question_answer

The geometry of \[{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}}\]ion is:

A)  trigonal bipyramidal      

B)  tetrahedral

C)  octahedral

D)  square planar

Correct Answer: D

Solution :

\[{}_{29}Cu=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}},4{{s}^{1}}\] \[C{{u}^{2+}}=[Ar]\,3{{d}^{9}},4{{s}^{0}}4{{p}^{0}}\] Hence, geometry of \[{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}}\] ion is square-planar and it is paramagnetic ion, as it has one unpaired electron.