JEE Main 26 Feb 2021 Shift 2 Solved Paper

Calgon is used to remove permanent hardness of water by removing cation $\mathrm{Mg}^{2+}$. Calgon is not an element, it is a compound. The 2 nd most abundant element by weight in the Earth's crust is silicon (Si). So, the statement in option (a) is not true. Calgon is sodium hexametaphosphate $(\mathrm{NaPO}_3)_6$ or $\mathrm{Na}_2[\mathrm{Na}_4(\mathrm{PO}_3)_6]$. It is a polymeric water soluble compound. It is also named as Graham's salt. $\mathrm{Mg}^{2+}$ of hard water is removed as $\underset{\text{Calgon}}{\mathrm{Na}_2[\mathrm{Na}_4(\mathrm{PO}_4)_3]} + 2\mathrm{Mg}^{2+}(\mathrm{aq}) \longrightarrow \mathrm{Na}_2[\mathrm{Mg}_2(\mathrm{PO}_3)_6](\mathrm{s})$ $+4\mathrm{Na}^{+}(\mathrm{aq})$ $\mathrm{Ca}^{2+}$ ion cannot be removed by calgon as $\mathrm{Ca}^{2+}$ is larger in size than $\mathrm{Mg}^{2+}$. So, the statement of option (b), (c) and (d) are true.