Grade XII DELHI SET 2
Chemistry (Theory)
[Time allowed: 3 hours]
[Maximum marks:70]
General Instructions:
(i) All questions are compulsory.
(ii) Marks for each question are indicated against it.
(iii) Question numbers 1 to 8 are very short-answer questions and carry 1 mark each.
(iv) Question numbers 9 to 18 are short-answer questions and carry 2 marks each.
(v) Question numbers 19 to 27 are also short-answer questions and carry 3 marks each.
(vi) Question numbers 28 to 30 are long-answer questions and carry 5 marks each.
(vii) Use Log Tables, if necessary. Use of calculators is not allowed.
Q19. Tungsten crystallizes in body centred cubic unit cell. If the edge of the unit cell is 316.5 pm,
what is the radius of tungsten atom?
3
OR
Iron has a body centred cubic unit cell with a cell dimension of 286.65 pm. The density of iron
is 7.874 g cm–3. Use this information to calculate Avogadro’s number (At. Mass of Fe = 55.845
u)
Ans.
It is body centred cubic cell for which radius
3
radius  r  
a
4
a = 316.5 = pm (given)
3
 316.5 pm
4
1.732

 316.5
4
 137.04 pm
Radius 
OR
a = 286.65 pm
= 286.65 × 10–10 cm
Density () = 7.874 g cm–3
At mass of Fe = 55.845 u
Z = 2 (For body centred cubic unit cell)
Avogadro Number (N0) = ?
Grade XII DELHI SET 2

Z×M
a3  N0
3
7.874 g cm 
2   55.845 g mol1 
 286.65 10 cm   N
2   55.845 g mol 

 286.65 10 cm    7.874 g cm 
3
10
0
1
N0
10
3
3
 6.022 1023 mol 1
Q20. A solution of glycerol (C3H8O3) in water was prepared by dissolving some glycerol in 500 g of
water. This solution has a boiling point of 100.42°C while pure water boils at 100°C. What mass
of glycerol was dissolved to make the solution?
(Kb = for water = 0.512 K kg mol–1
3
Ans.
WB = ?
WA = 500g
Kb = 0.512 Kkg mol–1
Tb  100.42C  100C
 0.42°C
M B  3  12  8  1  3  16
 36  8  48  92
K  WB  1000
Tb  b
M B  WA
0.512  WB 1000
92  500
0.42  92  500
 WB
0.512 1000
WB  37.73 g
0.42 
Q21. For the reaction
2NO(g) + Cl2(g) → 2 NOCl(g)
the following data were collected. All the measurements were taken at 263 K:
Experiment
Initial rate of disappearance
Initial [NO] (M) Initial [Cl2] (M)
No.
of Cl2 (M/min)
1
0.15
0.15
0.60
2
0.15
0.30
1.20
3
0.30
0.15
2.40
4
0.25
0.25
?
(a)
(b)
Write the expression for rate law.
Calculate the value of rate constant and specify its units.
Grade XII DELHI SET 2
(c)
Ans.
(a)
What is the initial rate of disappearance of Cl2 in exp. 4?
Rate law may be written as
Rate = k [NO]p [Cl2]q
The initial rate becomes
(Rate)0 = k [NO]p [Cl2]q
Comparing experiment 1 and 2
(Rate)1 = k (0.15)p (0.15)q = 0.60
(Rate)2 = k (0.15)p (0.30)q = 1.20
Dividing equation (ii) by (i)
 Rate 2 k  0.15 p  0.30 q 1.20


 Rate 1 k  0.15 p  0.15q 0.60
or 2q = 21
q=1
order with respect to Cl2 = 1
Comparing experiment 1 and 3
(Rate)1 = k (0.15)p (0.15)q = 0.60
(Rate)3 = k (0.30)p (0.15)q = 2.40
Dividing equation (ii) by (i)
 Rate 3 k  0.30  p  0.15q 2.40


