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

In a normal spinel types structure, the oxide ions are arranged in ccp whereas 1/8 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
ZnFe2O4
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
D
ZnFe2O2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is C ZnFe2O4
In one unit cell no. of O2=4

The no. of Zn2+=18×8=1

The no. of Fe3+=12×4=2

molecular formula of given spinel structure is ZnFe2O4

flag
Suggest Corrections
thumbs-up
1
similar_icon
Similar questions
View More
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Voids
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon