Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2022 (Term-II) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2022 (Term-II) with Solutions
Time Allowed: 2 hours
Maximum Marks: 35
General Instructions:
- This question paper contains 12 questions. All questions are compulsory.
- This question paper is divided into three Sections – Section A, B and C.
- Section A — Q. Nos. 1 to 3 are very short answer type questions carrying 2 marks each.
- Section B—Q. Nos. 4 to 11 are very short answers type questions carrying 3 marks each.
- Section C—Q. Nos. 12 is case-based question carrying 5 marks.
- Use of log tables and calculators is NOT allowed.
Section A
Question 1.
Predict the reagent for carrying out the following transformations : (Any two)
(i) Benzoyl chloride to Benzaldehyde
Answer:
Benzoyl chloride to Benzaldehyde
The reagent is H
2
/Pd – BaSO
4
(Lindler’s catalyst)
Rosenmund Reaction
(ii) Ethanal to 3-hydroxy butanal
Answer:
Ethanal to 3-hydroxy butanal
The reagent is Dilute Sodium hydroxide
Aldol condensation
(iii) Ethanoic acid to 2-chloroethanoic acid [1 × 2 = 2]
Answer:
Ethanoic Acid to 2-chloroethanoic acid
The reagent is Red phosphorus
Hell-Vohlard Zelinsky Reaction
Question 2.
(i) Why on dilution the Λ
m
of CH
3
COOH increases very fast, while that of CH
3
COONa increases gradually?
Answer:
Since CH
3
COOH is a weak electrolyte and CH
3
COONa is strong electrolyte, so on dilution the degree of dissociation increases for both electrolytes but more for CH
3
COOH. As a result, number of ions present also increases which leads to drastic
increase of Λ
m
for CH
3
COOH and gradual increase for CH
3
COONa.
(ii) What happens if external potential applied becomes greater than E°cell of electrochemical cell? [1 × 2 = 2]
Answer:
If external potential applied becomes greater than E°
cell
of electrochemical cell, then the reaction gets reversed, It now becomes an electrolytic cell and converts electrical energy into chemical energy.
Question 3.
An Organic compound (A) with molecular formula C
3
H
7
NO on heating with Br
2
and KOH forms a compound (B). Compound (B) on heating with CHCl
3
and alcoholic KOH produces a foul smelling compound (C) and on reacting with C
6
H
5
SO
2
Cl forms a compound (D) which is soluble in alkali.
Write the structures of (A), (B), (C) and (D). [2]
Answer:
The compound C
3
H
7
NO (A) is CH
3
CH
2
CONH
2
Section – B
Question 4.
Account for the following: [1 × 3 = 3]
(i) Cu
2+
salts are coloured while Zn
2+
salts are white.
Answer:
Cu
2+
= [Ar] 3d
9
4s°
Zn
2+
= [Ar] 3d
10
4s°
In case of Cu
2+
due to presence of one unpaired electrons, it forms coloured salts while in Zn
2+
due to absence of unpaired electrons all salts are white.
(ii) E° value for the Mn
3+
/Mn
2+
couple is much more positive than that for Cr
3+
/Cr
2+
.
Answer:
Mn
2+
= [Ar] 3d
5
4s° Cr
3+
= [Ar] 3d
3
Mn
3+
= [Ar] 3d
4
4s° Cr
2+
= [Ar] 3d
4
4s°
Since Mn
3+
has 3d
4
configuration, so it will have the tendency to gain one electron to become Mn
2+
with 3d
5
configuration which is more stable due to its symmetry and exchange energy while in case of Cr
3+
and Cr
2+
there is no such stable configuration has been found.
(iii) Transition metals form alloys. [1]
Answer:
Because transition metals can form homogeneous mixture with other transition metals due to their very similar atomic sizes as a result of which one metal can occupy the place of other metal from its crystal lattice and sometimes also occupies their interstitial spaces.
Question 5.
