The bond order of the N-O bonds in `NO_3^-` ion is

Options

A. 1.33
B. `1.50`
C. `1.00`
D. `0.33`

Answer

To determine the bond order of the N-O bonds in the nitrate ion (NO₃⁻), we can follow these steps:

### Step 1: Draw the Lewis Structure of NO₃⁻
- The nitrate ion has one nitrogen atom and three oxygen atoms.
- Nitrogen is the central atom, and it forms bonds with the three oxygen atoms.
- The total number of valence electrons is calculated as follows: Nitrogen has 5 valence electrons, and each oxygen has 6, plus one extra electron for the negative charge. Thus, the total is \(5 + (3 \times 6) + 1 = 24\) valence electrons.
- Distributing these electrons, we can draw the Lewis structure, showing that nitrogen forms one double bond with one oxygen and single bonds with the other two oxygens.

### Step 2: Identify Resonance Structures
- The nitrate ion has resonance structures. The double bond can be between nitrogen and any one of the three oxygen atoms while the others have single bonds.
- This means there are three resonance structures where the position of the double bond shifts among the three oxygen atoms.

### Step 3: Count Sigma and Pi Bonds
- In the Lewis structure of NO₃⁻, we have:
- 3 sigma bonds (one for each N-O bond).
- 1 pi bond (from the double bond).
- Therefore, we have a total of 3 sigma bonds and 1 pi bond.

### Step 4: Calculate Bond Order
- The bond order can be calculated using the formula:
\[ \text{Bond Order} = \frac{\text{Number of Sigma Bonds} + \frac{1}{2} \times \text{Number of Pi Bonds}}{\text{Number of Bonds}} \]
- Here, we have:
- Number of sigma bonds = 3
- Number of pi bonds = 1
- The total number of bonds is 3 (sigma) + 1 (pi) = 4.
- Plugging these values into the formula:
\[ \text{Bond Order} = \frac{3 + \frac{1}{2} \times 1}{3} = \frac{3 + 0.5}{3} = \frac{3.5}{3} \approx 1.33 \]

### Conclusion
- The bond order of the N-O bonds in the NO₃⁻ ion is approximately **1.33**.

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