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A crystal of Lead (II) sulphide has NaCl structure. In this crystal the shortest distance between the $Pb^{2+}$ ion and $S^{2−}$ ion is 297 pm. What is the length of the edge of the unit cell in lead sulphide? Also calculate the unit cell volume.

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$a=5.94\times 10^8cm,V=a^3=2.096\times 10^{-12}cm^3$
Hence (A) is the correct answer.
answered Jun 2, 2014 by sreemathi.v

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