In the cubic crystal of `CsCl (d = 3.97 g cm^(-3))`, the eight corners are occupied by `Cl^(Θ)` with a `Cs^(o+)` at the centre and vice versa. Calculate the distance between the neighbouring `Cs^(o+)` and `Cl^(Θ)` ions. What is the radius of the two ions? (`Aw` of `Cs = 132.91` and `Cl = 35.45)`

Explanation

In CsCl, `Cl^-` ions per unit cell =`8xx1/8`=1
`Cs^+` ions per unit cell =1
Thus, the unit cell of CsCl contains one CsCl molecule, i.e., Z=1
`rho=(ZxxM)/(a^3xxN_0)`
`therefore 3.97 =(1xx168.36)/(a^3xx6.023xx10^23)` [Mol. Mass of CsCl=132.91+35.45 =168.36]
or `a=4.13xx10^(-8)` cm or a=4.13 Å
For a cube of side length (a) equal to 4.13 Å, body diagonal `AE=sqrt3a=sqrt3xx4.13Å`=7.15 Å
As `Cl^-` ions A and E touch `Cs^+` ion, `therefore AE=2r_+ +2r_-`
`therefore 2r_+ + 2r_(-)`=7.15 Å or `r_+ + r_-` =3.57 Å
i.e., Distance between neighbouring `Cs^+` and `Cl^-` =3.57 Å
Assuming that the two `Cl^-` ions touch each other , length of the unit cell `=2r^(-)` =a=4.13 Å
`therefore r_(-)` =2.06Å `therefore r_+` =3.57-2.06=1.51 Å `therefore r_+//r_-` =1.51/2.06=0.73