RBSE 12th Chemistry Model Paper Set 9 with Answers in English

Students must start practicing the questions from RBSE 12th Chemistry Model Papers Model Paper Set 9 with Answers in English Medium provided here.

RBSE Class 12 Chemistry Model Paper Set 9 with Answers in English

Time: 2 Hours 45 Min
Maximum Marks: 56

General Instruction to the Examinees :

• Candidate must write first his/her roll No. on the question paper compulsory.
• All the questions are compulsory.
• Write the answer to each question in the given answer book only.
• For questions having more than one part the answers to those parts are to be written together in continuity.

Section – A
Multiple Choice Questions (MCQs):

Question 1.
Choose the correct option:
(i) The equivalent weight of KMnO₄ (formula weight: M) when it is used as an oxidant in neutral medium is: (1)
(a) M
(b) M/2
(c) M/3
(d) M/5
Answer:
(c) M/3

(ii) Fool’s gold is: (1)
(a) AS₂O₃
(b) Sb₂S₅
(c) FeS₂
(d) Alloy of Cu – Zn
Answer:
(c) FeS₂

(iii) Which of the following is strong base? (1)
(a) FCH₂NH₂
(b) FCH₂CH₂NH₂
(C) C₆H₅NH₂
(d) C₆H₅CH₂NH₂
Answer:
(d) C₆H₅CH₂NH₂

(iv) Anti-Markownikoff’sruleofHBrisnotshownby: (1)
(a) Propane
(b) 1-Butene
(c) But-2-ene
(d) Pent-2-ene
Answer:
(c) But-2-ene

(v) An oxygen-containing organic compound forms a carboxylic acid as the main product upon oxidation with its molecular mass higher by 14 units. The organic compound is: (1)
(a) an aldephyde
(b) a ketone
(c) a primary alcohol
(d) a secondary alcohol
Answer:
(c) a primary alcohol

(vi) Which of the following alkenes produces tertbutyl alcohol on acid dydration? (1)
(a) CH₃-CH=CH₂
(b) CH₃-CH=CH-CH₃
(c) CH₃-CH₂-CH=CH₂
(d) (CH₃)₂C=CH₂
Answer:
(d) (CH₃)₂C=CH₂

(vii) The number of moles of NaOH required for the preparation of 5 litre 2m NaOH solution is: (1)
(a) 10
(b) 1
(c) 5
(d) 2.5
Answer:
(a) 10

(viii) A metal is crystallised in face-centred cubic arrangement. If edge length of unit cell is a and radius of metal atom is r, then: (1)
(a ) r = \(\frac{\alpha}{2 \sqrt{2}}\)
(b) r = \(\frac{\alpha \sqrt{3}}{4}\)
(c) r = \(\frac{2 \alpha}{\sqrt{3}}\)
(d) None of these
Answer:
(a ) r = \(\frac{\alpha}{2 \sqrt{2}}\)

(ix) The number of octahedral voids for each sphere in face centred cubic structure will be: (1)
(a) 8
(b) 4
(c) 2
(d) 1
Answer:
(d) 1

Question 2.
Fill in the Blanks:
(i) For a first order of reaction, half life period is equal to ___________________ (1)
Answer:
0.693/K

(ii) Zirconium is best refined by ___________________ method. (1)
Answer:
Van Arkel

(iii) The most common mineral containing lanthanoids is ___________________. (1)
Answer:
Monazite

(v) [Ni(CN)₄]^2- is diamagnetic and has ___________________ shape. (1)
Answer:
Square planar

Question 3.
Very Short Answer Type Questions:
(i) Ortho nitrophenol has lower boiling point than p-nitrophenol. Why? (1)
Answer:
Ortho-nitrophenol has lower boiling point due to formation of intramolecular H-bonding whereas p-nitrophenol forms intermolecular H-bonding.

