CBSE Sample Papers for Class 12 Chemistry Set 3 with Solutions

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CBSE Sample Papers for Class 12 Chemistry Set 3 with Solutions

Time Allowed: 3 hours
Maximum Marks: 70

General Instructions :

1. There are 33 questions in this question paper with internal choice
2. SECTION A consists of 16 multiple-choice questions carrying 1 mark each.
3. SECTION A consists of 16 multiple-choice questions carrying 1 mark each.
4. SECTION C consists of 7 short answer questions carrying 3 marks each.
5. SECTION D consists of 2 case-based questions carrying 4 marks each.
6. SECTION E consists of 3 long answer questions carrying 5 marks each.
7. All questions are compulsory.
8. Use of log tables and calculators is not allowed.

SECTION – A (16 Marks)

The following questions are multiple-choice questions with one correct answer. Each question carries 1 mark. There is no internal choice in this section.

Question 1.
The number of Faradays required to reduce one mol of Cu⁺² to metallic copper is:
(a) One
(b) Two
(c) Three
(d) Four
Answer:
(b) Two
Explanation: Faraday’s first law state that “the amount of current passed through the electrode is directly proportional to the number of substances liberated from it.” The reduction of the one mole of Cu⁺² to metallic copper can be represented as: Cu⁺²+ + 2e^– → Cu
Since, there are two moles of electron involved in this reaction thus the amount of charge required is 2F.

Question 2.
Which of the following is the correct order of arrangement of the first five lanthanides according to atomic number?
(a) La, Ce, Pr, Nd, Pm
(b) La, Pr, Ce, Pm, Nd
(c) La, Pr, Ce, Nd, Pm
(d) La, Ce, Pr, Pm, Nd
Answer:
(a) La, Ce, Pr, Nd, Pm
Explanation: The first five elements of Lanthanides are:
Lanthanum (La) – 57
Cerium (Ce) – 58
Praseodymium (Pr) – 59
Neodymium (Nd) – 60
Promethium (Pm) – 61

Question 3.
Proteins are denatured in the:
(a) Mouth
(b) Stomach
(c) Small intestine
(d) Large intestine
Answer:
(b) Stomach
Explanation: Proteins are denatured in the stomach due to highly acidic medium.

Question 4.
The IUPAC name of [Pt(NH₃)₃(Br)(NO₂)Cl]Cl is :
(a) Triamminebromochloronitroplatinum(IV) chloride
(b) Triamminebromonitrochloroplatinum(lV) chloride
(c) Triamminechlorobromonitroplatinum(IV) chloride
(d) Triamminenitrochlorobromoplatinum(lV) chloride
Answer:
(a) Triamminebromochloronitroplatinum(IV) chloride
Explanation: The ligands are named in the alphabetic order according to the IUPAC nomenclature. For the given complex, the oxidation number of metal is +4. So, the name of [Pt(NH₃)₃(Br)(NO₂)Cl]Cl is Triamminebromochloronitroplatinum(IV) chloride.

Question 5.
Which of the following poisonous gas is formed when chloroform is exposed to light and air?
(a) Mustard gas
(b) Carbon monoxide
(c) Phosgene
(d) Chlorine
Answer:
(c) Phosgene
Explanation: When chloroform is exposed to air and sunlight, it undergoes oxidation and releases a poisonous gas called Phosgene gas (COCl₂) and Hydrochloric acid (HCl).

Question 6.
Which of the following statements about benzaldehyde is/are true?
(a) Reduces Tollen’s reagent
(b) Undergoes Aldol condensation
(c) Does not undergo Cannizzaro reaction
(d) Does not form an addition compound with sodium hydrogen sulphite
Answer:
(a) Reduces Tollen’s reagent
Explanation:: Benzaldehyde reduces Tollen’s reagent and undergoes Cannizzaro reaction. However, it forms an addition compound with NaHSOj but does not undergo aldol condensation. Thus, the statement which is correct about benzaldehyde is that it reduces Tollen’s reagent.

