Structure of `XeO_(2)F_(2)` is correctly represented by

Options

A.
B.
C.
D. Both (2) & (3)

Answer

To determine the correct structure of \( \text{XeO}_2\text{F}_2 \), we will follow these steps:

### Step 1: Calculate the total number of valence electrons.
1. **Identify the valence electrons for each atom:**
- Xenon (Xe) has 8 valence electrons.
- Oxygen (O) has 6 valence electrons, and since there are 2 oxygen atoms, the total contribution is \( 6 \times 2 = 12 \).
- Fluorine (F) has 7 valence electrons, and since there are 2 fluorine atoms, the total contribution is \( 7 \times 2 = 14 \).

2. **Total valence electrons:**
\[ \text{Total} = 8 (\text{Xe}) + 12 (\text{O}) + 14 (\text{F}) = 34 \text{ valence electrons} \]

### Step 2: Determine the number of bond pairs and lone pairs.
1. **Calculate the number of bond pairs:**
- Divide the total valence electrons by 8 (since each bond pair uses 2 electrons):
\[ \text{Bond pairs} = \frac{34}{2} = 17 \text{ pairs} \]

2. **Calculate the remainder:**
- The total number of pairs is 17, which means 34 electrons are used for bonding.
- Since we need to calculate the lone pairs, we find the remainder:
\[ \text{Remainder} = 34 - 32 = 2 \text{ electrons} \]
- Divide the remainder by 2 to find the number of lone pairs:
\[ \text{Lone pairs} = \frac{2}{2} = 1 \text{ lone pair} \]

### Step 3: Determine the steric number and geometry.
1. **Calculate the steric number:**
- The steric number is the sum of bond pairs and lone pairs:
\[ \text{Steric number} = 4 (\text{bond pairs}) + 1 (\text{lone pair}) = 5 \]

2. **Determine the geometry:**
- A steric number of 5 corresponds to a trigonal bipyramidal geometry.

### Step 4: Assign positions to the atoms.
1. **Positioning of atoms:**
- In a trigonal bipyramidal geometry, the lone pair will occupy an equatorial position (to minimize repulsion).
- The more electronegative atoms (in this case, fluorine) will occupy the axial positions, while the oxygen atoms will occupy the equatorial positions.

### Final Structure:
- The structure of \( \text{XeO}_2\text{F}_2 \) can be represented as:
- Axial positions: Fluorine (F), Fluorine (F)
- Equatorial positions: Oxygen (O), Oxygen (O), and one lone pair.

### Conclusion:
The correct representation of the structure of \( \text{XeO}_2\text{F}_2 \) is given by option B in the provided choices.

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