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 find 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⁻
1. Count the total number of valence electrons:
- Nitrogen (N) has 5 valence electrons.
- Each Oxygen (O) has 6 valence electrons, and there are three Oxygens: 3 × 6 = 18.
- The ion has an extra electron due to the negative charge: +1.
- Total = 5 + 18 + 1 = 24 valence electrons.

2. Arrange the atoms:
- Place Nitrogen in the center and surround it with three Oxygen atoms.

3. Distribute the electrons:
- Connect the Nitrogen to each Oxygen with single bonds initially.
- Distribute the remaining electrons to satisfy the octet rule for Oxygen.
- You will find that one of the Oxygens can form a double bond with Nitrogen to satisfy the octet rule for all atoms.

### Step 2: Count the Total Number of Bonds
1. In the Lewis structure, you will see:
- One N=O double bond (counts as 2 bonds).
- Two N-O single bonds (each counts as 1 bond).

2. Total bonds = 2 (from the double bond) + 1 + 1 (from the two single bonds) = 4 bonds.

### Step 3: Count the Number of Bonding Pairs Between Nitrogen and Oxygen
1. There are three N-O bonds in total:
- One double bond (N=O) counts as 1 bond between N and O.
- Two single bonds (N-O) each count as 1 bond.

2. Therefore, the total number of bonds between Nitrogen and Oxygen is 3.

### Step 4: Calculate the Bond Order
1. Use the formula for bond order:
\[ \text{Bond Order} = \frac{\text{Total number of bonds}}{\text{Number of bond pairs between individual atoms}} \]
\[ \text{Bond Order} = \frac{4}{3} \approx 1.33 \]

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