A) \[Ni{{(CO)}_{4}}\] and \[NiCl_{4}^{2-}\] are diamagnetic but \[{{[Ni{{(CN)}_{4}}]}^{2-}}\] is paramagnetic
B) \[Ni{{(CO)}_{4}}\]and \[{{[Ni{{(CN)}_{4}}]}^{2-}}\] are diamagnetic but \[NiCl_{4}^{2-}\]is paramagnetic
C) \[NiCl_{4}^{2-}\] and \[{{[Ni{{(CN)}_{4}}]}^{2-}}\]are diamagnetic but \[Ni{{(CO)}_{4}}\] is paramagnetic
D) \[Ni{{(CO)}_{4}}\] is diamagnetic but \[NiCl_{4}^{2-}\] and \[{{[Ni{{(CN)}_{4}}]}^{2-}}\] is paramagnetic
Correct Answer: B
Solution :
\[Ni\,\,(28)\,=[Ar]3{{d}^{8}},\,4{{s}^{2}}\] \[N{{i}^{2+}}=[Ar]3{{d}^{8}}\] \[Ni\] and \[N{{i}^{2+}}\] both have two unpaired electrons. \[CO\] and \[C{{N}^{-}}\] are strong field ligands and thus unpaired electrons get .paired. Hence, \[Ni{{(CO)}_{4}}\] and \[{{[Ni{{(CN)}_{4}}]}^{2-}}\] are diamagnetic. \[C{{l}^{-}}\]is a weak field ligand hence, \[NiCl_{4}^{2-}\] is paramagnetic.
You need to login to perform this action.
You will be redirected in
3 sec
