4
$\begingroup$

Given two ionic compounds, for instance:

$$\ce{CaS} \quad \mathrm{or} \quad \ce{KCl}$$

What is the procedure to predict which of the two have the highest lattice energy (in absolute value)?

(Suppose that I do not know the values of radii of the atoms but I can only look at the periodic table).


I know that I must look at the ratio

$$\frac{Z_1 Z_2}{r_1 + r_2} \propto E_\text{lattice},$$

where $r_1 + r_2$ is the distance between the two atoms in each molecule of the compound, $Z_1 Z_2$ is the product of atomic numbers of each element in the molecule.

But, for example, in the case above I have both $r_1 + r_2$ and $Z_1 Z_2$ bigger for $\ce{KCl}$ then $\ce{CaS}$, so I cannot say (without knowing the numerical values) if one of the two ratios would be higher.

Am I missing something?

$\endgroup$
  • 1
    $\begingroup$ The lattice energy is proportional to the charges of the atoms in the formula, not their atomic number! (There are no molecules in such simple ionic substances, btw.) $\endgroup$ – Karl Jul 24 '17 at 0:48
1
$\begingroup$

The Lattice Energy of Ionic Compounds is directly proportion to the Charge Density. As Ca, S have 2 units of charge while K, Cl have one and there is not much of a difference in size ( atleast not enough to overcome the doubling of the charge), the charge density of Ca and S ions would be more. Consequently, the lattice strength of CaS would be more.

$\endgroup$
  • $\begingroup$ So it depends mainly on how many electrons are exchanged! Thanks a lot! But, for istance, if I take $MgO$ and $MgS$, there are two electron exchanged in both, nevertheless the same problem showed in question remains, so is there a "secondary" criterium to select one of the two? $\endgroup$ – Sørën Jul 24 '17 at 8:38
  • 1
    $\begingroup$ @Sørën As the no. of e- exchanged are the same, we now have to focus on the size of the receiving atom (which will give us an idea about the strength of the electrostatic forces). $\endgroup$ – Ayushmaan Jul 24 '17 at 8:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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