[Rutherford]

No, there isn't just one lithium atom, then the empirical formula of lithium chloride would be LiCl₄. There is an octahedron Li in the center of the cube, but there are also 12 more at the center of each edge contributing 1/4: 1+12/4=4 Li atoms, or in other words: there are 4 octahedral holes in a fcc.

For the octahedron, interception with other octahedrons must be taken into consideration, that way it would be: 1Li and 6/6=1Cl.

[curiouscat]

No, there isn't just one lithium atom, then the empirical formula of lithium chloride would be LiCl₄. There is an octahedron Li in the center of the cube, but there are also 12 more at the center of each edge contributing 1/4: 1+12/4=4 Li atoms, or in other words: there are 4 octahedral holes in a fcc.

For the octahedron, interception with other octahedrons must be taken into consideration, that way it would be: 1Li and 6/6=1Cl.

You are right. That way I get a void fraction of ~0.2.{ Using 4(1.76) + 4(25.2). }

Still OK, I guess. If done the cubic unit cell way. Does this make sense?

I still have to do the calculation your way by using the octahedral cell. That ought to work too.

[curiouscat]

For the octahedron, interception with other octahedrons must be taken into consideration, that way it would be: 1Li and 6/6=1Cl.

On second thoughts, I suspect you cannot do the void fraction calculation with your octahedron. I don't think you can completely tile (rather fill) space with such octahedra. So you must keep losing volumes.

Not sure though. You can try.

[Rutherford]

The correct volume I got using the math formula V=sqrt(2)/3*a³ is 22.6ų. If I got a smaller volume using the apparently incorrect formula V=V_Cl+V_Li it would be because of voids present in the octahedron, but actually I got a bigger value by plugin in 1.82 and 0.75 Å, which I can't explain why. It is 27.0ų.

[curiouscat]

The correct volume I got using the math formula V=sqrt(2)/3*a³ is 22.6ų. If I got a smaller volume using the apparently incorrect formula V=V_Cl+V_Li it would be because of voids present in the octahedron, but actually I got a bigger value by plugin in 1.82 and 0.75 Å, which I can't explain why. It is 27.0ų.

I think you are making a mistake but not sure.

Ok, explain to me this part:

For the octahedron, interception with other octahedrons must be taken into consideration, that way it would be: 1Li and 6/6=1Cl.

Why the 6/6? How do you know 1/6 of the Cl atom comes to each octahedron.

[Rutherford]

Here is a package of octahedrons. Let's analyze the Cl atom I marked. I marked the 4 octahedrons which it belongs too. Obviously, two more octahedrons are missing there (if they were drawn, we couldn't see the chlorine atom): one on top of the Cl atom and one in front of Cl.

[curiouscat]

Here is a package of octahedrons. Let's analyze the Cl atom I marked. I marked the 4 octahedrons which it belongs too. Obviously, two more octahedrons are missing there (if they were drawn, we couldn't see the chlorine atom): one on top of the Cl atom and one in front of Cl.

What I'm saying is the green Cl atoms will always be peeking out. Some green volume won't be captured by the octahedra.  Hence you cannot do the usual 1/6 apportioning.

For a cube, say, all space is covered. So symmetric splits make sense.

[Rutherford]

I don't see how, but that must be it. Thanks.