## Chapter 12

## Q. 12.3.2

## Q. 12.3.2

Magnesium oxide crystallizes with the sodium chloride structure. If the edge length of the unit cell is determined to be 420 pm and the O^{2-} ion is assigned a radius of 126 pm, what is the radius of the Mg^{2+} ion?

## Step-by-Step

## Verified Solution

**You are asked** to calculate the radius of a cation in an ionic solid.

**You are given** the crystal structure of the solid, the edge length of the unit cell, and the ionic radius of the anion.

**Step 1.** Calculate the sum of the ionic radii from the fact that the edge length of the unit cell is twice the sum of the radii. The sum of the radii is the distance between the ion centers, which is a distance that can be determined from an x-ray diffraction experiment.

a = 2\left(r_{Mg^{2+}}+r_{O^{2-}}\right) = 420 pm

r_{Mg^{2+}}+r_{O^{2-}} = 210 pm

**Step 2.** Use the sum of the ionic radii to calculate the radius of the Mg^{2-} ion, taking the radius of the O^{2-} ion to be 126 pm and assuming that the ions are in contact.

r_{Mg^{2+}} = 210 pm – r_{O^{2-}} = 210 pm – 126 pm = 84 pm