Download Chemistry (Theory)

Grade XII Foreign SET 2
Chemistry (Theory)
[Time allowed: 3 hours]
marks:70]
[Maximum
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)
Complete and balance the following chemical equations.
(i)
Cr2O72 + I + H+ 
(ii)
(b)
MnO4 +SO32 + H+ 
Explain the following observation:
(i)
Transition elements and their compounds are known to act as catalysts.
(ii)
The higher oxidation states are usually exhibited by the members in the
middle of a series of transition elements.
(iii)
The metal-metal bonding is more frequently found with the second and
third series of transition elements.
OR
(a)
Calculate the number of unpaired electrons in the following gaseous state ions:
Mn3+, Cr3+, V3+ and Fe2+
Which one of these is the most stable in aqueous solutions?
(At nos, V = 23, Cr = 24, Mn = 25, Fe = 26)
(b)
Explain the following observation:
(i)
The transition metal ions are usually colored in aqueous solutions.
(ii)
Cu (I) is not stable in an aqueous solution.
(iii)
The highest oxidation state of a transition metal is exhibited in its oxide or
fluoride.
Ans: (a)
(i)
(ii)
Cr2O72 – +14H+ +6 I  2Cr 3+ +3I 2 +7 H 2 O
Grade XII Foreign SET 2
(b)
(a)
(i)
The catalytic activity of the transition elements can be explained by two
basic facts. Firstly, owing to their ability to show variable oxidation states
and form complexes, transition metals form unstable intermediate
compounds. Thus, they provide a new path with lower activation energy,
Ea, for the reaction and secondly transition metals also provide a suitable
surface for the reactions to occur.
(ii)
The higher oxidation states are usually exhibited by the members in the
middle of a series of transition elements due to involvement of all ns and
(n-1)d electrons in the bonding.
(iii)
The metals of the second and third series have greater enthalpies of
atomization due to this reason metal-metal bonding is more frequently
found with the second and third series of transition elements.
5
2
OR
Mn = 3d 4s
So, Mn3+ = 3d4
Number of unpaid electrons = 4
Cr = 3d44S2
Cr3+ = 3d3
Number of unpaired electrons = 3
V = 3d34S2
V3+ = 3d2
Number of unpaired electrons = 2
Fe = 3d64S2
Fe2+ = 3d6
Number of unpaired electrons = 4
Mn3+ will be most stable because smallest in size so it has maximum Mn3+ us
hydration energy and hence more stability.
(b)
(i) Most of the complexes of transition metals are colored. This is because of the
absorption of radiation from visible light region to promote an electron from one
Grade XII Foreign SET 2
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 metals absorb the radiation of a particular wavelength
and the rest is reflected, imparting color to the solution.
(ii)
Copper (I) compounds are unstable in aqueous solution and undergo
disproportionation.
The stability of Cu 2+ (aq) rather than Cu+ (aq) is due to the much more negative
ΔhydH of Cu2+ (aq) than Cu+, which more than compensates for the second
ionisation enthalpy of Cu.
(iii)
Fluorine is most electronegative element. The transition metals show
highest oxidation states with fluorine. The tendency of fluorine to stabilize
highest oxidation state is due to highest lattice enthalpy as in case of
CoF3 or higher bond enthalpy due to higher covalent bonds e.g. VF5 and
CrF6.
Ability of oxygen to stabilize these high oxidation states exceeds that of
fluorine. For example Mn forms highest fluoride as MnF4 whereas highest
oxide is Mn2O7. This is due to the tendency of oxygen to form multiple
bonds.
Q29. (a)
(b)
What type of battery is the lead storage battery? Write the anode and cathode
reactions and the overall reaction occurring in a lead storage battery sending out
an electric current.
A voltaic cell is set up at 25°C with the half-cells
Ag+ (0.001 M) |Ag and Cu2+ (0.10 M)| Cu.
What should be its cell potential? (E° = 0.46 V; Log105 = 5)
OR
(a)
Define the term molar conductivity and explain how molar conductivity changes
with solution concentration for weak and strong electrolytes.
(b)
A strip of nickel metal is placed in a 1-molar solution of Ni(NO3)2 and a strip of
silver metal is placed in a 1-molar solution of AgNO3. An electrochemical cell is
created when the two solutions are connected by a salt bridge and the two strips
are connected by wires to a voltameter.
(i) Write the balanced equations for the overall reactions occurring in the cell and
calculate the cell potential.
Grade XII Foreign SET 2
(ii) Calculate the cell potential, E, at 25° C for the cell if the initial concentration
of Ni(NO3)2 is 0.100 molar and the initial concentration of AgNO3 is 1.00 molar.
[ENi2+, Ni = -0.25 V; E+Ag+/Ag = 0.80 V; Log 10–1 = – 1]
Ans: (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:
Pb  SO
2
4

