The type of hybrid orbitals used by chlorine atom
in` ClO_(3)^(-)` is

Options

A. `sp^(3)`
B. `sp^(2)`
C. sp
D. none of these .

Explanation

To determine the type of hybrid orbitals used by the chlorine atom in the chlorate ion \( ClO_3^- \), we can follow these steps:

### Step 1: Determine the Valence Electrons of Chlorine
Chlorine is in Group 17 of the periodic table, which means it has 7 valence electrons.

### Step 2: Identify the Charge of the Ion
The chlorate ion \( ClO_3^- \) has a negative charge of -1.

### Step 3: Calculate the Steric Number
The steric number can be calculated using the formula:
\[ \text{Steric Number} = \frac{1}{2} \left( \text{Valence Electrons} + \text{Monovalent Atoms} + \text{Positive Charge} - \text{Negative Charge} \right) \]

For \( ClO_3^- \):
- Valence Electrons from Chlorine = 7
- Monovalent Atoms = 0 (there are no monovalent atoms attached to chlorine)
- Positive Charge = 0
- Negative Charge = -1 (since the overall charge is -1, we treat it as +1 in the formula)

Plugging in the values:
\[ \text{Steric Number} = \frac{1}{2} \left( 7 + 0 + 0 + 1 \right) = \frac{1}{2} \times 8 = 4 \]

### Step 4: Determine the Hybridization
According to VSEPR (Valence Shell Electron Pair Repulsion) theory, a steric number of 4 corresponds to \( sp^3 \) hybridization.

### Step 5: Identify the Hybrid Orbitals
The hybridization \( sp^3 \) indicates that the chlorine atom uses one \( s \) orbital and three \( p \) orbitals to form four equivalent hybrid orbitals.

### Conclusion
The type of hybrid orbitals used by the chlorine atom in \( ClO_3^- \) is \( sp^3 \).

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