The type of hybrid orbitals used by chlorine atom in `ClO_(2)^(-)` 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 ion ClO₂⁻ (chlorite ion), we can follow these steps:

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

### Step 2: Draw the Lewis Structure
1. Start by placing the chlorine atom in the center.
2. Attach two oxygen atoms to the chlorine atom.
3. Since ClO₂⁻ has a negative charge, we need to account for an additional electron.
4. Distribute the electrons: Chlorine will form two bonds with the two oxygen atoms and will have two lone pairs of electrons remaining.

### Step 3: Count Lone Pairs and Bond Pairs
- In the Lewis structure, we have:
- 2 bond pairs (from the Cl-O bonds)
- 2 lone pairs (remaining on the chlorine atom)

### Step 4: Calculate the Steric Number
The steric number is calculated as the sum of the number of bond pairs and lone pairs:
- Steric Number = Number of Bond Pairs + Number of Lone Pairs
- Steric Number = 2 (bond pairs) + 2 (lone pairs) = 4

### Step 5: Determine the Hybridization
For a steric number of 4, the hybridization is sp³. This means that one s orbital and three p orbitals from chlorine are mixed to form four equivalent sp³ hybrid orbitals.

### Step 6: Conclusion
The type of hybrid orbitals used by the chlorine atom in ClO₂⁻ is **sp³**.

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