I do not quite understand the following line of reasoning:

Why does Na not form $\ce{Na^{2+}}$ ions? To obtain $\ce{Na^{2+}}$ ions, the first and second ionization energy must be applied: 496 + 4563 = 5059 kJ/mol. The lattice energy of an imaginary "$\ce{NaCl2}$" would have to be in the same order of magnitude as for $\ce{MgCl2}$ (-2525 kJ/mol).

So far everything is understandable...

This amount would not be sufficient for ionization, the removal of an electron from an ion with noble gas configuration costs too much energy.

Why can magnesium form $\ce{Mg2+}$ ions? The magnesium atom reaches the noble gas configuration when two electrons are donated. The sum of the first and second ionization energies for magnesium is considerably lower than for sodium: 738 + 1450 = 2188 kJ/mol. The lattice energy of -2525 kJ/mol is high enough to provide the energy needed for ionization.

I don't understand why the lattice energy released has to be sufficient for ionization (at all why any ionization should even take place here). I wonder in general where the connection is.

One is the heat release when positive and negative ions in the gas state are combined to form a crystal and the other is just the energy to knock out electrons. Why does the lattice energy have to be high enough now to apply the first, second etc. ionization energy?

(In this section and before in the book the entropy or Gibbs energy was not introduced yet, the author himself says: "The correct quantity to be considered here is the free enthalpy instead of the enthalpy. However, we do not commit a big mistake if we calculate with enthalpy in this case".)

That's the information I'm taking out:

$$\ce{Na -> Na^{2+} + 2e-} ~~~ \Delta H = \pu{5059 kJ/mol} \\ \ce{Na^{2+}(g) +2Cl-(g) -> NaCl2 } ~~~ \Delta H \approx \pu{-2525 kJ/mol}$$

Then it is said that the amount of the lattice energy is not sufficient for the ionization of Na. This is true, after all 2525 kJ < 5059 kJ and now? Why should the lattice energy be used for this? I don't know what the author wants to make clear now, the jump to the original question why Na does not form $\ce{Na^{2+}}$ ions does not work for me. Why should the lattice energy even be responsible for the ionization of Na?

  • $\begingroup$ The enthalpies for the two steps, ionization and lattice formation, must add up to a sizable negative number for the atoms to associate from the gas phase and overcome an entropic penalty. $\endgroup$
    – Buck Thorn
    Nov 11, 2022 at 9:14
  • $\begingroup$ Have you read about Hess's law and the Born-Haber cycle? en.wikipedia.org/wiki/Born%E2%80%93Haber_cycle $\endgroup$
    – Buck Thorn
    Nov 11, 2022 at 9:16
  • $\begingroup$ Convenient reference for text/formula formatting: Notation basics / Formatting of math/chem expressions / upright vs italic // For more: Math SE MathJax tutorial. // Not to be applied in CH SE titles. $\endgroup$
    – Poutnik
    Nov 11, 2022 at 9:29
  • $\begingroup$ Note there was an error in second equation (Cl should be Cl-) and state of matter is not specified in the first (presumably all are gases). $\endgroup$
    – Buck Thorn
    Nov 11, 2022 at 10:00
  • $\begingroup$ It might help if you tell us what question you think is trying to be answered by the text you quote . Also where is it from? $\endgroup$
    – Ian Bush
    Nov 11, 2022 at 12:07

1 Answer 1


The author is trying to explain why $\ce{NaCl2}$ doesn't form given that $\ce{MgCl2}$ does, where Mg is just one electron away from Na on the periodic table. The reaction being considered is $$\ce{X(g) + Cl2(g) -> XCl2(s)} \tag{1}$$

Thermodynamic arguments (having to do with the stability in terms of energy of products relative to reagents) or kinetic arguments(having to do with energetic barriers to reaching the products from the reagents) can be used to explain why compounds don't form as expected.

