To determine the O-N-O bond angle, we need to analyze the hybridization and geometry of the central nitrogen atom in each species.

Step 1: Analyze NO₂⁺

The NO₂⁺ ion has a linear structure. The nitrogen atom is sp hybridized. It has two sigma bonds and no lone pairs. The O-N-O bond angle is $180^{\circ }$.

Step 2: Analyze NO₂

NO₂ is a bent molecule with an odd number of electrons. The nitrogen atom is sp² hybridized with one unpaired electron. The bond angle is approximately $134^{\circ }$.

Step 3: Analyze NO₂^-

NO₂^- has a bent structure. The nitrogen atom is sp² hybridized with one lone pair of electrons. The lone pair-bond pair repulsion is greater than bond pair-bond pair repulsion, reducing the bond angle from the ideal $120^{\circ }$. The O-N-O bond angle is approximately $115^{\circ }$.

Step 4: Analyze NO₃^-

NO₃^- has a trigonal planar structure. The nitrogen atom is sp² hybridized with no lone pairs. The O-N-O bond angle is $120^{\circ }$.

Step 5: Compare the bond angles

• NO₂⁺: $180^{\circ }$ (linear)
• NO₂: $134^{\circ }$ (bent)
• NO₂^-: $115^{\circ }$ (bent)
• NO₃^-: $120^{\circ }$ (trigonal planar)

Comparing these values, NO₂⁺ has the highest O-N-O bond angle.

Option Analysis:

• A) NO₂: Has an unpaired electron, leading to a bent structure with a bond angle of approximately $134^{\circ }$. Incorrect.
• B) NO₂⁺: Is linear with sp hybridization, resulting in a $180^{\circ }$ bond angle. Correct.
• C) NO₂^-: Has a lone pair on nitrogen, leading to a bent structure with a bond angle of approximately $115^{\circ }$. Incorrect.
• D) NO₃^-: Is trigonal planar with sp² hybridization and no lone pairs, resulting in a $120^{\circ }$ bond angle. Incorrect.

Correct Answer: (B)