I cannot find anywhere what the contribution of atoms situated at octahedral or tetrahedral voids in a HCP unit cell would be. I need to know this to be able to calculate the formula of a compound with HCP unit cell, when the positions of the atoms in the lattice are given. What I mean by 'contribution' is how much of the atom in that lattice point belongs to one unit cell. Eg:- in CCP packing the contribution of atoms at the corners is 1/8 th. I know that in a CCP unit cell the octahedral voids are there at the body center and the edge centers- so the contribution of an atom from both locations is different- 1 and 1/4 th. So I'm not sure whether the positions of OTs or TVs in a hcp unit cell are different (for the atoms of the same type) if so could someone tell how the contribution would be?


1 Answer 1


I cannot find anywhere..

Couldn't find? You can figure it out on your own, with me, right now.

Disclaimer: By "hexagonal unit cell" I assume you mean hexagonal prism, which comprises of 3 primitive hexagonal Bravais lattice.


Octahedral Voids (OhV)

The one on the left displays octahedral voids(visibly surrounded by 6 atoms). The OhV is at the center of the highlighted octahedron, and lies on the plane containing four of the six lattice points in the octahedron. Note that these lattice points belong to the unit cell, and the OhV must hence lie inside the unit cell.

This is the same for the other 5, amounting to a total of 6 OhV per HCP unit cell.

Being completely inside, its contribution is taken as 1.

  • BONUS: If you remember, No. of atoms per unit cell is 6, which tallies with the no. of OhV given by the formula $N_{OhV}=N$ where $N$ represents no. of atoms per unit cell.

Tetrahedral Voids (ThV)

I'm going to be frank: it IS not as easy as locating OhV. The figure above absolutely freaked me out when I tried to visualize ThV's. I'm not keen on recommending that. Instead, this 3D rendition should help immensely.

Type 1 ThV


The Type 1 Thv's are the ones inside in unit cell completely(so again, contribution=1). Notice that I have marked 2 tetrahedrons. Also note that each tetrahedron we form will tell us about the ThV its existence creates.

  • RED: Note that there will be another 2 like this one in the lower half. That makes a total of 6 Red-type ThV in the entire HCP unit cell.
  • GREEN: Note that there will be a total of 2 of the green-type ThV, the other one belonging to the inverted tetrahedron just below it.

That gives us a total of $6+2=8$ Type 1 ThV.

Type 2

Let's do a little recap to see which ThVs have we counted so far(displayed in red) ThV so far

I constructed a 3D model of 3 HCPs, which is required to visualize the remaining. 3d1

Now I will mark the ThV forming lattice points in red.

3d2 3d3

As you can see, lattice points from 3 unit cells are involved.

A quick analysis tells us that there will be 2 such Tetrahedrons for every vertical edge. A quicker analysis tells us that since each ThV so formed is present in 3 HCP unit cells, or is "shared" by them, Its contribution must be $\frac 13$.

We count the number of such tetrahedrons to be 12 per unit cell; with contributions, ThV Type 2 will be $12*\frac 13$ $=4$.

Ultimately, we get:

ThV type 1= 8

ThV type 2= 4

Total ThV= 12.

Which brilliantly tallies with the formula for number of ThV($N_{ThV}=2N$).

I hope managed to enlighten you.


  • OhV contribution is 1, no issues there.
  • ThV does not have a single "contribution", but rather could be 1 or $\frac 13$ depending on which ThV you're talking about.
  • 8
    $\begingroup$ The above answer is quite comprehensive, and i would like to complement that with this site, where the JSmol diagram of the HCP lattice helped me visualize it well. $\endgroup$
    – Keyenke
    Commented Sep 24, 2020 at 7:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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