Which of the following have identical bond order?

Options

A. `CN^(-)`
B. `NO^(+)`
C. `O_(2)^(+)`
D. `O_(2)^(2-)`

Explanation

To determine which of the given molecules have identical bond orders, we will follow these steps:

### Step 1: Identify the total number of electrons in each molecule.
- For CN⁻ (Cyanide ion):
- Carbon (C) has 6 electrons and Nitrogen (N) has 7 electrons. The negative charge adds 1 electron.
- Total electrons = 6 + 7 + 1 = 14 electrons.

- For NO⁺ (Nitric oxide ion):
- Nitrogen (N) has 7 electrons and Oxygen (O) has 8 electrons. The positive charge removes 1 electron.
- Total electrons = 7 + 8 - 1 = 14 electrons.

- For O₂⁺ (Oxygen ion):
- Each Oxygen has 8 electrons, and the positive charge removes 1 electron.
- Total electrons = 8 + 8 - 1 = 15 electrons.

- For O₂⁻ (Oxygen ion):
- Each Oxygen has 8 electrons, and the negative charge adds 1 electron.
- Total electrons = 8 + 8 + 1 = 17 electrons.

### Step 2: Determine the bond order using the formula.
The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2} \times (\text{Number of bonding electrons} - \text{Number of antibonding electrons}) \]

- For CN⁻ (14 electrons):
- The bond order is 3.

- For NO⁺ (14 electrons):
- The bond order is also 3.

- For O₂⁺ (15 electrons):
- The bond order is approximately 2.5.

- For O₂⁻ (17 electrons):
- The bond order is 1.

### Step 3: Compare the bond orders.
- CN⁻ has a bond order of 3.
- NO⁺ has a bond order of 3.
- O₂⁺ has a bond order of approximately 2.5.
- O₂⁻ has a bond order of 1.

### Conclusion:
The molecules CN⁻ and NO⁺ have identical bond orders of 3.