Chemical Bonding & Molecular Structure Practice Questions - DPP for JEE

 Page 1


1. (a) Greater the difference in electronegativity between the two atoms,
larger will be polarity and hence dipole moment. Thus (a) has
maximum dipole moment.
2. (b) BiCl
3
: ;  sp
2
 - Hybridisation (Trigonal
planar geometry); Bond angle = 120º
In these, order
of bond angle : BCl
3
 > PCl
3
 > AsCl
3
 > BiCl
3
3. (a) In NH
3
 the atomic dipole (orbital dipole due to lone pair) and bond
dipole are in the same direction whereas in NF
3
 these are in
Page 2


1. (a) Greater the difference in electronegativity between the two atoms,
larger will be polarity and hence dipole moment. Thus (a) has
maximum dipole moment.
2. (b) BiCl
3
: ;  sp
2
 - Hybridisation (Trigonal
planar geometry); Bond angle = 120º
In these, order
of bond angle : BCl
3
 > PCl
3
 > AsCl
3
 > BiCl
3
3. (a) In NH
3
 the atomic dipole (orbital dipole due to lone pair) and bond
dipole are in the same direction whereas in NF
3
 these are in
Species Number of lone pairs on
opposite direction so in the former case they are added up whereas
in the latter case net result is reduction of dipole moment. It has
been shown in the following figure :
4. (b) On changing N
2
 to N
+
? 2
, B.O. decreases from 3 to 2.5 whereas on
changing O
2 
to O
+
? 2
, B.O. increases from 2 to 2.5. In former case, the
bond dissociation energy decreases and in the latter case, it
increases.
5. (a) Hybridisation =  [No. of valence electrons of central atom + No.
of monovalent atoms attached to it +
Negative charge if any – Positive charge if any]
NO
2
? –
   
NO
3
? –
   
NH
2
? –
 
NH
4
? +
, 
SCN
–
  = sp
i.e., NO
2
? –
 and NO
3
? –
 have same hybridisation.
6. (c) ClO
3
? –
 and SO
3
? 2–
 both have same number of electrons (42) and central
atom in each being sp
3
 hybridised. Both are having one lone pair
on central atom hence they are pyramidal.
7. (a) The number of lone pairs of electrons on central atom in various
given species are
Page 3


1. (a) Greater the difference in electronegativity between the two atoms,
larger will be polarity and hence dipole moment. Thus (a) has
maximum dipole moment.
2. (b) BiCl
3
: ;  sp
2
 - Hybridisation (Trigonal
planar geometry); Bond angle = 120º
In these, order
of bond angle : BCl
3
 > PCl
3
 > AsCl
3
 > BiCl
3
3. (a) In NH
3
 the atomic dipole (orbital dipole due to lone pair) and bond
dipole are in the same direction whereas in NF
3
 these are in
Species Number of lone pairs on
opposite direction so in the former case they are added up whereas
in the latter case net result is reduction of dipole moment. It has
been shown in the following figure :
4. (b) On changing N
2
 to N
+
? 2
, B.O. decreases from 3 to 2.5 whereas on
changing O
2 
to O
+
? 2
, B.O. increases from 2 to 2.5. In former case, the
bond dissociation energy decreases and in the latter case, it
increases.
5. (a) Hybridisation =  [No. of valence electrons of central atom + No.
of monovalent atoms attached to it +
Negative charge if any – Positive charge if any]
NO
2
? –
   
NO
3
? –
   
NH
2
? –
 
NH
4
? +
, 
SCN
–
  = sp
i.e., NO
2
? –
 and NO
3
? –
 have same hybridisation.
6. (c) ClO
3
? –
 and SO
3
? 2–
 both have same number of electrons (42) and central
atom in each being sp
3
 hybridised. Both are having one lone pair
on central atom hence they are pyramidal.
7. (a) The number of lone pairs of electrons on central atom in various
given species are
central atom
IF
7
0
IF
5
1
ClF
3
2
XeF
2
3
Thus the correct increasing order is
IF
7
 < IF
5
 < ClF
3
 < XeF
2
8. (b)
The geometry of H
2
O should have been tetrahedral if there are all bond
pairs. But due to presence of two lone pairs the shape is distorted
tetrahedral. Hence bond angle reduced to 104.5° from 109.5°.
9. (a) In NH
3
 and  the hybridisation is sp
3
 and the bond angle is
almost 109º 28'.
10. (a) Hybridisation = 
(a) For AlH
3
,
Hybridisation of Al atom =  = 3 = sp
2
For AlH
4
? –
,
Hybridisation of Al atom =  = 4 = sp
3
(b) For H
2
O,
Page 4


1. (a) Greater the difference in electronegativity between the two atoms,
larger will be polarity and hence dipole moment. Thus (a) has
maximum dipole moment.
2. (b) BiCl
3
: ;  sp
2
 - Hybridisation (Trigonal
planar geometry); Bond angle = 120º
In these, order
of bond angle : BCl
3
 > PCl
3
 > AsCl
3
 > BiCl
3
3. (a) In NH
3
 the atomic dipole (orbital dipole due to lone pair) and bond
dipole are in the same direction whereas in NF
3
 these are in
Species Number of lone pairs on
opposite direction so in the former case they are added up whereas
in the latter case net result is reduction of dipole moment. It has
been shown in the following figure :
4. (b) On changing N
2
 to N
+
? 2
, B.O. decreases from 3 to 2.5 whereas on
changing O
2 
to O
+
? 2
, B.O. increases from 2 to 2.5. In former case, the
bond dissociation energy decreases and in the latter case, it
increases.
5. (a) Hybridisation =  [No. of valence electrons of central atom + No.
of monovalent atoms attached to it +
Negative charge if any – Positive charge if any]
NO
2
? –
   
