Question

Geometrical shapes of the complexes formed by the reaction of $N{i}^{2+}$ 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)

tetrahedral, square planar and octahedral

$N{i}^{2+}+4C{l}^{-}\to [NiC{l}_4{]}^{2-}$

$[NiC{l}_4{]}^{2-}=3d^8;C{l}^{-}$ being weak ligand doesn't compel (configuration $s{p}^3)$ for the pairing of electrons.
Hence, the complex has a tetrahedral geometry.

$\left[Ni\right(CN{)}_4{]}^{2-}\to 3d^8$ configuration. $(CN{)}^{-}$ being strong ligand compel for the pairing of the electrons. Hence, square planar geometry.

$N{i}^{2+}+6H_{2}O\to \left[Ni\right(H_{2}O{)}_6{]}^{2+}$

$\left[Ni\right(H_{2}O{)}_6{]}^{2+}\to 3d^8$ configuration. As with $3d^8$ configuration, two orbitals are not available for $d^{2}s{p}^3$ hybridisation.

So, it will have octahedral geometry.

Hence, option $B$ is correct.