Which of the following have identical bond order ?

Options

A. `CN^(-)`
B. `NO^(+)`
C. `O_(2)^(-)`
D. `O_(2)^(2-)`

Explanation

To determine which of the given species have identical bond orders, we will calculate the bond order for each species step by step.

### Step 1: Determine the number of electrons for each species
1. **CN⁻ (Cyanide ion)**:
- Carbon (C) has 6 electrons.
- Nitrogen (N) has 7 electrons.
- The negative charge indicates an additional electron.
- Total electrons = 6 + 7 + 1 = 14 electrons.

2. **NO⁺ (Nitric oxide cation)**:
- Nitrogen (N) has 7 electrons.
- Oxygen (O) has 8 electrons.
- The positive charge indicates one less electron.
- Total electrons = 7 + 8 - 1 = 14 electrons.

3. **O₂⁻ (Oxygen anion)**:
- Each oxygen atom has 8 electrons.
- Total for two oxygen atoms = 8 + 8 = 16 electrons.
- The negative charge indicates an additional electron.
- Total electrons = 16 + 1 = 17 electrons.

4. **O₂²⁻ (Dioxygen dianion)**:
- Each oxygen atom has 8 electrons.
- Total for two oxygen atoms = 8 + 8 = 16 electrons.
- The two negative charges indicate two additional electrons.
- Total electrons = 16 + 2 = 18 electrons.

### Step 2: Write the molecular orbital configuration for each species
1. **CN⁻**:
- Molecular orbital configuration:
- σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y²

2. **NO⁺**:
- Molecular orbital configuration:
- σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y²

3. **O₂⁻**:
- Molecular orbital configuration:
- σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x¹ π*2p_y¹

4. **O₂²⁻**:
- Molecular orbital configuration:
- σ1s² σ*1s² σ2s² σ*2s² σ2p_z² π2p_x² π2p_y² π*2p_x¹ π*2p_y¹

### Step 3: Count the bonding and antibonding electrons
1. **CN⁻**:
- Bonding electrons = 10 (from σ and π orbitals)
- Antibonding electrons = 4 (from σ* orbitals)
- Bond order = (Bonding - Antibonding) / 2 = (10 - 4) / 2 = 3.

2. **NO⁺**:
- Bonding electrons = 10 (same as CN⁻)
- Antibonding electrons = 4 (same as CN⁻)
- Bond order = (10 - 4) / 2 = 3.

3. **O₂⁻**:
- Bonding electrons = 10
- Antibonding electrons = 7 (π* orbitals)
- Bond order = (10 - 7) / 2 = 1.5.

4. **O₂²⁻**:
- Bonding electrons = 10
- Antibonding electrons = 8 (π* orbitals)
- Bond order = (10 - 8) / 2 = 1.

### Step 4: Compare bond orders
- CN⁻: Bond order = 3
- NO⁺: Bond order = 3
- O₂⁻: Bond order = 1.5
- O₂²⁻: Bond order = 1

### Conclusion
The species CN⁻ and NO⁺ have identical bond orders of 3.