Question 7.
Which one of the following aldehyde gives Cannizzaro reaction when heated with strong alkali?
(a) Benzaldehyde
(b) Acetaldehyde
(c) Propionaldehyde
(d) All of these
Answer:
(a) Benzaldehyde
Explanation: Aldehydes that do not have alpha hydrogen atom will undergo Cannizzaro reaction. Since, benzaldehyde does not have alpha hydrogen atom in its structure, therefore, it will undergo cannizzaro reaction. The reaction can be represented as:

Question 8.
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
Answer:
(c) 68.4
Explanation: ‘Cane Sugar’
W₁ = 5g
V₁ = 100 mL = 0.1 L
M₁ = 342
For isotonic solutions, C₁ = C₂
‘X’
W₂ = lg
V₂ = 100 mL = 0.1 L
M₂ = ?
\(\begin{aligned}
\frac{\mathrm{W}_1}{\mathrm{M}_1 \mathrm{~V}_1} & =\frac{\mathrm{W}_2}{\mathrm{M}_2 \mathrm{~V}_2} \Rightarrow \frac{5}{342 \times 0.1}=\frac{1}{\mathrm{M}_2 \times 0.1} \\
\mathrm{M}_2 & =\frac{342}{5}=68.4
\end{aligned}\)
Hence, the molecular weight of X is 68.4.

Question 9.
A graph of volume of hydrogen released vs time for the reaction between zinc and dil. HC1 is given in figure. On the basis of this, mark the correct option.

Hence, a catalyst does not increase or decrease the free energy change in the reaction.

Question 10.
Match the Columns:

Column I	Column II
1 Molal depression constant	(p) Henry’s law
2 Colligative property	(q) Osmotic pressure
3 Dilute solution	(r) Relative lowering of vapour pressure
4 Elevation of boiling point	(s) K kg mol-1
5 Solubility of gas in liquid	(t) Colligative property

(a) l-(s), 2-(r), 3-(q), 4-(t), 5-(p)
(b) l-(r), 2-(t), 3-(s), 4-(q), 5-(p)
(c) l-(p),2-(q),3-(t),4-(s),5-(r)
(d) l-(q), 2-(p), 3-(r), 4-(t), 5-(s)
Answer:
(a) l-(s), 2-(r), 3-(q), 4-(t), 5-(p)
Explanation: The unit of molal depression constant is K kg mol-1, relative lowering of vapour pressure is one of the type of colligative property, osmotic pressure is the external pressure that is applied on the solution to prevent the flow of solvent from a higher concentration area to the lower
concentration area. Elevation in boiling point is one of the type colligative property and solubility of gas in liquid is determined by Henry’s Law.

Question 11.

Identify Z.
(a) 2-lodoheptane
(b) Heptane-2-ol
(c) 2-lodohexane
(d) Heptanoic acid
Answer:
(d) Heptanoic acid
Explanation:

Question 12.
The effect of a catalyst in a chemical reaction is to change the:
(a) Activation energy
(b) Equilibrium concentration
(c) Heat of reaction
(d) Final products
Answer:
(a) Activation energy
Explanation: The effect of a catalyst in a chemical reaction is to change the activation energy because a catalyst provides an alternate pathway or reaction mechanism by reducing the activation energy of reactants.

Question No.13 to 16 consist of two statements- Assertion (A) and Reason (R). Answer these questions selecting the appropriate 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.

Question 13.
Assertion (A): Fibrous protein is a fibre-liked structure formed by the polypeptide chain. These proteins are held together by strong hydrogen and disulphide bonds.
Reason (R): It is usually soluble in water.
Answer:
(c) A is true, but R is false.
Explanation: The proteins which consist of linear, thread-like polypeptide chains arranged or twisted to form fibers are called fibrous proteins. The long chain in fibrous protein is held together by hydrogen bonds and some disulphide bonds. They are insoluble in water. Hence, assertion is true but reason is false.

Question 14.
Assertion (A): Transition metals and their many compounds act as good catalyst.
Reason (R): It is due to variable magnetic property.
Answer:
(c) A is true, but R is false.
Explanation: The transition metals and their compounds are known for their catalytic activity. This is because of their ability to adopt multiple oxidation states and to form complexes. Also, transition metals sometime form unstable intermediate complex which provide a new path with lower activation energy for the reaction. Transition elements provide a suitable surface for the reaction to take place. Since, the transition metal ions can change their oxidation states, they become more effective as catalysts. Thus, assertion is true but reason is false.

