CBSE Sample Papers for Class 12 Chemistry Term 2 Set 5 with Solutions

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CBSE Sample Papers for Class 12 Chemistry Standard Term 2 Set 5 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.

Answer:

Question 2.
At 291 K molar conductivities at infinite dilution of NH₄Cl, NaOH, NaCl are 129.8,217.4,108.9 ohm^-1cm² respectively. If molar conductivity of normal solution of NH₄OH is 9.33 ohm^-1cm². What is the degree of dissociation of NH₄OH solution?
Answer:
∧⁰_NH4OH = ∧⁰NH₄₊ +∧⁰_OH^–
∴ ∧⁰_NH4OH = ∧⁰_NH4OHCl + ∧⁰_NaOH – ∧⁰_Nacl
= 129.8 + 217.4-108.9
= 238.3 ohm^-1 cm²
Given ∧_c = 9.33 ohm^-1 cm²
∴ Degree of dissociation, α = \(\frac{\Lambda_{c}}{\Lambda}\)
= \(\frac{9.33}{238.3}\) = 0.039.

Question 3.
For a reaction,
Cl₂(g) + 2NO(g) → 2NOCl(g),
The rate law is expressed as; rate = k[Cl₂][NO]². What is the order of the reaction?
Answer:
Rate law; R = k [Cl₂] [NO]²
order 1 +2=3
Hence, It is a third-order reaction.

Section – B

Question 4.
Following are the transition metal ions of 3d series:
Ti³⁺, V²⁺, Mn³⁺, Cr³⁺
(Atomic numbers : Ti = 22, V = 23, Mn = 25, Cr = 24)
Answer the following:
(a) Which ion is most stable in an aqueous solution and why?
(b) Which ion is a strong oxidising agent and why?
(c) Which ion is colourless and why?
OR
Complete the following equation :
(a) 2MnO₄^– + 16H⁺ + 5S^2-
(b) KMnO₄ img
(c) What is the general electronic configuration of transition elements?
Answer:
Ti⁴⁺ = 1s²2s²2p⁶3s²3p⁶
V²⁺ = 1s²2s²2p⁶3s²3p⁶3d³
Mn³⁺ = 1s²2s²2p⁶3s²3p⁶3d⁴
Cr³⁺ = 1s²2s²2p⁶3s²3d³
(a) Ti⁴⁺ is most stable in an aqueous solution because of fully filled valence shell (3s²3p⁶) configuration
(noble gas configuration).
(b) Mn³⁺ is the strong oxidising agent as it oxidises other species it will reduce itself by taking an e and
will stabilise its configuration (3d⁵).
(c) Ti⁴⁺ is colourless due to absence of unpaired electrons (3s²3p⁶).
OR
(a) 2MnO₄^– + 16H ⁺+ 5S^2- → 2Mn²⁺ + 5S + 8H₂O
(b)

(c) The general electronic configuration of transition elements is (n – 1)d^1-10ns².

Question 5.
For the complex [NiCl₄]^2-, write.
(a) The IUPAC name (b) Flybridization state of metal atom.
(c) The shape of the complex (At. no. of Ni = 28).
OR
Write down the IUPAC name for each of the following complexes and indicate the oxidation state elec¬tronic configuration and coordination number.
Also give magnetic moment of the complex.
(a) K[Cr(H₂O)₂(C₂O₄)₂,3H₂O]
(b) [CO(NH₃)5Cl]Cl₂
(c) CrCl₃(py)₃
Answer:
(a) tetrachioridonickelate (II) ion.
(b) Hybridization of Ni in the complex [NiCl₄]^2- is sp³.
The hybridisation scheme is shown in the following diagram.

Hence the hybridisation of Ni²⁺ in the complex is sp³.
As the hybridisation of Ni is sp³, so, the shape of the complex is tetrahedral.

OR
(a) K[Cr(H₂O)₂(C₂O₄)₂,3H₂O] JUPAC Name: Potassium diaquadioxalatochromate (Ill) trthydrate.
Oxidation state of chromium =3
Electronic configuration : 3d³ (t²_2g g², e_g⁶)
Coordination number =6
Shape : Octahedral
Magnetic moment, µ = \(\sqrt{n(n+2)}\)
= \(\sqrt{3(3+2)}\)
= \(\sqrt{15}\) = 3.87 cm
~ 4BM

(b) [Co(NH₃)₅Cl]Cl₂
IUPAC name: Pentaamminechloridocobalt(lII) chloride
Oxidation state of CO = +3
Coordination number = 6
Shape: Octahedral.
Electronic configuration: d⁶: (t_2g⁶)
Magnetic Moment:
µ = \(\sqrt{n(n+2)}\)
µ = \(\sqrt{\mathrm{O}(\mathrm{O}+2)}\)
= 0.
Since there are no unpaired electron thus the magnetic moment will be O.

