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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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$BeF_3^-$

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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 of the central atom,$M$ is the number of monovalent atoms,$C$ is the cationic charge,and $A$ is the anionic charge.$1$. For $BeF_3^-$: $	ext{Steric Number} = \frac{1}{2} [2 + 3 - 0 + 1] = 3$. This corresponds to $sp^2$ hybridization.$2$. For $OH_3^+$: $	ext{Steric Number} = \frac{1}{2} [6 + 3 - 1 + 0] = 4$. This corresponds to $sp^3$ hybridization.$3$. For $NH_2^-$: $	ext{Steric Number} = \frac{1}{2} [5 + 2 - 0 + 1] = 4$. This corresponds to $sp^3$ hybridization.$4$. For $NF_3$: $	ext{Steric Number} = \frac{1}{2} [5 + 3 - 0 + 0] = 4$. This corresponds to $sp^3$ hybridization.Therefore,the central atom in $BeF_3^-$ does not use $sp^3$ hybrid orbitals.

In which of the following does the central atom $NOT$ use $sp^3$ hybrid orbitals in its bonding?

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