Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2021 (Term-I) to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2021 (Term-I) with Solutions
Time Allowed: 90 minutes
Maximum Marks: 35
General Instructions:
- This question paper contains 55 questions out of which 45 questions are to be attempted.
- All questions carry equal marks.
- This question paper consists of three Section – Section A, B and C.
- Section A contains 25 questions. Attempt any 20 questions from Q. No. 1 to 25.
- Section B contains 24 questions. Attempt any 20 questions from Q. No. 26 to 49.
- Section C contains 6 questions. Attempt any 5 questions from Q. No. 50 to 55.
- The first 20 questions each in Section A & Section B and first 5 questions attemped in Section C by a candidate will be evaluated.
- There is only one correct option for every Multiple Choice Question (MCQ). Marks will not be awarded for answering more than one option.
- There is no negative marking.
Section A
The section consists of 25 Multiple Choice Questions with overall choice to attempt any 20 questions. In case more than desirable number of questions are attempted, ONLY first 20 will be considered for evaluation.
Question 1.
In a crystal of an ionic compound, the ions Q form the ccp lattice and the ions P occupy all the tetrahedral voids. The formula of the compound is
(a) PQ
2
(b) P
2
Q
(C) PQ
(d) PQ
3
Answer:
(b) P
2
Q
There are 2 tetrahedral holes for every Q ion which are occupied by P ions.
Question 2.
Increasing the temperature of an aqueous solution will cause
(a) Increase in Molarity
(b) Increase in Molality
(c) Decrease in Molarity
(d) Decrease in Molality
Answer:
(c) Decrease in Molarity
As the temperature increases the volume of solution also increases which in turn decreases molarity.
Question 3.
Which of the following conditions is correct for an ideal solution?
(a) ∆H
mix
= 0 and ∆V
mix
= 0
(b) ∆H
mix
> 0 and ∆V
mix
> 0
(c) ∆H
mix
< 0 and ∆V
mix
< 0
(d) ∆H
mix
>0 and ∆V
mix
< 0
Answer:
(a) ∆H
mix
= 0 and ∆V
mix
= 0
Ideal solutions obey Raoult’s Law and the enthalpy of mixing and volume of mixing is zero for such solutions.
Question 4.
2-Bromo-2-methylpropane is allowed to react with alcoholic KOH solution. The major product formed is
Answer:
(d)
Question 5.
Which of the following intermediates is formed in the reaction shown below?
CH
3
-CH
2
-OH + HI → CH
3
-CH
2
-I + H
2
O
Answer:
(a)
Question 6.
Glucose is oxidized by Br
2
water to give
Answer:
(b)
Question 7.
Which of the following has the largest bond angle?
(a) NH
3
(b) H
2
O
(c) PH
3
(d) \(\mathrm{NH}_4^{+}\)
Answer:
(d) \(\mathrm{NH}_4^{+}\)
In N\(\mathrm{H}_4^{+}\) ions, all electrons are used up in formation of bond pairs and there is no Ione pair in it while in rest of the molecules there are both lone pair-lone pair and Ione-pair-bond pair repulsion.
Question 8.
Which of the following elements does not show an oxidation state higher than +2?
(a) Oxygen
(b) Sulphur
(e) Selenium
(d) Tellurium
Answer:
(a) Oxygen
Oxygen does not show an oxidation state more than +2 due to absence of vacant d-orbitals.
Question 9.
Which of the following is a molecular solid?
(a) KCl
(b) SiO
2
(c) Cu
(d) Ar
Answer:
(b) SiO
2
It is a discrete molecule held together by intermolecular forces.
Question 10.
For determination of molar mass of polymers and proteins, which colligative property is
(a) Relative lowering in vapour pressure
(b) Elevation in boiling point
(c) Osmotic pressure
(d) Depression in freezing point
Answer:
(c) Osmotic pressure
Question 11.
Which of the following molecular is chiral in nature?
Answer:
(b)
Question 12.
The reaction:
is an example of
(a) Reimer-Tieinami reaction
(b) Kolbe’s reaction
(c) Williamson’s synthesis
(d) Wurtz reaction
Answer:
(a) Reimer-Tieinami reaction
Question 13.
