The unit cell of #"CsCl"# (BCC) looks like this:

(Adapted from www.spaceflight.esa.int)

The #"Cs"# and #"Cl"# ions are touching along the red line connecting diagonally opposite corners of the cube.

Our job is to calculate #r_text(Cs) + r_text(Cl)#, which is half the length of this line.

**Step 1. Calculate the mass of the unit cell.**

The unit cell contains

#"1 Cs atom" + 8 × 1/8 "Cl atom" = "1 Cs atom + 1 Cl atom" = "1 CsCl formula unit"#

#"Mass of unit cell" = 1 color(red)(cancel(color(black)("FU"))) × (1 color(red)(cancel(color(black)("mol CsCl"))))/ (6.023 × 10^23 color(red)(cancel(color(black)("FU")))) ×"168.36 g"/(1 color(red)(cancel(color(black)("mol CsCl"))))#

#= 2.795 × 10^"-22"color(white)(l) "g"#

**Step 2. Calculate the volume of the unit cell.**

Density is an intensive property, so the density of the unit cell is the same as that of bulk #"CsCl"#.

#V = 2.795 × 10^"-22" color(red)(cancel(color(black)("g"))) × ("1 cm"^3)/(3.97 color(red)(cancel(color(black)("g")))) = color(white)(l)7.042 × 10^"-23"color(white)(l) "cm"^3#

**Step 3. Calculate the edge length of the unit cell.**

The volume of a cubic unit cell with edge length #l# is given by

#color(blue)(bar(ul(|color(white)(a/a)V = l^3color(white)(a/a)|)))" "#

∴ #l = root(3)V = root (3)(7.042 × 10^"-23" "cm"^3) = 4.130 × 10^"-8" color(red)(cancel(color(black)("cm"))) × (1 color(red)(cancel(color(black)("m"))))/(100 color(red)(cancel(color(black)("cm")))) × (10^12color(white)(l) "pm")/(1 color(red)(cancel(color(black)("m")))) = "413.0 pm"#

**Step 4. Calculate the internuclear distance in #"CsCl"#.**

(Adapted from Chemteam.info)

The length #d# of the diagonal on a face of the cell is given by

#d ^2 = l^2 + l^2 =2l^2#

The length #D# of the diagonal to opposite cornets of the cube is given by

#D^2 = l^2 + d^2 = l^2 + 2l^2 = 3l^2#

∴ #D = lsqrt3#

#D = 2r_text(Cs) + 2r_text(Cl) = lsqrt3#

#r_text(Cs) + r_text(Cl) = (lsqrt3)/2 = ("413.0 pm" ×sqrt3)/2 = "358 pm"#