RBSE 12th Chemistry Model Paper Set 5 with Answers in English

Students must start practicing the questions from RBSE 12th Chemistry Model Papers Model Paper Set 5 with Answers in English Medium provided here.

RBSE Class 12 Chemistry Model Paper Set 5 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 molality of water is:
(a) 18 mol L^-1
(b) 10^-7mol L^-1
(c) 55.5 mol L^-1
(d) None of these
Answer:
(c) 55.5 mol L^-1

(ii) Cryolite is an ore of:
(a) Fe
(b) Al
(c) Cu
(d) Ag
Answer:
(b) Al

(iii) The mole fraction of one molal aqueous solution is:
(a) 0.031
(b) 0.042
(c) 0.018
(d) 0.06
Answer:
(c) 0.018

(iv) Those natural occurring compounds from which element can be extracted profitably and conveniently known as: (1)
(a) minerals
(b) gangue
(c) ores
(d) flux
Answer:
(c) ores

(v) If two halogen atoms are attached with adjacent carbon atoms, then they are known as: ()1
(a) Polvmethylene dihalide
(b) Gem-dehalide
(c) Vic-dihaiide
(d) Alkylene halide
Answer:
(c) Vic-dihaiide

(vi) In Hinseberg’s test, the product of which amine does not dissolve in NaOH? (1)
(a) CH₃CH₂NH₂
(b) (CH₃)₂CHNH₂
(c) CH₃NHCH₃
(d) CH₃N(CH₃)₂
Answer:
(c) CH₃NHCH₃

(vii) General formula of aliphatic aldehyde is:
(a) C_nH_2n+2O
(b) C_nH_2nO
(c) C_nH_2n-2O
(d) C_nH_2n-1O
Answer:
(b) C_nH_2nO

(viii) Which of the following is strong base?
(a) FCH₂NH₂
(b) FCH₂CH₂NH₂
(c) C₆H₅NH₂
(d) C₆H₅CH₂NH₂
Answer:
(d) C₆H₅CH₂NH₂

(ix) Which compound produces acetaldehyde on heating with aqueous potassium hydroxide: (1)
(a) CH₃COCl
(b) CH₃CH₂Cl
(c) CH₂ClCH₂Cl
(d) CH₃CHCl₂
Answer:
(d) CH₃CHCl₂

Question 2.
Fill in the Blanks:
(i) Packing efficiency of fcc unit cell is _____________ .
Answer:
74%

(ii) Mixture of n-heptaneand n-hexaneisa _____________ soluble.
Answer:
Ideal

(iii) Oxalate is an example of _____________ dentate ligand.
Answer:
Bi

(iv) Carbon atom vinylic in halide is _____________ hybridised.
Answer:
sp²

Question 3.
Very Short Answer Type Questions:
(i) What type of interactions hold the molecules together in a polar molecular solid? (1)
Answer:
Dipole-dipole forces of attractions hold the molecules together in a polar molecular solid.

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

Answer:
IUPAC name: 4-bromo-4-methylpent-2-ene

(iii) Define the following: Schottky defect
Answer:
Schottky defect: If in an ionic crystal of type A⁺ B^–, equal number of cations and anions are missing from their lattice sites so that the electrical neutrality is maintained, it is called Schottky defect.

(iv) Write the IUPAC name of [Pt(NH₃)₄Cl₂]Cl₂. (1)
Answer:
The IUPAC name of the complex, [Pt(NH₃)₄Cl₂]Cl₂ is tetraamminedichlorido polatinum (IV) chloride.

(v) In some cases, it is found that a large number of colliding molecules have energy more than threshold energy, yet the reaction is slow. Why? (1)
Answer:
It is due to improper orientation, or due to another factor P called steric factor, which refers to the orientation of the colliding molecules. These are the two main factors which are responsible for a reaction to occur slowly.

(vi) Why is geometrical isomerism not possible in tetrahedral complexes having two different types of unidentate ligands coordinated with the central metal ion? (1)
Answer:
Tetrahedral complexes do not show geometrical isomerism because the relative positions of the unidentate ligands attached to the central metal atom or ion are the same with respect to each other.

(vii) For a reaction, (1)

(a) Write the order and molecularity of this reaction.
(b) Write the units of k.
Answer:
(a) Order of Reaction = Zero
Molecularity = Two
(b) Unit = Mol L^-1

(viii) Name the only vitamin which can be synthesized in our body. Name the disease caused due to the deficiency of this vitamin. (1)
Answer:
Vitamin which can be synthesized in our body:
Vitamin A Its deficiency causes xerophthalmia.

Section-B
Short Answer Type Questions:

Question 4.
A compound is formed by two elements X and Y. Atoms of the element ‘Y’ (as anions) make ccp and those of the element ‘X’ (as cation) occupy all the octahedral voids. What is the formula of the compound? (1½)
Answer:
Let number of atoms in ccp arrangement = N
Number of octahedral voids = N
As ‘Y’ anions occupy the ccp arrangement hence
Number of ‘Y’atoms N
Number of ‘X’ cation = All the octahedral voids = N
Thus, ratio X: Y
N : N
1 : 1
Hence, formula of the compound is XY.

