$ \mathrm{S}+\text { Conc. } \mathrm{H}_2 \mathrm{SO}_4 \longrightarrow X+Y $ Here $X$ is a gas and $Y$ is a…
$
\mathrm{S}+\text { Conc. } \mathrm{H}_2 \mathrm{SO}_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.
2, 1
1, 0
1, 2
2, 2
Solution
$\mathrm{S}+2 \mathrm{H}_2 \mathrm{SO}_4$ (conc.) $\longrightarrow 3 \mathrm{SO}_2+2 \mathrm{H}_2 \mathrm{O}$
(X) (Y)
As $\mathrm{SO}_2$ has one lone pair of electrons on central atom i.e., sulphur.
And $\mathrm{H}_2 \mathrm{O}$ has two lone pair of electrons on the central atom i.e., oxygen.