(c) CrCl₃(py)₃
IUPAC name: Trichioridotripyridinechromium (III)
Oxidation state of chromium = +3
Electronic configuration for d³ : t_2g³
Coordination number =6
Shape: Octahedral.
Magnetic moment, µ = \(\sqrt{n(n+2)}\)
= \(\sqrt{3(3+2)}\)
= \(\sqrt{15}\) = 3.87 cm
~ 4BM

Question 6.
Answer the following questions:
(a) What is zeta potential? Explain.
(b) Why colloidal solutions differ in colour?
(c) What is added to gasoline to decrease knocking?
Answer:
(a) As colloidal particles adsorb a specific charge on their surface, to acquire either negative or positive charge, they attract further oppositely charged ions from the dispersed medium which forms a second layer of mobile opposite charge which surrounds the first fix layer. Such a double layer of opposite charges is called the Helmholtz-double layer, as shown in figure.

The mobile layer differences into the bulk of the liquid. The potential difference between the fixed layer and the mobile diffused layer is called electrokinetic potential or zeta potential.

(b) Colour of a colloidal solution depends upon the size and shape of dispersed phase and nature of dispersion medium. Hence, different colloidal solutions transmit different light of spectrum depending upon the variations and hence have different colours.
(c) Tetraethyl lead (TEL).

Question 7.
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 ammonolysis of ethyl chloride?
Answer:
(a) The class of compounds here is ‘amines’. The 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.
(c)

Question 8.
Account for the following:
(a) p-Nitrobenzoic acid has higher K_a than benzoic acid.
(b) Carboxylic acids have higher boiling points than alcohols.
(c) Acetone is soluble in water but benzophenone is not.
Answer:
(a) Para nitro benzoic acid is more acidic than benzoic acid because of the NO₂ functional group which is an electron-withdrawing group. Due to which the electron density on the hydrogen atom of the carboxylic acid becomes low, and it can be easily removed. Hence p-Nitrobenzoic acid is more acidic and have higher K_a value.

(b) Carboxylic acids have higher boiling point than alcohols due to more extensive association of carboxylic acid molecules through intermolecular hydrogen bonding. The hydrogen bond formed by the carboxylic acids are stronger than those in alcohols because O-H bond in -COOH is more strongly polarised due to the presence of electron-withdrawing carboxyl group in adjacent position than the O-H bonds of alcohols. Therefore, the boiling points of carboxylic acids particularly lower members, are higher than alcohol of comparable molecular masses.

(c) Acetone is highly soluble in water because it can undergo hydrogen bonding with water molecules because of the presence of polar carbonyl group. In case of benzophenone the carbonyl
group is sterically hindered by two big phenyl group (C₆H₅COC₆H₅), hence, the carbonyl oxygen is masked and cannot participate in hydrogen bonding with water.

Question 9.
Following data are obtained for reaction:
N₂O₅ → 2NO₂+ \(\frac{1}{2}\) O₂

(a) Shows that it follows first order reaction.
(b) Calculate the half-life.
(Given log 2 = 0.3010, log 4 = 0.6021):
OR
(The decomposition of phosphine PH₃, proceeds according to the following equation:
4PH₃(g) → P_4(g) + 6H_2(g)
Its found that the reaction follows the following rate equation.
Rate = k[PH₃]
The half-life period of PH₃ is 37.9 s at 120°C.
(a) How much time is required for 3/4th of the PH₃ to decompose?
(b) What fraction of the original sample of PH₃ remains behind after 1 minute?
Answer:
(a) k= \(\frac{2.303}{t} \log \frac{\left[\mathrm{A}_{0}\right]}{\mathrm{A}}=\frac{2.303}{300} \log \frac{1.6 \times 10^{-2}}{0.8 \times 10^{-2}}\)
= \(\frac{2.303}{300} \log ^{2}=2.31 \times 10^{-3} \mathrm{~s}^{-1} \)
= At 600 s, K = \(\frac{2.303}{t} \log \frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)
= \(\frac{2.303}{300} \log \frac{1.6 \times 10^{-2}}{0.4 \times 10^{-2}} \)
= \(\frac{2.303}{600} \log 4=2.31 \times 10^{-3} \mathrm{~s}^{-1}\)
Since K is constant when using first-order equation therefore it follows order kinetics.
(b) t_1/2 = \(\frac{0.693}{k}=\frac{0.693}{2.31} \times 10^{-3}\) = 300 s
Thus, the half life of the reaction is 300s.