Question 5.
A compound forms hexagonal close-packed strucutre. What is the total number of voids in 0.5 mol of.it ? How many of these are tetrahedral voids? (1½)
Answer:
As we know that, number of atoms in the close packing = N = 0.5 mol
Number of tetrahedral voids
= 2N = 2 × 0.5 mol
Number of octahedral voids
= N = 0.5 mol
Hence, total number of atoms in close packing
= 0.5 mol
= 0.5 × 6.022 × 10²³
= 3.011 × 10²³ atoms
Number of tetrahedral voids
= 2 × 0.5 mol
= 2 × 0.5 × 6.022 × 10²³
= 6.022 × 023
= 0.5 mol
Number of octahedral voids
= 0.5 × 6.022 × 10²³
= 3.011 × 10²³
Total number of voids
= Octahedral voids + Tetrahedral voids
= 3.011 × 10²³ + 6.022 × 1o²³
= 9.033 × 10²³ voids.

Question 6.
Which compound in each of the following pairs will react faster in S_N2 reaction with – OH?
(a) CH³Br or CH³l
Answer:
CH₃l: Because Iodide is better leaving group than bromide.

(b) (CH³)³ CCl or CH³Cl (¾ + ¾ = 1½)
Answer:
CH₃Cl: Carbon atom leaving group is less hindered.

Question 7.
What criterion is followed for the selection of the stationary phase in chromatography? (1½)
Answer:
Criteria for the selection of stationary phase in chromatography are as follows:

• Impurities should be strongly adsorbed or are more soluble in stationary phase.
• Stationary phase should have high but selective adsorption power.
• The adsorbent should not react chemically with the solvent used for elution or with the components of the mixture under investigation.
• The impurities will be retained by stationary phase whereas the pure component should be easily removed.

Question 8.
Rearrange the compounds of each of the following sets in order of reactivity towards S_N2 ? displacement: (1½)
(a) 2-Bromo-2-methylbutane, 1-Bromopentane, 2-Bromopentane
Answer:
1-Bromopentane > 2-Bromo- pentane > 2-Bromo-2-methylbutane

(b) l-Bromo-3-methylbutane, 2-Bromo-2-methylbutane, 3-Bromo-2-methylbutane
Answer:
1 -Bromo-3-methylbutane > 3-Bromo-2-methylbutane

(c) 1-Bromobutane, l-Bromo-2-, 2-dimethyl-propane, l-Bromo-2-methylbutane.
Answer:
1-Bromobutane > 1-Bromo-2- methylbutane > 1-Bromo-2-, 2- dimethylpropane.

Question 9.
At a site, low grade copper ores are available and zinc and iron scraps are also available. Which of the two scraps would be more suitable for reducing the leached copper ore and why? (1½)
Answer:
Zinc being above iron in the electrochemical series (more reactive metal is zinc), the reduction will be faster in case zinc scrps are used. But zinc is costly metal than iron so using scraps will be advisable and advantageous.

Question 10.
For the reaction: 2N₂O₅(g) → 4NO₂(g) + O₂(g)
the rate of formation of NO₂(g) is 2.8 × 10^-3 Ms^-1. Calculate the rate of disappearance of N₂O₅(?). (1½)
Answer:
For the reaction: 2N₂O₅(g) → 4NO₂(g) + O₂(g) overall the rate of reaction is

Question 11.
How is the variability in oxidation states of transition metals different from that of the non transition metals ? Illustrate with examples. (1½)
Answer:
The oxidation states of transition metals differ from each other by unity due to incomplete filling of d-orbitals while oxidation state of non-transition elements normally differ by two units due to inert pair effect. For example:
Mn → + 2, + 3, + 4, + 5, + 6, + 7
Pb → +2, +4

Question 12.
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. (1½)
Answer:
Given, k = 5.5 × 10^-14 s^-1
For first order reaction,
t_1/2 = \(\frac{0.693}{k}\) = \(\frac{0.693}{5.5 \times 10^{-14}}\)
= 0.126 × 10¹⁴s
= 1.26 × 10¹³s

Question 13.
What are interstitial compounds ? Why are such compounds well known for transition metals ? (1½)
Answer:
Interstitial compounds : The compounds in which small atoms like H, C, N, He etc. occupy interstitial sites in the crystal lattice are called interstitial compounds.

These compounds are well known for transition metals because small atoms can easily occupy the positions in the voids present in the crystal lattices of transition metals.

