CBSE Sample Papers for Class 12 Chemistry Term 2 Set 4 with Solutions

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CBSE Sample Papers for Class 12 Chemistry Standard Term 2 Set 4 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.
Prove that for a first-order reaction the time required for 99.9 % completion is about 10 times its half-life period.
Answer:
We know that,
Formula: k = \(\frac{2.303}{t} \log \frac{a}{a-x}\)
t_1/2 = \(\frac{0.693}{k}\)
Now, t_99.9 = \(\frac{2.303}{t} \log \frac{100}{100-99.9}\)
= \(\frac{2.303}{k} \log \frac{100}{0.1}\)
= \(\frac{2.303}{k} \log 1000\)
t_99.9 = \(\frac{2.303 \times 3}{k}\)
t_99.9 = \(\frac{6.909}{k}\)
∴ t_1/2 = \(\frac{0.693}{k}\)
∴ k = \(\frac{0.693}{t_{1 / 2}}\)

Time required for 99.9% completion of the reaction = 10 t_1/2

Question 2.
Classify the following amines as primary, secondary or tertiary:

Answer:
Primary (a) and (c)
Secondary: (d)
Tertiary: (b).

Question 3.
Write IUPAC names for the following:

Answer:
(a) Ethyl4-Bromobenzoate
(b) Methyl 2-amine-butyrate/butanoate

Section – B

Question 4.
Answer the following questions:
(a) Why Zn, Cd and Hg are not regarded as transition metals?
(b) Explain why potassium dichromate is used as primary standard in volumetric analysis?
(c) What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements: 29, 59, 74, 95,102,104.
OR
Answer the following questions:
(a) Name two oxo-metal onions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.
(b) What are the different oxidation states exhibited by the lanthanoids?
(c) Complexes of which transition metals are useful in the polymerisation of alkynes?
Answer:
(a) Transition elements are characterised by partially filled (n-1) d subshells and Zn, Cd and Hg are
not regarded as transition elements as they have completely filled (n-1)d subshells. This can be
explained as follows:
1. The electronic configuration of zinc is [Ar]3d¹⁰4s².
2. The electronic configuration of cadmium is [Kr]4d¹⁰5s².
3. The electronic configuration of mercury is [Xe]5d¹⁰6s².
From the above electronic configuration, it is clear that zinc, cadmium and mercury have completely
filled (n-1)d sub-shell. Therefore, they are not considered as transition elements.

(b) K₂Cr₂O₇ is used as a primary standard in volumetric analysis because of the following reasons:
(i) It is available in pure state.
(ii) It is non-hygroscopic in nature.
(iii) It does not decolourise in solution.
(c) Inner transition metals are those elements in which the last electron enters the f-orbital. The elements in which the 4f and 5f orbitals are progressively filled are called f-block elements. The lanthanoids
from atomic number (Z = 58 to 71) and actinoids from atomic number (Z = 90 to 103). Among the given atomic numbers, the atomic numbers of the inner transition elements are 59,95 and 102.
OR
(a) MnO₄^–: Mn has +7 oxidation state and group number is 7.
CrO₄^2-: Cr has + 6 oxidation state and group number is 6.
(b) In the lanthanide series, +3 oxidation state is most common i.e., Ln(III) compounds are predominant. However, +2 and +4 oxidation states can also be found in the solution or in solid compounds.
(c) Nickel (Ni) metal is useful in the polymerisation of alkynes.

Question 5.
Answer the following:
(a) Write the formula of the following coordination compound – Iron (III) hexacyanoferrate (II)
(b) Predict the number of unpaired electrons in [MnBr₄]^2-.
(c) Write the hybridisation and number of unpaired electrons in the complex [CoF₆]^3-.
(Atomic No. of Co = 27).
OR
[Fe(H₂O)₆]³⁺ is strongly paramagnetic whereas [Fe(CN)₆]^3- is weakly paramagnetic. Explain.
Answer:
The molecular formula of Iron(III) hexacyanoferrate(II) is Fe₄[Fe(CN)₆]₃.
[MnBr₄]^2- Mn has (+ 2) oxidation state hence 4s⁰3d⁵ configuration. Br^– is a weak field ligand, five unpaired electrons. So, highly paramagnetic.

So highly paramagnetic.
(c) Electronic configuration of Co³⁺ ion is,

Electronic configuration of sp³d³ hybridized (as F^– is a weak field ligand) orbitais of Co³⁺, with six
pairs of electrons from six F^– ions:

There are 4 unpaired electrons in [COF₆]³.
OR
In both [Fe(H₂O)₆]³⁺ and [Fe(CN)₆]^3-, Fe exists in the +3 oxidation state i.e., in d⁵ configuration.

