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

An element crystallizes in fcc lattice having edge length 400 pm. The maximum diameter of an atom which can be placed in interstitial site without distorting the structure is (in pm):

Open in App
Solution

In fcc, there are octahedral voids and tetrahedral voids.

For tetrahedral voids, rR=0.225

For octahedral voids, rR=0.414

Where,
r= radius of the smaller ion
R= radius of the larger ion

For the maximum diameter of the atom in the interstitial site, octahedral voids must be taken because they can accommodate a larger atom.

rR=0.414

For FCC,

R=a22=40022
rR=0.414

r=0.414×40022=58.55 pm

Diameter is 2r=2×58.55=117.1 pm

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