To determine the hybridization of the central atom, we need to calculate the steric number (SN) for each molecule/ion. The steric number is the sum of the number of sigma bonds and the number of lone pairs around the central atom.

Steric Number and Hybridization:

• SN = 2: $sp$ hybridization
• SN = 3: $s{p}^2$ hybridization
• SN = 4: $s{p}^3$ hybridization

Let's analyze each option:

Option A: ${\mathrm{NO}}_2$ and ${\mathrm{NH}}_3$

• ${\mathrm{NO}}_2$: Nitrogen is the central atom. It forms two sigma bonds (one with each oxygen) and has one lone electron (odd electron molecule) and one lone pair (or can be considered as 2 sigma bonds + 1 lone electron + 1 lone pair, leading to SN = 3 for $s{p}^2$ hybridization, or more accurately, the odd electron is delocalized). However, for simplicity, considering it as 2 sigma bonds and 1 lone electron, it's often described as $s{p}^2$ hybridized with the odd electron in a p-orbital or hybrid orbital. Let's re-evaluate using a more standard approach for $N{O}_2$. The central N atom forms two sigma bonds with O atoms and has one unpaired electron. The geometry is bent, and the hybridization is $s{p}^2$.
• ${\mathrm{NH}}_3$: Nitrogen is the central atom. It forms three sigma bonds with H atoms and has one lone pair. SN = 3 + 1 = 4. Hybridization is $s{p}^3$.
• Since ${\mathrm{NH}}_3$ has $s{p}^3$ hybridization, option A is incorrect.

Option B: ${\mathrm{BF}}_3$ and ${\mathrm{NO}}_2^{-}$

• ${\mathrm{BF}}_3$: Boron is the central atom. It forms three sigma bonds with F atoms and has no lone pairs. SN = 3 + 0 = 3. Hybridization is $s{p}^2$.
• ${\mathrm{NO}}_2^{-}$: Nitrogen is the central atom. It forms two sigma bonds with O atoms and has one lone pair. SN = 2 + 1 = 3. Hybridization is $s{p}^2$.
• Both molecules/ions have $s{p}^2$ hybridization. Therefore, option B is correct.

Option C: ${\mathrm{NH}}_2^{-}$ and ${\mathrm{H}}_2\mathrm{O}$

• ${\mathrm{NH}}_2^{-}$: Nitrogen is the central atom. It forms two sigma bonds with H atoms and has two lone pairs (N has 5 valence electrons + 1 for negative charge = 6 electrons; 2 used in bonds, 4 remaining as 2 lone pairs). SN = 2 + 2 = 4. Hybridization is $s{p}^3$.
• ${\mathrm{H}}_2\mathrm{O}$: Oxygen is the central atom. It forms two sigma bonds with H atoms and has two lone pairs. SN = 2 + 2 = 4. Hybridization is $s{p}^3$.
• Since both have $s{p}^3$ hybridization, option C is incorrect.

Option D: ${\mathrm{BF}}_3$ and ${\mathrm{NH}}_2^{-}$

• ${\mathrm{BF}}_3$: As calculated above, it has $s{p}^2$ hybridization.
• ${\mathrm{NH}}_2^{-}$: As calculated above, it has $s{p}^3$ hybridization.
• Since ${\mathrm{NH}}_2^{-}$ has $s{p}^3$ hybridization, option D is incorrect.

Correct Answer: (B)