The hybridization of the central atom in `ClO_(2)F_(3)` is :

Options

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

Explanation

To determine the hybridization of the central atom in the molecule ClO₂F₃, we will follow a systematic approach using the formula for calculating the number of hybrid orbitals.

### Step-by-Step Solution:

1. **Identify the Central Atom**:
The central atom in ClO₂F₃ is chlorine (Cl).

2. **Determine the Valence Electrons of Chlorine**:
Chlorine is in group 17 of the periodic table and has 7 valence electrons.

3. **Count the Surrounding Atoms**:
- There are 2 oxygen (O) atoms. Oxygen is a bivalent atom, meaning it forms 2 bonds.
- There are 3 fluorine (F) atoms. Fluorine is a monovalent atom, meaning it forms 1 bond.

4. **Calculate the Total Number of Monovalent Atoms**:
Since each fluorine is monovalent, we count them as follows:
- 3 fluorine atoms contribute 3 to the count.
- The 2 oxygen atoms, being bivalent, do not contribute to the count of monovalent atoms.

5. **Apply the Formula**:
The formula to calculate the number of hybrid orbitals (X) is:
\[ X = \frac{1}{2} \left( \text{Number of valence electrons of central atom} + \text{Number of monovalent atoms} - \text{Charge on cation} + \text{Charge on anion} \right) \]
- Here, the charge on the cation and anion is 0 since ClO₂F₃ is a neutral molecule.
- Plugging in the values:
\[ X = \frac{1}{2} \left( 7 + 3 + 0 + 0 \right) = \frac{1}{2} \times 10 = 5 \]

6. **Determine the Hybridization**:
- If \(X = 5\), the hybridization is \(sp^3d\).

### Conclusion:
The hybridization of the central atom (chlorine) in ClO₂F₃ is **sp³d**.