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