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,

2R^{2} = (R+ r)^{2} + (R+ r)^{2}

4R^{2} = 2(R +r)^{2}

2R^{2} = (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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