Packing Efficiency (Old NCERT) - Free MCQ Practice Test with solutions,

The packing efficiency of a body-centred cubic unit cell is 68%.

Explanation: 

The packing efficiency is the fraction of the crystal (or unit cell) actually occupied by the atoms. It is usually represented by a percentage or volume fraction.

Mathematically Packing Efficiency is:

Number of atoms × volume occupied by 1 share × 100/Total volume of unit cell × 100

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

The constituent particles i.e. atoms, molecules and ions are assumed to be spheres.

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.

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%.

Packing efficiency is defined as the percentage of space occupied by constituent particles packed inside the lattice.

Packing efficiency = Number of atoms × volume occupied by one share × 100/total volume of the unit cell

• Packing efficiency in fcc and hcp structures is 74%

• The packing efficiency of the body-centred unit cell is 68%.

• The packing efficiency of the simple unit cell is 52.4%

Since HCP shows the maximum packing efficiency, the correct answer is A. 

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%.

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√2r.
Surface area (S.A.) = 6a²
From the relationship,
a² = 8r²
S.A. = 6a² = 48r²
When r = 166 pm, S.A. = 48(166pm) = 1.32 x 10⁶ pm². 

Density, d of unit cell is given by d
Given,
Density, d = 9.0 g/cm³
Atomic mass, M = 63.5 g/mole
Edge length = a
N_A = Avogadro’s number = 6.022 x 10²³
z = 4 atoms/cell
On rearranging the equation for density we get a³ 
Substituting the given values:

Therefore, a = 360.5 pm
The relation of edge length ‘a’ and radius of particle ‘r’ for FCC packing i.e. a = 2√2r.
On substituting the value of ‘a’ in the given relation,127.46 pm
Now, for spherical particles volume, V = 4πr³/3 and surface area, S = 4πr²
Required ratio = S/V=4πr²/(4πr³/3) = 3/r (after simplifying)
Thus, S/V = 3/127.46 = 0.0235.