Potassium fluoride `(KF)` has `NaCl` structure . Its density is `2.48g cm^(-3)` and its molar mass is `58g mol^(-1)`. Compute the distance between `K^(+)` and `F^(-) ions in `KF`.

Explanation

Since NaCl has fcc structure, therefore, KF has also fcc structure and thus, its rank is 4. Again, in a face centred cubic lattice the distance between the cation and anion is equal to the sum of their radii, which is equal to half of the edge length of the unit cell. That is
`r_(+)+r_(-)=(a)/(2) " " cdots(i)`
The edge length is calculated by the formula
`a^(3)= (ZxxM)/(rhoxx"AvNo.") " " cdots(ii)`
We know :
Z=4 , M = `58 "gmol"^(-1)`
`rho = 2.48 "gcm"^(-3) ` Av. No ` = 6.023xx 10^(23) "mol"^(-1)`
On substituting the known data in equation (ii), we get
`a^(3) = (4xx58)/(2.48xx6.023xx10^(23))`
=`155.318 xx10^(-24)`
a=` 5.375xx10^(-8) "cm" = 5.375 xx10^(-10) ` m
`:. = 537.5 xx10^(-12) m = 5375 ` pm
On putting the value of a in equation (i), we get,
`r_(+) + r_(-)= (a)/(2) = (537.5)/(2) = 269`pm .