Both $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$ and $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$ are…

1. $s p^3, s p^3$
2. $s p^3, d s p^2$
3. $d s p^2, s p^3$
4. $d s p^2, d s p^2$

Solution

In $\left[\mathrm{Ni}(\mathrm{CO})_4\right]$, the oxidation state of $\mathrm{Ni}$ is zero $(0)$. $ \mathrm{Ni}(28)[\mathrm{Ar}] 4 \mathrm{~s}^2, 3 d^8 $
$ \mathrm{CO} \text { is a strong ligand, causes coupling, thus } $
In $\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-}$, the oxidation state of $\mathrm{Ni}$ is $+2$ $\mathrm{Ni}^{2+}=[\mathrm{Ar}] 3 d^8, 4 s^0$
$ \mathrm{CN}^{-} \text {is strong ligand causes coupling. } $

Asked in: JEE Advanced 2008 (Paper 2)

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