Ferrous oxide has cubes structure and each edge of the unit cell is `5.0 Å` .Assuming of the oxide as `4.0g//cm^(3)` then the number of `Fe^(2+) and O^(2)` inos present in each unit cell will be

Options

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-)`

Answer

D

Explanation

Let the number of units of FeO present in the unit cell =n
Molecule mass of FeO `=56+16=72 g " mol"^(-1)`
Weight of n units `=((72 g " mol"^(-1))xxn)/((6.023xx10^(23) " mol"^(-1))`
(weights of cell.)
Density of cell `=("Weight of cell")/("Volume of cell " (a^(3)))`
`(4.09g " cm"^(-3))=((72g " mol"^(-1))xxn)/((6.023xx10^(23)" mol"^(-1))xx(125xx10^(-24)" cm"^(-3)))`
`n=4.27 approx 4`