Question 14.
Write reactions of the final alkylation product of aniline with excess of methyl iodide in the presence of sodium carbonate solution. (1½)
Answer:

Question 15.
Write chemical reaction of aniline with benzoyl chloride and write the name of product obtained. (1½)
Answer:

Section-C
Long Answer Type Questions:

Question 16.
Define the term solution. How many types of solutions are formed ? Write in brief about each type of solution with an example. (3)
Or
Boiling point of water at 750 mm Hg is 99.63°C. How much sugar is to be added to 500 g of water such that it boils at 100°C. (3)
Answer:
A homogeneous mixture of two or more than two components is called solution. A solution mainly has two components:
(i) Solvent: The component which is present in the largest quantity is called solvent. Solvent determines the physical state in which solution exists.
(ii) Solute: The component which is present in lesser quantity is called solute. A solution may contain one or more solutes.
Types of solutions: Depending upon the physical state of solute and solvent, solutions may be classified in following categories:

Question 17.
Give evidence that [Co(NH₃)₅Cl]SO₄ and [Co(NH₃)₅SO₄]Cl are ionisation isomers. (3)
Or
Explain on the basis of valence bond theory that [Ni(CN)₄]^2- ion with square planar structure is diamagnetic and the [NiCl₄]^2- ion with tetrahedral geometry is paramagnetic. (3)
Answer:
Ionisation isomers give different ions in aqueous solution. Therefore, they react differentially with different reagents.
[CO(NH₃)₅Cl]SO₄ gives sulphateions in aqueous solution which gives the precipitate of BaS04 on adding of BaCl₂.

However, it does not react with AgNO₃

[CO(NH₃)₅SO₄]Cl gives chloride ions in aqueous solution which gives the precipitate of AgCl on addition of AgNO₃.

But it does not react with BaCl₂ solution.

Question 18.
(a) Write the products formed when CH₃CHO reacts with the following reagents:
(i) HCN.
(ii) H₂N – OH
(iii) CH₃CHO in the presence of dilute NaOH
(b) Give simple chemical tests to distinguish between the following pairs of compounds: (1½ + 1½ = 3)
(i) Benzoic acid and Phenol
(ii) Propanal and Propanone.
Or
(a) Account for the following:
(i) Cl – CH₂COOH is a stronger acid than CH₃COOH.
(ii) Carboxylic acids do not give reactions of carbonyl group.
(b) Write the chemical equations to illustrate the following name reactions:
(i) Rosenmund reduction
(ii) Cannizzaro reaction
(c) Out of CH₃CH₂ – CO – CH₂ – CH₃ and CH₃CH₂ – CH₂ – CO – CH₃ which gives iodoform test? (1 + 1 + 1 = 3)
Answer:
(a)
(i)

(ii)

(iii)

(b) (i) Benzoic acid and Phenol: On addition of NaHCO₃ to both solutions, carBondioxide gas is evolved with benzoic acid while phenol does not form CO₂.

(ii) Propanal and Propanone: Propanal gives positive test with Fehling solution in which a red ppt. of cuprous oxide is obtained while propanone does not respond to test.

Section – D
Essay Type Questions:

Question 19.
(a) What happens when CH₃ – O – CH₃ is heated with HI? (2 + 2 = 4)
(b) Explain mechanism for hydration of acid catalyzed ethane:

Or
(a) Why phenol Is more acidic than ethanol? (2 + 2 = 4)
(b) Write the mechanism of acid dehydration of ethanol to yield ether:

Answer:
(a) Methly iodide (CH₃I) and Methonal (CH₃OH) are formed when CH₃ – O – CH₃ is heated with HI.

(b) Acid catalyzed hydration: Alkenes react with water in the presence of acid as catalyst to form alcohols.
Mechanism : It involves the following three steps:

Step 1: Protonation of alkene to form carbocation by electrophilic attack of H₃O⁺.

Step 2: Nucleophilic attack of water on carbonation

Step 3 : Deprotonation to form an alcohol

Question 20.
(a) Give reasons:
(i) CH₃ – CHO is more reactive than CH₃COCH₃ towards HCN.
(ii) 4-nitrobenzoic acid is more acidic than benzoic acid.
(b) Describe the following
(i) Acetylation
(ii) Cannizzaro reaction
(iii) Cross aldol condensation (2 + 2 = 4)
Or
(a) Complete the following equations:
(i)

(ii)

(iii)

(b) Distinguish between:

(ii)

Answer:
(a) (i) Because carbonyl carbon of CH₃ – CHO is more electrophilic than CH₃COCH₃, due to only one electron donating CH^–₃ group.
(ii) Because of electron withdrawing nature of -NO₂ group.

(b) (i) Acetylation: Introduction of an acetyl group/ CH₃CO^– by heating an organic compound with acetyl chloride/acetic anhydride.

(ii) Cannizzaro reaction: Aldehydes having no α-hydrogen atom when treated with conc. NaOH, undergoes self oxidation and self-reduction simultaneously.

(iii) Cros’s Aldol Condensation: When aldol condensation is carried out between two different aldehydes or ketones, it is called cross aldol condensation.