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

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$sp$,$sp^3d$,$sp^3d^2$

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(C) To determine the hybridization of the central atom,we use 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 $NO_2^+$:Central atom $N$ $(V=5)$,$M=0$,$C=1$,$A=0$.$	ext{Steric Number} = \frac{1}{2} [5 + 0 - 1 + 0] = 2$. Hybridization is $sp$.$2$. For $SF_4$:Central atom $S$ $(V=6)$,$M=4$,$C=0$,$A=0$.$	ext{Steric Number} = \frac{1}{2} [6 + 4 - 0 + 0] = 5$. Hybridization is $sp^3d$.$3$. For $PF_6^-$:Central atom $P$ $(V=5)$,$M=6$,$C=0$,$A=1$.$	ext{Steric Number} = \frac{1}{2} [5 + 6 - 0 + 1] = 6$. Hybridization is $sp^3d^2$.Thus,the correct hybridization modes are $sp$,$sp^3d$,and $sp^3d^2$.

What is the correct mode of hybridization of the central atom in the following compounds: $NO_2^+$,$SF_4$,$PF_6^-$?

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