According VSEPR theory, in the electronic geometry of the molecule(s), electron pair tend to minimise repulsion and follow the repulsion order in presence of lone pair of electrons with bond pair of electrons. lp-lpgt lp-bp gtbp-bp
and similarly double bond pair of electrons follow the repulsion order with angle bond pair of electrons. double bond -double bond gt double bond -single bond gtsingle bond -single bond
Which of the following statement is CORRECT about `XeO_(2)F_(2)`

Options

A. `XeO_(2)F_(2)` is `sp^(3)d` hybridized and has lone pair of electron in its electronic geometry
B. `XeO_(2)F_(2)` has maximum 10-lone pair of electrons
C. `XeO_(2)F_(2)` has distored see -saw shape.
D. None of these

Answer

To determine the correct statement about the molecule \( \text{XeO}_2\text{F}_2 \) according to VSEPR theory, we will follow these steps:

### Step 1: Determine the Valence Electrons
- **Xenon (Xe)** is in Group 18 and has 8 valence electrons.
- **Oxygen (O)** is in Group 16 and has 6 valence electrons. Since there are 2 oxygen atoms, they contribute \( 2 \times 6 = 12 \) valence electrons.
- **Fluorine (F)** is in Group 17 and has 7 valence electrons. Since there are 2 fluorine atoms, they contribute \( 2 \times 7 = 14 \) valence electrons.

Total valence electrons = \( 8 + 12 + 14 = 34 \) electrons.

### Step 2: Calculate the Hybridization
Using the formula for hybridization:
\[ \text{Hybridization} = \frac{1}{2} \left( \text{Valence Electrons} + \text{Monovalent Atoms} - \text{Charge on Cation} + \text{Charge on Anion} \right) \]
- Valence electrons = 8 (Xe) + 0 (monovalent atoms) - 0 (cation) + 0 (anion) = 8.
- Monovalent atoms = 2 (from 2 F atoms).

Plugging in the values:
\[ \text{Hybridization} = \frac{1}{2} (8 + 2 - 0 + 0) = \frac{1}{2} (10) = 5 \]
This indicates that the hybridization is \( \text{sp}^3\text{d} \).

### Step 3: Determine the Number of Lone Pairs
- Xenon has 8 valence electrons.
- It forms 2 bonds with fluorine (2 bond pairs) and 2 double bonds with oxygen (2 bond pairs).
- Total electrons used in bonding = \( 2 \text{ (F)} + 4 \text{ (2 double bonds with O)} = 6 \).
- Remaining electrons = \( 8 - 6 = 2 \) electrons, which corresponds to 1 lone pair.

### Step 4: Determine the Molecular Geometry
- With 5 hybrid orbitals, 2 are used for the fluorine atoms, 2 are used for the oxygen atoms, and 1 remains as a lone pair.
- The electronic geometry is trigonal bipyramidal due to 5 regions of electron density.
- The molecular shape, considering the lone pair, is distorted trigonal bipyramidal (or seesaw shape).

### Conclusion
The correct statement about \( \text{XeO}_2\text{F}_2 \) is:
- It is \( \text{sp}^3\text{d} \) hybridized and has 1 lone pair of electrons in its electronic geometry.

### Final Answer
The correct option is **1**: "XeO2F2 is sp3d hybridized and has a lone pair of electrons in its electronic geometry."

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