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

In a hypothetical ionic crystal, B is arranged in cubic close packing and A occupies all octahedral voids and alternative tetrahedral voids. The correct formula of the compound?

A
AB
No worries! We‘ve got your back. Try BYJU‘S free classes today!
B
A2B2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
AB2
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
A2B
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
Open in App
Solution

The correct option is C A2B

No. of atoms ina ccp unit cell is fcc=4

Therefore no. of B atom present =4

No of A atoms= Alternate tetrahedral voids + octahedral voids

=(82+4)=8

The formula of the compound=A8B4 or A2B


flag
Suggest Corrections
thumbs-up
0
Join BYJU'S Learning Program
similar_icon
Related Videos
thumbnail
lock
Voids
CHEMISTRY
Watch in App
Join BYJU'S Learning Program
CrossIcon