Iron forms iron halides salts by reacting the metal directly with halogen. `FeI_(3)` does not exist. `FeF_(3)` is white soled inspite of five unpaired electrons with `d^(5)` configuration . `FeCl_(3)` is soluble in water and is used as a mordant in dyeing industry.
Q.`FeI_(3)` does not exist because:

Options

A. of large size
B. `Fe^(3+)` oxidise `I^(-)` to `I_(2)`
C. of low lattice energy
D. iodine is not highly electronegative enough to oxidise Fe to `Fe^(3+)`

Answer

To explain why `FeI_(3)` does not exist, we can break down the reasoning into several steps:

### Step-by-Step Solution:

1. **Understanding the Oxidation States**:
- Iron (Fe) can exist in multiple oxidation states, primarily +2 (Fe²⁺) and +3 (Fe³⁺). In `FeI_(3)`, iron would be in the +3 oxidation state.

2. **Behavior of Iodide Ion**:
- The iodide ion (I⁻) is a strong reducing agent. This means it has the ability to donate electrons and, in the process, can reduce other species.

3. **Reduction of Iron**:
- When `FeI_(3)` is formed, the Fe³⁺ ion would need to coexist with I⁻ ions. However, the iodide ion can reduce Fe³⁺ back to Fe²⁺. This means that instead of forming `FeI_(3)`, the reaction would favor the formation of `FeI₂` (where Fe is in the +2 oxidation state).

4. **Formation of Iodine**:
- As the iodide ions reduce Fe³⁺ to Fe²⁺, they themselves get oxidized to form iodine (I₂). Thus, the reaction can be represented as:
\[ 2Fe^{3+} + 6I^{-} \rightarrow 2Fe^{2+} + 3I_{2} \]
- This indicates that `FeI_(3)` cannot be stable because the iodide ions will not remain in the +1 state.

5. **Conclusion**:
- Therefore, `FeI_(3)` does not exist because the iodide ion reduces Fe³⁺ to Fe²⁺, leading to the formation of `FeI₂` instead.