RBSE 12th Chemistry Model Paper Set 3 with Answers in English

Students must start practicing the questions from RBSE 12th Chemistry Model Papers Model Paper Set 3 with Answers in English Medium provided here.

RBSE Class 12 Chemistry Model Paper Set 3 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) Solid AB has NaCl type of structure. If radius is 100 pm, then radius of B^– will be: (1)
(a) 241pm
(b) 100 pm
(c) 50 pm
(d) 200 pm
Answer:
(a) 241pm

(ii) The solution of solute per kilogram solvent is called: (1)
(a) Normality
(b) Molarity
(c) Mole fraction
(d) Molality
Answer:
(d) Molality

(iii) The rate of reaction of a substance depends on: (1)
(a) active mass
(b) molecular weight
(c) atomic weight
(d) equivalent weight
Answer:
(a) active mass

(iv) The most abundant metal in the earth’s crust is: (1)
(a) Fe
(b) Al
(c) Ca
(d) Na
Answer:
(b) Al

(v) The species/elemental atom that has minimum number of unpaired electrons is: (1)
(a) Kr⁺
(b) Mn²⁺
(c) Fe³⁺
(d) Fe¹⁺
Answer:
(a) Kr⁺

(vi) Denatured alcohol is:
(a) Ethanol + Methanol
(b) Rectified spirit
(c) Ethanol
(d) Rectified spirit + methanol + naphtha
Answer:
(d) Rectified spirit + methanol + naphtha

(vii) Which of the following is chloral?
(a) CCl₃CH₃
(b) Cl₂CHCOOH
(c) CCl₃CHO
(d) Cl₃CCOOH
Answer:
(c) CCl₃CHO

(viii)

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

(ix) The end product of hydrolysis of starch is:
(a) Glucose
(b) Fructose
(c) Glucose and Frustose
(d) Sucrose
Answer:
(a) Glucose

Question 2.
Fill in the Blanks:
(i) The total number of tetrahedral voids in the face centred unit cell is ______________ .
Answer:
8

(ii) Rate constant and rate of a reaction have the same units for reaction of _______________ order.
Answer:
zero

(iii) CrO₃ is an oxide.
Answer:
acidic

(iv)

Answer:

Question 3.
Very Short Answer Type Questions:
(i) Suppose a solid solution is formed between two substances, one whose particles are very large and the other whose particles are very small. What kind of solid solution is this likely to be? (1)
Answer:
Interstitial solid solution.

(ii) What is the role of zinc metal in the extraction of silver? (1)
Answer:
In the extraction of silver, sine metal acts as a reducing agent.

(iii) Arrange the following compounds in an increasing order to basic strengths in their aqueous solutions: (1)
NH₃, CH₃NH₂, (CH₃)₂NH, (CH₃)₃N
Answer:
Basicity order (due to stability of ammonium cation)

(iv) What is the function of SiO₂ in the metallurgy of copper? (1)
Answer:
Silica (SiO₂) is added in the reverberatory furnace during the extraction of Cu to remove impurities of iron oxide (FeO) present in the ore. Silica here acts as flux and reacts with iron oxide gangue to remove it as slag, iron silicate.

(v) What is the composition of copper matte? (1)
Answer:
Copper matte contains Cu₂S and FeS.

(vi) What is a glycosidic linkage? (1)
Answer:
The two monosaccharide units are joined together through an etheral or oxide linkage formed by loss of a molecule of water. Such a linkage between two monosaccharide units through oxygen atom is called gly- cosidic linkage.

(vii) What are the products of hydrolysis of sucrose? (1)
Answer:
Invert sugar: An equimolar mixture of glucose and fructose is obtained by hydrolysis of sucrose in presence of an acid such as dil. HCI or the enzyme invertase or sucrase and is called invert sugar.

(viii) How is the vapour pressure of a solvent affected when a non-volatile solute is dissolved in it? (1)
Answer:
When a non-volatile solute is added to a solvent, its vapour pressure decreases because some of the surface sites are occupied by solute molecules. Thereby, the fraction of the surface covered by the solvent molecules gets reduced. Thus, less space is available for the solvent molecules to vaporise. Therefore, vapour pressure is also reduced.

Section – B
Short Answer Type Questions:

Question 4.
Define the following:
(a) Ideal solution
Answer:
Ideal solution : The solutions which’ obey Raoult’s law over, the entire range of concentration are known as ideal solutions.

(b) Molarity (M)
Answer:
Molarity (M): Molarity is defined as the number of moles of solute dissolved in 1L or 1 dm³ of the solution.
Molarity =

e.g. 0.25 mol L^-1 (or 0.25 M) solution of NaOH means that 0.25 mole of NaOH has been dissolved in one litre (1 dm³) of solution.