Question 15.
Assertion (A): To obtain maximum work from a galvanic cell, charge has to be passed reversibly.
Reason (R): The reversible work done by a galvanic cell is equal to decrease in its Gibbs energy.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
Explanation: Electrical work done in one second is equal to electrical potential multiplied by total charge. If we want to obtain maximum work from a galvanic cell then charge has to be passed reversibly. The reversible work done by a galvanic cell is equal to decrease in its Gibbs energy and therefore, if the emf of cell is E and «F is the amount of charge passed and AG is the Gibbs energy of reaction then,
ΔG = -nFE_cell

Question 16.
Assertion (A): Ethanal is soluble in water.
Reason (R): Lower aldehydes form effective hydrogen bond with water molecules.
Answer:
(a) Both A and R are true and R is the correct explanation of A.
Explanation: : Ethanal is soluble in water in all proportions. The reason for the solubility is that although aldehydes cannot hydrogen bond with themselves, but lower aldehydes can hydrogen bond with water molecules. Thus, both assertion and reason is correct and reason is the correct explanation of assertion.

SECTION – B (10 Marks)

This section contains 5 questions with internal choice in one questions. The following questions are very short answer type and carry 2 marks each.

Question 17.
(a) What is meant by ‘limiting molar conductivity’?
(b) Express the relation between conductivity and molar conductivity of a solution held in a cell.
Answer:
(a) The molar conductivity of a solution at infinite dilution is called limiting molar conductivity and is
represented by the symbol ∧_m.
(b) The molar conductivity of a given solution is related to conductivity as:
∧_m = \( \frac{\mathrm{K}}{\mathrm{C}}=\frac{\text { Conductivity }}{\text { Concentration }}\)

Question 18.
Write the IUPAC name of :

Answer:

Answer the following questions:

Question 19.
(a) La (5d¹6s²), Gd (4f⁷d¹6s²) and Lu (4f¹⁴5d¹6s²) have abnormally low third ionisation enthalpies. Why?
(b) Which element among 3d transition elements exhibits the highest oxidation state?
Answer:
(a) La³⁺, Gd³⁺ and Lu³⁺ have stable configuration 4f⁰, 4f⁷ and 4f¹⁴ respectively. So, they prefer the
formation of low energy +3 states and hence third ionisation enthalpies are low.
(b) Among 3d transition elements Manganese (Mn) exhibits the highest oxidation state of (+7).

Question 20.
Write any three differences between potential difference and EMF.

S.No	Potential Difference	EMF
1.	It is difference between electrode potential of two electrodes when no current is flowing through circuit.	It is difference of potential between electrode in a closed circuit.
2.	It is the maximum voltage obtained from a cell.	It is less than maximum voltage obtained from a cell.
3.	It is responsible for a steady flow of current.	It is not responsible for a steady flow of current.

Question 21.
A 6.90 M solution of KOH in water contains 30% by mass of KOH. Calculate the density of the KOH solution. (Molar mass of KOH = 56 g/mol).
Answer:
6.90 M = 6.9 moles of KOH in 1000 ml solution.
i.e., 6.9 x 56 g = 386.4 of KOH in 1000 ml solution.
30% of mass of KOH means, 30 g of KOH in 100 g of solution. 100
Hence, 386.4 g of KOH in \( \frac{100}{30}\) x 386.4 g= 1288 g of solution.
Density = \( \frac{\text { Mass }}{\text { Volume }}=\frac{1288 \mathrm{~g}}{1000 \mathrm{ml}}\) = 1.288 = g/ml of solution

SECTION – C (21 Marks)

This section contains 7 questions with internal choice in one questions. The following questions are short answer type and carry 3 marks each.

