Four diatomic species are listed below Identify the correct order in which the bond order is increasing in then :

Options

A. NO `lt O_(2)^(-) lt C_(2)^(2-) lt He_(2)^(+)`
B. `O_(2)^(-) lt NO lt C_(2)^(2-) lt He_(2)^(+)`
C. `C_(2)^(2-) lt He_(2)^(+) lt O_(2)^(-) lt NO `
D. `He_(2)^(+) lt O_(2)^(-) lt NO lt C_(2)^(2-)`

Answer

To determine the correct order of bond order for the given diatomic species (NO, O2-, C2 2-, and He2+), we will follow these steps:

### Step 1: Calculate the bond order for each species.

**1. For NO:**
- Total electrons: 15 (7 from N + 8 from O)
- Molecular orbital configuration:
- σ1s² σ1s*² σ2s² σ2s*² σ2p z² π2p x² = π2p y² (2 electrons in bonding orbitals, 3 in anti-bonding)
- Bond order formula:
\[ \text{Bond Order} = \frac{1}{2} \left( \text{Number of electrons in bonding orbitals} - \text{Number of electrons in anti-bonding orbitals} \right) \]
- Calculation:
\[ \text{Bond Order} = \frac{1}{2} (10 - 5) = 2.5 \]

**2. For O2-:**
- Total electrons: 17
- Molecular orbital configuration:
- σ1s² σ1s*² σ2s² σ2s*² σ2p z² π2p x² = π2p y² (2 electrons in bonding orbitals, 3 in anti-bonding)
- Calculation:
\[ \text{Bond Order} = \frac{1}{2} (10 - 7) = 1.5 \]

**3. For C2 2-:**
- Total electrons: 14
- Molecular orbital configuration:
- σ1s² σ1s*² σ2s² σ2s*² σ2p z² π2p x² = π2p y² (10 in bonding, 4 in anti-bonding)
- Calculation:
\[ \text{Bond Order} = \frac{1}{2} (10 - 4) = 3 \]

**4. For He2+:**
- Total electrons: 3
- Molecular orbital configuration:
- σ1s² σ1s*¹ (2 in bonding, 1 in anti-bonding)
- Calculation:
\[ \text{Bond Order} = \frac{1}{2} (2 - 1) = 0.5 \]

### Step 2: List the bond orders calculated.
- NO: 2.5
- O2-: 1.5
- C2 2-: 3
- He2+: 0.5

### Step 3: Arrange the bond orders in increasing order.
- He2+: 0.5
- O2-: 1.5
- NO: 2.5
- C2 2-: 3

### Final Order of Bond Order:
The correct order of bond order increasing is:
1. He2+ (0.5)
2. O2- (1.5)
3. NO (2.5)
4. C2 2- (3)