Class 12 Chemistry The Solid State Placing second layer over the first layer

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

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