This chapter belongs to a previous NCERT/CBSE curriculum and is not part of the current 2026-27 chapter list. Removed in 2022-23.
Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter The Solid State. This page offers a step-by-step solution to the specific question from Exercise 2, Question 10:
Calculate the efficiency of packing in case of a metal crystal for
(i) simple cubic
(ii) body-centred cubic
(iii) face-centred cubic (with the assumptions that atoms are touching each other).
. With detailed answers and explanations for each chapter, students can strengthen their understanding and prepare confidently for exams. Ideal for CBSE and other board students, this resource will simplify your study experience.Calculate the efficiency of packing in case of a metal crystal for
(i) simple cubic
(ii) body-centred cubic
(iii) face-centred cubic (with the assumptions that atoms are touching each other).
(i) Simple cubic:
Suppose the edge length of the unit cell = a
&
Radius of the sphere = r
Then,since the sphere are touching each other along the edge,therefore a = 2r
Now there are 8 spheres at the corners of the cube & each sphere at the corner is shared by 8 unit cells & the contribution per unit cell is 1/8 so that
Number of spheres per unit cell is 8 x 1/8 = 1
Volume of sphere =4/3πr3 & volume of cube = a3 = (2r)3 = 8r3
Now packing efficiency = (volume of one sphere / total volume of cubic unit cell) x 100
Or
(4/3 πr3 / 8r3) x 100 = 52.4%
Therefore the volume occupied in simple cubic arrangement = 52.4%
(ii) Body centered cubic:
Let us suppose the edge leght = a & radius of each sphere = r then there are 8 spheres at the corners & 1 in the body of unit cell
Therefore number of spheres per unit cell = (8 x1/8) + 1 = 2
Now volume of unit cell = a3 = (4r / √3)3
and volume of a sphere = 4 / 3πr3
Total volume of two spheres = 2 x 4/3πr3
Packing efficiency = (volume of two spheres in unit cell/total volume of unit cell ) x 100
= (2 x 4/3πr3 / (4r/√3)3 ) x 100 = 68%
Therefore volume occupied in bcc arrangement = 68%
(iii) Face centered:
let us suppose the edge length of the unit cell = a
Radius of each sphere = r
Now there are 8 spheres at the corner & 6 at the faces
Therefore number of spheres in unit cell = (8 x 1/8 + 6 x1/2) = 4
From the arrangement of fcc, we get a = 2√2r
Now volume of a unit cell = a3 = (2√2r)3 = 16√2r3
Total volume of 4 spheres = 4 x 4/3 πr3 = 16/3 πr3
Packing efficiency = (volume of 4 spheres in the unit cell/total volume of unit cell) x 100
= (16/3 πr3 /16√2r3) x 100 = 74%
Therefore volume occupied in fcc = 74%
The reaction between A and B is first order with respect to A and zero order with respect to B. Fill in the blanks in the following table:
| Experiment |
A/ mol L - 1 |
B/ mol L - 1 |
Initial rate/mol L - 1 min - 1 |
| I | 0.1 | 0.1 |
2.0 × 10 - 2 |
| II | -- | 0.2 |
4.0 × 10 - 2 |
| III | 0.4 | 0.4 | -- |
| IV | -- | 0.2 |
2.0 × 10 - 2 |
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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Welcome to the NCERT Solutions for Class 12 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 2 , Question 10: Calculate the efficiency of packing in case of a metal crystal for (i) simple cubic (ii) body-....
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