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

1. For ${\mathrm{NO}}_2^{+}$:

• Lewis structure: ${\mathrm{O}=\mathrm{N}=\mathrm{O}}^{+}$.
• Nitrogen forms two sigma bonds (one with each oxygen).
• Number of lone pairs on N = 0.
• Steric number (SN) = 2 (sigma bonds) + 0 (lone pairs) = 2.
• Hybridization for SN = 2 is $sp$.

2. For ${\mathrm{NO}}_3{}^{-}$:

• Lewis structure: Nitrogen is bonded to three oxygen atoms. One N-O bond is a double bond, and two are single bonds (with resonance structures).
• Nitrogen forms three sigma bonds (one with each oxygen).
• Number of lone pairs on N = 0.
• Steric number (SN) = 3 (sigma bonds) + 0 (lone pairs) = 3.
• Hybridization for SN = 3 is $s{p}^2$.

3. For ${\mathrm{NH}}_4^{+}$:

• Lewis structure: Nitrogen is bonded to four hydrogen atoms.
• Nitrogen forms four sigma bonds (one with each hydrogen).
• Number of lone pairs on N = 0.
• Steric number (SN) = 4 (sigma bonds) + 0 (lone pairs) = 4.
• Hybridization for SN = 4 is $s{p}^3$.

Therefore, the hybridizations are $sp$, $s{p}^2$, and $s{p}^3$ respectively.

Option Analysis:

• A) $sp,s{p}^3$, and $s{p}^2$: Incorrect. The hybridization for ${\mathrm{NO}}_3{}^{-}$ is $s{p}^2$, not $s{p}^3$, and for ${\mathrm{NH}}_4^{+}$ is $s{p}^3$, not $s{p}^2$.
• B) $sp,s{p}^2$, and $s{p}^3$: Correct. This matches our calculated hybridizations for ${\mathrm{NO}}_2^{+}$, ${\mathrm{NO}}_3{}^{-}$, and ${\mathrm{NH}}_4^{+}$ respectively.
• C) $s{p}^2,sp$, and $s{p}^3$: Incorrect. The hybridization for ${\mathrm{NO}}_2^{+}$ is $sp$, not $s{p}^2$, and for ${\mathrm{NO}}_3{}^{-}$ is $s{p}^2$, not $sp$.
• D) $s{p}^2,s{p}^3$, and $sp$: Incorrect. All three hybridizations are mismatched with the respective species.

Correct Answer: (B)