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Byju's Answer
Standard XII
Chemistry
Introduction of VSEPR
apply VSEPR ...
Question
apply VSEPR postulates to describe the difference in the shapes of polyvalent species SF
4
,CF
4
and XeF
4
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Solution
Dear Student,
SF
4
:
Number
of
bond
pairs
=
4
Number
of
lone
pairs
=
1
Total
number
of
electron
pairs
=
4
+
1
=
5
Therefore
,
the
geometry
of
SF
4
should
be
triagonal
bipyramidal
but
due
to
the
presence
of
1
lone
pair
,
there
occurs
lone
pair
-
bond
pair
repulsion
which
distorts
it
geometry
to
see
-
saw
.
Therefore
,
the
shape
of
SF
4
is
see
-
saw
.
CF
4
:
Number
of
bond
pairs
=
4
Number
of
lone
pairs
=
0
Total
number
of
electron
pairs
=
4
+
0
=
4
So
,
its
geometry
is
tetrahedral
.
Since
there
is
no
lone
pair
present
,
so
there
will
be
no
distortion
and
its
shape
will
be
tetrahedral
.
XeF
4
:
Number
of
bond
pairs
=
4
Number
of
lone
pairs
=
2
Total
number
of
electron
pairs
=
4
+
2
=
6
So
,
its
geometry
must
be
octahedral
.
But
due
to
the
presence
of
two
unpaired
electrons
,
there
occurs
lone
pair
-
bond
pair
as
well
as
lone
pair
-
lone
pair
repulsion
which
distorts
its
shape
to
square
planar
.
Therefore
,
the
shape
of
XeF
4
is
square
planar
.
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