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

Answer

4

Explanation

Volume of the unit cell `a^(3)=(5xx10^(-8))^(3)`
`=1.25xx10^(-22)cm^(3) ( :. a=5Å)`
Density `(rho)=4.0gcm^(3)`
Mass of unit cell `=Vxxrho=1.25xx10^(-22)xx4.0`
`=5.0xx10^(-22)g`
Number of `FeO` molecules per unit cell `=(5.0xx10^(-22)g)/(1.195xx10^(-22)g)`
`=4.19~~4`
Thus there are four `Fe^(2+)` ions and four `O^(2-)` ions and four `O^(2-)` ions in each unit cell.