Animated Solution for Chemistry - Coordination Compounds: According to the valence bond theory the hybridisation of central metal atom is $ds{p}^2$ for which one of the following compounds?

Welcome to a classic exploration of Coordination Compounds! Today, we are diving deep into the Valence Bond Theory (VBT) to determine which of the given Nickel complexes exhibits a $ds{p}^2$ hybridization. This is a fundamental concept that tests your understanding of oxidation states, electronic configurations, and the profound influence of ligands on the central metal atom.

The VBT Approach

To solve any VBT problem, we follow a strict, logical sequence: 1. Identify the central metal atom and calculate its oxidation state. 2. Write the electronic configuration of the metal ion. 3. Analyze the ligands attached to the metal. Are they Strong Field Ligands (SFL) or Weak Field Ligands (WFL)? 4. Determine the pairing of electrons. Strong field ligands force unpaired electrons to pair up against Hund's rule, while weak field ligands do not. 5. Find the hybridization based on the coordination number and the available empty orbitals.

Let's break down each option using this powerful framework.

Option (a): ${\text{NiCl}}_2\cdot 6{\text{H}}_2\text{O}$ In this complex, Nickel is in a $+2$ oxidation state. The ground state configuration of Nickel is $[\text{Ar}]3d^{8}4s^2$, so ${\text{Ni}}^{2+}$ becomes $[\text{Ar}]3d^{8}4s^0$. The ligands here are water (${\text{H}}_2\text{O}$) and chloride (${\text{Cl}}^{-}$), both of which are Weak Field Ligands. Because they are weak, they cannot force the two unpaired electrons in the $3d$ subshell to pair up. With a coordination number of 6, the metal must utilize the outer $4d$ orbitals to accommodate the incoming electron pairs. This results in an $s{p}^3{d}^2$ hybridization, giving the complex an octahedral geometry.

Option (b): ${\text{K}}_2[\text{Ni}(\text{CN}{)}_4]$ Here again, Nickel is in a $+2$ oxidation state, giving us the $[\text{Ar}]3d^{8}4s^0$ configuration. However, the game changes completely because Cyanide (${\text{CN}}^{-}$) is a Strong Field Ligand. The strong ligand field forces the two unpaired electrons in the $3d$ orbitals to pair up. This brilliant move leaves one inner $3d$ orbital completely empty! Since the coordination number is 4, the metal uses this empty inner $3d$ orbital, the $4s$ orbital, and two $4p$ orbitals. This perfectly yields the $ds{p}^2$ hybridization, which corresponds to a square planar geometry. We have found our answer!

Option (c): $[\text{Ni}(\text{CO}{)}_4]$ In Nickel tetracarbonyl, Nickel is in a rare $0$ oxidation state. Its configuration remains $[\text{Ar}]3d^{8}4s^2$. Carbon monoxide ($\text{CO}$) is an exceptionally strong field ligand. It exerts such a strong influence that it forces the two $4s$ electrons to jump down and pair up with the electrons in the $3d$ orbitals, completely filling the $3d$ subshell as $3d^{10}$. Now, for a coordination number of 4, the metal has to use the empty $4s$ and three $4p$ orbitals. This results in an $s{p}^3$ hybridization, giving a tetrahedral geometry.

Option (d): ${\text{Na}}_2[{\text{NiCl}}_4]$ Finally, in this complex, Nickel is back to a $+2$ oxidation state ($3d^8$). Chloride (${\text{Cl}}^{-}$) is a Weak Field Ligand, so no pairing of electrons occurs. With a coordination number of 4, the metal must use the outer $4s$ and three $4p$ orbitals, resulting in an $s{p}^3$ hybridization, which is also tetrahedral.

By systematically applying the principles of Valence Bond Theory, we can confidently conclude that only ${\text{K}}_2[\text{Ni}(\text{CN}{)}_4]$ exhibits the $ds{p}^2$ hybridization. Always remember to check the oxidation state and the strength of the ligand—that is the ultimate key to mastering coordination chemistry!