According to MO theory the bond orders for `O_(2)^(-2),CO` and `NO^(+)` respectively, are

Options

A. `1,3` and `3`
B. `2,3` and `3`
C. `1, 2 and 3`
D. `1, 3 and 2`

Answer

To determine the bond orders for \( O_2^{2-} \), \( CO \), and \( NO^+ \) using Molecular Orbital (MO) theory, we can follow these steps:

### Step-by-Step Solution:

1. **Identify the total number of electrons**:
- For \( O_2^{2-} \): Oxygen has an atomic number of 8. Therefore, for two oxygen atoms, the total number of electrons is \( 8 + 8 = 16 \). Since it has a \( 2- \) charge, we add 2 more electrons: \( 16 + 2 = 18 \) electrons.
- For \( CO \): Carbon has an atomic number of 6 and oxygen has an atomic number of 8. Therefore, the total number of electrons is \( 6 + 8 = 14 \) electrons.
- For \( NO^+ \): Nitrogen has an atomic number of 7 and oxygen has an atomic number of 8. Therefore, the total number of electrons is \( 7 + 8 = 15 \). Since it has a \( +1 \) charge, we subtract 1 electron: \( 15 - 1 = 14 \) electrons.

2. **Determine the arrangement of electrons in molecular orbitals**:
- The molecular orbital filling order for diatomic molecules (for \( O_2 \) and below) is:
- \( \sigma_{1s}^2 \)
- \( \sigma_{1s}^*^2 \)
- \( \sigma_{2s}^2 \)
- \( \sigma_{2s}^*^2 \)
- \( \sigma_{2p_z}^2 \)
- \( \pi_{2p_x}^2 \)
- \( \pi_{2p_y}^2 \)
- \( \pi_{2p_x}^* \)
- \( \pi_{2p_y}^* \)

3. **Fill the molecular orbitals for each species**:
- For \( O_2^{2-} \) (18 electrons):
- Filling: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- Total electrons in bonding orbitals = 10 (from \( \sigma_{1s}, \sigma_{2s}, \sigma_{2p_z}, \pi_{2p_x}, \pi_{2p_y} \))
- Total electrons in antibonding orbitals = 8 (from \( \sigma_{1s}^*, \sigma_{2s}^* \))
- Bond order = \( \frac{(10 - 8)}{2} = 1 \)

- For \( CO \) (14 electrons):
- Filling: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \)
- Total electrons in bonding orbitals = 10
- Total electrons in antibonding orbitals = 4
- Bond order = \( \frac{(10 - 4)}{2} = 3 \)

- For \( NO^+ \) (14 electrons):
- Filling: \( \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^1 \)
- Total electrons in bonding orbitals = 9
- Total electrons in antibonding orbitals = 4
- Bond order = \( \frac{(9 - 4)}{2} = 2.5 \)

4. **Summarize the bond orders**:
- \( O_2^{2-} \): Bond order = 1
- \( CO \): Bond order = 3
- \( NO^+ \): Bond order = 2.5

### Final Answer:
The bond orders for \( O_2^{2-}, CO, \) and \( NO^+ \) are \( 1, 3, \) and \( 2.5 \) respectively.