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Q.

What is the hybridization of $K_{3} [Cu {(CN)}_{4}]$

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a

${sp}^{2}$

b

${dsp}^{2}$

c

$d^{3} s$

d

${sp}^{3}$

answer is B.

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Detailed Solution

$K_{3} [Cu {(CN)}_{4}] has {Cu}^{+} : 3 d^{10}$

Thus ${sp}^{3}$ hybrid regardless of strength of ligand.

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