Oswal Specimen Papers ISC Class 12 Chemistry Solutions (Specimen Paper - 7)


1. 1. (A) (i) ethanol
(ii) above, below
(iii) cannot, rate
(iv) ethanenitrile

(B) (i) (a) Lipase
(ii) (c) 16 times
(iii) (a) α-amino acids
(iv) (b) ClO3-
(v) (a) Alcohols 

Explanation :    

Lucas test in alcohols is conducted to distinguish between primary, secondary, and tertiary alcohols. It is based on the difference in reactivity of the three classes of alcohols with hydrogen halides through a substitution reaction:

ROH + HCl → RCl + H2O

(vi) (d) Assertion is false but reason is true.

Explanation :    

Nitration of chlorobenzene leads to the formation of m-nitrochlorobenzene but the chlorination of nitrobenzene leads to the formation of m-nitrochlorobenzene. Also, -NO2 is a m-directing-group. Thus, assertion is false but reason is true.

(vii) (a) Both assertion and reason are true and reason is the correct explanation of assertion.

Explanation :    

A transition metal exhibits higher oxidation states in oxides and fluorides because oxygen and fluorine are highly electronegative elements, small in size and strongest oxidising agents. For example; Osmium shows an oxidation state of +6 in OsF6 and vanadium shows an oxidation state of +5 in V2O5. Thus, both assertion and reason are true and reason is the correct explanation of assertion. 

(C) (i) One or more polypeptide chains can be found in proteins. Each polypeptide is a protein with amino acids linked together in a certain sequence, and this sequence of amino acids is referred to as the protein’s fundamental structure. Any modification in this core structure, i.e. the amino acid sequence, results in a different protein.
(ii) Protein denaturation is an irreversible alteration that occurs when proteins are heated with alcohol, concentrated inorganic acids, or heavy metal salts. Protein uncoils and loses its form and typical biological activity is lost.
(iii) Denaturation breaks down the protein’s secondary and tertiary structures while leaving the fundamental (primary) structure intact. Only the primary structure of the protein remains intact during denaturation.


2. Given: wB = 250 g, wA = 60 g, MA = 180g mol–1

$$K_f=1.86Kkg\space mol^{–1}\\\Delta T_f=K_fm\\=Kf×\frac{W_A×1000}{M_A×W_B}\\=1.86×\frac{60×1000}{180×250}\\=\frac{1.86×600}{18×25}\\=\frac{1116}{450}\\=2.48 K\\\Delta T_f=T_{solvent}– T_{solution}\\T_{solution}=T_{solvent} – \Delta T_f\\= 273.15 – 2.48 = 270.67 K\\\text{Hence, the freezing point of a solution is 270.67 K.}$$

3. (i) In Friedel - Crafts reaction, AlCl3 is added as a catalyst which is a Lewis acid. It forms a salt with aniline due to which the nitrogen of aniline acquires positive charge. This positively charged nitrogen acts as a strong deactivating group, hence, aniline does not undergo Friedel - Crafts reaction.

Friedel - Crafts

(ii) (CH3)2NH is more basic than (CH3)3N in an aqueous solution. This is because (CH3)2NH2 + is hydrated to a greater extent than (CH3)3NH+. As the number of methyl groups increases, the extent of hydration decreases due to steric hindrance. Greater is the extent of hydration, greater is the stability of ion and greater is the basic strength of amine. 


Therefore, with the increase in methyl group, hydrogen bonding, and stabilisation by solvation decreases. This net effect results in a decrease of basic strength from secondary to the tertiary amine.



