CBSE Previous Year Question Papers Class 12 Chemistry 2016 Outside Delhi

CBSE Previous Year Question Papers Class 12 Chemistry 2016 Outside Delhi

Time allowed: 3 hours
Maximum Marks: 70

General Instructions

• All questions are compulsory.
• Section A: Questions number 1 to 5 are very short answer questions and carry 1 mark each.
• Section B: Questions number 6 to 12 are short answer questions and carry 2 marks each.
• Section C: Questions number 13 to 24 are also short answer questions and carry 3 marks each.
• Section D: Questions number 25 to 27 are long answer questions and carry 5 marks each.
• There is no overall choice. However, an internal choice has been provided in two questions of one mark, two questions of two marks, four questions of three marks and all the three questions of five marks weightage. You have to attempt only one of the choices in such questions
• Use of log tables, if necessary. Use of calculators is not allowed.

CBSE Previous Year Question Papers Class 12 Chemistry 2016 Outside Delhi Set I

Section – A

Question 1.
What type of magnetism is shown by a substance if moments of domains are arranged in the same direction? [1]

Question 2.

which is more reactive towards S_N1 reaction and why? [1]
Answer:

Question 3.
On adding NaOH to ammonium sulphate a colourless gas with pungent odour is evolved which forms a blue coloured complex with Cu²⁺ ion. Identify the gas. [1]
Answer:
The gas evolved is ammonia
(NH₃).(NH₄)₂SO₄(aq) + 2NaOH → Na₂SO₄ (aq) + 2H₂O(l) + 2NH₃(g)
Ammonia reacts with a solution of Cu²⁺ ion to form a deep blue coloured complex, [Cu(NH₃)₄]²⁺.

Question 4.
Write the main reason for the stability of colloidal solutions. [1]
Answer:
All the colloidal particles in a given solution carry the same charge and the dispersion medium has an opposite and equal charge; the system as a whole being electrically neutral. This is the main reason for the stability of the colloidal solution.

Question 5.
Write the IUPAC name of the given compound. [1]

Answer:

Question 6.
When a coordination compound CrCl₃.6H₂O is mixed with AgNO₃, 2 moles of AgCl are precipitated per mole of the compound. Write [2]

1. The structural formula of the complex.
2. IUPAC name of the complex

Answer:

1. The structural formula of the complex is [Cr(H₂O)₅ Cl] Cl₂.H₂O because two moles of chlorine are outside coordination entity to form two moles of AgCl from per mole compound.
2. Pentaaquachlorido Chromium(III) chloride monohydrate.

Question 7.
From the given cells: [2]
Lead storage cell, Mercury cell, Fuel cell and Dry cell
Answer the following:

1. Which cell is used in hearing aids?
2. Which cell was used in the Apollo Space Programme?
3. Which cell is used in automobiles and inverters?
4. Which cell does not have a long life?

Answer:

1. Hearing aid-Mercury cell.
2. Apollo Space Programme-Fuel cell.
3. Automobile and inverters-Lead storage cell.
4. A cell does not have a long life-Dry cell.

Question 8.
When chromite ore FeCr₂O₄ is fused with NaOH in presence of air, a yellow coloured compound (A) is obtained which on acidification with dilute sulphuric acid gives a compound (B), compound (B) on reaction with KCl forms an orange coloured crystalline compound (C). [2]
(i) Write the formulae of the compounds (A), (B) and (C).
(ii) Write one use of the compound (C).
OR
Complete the following chemical equations:
(i) 8MnO₄^– + 3S₂O₃^2- + H₂O →
(ii) Cr₂O₃^2- + 3Sn²⁺ + 14 H⁺ →
Answer:
(i) On fusing chromite ore with Sodium hydroxide in presence of air, yellow coloured “sodium chromate” (A) is formed.
FeCr₂O₄ + 16 NaOH + 7O₂ → 8Na₂CrO₄(A) + 2Fe₂O₃ + 8H₂O
On acidification with dil. H₂SO₄ it forms sodium dichromate (B).
2Na₂CrO₄(A) + H₂SO₄ → Na₂Cr₂O₇(B) + Na₂SO₄ +H₂O.
Compound (B) i.e. sodium dichromate forms potassium dichromate, orange coloured crystals (C) on treating with KCl.
Na₂Cr₂O₇ (B) + 2KCl → K₂Cr₂O₇ (C) + 2NaCl.
The formula of compounds are:
(A) Sodium chromate – Na₂CrO₄
(B) Sodium dichromate – Na₂Cr₂O₇
(C) Potassium dichromate – K₂Cr₂O₇

(ii) Potassium dichromate, K₂Cr₂O₇ is most commonly used as an oxidizing agent in various laboratory and industrial applications.
OR
(i) 8MnO₄^– + 3S₂O₃^2- + H₂O → 8MnO₂ + 6SO₄^2- + 2OH^–
(ii) Cr₂O₇^2- + 3Sn²⁺ + 14H⁺ → 3Sn⁴⁺ + 2Cr³⁺ + 7H₂O

Question 9.
Write the mechanism of the following reaction: [2]

Answer:
Formation of ether from alcohol is a nucleophilic bimolecular reaction (S_N2). A protonated alcohol is attacked by another alcohol molecule.
Reaction Steps:

Question 10.

