Question

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

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

Answer 1

Answer: (A)

${}_{29}Cu=[Ar]3d^{10}4s^1$

$C{u}^{2+}=[Ar]3d^9$

$C{u}^{2+}=$ Ground state

${\left[Cu{\left(N{H}_3\right)}_4\right]}^{2+}=$