Cl-O bond order in perchlorate ion is

Options

A. 1.33
B. 1.5
C. 1.75
D. 1.9

Explanation

To find the bond order of the Cl-O bond in the perchlorate ion (ClO₄⁻), we can follow these steps:

### Step 1: Determine the Lewis Structure of ClO₄⁻
- **Valence Electrons Calculation**:
- Chlorine (Cl) has 7 valence electrons.
- Each Oxygen (O) has 6 valence electrons, and there are 4 oxygen atoms, contributing a total of 24 electrons.
- The ion has an additional electron due to the -1 charge.
- Total valence electrons = 7 (Cl) + 24 (4 O) + 1 (charge) = 32 electrons.

### Step 2: Draw the Lewis Structure
- Start by placing Cl in the center and surrounding it with the four O atoms.
- Connect each O atom to Cl with single bonds initially.
- This uses 8 electrons (4 single bonds).
- Place the remaining 24 electrons around the O atoms to satisfy the octet rule.
- To minimize formal charges, convert some single bonds to double bonds.

### Step 3: Identify Resonance Structures
- The Lewis structure can have multiple resonance forms.
- In the case of ClO₄⁻, there are four resonance structures where:
- Three O atoms are double bonded to Cl, and one O atom is single bonded with a negative charge.
- Each resonance structure contributes to the overall bonding character.

### Step 4: Calculate the Bond Order
- **Bond Order Formula**:
\[ \text{Bond Order} = \frac{\text{Total number of bonds}}{\text{Number of resonance structures}} \]
- In the resonance structures, we have:
- 3 double bonds (Cl=O) and 1 single bond (Cl-O⁻).
- Total number of bonds = 3 (double bonds) + 1 (single bond) = 7 bonds.
- Number of resonance structures = 4.

### Step 5: Plug into the Bond Order Formula
- Using the formula:
\[ \text{Bond Order} = \frac{7}{4} = 1.75 \]

### Conclusion
The bond order of the Cl-O bond in the perchlorate ion (ClO₄⁻) is **1.75**.

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