(ii) Write the structure of 3-oxopentanal. (1)
Answer:

(iii) Arrange the following in increasing order of basic strength (1)
Aniline, p-Nitroaniline and p-Toluidine
Answer:
p-Nitroaniline< Aniline

(iv) Arrange the following in the decreasing order of their basic strength in aqueous solutions: (1)
CH₃NH₂, (CH₃)₂, NH,(CH₃)₃N and NH₃
Answer:
(CH₃)₂ NH > CH₃NH₂ > (CH₃)₃ N > NH₃

(v) A first order reaction is found to have a rate constant, k = 5.5 × 10^-14 s^-1. Find the half-life of the reaction. (1)
Answer:
Given, k = 5.5 × 10^-14 s^-1
For first order reaction,
t_½ = \(\frac{0.693}{k}=\frac{0.693}{5.5 \times 10^{-14}}\)
= 0.126 × 10¹⁴ s
= 1.26 × 10¹³ s

(vi) What type of isomerism is shown by the following complex [Co(NH₃)₆] [Cr(CN)₆]? (1)
Answer:
The complex, [Co(NH₃)₆] [Cr(CN)₆] shows coordination isomerism which is caused by the interchange of ligands between the two complex ions. Its another coordination isoner is [Cr(NH₃)₆] [CO(CN)₆].

(vii) Some liquids on mixing form azeotropes. What are azeotropes? (1)
Answer:
Azeotropes are binary mixture having the same composition in liquid and vapour phase and boil at constant temperatures.

(viii) What is the role of CO₂, in the extractive metallurgy of aluminium from its ore? (1)
Answer:
The aluminate in solution is neutralised by passing CO₂ and hydrated Al₂O₃ is precipitated.
2Na[Al (OH)₄](aq) + C0₂(g) → Al₂O₃.xH₂O(s) + 2NaHCO₃(a<7)

Section – B
Short Answer Type Questions:

Question 4.
Explain the Henry’s law about dissolution of a gas in a liquid. (1½)
Answer:
Henry’s law states that, the partial pressu re of the gas in vapour phase (p)is directly proportional to the mole fraction of the gas (χ) in the solution.
P = K_H × χ
Here, K_H = Henry’s law constant. Different gases have different K_H values at the same temperature.

Question 5.
Define azeotropes. What type of azeotTope is formed by negative deviation from Raoult’s law? Give an example. (1½)
Answer:
Azeotropes: Azeotoropes are binary mixture having the same composition in liquid and vapour phase and boil at constant temperatures.
Azeotropes showing negative deviation from Raoult’s law form maximum boiling azeotropes at a specific composition, e.g. azeotrope formed from nitric acid and water. Azeotropes showing positive deviation from Raoult’s law form mini¬mum boiling azeotropes at a specific composition, e.g. ethonol water,mixture.

Question 6.
What is meant by crystal field splitting energy? On the basis of crystal field theory, write die electronic configuration of d⁴ in terms of t_2g and e_g in an octahedral field when (1½)
(i) D₀ > P
(ii) D₀ < P
Answer:
Crystal field splitting energy: In an octahedral field, when ligands approach to the central metal ion to form coordination bonds, the five degenerated d-obitals of central metal ion splits in two sets of orbitals having different energies. This splitting of degenerated d-orbitals is called central field splitting and the energy difference between two sets of orbitals is called crystal field splitting energy (CFSE, ∆₀)

• ∆ > P, (pairing energy), the 4th electron pair up in one of the t_2g or bitals giving the configuration t_2g⁴ e_g⁰ J.
• If ∆₀ > P the 4th electron enters one of the eg orbitals giving the configuration t_2g³ e_g¹.

Question 7.
Ionic solids, which have anionic vacancies due to metal excess defect, develop colour. Explain with the help of a suitable example. (1½)
Answer:
When crystals of alkali metal halides, such as NaCl are heated in the atmosphere of sodium vapour, then the excess of sodium atoms are deposited on the surface of the crystal. The Cl^– ions diffuse to the surface and combine with sodium atoms to form NaCl. This happens by the loss of electrons by Na atom to form Na⁺ ions. These released electrons diffuse into the crystal and occupy anionic sites. It is known as F-centre. F-centre is responsible to colour. Due to this F-centre, they impart yellow colour to the crystal because they absorb energy from the visible light and get excited. Similarly excess of Li makes LiCl crystal pink in colour and excess of K makes KCl crystal violet.

Question 8.
What are semiconductors? Decribe thetwo main types of semiconductors. (1½)
Answer:
Semiconductor: The solid materials whose electrical conductivity lies between those of the typical metallic conductors and insulators are termed as seniconductors. The semiconductors possess conduc¬tivity in the range of 10² to 10^-9 ohm^-1 cm^-1.

