Match the following
`{:(,"List-I",,"List-II"),(A,ClO^(-),(1),"one lone pair on chlorine"),(B,ClO_(2)^(-),(2),"two lone pairs on chlorine"),(C,ClO_(3)^(-),(3),"there lone pairs on chlorine"),(,,(4),"Cl is in "sp^(3)" hybridization"),(,,(5),"most stable"):}`
The correct match is

Options

A. A-3, 4 B-2,4,5, C-1, 4
B. A-3,4 B-2,4, C-1, 4, 5
C. A-3,4,5 B-2, 4 C-1,4
D. A-3, 4 B-2, 4 C-1, 4

Explanation

To solve the matching question, we need to analyze the given compounds in List-I (ClO^(-), ClO2^(-), and ClO3^(-)) and determine their characteristics, such as the number of lone pairs on chlorine and hybridization. We will then match them with the appropriate descriptions in List-II.

### Step-by-Step Solution:

1. **Analyze ClO^(-)**:
- **Valence Electrons Calculation**:
- Chlorine (Cl) has 7 valence electrons.
- Oxygen (O) has 6 valence electrons, and since there is one oxygen, we take 6.
- The negative charge adds 1 electron.
- Total = 7 (Cl) + 6 (O) + 1 (negative charge) = 14 electrons.
- **Hybridization**:
- Using the formula:
\[ \text{Hybridization} = \frac{\text{(Number of valence electrons on central atom)} + \text{(Number of monovalent atoms)} + \text{(Charge)}}{2} \]
Here, Cl has 7 valence electrons, and there is 1 monovalent atom (the negative charge is counted as +1).
\[ \text{Hybridization} = \frac{7 + 1}{2} = 4 \quad \text{(sp}^3\text{)} \]
- **Lone Pairs**:
- There is 1 bond pair (Cl-O bond), so the number of lone pairs = 4 - 1 = 3.
- **Match**:
- ClO^(-) has 3 lone pairs and sp^3 hybridization. So, it matches with (3) "three lone pairs on chlorine" and (4) "Cl is in sp^3 hybridization".

2. **Analyze ClO2^(-)**:
- **Valence Electrons Calculation**:
- Cl: 7, O: 6 (2 O atoms), -1 charge.
- Total = 7 + 12 + 1 = 20 electrons.
- **Hybridization**:
- Using the same formula:
\[ \text{Hybridization} = \frac{7 + 2 + 1}{2} = 5 \quad \text{(sp}^3\text{d)} \]
- **Lone Pairs**:
- There are 2 bond pairs (Cl-O bonds), so the number of lone pairs = 5 - 2 = 3.
- **Match**:
- ClO2^(-) has 3 lone pairs and sp^3d hybridization. So, it does not match with any options in List-II.

3. **Analyze ClO3^(-)**:
- **Valence Electrons Calculation**:
- Cl: 7, O: 6 (3 O atoms), -1 charge.
- Total = 7 + 18 + 1 = 26 electrons.
- **Hybridization**:
- Using the same formula:
\[ \text{Hybridization} = \frac{7 + 3 + 1}{2} = 5 \quad \text{(sp}^3\text{d)} \]
- **Lone Pairs**:
- There are 3 bond pairs (Cl-O bonds), so the number of lone pairs = 5 - 3 = 2.
- **Match**:
- ClO3^(-) has 2 lone pairs and sp^3d hybridization. So, it matches with (2) "two lone pairs on chlorine" and (4) "Cl is in sp^3 hybridization".

### Final Matches:
- A: ClO^(-) → (3) "three lone pairs on chlorine" and (4) "Cl is in sp^3 hybridization".
- B: ClO2^(-) → (2) "two lone pairs on chlorine" and (4) "Cl is in sp^3 hybridization".
- C: ClO3^(-) → (1) "one lone pair on chlorine" and (5) "most stable".

### Summary of Matches:
- A → (3, 4)
- B → (2, 4)
- C → (1, 5)