Question

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

• A. $a=2.97\times {10}^{-9}$ cm, $V=1.55\times {10}^{-22}$ cm${}^{-3}$
• B. $a=5.94\times {10}^{-8}$ cm, $V=2.096\times {10}^{-22}$ cm${}^{-3}$
• C. $a=11.88\times {10}^{-9}$ cm, $V=4.1\times {10}^{-22}$ cm${}^{-3}$
• D. None of the above

Answer 1

Answer: (B)

$a=5.94\times {10}^{-8}$ cm, $V=2.096\times {10}^{-22}$ cm${}^{-3}$

the shortest distance between $P{b}^{+2}$ ion and $S^{2-}$ ions is $297$ pm. This corresponds to one half the edge length. This is because along the edge, the spheres touch each other. If a is the edge length and r and r' are the radii of cation and anion respectively, then
$a=2(r+r^{\prime })$
The shortest distance between the cation and anion is $r+r^{\prime }=297pm$
Hence, $a=2(r+r^{\prime })=2\times 297pm=594pm=5.94\times {101}^{-8}cm$
The volume $V=a^3=(5.94\times {10}^{-8}cm{)}^3=2.096\times {10}^{-22}c{m}^3$