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Ferrous oxide has cubes structure and ea...

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

A

two `Fe^(2+)` and four `O^(2-)`

B

three `Fe^(2+)` and three `O^(2-)`

C

four `Fe^(2+)` and two `O^(2-)`

D

four `Fe^(2+)` and four `O^(2-)`

Text Solution

Verified by Experts

The correct Answer is:
D
Let the number of units of FeO present in the unit cell =n
Molecule mass of FeO `=56+16=72 g " mol"^(-1)`
Weight of n units `=((72 g " mol"^(-1))xxn)/((6.023xx10^(23) " mol"^(-1))`
(weights of cell.)
Density of cell `=("Weight of cell")/("Volume of cell " (a^(3)))`
`(4.09g " cm"^(-3))=((72g " mol"^(-1))xxn)/((6.023xx10^(23)" mol"^(-1))xx(125xx10^(-24)" cm"^(-3)))`
`n=4.27 approx 4`
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