Animated Solution for Chemistry - Coordination Compounds: The molecule in which hybrid MOs involve only one $d$-orbital of the central atom is

To determine which molecule utilizes exactly one $d$-orbital in its hybridisation, we must systematically evaluate the steric number and coordination geometry of each option provided. The hybridisation of a central atom is dictated by the number of electron domains (bond pairs and lone pairs) surrounding it, or in the case of coordination complexes, by the coordination number and the nature of the ligands.

Let's break down the first three options:

1. ${\text{XeF}}_4$ (Xenon tetrafluoride): Xenon, a noble gas, has 8 valence electrons. In ${\text{XeF}}_4$, it forms 4 single bonds with fluorine atoms, utilizing 4 electrons. The remaining 4 electrons form 2 lone pairs.

The steric number is $4\text{ (bond pairs)}+2\text{ (lone pairs)}=6$. A steric number of 6 corresponds to an octahedral electron geometry, which requires $s{p}^3{d}^2$ hybridisation. This involves two $d$-orbitals.

2. $[{\text{CrF}}_6{]}^{3-}$ (Hexafluorochromate ion): This is a classic octahedral coordination complex with a coordination number of 6. Regardless of whether it forms an inner orbital ($d^{2}s{p}^3$) or an outer orbital ($s{p}^3{d}^2$) complex, an octahedral geometry inherently requires the participation of two $d$-orbitals ($d_{z^2}$ and $d_{x^2-y^2}$).

3. ${\text{BrF}}_5$ (Bromine pentafluoride): Bromine has 7 valence electrons. It forms 5 single bonds with fluorine, leaving 2 electrons to form 1 lone pair.

The steric number is $5\text{ (bond pairs)}+1\text{ (lone pair)}=6$. Similar to ${\text{XeF}}_4$, this results in $s{p}^3{d}^2$ hybridisation, which again utilizes two $d$-orbitals.

Analyzing $[{\text{Ni(CN)}}_4{]}^{2-}$

Now, let's focus our attention on the tetracyanonickelate ion, $[{\text{Ni(CN)}}_4{]}^{2-}$.

First, we determine the oxidation state of Nickel. Since each cyanide (${\text{CN}}^{-}$) ligand carries a $-1$ charge and the overall complex has a $-2$ charge, Nickel must be in the $+2$ oxidation state.

The ground state electronic configuration of neutral Nickel is $[\text{Ar}]3d^{8}4s^2$. Therefore, the ${\text{Ni}}^{2+}$ ion has a configuration of $[\text{Ar}]3d^8$. According to Hund's rule, these 8 electrons will fill the five $3d$ orbitals such that there are 3 pairs and 2 unpaired electrons.

The Role of the Ligand

Here is where the magic happens. The cyanide ion (${\text{CN}}^{-}$) is a strong field ligand. In the presence of a strong field ligand, the energy gap between the $d$-orbitals increases significantly, forcing the unpaired electrons in the $3d$ subshell to pair up against Hund's rule.

When the two unpaired electrons pair up, they leave exactly one $3d$ orbital completely vacant.

The coordination number of the complex is 4, meaning the ${\text{Ni}}^{2+}$ ion needs 4 empty orbitals to accept the electron pairs donated by the four ${\text{CN}}^{-}$ ligands. It utilizes the newly vacated $3d$ orbital, the empty $4s$ orbital, and two of the empty $4p$ orbitals.

This specific combination results in $ds{p}^2$ hybridisation. As the notation clearly shows, this hybridisation scheme involves exactly one $d$-orbital (specifically, the $d_{x^2-y^2}$ orbital). This perfectly matches the condition asked in the question, making $[{\text{Ni(CN)}}_4{]}^{2-}$ the correct answer.