Question

What type of hybridization is present in \([\text{CoF}_{6}]^{3-}\) ?

• A. \(d^{2}sp^{3}\)
• B. \(sp^{3}d^{2}\)
• C. \(dsp^{2}\)
• D. \(sp^{3}\)

Hint:

Fluoride is a weak field ligand, so the 3d electrons do not pair and outer d orbitals are used.

Answer 1

The Correct Option is B

Step 1: Understanding the Concept
The hybridisation depends on the oxidation state of the metal and whether the ligand is strong field or weak field.

Step 2: Oxidation State and Configuration
In \([\text{CoF}_{6}]^{3-}\), \(x+6(-1)=-3\), so \(x=+3\).
\(\text{Co}^{3+}\) is \(3d^{6}\).

Step 3: Effect of the Ligand
\(\text{F}^{-}\) is a weak field ligand and does not pair the \(3d\) electrons.
The inner \(3d\) orbitals are therefore not free.

Step 4: Hybridisation
Cobalt uses one \(4s\), three \(4p\) and two \(4d\) orbitals.
This gives \(sp^{3}d^{2}\) (outer orbital, octahedral).

Final Answer:
The complex has \(sp^{3}d^{2}\) hybridisation, option (B). \[ \boxed{\text{(B) }sp^{3}d^{2}} \]