Ferrous oxide has a cubic structure and eged length of the unit cell is 5.0 A. Assuming the density of ferrous oxide to be `3.84g//cm^(3)` the no. of `Fe^(2+) and O^(-2)`ions present in each unit cell are `("use "N_(A) = 6 xx 10^(23))`

Options

A. `4Fe^(2+) and 4O^(2-)`
B. `2Fe^(2+) and 2O^(2-)`
C. `1Fe^(2+) and 1O^(2-)`
D. `3Fe^(2+) and 4O^(2-)`

Answer

A

Explanation

`d =(Zm)/(a^(3)N), Zn = (3.84 xx 5 xx 5 xx 10^(-24) xx 6 xx 10^(23))/72 =4` no. of `Fe^(+2)4` & `4O^(-2)` ions.