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Chemistry Question on coordination compounds

The geometry of [Cu(NH3)4]2+{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}} ion is:

A

trigonal bipyramidal

B

tetrahedral

C

octahedral

D

square planar

Answer

square planar

Explanation

Solution

29Cu=1s2,2s22p6,3s23p63d10,4s1{}_{29}Cu=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}},4{{s}^{1}} Cu2+=[Ar] 3d9,4s04p0C{{u}^{2+}}=[Ar]\,3{{d}^{9}},4{{s}^{0}}4{{p}^{0}} Hence, geometry of [Cu(NH3)4]2+{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}} ion is square-planar and it is paramagnetic ion, as it has one unpaired electron.