JEE Main · 2020 · 09 Jan 2020 · Shift 2 · Classification of Elements and Periodicity

The large gap between the first ionization enthalpy, $496\text{kJ/mol}$, and the second, $4560\text{kJ/mol}$, shows that removing a second electron is prohibitively costly, the signature of a Group 1 metal that forms a stable ${\mathrm{M}}^{+}$ ion, whose hydroxide is $\mathrm{MOH}$. Neutralizing one mole of $\mathrm{MOH}$ needs one mole of $\mathrm{HCl}$, from $\mathrm{MOH}+\mathrm{HCl}\to \mathrm{MCl}+{\mathrm{H}}_2\mathrm{O}$, but only half a mole of the diprotic ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$, from $2MOH+{\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4\to {\mathrm{M}}_2\mathrm{S}{\mathrm{O}}_4+2{\mathrm{H}}_2\mathrm{O}$. The amounts are therefore $1$ mole of $\mathrm{HCl}$ and $0.5$ mole of ${\mathrm{H}}_2\mathrm{S}{\mathrm{O}}_4$.