Q. Ferrous oxide has a cubic structure and each edge of the unit cell is $5.0$ $\mathring{\mathrm{A}}$. Assuming density of the oxide as 4.0g/$cm^{- 3}\$then the number of $Fe^{2 +}$ and $O^{2 -}$ ions present in each unit cell will be
A) Two $Fe^{2 +}$ and four $O^{2 -}$
B) Three $Fe^{2 +}$ and three $O^{2 -}$
C) four $Fe^{2 +}$ and two $O^{2 -}$
D) four $Fe^{2 +}$ and four $O^{2 -}$
Explanation
(d)
Let the units of ferrous oxide in a unit cell $= \ n$. Molecular
Weight of ferrous oxide ($FeO$)
$= 56\ + \ 16\ = 72g\ mol^{- 1}$
Weight of $n$ units $= \frac{72\ \times \ n}{6.023\ \times \ 10^{23}}$
Volume of one unit $= (length\ of\ corner)^{3}$
$= (5\mathring{\mathrm{A}})^{3} = 125\ \times \ 10^{- 24}cm^{3}$
Density $= \frac{wt.of\ cell}{volume}$
$\therefore 4.09 = \frac{72\ \times \ n}{6.023 \times 10^{23}\ \times \ 125\ \times \ 10^{- 24}}$
Hence, $n = \ \frac{3079.2\ \times \ 10^{- 1}}{72} = 42.7 \times \ 10^{- 1}$
$= 4.27\ \approx 4$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.