To determine the hybridization of the central atom in each compound, we can use the steric number method or by drawing the Lewis structure and counting electron domains.

Concept: Steric Number Method
Steric Number = (Number of sigma bonds) + (Number of lone pairs on the central atom)

• Steric Number = 2 $\to$ sp hybridization
• Steric Number = 3 $\to$ sp\u00b2 hybridization
• Steric Number = 4 $\to$ sp\u00b3 hybridization

Why (B) is correct:

Option B) ${\mathrm{SO}}_2$
The central atom is Sulfur (S).
Lewis structure of ${\mathrm{SO}}_2$: S is double-bonded to one O, single-bonded to another O (with a negative charge on O), and has one lone pair. Or, more commonly, two S=O double bonds and one lone pair on S (resonance structures). In either case, the sulfur atom forms two sigma bonds and has one lone pair.
Steric Number = 2 (sigma bonds) + 1 (lone pair) = 3.
Therefore, the hybridization of S in ${\mathrm{SO}}_2$ is ${\mathrm{sp}}^2$.

Option Analysis:

A) ${\mathrm{CO}}_2$
The central atom is Carbon (C).
Lewis structure: O=C=O. Carbon forms two sigma bonds (one in each double bond) and has no lone pairs.
Steric Number = 2 (sigma bonds) + 0 (lone pairs) = 2.
Hybridization of C in ${\mathrm{CO}}_2$ is $\mathrm{sp}$.

C) ${\mathrm{N}}_2\mathrm{O}$
The central atom is Nitrogen (N) in the N-N-O arrangement.
Lewis structure: N$\equiv$N-O (with charges). The central N forms one sigma bond with the terminal N and one sigma bond with the O. It has no lone pairs.
Steric Number = 2 (sigma bonds) + 0 (lone pairs) = 2.
Hybridization of central N in ${\mathrm{N}}_2\mathrm{O}$ is $\mathrm{sp}$.

D) $\mathrm{CO}$
The central atom can be considered either C or O, but it's a diatomic molecule.
Lewis structure: C$\equiv$O (with charges). Carbon forms one sigma bond and has one lone pair. Oxygen forms one sigma bond and has one lone pair.
For C: Steric Number = 1 (sigma bond) + 1 (lone pair) = 2. Hybridization is $\mathrm{sp}$.
For O: Steric Number = 1 (sigma bond) + 1 (lone pair) = 2. Hybridization is $\mathrm{sp}$.

Correct Answer: (B)