CBSE Class 12 Chemistry Question Paper 2015 Outside Delhi with Solutions

Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2015 Outside Delhi to familiarize themselves with the exam format and marking scheme.

CBSE Class 12 Chemistry Question Paper 2015 Outside Delhi with Solutions

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.
  † Deleted from Syllabus.

Question 1.
A delta is formed at the meeting point of sea water and river water. Why? [1]
Answer:
Delta is formed at the meeting point of sea water and river water due to coagulation of colloidal clay particles.

tQuestion 2.
What is the formula of a compound in which the element Y forms ccp lattice and atoms of X occupy 2/3^rd of tetrahedral voids? [1]
Answer:
Y atoms are N (No. of tetrahedral voids are 2N),
No. of tetrahedral voids occupied by X are \(\frac{2}{3}\) × 2N = \(\frac{4 N}{3}\)
X : Y = 4N : 3N Formida : X₄Y₃

Question 3.
Write the formulae of any two oxoacids of sulphur. [1]
Answer:
H₂SO₃ and H₂SO₄

Question 4.
Write the IUPAC name of the given compound: [1]

Answer:
IUPAC name : l-Ethoxy-2-methylpropane

Question 5.
Which would undergo SN1 reaction faster in the following pair : [1]

Answer:
, because the secondary carbo cation formed is more stable than primary I carbo cation.

Question 6.
Write the reagents required in the following reactions : [2]

Answer::

Or

Arrange the following compounds in increasing order of their property as indicated :
(i) CH₃COCH₃, C₆H₅COCH₃, CH₃CHO
(reactivity towards nucleophilic addition reaction)

(ii) Cl-CH₂-COOH, F-CH₂-COOH, CH₃-COOH (acidic character)
Answer:
(i) C₆H₅COCH₃ < CH₃COCH₃ < CH₃CHO
(Reactivity towards nucleophilic addition)

(ii) CH₃-COOH < C₆H₅ < F-CH₂-COOH
(Increasing acidic character)

Question 7.
(i) On mixing liquid X and liquid Y, volume of the resulting solution decreases. What type of deviation from Raoult’s law is shown by the resulting solution? What change in temperature would you observe after mixing liquids X and Y?
(ii) What happens when we place the blood cell in water (hypotonic solution)? Give reason. [2]
Answer:
(i) Volume decreases by mixing X and Y. It shows negative deviations from Raoult’s law.
There will be rise in temperature. (∆H_mix < 0)

(ii) Blood cell will swell due to osmosis as water enters the cell.

Question 8.
(i) Write down the IUPAC name of the following complex :
[CO(NH₃)₅Cl]²⁺
(ii) Write the formula for the following complex :
Potassium tetrachloridonickelate (II) [2]
Ans
(i) [CO(NH₃)₅Cl]²⁺
IUPAC name : Pentaammine chlorido cobalt (III) ion

(ii) Formula of the complex Potassium tetrachloridonickelate (II) K₂[NiCl₄]

Question 9.
Calculate the time to deposit 1.27 g of copper at cathode when a current of 2A was passed through the solution of CuSO₄.
(Molar mass of Cu = 63.5 g mol^-1, 1 F = 96500 C mol^-1)
Answer:
CuSO₄ → Cu+ + SO₄^2-
Cu²⁺ + 2e^– → Cu

63.5 gram of copper is deposited = 2 × 96500 C
1.27 gram of Cu is deposited = \(\frac{2 \times 96500}{63.5}\) × 1.27 = I × t

t = \(\frac{2 \times 96500 \times 1.27}{63.5 \times 2}\) = 1930 seconds

Question 10.
Write one similarity and one difference between the chemistry of lanthanoids and that of actinoids. [2]
Answer:
Similarity : Both lanthanoids and actinoids show contraction in size and irregularity in their electronic configuration.
Difference : Actinoids show wide range of oxidation states but lanthanoids do not.

Question 11.
How can the following conversions be carried out :
(i) Aniline to bromobenzene
(ii) Chlorobenzene to 2-chloroacetophenone
(iii) Chloroethane to butane
Answer:
(i) Aniline to bromobenzene

Or

What happens when
(i) Chlorobenzene is treated with Cl₂/FeCl₃,
(it) Ethyl chloride is treated with AgNO₂,
(iii) 2-bromopentane is treated with alcoholic KOH?
Write the chemical equations in support of your answer.
Answer:

Question 12.
Examine the given defective crystal :

Answer the following questions :
+(i) Is the above defect stoichiometric or non-stoichiometric?
+(ii) Write the term used for this type of defect. Give an example of the compound which shows this type of defect.
+(iii) How does this defect affect the density of the crystal? [3]
Answer:
(i) It is stoichiometric defect.
(ii) Schottky defect, e.g. NaCl.
(iii) Density of crystal decreases.

Question 13.
Conductivity of 2.5 × 10^-4 M methanoic acid is 5.25 × 10^-5 Scm^-1. Calculate its molar conductivity and degree of dissociation.
Given : λ°(H⁺) = 349.5 Scm² mol^-1 and λ°(HCOO) = 50.5 Scm^-1. [3]
Answer:
Concentration is 2.5 × 10^-4 M
K = 5.25 × 10^-5 Scm^-1

Question 14.
Predict the products of the following reactions : [3]

Answer:

Question 15.
(a) Account for the following : [3]
(i) Cu⁺ is unstable in an aqueous solution.
(ii) Transition metals form complex compounds.
(b) Complete the following equation :
Cr₂O₇^2- + 8H⁺ + 3NO₂^– →
Answer:
(a) (i) Cu⁺ is unstable in an aqueous solution because Cu⁺ undergoes disproportionation reaction as follows :
2Cu⁺ → Cu²⁺ + Cu

(ii) Transition metals form complex compounds due to small size of metal, higher nuclear (ionic) charge and availability of vacant or incompletely filled d-orbitals.

(b) Cr₂O₇^2- + 8H+ + 3NO₂^– → 2Cr³⁺ + 3NO₃⁺ + 4H₂O

Question 16.
Write the names and structures of the monomers of the following polymers :
(i) Terylene
(ii) Buna-S
(iii) Neoprene [3]
Answer:
(i) Terylene :
Terephthalic acid and ethylene glycol

Question 17.
A solution is prepared by dissolving 10 g of non-volatile solute in 200 g of water. It has a vapour pressure of 31.84 mm Hg at 308 K. Calculate the molar mass of the solute. (Vapour pressure of pure water at 308 K = 32 mm Hg)
Answer:

Question 18.
(i) Name the method of refining to obtain silicon of high purity.
(ii) What is the role of SiO₂ in the extraction of copper?
(iii) What is the role of depressants in froth floation process? [3]
Answer:
(i) Silicon of high purity can be obtained by zone refining.
(ii) SiO₂ acts as acidic flux to remove the impurities of iron oxide.
(iii) Depressants prevent the formation of froth with air bubbles of other sulphide ore.

Question 19.
Write any three differences between Physisorption and Chemisorption. [3]
Answer:
Tablee 1

Physisorption
It is not specific in nature i.e. all gases are adsorbed on all solids to some extent.
Adsorbate is held by weak van der Waals’ force.
It is reversible i.e. desorption of gas takes place by increasing the temperature or decreasing the pressure.
It forms multimolecular layers.

Chemisorption
It is highly specific in nature and occurs only when there is some possibility of compound formation between the gas being adsorbed and the solid being adsorbent.
Adsorbate molecules are held by strong force like a chemical bond.
It is irreversible in nature as it involves formation of compound instead of release of gas.
It forms unimolecular layer.

Question 20.
Give reasons for the following :
(i) Phenol is more acidic than methanol.
(ii) The C-O-H bond angle in alcohols is slightly less than the tetrahedral angle (190°28′).
(iii) (CH₃)₃C-O-CH, on reaction with HI gives (CH₃)₃C-I and CH₃-OH as the main products and not (CH₃)₃C-OH and CH₃-I.
Answer:
(i) Phenol is more acidic than methanol because in phenol, phenoxide ion formed is more stabilized by resonance than phenol, There is no resonance in methanol.
(ii) The C-O-H bond angle in alcohols is slightly less than tetrahedral due to repulsion between the lone pairs of electrons of oxygen.
(iii) (CH₃)₃C⁺ is 3° carbo-cation which is more stable than CH₃⁺ for S_N1 reaction.

