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`.

Options

A. 134.35pm
B. 268.7 pm
C. 403.05 pm
D. 201.52 pm

Answer

B

Explanation

Since NaCl has f.c.c. structure, KF has also fcc structure and Z=4.
Further 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 i.e.,`r^(+)+r^(-)`=a//2.
Now edge length `a^(3)=(ZtimesM)/(rhotimesN_(A))`
`=(4times58)/(2.48times6.023times10^(23))=155.318times10^(-24)cm^(3)`
`thereforea=5.375times10^(-8)cm=5.375times10^(-10)m`
`=537.5times10^(-12)=537.5 "pm"`
`"Now" r^(+)+r^(-)=a//2=(537.5)/2=268.7"pm"`