NO
3
? –
   
NH
2
? –
 
NH
4
? +
, 
SCN
–
  = sp
i.e., NO
2
? –
 and NO
3
? –
 have same hybridisation.
6. (c) ClO
3
? –
 and SO
3
? 2–
 both have same number of electrons (42) and central
atom in each being sp
3
 hybridised. Both are having one lone pair
on central atom hence they are pyramidal.
7. (a) The number of lone pairs of electrons on central atom in various
given species are
central atom
IF
7
0
IF
5
1
ClF
3
2
XeF
2
3
Thus the correct increasing order is
IF
7
 < IF
5
 < ClF
3
 < XeF
2
8. (b)
The geometry of H
2
O should have been tetrahedral if there are all bond
pairs. But due to presence of two lone pairs the shape is distorted
tetrahedral. Hence bond angle reduced to 104.5° from 109.5°.
9. (a) In NH
3
 and  the hybridisation is sp
3
 and the bond angle is
almost 109º 28'.
10. (a) Hybridisation = 
(a) For AlH
3
,
Hybridisation of Al atom =  = 3 = sp
2
For AlH
4
? –
,
Hybridisation of Al atom =  = 4 = sp
3
(b) For H
2
O,
Hybridisation of O atom
=  = 4  = sp
3
For  H
3
O
+
,  Hybridisation of O atom
=  = 4  = sp
3
(c) For NH
3
Hybridisation  of N atom
=  = 4  = sp
3
For Hybridisation
 
of N atom
=  = 4  = sp
3
Thus hybridisation changes only in option (a).
11. (b)
12. (b) Hybridisation in = ½ [5 + 0 + 3 –0] = 4 sp
3
. In bonding
only d orbital of P, p orbital of O can be involved. Since hybrid
atomic orbitals do not  form bond.
13. (a) For any species to have same bond order we can expect them to
Page 5


1. (a) Greater the difference in electronegativity between the two atoms,
larger will be polarity and hence dipole moment. Thus (a) has
maximum dipole moment.
2. (b) BiCl
3
: ;  sp
2
 - Hybridisation (Trigonal
planar geometry); Bond angle = 120º
In these, order
of bond angle : BCl
3
 > PCl
3
 > AsCl
3
 > BiCl
3
3. (a) In NH
3
 the atomic dipole (orbital dipole due to lone pair) and bond
dipole are in the same direction whereas in NF
3
 these are in
Species Number of lone pairs on
opposite direction so in the former case they are added up whereas
in the latter case net result is reduction of dipole moment. It has
been shown in the following figure :
4. (b) On changing N
2
 to N
+
? 2
, B.O. decreases from 3 to 2.5 whereas on
changing O
2 
to O
+
? 2
, B.O. increases from 2 to 2.5. In former case, the
bond dissociation energy decreases and in the latter case, it
increases.
5. (a) Hybridisation =  [No. of valence electrons of central atom + No.
of monovalent atoms attached to it +
Negative charge if any – Positive charge if any]
NO
2
? –
   
NO
3
? –
   
NH
2
? –
 
NH
4
? +
, 
SCN
–
  = sp
i.e., NO
2
? –
 and NO
3
? –
 have same hybridisation.
6. (c) ClO
3
? –
 and SO
3
? 2–
 both have same number of electrons (42) and central
atom in each being sp
3
 hybridised. Both are having one lone pair
on central atom hence they are pyramidal.
7. (a) The number of lone pairs of electrons on central atom in various
given species are
central atom
IF
7
0
IF
5
1
ClF
3
2
XeF
2
3
Thus the correct increasing order is
IF
7
 < IF
5
 < ClF
3
 < XeF
2
8. (b)
The geometry of H
2
O should have been tetrahedral if there are all bond
pairs. But due to presence of two lone pairs the shape is distorted
tetrahedral. Hence bond angle reduced to 104.5° from 109.5°.
9. (a) In NH
3
 and  the hybridisation is sp
3
 and the bond angle is
almost 109º 28'.
10. (a) Hybridisation = 
(a) For AlH
3
,
Hybridisation of Al atom =  = 3 = sp
2
For AlH
4
? –
,
Hybridisation of Al atom =  = 4 = sp
3
(b) For H
2
O,
Hybridisation of O atom
=  = 4  = sp
3
For  H
3
O
+
,  Hybridisation of O atom
=  = 4  = sp
3
(c) For NH
3
Hybridisation  of N atom
=  = 4  = sp
3
For Hybridisation
 
of N atom
=  = 4  = sp
3
Thus hybridisation changes only in option (a).
11. (b)
12. (b) Hybridisation in = ½ [5 + 0 + 3 –0] = 4 sp
3
. In bonding
only d orbital of P, p orbital of O can be involved. Since hybrid
atomic orbitals do not  form bond.
13. (a) For any species to have same bond order we can expect them to
have same number of electrons. Calculating the number of
electrons in various species.
; 
NO
+
(7 + 8 – 1= 14); CN
+
 (6 + 7 –1 = 12)
We find CN
–
 and NO
+
 both have 14 electrons so they have same bond
order. Correct answer is (a).
14. (d)
it contains  one unpaired electron hence paramagnetic.
15. (b) The delocalised  bonding between filled p-orbital of F and
vacant p-orbital of B leads to shortening of B–F bond length which
results in higher bond dissociation energy of the B–F bond.
16. (b) Compounds involved in chelation become non-polar.