Daily Practice Problems
Overall
Score Total no of Correct Incorrect Unattempted
question
2 20 1 2 17
Solution
1 Which of the following statements regarding the 1m
structure of SOCl is incorrect ?
2
41 s
A The sulphur is sp hybridised and it has a
3
tetrahedral shape
B The sulphur is sp hybridised and it has a
3
trigonal pyramid shape
C The oxygen-sulphur bond has pπ − dπ bond
D It contain one lone pair of electron in sp hybrid
3
orbital of sulphur.
Solution : A
1 mins
In SOCl , the sulphur atom is sp hybridised. Due
2
3
to the presence of lone pair of electrons on the
S atom, the molecule has distorted tetrahedral
′ ′
geometry and trigonal pyramidal shape. The
shape contains two chlorine and one oxygen
atoms in a triangle. The S − O π bond is dπ − pπ
type.
2 Which of the following statements regarding the 0s
structure of SOCl is incorrect ?
2
A The sulphur is sp hybridised and it has a
3
tetrahedral shape
B The sulphur is sp hybridised and it has a
3
trigonal pyramid shape
C The oxygen-sulphur bond has pπ − dπ bond
D It contain one lone pair of electron in sp hybrid
3
orbital of sulphur.
Solution : A
1 mins
In SOCl , the sulphur atom is sp hybridised. Due
2
3
to the presence of lone pair of electrons on the
S atom, the molecule has distorted tetrahedral
′ ′
geometry and trigonal pyramidal shape. The
shape contains two chlorine and one oxygen
atoms in a triangle. The S − O π bond is dπ − pπ
type.
3 Which of the following molecules are planar in 36 s
shape?
A N (SiH3 )3
B N (CH3 )3
C P (SiH3 )3
D Both (a) and (c)
Solution : A
2 mins
a) In N (SiH ) , N has one lone pair and three
3 3
bond pairs. Hence, it should posses tetrahedral
geometry and pyramidal shape. But it forms a
planar structure because of the presence of back
bonding between the filled 2p orbital of
the central atom (N ) and the empty 3d orbital of
Si.
b) N (CH ) has tetrahedral geometry
3 3
and pyramidal shape due to the absence of back
bonding as C does not have vacant d-orbital.
c) In P (SiH ) , the shape is pyramidal as there is
3 3
very less extent or no back bonding between P
and Si. This is due to the inefficient overlaping of
orbitals.
4 Which of the following molecules are planar in 0s
shape?
A N (SiH3 )3
B N (CH3 )3
C P (SiH3 )3
D Both (a) and (c)
Solution : A
2 mins
a) In N (SiH ) , N has one lone pair and three
3 3
bond pairs. Hence, it should posses tetrahedral
geometry and pyramidal shape. But it forms a
planar structure because of the presence of back
bonding between the filled 2p orbital of
the central atom (N ) and the empty 3d orbital of
Si.
b) N (CH ) has tetrahedral geometry
3 3
and pyramidal shape due to the absence of back
bonding as C does not have vacant d-orbital.
c) In P (SiH ) , the shape is pyramidal as there is
3 3
very less extent or no back bonding between P
and Si. This is due to the inefficient overlaping of
orbitals.
5 The correct order of bond angles among the 39 s
following:
A OF2 > H2 O
B OCl2 > OBr2
C CH4 > N H3
D All of the above
Solution : C
3 mins
a) Since fluorine is more electronegative than
oxygen, the bond pairs are close to fluorine and
hence there is minimum repulsion between and
the bond pairs compared to H O where the bond
2
pairs are close to oxygen and hence, the bond
angle of OF is smaller than that of OH
2 2
b) Since Br is larger than Cl the bond angle of
OBr is larger than OCl
2 2
C) Since there are no lone pairs in methane, it will
not have any distortion and hence a greater bond
angle than N H 3
6 The correct order of bond angles among the 0s
following:
A OF2 > H2 O
B OCl2 > OBr2
C CH4 > N H3
D All of the above
Solution : C
3 mins
a) Since fluorine is more electronegative than
oxygen, the bond pairs are close to fluorine and
hence there is minimum repulsion between and
the bond pairs compared to H O where the bond
2
pairs are close to oxygen and hence, the bond
angle of OF is smaller than that of OH
2 2
b) Since Br is larger than Cl the bond angle of
OBr is larger than OCl
2 2
C) Since there are no lone pairs in methane, it will
not have any distortion and hence a greater bond
angle than N H 3
7 Which compound has a planar structure? 0s
A XeF4
B XeO3
C XeO2 F2
D XeO4
Solution : A
6 mins
In XeF molecule, the central atom (Xe) has 6
4
electron pairs around it in which 4 electron pairs
are bonding pairs and 2 electron pairs are lone
pairs.
