Question

Iron$\left(\text{II}\right)$oxide crystallise in cubic structure with unit cell edge length of $5A^{\circ }$. If the density of oxide is $3.8{\text{g cm}}^{-3}$. Calculate the number of $F{e}^{2+}$ ions present in each unit cell.
[ Molar mass of $Fe=56\text{g/mol}$ ]

• A. $1$
• B. $2$
• C. $3$
• D. $4$

Answer 1

The correct option is D $4$

Given:
Edge length, $a$= $5A^{\circ }=5\times {10}^{-8}\text{cm}$
Density, $D$ = $3.8{\text{g cm}}^{-3}$
Molecular mass of $FeO=56+16=72\text{u}$
We have the equation,
$D=\frac{ZM}{N_A{a}^3}$
$N_A\to 6.023\times {10}^{23}$
$Z\to \text{Number of}FeO\text{per unitcell}$
Substituting the values,
$Z=\frac{D{N}_A{a}^3}{M}=\frac{3.8\times 6.023\times {10}^{23}\times (5\times {10}^{-8}{)}^3}{72}=4$
So there are four $FeO$ per unit cell. Hence $4F{e}^{2+}$ per unit cell.