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In the cubic crystal of CsCl (d = 3.97 g...

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)`

Text Solution

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In a unit cell, there is one `Cl^(o+)` ions and one `Cl^(Θ)` ion.
`rho = (1 xx 168.36)/(a^(3) xx 6.023 xx 10^(23))`
`3.97 g cm^(-3) = (1 xx 168.36)/(a^(3) xx 6.023 xx 10^(23))`
`implies a= 4.13 Å`
For `bc c`,
`2r_(o+) + 2r_(Θ) = sqrt3a`
`r_(o+) + r_(Θ) = (sqrt3 xx 4.13)/(2) = 3.57 Å`
`2r_(Θ) = 4.13 Å implies r_(Θ) = 2.065`
`r_(o+) = 3.57 - 2.065 = 1.505`
Hence, `(r_(o+)) /(r_(Θ)) = 0.729`.
(The radius ratio suggests `bc c` structure)
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