question_answer

The ion which is not tetrahedral in shape is:

A) \[{{\operatorname{BF}}^{-}}_{4}\]                              

B) \[{{\operatorname{NH}}^{+}}_{4}\]

C) \[{{\left[ \operatorname{Cu}{{\left( N{{H}_{3}} \right)}_{4}} \right]}^{2+}}\]    

D) \[{{\operatorname{NiCl}}^{2-}}_{4}\]

Correct Answer: C

Solution :

\[{{\left[ \operatorname{Cu}{{\left( N{{H}_{3}} \right)}_{4}} \right]}^{2+}}\] It is square planar in shape.