Solid State Question 197
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-c{{m}^{-3}} $ , then the number of $ F{{e}^{2+}} $ and $ {{O}^{2-}} $ ions present in each unit cell will be
[MP PET 2000]
Options:
A) Four $ F{{e}^{2+}} $ and four $ {{O}^{2-}} $
B) Two $ F{{e}^{2+}} $ and four $ {{O}^{2-}} $
C) Four $ F{{e}^{2+}} $ and two $ {{O}^{2-}} $
D) Three $ F{{e}^{2+}} $ and three $ {{O}^{2-}} $
Show Answer
Answer:
Correct Answer: A
Solution:
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 $ ={{(lengthofcorner)}^{3}} $
$ ={{(5{\AA})}^{3}}=125\times {{10}^{-24}}cm^{3} $ Density $ =\frac{wt\text{.}ofcell}{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 $
BETA
