Chemistry - Complete P-block Question with Solution | TestHub

Yellow non-metal insoluble non-metal $[\mathrm{X}]$ is Sulfur.

On heating sulfur with concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$, a colourless, pungent gas $[\mathrm{A}]$ is formed, which is sulfur dioxide.

$\mathrm{S}+2{\mathrm{H}}_2{\mathrm{SO}}_4\to 3{\mathrm{SO}}_2+2{\mathrm{H}}_2\mathrm{O}$ Reaction of ${\mathrm{SO}}_2$ with ${\mathrm{Cl}}_2$ and water produces Sulfuric acid, a strong dibasic oxoacid. Hence $[\mathrm{B}]={\mathrm{H}}_2{\mathrm{SO}}_4$.

On boiling sulfur with aqueous ${\mathrm{Na}}_2{\mathrm{SO}}_3$,

Sodium thiosulfate is formed, so $[\mathrm{C}]={\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3$. ${\mathrm{Na}}_2{\mathrm{S}}_2{\mathrm{O}}_3+2\mathrm{HCl}\to \mathrm{S}\downarrow +{\mathrm{SO}}_2\uparrow +2\mathrm{NaCl}+{\mathrm{H}}_2\mathrm{O}$

Oxidation of ${\mathrm{SO}}_2$ in presence of ${\mathrm{V}}_2{\mathrm{O}}_5$ gives Sulfur trioxide, so $[\mathrm{D}]={\mathrm{SO}}_3$. ${\mathrm{SO}}_2+\frac{1}{2}{\mathrm{O}}_2\stackrel{{\mathrm{V}}_2{\mathrm{O}}_5}{\to }{\mathrm{SO}}_3$

Finally, ${\mathrm{SO}}_3$ reacts with concentrated ${\mathrm{H}}_2{\mathrm{SO}}_4$ to form Oleum (pyrosulfuric acid), hence $[\mathrm{E}]={\mathrm{H}}_2{\mathrm{S}}_2{\mathrm{O}}_7$.