wiz-icon
MyQuestionIcon
MyQuestionIcon
1
You visited us 1 times! Enjoying our articles? Unlock Full Access!
Question

In a normal spinel type structure, the oxide ions are arranged in ccp whereas 18 tetrahedral holes are occupied by Zn2+ ions and 50% of octahedral holes are occupied by Fe3+ ions. The formula of the compound is ?


A

Zn2Fe2O4

No worries! We‘ve got your back. Try BYJU‘S free classes today!
B

ZnFe2O3

No worries! We‘ve got your back. Try BYJU‘S free classes today!
C

Zn2Fe2O4

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D

Zn2Fe2O2

No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C

Zn2Fe2O4


Number of O - atoms per unit cell =18×8+12×6=4
Number of octahedral holes per unit cell =1×4=4
Number of Fe3+ ions per unit cell =50×4100=2
Number of Zn2+ ions per unit cell =18×8=1
Hence, formula : ZnFe2O4


flag
Suggest Corrections
thumbs-up
0
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Examples of Ionic Compounds
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon