Find hybridisation of N in $\mathrm{NO_2^-}$

• Total valence electrons: $5 + 2(6) + 1 = 18$
• The nitrite ion has resonance structures with N as the central atom.
• Around nitrogen, there are:
  • two $\sigma$ bonds with O atoms
  • one lone pair on N
• So, steric number of N = $2 + 1 = 3$.
• Hybridisation corresponding to steric number 3 is: $\boxed{sp^2}$

Find hybridisation of N in $\mathrm{NO_2^+}$

• Total valence electrons: $5 + 2(6) - 1 = 16$
• In nitronium ion, structure is linear: $\mathrm{O=N=O}$
• Around nitrogen, there are:
  • two $\sigma$ bonds
  • no lone pair
• Steric number of N = 2.
• Hybridisation corresponding to steric number 2 is: $\boxed{sp}$

Find hybridisation of N in $\mathrm{NH_4^+}$

• Around nitrogen, there are:
  • four $\sigma$ bonds with H
  • no lone pair
• Steric number of N = 4.
• Hybridisation corresponding to steric number 4 is: $\boxed{sp^3}$

Arrange in the asked order

For $\mathrm{NO_2^-}$, $\mathrm{NO_2^+}$, $\mathrm{NH_4^+}$ respectively: $sp^2,\ sp,\ sp^3$

Match with options

This corresponds to: $\boxed{\text{D}}$

Comparison with stored answer

Stored correct answer = D, and our derived answer is also D. Hence they agree.