CBSE Sample Papers for Class 12 Chemistry Term 2 Set 3 with Solutions

Students can access the CBSE Sample Papers for Class 12 Chemistry with Solutions and marking scheme Term 2 Set 3 will help students in understanding the difficulty level of the exam.

CBSE Sample Papers for Class 12 Chemistry Standard Term 2 Set 3 with Solutions

Max. Marks: 35
Time: 2 Hours

General Instructions:
Read the following instructions carefully:

• There are 12 questions in this question paper with internal choice.
• Section A – Q. No. 1 to 3 are very short answer questions carrying 2 marks each.
• Section B – Q. No. 4 to 11 are short answer questions carrying 3 marks each.
• Section C- Q. No. 12 is case-based question carrying 5 marks.
• All questions are compulsory.
• Use of log tables and calculators is not allowed

Section – A

Question 1.
Explain Perkin’s reaction.
Answer:
The Perkin reaction is an organic reaction developed by English Chemist William Henry Perkin in the year 1868. i.e., used to make cinnamic acids. The reaction belongs to the Carbonyl compounds and it gives an α,β-unsaturated aromatic acid by the aldol condensation of an aromatic aldehyde and an acid anhydride, in the presence of an alkali salt of the acid. Perkin used this reaction to make cinnamic acids. In the reaction to make cinnamic acid, benzaldehyde (aromatic aldehyde) was treated with acetic anhydride in the presence of sodium acetate (weak base) to give cinnamic acid. The reaction can be represented as follows:

Question 2.
Answer the following questions:
(a) What is the relationship between cell potential and equilibrium constant?
(b) Why hydrogen is used as a Standard electrode?
(c) What is the effect of catalyst on Gibbs energy (ΔG)?
Answer:
(a) At equilibrium, the value of cell potential is zero.
(b) Hydrogen is used as a standard electrode because it is an inert electrode and its reduction potential is 0.
(c) There will be no effect of catalyst on Gibbs energy.

Question 3.
Answer the following questions:
(a) Tertiary amines do not undergo acylation reaction.
(b) o-Toluidine is less basic than aniline
Answer:
(a) Tertiary amines cannot undergo acylation as they do not have any replaceable hydrogen atom. Tertiary amines thus are non-reactive towards acylation.

(b) o-Toluidine is less basic than aniline because of ortho effect. The ortho position substitution makes it difficult for the – NH₂ group to donate electrons in spite of the electron-donating tendency of (CH₃) methyl group as a substituent.

Section – B

Question 4.
(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)
OR
What is the electronic configuration of Co in [CoF₆]^3- according to the valence bond theory of coordination compounds. Also, predict the magnetic character of the [CoF₆]^3-.
Answer:
(a) On the basis of crystal field theory, for a d₄ ion, if Δo < P, then the complex is a high spin complex formed by the association of weak field ligands with the metal ion. As a result, the fourth electron enters one of the eg orbitals, thereby, exhibiting the electronic configuration t_2g3e_g¹.

(b) In [NiCl₄]^2-, Ni is in the +2 state and Cl^– which is a weak field ligand does not cause pairing of unpaired 3d electrons. Hence, it is paramagnetic. On the other hand, in [Ni(CO)₄], Ni has 0 oxidation state and CO is a strong field ligand which causes pairing of unpaired 3d electrons. Since no unpaired electrons are present, it is diamagnetic.
OR
Electronic configuration of Co³⁺ ion is: 3d

Electronic configuration of sp³d² hybridised (as F- is a weak field ligand) orbitals of Co³⁺’, with six pairs of electrons from six F- ions:

As there are 4 unpaired electrons, the [CoF₆]^3- compound is paramagnetic.

Question 5.
From the following ions which are given:
Cr²⁺, Cu²⁺, Cu⁺, Fe²⁺, Fe³⁺, Mn³⁺
Identify the ion which is:
(a) A strong reducing agent.
(b) Unstable in aqueous solution.
(c) A strong oxidising agent.
OR
Write down the electronic configuration o(:
(a) Cr³⁺
(b) Cu⁺
(c) Ce ⁴⁺
(d) Co²⁺
(e) Mn²⁺
(f) Lu²⁺
Answer:
(a) Cr²⁺ is a strong reducing agent as it is very easily oxidized to more stable Cr³⁺ (d³) where it attains a stable half-filled t_2g configuration.
(b) Cu⁺ is unstable in an aqueous solution because it disproportionates in water to form Cu²⁺ and Cu.
(c) Mn³⁺ is a strong oxidising agent because it gets reduced to Mn²⁺ (d⁵) in the process and attains an
extra stable half-filled d orbital configuration.
Or

