Which is incorrect statement for `XeF_2` ?

Options

A. It has linear structure
B. It is hydrolysed rapidly in aqueous solution of a base
C. It oxidizes `Cl^(-) and I^(-) " to " Cl_2 and I_2 ` respectively
D. It cannot act as `F^(-)` donor

Answer

To determine which statement about \( \text{XeF}_2 \) is incorrect, we will analyze each statement provided in the question.

### Step-by-Step Solution:

1. **Understanding the Structure of \( \text{XeF}_2 \)**:
- The central atom in \( \text{XeF}_2 \) is xenon (Xe), which has an atomic number of 54 and an electronic configuration of \( [Kr] 4d^{10} 5s^2 5p^6 \).
- In the excited state, one of the 5p electrons is promoted to the 5d orbital, leading to the hybridization of \( sp^3d \).
- The geometry of \( \text{XeF}_2 \) is trigonal bipyramidal, but due to the presence of three lone pairs of electrons, the molecular shape is linear.

**Conclusion**: The first statement that \( \text{XeF}_2 \) has a linear structure is **true**.

2. **Hydrolysis of \( \text{XeF}_2 \)**:
- When \( \text{XeF}_2 \) is treated with water, it undergoes hydrolysis to produce xenon, hydrofluoric acid (HF), and oxygen.
- The balanced equation for this reaction is \( 2 \text{XeF}_2 + 2 \text{H}_2O \rightarrow 2 \text{Xe} + 4 \text{HF} + \text{O}_2 \).
- This indicates that \( \text{XeF}_2 \) hydrolyzes rapidly in aqueous solution.

**Conclusion**: The second statement about hydrolysis is **true**.

3. **Oxidizing Chloride and Iodide**:
- \( \text{XeF}_2 \) can oxidize \( \text{Cl}^- \) to \( \text{Cl}_2 \) and \( \text{I}^- \) to \( \text{I}_2 \).
- The reactions can be represented as:
- \( \text{XeF}_2 + 2 \text{HCl} \rightarrow \text{Cl}_2 + \text{Xe} + 2 \text{HF} \)
- \( \text{XeF}_2 + 2 \text{KI} \rightarrow \text{I}_2 + \text{Xe} + 2 \text{KF} \)
- In both cases, the oxidation states change, confirming that \( \text{XeF}_2 \) acts as an oxidizing agent.

**Conclusion**: The third statement regarding oxidation is **true**.

4. **Fluoride Ion Donor**:
- The statement claims that \( \text{XeF}_2 \) cannot act as a fluoride ion donor.
- However, \( \text{XeF}_2 \) can react with \( \text{PF}_5 \) to form \( \text{XeF}^+ \) and \( \text{PF}_6^- \), indicating that it can donate a fluoride ion.

**Conclusion**: The fourth statement is **incorrect**.

### Final Answer:
The incorrect statement about \( \text{XeF}_2 \) is that it **cannot act as a fluoride ion donor**.