Energy Required to Convert Lithium to Divalent Lithium Ion

To determine the energy required to convert $3.5\text{ mg}$ of gaseous lithium ($\text{Li}$) into divalent gaseous lithium ions ($\text{Li}^{2+}$), we follow these steps:

Step 1: Calculate the total molar energy required for the transformation

The process of converting gaseous lithium atoms to divalent gaseous lithium ions involves two successive ionization steps:

1. $\text{Li}(\text{g}) \rightarrow \text{Li}^+(\text{g}) + e^-$ with first ionization enthalpy $IE_1 = 520\text{ kJ mol}^{-1}$
2. $\text{Li}^+(\text{g}) \rightarrow \text{Li}^{2+}(\text{g}) + e^-$ with second ionization enthalpy $IE_2 = 7297\text{ kJ mol}^{-1}$

The net reaction is: $\text{Li}(\text{g}) \rightarrow \text{Li}^{2+}(\text{g}) + 2e^-$

The total energy required per mole ($\Delta H$) is the sum of the first and second ionization enthalpies: $\Delta H = IE_1 + IE_2 = 520\text{ kJ mol}^{-1} + 7297\text{ kJ mol}^{-1} = 7817\text{ kJ mol}^{-1}$

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Step 2: Calculate the number of moles of lithium

The given mass of lithium is $m = 3.5\text{ mg} = 3.5 \times 10^{-3}\text{ g}$. Given the molar mass of lithium $M = 7\text{ g mol}^{-1}$, the number of moles ($n$) is: $n = \frac{m}{M} = \frac{3.5 \times 10^{-3}\text{ g}}{7\text{ g mol}^{-1}} = 0.5 \times 10^{-3}\text{ mol}$

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Step 3: Calculate the total energy required

The total energy required ($E$) to convert $3.5\text{ mg}$ of $\text{Li}(\text{g})$ to $\text{Li}^{2+}(\text{g})$ is: $E = n \times \Delta H = (0.5 \times 10^{-3}\text{ mol}) \times (7817\text{ kJ mol}^{-1})$ $E = 3.9085\text{ kJ}$

Rounding off to the nearest integer gives: $E \approx 4\text{ kJ}$

Final Answer: The energy required is 4 (rounded to the nearest integer).