(a) Calculate ∆
r
G° and log K
c
for the following cell:
Ni(s) + 2 Ag
+
(aq) → Ni
2+
(aq) + 2Ag(s)
Given that E°
cell
= 1.05 V, IF = 96,500 C mol
-1
. [3]
Answer:
Given. E°
cell
= 1.05 V
IF = 96,500 C mol
-1
Ni(s) + 2Ag
+
(aq) → Ni
2+
(aq) + 2Ag(s) ……[(n = 2)
Using formula, ∆G° = -nFE°
∆G = -2 × 96500 × 1.05 = -202650 J
∆G° = -2.303 RT × log K
c
On substituting the values,
-202650J = -2.303 × 8.314 × 298 × log K
c
= log K
c
= \(\frac{-202650 \mathrm{~J}}{-5705.85}\)
⇒ K
c
= Antilog 35.5
∴ K
c
= 3.16 × 10
-36
Or
(b) Calculate the e.m.f. of the following cell at 298K:
Fe (s) | Fe
2+
(0.001 M) | | H
+
(0.01M) | H
2
(g) (1 bar) | Pt(s)
Given that E°
cell
= +0.44 V
[log 2 = 0.3010 log 3 = 0.4771 log 10 = 1] [3]
Answer:
Fe(s) + 2H
+
(aq) → Fe
2+
(aq) + H
2
(g)
Given. E°cell = +0.44 V, n = 2
Using Nernst equation,
E°cell = E°
cell
– \(\frac{0.0591}{n}\)\(\log \frac{[\text { Product }]}{[\text { Reactant }]}\)
⇒ E°
cell
= E°
right
E°
left
⇒ E°
cell
= 0 – (-0.44) V ⇒ E°
cell
= 0.44 V
∵ Concentration of solid substance is always 1.
∴ [Fe] = 1 and for [H
2
] = 1 (gaseous)
E
cell
= o.44 – \(\frac{0.0591}{2}\)log\(\frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^2}\)
E
cell
= 0.44 – \(\frac{0.0591}{2}\)log\(\frac{\left[\mathrm{Fe}^{2+}\right]}{\left[\mathrm{H}^{+}\right]^2}\)
= +0.44 – \(\frac{0.0591}{2}\)log\(\frac{[0.001]}{[0.01]}\)
= 0.44 – \(\frac{0.0591}{2}\)log\(\frac{1}{10}\)
= 0.44 – \(\frac{0.0591}{2}\)log10
-1
⇒ E°
cell
= +0.44 – \(\frac{0.0591}{2}\) × (-1)
⇒ E°
cell
= 0.44 + 0.02955
⇒ E°
cell
= 0.46955
∴ E°
cell
= 0.47 V
Question 6.
(a) Using valence bond theory, predict the hybridization and magnetic character of following:
[CoF
6
]
3-
[Atomic number of Co = 27]
Answer:
It is a paramagnetic in nature due to presence of four unpaired electrons.
(b) Write IUPAC name of the following complex :
[CoBr
2
(en)
2
]
+
Answer:
[CoBr
2
(en)
2
]
+
IUPAC Name. Dibromidobis (ethane-1,2 -diamine) Cobalt (III) ion
(c) How many ions are produced from the complex [Co(NH
3
)
6
]Cl
2
in solution? [1 × 3 = 3]
Answer:
[Co(NH
3
)]Cl
2
→ [Co(NH
3
)
6
]
2+
+ 2 Cl
–
Number of ions produced form the complex are three.
Question 7.
(a) Differentiate between the following: [1 × 3 = 3]
(i) Adsorption and Absorption
Answer:
Adsorption and Absorption
| Adsorption | Absorption |
| It is surface phenomena. | It is a bulk phenomenon. |
| Molecules undergoing adsorption are soaked up by the air. | Molecules undergoing absorption are soaked up by the length. |
| Film of adsorbate is developed on the surface. | It includes complete volume of the absorbing agent. |
| It is the phenomenon by which one substance gets concentrated mainly on the surface of the other substance rather than in the bulk of a solid or liquid. | It is the phenomenon by which one substance gets uniformly distributed throughout the body of the other substance. |
| Its concentration is different at surface from the bulk. | Its concentration is same throughout the bulk. |
(ii) Lyophobic Sol and Lyophilic Sol
Answer:
Lyophobic Sol and Lyophilic Sol
| Lyophobic Sol | Lyophilic Sol |
| They are irreversible in nature. | They are reversible in nature. |
| They cannot be disintegrated into dispersed phase and medium. | They can be disintegrated into dispensed phase and medium. |
| They are less stable. | They are more stable. |
| No affinity, Solvent repelling. | High affinity for dispersion medium and solvent attracting. |
|
Lyophobic sols substances like-metals, their sulphides, etc., when simply mixed with the dispersion medium do not form the colloidal sol. Their colloidal sols can only be prepared by specific methods. They are not much hydrated and are irreversible in nature. They are also called extrinsic colloids.
