-1
$\begingroup$

My chemistry lecturer taught us that lattice energy is the energy required to break an ionic bond between two oppositely charged ions in a crystal lattice.

He then told us that lattice energy is also the energy released when separate oppositely charged ions come together to form a bond. He told us that the ions "must" be in gaseous state to combine.

I thought that the reason was that there is no way for the ions to combine other than in the gaseous state. But the reason he gave appeared quite different than mine. I couldn't grasp the reason he told us.

Can you please help in making me understand the reason properly?

$\endgroup$
  • 1
    $\begingroup$ Ions may combine in pretty much any state; if you have the reason to think otherwise, you may just as well throw that reason away as nonsense. Gaseous state, though, surely gets mentioned in the definition of lattice energy, and that for a good reason. $\endgroup$ – Ivan Neretin May 28 '19 at 17:06
  • $\begingroup$ Can you share the reason Ivan? $\endgroup$ – NightKruger May 28 '19 at 17:25
  • 2
    $\begingroup$ Well, it's simple: you need to measure the energy released when the bonds form, hence you start from the state with no bonds and go to the state with bonds. The latter is the crystal; now what about the former? $\endgroup$ – Ivan Neretin May 28 '19 at 17:34
  • 1
    $\begingroup$ "Easier" is not a concern at all. The process is imaginary anyway. $\endgroup$ – Ivan Neretin May 28 '19 at 17:51
  • 2
    $\begingroup$ You should read about the Madelung constant and the Born–Haber cycle. $\endgroup$ – MaxW May 28 '19 at 17:54
1
$\begingroup$

The actual, "real" state of the atoms (and then ions) that combine to form an ionic compound is not particularly significant to the question at hand.

When we want to calculate the lattice energy of an ionic compound, one method used is a Born-Haber Cycle. In essence, we use Hess' Law to combine a series of elementary steps that results in the overall transformation we are interested in. The enthalpies of each of these steps can be combined and used alongside the enthalpy of the overall reaction to determine the lattice energy.

Let's take the formation of sodium chloride from its elements as an example. The overall reaction is:

$\ce{Na(s) + 1/2Cl2(g) -> NaCl(s)}$ $\Delta H_f = -411 kJ/mol$ .[ 1 ]

Remember that it's OK to use fractions in chemical equations when we are doing energy calculations.

In many texts, the lattice energy is defined as the energy released when gas-phase ions are combined to form a solid compound. $^{[2]}$

One of the reasons that we have to have gas-phase ions is that the thermodynamic data for things ionization energy and electron affinity are for gas-phase particles. This also means that I have to make sure that all the particles in the systems are transformed into gas-phase atoms which can then be ionized.

The complete cycle for the formation of sodium chloride would be:

$\ce{Na(s) ->Na(g)\quad \Delta H_{sublimation}}$

$\ce{1/2Cl_2(g) -> Cl(g) \quad \Delta H_{bond \quad energy}}$

$\ce{Na(g) -> Na^+(g) + e^- \quad \Delta H_{ionization}}$

$\ce{Cl(g) + e^- -> Cl^-(g) \quad \Delta H_{electron \quad affinity}}$

$\ce{Na^+(g) + Cl^-(g) -> NaCl(s) \quad \Delta H_{lattice \quad energy}}$

To sum this up; we can think of the lattice energy as holding ions to each other. When the ions come together to form the compound, this energy is released. If we want to separate the ions (for example, if we want to vapourize the compound) we need to add the same (slightly more really) amount of energy. The requirement that the ions be in the gas phase comes from the mathematical model that we use.


Footnotes

[2]:Atkins; et al. (2010). Shriver and Atkins' Inorganic Chemistry (Fifth ed.). New York: W. H. Freeman and Company. ISBN 978-1-4292-1820-7

$\endgroup$
  • $\begingroup$ Can you explain what you mean by the thermodynamic Data for things ionisation energy and electron affinity? $\endgroup$ – NightKruger May 29 '19 at 1:49

Not the answer you're looking for? Browse other questions tagged or ask your own question.