RBSE 12th Chemistry Model Paper Set 6 with Answers in English

Students must start practicing the questions from RBSE 12th Chemistry Model Papers Model Paper Set 6 with Answers in English Medium provided here.

RBSE Class 12 Chemistry Model Paper Set 6 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) Which of the following is an outer orbital complex? (1)
(a) [Fe(CN)₆]^4-
(b) [FeF₆]3^–
(c) [Co(NH₃)₆]²⁺
(d) [Cu(CN)₆]²⁺
Answer:
(b) [FeF₆]3^–

(ii) The primary and secondary valencies of chromium in the complex ion dichloridoxalato chromium (III) are respectively? (1)
(a) 3,4
(b) 4,3
(c) 3,6
(d) 6,3
Answer:
(c) 3,6

(iii) The “spin-only” magnetic moment of Ni2 in aqueous solution would be (At No. Ni = 28)? (1)
(a) 0
(b) 1.73
(c) 2.84
(d) 4.90
Answer:
(c) 2.84

(iv) Which of the following does not show by molecular formula C₃H₉N? (1)
(a) 1° amine
(b) 2° amine
(c) 3° amine
(d) quaternary ammonium salt
Answer:
(d) quaternary ammonium salt

(v) How many isomers will be possible for compound having molecular formula C₇H₉N? (1)
(a) 4
(b) 5
(c) 6
(d) 7
Answer:
(b) 5

(vi) The rate constant of reaction depends on: (1)
(a) temperature
(b) mass
(c) weight
(d) catalyst
Answer:
(a) temperature

(vii) \(\frac{1}{2} \frac{d\left[\mathrm{NH}_{3}\right]}{d t}\) in the reaction N₂ + 3H₂ →
(a) Rate of dissociation of ammonia.
(b) Rate of combination of N₂.
(c) Rate of combination of H₂.
(d) Rate of formation of ammonia.
Answer:
(d) Rate of formation of ammonia.

(viii) A molal solution is that in which 1 mole solute is dissolved in:
(a) 1000 g solvent
(b) 1 L solvent
(c) 1 L solution
(d) 22.4 L solution
Answer:
(a) 1000 g solvent

(ix) The total volume of atoms present in face-centred cubic unit cell is:
(a) \(\frac{24}{3}\) πr³
(b) \(\frac{12}{3}\) πr³
(c) \(\frac{16}{3}\) πr³
(d) \(\frac{20}{3}\) πr³
Answer:
(d) \(\frac{20}{3}\) πr³

Question 2.
Fill in the Blanks:
(i) Unit of rate constant for first order reaction is
Answer:
s^-1

(ii) Chemical formula of copper glance is
Answer:
Cu₂S,

(iii) CH₃ – CH₂ – Cl + Ag₂0 (wet)
Answer:
CH₃CH₂ – OH

(iv)

Answer:
picric acid.

Question 3.
Very Short Answer Type Questions:
(i) Define rate of a reaction. (1)
Answer:
Rate of a reaction: Change in concentration of reactants or products in unit time is known as rate of a reaction. Its unit is (mol L^-1)^1-n s^-1.

(ii) What is the number of atoms per unit cell (Z) in a body-centred cubic structure? (1)
Answer:
Contribution by the atoms present at eight corners.
= 8 × \(\frac{1}{2}\) = 1
Contribution by the atoms present at centre = 1
Total number of atoms present in unit cell = 1 + 1 = 2

(iii) Define activation energy. (1)
Answer:
It is the extra energy contained by reactant molecules that results into effective collisions between them to form the products. It is denoted by E_a.

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

Answer:
IUPAC name: 2,5-dichlorotoluene.

(v) State clearly what are known as nucleosides and nucleotides. (1)
Answer:
Nucleosides: A nucleoside contains only two basic components of nucleic acids i.e. a pentose sugar and a nitrogenous base. During their formation 1-position of the pyrimidine or 9-position of the purine moitey is linked to C₁ of the sugar (ribose or deoxyribose) by a β- linkage.

Nucleotide: A nucleotide contains all the three basic components of nucleic acids, i.e. a phosphoric acid group, a pentose sugar and a nitrogenous base. These are formed by esterification of C₅ – OH of the sugar of the nucleoside with phosphoric acid.

