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 ` = 8 xx 1/8 =1`
` Cs^(+)` ions per unit cell = 1
Thus, the unit cell of CsCl contains one CsCl molecule, i.e, Z =1
` p = ( Zxx M)/(a^(3) xx N_(0))`
` 3.97 = ( 1 xx 168.36)/ (a^(3) xx 6.023 xx 10^(23)) `
or a = `4.13 xx 10^(-18) m or a = 4.13 Å`
For a cube of side length (a) equal to 4.13 Å , body diagonal AE =` sqrt3 a = sqrt3 xx 4.13 Å = 7.15 Å`
As `Cl^(-)` ions A and E touch `Cs^(+) and Cl^(-) = 3.57 Å`
` 2r^(+) + 2r^(-) = 7.15 Å or r_(+) + r_(-) = 3.15 Å`
i.e, Distance between neighboring `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`
` r_(-) = 2.06 Å therefore r_(+)= 3.57 -2.06 = 1.51 Å therefore r_(+)//r_(-) = 1.51// 2.06 = 0.73`