JEE MainChemistrySolid State
A metal crystallises in a face-centred cubic (fcc) structure. If the radius of the metal atom is r , what is the shortest distance between the centres of two nearest tetrahedral voids in the lattice?
Options
- A2 r
- B2r
- C2 2 r
- Dr 2
Correct answer
A. 2 r
Step-by-step solution
In an fcc unit cell, there are 8 tetrahedral voids located on the body diagonals. These voids form a smaller simple cube of edge length a 2 within the unit cell. Thus, the shortest distance between the centres of two nearest tetrahedral voids is a 2 . For an fcc lattice, the relationship between the edge length a and the atomic radius r is given by: 2 a = 4r a = 2 2 r Substituting the value of a into the distance expression: Shortest distance = a 2 = 2 2 r 2 = 2 r Answer: 2 r