# Why is the common magnesium ion Mg(II) and not Mg(I) when the second ionization energy is higher than the first ionization energy?

The first ionization energy for magnesium is given as $$\pu{737.7 kJ/mol}$$, and the second ionization energy is $$\pu{1450.7 kJ/mol}$$.

Given this information, doesn't it take $$\pu{737.7 kJ/mol}$$ to form $$\ce{Mg^1+}$$, while $$\pu{737.7kJ/mol} + \pu{1450.7kJ/mol} = \pu{2188.4 kJ/mol}$$ is required to form $$\ce{Mg^2+}$$?

Why is it then, that the $$\ce{Mg^2+}$$ ion is more common than the $$\ce{Mg^1+}$$ ion, while more energy is required to form the $$\ce{Mg^2+}$$ ion?

• That depends on what else is accepting electrons. Just because it is less favorable did not mean that the sum when you account for the oxidant is not overall favorable. Also, you're looking at gas phase ionization energy? In solution, it's a different game. – Zhe Apr 15 '17 at 15:24

As mentioned by Zhe, we have to look at the entire process by which an ionic compound is formed, not just the energy for a single part.

For example, consider the formation reaction of $$\ce{MgO}$$ $$\ce{Mg(s) +1/2O2(g)->MgO}$$ To find the enthalpy of this process, we could follow a Born-Haber cycle:

This demonstrates that while the ionization of $$\ce{Mg}$$ to $$\ce{Mg^{2+}}$$ might be unfavorable, the overall process to form the compound is not. So the ionization of $$\ce{Mg}$$ is driven by the favorability of the overall process.2

It is also worth noting why we do not instead form $$\ce{Mg2O}$$. If we were to construct a Born-Haber cycle for this compound, it would take less energy to form $$\ce{2Mg^{+}(g) +O^{2-}(g)}$$ than it would to form $$\ce{Mg^{2+} (g) + O^{2-}(g)}$$. However, the lattice energy of $$\ce{Mg2O}$$ would be sufficiently smaller than the lattice energy of $$\ce{MgO}$$ that $$\ce{MgO}$$ is still more favorable to form. Lattice energies are heavily dependent on the charge of the ions involved and having a $$+2$$ ion will lead to twice as large a lattice energy as an ion with a $$+1$$ charge all else being equal.

Even in this simple case, all else might not be equal. To make a more direct comparison between we can utilize the Born-Lande or Born-Mayer to obtain a calculated value of the lattice energy.

• Also, in solution, the hydration enthalpy of $\ce{Mg^2+}$ is sufficiently large to compensate for the 2nd IE. The same cannot be said for, for example, Na, which has a huge 2nd IE. – orthocresol Apr 15 '17 at 17:34
• @orthocresol based on this data table from the RSC, the enthalpy of hydration of $\ce{Mg^{2+}}$ would be insufficient to overcome the two ionization energies. – Tyberius Apr 15 '17 at 18:06
• Well, my statement was rather simplified, there are certainly more factors that go into a proper energy cycle (hydration of the counterion for example) however the larger hydration enthalpy of $\ce{Mg^2+}$ compared to that of $\ce{Mg+}$ was the factor I thought would tilt the balance in favour of Mg(II). – orthocresol Apr 15 '17 at 18:07
• @orthocresol I would agree. I just realized I sidestepped around the OPs actual question. – Tyberius Apr 15 '17 at 18:11