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 species have identical bond orders, we need to calculate the bond order for each species based on the number of electrons present. Here’s a step-by-step solution:

### Step 1: Identify the species
Let's assume the species in question are CN⁻ (cyanide ion), NO⁺ (nitric oxide cation), O₂ (dioxygen), and O₂²⁻ (peroxide ion).

### Step 2: Calculate the total number of electrons for each species
1. **CN⁻**:
- Carbon (C) has 4 electrons.
- Nitrogen (N) has 5 electrons.
- The negative charge (-1) adds 1 electron.
- Total = 4 (C) + 5 (N) + 1 = 10 electrons.

2. **NO⁺**:
- Nitrogen (N) has 5 electrons.
- Oxygen (O) has 6 electrons.
- The positive charge (+1) removes 1 electron.
- Total = 5 (N) + 6 (O) - 1 = 10 electrons.

3. **O₂**:
- Each oxygen (O) has 6 electrons.
- Total = 6 (O) + 6 (O) = 12 electrons.

4. **O₂²⁻**:
- Each oxygen (O) has 6 electrons.
- The two negative charges (-2) add 2 electrons.
- Total = 6 (O) + 6 (O) + 2 = 14 electrons.

### Step 3: Compare the total number of electrons
- CN⁻ has 10 electrons.
- NO⁺ has 10 electrons.
- O₂ has 12 electrons.
- O₂²⁻ has 14 electrons.

### Step 4: Identify species with identical bond orders
From the calculations:
- CN⁻ and NO⁺ both have 10 electrons.
- O₂ has 12 electrons.
- O₂²⁻ has 14 electrons.

Since CN⁻ and NO⁺ have the same number of electrons, they will have identical bond orders.

### Conclusion
The species that have identical bond orders are **CN⁻ and NO⁺**.

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