# Regarding comparison of ionization energies [closed]

Why is the ionization energy of $$\ce{Mg^2+}$$ greater than the ionization energy of $$\ce{Ne}$$ (neon)?

My teacher said the answer was $$\ce{Mg^2+}$$ but I have no idea why as my general knowledge tells me that $$\ce{Ne}$$ has the greatest ionization energy since ionization energy increases across the group? Would someone explain why $$\ce{Mg^2+}$$ has a greater ionization energy?

• How many electrons have to be removed, and what will be the net charge when done? – DrMoishe Pippik Sep 14 '20 at 2:28

## 1 Answer

When you talk about the ionization energy of $$\ce{Mg^2+}$$, it is the third ionization of $$\ce{Mg}$$, which is equal to $$\pu{7734 kJ/mol}$$. When you talk about the ionization energy of $$\ce{Ne}$$, it is the first ionization of $$\ce{Ne}$$, which is $$\pu{2081 kJ/mol}$$. Thus, actually, $$IE_\ce{Mg^2+} \gt IE_\ce{Ne}$$.

Keep in mind that the first and second ionization energies of $$\ce{Mg}$$ are equal to $$\pu{738 kJ/mol}$$ and $$\pu{1451 kJ/mol}$$, respectively, both of which are smaller than the first ionization of $$\ce{Ne}$$ ($$\pu{2081 kJ/mol}$$).

Simplified reason: Both $$\ce{Mg^2+}$$ ion and $$\ce{Ne}$$ atom have the electronic configuration ($$\mathrm{1s^22s^22p^6}$$). Yet, $$\ce{Mg^2+}$$ has extra two protons in its nucleus than that of $$\ce{Ne}$$, thus attraction forces on electrons in $$\ce{Mg^2+}$$ are higher than that of $$\ce{Ne}$$ (e.g., $$F_{q_+q_-} = \dfrac{1}{4 \pi \epsilon_\circ} \cdot \dfrac{q_+q_-}{d^2}$$; also keep in mind that $$r_\ce{Mg^2+} \lt r_\ce{Ne}$$). By that reason alone, one can assume that the ionization energy of $$\ce{Mg^2+}$$ is higher than that of $$\ce{Ne}$$.

Note that your statement of "$$\ce{Ne}$$ has the greatest ionization energy" is incorrect. In that concept, the first ionization of $$\ce{He}$$ ($$\pu{2372 kJ/mol}$$) is greater than that of $$\ce{Ne}$$ ($$\pu{2081 kJ/mol}$$).