A) \[[Ni{{(CO)}_{4}}]\]
B) \[{{[NiC{{l}_{4}}]}^{2-}}\]
C) \[{{[Ni{{({{H}_{2}}O)}_{6}}]}^{2+}}\]
D) \[{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}}\]
Correct Answer: D
Solution :
\[{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2+}}\]ion has square-planar structure. \[{}_{29}Cu=1{{s}^{2}},2{{s}^{2}}2{{p}^{6}},3{{s}^{2}}3{{p}^{6}}3{{d}^{10}},4{{s}^{1}}\] \[C{{u}^{2+}}=[Ar]\,3{{d}^{9}}4s\] \[\underset{3d}{\mathop{}}\,\underset{4s}{\mathop{}}\,\,\,\,\,\underset{4p}{\mathop{}}\,\] \[C{{u}^{2+}}\]in presence of \[N{{H}_{3}}=\] \[\underset{{}}{\mathop{}}\,\underbrace{\underset{{}}{\mathop{}}\,\,\,\,\,\underset{{}}{\mathop{\,\,\,\,\,}}\,}_{\text{ds}{{\text{p}}^{\text{2}}}\text{-hybridisation}}\] \[{{[Cu{{(N{{H}_{3}})}_{4}}]}^{2-}}=\] \[\underset{\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{N}{{\text{H}}_{\text{3}}}}{\mathop{\,\,}}\,\underset{\text{N}{{\text{H}}_{\text{3}}}}{\mathop{}}\,\,\,\underset{\,\,\,\,\,\text{N}{{\text{H}}_{\text{3}}}\,\,\,\,\text{N}{{\text{H}}_{\text{3}}}}{\mathop{\,\,}}\,\] As, \[C{{u}^{2+}}\] undergoes \[ds{{p}^{2}}\text{-}\]hybridisation, this ion is square planar.
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