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

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

Ferrous oxide has a cubic structure and each edge of the unit cell is 5.0 Å. Assuming density of the oxide as 4.0gcm^(-3) , then the number of Fe^(2+)andO^(2-) ions present in each unit cell will be

Ferrous oxide (FeO) crystal has a cubic structure and each edge of the unit cell is 5.0 A^(@) Taking density of the oxide as 4.0gcm^(-3) . The number of Fe^(2+) and O^(2-) ions present in each unit cell is:

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))