1. Write the order and molecularity of this reaction.
2. Write the unit of k. [2]

Answer:

1. For any reaction; Rate = K [A] order [A] = concentration of reactant. Hence its zero-order reactions and its molecularity are two.
2. Unit of K for a zero-order reaction is mol L^-1 sec^-1.

Question 11.
The rate constant for the first-order decomposition of H₂O₂ is given by the following equation:

Calculate Ea for this reaction and rate constant K if its half-life period is 200 minutes.
(Given: R = 8.314 JK^-1 mol^-1) [3]
Answer:
According to the Arrhenius equation

Question 12.
(i) Differentiate between adsorption and absorption.
(ii) Out of MgCl₂ and AlCl₂, which one is more effective in causing coagulation of negatively charged sol and why?
(iii) Out of sulphur sol and proteins, which one forms multimolecular colloids? [3]
Answer:
(i)

S.No.	Adsorption	Absorption
1.	It is a surface phenomenon, Adsor­bate molecules are held at the surface of the adsorbent.	Absorption occurs in the bulk of absorbing sub­stance.
2.	Initially, the rate of adsorption is rapid.	Absorption occurs at a uniform rate.
3.	The concentration of the adsorbent surface is much more than that in the bulk.	Absorbing material is uniformly distributed throughout the bulk means the concentration is the same throughout and it is essentially a bulk phe­nomenon.
4.	E.g., water vapours on silica gel.	E.g., Water vapours are absorbed by anhydrous CaCl2.

(ii) According to Hardy-Schulze law, the ions carrying the opposite charge to that on sol are responsible for coagulation of the sol. These are called active ions. Hence as the sol is negative, Mg²⁺ and Al³⁺ ions will cause coagulation.
As coagulation power of electrolyte is proportional to the valency of oppositely charged ion, so AlCl₃ will be more effective than MgCl₂.

(iii) Sulphur sol will form the multimolecular colloid. A sol of sulphur consists of colloidal particles which are aggregates of S₈ molecules.

Question 13.
Give reasons: [3]
(i) C-Cl bond length in chlorobenzene is shorter than C-Cl bond length in CH₃-Cl.
(ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(iii) S_N1 reactions are accompanied by racemization in optically active alkyl halides.
Answer:
(i) C-Cl bond length in chlorobenzene is shorter than CH₃-Cl, C-Cl bond as in chlorobenzene due to resonance C-Cl bond has a partial double bond character which reduces the bond length.

(ii) In cyclohexyl chloride, the carbon in C-Cl bond is sp³ hybridised whereas in chlorobenzene C-Cl bond carbon is sp² hybridised, sp² is more electronegative than sp³ carbon. Hence C-Cl bond of chlorobenzene is less polar.

(iii) In S_N1 reaction a carbocation intermediate is formed. In the case of optically active alkyl halide, the attack of nucleophile in the next step to carbocation can occur from both the faces of the trigonal planar species in equal probability. Thus 50 : 50 racemic mixture is obtained.

Question 14.
An element crystallizes in a f.c.c. lattice with a cell edge of 250 pm. Calculate the density if 300 g of this element contain 2 × 10²⁴ atoms. [3]

Question 15.
Give reasons: [3]

1. Mn shows the highest oxidation state of +7 with oxygen but with fluorine, it shows the highest oxidation state of +4
2. Transition metals show variable oxidation states.
3. Actinoids show irregularities in their electronic configurations.

Answer:

1. Mn shows the highest oxidation state of +7 with oxygen because it can form pπ-dπ multiple bonds. On the other hand, Mn shows the highest oxidation state of +4 with fluorine because it can form only a single bond.
2. Transition metals show variable oxidation state because of use of ns and (n-1) d shell electrons while bonding as the shells have similar energy.
3. Actinoids show irregularities in their electronic configurations because 6d, 7s and 5f electrons or shells have less energy difference and electrons can be accommodated in any of them.

