CsCl had bcc arrangement and its uint cell edge length is 400 pm. Calculate the interionic distance in CsCl.

CsCl had bcc arrangement and its uint cell edge length is 400 pm. Calc

1.	CsCl had bcc arrangement and its uint cell edge length is 400 pm. Calculate the interionic distance in CsCl.
Answer» Solution :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