| A crystal of lead(II) sulphide has NaCl structure. In this crystal the shortest distance between Pb+2ion and S2–ion is 297 pm. What is the length of the edge of the unit cell in lead sulphide? |
| (1) | a = 9 × 10–5cm |
| (2) | a = 6.25 × 10–8cm |
| (3) | a = 4.25 × 10–8cm |
| (4) | a = 5.94 × 10–8cm |
We are given that PbS has NaCl structure. As we know that NaCl has fcc structure, we should expect an arrangement similar to that in NaCl. Therefore in a unit cel, the sulphide ions would be present at the corners and on the faces of the cube and the Pb+2 ion would be present in the void on the edge such that all the spheres touch each other.
Let r+ be the radius of the Pb2+ ion and r- be the radius of the S2- ion. From the figure, it can be seen that
edge length = r- + 2 X r+ + r- = 2 [ r+ + r-]
= 2 X 297 pm (given that the shortest distance between Pb2+ ion and S2- ion is 297 pm)
= 594 pm = 5.94 X 10-10m
= 5.94 X 10-8 cm
Therefore answer is option (4)