These are of two types:
(a) w-type semiconductors: Doping of higher group element impurity forms w-type semiconductors, e.g. when ‘As’ is doped in ‘Ge’.
(b) p-type semiconductors: Impurity of lower groups forms election deficient bond in the structure. Electron deficiency develops to p-hole.

Question 9.
What happens when phenol is heated with zinc dust? (1½)
Answer:
Benzene is formed when phenol is heated with zinc dust.

Question 10.
What is ambidentate ligand? Give an example. (1½)
Answer:
Ambidentate ligand: Ligands which can ligate through two different atoms (with any one at a time) present in it are called ambidenatate ligands.
e.g. NO₂^–, SCN^–, CNO^– etc

NO₂^– can ligate through two sites,
eg:

Question 11.
Illustrate the following reactions giving a chemical equation for each : (1½)
(a) Kolbe’s reaction
Answer:
Kolbe’s reaction: Phenol reacts with CO₂ in presence of sodium hydroxide (NaOH) at 4-7 atm and 390-410 K giving salicylic acid.

(b) Williamson synthesis of an ether.
Answer:
Williamson synthesis of an ether: The reaction involves the nucleophilic substitution of the halide ion from the alkyl halide by the alkoxide ion by S_N2 mechanism. Example:

Question 12.
What is Tollen’s reagent? Write one usefulness of this reagent. (1½)
Answer:
Ammonical silver nitrate solution is called Tollen’s reagent.

Uses: It is used to test aldehydes. Both aliphatic and aromatic aldehydes reduce Tollen’s reagent to shining silver mirror. It is also used to distinguish aldehydes from ketones.

Question 13.
Give the structure and IUPAC name of the product formed when propanone is reacted with methylmagnesium bromide followed by hydrolysis. (1½)
Answer:

IUPAC name: 2-methylpropan-2-ol

Question 14.
How can you convert aniline to iodobeazene? (1½)
Answer:

Question 15.
Write structures and IUPAC name of: (1½)
(a) the amide which gives propanamine by Hoffmann bromamide reaction.
Answer:
Propanamine contains three carbon atoms. Hence the amide molecule must contain four carbon atoms. Structure and IUPAC name of the starting amide with four carbon atoms are given below:

The complete reaction is as follows:

(b) the amine produced by the Hoffmann degradation of benzamide.
Answer:
Benzamide is an aromatic amide containing seven carbon atoms. Hence, the amine formed from benzamide is aromatic primary amine containing six carbon atoms.

The complete reaction is as follows :

Section – C
Long Answer Type Questions:

Question 16.
(a) What type of linkage is present in proteins? (1 + 1 + 1 = 3)
(b) Give one example each of water-soluble and fat-soluble vitamins.
(c) Draw pyranose structure of glucose
Or
Define the following terms: (1 + 1 + 1 = 3)
(a) Glycosidic linkage
(b) Oligosaccharides
(b) Invert sugar
Answer:
(a) Peptide linkage is present in proteins.
(b) Vitamin C is water soluble and Vitamin D is fat soluble vitamin.
(c) Pyranose structure of glucose

Question 17.
Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate? (3)
Or
Give example and suggest reasons for the following features: (1 + 1 + 1 = 3)
(a) The lower oxide of the transition metal are basic and the highest oxides are amphoteric/acidic.
(b) A transition metal exhibits highest oxidation state in oxides and fluorides.
(c) The highest oxidation state is exhibited in oxo-anions of a metal.
Answer:
Preparation: Potassium dichromate can be prepared from iron chromite ore with the help of the following steps:
(i) Conversion of chromite ore to sodium chromate: When ore is fused with sodium or potassium carbonate or sodium or potassium hydroxide in presence of air then sodium chromate is formed.

(ii) Conversion of sodium chromate into sodium dichromate: The yellow solution of sodium chromate is filtered and acidified with sulphuric acid to give a solution from which average sodium dichromate Na₂Cr₂O₇.2H₂O can be crystallised.

(iii) Conversion of sodium dichro-mate into potassium dichromate:
Na₂Cr₂O₇ is more soluble when it is treated with KCl then K₂Cr₂O₇ is formed.

Effect of pH: Chromate and dichromate ions exist in equilibrium. Depending on pH, they can be interconverted. On increasing pH i.e., basic medium dichromate ion is converted into chromate ion. While in acidic medium i.e., on decreasing pH, chromate ion is converted into dichromate ion.

