I'm trying to understand what's behind the ionisation of atoms in a redox reaction. I apologise if the answer to this question is obvious. I've searched around, but, surprisingly, I haven't found a clear answer on this.

Many many years ago, in secondary school, I remember watching a demonstration of a redox reaction of magnesium and oxygen (Mg2(s) + O2(g) --> MgO).

This involved a piece of magnesium being placed over a flame, after which a rather intense light would be emitted as the magnesium reacted with the oxygen, leaving behind MgO.

My understanding of this reaction is as follows: two valence electrons break free from the Mg atom and become part of the O atom, resulting in two ions, both of which have a stable octet configuration (8 electrons in their valence energy level). The electrostatic forces of the two ions result in an ionic bond, MgO.

Assuming the above is correct:

  1. How did the two electrons break free from the Mg atom? What caused that ionisation? Was it the heat from the flame? I've read somewhere that the heat from the flame only speeds up the reaction, and that the reaction would take place even in the absence of the flame, but at a considerably slower rate. If this is true, then again, what's causing the ionisation? Is it the heat from the ambient environment? Is that heat enough to ionise an Mg atom?

  2. What forces are behind the two "escaped" electrons joining the O atom? The O atom is neutral, so there should be no electrostatic forces pulling in those two electrons. So what's behind the ionisation of the O atom?

  • 1
    $\begingroup$ They simply bind to each other, that's it. $\endgroup$
    – Mithoron
    Nov 6, 2023 at 1:37
  • $\begingroup$ Who says both electrons from one Mg end up on the same oxygen? $\endgroup$
    – Ian Bush
    Nov 6, 2023 at 7:46
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    $\begingroup$ The step-wise mechanism of the reaction might well depend on whether a flame is applied (what is the source and composition of this flame?). This question would largely be answered with an explanation of the location and mechanism of the reaction. $\endgroup$
    – Buck Thorn
    Nov 6, 2023 at 10:35
  • $\begingroup$ @Mithoron The ions bind to each other due to the electrostatic forces, but what I don't understand is how they become ions. Some amount of energy is required for the two electrons to break free from the Mg atom, right? Where does that energy come from? Is it the heat from the flame? And what forces attract the two electrons to the O atom? The O atom is electrically neutral, so what forces pull the electrons towards the O atom? $\endgroup$ Nov 6, 2023 at 12:31
  • 1
    $\begingroup$ I'm saying your "understanding" is thoroughly wrong - premise is false, and you should have looked for earlier questions with this annoying misconception. $\endgroup$
    – Mithoron
    Nov 6, 2023 at 12:38

1 Answer 1


The fundamental mistake you're making here is treating each atom in the reaction as though it's isolated. If you do that, the reaction will make no sense. You have to look at them together.

The energy to remove the electrons does come from heat. Yes, magnesium does oxidize at room temperature, but no, there is not enough heat at room temperature to ionize the magnesium atom. That is, room temperature does not provide enough energy to rip the electrons completely away from the magnesium atom and fling them out into space. We know this because magnesium does not spontaneously emit electrons at room temperature. (But if you heat metals to high enough temperatures, there will be enough thermal energy to start emitting electrons; that's how vacuum tubes work.)

When magnesium reacts with oxygen, the electrons do not get flung out into space. In fact, the electrons barely go anywhere. They are still right next door to the magnesium atom, attached to the oxygen. That requires a lot less energy than completely ripping them away; that's why the reaction can proceed at room temperature. In other words, it is the formation of the ionic bond that makes all of this possible.

As for oxygen, it's easy to rationalize how the first electron attaches. This is similar to how a statically charged object, like a balloon, can cling to a neutral object. Atoms are neutral, but they are polarizable. As an electron approaches a neutral oxygen atom, it repels the electron cloud slightly, exposing a little bit of juicy positive charge that is enough to make it attach. This is a relatively weak effect. Attaching an additional electron to oxygen gets you 141 kJ/mol. By contrast, removing an electron from neutral oxygen costs 1300 kJ/mol. But 141 kJ/mol is still substantial on a chemistry scale.

The second electron is harder to understand, because now there is an excess negative charge on the atom, and it's hard to see how you could attach another electron. In fact, it does not happen. The oxygen dianion is not stable in a vacuum; it will emit an electron. But once again, the reaction is not occurring in a vacuum. There is a positively charged magnesium atom right next door, which provides an attractive force that prevents the second electron from flying away. Once again the ionic bond is critical; the reaction doesn't work without it.

  • $\begingroup$ Thanks for this! So just to be clear, the heat at room temperature is what allows for the two valence electrons to break free from the Mg, but only when they're in close proximity to the O atom? So if we attempted the reaction in an environment with a much lower temperature, nothing would happen? $\endgroup$ Nov 6, 2023 at 13:25
  • $\begingroup$ @JohnO'brien Electrons don't ever "break free" in chemistry, unless it's plasma chemistry. Your atoms just bump into each other creating highly polarised bond that is called "ionic". That does not mean there are separate ions in a molecule, just that ionic component of bonding is greater than covalent. $\endgroup$
    – Mithoron
    Nov 6, 2023 at 15:04
  • $\begingroup$ Yes, reducing temperature slows reactions. The atoms can only bond if they collide with enough energy to overcome the activation energy. This becomes less likely at low temperatures. Eventually it will essentially stop. I don't know how cold you have to make magnesium before oxidation becomes negligible... probably pretty cold. $\endgroup$
    – anon
    Nov 6, 2023 at 15:39

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