To determine the hybridization of the central atom (nitrogen) in each species, we need to calculate the steric number (SN) for nitrogen. The steric number is the sum of the number of sigma bonds and the number of lone pairs around the central atom.

1. For NO₂⁺ (Nitronium ion):

• Nitrogen is the central atom.
• Valence electrons of N = 5.
• Oxygen forms double bonds. Each oxygen contributes 2 electrons to bonding.
• The +1 charge means one electron is removed from nitrogen.
• Total valence electrons = 5 (N) - 1 (charge) + 2 * 6 (O) = 16 electrons.
• The Lewis structure is O=N=O⁺.
• Nitrogen forms two sigma bonds (one with each oxygen) and has no lone pairs.
• Steric Number (SN) = 2 (sigma bonds) + 0 (lone pairs) = 2.
• SN = 2 corresponds to sp hybridization.

2. For NO₃^- (Nitrate ion):

• Nitrogen is the central atom.
• Valence electrons of N = 5.
• Valence electrons of O = 6.
• The -1 charge means one extra electron.
• Total valence electrons = 5 (N) + 3 * 6 (O) + 1 (charge) = 24 electrons.
• The Lewis structure shows nitrogen forming one double bond and two single bonds with oxygen atoms, with a delocalized negative charge.
• Nitrogen forms three sigma bonds (one with each oxygen) and has no lone pairs.
• Steric Number (SN) = 3 (sigma bonds) + 0 (lone pairs) = 3.
• SN = 3 corresponds to sp² hybridization.

3. For NH₄⁺ (Ammonium ion):

• Nitrogen is the central atom.
• Valence electrons of N = 5.
• Valence electrons of H = 1.
• The +1 charge means one electron is removed.
• Total valence electrons = 5 (N) + 4 * 1 (H) - 1 (charge) = 8 electrons.
• The Lewis structure shows nitrogen forming four single bonds with hydrogen atoms.
• Nitrogen forms four sigma bonds and has no lone pairs.
• Steric Number (SN) = 4 (sigma bonds) + 0 (lone pairs) = 4.
• SN = 4 corresponds to sp³ hybridization.

Therefore, the hybridization of nitrogen in NO₂⁺, NO₃^-, and NH₄⁺ are sp, sp², and sp³ respectively.

Option Analysis:

• A) sp², sp³ and sp² respectively: Incorrect. NO₂⁺ is sp, NO₃^- is sp², NH₄⁺ is sp³.
• B) sp, sp² and sp³ respectively: Correct, as determined above.
• C) sp², sp, and sp³ respectively: Incorrect. NO₂⁺ is sp, NO₃^- is sp².
• D) sp², sp³ and sp respectively: Incorrect. NO₂⁺ is sp, NO₃^- is sp², NH₄⁺ is sp³.

Correct Answer: (B)