Oswal Practice Papers CBSE Class 12 Chemistry Solutions (Practice Paper - 10)

Section-A

1. (ii) Dehydrohalogenation

Explanation :

Dehydrohalogenation is a type of elimination reaction in which a hydrogen atom and a halogen are removed together. Usually, this reaction is used for making alkenes from alkyl halides. The conversion of ethyl bromide to ethylene is an example of dehydrohalogenation reaction. During the reaction, ethyl bromide is treated with a strong base like KOH to convert it into ethylene. The base attacks the Hydrogen on the β-Carbon, forming H₂O. This leads to bromine leaving the α-Carbon to combine with K⁺. The leftover electrons from both carbons make another bond, so ethylene is formed.
$$\underset{(Ethyl bromide)}{CH_3—CH_2—Br}+\underset{(Base)}{KOH}\xrightarrow{}\underset{(Ethylene)}{CH_2 ≡ CH_2}+ H_2O + KBr$$

2. (b) water will move from side (B) to side (A) if a pressure greater than osmotic pressure is applied on piston (B).

Explanation :

Water will move from side B (concentrated sodium chloride solution) to side A (fresh water) if a pressure greater than osmotic pressure is applied on piston B.

3. (d) I₂ and NaOH

Explanation :

Acetophenone reacts with NaOH and I₂ to give yellow ppt of CHI₃ but benezophenone (C₆H₅COC₆H₅) does not. Hence, it can be used to distinguish between them.
$${C_6H_5COCH_3}\xrightarrow{I_2,NaOH}\underset{Yellow ppt.}{CHI_3}+C_6H_5COONa$$

4. (b) 3, 5-Dibromo-4-hydroxybenzene sulphonic acid

Explanation :

The compound ‘X’ will be 3, 5-Dibromo-4-hydroxybenzene sulphonic acid.

5. (a) But-2-enal

Explanation :

The carbon atom of the CHO group is assigned the first number. So, in this case, the double bond occurs at second carbon, the prefix 2- is used for the naming of double bond. Hence, the correct IUPAC name of the given compound is But-2-enal.

6. (c) 0.5 mol L^–1

Explanation :

According to equation,
$$\text{No. of moles of NaOH =}\frac{10}{40}=0.25\space\text{mole}\\\text{Molarity =}\frac{(0.25×1000)}{500}= 0.5 mol L^{-1}$$

7. (a) 1-(p), 2-(q), 3-(r), 4-(s), 5-(t)

Explanation :

Ammonical silver nitrate is Tollen’s reagent. DDT (dichloro diphenyl trichloroethane) is an insecticide. Freon are low toxicity gases or liquids which are generally used as refrigerants and as aerosol. Iodoform is used as an antiseptic. RMgX is a grignard reagent.

8. (b) Lanthanoid contraction

Explanation :

Due to lanthanide contraction Zr and Hf have almost similar atomic radii. It can be explained on the basis of shielding effect. In case of post lanthanide elements like Hf, 4f subshell is filled and it is not very effective at shielding the outer shell electrons. Therefore, Zr and Hf have almost similar atomic radii.

9. (c) Fehling’s solution

Explanation :

Aromatic aldehydes such as benzaldehyde and ketones such as acetone does not respond to Fehling’s test. This is because benzaldehyde does not contain alpha hydrogen so intermediate enolate formation does not take place. Thus, it does not react with Fehling’s solution. On the other hand, ketone does not react with the Fehling’s solution unless they are alpha-hydroxy ketones. Acetone is not alpha-hydroxy ketone so it will also not reduce the Fehling’s solution.

10. (b) Benzaldehyde

Explanation :

The reduction of benzoyl chloride with Pd and BaSO₄ produces benzaldehyde. This reaction is called Rosenmund reaction.

11. (d) Denaturation of protein

Explanation :

The major components of milk are lactose, protein, and fats. During the curdling process, the lactose present in the milk is fermented by lactic acid bacteria and produces lactic acid which causes a decrease in the pH of milk. At low pH, the proteins present in the milk are denatured and get precipitated

12. (d) CH₃CHO

Explanation :

CH₃CHO will not give Cannizzaro’s reaction because it contains a-hydrogen while other three compounds have no a-hydrogen. Hence, they will give Cannizzaro’s reaction.

13. (b) Both A and R are true and R is not the correct explanation of A.

Explanation :

Azeotropic mixtures are formed only by non-ideal solutions and they may have boiling points either greater than both the components or lesser than both the components. The composition of the vapour phase is same as that of the liquid phase of an azeotropic mixture. Thus, both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.

14. (c) A is true, but R is false.

