🤓 Based on our data, we think this question is relevant for Professor Saitta's class at UCF.

The lattice energy can be estimated using:

$\overline{){\mathbf{lattice}}{\mathbf{}}{\mathbf{energy}}{\mathbf{=}}\frac{\left|\mathbf{cation}\mathbf{}\mathbf{charge}\mathbf{\times}\mathbf{anion}\mathbf{}\mathbf{charge}\right|}{\mathbf{cation}\mathbf{}\mathbf{period}\mathbf{}\mathbf{\#}\mathbf{}\mathbf{+}\mathbf{}\mathbf{anion}\mathbf{}\mathbf{period}\mathbf{}\mathbf{\#}}}$

**For KCl:**

K^{+} is from Period 4 with a charge of +1 while Cl^{–} is from Period 3 with a charge of –1.

The lattice energy of KCl is:

$\mathbf{lattice}\mathbf{}\mathbf{energy}\mathbf{=}\frac{\left|\left(\mathbf{+}\mathbf{1}\right)\mathbf{\times}\left(\mathbf{-}\mathbf{1}\right)\right|}{\mathbf{4}\mathbf{+}\mathbf{3}}$** = 0.143**

**For RbBr:**

Rb^{+} is from Period 5 with a charge of +1 while Br^{–} is from Period 4 with a charge of –1.

The lattice energy of RbBr is:

Arrange these compounds in order of increasing magnitude of lattice energy: KCl, SrO, RbBr, CaO.