The correct order of bond order values among the following
(i) `NO^(-)` (ii) `NO^(+)`
(iii) `NO` (iv) `NO^(2+)`
(v) `NO^(2-)`

Options

A. I lt IV lt III lt II lt V
B. IV = II lt I lt V lt III
C. V lt I lt III lt IV = II
D. II lt III lt IV lt I lt V

Explanation

To determine the correct order of bond order values among the given species (i) `NO^(-)`, (ii) `NO^(+)`, (iii) `NO`, (iv) `NO^(2+)`, and (v) `NO^(2-)`, we will calculate the bond order for each species using the formula:

\[ \text{Bond Order} = \frac{1}{2} \times (\text{Number of bonding electrons} - \text{Number of antibonding electrons}) \]

### Step 1: Calculate Bond Order for `NO^(-)`

1. **Determine the number of electrons**:
- `NO` has 11 electrons. The negative charge adds 1 more electron, so `NO^(-)` has 12 electrons.

2. **Molecular orbital configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², π2p_x², π2p_y²

3. **Count bonding and antibonding electrons**:
- Bonding electrons = 2 (σ1s) + 2 (σ2s) + 4 (π2p) = 8
- Antibonding electrons = 2 (σ*1s) + 2 (σ*2s) = 4

4. **Calculate bond order**:
\[ \text{Bond Order} = \frac{1}{2} \times (8 - 4) = \frac{1}{2} \times 4 = 2 \]

### Step 2: Calculate Bond Order for `NO^(+)`

1. **Determine the number of electrons**:
- `NO` has 11 electrons. The positive charge removes 1 electron, so `NO^(+)` has 10 electrons.

2. **Molecular orbital configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², π2p_x¹, π2p_y¹

3. **Count bonding and antibonding electrons**:
- Bonding electrons = 2 (σ1s) + 2 (σ2s) + 2 (π2p) = 6
- Antibonding electrons = 2 (σ*1s) + 2 (σ*2s) = 4

4. **Calculate bond order**:
\[ \text{Bond Order} = \frac{1}{2} \times (6 - 4) = \frac{1}{2} \times 2 = 1 \]

### Step 3: Calculate Bond Order for `NO`

1. **Determine the number of electrons**:
- `NO` has 11 electrons.

2. **Molecular orbital configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², π2p_x², π2p_y¹

3. **Count bonding and antibonding electrons**:
- Bonding electrons = 2 (σ1s) + 2 (σ2s) + 4 (π2p) = 8
- Antibonding electrons = 2 (σ*1s) + 2 (σ*2s) = 4

4. **Calculate bond order**:
\[ \text{Bond Order} = \frac{1}{2} \times (8 - 4) = \frac{1}{2} \times 4 = 2 \]

### Step 4: Calculate Bond Order for `NO^(2+)`

1. **Determine the number of electrons**:
- `NO` has 11 electrons. The positive charge removes 2 electrons, so `NO^(2+)` has 9 electrons.

2. **Molecular orbital configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², π2p_x¹

3. **Count bonding and antibonding electrons**:
- Bonding electrons = 2 (σ1s) + 2 (σ2s) + 2 (π2p) = 6
- Antibonding electrons = 2 (σ*1s) + 2 (σ*2s) = 4

4. **Calculate bond order**:
\[ \text{Bond Order} = \frac{1}{2} \times (6 - 4) = \frac{1}{2} \times 2 = 1 \]

### Step 5: Calculate Bond Order for `NO^(2-)`

1. **Determine the number of electrons**:
- `NO` has 11 electrons. The negative charge adds 2 electrons, so `NO^(2-)` has 13 electrons.

2. **Molecular orbital configuration**:
- The configuration is:
- σ1s², σ*1s², σ2s², σ*2s², π2p_x², π2p_y², σ2p_z¹

3. **Count bonding and antibonding electrons**:
- Bonding electrons = 2 (σ1s) + 2 (σ2s) + 4 (π2p) + 2 (σ2p) = 10
- Antibonding electrons = 2 (σ*1s) + 2 (σ*2s) = 4

4. **Calculate bond order**:
\[ \text{Bond Order} = \frac{1}{2} \times (10 - 4) = \frac{1}{2} \times 6 = 3 \]

### Summary of Bond Orders
- `NO^(-)` = 2
- `NO^(+)` = 1
- `NO` = 2
- `NO^(2+)` = 1
- `NO^(2-)` = 3

### Final Order of Bond Orders
From highest to lowest:
1. `NO^(2-)` (3)
2. `NO^(-)` (2)
3. `NO` (2)
4. `NO^(+)` (1)
5. `NO^(2+)` (1)