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 (iv) = (iii) lt (ii)
D. (ii) lt (iii) lt (iv) lt (i) lt (v)

Explanation

To determine the correct order of bond order values among the species \( NO^{-} \), \( NO^{+} \), \( NO \), \( NO^{2+} \), and \( NO^{2-} \), we will calculate the bond order for each species using the molecular orbital theory. The bond order is calculated using the formula:

\[ \text{Bond Order} = \frac{N_b - N_a}{2} \]

where \( N_b \) is the number of electrons in bonding molecular orbitals and \( N_a \) is the number of electrons in anti-bonding molecular orbitals.

### Step 1: Calculate the bond order for \( NO \)

1. **Count the total number of electrons**:
- Nitrogen (N) has 7 electrons and Oxygen (O) has 8 electrons.
- Total = \( 7 + 8 = 15 \) electrons.

2. **Molecular orbital configuration**:
- The configuration is:
\[ \sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2p_z}^{2} \pi_{2p_x}^{2} \pi_{2p_y}^{2} \pi_{2p_x}^{*1} \pi_{2p_y}^{*0} \]
- Bonding electrons \( N_b = 10 \) (from \( \sigma_{1s}^{2}, \sigma_{2s}^{2}, \sigma_{2p_z}^{2}, \pi_{2p_x}^{2}, \pi_{2p_y}^{2} \))
- Anti-bonding electrons \( N_a = 5 \) (from \( \pi_{2p_x}^{*1}, \pi_{2p_y}^{*0} \))

3. **Calculate bond order**:
\[ \text{Bond Order} = \frac{10 - 5}{2} = 2.5 \]

### Step 2: Calculate the bond order for \( NO^{-} \)

1. **Total electrons**:
- \( 15 + 1 = 16 \) electrons.

2. **Molecular orbital configuration**:
- The configuration is:
\[ \sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2p_z}^{2} \pi_{2p_x}^{2} \pi_{2p_y}^{2} \pi_{2p_x}^{*1} \pi_{2p_y}^{*1} \]
- Bonding electrons \( N_b = 10 \)
- Anti-bonding electrons \( N_a = 6 \)

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

### Step 3: Calculate the bond order for \( NO^{+} \)

1. **Total electrons**:
- \( 15 - 1 = 14 \) electrons.

2. **Molecular orbital configuration**:
- The configuration is:
\[ \sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2p_z}^{2} \pi_{2p_x}^{2} \pi_{2p_y}^{2} \pi_{2p_x}^{*0} \pi_{2p_y}^{*1} \]
- Bonding electrons \( N_b = 10 \)
- Anti-bonding electrons \( N_a = 4 \)

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

### Step 4: Calculate the bond order for \( NO^{2+} \)

1. **Total electrons**:
- \( 15 - 2 = 13 \) electrons.

2. **Molecular orbital configuration**:
- The configuration is:
\[ \sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2p_z}^{2} \pi_{2p_x}^{2} \pi_{2p_y}^{1} \pi_{2p_x}^{*0} \pi_{2p_y}^{*0} \]
- Bonding electrons \( N_b = 9 \)
- Anti-bonding electrons \( N_a = 4 \)

3. **Calculate bond order**:
\[ \text{Bond Order} = \frac{9 - 4}{2} = 2.5 \]

### Step 5: Calculate the bond order for \( NO^{2-} \)

1. **Total electrons**:
- \( 15 + 2 = 17 \) electrons.

2. **Molecular orbital configuration**:
- The configuration is:
\[ \sigma_{1s}^{2} \sigma_{1s}^{*2} \sigma_{2s}^{2} \sigma_{2s}^{*2} \sigma_{2p_z}^{2} \pi_{2p_x}^{2} \pi_{2p_y}^{2} \pi_{2p_x}^{*1} \pi_{2p_y}^{*1} \]
- Bonding electrons \( N_b = 10 \)
- Anti-bonding electrons \( N_a = 7 \)

3. **Calculate bond order**:
\[ \text{Bond Order} = \frac{10 - 7}{2} = 1.5 \]

### Summary of Bond Orders:
- \( NO^{+} \): 3
- \( NO \): 2.5
- \( NO^{2+} \): 2.5
- \( NO^{-} \): 2
- \( NO^{2-} \): 1.5

### Final Order:
The correct order of bond order values is:
1. \( NO^{+} \) (3)
2. \( NO \) = \( NO^{2+} \) (2.5)
3. \( NO^{-} \) (2)
4. \( NO^{2-} \) (1.5)