In basic, medium:
Cr₂O₄^2- + 2OH^– → 2CrO₄^2- + H₂O

In acidic medium:

Question 18.
How.will you convert ethanal into the following compounds: (1 + 1 + 1 = 3)
(a) Butane-1-3-diol,
(b) But-2-enal
(c) Biut-2-enoic acid
Or
An organic compound A (molecular formula C₈H₁₆O₂) was hydrolysed with dilute sulphuric acid to give a carboxylic acid (B) and an alcohol (C). Oxidation of (C) with chromic acid produced (B). (C) on dehydration gives but-1-ene as the major product. Write equations for the reaction involved. (3)
Answer:

Section – D
Essay Type Questions:

Question 19.
(a) For a reaction, A + B → P, the rate is given by Rate = k [A] [B]²
(i) How is the rate of reaction affected if the concentration of B is doubled?
(ii) What is the overall order of reaction if A is present in large excess?
(b) A first order reaction takes 30 min for 50% completion. Calculate the time required for 90% completion of this reaction, (log 2 = 0.3010) (1 + 1 + 1 + 1 = 4)
Or
(a) For a chemical reaction R → P, the variation in the concentration, In [R] vs time(s) plot is given as:

(i) Predict the order of the reaction.
(ii) What is the slope of curve?
(iii) Write the unit of rate constant for this reaction.
(b) Show that the time required for 99% completion is double of the time required for the completion of 90% reaction. (2 + 2 = 4)
Answer:
(a) A + B → P
Rate = k[A] [B]² (given)
(i) If concentration of B is doubled, then rate of reaction
= k[A][2B]² = 4k[A][B]²
∴ Rate becomes 4 times the original rate.

(ii) If A is present in large excess, then the reaction will be independent of the concentration of A and will be dependent only on the concentration of B. As [B]² will be the only determining factor in the order of reaction, the overall order of the reaction will be two.

(b) For the given first order reaction, the rate constant for 50% completion is given by
k = \(\frac{2.303}{t}\)log\(\frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}\) …………(i)
Here, t = time taken for 50% completion = 30 min

[R]₀ = initial concentration of reactant
[R] = final concentration of reactant

Let [R]₀ be 100 and due to 50% completion of reaction, [R] will be 100 – 50, i.e. 50
Putting values in (i), we get
k = \(\frac{2.303}{30}\) log\(\frac{100}{50}\)
= \(\frac{2.303}{30}\)log 2
= 0.023 min^-1

For same reaction, the time required for 90% completion of reaction can be computed using the expression,
k = \(\frac{2.303}{t}\)log\(\frac{[\mathrm{R}]_{0}}{[\mathrm{R}]}\)

Here, [R] = final concentration of reactant = 100 – 90 = 10
2.303 = \(\frac{2.303}{t}\)log\(\frac{100}{10}\)
⇒ t = \(\frac{2.303}{0.023}\)log 10
= 100.13 min
Therefore, the time required for 90% completion of the given first-order reaction is 100.13 min.

Question 20.
Primary alkyl halides C₄H₉Br (a) reacted with alcoholic KOH to give compound (b). Com-pound is reacted with HBr to give (c) which is an isomer of (a). When (a) is reacted with Na metal it gives compound (d), C₈H₁₈ which is different from the compound formed when n- butyl bromide is reacted with sodium. Give the structural formula of (a) and write the equations for all the reactions. (4)
Or
Write the major product(s) in the following: (4)

Answer:
(a) The two primary alkyl bromides are possible from the molecular formula (a) C₄H₉Br. These are:

Thus compound (a) is either n-butyl bromide or isobutyl bromide.

(b) Since compound (a) when reacted with Na gives a compound (d), C₈H₁₈ which is different from the compound formed when n-butyl bromide is reacted with Na, therefore compound (a) must be isobutyl bromide and compound (d) must be 2,5-dimethyl hexane.

(c) If compound (a) is isobutyl bromide then compound (d) which it gives on treatment with alcoholic KOH must be 2-methyl-1-propene.

(d) The compound (b) on treatment with HBr gives compound (c) according to Markownikov’s rule. Therefore, compound (c) is tert-butyl bromide which is an isomer of compound (a).

Thus, compound (a) is isobutyl bromide, (b) is 2-methylprop-1-ene, (c) is tert-butyl bromide and (d) is 2, 5-dimethyl hexane.