RBSE 12th Chemistry Model Paper Set 7 with Answers in English

Students must start practicing the questions from RBSE 12th Chemistry Model Papers Model Paper Set 7 with Answers in English Medium provided here.

RBSE Class 12 Chemistry Model Paper Set 7 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 normality of 2M sulphuric acid is: (1)
(a) 2N
(b) 4N
(c) N/2
(d) N/4
Answer:
(b) 4N

(ii) 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

(iii) The vapour pressure of liquid: (1)
(a) increases with increase in temperature
(b) does not depend on temperature
(c) decreases with increase in temperature
(d) None of the above
Answer:
(a) increases with increase in temperature

(iv) Which of the following compound has tetrahedral geometry? (1)
(a) [Ni(CN)₄]^2-
(b) [Pd[CN)₄]^2-
(c) [PdCl₄]^2-
(d) [NiCl₄]^2-
Answer:
(d) [NiCl₄]^2-

(v) The hybridisation of Fe in K4[Fe(CN)₆] is: (1)
(a) dsp¹
(b) sp³
(c) d²sp³
(d) sp³d²
Answer:
(c) d²sp³

(vi) Which gives iodine test? (1)
(a) Polypeptide
(b) Glucose
(c) Starch
(d) Glycogen
Answer:
(c) Starch

(vii)

has IUPAC name: (1)
(a) 1,2,3-Tricyano propane
(b) Propane-1,2,3-tricarbonitrile
(c) 3-Cyanopentane-l, 5-dinitrile
(d) 1,2,3-Propane trinitrile
Answer:
(c) 3-Cyanopentane-l, 5-dinitrile

(viii) The reaction given below know as: (1)

(a) Kharash effect
(b) Darjen reactions
(c) Williamson’s synthesis
(d) Wurtz reaction
Answer:
(b) Darjen reactions

(ix) Which of the following can differentiate between aldehyde and ketone? (1)
(a) Fehling solution
(b) Hydrazine
(c) Pyrogallol
(d) Grignard Reagent
Answer:
(a) Fehling solution

Question 2.
Fill in the Blanks:
(i) Molecular formula of bauxite is ______________. (1)
Answer:
Al₂O₃.H₂O

(ii) In KMnO₄ oxidation state of Mn is ______________. (1)
Answer:
+ 7

(iii) Solution of ammonical silver nitrate is known as ______________. (1)
Answer:
Tollen’s reagent

(iv) The bond formed between amino adds is known as ______________. (1)
Answer:
peptide bond.

Question 3.
Very Short Answer Type Questions:
(i) Which point defect in crystals not does alter the density of the relevant solid? (1)
Answer:
Frenkel defect.

(ii) Name the products of hydrolysis of lactose. (1)
Answer:
Lactose on hydrolysis with dilute acids gives an equimolar mixture of D-glucose and D-galactose

(iii) Arrange the following compounds in increasing order of solubility in water: (1)
C₆H₅NH₂,(C₂H₃)₂NH,C₂H₅NH₂
Answer:
C₆H₅NH₂ < (C₂H₅)₂NH < C₂H₅NH₂

(iv) Write the IUPAC name of the given compound : (1)

Answer:
IUPAC name : 1-Ethoxy-2-methyl- propane

(v) Write the IUPAC name of the given compound : (1)

Answer:
2-Phenylethanol

(vi) Draw the structural formula of 1-phenyl propan-1 -one molecule. (1)
Answer:
1 phenyl propan-1-one

(vii) Write the coordination number and oxidation state of platinum in the complex [Pt(en)₂ Cl₂].
Answer:
Coordination number of platinum in the complex, [Pt (en)₂ Cl₂] is 6 as en is a bidentate ligand. Let the oxidation state of Pt is x.
x + 0 + 2(-2) = 0 ⇒ t = + 2
Thus, oxidation state of Pt be + 2

(viii) Name a transition element which does not exhibit variable oxidation states. (1)
Answer:
Scandium (Z = 21) does not exhibit variable oxidation states.

Section – B
Short Answer Type Questions:

Question 4.
Why is zinc not extracted from zinc oxide through reduction using CO?
Answer:
According to Effingham diagram the value of ∆G° for the oxidation of CO into C0₂ is higher than the value of ∆G° for the oxidation of Zn into ZnO. Hence CO can not reduce ZnO into Zn. Further more the reduction may require very high temperature if CO is used as a reducing agent in this case.