Since CN^– is a strong field ligand, it causes the pairing of unpaired electrons. Therefore, there is only
one unpaired electron left in the d-orbital.

Therefore,
µ = \(\sqrt{n(n+2)}\)
= \( \sqrt{1(1+2)}\)
= \(\sqrt{3}\)
= 1.732 BM
On the other hand, H₂O is a weak field ligand. Therefore, it cannot cause the pairing of electrons.
This means that the number of unpaired electrons is 5.

Therefore,
µ = \(\sqrt{n(n+2)}\)
= \(\sqrt{5(5-2)}\)
= \(\sqrt{35}\)
= 5.916 BM
Hence, [Fe(H₂O)₆]³⁺ is strongly paramagnetic, while [Fe(CN)₆]^3- is weakly paramagnetic.

Question 6.
Explain giving reasons:
(a) Transition metals and many of their compounds show paramagnetic behaviour.
(b) The enthalpies of atomisation of the transition metals are high.
(c) The transition metals generally form coloured compounds.
Answer:
(a) Transition metals show paramagnetic behaviour due to the presence of unpaired electrons in the d-orbitals. Each electron having a magnetic moment associated with its spin angular momentum. However, in the first transition series, the orbital angular momentum is quenched. Therefore, the resulting paramagnetism is only because of the unpaired electron.

(b) Transition elements have large number of d-shell valence electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of atomisation of transition metals is high.

(c) Most of the complexes of transition metals are coloured. This is because of the absorption of radiations from visible light region to promote an electron from one of the d-orbitals to another i.e., d-d transition. In the presence of ligands, the d-orbitals split up into two sets of orbitals having different energies. Therefore, the transition of electrons can take place from one set to another. The energy required for these transitions is quite small and falls in the visible region of radiation. Hence reflected radiation is visible in the form of coloured solutions.

Question 7.
What are lyophilic and lyophobic sols? Give one example of each type. Why hydrophobic sols are easily coagulated?
Answer:
Lyophiic sols: Colloidal sols that are formed by mixing substances such as gum, gelatin, starch, etc., with a suitable liquid (dispersion medium) are called lyophilic sols. These sols are reversible in nature i.e., if two constituents of the sol are separated by any means (such as evaporation), then the sol can be prepared again by simply mixing the dispersion medium with the dispersion phase and shaking the mixture.

Lyophobic sols: When substances such as metals and their sulphides etc., are mixed with the dispersion medium, they do not form colloidal sols. Their colloidal sols can be prepared only by special methods. Such sols are called lyophobic sols. These sols are irreversible in nature. For example: SoIs of metals, sols of Au and Ag.

Now, the stability of hydrophilic sols depends on two things the presence of a charge and the solvation of colloidal particles. On the other hand, the stability of hydrophobic sols is only because of the presence of a charge. Therefore, the latter are much less stable than the former. If the charge of hydrophobic sols is removed (by addition of electrolytes), then the particles present in them come closer and forms aggregates, leading to precipitation or coagulation.

Question 8.
A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed. (Given: log 2 = 0.3010, log 3= 0.4771, log 4= 0.6021)
Answer:
For first order reaction,
k = \(\frac{2.303}{t} \log \frac{a}{a-x}\)
where, a = initial amount, a — x = amount left after time t for 25% decomposition:
k = \(\frac{2.303}{20} \log \frac{100}{75}\) ……………… (1)
for 75% decomposition:
k= \(\frac{2.303}{t} \log \frac{100}{25}\) ………………………… (2)
k is constant throughout the process equation (1) = equation (2)
Thus, on comparing equation (1) and equation (2), we have

t = \(\frac{1.204}{0.125}\)
t = 9.632 minutes.

Question 9.
(a) Write the chemical reaction involved in Cannizzaro reaction.
(b) Write the reaction where aldehydes and ketones can be converted to alkanes on treatment with zinc
and hydrochloric acid. What is the reaction called?
(c) Predict the product of following reaction:
C₂H₅COCH₃ + NaOl →
OR
(a) What is aldol condensation?
(b) Arrange in decreasing order of acidity:
Benzoic acid, 4-Methoxy benzoic acid and 4-Nitrobenzoic acid
(c) Give IUPAC names of following compounds:
(i) OHCC₆H₄CHO
(ii) CH₃CO(CH₂)₄CH₃.
Answer:
(a) Cannizzaro’s reaction: In this reaction those aldehydes which do not have an a-hydrogen undergo self-oxidation and reduction (disproportionation) reaction on treatment with concentrated alkali. In this reaction one molecule of aldehyde undergoes reduction to alcohol while another molecule of aldehyde undergoes oxidation to carboxylic acid salt.

