Question

What type of hybridization is present in $\mathrm{Ni}$ of $[\mathrm{Ni}(\mathrm{Cl})_4]^{2-}$ and $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$ respectively?

• A. $\mathrm{dsp}^2$ and $\mathrm{ddp}^2$
• B. $\mathrm{sp}^3$ and $\mathrm{dsp}^2$
• C. $\mathrm{dsp}^2$ and $\mathrm{sp}^3$
• D. $\mathrm{sp}^3$ and $\mathrm{sp}^3$

Hint:

Remember this classic rule of thumb for $\mathrm{Ni}^{2+}$ ($d^8$) complexes with a coordination number of 4: a weak field ligand like $\mathrm{Cl}^-$ always results in a tetrahedral geometry ($\mathrm{sp}^3$), whereas a strong field ligand like $\mathrm{CN}^-$ always forces a square planar structure ($\mathrm{dsp}^2$).

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
The question asks to find the orbital hybridization of the central Nickel ($\mathrm{Ni}$) ion in two coordination complexes: $[\mathrm{Ni}(\mathrm{Cl})_4]^{2-}$ and $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$.

Step 2: Key Formula or Approach:
According to Valence Bond Theory (VBT) and Crystal Field Theory (CFT): 1. Identify the oxidation state of the central metal ion. 2. Determine its $d$-electron configuration. 3. Check the nature of the ligands: strong field ligands cause electron pairing, while weak field ligands leave spins unpaired.

Step 3: Detailed Explanation:
In both coordination complexes, Nickel is in the $+2$ oxidation state ($\mathrm{Ni}^{2+}$). The atomic number of $\mathrm{Ni}$ is 28 ($[\mathrm{Ar}]\,3d^8\,4s^2$), so the configuration of $\mathrm{Ni}^{2+}$ is $[\mathrm{Ar}]\,3d^8$.
Case 1: $[\mathrm{Ni}(\mathrm{Cl})_4]^{2-}$
The chloride ion ($\mathrm{Cl}^-$) is a weak field ligand. It cannot overcome the pairing energy, so the eight $3d$ electrons remain distributed across the five orbitals with two unpaired electrons. Because the inner $3d$ orbitals are completely unavailable, the four incoming lone pairs occupy the outer $\mathrm{4s}$ and three $\mathrm{4p}$ orbitals, leading to $\mathrm{sp}^3$ hybridization (tetrahedral geometry).
Case 2: $[\mathrm{Ni}(\mathrm{CN})_4]^{2-}$
The cyanide ion ($\mathrm{CN}^-$) is a strong field ligand. It forces the electrons in the $3d$ subshell to pair up completely, leaving one internal $3d$ orbital vacant. The four coordinate bonds are formed by utilizing this empty internal $3d$ orbital along with the $\mathrm{4s}$ and two $\mathrm{4p}$ orbitals, resulting in $\mathrm{dsp}^2$ hybridization (square planar geometry).

Step 4: Final Answer:
The hybridizations are $\mathrm{sp}^3$ and $\mathrm{dsp}^2$ respectively, which corresponds perfectly to option (B).