Example: As 2 S 3 sol. |
Lyophilic sols Liquid loving colloids in which there is affinity between disperse phase and dispersion medium.
Example : Starch sol, Gum sol, Gelatin sol |
(iii) Multimolecular Colloid and Macromolecule colloid.
Answer:
Multimolecular Collid and Macromolecular Colloid
| Multimolecular Colloid | Macromolecular Colloid |
| They are formed by aggregation of smaller molecules when they are dissolved in a solvent. | They are individual particles that are large enough to be considered as colloids. |
| Formed form particles having low molecular masses and diameter less than 1 nm. | Formed from particles of high molecular masses. |
| They are lyophobic in nature. | They are lyophillic in nature. |
| Aggregates are held by weak Vander Waal forces | Particles are held by strong forces |
Or
(b) (I) Define the following terms : [3]
(i) Zeta Potential
Answer:
Zeta Potential. When one type of ions of the electrolyte are adsorbed on the surface of colloidal particles, it forms a fixed layer which attracts another layer of opposite ions thus forming a Helmholtz electrical double layer whose potential difference between the two layers is termed as Zeta Potential.
(ii) Coagulation
Answer:
Coagulation: The process of settling of colloidal particles is called coagulation or precipitation of the solution.
(II) Why a negatively charged sol is obtained when AgNO
3
solution is added to KI solution? [3]
Answer:
When AgNO
3
solution is added to KI solution, a negatively charged sol is obtained because the precipitated Agl adsorbs I
–
ions from the dispersion medium.
Question 8.
Define transition metals. Why Zn, Cd and Hg are not called transition metals? How is the variability in oxidation states of transition metals different from that of p-block elements? [3]
Answer:
Elements which have partially filled d-orbitals in their ground states or of its any oxidation states are called transition metals.
Zn, Cd and Hg are not called transition metals because the orbitals of these elements are completely filled in their ground state as well as in their general oxidation state.
In transistion metals the oxidation states can vary from +1 to highest oxidation state by remaining all its valence electrons and the oxidation stale differ by 1. For example, Fe
2+
and Fe
3+
, Cu
2+
and Cu
+
while in p-block elements. The oxidation states differ by 2 due to inert pair effect For example, Pb
2+
and P
4+
, As
3+
and As
5+
, Bi
3+
and Bi
+5
.
Question 9.
(a) What happens when [1 × 3 = 3]
(i) Propanone is treated with CH
3
MgBr and then hydrolysed?
Answer:
Propanone is treated with CH
3
MgBr and then hydrolysed
(ii) Ethanal is treated with excess ethanol and acid?
Answer:
Ethanal is treated with excess ethanol and acid
(iii) Methanal undergoes Cannizzaro reaction?
Answer:
Methanal undergoes Cannizzaro reaction
Or
(b) Write the main product in the following reactions : [1 × 3 = 3]
(i) 2CH
3
COCl + (CH
3
)
2
Cd →
(ii)
(iii)
Answer:
Question 10.
Give reasons : [1 × 3 = 3]
(i) Ammonolysis of alkyl halides is not a good method to prepare pure primary amines.
Answer:
Because the primary amine formed by ammonolysis itself acts as a nucleophile and produces further 2° and 3° alkyl amine. Ammonoylsis fields a mixture of 1°, 2° and 3° amines and their separation in a complicated process as the formed 1° amine itself acts as a nucleophile and produces further 2° and 3° alkyl amine.
(ii) Aniline does not give Friedel-Crafts reaction.
Answer:
Aniline being a Lewis base reacts with Lewis acid AlCl
3
to form a salt.
As a result, N of aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Freidel Craft reaction.
(iii) Although -NH
2
group is o/p directing in electrophilic substitution reactions, yet aniline on nitration gives good yield of m-nitroaniline.
Answer:
Because of nitration in an acidic medium, aniline gets protonated to give anilinium ion which is m-directing.
Question 11.
(a)
(i) Which acid of the following pair would you expect to be stronger?
F – CH
2
– COOH or CH
3
– COOH
Answer:
F —CH
2
—COOH is a stronger acid than CH
3
COOH because due to -I effect of F electron density in the O — H bond decreases thereby making release of a proton easier.
