Question

Geometrical shapes of the complexes formed by the reaction of ${Ni}^{2+}$ with $C{l}^{-},C{N}^{-}$ and $H_{2}O$ , respectively, are

• A. octahedral, tetrahedral and square planar
• B. tetrahedral, square planar and octahedral
• C. square planar, tetrahedral and octahedral
• D. octahedral, square planar and octahedral

Answer 1

Answer: (B)

$N{i}^{2+}+4CN\to {\left[Ni(CN{)}_4\right]}^{2-}$
Here, $N{i}^{2+}$ has $d^8$-configuration with $C{N}^{-}$as strong ligand.

$d^8$-configuration in strong ligand field gives $ds{p}^2$-hybridisation, hence square planar geometry.
$N{i}^{2+}+4C{l}^{-}⟶{\left[NiC{l}_4\right]}^{2-}$
Here, $N{i}^{2+}$ has $d^8$-configuration with $C{l}^{-}$as weak ligand.

$d^8$-configuration in weak ligand field gives $s{p}^3$-hybridisation, hence tetrahedral geometry.
$N{i}^{2+}$ with $H_{2}O$ forms ${\left[Ni{\left(H_{2}O\right)}_6\right]}^{2+}$ complex and $H_{2}O$ is a weak ligand.

Therefore, ${\left[Ni{\left(H_{2}O\right)}_6\right]}^{2+}$ has octahedral geometry.