Question 21.
(i) Which one of the following is a polysaccharide :
Starch, Maltose, Fructose, Glucose?
(ii) What one difference between α-helix and ß-pleated sheet structure of protein.
(iii) Write the name of the disease caused by the deficiency of Vitamin B12. [3]
Answer:
(i) Starch is a polysaccharide.
(ii) α-Helix structure : The polypeptide chains are held together (stabilized) by intramolecular H-bonding.
ß-Pleated sheet structure : The two neighbouring polypeptide chains are held together by intermolecular H-bonding.
(iii) Disease caused by the deficiency of Vitamin B₁₂ is Pernicious anaemia.

Question 22.
(i) What type of isomerism is shown by the complex [Cr(H₂O)₆]Cl₃?
(ii) On the basis of crystal field theory, write the electronic configuration for d4 ion if Ao > P.
(iii) Write the hybridization and shape of [CoF₆]^3-.
(Atomic number of Co = 27) [3]
Answer:
(i) [Cr(H2O)6]Cl3 shows Hydration isomerism.
(ii) Electronic configuration for d⁴ ion if A0 > P is t_2g \({ }^4 e_{\mathrm{g}}^0\)
(iii) [CoF₆]^3- has sp³d² hybridization and octahedral shape.

Question 23.
Seeing the growing cases of diabetes and depression among young children, Mr. Chopra, the principal of one reputed schoool organized a seminar in which he invited parents and principals. They all resolved this issue by strictly banning junk food in schools and introducing helathy snacks and drinks like soup, lassi, milk, etc. in school canteens. They also decided to make compulsory half an hour of daily physical activities for the students in the morning assembly. After six months, Mr. Chopra conducted the health survey in most of the schools and discovered a tremendous improvement in the health of the students.
After reading the above passage, answer the following questions :
*(i) What are the values (at least tivo) displayed by Mr. Chopra?
(ii) As a student, how can you spread awareness about this issue?
(iii) Why should antidepressant drugs not be taken without consulting a doctor?
(iv) Give two examples of artificial sweeteners. [4]
Answer:
*(i) As per latest CBSE curriculum, Value Based Questions will not be asked in the examination.
(ii)

• As a student, I will advise my friends and people of society to undertake daily physical activitiy like morning walk, cycling etc.
• I can also spread awareness about this issue through social media, debates, discussions in school, school plays, street plays and other such means.

(iii) Proper dose can be prescribed by the doctor only, as wrong choice and overdose of medicine may be harmful.

(iv) (a) Aspartame, (b) Saccharin.

Question 24.
An aromatic compound ‘A’ of molecular formula C7H602 undergoes a series of reactions as shown below. Write the structures of A4 B, C, D and E in the following reactions : [5]

Answer:

OR

(a) Write the structures of main products when benzene diazonium chloride reacts with the following reagents :
(i) H₃PO₂ + H₂O
(ii) CuCN/KCN
(iii) H₂O
(b) Arrange the following in the increasing order of their basic character in an aqueous solution :
C₂H₅NH₂, (C₂H₅)₂NH, (C₃H₅)₃N
(c) Give a simple chemical test to distinguish between the following pair of compounds :
C₆H₅-NH₂ and C₆H₅-NH-CH₃
Answer:
(a) The structure of main products when aniline (benzene diazonium chloride) reacts with

(b) C₂H₅NH₂ < (C₃H₅)₃N < (C₂H₅)₂NH
(c) Aniline and Benzylamine can be distinguished by the Nitrous acid test. Benzylamine reacts with HNOz to form a diazoniurrt salt which being unstable even at low temperature, decomposes with evolution of N₂ gas

Aniline on the other hand, reacts with HNO₂ to form benzenediazonium chloride which is stable at 273-278 K and hence does not decompose to evolve N₂ gas.

Question 25.
For the hydrolysis of methyl acetate in aqueous solution, the following results were obtained :

(i) Show that it follows pseudo first order reaction, as the concentration of water remains constant.
(it) Calculate the average rate of reaction between the time interval 10 to 20 seconds.
(Given : log 2 = 0.3010, log 4 = 0.6021)
Answer:

As k₁ and k₂ are equal, hence pseudo rate constant is same. It follows the pseudo first order reaction.

(ii) Average rate of reaction between 10 to 20 seconds
= \(\frac{-\Delta[\mathrm{R}]}{\Delta t}\)
= \(\frac{-(0.025-0.05)}{(20-10)}\)
= \(\frac{0.025}{10}\)
= 0.0025 mol lit^-1 sec^-1