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Which of the following is not isostructural with $SiCl_4$ :-
A
$SCl_4$
B
$SO_4^{2-}$
C
$PO_4^{3-}$
D
$NH_4^+$
Detailed Solution
$SiCl_4$: silicon has four bond pairs and no lone pair. It is $sp^3$ hybridised and tetrahedral.
$SO_4^{2-}$: sulphur has four sigma bond pairs and no lone pair; $sp^3$, tetrahedral.
$PO_4^{3-}$: phosphorus has four sigma bond pairs and no lone pair; $sp^3$, tetrahedral.
$NH_4^+$: nitrogen has four bond pairs and no lone pair; $sp^3$, tetrahedral.
$SCl_4$: sulphur has six valence electrons; four are used in the four S-Cl bonds and two remain as one lone pair. With five electron pairs it is $sp^3d$ hybridised, and the shape is see-saw (distorted tetrahedral), like $SF_4$.
So $SCl_4$ is not isostructural with $SiCl_4$.
$SO_4^{2-}$: sulphur has four sigma bond pairs and no lone pair; $sp^3$, tetrahedral.
$PO_4^{3-}$: phosphorus has four sigma bond pairs and no lone pair; $sp^3$, tetrahedral.
$NH_4^+$: nitrogen has four bond pairs and no lone pair; $sp^3$, tetrahedral.
$SCl_4$: sulphur has six valence electrons; four are used in the four S-Cl bonds and two remain as one lone pair. With five electron pairs it is $sp^3d$ hybridised, and the shape is see-saw (distorted tetrahedral), like $SF_4$.
So $SCl_4$ is not isostructural with $SiCl_4$.