Magnesium has an HCP crystal structure, a \(c / a\) ratio of \(1.624,\) and a density of \(1.74 \mathrm{g} / \mathrm{cm}^{3}\) Compute the atomic radius for \(\mathrm{Mg}.\)

Hexagonal Close Packed (HCP) structure is a type of crystal lattice structure with layers of atoms packed closely together in repeating hexagonal patterns. In an HCP structure, the ratio of the height \(c\) to the side length \(a\) of the hexagonal cell is \(1.624\). Each unit cell in the HCP structure has 6 atoms.

We have the following information given in the exercise: - \(c/a = 1.624\) - Density of Magnesium (\(\rho\)) = 1.74 g/cm³ - Molar mass of Magnesium = 24.305 g/mol - Avogadro's Number (N_A) = \(6.022 \times 10^{23}\) atoms/mol

The formula for the volume of a unit cell in an HCP structure is given by: $$V = \frac{3\sqrt{2}}{2} a^2c$$ Given that the ratio \(c/a = 1.624\), we can express \(c\) as \(c = 1.624a\). Plugging this into the volume formula, we get: $$V = \frac{3\sqrt{2}}{2} a^2(1.624a) = 2.489a^3$$

In an HCP unit cell, there are 6 atoms. Therefore, the number of atoms in Mg crystal structure is 6.

Using the value of the volume of the unit cell, we can compute the volume occupied by one Mg atom: $$V_{atom} = \frac{V}{6} = \frac{2.489a^3}{6}$$

We can determine the mass of magnesium in the unit cell by dividing the molar mass of magnesium by Avogadro's number, which will give the mass of one magnesium atom. Then we multiply it by 6 (atoms per unit cell): $$m_{Mg} = 6 \cdot \frac{M_{Mg}}{N_A} = 6 \cdot \frac{24.305 \mathrm{g/mol}}{6.022 \times 10^{23} \mathrm{atoms/mol}}$$

The density formula for a crystal structure is: $$\rho = \frac{m}{V}$$ Plugging in the given density of magnesium and the derived expressions for mass and volume, we get: $$1.74 \mathrm{g/cm^3} = \frac{6 \cdot \frac{24.305 \mathrm{g/mol}}{6.022 \times 10^{23} \mathrm{atoms/mol}}}{\frac{2.489a^3}{6}}$$ Solve this equation for \(a^3\): $$a^3 = \frac{6 \times 24.305 \mathrm{g/mol}}{1.74 \mathrm{g/cm^3} \times 6.022 \times 10^{23} \mathrm{atoms/mol} \times 2.489}$$ Hence, \(a^3 = 3.68197 \times 10^{-23} \mathrm{cm^3}\), and \(a = 3.3250 \times 10^{-8} \mathrm{cm}\). Finally, the atomic radius for Magnesium can be calculated from the value of \(a\) we just found: $$r = \frac{\sqrt{2}a}{4} = \frac{\sqrt{2}(3.3250 \times 10^{-8} \mathrm{cm})}{4} = 1.1805 \times 10^{-8} \mathrm{cm}$$ So, the atomic radius for Magnesium is \(1.1805 \times 10^{-8}\) cm.