Arrange the `XeF_(2),XeF_(4)` in decreasing order of Xe-F bond length, give reason also.

Answer

To arrange the compounds \( \text{XeF}_2 \) and \( \text{XeF}_4 \) in decreasing order of Xe-F bond length, we need to analyze their structures and hybridization.

### Step 1: Identify the Hybridization
- **For \( \text{XeF}_2 \)**: The hybridization is \( \text{sp}^3\text{d} \). This molecule has a linear geometry due to the presence of three lone pairs of electrons on the central xenon atom. The two fluorine atoms occupy the axial positions.

- **For \( \text{XeF}_4 \)**: The hybridization is \( \text{sp}^3\text{d}^2 \). This molecule adopts a square planar geometry with four fluorine atoms in the equatorial plane and two lone pairs in the axial positions.

### Step 2: Analyze Bond Lengths
- In \( \text{XeF}_2 \), the bond lengths are influenced by the presence of lone pairs. The axial bonds (Xe-F) are longer due to the repulsion between the lone pairs and the bonded pairs.

- In \( \text{XeF}_4 \), the bond lengths are shorter than those in \( \text{XeF}_2 \) because the square planar arrangement minimizes the lone pair-bond pair repulsion, leading to a more compact structure.

### Step 3: Compare the Bond Lengths
- Since the bond length in \( \text{XeF}_2 \) is affected by the presence of lone pairs and the linear arrangement, it results in a longer bond length compared to \( \text{XeF}_4 \).

- Therefore, we can conclude that the bond length of Xe-F in \( \text{XeF}_2 \) is greater than that in \( \text{XeF}_4 \).

### Final Arrangement
Thus, the decreasing order of Xe-F bond lengths is:
\[ \text{XeF}_2 > \text{XeF}_4 \]

### Reason
The reason for this arrangement is that in \( \text{XeF}_2 \), the axial bond lengths are longer due to the lone pair-bond pair repulsion, while in \( \text{XeF}_4 \), the square planar geometry allows for shorter bond lengths due to minimized repulsion.

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