RBSE 12th Chemistry Model Paper Set 8 with Answers in English

Students must start practicing the questions from RBSE 12th Chemistry Model Papers Model Paper Set 8 with Answers in English Medium provided here.

RBSE Class 12 Chemistry Model Paper Set 8 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) Amongst the following ion which one has the highest paramagnetism? (1)
(a) [Cr(H₂O)₆]³⁺
(b) [Fe(H₂O)₆]²⁺
(c) [CU(H₂O)₆]³⁺
(d) [Zn(H₂O)₆]²⁺
Answer:
(b) [Fe(H₂O)₆]²⁺

(ii) Which forms interstitial compounds? (1)
(a) Fe
(b) Co
(c) Ni
(d) All of these
Answer:
(d) All of these

(iii) Which of the following has highest molar conductivity? (1)
(a) Diamminedichloro platinum (II)
(b) Tetraamminedichlorocobalt (IV) chloride
(c) Potassium hexacyanoferrate (II)
(d) Hexaaquochromiuih (III) chloride
Answer:
(b) Tetraamminedichlorocobalt (IV) chloride

(iv) On treatment with methyl alcohol glucose gives: (1)
(a) α-methylglucoside
(b) β-methyl glucoside
(c) Bothe (a) and (b)
(d) None of these
Answer:
(c) Bothe (a) and (b)

(v) Which compound is obtained during the complete hydrolysis of cellulose? (1)
(a) D-glucose
(b) L-glucose
(c) D-fructose
(d) D-ribose
Answer:
(a) D-glucose

(vi) Which of the following is gem-di halide. (1)
(a) CH₃CHBr₂
(b) BrCH₂CH₂Br
(c) CH₃CHBrCH₂Br
(d) CH₃CHBrCH₂CH₂Br
Answer:
(a) CH₃CHBr₂

(vii) Formaldehyde on reacting with ammonia gives: (1)
(a) Paraformaldehyde
(b) Urotropin
(c) Formose
(d) Bakelite
Answer:
(b) Urotropin

(viii) Formaldehyde on heating with KOH gives: (1)
(a) Acetylene
(b) Methane
(c) Methanol
(d) Ethyl formate
Answer:
(c) Methanol

(ix) In Hinesberg’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

Question 2.
Fill in the Blanks:
(i) Packing efficiency, of ccp is _____________ . (1)
Answer:
74%

(ii) [Pt(NH₃)Cl₂] complex is known as _____________ (1)
Answer:
platin

(iii) _____________ compounds are used as refrigrents. (1)
Answer:
freons

(iv) Vitamin C is a _____________ soluble vitamin. (1)
Answer:
water

Question 3.
Very Short Answer Type Questions:
(i) Name the chief ores of aluminium and zinc. (1)
Answer:
Chief ore of aluminium is bauxite (AlO_x (OH)_{3 – 2x}) where, 0 < x < 1 and of zinc is zinc blende or sphalerite (Zns).

(ii) Write the IUPAC name of Ph-CH=CH-CHO.
Answer:
IUPAC name: 3-phenylprop-2-enal.

(iii) Silver atom has completely filled d-orbitals (4d¹⁰) in its ground state. How can you say that it is a transition element ? (1)
Answer:
Silver can exhibit + 2 O.S. wherein it will have incompletely filled d- orbitals (4d⁹), hence silver is a transition element.

(iv) Write the structural formula of 1-phenylpentan-1-one. (1)
Answer:
1-Phenylpentan-l-one

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

Answer:
2,5-dinitrophenol.

(vi) What is meant by reverse osmosis? (1)
Answer:
If a pressure higher than the osmotic pressure is applied in the solution, the solvent will flow from the solution into the pure solvent through the semi-permeable membrane and the process is called reverse-osmosis.

(vii) State the main advantage of molality over molarity as the unit of concentration. (1)
Answer:
Molality does not change with change in temperature while molarity decrease with rise in temperature.

(viii) Explain the following terms with suitable examples: (1)
(a) Frenkel defect
(b) F-centres
Answer:
(a) Frenkel defect: The defect in which the smaller ion/cation is dislocated to a nearby interstitial site. Example: Silver halides, ZnS.
(b) F-centres: The anion vacancy occupied by an electron is called F-centre in alkali metal halides.
Examples: NaCl, KCl, LiCl

Section-B
Short Answer Type Questions:

Question 4.
Differentiate between ‘minerals’ and ‘ores’. (1½)
Answer:
Minerals: These are the combined states of metal in which metal accurs naturally in the earth’s crust. For example : clay, bauxite, mica, dolomite, zinc blende etc.
Ores: The minerals from which a metal can profitably and conveniently extracted is called ores.
All the ores are minerals but all the minerals are not ores.
For example : Bauxite is an ore but clay and mica are not because aluminium metal can be extracted economically of profitably from bauxite not from clay or mica.

