In $NO_{3}^{-}$ ion the number of bond pairs and lone pairs of electrons on nitrogen atom are respectively:

Question

Accepted Answer

In the nitrate ion, $NO_3^-$, the nitrogen atom is involved in forming bonds with three oxygen atoms. Let's analyze the bonding and electron arrangement to understand why Option 4 is the correct answer.

### Explanation for Option 4 (Correct Answer: 4 bond pairs, 0 lone pairs):

1. **Bond Formation**: In the $NO_3^-$ ion, nitrogen forms three sigma bonds with three oxygen atoms. Additionally, there is resonance in the nitrate ion, which involves the delocalization of electrons. This results in one of the nitrogen-oxygen bonds being a double bond at any given time, while the other two are single bonds. However, due to resonance, the bond order is effectively 1.33 for each N-O bond.

2. **Bond Pairs**: Since nitrogen forms three sigma bonds and participates in resonance, it effectively uses all its valence electrons in bonding. Therefore, there are 4 bond pairs around the nitrogen atom (considering the resonance structure).

3. **Lone Pairs**: Nitrogen, in this configuration, does not have any lone pairs of electrons. All its valence electrons are involved in bonding with oxygen atoms.

### Explanation for Why Other Options are Incorrect:

- **Option 1 (2 bond pairs, 2 lone pairs)**: This option suggests that nitrogen has two bond pairs and two lone pairs. However, in $NO_3^-$, nitrogen forms bonds with all three oxygen atoms, leaving no lone pairs on nitrogen. Thus, this option is incorrect.

- **Option 2 (3 bond pairs, 1 lone pair)**: This option implies that nitrogen has one lone pair of electrons. As explained, nitrogen uses all its valence electrons to form bonds with oxygen atoms, leaving no lone pairs. Therefore, this option is incorrect.

- **Option 3 (1 bond pair, 3 lone pairs)**: This option is incorrect because it suggests that nitrogen forms only one bond and has three lone pairs. In reality, nitrogen forms bonds with all three oxygen atoms, and there are no lone pairs on nitrogen in the $NO_3^-$ ion.

In conclusion, Option 4 is the correct answer because it accurately reflects the bonding situation in the $NO_3^-$ ion, where nitrogen forms four bond pairs and has no lone pairs.