r_{C+}/ r_{A-}
= 0.98 ×10^{−10} / 1.81 ×10^{−10} = 0.54

it is in the range of 0.
414 - 0.732 , hence the coordination number of each ion is 6

**Solution**

√3 a = r_{Cs+} +
2r_{Cl-} + r_{Cs+}

(√3/2) a = ( r_{Cl-}
+ r_{Cs+})

(√3/2) x 400 = inter
ionic distance

for an fcc structure r_{x+} / r_{y−} = 0.414

given that r_{X+} = 100 *pm*

r _{y−} = 100 *pm*
/ 0.414

packing efficiency = 68%

therefore empty space
percentage = (100-68) = 32%

let edge length = a

√2a = 4r

a = [4 × 300] / √2

a = 600 ×1.414

a = 848.4 pm

No Explaination.

if a is the length of the side, then the length of the leading
diagonal passing through the body centered atom is √3a

Required distance (√3/2) a

ρ = n × M / a^{3}N_{A}

for bcc

n = 2

M=39

nearest distance 2r =
4.52