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MHT-CET Chemistry · Solid State

Packing Efficiency and Voids

Packing efficiency is the fraction of the unit cell that the atoms actually fill — 52.4% simple cubic, 68% bcc, 74% fcc and hcp — and the rest is void; in close packing every atom brings one octahedral and two tetrahedral voids.

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Why this matters

49 PYQs, 6 of them HARD — the chapter's densest HARD cluster. Three question shapes: the occupied or void volume of a cell from its volume (multiply by 0.74, 0.68, 0.524 or their complements), the number of voids in a given number of moles (N and 2N), and the formula of a compound from which voids the cations fill. The HARD rows from 2021 to 2025 are the last shape and a lost-digit stem; the arithmetic is never hard.

Concept 1 of 4: Packing Efficiency: 52.4%, 68%, 74%

Divide the volume of the atoms in the cell by the cell volume. Because the edge–radius relation fixes rr as a multiple of aa, the ratio has no units and no numbers in it — it is a pure property of the cell type.

Definition

  • Simple cubic: 43πr3(2r)3=π6=52.4%\dfrac{\tfrac{4}{3}\pi r^3}{(2r)^3} = \dfrac{\pi}{6} = 52.4\%. Void 47.6%.
  • bcc: 2⋅43πr3(4r/3)3=3π8=68%\dfrac{2 \cdot \tfrac{4}{3}\pi r^3}{(4r/\sqrt{3})^3} = \dfrac{\sqrt{3}\pi}{8} = 68\%. Void 32%.
  • fcc / ccp and hcp: 4⋅43πr3(22r)3=π32=74%\dfrac{4 \cdot \tfrac{4}{3}\pi r^3}{(2\sqrt{2}r)^3} = \dfrac{\pi}{3\sqrt{2}} = 74\%. Void 26%.
  • Silver, copper, gold, aluminium are fcc — packing efficiency 74%.

Packing efficiency

PE=n⋅43πr3a3×100:π6, 3π8, π32\text{PE} = \frac{n \cdot \tfrac{4}{3}\pi r^3}{a^3} \times 100:\quad \tfrac{\pi}{6},\ \tfrac{\sqrt{3}\pi}{8},\ \tfrac{\pi}{3\sqrt{2}}

Worked example

Derive the packing efficiency of the simple cubic cell.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2022 · Shift 1 · Q61Easy

Example 1 · Solid State · Packing Efficiency and Voids

What is the total volume occupied by atoms in bcc unit cell?

Giving 74% for bcc

bcc is the middle value, 68%. 74% belongs to the two close-packed structures (fcc and hcp), which is why they are called close packed.
Drill 1 more on packing efficiency: 52.4%, 68%, 74%

Concept 2 of 4: Occupied Volume, Void Volume and the Volume Per Particle

Given the cell volume, the atoms occupy PE × V and the void is (1 − PE) × V. The volume of ONE particle is the occupied volume divided by the particles per cell. Run the same three numbers in reverse when the void volume is given and the cell volume is asked.

Definition

  • Occupied volume =PE×Vcell= \text{PE} \times V_{\text{cell}}: fcc 0.74 V0.74\,V, bcc 0.68 V0.68\,V, sc 0.524 V0.524\,V.
  • Void volume =(1−PE)×Vcell= (1 - \text{PE}) \times V_{\text{cell}}: fcc 0.26 V0.26\,V, bcc 0.32 V0.32\,V, sc 0.476 V0.476\,V.
  • Volume of one particle =PE×Vn= \dfrac{\text{PE} \times V}{n}: fcc 0.185 V0.185\,V, bcc 0.34 V0.34\,V, sc 0.524 V0.524\,V.
  • Reverse: Vcell=void volume1−PEV_{\text{cell}} = \dfrac{\text{void volume}}{1 - \text{PE}} or volume of one particlePE/n\dfrac{\text{volume of one particle}}{\text{PE}/n}.