Question 5.
Describe the method for refining nickel. (1½)
Answer:
Mond’s Process : Nickel can be refined by Mond’s process. In this process nickel is first heated with carbon mono oxide at about 330-350 K temperature then it is converted into volatile complex leaving behind non volatile impurities. This volatile complex is further heated at high temperature then it decomposes into pure nickel metal and carbon mono oxide gas.

Question 6.
Define Raoult’s law in its general form in references to solutions. (1½)
Answer:
Raoult’s law: For a solution of volatile liquids, the partial vapour pressure of each component of the solution is directly proportional to its mole fraction present in solution. Thus, for any component, partial vapour pressure.
P ∝ χ ⇒ P = P°χ
where, p° is the vapour pressure of pure component and χ is the mole fraction of that component.

Question 7.
(a) Based on the nature of intermolecular forces, classify the following solids : Benzene, Silver. (½)
Answer:
Benzene: Molecular solid (non-polar)
Silver: Metallic solid

(b) AgCl shows Frenkel defect while NaCl does not Give reason. (½)
Answer:
Due to intermediate radius of AgCl, the size of Ag⁺ is smaller than larger Na⁺ ion of NaCl so it can easily occupy interstitial spaces and shows Frenkel defect.

(c) What type of semiconducfor is formed when Ge is doped with Al? (½)
Answer:
p-type semiconductor is formed when Ge is doped with Al.

Question 8.
(a) For a reaction A + B → Product, the rate law is given by,
Rate = k[A]¹[B]². What is the order of the reaction?
(b) Write the unit of rate constant ‘k’ for the first order reaction. (¾ + ¾ = 1½)
Answer:
(a) The sum of the powers of the concentrations of reactants in the rate law expression is called order of that chemical reaction.
Rate = k[A]¹[B]²
Order of reaction = (1 + 2) = 3

(b) Unit of rate constant for first order reaction is s^-1 or min^-1 or time^-1.

Question 9.
A reaction is of first order with respect to reactant A and of second order with respect to reactant B. How is the rate of this reaction affected when:
(a) the concentration of B alone is increased to three times?
(b) the concentrations of A as well as B are doubled? (¾ + ¾ = 1½)
Answer:
Since, the reaction is of first order w. r. t. A and second order wrt B, then the rate law can be given as, (Rate)₁ = k [A][B]².
(a) When the concentration of B is increased to three times (3B), the fate would be
(Rate)₂ = k [A] [3B]²
(Rate)₂ = 9k [A] [B]² = 9 × (Rate)₁
∴ Rate is increased by 9 times.

(b) (Rate)₃ = k [2A] [2B]²
(Rate)₃ = 2 × 2 × 2 × k [A] [B]²
= 8 × (Rate)₁
∴ Rate is increased by 8 times.

Question 10.
Write the IUPAC nine of the complex [Cr (NH₃)₄Cl₂]⁺. What type of Isomerism does It exhibit? (1½)
Answer:
IUPAC name is tetraamminedichlorido chromium (III) ion.
Isomerism: It shows geometrical isomerism and has two isomers as and trans that can be represented as:

Question 11.
What are transition elements? Write two characteristics of the transition elements. (1½)
Answer:
The elements which lie in between s-and p-block elements in the long form of periodic table belonging to groups 3-12 in which different electrons of d-orbitals are progressively filled in each of the four long periods are called transition elements. Their general electronic configuration is (n – 1)d^1-10ns^1-2. The general charac-teristics of transition elements are high melting and boiling points paramagnetic behaviour, variable oxidation states, catalytic properties, etc.

Question 12.
Define the term mole fraction. (1½)
Answer:
The mole fraction of a component is the ratio of the number of moles of the component to the total number of moles of all the components present in the solution.
Mole fraction of a component

Number of moles of = \(\frac{\text { the component }}{\text { total number of moles of all the components}}\)
For a binary solution, mole fraction of component A,
χ_A = \(\frac{n_{\mathrm{A}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}\)
Similarly, for B,
χ_B = \(\frac{n_{\mathrm{B}}}{n_{\mathrm{A}}+n_{\mathrm{B}}}\)
and
χ_A + χ_B = 1

Question 13.
What is the role of NaOH in the metallurgy of aluminium? (1½)
Answer:
Role of NaOH in the metallurgy of aluminium: The ore of aluminium (bauxite) usually contains SiO₂, iron oxides and titanium oxide (TiO₂) as impurities. Concentration of bauxite ore is carried out by digesting the powdered ore with a concentrated solution of NaOH at 473-523 K and 35-36 bar pressure. This way, Al₂O₃ is leached out as sodium aluminate (and SiO₂ too as sodium silicate) leaving behind the impurities.
Al₂O₃(S) + 2 NaOH(flfl) + 3H₂O(2) → 2Na [AI (OH)₄](aq)

Question 14.
Which is a stronger reducing agent Cr²⁺ or Fe²⁺and why? (1½)
Answer:
Cr²⁺is stronger reducing agent than Fe²⁺.
Reason: In case of Cr²⁺, the configuration changes from d⁴ to d³ and it is oxidised, into Cr³⁺. It is due to extra stability of (t_2g)³ configuration.
But in case of Fe²⁺ ion, the configuration changes from d₆ to d₅ and it changes into Fe³⁺ ion.
(Note : In aqueous medium, d³ configuration is more stable as compared to d⁵ configuration.)

