The hybridisations of atomic orbitals of nitrogen in NO$^+$, NO$_3^-$ and NH$_3$ respectively are:

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1. **Formula for Hybridisation:** The steric number (H) or number of hybrid orbitals can be calculated using the formula: $H = \frac{1}{2} [V + M - C + A]$, where V = valence electrons of central atom, M = number of monovalent atoms, C = positive charge, A = negative charge.
2. **Analysis of NO$^+$ (Nitrosonium ion):**
   - Central atom N ($V=5$). No monovalent atoms ($M=0$). Cationic charge ($C=1$).
   - $H = \frac{1}{2} [5 + 0 - 1 + 0] = 2$.
   - $H=2$ corresponds to **sp** hybridisation .
3. **Analysis of NO$_3^-$ (Nitrate ion):**
   - Central atom N ($V=5$). No monovalent atoms ($M=0$). Anionic charge ($A=1$).
   - $H = \frac{1}{2} [5 + 0 - 0 + 1] = 3$.
   - $H=3$ corresponds to **sp$^2$** hybridisation .
4. **Analysis of NH$_3$ (Ammonia):**
   - Central atom N ($V=5$). Three monovalent H atoms ($M=3$).
   - $H = \frac{1}{2} [5 + 3 - 0 + 0] = 4$.
   - $H=4$ corresponds to **sp$^3$** hybridisation .
5. **Conclusion:** The correct order is sp, sp$^2$, and sp$^3$.

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