Question

What type of hybridization is exhibited by $[\text{CoF}_6]^{3-}$?

• A. $\text{sp}^3$
• B. $\text{sp}^3\text{d}^2$
• C. $\text{dsp}^2$
• D. $\text{d}^2\text{sp}^3$

Hint:

For coordination number 6 complexes involving $\text{Co}^{3+}$ ($\text{d}^6$):
If the ligand is a strong field ligand (like $\text{CN}^-$, $\text{NH}_3$), it causes pairing leading to inner orbital $\text{d}^2\text{sp}^3$ hybridization. If the ligand is a weak field ligand (like $\text{F}^-$, $\text{Cl}^-$), no pairing happens, leading directly to outer orbital $\text{sp}^3\text{d}^2$ hybridization.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the specific orbital hybridization of the central Cobalt (Co) ion in the octahedral coordination complex ion $[\text{CoF}_6]^{3-}$ using Valence Bond Theory (VBT).

Step 2: Detailed Explanation:
Let's systematically evaluate the electronic state of the central metal atom:
1.

Determine the oxidation state of Co: Let the oxidation state of Co be $x$. Flouride ($\text{F}^-$) is a monodentate anionic ligand with a $-1$ charge. $$ x + 6(-1) = -3 \implies x = +3 $$ Thus, Cobalt exists as a $\text{Co}^{3+}$ ion.
2.

Electronic Configuration: Atomic number of $\text{Co} = 27$: $[\text{Ar}] 3\text{d}^7 4\text{s}^2$ For $\text{Co}^{3+}$ ion: $[\text{Ar}] 3\text{d}^6 4\text{s}^0 4\text{p}^0 4\text{d}^0$
3.

Nature of the Ligand: Fluoride ($\text{F}^-$) is a well-known

weak field ligand. According to spectrochemical series behavior, it does not possess enough field energy to force pairing of the electrons inside the $3\text{d}$ subshell. Therefore, the six $3\text{d}$ electrons remain distributed according to Hund's rule (one paired orbital and four unpaired electrons).
4.

Orbital Hybridization: Since the inner $3\text{d}$ orbitals are unavailable for bonding, the complex must utilize outer shell empty valence orbitals to accommodate 6 pairs of incoming ligand electrons.
The metal ion mixes one $4\text{s}$, three $4\text{p}$, and two $4\text{d}$ atomic orbitals:
$$ 4\text{s} + 4\text{p}_x + 4\text{p}_y + 4\text{p}_z + 4\text{d}_{z^2} + 4\text{d}_{x^2-y^2} \rightarrow \text{sp}^3\text{d}^2 $$ This configuration gives six empty $\text{sp}^3\text{d}^2$ outer-orbital hybrid paths to form an outer-orbital high-spin complex.

Step 3: Final Answer: The hybridization exhibited is $\text{sp}^3\text{d}^2$, which corresponds directly to option (B).