Question 5.
Explain the following terms with suitable examples: Ferromagnetism and Ferrimagnetism. (1½)
Answer:
Ferromagnetic solids: The solids which are strongly attracted by external magnetic field and do not lose their magnetism When (he external field is removed are called ferromagnetic solids. The property, thus exhibited, is termed as ferromagnetism.
Example: Fe, Co and Ni show ferromagnetism at room temperature.

Ferrimagnetic solids: The solids which are expected to show large magnetism due to the presence of impaired electrons but in fact have small net magnetic moment, are called ferrimagnetic solids.
Example: Fe₃O₄ and ferrites.

Question 6.
The initial concentration of N₂O₅ in the following first order reaction N₂O_g(g) → 2NO_2(g) + \(\frac{1}{2}\)O₂(g) was 1.24 × 10^-2 mol L^-1 at 318 K. The concentration of N₂O₅ after 60 minutes was 0.20 × 10^-2 mol L^-1. Calculate the rate constant of the reaction at 318 K. (1½)
Answer:
Initial concentration [A]₀
= 1.24 × 10^-2 mol L^-1
Final concentration [A]
= 0.20 × 10^-2 mol L^-1
time (t) = 60 minutes
For first order reaction,

Question 7.
Name the following coordination, compounds according to IUPAC system of nomenclature.
(a) (CO(NH₃)₄ (H₂O) Cl] Cl₂ (1½)
(b) [CrCl₂(en)₂]Cl.
(where, en = ethane-1,2-diamine)
Answer:
IUPAC name of the following coordination compounds are:
(a) [Co (NH₃)₄ (H₂O) Cl] Cl₂
Tetraamirtiheaquachloridocobalt
(III) chloride

(b) [CrCl₂(en)₂] Cl:
Dichlorido bis-(ethane-1, 2-diamine) chromium (III) chloride.

Question 8.
Draw the structures of isomers, if any and write the names of the following complexes: (1½)
(a) [Cr(NH₃)₄Cl₂]⁺
(b) [Co(en)₃]³⁺
(Atomic no. of Cr = 24, Co = 27)
Answer:
(a) IUPAC name of [Cr(NH₃)₄Cl₂]⁺ is tetraammine dichloride chormium (III)

(b) [Co(en)₃]³⁺

IUPAC name
Tris-(ethane-l, 2-diamine) cobalt (III) ion.
Structure: There are two optical isomers of [Co(en)₃]³⁺, one is dextro and’other is laevo whose structures are:

Question 9.
What will be the effect of temperature on rate constant?
Answer:
It has been observed that rate constant almost becomes double for every 10 K rise in temperature. The ratio of rate constants at the two temperatures differ by 10 K is called temperature coefficient and is generally lies between 2 and 3.
Temperature Coefficient = \(\frac{(\text { Rate constant })_{t+10}}{(\text { Rate constant })_{t}}\)
= \(\frac{k_{t+10}}{k_{t}}\)
≈ 2 or 3
The temperature dependence of the rate of a chemical reaction is explained by Arrhenius equation.
According to Arrhenius equation,
k = Ae^-E_a/RT
where, A= Arrhenius factor or frequency factor or pre-exponential factor
R = Gas constant
T = Temperature
E_a = Activation energy

Question 10.
Name the reagents used in the following reactions:

Answer:
(a) LiAIH₄ (Lithium Aluminium hydride)
(b) KMnO₄ (Alkaline Potassium permanganate).

Question 11.
Arrange the following compounds in increasing order of their property as indicated: (1½)
(a) CH₃COCH₃, C₆H₅COCH₃, CH₃CHO
(reactivity towards nucleophilic addition)
Answer:
C₆H₅COCH₃ < CH₃COCH₃ < CH₃CHO
(Reactivity towards nucleophilic addition)

(b) CI – CH₃ – COOH, F – CH₂ – COOH, CH₃ – COOH
(acidic character)
Answer:

Question 12.
A compound is formed by two elements M and N. The element N forms ccp and atoms of M occupies \(\frac{1}{3}\)rd of tetrahedral voids. What is the formula of the compound? (1½)
Answer:
Let, number of atoms in ccp = x
Number of tetrahedral voids = 2x
Number of octahedral voids = x
As element M occupies \(\frac{1}{3}\)rd of tetrahedral voids hence,
Number of M atoms = \(\frac{1}{3}\) × 2x
= \(\frac{2 x}{3}\)
As element N occupies cap array hence,
Number of N atoms = x
Ratio of M and N
M : N
\(\frac{2 x}{3}\) : x
2 : 3
Formula of compound = M₂N₃.

