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My textbook has given data for third and fourth nearest neighbours to be 12 and 8 with distances $\sqrt{2}a$ and $\frac{\sqrt{11}a}{2}$.

I have been able to calculate for the first and second nearest neighbour but it has become difficult to visualise for the other two to calculate. Can you help me with hints on how to proceed preferably with a diagram.

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With $\sqrt3\over2$ being that close to $1$, BCC packing is better not looked at in terms of coordination spheres. But if you insist...

Say you are sitting in the center of a cell. Then:

  • Your first neighbours are at the corners of the same cell.
  • Second neighbours are at the centers of the nearest adjacent cells.
  • Third neighbours: centers of the next adjacent cells (those which share two corners with your cell).
  • Fourth neighbours: far corners of the nearest adjacent cells.

On a side note, there are more than just 8 of the latter.

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  • $\begingroup$ "BCC packing is better not looked at in terms of coordination spheres" then in what terms should we look at it instead? $\endgroup$ – Gaurang Tandon Jul 3 '18 at 13:04
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    $\begingroup$ @GaurangTandon Two interpenetrating cubic lattices would do. $\endgroup$ – Ivan Neretin Jul 3 '18 at 13:07
  • $\begingroup$ ...as the one mentioned in Oscar's answer? ("two simple cubic lattice, one with points at coordinates $(ma,na,pa)$ where $m,n,p$ are integers, the other with points at $((m+(1/2))a,(n+(1/2))a,(p+(1/2))a)$") $\endgroup$ – Gaurang Tandon Jul 3 '18 at 13:08
  • $\begingroup$ Yeah, that one. $\endgroup$ – Ivan Neretin Jul 3 '18 at 13:24
  • $\begingroup$ Although it is quite common to look at the radial distribution function from x-ray diffraction and find the peaks corresponding to the nth nearest neighbors. From that perspective, 'coordination spheres' make a lot of sense. $\endgroup$ – Jon Custer Jul 3 '18 at 18:52
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You can think of the body centered cubic lattice as two simple cubic lattice, one with points at coordinates $(ma,na,pa)$ where $m,n,p$ are integers, the other with points at $((m+(1/2))a,(n+(1/2))a,(p+(1/2))a)$. If you work out increasing distances for both omponent cubic lattices you get, in units of $a$:

$\sqrt{0^2+0^2+1^2}=1$

$\sqrt{0^2+1^2+1^2}=\sqrt{2}$

$\sqrt{1^2+1^2+1^2}=\sqrt{3}$

$\sqrt{0^2+0^2+2^2}=2$ -- generally square roots of whole numbers

and also

$\sqrt{(1/2)^2+(1/2)^2+(1/2)^2}=\sqrt{3}/2$

$\sqrt{(1/2)^2+(1/2)^2+(3/2)^2}=\sqrt{11}/2$

$\sqrt{(1/2)^2+(3/2)^2+(3/2)^2}=\sqrt{19}/2$

$\sqrt{(3/2)^2+(3/2)^2+(3/2)^2}=3\sqrt{3}/2$ -- generally half the square roots of numbers that are 3 more than multiples of 8. Each component square under the radical is one fourth of an odd square, and each odd square is one more than a multiple of 8.

Incidentally, there are not eight fourth-nearest neighbors. There are 24. The $3/2$ coordinate may occur along any one of three axes and each nonzero coordinate can independently have a negative sign.

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