S + Conc. H_2SO_4 longrightarrow X + YHere X | vedclass

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(C) The chemical reaction is:$S + 2H_2SO_4 	ext{ (conc.)} \longrightarrow 3SO_2 + 2H_2O$Here,$X = SO_2$ (gas) and $Y = H_2O$ (liquid).Both $SO_2$ and

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$1, 2$

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(C) The chemical reaction is:$S + 2H_2SO_4 	ext{ (conc.)} \longrightarrow 3SO_2 + 2H_2O$Here,$X = SO_2$ (gas) and $Y = H_2O$ (liquid).Both $SO_2$ and $H_2O$ are triatomic molecules.In $SO_2$,the central atom is sulphur $(S)$. Sulphur has $6$ valence electrons. It forms two double bonds with two oxygen atoms,using $4$ electrons. Thus,it has $6 - 4 = 2$ electrons remaining,which form $1$ lone pair.In $H_2O$,the central atom is oxygen $(O)$. Oxygen has $6$ valence electrons. It forms two single bonds with two hydrogen atoms,using $2$ electrons. Thus,it has $6 - 2 = 4$ electrons remaining,which form $2$ lone pairs.Therefore,the number of lone pairs on the central atoms of $X$ and $Y$ are $1$ and $2$ respectively.

$S +$ Conc. $H_2SO_4 \longrightarrow X + Y$
Here $X$ is a gas and $Y$ is a liquid and both are triatomic molecules. The number of electron lone pairs present on the central atoms of $X$ and $Y$ are respectively.

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$S +$ Conc. $H_2SO_4 \longrightarrow X + Y$Here $X$ is a gas and $Y$ is a liquid and both are triatomic molecules. The number of electron lone pairs present on the central atoms of $X$ and $Y$ are respectively.