Concept: Hybridization is the concept of mixing atomic orbitals to form new hybrid orbitals suitable for the pairing of electrons to form chemical bonds in valence bond theory. The state of hybridization can be determined by counting the number of sigma bonds and lone pairs around the central atom (steric number).

Why (B) is correct:

Solid ${\mathrm{N}}_2{\mathrm{O}}_5$ exists as an ionic compound, nitronium nitrate, ${\left[{\mathrm{NO}}_2\right]}^{+}{\left[{\mathrm{NO}}_3\right]}^{-}$. We need to determine the hybridization of the anionic part, which is the nitrate ion, ${\mathrm{NO}}_3^{-}$.

1. Lewis Structure of ${\mathrm{NO}}_3^{-}$: The central atom is Nitrogen (N). It forms three sigma bonds with three oxygen atoms and has no lone pairs. One of the N-O bonds is a double bond, and the other two are single bonds, but due to resonance, all N-O bonds are equivalent and have partial double bond character.

2. Steric Number Calculation: For the central nitrogen atom in ${\mathrm{NO}}_3^{-}$, there are 3 sigma bonds and 0 lone pairs. Therefore, the steric number = 3 + 0 = 3.

3. Hybridization: A steric number of 3 corresponds to ${\mathrm{sp}}^2$ hybridization.

Option Analysis:

• A) $\mathrm{sp}$: This hybridization corresponds to a steric number of 2 (e.g., linear geometry). This is incorrect for ${\mathrm{NO}}_3^{-}$.
• B) ${\mathrm{sp}}^2$: This hybridization corresponds to a steric number of 3 (e.g., trigonal planar geometry). This is correct for the nitrate ion.
• C) ${\mathrm{sp}}^3$: This hybridization corresponds to a steric number of 4 (e.g., tetrahedral geometry). This is incorrect for ${\mathrm{NO}}_3^{-}$.
• D) Not applicable: Hybridization is applicable to the central atom in the nitrate ion. This is incorrect.

Correct Answer: (B)