Xenon flurides are very good oxidisng and fluorinating agents. They also act as `F^(-)` donors and acceptors. When `XeF_(4)` donates its fluoride to `SbF_(5)`, then the states of hybridisation of central atom of cationic part ad anionic part of product are:

Options

A. `sp^(3)d,sp^(3)d^(2)`
B. `sp^(3)d^(2),sp^(3)d`
C. `sp^(3)d^(2),sp^(3)d^(2)`
D. `dsp^(2),sp^(3)`

Explanation

To solve the problem, we need to determine the hybridization states of the central atoms in the cationic part (Xenon trifluoride, \(XeF_3^+\)) and the anionic part (Antimony hexafluoride, \(SbF_6^-\)) after the reaction between \(XeF_4\) and \(SbF_5\).

### Step-by-Step Solution:

1. **Identify the Reaction**:
- Xenon tetrafluoride (\(XeF_4\)) donates a fluoride ion to antimony pentafluoride (\(SbF_5\)).
- The products formed are \(XeF_3^+\) (cation) and \(SbF_6^-\) (anion).

2. **Determine the Oxidation States**:
- In \(XeF_4\), Xenon has an oxidation state of +4.
- After donating a fluoride ion, in \(XeF_3^+\), the oxidation state of Xenon becomes +3.
- In \(SbF_5\), Antimony has an oxidation state of +5.
- After accepting a fluoride ion, in \(SbF_6^-\), the oxidation state of Antimony becomes +6.

3. **Calculate the Hybridization of \(XeF_3^+\)**:
- Xenon starts with 8 valence electrons.
- It loses 1 electron due to the +3 oxidation state: \(8 - 1 = 7\).
- It is bonded to 3 fluoride ions (3 bond pairs).
- The total electron pairs = Bond pairs + Lone pairs = 3 + 2 (from the remaining 4 electrons).
- The hybridization can be determined using the formula:
\[ \text{Hybridization} = \text{Number of electron pairs} \]
- Total pairs = 5 (3 bond pairs + 2 lone pairs) → \(sp^3d\).

4. **Calculate the Hybridization of \(SbF_6^-\)**:
- Antimony has 5 valence electrons.
- It gains 1 electron due to the -1 charge: \(5 + 1 = 6\).
- It is bonded to 6 fluoride ions (6 bond pairs).
- The total electron pairs = 6 bond pairs + 0 lone pairs = 6.
- The hybridization can be determined as:
\[ \text{Hybridization} = \text{Number of electron pairs} \]
- Total pairs = 6 → \(sp^3d^2\).

5. **Final Answer**:
- The hybridization of the central atom in the cationic part (\(XeF_3^+\)) is \(sp^3d\).
- The hybridization of the central atom in the anionic part (\(SbF_6^-\)) is \(sp^3d^2\).

### Conclusion:
The states of hybridization of the central atom of the cationic part and anionic part of the products are \(sp^3d\) for \(XeF_3^+\) and \(sp^3d^2\) for \(SbF_6^-\) respectively.