Which of the following is fibrous protein?
(a) Albumin
(b) Keratin
(c) Insulin
(d) Globin
Answer:
(b) Keratin
Keratin is a fibrous protein.
Question 14.
The acid strength of HF, HCl, HBr and HI increases in the order
(a) HF < HCl < BHr < HI
(b) HI < HBr < HCl < HF
(c) HBr < HI < HCl < HF
(d) HF < HBr < HI < HCl
Answer:
(a) HF < HCl < BHr < HI
With increase in size and decrease in electronegatiity down the group, the ease to lose H + ion increases.
Question 15.
Which is not correct about concentrated H
2
SO
4
?
(a) Dehydrating agent
(b) Oxidising agent
(c) Ka
2
» Ka
1
(d) It forms two series of salts
Answer:
(c) Ka
2
» Ka
1
H 2 SO 4 has larger value of Ka 1 because it is largely dissociated into H + and HS\(\mathrm{O}_4^{-}\) ions and acts as a strong acid.
Question 16.
Pure water boils at 373.15 K and nitric acid boils at 359.15 K. An azeotropic mixture of H
2
O and HNO
3
boils at 393.55 K. Distilling the azeotropic mixture will cause
(a) Pure nitric acid to distil over first.
(b) Pure water to distil over first.
(c) One of them to distill over with a small amount of the other.
(d) both of them to distil over in the same composition as that of the mixture being distilled.
Answer:
(d) both of them to distil over in the same composition as that of the mixture being distilled.
Azeotropic mixture of HNO 3 (68%) + H 2 O (32%) shows that composifion of water is less than HNO 3 , so the distillation of the mixture will distil over in the saine composition as they are being distilled.
Question 17.
A 5% (by mass) solution of glucose (molar mass = 180 g mol
-1
) is isotonic with 1% solution
(by mass) of a substance ‘X’. The molar mass of ‘X’ is
(a) 36 g mol
-1
(b) 18 g mol
-1
(c) 72 g mol
-1
(d) 900 g mol
-1
Answer:
(a) 36 g mol
-1
\(\frac{5}{180}\) = \(\frac{1}{x}\)
⇒ x = \(\frac{180}{5}\)
∴ x = 36 g mol
-1
Question 18.
An S
N
1 reaction of an enantiomerically sure chiral alkyl halide gives a product
(a) with retention of configuration
(b) with inversion of configuration
(c) with racemisation
(d) with partial racemisation
Answer:
(c) with racemisation
Question 19.
In the following reaction
The product formed is
Answer:
PhCH
2
-CH
2
-OH
Question 20.
α-D-Glucopyranose and β-D-Glucopyranose are
(a) Isomers which differ in configuration at C-5
(b) Geometrical isomers
(c) Functional isomers
(d) Anomers
Answer:
(d) Anomers
Anomers, because they both differ in the configuration at anorneric carbon atom.
Question 21.
Fluorine does not exhibit variable oxidation states due to
(a) Non-availability of d-orbitals in valence shell
(b) Low bond dissociation enthalpy
(c) High electronegativity
(d) Small size
Answer:
(a) Non-availability of d-orbitals in valence shell
Question 22.
The formation of \(\mathrm{O}_2^{+}\)[PtF
6
]
–
is the basis for the formation of Xenon fluoride. This is because
(a) O
2
and Xe have comparable electronegativities.
(b) O
2
and Xe have comparable sizes.
(c) O
2
and Xe have comparable ionization enthalpies.
(d) O
2
and Xe have comparable electron gain enthalpies.
Answer:
(c) O
2
and Xe have comparable ionization enthalpies.
Question 23.
Chlorobenzene reacts with Cl
2
in the presence of FeCl
3
giving ortho and para chloro compounds. The reaction is
(a) Nucleophilic substitution reaction
(b) Nucleophilic addition reaction
(c) Electrophilic addition reaction
(d) Electrophilic substitution reaction
Answer:
(d) Electrophilic substitution reaction
Question 24.
Phenol is more acidic than ethanol because
(a) Ethoxide ion is more stable than Phenoxide ion.
(b) Phenoxide ion is more stable than Ethoxide ion.
(c) Phenol undergoes electrophilic substitution reaction.
