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    JEE Mains 2026 April 6 Shift 2 Question Paper with Solutions PDF
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    Packing Efficiency Of A Unit Cell - Practice Questions & MCQ

    Edited By admin | Updated on Sep 18, 2023 18:35 AM | #JEE Main

    Quick Facts

    • Packing Efficiency is considered one the most difficult concept.

    • 128 Questions around this concept.

    Solve by difficulty

    QuestionEASY

    Total volume of atoms present in a face centred cubic unit cell of a metal is ( r is atomic radius ) :

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    What is the length of side of a cell when element having an atomic radius 2.68 nm crystallizes in an face centered cubic unit cell

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    Lithium forms body centred cubic structure. The length of the side of its unit cell is 351 pm. Atomic radius of the lithium will be :

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    Which of the following conformations will be the most stable ?

     

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    Percentages of free space in cubic close packed structure and in body centred packed structure are respectively

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    Calculate the density of solid if the atomic solid is crystallises in body centered cubic lattice and the inner surface of atom at adjacent corner are separated by 60.3 pm and atomic mass of A is.

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    Calculate the ratio of the side of the unit cell of KCl and NaCl if $\mathrm{r}_{\mathrm{Na}^{+}} / \mathrm{r}_{\mathrm{Cl}^{-}}=0.55$ and $\mathrm{r}_{\mathrm{K}+} / \mathrm{r}_{\mathrm{Cl}^{-}}=0.72$

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    Copper crystallizes in fcc with a unit cell length of 361 pm. What is the radius of copper atom ?

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    In metallic solids, the number of atoms for the face-centered and the body-centered cubic unit cells, are, respectively

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    In the cubic close packing, the unit cell has ________.

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    Concepts Covered - 0

    Packing Efficiency

    Packing Efficiency in HCP and CCP Structures

    Both types of close packing (hcp and ccp) are equally efficient. Let us calculate the efficiency of packing in ccp structure. In the figure, let the unit cell edge length be 'a' and face diagonal AC = b.

    \begin{array}{l}{\text {In } \Delta \mathrm{ABC}} \\\\ {\mathrm{AC}^{2}=\mathrm{b}^{2}=\mathrm{BC}^{2}+\mathrm{AB}^{2}} \\\\ {=\mathrm{a}^{2}+\mathrm{a}^{2}=2 \mathrm{a}^{2} \text { or }} \\\\ {\mathrm{b}=\sqrt{2} \mathrm{a}} \\\\ {\text { If } \mathrm{r} \text { is the radius of the sphere, we find }} \\\\ {\mathrm{b}=4 \mathrm{r}=\sqrt{2 \mathrm{a}}} \\\\ {\text { or } \mathrm{a}=\frac{4 \mathrm{r}}{\sqrt{2}}=2 \sqrt{2 \mathrm{r}}} \\\\ {\left.\text { (We can also write, } \mathrm{r}=\frac{\mathrm{a}}{2 \sqrt{2}}\right)}\end{array}

    \begin{array}{l}{\text {As we know, that each unit cell in ccp structure, has }} {\text {effectively } 4 \text { spheres. Total volume of four }} \\ {\text {spheres is equal to } 4 \times(4 / 3) \pi \mathrm{r}^{3} \text { and volume of the cube is }} {\mathrm{a^{3} or\: }(2 \sqrt{2} \mathrm{r})^{3} \text { . }} \\\\ {\text {Therefore, }}\end{array}

    \begin{array}{l}{\text { Packing efficiency = }} {\frac{\text { Volume occupied by four spheres in the unit cell } \times 100 \%}{\text {Total volume of the unit cell }}}\end{array}

    \begin{array}{l}{=\frac{4 \times(4 / 3) \pi \mathrm{r}^{3} \times 100}{(2 \sqrt{2})^{3}} \%} \\\\ {=\frac{(16 / 3) \pi \mathrm{r}^{3} \times 100}{16 \sqrt{2} \mathrm{r}^{3}}=74 \%}\end{array}
     

    Efficiency of Packing in Body Centred Cubic Structures


    From figure, it is clear that the atom at the centre is in touch with the other two atoms diagonally arranged.

    \begin{array}{l}{\text {In } \Delta \mathrm{EFD}} \\\\ {\mathrm{b}^{2}=\mathrm{a}^{2}+\mathrm{a}^{2}=2 \mathrm{a}^{2}} \\\\ {\mathrm{b}=\sqrt{2 \mathrm{a}}} \\\\ {\text { Now in } \Delta \mathrm{AFD}} \\\\ {\mathrm{c}^{2}=\mathrm{a}^{2}+\mathrm{b}^{2}=\mathrm{a}^{2}+2 \mathrm{a}^{2}=3 \mathrm{a}^{2}} \\\\ {\mathrm{c}=\sqrt{3} \mathrm{a}}\end{array}
    The length of the body diagonal is equal to 4r, here is the radius of the sphere (atom), as all the three spheres along the diagonal touch each other.