Explanation :

Halogen acids react with alcohols to form haloalkanes. For a given alcohol the order of reactivity of halogen acids follows the sequence HI > HBr > HCl. It is because of the fact that I^– is a stronger nucleophile than Br– which in turn is a stronger nucleophile than Cl^– . Thus, assertion is true but reason is false.

15. (b) Both A and R are true and R is not the correct explanation of A.

Explanation :

Aldehydes which do not contain a-hydrogen undergo Cannizzaro reaction. 2, 2-Dimethylpropanal undergoes Cannizzaro reaction with concentrated NaOH because it does not contain a-hydrogen atom.

Cannizzaro reaction is a disproportionation reaction because it involves the base-induced disproportionation of two molecules of a non-inolizable aldehyde to give a primary alcohol and carboxylic acid. Thus both assertion and reason are correct but reason is not the correct explanation of
the assertion.

16. (b) Both A and R are true and R is not the correct explanation of A.

Explanation :

When a solution is separated from the pure solvent by a semipermeable membrane, the solvent molecules pass through it from pure solvent side to the solution side. Diffusion is the process of movement of solvent molecules from the region of higher concentration to region of low concentration. Thus, both assertion and reason are correct statements but reason is not the correct explanation of assertion.

Section-B

17. Crystal field splitting energy (Δ) is the energy difference between two sets of d-orbitals (of metal atom) formed after splitting in presence of ligands. For example, in octahedral complexes d-orbitals split in two energy levels e_g (higher than degenerate level) and t_2g (lower than degenerate level). The e_g level has d_{x² −y²}and d_z² (pointing towards the axes along the direction of the ligand) and t_2g level has d_xy, d_yz and d_xz orbitals (directed between the axes).

(a) Hexaammine cobalt (III) Sulphate

(b) Potassium hexacyanide ferrate (II)

18. Initial amount = 5 g
Final concentration = 3 g
Rate constant = 1.15 × 10^–3 s^–1
We know that for a first order reaction,
$$t=\frac{2.303}{k}log\frac{[R_0]}{[R]}\\=\frac{2.303}{1.15×10^{-3}}log\frac{[5]}{[3]}\\=\frac{2.303}{1.15×10^{-3}}×0.2219=444.28s=444s\text{(approx.)}$$

19. The half cell reactions can be written as :
$$Mg^{2+}+2e^{–}→Mg;E_{cell}=-36+\frac{0.0591}{2}log[Mg^{2+}]=-2.448...(1)\\Cu^{2+}+2e^–→Cu;E_{cell}=0.34+\frac{0.0591}{2}log[Cu^{2+}]...(2)\\\text{Considering:}Cu^{2+}(aq)+Mg(s)→Mg^{2+}(aq)+Cu(s),\text{as the cell reaction}\\So,E_{cell} will be:E_{cell}=0.251V–(–2.448)V\\=2.69 V$$

20. (a) Fehling’s solution is a tartaric acid complex of cupric ions. When acetaldehyde is heated with Fehling’s solution, it gives a red precipitate of Cu₂O. During the reaction, acetaldehyde is oxidized to acetate ion and cupric ions are reduced to cuprous oxide.
$$CH_3CHO+2Cu^{+2}+5OH^{–}\xrightarrow{\Delta}\underset{(Red ppt.)}{CH_3COO^{–}}+Cu_2O↓+3H_2O.$$

(b) When acetaldehyde is treated with phenyl hydrazine, the Nitrogen atom of the phenylhydrazene serves as a nucleophile and attacks the electrophilic carbon of the acetaldehyde and removes water to give acetaldehyde phenylhydrazone.

21. (a)

(b)

Section-C

22. The given information is:
P₁ (Vapour pressure of solvent) = 54.2 mm Hg;
Molecular weight of glucose (C₆H₁₂O₆), M₂ = 180 g mol^–1;
Lowering of vapour pressure (i.e. P₁ – p₁) = 0.23 mm Hg;
M₁ = 18 g; w₁ = 100 g
Relative lowering of vapour pressure can be given the below formula :
$$\frac{P_1-wf_1}{p_1}=\frac{w_2×M_1}{M_2×w_1}$$

Here, P₁ and p₁ are vapour pressure of solvent and solution respectively, w₁ and w₂ are masses and M₁ and M₂ are molar masses of the solute and solvent respectively.
Now, putting the values in equation:
$$\frac{0.23mmHg}{54.2mmHg}=\frac{w_2×18gmol^{-1}}{18gmol^{-1}×100g}\\w_2=4.24g\\\text{Hence, weight of glucose is 4.24 g.}$$

23. Nucleoside: A nucleoside contains only two basic components of nucleic acids i.e., a pentose sugar and a nitrogenous base. During their formation 1-position of the pyrimidine or 9-position of the purine moitey is linked to C₁ of the sugar (ribose or deoxyribose) by a b-linkgae.

