Which complex has $d {s p}^{2}$ hybridization and square planar shape?

{[N i {(C N)}_{4}]}^{2 -}

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{[C u {({N H}_{3})}_{4}]}^{2 +}

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{[P t {(C l)}_{4}]}^{2 -}

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Solution

The correct option is A ${[N i {(C N)}_{4}]}^{2 -}$
$[N i (C N)_{4}]^{2 -}$
$_{28} N i = 4 s^{2} 3 d^{8}$
$N i^{2 +} = 4 s^{0} 3 d^{8}$
(Refer to Image 1)
$[N i (C N)_{4}]^{2 -}$
(Refer to Image 2)

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Similar questions

The type of hybridization of 'Pt' in ${[P t C l_{4}]}^{- 2}$ and geometry of ${[P t C l_{4}]}^{- 2}$ are

\begin{matrix} L i s t - I & L i s t - I I (p) [N i (C N)_{4}]^{2 -} & (1) s p^{3} h y b r i d i z a t i o n (Q) [N i (C O)_{4}] & (2) d s p^{2} h y b r i d i z a t i o n (R) [C u (N H_{3})_{4}]^{2 +} & (3) μ = 0 B M (S) [P d (C l)_{4}]^{2 -} & (4) μ = 1.732 B M \end{matrix}

[N i (C N)_{4}]^{2 -}

d -

{[C u {(N H_{3})}_{4}]}_{4}^{2 +}

Which of the following complexes is/are square planar in shape ?
(I) [Ni(CN)4]2− (II) [MnCl4]2− (III) [ZnCl4]2− (IV) [Ni(NH3)4]2+

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