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I was going through my chemistry textbook (Chemistry, 10th Ed. by Raymond Chang) when I encountered this explanation of lattice energy.

9.3 Lattice Energy of Ionic Compounds

We can predict which elements are likely to form ionic compounds based on ionization energy and electron affinity, but how do we evaluate the stability of on ionic compound? Ionization energy and electron affinity are defined for processes occurring in the gas phase, but at 1 atm and 25 °C all ionic compounds are solids. The solid state is a very different environment because each cation in a solid is surrounded by a specific number of anions, and vice versa. Thus, the overall stability of a solid ionic compound depends on the interactions of all these ions and not merely on the interaction of a single cation with a single anion. A quantitative measure of the stability of any ionic solid is its lattice energy, defined as the energy required to completely separate one mole of a solid ionic compound into gaseous ions (see Section 6.7).

In this text, it seems like the writer is implying that for an ion pair in the gaseous state, ionisation energy (IE) and electron affinity (EA) can be used to measure it's stability. But even an ion pair in the gaseous state is held by an electrostatic force that gives it more stability, so more energy will be needed to separate the ion pair and that energy would be more than IE and EA. What does the writer mean, and what is the correct definition of lattice energy?

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  • $\begingroup$ "Gas phase" is supposed to mean "zero contact, infinite distance between ions" in this context. $\endgroup$
    – Karl
    Aug 5, 2020 at 12:36
  • $\begingroup$ In other words, that sentence about the gas phase and 1 atm/25°C is utter bullshit. Can you please give the book title and author? $\endgroup$
    – Karl
    Aug 5, 2020 at 19:52
  • $\begingroup$ It begins by explaining that the relative EA and IE of two atoms can be used to estimate whether they form an ionic compound. The EA and IE are often combined to compute an electronegativity parameter, which is compared between atoms. In this case the question is about the nature of the bond between the atoms, not about the stability of the resulting solid. $\endgroup$
    – Buck Thorn
    Aug 6, 2020 at 9:27
  • $\begingroup$ @Karl so are you implying that the text is incoherent? $\endgroup$
    – Taofeek
    Aug 6, 2020 at 15:24
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    $\begingroup$ The last sentence is just as wrong. The lattice energy is the one needed to separate the ions to infinite distance. That´s not the same as vaporising them. The difference won´t be large, but Mr. Chang is still lying to students for no good reason. Perhaps he thought they were too stupid to grasp the idea of a theoretical value, or was trying to be practical, whatever. Schoolbook authors often do that, and its also disgusting there, but a university professor? $\endgroup$
    – Karl
    Aug 6, 2020 at 15:57

2 Answers 2

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In summary: IE and EA are just partial terms in the formation enthalpy of solid ionic compounds, that includes also LE and atomization enthalpy of elements. Their values alone can give a hint, but do not provide the final word.


Notice that the text says nothing explicit about ion pairs.

IE + EA relate to free gaseous unbound ions. Like:

$$\ce{M(g) -> M+(g) + e-(g)}\tag{IE}$$

resp.

$$\ce{X(g) + e-(g) -> X-(g)}\tag{EA}$$

The lattice energy (LE) is the energy released by forming solid ionic compounds from free ions:

$$\ce{ M+(g) + X-(g) -> MX(s)}\tag{LE}$$

Gaseous ion pairs have already released a part of the lattice energy.

Note that for ion pairs, the pairing energy may be smaller than the sum of IE+EA, especially for big ions, so forming pairs may not be favoured.

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A picture is worth a thousand words.

enter image description here

In the picture above, start with a mole of solid sodium and half a mole of Cl$_2$ gas, in the ground state. Then gasify the sodium with 107.3 kJ of heat, then ionize it with 496 kJ of energy, then dissociate Cl$_2$ to Cl radicals with 121.7 kJ, then allow the electron from the sodium to combine with the chlorine radical. You get heat released: 384.8 kJ.

The picture shows only one NaCl pair, so understanding how this energy is "released", and what it does to the system becomes a little mysterious, but if you realize that 6 x 10$^{23}$ ion pairs are all doing this at the same time, it's a bit easier. This is called the Born-Haber cycle.

One ion-pair doesn't have a lattice energy, but the mole of NaCl does, because one ion-pair doesn't have an extended lattice. When you add everything up, the lattice energy (the blue arrow on the right, pointing downward) comes out to -787.8 kJ/mol. http://derekcarrsavvy-chemist.blogspot.com/2016/12/lattice-energy-1-calculating-lattice.html

The calculation suggests that NaCl is more stable, energetically, than a mixture of solid sodium and chlorine gas, which is in agreement with observation. Also less reactive.

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