The atom at <1, 1, 1> also has four neighbors including <0, 0, 0>. I claim that the other three are at <2, 2, 0>, <2, 0, 2> and <0, 2, 2>.

All atoms have integer coordinates and there are two populations:

- C0
- atoms whose coordinates are all even and whose sum is 0 mod 4.
- C1
- atoms whose coordinates are all odd and whose sum is 3 mod 4.

Cubic ice has a similar structure.

- H0
- holes whose coordinates are all even and whose sum is 2 mod 4.
- H1
- holes whose coordinates are all odd and whose sum is 1 mod 4.

The density of diamond is (atoms / cell)(mass of carbon atom)/(cell edge)

There are these 8 atoms in the half open cell [0, 4)^{3}:

<0, 0, 0>, <2, 2, 0>, <2, 0, 2>, <0, 2, 2>, <1, 1, 1>, <3, 3, 1>, <3, 1, 3>, <1, 3, 3>.

The length units used above are 0.154/√(3) nm and the cell edge is 4 times that.
Thus density =

(8(atoms/cell))(19.925 10^{−24}g/atom)/((4(0.154)/√(3))^{3} (nm^{3}/cell))
= 3.54 10^{−21} g/nm^{3}
= 3.54 g/cm^{3}

(It is reported as 3.52 g/cm.)

Lets work out the Voronoi zone of <0, 0, 0>. We seek the facets. Each nearest neighbor contributes a facet. x+y+z = 3/2 is the plane midway between <0, 0, 0> and <1, 1, 1> and x−y−z = 3/2 is half way to <1, −1, −1>, etc. The interior is defined by x+y+z<3/2 and x−y−z<3/2 etc. These four planes belong to a regular tetrahedron. It is tempting to guess that the tetrahedron is the zone, but (Voronoi) zones fill space and regular tetrahedra can’t. The vertices of the tetrahedron are <−3/2, −3/2, −3/2>, <3/2, 3/2, −3/2> etc.

The neighbor’s 12 neighbors are at <±2, ±2, 0>, etc. The zone facets due to those are |x|+|y|<2, |x|+|z|<2, and |y|+|z|<2. These are facets for the Rhombic dodecahedron. The vertices of this are at <±1, ±1, ±1> and <±2, 0, 0> etc. The short diagonals of the rhombi are the edges of the cube |x|, |y|, |z| < 1.

Of the vertices of the dodecahedron, only <−1, −1, −1> and <1, 1, −1> etc. lie inside the tetrahedron. None of the tetrahedron vertices lie in the dodecahedron.

From the center of a cavity there are four closest atoms. The zone of each of these four must reach the center. These four zones come together, each touching the other three−each with three facets coming together at the center point. If we removed population of C1 atoms from the crystal, the rhombic dodecahedron would be the Voronoi zone. As it is, it is twice too big.

The intersection between x+y+z = 3/2 and x−y−z = 3/2 is the line x=3 & y=−z. <3, 0, 0> is on that line but that point is closer to the atom <4, 0, 0>. This suggests that <4, 0, 0> contributes a facet. But first the 12 neighbor’s neighbors, such as <2, 2, 0>, the facet for which is the plane x+y=2. That plane and x+y+z = 3/2 intersect at z=−1/2 & x+y=2. Now I will guess and try to verify. The complementary diamond crystal has atoms in the cavities of the real crystal. The bonds of the complementary crystal form the edges of the Voronoi zones of the real crystal. The zones cover the space and thus cover the center of each cavity. Indeed the zones of four atoms meet at that center.

<−1, −1, −1> is an H1 cavity center adjoining <0, 0, 0>. <1, −1, −1> is C1 atom adjoining the same cavity. The bisecting plane between <0, 0, 0> and <1, −1, −1> is x−y−z=3/2.

The other neighbors of <1, 1, 1> are <2, 2, 0>, <2, 0, 2> and <0, 2, 2> (see above), and the planes midway between these and <0, 0, 0> are x+y=2, x+z=2 and y+z=2.

See Repasky’s page for geometry of many familiar molecules.