Jake--It seems to me that the unit cell edge of 564.0 pm is the radius of 1 Na ion + the diameter of the chloride ion + the radius of a second Na ion with the edge of each just touching. So wouldn't the diameter of the chloride ion just be 564.0 - 2*(radius Na ion) = ?? about 374 pm. Check my thinking.
You can check that out by looking at the diagonal of the face (where radius Cl + space + diameter Cl + space + radius Cl) DON'T touch. The diagonal will be sqrt(a^2+ a^2) = 797.6 pm. So if diameter is 374, then radius is 374/2 = 187. That means that 187 for the radius of the first Cl at the lower corner + 374 for the diameter of the Cl in the face + 187 for the Cl at the upper diagonal corner = 187 + 374 + 187 = 748 pm. That leaves 797.6 - 748 = 49.6 pm for the spaces between the Cl ions. We have two spaces which means the Cl atoms on the diagonal of the face are separated by 49.6/2 pm. Again, CHECK MY THINKING.
The unit cells for most alkali metal halides are face-entered for both the cations and anions. In the case of NaCl the sodium ion is just a little too large for the chloride ions to touch each other along the face diagonal. The unit cell edge length is 564.0 pm. The sodium ion has a radius of 95.0 pm. What is the DIAMETER of the chloride ion in picometers?
how would i work out this question?
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The purpose of that exercise on the diagonal was just to prove that the Cl ions were NOT touching and that there was, indeed, a little space between Cl ions. That gave some credibility, I thought, to the value of 374 for the diameter of the Cl ion.
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