5. (i) Differential rate equation of reaction is rate (r) = k[A]1[B]2 = k[A][B]2
(ii) When concentration of B is tripled, it means concentration of B becomes [3 × B]
New rate of reaction, r′ = k[A][3B]

$$\frac{r'}{r}=\frac{k[A]^13[B]^2}{k[A]^1[B]^2}= 9 or\space r′ = 9r\\\text{Thus, the reaction rate will increase to 9 times.}$$

6. Cu2++2eCu
63.5 g of copper deposited by 2 × 96500 C
1.27 g of copper will be deposited by

$${r}=\frac{2×96500×1.27}{63.5}C=3860 C\\\text{We know},Q=I_t\\\text{Here}, I=2A\text{and} Q=3860C\\t=\frac{Q}{t}=\frac{3860}{2}= 1930s\\\text{Thus, the item to deposit 1.27 g of copper at cathode is 1930s.}$$

7. (i)


Benzamide on heating with a mixture of Br2 in presence of NaOH or KOH (i.e., NaoBr or KOBr) will a give aniline.


Aniline first converts to benzene diazonium chloride using NaNO2/HCl at 0-5°C. Then diazonium ion reacts with water at 100°C to produce phenol.

8. (i)



(ii) The completed sequence is as follows:

9. Given:
MB = 256 g mol−1, WA = 75 g
Kf = 5.12 kg mol−1, ΔTf = 0.48K 

From the formula,

$$\Delta T_f=K_f×\frac{W_B}{M_B}×\frac{1000}{W_A}\\\text{Mass of solute,}\\W_B=\frac{\Delta T_f×M_B×W_A}{K_f×1000}\\=\frac{0.48×256×75}{5.12×1000}\\=\frac{(0.48K)(256 gmol^{-1})(0.075kg)}{(5.12K\space kgmol^{-1})}=1.8g\\\text{Hence, the mass of compound is 1.8 g.}$$

10. (i) Primary and secondary amines are engaged in intermolecular association due to hydrogen bonding between nitrogen of one and hydrogen of another molecule. Due to the presence of three hydrogen atoms, the intermolecular association is more in primary amines than in secondary amines as there are two hydrogen atoms available for hydrogen bond formation in it.
(ii) Ethylamine is soluble in water whereas aniline is not. This is because: Ethylamine when added to water forms intermolecular H—bonds with water. Hence, it is soluble in water.


But aniline does not undergo H—bonding with water to a very extent due to the presence of a large hydrophobic—C6H5 group. Hence, aniline is insoluble in water.


11. Transition metals have partly filled d-orbitals. So, the single electrons available in d-orbitals absorb energy and go to higher unoccupied electronic energy levels due to d-d transitions. The energy of excitation corresponds to the frequency of light absorbed and the colour observed corresponds to the complementary colour of the light absorbed (whose frequency lies in the visible region).


12. Arrhenius equation is given by:

$$k=Ae^{–E_a/RT}\\k=(2.5×10^{14}s^{–1})e^{(–25000K/T)}\\\text{Comparing the two equations :}\\E_a/RT=\frac{25000 k}{T}\\E_a=25000K×8.314JK^{–1}mol^{–1}\\=207850Jmol^{–1}\\E_a=207.9kJmol^{–1}\\\text{Now, for first order reactions, rate constant is given by :}\\k=\frac{0.693}{t_{1/2}}\\\text{Substituting the value of}\space t_{1/2}:\\k=\frac{0.693}{300}\\k=2.3×10^{–3}s^{–1}\\\text{Hence, Ea and rate constant for this reaction is} 207.9kJmol^{–1} and  2.3×10^{–3}s^{–1}.$$

13. As the chemical combination Br2 and KOH are used for Hofmann bromamide reaction where an amide is reduced to amine, the compound B, C6H7N seems to be aniline. The reaction can be suggested as follows: 

Then aniline undergoes carbylamine reaction to give the foul smelling compound C.

Phenyl isocyanide

IUPAC names are as follows:
Compound A – Benzamide
Compound B – Aniline or Benzenamine
Compound C – Isocyanobenzene or phenylisocyanide.

