JEE Main 2023 — General Principles and Processes of Isolation of Metals Question with Solution

For line II $∆\mathrm{S}$ is more negative than line I. In the Ellingham diagram, a sharp change in slope of the line is observed for $\mathrm{Mg}-\mathrm{MgO}$ at $~{1120}^{\mathrm{o}}\mathrm{C}$ because that is the boiling point of magnesium.

From diagram, $$\begin{gathered} ∆{\mathrm{S}}_{\mathrm{I}}=\left({\mathrm{S}}_{\mathrm{solid}}\right)-({\mathrm{S}}_{\mathrm{solid}}+{\mathrm{S}}_{\mathrm{gas}}) \\ ∆{\mathrm{S}}_{\mathrm{II}}=\left({\mathrm{S}}_{\mathrm{liquid}}\right)-({\mathrm{S}}_{\mathrm{solid}}+{\mathrm{S}}_{\mathrm{gas}}) \end{gathered}$$

There is a large increase in entropy associated with the change of state of magnesium.

So, both Assertion (A) and Reason (R) are true and (R) is the correct explanation of (A).