(ii) Arrange the following compound in increasing order of their boiling points :
CH
3
CH
2
OH, CH
3
-CHO, CH
3
-COOH
Answer:
CH
3
CH
2
OH > CH
3
-COOH > CH
3
-CHO
(iii) Give simple chemical test to distinguish between Benzaldehyde and Acetophenone. [1 × 3 = 3]
Answer:
Benzaldehyde and Acetophenone :
By Iodoform test : Acetophenone being a methyl ketone on treatment with I
2
/NaOH (NaOI) undergoes iodoform test to give yellow ppt. of iodoform but benzaldehyde does not.
Or
(b)
(i) Which will undergo faster nucleophilic addition reaction?
Acetaldehyde or Propanone
Answer:
Acetaldehyde undergoes faster nucleophilic addition reaction than propanone due to electronic and steric effect.
(ii) What is the composition of Fehling’s reagent?
Answer:
Fehling’s reagent consists of two Solutions — Solution A and Solution B
Fehling’s Solution A. 7% CuSO
4
Solution (blue)
Fehling’s Solution B. 25% KOH and 35% Sodium potassium tartrate (Colourless)
(iii) Draw structure of the semicarbazone of Ethanol. [1 × 3 = 3]
Answer:
CH
3
CH = N-NH-CONH
2
Semicarbazone of ethanol
Section – C
Question 12.
Read the following passage and answer the questions that follow :
The rate of reaction is concerned with decrease in concentration of reactants or increase in the concentration of products per unit time. It can be expressed as instantaneous rate at a particular instant of time and average, rate over a large interval of time. A number of factors such as temperature, concentration of reactants, catalyst affect the rate of reaction. Mathematical representation of rate of a reaction is given by rate law :
Rate = k[A]
x
[B]
y
x and y indicate how sensitive the rate is to the change in concentration of A and B. Sum of x + y gives the overall order of a reaction.
When a sequence of elementary reactions gives us the products, the reactions are called complex reactions. Molecularity and order of an elementary reaction are same. Zero order reactions are relatively uncommon but they occur under special conditions. All natural and artificial radioactive decay of unstable nuclei take place by first order kinetics.
(a) What is the effect of temperature on the rate constant of a reaction?
Answer:
With increase in temperature, both rate of reaction and rate constant increase and it gets doubled when the temperature is increased by 10°C.
(b) For a reaction A + B → Product, the law is given by, Rate = k[A]
2
[B]
1/2
. What is the order of the reaction?
Answer:
According to the formula
r = k [A]
1/2
[B]
2
Order w.r.t. A = \(\frac{1}{2}\)
Order w.r.t. B = 2
∴ Overall order = \(\frac{1}{2}\) + 2 = \(\frac{5}{2}\)
(c) How order and molecularity are different for complex reactions?
Answer:
Order of reaction is the sum of powers of the molar concentrations of reacting species while molecularity is the total number of reacting species which belong to a chemical change. For complex reactions they are meaningless as they are completed in multiple steps.
(d) A first order reaction has a rate constant 2 × 10
-3
s
-1
. How long will 6g of this reactant take to reduce to 2g?
Answer:
Given:
K = 2 × 10
-3
s
-1
[R
0
] = 6 g
[R] = 2 g
Using the formula,
t = \(\frac{2.303}{\mathrm{~K}}\)log\(\frac{\left[R_0\right]}{[R]}\)
= \(\frac{2.303}{2 \times 10^{-3}}\)log\(\frac{6}{2}\)
= 1.1515 × 10
3
log 3
= 1.1515 × 10
3
× 0.4771
= 0.54938 × 10
3
= 549 seconds
Or
The half life for radioactive decay of
14
C is 6930 years. An archaeological artifact containing wood had only 75% of the
14
C found in a living tree. Find the age of the sample.
[log 4 = 0.6021 log 3 = 0.4771 log 2 = 0.3010 log 10 = 1] [1 + 1 + 1 + 2]
Answer:
Given: t
1/2
= 6930 years
Using the formula,
k = \(\frac{0.693}{t_{1 / 2}}\)
= \(\frac{0.693}{6930}\) years
= 10
-4
years
Again, t = \(\frac{2.303}{\mathrm{~K}}\) log \(\frac{[R]_0}{[R]}\)
t = \(\frac{2.303 \times 6930}{0.693}\) log\(\frac{100}{75}\)
t = 23030[log 100 – log 75]
t = 23030 (2 – 1.8751)
= 23030 × 0.1 249
∴ t = 2876 years (approximately)
∴ The age of the sample is 2876 years