14. (i) Primary structure of protein – Peptide bond (or linkage)
(ii) Cross linkage of polypeptide chain – Polypeptide linkage
(iii) α-helix formation – Hydrogen bond

15. Given:
$$\text{Molar conductivity}(\lambda m)\text{for acetic acid}=39.05 S cm^2 mol^{−1}\\\lambda°(H^+)=349.6 S cm^2mol^{−1}\\\lambda°(CH_3COO^–)=40.95 Scm^2 mol^{−1}\\\text{We know that}:\\\lambda°m(CH_3COOH)=\lambda°(H^+)+\lambda(CH_3COO^−)\\39.05=349.6 + 40.9\\39.05=390.5Scm^2/mol\\also;\\\alpha=\frac{\lambda_m}{\lambda_m°}\\\alpha=\frac{39.05}{390.5}\\α = 0.1\\\text{Thus, the degree of dissociation of acetic acid is 0.1.}$$

16. Proteins are large, complex molecules that play many critical roles in the body. They do most of the work in cells and are required for the structure, function, and regulation of the body’s tissues and organs.

  1. Primary structure : The specific sequence in which the various a-amino acids are present in a protein and are linked to one another is called its primary structure. Any change in primary structure creates a different proteins.
  2. Secondary structure : The conformation in which the polypeptide chain assembles as a result of hydrogen bonding is known as secondary structure. The two types of secondary structure are a-helix and b-pleated sheet structure.

(a) In a-helix structure, the polypeptide chain form all the possible hydrogen bonds by twisting into right handed screw (helix) with NH group of each amino acids residue hydrogen bonded to the C = O group of an adjacent turn of the helix.

(b) In b-pleated structure, all peptide chains are streched out to nearly maximum extention then laid side by side and are held together by hydrogen bonds.

17. (i)



(ii) (a) But-1-ene to 1-Iodobutane conversion


(b) Benzene to Acetophenone conversion

(c) Ethanol to propane nitrile conversion:
This can be carried out in two steps, first, conversion of ethanol to ethyl chloride by reacting with zinc chloride and then reacting ethyl chloride with KCN in aq. ethanol.

$$1.\underset{Ethanol}{C_2H_5OH}\xrightarrow{Zncl_2}\underset{Ethyl chloride}{C_2H_5CI}\\2.\underset{Ethyl chloride}{CH_3-CH_2}\xrightarrow[-KCL]{KCN/aq.ethanol}\underset{Propane\space nitrile}{CH_3-CH_2-CN}$$

18. (i) As step (1) is slow, it is the rate – determining step as it dictates the overall pace of the reaction. Hence, Rate law of the reaction is given by:
Rate = – d [H2O]/dt = k [H2O2][I–]

(ii) As the rate law shows that the reaction is first order with H2O2 and I- both, hence the overall order of the reaction becomes 1 + 1 = 2. 

(iii) Step number (1) is the slowest among two steps, hence it is the rate determining step.


19. (i) (a) Electronic configuration: In the 1st, 2nd and 3rd transition series, the 3d, 4d and 5d orbitals are respectively filled.
We know that elements in the same vertical column generally have similar electronic configurations. In the first transition series, two elements show unusual electronic configurations:
Cr (24) = 3d5 4s1
Cu (29) = 3d10 4s1
Similarly, there are exceptions in the second transition series. These are:
Mo (42) = 4d5 5s1
Tc (43) = 4d6 5s1
Ru (44) = 4d7 5s1
Rh (45) = 4d8 5s1
Pd (46) = 4d10 5s0
Ag (47) = 4d10 5s1
There are some exceptions in the third transition series as well. There are:
W (74) = 5d4 6s2
Pt (78) = 5d9 6s1
Au (79) = 5d10 6s1
As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar.

(b) Oxidation states: The elements in the same vertical column generally shows similar oxidation state. The oxidation state shown by the elements in the middle of each series is maximum and minimum at the extreme ends. +2 and +3 oxidation states of first transition series elements are more stable
than +2 and +3 oxidation states of second and third transition series elements. Cobalt forms several complexes with +2 and +3 oxidation states but no such complexes are known for Rh,Ir which belong to the same group to which Co belongs. The maximum oxidation state in a group increases from
first transition series to third transition series. In group 8, iron shows +2 and +3 oxidation states and ruthenium and osmium show +4, +6 and +8 oxidation states.

