Question

3. X-ray diffraction studies show that the edge length of the unit cell in crystal- line sodium chloride is 5.638 A. Calculate the internuclear Na*-Cl distance. Calculate the density of crystalline sodium chloride. Compare your answer with the experimental value obtained from the Handbook of Chemistry and Physics.

Answer 1

Hi,

NaCl crystallizes as an interpenetrated fcc packing of Na⁺ and Cl^- ions.

Hence, edge length = 2*r_cation + 2*r_anion

=> Since interionic Na⁺--Cl^- distance = r_cation + r_anion? = edge length / 2 = 5.638/2 = 2.819

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Density of NaCl lattice will be: mass of all atoms in one unit cell / volume of unit cell

Since 1 unit cell of NaCl consists 4 Na⁺ and 4 Cl^- ions,

mass of one unit cell = 4*23 + 4*35.5 = 234 atomic unit = 234*1.66*10^-24 grams

Volume of 1 unit cell (since its a cubic lattice) = (edge length)³

= (5.638*10^-8 cm)³ = 179.22*10^-24 cm³  

Hence the density will be = 234*1.66/179.22 g/cm³  

= 2.167 g/cm³

reported value from literature = 2.165 g/cm³.

The value calculated here is <0.1 % away from the actual value !