Determine the hybridization of nitrogen in each species using steric number:

$\text{Steric number} = \text{number of } \sigma\text{-bonds} + \text{number of lone pairs on central atom}$

Option A: $\mathrm{NO_2^+}$ (nitronium ion)

Structure: $\mathrm{O=N=O}$ Nitrogen is the central atom.

• Number of $\sigma$-bonds on N = 2
• Number of lone pairs on N = 0

So, $\text{Steric number} = 2 + 0 = 2$ Hybridization for steric number 2 is: $sp$

Option B: $\mathrm{NO_2^-}$ (nitrite ion)

Resonance structure has N central with two O atoms and one lone pair on N.

• Number of $\sigma$-bonds on N = 2
• Number of lone pairs on N = 1

So, $\text{Steric number} = 2 + 1 = 3$ Therefore, hybridization is: $sp^2$

Option C: $\mathrm{NO_3^-}$ (nitrate ion)

Nitrogen is bonded to three oxygen atoms, with no lone pair on N.

• Number of $\sigma$-bonds on N = 3
• Number of lone pairs on N = 0

So, $\text{Steric number} = 3 + 0 = 3$ Therefore, hybridization is: $sp^2$

Option D: $\mathrm{NO_3^-}$

This is identical to option C, so again the hybridization is: $sp^2$

Conclusion The species in which nitrogen is $sp$ hybridized is: $\boxed{\mathrm{NO_2^+}}$

Comparison with stored answer Stored correct answer: A

Our derived answer is also A, so they agree.