[VISUAL] 6. The first ionization energy of an element is the energy required to remove an electron f — Atomic Structure Chemistry Question

STEPS:

1. Understand first ionization energy ($\text{IE}_1$) equations: The first ionization energy is defined as the energy required to remove the most loosely bound electron from a gaseous atom:
$$\text{X}(g) \rightarrow \text{X}^+(g) + e^- \quad \Delta H = \text{IE}_1 > 0$$
Because energy must be absorbed to overcome the electrostatic attraction between the nucleus and the electron, ionization is always an endothermic process ($\Delta H > 0$).
2. Deconstruct the given net reactions into individual steps: Every reaction in the options has the general form:
$$\text{Y}(g) + \text{Mg}^+(g) \rightarrow \text{Y}^+(g) + \text{Mg}(g)$$
This chemical change is a combination of two separate gas-phase processes:
* Step 1 (Ionization of Y): $\text{Y}(g) \rightarrow \text{Y}^+(g) + e^-$
$$\Delta H_1 = \text{IE}_1(\text{Y})$$
* Step 2 (Reduction of $\text{Mg}^+$): $\text{Mg}^+(g) + e^- \rightarrow \text{Mg}(g)$
This is the exact reverse of the first ionization of magnesium, meaning its enthalpy change is equal in magnitude but opposite in sign:
$$\Delta H_2 = -\text{IE}_1(\text{Mg})$$
3. Set up the expression for the overall enthalpy change ($\Delta H_{\text{rxn}}$):
By Hess's Law, adding these two steps together yields the overall reaction and allows us to sum their enthalpy changes:
$$\Delta H_{\text{rxn}} = \Delta H_1 + \Delta H_2 = \text{IE}_1(\text{Y}) - \text{IE}_1(\text{Mg})$$
4. Determine the condition required for an exothermic reaction:
For a reaction to be exothermic, the overall enthalpy change must be negative ($\Delta H_{\text{rxn}} < 0$):
$$\text{IE}_1(\text{Y}) - \text{IE}_1(\text{Mg}) < 0 \implies \text{IE}_1(\text{Y}) < \text{IE}_1(\text{Mg})$$
Therefore, the reaction is exothermic if and only if the first ionization energy of the reactant element $\text{Y}$ is less than the first ionization energy of magnesium.
5. Analyze the first ionization energy graph:
Locate magnesium ($\text{Mg}$, atomic number 12) on the provided graph:
* $\text{IE}_1(\text{Mg})$ is approximately $738\text{ kJ/mol}$.
* Now compare this value to the other elements listed as reactants in the choices:
* Chlorine ($\text{Cl}$, atomic number 17): $\text{IE}_1(\text{Cl}) \approx 1251\text{ kJ/mol}$
* Aluminum ($\text{Al}$, atomic number 13): $\text{IE}_1(\text{Al}) \approx 578\text{ kJ/mol}$
* Phosphorus ($\text{P}$, atomic number 15): $\text{IE}_1(\text{P}) \approx 1012\text{ kJ/mol}$
* Sulfur ($\text{S}$, atomic number 16): $\text{IE}_1(\text{S}) \approx 1000\text{ kJ/mol}$
6. Identify the exothermic reaction:
Since aluminum is the only element among the choices with a first ionization energy lower than magnesium ($578\text{ kJ/mol} < 738\text{ kJ/mol}$), the transfer of an electron from $\text{Al}$ to $\text{Mg}^+$ is thermodynamically favorable. This confirms Option B is the correct answer.

*

WHY_OTHERS_WRONG:

* Option A is incorrect: Because chlorine has a much higher first ionization energy than magnesium ($1251\text{ kJ/mol} > 738\text{ kJ/mol}$), removing an electron from $\text{Cl}$ to give it to $\text{Mg}^+$ is highly unfavorable and endothermic:
$$\Delta H_{\text{rxn}} \approx 1251 - 738 = +513\text{ kJ/mol}$$
* Option C is incorrect: Phosphorus has a higher first ionization energy than magnesium ($1012\text{ kJ/mol} > 738\text{ kJ/mol}$). This is due to phosphorus's stable, half-filled $3\text{p}$ subshell ($3\text{p}^3$) and greater effective nuclear charge, making this reaction endothermic:
$$\Delta H_{\text{rxn}} \approx 1012 - 738 = +274\text{ kJ/mol}$$
* Option D is incorrect: Sulfur has a higher first ionization energy than magnesium ($1000\text{ kJ/mol} > 738\text{ kJ/mol}$). Although sulfur's ionization energy is slightly lower than phosphorus's due to electron-electron repulsion in its paired $3\text{p}$ orbital, it is still higher than magnesium's, making the reaction endothermic:
$$\Delta H_{\text{rxn}} \approx 1000 - 738 = +262\text{ kJ/mol}$$