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Test: Packing Efficiency (Old NCERT) - NEET MCQ


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10 Questions MCQ Test Chemistry Class 12 - Test: Packing Efficiency (Old NCERT)

Test: Packing Efficiency (Old NCERT) for NEET 2024 is part of Chemistry Class 12 preparation. The Test: Packing Efficiency (Old NCERT) questions and answers have been prepared according to the NEET exam syllabus.The Test: Packing Efficiency (Old NCERT) MCQs are made for NEET 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Packing Efficiency (Old NCERT) below.
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Test: Packing Efficiency (Old NCERT) - Question 1

Packing efficiency of body centred cubic unit cell is:

Detailed Solution for Test: Packing Efficiency (Old NCERT) - Question 1

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

Test: Packing Efficiency (Old NCERT) - Question 2

Packing efficiency in a unit cell is never 100% because constituent particles are assumed to be:

Detailed Solution for Test: Packing Efficiency (Old NCERT) - Question 2

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

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Test: Packing Efficiency (Old NCERT) - Question 3

The packing efficiency of the two-dimensional square unit cell shown in the given figure is:

Detailed Solution for Test: Packing Efficiency (Old NCERT) - Question 3

Let us take the radius of each atom to be ‘r’. So, from the structure we can say that the diagonal of the square unit is ‘4r’. …….(1)

Let the side of the square unit be ‘L’. So, the diagonal of the square will be:

 

Now since we have 2 values for the diagonal, let’s equate them to each other.
(1) = (2)

Total area (A) of the square unit can be written as:
A = L2
In the above equation put the value of L from equation (3):


= 16r2/2
= 8r2 ……….(4)
Total number of spheres inside the square unit = 1 (the one at the centre) + 4 (1/4) (from the 4 spheres at the corners, each contributing 1 / 4 to the square unit cell)
= 1 + 4 ( 1/4 ) = 2
Area of 1 sphere = πr2
Since there are 2 spheres, the total area of spheres will be = 2(πr2) ………...(5)
Packing fraction or packing efficiency of the square unit cell is defined as the ratio of area occupied by the spheres and total area of the square unit cell
Use the values of the area occupied by the sphere from equation (5) and total area of square unit cell from equation (4).
Packing efficiency = total area of spheres / total area of the square unit cell
= 2(πr2)/8r2
= π/4 (π = 3.14)
= 0. 785

So, the percentage packing efficiency will be 78.5%.
The correct option is: (D) 78.54%

Test: Packing Efficiency (Old NCERT) - Question 4

Packing efficiency in a unit cell is never 100% because constituent particles are assumed to be:

Detailed Solution for Test: Packing Efficiency (Old NCERT) - Question 4

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

Test: Packing Efficiency (Old NCERT) - Question 5

What is the dimensional formula of packing fraction?

Detailed Solution for Test: Packing Efficiency (Old NCERT) - Question 5

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.

Test: Packing Efficiency (Old NCERT) - Question 6

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

Detailed Solution for Test: Packing Efficiency (Old NCERT) - Question 6

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%.

Test: Packing Efficiency (Old NCERT) - Question 7

Which of the following shows maximum packing efficiency?

Detailed Solution for Test: Packing Efficiency (Old NCERT) - Question 7

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. 

Test: Packing Efficiency (Old NCERT) - Question 8

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

Detailed Solution for Test: Packing Efficiency (Old NCERT) - Question 8

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%.

Test: Packing Efficiency (Old NCERT) - Question 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?

Detailed Solution for Test: Packing Efficiency (Old NCERT) - Question 9

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

Test: Packing Efficiency (Old NCERT) - Question 10

If copper, density = 9.0 g/cm3 and atomic mass 63.5, bears face-centered unit cells then what is the ratio of surface area to volume of each copper atom?

Detailed Solution for Test: Packing Efficiency (Old NCERT) - Question 10

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
NA = 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.

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