How many of the following species have bond order of `2.5` ?
`N_(2)^(o+)` (ii) `N_(2)^(Theta) ` (iii) `O_(2)^(o+)` (iv) `O_(2)^(Theta)` (v) `NO` (vi) `CN` .

Explanation

To determine how many of the given species have a bond order of 2.5, we will analyze each species step by step, calculating their bond orders based on their electronic configurations.

### Step 1: Analyze `N2^o+`
1. **Electronic Configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², π2p_y², π2p_z², σ2p_x¹
2. **Count Bonding and Antibonding Electrons**:
- Bonding electrons (nb) = 2 (σ1s) + 2 (σ2s) + 2 (π2p_y) + 2 (π2p_z) + 1 (σ2p_x) = 9
- Antibonding electrons (na) = 2 (σ*1s) + 2 (σ*2s) = 4
3. **Bond Order Calculation**:
- Bond Order = ½(nb - na) = ½(9 - 4) = ½(5) = 2.5

### Step 2: Analyze `N2^Θ`
1. **Electronic Configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², π2p_y², π2p_z², σ2p_x²
2. **Count Bonding and Antibonding Electrons**:
- Bonding electrons (nb) = 2 + 2 + 2 + 2 + 2 = 10
- Antibonding electrons (na) = 2 + 2 = 4
3. **Bond Order Calculation**:
- Bond Order = ½(10 - 4) = ½(6) = 3.0 (not 2.5)

### Step 3: Analyze `O2^o+`
1. **Electronic Configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², σ2p_x², π2p_y², π2p_z², π*2p_y¹
2. **Count Bonding and Antibonding Electrons**:
- Bonding electrons (nb) = 2 + 2 + 2 + 2 + 2 = 10
- Antibonding electrons (na) = 2 + 2 + 1 = 5
3. **Bond Order Calculation**:
- Bond Order = ½(10 - 5) = ½(5) = 2.5

### Step 4: Analyze `O2^Θ`
1. **Electronic Configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², σ2p_x², π2p_y², π2p_z², π*2p_y², π*2p_z¹
2. **Count Bonding and Antibonding Electrons**:
- Bonding electrons (nb) = 10
- Antibonding electrons (na) = 2 + 2 + 2 + 1 = 7
3. **Bond Order Calculation**:
- Bond Order = ½(10 - 7) = ½(3) = 1.5 (not 2.5)

### Step 5: Analyze `NO`
1. **Electronic Configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², σ2p_x², π2p_y², π2p_z², π*2p_y¹
2. **Count Bonding and Antibonding Electrons**:
- Bonding electrons (nb) = 10
- Antibonding electrons (na) = 2 + 2 + 1 = 5
3. **Bond Order Calculation**:
- Bond Order = ½(10 - 5) = 2.5

### Step 6: Analyze `CN`
1. **Electronic Configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², π2p_y², π2p_z², σ2p_x¹
2. **Count Bonding and Antibonding Electrons**:
- Bonding electrons (nb) = 9
- Antibonding electrons (na) = 2 + 2 = 4
3. **Bond Order Calculation**:
- Bond Order = ½(9 - 4) = ½(5) = 2.5

### Conclusion:
The species with a bond order of 2.5 are:
1. `N2^o+`
2. `O2^o+`
3. `NO`
4. `CN`

Thus, **4 species** have a bond order of 2.5.