(b) Clemmensen’s reduction: When the carbonyl group of aldehyde and ketone is reduced to CH₂ group on treatment with zinc amalgam and concentrated hydrochloric acid then reaction is called Clemmensen’s reduction.


C₂H₅COCH₃ + NaOl→ C₂H₅COONa + CH₃I
OR
(a) Aldol condensation: When two molecules of aldehydes or ketones containing at least one α-hydrogen atom on treatment with dilute alkali undergoes condensation to form β-hydroxy aldehyde (aldol) or
β-hydroxy ketone (ketol) is known as aldol condensation.

(b) 4-Nitrobenzoic acid > Benzoic acid > 4-Methoxy benzoic acid.
(c) (i) Benzene-1, 4-dicarbaldehyde (ii) Heptan-2-one.

Question 10.
Complete the following reactions:
(a) C₆H₅ – NH₂ + CHCl₃ + alc.KOH →
(b) C₆H₅ – N₂Cl + H₃PO₂ + H₂O →
(c) C₆H₅NH₂ + (CH₃CO)₂O →
Answer:
(a) C₆H₅ – NH₂ + CHCl₃ + 3KOH → C₆H₅ – NC + 3KCl + 3H₂O
(b) C₆H₅N₂Cl + H₃PO₂ + H₂O → C₆H₅ – N₂ + H₃PO₃ + HCl.

Question 11.
Write the product formed when benzaldehyde reacts with the following reactants:
(a) CH₃CHO in presence of dilute NaOH
(b)

(c) Conc.NaOH
OR
An organic compound A(C₇H₆C₁₂) on treatment with NaOH solution gives another compound B
(C₇H₆O). B on oxidation gives an acid C(C₇H₆O₂) which on treatment with a mixture of conc. HNO₃ and
H₂SO₄ gives compound D (C₇H₅NO₄). B on treatment with conc. NaOH gives a compound E (C₇H₈O) and C₆H₅COONa. Deduce the structure of [A], [B], [C], [D] and [E].
Answer:
(a) Cinnamaldehyde is formed when benzaldehyde reacts with acetaldehyde in presence of dii. NaOH by aldol condensation.

(b) Benzaldehyde reacts with phenylhydrazine to give phenylhydrazine.

(c) Benzaldehyde does not have α-hydrogen so on heating with concentrated KOH solution a mixture of alcohol and salt is formed, this is Cannizzaro’s reaction.

OR
[A] seems like an organic compound of benzene from formula C₇H₆C₁₂.

Section – C

Question 12.
Read the passage given below and answer the questions that follow:
The calculation of cell potential for emf requires only the addition of the emf. values for each half-reaction. while the same cell potential calculation using standard potentials requires the usage of the following
convention:
E°cell = E°cathode – E°ariode
Each half-cell reaction has a specific standard potential reported as the potential of the reduction reaction vs. the Normal Hydrogen Electrode (NHE). In an electrochemical cell, there is a half-cell corresponding to the working electrode (WE), where the reactions understudy take place, and a reference half-cell.

Experimentally the cell potential is measured as the difference between the potentials of the WE half-cell and the reference electrode/reference half-cell. The archetypal reference electrode is the NHE, also known as the Standard Hydrogen Electrode (SHE) and is defined, by convention, as 0.000V for any temperature.
(a) What is the role of salt bridge?
(b) What is EMF of a cell?
(c) How electrical conductance changes with temperature?
(d) EMF of a cell depends, on which factors.
Or
(d) Predict the value of EMF of a cell in which the chemical reactions achieve equilibrium.
Answer:
(a) A salt bridge maintains the electrical neutrality between solutions of both the half cells.
(b) Cell potential is called EMF of the cell when no current is drawn through the cell.
(c) Electrical conductance increases with increase in temperature.
(d) EMF of a cell depends on:
1. Temperature
2. Nature of electrolyte
3. Concentration of electrolyte in two half cells.
OR
(d) At equilibrium, ΔG = O.
∴ ΔG = -nFE⁰
0 = -nFE⁰
E⁰ = 0.