Nucleotides: A nucleotide contains all the three basic components of nucleic acids. i.e., a phosphoric acid group, a pentose sugar and nitrogenous base. These are formed by esterification of C₅′—OH of the sugar of the nucleoside with phosphoric acid.

24. Here, n = 2, E^o _cell= 0.236 V, T = 298 K
We know that :
ΔrG° = -nFE°_cell
= – 2 × 96487 × 0.236
= – 45541.864 J mol^–1
= – 45.54 kJ mol^–1

Note : Generally we take value of F as 96500 C. Then

ΔrG° = – 2 × 96500 × 0.236
= – 45548.00 J mol^–1
= – 45.548 kJ mol^–1.

$$\text{Again},\Delta_rG°=–2.303RT\space logK_c\\⇒log\space K_c=-\frac{\Delta_rG°}{2.303 RT}\\=-\frac{-45.54×10^3}{2.303×8.314×294}\\=7.981\\K_c=\text{Antilog}(7.981)=9.57×10^7.$$

25. (a) The ‘silver’ UK coins are made up of Cu/Ni alloy.
(b) The electronic configuration of Uranium (Z = 92) is : [Rn]5f ³6d¹⁷s² The electronic configuration of U³⁺ ion is : [Rn]5f³
(c) Although the 5f orbitals resemble the 4f orbitals in their angular part of wave function, they are not as buried as 4f orbitals hence, 5f electrons can participate in bonding to a far great extent compared to 4f electrons.

26. (a) Due to lanthanide contraction, the change in the atomic or ionic radii of these elements is very small. Hence, their chemical properties are similar. This makes their separation difficult.
(b) The greater the number of unpaired electrons greater is interatomic interaction and greater will be the enthalpy of atomisation. Since transition elements have unpaired electrons they have greater interatomic interaction and exhibit higher enthalpies of atomisation.
(c) Transition elements have a high enthalpy of hydration because of the presence of strong metallic bonds which is a result of the unpaired electron in the (n – 1) d subshell. The atoms in these elements are tightly packed. The strong bonds basically attribute to the elements having high enthalpies

27. (a) Amino acids behave like salt due to the pressure of both acidic and basic groups in the same molecule.
In aqueous solution, the carbonyl group loses a proton and amino accepts a proton. Thus, the carboxyl and amino group neutralize each other to form a salt like structure called dipolar ion or switter ion.

(b) A peptide is two or more amino acids joined together by peptide bonds, and a polypeptide is a chain of many amino acids. A protein contains one or more polypeptides. Therefore, proteins are long chains of amino acids held together by peptide bonds.
(c) Glycine is optically inactive amino acid.

28.

(C₈H₈Br₂) must be compound (I). As from the given information.
Compound (I) reacts with aq. KOH to give (II) (C₈H₉BrO), hence it must be undergoing nucleophilic substitution reaction. But only one (Br) group is substituted. Hence only one alkyl halide (1°, 2° or 3°) and another (Br) must be an aryl halide group.
Also, heating (I) with alcoholic KOH provides (III) (C₈H₇Br) confirming the aryl halide theory. The compound (III) must be an alkene as it undergoes addition reaction to form compound (IV).

Hence compound (III) and (IV) are similar.
Bromine mentioned is in para position to offer least or no steric hindrance to the above reactions.

Section-D

29. (a) Functional isomerism is found in between methyl cyanide and methyl isocyanide as both have same molecular formula but different functional groups.
$$(b)\space\space\space\underset{Methylamine}{CH_3–NH_2}\xrightarrow[Ethanol]{CHCl_3/KOH,\Delta}\underset{Methylisocyanide}{CH_3NC}$$

(c) N-atom in the molecule of methyl cyanide in partially negative charged. So, it can easily participate in the hydrogen bond formation with the water molecules as follow:

That is why, methyl cyanide shows significant solubility in water.

N-atom in the molecule of methyl isocyanide becomes partially positive charged and hence it fails to form H-bond with water molecules. So, it is almost insoluble in water.

30. (a) Dry cell is called a primary cells a is in these cells, the electrode reactions can not be reversed by an external electric energy source.
(b) Mercury cell is the example of primary cell.
(c) Dry cell does not have an indefinite life because NH₄Cl being acidic corrodes the zinc container even when not in use.

