question_answer

Amongst \[Ni{{(CO)}_{4}},{{[Ni{{(CN)}_{4}}]}^{2-}}\] and \[NiC{{l}_{4}}^{2-}\]

A) \[Ni{{(CO)}_{4}}\] and \[NiC{{l}_{4}}^{2-}\] are diamagnetic and \[{{[Ni{{(CN)}_{4}}]}^{2-}}\] is paramagnetic

B) \[{{[Ni{{(CN)}_{4}}]}^{2-}}\] and \[{{[Ni{{(CN)}_{4}}]}^{2-}}\] are diamagnetic and \[Ni{{(CO)}_{4}}\] is paramagnetic

C) \[Ni{{(CO)}_{4}}\] and \[{{[Ni{{(CN)}_{4}}]}^{2-}}\] are diamagnetic and \[NiC{{l}_{4}}^{2-}\]is paramagnetic

D) \[Ni{{(CO)}_{4}}\] is diamagne and \[NiC{{l}_{4}}^{2-}\] and \[{{[Ni{{(CN)}_{4}}]}^{2-}}\] are paramagnetic

Correct Answer: C

Solution :

[c] \[[Ni{{(CO)}_{4}}]:\]

Oxidation state of \[Ni\]is 0

\[Ni=(Z=28)\Rightarrow 3{{d}^{8}}4{{s}^{2}}\]

Co is a strong field ligand, it causes pairing.

\[s{{p}^{3}}\]-Hybridisation (tetrahedral)

There are no unpaired electrons, so the complex is diamagnetic. Spin magnetic moment = Zero.

\[{{[NiC{{l}_{4}}]}^{2-}}:\]

\[N{{i}^{2+}}=3{{d}^{8}}\]

Cyano is a strong field ligand, it causes pairing.

\[ds{{p}^{2}}\]-hybridsation (square planar)

There are no unpaired electrons so, the complex is diamagnetic. Spin magnetic moment = zero.

\[{{[NiC{{l}_{4}}]}^{2-}}:\]

\[N{{i}^{2+}}=3{{d}^{8}}\].

Chloride is a weak field ligand, no pairing.

\[s{{p}^{3}}\]-hybridisation (tetrahedral)

There are two unpaired electrons, so the complex is paramagnetic.

Spin magnetic moment

\[\mu =\sqrt{n(n+2)}BM=\sqrt{2(2+2)}BM=\sqrt{8}BM\]