Ferrous oxide has a cubie structure and edge length of the uint cell is ` 5.0Å` . Assuming the density o ferrous oxide to be `3.84 g//cm^(3)` , the no. Of `Fe^(2+)` and `O^(2-)` ions present in each unit cell be : (use `N_(A)= 6xx 10^(23)`):
A
`4 Fe^(2+) and 4 O^(2-)`
B
`2 Fe^(2+) and 2 O^(2-)`
C
`1 Fe^(2+) and 1 O^(2-)`
D
`3 Fe^(2+) and 4 O^(2-)`
Text Solution
Verified by Experts
The correct Answer is:
A
Topper's Solved these Questions
SOLID STATE
NARENDRA AWASTHI|Exercise Level 1 (Q.33 To Q.62)|1 Videos
SOLID STATE
NARENDRA AWASTHI|Exercise Level 3 - One Or More Answers Are Correct|1 Videos
Ferrous oxide (FeO) crystal has a cubic structure and each edge of the unit cell is 5.0 A^(@) Taking density of the oxide as 4.0gcm^(-3) . The number of Fe^(2+) and O^(2-) ions present in each unit cell is:
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
FeO crystal has a simple cubic structure and each edge of the unit cell is 5 A^(@) . Taking density of crystal as 5gm//"cc" the number of Fe^(2+) and O^(2-) ions present in each unit cell are: