Select the CORRECT statement about `ClO_(3)-`,

Options

A. `ClO_(3)-` has unpaired electron
B. `ClO_(3)-` has pyramidal structure
C. In `ClO_(3)-,Cl` atom is `sp^(3)` hybridised
D. In `ClO_(3)-,Cl` atom is `sp^2` hybridised

Answer

To solve the question regarding the correct statement about \( \text{ClO}_3^- \), we will analyze the hybridization, structure, and electron configuration of the chlorate ion step by step.

### Step 1: Determine the Valence Electrons
Chlorine (Cl) has 7 valence electrons, and each oxygen (O) has 6 valence electrons. Since there are three oxygen atoms in \( \text{ClO}_3^- \) and the ion has a negative charge, we need to account for that as well.

- Valence electrons from Cl: 7
- Valence electrons from 3 O: \( 3 \times 6 = 18 \)
- Total valence electrons: \( 7 + 18 + 1 = 26 \) (adding 1 for the negative charge)

### Step 2: Calculate the Hybridization
Using the formula for hybridization:
\[ \text{Number of hybrid orbitals} = \frac{1}{2} \times (\text{Valence electrons} + \text{Monovalent atoms} - \text{Charge on cation} + \text{Charge on anion}) \]
In this case:
- Valence electrons = 7 (Cl) + 18 (O) = 26
- Monovalent atoms = 0 (since there are no monovalent atoms)
- Charge on cation = 0
- Charge on anion = -1

Substituting these values:
\[ \text{Number of hybrid orbitals} = \frac{1}{2} \times (26 + 0 - 0 + 1) = \frac{1}{2} \times 27 = 13.5 \]
Since this calculation seems incorrect, let's use the simpler method of counting the regions of electron density around the chlorine atom.

### Step 3: Count the Regions of Electron Density
In \( \text{ClO}_3^- \):
- Chlorine is bonded to three oxygen atoms.
- There is one lone pair on chlorine.

This gives us a total of 4 regions of electron density (3 bonds + 1 lone pair).

### Step 4: Determine the Hybridization Type
With 4 regions of electron density, the hybridization is \( \text{sp}^3 \).

### Step 5: Determine the Molecular Geometry
The molecular geometry of \( \text{ClO}_3^- \) with 4 regions of electron density (3 bonding pairs and 1 lone pair) is trigonal pyramidal.

### Step 6: Identify Unpaired Electrons
In the \( \text{ClO}_3^- \) structure:
- Chlorine has one lone pair and forms three bonds with oxygen.
- The lone pair contains unpaired electrons.

### Conclusion
Based on the analysis:
1. The hybridization is \( \text{sp}^3 \) (correct).
2. The structure is trigonal pyramidal (correct).
3. There are unpaired electrons in the lone pair (correct).

Thus, the correct statements about \( \text{ClO}_3^- \) are that it has \( \text{sp}^3 \) hybridization, a pyramidal structure, and contains unpaired electrons.

### Final Answer
The correct statements about \( \text{ClO}_3^- \) are 1, 2, and 3.

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