Casium may be considered to form interpentrating simple primitice cubic crystal. The edge length of unit cell is `412` pm. Determine
a. The density of `CsCl`.
b. The inoic radius of `Cs^(o+)` if the ionic radius of `Cl^(Θ)` is `181` pm. Given: `Aw (Cs) = 133 g mol^(-1)`

Explanation

The interpenetrating simple primitive cubic crystals mean the unit cell is body-centred where `Cl^(Θ)` ions occupy corners and `Cs^(o+)` ion and one `Cl^(Θ)` ion (or one molecule of `CsCl`) per unit cell.
From the expression, `rho = (Z_(eff) xx Aw)/(a^(3) xx N_(A))`
`= (1)/((412 xx 10^(-12) m)^(3)) ((168.5 g mol^(-1))/(6.023 xx 10^(23) mol^(-1)))`
`= 0.4 xx 10^(7) g m^(-3) -= 4.0 g cm^(-3)`
Now since `Cs^(o+)` ions touch the two chlorides along the cross diagonal of the cube, we will have
`2r_(c) + 2r_(a) = sqrt3a`
or `r_(c) = ((sqrt3)/(2)) a - r_(a) = ((sqrt3)/(2)) (412 "pm") - (181 "pm")`
`= 175.8` pm