Grade XII DELHI SET 1
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.
Q28. (a)
(b)
What type of a battery is the lead storage battery? Write the anode and the
cathode reactions and the overall reaction occurring in a lead storage battery when
current is drawn from it.
In the button cell, widely used in watches, the following reaction takes place
Zn s  Ag 2Os  H2Ol  
 Zn 2aq   2Ags  2OHaq 
Determine E° and G° for the reaction.
 given: E°
Ag + /Ag
  0.80V, E°Zn 2 /Zn  0.76V

5
OR
Ans.
(a)
Define molar conductivity of a solution and explain how molar conductivity
changes with change in concentration of solution for a weak and a strong
electrolyte.
(b)
The resistance of a conductivity cell containing 0.001 M KCl solution at 298 K is
1500. What is the cell constant if the conductivity of 0.001 M KCl solution at
298 K is 0.146 × 10–3 S cm–1?
(a)
A lead storage battery has a secondary cell. Thus, it can be recharged by passing
direct current through it. Therefore, it can be reused.
It is used in automobiles.
In a lead storage cell, the anode is made of spongy lead and the cathode is a grid
of lead packed with lead dioxide. The electrolyte used is H2SO4.
At anode:
2
Pb  SO 4 
 PbSO4  2e
At cathode:
PbO2  SO 24  4H  2e 
 PbSO4  2H2O
Overall cell reaction:
Grade XII DELHI SET 1
(b)
Pb  PbO2  4H  2SO42 
 2PbSO4  2H2O
From the given reaction, it is known that zinc is oxidised and silver is reduced in
the button cell.
The reactions occurring at the anode and cathode are –


 Zn (2aq
Anode: Zn ( s ) 
)  2e
Cathode: Ag 2O( s )  H2O(l )  2e  2Ag( s )  2OH(aq )
0
0
Now, E 0cell  E cathode
 E anode
 E 0Ag /Ag  E 0Zn 2 / Zn
 0.80  (0.76)
 1.56 V
We know that
r G 0  nE0 F
Here,
n=2
F = 96500 C mol–1
 r G 0  2 1.56  96500 J mol1
 301080 J mol1
OR
(a)
Molar conductivity of a solution at a given concentration is the conductance of
volume V of a solution containing 1 mole of the electrolyte kept between two
electrodes with the area of cross-section A and distance of unit length.
A
m 
l
Now, l = 1 and A = V (volume containing 1 mole of the electrolyte).
 m  V
Molar conductivity increases with a decrease in concentration. This is because the
total volume V of the solution containing one mole of the electrolyte increases on
dilution.
Variation of molar conductivities with dilution:
For strong electrolytes, molar conductivity slowly increases with dilution.
For weak electrolytes, molar conductivity increases steeply on dilution, especially
near lower concentrations.
The variation of  m with c for strong and weak electrolytes is shown in the
following plot:
Grade XII DELHI SET 1
(b)
Given,
Conductivity, κ = 0.146 × 10–3 S cm–1
Resistance, R = 1500 
 Cell constant = κ × R
= 0.146 × 10–3 × 1500
= 0.219 cm–1
Q29. (a)
Complete the following chemical reactions equations:
(i)
P4  SO2Cl2 

(ii)
(b)
XeF6  H2O 

Predict the shape and the asked angle (90° or more or less) in each of the
following cases:
SO32 and the angle O – S – O
(i)
(ii)
ClF3 and the angle F – Cl – F
(iii)
XeF2 and the angle F – Xe – F
OR
(a)
Complete the following chemical equations:
NaOH + Cl2 

