Ferrous oxide has a cubic structure and each edge of the unit cell is 5.0$\stackrel{°}{\mathrm{A}}$. Assuming density of the oxide as 4.0g cm^–3, then the number of Fe²+ and O²– ions present in each unit cell will be

Let the units of ferrous oxide in a unit cell be n.
The molecular weight of ferrous oxide (FeO) = 56 + 16 = 72 g mol⁻¹

Weight of n units = $\frac{72\times \mathrm{n}}{6.023\times {10}^{23}}$

Volume of one unit =(length of corner)³

⇒ Volume of one unit = ${\left(5\stackrel{°}{\mathrm{A}}\right)}^3=125\times {10}^{-24}{\mathrm{cm}}^3$

$$\begin{gathered} \mathrm{Density}=\frac{\mathrm{wt}. \mathrm{of} \mathrm{cell}}{\mathrm{volume}} \\ \Rightarrow 4.0=\frac{72\times \mathrm{n}}{6.023\times {10}^{23}\times 125\times {10}^{-24}} \\ \Rightarrow \mathrm{n}=\frac{3079.2\times {10}^{-1}}{72}=42.7\times {10}^{-1}=4.27\approx 4 \\ \mathrm{Therefore}, \mathrm{the} \mathrm{number} \mathrm{of} {\mathrm{Fe}}^{2+} \mathrm{and} {\mathrm{O}}^{2-} \mathrm{ions}=4 \end{gathered}$$