(c) Ionisation enthalpies: The first ionisation enthalpies in each series generally increases gradually as we move from left to right though some exceptions are observed in each series. The first ionisation enthalpies of some elements in the second (4d) series are higher while some of them have lower value than the element in 3d series in the same vertical column. However, 5d series elements have higher ionisation enthalpies than 3d and 4d series elements. In 5d series, 4f orbitals are filled which have poor shielding effect and valence electrons experience higher effective nuclear charge. This results in higher ionisation enthalpies for 5d series elements.

(ii) (a) Ni(CO)4 :

Shape - Tetrahedral

(b) Fe(CO)5 :

Shape - Trigonal bipyramidal

20. (i) (a) Ionisation isomerism
(b) Optical isomerism
(c) Coordination isomerism

(ii) Electronic configuration of CO3+ ion
[Co(C2O4)3]3– has d2sp3 hybridisation (low spin complex) and [CoF6]3– has sp3d2 hybridisation (high spin complex).
(a) [Co(C2O4)3]3– is diamagnetic
(b) [Co(C2O4)3]3– is more stable.
(c) [CoF6]3– is outer orbital complex
(d) [Co(C2O4)3]3– is low spin complex. 

21. (i) (a) (1) Hearing aid — Mercury cell.
(2) Apollo Space Programme — Fuel cell.
(3) Automobile and inverters — Lead storage cell.
$$log=\frac{K_2}{K_1}=\frac{E_a}{2.203R}[(\frac{1}{T_1})-(\frac{1}{T_2})]\\Here,T_1=298K,T_2=308K,R=8.314K^{–1}mol^{–1}\\\frac{K_2}{K_1}=2\\log2=\frac{E_a}{2.303}×8.314[(\frac{1}{298})-(\frac{1}{308})]\\0.3010=\frac{E_a}{2.303}×8.314[\frac{10}{298×308}]\\E_a=\frac{0.3010×2.303×8.314×298×308}{10}\\= 52898 Jmol^{–1}=52.9KJ mol^{–1}.\\\text{Hence, the value of activation energy will be 52.9kJ} mol^{–1}.$$


(ii) (a) For the given cell representation, the cell reaction will be :
Fe(s) + 2H+(aq) Fe2+(aq) + H2(g)
The standard emf of the cell will be :
cell = E°H+/H2 – E°Fe2+/Fe
cell = (0 – 0.44) = 0.44 V
The Nernst equation for the cell reaction at 25°C will be
$$E_{cell}=E°_{cell}–\frac{0.0591}{n}log\frac{[Fe^{2+}]}{[H^+]^2}\\= 0.44–\frac{0.0591}{2}log\frac{0.001}{(0.01)^2}\\= 0.44–\frac{0.0591}{2}log 10\\=0.44–\frac{0.0591}{2}\\E_{cell}=0.4105V≈0.41V\\\text{Hence, the e.m.f. of the cell will be 0.41 V.}$$

(b) Given: E°cell = + 0.46 V and log10n = n.
Ag(s) | Ag+(10–3) || Cu2+(10–1M) | Cu(s)
E°cell = 0.46 V
T = 298 K

$$E_{cell}=E°_{cell}–\frac{RT}{nF}in\frac{\text{[Oxidised species]}}{\text{[Reduced species]}}\\\text{Cell reaction:}\\2Ag(s)+Cu^{2+}(aq)→2Ag^+(aq)+Cu(s)\\n=2\\E_{cell}=0.46–\frac{0.0591}{2}log\frac{[10^{-3}]^2}{[10^{-1}]}\\=0.46–\frac{0.0591}{2}log\space 10^{-5}\\=0.46+\frac{0.0591×5}{2}\\E_{cell}=0.61V\\\text{Thus, e.m.f. of the cell is 0.61 V.}$$

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