Which of the following has the highest value for bond order `-C_(6)H_(6),CO_(3)^(2-) ,NO^(-) and SO_(4)^(2-)` ?

Options

A. `C_6H_6`
B. `CO_3^(2-)`
C. `NO^(-)`
D. `SO_(3)`

Answer

To determine which of the given compounds has the highest bond order among \( C_6H_6 \), \( CO_3^{2-} \), \( NO^{-} \), and \( SO_4^{2-} \), we will calculate the bond order for each compound step by step.

### Step 1: Calculate the bond order of \( C_6H_6 \) (Benzene)

1. **Identify the structure**: Benzene has a cyclic structure with alternating single and double bonds, represented as:
\[ C_6H_6 \text{ (Benzene)} \]
It can be represented with resonance structures.

2. **Count the bonds**: In benzene, each carbon-carbon bond can be considered as having a bond order of 1.5 due to resonance. There are 6 carbon-carbon bonds in total.

3. **Bond order formula**:
\[ \text{Bond Order} = \frac{\text{Number of bonds}}{\text{Number of resonance structures}} = \frac{6 \text{ bonds}}{2 \text{ resonance structures}} = 1.5 \]

### Step 2: Calculate the bond order of \( CO_3^{2-} \) (Carbonate ion)

1. **Identify the structure**: The carbonate ion has three resonance structures.

2. **Count the bonds**: In each resonance structure, there are 4 bonds (1 double bond and 2 single bonds).

3. **Bond order formula**:
\[ \text{Bond Order} = \frac{\text{Total number of bonds}}{\text{Number of resonance structures}} = \frac{4 \text{ bonds}}{3 \text{ resonance structures}} \approx 1.33 \]

### Step 3: Calculate the bond order of \( NO^{-} \) (Nitroxide ion)

1. **Identify the structure**: The nitroxide ion has a total of 15 electrons.

2. **Count the bonds**: The molecular orbital theory suggests that there are 10 electrons in bonding molecular orbitals and 4 in antibonding orbitals.

3. **Bond order formula**:
\[ \text{Bond Order} = \frac{\text{Number of bonding electrons} - \text{Number of antibonding electrons}}{2} = \frac{10 - 4}{2} = 3 \]

### Step 4: Calculate the bond order of \( SO_4^{2-} \) (Sulfate ion)

1. **Identify the structure**: The sulfate ion has four resonance structures.

2. **Count the bonds**: Each structure has 6 bonds (1 double bond and 3 single bonds).

3. **Bond order formula**:
\[ \text{Bond Order} = \frac{\text{Total number of bonds}}{\text{Number of resonance structures}} = \frac{6 \text{ bonds}}{4 \text{ resonance structures}} = 1.5 \]

### Conclusion: Compare the bond orders

- \( C_6H_6 \): Bond order = 1.5
- \( CO_3^{2-} \): Bond order ≈ 1.33
- \( NO^{-} \): Bond order = 3
- \( SO_4^{2-} \): Bond order = 1.5

The compound with the highest bond order is \( NO^{-} \) with a bond order of 3.

### Final Answer:
**The compound with the highest bond order is \( NO^{-} \).**