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Byju's Answer
Standard XII
Chemistry
AVSEPR
Arrange the m...
Question
Arrange the molecule in the increasing order of bond angle.
A
H
2
O
<
O
F
2
<
O
C
l
2
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B
O
C
l
2
<
H
2
O
<
O
F
2
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C
O
F
2
<
O
C
l
2
<
H
2
O
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D
O
F
2
<
H
2
O
<
O
C
l
2
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Solution
The correct option is
A
H
2
O
<
O
F
2
<
O
C
l
2
H
2
O
<
O
F
2
<
O
C
l
2
O
C
l
2
have a maximum bond angle as the size of
C
l
atom is bigger due to which
2
C
l
atom repel each other, thus have bigger bond angle than
H
2
O
and
O
F
2
.
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Similar questions
Q.
The sequence that correctly represents the increasing order of bond angles in
O
F
2
,
O
C
l
2
,
C
l
O
2
and
H
2
O
is
:
Q.
I
II
O
C
l
2
O
F
2
The bond angle is greater in
O
C
l
2
than in
O
F
2
because there is some
π
interactions in
O
C
l
2
involving donation from the filled p-orbitals on oxygen into the empty d-orbitals on chlorine No
π
-bonding is possible for
O
F
2
because all orbitals are filled on both atoms.
If true enter 1, else enter 0.
Q.
The bond angle in
O
F
2
out of
O
F
2
,
C
l
2
O
,
B
r
2
O
is minimum. It is because in case of
O
F
2
:
Q.
Δ
H
for the reaction,
O
F
2
+
H
2
O
→
O
2
+
2
H
F
:(B.E. of O - F, O - H, H - F and O = O are 44, 111, 135 and 119 kcal
m
o
l
−
1
respectively)
Q.
Consider the following molecules :
(I)
H
2
O
(II)
H
2
S
(III)
H
2
S
e
(IV)
H
2
T
e
Arrange these molecules in increasing order of bond angle.
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