In the cubic crystal of CsCl (d = 3.97 g cm–3), the eight corners are occupied by CI–with a Cs+ at the centre and vice versa. Calculate the distance between the neighbouring Cs+ and CI– ions. What is the radius ratio of the two ions ? (At. mass of Cs = 132.91 and CI = 35.45). from Chemistry The Solid State Class 12 Nagaland Board
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Class 10 Class 12
In the cubic crystal of CsCl (d = 3.97 g cm–3), the eight corners are occupied by CIwith a Cs+ at the centre and vice versa. Calculate the distance between the neighbouring Cs+ and CI ions. What is the radius ratio of the two ions ? (At. mass of Cs = 132.91 and CI = 35.45).

In a unit cell there are one Cs and 1 x 8/8 =1 chlorine  (Cl-) such that one CsCl molecule 

therefore 
As we have given 
density = 3.97 g cm-3
Mass of CsCl = 168.36g
Number of unit cell(Z) = 1

apply formula d= Z xMa3 x NA3.97 = 1 x 168.36a3 x 6.023 x 1023a = 4.13 x 10-8 cma =4.13 A0

for a cube of side  length 4.13A0 diagonal

                                    =3a i.e.,3 x 4.13 = 7.15A0

as it is a BCC with  Cs+ at centre radius r+ and Cl- at corner radius r- so, 

2r+ +2r- =7.15 or r+ +r- =3.57A0

such that distance between neighbouring Cs+ and Cl- =3.57A0

now assume two Cl- ion touch with each other so length of unit cell = 2r- =4.13
r- =2.06A0
r+ =3.57 -2.06 =1.51

r+/r-=1.51/2.06 =0.73

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