The bond energies `NO,NO^(+)` and `NO^(Theta)` ion follows the order

Options

A. `NO^(o+) gt NO gt NO^(Theta)`
B. `NO gt NO^(o+) gt NO^(Theta)`
C. `NO^(Theta) gt NO gt NO^(o+)`
D. `NO^(o+) gt NO^(Theta) gt NO`

Answer

To determine the order of bond energies for the species NO, NO⁺, and NO⁻, we need to analyze their bond orders. The bond energy is directly related to the bond order; higher bond order indicates stronger bonds and thus higher bond energy.

### Step-by-Step Solution:

1. **Identify the number of electrons in each species**:
- For NO: 15 electrons
- For NO⁺: 14 electrons (1 electron removed)
- For NO⁻: 16 electrons (1 electron added)

2. **Determine the electronic configuration for each species**:
- **NO (15 electrons)**:
- Configuration: σ1s², σ*1s², σ2s², σ*2s², σ2p_x², π2p_y², π2p_z², π*2p_y¹, π*2p_z⁰
- **NO⁺ (14 electrons)**:
- Configuration: σ1s², σ*1s², σ2s², σ*2s², σ2p_x², π2p_y², π2p_z²
- **NO⁻ (16 electrons)**:
- Configuration: σ1s², σ*1s², σ2s², σ*2s², σ2p_x², π2p_y², π2p_z², π*2p_y¹, π*2p_z¹

3. **Calculate the bond order for each species**:
- **Bond Order Formula**: Bond Order = ½ (Nb - Na)
- Where Nb = number of bonding electrons and Na = number of antibonding electrons.

- **For NO**:
- Nb = 10 (σ1s² + σ2s² + σ2p_x² + π2p_y² + π2p_z²)
- Na = 5 (σ*1s² + σ*2s² + π*2p_y¹ + π*2p_z⁰)
- Bond Order = ½ (10 - 5) = 2.5

- **For NO⁺**:
- Nb = 10 (same as NO)
- Na = 4 (σ*1s² + σ*2s² + π*2p_y² + π*2p_z²)
- Bond Order = ½ (10 - 4) = 3

- **For NO⁻**:
- Nb = 10 (same as NO)
- Na = 6 (σ*1s² + σ*2s² + π*2p_y¹ + π*2p_z¹)
- Bond Order = ½ (10 - 6) = 2

4. **Order the bond energies based on bond orders**:
- Bond Order for NO⁺ = 3 (highest)
- Bond Order for NO = 2.5
- Bond Order for NO⁻ = 2 (lowest)

5. **Conclusion**:
- The order of bond energies is: NO⁺ > NO > NO⁻.

### Final Answer:
The bond energies follow the order: **NO⁺ > NO > NO⁻**.