In $C u$ -ammonia complex, the state of hybridization of $C u^{2 +}$ is :

s p^{3}

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d^{3} s

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s p^{2}

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d s p^{2}

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Solution

The correct option is D $d s p^{2}$
in presence of strong ligand $N H_{3}$ low spin complex forms and $d s p^{2}$ hybridization takes place.
And shape is square planar.
Also.
$C u^{+ 2} = d^{9}$
so it have one unpaired electron and its magnetic moment is
$μ = \sqrt{n (n + 2)} = \sqrt{3} = 1.73 B M$

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