So the geometry is octahedral but the shape is
square planar due to the two lone pairs on it. The
two lone pairs repel each other, so they are
placed at an angle of 180 and the shape is
o
square planar.
8 Which compound has a planar structure? 13 s
A XeF4
B XeO3
C XeO2 F2
D XeO4
Solution : A
6 mins
In XeF molecule, the central atom (Xe) has 6
4
electron pairs around it in which 4 electron pairs
are bonding pairs and 2 electron pairs are lone
pairs.
So the geometry is octahedral but the shape is
square planar due to the two lone pairs on it. The
two lone pairs repel each other, so they are
placed at an angle of 180 and the shape is
o
square planar.
-1.0/4.0
9 Which of the following pairs have identical shapes? 38 s
A CF4 , SF4
B XeF2 , CO2
C BF3 , P Cl3
D P F5 , BrF5
Solution : B
3 mins
The molecules assume different geometry based
on the number of bond pairs/lone pairs in the
valence shell of the central atom.
However, the lone pairs are ignored in assigning
the shape.
(a) CF4 having 4 sigma bond pairs and no lone
pairs is sp hybridised i.e., tetrahedral in shape.
3
Whereas, SF having 4 sigma bond pairs and 1
4
lone pair is sp d hybridised i.e., see-saw in shape.
3
(b) XeF2 having 2 sigma bond pairs and 3 lone
pairs is sp d hybridised i.e., linear in shape. CO
3
2
having 2 sigma bond pairs and no lone pairs is sp
hybridised i.e linear in shape.
(c) BF3 having 3 sigma bond pairs and no lone
pairs is sp hybridised i.e., trigonal planar in
2
shape. P Cl having 3 sigma bond pairs and 1 lone
3
pair is sp hybridised i.e., pyramidal in shape.
3
(d) P F5 having 5 sigma bond pairs and no lone
pairs is sp d hybridised i.e., trigonal bipyramidal in
3
shape. BrF having 5 sigma bond pairs and 1 lone
5
pair is sp d hybridised i.e., square pyramidal in
3 2
shape.
Thus, XeF and CO are the species having
2 2
identical shape i.e., linear.
10 Which of the following pairs have identical shapes? 0s
A CF4 , SF4
B XeF2 , CO2
C BF3 , P Cl3
D P F5 , BrF5
Solution : B
3 mins
The molecules assume different geometry based
on the number of bond pairs/lone pairs in the
valence shell of the central atom.
However, the lone pairs are ignored in assigning
the shape.
(a) CF4 having 4 sigma bond pairs and no lone
pairs is sp hybridised i.e., tetrahedral in shape.
3
Whereas, SF having 4 sigma bond pairs and 1
4
lone pair is sp d hybridised i.e., see-saw in shape.
3
(b) XeF2 having 2 sigma bond pairs and 3 lone
pairs is sp d hybridised i.e., linear in shape. CO
3
2
having 2 sigma bond pairs and no lone pairs is sp
hybridised i.e linear in shape.
(c) BF3 having 3 sigma bond pairs and no lone
pairs is sp hybridised i.e., trigonal planar in
2
shape. P Cl having 3 sigma bond pairs and 1 lone
3
pair is sp hybridised i.e., pyramidal in shape.
3
(d) P F5 having 5 sigma bond pairs and no lone
pairs is sp d hybridised i.e., trigonal bipyramidal in
3
shape. BrF having 5 sigma bond pairs and 1 lone
5
pair is sp d hybridised i.e., square pyramidal in
3 2
shape.
Thus, XeF and CO are the species having
2 2
identical shape i.e., linear.
11 Which of the following is the correct geometry for 0s
P OCl3 molecule?
A Tetrahedral
B Bent
C Trigonal pyramidal
D Linear
Solution : A
1 mins
P OCl3 has 4 bond pair and zero lone pair of
electron.
Phosphorus atom makes a double bond with
oxygen and 3 single bond with chlorine.
It has a hybridization of sp and has a tetrahedral
3
geomety
12 Which of the following is the correct geometry for 2m
P OCl3 molecule? 12 s
A Tetrahedral
B Bent
C Trigonal pyramidal
D Linear
Solution : A
1 mins
P OCl3 has 4 bond pair and zero lone pair of
electron.
