Question

$\left[Cu\right(N{H}_3{)}_4{]}^{2+}$ shows the following hybridization

• A. $ds{p}^2$
• B. $s{p}^{3}d$
• C. $ds{p}^3$
• D. $s{p}^3$

Answer 1

The correct option is A $ds{p}^2$

$Cu=\left[Ar\right]3d^{10}4s^1$
$C{u}^{2+}=\left[Ar\right]3d^9$

The electron pair donate by $N{H}_3$ enters into $4s,4p$ and one $3d$ subshell of Cu to form $ds{p}^2$ hybrid orbital.