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In which of the following molecules all the bonds are not equal :-
A
$ClF_3$
B
$BF_3$
C
$AlF_3$
D
$NF_3$
Detailed Solution
In $ClF_3$ the Cl atom is $sp^3d$ hybridised, with three bond pairs and two lone pairs, in a trigonal bipyramidal arrangement of electron pairs.
The two lone pairs occupy equatorial positions, so the molecule is T-shaped, with two axial Cl-F bonds and one equatorial Cl-F bond.
The axial bonds are longer than the equatorial bond, because the axial bond pairs suffer greater repulsion. So all the bonds in $ClF_3$ are not equal.
$BF_3$ is $sp^2$ hybridised and trigonal planar, with three equal B-F bonds.
$AlF_3$ has three equivalent Al-F bonds.
$NF_3$ is $sp^3$ hybridised and pyramidal, with three equal N-F bonds.
So the molecule in which all the bonds are not equal is $ClF_3$.
The two lone pairs occupy equatorial positions, so the molecule is T-shaped, with two axial Cl-F bonds and one equatorial Cl-F bond.
The axial bonds are longer than the equatorial bond, because the axial bond pairs suffer greater repulsion. So all the bonds in $ClF_3$ are not equal.
$BF_3$ is $sp^2$ hybridised and trigonal planar, with three equal B-F bonds.
$AlF_3$ has three equivalent Al-F bonds.
$NF_3$ is $sp^3$ hybridised and pyramidal, with three equal N-F bonds.
So the molecule in which all the bonds are not equal is $ClF_3$.