA (t-RNA
(c) Two main functions of nucleic acids are:
(i) DNA is responsible for the transmission of inherent characters from one
generation to the
next. This process of transmission is called heredity.
(ii) Nucleic acids (both DNA and RNA) are responsible for protein synthesis in a cell.
RNA
34 DNA (FUNCTIONAL DIFFERENCE )
DNA is the chemical basis of heredity
RNA is not responsible for heredity.
DNA molecules do not synthesize
Proteins are synthesised by RNA
proteins, but transfer coded message
molecules in the cells
for the synthesis of proteins in the
cells
(b)Starch consists of two components - amylose and amylopectin. Amylose is a long
linear chain of α-D-(+)-glucose units joined by C1-C4 glycosidic linkage ( α-link).
Amylopectin is a branched-chain polymer of α-D-glucose units, in which the chain is
formed by C1-C4-glycosidic linkage and the branching occurs by C1-C6glycosidic linkage.
cellulose is a straight-chain polysaccharide of β-D-glucose units joined by C1-C4glycosidic linkage ( β-link).
OR
(a) β-form
(b) Glyceraldehyde is C2H6O3
1
2
2
1
1
1
1
1
1
( c) Carbohydrates that yield two to ten monosaccharide units, on hydrolysis, are
called oligosaccharides
(a) C2(H2O)2 is not a carbohydrate. Chemically, the carbohydrates may be defined
as optically active polyhydroxy aldehydes or ketones.
Page 5 of 6
2
(a) The various sugars present in nucleic acids are Ribose in RNA and deoxyribose in
35 DNA.
(b)
(i) Fructose also has the molecular formula C6H12O6
(ii) it contains a ketonic functional group at carbon number- 2
(iii) Fructose exists in two cyclic forms which are obtained by the addition of —OH at
C5 to the carbonyl group
(iv)The ring, thus formed is a five membered ring and is named as furanose with
analogy to the compound furan.
(v)Furan is a five membered cyclic compound with one oxygen and four carbon atoms.
(vi) The cyclic structures of two anomers of fructose are represented by Haworth
structures as given.
1
½ *6=3
1
OR
(a)(i) In cyclic structure of glucose one of the —OH groups may add to the —CHO
group and form a cyclic hemiacetal structure.
(ii) glucose forms a six-membered ring in which —OH at C-5 is involved in ring
formation. This explains the absence of —CHO group.
(iii) The two cyclic hemiacetal forms of glucose differ only in the configuration of the
hydroxyl group at C1, called anomeric carbon
3
(iv) The six membered cyclic structure of glucose is called pyranose structure (α– or β),
in analogy with pyran
(v)
(b) Nucleotides are joined together by phosphodiester linkage between 5′ and 3′
carbon atoms of the pentose sugar
2
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