Understanding Close Packing in 3D with Solid State: Mr V's Homework Puzzle

[Mr Virtual]

Homework Statement

Chapter name: Solid state
Topic: close packed structures in crystalline solids
Sub-topic: close packing in 3D
In my Chemistry book I read:

Let the number of closed packed spheres be N, then:
The number of octahedral voids generated=N
The number of tetrahedral voids generated=2N

Now let us suppose there is 1 sphere only. Can anybody explain to me where the one octahedral void and two tetrahedral voids are? I cannot understand how they deduced the above formula. There is no explanation given in the book. It just says: "It is found that such and such thing happens". No illustration is provided. I thought that I could find the answer to my problem on this forum.

Homework Equations

If N is the number of spheres, then
number of octahedral voids=N
number of tetrahedral voids=2N

The Attempt at a Solution

No clue. If possible, please answer with pictures/graphics.

Thanks

Eagerly waiting!
Mr V

 

[Gokul43201]

Now let us suppose there is 1 sphere only.

You can't do that. The ratio of the number of octahedral and tetrahedral voids to the number of lattice sites is accurate only for large N. With a single lattice site, it is meaningless to even speak of an interstitial.

For visualization: http://www.iucr.org/iucr-top/comm/cteach/pamphlets/5/node1.html

PS: Next time, please include the name of the source (your Chemistry book), and the location within it, where the quoted text comes from.

 

[Mr Virtual]

PS: Next time, please include the name of the source (your Chemistry book), and the location within it, where the quoted text comes from.

I don't know if this book is available outside my country.
Name: Chemistry Part 1 Textbook for class XII NCERT
Page number: 15

This is what I found on the link you gave me:

To determine the number of tetrahedral and octahedral voids in a three-dimensional close-packing of spheres, we note that a sphere in a hexagonal close-packed layer A is surrounded by three B voids and three C voids (Fig. 1). When the next layer is placed on top of this, the three voids of one kind (say B) are occupied and the other three (say C) are not. Thus the three B voids become tetrahedral voids and the three C voids become octahedral voids. A single sphere in a three-dimensional close-packing will have similar voids on the lower side as well. In addition, the particular sphere being considered covers a triangular void in the layer above it and another in the layer below it. Thus two more tetrahedral voids surround the spheres. This results in 2 3 + 1 + 1 = 8 tetrahedral voids and 2 3 = 6 octahedral voids surrounding the sphere. Since a tetrahedral void is shared by four spheres, there are twice as many tetrahedral voids as there are spheres. Similarly, since an octahedral void is surrounded by six spheres, there are as many octahedral voids as there are spheres.

So, N is actually the number of spheres present in a crystal lattice. An isolated sphere cannot have any voids, but if we consider a single sphere in a lattice, then it makes two tetrahedral voids (by covering the triangles of the above and below layer). As regards to the single octahedral void, I am a little confused. I know that if three spheres are arranged in such a way that they form a triangular void pointing north, and another three spheres form a void pointing south, then if we put one set of three spheres on top of the other set, such that the individual void of each set is not covered, then we say that the two-triangle void is an octahedral void.
Maybe the logic involved is this: suppose the upper layer has only 2 spheres (the lower layer contains its 3 spheres). Now until and unless the third sphere is also there in the first layer, the octahedral void cannot be formed. So each sphere in the lattice helps in completing an octahedral void. Perhaps this is why it is said that a single sphere generates a single octahedral void. Am I right?

Anyway, thanks a lot for your reply and for clearing my confusion!

warm regards
Mr V