A) \[{{[Ni{{(CN)}_{4}}]}^{2-}}\]-tetrahedral;\[[Ni{{(CO)}_{4}}]\]-Paramagnetic
B) \[{{[NiC{{l}_{4}}]}^{2-}}\] -paramagnetic; \[[Ni{{(CO)}_{4}}]\]-tetrahedral
C) \[{{[NiC{{l}_{4}}]}^{2-}}\] -diamagnetic; \[[Ni{{(CO)}_{4}}]\] -square-planar
D) \[{{[NiC{{l}_{4}}]}^{2-}}\] -Square-planar; \[{{[Ni{{(CN)}_{4}}]}^{2-}}\] -paramagnetic
Correct Answer: B
Solution :
\[{{[Ni{{(CN)}_{4}}]}^{2-}}\] is square planar, diamagnetic (0 unpaired electrons) with \[ds{{p}^{2}}\] hybridisation. \[[Ni{{(CO)}_{4}}]\] -is tetrahedral, diamagnetic (0 unpaired electrons) with \[s{{p}^{3}}\] hybridisation. \[{{[NiC{{l}_{4}}]}^{2-}}\] is tetrahedral, paramagnetic (2 unpaired electrons) with \[s{{p}^{3}}\] hybridisation. Hence, the option (B) is the correct answer.
You need to login to perform this action.
You will be redirected in
3 sec