Question 22.
(a) On the basis of crystal field theory, write the electronic configuration for d⁴ ion if Δ₀ < P.
(b) [NiCl₄]^2- is paramagnetic, while [Ni(CO)₄] is diamagnetic, though both are tetrahedral. Why? (Atomic number of Ni = 28)
Answer:
(a) On the basis of crystal field theory, for a d⁴ ion, if A0 < P, then the complex is a high spin complex formed by association of weak field ligands with the metal ion. As a result, the fourth electron enters one of the e₈ orbitals, thereby, exhibiting the electronic configuration t_2g³e₈¹.
(b) Though both [NiCl₄]^2- and [Ni(CO)₄] are tetrahedral, their magnetic characters are different. This is due to a difference in the nature of ligands. Cl^– is a weak field ligand and it does not cause the pairing of unpaired 3d electrons. Hence, [NiCl₄]^2- a paramagnetic.

But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electons. Also, it causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp³ hybridization. Since no unpaired electrons are present in this case. [Ni(CO)⁴] is diamagnetic.

Question 23.
Attempt any two.
(a) Which type of compounds show linkage isomerism? What would be the structural formula for linkage isomer of [CO(NH₃)₅(NO₂)]Cl₂ ?
(b) How are the inner orbital complexes formed?
(c) What is a chelate ligand? Give one example.
Answer:
(a) Linkage isomerism arises in a coordination compound containing ambidentate ligand. Thestructural formula for linkage isomer of [CO(NH₃)₅(NO₂)]Cl₂ would be [CO(NH₃)₅(ONO)]Cl₂.
(b) Inner orbital complexes are coordination compounds composed of a central metal atom having hybridization of the atomic orbitals including d-orbitals of inner shell and s, p-orbitals from the outer shell. In other words, the central metal atom of these complexes uses inner shell d-orbitals for the hybridization of atomic orbitals. Therefore, these d-orbitals are in a lower energy level than s and
p-orbitals. The most common hybridization of the metal atom in inner orbital complexes is d²sp³. But there can be some other hybridizations as well, such as d²sp².
(c) The ligands which can form two co-ordinate covalent bonds through two donor atoms are called
bidentate ligands. These bidentate ligands are also called chelate ligands.
E.g.: C₂O₄^-2,CO₃^-2 etc.

Question 24.
(a) What is Hinsberg reagent?
(b) Complete the following equations:

Answer:

Question 25.
Answer following questions :
(a) What is Lanthanoid contraction? What is the cause of Lanthanoid contraction?
(b) Explain why Cu²⁺ ion is stable in aqueous solution.
(c) Trivalent Lanthanoid ions are coloured, why?
Answer:
(a) The gradual decrease in atomic radii from La to Lu due to poor screening effect of 4f-electrons is called Lanthanoid contraction.
1. Lanthanoid contraction of 4f-electron.
2. Increased effective nuclear charge.
(b) In an aqueous solution, Cu²⁺ is more stable than Cu⁺. Hydration enthalpy of Cu²⁺ ion is more negative which more than compensates for the second ionisation enthalpy of Cu.
(c) Trivalent lanthanoids are coloured due to the presence of f -electrons which causes f-felectronic transition.

Question 26.
The electrical resistance of a column of 0.05 M NaOH solution of diameter 1 cm and length 50 cm is 5.55 x 10³ ohm. Calculate its:
(a) Resistivity
(b) Conductivity
(c) Molar conductivity
Answer:
(a) Resistivity:
A = nr² = 3.14 (0.5)² cm² = 0.785 cm²
l = 50 cm

Question 27.
This important class of organic compounds are derived by replacing one or more hydrogen atoms of ammonia molecule by alkyl or aryl group(s). The nitrogen atom in this class of compound is trivalent. There are many important natural and synthetic compounds of this class. A lot of important drugs also belong to this class of compounds.
(a) Name the class of compounds being discussed. What is the hybridisation state of nitrogen in these class of compounds?
(b) What is ammonolysis?
(c) What are the compounds obtained on?
Answer:
(a) The class of compounds here is ‘amines’. Hybridisation state of nitrogen in these class of compounds is sp³.
(b) The process of cleavage of C-X bond of an alkyl halide molecule by ammonia molecule is known as ammonolysis. The halogen atom of alkyl or benzyl halide gets replaced by an amino (-NH₂) group. The primary amine thus obtained behaves as a nucleophile and in subsequent reactions, secondary amine, tertiary amine and quaternary ammonium salts are produced.

