`MF + XeF_(4) rarr M^(+) A^(-) (M^(+)-` alkali metal cation) The state of hybridisation of the central atom in A and sphere of the species are:

Options

A. `sp^(3)d , TBP`
B. `sp^(3)d^(3)`, distorted octahedral
C. `sp^(3)d^(3)`, pentagonal planar
D. No compound formed at all

Answer

To solve the problem, we need to analyze the reaction and determine the hybridization of the central atom in the resulting species A and the geometry of the species formed.

### Step-by-Step Solution:

1. **Identify the Reactants and Products**:
The reaction given is:
\[ \text{MF} + \text{XeF}_4 \rightarrow \text{M}^+ + \text{A}^- \]
Here, MF is an alkali metal fluoride, and XeF4 is xenon tetrafluoride.

2. **Determine the Nature of M**:
Since M is an alkali metal cation (M+), it can be K+, Rb+, or Cs+. This indicates that M is likely to be one of these alkali metals.

3. **Formation of the Product**:
When MF reacts with XeF4, it forms a complex:
\[ \text{M}_2\text{XeF}_6 \]
This suggests that the anion A is \(\text{XeF}_6^{2-}\).

4. **Analyze the Central Atom in A**:
The central atom in the anion A (\(\text{XeF}_6^{2-}\)) is xenon (Xe).

5. **Count the Electron Pairs**:
- Xenon (Xe) in \(\text{XeF}_6^{2-}\) has 6 fluorine atoms bonded to it, which means there are 6 bond pairs.
- The -2 charge indicates that there are 2 additional electrons, which will form 1 lone pair.

6. **Determine the Hybridization**:
- Total electron pairs = Bond pairs + Lone pairs = 6 + 1 = 7.
- The hybridization corresponding to 7 electron pairs is \(sp^3d^3\).

7. **Determine the Geometry**:
- With 6 bond pairs and 1 lone pair, the geometry is pentagonal bipyramidal, but since we have a lone pair, the effective geometry is pentagonal planar.

### Final Answer:
- The state of hybridization of the central atom in A (\(\text{XeF}_6^{2-}\)) is \(sp^3d^3\).
- The geometry of the species is pentagonal planar.