Sigma bond between `CI` and `O` in `CIO_4^(-)` is formed by ---- overlapping

Options

A. `sp^2 - p`
B. `sp^2 - s`
C. `sp^3 - s`
D. `sp^3 - p`

Answer

To determine the type of overlapping that forms the sigma bond between chlorine (Cl) and oxygen (O) in the chlorate ion \( ClO_4^- \), we can follow these steps:

### Step 1: Identify the central atom and its valence electrons
Chlorine (Cl) is the central atom in \( ClO_4^- \). Chlorine has 7 valence electrons.

### Step 2: Account for the negative charge
The negative charge on the ion indicates that we need to add one more electron to the total count. Therefore, the total number of valence electrons becomes:
\[ 7 \, (\text{from Cl}) + 4 \, (\text{from 4 O}) + 1 \, (\text{for the negative charge}) = 12 \, \text{valence electrons} \]

### Step 3: Determine the hybridization of chlorine
To find the hybridization, we need to determine the number of hybrid orbitals required. Chlorine is bonded to four oxygen atoms, which means it will need four hybrid orbitals. The hybridization that accommodates four orbitals is \( sp^3 \).

### Step 4: Identify the orbitals involved in the sigma bond formation
In the sigma bond formation between Cl and O, the \( sp^3 \) hybrid orbitals from chlorine overlap with the \( p \) orbitals of oxygen.

### Step 5: Conclusion
Thus, the sigma bond between Cl and O in \( ClO_4^- \) is formed by the overlapping of \( sp^3 \) hybrid orbitals from chlorine and \( p \) orbitals from oxygen.

### Final Answer
The sigma bond between Cl and O in \( ClO_4^- \) is formed by \( sp^3 \) and \( p \) overlapping.

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