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

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$CO_2$

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(D) To determine the hybridization,we calculate the steric number $(SN)$ using the formula: $SN = \frac{1}{2} (V + M - C + A)$,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.$1$. For $SO_4^{2-}$: $SN = \frac{1}{2} (6 + 0 - 0 + 2) = 4$,which corresponds to $sp^3$ hybridization.$2$. For $CH_4$: $SN = \frac{1}{2} (4 + 4 - 0 + 0) = 4$,which corresponds to $sp^3$ hybridization.$3$. For $H_2O$: $SN = \frac{1}{2} (6 + 2 - 0 + 0) = 4$,which corresponds to $sp^3$ hybridization.$4$. For $CO_2$: $SN = \frac{1}{2} (4 + 0 - 0 + 0) = 2$,which corresponds to $sp$ hybridization.Therefore,$CO_2$ does not contain $sp^3$ hybridization.

In the following,which species does not contain $sp^3$ hybridization?

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