Question

Nickel $\left( {Z = 28} \right)$ combines with uninegative monodentate ligand to form a diamagnetic complex \[{\left[ {{\text{Ni}}{{\text{X}}_{\text{4}}}} \right]^{ - 2}}\]. The hybridization involved and the number of unpaired electrons present in the complex is respectively:
A. ${\text{s}}{{\text{p}}^2}$, two
B. ${\text{ds}}{{\text{p}}^2}$, zero
C. ${\text{ds}}{{\text{p}}^2}$, one
D. ${\text{s}}{{\text{p}}^3}$, zero

Answer 1

Hint:The orbitals of metal combine to form the hybrid orbitals. Ligand donates electron pairs in these hybrid orbitals. The hybridization can be determined by counting the number of sigma bonds. After the formation of hybrid orbitals, the electrons get arranged in the orbitals. The presence of unpaired electrons in the metal complex tells the paramagnetic nature of the complex. . The presence of paired electrons in the metal complex tells the diamagnetic nature of the complex.

Complete answer:
According to the valence bond theory, the orbitals of metals combine to form orbitals of the same energy. These orbitals are known as hybrid orbital. Each ligand donates an electron pair to a hybrid orbital. Based on the number of electron pairs, the hybridization is determined.
First, determine the oxidation number of the metal as follows:
Suppose the charge of the metal is x so,
The charge of the coordination entity is $ - 2$ . The ligands are uninegative so the oxidation number of ligands is $ - 1$.
$\Rightarrow x + \,\left( { - 1 \times 4} \right) = - 2$
$\Rightarrow x = - 2 + 4$
$\Rightarrow x = + 2$
So, the oxidation number of nickel-metal in \[{\left[ {{\text{Ni}}{{\text{X}}_{\text{4}}}} \right]^{ - 2}}\] complex as $ + 2$.
The hybridization in \[{\left[ {{\text{Ni}}{{\text{X}}_{\text{4}}}} \right]^{ - 2}}\] is as follows
Valence electronic configuration of ${\text{N}}{{\text{i}}^{2 + }} = 3{d^8}$.
A total of four electron pairs are present in four hybridized orbitals so, the hybridization of \[{\left[ {{\text{Ni}}{{\text{X}}_{\text{4}}}} \right]^{ - 2}}\]complex is ${\text{ds}}{{\text{p}}^2}$. The number of unpaired electrons is zero.
So, the hybridization involved and the number of unpaired electrons present in the complex is ${\text{ds}}{{\text{p}}^2}$, zero respectively.

Therefore, option (B) ${\text{ds}}{{\text{p}}^2}$, zero.

Note:
The number of hybrid orbitals also represents the type of hybridization. It is given that the complex is diamagnetic so, it can be said without drawing the hybridization, that the complex has zero unpaired electrons. The complex in coordination number four shows two types of hybridization; ${\text{ds}}{{\text{p}}^{\text{2}}}$ and ${\text{s}}{{\text{p}}^{\text{3}}}$. The ${\text{ds}}{{\text{p}}^{\text{2}}}$ complexes are diamagnetic and complexes having ${\text{s}}{{\text{p}}^{\text{3}}}$ hybridization are paramagnetic generally.