Acidified potassium dichromate oxidizes sulphides (S²⁻) to ………………

The correct option/answer is Sulphur (S).

Acidified potassium dichromate (${\mathrm{K}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7$) is a powerful oxidizing agent. In an acidic medium, the dichromate ion (${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$) undergoes reduction to form chromium(III) ions (${\mathrm{Cr}}^{3+}$). The reduction half-reaction is given by:
$${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+14{\mathrm{H}}^{+}+6{\mathrm{e}}^{-}\to 2{\mathrm{Cr}}^{3+}+7{\mathrm{H}}_2\mathrm{O}$$

During this process, the dichromate oxidizes sulfide ions (${\mathrm{S}}^{2-}$) present in the solution by accepting electrons from them.

Sulfide ions are oxidized to elemental sulfur (S) by losing two electrons. The oxidation half-reaction is:
$${\mathrm{S}}^{2-}\to \mathrm{S}+2{\mathrm{e}}^{-}$$

To get the overall balanced chemical equation, we multiply the oxidation half-reaction by 3 and add it to the reduction half-reaction:
$${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}+3{\mathrm{S}}^{2-}+14{\mathrm{H}}^{+}\to 2{\mathrm{Cr}}^{3+}+3\mathrm{S}\downarrow +7{\mathrm{H}}_2\mathrm{O}$$

This reaction is visually characterized by a color change from orange (due to the presence of ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ ions) to green (due to the formation of ${\mathrm{Cr}}^{3+}$ ions) along with the precipitation of yellow colloidal sulfur (S).