# Chemistry - Calculate the third and fourth nearest neighbours in bcc

## Solution 1:

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.

## Solution 2:

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.