(i)
 hot and conc.
(ii)
Ans.
XeF4  O2 F2 

(b)
Draw the structures of the following molecules:
(i)
H3PO2
(ii)
H2 S2 O7
(iii)
XeOF4
(a)
(i)
(ii)
P4 + 10 SO2Cl2  4 PCl5 + 10 SO2
Equation for complete hydrolysis:
XeF6  3H2O 
 XeO3  6HF
5
Grade XII DELHI SET 1
Equations for partial hydrolysis
XeF6  H2O 
 XeOF4  2HF
XeF6  2H2O 
 XeO2 F2  4HF
(b)
(i)
Shape: Pyramidal
Angle O – S – O: More than 90o
(ii)
Shape: Bent T-shaped
Angle F – Cl – F: Less than 90o
(iii)
Shape: linear
Angle F – Xe – F : 180o
OR
(a)
(i)
6NaOH + 3Cl2 
 5NaCl  NaClO3 + 3H2O
(hot and
conc.)
(ii)
XeF4  O2 F2 
 XeF6  O2
Grade XII DELHI SET 1
(b)
(i)
(ii)
(iii)
Q30. (a)
(b)
Illustrate the following name reaction giving suitable example in each case:
(i)
Clemmensen reduction
(ii)
Hell-Volhard-Zelinsky reaction
How are the following conversions carried out?
(i)
Ethylcyanide to ethanoic acid
(ii)
Butan-1-ol to butanoic acid
(iii)
Benzoic acid to m-bromobenzoic acid
OR
Ans.
(a)
Illustrate the following reactions giving a suitable example for each.
(i)
Cross aldol condensation
(ii)
Decarboxylation
(b)
Give simple tests to distinguish between the following pairs of compounds
(i)
Pentan-2-one and Pentan-3-one
(ii)
Benzaldehyde and Acetophenone
(iii)
Phenol and Benzoic acid
(a)
(i)
Clemmensen Reduction
5
Grade XII DELHI SET 1
The carbonyl group of aldehydes and ketones is reduced to CH2 group on
treatment with zinc-amalgam and concentrated hydrochloric acid. This is known
as Clemmensen reduction.
(ii)
(b)
Hell-Volhard-Zelinsky (HVZ )reaction.
Carboxylic acids having an α-hydrogen are halogenated at the α-position on
treatment with chlorine or bromine in the presence of small amount of red
phosphorus to give α-halocarboxylic acids. The reaction is known as HellVolhard-Zelinsky reaction.
(i)
(ii)
(i) K 2 Cr2 O7 /H 2SO4
CH3CH 2CH 2CH 2OH 
 CH3CH 2CH 2COOH
(ii) Dil. H2SO4
Butanol
(iii)
Butanoic acid
Grade XII DELHI SET 1
OR
(a)
(b)
(i)
Cross aldol condensation:
When aldol condensation is carried out between two different aldehydes and / or
ketones, it is called cross aldol condensation. If both of them contain α-hydrogen
atoms, it gives a mixture of four products.
(ii)
Decarboxylation refers to the reaction in which carboxylic acids lose
carbon dioxide to form hydrocarbons when their sodium salts are heated
with soda-lime.
(i)
Pentan-2-one and pentan-3-one can be distinguished by iodoform test.
Pentan-2-one is a methyl ketone. Thus, it responds to this test. But pentan3-one not being a methyl ketone does not respond to this test.
(ii)
Benzaldehyde (C6H5CHO) and acetophenone (C6H5COCH3) can be
distinguished by iodoform test.
Acetophenone, being a methyl ketone on treatment with I2/NaOH
undergoes iodoform reaction to give a yellow ppt. of iodoform. On the
other hand, benzaldehyde does not give this test.
Grade XII DELHI SET 1
C6 H 5COCH 3  3NaOI
Acetophenone

C6 H 5COONa  CHI3   2NaOH
Iodoform
NaOI
C6 H 5CHO 
 No yellow ppt of iodoform
(iii)
Benzaldehyde
Phenol and benzoic acid can be distinguished by ferric chloride test.
Ferric chloride test:
Phenol reacts with neutral FeCl3 to form ferric phenoxide complex giving violet
colouration.
But benzoic acid reacts with neutral FeCl3 to give a buff-coloured precipitate of
ferric benzoate.