Question

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

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

Answer 1

Answer: (D)

$Cu$-ammonia complex is ${\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}$

$N{H}_3$, being strong field ligand, shifts the unpaired electron to higher level.