(vi) (a) Give two differences between globular and fibrous proteins. (½ + ½ = 1)
Answer:

Globular Proteins	Fibrons Proteins
1. Globular proteins have almost spheroidal shape due to folding of the polypeptide chain.	1. Polypeptide chains of fibrous proteins consist of thread like molecules which tend to lie side by side to form fibres.
2. Globular proteins are soluble in water.	2. Fibrous proteins are insoluble in water.

(b) What change occurs in the nature of egg protein on boiling?
Answer:
Because the egg comes in contact with a solution of higher osmotic pressure, the egg will shrink due to going out of water. This shrinking of egg is called plasmolysis.

(vii) Name the alcohol that is used to make the following ester: (1)

Answer:
Alcohol used: Propan-2-ol

(viii) Write the IUPAC names of the following compounds: (1)
(a) CH₂ = CHCH₂Br
(b) (CCl₃)₃CCl
Answer:
(i) CH₂ = CHCH₂ – Br
IUPAC name: 3-Bromopropene:
(ii) (CCl₃)₃ – C – Cl:
IUPAC name: 2-(Trichloromethyl) – 1,1,1,2, 3, 3, 3-heptachloropropane

Section – B
Short Answer Type Questions

Question 4.
Which compound in each of the following pairs will reacts faster in S_N2 reaction with -OH ?
(i) CH₃Br or CH₃I
(ii)(CH₃)₃CCl or CH₃a (1½)
Answer:
(a) CH₃ – I will reacts faster in S_N2 reaction than CH₃-Br because I- ion is a better leaving group than Br^–ion

(b) CH₃Cl will reacts faster in S_N2 reaction than (CH₃)₃CCl because of less steric hindrance.

Question 5.
Write the equations for the preparation of 1-bromobutane from: (¾ + ¾ = 1½)
(a) 1-butanol
Answer:
Preparation of 1-bromobutane from 1-butanol

(b) but-l-ene
Answer:
Preparation of 1-bromobutane from but-l-ene

Question 6.
State Raoult’s law for a solution containing volatile components. How does Raoult’s law become a special case of Henry’s law? (1½)
Answer:
Raoult’s law as a special case of Henry’s law:
In the solution of a gas in a liquid, one of the components is so volatile that it exists as a gas and its solubility is given by Henry’s law which states that, p = K_Hχ

i.e. partial pressure of the volatile component (gas) is directly proportional to the mole fraction of that component (gas) in the solution.

When the equations of Raoult’s law and Henry’s law are compared, it can be seen that the partial pressure of the volatile component or gas is directly proportional to its mole fraction in solution. Only the proportionality constant KH differs from p°.
Hence, Raoult’s law and Henry’s law has been idential except that their proportionality constants are different.

Therefore, Raoult’s law becomes a special case of Henry’s law in which K_H becomes equal to vapour pressure of pure component p°.

Question 7.
An aqueous solution of sodium chloride freezes below 273 K. Explain the lowering in freez¬ing point of water with the help of a suitable diagram. (1½)
Answer:
When a non-volatile solute is added to a solvent, the freezing point of the solution is always lower than that of pure solvent as the vapour pressure of the solvent decreases in the presence of non-volatile solute.
Plot for the lowering in freezing point of water when NaCl is added to it is shown below.

Question 8.
Explain how vacancies are introduced in an ionic solid when a cation of higher valence is added as an impurity in it (1½)
Answer:
When a Cation of higher valence is added to an ionic solid then impurity defect is produced. In impurity defect, two or more cations of lower valency are replaced by a cation of higher valency in order to maintain the electrical neutrality. Hence, some vacancies are created.For example : in ionic crystal of NaCl, impurity of Sr²⁺ is added as SrCl₂. Then two Na⁺ ions left its lattice site leaving behind holes. One of these hole will be occupied by one Sr²⁺ ion to maintain electrical neutrality and other hole will remain vacant.

Question 9.
Atoms of element B form hep lattice and those of the element A occupy 2jyd of tetrahedral voids. What is the formula of the compound formed by the element A and B? (1½)
Answer:
Let,
Number of atoms in hep lattice = N
Number of atoms in tetrahedral voids =2N
As element B is arranged in hep, lattice, hence the number of atoms of B = N
As element A occupies
= \(\frac{2}{3}\) × Tetrahedral voids
= \(\frac{2}{3}\) × 2N
= \(\frac{4 \mathrm{~N}}{3}\)
Ratio of elements = A : B
= \(\frac{4 \mathrm{~N}}{3}\) : N 4 : 3

Question 10.
Write the reactions involved in the following: (¾ + ¾ = 1½)
(a) Hell-Volhard Zelinsky reaction
Answer:
R – CH₂COOH

(b) Decarboxylation reaction
Answer:

Question 11.
Write the reactions involved in the following reactions : (¾ + ¾ = 1½)
(a) Clemmensen reduction
Answer:
Clemmensen reduction. The carbonyl group of aldehydes and ketones is reduced to CH₂ group on treatment with zinc amalgam and concentrated hydrochloric acid.

