Among the compounds,BF_3,NCl_3,H_2S,SF_4 and | vedclass

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(D) To determine the hybridization of the central atom,we use the formula: $H = \frac{1}{2}(V + M - C + A)$,where $V$ is the number of valence electro

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$NCl_3$ and $H_2S$

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(D) To determine the hybridization of the central atom,we use the formula: $H = \frac{1}{2}(V + M - C + A)$,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 $BF_3$: $H = \frac{1}{2}(3 + 3) = 3$ ($sp^2$ hybridization).$2$. For $NCl_3$: $H = \frac{1}{2}(5 + 3) = 4$ ($sp^3$ hybridization).$3$. For $H_2S$: $H = \frac{1}{2}(6 + 2) = 4$ ($sp^3$ hybridization).$4$. For $SF_4$: $H = \frac{1}{2}(6 + 4) = 5$ ($sp^3d$ hybridization).$5$. For $BeCl_2$: $H = \frac{1}{2}(2 + 2) = 2$ ($sp$ hybridization).Comparing the results,both $NCl_3$ and $H_2S$ have $sp^3$ hybridization. Therefore,the correct option is $(D)$.

Among the compounds,$BF_3$,$NCl_3$,$H_2S$,$SF_4$ and $BeCl_2$,identify the ones in which the central atom has the same type of hybridization.

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