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Both $[Ni(CO)_4]$ and $[Ni(CN)_4]^{2-}$ are diamagnetic.The hybridisation of nicked in these complexes ,respectively are

$\begin{array}{1 1}(a)\;sp^3,sp^3&(b)\; sp^3,dsp^2\\(c)\;dsp^2,sp^3&(d)\;dsp^2,dsp^2\end{array}$

1 Answer

According to valency bond theory (VBT),in $[Ni(CO)_4]$ Ni is in zero oxidation state due to CO strong ligand,the electrons are paired up.
$Ni :3d^84s^2$
In $[Ni(CN)_4]^{2-}$ Ni is in $+2$ oxidation state.
$Ni^{+2} : 3d^84s^0$
$\therefore [Ni(Co)_4]\rightarrow sp^3$ hybridisation
$[Ni(Co)_4]\rightarrow dsp^2$ hybridisation
Hence (b) is the correct option.
answered Mar 6, 2014 by sreemathi.v
 

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