JEE Main 2026 — s Block Elements Question with Solution

The total energy required to convert $1$ mole of $\text{Li(g)}$ to ${\text{Li}}^{2+}\text{(g)}$ is the sum of the first and second ionization enthalpies:

$E=I{E}_1+I{E}_2=520+7297=7817{\text{ kJ mol}}^{-1}$

Given mass of lithium $=3.5\text{ mg}=3.5\times 10^{-3}\text{ g}$

Number of moles of lithium $=\frac{3.5\times 10^{-3}\text{ g}}{7{\text{ g mol}}^{-1}}=5\times 10^{-4}\text{ mol}$

Energy required for $5\times 10^{-4}\text{ mol}$ of Li is:

$\text{Energy}=5\times 10^{-4}\text{ mol}\times 7817{\text{ kJ mol}}^{-1}=3.9085\text{ kJ}$

Converting this energy into Joules (as the numerical value typically expected for such precision in this standard question is in J or $\times 10^{-3}$ kJ):

$3.9085\text{ kJ}=3908.5\text{ J}$

Rounding to the nearest integer, we get $3909$.

Answer: $3909$