Question 5.
How is ‘cast iron’ different from ‘Pig iron’?
Answer:

• Cast iron is different from Pig iron with respect to the carbon contents. Pig iron is obtained from blast fumance and have about4% carbon and other impurities like S, P, As, Mn etc. It can be mould into different shapes.
• Cast iron is obtained from Pig iron when Pig iron is mixed with scrap of carbon and oxidised by passing hot air through it. Cast iron has 3% impurities of carbon and other elements.

Question 6.
Define the following terms.
(a) Half-life of a reaction (t_1/2)
(b) Rate constant (k) (1½)
Answer:
(a) Half-life (t_1/2) of a reaction is the time in which the concentration of a reactant is reduced to one half of its initial concentration.
(b) Rate constant (k) is equal to the rate of reaction when molar concentration of reactants is unity. Its unit depends upon the order of reaction.

Question 7.
A reaction is of second order with respect to a reactant. How is its rate affected if the concentration of the reactant is
(a) doubled?
(b) reduced to half? (1½)
Answer:
(a) Write the rate law for initial concentration of reactant and for the conditions, when concentration is doubled or reduced to half.
(b) Now, compare the initial concen-tration condition with changed con-centration to find the effect on rate.

Question 8.
List two main differences between order of a reaction and molecuiarity of a reaction, (1½)
Answer:

S. Molecularity No. of a reaction	Order of reaction
(i) The number of reacting species which must collide simultaneously in order to bring about a chemical reaction is called the molecularity of a reaction.	(ii) The sum of powers of concentrations of the reactants in the rate law expression is called the order of that chemical reaction.
(ii) Molecularity is always a whole	(ii) Order may have zero, whole number,

Question 9.
Find the boiling point of a solution containing 0.520 g of glucose (C₆H₁₂O₆) dissolved in 80.2 g of water. [Given, K_b for water = 0.52 Km^-1]. (1½)
Answer:
Given Kb =0.52 Km^-1

Mass of solute (W₂)
= 0.520 g (glucose)

Mass of solvent (W₁)
= 80.20 g (water)

Molar mass of solute (M₂)
= 180 g mol^-1
∵ ∆T_b = K_b × \(\frac{W_{1}}{M_{2}} \times \frac{1000}{W_{1}}\)
∆T_b = \(\frac{0.52 \times 0.52 \times 1000}{180 \times 80.2}\)
∆T_b = 0.0187 ≈ 0.019

∴ Boiling Point
T_b = 373 + 0.019
= 373.019 K
Hence b.p = 373.02 K

Question 10.
The partial pressure of ethane over a saturated solution containin 6.56 × 10^-2 g of ethane is 1 bar. If the solution contains 5.0 × 10^-2 of ethane, then what will be the partial pressure of the gas? (1½)
Answer:

Question 11.
Write units of rate constants for zero order and for the second order reactions if the concentration is expressed in mol L^-1 and time in seconds. (1½)
Answer:
Unit of rate constant (k)

Question 12.
In the series Sc (Z = 21) to Zn (Z = 30), the enthalpy of atomisation of zinc is the lowest,126 kJ mol^-1. Why? (1½)
Answer:
The enthalpy of atomisation of zinc is lowest because in the formation of metallic bond, no electrons from 3d- orbitals are involved in case of zinc, formation which increases the strength of bond.

Question 13.
How many ions are produced from the complex CO(NH₃)₆C₁₂ in solution? Explain. (1½)
Answer:
Oxidation number of cobalt = 6
The complex [Co(NH₃)₆] Cl₂ ionises as

Hence three ions are produced.

Question 14.
The E° (M²⁺/M) value for copper is positive (+ 0.34 V). What is possible reason for this ? (1½)
Answer:
For the calculation of E° (M²⁺/M) values, three energies are responsible. They are ionisation energies, sublimation energies and hydration energies. In case of copper, the values of ionisation energies and sublimation energies are higher while the value of hydration energy is low hence E° (M²⁺/M) value for copper is positive (+ 0.34 V).

Question 15.
What is meant by chelate effect ? Give an example. (1½)
Answer:
When a di-or polydentate ligand coordinates to the central metal atom or ion using its to as more donor atoms, then a ring like structure is formed which increase the stability of the complex. This enhanced stability of the complexes containing bidentate of polydentate ligands (chelating ligands) in called chelate effect:
Example – [Cu(CN)₂]²⁺

Section – C
Long Answer Type Questions

Question 16.
An element crystallises in bcc lattice with cell edge of400pm. Calculate its density if 500 g of this element contains 2.5 × 10²⁴ atoms. (3)
Or
An element crystallizes in a bcc lattice with cell edge of 500 pin. The density of the element is 7.5g cm^-3. How many atoms are present in 300 g of the element? (3)
Answer:
Given: a = 400pm = 400 × 10¹⁰ cm
Z = 2(for bcc) M = ? d = ?

