Step 1: Understanding the Concept:
Hybridization of the central atom dictates the molecular geometry and is determined by the number of sigma bonds and lone pairs around the atom.
Step 2: Key Formula or Approach:
Use the Steric Number (SN) formula: $SN = (\text{Number of atoms bonded}) + (\text{Number of lone pairs})$. SN = 2 ($sp$), 3 ($sp^2$), 4 ($sp^3$). Adjust valence electrons for ionic charges.
Step 3: Detailed Explanation:
1. NO$_{2^{+}$ (Nitryl cation):} Nitrogen group 15 has 5 valence electrons. Subtract 1 for the positive charge = 4 electrons. It forms 2 double bonds with 2 oxygen atoms.
$SN = 2$ (bonded atoms) $+ 0$ (lone pairs) $= 2$. Hybridization: sp. Shape: Linear.
2. NO$_{3^{-}$ (Nitrate ion):} Nitrogen has 5 valence electrons. Add 1 for the negative charge = 6 electrons available for bonding. It bonds to 3 oxygen atoms (one double bond, two single coordinate/resonance bonds).
$SN = 3$ (bonded atoms) $+ 0$ (lone pairs) $= 3$. Hybridization: sp$^{2$}. Shape: Trigonal planar.
3. NH$_{4^{+}$ (Ammonium ion):} Nitrogen has 5 valence electrons. Subtract 1 for positive charge = 4 electrons. It forms 4 single bonds with 4 hydrogen atoms.
$SN = 4$ (bonded atoms) $+ 0$ (lone pairs) $= 4$. Hybridization: sp$^{3$}. Shape: Tetrahedral.
Step 4: Final Answer:
The order is sp, sp$^{2}$, sp$^{3}$.