NCERT Class XII Chemistry
Chapter - The Solid State
Questions with Solutions

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Question : 10
Total: 50
Calculate the efficiency of packing in case of a metal crystal for
(i) simple cubic
(ii) face-centred cubic (with the assumptions that atoms are touching each other).
(iii) body-centred cubic
Solution:  
(i) In a simple cubic unit cell :
Suppose the edge length of the unit cell = a and radius of the sphere = r
As spheres are touching each other, evidently, a=2r
No. of spheres per unit cell =
1
8
×8
=1

Volume of the sphere =
4
3
π
r3

Volume of the cube =a3=(2r)3=8r3
therefore Fraction occupied. ,i.e., packing fraction =(
4
3
π
r3
)
/ 8r3=0.524
or % occupied i.e., packing efficiency = 52.4%
(ii) In face-centred cubic structure : As sphere on the face-centre is touching the spheres at the corners, evidently AC=4r.
But from right angled triangle ABC, AC=AB2+BC2
=a2+a2=2a
2a=4r or a=
4
2
r

Volume of the unit cell =a3= (
4
2
r
)
3
=
32
2
r3

No. of spheres in the unit cell =8×
1
8
+6×
1
2
=4

Volume of four spheres =4×
4
3
π
r3
=
16
3
π
r3

∴ Fraction occupied i.e., packing fraction =
16πr3
3
32r3
2
=0.74

or % occupied i.e., packing efficiency = 74%
(iii) In body-centred cubic structure : As the sphere at the body-centre touches the spheres at the corners, body diagonal, AD=4r.
Further, face diagonal,
AC=AB2+BC2=a2+a2=2a
and body diagonal,
AD=AC2+CD2=2a2+a2=3a
3a=4r
or a=
4r
3

Volume of the unit cell =a3=(
4r
3
)
3
=
64r3
33

No. of spheres per unit cell =8×
1
8
+1
=2

Volume of two spheres =2×
4
3
π
r3
=
8
3
π
r3

therefore Fraction occupied i.e., packing fraction =
8
3
π
r3
64r3
33
=0.68

or % occupied i.e ., packing efficiency = 68%
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