Geometrical shapes of the complexes formed by the reaction of $\mathrm{Ni}^{2+}$ with $\mathrm{Cl}^{-}$,…

1. octahedral, tetrahedral and square planar
2. tetrahedral, square planar and octahedral
3. square planar, tetrahedral and octahedral
4. octahedral, square planar and octahedral

Solution

$ \text { } \mathrm{Ni}^{2+}+4 \mathrm{CN}^{-} \longrightarrow\left[\mathrm{Ni}(\mathrm{CN})_4\right]^{2-} \text {. Here } \mathrm{Ni}^{2+} \text { has } d^8 \text { configuration with } \mathrm{CN}^{-} \text {as strong ligand. } $
$d^8$-configuration in strong ligand field gives $d s p^2$-hybridisation, hence square planar geometry. $ \mathrm{Ni}^{2+}+4 \mathrm{Cl}^{-} \longrightarrow\left[\mathrm{NiCl}_4\right]^{2-} $ Here $\mathrm{Ni}^{2+}$ had $d^8$-configuration with $\mathrm{Cl}^{-}$as weak ligand.
$d^8$-configuration in weak ligand field gives $s p^3$ hydridisation, hence tetrahedral geometry. $\mathrm{Ni}^{2+}$ with $\mathrm{H}_2 \mathrm{O}$ forms $\left[\mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}$ complex and $\mathrm{H}_2 \mathrm{O}$ is a weak ligand.
$ \text { Therefore, } \mathrm{Ni}\left(\mathrm{H}_2 \mathrm{O}\right)^{2+} \text { has octahedral geometry. } $

Asked in: JEE Advanced 2011 (Paper 1)

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