Placing second layer over the first layer
- Consider a two dimensional hexagonal close packed layer of type ‘A’ and place a similar layer above it in such a way that the spheres of the second layer lies in the depressions of the first layer.
- Let the second layer be of type ‘B’ due to its different alignment.
- All the triangular voids of the first layer are not covered by the spheres of the second layer which gives rise to a different arrangement.
- The area where a sphere of the second layer is above the void of the first layer (or vice versa) a tetrahedral void is formed.
- These voids are called tetrahedral voids because a joining of the centers of four spheres gives rise to tetrahedron.
- There are areas where the triangular voids in the second layer lies above the triangular voids of the first layer, and the triangular shapes of these do not overlap.
- One of them has the apex of the triangle pointing upwards and the other downwards.
- Such voids are surrounded by six spheres and are called octahedral voids (O).
Let the number of close packed spheres be N, then:
The number of octahedral voids generated in the structure = N.
The number of tetrahedral voids generated in the structure = 2N.
Question : If the radius of the octachedral void is r and radius of the atoms in close packing is R, derive relation between r and R.
Answer: Given that
Radius of octahedral void in close packing = r
Radius of atom in close packing = R
Using Pythagoras theorem,
2R2 = (R+ r)2 + (R+ r)2
4R2 = 2(R +r)2
2R2 = (R+ r)2
R√2 = R +r
R√2 – R = r
r = R√2 – R
Putting the value of √2, we derive
r = R (1.414-1)
r = 0.414R