Pick out the incorrect statement for `XeF_4`

Options

A. `XeF_4` disproportionates violently with water
B. It is used as fluorinating agent
C. It has octahedral shape
D. It oxidizes `I^(-)" or " I_2`

Answer

To solve the question of identifying the incorrect statement for \( \text{XeF}_4 \), let's analyze each of the provided statements step by step.

### Step 1: Analyze the first statement
**Statement 1:** \( \text{XeF}_4 \) disproportionates with water.

- **Reaction:** The reaction can be represented as:
\[ \text{XeF}_4 + \text{H}_2\text{O} \rightarrow \text{XeO}_3 + \text{HF} + \text{O}_2 \]
- **Oxidation States:** In \( \text{XeF}_4 \), xenon (Xe) is in the +4 oxidation state. In \( \text{XeO}_3 \), xenon is in the +6 oxidation state, and in elemental xenon (Xe), it is 0. This indicates that \( \text{XeF}_4 \) is being oxidized to \( \text{XeO}_3 \) and reduced to elemental xenon.
- **Conclusion:** This statement is **true** as disproportionation involves both oxidation and reduction.

### Step 2: Analyze the second statement
**Statement 2:** \( \text{XeF}_4 \) is used as a fluorinating agent.

- **Reaction:** When \( \text{XeF}_4 \) reacts with \( \text{SbF}_5 \), it forms \( \text{XeF}_3^+ \) and \( \text{SbF}_6^- \).
- **Conclusion:** This statement is also **true** as \( \text{XeF}_4 \) acts as a fluorinating agent.

### Step 3: Analyze the third statement
**Statement 3:** \( \text{XeF}_4 \) has an octahedral shape.

- **Hybridization:** The hybridization of xenon in \( \text{XeF}_4 \) is \( \text{sp}^3\text{d}^2 \) due to 4 bonding pairs (with fluorine) and 2 lone pairs.
- **Geometry vs. Shape:** The geometry is octahedral due to the 6 regions of electron density (4 bonds + 2 lone pairs), but the actual shape is square planar because the lone pairs occupy the axial positions.
- **Conclusion:** This statement is **false** because while the geometry is octahedral, the shape is square planar.

### Step 4: Analyze the fourth statement
**Statement 4:** \( \text{XeF}_4 \) oxidizes \( \text{I}^- \) to \( \text{I}_2 \).

- **Reaction:** The reaction can be represented as:
\[ \text{XeF}_4 + \text{KI} \rightarrow \text{KF} + \text{I}_2 + \text{Xe} \]
- **Oxidation States:** Here, \( \text{I}^- \) (with an oxidation state of -1) is oxidized to \( \text{I}_2 \) (with an oxidation state of 0).
- **Conclusion:** This statement is **true** as \( \text{XeF}_4 \) indeed oxidizes iodide ions.

### Final Conclusion
The incorrect statement among the four provided is **Statement 3**, which claims that \( \text{XeF}_4 \) has an octahedral shape. The correct shape is square planar.

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