-3
$\begingroup$

Can anyone explain me that how the coordination number of corner sphere of this 3-D packing is 6 ? it should be 3

3D close packing from 2-D square close packed layers

$\endgroup$
1
  • 2
    $\begingroup$ The 3D packing is considered infinite, so there is no corner sphere. $\endgroup$
    – Poutnik
    Apr 4, 2021 at 5:11

2 Answers 2

1
$\begingroup$

As @Poutnik said, the packing is infinite, so there are no corner spheres.

Consider it this way-

  1. Each sphere is in contact with four other spheres in it's own layer, as shown- enter image description here
  1. Also note that each sphere is in contact with one sphere on the layer just above it and one sphere on the layer just below it. A total of two additional spheres thus.

Hence the co-ordination number is 4+2=6.

$\endgroup$
0
0
$\begingroup$

As @Poutnik mentioned and @Meta xylene confirmed, the packing is infinite, so there are no corners. In case you have any difficulty in understanding this, consider it this way: the coordination number will be the maximum number of adjacent spheres...So now if you take the apparent corner sphere, the number of adjacent spheres is 3, taking one on the edge, the answer will be 4 and for a central sphere, it will be 6. The maximum out of 3,4 and 6 is obviously 6, so that is the coordination number. (Just remember that this is just a simple hack, not the correct theory or explanation!)

enter image description here

Here, the first figure is Square Close Packing and the second one is Hexagonal Close Packing of spheres in 2D. Now, stacking two or more of the square close packing (AA type) one on top of the other, we get ccp, which stands for cubic close packing (AAA type)

enter image description here

Consider a sphere in ccp, it will have 6 adjacent spheres, i.e, one on the top, one on the bottom, and 4 in the same plane (see it as North, South, East and West of the sphere you are considering maybe) [the spheres are marked in the square close packing figure]

So, the coordination number is 6. I hope my answer helps you, the "hack" I mentioned worked for me and is really simple! :)

$\endgroup$
1
  • $\begingroup$ Please note that spheres diagonal to each other are NOT considered adjacent. Also, we have only taken into account Cubic close packing here... $\endgroup$ Apr 4, 2021 at 11:03

Not the answer you're looking for? Browse other questions tagged or ask your own question.