Question

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

The correct option is B Tetrahedral, square planar and octahedral.

Nickel reacts with $C{l}^{-},C{N}^{-}$ and $H_{2}O$ to form $[NiC{l}_4{]}^{2-},[Ni(CN{)}_4{]}^{2-}$ and $\left[Ni\right(H_{2}O{)}_6{]}^{2+}$.

Electronic configuration of Nickel is $\left[Ar\right]3d^{8}4s^2$.

In $[NiC{l}_4{]}^{2-}$,

$N{i}^{2+}:\left[Ar\right]3d^{8}4s4p$

Chloride ion is weak field ligand due to which electron pairing does not takes place. And electrons from ligand occupy $4s$ and $4p$ orbitals of metal. Therefore, hybridization is $s{p}^3$ and geometry of $[NiC{l}_4{]}^{2-}$ is tetrahedral.

In $\left[Ni\right(CN{)}_4{]}^{2-}$,

$N{i}^{2+}:\left[Ar\right]3d^{8}4s4p$

Cyanide ion is a strong field ligand due to which electron pairing takes place. And electrons from ligand occupy one $3d$, $4s$ and two $4p$ orbitals of metal. Therefore, hybridization is $ds{p}^2$ and geometry of $\left[Ni\right(CN{)}_4{]}^{2-}$ is square planar.

In $\left[Ni\right(H_{2}O{)}_6{]}^{2+}$,

$N{i}^{2+}:\left[Ar\right]3d^{8}4s4p3d$

Water is a weak ligand due to which electron pairing does not take place. And electrons from ligand occupy $4s$, $4p$ and two $4d$ orbitals of metal. Therefore, hybridization is $s{p}^3{d}^2$ and geometry of $\left[Ni\right(H_{2}O{)}_6{]}^{2+}$ is octahedral.

Therefore, geometrical shape of $[NiC{l}_4{]}^{2-},[Ni\left(CN{)}_4{]}^{2-}and\right[Ni(H_{2}O{)}_6{]}^{2+}$ is tetrahedral, square planar and octahedral.

Hence, the correct option is (B).