Question 28.
Given below is the sketch of a plant for carrying out a process.

(a) Name the process occurring in the above plant.
(b) Name one SPM which can be used in this plant.
(c) Give one practical use of the plant.
Answer:
(a) Reverse osmosis
(b) Film of cellulose acetate
(c) This can be used as a desalination plant to meet potable water requirements.

SECTION – D (8 Marks)

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.

Question 29.
The reaction of ethanol with cone. H₂SO₄ was carried out under different thermal conditions and following results are obtained as furnished below:

Reactant	Temperature	Product
Ethanol + Cone. H2SO4	25°C (Room temperature)	CH3CH2OSO3H (Ethyl hydrogen sulphate)
Ethanol + Cone. H2SO4	140°C	CH3-CH2-O-CH2-CH3 (diethyl ether)
Ethanol + Cone. H2SO4	170°C	CH2 = CH2 (Ethylene)

Answer the following questons:
(a) Predict the product when ethanol is treated with cone. H₂SO₄ at 0°C.
(b) At 170°C, the product formed in the table follow substitution or elimination pathway?
(c) Write the reaction mechanism involved in the above mentioned reaction at 170°C.

OR

Write the reaction mechanism involved in the above mentional reaction at 140°C.
Answer:
(a) CH₃CH₂OH₂⁺HSO₄^–(Ethyl oxonium hydrogen sulphate)
(b) Elimination pathway is followed by the product formed at 170°C
(c) H₂SO₄ → H⁺ + HSO₄^–

Question 30.
Water solubility of some organic compounds alongwith some carbohydrates were investigated and observations are furnished as follow:

Organic compound	Solubility in H2O
Glucose	✓
Sucrose	✓
Benzene	✗
Cyclohexane	✗

Answer the following questions:
(a) How many -OH groups are present in a sucrose molecule?
(b) What happens when glucose is treated with dilute NaOH?
(c) Why Glucose or Sucrose is water soluble?

OR

Why cyclohexane or benzene is water insoluble?
Answer:
(a) Eight -OH groups are present in the sucrose molecule.
(b) When glucose is treated with dilute NaOH, it forms a mixture of D-glucose, D-Fructose and D-Mannose due to reversible isomerisation.
(c) There are five -OH groups present in the glucose molecule and right -OH groups in the sucrose molecule. These -OH groups link to water molecules strongly by hydrogen bonds. That is why glucose and sucrose are water soluble.

OR

Cylohexane and benzene are hydrocarbons lacking any -OH group. So, these compounds can not form hydrogen bonds with water and hence, are water insoluble.

SECTION – E (15 Marks)

The following questions are long answer type and carry 5 marks each. All questions have an internal choice.

Question 31.
What happens when:
(a) Bromobenzene is treated with Mg in the presence of dry ether.
(b) Chlorobenzene is subjected to hydrolysis.
(c) Ethyl chloride is treated with aqueous KOH.
(d) Methyl bromide is treated with sodium in the presence of dry ether.
(e) Methyl chloride is treated with KCN.
(f) n-butyl chloride is treated with alcoholic KOH.
(g) Chloroform is heated with Ag powder.
Answer:
(a) When bromobenzene is treated with Mg in the presence of dry ether, phenylmagnesium bromide a Grignard’s reagent is formed.

Question 32.
An acid [A] C₈H₇O₂Br on bromination in the presence of FeBr₃ gives two isomers [B] and [C] of the formula C₈H₆O₂Br₂. Vigorous oxidation of [A], [B] and [C] gives acids [D], [E] and [F] respectively. The acid [D] C₇H₅O₂Br is the strongest acid among all the isomers whereas [E] and [F] each has molecular formula of C₇H₄O₂Br₂. Give structure of [A] to [F] with justification.

