Question

Which of the following have tetrahedral structures?

• A. $[\text{Ni}(\text{CN})_4]^{2-}$
• B. $[\text{Ni}(\text{CO})_4]$
• C. $[\text{NiCl}_4]^{2-}$
• D. $\text{CrO}_4^{2-}$

Hint:

For $d^8$ configuration of central metal ions like $\text{Ni}^{2+}$:
- Strong-field ligands (e.g., $\text{CN}^-$) result in square planar geometry ($\text{dsp}^2$).
- Weak-field ligands (e.g., $\text{Cl}^-$) result in tetrahedral geometry ($\text{sp}^3$).

Answer 1

The Correct Option is B, C, D

Step 1: Understanding the Concept:
We need to determine the geometry of several coordination complexes and polyatomic ions using Valence Bond Theory (VBT) and VSEPR. Geometry depends on the hybridization of the central atom.
Step 2: Detailed Explanation:
- (A) $[Ni(CN)_4]^{2-$:} $Ni^{2+}$ ($d^8$ configuration). Cyanide ($CN^-$) is a strong field ligand. It causes pairing of d-electrons, leaving one empty $3d$ orbital. Hybridization is $dsp^2$. The structure is square planar.
- (B) $[Ni(CO)_4]$: $Ni^0$ ($3d^8 4s^2$ configuration). $CO$ is a strong field ligand and causes the $4s$ electrons to shift into the $3d$ subshell. This gives a filled $3d^{10}$ configuration. The empty $4s$ and three $4p$ orbitals hybridize to $sp^3$. The structure is tetrahedral.
- (C) $[NiCl_4]^{2-$:} $Ni^{2+}$ ($d^8$ configuration). Chloride ($Cl^-$) is a weak field ligand and cannot cause pairing. $Ni^{2+}$ uses its $4s$ and three $4p$ orbitals for hybridization ($sp^3$). The structure is tetrahedral.
- (D) $CrO_4^{2-$:} Chromate ion. $Cr$ is in the $+6$ oxidation state ($d^0$ configuration). According to VSEPR, 4 bonds around the central atom with 0 lone pairs result in $sp^3$ (or $d^3s$) hybridization. The structure is tetrahedral.
Step 3: Final Answer:
Species (B), (C), and (D) have tetrahedral structures.