Question 6.
Complete the following equations:

Answer:

Question 7.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained:

(a) Show that it follows a 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: log2 = 0.3010, log4 = 0.6021)
(c) A first-order reaction takes 30 minutes for 50% completion. Calculate the time required for 90% completion of this reaction.
Answer:
(a) [A]₀ = 0.01 mol/L
[A] = 0.05 mol/L at time t = 10 s.
k = \(\frac{2.303}{t} \log \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]}\)
We know,
k = \(\frac{2.303}{10} \log \frac{0.10}{0.05}\)
k = 0.0693 s^-1
t =20 s
And k = \(\frac{2.303}{t} \log \frac{[\mathrm{A}]_{0}}{[\mathrm{~A}]}=\frac{2.303}{20} \log \frac{0.10}{0.025} \)
k = 0.0693 s^-1
Thus, its pseudo first order reaction.

(b) The average rate constant is
\(\frac{-\Delta[\mathrm{R}]}{\Delta t}=\frac{\Delta\left[\mathrm{CH}_{3} \mathrm{COOCH}_{3}\right]}{\Delta t}\)
= \( \frac{\left[0.025 \mathrm{~mol} \mathrm{~L}^{-1}-0.05 \mathrm{~mol} \mathrm{} \mathrm{L}^{-1}\right]}{20 \mathrm{~s}-10 \mathrm{~s}}\)
= \(\frac{0.025}{10 s}\) mol L^-1 s^-1.
(c) We have t_1/2 =30 min.
[R] = [R]₀ – 90% of [R]₀
=[R]₀ – \(\frac{90[\mathrm{R}]_{0}}{100}\)

[R] = \(\frac{[\mathrm{R}]_{0}}{10}\)
According to the first order, k =\(\frac{0.693}{t_{1 / 2}}=\frac{0.693}{30}\) =0.0231 min^-1
t = \(\frac{2.303}{\mathrm{~K}} \log \frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}\)
t = \(\frac{2.303}{0.0231} \log \frac{[\mathrm{R}]_{0}}{\frac{[\mathrm{R}]_{0}}{10}}=\frac{2.303}{0.0231} \log 10\)
t =99.7mm.

Question 8.
Write the complete reaction for the following conversions mentioning the necessary conditions:
(i) Toluene to benzaldehyde
(ii) Aldelyde to acetal
Carbonyl carbon is an electrophilic centre whereas carbonyl oxygen is a nucleophilic centre. Example.
Predict the product.

(a)

(ii)

(b) Carbonyl carbon is an electrophilic centre where as carbonyl oxygen is a nucleophilic centre because of high electronegativity, electron density shifts towards oxygen and carbon becomes electron-deficient and acts as electrophilic centre whereas oxygen acts as nucleophilic centre.
(c)

Question 9.
(a) Write the reaction indicating the conditions for the following conversions.
(i) Benzaldelyde to Benzyl alcohol
(ii) Acetaldelyde to acetone
(iii) Propyne to propanone.
(b) Explain the following facts:
Treatment of benzaldehyde with HCN gives a mixture of two isomers which cannot be separated even by careful fractional distillation.
OR
An organic compound [A] having molecular formula C₃H₆O, gives iodoform reaction and forms a compound [B]. [B] on heating with Ag powder, gives compound [C]. [C] reacts with dil. H₂SO₄ and mercuric sulphate to obtain compound [D]. Compound [D] undergoes Aldol condensation. Write down the names and structures of all the compounds starting from [A] to [D] with the help of chemical equations.
Answer:
(a)
(i)


(b) Benzaldehyde on treatment with HCN gives benzaldehyde cyanohydrin which is a chiral molecule and exists as two optical isomers which can not be separated by fractional distillation.

