Ferrous oxide has a cubic structure. The length of edge of the unit cell is 5 Å. The density of the oxide is `4.0 g cm^(-3)`. Then the number of `Fe^(2+) and O^(2-)` ions present in each unit cell will be

Options

A. four `Fe^(2+) " and four " O^(2-)`
B. two `Fe^(2+) " and two " O^(2-)`
C. four `Fe^(2+) " and four " O^(2-)`
D. two `Fe^(2+) " and two " O^(2-)`

Answer

A

Explanation

`d= (ZxxM)/(N_A xxV) or Z = (d xx N_A xxV)/M`
`a = 5 Å= 5 xx 10^(-8) cm`
Molecular mass, `M=56 + 16 = 72 g`
`Z =(4xx 6.023 xx 10^(24) xx 125 xx 10^(-24))/72 = 4.18 ~~ 4`
Four molecules of ferrous oxide, FeO are there per unit cell, it means `4Fe^(2+) and 4O^(2-)` ions are present per unit cell.