To solve the question regarding the geometrical differences between `[XeO_(6)]^(4-)` and `XeF_(6)`, we will analyze the hybridization and molecular geometry of both compounds step by step.
### Step 1: Determine the oxidation state of xenon in both compounds.
- For `[XeO_(6)]^(4-)`, the oxidation state of xenon can be calculated as follows:
- Let the oxidation state of xenon be \( x \).
- Each oxygen has an oxidation state of -2, and there are 6 oxygen atoms.
- The overall charge of the ion is -4.
- Therefore, the equation is:
\[
x + 6(-2) = -4 \implies x - 12 = -4 \implies x = +8
\]
- For `XeF_(6)`, similarly:
- Let the oxidation state of xenon be \( x \).
- Each fluorine has an oxidation state of -1, and there are 6 fluorine atoms.
- The overall charge is neutral (0).
- Therefore, the equation is:
\[
x + 6(-1) = 0 \implies x - 6 = 0 \implies x = +6
\]
### Step 2: Calculate the total number of valence electrons for xenon in both compounds.
- For `[XeO_(6)]^(4-)`:
- Xenon has 8 valence electrons.
- The charge of -4 adds 4 more electrons.
- Total = \( 8 + 4 = 12 \) electrons.
- For `XeF_(6)`:
- Xenon has 8 valence electrons.
- There is no additional charge.
- Total = \( 8 \) electrons.
### Step 3: Determine the number of sigma bonds and lone pairs.
- For `[XeO_(6)]^(4-)`:
- There are 6 oxygen atoms, each forming a sigma bond with xenon.
- Thus, there are 6 sigma bonds and no lone pairs (since all 12 electrons are used in bonding).
- For `XeF_(6)`:
- There are 6 fluorine atoms, each forming a sigma bond with xenon.
- Thus, there are 6 sigma bonds.
- Since xenon has 8 valence electrons and 6 are used for bonding, 2 electrons remain, which means there is 1 lone pair.
### Step 4: Determine the hybridization and geometry.
- For `[XeO_(6)]^(4-)`:
- The hybridization is determined by the number of sigma bonds:
- 6 sigma bonds → \( sp^3d^2 \).
- The geometry is octahedral.
- For `XeF_(6)`:
- The hybridization is determined by the number of sigma bonds and lone pairs:
- 6 sigma bonds + 1 lone pair → \( sp^3d^3 \).
- The geometry is distorted due to the presence of the lone pair, making it disordered.
### Conclusion
- `[XeO_(6)]^(4-)` is octahedral because it has 6 sigma bonds and no lone pairs, leading to a stable octahedral geometry.
- `XeF_(6)` is disordered because it has 6 sigma bonds and 1 lone pair, leading to a distorted geometry.
### Final Answer
`[XeO_(6)]^(4-)` is octahedral whereas `XeF_(6)` is disordered because xenon in `[XeO_(6)]^(4-)` has no lone pairs and 6 sigma bonds, leading to an octahedral geometry, while in `XeF_(6)`, xenon has one lone pair and 6 sigma bonds, resulting in a distorted geometry.