Using formula d = \(\frac{\mathrm{Z} \times \mathrm{M}}{a^{3} \times \mathrm{N}_{\mathrm{A}}}\)
∴2.5 × 10²⁴ atoms of nelement have mass = 500 g
∴ 6.022 × 20²³ atoms of an element have mass
= \(\frac{500 \times 6.022 \times 10^{23}}{2.5 \times 10^{24}}\)
∴ M = 120.44 g
Substituting all values in the formula:

Question 17.
Show how will you synthesize:
(a) 1-phenylethanol from a suitable alkene?
(b) Cydohexylmethanol using an alkyl halide by an S*[2 reaction?
(c) Pentan-l-ol using a suitable alkyl halide? (1 + 1 + 1 = 3)
Answer:
(a) Addition of H₂O to ethenyl- benzene (or styrene) in presence of dil.H₂SO₄ gives 1-phenylethanol.

(b) Hydrolysis of cyclohexylmethyl bromide by aqueous NaOH gives cyclohexylmethanol.

(c) Hydrolysis of 1-bromopentaneby aqueous NaOH gives pentan-l-ol.

Question 18.
Give reasons:
(a) Aniline is a weaker base than cyclohexylamine..
(b) It is difficult to prepare pure amines by ammonolysis of alkyl halides. (3)
01
Give reasons:
(a) Electrophilic substitution in aromatic amines takes place more readily than benzene.
(b) CH₃CONH₂ is a weaker base than CH₃CH₂NH₂. (3)
Answer:
(a) In aniline, the lone pair of electrons on the N-atom is delocalised over the benzene ring. As a result, the electron density on the nitrogen decreases.
But in cyclohexylamine, the lone pair of electrons on N-atom is readily available due to absence of π-electrons. Hence aniline is weaker base than cyclohexylamine.

(b) Because the primary amine formed by ammonolysis itself acts as a nucleophile and produces further , 2° and 3° alkyl amine.

Section – D
Essay Type Questions:

Question 19.
Explain why:
(a) the dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(b) alkyl halides, though polar are immiscible with water.
(c) Grignard reagents should be prepared under anhydrous conditions. (4)
Or
Primary alkyl halides C₄H₉Br (a) reacted with alcoholic KOH to give compound (b). Compound (b) 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 equa-tions for all the reactions., (4)
Answer:
(a)

The polarity of C — Cl bond in chlorobenzene is less than that of same bond in cyclohexyl chloride because of carbon atom involved in chlorobenzene is more electronegative due to greater s- character as compared to the carbon atom in cyclohexyl chloride with lesser s-character. Therefore, the dipole moment of chlorobenzene is less than cyclohexyl chloride.

(b) Alkyl halides or haloalkanes are polar molecules, therefore their molecules are held together with dipole-dipole interactions. In H20 the molecules are held together by intermolecular H-bonding. Since the new forces of attraction between H₂O and haloalkane molecules are weaker than the forces of attraction already existing between haloalkane haloalkane molecules and water- water molecules, therefore haloalkanes or alkyl halides are immiscible with water.

(c) Grignard reagents (R—Mg—X) should be prepared under anhydrous conditions because these are readily decomposed with water to give alkanes.

That is why ether used as solvent in the preparation of Grignard reagents in completely anhydrous conditions.

Question 20.
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-l-ene as the major product. Write equations for the reaction involved. (4)
Or
Arrange the following compounds in increasing order of their property as indicated:
(a) Acetaldehyde, Acetone, Di-tert-butylketone, Methyl tert-butylketone (Reactivity towards HCN).
(b) CH₃CH₂CH (Br) COOH, CH₃CH (Br) CH₂COOH, (CH₃)₂CHCOOH, CH₃CH₂CH₂COOH (Acid strength)
(c) Benzoic acid, 4-Nitrobenzoic acid, 3,4-Dinitro-benzoic acid, 4-methoxybenzoic acid (Acid strength). (4)
Answer:
(a) The given informations show that the compound [A] upon hydrolysis gives a carboxylic acid [B] and alcohol [C]. It must be an ester.
(b) Since the alcohol [C] upon oxidation with chromic acid gives back the carboxylic acid [B]., both the carboxylic acid and alcohol must have the same number of C-atom i.e., four each.
(c) On dehydration, the alcohol [C] gives an alkene.

The relevant equations for all the; reactions involved may be explained as follows: