wiz-icon
MyQuestionIcon
MyQuestionIcon
5
You visited us 5 times! Enjoying our articles? Unlock Full Access!
Question

A crystal of lead (II) sulphide has NaCl structure. In this crystal the shortest distance between a Pb2+ ion and S2
ion is 300 pm. What is the volume of a unit cell of lead sulphide ?

A
216×1024cm3
Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses
B
2.7×1023cm3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
C
216×1030cm3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
D
27×1030cm3
No worries! We‘ve got your back. Try BYJU‘S free classes today!
Open in App
Solution

The correct option is A 216×1024cm3
Side length of unit cell of crystal = 2×300=600 pm

vol = (side length)3=(600)3pm3=216×106×1030cm3=216×1024cm3

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