    \begin{array}{l}{\text { So } \sqrt{3} a=3 r} \\ {\qquad a=\frac{4 r}{\sqrt{3}}} \\ {\text { Hence we can write, } r=\frac{\sqrt{3}}{4} \text { a }} \\ {\text { In this type of structure, total number of atoms is } 2} \\ {\text { and their volume is } 2 \times \frac{(4)}{3} \pi r^{3}}\end{array}
    \\\mathrm{Volume \: of \: the\: cube,\: a^{3}\: will\: be\: equal\: to\: \frac{(4r)^{3}}{\sqrt{3}} \: or\: a^{3}}\\\\\mathrm{Therefore,\: Packing\: efficiency\: =\: \frac{\text { Volume occupied by two spheres in the unit } \times 100 \%}{\text { Total volume of the unit cell }}}\\\\=\frac{4 \times(4 / 3) \pi \mathrm{r}^{3} \times 100}{[(4 \sqrt{3}) \mathrm{r}]^{3}} \%\\\\=\frac{4 \times(4,3) \pi \mathrm{r}^{3} \times 100}{64 /(3 \sqrt{3}) \mathrm{r}^{3}} \%\\\\=68 \%

    Mathematical Analysis of Cubic System

    Coordination Number (C. No.)

    • In Simple Cubic (SC): 6
    • In Face Centered Cubic (FCC): 12
    • In Body Centered Cubic (BCC): 8

    Density of Lattice Matter (d)
    It is the ratio of mass per unit cell to the total volume of a unit cell and it is found out as follows.

    \mathrm{d}=\frac{\mathrm{Z} \times \text { Atomic weight }}{\mathrm{N}_{0} \times \text { Volume of unit cell }\left(\mathrm{a}^{3}\right)}
    Here, d = Density
             Z = Number of atoms
             N0 = Avogadro number
             a3 = Volume
             a = Edge length
    Here in order to find density of unit cell in cm3, m must be taken in g/mole and should be in cm.

    Radius Ratio
    It is the ratio of radius of an octahedral void to the radius of sphere-forming the close-packed arrangement. Normally, ionic solids are more compact as voids are also occupied by cation (smaller in size) pattern of arrangements and type of voids both depend upon relative size (ionic size) of two ions in solid. 
    Example, when r+ = r- the most probable and favourable arrangement is BCC type.
    With the help of relative ionic radii, it is easier to predict the most probable arrangement. This property is expressed as radius ratio.

    \mathrm{\text { Radius ratio }=\frac{r^{+} \text {(radius of cation) }}{r^{-} \text { (Radius of anion) }}}
    From the value of radius ratio, it is clear that larger the radius ratio larger is the size of cation and more will be the number of anions needed to surround it, that is, more co-ordination number.

    • Radius ratio for tetrahedron

      \begin{array}{l}{\text { Angle } A B C \text { is the tetrahedral angle of } 109.5^{\circ}} \\\\ {\angle A B D=\frac{109.5}{2}=54.75^{\circ}} \\\\ {\text { In triangle } A B D} \\\\ {\text { Sin } A B D=0.8164=A D / A B} \\\\ {\text { or } \frac{r^{*}+r}{r}=\frac{1}{0.8164}=1.225} \\\\ {\text { or } \frac{r^{*}}{\Gamma}=0.225}\end{array}


       
    •  Radius ratio for octahedron
      \begin{array}{l}{\mathrm{AB}=\mathrm{r}^{+}+\mathrm{r}} \\\\ {\mathrm{BD}=\mathrm{r}} \\\\ {\angle \mathrm{ABC}=45^{\circ}} \\\\ {\text { In triangle } \mathrm{ABD}}\end{array}\\\\\\\begin{array}{l}{\operatorname{Cos} A B D=0.7071=B D / A B} \\\\ {=\frac{r}{r^{+}+r}} \\\\ {\text { or } \frac{r^{+}+r}{r}=\frac{1}{0.707}=1.414} \\\\ {\text { or } \frac{r}{r^{4}}=0.414}\end{array}

    Study other Related Concepts

    Packing Efficiency Of A Unit Cell Current Topic

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    Charles’ Law
    Gay Lussac’s Law
    Avogadro's Law
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    Dalton's Law of Partial Pressure
    Graham's Law: Diffusion And Effusion

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