The difference between bond orders of `CO` and `NO^(o+)` is `(x)/(2)` where x = _________ .
(Round off to the Nearest Integer)

Answer

To solve the problem, we need to find the bond orders of carbon monoxide (CO) and the nitric oxide ion (NO^+), and then calculate the difference between these bond orders.

### Step-by-Step Solution:

1. **Determine the Total Number of Electrons:**
- CO consists of Carbon (C) and Oxygen (O).
- Carbon (C) has 6 electrons.
- Oxygen (O) has 8 electrons.
- Total electrons in CO = 6 + 8 = **14 electrons**.
- NO^+ consists of Nitrogen (N) and Oxygen (O).
- Nitrogen (N) has 7 electrons.
- Oxygen (O) has 8 electrons.
- Since NO^+ has a positive charge, it loses one electron.
- Total electrons in NO^+ = 7 + 8 - 1 = **14 electrons**.

2. **Molecular Orbital Configuration:**
- Both CO and NO^+ have 14 electrons, so their molecular orbital configurations will be similar.
- The filling of molecular orbitals follows this order:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p_z)², π(2p_x)², π(2p_y)²
- For both CO and NO^+, the configuration will be:
- σ(1s)², σ*(1s)², σ(2s)², σ*(2s)², σ(2p_z)², π(2p_x)², π(2p_y)²
- This accounts for 10 bonding electrons and 4 antibonding electrons.

3. **Calculate Bond Order:**
- The bond order (BO) is calculated using the formula:
\[ \text{Bond Order} = \frac{\text{Number of electrons in bonding MOs} - \text{Number of electrons in antibonding MOs}}{2} \]
- For both CO and NO^+:
- Number of bonding electrons = 10
- Number of antibonding electrons = 4
- Bond Order = \(\frac{10 - 4}{2} = \frac{6}{2} = 3\)

4. **Calculate the Difference in Bond Orders:**
- Bond Order of CO = 3
- Bond Order of NO^+ = 3
- Difference in bond orders = \(3 - 3 = 0\)

5. **Relate the Difference to x:**
- According to the problem, the difference in bond orders is given as \(\frac{x}{2}\).
- We have found that the difference is 0, so:
\[ 0 = \frac{x}{2} \]
- Solving for x gives:
\[ x = 0 \times 2 = 0 \]

### Final Answer:
The value of x is **0**.