`[XeO_(6)]^(4-)` is octahedral whereas `XeF_(6)` is a disordered one, because

Options

A. fluorine is more electronegative than oxygen
B. Xe has a lone pair in `XeF_(6)`
C. `XeF_(6)` is neutral whereas `[XeO_(6)]^(4-)` anionic
D. `Xe-F` bond has more ionic characters

Answer

To understand why `[XeO_(6)]^(4-)` is octahedral while `XeF_(6)` is distorted, we need to analyze the hybridization and the presence of lone pairs in these compounds.

### Step-by-Step Solution:

1. **Determine the Hybridization of `[XeO_(6)]^(4-)`:**
- Xenon (Xe) has 8 valence electrons (as it is a noble gas).
- The formula for hybridization is:
\[ n = \frac{1}{2} \left( \text{Number of valence electrons} + \text{Number of monovalent atoms} - \text{Charge} \right) \]
- For `[XeO_(6)]^(4-)`, we have:
- Valence electrons from Xe = 8
- Oxygen is bivalent and forms double bonds, so it does not contribute additional monovalent atoms.
- Charge = -4
- Thus, substituting into the formula:
\[ n = \frac{1}{2} \left( 8 + 0 + 4 \right) = \frac{12}{2} = 6 \]
- This indicates that the hybridization is `sp^3d^2`, which corresponds to an octahedral geometry.

2. **Determine the Hybridization of `XeF_(6)`:**
- For `XeF_(6)`, we again use the hybridization formula:
- Valence electrons from Xe = 8
- There are 6 fluorine atoms (monovalent).
- Charge = 0
- Thus, substituting into the formula:
\[ n = \frac{1}{2} \left( 8 + 6 + 0 \right) = \frac{14}{2} = 7 \]
- This indicates that the hybridization is `sp^3d^3`, which typically corresponds to a pentagonal bipyramidal geometry.

3. **Analyze the Presence of Lone Pairs:**
- In `[XeO_(6)]^(4-)`, there are no lone pairs on the xenon atom since all valence electrons are involved in bonding with oxygen atoms.
- In `XeF_(6)`, we have:
- 8 valence electrons from Xe and 6 from 6 fluorine atoms, leading to 2 remaining electrons.
- These 2 electrons form a lone pair on the xenon atom.
- The presence of this lone pair in `XeF_(6)` causes repulsion between the lone pair and the bonding pairs, leading to a distortion in the geometry from the ideal pentagonal bipyramidal shape.

4. **Conclusion:**
- `[XeO_(6)]^(4-)` is octahedral because it has no lone pairs, allowing for a symmetrical arrangement of the oxygen atoms.
- `XeF_(6)` is distorted due to the presence of a lone pair, which causes repulsion and alters the expected geometry.