OR
Compound [A] gives iodoform reaction hence it must have a must be [A].
Hence, the reactions are:

Question 10.
Answer the following questions:
(a) What is the use of emulsion in photography?
(b) Colloidal medicines are more effective. Why?
(c) What is the difference between “Sol” and “gel”?
Answer:
(a) Emulsion is a light-sensitive material which is used to coat film and photographic paper so that images can be made to appear. The emulsion consists of light-sensitive crystals which are suspended in gelatin, creating a mixture which can be evenly applied to a base such as paper, glass, celluloid, or fabric.
(b) Medicines are more effective in colloidal state because colloids have a larger surface area. Thus, they easily gets assimilated, absorbed and digested in our body.
(c)

Sol	Gel
1. The liquid state of a colloidal solution is called sol.	The solid or semi solid-state of a colloidal solution is called gel.
2. The sol does not have a definite structure.	The gel possesses honeycomb-like structure.
3. The dispersion medium of the sol may be water.	The dispersion medium of gel will be hydrated colloid particles.
4. The sol can be converted to gel by cooling.	The gel can be converted to sol by heating.
5. The sol can be easily dehydrated.	The gel cannot be dehydrated.
6. The viscosity of the sol is very low.	The viscosity of the gel is very high.
7. Sol is categorized into lyophobic and lyophilic sols. Example: Blood	There is no such classification of gel Example: Fruit jelly, cooked gelatin gelly.

Question 11.
FeSO₄ solution mixed with (NH₄)₂SO₄ solution in 1: 1 molar ratio gives the test of Fe²⁺ ion but CuSO₄ solution mixed with aqueous ammonia in 1: 4 molar ratio does not gives the test of Cu²⁺ ion. Explain why?
OR
The crystal field theory assumes that the interaction between the metal ion and ligand is purely electrostatic. When ligands approach central metal atom/ion the five degenerated orbitals of the central metal atom becomes differential. In a complex, the central metal atom or ion is surrounded by various atoms or groups of atoms called ligands.
(a) What is the type of bond in metal complex according to crystal field theory?
(b) What is the number of ligands in Cu₂ [Fe(CN)₆].
(c) Which one is lower energy set between t_2g and eg, after splitting of five degenerate orbital or metal atom/ion on complexation in tetrahedral complex?
(d) Define crystal field splitting.
(e) What do you mean by weak field ligands?
Answer:

Both the compounds i.e., FeSO₄.(NH₄)₂SO₄.6H₂O and [Cu(NH₃)₄]SO₄.5H₂O fall under the category of
addition compounds. But Mohr’s salt is an example of a double salt, while the latter is a coordination compound.

A double salt is an addition compound that is stable in the solid-state but it dissociates up into its constituent ions in the dissolved state. Hence will give test of each individual constituent ions. For e.g., FeSO₄.(NH₄)₂SO₄.6H₂O breaks into Fe²⁺, NH⁴⁺ and SO₄^2- ions. Hence, it gives a positive test for Fe²⁺ ions.

A coordination compound is an addition compound which retains its identity in the solid as well as in the dissolved state. However, the individual properties of the constituents are lost. In [Cu(NH₃)₄]SO₄.5H₂O
does not show the test for Cu²⁺ because Cu²⁺ ion is present inside the complex entity, [Cu(NH₃)₄]²⁺’.
OR
(a) Ionic bond or electrostatic bond.
(b) Number of ligands 6.
(c) Lower energy set = cg.
(d) The splitting of degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting.
(e) If energy separation for octahedral crystal field is less than energy required for electron pairing
energy in single orbital, then fourth electron enters eg orbital. So, ligands whose D₀ < P are weak field ligands.

Section – C

Question 12.
Read the passage given below and answer the questions that follow:
The potential of an electrode at a given temperature depends upon the concentration of the ions in the surrounding solution.
If the concentration of ions is unity and the temperature is 25°C, the potential of the electrode is termed as the “Standard Electrode Potential (E°cell)”. For example, the standard electrode reduction potential for a substance having half cell reaction, Cu⁺² + 2e^– → Cu, has the value = + 0.34 V.
(a) What is electrode?
(b) What is the relation between free energy change and emf of a cell?
(c) Which type of reaction occurs at anode and cathode during electrolysis?
(d) Which substance is used as electrode in a dry cell?
OR
(d) Mention the applications of electrochemical series.
Answer:
(a) An electrode is a solid electric conductor that carries electric current into non-metallic solids, or liquids, or gases.
(b) The maximum cell potential is directly related to the free energy difference between the reactants and the products in the cell.
ΔG° = -nFε⁰
ε⁰ = Volts = Work (J)/ charge (c)
F = Faraday = 96,485 coulombs of charge per mole of electrons
n = number of moles transferred per mole of reactant and product
(c) Oxidation occurs at anode and reduction occurs at cathode during electrolysis.
(d) Carbon rod as cathode and zinc (Zn) cup as anode.
OR
(d) Applications of electrochemical series are:
1. Predicting feasibility of redox reaction.
2. Predicting the capability of a metal to displace H₂ gas from acid.
3. Higher reduction potentials are strong oxidising agents while lower reduction potentials are strong reducing agents.