Hybridisation of chlorine atom in `ClO^(-), ClO_(2)^(-),ClO_(3)^(-) and ClO_(4)^(-)`, respectively

Options

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

Explanation

To determine the hybridization of the chlorine atom in the species ClO^(-), ClO2^(-), ClO3^(-), and ClO4^(-), we will follow a systematic approach.

### Step-by-Step Solution:

1. **Identify the Valence Electrons of Chlorine:**
- Chlorine (Cl) has an atomic number of 17, which gives it an electronic configuration of 2, 8, 7. Therefore, chlorine has 7 valence electrons.

2. **Analyze ClO^(-):**
- In ClO^(-), chlorine forms one single bond with oxygen and carries a negative charge.
- The single bond uses 1 electron from chlorine, leaving 6 electrons.
- These 6 electrons can be arranged as 3 lone pairs.
- Total: 1 sigma bond (from Cl-O) + 3 lone pairs = 4 regions of electron density.
- Hybridization = sp³.

3. **Analyze ClO2^(-):**
- In ClO2^(-), chlorine forms one double bond with one oxygen and a single bond with another oxygen.
- The double bond uses 2 electrons, and the single bond uses 1 electron, which totals 3 electrons used.
- This leaves 4 electrons, which can be arranged as 2 lone pairs.
- Total: 2 sigma bonds (1 from Cl=O and 1 from Cl-O) + 2 lone pairs = 4 regions of electron density.
- Hybridization = sp³.

4. **Analyze ClO3^(-):**
- In ClO3^(-), chlorine forms one double bond with one oxygen and two single bonds with the other two oxygens.
- The double bond uses 2 electrons, and the two single bonds use 2 electrons (1 each), totaling 4 electrons used.
- This leaves 3 electrons, which can be arranged as 1 lone pair.
- Total: 3 sigma bonds (1 from Cl=O and 2 from Cl-O) + 1 lone pair = 4 regions of electron density.
- Hybridization = sp³.

5. **Analyze ClO4^(-):**
- In ClO4^(-), chlorine forms four double bonds with four oxygen atoms.
- Each double bond uses 2 electrons, totaling 8 electrons used.
- Since chlorine only has 7 valence electrons, it shares all its electrons and has no lone pairs left.
- Total: 4 sigma bonds (1 from each Cl=O) + 0 lone pairs = 4 regions of electron density.
- Hybridization = sp³.

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