The number of lone pairs on chlorine atom in `CIO^(-),CIO_(2)^(-),CIO_(3)^(-),CI_(4)^(-)`

Options

A. 0,1,2,3
B. 1,2,3,4
C. 4,3,2,1
D. 3,2,1,0

Explanation

To determine the number of lone pairs on the chlorine atom in the species ClO^(-), ClO2^(-), ClO3^(-), and ClO4^(-), we will analyze each species step by step.

### Step 1: Analyze ClO^(-)
1. **Valence Electrons**: Chlorine (Cl) has an atomic number of 17, with an electronic configuration of 2, 8, and 7. Thus, it has 7 valence electrons.
2. **Charge Consideration**: The ClO^(-) ion has an extra electron due to the negative charge, giving chlorine a total of 8 electrons to consider.
3. **Bond Formation**: In ClO^(-), chlorine forms one bond with oxygen. This uses 1 of its electrons.
4. **Lone Pairs Calculation**:
- Total electrons = 8
- Electrons used in bonding = 1
- Remaining electrons = 8 - 1 = 7
- Lone pairs = 7 / 2 = 3.5, which means there are 3 lone pairs and 1 unpaired electron.

**Lone pairs on Cl in ClO^(-)**: 3

### Step 2: Analyze ClO2^(-)
1. **Valence Electrons**: Chlorine has 7 valence electrons.
2. **Charge Consideration**: ClO2^(-) has an extra electron, giving chlorine a total of 8 electrons.
3. **Bond Formation**: In ClO2^(-), chlorine forms one double bond with one oxygen and a single bond with another oxygen. This uses 2 electrons for the double bond and 1 electron for the single bond.
4. **Lone Pairs Calculation**:
- Total electrons = 8
- Electrons used in bonding = 3 (2 for double bond + 1 for single bond)
- Remaining electrons = 8 - 3 = 5
- Lone pairs = 5 / 2 = 2.5, which means there are 2 lone pairs and 1 unpaired electron.

**Lone pairs on Cl in ClO2^(-)**: 2

### Step 3: Analyze ClO3^(-)
1. **Valence Electrons**: Chlorine has 7 valence electrons.
2. **Charge Consideration**: ClO3^(-) has an extra electron, giving chlorine a total of 8 electrons.
3. **Bond Formation**: In ClO3^(-), chlorine forms two double bonds with two oxygens and one single bond with one oxygen. This uses 2 electrons for each double bond (4 total) and 1 for the single bond.
4. **Lone Pairs Calculation**:
- Total electrons = 8
- Electrons used in bonding = 5 (4 for double bonds + 1 for single bond)
- Remaining electrons = 8 - 5 = 3
- Lone pairs = 3 / 2 = 1.5, which means there is 1 lone pair and 1 unpaired electron.

**Lone pairs on Cl in ClO3^(-)**: 1

### Step 4: Analyze ClO4^(-)
1. **Valence Electrons**: Chlorine has 7 valence electrons.
2. **Charge Consideration**: ClO4^(-) has an extra electron, giving chlorine a total of 8 electrons.
3. **Bond Formation**: In ClO4^(-), chlorine forms four double bonds with four oxygens. This uses 2 electrons for each bond (8 total).
4. **Lone Pairs Calculation**:
- Total electrons = 8
- Electrons used in bonding = 8
- Remaining electrons = 8 - 8 = 0
- Lone pairs = 0 / 2 = 0.

**Lone pairs on Cl in ClO4^(-)**: 0

### Summary of Lone Pairs:
- ClO^(-): 3 lone pairs
- ClO2^(-): 2 lone pairs
- ClO3^(-): 1 lone pair
- ClO4^(-): 0 lone pairs