Question

Ferrous oxide has a cubic structure and each edge of the unit cell is 5.0 Å. 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. Four $Fe^{2 +}$ and four $O^{2 -}$
• B. Two $Fe^{2 +}$ and four $O^{2 -}$
• C. Four $Fe^{2 +}$ and two $O^{2 -}$
• D. Three $Fe^{2 +}$ and three $O^{2 -}$

Answer 1

Answer: (A)

Four $Fe^{2 +}$ and four $O^{2 -}$

Answer 2

Let the units of ferrous oxide in a unit cell $= n$, molecular weight of ferrous oxide

$(FeO) = 56 + 16 = 72gmol^{- 1},$

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

Volume of one unit $= (\text{lengthofcorner})^{3}$

$= (5Å)^{3} = 125 \times 10^{- 24}cm^{3}$

Density$= \frac{\text{wt.ofcell}}{\text{volume}},4.09 = \frac{72 \times n}{6.023 \times 10^{23} \times 125 \times 10^{- 24}}$

$n = \frac{3079.2 \times 10^{- 1}}{72} = 42.7 \times 10^{- 1} = 4.27 \approx 4$