Solid State Question 74
Question: Ferrous oxide has a cubic structure and each edge of the unit cell is $ 5.0A^{o} $ . Assuming density of the oxide as $ 4.0gc{{m}^{-3}} $ the number of $ F{{e}^{2+}} $ and $ {{O}^{2-}} $ ions present in each unit cell will be
Options:
A) two $ F{{e}^{2+}} $ and four $ {{O}^{2-}} $
B) three $ F{{e}^{2+}} $ and three $ {{O}^{2-}} $
C) four $ F{{e}^{2+}} $ and two $ {{O}^{2-}} $
D) four $ F{{e}^{2+}} $ and four $ {{O}^{2-}} $
Show Answer
Answer:
Correct Answer: D
Solution:
[d] Let the units of ferrous oxide in a unit cell = n Molecular weight of ferrous oxide $ (FeO) $ $ =56+16=72gmo{{l}^{-1}} $
Weight of n units $ =\frac{72\times n}{6.023\times 10^{23}} $
Volume of one unit $ \text{= (length of corner}{{)}^{3}} $ $ ={{(5A^{0})}^{3}}=125\times {{10}^{-24}}cm^{3} $ $ \text{Density =}\frac{wt\text{.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=4 $
[d] They are comparatively soft and not very rigid.
BETA