Question 15.
Write down the IUPAC name of the complex [Pt (en)₂Cl₂]²⁺. What type of isomerism is shown by this complex? (1½)
Answer:
IUPAC name of [Pt (en)₂Cl₂]²⁺ is dichlorobis (ethane 1, 2-diamine) platinum (IV) ion. The given complex shows geometrical isomerism and optical-isomerism as follows.

Section-C
Long Answer Type Questions

Question 16.
With the help of diagram, explain the role of activated complex in a reaction. (3)
Or
Mention the factors that affect the rate of a chemical reaction. (3)
Answer:
When the colliding molecules possess the kinetic energy less than the threshold Energy (E_t), they require the energy to achieve E_t. This amount of energy (energy required) is known as activation energy (E_a). After achieving the activation energy (i.e. E_a), they reach to an activated stage. This stage is different from the reactant as well as from the product, e.g. in the reaction between H₂(g) and I₂(g), activated complex has configuration in which H – H and I – I bonds are breaking and H-I bonds are forming as shown below:

Potential energy diagram of this action is shown below

Question 17.
Out of C₆H₅CH₂Cl and C₆H₅CH(Cl)C₆H₅ which is more easily hydrolysed by aqueous KOH?
Or
p-Dichlorobenzene has higher melting point and lower solubility than those of o- and m- isomers. Discuss. (3)
Answer:
C₆H₅CH₂Cl is a primary (1°) aralkyl halide while. C₆H₅CH(Cl)C₆H₅ is a secondary (2°) aralkyl halide.

In S_N1 reactions, the reactivity depends upon the stability of carbocations. Since the carbocation

C₆H₅ gets hydrolysed more easily than C₆H₅CH₂Cl under S_N1 conditions.

Question 18.
(a) Give chemical tests to distinguish between the following pairs of compounds:
(i) Ethanal and Propanal
(ii) Phenol and Benzoic acid
(b) How will you bring about the following conversions ?
(i) Benzoic acid to benzaldehyde
(ii) Ethanal to but-2-enal
(iii) Propanone to propene
Give complete reaction in each case. (4)
Or
(a) How will you bring about the following conversions:
(i) Ethanol to 3-hydroxybutanal
(ii) Benzaldehyde to Benzophenone
(b) An organic compound A has the molecular formula C8H1602. It gets hydrolysed with dilute sulphuric acid and gives a carboxylic acid B and an alcohol C. Oxidation of C with chromic acid also produces, B. C on dehydration reaction gives but-l-ene. Write equations for the reactions involved. (4)
Answer:
(a) (i) Ethanal and Propanal: Ethanal and propanal can be distinguished by iodoform test. Warm each compound with iodine and sodium hydroxide solution in water. Ethanal gives yellow crystals of iodoform while propanal does not respond to iodoform test.

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

(b) (i) Benzoic add to benzaldehyde:

(ii) Ethanal to but-2-enal

(iii) Propanone to propene

Section-D
Essay Type Questions:

Question 19.
Explain the mechanism of the following reactions:
(a) Addition of Grignard’s reagent to the carbonyl group of a compound forming an adduct followed by hydrolysis.
(b) Acid catalysed dehydration of an alcohol forming an alkene.
(c) Acid catalysed hydration of an alkene forming an alcohol. (4)
Or
How are the following conversions carried out?
(a) Propene to propan-2-ol
(b) Benzly chloride to Benzyl alcohol
(c) Anisole top-Bromoanisole. (4)
Answer:
(a) Carbonyl group undergoes nucleophilic addition reaction with Grignard reagent to form an adduct which undergoes hydrolysis to give alcohol in the following manner:

(b) The mechanism of dehydration of ethanol involves the following steps:
Mechanism: It involves the following three step:
Step 1 : Formation of protonated alcohol H H
H-C-C-O-H+H

Step 2: Formation of Carbocation: It is the slowest and rate determining step.

Step 3: Formation ofethene by elimination of a proton

(c) Acid catalysed 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 from carbocation by electrophilic attack of

Step 2: Nucleophilic attack of water on carbocation

Step 3: Deprotonation to form an alcohol

Question 20.
Write chemical equations for the following conversions:
(a) Nitrobenzene to benzoic acid.
(b) Benzyl chloride to 2-phenylethanamine
(c) Aniline to benzyl alcohol. (4)
Or
Give the structure of A, B and C in the following reactions:

Answer:
(a) Nitrobenzene to benzoic acid.

(b) Benzyl chloride to 2-phenylethanamine

(c) Aniline to benzyl alcohol.