Question

Iron (II) oxide has a cubic structure and each unit cell has side 5 Å. If the density of the oxide is `4 g cm^(-3)`, the number of oxide ions present in each unit cell is (Molar mass of FeO =`72 "g mol"^(-1) , N_A=6.02xx10^23 mol^(-1)`)

Answer 1

Correct Answer - 4
`Z=(rhoxxa^3xxN_0)/M`
`=((4 g cm^(-3))(5xx10^(-8)cm)^3(6.02xx10^23 mol^(-1)))/(72 g mol^(-1))approx 4`
Thus, there are 4 formula units (FeO) per unit cell.Hence, `O^(2-)` ions =4