Calculation of Ionic Radii

Calculation of Ionic Radii

The edge length of the unit cell of LiCl (NaCl-like structure, FCC) is 0.514 nm or 5.14 Å. Assuming that the lithium ion is small enough so that the chloride ions are in contact, as in the figure below, calculate the ionic radius for the chloride ion.

Ionic compounds with anions that are much larger than cations, such as NaCl, usually form an FCC structure. They can be described by FCC unit cells with cations in the octahedral holes.

Note: The length unit angstrom, Å, is often used to represent atomic-scale dimensions and is equivalent to 10⁻¹⁰ m.

Solution

On the face of a LiCl unit cell, chloride ions contact each other across the diagonal of the face:

Drawing a right triangle on the face of the unit cell, we see that the length of the diagonal is equal to four chloride radii (one radius from each corner chloride and one diameter—which equals two radii—from the chloride ion in the center of the face), so d = 4r. From the Pythagorean theorem, we have:

\({a}^{2}+{a}^{2}={d}^{2}\)

which yields:

\({\left(0.514\phantom{\rule{0.2em}{0ex}}\text{nm}\right)}^{2}+{\left(0.514\phantom{\rule{0.2em}{0ex}}\text{nm}\right)}^{2}={\left(4r\right)}^{2}=16{r}^{2}\)

Solving this gives:

\(r=\phantom{\rule{0.2em}{0ex}}\sqrt{\cfrac{{\left(0.514\phantom{\rule{0.2em}{0ex}}\text{nm}\right)}^{2}+{\left(0.514\phantom{\rule{0.2em}{0ex}}\text{nm}\right)}^{2}}{16}}\phantom{\rule{0.2em}{0ex}}=\text{0.182 nm}\phantom{\rule{0.2em}{0ex}}\left(\text{1.82 Å}\right)\phantom{\rule{0.2em}{0ex}}{\text{for a Cl}}^{\text{−}}\phantom{\rule{0.2em}{0ex}}\text{radius}.\)