(b) Cannizzaro reaction
Answer:
Cannizzaro reaction. Aldehydes, which do not have an a hydrogen atom undergo self oxidation and reduction on treatment with cone, alkali and produce alcohol and carboxylic acid salt.

Question 12.
In the following cases rearrange the compounds as directed:
(a) In an increasing order of basic strength:
C₆H₅NH₂, C₆H₅, N(CH₃)₂, (C₂H₅)₂NH and CH₃NH₂
Answer:
(a) Order of basic strength:

(b) In a decreasing order of basic strength: Aniline, p-nitroaniline and p-toluidine
Answer:
The decreasing order of basic strength:

(c) In an increasing order of pKb values:
C₂H₅NH₂, C₆H₅, NHCH₃, (C₂H₅)₂NH and C₆H₅NH₂ (1½)
Answer:
Increasing order of pK_b values:
(C₂H₅)₂NH < C₂H₅NH₂ < C₆H₅NHCH₃ < C₆H₅NHCH₂
Since, a stronger base has a lower pk_b value therefore basic strength order.
(C₂H₅)₂NH > C₂H₅NH₂ > C₆H₅NHCH₃ > C₆H₅NH₂

Question 13.
Complete the following chemical equations: (1½)
(a)

Answer:

(b) C₆H₅N₂CT + CH₃CH₂OH →
Answer:

(c) RNH₂ + CHCl₃ + KOH →
Answer:

Question 14.
What is essentially the difference between a-form of glucose and j3-form of glucose? Explain. (1½)
Answer:
In α-glucose, the OH group at C₁ is towards right while in β-D-glucose, the OH group at C₁ is towards left.

Question 15.
What are essential and non-essential amino acids in human food? Give one example of each type. (1½)
Answer:
Essential amino acids: Amino acids which the body cannot synthesize are called essential amino acids.
Example: Valine, leucine etc. There-fore they must be supplied in diet.
Non-essential amino acids: Amino acids which the body can synthesize are called non-essential amino acids. Therefore, they may or may not be present in diet.
Example: Glycine, alanine etc.

Section-C
Long Answer Type Questions:

Question 16.
Name the process from which chlorine is obtained as a by-product. What will happen if an aqueous solution of NaCl is subjected to electrolysis ? (3)
Or
Give reasons for the following:
(a) Alumina is dissolved in cryolite instead of being electrolysed directly.
(b) Extraction of copper directly from sulphide ores is less favourable than that from its oxide ore through reduction. (3)
Answer:
Chlorine can be obtained as by¬product in Down’s process, – In Down’s process molten NaCl is subjected to electrolysis.

At Cathode: Na⁺ + e^– → Na
At Anode: Cl^– → \(\frac{1}{2}\)Cl₂ + e^–
Here sodium metal is obtained as main product while chlorine is obtained as by product.
Electrolysis of aqueous NaCl : Chlorine can also be obtained by the electrolysis of aqueous NaCl in Nelson’s Cell. The related reactions are as follow:

At Cathode: Na⁺ and H⁺ both ions migrate towards cathode but H + ions are dicharged preferenlially.
At cathode: H⁺ + e^– → \(\frac{1}{2}\)H₂(g)
At anode : Both Cl^– and OH^– ions migrate towards the anode but Cl^– ions are discharged preferentially.
At cathode: Cl^– → \(\frac{1}{2}\)Cl₂ + e^–
Thus remaining ions i.e. Na+ and OH’ combines with each other and forms NaOH.
Na⁺ + OH^– → NaOH

Question 17.
Account for the following:
(a) The enthalpies of atomisation of the transition metals are high.
(b) The lowest oxide of a transition metal is basic, the highest is amphoterie or acidic.
(c) Cobalt (II) is stable in aqueous solution but in the presence of complexing agents, it is easily oxidised. (3)
Or
Give reasons:
(a) E° value for Mn³⁺/Mn²⁺ couple is much more positive than that for Fe³⁺/Fe²⁺.
(b) Iron has higher enthalpy of atomisation than that of copper.
(c) Sc³⁺ is colourless in aqueous solution whereas Ti³⁺ is coloured. (3)
Answer:
(a) Transition metals have high enthalpies of atomisation due to strong metallic bonding and additional covelent bonding. Metallic bonding is due to their smaller size while covalent bonding is due to d-d overlapping.