Here the author uses a thermodynamic explanation. Given the apparent similarity of $\ce{NaCl2}$ and $\ce{MgCl2}$ the failure to form $\ce{NaCl2}$ is traced to the energetic cost of forming it.

And why is $\ce{Na^{2+}}$ so costly to form? Because sodium has a large second ionization energy.

How do you do energy accounting to show this? You use Hess's law, which allows splitting a reaction into smaller steps through intermediates. Since enthalpy H is a state function, the path that takes you from reagents to products doesn't matter. To obtain the change in enthalpy for the overall process just add the changes for the individual steps.

In this case the steps are

$$\ce{X(g) + Cl2(g) -> X(g) + 2Cl(g)}\\\ce{X(g) + 2Cl(g) -> X^{2+}(g) + 2Cl-(g)}\\\ce{X^{2+}(g) + 2Cl-(g) -> XCl2(s)}$$

The first step is the atomization of $\ce{Cl2(g)}$, the second the ionization of the atoms in the gas phase, the third the formation of the lattice. The total enthalpy change for reaction (1) at top is just the sum of the enthalpies for these individual steps.

The enthalpies for the steps, atomization, ionization and lattice formation, must add up to a sizable negative number for the atoms to associate from the gas phase and overcome an entropic penalty.

There is some subtlety to this. An ionization step is introduced as a step on the way from gas phase atoms to the solid as a way to explain the stability of ionic solids and an additional justification for the proposed structure of such solids as composed of lattices of ions. You might have skipped that step and gone from the gas phase atoms to the lattice, without incorporating an ionization step, but that would not have provided much insight into the fundamental question, why does sodium behave differently?

Born-Haber cycles are a method of accounting (a "book-keeping" tool) but also have explanatory power, in that like an accounting sheet they can point out where there might be important steps responsible for differences between processes. Ionization does occur, and a lattice does form, but the events might not necessarily occur in the order suggested by a Born-Haber cycle. These events may occur simultaneously, for instance. In that case, the energy for the simultaneous event is the sum of the energies for the step-wise events.

There is also the issue of assuming that the lattice enthalpy for formation of the hypothetical $\ce{NaCl2}$ and $\ce{MgCl2}$ are equivalent. You have no way of proving this without further experiments (or computational chemistry), but it is consistent with the fact that in practice $\ce{NaCl2}$ crystals are not observed.

In practice the formation of salts by mixing gaseous ions would be difficult to carry out (although clusters might be studied this way). So this is primarily a conceptual problem that demonstrates the importance of ionization energies in determining the relative stability of ionic compounds, and it also illustrates the usefulness of Hess's law.

It might just be possible to make $\ce{NaCl2}$ under the right circumstances, just not at room temperature and pressure (or maybe not?).

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – andselisk
    Nov 13, 2022 at 13:37
  • $\begingroup$ NaCl2 would be unstable to NaCl + Cl $\endgroup$
    – jimchmst
    Nov 14, 2022 at 22:31
  • $\begingroup$ @jimchmst Yes. The question is how this can be explained in terms of the 2nd IE of sodium. $\endgroup$
    – Buck Thorn
    Nov 15, 2022 at 8:10
  • $\begingroup$ I suppose deltaH, deltaS, delta G, Keq could all be calculated, and one could determine if there were enough Na and Cl on Earth for a molecule to exist at equilibrium. The answer to the ultimate question [last sentence in the post]is that the lattice energy has nothing to do with the ionization energy. It's responsible for what happens when the positive and negative ions are brought together to form a compound. An Na++ ion and electron at infinity are perfectly stable as long as nothing comes close to the Na++. Think of Na++ as an almost alpha particle. Born-Haber cycle is not a mechanism. $\endgroup$
    – jimchmst
    Nov 20, 2022 at 3:54
  • $\begingroup$ @jimchmst Yes, I added more to emphasize that the Born-Haber method is a book-keeping tool. $\endgroup$
    – Buck Thorn
    Nov 20, 2022 at 10:41

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