 Rate 1 k  0.15 p  0.15q 0.60
or 2p = 4
2p = 22
p=2
Thus order with respect to NO is 2.
Rate law = k [NO]2 [Cl2]1
(b)
...
...
(i)
(ii)
...
...
(i)
(ii)
The rate law for the reaction Rate = k[NO]2 [Cl2]
Rate constant can be calculated by substituting the value of rate,[NO] and [Cl2] for any
experiment
Rate
0.60
k=
=
2
2
 NO Cl2   0.15  0.15
0.60
0.003375
=177.77 mol 2 L2 min 1
=
(c)
3
Let initial rate of disappearance of Cl2 in exp. 4 is r4.
 r4 = k  NO  Cl2 
2
=177.77×  0.25   0.25 
2
= 2.78 M/min
Q22. State a reason for each of the following situations:
Grade XII DELHI SET 2
(i) Co2+ is easily oxidized to Co3+ in presence of a strong ligand.
(ii) Co is a stronger complexing reagent than NH3.
(iii) The molecular shape of Ni(CO)4 is not the same as that of [Ni(CN)4]2–
Ans.
(i) The electronic configuration of Co3+ is 3d6 4sº. So, pairing occurs in the presence of a strong
ligand. Thus, there are no unpaired electrons and it is stable.
Co3+ – 3d6
In the presence of a strong ligand:
However, in Co2+, whose electronic configuration is 3d7, there is one unpaired electron even
after pairing occurs in the presence of a strong ligand. Hence, Co2+ is oxidised to Co3+.
(ii) Co(II) is stable in aqueous solutions. However, in the presence of strong field complexing
reagents, it is oxidized to Co (III). Although the 3rd ionization energy for Co is high, but the
higher amount of crystal field stabilization energy (CFSE) released in the presence of strong
field ligands overcomes this ionization energy.
(iii) In Ni(CO)4, Ni is in the zero oxidation state i.e., it has a configuration of 3d8 4s2.
But CO is a strong field ligand. Therefore, it causes the pairing of unpaired 3d electrons. Also, it
causes the 4s electrons to shift to the 3d orbital, thereby giving rise to sp3 hybridization.
In [Ni(CN)4]2–, Ni exists in the +2 oxidation state i.e., d8 configuration.
Ni2+:
CN– is a strong field ligand. It causes the pairing of the 3d orbital electrons. Then, Ni2+
undergoes dsp2 hybridization.
As there are no unpaired electrons, it is diamagnetic.
Grade XII DELHI SET 2
Q23. How would you account for the following?
(i) With the same d-orbital configuration (d4) Cr2+ is a reducing agent while Mn3+ is an
oxidizing agent.
(ii) The actinoids exhibit a larger number of oxidation states than the corresponding members in
the lanthanoid series.
(iii) Most of the transition metal ions exhibit characteristic in colours in aqueous solutions.
Ans.
(i) Cr2+ is strongly reducing in nature. It has a d4 configuration. While acting as a reducing
agent, it gets oxidized to Cr3+ (electronic configuration, d3). This d3 configuration can be written
3
as t2g
configuration, which is a more stable configuration. In the case of Mn3+ (d4), it acts as an
oxidizing agent and gets reduced to Mn2+ (d5). This has an exactly half-filled d-orbital and is
highly stable.
(ii) The wide range of oxidation states of actinoids is attributed to the fact that the 5f, 6d and 7s
energy levels are comparable energies. Therefore all these three sub shells can participate. But
the most common oxidation state of actinoids is +3.
(iii) Most of the complexes of transition elements are coloured. This is because of the
absorption of radiation from visible light region to promote an electron from one of the d–
orbitals to another. In the presence of ligands, the d-orbitals split up into two sets of orbitals
having different energies. Therefore, the transition of electrons can take place from one set to
another. The energy required for these transitions is quite small and falls in the visible region of
radiation. The ions of transition elements absorb the radiation of a particular wavelength and the
rest is reflected, imparting colour to the solution.
Q24. Write chemical equation for the following conversions:
(i) Nitrobenzene to benzoic acid.
(ii) Benzyl chloride to 2-phenylethanamine.
(iii) Aniline to benzyl alcohol.
Ans.
(i)
Nitrobenzene to benzoic acid
(ii)
Benzyl chloride to 2-phenylethanamine
3
Grade XII DELHI SET 2
(iii)
Aniline to benzyl alcohol
Q25. What are the following substances? Give one example of each one of them.
(i) Tranquilizers
(ii) Food preservatives
(iii) Synthetic detergents
Ans.
3
(i) Tranquilizers: The chemical substances used for the treatment of stress, mild and severe
mental diseases are called tranquilizer.
Example : Noradrenaline
(ii) Food Preservatives: These are the chemical substances which are added to the food
materials to prevent their spoilage and to retain their nutritive value for long periods.
Example: Vinegar
(iii) Synthetic detergents: Synthetic detergents are cleansing agents which have all the
properties of soaps but actually do not contain any soap. These can be used both in soft and
hard water.
Example: Sodium p-dodecylbenzenesulphonate.
Q26. Draw the structure and name the product formed if the following alcohols are oxidized. Assume
that an excess of oxidizing agent is used.
3
(i) CH3CH2CH2CH2OH
(ii) 2-butenol
(iii) 2-methyl-1-proponal
Grade XII DELHI SET 2
Ans.
Q27. Although chlorine is an electron withdrawing group, yet it is ortho-, para-directing in
electrophilic aromatic substitution reactions. Explain why it is so?
Ans.
3
Chlorine withdraws electrons through inductive effect and releases electrons through resonance.
Through inductive effect, chlorine destabilises the intermediate carbocation formed during the
electrophilic substitution.
Grade XII DELHI SET 2
Through resonance, halogen tends to stabilise the carbocation and the effect is more pronounced
at ortho-and para-position. The inductive effect is stronger than resonance and causes net
electron withdrawal and thus causes net deactivation. The resonance effect tends to oppose the
inductive effect for the attack at ortho-and para-position and hence makes the deactivation less
for ortho- and para-attack. Reactivity is thus controlled, by the stronger inductive effect and
orientation is controlled by resonance effect.