(b) In lower oxidation states, transition metals behave like metals and metal oxides are basic in nature. Thus, in lower oxidation states, transition metal oxides are basic.
As the oxidation state increases, its metallic character decreases due to decrease in size, thus, it becomes less metallic or more non-metallic. Oxides of a non-metal may be acidic or neutral. Thus, in higher oxidation states, transition metal oxides are amphoteric or acidic.

(c) In the presence of complexing agents, cobalt gets oxidised from + 2 to + 3 state because it provides energy to remove an electron from Co²⁺. Moreover, Co (III) is more stable than Co(II).

Question 18.
Illustrate the following reactions giving a chemical equation for each:
(a) Kolbe’s reaction
(b) Williamson synthesis of an ether. (3)
Or
Explain the following reactions with an example for each:
(a) Reimer-Tiemann reaction
(b) Friedel-Crafts reaction. (3)
Answer:
(a) Kolbe’s reaction: Phenol reacts with C02 in presence of sodium hydroxide (NaOH) at 4-7 atm and 390-410 K giving salicylic acid.

(b) 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:

Section-D
Essay Type Questions:

Question 19.
Indicate the types of isomerism exhibited by the following complexes and draw the structures of these isomers:
(a) K[Cr(H₂O)₂ (C₂O₄)₂]
(b) [Co(en)₃)Cl₃
(c) [Co(NH₃)₅ (NO₂)] (NO₃)₂
(d) [Pt(NH₂) (H₂O)Cl₂] (4)
Or
Discuss the nature of bonding in the following coordination entites on the basis of valence bond theory:
(a) [Fe(CN)₆]^4-
(b) [FeF₆]_3-
Answer:
(a) The complex K[Cr(H₂O)₂ (C₂O₄)₂ shows geometrical isomerism structure of its ds and trans isomers are as follows:

Optical isomers of ds form are:

Trans isomer does not show geometrical isomerism.
(b) The complex [Co(en)₃]Cl₃ shows optical isomerism and it has two optical isomers.

(c) The complex [Co(NH₃)₅ (NO₂)] (NO₃)₂ shows ionisation isomerism and linkage isomerism.
(i) Ionisation isomerism: [Co(NH₃)₅
(NO₂)] (NO₃)₂ and [Co(NH₃)₅ (NO₃)] (NO₂).(NO₃)₂
(ii) Linkage isomerism: [Co(NH₃)₅ (NO₂)](NO₃)₂ and [Co(NH₃)₅ (ONO) (N0₃)₂
(d) The complex [Pt(NH₃) (H₂O)Cl₂] shows geometrical isomerism.

Question 20.
Give a plausible explanation for each one of the following:
(i) There are two -NH₂ groups in semicarbazide. However, only one such group is involved in the formation of semicarbazones.
(ii) Cyclohexanone forms cyanohydrin in good yield but 2 4,6-taimethylcyclohexanone does not
(b) An organic compound with molecular formula C₉H₁₀O forms 2,4, – DNP derivative, reduces Tollens reagent and undergoes Cannizzaro reaction. On vigorous oxidation it gives 1,2-benzene dicarboxylic acid. Identify the compound. (4)
Or
What is meant by following terms ? Give an example of the reaction in each case: (4)
(a) Cyanohydrin
(b) Acetal
(c) Semicar bazone
(d)Aldol
(e) Hemiacetal
(f) Oxime
(g) Ketal
(h)Imine
(i) 2,4-DNP derivative
(j) Schiff’s base
Answer:
(a) (i) Because one of the – NH₂ in semicarbazide is involved in the resonance with group.

(ii) Due to steric hinderance in 2. 4. 6-trimethylcyclohexanone, it doesn’t react with HGN easily.

(b) Compound C₉H₁₀O, forms 2,4- DNP derivative, so it contains a carbonyl group. Also it reduces Tollens reagent therefore carbonyl group is an aldehyde group. Since it undergoes Cannizzaro’s reaction, aldehyde has no α-hydrogen atom, so compound is C₈H₉ CHO. On vigorous oxidation, compound gives 1,2-benzene dicarboxylic acid hence the compound is