Question

If the distance between \[N{{a}^{+}}\]$and $\[C{{l}^{-}}\] ions in sodium chloride crystal is y pm, the length of the edge of the unit cell is:
(a) 4y pm
(b) \[\dfrac{y}{4}pm\]
(c) \[\dfrac{y}{2}pm\]
(d) 2y pm

Answer 1

Hint: The radius of a mono-atomic ion in an ionic crystal is called ionic radius. The ions are treated as hard spheres such that the sum of radius of a cation and radius of an anion will give its ionic radii.

Complete step by step solution: \[NaCl\] has a fcc unit cell. The chloride ions are present in the corners while the sodium ions are in the face center.
For fcc of \[NaCl\]edge length \[a=2{{r}_{N{{a}^{+}}}}+2{{r}_{C{{l}^{-}}}}\]= \[2({{r}_{N{{a}^{+}}}}+{{r}_{C{{l}^{-}}}})\]
Here \[2({{r}_{N{{a}^{+}}}}+{{r}_{C{{l}^{-}}}})\] is the ionic radius, the distance between chlorine and sodium ions. According to the question it is y pm.
Hence, from the above equation of edge length, we get that length of edge is twice the distance between the chloride and sodium ions. The correct answer to the question is option (d) 2y pm.

Additional Information: The measure of size of an atom’s ion in a crystal lattice is called ionic radius. In the past, X-ray diffraction technique was used to determine how much the radius of an ion increased or decreased. Nowadays, the Hard sphere model is used, where an ion is considered as a hard sphere. In this model these spheres do not overlap each other. Hard sphere models can be applied to metallic as well as ionic compounds. In general, the internuclear distance equation is used to obtain the value of ionic radius.

Note: The unit cell of \[NaCl\] is of fcc arrangement. Face centered unit cells have ions in its face centres as well as in the corners. Chloride ions are in the corners and sodium ions occupy the face centre.