KF has NaCl structure. What is the distance between `K^(+)` and `F^(-)` in KF, if the density is `2.48g cm^(-3)`?

Explanation

KF has NaCl (fcc) structure , therefore , the number of KF molecules per unit cell, Z = 4
Molar mass of KF = `58 g mol^(-1)`
Let the edge of unit cell = a cm
Density, `rho = 2.48 g cm^(-3)`
Density , `rho = (Z xx M)/(a^3 xx N_A)`

`2.48 g cm^(-3) = (4 xx 58 g mol^(-1))/(a^3 xx (6.022 xx 10^(23) mol^(-1)))`
`a^3 = (4 xx 58 g mol^(-1))/((2.48 g cm^(-3)) xx (6.022 xx 10^(23) mol^(-1)))`
`= 155.3 xx 10^(-24) cm^(3)`
`:. a = (155.3 xx 10^(-24))^(1//3)`
or `a = 5.375 xx 10^(-10) cm " or " 537.5 p m`.
Now, if radius of `F^(-)` is `r_(F^-)` and radius of `K^(+)` is `r_(K^+)`, then according to the figure, edge of unit cell,
`a = r_(F^-) + 2r_(K^+) + r_(F^-)`
`= 2(r_(K^+) + r_(F^-))`
or `r_(K^+) + r_(F^-) = a/2`
Thus, distance between `K^(+)`and `F^(-)` ions will be half the edge length in the unit cell.
Thus, the distance between `K^(+)` and `F^(-)` ions = `(537.5)/2`
`= 268.8 p m`.