OR

(a) Account for the following:
(i) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction.
(ii) pK_a value of 4-nitrobenzoic acid is lower than that of benzoic acid.
(b) (i) Arrange the following compounds in increasing order of their boiling points.
CH₃CHO, CH₃CH₂OH, CH₃OCH₃, CH₃CH₂CH₃
(ii) Arrange the following in decreasing order of acid catalysed esterification:

Answer:
(i) From the formula C₈H₇O₂Br seems like a carboxylic acid. Formation of two isomers [B] and [C] of formula C₈H₆O₂Br₂ must have given an ortho and para isomer.

Compound [A] can be any of these I, II or III structures.
(ii) As vigorous oxidation of [A] along with acid [B] and [C] produces strongest acid [D] from structure [A] hence compound [A] must be structure [I].

[D] is strongest acid because of ortho-effect whereas presence of Br molecule at the meta position weakens the acidic strength.

OR

(a) (i) Aromatic carboxylic acids do not undergo Friedel-Crafts reaction because the carboxyl group is deactivating for electrophilic substitution reaction, secondarily, the catalyst aluminium chloride gets bonded to the carboxyl group.
(ii) pK_a value of 4-Nitrobenzoic acid is lower than benzoic acid, which means 4-nitrobenzoic acid is more acidic than benzoic acid. Being an electron with drawing group, the -NO₂ group withdraws electrons towards itself resulting in ease of carboxylic proton release, hence increasing the acidity.

(b) (i) The molecular masses of the given compounds are comparable CH₃CH₂OH undergoes extensive intermolecular H-bonding, resulting in the association of molecules. Therefore, it has the highest boiling point. CH₃CHO is more polar than CH₃OCH₃ and so CH₃CHO has stronger intermolecular dipole-dipole interaction than CH₃OCH₃.CH₃CH₂CH₃ has only weak Van der Waals force. Therefore, the increasing order of their boiling points is as:
CH₃CH₂CH₃ < CH₃OCH₃ < CH₃CHO < CH₃CH₂OH
(ii) As we know that in esterification, alcohol molecule attacks the carbonyl carbon protonated by acid. Presence of substitution in ortho position will cause steric hindrance to such attack in benzoic acid.
Hence, increasing order of reactivity is,
2, 6-Dimethyl benzoic acid < 2, 4-Dimethylbenzoic acid < benzoic acid.

(iii) The rate of hydrolysis depends on the electron deficiency of the carbonyl carbon of the ester group. Hence presence of electron withdrawing groups increases the rate of reaction whereas electron donating group decreases the rate of hydrolysis. Hence the decreasing order of rate for the given esters is.

Question 33.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:

t/s	0	10	20
[CH3COOCH3]/mol L-1	0.10	0.05	0.025

(a) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(b) Calculate the average rate of reaction between the time interval 10 to 20 seconds.
(Given: log 2 = 0.3010, log 4 = 0.6021)
(c) For a reaction A + B → P, the rate is given by Rate = k [A][B]².
1. How is the rate of reaction affected if the concentration of B is double?
2. What is the overall order of reaction if A is present in large excess?
(d) A first order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.

OR

(a) Why molecularity of a reaction cannot be zero?
(b) The decomposition of NH₃ on platinum surface is zero order reaction. What are the rates of production of N₂ and H₂ if k = 2.5 × 10^-4 mol L^-1 s^-1?
(c) For the reaction R P, the concentration of a reactant changes from 0.03 M to 0.02 M in 25 minutes.
Calculate the average rate of reaction using units of time both in minutes and seconds.
Answer:
(a) [A]₀ = 0.01 mol/L
[A] = 0.05 mol/L at time t = 10 s.

(c) 1. Since the given reaction has order two with respect to reactant B, thus if the concentration of B is doubled in the given reaction, then the rate of reaction will become four times.
2. If A is present in large excess, then the reaction will be independent of the concentration of A and will be dependent only on the concentration of B. Thus, the order of reaction will be 2.

OR

(a) Molecularity is the number of given reactant molecules or atoms that are colliding in the elementary reaction. Hence a minimum of one reactant molecule, atom or ion is required to initiate a chemical reaction. Hence, molecularity cannot be zero.
(b) The decomposition of NH₃ on platinum surface is represented by the following equation: