RBSE 12th Chemistry Model Paper Set 2 with Answers in English

Students must start practicing the questions from RBSE 12th Chemistry Model Papers Model Paper Set 2 with Answers in English Medium provided here.

RBSE Class 12 Chemistry Model Paper Set 2 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 value of coordination number in hexagonal close packing (hep) is: (1)
(a) 12
(b) 8
(c) 6
(d) 4
Answer:
(a) 12

(ii) What will be the edge length of unit cell of a face-centred cubic lattice if radius of sphere present in it is 500 pm ? (1)
(a) 1414 pm
(b) 1000 pm
(c) 500 pm
(d) 250 pm
Answer:
(a) 1414 pm

(iii) The rate of reaction becomes twice on increasing the concentration of A four times for the reaction A → B. The order of reaction is: (1)
(a) two
(b) one
(c) half
(d) zero
Answer:
(c) half

(iv) The basic character of the transition metal monoxides follows the order: (1)
(a) CrO > VO > FeO > TiO
(b) TiO > FeO > VO > CrO
(c) TiO > VO > CrO > FeO
(d) VO > CrO > TiO > FeO
Answer:
(c) TiO > VO > CrO > FeO

(v) The number of possible isomers for the complex [Co(C₂O₄)₂ (NH₃)₂] is: (1)
(a) 1
(b) 2
(c) 3
(d) 4
Answer:
(d) 4

(vi) The complexes [Co(NH₃)₆] [Cr(CN)₆] and [Cr(NH₃)₆] [Co(CN)₆] are the examples of which type of isomerism? (1)
(a) Geometrical isomerism
(b) Linkage isomerism
(c) Ionisation isomerism
(d) Coordination isomerism
Answer:
(d) Coordination isomerism

(vii) An organic! compound X having C, H and O possesses a pleasant odour with boiling point of 78°C is: (1)
(a) C₂H₅Cl
(b) C₂H₅OH
(C) CH₃COOC₂H₅
(d) C₂H₆
Answer:
(b) C₂H₅OH

(viii) Molecular formula of simple disaccharide is: (1)
(a) C₁₀O₁₈O₉
(b) C₁₀H₂₀O₁₁
(c) C₁₈H₃₂O₁₆
(d) C₁₂H₂₂O₁₁
Answer:
(d) C₁₂H₂₂O₁₁

(ix) Which of the following is laevorotatoiy? (1)
(a) Glucose
(b) Sucrose
(c) Fructose
(d) None of these
Answer:
(c) Fructose

Question 2.
Fill in the Blanks:
(i) IUPAC name of isobutyric add is _______________ (1)
Answer:
2-methylpropanoic acid,

(ii) _______________ are building blocks of proteins. (1)
Answer:
Amino acids

(iii) Mixture of _______________ and _______________ form a non-ideal solution. (1)
Answer:
Alcohol and water

(iv) Formation of HCl in presence of light is a _______________ order reaction. (1)
Answer:
zero.

Question 3.
Very Short Answer Type Questions:
(i) What happens when bromine attacks CH₂ = CH – CH₂ – C ≡ CH? (1)
Answer:

(ii) Write the IUPAC name of the following: (1)

Answer:
IUPAC name: 3 Bromo-2-methyl propene.

(iii) Draw the structure of 2,6-dimethylphenol. (1)
Answer:

(iv) Name the method that is used for refining of nickel. (1)
Answer:
Mond’s process is used for refining of nickel metal. It involves the following reactions:

(v) Explain why Cu⁺ ion is not stable in aqueous solution ? (1)
Answer:
Cu⁺ ion in aqueous solution undergoes disproportionation i.e.,
2Cu⁺ (aq) → Cu²⁺_{(aq) + Cu(s)
Becuase the E° value of this reaction is favourable as Cu²⁺ has higher hydration enthalpy which compensates I.E. and sublimation energies.}

(vi) What is the difference between a complex and a double salt? (1)
Answer:
Double salt is combination of (+) ve and (-) ye ions, which completely dissociates into its ions, when dissolved in water. Whereas complex is a salt, in which molecular structure of complex ion retain itself in aqueous solution, i.e., do not dissociates into its ions completely.

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

(viii) Why is an alkylamine more basic than ammonia? (1)
Answer:
Due to electron releasing inductive effect (+1) of alkyl group, the electron density on the nitrogen atom increases and thus, it can donate the lone pair of electrons more easily than ammonia.

Section-B
Short Answer Type Questions:

Question 4.
Identify all the possible monochloro structural isomers expected to be formed on free radical monochlorination of (CH₃)₂CHCH₂CH₃. (1½)
Answer:
In the given molecule, there are four different types of hydrogen atoms. Replacement of these hydrogen atoms will give the following
(CH₃)₂CHCH₂CH₂Cl
(CH₃)₂CHCH(Cl)CH₃
(CH₃)₂C(Cl)CH₂CH₃
CH₃CH(CH₂Cl)CH₂CH₃

Question 5.
Which one in the following pairs of substances undergoes S_N2 substitution reaction faster and why? (¾ + ¾ = 1½)
(a)

Answer:

CH₂Cl is a primary halide and therefore undergoes S_N2 reaction faster.

(b)

Answer:

As iodine is a better leaving group because of its large size, therefore undergoes S_N2 reaction faster.

Question 6.
Name the parameters that characterise a unit cell. (1½)
Answer:
A unit cell has following parameters:

(i) Three edge length a, b, c
(ii) Three axial angle α, β, γ.

Question 7.
Explain how much portion of an atom located at (a) comer and (b) body-centre of a cubic unit cell is part of its neighbouring unit cell ? (1½)
Answer:
(a) Portion of an atom at comer: An atom at corner is shared by eight adjacent unit cell. Four unit cells on one layer while four unit cells either above or below the layer, Hence, the portion of the atom at the comer that belongs to one unit cell is equal to \(\frac{1}{8}\).

(b) Portion of an atom at body centre: An atom at the centre of body of a cubic unit cell is not shared by any other unit cell. Hence, it belongs the whole part or fully to the unit cell.

Question 8.
Suggest a condition under which magnesium could reduce alumina. (1½)
Answer:
The two equations involved in this processes are:
(a) \(\frac{4}{3}\)Al + O₂ → \(\frac{2}{3}\)Al₂O₃
and
(b) 2Mg + O₂ → 2MgO
In Ellingham diagram, at the point of intersection of the Al₂O₃ and MgO curves, the ΔG° becomes zero for the reaction.
\(\frac{2}{3}\)Al₂O₃ + 2 Mg → 2MgO + \(\frac{4}{3}\)Al
Below the point magnesium can reduce alumina.

Question 9.
Give the IUPAC names of the following compounds: (1½)
(a)

Answer:
4-Chloro-2, 3-dimethylpentan-1-ol.

(b)

Answer:
(b) 2-Ethoxypropane

(c)

Answer:
2, 6-Dimethylphenol

(d)

Answer:
1-Ethoxy-2-nitrocyclohexane.

Question 10.
Although thermodynamically feasible, in practice, magnesium metal is not metal for the reduction of alumina in the metallurgy of aluminium. Why? (1½)
Answer:
Temperatures above the point of intersection of Al₂O₃ are MgO curves, magnesium can reduce alumina. But the temperature required would be so high that the process will be uneconomic and technologically difficult.

Question 11.
Show how are the following alcohols prepared by the reaction of suitable Grignard reagent with methanol?
(a)

Answer:
The structure of given alcohol suggests that the Grignard reagent that reacts with methanol is isopropyl magnesium halide.

(b)

Answer:
The structure of given alcohol suggests that the Grignard reagent that reacts with methanol is cyclohexyl magnesium halide.

Question 12.
Write the agents required in the following reactions :
Answer:

Question 13.
On ground you can say that scandium (Z = 21) is a transition element but zinc (Z = 30) is n. (1½)
Answer:
On the basis of incompletely filled 3d orbitals in cage of scandium atom in its ground state (3d¹), it is regarded as a transition element. On the other hand, zinc atom has completely filled d-orbitais (3d¹⁰) in its ground state as well as in its excited state, hence it is not regarded as a transition element.

Question 14.
Write the equations involved in the following reactions: (¾ + ¾ = 1½)
(a) Wolff-Kishnerreduction
Answer:

(b) Etard reaction
Answer:

Question 15.
Why. do the transition element exhibit higher enthalpies of atomisation ? (1½)
Answer:
Because of large number of unpaired electrons in their atoms they have stronger interatomic interaction and hence stronger bonding between atoms resulting in higher enthalpies of atomisation.

Section – C
Long Answer Type Questions:

Question 16.
Explain with two examples each of the following: (3)
Coordination entity, ligand, coordintion number, coordination polyhedron, hornoleptic and heteroleptic complexes.
Or
What is meant by unidentate, bidentate and ambidentate ligands ? Give two examples for each. (3)
Answer:
(i) Coordination entity : A coordination entity constitutes a central metal atom or ion bonded to a fixed number of ions or molecules.
Example : [CO(NH₃)₃Cl₃] is a coordination entity in which cobalt ion is surrounded by three ammonia molecules and three chloride ions. Other examples are: (a)[Pt(NH₃)₂Cl₂]
(b)[Fe(CN)₆]^4-
(c) [CO(NH₃)₆]³⁺ etc.

(ii) Ligand: The ions or molecules bound to the central metal atom/ion in the coordination entity are called ligands. These may be simple ions as Cl^– and may be large molecule as NH₂CH₂CH₂NH₂ or even macro-molecules like proteins.

(iii) Coordination number: It is the number of ligand donor atoms to which the metal is directly bonded’. For example, [PtCl₆]^2-, the coordination number of Pt is 6 while in [Ni(NH₃)₄]²⁺, the C.N. of Ni is 4.

