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 `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 oxide-
`(FeO) = 56 + 16 = 72 g "mol"^(-1)`
weight of n units `=(72 xx n)/(6.0222 xx 10^(23))`
Volume of one unit = `("length of corner")^(3)`
`=(5A)^(3) = 125 xx 10^(-24) cm^(3)`
Density `=("wt. of cell")/("volume"), 4.09 = (72 xx n)/(6.022 xx 10^(23) xx 125 xx 10^(-24))`
`n = (3079.2 xx 10^(-1))/72 = 42.7 xx 10^(-1) = 4.27 = 4`