Students can use CBSE Previous Year Question Papers Class 12 Chemistry with Solutions and CBSE Class 12 Chemistry Question Paper 2016 Outside Delhi to familiarize themselves with the exam format and marking scheme.
CBSE Class 12 Chemistry Question Paper 2016 Outside Delhi with Solutions
Time allowed: 3 hours
Maximum marks : 70
General Instructions:
- All questions are compulsory.
- Questions number 1 to 5 are very short-answer questions and carry 1 mark each.
- Questions number 6 to 10 are short-answer questions and carry 2 marks each.
- Questions number 11 to 22 are also short-answer questions and carry 3 marks each.
- Question number 23 is a value based question and carries 4 marks.[*]
- Questions number 24 to 26 are long-answer questions and carry 5 marks each.
-
Use log tables, if necessary, Use of calculators is not allowed.
† Deleted from Syllabus.
Question 1.
Write the structure of an isomer of compound C
4
H
9
Br which is most reactive towards S
N
1 reaction. [1]
Answer:
(2-Bromo-2-methyI propane) or tert-butyl bromide is most reactive towards S
N
1 reaction as it can form 3° carbocation.
+Question 2.
Pb(N0
3
)
2
on heating gives a brown gas which undergoes dimerization on cooling.
Identify the gas. [1]
Answer:
The brown gas is nitrogen dioxide (NO
2
) which can dimerize to N
2
O
4
+Question 3.
Give an example each of a molecular solid and an ionic solid. [1]
Ans.
Molecular solid → Iodine (I
2
)
Ionic solid → Sodium chloride (NaCl)
Question 4.
Write the IUPAC name of the given compound: [1]
Answer:
2-Phenylethanol
Question 5.
What is the reason for the stability of colloidal sols? [1]
Answer:
The stability of the colloidal solution is becuase of solvation and the presence of charge on the dispersed phase particles.
Question 6.
(i) Gas (A) is more soluble in water than Gas (B) at the same temperature. Which one of the two gases will have the higher value of KH (Henry’s constant) and why?
(ii) In non-ideal solution, what type of deviation shows the formation of maximum boiling azeotropes? [2]
Answer:
(i) Gas (B) will have higher value of K
H
(Henry’s constant) than Gas (A) at the same temperature because lesser the solubility of a gas in a given solvent, higher will be the value of K
H
for a gas.
(ii) Negative deviations from Raoult’s Taw show the formation of maximum boiling azeotropes.
Question 7.
Write the structures of the following:
(i) BrF
3
(ii) XeF
4
Ans
(i) BrF
3
:
(ii) XeF
4
:
Or
What happens when:
(i) SO
2
gas is passed through an aqueous solution of Fe
2+
salt?
(ii) XeF
4
reacts with SbF
5
?
Answer:
(i) In this sulphur dioxide acts as a reducing agent and reduces Fe
2+
to Fe
2+
.
2Fe 2+ + SO 2 + 2H 2 O → 2Fe 2+ + SO 4 2- + 4H +
(ii) XeF 4 + SbF 5 → [XeF 3 ] + + [SbF 6 ] –
Question 8.
When a coordination compound COCl
3
.6NH
3
is mixed with AgNO
3
, 3 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) Complex so formed is:
COCl
3
.6NH
3
+ AgNO
3
→ [Co(NH
3
)
6
]Cl
3
(ii) IUPAC name of above complex is:
Hexaamminecobalt (III) chloride
Question 9.
For a reaction:
Rate = k
(i) Write the order and molecularity of this reaction.
(ii) Write the unit of k.
Answer:
(i) H, + Cl2 2HC1
This reaction is zero order reaction and molecularity is two.
(ii) Unit of k = mol L -1 s -1
Question 10.
Write the chemical equations involved in the following reactions: [2]
(i) Hoffmann-bromamide degradation reaction
(ii) Carbylamine reaction
Answer:
(i) Hoffmann’s bromamide reaction : In this reaction, migration of an alkyl or aryl group takes place from carbonyl carbon of the amide to the nitrogen atom. Therefore the amine so formed has one carbon atom less than that of amide.
(ii) Carbylamine reaction. This reaction used to distinguish primary amines from 2° and 3° amines as it is only given by 1° amines with the production of a very bad smelling organic compound.
For example :
Question 11.
An element crystallizes in a b.c.c. lattice with .cell edge of 500pm. The density of the element is 7.5g cm
-3
. How many atoms are present in 300 g of the element? [3]
Answer:
Given: For b.c.c. structure, z = 2
Edge of the unit cell, a = 500 pm = 500 × 10
-10
cm
Density d = 7.5 g cm
-3
Using the formula,
Question 12.
For the first order thermal decomposition reaction, the following data were obtained: [3]
Calculate the rate constant.
(Given: log 2 = 0.301, log 3 = 0.4771, log 4 = 0.6021)
Answer:
Given: Initial pressure, P
0
= 0.30 atm
P
t
= 0.50 atm
t = 300 s
Question 13.
