Question

In the cubic crystal of CsCl (d = 3.97 g/cm³), the eight corners are occupied by Cl⁻ with a Cs⁺ at the centre and vice-versa. Calculate the distance between the neighbouring Cs⁺ and Cl⁻ ions. What is the radius ratio of the two ions? (At. masses: Cs = 132.91, Cl = 35.45)

Answer 1

Number of Cl⁻ ions per unit cell = `8 xx 1/8 = 1`

Number of Cs⁺ ions per unit cell = 1 × 1 = 1

Therefore, the unit cell of caesium chloride contains one Cs⁺ Cl⁻ unit, i.e., Z = 1

Given that ρ = 3.97 g/cm³,

M = 132.91 + 35.45 = 168.36

∴ `rho = (Z xx M)/(a^3 xx N_A)`

∴ `a = ((Z xx M)/(rho xx N_A))^(1/3)`

= `((1 xx 168.36)/(3.97 xx 6.022 xx 10^23))^(1/3)`

= `((1 xx 168.36)/(23.90 xx 10^23))^(1/3)`

= `(7.044 xx 10^(-23))^(1/3)`

= 4.13 × 10⁻⁸ cm

Since the crystal lattice is body-centred,

body diagonal AD = `sqrt 3  a`

= `sqrt 3 xx 4.13`

= `1.73 xx 4.13`

= 7.15 Å

Also, AD = 2r₊ + 2r₋ = 7.15

or `r_+ + r_- = 7.15/2`

= 3.57 Å

Hence, the distance between the neighbouring Cs⁺ and Cl⁻ ions is 3.57 Å.

If two Cl⁻ ion touch each other,

2r₋ = a

= 4.13 Å

or, `r_- = 4.13/2 = 2.06` Å

r₊ = 3.57 − 2.06

= 1.51 Å

Hence, the radius ratio

`r_+/r_- = 1.51/2.05`

= 0.73 Å