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 will follow these steps:

### Step 1: Identify the number of electrons in each species
- **CN⁻**: Carbon (C) has 6 electrons, Nitrogen (N) has 7 electrons, and CN⁻ has one additional electron due to the negative charge. Thus, total electrons = 6 + 7 + 1 = **14 electrons**.
- **NO⁺**: Nitrogen (N) has 7 electrons, Oxygen (O) has 8 electrons, and NO⁺ has one less electron due to the positive charge. Thus, total electrons = 7 + 8 - 1 = **14 electrons**.
- **O₂⁻**: Oxygen (O) has 8 electrons, and O₂ has 16 electrons. Adding one more electron gives us O₂⁻ a total of 17 electrons.
- **O₂²⁻**: O₂ has 16 electrons, and adding two more electrons gives us O₂²⁻ a total of 18 electrons.

### Step 2: Fill the molecular orbitals according to the molecular orbital theory
The order of filling for the molecular orbitals for these species is:
- σ(1s), σ*(1s), σ(2s), σ*(2s), π(2p_z), π(2p_x), π(2p_y), π*(2p_x), π*(2p_y), π*(2p_z)

### Step 3: Calculate the bond order for each species
The bond order is calculated using the formula:
\[ \text{Bond Order} = \frac{1}{2} (N_b - N_a) \]
where \( N_b \) is the number of bonding electrons and \( N_a \) is the number of anti-bonding electrons.

#### For CN⁻:
- **Electrons filled**:
- σ(1s): 2
- σ*(1s): 2
- σ(2s): 2
- σ*(2s): 2
- π(2p_z): 2
- π(2p_x): 2
- π(2p_y): 2
- **Total bonding electrons (N_b)** = 10
- **Total anti-bonding electrons (N_a)** = 4
- **Bond Order** = \( \frac{1}{2} (10 - 4) = \frac{6}{2} = 3 \)

#### For NO⁺:
- **Electrons filled**: Same as CN⁻ since it also has 14 electrons.
- **Bond Order** = \( \frac{1}{2} (10 - 4) = \frac{6}{2} = 3 \)

#### For O₂⁻:
- **Electrons filled**:
- σ(1s): 2
- σ*(1s): 2
- σ(2s): 2
- σ*(2s): 2
- π(2p_z): 2
- π(2p_x): 2
- π(2p_y): 1 (one additional electron)
- **Total bonding electrons (N_b)** = 10
- **Total anti-bonding electrons (N_a)** = 7
- **Bond Order** = \( \frac{1}{2} (10 - 7) = \frac{3}{2} = 1.5 \)

#### For O₂²⁻:
- **Electrons filled**:
- σ(1s): 2
- σ*(1s): 2
- σ(2s): 2
- σ*(2s): 2
- π(2p_z): 2
- π(2p_x): 2
- π(2p_y): 2 (two additional electrons)
- **Total bonding electrons (N_b)** = 10
- **Total anti-bonding electrons (N_a)** = 8
- **Bond Order** = \( \frac{1}{2} (10 - 8) = \frac{2}{2} = 1 \)

### Conclusion:
The species with identical bond orders are **CN⁻** and **NO⁺**, both having a bond order of **3**.