Define the following terms:[3]
(i) Lyophilic colloid
(ii) Zeta potential
(iii) Associated colloids
Answer:
(i) Lyophilic colloid : Liquid loving colloids in which there is affinity between disperse phase and dispersion medium.
Example ; Starch sol, Gum sol, Gelatin sol
(ii) Zeta potential. When one type of the ions of the electrolyte are adsorbed on the surface of colloidal particles it forms a fixed layer which attracts another layer of opposite ions thus forming a Helmholtz electrical double layer whose potential difference between the two layers is termed as zeta potential.
(iii) Associated colloids :
1. They are formed by aggregation of a large number of ions in concentrated solution Example : soap sol
2. Their molecular masses are generally high.
3. Higher is the concentration, greater are the Vander Waals’ forces.
Question 14.
(i) Name the method of refining of nickel.
(ii) What is the role of cryolite in the extraction of aluminium?
(iii) What is the role of limestone in the extraction of iron from its oxides? [3]
Answer:
(i) Mond’s Process.
(ii) The melting point of alumina is very high. It is dissolved in cryolite which lowers the melting point and brings conductivity.
(iii) Limestone decomposes in the blast furnace into CaO and C02 where CaO acts as basic flux which combines with acidic impurities of the ore like silica i.e., Si02 to form calcium silicate in the form of slag.
Question 15.
Calculate the boiling point of solution when 4 g of MgSO
4
(M =120 g mol
-1
) was dissolved in 100 g of water, assuming MgSO
4
undergoes complete ionization.
(K
b
for water = 0.52 K kg mol
-1
) [3]
Answer:
Since MgSO
4
is an ionic compound, so undergoes complete ionisation in the following way:
∴ Total number of moles = 1 + 1 = 2
Thus i = \(\frac{2}{1}\) = 2
Using formula:
∆T
b
= iK
b
m
Question 16.
Give reasons: [3]
(i) SO
2
is reducing while TeO
2
is aft oxidizing agent.
(ii) Nitrogen does not form pentahalide.
(iii) ICl is more reactive than I
2
.
Answer:
(i) SO
2
is reducing while TeO
2
is an oxidising agent because sulphur can expand its covalency upto +6 from +4 due to presence of empty d-orbital but as we move down the group the stability of +6 oxidation state decreases and of +4 oxidation state increases due to inert pair effect. Hence SO
2
acts as reducing agent while TeO
2
acts as an oxidising agent.
(ii) Due to absence of empty d-orbitals, N
2
does not form pentahalides.
(iii) Because ICl bond is weaker than I -1 bond as a result of which ICl breaks easily to form halogen atoms which readily bring about the reaction, hence more reactive.
Question 17.
Write the final product(s) in each of the following reactions: [3]
Answer:
Question 18.
Give reasons for the following:
(i) Aniline does not undergo Friedal-Crafts reaction.
(ii) (CH
3
)
2
NH is more basic than (CH
3
)
3
N in an aqueous solution.
(iii) Primary amines have higher boiling point than tertiary amines.
Answer:
(i) Aniline being a Lewis base reacts with Lewis acid AlCl
3
tp form a salt.
As a result, N of aniline acquires positive charge and hence it acts as a strong deactivating group for electrophilic substitution reaction. Consequently, aniline does not undergo Freidel Crafts reaction.
(ii) In (CH 3 )N there is maximum steric hindrance and least solvation but in (CH 3 ) 2 NH the solvation is more and the steric hindrance is less than in (CH 3 ) 3 NH; although + I effect is less, since there are two methyl groups; di-methyl amine is still a stronger base than tri-methyl.
(iii) Due to presence of two H-atoms on N-atom of primary amines, they undergo extensive intermolecular H-bonding while tertiary amines due to the absence of a H-atom on the N-atom, do not undergo H-bonding. As a result, primary amines have higher boiling points than 3° amines.
Question 19.
How do you convert: [3]
(i) Chlorobenzene to biphenyl
(iii) 2-bromobutane to but-2-ene
Answer:
OR
Answer:
Question 20.
(i) What is the role of Sulphur in the vulcanization of rubber?
(ii) Identify the monomers in the following polymer:
(iii) Arrange the following polymers in the increasing order of their intermolecular forces:
Terylene, Polythene, Neoprene
Answer:
(i) Sulphur introduces sulphur bridges or cross-links between polymer chains thereby imparting more tensile strength, elasticity and resistance to abrasion.
(ii) The monomers in the given polymer are:
Ethylene glycol HO-CH
2
CH
2
-OH
(iii)Neoprene < Polyethene < Terylene
Question 21.
(i) Write the structural difference between starch and cellulose.
(ii) What type of linkage is present in Nucleic acids?
(iii) Give one example each for fibrous protein and globular protein. [3]
Answer:
(i) Starch contains the α-D-glucose as its monomer units while cellulose contains ß-D- glucose as its monomer units.
(ii) Phosphodiester linkages are present in Nucleic Acids
(ii) (a) Globular protein : All enzymes and hormones like insulin
(b) Fibrous protein : Keratin in skin, nails etc.