Phosphorus atom makes a double bond with
oxygen and 3 single bond with chlorine.
It has a hybridization of sp and has a tetrahedral
3
geomety
13 Which of the following species contains equal 18 s
number of σ- and π- bonds?
A H CO
−
B XeO4
C (CN )2
D CH2 (CN )2
Solution : B
2 mins
14 Which of the following species contains equal 0s
number of σ- and π- bonds?
A H CO
−
B XeO4
C (CN )2
D CH2 (CN )2
Solution : B
2 mins
15 A sp hybridized orbital contains:
3 0s
1
A s character
4
1
B s character
2
2
C s character
3
3
D s character
4
Solution : A
1 mins
In a sp hybridized orbital number of s orbital = 1
3
and number of p orbitals = 3.
1
% s - character = s - character
4
16 A sp hybridized orbital contains:
3 34 s
1
A s character
4
1
B s character
2
2
C s character
3
3
D s character
4
Solution : A
1 mins
In a sp hybridized orbital number of s orbital = 1
3
and number of p orbitals = 3.
1
% s - character = s - character
4
4.0/4.0
17 Calculate the percentage of p − character of the 0s
orbitals of ‘B atom involved in bonding with H -
′
atom in BH molecule.
3
A 66.66%
B 33.33%
C 50%
D 75%
Solution : A
2 mins
The ground state electronic configuration of B
atom is: 1s 2s 2p 2 2 1
The excited state electronic configuration of B
atom is: 1s 2s 2p 2 1 2
Thus, here 2s orbital and two 2p orbitals
undergoes hybridisation to give three sp 2
hybridised orbitals. Hence, boron in BH is sp
3
2
hybridised.
Thus, the percentage of p − character of the
hybridised orbital of ‘B in BH molecule is
′
3
2
× 100 = 66.66 %
3
18 Calculate the percentage of p − character of the 56 s
orbitals of ‘B atom involved in bonding with H -
′
atom in BH molecule.
3
A 66.66%
B 33.33%
C 50%
D 75%
Solution : A
2 mins
The ground state electronic configuration of B
atom is: 1s 2s 2p 2 2 1
The excited state electronic configuration of B
atom is: 1s 2s 2p 2 1 2
Thus, here 2s orbital and two 2p orbitals
undergoes hybridisation to give three sp 2
hybridised orbitals. Hence, boron in BH is sp
3
2
hybridised.
Thus, the percentage of p − character of the
hybridised orbital of ‘B in BH molecule is
′
3
2
× 100 = 66.66 %
3
-1.0/4.0
19 Statement 1: N M e and N (SiM e
3 3) are not 0s
3
isostructural.
Statement 2: N M e and N (SiM e
3 3
)
3
have the same
steric number.
Regarding the above statements which of the
following is/are correct?
A Both the statements are true
B Both the statements are false
C Only statement 1 is true
D Only statement 2 is true
Solution : C
2 mins
N (SiM e3 ) is trigonal planar in shape because of
3
pπ − dπ back bonding between the vacant d-
orbital of Si and the filled 2p orbital (or lone pair)
of N atom. All the bond lenghts are equal having
a partial double bond character and the structure
is shown below.
So, the steric number ofN (SiM e ) is 3.
3 3
In N M e , there is no such pπ − dπ back bonding
3
as carbon does not have vacant d-orbitals.
The steric number of N M e is 4 and its structure
3
is pyramidal with lone pair at its apex.
(i) (ii)
20 Statement 1: N M e and N (SiM e
3 3) are not 46 s
3
isostructural.
Statement 2: N M e and N (SiM e
3 3
)
3
have the same
steric number.
Regarding the above statements which of the
following is/are correct?
A Both the statements are true
B Both the statements are false
C Only statement 1 is true
D Only statement 2 is true
Solution : C
2 mins
N (SiM e3 ) is trigonal planar in shape because of
3
pπ − dπ back bonding between the vacant d-
orbital of Si and the filled 2p orbital (or lone pair)
of N atom. All the bond lenghts are equal having
a partial double bond character and the structure
is shown below.
So, the steric number ofN (SiM e ) is 3.
3 3
In N M e , there is no such pπ − dπ back bonding
3
as carbon does not have vacant d-orbitals.
The steric number of N M e is 4 and its structure
3
is pyramidal with lone pair at its apex.
(i) (ii)
