CBSE Class 12 Chemistry Question Paper 2016 (Delhi) with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2016 (Delhi) to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Chemistry Question Paper 2016 (Delhi) with Solutions

General Instructions:

Time allowed : 3 hours
Maximum marks : 70

General Instructions:

1. All questions are compulsory.
2. Questions number 1 to 8 are very short answer questions and carry 1 mark each.
3. Questions number 9 to 18 are short answer questions and carry 2 marks each.
4. Questions number 19 to 27 are also short answer questions and carry 3 marks each.
5. Questions number 28 to 30 are long answer questions and carry 5 marks each.
6. Use log tables if necessary. Use of calculators is not allowed.
   Deleted from Syllabus.

Question 1.
Out of and which is more reactive towards SN1 reaction and why? [1]
Answer:
(2-Chlorobutane) is more reactive than
(1-Chloro-2-methylpropane) because 2-Chlorobutane forms a more stable 2° carbocation.

Question 2.
On adding NaOH to ammonium sulphate, a colourless gas with pungent odour is evolved which forms a blue coloured complex with Cu ²⁺ ion. Identify the gas. [1]
Answer:
The gas with a pungent odour is Ammonia (NH ₃ ) and the blue coloured complex is Tetra- ammine copper (II) sulphate monohydrate.

Question 3.
What type of magnetism is shown by a substance if magnetic moments of domains are arranged in same direction? [1]
Answer:
Ferromagnetism is shown by a substance if magnetic moments of domains are arranged in same direction.

Question 4.
Write the IUPAC name of the given compound: NH ₂

Answer:

Question 5.
Write the main reason for the stability of colloidal sols. [1]
Answer:
The stability of the colloidal solution is because of solvation and the presence of charge on the dispersed phase particles.

Question 6.
From the given cells: [2]
Lead storage cell, Mercury cell, Fuel cell and Dry cell Answer the following:
(i) Which cell is used in hearing aids?
(ii) Which cell was used in Apollo Space Programme?
(iii) Which cell is used in automobiles and inverters?
(iv) Which cell does not have long life?
Answer:
(i) Mercury cell is used in hearing aids.
(ii) Fuel cell was used in Apollo Space Programme.
(iii) Lead storage cell is used in automobiles and inverters.
(iv) Dry cell does not have a long life.

Question 7.
When chromite ore FeCr ₂ O ₄ 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).
(i) Write the formulae of the compounds (A), (B) and (C).
(it) Write one use of compound (C). [2]
Answer:
The chromite ore FeCr ₂ O ₄ on fusion with NaOH in presence of air, forms a yellow coloured compound (A) i.e. Sodium chromate.

Sodium chromate (A) upon acidification with dilute sulphuric acid gives Sodium dichromate (B).

Sodium dichromate (B) on reaction with KCl forms orange coloured compound Potassium dichromate (C).

(i) Thus (A) → Sodium chromate Na ₂ CrO ₄
(B) → Sodium dichromate Na ₂ Cr ₂ O ₇
(C) → Potassium dichromate K ₂ Cr ₂ O ₇
(ii) (C) is used as a strong oxidising agent in acidic medium in volumetric analysis.

Or

Complete the following chemical equations:

Answer:

Question 8.
When a co-ordination compound CrCl ₃ .6H ₂ O is mixed with AgN0 ₃ , 2 moles of AgCl are precipitated per mole of the compound. Write
(i) Structural formula of the complex.
(ii) IUPAC name of the complex. [2]
Answer:
(i) The complex formed on mixing a coordination compound CrCl ₃ .6H ₂ 0 with AgN0 ₃ is as follows

CrCl ₃ .6H ₂ 0 + AgNO ₃ → [Cr(H ₂ O) ₅ Cl]Cl ₂ . H ₃ 0

(ii) Pentaaquachloridochromium (III) chloride monohydrate

Question 9.
For a reaction:

(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k.
Answer:

(i) It is a zero order reaction and its molecularity is two.
(ii) Unit of k is mol L ^{-1 s -1 .}

Question 10.
Write the mechanism of the following reaction:

Answer:
Mechanism:

Question 11.
Give reasons: [3]
(i) C-Cl bond length in chlorobenzene is shorter than C-Cl bond length in CH ₃ -Cl.
(ii) The dipole moment of chlorobenzene is lower than that of cyclohexyl chloride.
(iii) SN1 reactions are accompanied by racemization in optically active alkyl halides.
Answer:
(i) In haloalkanes, the halogen atom rs attached to sp ³ -hvhrid lied carbon while in haloarenes it is attached to sp-hvhridized carbon whose size is smaller than sp ³ orbital carbon. Therefore, C-Cl bond in chlrobenzene is shorter than alkvl chloride.

