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

Options

A. `Four Fe^(2+) and four O^(2-)`
B. `Two Fe^(2+) and four O^(2-)`
C. `Four Fe^(2+) and two O^(2-)`
D. ` Three Fe^(2+) and three O^(2-)`

Answer

a

Explanation

Let the units of ferrous oxide in a unit cell `= n`, molecular weight of ferrous `(FeO) = 56 + 16 = 72 g mol^(-1)`
Weight of n unit`= (72 xx n)/(6.025 xxx 10^(23))`
Volume of one unit `= ("length of corner")^(3)`
`= (5 Å)^(3) = 125 xx 10^(-26) cm^(3)`
Density `= ("wt, of cell" )/("volume")`
`4.09 = (72 xx n)/(6.023 xx 10^(23) xx 125 xx 10^(-24))`
`n = (3079.2 xx 10^(-1))/(72)`
`= 42.7 xx 10^(-1) = 4.27 = 4`