KF has NaCl structure. What is the distance between K+ and F– in KF if density is 2.48 g cm-3?
KF has NaCl structure. What is the distance between K+ and F– in KF if density is 2.48 gm cm-3 ?
Answer:
Given Values
Density = 2.48 gm/cm3
Formula mass (M) = KF = 39+19=58
KF has NaCl structure : Means, KF has FCC unit cell. So, Z = 4
Avogadro Number (NA) = 6.02 × 1023
Formula used : \( D=\frac{Z \times M}{\left ( a \right )^{3}\times N_{A}} \)
Calculations
Let us first find the edge length (a) using the density of the unit cell\( \left ( a \right )^{3}=\frac{Z \times M}{D \times N_{A}} \)
\( \left ( a \right )^{3}=\frac{4 \times 58}{2.48 \times 6.02 \times 10^{23}} \) = 155.4 × 10-24 cm3
a = 5.38 × 10-8 cm = 538 pm
Let us first find the distance between Potassium ion and Fluoride ion\( d = \frac{a}{2} = \frac{358~pm}{2} = 269~pm \)
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