`Cl` in `ClO^(-)` undergoes ....... hybridisation 

Options

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

Answer

To determine the hybridization of chlorine (Cl) in the hypochlorite ion (ClO⁻), we can follow these steps:

### Step 1: Identify the Valence Electrons
Chlorine (Cl) is in group 17 of the periodic table and has 7 valence electrons.

### Step 2: Count the Monovalent Atoms
In the hypochlorite ion (ClO⁻), there is one oxygen atom (O), which is bivalent (it can form two bonds). Since there are no monovalent atoms (atoms that can form only one bond), we do not add any for monovalent atoms.

### Step 3: Account for the Negative Charge
The hypochlorite ion has a negative charge (-1). To account for this charge, we add 1 to our total count.

### Step 4: Calculate the Steric Number
The steric number (s) can be calculated using the formula:
\[ s = \text{Number of valence electrons} + \text{Number of monovalent atoms} + \frac{\text{Charge}}{2} \]

Substituting the values we have:
- Number of valence electrons in Cl = 7
- Number of monovalent atoms = 0
- Charge = -1

So,
\[ s = 7 + 0 + \frac{-1}{2} = 7 + 0 + 0.5 = 7 + 0.5 = 8 \]

### Step 5: Determine the Hybridization
The steric number of 4 indicates that the hybridization is \( sp^3 \). This means that chlorine in ClO⁻ undergoes \( sp^3 \) hybridization.

### Final Answer
The hybridization of Cl in ClO⁻ is \( sp^3 \).

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