(d) Phenol undergoes protonation easily.
Answer:
(b) Phenoxide ion is more stable than Ethoxide ion.
Phenoxide is stabilized by resonance to a greater extent than ethoxide ion.
Question 25.
Which of the following statements is not correct about amorphous solids?
(a) Amorphous solids are anisotropic.
(b) Amorphous solids have a tendency to flow.
(c) Amorphous solids have short range order.
(d) Amorphous solids have irregular shape.
Answer:
(a) Amorphous solids are anisotropic.
Amorphous solids are isotropic in nature because they show thermal arid optical properties which are same in all directions.
Section B
This section consists of 24 Multiple Choice Questions with overall choice to attempt any 20 questions. In case more than desirable number of questions are attempted, ONLY first 20 will be considered for evaluation.
Question 26.
An element with molar mass 96 g mol-1 forms a cubic unit cell with edge length 4 × 10
-8
cm. If density is 10 g cm
-3
, the nature of unit cell is (N
A
= 6 × 10
23
mol
-1
).
(a) simple cubic
(b) bcc
(c) fcc
(d) End centered cubic
Answer:
(c) fcc
Question 27.
When 2.5 g of a non-volatile solute was dissolved in 50 mL of water, it gave boiling point elevation of 0.52 °C. The molar mass of the solute is (K
b
for water = 0.52 km
-1
)
(a) 100 g mol
-1
(b) 50 g mol
-1
(c) 25 g mol
-1
(d) 75 g mol
-1
Answer:
(b) 50 g mol
-1
Question 28.
Which of the following gas is released on heating ammonium dichromate (NH
4
)
2
Cr
2
O
7
?
(a) NO
2
(b) N
2
O
(C) NO
(d) N
2
Answer:
(d) N
2
(NH 4 ) 2 Cr 2 O 7 (s) → N 2 (g) + 4H 2 O (g) + Cr 2 O 3 (g)
Question 29.
On its reaction with water and alkalies, the behaviour of SO
2
(g) is very similar to that of which gas?
(a) NO
2
(b) CO
2
(c) NH
3
(d) N
2
O
Answer:
(b) CO
2
Question 30.
Which of the following is most reactive towards nucleophilic substitution reaction?
Answer:
(a)
The electro-withdrawing -NO 2 group decreases the electron density and increases the reactivity, hence undergoes S N reaction easily.
Question 31.
In the following reaction.
Answer:
(c)
Wurtz reaction
Question 32.
The structure(s) of the major product(s) expected from the mononitration of 3-methyiphenol will be
Answer:
(a)
Question 33.
1-Phenylethanol may be prepared by the reaction of C
6
H
5
MgBr with
Answer:
(b)
CH
3
-CHO
Question 34.
Two among the three components of DNA are 2-deoxyribose and a nitrogen containing heterocyclic base. The third component is
(a) D-ribose
(b) Thymine
(c) Guanine
(d) Phosphoric acid
Answer:
(d) Phosphoric acid
Question 35.
Which one among the following bases is usually not present in RNA?
(a) Uracil
(b) Thymine
(c) Adenine
(d) Guanine
Answer:
(b) Thymine
Thymine is present in DNA and binds to adenine.
Question 36.
Glucose on reaction with (CH
3
CO)
2
O forms glucose pentaacetate which confirms the presence of
(a) — CHO group
(b) — COOH group
(c) Five—OH groups
(d) A straight chain
Answer:
(c) Five—OH groups
Five -OH groups
Question 37.
To increase the solubility of CO
2
gas in soft drinks, the bottle is sealed under
(a) Low pressure
(b) High temperature
(c) Constant pressure
(d) High pressure
Answer:
(d) High pressure
Solubility of gas in liquid is directly proportional to the pressure of the gas.
Question 38.
A solution of a pair of volatile liquids A and B shows negative deviation from Raoult’s law. This is because
(a) p
A
> p°
A
x
A
and p
B
> P°
B
x
B
(b) the intermolecular forces A —A, B—B < A—B.
(c) Both ∆H
mixing
and ∆V
mixing
are positive.
(d) All of the above
Answer:
(b) the intermolecular forces A —A, B—B < A—B.
Question 39.
