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(A) To determine the hybridization,we calculate the steric number using the formula: $	ext{Steric Number} = \frac{1}{2} [V + M - C + A]$,where $V$ is

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$BF_3$ and $NO_2^{-}$

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(A) To determine the hybridization,we calculate the steric number using the formula: $	ext{Steric Number} = \frac{1}{2} [V + M - C + A]$,where $V$ is the number of valence electrons,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1. BF_3$: Steric number = $\frac{1}{2} [3 + 3] = 3$ ($sp^2$ hybridization).$2. NO_2^{-}$: Steric number = $\frac{1}{2} [5 + 0 + 1] = 3$ ($sp^2$ hybridization).$3. NH_2^{-}$: Steric number = $\frac{1}{2} [5 + 2 + 1] = 4$ ($sp^3$ hybridization).$4. H_2O$: Steric number = $\frac{1}{2} [6 + 2] = 4$ ($sp^3$ hybridization).$5. NO_2$: The nitrogen atom has one unpaired electron,but for hybridization purposes,it is considered $sp^2$ due to the presence of one double bond and one lone electron.Thus,both $BF_3$ and $NO_2^{-}$ exhibit $sp^2$ hybridization.

In which one of the following pairs do the central atoms exhibit $sp^2$ hybridization?

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