Second question:
In the LiCl structure shown in 9.18 [LiCl's lattice, where the "Cl- ions are in contact with each other, forming a face centered cubic lattice."], the chloride ions form a face-centered cubic unit cell 0.513 nm on an edge. The ionic radius of Cl- is 0.181 nm.
A) Along a cell edge, how much space is in between the Cl- ions?
B) Would an Na+ ion (r=0.095 nm) fit into this space? A K+ ion (r=0.133 nm)?
I guess I just need help with letter A, since B will come as a result of that. I don't know how to do it though. Any help would be appreciated
According to the pictures I can find, we have one atom at the left corner with its radius and one atom at the right corner with its radius and a space between. The radius + radius + space = a, the cell edge. Therefore, if we take (cell edge - 2*radius) = 0.513 nm -(2*1.81 nm) = the distance of the space. Check my thinking.
3 answers