What is the CFSE for [FeF₆]^3- ? 

0 Dq

The correct answer is option 1. 0 Dq.

To calculate the Crystal Field Stabilization Energy (CFSE) for the complex \([FeF_6]^{3-}\), follow these steps:

Iron in \([FeF_6]^{3-}\) is in the +3 oxidation state.

The electronic configuration of Fe(III) is \( [Ar] \ 3d^5 \).

Fluoride is a weak field ligand and will not cause significant splitting of the d-orbitals, leading to an octahedral geometry with low crystal field splitting energy (\(\Delta_0\) or \(Dq\)).

In an octahedral field, the \(d\)-orbitals split into two sets: \( t_{2g} \) (lower energy) and \( e_g \) (higher energy).

For Fe(III) with \(d^5\) configuration:

In an octahedral field with weak field ligands, Fe(III) will have a high-spin configuration: \( t_{2g}^3 \ e_g^2 \). The electron configuration will be \( t_{2g}^3 \ e_g^2 \).

CFSE is calculated using the formula:

\(\text{CFSE} = \Delta_0 \times (\text{number of electrons in } t_{2g} \text{ orbitals}) - 0.6 \Delta_0 \times (\text{number of electrons in } e_g \text{ orbitals})\)

For \( t_{2g}^3 e_g^2 \) configuration:

\(\text{CFSE} = 0.4 \Delta_0 \times 3 - 0.6 \Delta_0 \times 2\)

\(\text{CFSE} = 1.2 \Delta_0 - 1.2 \Delta_0\)

\(\text{CFSE} = 0\)

Therefore, the CFSE for \([FeF_6]^{3-}\) is: 0 Dq.