(ii) In chlorobenzene, the chlorine atom is attached to sp ² hybridized carbon atom (more ‘s’ character). In cyclohexvl chloride, the chlorine atom is attached to sp ³ hybridized carbon atom (less ‘s’ character than sp ² hybridized carbon atom). Hence chlorobenzene is more electronegative than cyclohexyl chloride. Therefore, the density of electrons C-Cl bond at chlorine atom is less in chlorobenzene than cyclohexyl chloride.

The density of electrons C-Cl bond at chlorine atom in chlorobenzene decreases due to the -R effect of the benzene ring which is not in cyclohexyl chloride. Due to this polarity of the C-Cl bond decreases and hence dipole moment of chlorobenzene is lower than cyclohexyl chloride.

(iii) S _N 1 reactions are accompanied by racemization in optically active alkyl halides because the nucleophile will have an equal opportunity to attack on sp ² hybridised carbocation from either sides to give a racemic mixture.
Example:

Question 12.
An element crystallizes in a f.c.c. lattice with cell edge of 250 pm. Calculate the density if 300 g of this element contains 2 × 10 ²⁴ atoms. [3]
Answer:
Given: a = 250 pm = 250 × 10 ^-10 cm
z = 4 (for fee)
M = ? d = ?
Using formula : d = \(\frac{z \times \mathrm{M}}{a^3 \mathrm{~N}_{\mathrm{A}}}\)
∵ 2 × 10 ²⁴ atoms of an element have mass = 300 g
∴ 6.022 × 10 ²³ atoms of an element have mass = \(\frac{300 \times 6.022 \times 10^{23}}{2 \times 10^{24}}\) = 90.33 g

Now M = 90.33 g
Substituting all values in formula

d = \(\frac{4 \times 90.33}{\left(250 \times 10^{-10}\right)^3 \times N_0}\)

= \(\frac{4 \times 90.33}{(250)^3 \times 6.022 \times 10^{23} \times 10^{-30}}\) [∵ N = 6.022 × 10 ²³ ]

= 38.4 gm cm ^-3

Question 13.
The rate constant for the first order decomposition of H2Oz is given by the following equation: [3]

log k = 14.2 – \(\frac{1.0 \times 10^4}{T}\) K

Calculate Ea for this reaction and rate constant k if its half-life period be 200 minutes.
(Given: R = 8.314 J K ^-1 mol ^-1 )
Answer:
Given: t _1/2 = 200 min
E _a = ?, T = ?
Using Arhenius equation
log k = log A – \(\frac{\mathrm{E}_{\mathrm{a}}}{2.303 \mathrm{RT}}\)

On comparing above equation with the given equation

Question 14.
(i) Differentiate between adsorption and absorption.
(ii) Out of MgCl2 and AlCly 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) Tablee 1
Adsorption
(a) It is the phenomenon by which one substance gets concentrated mainly on the surface of the other substance rather than in the bulk of a solid or liquid.
(b) It is a surface phenomenon.
(c) Its concentration is different at surface from the bulk.

Absorption
(a) It is the phenomenon by which one substance gets uniformly distributed thoughout the body of the other substance.
(b) It is a bulk phenomenon.
(c) Its concentration is same throughout th- bulk.

(ii) AlCl ₃ is more effective than MgCl ₂ in causing coagulation of negatively charged sol because coagulating power of an electrolyte is directly proportional to the valency of the active ion i.e., Al ³⁺ > Mg ²⁺ .

(iii) Sulphur sol forms multimolecular colloids.

Question 15.
(i) Name the method of refining of metals such as Germanium.
(ii) In the extraction of Al, impure Al ₂ O ₃ is dissolved Hn cone. NaOH to form sodium aluminate and leaving impurities behind. What is the name of this process?
(iii) What is the role of coke in the extraction of iron from its oxides? [3]
Answer:
(i) Zone refining is used for refining of Germanium.
(ii) Leaching of Alumina or Bayer’s process.
(iii) Coke i.e., CO acts as a reducing agent in the extraction of iron from its oxides in the zone of reduction in blast fumance at 1123 K.

FeO + CO → Fe + CO ₂

Question 16
Calculate e.m.f. of the following cell at 298 K: [3]

2Cr(s) + 3Fe ²⁺ (0.1 M) → 2Cr ³⁺ (0.01 M) + 3 Fe(s)

Given: E°(Cr ³⁺ | Cr) = -0.74 V; E°(Fe ²⁺ | Fe) = -0.44 V

Cell reaction:

2Cr(s) + 3Fe ²⁺ (0.1 M) → 2Cr ³⁺ (0.01 M) + 3 Fe(s)

Given: E°(Cr ³⁺ | Cr) = -0.74 V
E°(Fe ²⁺ | Fe) = -0.44 V
E _cell = ?
E° _cell = E° _cathode – E° _anode
= (-0.44) – (-0.74)
= -0.44 + 0.74 = 0.30 V

Using Nernest equation

Question 17.
Give reasons: [3]
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4. (2011 O.D.)
(ii) Transition metals show variable oxidation states. (2009 Delhi)
(iii) Actinoids show irregularities in their electronic configurations.
Answer:
(i) Because oxygen stabilizes the highest oxidation state (+7 of Mn) even more than fluorine i.e., +4 since oxygen has the ability to form multiple bonds with metal atoms.
(ii) The variability of oxidation state of transition elements is due to incompletely filled d-orbitals and presence of unpaired electrons.
(iii) This happens because the energy difference between 5f, 6d and 7s subshells of the actinoids is very small and hence electrons can be accomodated in any of them.

