1. No category

Class XII_All India Chemistry_Set-2
9.
Ans.
10.
Ans.
(a)
Why does presence of excess of lithium makes LiCl crystals pink?
(b)
A solid with cubic crystal is made of two elements P and Q. Atoms of Q
are at the corners of the cube and P at the body-centre. What is the formula
of the compound?
2
(a)
When crystals of LiCl is heated in presence of excess of lithium, Cl- ions from
crystal diffuse on surface and combine with ionised Li to form LiCl. The
released unpaired electrons occupy the anionic sites known as F-centres. The pink
colour results by excitation of these electrons when they absorb energy from
visible light falling on them.
(b)
It is given that the atoms of Q are present at the corners of the cube. Therefore,
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number of atoms of Q in one unit cell =  8  1
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Also, It is also given that the atoms of P are present at the body-centre.
Therefore, number of atoms of P in one unit cell = 1
This means that the ratio of the number of P atoms to the number of Q atoms, P:
Q = 1:1
Hence, the formula of the compound is PQ.
Write the equations involved in the following reactions:
(i)
Reimer – Tiemann reaction
(ii)
Williamson’s ether synthesis
(i)
(ii)
2
Reimer-Tiemann reaction :
Williamson’s ether synthesis:
R — X  R — ONa  R — O — R  NaX
Alkyl halide
Sodium alkoxide
Ether
For e.g.
CH3I  CH3CH2ONa  CH3O —CH2CH3  NaI
Methyl iodide
Sodium ethoxide
ethyl methylether
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Class XII_All India Chemistry_Set-2
11.
Define thermoplastic and thermosetting polymers. Give one example of each.
2
OR
What is a biodegradable polymer? Give an example of a biodegradable aliphatic
polyester.
Ans.
Thermoplastic Polymers: These are the linear or slightly branched long chain molecules
capable of repeatedly softening on heating and hardening on cooling. These possess
intermolecular forces of attraction intermediate between elastomers and fibres. For e.g.
Polythene, Polyvinyls etc.
Thermosetting Polymers: These are cross-linked or heavily branched molecules which
on heating undergo extensive cross linking in moulds and again become influsible. These
cannot be reused. For e.g. bakalite, urea-formaldehyde resins etc.
OR
Biodegradable Polymers: In view of the disposable problems of the polymer wastes and
for preparing polymers for other safe uses in human system, biodegradable synthetic
polymers have been developed. They mostly have functional groups similar to functional
groups present in biopolymers.
An example of biodegradable aliphatic polyester is PHBV i.e., Poly β-hydroxybutyrate co-β-hydroxy valerate
12.
Explain the mechanism of the following reaction:
H
CH3 — CH 2 — OH 
 CH 2  CH 2  H 2O
443K
Ans.
Mechanism of the following reaction is given below:
H
CH3 — CH2 — OH 
 CH2  CH2  H2O
443 K
2
Step 1: Formation of protonated alcohol
Step 2: Formation of carbocation
2
Class XII_All India Chemistry_Set-2
Step 3: Formation of ethene by removal of proton
13.
How are interhalogen compounds formed? What general compositions can be
assigned to them?
2
Ans.
Halogens can react with each other to form a number of compounds called interhalogen
compounds. Their general formula is AXn where A is a less electronegative halogen
while X is more electronegative halogen and n is number of X atoms in compound.
They can be assigned general compositions as AX, AX3, AX5 and AX7.
14.
Draw the structures of the following molecules:
(i)
XeF6
(ii)
H2S2O7
Ans.
15.
Ans.
(i)
Structure of XeF6 :
(ii)
H2S2O7
Aluminium crystallizes in an fcc structure. Atomic radius of the metal is 125 pm.
What is the length of the side of the unit cell of the metal?
2
2
Aluminium crystallizes in FCC structure atomic radius (r) = 125 pm
4
Length of the side of the unit cell in fcc structure, a 
r =2 2r
2
a = 2x 1.414x 125
a = 353.5 pm
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Class XII_All India Chemistry_Set-2
16.
The standard electrode potential (E°) for Daniell cell is + 1·1 V. Calculate the
ΔG° for the reaction
Zn  s   Cu 2  aq  
 Zn 2  aq   Cu  s 
(1 F = 96500 C mol–1).
Ans.
E° for Daniel cell = 1.1 V
Zn(s) + Cu2+ (aq) → Zn2+ (aq) + Cu(s)
1F = 96500 C mol–1
∆G ° = ?
n = 2 (no. of e–s exchanged)
Since,
∆G ° = –nFE°
Therefore, ∆G ° = –2 × 96500 × 1.1
∆G ° = –212300 J mol–1
∆G ° = –212.3 kJ mol–1
17.
(a)
For a reaction A + B → P, the rate law is given by,
r = k [A]1/2 [B]2.
What is the order of this reaction?
(b)
A first order reaction is found to have a rate constant k = 5·5  10–14 s–1.
Find the half life of the reaction.
Ans.
(a)
2
For A + B → P
r = k[A]1/2 [B]2
The order of the reaction = 2 
(b)
2
1 5
  2.5
2 2
For first order reaction
k = 5.5 × 10–14 s–1
t
Half life period ( 1/ 2 ) for the first order reaction
0.693
0.693
t1/2 

k
5.5 1014
t1/2  1.261013 s
18.
Outline the principles of refining of metals by the following methods:
(i)
Zone refining
(ii)
Vapour phase refining
2
Ans.
Principles of refining of metals by:
(i)
Zone refining: The impurities are more soluble in the melt than in the solid state
of the metal. Hence, the pure metal crystallizes out of the melt and impurities pass
on into the adjacent molten zone and like this at one end impurities get
concentrated.
(ii)
Vapour phase refining: The metal to be refined should form volatile
compound with the suitable reagent and the compound so formed should
be easily decomposable to the recovery of pure metal.
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