First and second ionization enthalpies of lithium are 520 kJ mol ^...

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:

$E = IE_1 + IE_2 = 520 + 7297 = 7817 \text{ kJ mol}^{-1}$

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

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

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

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

Rounding to the nearest integer, we get $4$ KJ.

Answer: $4$