Overall reaction: Zn + 2NH₄ ⁺(aq) + 2MnO₂(s) → Zn⁺² + 2NH₃ + 2MnO(OH)

Section-E

31. (a)

The splitting of the d-orbitals in an octahedral field takes place in such a way that d_x²–y², d_z2 experiences a rise in energy and forms the e_g level, d_xy, d_yz and d_zx experience a fall in energy and forms the t_2g level.

(b) (b) 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,
$$μ=\sqrt{n(n+2)}=\sqrt{1(1+2)}\\=\sqrt{3}=1.732BM\\\text{On the other hand,}H_2O\text{is a weak field ligand. Therefore, it cannot cause the pairing of electrons.}\\\text{This means that the number of unpaired electrons is 5. Therefore,}\\μ=\sqrt{n(n+2)}\\=\sqrt{5(5-2)}\\=\sqrt{35}=5.916 BM\\Hence, [Fe(H_2O)_6]^{3+}\text{is strongly paramagnetic, while}[Fe(CN)_6]^{3–}\text{is weakly paramagnetic}.$$

(a) [Co(H₂O)(CN)(en)₂]²⁺
Let the oxidation number of Co be x.
The charge on the complex is + 2.

(d) K₃[Fe(CN)₆]

32. (a) The plot of ln k Vs. 1/T gives a straight line according to the equation given below :
$$ln\space k=\frac{-E_a}{RT}+In\space A$$

Activation energy can be calculated by the value of slope (– E_a/R) of the graph.

Given : Slope = – 5841 K and R = 8.314 J thin K ^–1 mol^–1
Putting the values

$$Slope=\frac{-E_a}{R}\\–Ea=–5841K×8.314J\text{thin}K^{– 1}mol^{– 1}\\= 48562 J mol^{–1}$$

(b) The action of the catalyst can be explained by intermediate complex theory. According to this theory, a catalyst participates in a chemical reaction by forming temporary bonds with the reactants resulting in an intermediate complex. This has a transitory existence and decomposes to yield
products and catalysts. This explains that catalysts are obtained as such from reaction mixture.
(c) Reactions which do not take place in a single step but take place in a sequence of a number of elementary steps are called as complex reactions. In some complex reactions, products are not formed in steps directly involving the reactants.
(d) Order of a reaction can be zero but molecularity of a reaction can never be equal to zero because number of molecules colliding with each other in a reaction can never be zero.
(e) (i) If reaction is 20% completed then the remaining reactant is 80%, Hence, R₁ = 100; R₂ = 80
For a first order reaction, rate constant is given by :
$$\text{Rate constant k=}\frac{2.303}{(t_2-t_1)}log(\frac{[R]_1}{[R]_2})\\\text{Putting the given values :}\\k=\frac{2.303}{10\space min}×log[\frac{100}{(100-20)}]\\=0.2303×0.1=0.023\\=2.3×10–2min^{–1}\\ (ii)\text{Time, t=}\frac{2.303}{k}log(\frac{[R]_1}{[R]_2})\\=\frac{2.303}{2.3×10^{-2}min^{–1}}log[\frac{100}{(100-75)}]\\=0.6×10^2\text{min}=60\text{min}.$$

(f) The order of all photochemical reactions is zero as it does not depend upon the concentration of reactants.
(g) Temperature, concentration of reactants and catalyst affect the rate of reaction.

33. (a) The structures of all isomeric alcohols of molecular formula, C₅H₁₂O are shown below :

(b) Primary alcohol : Pentan-1-ol; 2-Methylbutan-1-ol; 3-Methylbutan-1-ol; 2, 2-Dimethylpropan-1-ol Secondary alcohol : Pentan-2-ol; 3-Methylbutan-2-ol; Pental-3-ol Tertiary alcohol : 2-methylbutan-2-ol.

$$(a)\space\space\space(i)C_2H_5O–OC_2H_5+HI\xrightarrow{373k}\underset{Ethanol}{C_2H_5–OH}+\underset{Ethyl iodide}{C_2H_5I}$$

Ethanol and ethyl iodide is formed. In case when HI is in excess then the remaining ethanol will also get converted to ethyl iodide.

$$C_2H_5–OH+HI\xrightarrow{}C_2H_5–I+H_2O$$

(iii)$$\space\space\space CH_3CH_2OH\xrightarrow[443 K]{conc.H_2SO_4}\underset{Ethene}{CH_2=CH_2+H_2O}\\\text{Dehydration of alcohol results in the formation of ethene.}$$

(b) (i) Phenol will give yellowish white ppt. on treating with bromine water whereas cyclohexanol will not.

(ii) Propan-2-ol