OR
(a) As we can see from the rate equation that its a first-order reaction.
Initial concentration, let be a
Then concentration after time, t = \(\frac{3}{4}\)a = x
∴ t = \(\frac{2.303}{k} \log \frac{a}{a-x}\) also t_1/2 = \(\frac{0.693}{k}\)
= \(\frac{2.303}{\frac{0.693}{37.9}} \log \frac{a}{a-\frac{3}{4} a}\)
⇒ k = \(\frac{0.693}{37.9}=0.0183 \mathrm{~s}^{-1}\)
= \(\frac{2.303 \times 37.9}{0.693} \log 4\) [log 4 = 0.6021]
= \(\frac{2.303 \times 37.9}{0.693} \times 0.6021\) = 75.83 sec

(b) k = \(\frac{2.303}{t} \log \frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\)
t =60 s

Thus, \(\frac{\left[\mathrm{A}_{0}\right]}{[\mathrm{A}]}\) = 2.99.

Question 10.
(a) State the products of electrolysis obtained on the cathode and anode in the following:
(a) Dilute solution of H₂SO₄ with Pt electrodes.
(b) Aqueous solution of AgNO₃ with Ag electrodes.
(b) Write the cell formation and calculate the standard cell potential of the galvanic cell for the following
reaction:
Fe²⁺(aq) + Ag⁺(aq) → Fe³⁺(aq) + Ag(s)
Also, calculate ΔG° for the above reaction.
Given: [E°_Ag⁺/Ag = + 0.80 V, E°_{Fe³⁺/Fe²⁺} = 0.77 V] 1 F = 96500 C mol^-1.
Answer:
(a)
(i) Dilute solution of H₂SO₄ with Pt electrode have ions, H⁺, SO₄^2- and OH^– in it as Pt is an inert electrode.
At cathode (Reduction) : 2H⁺(aq) + 2e^– → H_2(g)
At anode (Oxidation) : 2OH^–_(aq) → O_2(g) + 2H⁺_(aq) + 4e^–
SO₄^2- will remain in solution as it has a higher discharge potential than OH^–.
So, H₂ gas is evolved at cathode and O² gas is evolved at anode.

(ii) Aqueous solution of AgNO₃ with Ag electrode.
Ions present are: Ag⁺, H⁺, OH^–, NO₃^– and Ag atom.
At cathode (Reduction): Ag⁺ _(aq) + e^– → Ag_(s)
At anode (Oxidation): Ag_(s) → Ag⁺_(aq) + e^–
As Ag has highest discharge potential compared to all other ions hence it gets dissolved as Ag⁺_(aq) ion.
Hence at cathode we get solid Ag deposit and at anode the silver electrode dissolves.

(b) pt, Fe²⁺| Fe³⁺_(aq) || Ag⁺_(aq)|Ag_(s)
E°cell = E°_Ag⁺/Ag – E° _{Fe³⁺/Fe²⁺}
= + 0.80 V — (+ 0.77 V)
ΔG°= — nFE°_cell n = 1
ΔG° = 1 x 96500 C mol^-1 x 0.03 V
= -2895 J mol^-1
= – 2.895 kJ/mol^-1.

Question 11.
(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)

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

OR
Compound [A] ves iodoform reaction henœ it must have a (CH₃CO) group. must be [A].
Hence, the reactions are:

Aldol condensation of [D] i.e., ethanol occurs.

Section – C

Question 12.
Read the passage given below and answer the questions that follow:
Both aldehydes and ketones are collectively termed as carbonyl compounds as they contain carbonyl
group.
Aliphatic aldehydes and ketones both are expressed by the general formula C_nH_2nO. For example both propionaldehyde (CH₃CH₂CHO) and action (CH₃COCH₃) can be represented by the molecular formula C₃H₆O.
The simplest aromatic aldehyde is benzaldehyde.
(a) What happens when acetaldehyde is reacted with a trace of H₂SO₄.
(b) Which type of aldehydes undergo aldol condensation?
(c) Write one chemical reaction to exemplify Cannizzaro reaction.
(d) Which aldehyde is a gas? Why it is soluble in water?
OR
(d) Write the mechanism of addition of HCN to >C = O group.
Answer:
(a) Acetaldehyde on reaction with trace of H₂SO₄ gives a cyclic trimer compound known as
paraldehyde.

(b) All the aldehydes which have a-hydrogen atom undergoes aldol condensation.

(d) Methanal is a gas. It is soluble in water because it form H-bond with water molecules.
OR

Mechanism for the addition of HCN to ethanal can be represented as follows:
Step I: Nucleophilic addition of CN^–

Step II: Addition of H⁺ to O^–