This set of Chemistry MCQs for Schools focuses on “Solid State – Packing Efficiency”.

1. What does the ratio ‘space occupied/total space’ denote?

a) Packing factor

b) Packing efficiency

c) Particle fraction

d) Packing unit

View Answer

Explanation: Packing factor is a fraction of total space of the unit cell occupied by the constituent particles.

2. What is the dimensional formula of packing fraction?

a) M^{0}L^{3}T^{0}

b) M^{0}L^{0}T^{0}

c) ML^{0}T^{0}

d) M^{0}L^{2}T^{0}

View Answer

Explanation: Packing fraction is a dimensionless quantity which is the ratio of space occupied to total crystal space available. Since both the quantities have the same units the ratio renders dimensionless.

3. Arrange the types of arrangement in terms of decreasing packing efficiency.

a) BCC < Simple cubic < CCP

b) HCP < CCP < BCC

c) HCP < BCC < Simple cubic

d) CCP < BCC < HCP

View Answer

Explanation: HCP and CCP have the highest packing efficiency of 74% followed by BCC which is 68%. The simple cubic structure has a packing efficiency of 54%.

4. If the body-centered unit cell is assumed to be a cube of edge length ‘a’ with spherical particles of radius ‘r’ then how is the diameter, d of particle and surface area, S of the cell related?

a) S = 32d^{4}/3

b) S = 2d^{2}

c) S = 4d^{2}

d) S = 8d^{2}

View Answer

Explanation: For BCC unit cell the relation between radius of a particle ‘r’ and edge length of unit cell, a, is r = \(\frac{\sqrt{3}}{4}\)a.

We know that diameter, d = 2r = \(\frac{\sqrt{3}}{2}\)a

Implying d

^{2}= \(\frac{3}{4}\)a

^{2}

Therefore, 4d

^{2}/3=a

^{2}

Multiplying by 6 on both sides gives S = 6a

^{2}= 8d

^{2}, where S is the surface area of the cube = 6a

^{2}.

5. Which of the following metals would have the highest packing efficiency?

a) Copper

b) Potassium

c) Chromium

d) Polonium

View Answer

Explanation: Copper metal bears face-centered unit cells in its crystal structure. Potassium and chromium both have body-centered unit cells whereas polonium is the only known metal to bear a simple cubic structure. FCC structure has the highest efficiency.

6. “The packing efficiency can never be 100%”. Is this true or false?

a) False

b) True

View Answer

Explanation: Packing efficiency can never be 100% because in packing calculations all constituent particles filling up the cubical unit cell are assumed to be spheres.

7. What are the percentages of free space in a CCP and simple cubic lattice?

a) 52% and 74%

b) 48% and 26%

c) 26% and 48%

d) 74% and 52%

View Answer

Explanation: The packing efficiency in CCP and simple cubic lattice are 74% and 52%, respectively. Hence the corresponding free spaces will be 100% – 74% = 26% and 100% – 52% = 48%.

8. How many atoms surround the central atom present in a unit cell with the least free space available?

a) 4

b) 6

c) 8

d) 12

View Answer

Explanation: FCC, CCP and HCP are unit cells with least free space available i.e. highest packing efficiency. The coordination number of given cells are 12.

9. If metallic atoms of mass 197 and radius 166 pm are arranged in ABCABC fashion then what is the surface area of each unit cell?

a) 1.32 × 10^{6} pm^{2}

b) 1.32 × 10^{-18}pm^{2}

c) 2.20 × 10^{5} pm^{2}

d) 2.20 × 10^{-19} pm^{2}

View Answer

Explanation: ABCABC arrangement is found in CCP.

In closed cubic packing, relation between edge length of unit cell, a, and radius of particle, r, is given as a=2\(\sqrt{2}\)r.

Surface area (S.A.) = 6a

^{2}

From the relationship,

a

^{2}= 8r

^{2}

S.A. = 6a

^{2}= 48r

^{2}

When r = 166 pm, S.A. = 48(166pm) = 1.32 x 10

^{6}pm

^{2}.

10. If copper, density = 9.0 g/cm^{3} and atomic mass 63.5, bears face-centered unit cells then what is the ratio of surface area to volume of each copper atom?

a) 0.0028

b) 0.0235

c) 0.0011

d) 0.0323

View Answer

Explanation: Density, d of unit cell is given by d = \(\frac{zM}{a^3N_A}\)

Given,

Density, d = 9.0 g/cm

^{3}

Atomic mass, M = 63.5 g/mole

Edge length = a

N

_{A}= Avogadro’s number = 6.022 x 10

^{23}

z = 4 atoms/cell

On rearranging the equation for density we get a

^{3}= \(\frac{zM}{dN_A}\)

Substituting the given values:

a

^{3}= \(\frac{4 \times 63.5}{9 \times 6.022 \times 10^{23}}\)

Therefore, a = 360.5 pm

The relation of edge length ‘a’ and radius of particle ‘r’ for FCC packing i.e. a = 2\(\sqrt{2}\)r.

On substituting the value of ‘a’ in the given relation, r = \(\frac{360.5}{2\sqrt{2}}\)=127.46 pm

Now, for spherical particles volume, V = 4πr

^{3}/3 and surface area, S = 4πr

^{2}

Required ratio = S/V=4πr

^{2}/(4πr

^{3}/3) = 3/r (after simplifying)

Thus, S/V = 3/127.46 = 0.0235.

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