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

Answer

269 pm

Explanation

Step I. Calculate of Edge length of unit cell.
No. of particles in fcc type unit cell (Z) = 4
Gram molar mass of KF (M) = `39+19=58.9"g mol"^(-1)`
Mass of unit cell = `(ZxxM)/(N_(0))=4xx((58.0"g mol"^(-1)))/((6.022xx10^(23)mol^(-1)))=38.5xx10^(-24)cm^(3)`
Volume of unit cell `(a^(3))=("Mass of unit cell")/("Density of unit cell")=((38.5xx10^(-23)g))/((2.48gcm^(-3)))=155.2xx10^(-24)cm^(3)`
Edge length of unit cell (a) `=(155.2xx10^(-24)cm^(3))^(1//3)=5.37xx10^(-8)cm=537xx10^(-8)xx10^(10)=537"pm."`
Step II. Calculation of distance between `K^(+)` and `F^(-)` ions.
Edge length `K^(+)F^(-)` unit cell `=2(r_(K^(+))+r_(F^(-)))`
or `2(r_(K^(+))+r_(F^(-)))="537pm"`
or `(r_(K^(+))+r_(F^(-)))=(("537pm"))/(2)="295pm."`