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Q.
Geometrical shapes of the complexes formed by the reaction of $N i^{2 +} \, with \, Cl^{-},CN^{-} \, and \, H_{2}O$ , respectively, are
NTA AbhyasNTA Abhyas 2022
Solution:
$Ni^{2 +}+4CN^{-} \rightarrow \left(\left[N i \left(C N\right)_{4}\right]\right)^{2 -}$
Here $Ni^{2 +}$ has $\text{d}^{\text{8}}$ -configuration with $CN^{-}$ as strong ligand.
$\text{d}^{\text{8}}$ -configuration in strong ligand field gives $\text{dsp}^{\text{2}}$ hybridisation, hence square planar geometry.
$Ni^{2 +}+4Cl^{-} \rightarrow \left[N i C l_{4}\right]^{2 -}$
Here $Ni^{2 +}$ has $\text{d}^{\text{8}}$ -configuration with $Cl^{-}$ as weak ligand.
$d^{8}$ -configuration in weak ligand field gives $sp^{3}$ hybridisation, hence tetrahedral geometry.
$Ni^{2 +}$ with $H_{2}O$ forms $\left(\left[N i \left(H_{2} O\right)_{6}\right]\right)^{2 +} \, $ complex and $H_{2}O$ is a weak ligand.
Therefore, $\left(\left[N i \left(H_{2} O\right)_{6}\right]\right)^{2 +}$ has octahedral geometry.
In octahedral complexes with $\text{d}^{\text{8}}$ configuration, there is always $\text{sp}^{\text{3}} \text{d}^{\text{2}}$ hybridisation (outer orbital complex) irrespective of strong ligand field or weak ligand field.