Question 18.
Write the main product(s) in each of the following reactions: [3]

Answer:

Question 19.
Write the structures of A, B and C in the following:

Answer:

Question 20.
(i) What is the role of t-butyl peroxide in the polymerization of ethene?
(ii) Identify the monomers in the following polymer.

[-NH-(CH ₂ ) ₆ -NH-CO-(CH ₂ ) ₄ -CO-] _n

(iii) Arrange the following polymers in the increasing order of their intermolecular forces: Polystyrene, Terylene, Buna-S [3]
Answer:
(i) t-butyl peroxide acts as a radical initiator or chain initiator in the polymerization of ethene. This initiator readily decomposes on mild heating to form initiator free radicals which adds to double bond of ethene molecule to form a new and larger free radical,

(ii) Adipic acid HOOC(CH2)4—COOH and Hexamethylene diamine NH ₂ (CH ₂ )6NH ₂ .

(iii) The increasing order of polymers on the basis of their intermolecular forces:
Buna-S < Polystyrene < Terylene

Or

Write the mechanism of free radical polymerization of ethene.
Answer:
Free radical polymerization of ethene:
This process normally occurs at high temperature and under high pressure in presence of small amounts of benzoyl peroxide as the initiator in three steps:
(i) Chain initiation step. Benzoyl peroxide undergoes homolysis and works as chain initiator.

(ii) Chain Propagating step:

(iii) Chain Terminating step:

Question 21.
(i) Write the name of two monosaccharides obtained on hydrolysis of lactose sugar. (2012 Comptt. Delhi)
(ii) Why Vitamin C cannot be stored in our body?
(iii) What is the difference between a nucleoside and nucleotide? (2010 Delhi) [3]
Answer:
(i) On hydrolysis, lactose gives ß-D-galactosa and ß-D-glucose.

(ii) Vitamin C is mainly ascorbic acid which is water soluble and is readily excreted through urine and thus cannot be stored in the body.
Nucleoside = Sugar + Base

(iii) Nucleoside. A nucleoside contains only two basic components of nucleic acids, i.e., a pentose sugar and a nitrogenous base. It is formed by the attachment of a base to V position of sugar.

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

Question 22.
(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 Cl ₂ ] ²⁺ which is optically active. [3]
Answer:
(a) [Fe(CN) ₆ ] ^3-
The element Fe is in +3 oxidation state. As CN ^– ion is a strong field ligand, the electron pairing is possible in this case.

(i) The hybridisation is of d ² sp ³ type.
(ii) There is one unpaired electron so it is paramagnetic in nature.
μ = \(\sqrt{n(n + 2)}\) BM
= \(\sqrt{1(1 + 2)}\) BM
∴ μ = 1.73 BM

(iii) It is inner orbital or low spin complex.

(b) cis form of [Pt(en) ₂ Cl ₂ ] ²⁺ will be optically active

Question 23.
Due to hectic and busy schedule, Mr. Angad made his life full of tensions and anxiety. He started taking sleeping pills to overcome the 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 few days he started feeling better. 4 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 tanquilizers? Give two examples.
Answer:
*(i) As per latest CBSE Curriculum, Value Based Questions will not be asked in the examination.
(ii) The sleeping pills have many adverse side effects which are not good for the body. Moreover, these are habit forming and cause addiction. So it is always advisable to take them under strict supervision of a doctor.
(iii) Tranquilizers are chemical substances used for the treatment of anxiety, stress or mild and severe mental disorders etc. For example, Equanil, valium etc.

Question 24.
(a) Account for the following:
(i) Ozone is thermodynamically unstable.
(ii) Solid PCl ₅ is ionic in nature.
(iii) Fluorine forms only one oxoacid HOF.

(b) Draw the structure of (i) BrF ₅ (ii) XeF ₄
Answer:
(a) (i) Ozone is thermodynamically very unstable because:

• The decomposition of ozone into oxygen is exothermic in nature. (∆H = -ve)
• There is also increase in entropy which in turn makes AG -ve and reaction spontaneous.

