Which molecule has sp^3 hybridization of the | vedclass

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

Vedclass · Accepted Answer

$PCl_3$

Vedclass · Answer

(A) To determine the hybridization,we use 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.$A$. For $PCl_3$: $	ext{Steric Number} = \frac{1}{2} [5 + 3] = 4$. This corresponds to $sp^3$ hybridization.$B$. For $SO_3$: $	ext{Steric Number} = \frac{1}{2} [6 + 0] = 3$. This corresponds to $sp^2$ hybridization.$C$. For $BF_3$: $	ext{Steric Number} = \frac{1}{2} [3 + 3] = 3$. This corresponds to $sp^2$ hybridization.$D$. For $NO_3^-$: $	ext{Steric Number} = \frac{1}{2} [5 + 0 - 0 + 1] = 3$. This corresponds to $sp^2$ hybridization.Therefore,$PCl_3$ is the correct answer.

Which molecule has $sp^3$ hybridization of the central atom?

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