Question 13.
How will you convert ethanal into the following compounds: (1½)
(a) Butane-1-3-diol,
Answer:

(b) But-2-enal
Answer:

(c) But-2-enoic acid
Answer:

Question 14.
Describe the following giving the relevant chemical equation in each case: (1½)
(a) Carbylamine reaction
Answer:
Carbylamine reaction: Aliphatic and aromatic primary amines on heating with chloroform and ethanolic KOH form isocyanides or carbylamines which are foul smelling substances. This reaction is known as carbylamines reaction.

(b) Hofmann’s bromamide reaction
Answer:
Hofmann’s bromamide reaction: Primary amines can be prepared by treating an amide with Br2 in an aqueous or alcoholic solution of NaOH.

Question 15.
Give a chemical test to distinguish between ethylamine and aniline. (1½)
Answer:
Ethylamine and aniline:
By Azo dye test: It involves the reaction of any aromatic primary amine with HNO₂ (NaNO₂ + dil. HCl) at 273-278 K followed by treatment with an alkaline solution of 2-naphthol when a brilliant yellowy orange or red coloured dye is obtained.

Aliphatic 1° amines under these conditions give a brisk evolution of N₂, gas with the formation of 1° alcohol i.e., solution remains clear.

Section-C
Long Answer Type Questions:

Question 16.
What are the characteristics of the transition elements and why are they called transition elements ? Which of the d-block elements may not be regarded as the transition element ? (3)
OR
Explain the following by giving reasons:
(a) Transition metals and many of their compounds show paramangetic behaviour.
(b) The enthalpies of atomisation of the transition metals are high.
(c) The transition metals generally form coloured compounds.
(d) Transition metals and their many compounds act as good catalyst. (3)
Answer:
General characteristics of transition elements:

• Except mercury which is liquid, all the other transition elements have typical metallic structure and’show all the properties of metals like conductivity, malleability, ductility, metallic luxture, high tensile strength etc.
• They have high melting and boiling points, high enthalpies of vaporisation, high enthalpies of atomisation and high enthalpies of hydration of their ions.
• They are electropositive in nature.
• They show variable oxidation states.
• A number of transition metals and their compounds show catalytic properties.
• Most of the transition metals form coloured compounds.
• These compounds are generally paramagnetic in nature.
• They have a great tendency to form complexes.
• They form interstitial compounds with elements like H, C, B and N.
• They form alloys.
  The d-block elements are called transition elements because these elments have their properties intermediate between those of s and p block elements and represent a change from the most electropositive s-block to the least electropositive p- block elements. Thus acting as transition elements.
  Zn, Cd and Hg have general formula (n – 1) d¹⁰ ns² i.e., they have completely filled electronic configuration for d-shell in ground state as well as in exited state. So they are not regarded as transition elements.

Question 17.
What is crystal field splitting energy ? How does the magnitude of Δ₀ decide the actual configuration of d-orbitals in a coordination entity? (3)
OR
Aqueous copper sulphate solution (blue in colour) gives:
(a) a green precipitate with aqueous potassium fluoride and
(b) a bright green solution with aqueous potassium chloride. Explain these experimental results. (1½ + 1½ = 3)
Answer:
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, Δ₀)

The configuration of d-orbitals in a coordination complex depends on the value of Δ₀ < p, the fourth electron enters one of the orbitals hence configuration becomes (t_2g)^z (e_g)¹ and high spin complex is formed. The ligands if Δ₀ > p, fourth electron get paired up in one of the t_2g orbitals hence configuration becomes (t_2g)⁴ (e_g)⁰ and low spin complex is formed. The ligands for which Δ₀ > p are called strong field ligands.

Question 18.
(a) How are the following obtained? (3)
(i) Benzoic acid from ethyl benzene.
(ii) Benzaldehyde from toluene.
(b) Complete each synthesis by giving the missing material, reagent or products:

(a) Write chemical equations to illustrate the following name reactions:
(i) Cannizzaro reaction
(ii) Hell-Volhard-Zelinsky reaction

(b) Give chemical tests to distinguish between the following pairs of compounds:
(i) Propanal and Propanone
(ii) Acetophenone and Benzophenone
(iii) Phenol and Benzoic acid (1½ + 1½ = 3)
Answer:

Section-D
Essay Type Questions:

