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Question

A crystal of lead (II) sulphide has NaCl structure. In this crystal, the shortest distance between Pb+2 ion and S2 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=2.97×109 cm, V=1.55×1022 cm3
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B
a=5.94×108 cm, V=2.096×1022 cm3
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C
a=11.88×109 cm, V=4.1×1022 cm3
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D
None of the above
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Solution

The correct option is C a=5.94×108 cm, V=2.096×1022 cm3
the shortest distance between Pb+2 ion and S2 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)
The shortest distance between the cation and anion is r+r=297pm
Hence, a=2(r+r)=2×297pm=594pm=5.94×1018cm
The volume V=a3=(5.94×108cm)3=2.096×1022cm3
421074_264544_ans_da87be9d7c8848c1a3f24773c2359456.png

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