Question 16.
Write the main product(s) in each of the following reactions: [3]

Answer:

Question 17.

1. Name the method of refining of metals such as Germanium.
2. In the extraction of Al, impure Al₂O₃ is dissolved in conc. NaOH to form sodium aluminate and leaving impurities behind. What is the name of this process?
3. What is the role of coke in the extraction of iron from its oxides? [3]

Answer:

1. Zone refining method is used for refining of metals such as germanium which is based on the principle that the impurities are more soluble in the molten state (melt) than in the solid-state of the metal.
2. Leaching: This method consists of treating the powdered ore with a suitable reagent which can selectively dissolve the ore but not the impurities.
3. Coke acts as a reducing agent and it reduces the iron ore hematite.

Question 18.
Calculate e.m.f. of the following cell at 298 K. [3]

Answer:

Question 19.

1. Write the name of two monosaccharides obtained on hydrolysis of lactose sugar.
2. Why Vitamin C cannot be stored in our body?
3. What is the difference between a nucleoside and nucleotide? [3]

Answer:

1. The two monosaccharides are β-D-galactose and β-D-glucose.
2. Vitamin C is a water-soluble vitamin and hence get excreted by the urine. So it cannot be stored in the body and needs to be supplemented regularly.
3. When a base (purine or pyrimidine) get attached to 1′ position of a pentose sugar a nucleoside is formed. When a nucleoside is further linked to phosphoric acid at 5′ position of the sugar moiety, we get a nucleotide.
   

Question 20.
(a) For the complex [Fe(CN)₆]^3-, write the hybridization type, magnetic character and spin nature of the complex. (At number Fe = 26).
(b) Draw one of the geometrical isomers of the complex [Pt (en)₂ Cl₂]²⁺, which is optically active? [3]
Answer:

Hence the complex has d²sp³ hybridization.
Type: Octahedral complex, inner orbital complex.
Magnetic character: One impaired electron hence paramagnetic.
It will have a total electron spin moment of 1 electron:
(µ) = √1(1+2) = √3 BM, Low spin complex

(b) [Pt (en)₂Cl₂]²⁺, that is isomer is optically active.

Question 21.
Write the structure of A, B and C in the following:

Answer:

Question 22.
(i) What is the role of r-butyl peroxide in the polymerization of ethane?
(ii) Identify the monomers in the following polymer:
—[NH—(CH₂)₆—NH—CO—(CH₂)₄—CO—]_n
(iii) Arrange the following polymers in the increasing order of their intermolecular forces.
Polystyrene, Terylene, Buna-S. [3]
OR
Write the mechanism of free radical polymerisation of ethene.
Answer:
(i) The polymerisation of ethene to low-density polyethene (L.D.P.) needs the presence of a free radical generating initiator (catalyst), t-butyl peroxide helps in starting the chain of radical formations.

This is Nylon-6,6 and its monomers are:
hexamethylene diamine
H₂N—(CH₂)₆—NH₂ and adipic acid
HOOC—(CH₂)₄—COOH.
(iii) Buna-S (Elastomer) < Polystyrene (Thermoplast) < Terylene (Fibre)
OR
Mechanism of Polymerisation of ethene.
1. Initiation: The process starts with the formation of a free radical by addition of catalyst-free radical like phenyl or benzoyl etc., generating a new and larger free radical.

2. Propagation: The radical reacts with another molecule of ethene thus forming a bigger radical molecule. The process continues until the required length of chain we need.

3. Chain Termination: Where free radical combine with each other the chain terminates resulting in the formation of a polymer.

Question 23.
Due to a hectic and busy schedule, Mr Angad made his life full of tensions and anxiety. He started taking sleeping pills to overcome depression without consulting the doctor. Mr Deepak a close friend of Mr Angad advised him to stop taking sleeping pills and suggested to change his lifestyle by doing Yoga, meditation and some physical exercise. Mr Angad followed his friend’s advice and after a few days, he started feeling better.
After reading the above passage answer the following: [4]
(i) What are the values (at least two) displayed by Mr Deepak?
(ii) Why is it not advisable to take sleeping pills without consulting doctor?
(iii) What are tranquilizers? Give two examples.
Answer:
(ii) Sleeping pills are tranquilizers and may cause harmful side effects as they slow down the working of the brain and nervous system. Hence a doctor must be consulted to regularise the doses of such drugs.

(iii) Tranquilizers are a class of drugs or chemicals which are used to treat stress and mental disease.
Example: Iproniazid and Equanil.

