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Which compound has planar structure :
A
$XeF_4$
B
$XeOF_2$
C
$XeO_2F_2$
D
$XeO_4$
Detailed Solution
$XeF_4$: Xe has 4 bond pairs and 2 lone pairs ($sp^3d^2$). The lone pairs occupy opposite (axial) positions, giving a square planar shape.
$XeO_4$ is tetrahedral, $XeO_2F_2$ is see-saw, and $XeOF_2$ is T-shaped; these are not planar in the same way.
So $XeF_4$ has a planar structure.
$XeO_4$ is tetrahedral, $XeO_2F_2$ is see-saw, and $XeOF_2$ is T-shaped; these are not planar in the same way.
So $XeF_4$ has a planar structure.
