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(B) 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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$NF_3, H_2O$

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(B) 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.For $NF_3$: $	ext{Steric Number} = \frac{1}{2} (5 + 3) = 4$,which corresponds to $sp^3$ hybridization.For $H_2O$: $	ext{Steric Number} = \frac{1}{2} (6 + 2) = 4$,which corresponds to $sp^3$ hybridization.Thus,both $NF_3$ and $H_2O$ exhibit $sp^3$ hybridization.

In which of the following pairs do both species have $sp^3$ hybridization?

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