The structure of XeF
6
is
(a) Distorted Octahedral
(b) Regular Octahedral
(c) Square Pyramidal
(d) Square Planar
Answer:
(a) Distorted Octahedral
Hybridisation is sp
3
d
3
(XeF
6
)
Question 40.
On the basis of ∆H°
bond
, which of the following has the strongest bond?
(a) H – I
(b) H – Cl
(c) H – F
(d) H — Br
Answer:
(c) H – F
H — F has highest bond dissociation enthaiphy due to large electronegativity difference
between H & F and due to strongest bond. But it decreases down the group with
increase in size.
H — F > H — Cl > H — Br > H — I
Question 41.
Which of the following oxoacids of Sulphur contains peroxide linkage?
(a) H
2
SO
4
(b) H
2
S
2
O
7
(C) H
2
SO
3
(d) H
2
S
2
O
8
Answer:
(d) H
2
S
2
O
8
Question 42.
Which of the following reactions is not correct?
(a) 2F
2
+ 2H
2
O → 4H
+
+ 4F
–
+ O
2
(b) 2I
2
+ 2H
2
O → 4H
+
+ 4I
–
+ O
2
(c) Cl
2
+ H
2
O → Hcl + HOCl
(d) Br
2
+ H
2
O → HBr + HOBr
Answer:
(b) 2I
2
+ 2H
2
O → 4H
+
+ 4I
–
+ O
2
2I
2
+ 2H
2
O → 4H
+
+ 4I
–
+ O
2
The correct reaction is
I
2
+ H
2
O → HI + HOI
Only fluorine reacts differently due to its higher electronegatively from other members and also from oxygen which allows fluorine to oxidize oxygen while other halogens cannot do so.
Question 43.
Which of the following belongs to the class of allylic halidep?
Answer:
(d)
Question 44.
Which reagent will be required for one step conversion of benzenediazonium chloride to phenol?
(a) Cu
2
Cl
2
(b) NaOH(aq)
(c) H
2
O
(d) Alcoholic KOH
Answer:
(c) H
2
O
Given below are the questions (45-49) labelled as Assertion (A) and Reason (R). 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.
Question 45.
Assertion (A): Relative lowering in vapour pressure is a colligative property.
Reason (R): Relative lowering in vapour pressure depends upon mole fraction of pure solvent.
Answer:
(c) (A) is true, but (R) is false.
Relative lowering in vapour pressure depends only on the number of solute particles and not on its nature. It depends on the mole fraction of solute.
Question 46.
Assertion (A): ZnO on heating turns yellow.
Reason (R): Excess Zn
2+
ions move to interstitial sites and the electron to neighbouring
interstitial sites.
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Because ZnO loses Oxygen reversibly at high temperature.
Question 47.
Assertion (A): F
2
is a powerful oxidizing agent.
Reason (R): Fluorine shows anomalous behaviour.
Answer:
(b) Both (A) and (R) are true, but (R) is not the correct explanation of (A).
F
2
is a powerful oxidizing agent. Due to its high electronegativity it attracts shared pair
of electrons towards itself strongly and oxidizes others.
F
2
shows anomalous behaviour due to its small size, highest electronegativity, low
bond dissociation enthalpy (F—F) and non-availability of d-orbitais in valence shell.
Question 48.
Assertion (A): Monoclinic sulphur is stable at room temperature.
Reason (R): Both Rhombic Sulphur and Monoclinic Sulphur have S8 molecules.
Answer:
(d) (A) is false, but (R) is true.
Monoclinic sulphur is less stable than Rhombic sulphur at room temperature because Rhombic sulphur converted into monoclinic sulphur on heating.
Question 49.
Assertion (A): Reaction of (CH
3
)
3
C – Br with CH
3
ONa gives majorly 2-methylpropene.
Reason (R): CH
3
ONa acts as a strong base.
Answer:
(a) Both (A) and (R) are true and (R) is the correct explanation of (A).
Because here CH
3
ONa acts as a strong nucleophile and base and causes elimination
reaction.
Section C
This section consists of 6 Multiple Choice Questions with an overall choice to attempt any 5. In case more than desirable number of questions are attempted, ONLY first 5 will be considered for evaluation.
