Question

Tetraaminecopper(II) ion is a square planar complex with one unpaired electron. According to valence bond theory the hybrid state of copper should be

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

Answer 1

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

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

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.