[Rutherford]

The 5. part of the second problem is problematic for me: http://icho2014.hus.edu.vn/document/01-24-2014%20IChO46-Preparatory.pdf so I would like if someone can check whether the radii are 0.9Å and 1.67Å for Li⁺ and Cl^- respectively. I am unsure because of part 5.2. as both values are far away from experimental ones.

[curiouscat]

I get 1.81 Å for Cl-. Kind of close to the expt. value.

For Li+ I get 0.75 Å.  That's 20% off.

Did you assume fcc or bcc? That might be the crux.
 

[CrazyAssasin]

I got 1.81 and 0.75 too

[Rutherford]

I assumed fcc, shouldn't I?

[curiouscat]

I assumed fcc, shouldn't I?

fcc is good. I did too. Wonder where you went wrong.

Show your working maybe?

[Rutherford]

Okay, the 5.14Å is the diagonal of the octahedron which is equal to:
1° d=2(r_Cl+r_Li)
Volume of the octahedron is:
V=sqrt(2)/3*a³ (a is the edge of the octahedron which is equal to d/sqrt(2)). I calculated V to be 22.6ų. V can be calculated also from:
2° V=V_Cl+V_Li=4/3π(r_Cl³+r_Li³)
Solving 1° and 2°, I got the answers I wrote  .

[curiouscat]

V can be calculated also from:
2° V=V_Cl+V_Li

I don't think you can use this 2°. That's your error (I think).

This assumes a void fraction of zero. Perfect packing is impossible in crystals. Ion volumes never add up to a unit cell volume. Assuming close contact (like you did in 1° is more reasonable)

[CrazyAssasin]

5.14Å is the length of the edge, not of the diagonal

[curiouscat]

5.14Å is the length of the edge, not of the diagonal

No that's fine. Edge of cube is diagonal of octahedron.

[Rutherford]

V can be calculated also from:
2° V=V_Cl+V_Li

I don't think you can use this 2°. That's your error (I think).

This assumes a void fraction of zero. Perfect packing is impossible in crystals. Ion volumes never add up to a unit cell volume. Assuming close contact (like you did in 1° is more reasonable)

Yes. What's the other equation? I can think only of 1°.

[curiouscat]

Yes. What's the other equation? I can think only of 1°.

Think harder. 

Hint: Along the cubes face-diagonal Cl- ions must contact. Right?

[Rutherford]

Right. Thanks, I got the right answer.

[curiouscat]

Right. Thanks, I got the right answer.

Glad.

I like these questions you've been posting. They are fun but not drudgery.

[Rutherford]

I plugged the correct values 1.82 and 0.75 Å into equation 2° and I got a volume of 27ų which is bigger than the one calculated using the edge length of the octahedron. Seems like the error where I overview the packing efficiency isn't the only one. Something else was wrong there, too and I don't know what .

[curiouscat]

I plugged the correct values 1.82 and 0.75 Å into equation 2° and I got a volume of 27ų which is bigger than the one calculated using the edge length of the octahedron. Seems like the error where I overview the packing efficiency isn't the only one. Something else was wrong there, too and I don't know what .

Of course. Your octahedron doesn't really contain one full atom of both. That's why. Look at this figure.



Why would you think one full atom of Li & Cl is fully inside the octahedron?

If you did want to do this try for the full cube. It ought to have one full Li atom & 8 vertex Cl- atoms (each contributing 1/8th ) plus 6 face center Cl- atoms (each contributing 1/2).

That's a total of 1 Li & 1+3 = 4 Cl atoms. The volume together is 1(1.76) + 4(25.2)≈102 ų

Compare to cube volume = 5.14³ =136 ų

That'd yield a void fraction of 34/136 = 0.25.  Pretty close to what's expected I'd hope?