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Q.
Potassium fluoride $KF$ has $NaCl$ structure. Its density is $248 \,g\, cm^{-3}$ and its molar mass is $58 \,g\, mol^{-1}$. The distance between K+ and F ions in KF is
The Solid State
Solution:
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} = \frac{Z\times M}{\rho\times N_{A}} $
$ =\frac{ 4\times 58}{2.48\times 6.023\times 10^{23}} $
$ = 155.318 \times 10^{-24} $
$ \therefore a= 5.375\times10^{-8}cm $
$ =5.375 \times10^{-10} m$
$= 537.5\times 10^{-12} $
$ = 537.5 pm$
Now $r^{+} + r^{-} = a/2 $
$ = \frac{537.5}{2} = 268.7 \,pm $