Consider the following species :
` CN^(+) ,CN^(-) ,NO and CN`
Which one of these will have the highest bond order?

Options

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

Explanation

To determine which of the given species (CN⁺, CN⁻, NO, and CN) has the highest bond order, we will calculate the bond order for each species using their molecular orbital configurations.

### Step 1: Calculate the Bond Order for NO
1. **Determine the total number of electrons**:
- Nitrogen (N) has 7 electrons, and Oxygen (O) has 8 electrons.
- Total = 7 + 8 = 15 electrons.

2. **Write the molecular orbital configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², σ2p_z², π2p_x², π2p_y², π*2p_x¹
- This accounts for 15 electrons.

3. **Count bonding and anti-bonding electrons**:
- Bonding electrons = 10 (σ1s + σ2s + σ2p)
- Anti-bonding electrons = 5 (σ*1s + σ*2s + π*2p)

4. **Calculate bond order**:
\[ \text{Bond Order} = \frac{\text{Bonding Electrons} - \text{Anti-bonding Electrons}}{2} = \frac{10 - 5}{2} = 2.5 \]

### Step 2: Calculate the Bond Order for CN⁻
1. **Determine the total number of electrons**:
- Carbon (C) has 6 electrons, and Nitrogen (N) has 7 electrons.
- With an extra electron due to the negative charge:
- Total = 6 + 7 + 1 = 14 electrons.

2. **Write the molecular orbital configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², π2p_x², π2p_y², σ2p_z²
- This accounts for 14 electrons.

3. **Count bonding and anti-bonding electrons**:
- Bonding electrons = 8 (σ1s + σ2s + π2p)
- Anti-bonding electrons = 2 (σ*1s + σ*2s)

4. **Calculate bond order**:
\[ \text{Bond Order} = \frac{8 - 2}{2} = 3 \]

### Step 3: Calculate the Bond Order for CN⁺
1. **Determine the total number of electrons**:
- Total = 6 (C) + 7 (N) - 1 (due to positive charge) = 12 electrons.

2. **Write the molecular orbital configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², π2p_x², π2p_y²
- This accounts for 12 electrons.

3. **Count bonding and anti-bonding electrons**:
- Bonding electrons = 6 (σ1s + σ2s + π2p)
- Anti-bonding electrons = 2 (σ*1s + σ*2s)

4. **Calculate bond order**:
\[ \text{Bond Order} = \frac{6 - 2}{2} = 2 \]

### Step 4: Calculate the Bond Order for CN
1. **Determine the total number of electrons**:
- Total = 6 (C) + 7 (N) = 13 electrons.

2. **Write the molecular orbital configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², π2p_x², π2p_y², σ2p_z¹
- This accounts for 13 electrons.

3. **Count bonding and anti-bonding electrons**:
- Bonding electrons = 7 (σ1s + σ2s + π2p)
- Anti-bonding electrons = 2 (σ*1s + σ*2s)

4. **Calculate bond order**:
\[ \text{Bond Order} = \frac{7 - 2}{2} = 2.5 \]

### Conclusion
After calculating the bond orders:
- NO: 2.5
- CN⁻: 3
- CN⁺: 2
- CN: 2.5

The species with the highest bond order is **CN⁻** with a bond order of **3**.

### Final Answer
The species with the highest bond order is **CN⁻**.

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