The planar compounds is/are:

¨ N (S i H_{3})_{3}

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{I F}_{4}^{-}

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B F_{3}

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B r F_{3}

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Solution

The correct options are
A $¨ N (S i H_{3})_{3}$
B ${I F}_{4}^{-}$
C $B F_{3}$
D $B r F_{3}$
In $(S i H_{3})_{3} N$ , $S i$ has vacant $d$ orbitals which provides $p π - d π$ bonding and so, nitrogen has $s p^{2}$ hybridization with trigonal planar shape.
$I F_{4}^{-}$ is square planar.
$B F_{3}$ trigonal planar.
$B r F_{3}$ trigonal planar.

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