Question 24.
(a) Write the structures of A, B, C, D and E in the following reactions: [5]

OR
(a) Write the chemical equation for the reaction involved in the Cannizzaro reaction.
(b) Draw the structure of the semicarbazone of ethanal.
(c) Why the pK_a of F-CH₂-COOH is lower than that of Cl-CH₂-COOH?
(d) Write the product in the following reaction

(e) How can you distinguish between propanal and propanone?
Answer:


OR
(a) For aldehydes which do not have α-hydrogen atom self oxidation and reduction takes place in the presence of concentrated alkali. This produces one mole of alcohol and one mole of salt of carboxylic acid. This is called Cannizzaro’s reaction.

(c) pK_a of F-CH₂-COOH is lower than that of pK_a of Cl-CH₂-COOH as F-CH₂-COOH is a stronger acid. This is because of higher electronegativity of F atom than Cl atom.

(e) Tollen’s reagent will give a positive test of silver mirror formation with propanal, while propane does not give this test since aldehydes can oxidise Tollens’ reagent to metallic silver but ketones cannot.

Question 25.
(a) Calculate the freezing point of a solution when 1.9 g (of MgCl₂ (M = 95 g/mol) was dissolved in 50 g of water, assuming MgCl₂. Undergoes complete ionization. [5]
(K_f for water = 1.86 K kg mol^-1)
(b) (i) Out of 1 M glucose and 2 M glucose, which one has a higher boiling point and why?
(ii) What happens when the external pressure applied becomes more than the osmotic pressure of a solution?
OR
(a) When 2.56 g of sulphur was dissolved in 100 g of CS2, the freezing point lowered by 0.383 K. Calculate the formula of sulphur (S_x).[5] (K_f for CS₂ = 3.83 K kg mole^-1, Atomic mass of sulphur = 32 g/mol^-1).
(b) Blood cells are isotonic with 0.9% sodium chloride solution what happens if we place blood cells in a solution containing.
(i) 1.2% sodium chloride solution?
(ii) 0.4% sodium chloride solution?
Answer:
(a) MgCl2 on ionisation gives 3 ions each mole.

(b) (i) 2M glucose will have a higher boiling point because the boiling point of a solution of a non-volatile liquid increases with increase in concentration
(ii) When the external pressure exerted on the solution is higher than the osmotic pressure, pure solvent starts flowing out of the solution through the semi-permeable membrane. This process is known as reverse osmosis.

(b) (i) 1.2 % Sodium chloride is hypertonic than blood cells, hence cells will shrink. Plasmolysis will take place.
(ii) 0.4% Sodium chloride solution is hypotonic than a blood cell, so cells will swell. Endo osmosis will take place.

Question 26.
(a) Account for the following: [5]
(i) Ozone is thermodynamically unstable.
(ii) Solid PCl₅ is ionic in nature.
(iii) Fluorine forms only one oxoacid HOF.
(b) Draw the structure of:
(i) BrF₅
(ii) XeF₄
OR
(i) Compare the oxidising action of F₂ and Cl₂ by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
(ii) Write the conditions to maximize the yield of H₂SO₄ by contact process.
(iii) Arrange the following in the increasing order of property mentioned.
(a) H₃PO₃ / H₃PO₄, H₃PO₂ (Reducing Characters)
(b) NH₃ / PH₃, AsH₃, SbH₃ / BiH₃ (Base strength)
Answer:
(a) (i) Ozone easily decomposes to give nascent oxygen:
O₃ → O₂ + [O]
because the reaction is exothermic, (∆H = negative), and results in the increases in entropy (∆S = positive). Overall Gibb’s energy change is quite high and negative.
(iii) Due to high electronegativity and small size fluorine forms only one oxoacid, HOF.

(i) Fluorine is a much stronger oxidizing agent than chlorine. The oxidizing power depends on three factors.
(a) Bond dissociation energy.
(b) Electron gain enthalpy.
(c) Hydration enthalpy.

The electron gain enthalpy of chlorine is more negative than that of Fluorine. However, the bond dissociation energy of fluorine is much lesser than that of chlorine. Also, because of its small size, the hydration energy of fluorine is much higher than that of chlorine. Therefore, the latter two factors compensate more than for the less negative electron gain enthalpy of fluorine. Thus, fluorine is a much stronger oxidizing agent than chlorine.

(ii) The condition necessary to maximize the yield of H₂SO₄ by contact process is.
(a) A moderately low temperature of about 720 K and a high pressure of about 2 bar yields maximum H₂SO₄ acid.
(b) It’s an exothermic reaction and the forward reaction causes a decrease in pressure.

Note: All questions are same in Outside Delhi Set II and III and Delhi Set-I, II and III

CBSE Previous Year Question Papers