The unit cell has 1/8 of an anion in each corner and 1/2 of an anion in each face giving 8(1/8) + 6(1/2) = 4 anions per cell. Charge balance says there must be four cations per cell, as well. thus, the mass of NaSbF6 in each cell is
Problem 5.14
NaSbF6 has the NaCl structure. The density is 4.37 g/cm3.
Calculate the radius of SbF6- using the radius of
Na+, the density, the formula weight, and the number of formula
units per unit cell.
Answer:
The NaCl cell looks something like:
The unit cell has 1/8 of an anion in each corner and 1/2 of an anion
in each face giving 8(1/8) + 6(1/2) = 4 anions per cell. Charge balance
says there must be four cations per cell, as well. thus, the mass of NaSbF6
in each cell is
The volume of the cell is a3, where a is the length of the unit cell dimension.
From the density, then
Assuming the ions are "touching" along the cell edge, so a = 2r- + 2r+ and r+ = 116 pm for Na+ in a six-coordinate environment. Thus,
733 = 2r- + 2(116)
r- = 251 pm = the anionic radius of SbF6-.