Question

Chromium metal crystallises with a body centered cubic lattice. The length of the unit cell is found to be 287pm. Calculate the atomic radius. What would be the density of chromium in g/cc?

Answer 1

This numerical can be solved using density of the unit cell=(Mass of the unit cell/Volume of the unit cell). In a body centered cubic structure, the atomic radius(r) is equal to \sqrt{3} \mathrm{a} where a is the edge length.

Complete step by step answer:
In this numerical given, the chromium metal is used.
We know the atomic mass of the chromium is $52 \mathrm{g} / \mathrm{mol}$.
Edge of unit cell $=287 \mathrm{pm}=287 \mathrm{x} 10^{-10} \mathrm{cm}$
Volume of the unit cell $=\left(287 \times 10^{-10}\right)^{3} \mathrm{cm}^{3}=23.9 \mathrm{x} 10^{-24} \mathrm{cm}^{3}$
In a body centered cubic structure, there are two atoms in one unit cell.
Mass of one atom $=($ Molar mass/Avogadro Number $)=\left(52 / 6.02 \times 10^{23}\right)$
Mass of unit cell = Number of atoms in unit cell x mass of each atom

$$=2 \times\left(52 / 6.02 \times 10^{23}\right) \mathrm{g}$$

Density of unit cell $=($ Mass of the unit cell/Volume of the unit cell)

$$\begin{array}{l} =(\mathrm{ii}) /(\mathrm{i}) \\\ =7.2 \mathrm{g} / \mathrm{cc} \end{array}$$

Now,
Given edge $(\mathrm{a})=287 \mathrm{pm}$
In bec structure, $r=\sqrt{3} a=1.732 \times 287$

$$=497.08 \mathrm{pm}$$

The answer is the atomic radius is 497.08 pm and density is $7.2 \mathrm{g} / \mathrm{cc}$.

Note: Always calculate the Mass of the unit cell as per given formula. Take the avogadro number as $6.02 \times 10^{23}$. Chromium always forms bcc structure. Do not confuse this with the fcc structure, that is face centered cubic structure.