(iv) Coordination polyhedron: The spatial arrangement of the ligand atoms which are directly attached to the central atom/ion defines a coordination polyhedron about the central atom, e.g., octahedral, square planar and tetrahedral polyhedron.

(v) Homoleptic complexes : Complexes in which a metal is bounded to only one kind of donor groups e.g., [CO(NH₃)₆]³⁺, [Cr(H₂O)₆]³⁺ etc.

(vi) Heteroleptic complexes: Complexes in which a metal is bound to more than one kind of donor groups, e.g., [Co(NH₃)₄Cl₂]⁺, [Cr(H₂O)₄Br₂]⁺ etc.

Question 17.
Arrange the following compounds in the increasing order of their boiling points: (3)
CH₃CH₂CH₂CHO, CH₃CH₂CH₂CH₂OH, H₃C₂-O-C₂H₃, CH₃CH₂CH₂CH₃
Or
Predict the products of the following reactions :
(a)

(b)

(c)

Answer:
The molecular masses of these compounds are in the range of 72 to 74. Since only butan-l-ol molecules are associated due to extensive intermolecular hydrogen bonding, therefore, the boiling point of butan- l-ol would be the highest. Butanal is more polar than ethoxyethane. Therefore, the intermolecular dipole-dipole attraction is stronger in the former. n-Pentane molecules have only weak van der Waals forces. Hence increasing order of boiling points of the given compounds is as follows:
CH₃CH₂CH₂CH₃<H5C₂-0-C₂H5 < CH₃CH₂CH₂CHO < CH₃CH₂CH₂CH₂OH

Question 18.
How would you account for the following:
(a) Electrophilic substitution in case of aromatic amines takes place more readily than benzene.
(b) Ethanamide is a weaker base than ethanamine. (½ + ½ = 3)
Or
Illustrate the following reactions:
(a) Sandmeyer’s reaction
(b) Coupling reaction (½ + ½ = 3)
Answer:
(a) Aniline exists as a resonance hybrid of the following” five structures:

The electron density is maximum at ortho and para positions to the NH₂ group. But in benzene there is no delocalisation of electron at any position and hence electrophilic substitution in case of aromatic amines takes place more readily than benzene.

(b) In ethanamide the lone pair of electron of N-atom is not available due to resonance structure

Section-D
Essay Type Questions:

Question 19.
Vapour pressure of chloroform (CHCl₃) and dichloromethane (CH₂Cl₂) at 298 K are 200 mm Hg and 415 mm Irrespectively, (i) Calculate the vapour pressure of the solution prepared by mixing.25- 5 g of CHCl₃ and 40 gof CH₂Cl₂ at 298 K and (ii) mole fractions of each component in vapour phase. (4)
Or
Calculate: (a) molality, (b) molarity and (c) mole fraction of KI if density of 20% (mass/mass) aqueous solution of KI is 1-202 gmL^-1. (4)
Answer:
Let CH₂Cl₂ = A
CHCl₃ = B
Molar mass of A,
M_a = 12 × 1 + 1 × 2 + 2 × 35.5
= 85 g mol^-1
Molar mass of B,
MB = 12 × 1 + 1 × 1 + 3 × 35-5
= 119.5 g mol^-1
Mass of A,
W_A = 40 g Mass of B,
W_B = 25.5 g
Vapour pressure of A,
p°_A = 415 mm of Hg
Vapour pressure of B,
p°<subB = 200 mm of Hg

x_B = 1 – x_A
= 1 – 0.688
= 0.31
p = p_A + P<subB
= p°_A x_A + p°_B x_B
= 415 × o.688 + 200 × 0.312
= 285.52 + 62.4
= 347.92 mmof Hg
Mole fraction of components in gas phase,
PA = y_A.p
p°_A x_A = y_A.P
415 × 0.688 = y_A × 347.92
y_A = \(\frac{285 \cdot 52}{347 \cdot 92}\) = 0.82
= 1 – 0.82
= 0.18

Question 20.
In a pseudo first order hydrolysis of ester in water, the following results were obtained:

(a) Calculate the average rate of reaction between the time interval 30 to 60 seconds.
(b) Calculate the pseudo first order rate constant for the hydrolysis of ester. (2 + 2 = 4)
Or
A reaction is first order with respect to A and second order in with respect to B. (4)
(a) Write the differential rate equation.
(b) How is the rate affected on increasing the concentration of B three times.
(c) How is the rate affected when the concentration of both A and B are doubled ?
Answer:
(a) Average rate during the time interval 30-60 seconds
Average rate
= \(\frac{C_{2}-C_{1}}{t_{2}-t_{1}}\)
= \(\frac{0.17-0.31}{60-30}\)
= \(\frac{-0.14}{30}\)
= – 4.67 × 10^-3
= mol L^-1 s^-1

(b) Pseudo first order rate constant

Here the value of rate constant is 1.98 × 10^-2s^-1.