Question 50.
Match the following:
Which of the following is the best matched option?
(a) (i)-B, (ii)-A, (iii)-C, (iv)-D
(b) (i)-D, (ii)-B, (v)-A, (iii)-C
(c) (i)-C, (ii)-D, (iii)-A, (v)-B
(d) (i)-A, (ii)-D, (iii)-C, (iv)-B
Answer:
(c) (i) – C, (ii) — D, (iii) — A, (v) — B
- Cl 2 act as bleaching agent as it decolourises clothes due to liberation o nascent of Chlorine.
- He has low solubility in blood and therefore used in diving apparatus to remove excessive N 2 from blood.
- N 2 is inert at room temperature due to its high bond dissociation energy, it requires a lot of energy to break these bonds and take part in a reaction.
- Moist SO 2 acts as a reducing agent as it liberates nascent hydrogen which reduces H 2 O 2 into H.
Question 51.
Which of the following analogies is correct?
(a) Chloroform-acetone : Positive deviation :: Ethanol-H
2
O : Negative deviation.
(b) pA = p°
A
. x
A
: Henry’s law :: p = K
H
. x : Raoult’s law.
(c) P
Total
= P
A
+ P
B
: Non-ideal solution :: P
Total
> p
A
+ p
B
: Ideal solution
(d) π = CRT : Osmotic pressure : : P π : Reverse osmosis.
Answer:
(d) π = CRT : Osmotic pressure : : P π : Reverse osmosis.
- CHCl 3 and Acetone shows negative deviation due to formation of H-bonding while Ethanol -H 2 O shows positive deviation
-
Henry Law is P = K
H
× x
Raoult’s law is A A = P° A . x A -
P
T
= P
A
+ P
B
Ideal solution
P T > P (Raoult’s law) Show deviations – Non ideal solution . - Reverse osmosis occurs when a pressure higher than osmotic pressure is applied on the solution (P > π)
Question 52.
Complete the following analogy: [1]
o-nitrophenol: A :: o-cresol: B
(a) A : more acidic than phenol B : less acidic than phenol
(b) A : less acidic than phenol B : more acidic than phenol
(c) A : more acidic than phenol B : more acidic than phenol
(d) A : less acidic than phenol B : less acidic than phenol
Case : Read the passage given below and answer the following questions 53-55.
No crystal is found to be perfect at room temperature. These defects are basically irregularities in the arrangement of constituent particles. These defects can be stoichiometric or Non- stoichiometric. Stoichiometric defects are of two types : Schottky and Frenkel defect. Schottky is basically a vacancy defect while Frenkel is an interstitial defect. Due to non-stoichiometric defects, the formula of the ionic compound is different from the ideal formula. These defects are also of two types :
(i) Metal excess defect and
(ii) Metal deficiency defect.
Answer:
(a) A : more acidic than phenol B : less acidic than phenol
Acidic strength α stability of anion α \(\frac{\text { E.W.G }}{\text { E.D.G }}\)
So acidic strength order will be
Question 53.
What type of defect is shown by AgCl?
(a) Schottky defect
(b) Frenkel defect
(c) Metal excess defect
(d) Metal deficiency defect
Answer:
(b) Frenkel defect
Size of Ag + < Cl – . Here Ag + occupies interstial spaces in its crystal lattic by leaving their position in lattice.
Question 54.
Which of the following defects lowers the density of the crystal but does not affect the stoichiometry?
(a) Schottky defect
(b) Frenkel defect
(c) Metal excess defect
(d) Metal deficiency defect
Answer:
(a) Schottky defect
In this defect equal number of cation and anion are found missing from its lattice, thus it deceases density but does not alter stoichiometry.
Question 55.
Excess of potassium makes KCl crystals violet due to the formation of
(a) Cation vacancies
(b) Anion vacancies
(c) F-centres
(d) Interstitial defect
Answer:
(c) F-centres
Because the Cl – ions diffuse into the surface and combine with K atoms while the e – ’s produced by ionization of K atoms then diffuse into the crystals and are trapped into anion vacancies (F-centres). When visible light falls on the crystal, these e – ’s gain energy and get excited to higher level and shows corresponding violet colour when they come back to the ground state.