(ii) In solid state, PCl ₅ exists as [PCl ₄ ] ⁺ [PCl ₆ ] ^– in which the cation is tetrahedral and anion is octahedral. Because of the precence of strong attractive forces, it is a solid.
(iii) Due to absence of d-orbitals in fluorine, it can only form one oxoacid i.e., HOF

(b) (i) structure of BrF ₅

(ii) XeF ₄
Shape → Square plannar

Or

(i) Compare the oxidizing action of F ₂ and Cl ₂ by considering parameters such as bond dissociation enthalpy, electron gain enthalpy and hydration enthalpy.
(ii) Write the conditions to maximize the yield of H ₂ SO ₄ by contact process. (2008 Delhi)
(iii) Arrange the following in the increasing order of property mentioned:
(a) H ₃ PO ₄ < H ₃ PO ₃ < H ₃ PO ₂ (Reducing character)
(b) BiH ₃ < SbH ₃ < AsH ₃ < PH ₃ < NH ₃ (Basic strength)
Answer:
(i) Since the standard reduction potential value of fluorine is more than that of chlorine, so fluorine is a stronger oxidising agent.

• Bond dissociation enthalpy of F ₂ is less as compared to that of chlorine.
• The negative electron gain enthalpy of fluorine is slightly less than that of chlorine.
• The hydration enthalpy of flouride ion is much higher than that of Cl- ion due to smaller size.

(ii) Favourable conditions for manufacturing sulphuric acid by contact process are:

• high pressure of about 2 bars
• low temperature 720 K
• presence of V ₂ Os catalyst

(iii) (a) H ₃ PO ₄ < H ₃ PO ₃ < H ₃ PO ₂ (Reducing character)
(b) BiH ₃ < SbH ₃ < AsH ₃ < PH ₃ < NH ₃ (Basic strength)

Question 25.
(a) Write the structures of A, B, C, D and E in the following reactions:

Answer:

Or

(a) Write the chemical equation for the reaction involved in Cannizzaro reaction.
(b) Draw the structure of the semicarbazone of ethanal.
(c) Why pKa of F-CH ₂ -COOH is lower than that of Cl-CH ₂ -COOH?
(d)Write the product in the following reaction:

(e) How can you distinguish between propanal and propanone? (2008 Delhi)
Answer:
(a) Cannizzaro’s 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.

(b) semicarbazone of ethanal

(c) In FCH ₂ -COOH fluorine is more electron withdrawing than chlorine in ClCH ₂ – COOH so FCH ₂ -COOH is more acidic than ClCH ₂ -COOH, hence its pKa value is lesser than ClCH ₂ COOH.

(d)

(e) Propanal and Propanone: Propanal being an aldehyde gives positive test with Fehling solution in which a red brown ppt. of cuprous oxide is obtained while propanone being a ketone does not respond to this test.

Question 26.
(a) Calculate the freezing point of solution when 1.9 g of MgCl ₂ (M = 95 g mol ^-1 ) was dissolved in 50 g of water, assuming MgCl ₂ undergoes complete ionization.
(Kf for water = 1.86 K kg mol ^-1 ) [5]
(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 solution?
Answer:
(a) Since MgCl ₂ is an ionic compound, so it undergoes complete dissociation.

(b) (i) 2M glucose will have a higher boiling point than 1M glucose because elevation in boiling point is a colligative property which depends upon the number of particles in the solution which is more in the case of 2M glucose solution.
(ii) When the external pressure applied becomes more than the osmotic pressure of the solution, then the solvent will flow from the solution into the pure solvent through the semi-permeable membrane. The process is called reverse osmosis (RO).

Or

(a) When 2.56 g of sulphur was dissolved in 100 g of CS ₂ , the freezing point lowered by 0.383 K. Calculate the formula of sulphur (S _X ).
(K _f for CS ₂ = 3.83 K kg mol ^-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:
Given: wb = 2.56 g
wa = 100 g = 0.1 kg ATf = 0.383 K
K _f = 3.83 K kg mol ^-1
Atomic mass of sulphur = 32 g mol ^-1
M _b = ?
Using Formula

M _b = \(\frac{\mathrm{K}_{\mathrm{f}} \times w_b}{\Delta \mathrm{T}_{\mathrm{f}} \times w_a}\) = \(\frac{3.83 \times 2.56}{0.383 \times 0.1}\)

∴ M _b = 256 g mol ^-1

Hence the no. of atoms present in one molecule of sulphur = \(\frac{256}{32}\) = 8
∴ the formula is S ₈ .

(b) (i) If RBCs are placed in contact with 1.2% NaCl solution, then the osmotic pressure of 1.2% NaCl becomes higher than that of RBCs due to which water present inside the cells moves into the NaCl solution which results in shrinkage of RBCs.

(ii) Reverse process will take place if RBCs are kept in contact with 0.4% NaCl solution which has less osmotic pressure due to which water moves into RBCs and they will swell.