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Standard X
Chemistry
Covalent Bond
The number of...
Question
The number of lp-bp repulsion present in
C
l
F
3
at nearly 90 degree angle are_______.
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Solution
From figure it can be seen that
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Similar questions
Q.
The number of lp-bp repulsion present in
C
l
F
3
at nearly 90 degree angle are
:
Q.
Assertion :
N
-atom in
N
H
3
is
s
p
3
hybridised and bond angle is
107
o
. Reason:
l
p
−
b
p
repulsion (VSEPR) decreases bond angle to
107
o
.
Q.
Give the correct order of initials
T
or
F
for the following statements. Use
T
if statement is true and use
F
if the statement is false:
(I) The order of repulsion between pair of electrons is
l
p
−
l
p
>
l
p
−
b
p
>
b
p
−
b
p
where
l
p
=
lone pair
,
b
p
=
bond papir
(II) In general, as the number of lone pair of electrons on central atom increases , value of bond angle from normal bond angle also increases.
(III) The number of lone pair on
O
in
H
2
O
is
2
while on
N
in
N
H
3
is 1.
(IV) The structures of xenon flourides and xenon oxyflourides could not be explained on the basis of VSEPR theory.
Q.
Hybridization is the chemist's attempt to explain the observed molecular shape by constructing hybridized atomic orbitals with the appropriate inter orbital angles. The molecule for which deviation from a normal bond angle is observed,
V
S
E
P
R
theory suggests electron pair repulsive interaction (
l
p
−
l
p
>
l
p
−
b
p
>
b
p
). While from hybridization point of view that is a departure from normal hybridization because the angle between any equivalent hybrid orbitals determines the fraction of s and p characters of the hybrid and vice-versa.
In which species number of lone pairs on iodine and number of d-orbitals used in hybridization by iodine are the same?
Q.
Number of lone pair-bond pair repulsion at
90
∘
are (P) in
I
−
3
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