Question

Which of the following hybridisation is correct for \([\text{FeF}_6]^{3-}\)?

• A. \(sp^3d^2\) with 1 unpaired electrons
• B. \(sp^3d^2\) with 5 unpaired electrons
• C. \(d^2sp^3\) with 1 unpaired electrons
• D. \(d^2sp^3\) with 5 unpaired electrons

Hint:

Comparing \(\text{Fe}^{3+}\) (\(3d^5\)):
- With weak field ligand \(\text{F}^-\): no pairing \(\rightarrow sp^3d^2\) (outer orbital) with 5 unpaired electrons.
- With strong field ligand \(\text{CN}^-\): pairing occurs \(\rightarrow d^2sp^3\) (inner orbital) with 1 unpaired electron.

Answer 1

The Correct Option is B

Concept:
According to Valence Bond Theory (VBT) and Crystal Field Theory (CFT), the hybridization and magnetic properties of an octahedral complex depend on the oxidation state of the metal and the field strength of the ligands.
Weak field ligands produce small crystal field splitting (\(\Delta_o < P\)), resulting in high-spin complexes where electrons remain unpaired and outer \(d\)-orbitals (\(4d\)) are utilized for hybridization.

Step 1: Determining the Oxidation State of Iron:
Let the oxidation state of iron be \(x\).
Fluoride is a monoanionic ligand (\(\text{F}^-\), charge = \(-1\)):
\[ x + 6(-1) = -3 \implies x = +3 \] The iron atom is in the \(+3\) oxidation state.

Step 2: Electronic Configuration of \(\text{Fe}^{3+}\):
Neutral iron (\(Z = 26\)) has the ground-state configuration:
\[ \text{Fe}: [\text{Ar}] \, 3d^6 4s^2 \] Removing two \(4s\) electrons and one \(3d\) electron gives:
\[ \text{Fe}^{3+}: [\text{Ar}] \, 3d^5 4s^0 4p^0 4d^0 \]

Step 3: Determining Hybridization and Unpaired Electrons:
The fluoride ion (\(\text{F}^-\)) is a weak field ligand positioned low in the spectrochemical series.
Because \(\Delta_o\) is less than the pairing energy \(P\), the crystal field is too weak to force the pairing of \(3d\) electrons.
Therefore, all five electrons remain unpaired, occupying the five \(3d\) orbitals singly:
\[ 3d: \quad \uparrow \quad \uparrow \quad \uparrow \quad \uparrow \quad \uparrow \quad (5 \text{ unpaired electrons}) \] To accommodate six lone pairs donated by the six fluoride ligands, the \(\text{Fe}^{3+}\) ion uses:
- One empty \(4s\) orbital,
- Three empty \(4p\) orbitals,
- Two empty outer \(4d\) orbitals.
These six atomic orbitals hybridize to form six equivalent \(sp^3d^2\) hybrid orbitals, forming an outer-orbital, high-spin octahedral complex with \(5\) unpaired electrons.
Final Answer:
The complex has \(sp^3d^2\) hybridization with 5 unpaired electrons, which corresponds to option (B).