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Question

For a NaCl unit cell, the Na+ occupies the edge centres and body centred positions of cube while Cl− occupies the corners and face centres of the cube.

Which is the correct expression for determining packing fraction (P.F.) of NaCl unit cell, if ions along an edge diagonal are absent?

A
P.F.=43π(r3++r3)162r3
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B
P.F.=43π(52r3++4r3)162r3
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C
P.F.=43π(52r3++r3)162r3
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D
P.F.=43π(72r3++4r3)162r3
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Solution

The correct option is B P.F.=43π(52r3++4r3)162r3
Given, in NaCl
Na+ occupies the 12 edge centres and 1 body centre
Cl occupies 8 corners and 6 face centers in the cube which is an fcc lattice arrangement.
Along an edge diagonal, one body centred atom and two opposite edge centre atoms are present.
Therefore, if the ions present along edge diagonal is removed
Effective number of Na+ ions in an unit cell =10×14
(Here 2Na+ from edges and Na+ from body centred is removed.)

ZNa+=52

No Cl ion numbers are affected during the removal of edge diagonal. So,
Effective no. of Clions in one unit cell =8×18+6×12=4
ZCl=4

For fcc lattice arrangement,
We know,
2 a=4r
a=4r2
Volume of unit cell =a3=162 r3
Packing fraction =(ZNa+×43πr3+)+ZCl×43πr3a3

Packing fraction =43π(52r3++4r3)162r3

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