Question

Ferrous oxide has a cubic structure and the edge length of its unit cell is $5.0A^{\circ }$. Assuming the density of ferrous oxide to be $3.84{\text{g/cm}}^3$ , the number of $F{e}^{2+}$ and $O^{2-}$ ions present in each unit cell should be:
$(\text{use}{N}_A=6\times {10}^{23})$:

• A. $4F{e}^{2+}\text{and}4O^{2-}$
• B. $2F{e}^{2+}\text{and}2O^{2-}$
• C. $1F{e}^{2+}\text{and}1O^{2-}$
• D. $3F{e}^{2+}\text{and}4O^{2-}$

Answer 1

The correct option is A $4F{e}^{2+}\text{and}4O^{2-}$

Given, $a=5A^o=5\times {10}^{-8}cm$
Mass = density $\times$ volume of unit cell
Gram molar mass of FeO = 56 + 16 = 72g

To get the number of formula units per unit cell,
$\text{Number of formula units}=\frac{\text{Mass of a mole of unit cells}}{\text{gram molar mass of FeO}}$
$⟹n=\frac{3.84\times (5\times {10}^{-8}{)}^3\times 6\times {10}^{23}}{72}=4$

Hence, there are $4FeO$ units per unit cell, i.e. 4 each of $F{e}^{2+}$ and $O^{2-}$ ions