Question 19.
(a) State one use each of DDT and iodoform. (1 + 3 = 4)
(b) Which compound in the following couples will react faster in S_N2 displacement and why?
(i) 1-bromopentane or 2-bromopentane
(ii) l-bromo-2-rhethylbutane or 2-bromo-2-methylbutane.
(iii) How would you differentiative between S_N1 and S_N2 mechanisms of substitution reactions? Give one example of each.
OR
Answer the following:
(a) Haloalkanes easily dissolve in organic solvents, why?
(b) What is known as a recemic mixture? Give an example.
(c) Of the two bromoderivatives, C₆H₅CH (CH₃) Br and C₆H₅CH (C₆H₅) Br, which one is more reactive in S_N1 substitution reaction and why?
(d) Although chlorine is an electron withdrawing group, yet it is ortho, para-directing in electrophilic aromatic substitution reactions. Explain why it is so? [1 + 1 + 1 + 1 = 4]
Answer:
(a) Use of DDT (Dichlorodiphenyl Trichloroethane): As a powerful insecticide, it is widely used for sugar¬cane and fodder crops to kill mosquitoes, lice which carry pathogens.

Use of iodoform (CHI₃): It is used as an antiseptic for dressing wounds. Its antiseptic action is due to liberation of iodine when iodoform comes in contact with skin but not due to iodoform itself.

(b) In S_N2 reactions, reactivity depends upon steric hindrance
(i) 1-Bromopentane (1° halogen) having less steric hindrance therefore is more reactive than 2- Bromopentane hence undergoes S_N2 reactions faster.

(ii) 1-Bromo-2-methylbutane having less steric hindrance, is thus more reactive towards S_N2 reaction than 2-bromo-2-methyl butane (more steric hindrance).

(iii)

SN1 (Substitution Nucleophilic Umimolecular)	SN2 (Substitution Nucleophilic Biomolecular)
1. It takes place in 2 steps.	1. It takes place in single step.
2. It follows first order kinetics.	2. It follows second order kinetics.
3. The rate of reaction depends upon the concentration of 3° alkyl halide only and is independent of the concentration of OH– ion Rate = K [3° Alkyl halide]	3. The rate of reaction depends upon the concentration of both the reactants Rate = K [RX] [OH–]
4. The Nu– attacks from front side.	4. The Nu– attacks from back side.
5. The reaction occurs occurs through a stable 3° carbocation.	5. The reaction occurs through an unstable transition state.
6. There is no Walden inversion. Example : Rxn of 3° Butyl bromide with aq KOH step 1:	6. The inversion of configuration occurs which is known as Walden inversion. Example:

Question 20.
(a) Arrange the following compounds in increasing order of boiling points :
(i) Pentan-l-ol, butan-l-ol, butan-2-ol, ethanol, propan-l-ol, methanol.
(ii) pentan-l-ol, n-butane, pentanal, ethoxyethane.
(b) Write the structures of the major products expected from the following reactions:
(i) Mononitration of 3-methylphenol (ii) Dinitration of 3-methylphenol (iii) Mononitration of phenyl ethanoate. (2 + 2 = 4)
OR
Predict the products of the following reactions:
(a) CH₃ – CH₂ – CH₂ – O – CH₃ + HBr→
(b)

(c)

(d)

Answer:
(a) (i) Boiling points increase regularly as the molecular mass increases due to the corresponding increase in their vander Waals forces of attraction. Thus, boiling points increase in the order:
methanol, ethanol, propan-l-ol, butan-l-ol and pentan-l-ol.
Among isomeric alcohols, 2° alcohols have lower boiling points than 1° alcohols due to a corresponding decrease in the extent of H-bonding because of steric hindrance. Thus, b.p. of butan-2-ol is lower than that of butan-l-ol. Thus, the overall increasing order of boiling point is:
methanol < ethanol < propan-l-ol < butan-2-ol < butan-l-ol < pentan-l-ol

(ii) Boiling points of ethers having molecular mass equal to or less than K-butane (i.e., 58 g mol^-1) are higher than those of the corresponding n-alkanes. However, the boiling points of ethers having molecular mass equal to or higher than n-pentane (i.e., 72 g mol^-1) are lower than the corresponding n-alkanes. Thus, b.p. of n-butane is lower than that of ethoxyethane. Further, the b.ps of alcohols are much higher than those of alkanes, and ethers of comparable molecular masses due to intermolecular H-bonding. However, boiling points of aldehydes are lower than those of corresponding alcohols due to absence of H-bonding but are still higher than those of alkanes and ethers of comparable molecular masses. Thus, the overall increasing order of boiling point is:
n-butane < ethoxyethane < pentanal < pentan-l-ol.

(b) Both – OH and – CH₃ groups are o, p-directing. Therefore, positions 2, 4 aftd 6 are activated but due to steric hindrance, substitution does not occur at a position in between the two groups, i.e., position 2. Consequently, electrophilic substitution occurs at 4 and 6 positions giving a mixture of products. .

(i)

(ii)

(iii) -OCOCH₃ is 0, p-directing. However, due to steric hindrance, p- product predominates. Thus,