Question

PbS has rock salt type structure. If its density is $12.7$ g cm${}^{-3}$. The distance between $P{b}^{2+}$ and $S^2$ ions in PbS is :

(Given that the atomic weight of Pb $=207$ amu and of S $=32$ amu)

• A. $500$
• B. $125$
• C. $250$
• D. None of the above

Answer 1

Answer: (C)

$250$

PbS has an fcc structure.

The expression for the density of the unit cell is:

$d=\frac{Z\times M}{N_0\times a^3}$

Where Z= 4

M= 207 + 32 = 239

d= 12.7 g cm${}^{-3}$

$12.7=\frac{4\times 239}{6.023\times {10}^{23}\times a^3}$

$a=500\times {10}^{-10}cm=500$ pm

Distance between two $=\frac{a}{2}=\frac{500}{2}=250$ pm

Hence, the correct option is $\text{C}$