`CsCl` has `bc c` arrangement and its unit cell edge length is `400` pm. Calculate the interionic distance in `CsCl`.

Explanation

The bcc arrangement iof CsCl , wehre balck cricle is ` Cs^(+)` ion and colured circles are` Cl^(-) ` ions. The aim is to find half of the body diagonal AE. If the edge of the unit cell is 'a' then,
` CE = sqrt(a^(2) +a^(2)) = sqrt2 a`
` AE = sqrt((sqrt2a)^(2) + a^(2)) = sqrt3 a = sqrt3 xx 400`
Interionic distanace = `1/2 AE = sqrt 3 xx 200 = 346.4 ` pm