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: Understand the target.
We must find the hybridisation of the cobalt atom in the complex $[\text{CoF}_6]^{3-}$. There are six fluoride groups around cobalt, so the shape is octahedral.

Step 2: Find the charge on cobalt.
Each fluoride is $\text{F}^-$ with charge $-1$, and there are six of them. Let cobalt be $x$.
\[ x + 6(-1) = -3 \;\Rightarrow\; x = +3 \]
So cobalt is $\text{Co}^{3+}$.

Step 3: Write the electron arrangement.
Neutral cobalt (atomic number 27) is $[\text{Ar}]3d^7 4s^2$. Removing three electrons to make $\text{Co}^{3+}$ leaves $[\text{Ar}]3d^6$, with the 4s and 4p empty.

Step 4: Decide if the ligand is strong or weak.
Fluoride is a weak field ligand. A weak ligand cannot push the d electrons to pair up. So the six 3d electrons stay spread out and the inner 3d orbitals are not free for bonding.

Step 5: Choose the orbitals for bonding.
Since the inner 3d orbitals are blocked, cobalt must use outer orbitals to accept the six lone pairs from fluoride. It uses one 4s, three 4p, and two 4d orbitals.
\[ 4s + 4p_x + 4p_y + 4p_z + 4d + 4d \rightarrow sp^3d^2 \]

Step 6: State the result.
Using outer d orbitals gives $sp^3d^2$ hybridisation, an outer orbital high spin complex. This is option 2. If fluoride had been a strong ligand the answer would have been $d^2sp^3$ instead.
\[ \boxed{sp^3d^2} \]