The hybridisation of chlorine in `ClO^(-), ClO_(2)^-, ClO_(3)^(-)`, and `ClO_(4)^(-)` ions are 

Options

A. `sp^3, sp^3d, sp^3 d^2, sp^3 d^3`
B. `sp^3 d^3, sp^3 d^2, sp^3 d, sp^3`
C. `sp^2, sp^2, sp^2, sp^2`
D. `sp^3, sp^3, sp^3, sp^3`

Answer

To determine the hybridization of chlorine in the ions ClO^(-), ClO2^(-), ClO3^(-), and ClO4^(-), we will follow these steps:

### Step 1: Determine the number of valence electrons for chlorine in each ion.
- Chlorine (Cl) has 7 valence electrons.

### Step 2: Analyze each ion to find the number of lone pairs and bond pairs.

#### For ClO^(-):
1. **Valence Electrons**: Cl has 7, and the negative charge adds 1 more electron, giving us 8 total electrons.
2. **Bonding**: Cl forms a single bond with O (using 2 electrons).
3. **Lone Pairs**: 8 total electrons - 2 (bonding) = 6 electrons left, which means there are 3 lone pairs.
4. **Hybridization**: Total pairs = 3 lone pairs + 1 bond pair = 4 pairs. Thus, hybridization is **sp³**.

#### For ClO2^(-):
1. **Valence Electrons**: Cl has 7, and the negative charge adds 1 more electron, giving us 8 total electrons.
2. **Bonding**: Cl forms one double bond with O (using 4 electrons) and one single bond with another O (using 2 electrons).
3. **Lone Pairs**: 8 total electrons - 4 (double bond) - 2 (single bond) = 2 electrons left, which means there are 2 lone pairs.
4. **Hybridization**: Total pairs = 2 lone pairs + 2 bond pairs = 4 pairs. Thus, hybridization is **sp³**.

#### For ClO3^(-):
1. **Valence Electrons**: Cl has 7, and the negative charge adds 1 more electron, giving us 8 total electrons.
2. **Bonding**: Cl forms three double bonds with three O atoms (using 6 electrons).
3. **Lone Pairs**: 8 total electrons - 6 (bonding) = 2 electrons left, which means there is 1 lone pair.
4. **Hybridization**: Total pairs = 1 lone pair + 3 bond pairs = 4 pairs. Thus, hybridization is **sp³**.

#### For ClO4^(-):
1. **Valence Electrons**: Cl has 7, and the negative charge adds 1 more electron, giving us 8 total electrons.
2. **Bonding**: Cl forms four double bonds with four O atoms (using 8 electrons).
3. **Lone Pairs**: 8 total electrons - 8 (bonding) = 0 electrons left, which means there are no lone pairs.
4. **Hybridization**: Total pairs = 0 lone pairs + 4 bond pairs = 4 pairs. Thus, hybridization is **sp³**.

### Summary of Hybridization:
- ClO^(-): sp³
- ClO2^(-): sp³
- ClO3^(-): sp³
- ClO4^(-): sp³

### Final Answer:
The hybridization of chlorine in ClO^(-), ClO2^(-), ClO3^(-), and ClO4^(-) is **sp³** for all.