 PbSO4  2e
At cathode:
PbO2  SO 24  4H  2e 
 PbSO4  2H2O
Overall cell reaction:
Pb  PbO2  4H  2SO42 
 2PbSO4  2H2O
(b)
Cu + 2Ag+  Cu2+ + 2Ag
Half-cell reactionsCathode (reduction): 2Ag+ (0.001 M) + 2e–  2Ag(s)
Anode (oxidation): Cu(s)  Cu2+ (0.10 M) + 2e–
n=2
E0cell = 0.46 V (Given)
Cu 2  
0.059
0
E Cell = E cell 
log
2
n
 Ag  
 0.1
0.059
 0.46 
log
2
2
 0.001
0.059
log105
2
0.059
 0.46 
 (5)
2
 0.46 – 0.1475
 0.3125 V
 0.46 
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.
Grade XII Foreign SET 2
A
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.
m 
Variation of molar conductivities with dilution:
In case of weak electrolyte: On increasing the dilution, undissociated weak
electrolytes dissociate in small extent further so, the number of conducting ions
increases resulting increase in the molar conductivity.
In case of strong electrolyte: In the concentrated solution, there is more
interionic force of attraction so, they are not much free for the conduction. On
dilution, the distance between the ions increases so, interionic interactions
decreases. Hence mobility of ions increases resulting in increase in molar
conductivity.
The variation of  m with c for strong and weak electrolytes is shown in the
following plot:
(b)
Half cell reactions:
Anode (oxidation): Ni(s)  Ni2+(1m) + 2e–
Cathode (reduction): 2Ag+(1m) + 2e  2Ag(s)
Overall reactions:
Ni + 2Ag +  Ni 2+ + 2Ag
....(1)
0
0
E 0cell = E (cathode)
 E (anode)
= 0.80  (– 0.25)
=1.05 V
From equation (1): n = 2
Grade XII Foreign SET 2
 
Ni 2 +  Ag
0.059
E cell = E 
log
2
n
Ni Ag + 
0.059
E cell =1.05 
log (1)
2
0.059
=1.05 
(0)
2
=1.05 V
0
cell
 
2
(ii) E 0cell =1.05 V
n=2
Ni 2+  = 0.100 M,
Ag +  =1.00 M
Ni 2 + 
0.059
log
2
n
Ag + 
 0.1
0.059
=1.05 
log
2
2
1.0
0
E cell = E cell

=1.05  0.0295  1
=1.05 + 0.0295
=1.0795 V  1.08 V
Q30. (a)
(c)
Describe the mechanism of the addition of Grignard reagent to the carbonyl group
of a compound a form an adduct which on hydrolysis yield an alcohol.
Draw the structure of the following compounds:
(i)
3-Methylbutanal
(ii)
Hexane-1, 6-dioic acid
(iii)
p-Nitropropiophenone
OR
(a)
Illustrate the following reactions giving suitable chemical equations for each:
(i)
Cannizzaro reaction
(ii)
Hell-Volhard-Zelinsky reaction.
(d)
How would you bring about the following conversions? Write the complete
equations in each case:
(i)
Ethanal to 3-hydroxybutanal
(ii)
Benzoic acid to m-nitrobenzyl alcohol
(iii)
Benzaldehyde to benzophenone
Grade XII Foreign SET 2
Ans: (a)
Alcohols are produced by the addition reaction of Grignard reagents to the
carbonyl group of aldehydes and ketones to give alcohol.
Mechanism:
Step1: The first step of the reaction is the nucleophilic addition of Grignard
reagent to the carbonyl group to form an adduct.
Step2: Hydrolysis of the adduct yields an alcohol.
(b)
(i)
3-Methylbutanal
(ii)
Hexane-1, 6-dioic acid
(iii)
p-Nitropropiophenone
OR
Ans:
(a)
(i)
Cannizzaro reaction
Aldehydes which do not have an α-hydrogen atom, undergo self oxidation
and reduction (disproportionation) reaction on treatment with concentrated
alkali. In this reaction, one molecule of the aldehyde is reduced to alcohol
while another is oxidised to carboxylic acid salt.
Example-
Grade XII Foreign SET 2
(ii)
Hell-volhard-Zelinsky 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
Hell-Volhard-Zelinsky reaction.
Example-
(b)
(i)
Ethanal to 3-hydroxybutanal
(ii)
Benzoic acid to m-nitrobenzyl alcohol
(iii)
Benzaldehyde to benzophenone.
Grade XII Foreign SET 2