Question

KF has NaCl type structure. What is the distance (in $\mathring{A}$) between K${}^{+}$ and F${}^{-}$ in KF if density of KF is $2.48$ g cm${}^{-3}$?

• A. $2.685$
• B. $4.632$
• C. $6.235$
• D. None of the above

Answer 1

The correct option is A $2.685$

As we know,

Density, $d=\frac{n\times M}{V\times N_A}$

The structure of NaCl is formed by repeating the face-centred cubic unit cell.

Hence, $n=4$

Density, $d=2.48gc{m}^{-3}$

$2.48=\frac{4\times 58}{a^3\times 6.02\times {10}^{23}}$

$\Rightarrow a=5.37\times {10}^{-8}cm=5.37A^0$

Distance between $K^{+}$ and $F^{-}=\frac{a}{2}=2.685A^0$