Which of the following have same bond order :-
(I) CO (II)`CN^(-)` (III) `O_(2)^(+)` (IV) `NO^(+)`

Options

A. I,II,III
B. I,II,IV
C. I,III,IV
D. II,III,IV

Answer

To determine which of the given species (CO, CN^(-), O2^(+), and NO^(+)) have the same bond order, we need to calculate the bond order for each molecule. The bond order can be calculated using the formula:

\[ \text{Bond Order} = \frac{(\text{Number of bonding electrons} - \text{Number of antibonding electrons})}{2} \]

### Step 1: Calculate the bond order for CO

1. **Determine the total number of electrons:**
- Carbon (C) has 6 electrons.
- Oxygen (O) has 8 electrons.
- Total = 6 + 8 = 14 electrons.

2. **Electron configuration:**
- The molecular orbital configuration for CO is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^0 \pi_{2p_y}^*^0 \]
- Bonding electrons = 10 (from σ and π orbitals).
- Antibonding electrons = 4 (from σ* and π* orbitals).

3. **Calculate bond order:**
- Bond Order = (10 - 4) / 2 = 3.

### Step 2: Calculate the bond order for CN^(-)

1. **Determine the total number of electrons:**
- Carbon (C) has 6 electrons.
- Nitrogen (N) has 7 electrons.
- CN^(-) has an additional electron due to the negative charge.
- Total = 6 + 7 + 1 = 14 electrons.

2. **Electron configuration:**
- The molecular orbital configuration for CN^(-) is the same as CO since they have the same number of electrons.
- Bonding electrons = 10.
- Antibonding electrons = 4.

3. **Calculate bond order:**
- Bond Order = (10 - 4) / 2 = 3.

### Step 3: Calculate the bond order for O2^(+)

1. **Determine the total number of electrons:**
- Oxygen (O) has 8 electrons.
- O2 has a total of 16 electrons.
- O2^(+) has one less electron due to the positive charge.
- Total = 16 - 1 = 15 electrons.

2. **Electron configuration:**
- The molecular orbital configuration for O2^(+) is:
\[ \sigma_{1s}^2 \sigma_{1s}^*^2 \sigma_{2s}^2 \sigma_{2s}^*^2 \sigma_{2p_z}^2 \pi_{2p_x}^2 \pi_{2p_y}^2 \pi_{2p_x}^*^1 \pi_{2p_y}^*^0 \]
- Bonding electrons = 10.
- Antibonding electrons = 5 (4 from σ* and π* orbitals).

3. **Calculate bond order:**
- Bond Order = (10 - 5) / 2 = 2.5.

### Step 4: Calculate the bond order for NO^(+)

1. **Determine the total number of electrons:**
- Nitrogen (N) has 7 electrons.
- Oxygen (O) has 8 electrons.
- Total = 7 + 8 = 15 electrons.
- NO^(+) has one less electron due to the positive charge.
- Total = 15 - 1 = 14 electrons.

2. **Electron configuration:**
- The molecular orbital configuration for NO^(+) is the same as CO and CN^(-).
- Bonding electrons = 10.
- Antibonding electrons = 4.

3. **Calculate bond order:**
- Bond Order = (10 - 4) / 2 = 3.

### Conclusion

From the calculations:
- The bond order for CO = 3.
- The bond order for CN^(-) = 3.
- The bond order for O2^(+) = 2.5.
- The bond order for NO^(+) = 3.

Thus, CO, CN^(-), and NO^(+) have the same bond order of 3.

### Final Answer
The species that have the same bond order are CO, CN^(-), and NO^(+).