`FeO` crystal has a simple cubic structure and each edge of the unit cell is `5 A^(@)`. Taking density of crystal as `5gm//"cc"` the number of `Fe^(2+)` and `O^(2-)` ions present in each unit cell are:

Options

A. `4Fe^(2+)` and `4O^(2-)`
B. `6Fe^(2+)` and `6O^(2-)`
C. `2Fe^(2+)` and `2O^(2-)`
D. `1Fe^(2+)` and `1O^(2-)`

Answer

A

Explanation

Volume of unit cell `=1.25xx10^(-22)"cc"`
Mass of unit cell `=1.25xx10^(-22)xx4`
`=5xx10^(-22)` gm
For one molecle `=72/(6.022xx10^(23))=1.195xx10^(-22)g`
Hence number of `FeO` molecules per unit cell `=(5xx10^(-22))/(1.195xx10^(-22))=4.19~~4`
Hence there are `4Fe^(2+)` and `4O^(2-)` in each unit cell.