$[C u (N H_{3})_{4}]^{2 +}$ shows the following hybridization

d s p^{2}

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s p^{3} d

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d s p^{3}

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s p^{3}

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Solution

The correct option is A $d s p^{2}$
$C u = [A r] 3 d^{10} 4 s^{1}$
$C u^{2 +} = [A r] 3 d^{9}$

The electron pair donate by $N H_{3}$ enters into $4 s, 4 p$ and one $3 d$ subshell of Cu to form $d s p^{2}$ hybrid orbital.

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Similar questions

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

\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}

P C l_{5}

[C u {({N H}_{3})}_{4}] [P t {C l}_{4}]

{[F e {({N H}_{3})}_{6}]}_{2} {[P t {(C N)}_{6}]}_{3}

[C o {({N H}_{3})}_{6}] [C r {(C_{2} O_{4})}_{3}]

[P t {(e n)}_{3}] {({S O}_{4})}_{2}

P C l_{5}

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