Question

Lead (II) fluoride, PbF2, crystallizes in a cubic unit cell in which the lead ions occupy a face-centered cubic arrangement in the unit cell, and the fluoride ions are located in certain spaces between the lead ions. How many of each ion are contained within a single unit cell? # of Pb2+ ions = ? # of F– ions =?

Answer 1

Lead (II) fluoride crystallizes in Face Centered Cubic arrangements like the following figure.

Here Pb²⁺ ions occupy the faces of the cube and F^- ions occupy the corners.

Now since an atom in face contribute 1/2 to the unit cell atom contribution, and an atom in corner contribute 1/8 to the unit cell atom contribution; the 6 Pb²⁺ atom at 6 faces contribute 6 X (1/2) = 3 and 8 F^- atoms at 8 corners contribute 8 X (1/8) = 1 to the unit cell.

So total unit cell atom contribution is 3 + 1 = 4.

Thus, Each unit cell contain 3 Pb²⁺ atom and 1 F^- atom.