Question 22.
For the complex [Fe(H
2
O)
6
]3+, write the hybridization, magnetic character and spin of the complex. (At. number: Fe = 26)
Draw one of the geometrical isomers of the complex [Pt(en)2Cl2]2+ which is optically inactive. [3]
Answer:
(a) [Fe(H
2
O)
6
]
3+
: The element Fe is in +3 oxidation state. As H
2
O is a weak field ligand, so electron pairing is not possible in this case.
(i) Hybridisation is sp
3
d
2
(ii) Since 5 unpaired electrons are present so it is paramagnetic in nature.
μ = \(\sqrt{n(n +1)}\) BM = \(\sqrt{5(5 +1)}\) = 5.92 BM
(iii) It forms outer orbital or high spin complexes.
(b) The optically inactive isomer of complex [Pt(en)
2
Cl
2
]
2+
is:
trans form of the complex
Question 23.
Due to hectic and busy schedule, Mr. Singh started taking junk food in the lunch break and slowly became habitual of eating food irregularly to excel in his field. One day during a meeting he felt severe chest pain and fell down. Mr. Khanna, a close friend of Mr. Singh took him to the doctor immediately. The doctor diagnosed that Mr. Singh was suffering from acidity and prescribed some medicines. Mr. Khanna advised him to eat home made food and change his lifestyle by doing yoga, meditation and some physical exercise. Mr. Singh followed his friend’s advice and after 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. Khanna?
(ii) What are antacids? Give one example.
(iii) Would it be advisable to take antacids for a long period of time? Give reason.
Ans.
*(i) As per latest CBSE Curriculum, Value Based Questions will not be asked in the examination.
(ii) The chemical substances which can reduce or neutralise the acidity in stomach and raise the pH to some appropriate level are called antacids.
Example: Magnesium hydroxide, sodium hydrogen carbonate etc.
(iii) No, because excessive use of antacids may lead to constipation and can even make the stomach alkaline leading to production of more acids.
Question 24.
(a) Calculate E°
cell
for the following reaction at 298 K:
2Al(s) + 3Cu
2+
(0.01M) → 2Al
3+
(0.01M) + 3Cu(s)
Given: E
cell
= 1.98 V
(b) Using the E° values of A and B, predict which is better for coating the surface of iron [E°(Fe
2+
/Fe) = -0.44 V] to prevent corrosion and why?
Given: E°(A
2+
/A) = -2.37 V; E°(B
2+
/B) = -0.14 V [5]
Answer:
(a) For the reaction
(b) Element A will be better for coating thfe surface of iron than element B because its E° value is more negative.
OR
(a) The conductivity of 0.001 mol L
-1
solution of CH3COOH is 3.905 × 10
-5
S cm
-1
. Calculate its molar conductivity and degree of dissociation (a).
Given: λ°(H
+
) = 349.6 S cm
2
mol
-1
and λ° (CH
3
COO
–
) = 40.9 S cm
2
mol
-1
(b) Define electrochemical cell. What happens if external potential applied becomes greater than E°
cell
of electrochemical cell?
Answer:
(a) Concentration = 0.001 mol L
-1
K = 3.905 × 10
-5
S cm
-1
\(\Lambda_{\mathrm{m}}^{\mathrm{c}}\) = ?
Using formula,
(b) Electrochemical cell: It is a device which converts chemical energy into electrical energy i.e., produced as a result of redox reaction taking place in the electrolyte.
The reaction gets reversed and it becomes non-spontaneous. It starts acting as an electrolytic cell.
Question 25.
(a) Account for the following:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
(ii) Cr
2+
is a strong reducing agent.
(iii) Cu
2+
salts are coloured while Zn
2+
salts are white.
(b) Complete the following equations:
Answer:
(a)
(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) Cr 2+ exists in the d4 system and is easily oxidized to Cr 3+ by loosing one electron which has the stable d 3 [t 2g 2 ] orbital configuration. So, Cr 2+ is a strong reducing agent.
(iii) Cu 2+ has the configuration 3d 9 with one unpaired electron which gets excited in the visible region to impart its colour while Zn 2+ has 3d 10 configuration without any unpaired electron so no d – d transition possible and hence colourless.
Or
The elements of 3d transition series are given as:
Sc Ti V Cr Mn Fe Co Ni Cu Zn
Answer the following:
(i) Write the element which shows maximum number of oxidation states. Give reason.
(ii) Which element has the highest m.p.?
(iii) Which element shows only +3 oxidation state?
(iv) Which element is a strong oxidizing agent in +3 oxidation state and why?
Answer:
(i) Mn shows maximum number of oxidation states upto +7. It has the maximum number of unpaired electrons.
(ii) Cr has the highest melting point.
(iii) Sc shows only +3 oxidation state.
(iv) Mn is a strong oxidizing agent in +3 oxidation state because after reduction it attains +2 oxidation state in which it has the most stable half filled (d
5
) configuration.