Question

Ferrous oxide has a cubic structure and each edge of the unit cell is $5.0\stackrel{\circ }{A}$. Assuming the density of the oxide as $4.0gc{m}^{-3}$, find the number of $F{e}^{2+}$ and $O^{2-}$ ions present in each unit cell.

• A. Four $F{e}^{2+}$ and two $O^{2-}$ ions
• B. Four $F{e}^{2+}$ and four $O^{2-}$ ions
• C. Two $F{e}^{2+}$ and four $O^{2-}$ ions
• D. Two $F{e}^{2+}$ and two $O^{2-}$ ions

Answer 1

The correct option is B Four $F{e}^{2+}$ and four $O^{2-}$ ions

The length of edge of the unit cell is 5 $A^o$ or $5\times {10}^{-8}cm$
The volume of the unit cell is $(5\times {10}^{-8}cm{)}^3=1.25\times {10}^{-22}c{m}^3$
The density is 4.0 g $c{m}^{-3}$
The mass of the unit cell is $4.0gc{m}^{-3}\times 1.25\times {10}^{-22}c{m}^3=5.0\times {10}^{-22}g$......(1)
The mass of one mole of $FeO$ is $55.8+16=71.8$ g.

The mass of one molecule of $FeO$ will be $\frac{71.8}{6.023\times {10}^{23}}=1.192\times {10}^{-22}$ g.....(2)
Divide (1) with (2) to get number of $FeO$ formula units per unit cell.
$\frac{5.0\times {10}^{-22}g}{1.192\times {10}^{-22}}\simeq 4$
Then the number of $F{e}^{2+}$ and $O^{2-}$ ions
present in each unit cell will be four $F{e}^{2+}$ and four $O^{2-}$