# Comparing the Lattice energy of different compounds

I am confused about some concepts related to Lattice energy.

First of all,how to compare lattice energy between $$2$$ compounds belonging to different groups and periods?

1.Between $$\ce{AlF3}$$ and $$\ce{MgO}$$ (In $$\ce{AlF3}$$ product of ionic charge is $$3\times1$$ and in MgO it is $$2\times 2$$. So,according to me $$\ce{MgO}$$ should have higher lattice energy but the opposite is true.

2.Between $$\ce{KF}$$ and $$\ce{LiF}$$ (According to Fajan's rule $$\ce{KF}$$ is more ionic than $$\ce{LiF}$$ and lattice energy depends directly on ionic character).But, value of lattice energy of $$\ce{KF}$$ is $$\pu{808 kJ mol-1}$$ and $$\ce{LiF}$$ is $$\pu{1030 kJ mol-1}$$. How is this possible?

So,can you please explain me how to compare lattice energies of different compounds and which factor is more dominating size of cation and anion or charge on ions.

• I don't think this question is too broad - see answer.
– A.K.
Nov 26 '18 at 17:50

You might want to look at

https://scilearn.sydney.edu.au/fychemistry/prelab/e12.shtml

and the related page

https://scilearn.sydney.edu.au/fychemistry/calculators/lattice_energy.shtml

which has a nice lattice energy calculator that allows you to play with the parameters and see how the lattice energy varies, but I'll try to summarise those the argument below.

Here we are dealing with the ionic model - everything is totally ionic, there is total charge separation, all binding is electrostatic. Thus our system consists of a set of point charges. We can write the energy of such a system as

$$E=N_AM \frac{z_1 z_2 e^2}{4 \pi \epsilon_0 r}$$

where $$N_A$$ is the Avogardro constant, $$z_1$$ the charge on the first ion, $$z_2$$ the charge on the second, r is the closest separation of the ions and M the Madelung constant. Now note the Madelung constant only depends upon the structure of the crystal - all Sodium Chloride type crystals, whether it be NaCl, KF, or FeO, all use the same value off M. Looking at the formula we can see that FOR A GIVEN STRUCTURE (i.e. with the same M)

1. High charges on the ions mean high lattice energy
2. Small separation means high lattice energy

Thus point 2 addresses point 2 in your question, K is bigger than Li, hence the separation is bigger in KF than LiF, hence KF has a lower lattice energy than LiF. The first point explains why MgO has a higher lattice energy than NaF. However it doesn't cover your first example, as $$\ce{AlF3}$$ and MgO have different structures (and actually I don't think it is a very good question).

Comparing different structures is very difficult. About the best we can do is the electrostatic term from the Kapustinskii equation (https://en.wikipedia.org/wiki/Kapustinskii_equation) $$U_{L}={K}\cdot {\frac {\nu \cdot |z^{+}|\cdot |z^{-}|}{r^{+}+r^{-}}}$$ This is an approximation to the above. Here K is a constant which is INDEPENDENT of structure and $$\nu$$ is the number of ions in the empirical formula, 2 for MgO, 4 for $$\ce{AlF_3}$$. Thus assuming $${r^{+}+r^{-}}$$ is the same for both of these $$\ce{AlF3}$$ has a bigger lattice energy because $$\nu\cdot |z^{+}|\cdot |z^{-}|$$ is bigger - For $$\ce{AlF3}$$ it is 4*3*1=12, while for MgO it is 2*2*2=8. What this is really saying is that the $$\ce{AlF3}$$ structure has a much larger Madelung constant than the NaCl one, enough to overcome the charge differences, and thus it has a higher lattice energy.

in AlF3 in case of each AlF bond, charge of Al =3 => product of charges in case of each bond=3 =>sum of products of charges =3x3=9>4 so, AlF3 has lattice energy more than MgO.