To solve the problem of identifying which pairs of molecules have a bond order of three and are isoelectronic, we can follow these steps:
### Step 1: Identify the pairs of molecules
The question provides several pairs of molecules. We need to analyze each pair to determine their bond order and whether they are isoelectronic.
### Step 2: Calculate the total number of electrons for each molecule
1. **For NO⁺ (Nitric Oxide cation)**:
- Nitrogen (N) has 7 electrons.
- Oxygen (O) has 8 electrons.
- Since it is NO⁺, we subtract 1 electron.
- Total electrons = 7 (N) + 8 (O) - 1 = 14 electrons.
2. **For CO⁺ (Carbon Monoxide cation)**:
- Carbon (C) has 6 electrons.
- Oxygen (O) has 8 electrons.
- Since it is CO⁺, we subtract 1 electron.
- Total electrons = 6 (C) + 8 (O) - 1 = 13 electrons.
Since NO⁺ has 14 electrons and CO⁺ has 13 electrons, they are not isoelectronic.
### Step 3: Analyze the second pair (CN⁻ and CO)
1. **For CN⁻ (Cyanide anion)**:
- Carbon (C) has 6 electrons.
- Nitrogen (N) has 7 electrons.
- Since it is CN⁻, we add 1 electron.
- Total electrons = 6 (C) + 7 (N) + 1 = 14 electrons.
2. **For CO (Carbon Monoxide)**:
- Carbon (C) has 6 electrons.
- Oxygen (O) has 8 electrons.
- Total electrons = 6 (C) + 8 (O) = 14 electrons.
Since both CN⁻ and CO have 14 electrons, they are isoelectronic.
### Step 4: Calculate the bond order for CN⁻ and CO
To calculate the bond order, we can use the formula:
\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]
#### For CN⁻:
- The electron configuration can be determined using Molecular Orbital Theory:
- Total electrons = 14
- Fill the orbitals:
- 1σ (2), 1σ* (2), 2σ (2), 2σ* (0), 2π (4), 2π* (2), 3σ (2)
- Bonding electrons = 10 (1σ + 2σ + 2π)
- Antibonding electrons = 4 (1σ* + 2π*)
Thus, the bond order for CN⁻:
\[ \text{Bond Order} = \frac{10 - 4}{2} = \frac{6}{2} = 3 \]
#### For CO:
- The electron configuration is similar:
- Total electrons = 14
- Fill the orbitals:
- 1σ (2), 1σ* (2), 2σ (2), 2σ* (0), 2π (4), 2π* (2), 3σ (2)
- Bonding electrons = 10
- Antibonding electrons = 4
Thus, the bond order for CO:
\[ \text{Bond Order} = \frac{10 - 4}{2} = \frac{6}{2} = 3 \]
### Conclusion
Both CN⁻ and CO have a bond order of 3 and are isoelectronic. Therefore, the correct answer is the second option.