Occupied and void volume

Vocc=PE⋅V,Vvoid=(1−PE) V,Vparticle=PE⋅VnV_{\text{occ}} = \text{PE}\cdot V,\qquad V_{\text{void}} = (1-\text{PE})\,V,\qquad V_{\text{particle}} = \frac{\text{PE}\cdot V}{n}

Worked example

The volume of a bcc unit cell is 2.0×10−22 cm32.0 \times 10^{-22}\ \text{cm}^3. Find the void volume and the volume of one particle.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 1 · Q73Easy

Example 2 · Solid State · Packing Efficiency and Voids

Find the void volume of fcc unit cell in cm3^3 if the volume of fcc unit cell is 1.25×10−22 cm31.25 \times 10^{-22}\,\text{cm}^3.

Reporting the occupied volume when ONE particle is asked

'Volume occupied by a particle in fcc' means one of the four: 0.74 V/4=0.185 V0.74\,V/4 = 0.185\,V. The option 0.74 V0.74\,V is always there for the student who skips the division.
Drill 10 more on occupied volume, void volume and the volume per particle

Concept 3 of 4: Tetrahedral and Octahedral Voids: 2N and N

Close packing (ccp or hcp) of N spheres leaves N octahedral holes (each ringed by 6 spheres) and 2N tetrahedral holes (each capped by 4 spheres). For a mole count, N is moles × Avogadro's number.

Definition

  • Tetrahedral void: formed by 4 spheres (three in a layer, one on top); 2 per atom. Per fcc cell: 8.
  • Octahedral void: surrounded by 6 spheres; 1 per atom. Per fcc cell: 4.
  • Total voids per atom =3= 3; in xx mol: tetrahedral 2xNA2xN_A, octahedral xNAxN_A, total 3xNA3xN_A.
  • Octahedral voids are larger than tetrahedral (radius ratio 0.414 against 0.225).

Void counts

Ntet=2N,Noct=N,N=nmol×6.022×1023N_{\text{tet}} = 2N,\qquad N_{\text{oct}} = N,\qquad N = n_{\text{mol}} \times 6.022 \times 10^{23}

Worked example

Find the number of octahedral and tetrahedral voids in 0.5 mol of a metal packed in hcp.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 16th May Shift 2 · Q77Easy

Example 3 · Solid State · Packing Efficiency and Voids

What is the total number of tetrahedral voids in 0.6 mole of compound that forms hcp structure?

Swapping the two counts

Tetrahedral is the SMALLER hole and the MORE numerous: 2 per atom. Octahedral is larger and fewer: 1 per atom. The option with the numbers reversed is always offered.
Drill 7 more on tetrahedral and octahedral voids: 2n and n

Concept 4 of 4: Formula of a Compound From the Voids the Cations Fill

Let the close-packed ions number N. A cation that fills a fraction ff of the tetrahedral voids numbers 2fN2fN; of the octahedral voids, fNfN. The ratio of the two counts, reduced to whole numbers, is the formula. For corner-and-face arrangements count the sharing fractions instead.

Definition

  • B in ccp/hcp, A in 13\tfrac{1}{3} of tetrahedral voids: A =23N= \tfrac{2}{3}N, ratio A:B =2:3= 2:3 → A2B3\text{A}_2\text{B}_3.
  • B in ccp, A in half the tetrahedral voids: A =N= N → AB (zinc blende).
  • B in ccp, A in all octahedral voids: AB (rock salt); in all tetrahedral voids: A2B\text{A}_2\text{B} (fluorite-type antistructure).
  • A at corners, B at face centres: 1:31 : 3 → AB3\text{AB}_3. A at corners, B at body centre: AB.
  • Removing an atom: A at corners with one corner missing → 78\tfrac{7}{8} A per cell.

Cations from void fraction

nA=ftet⋅2N+foct⋅N,formula=AnABNn_{\text{A}} = f_{\text{tet}} \cdot 2N + f_{\text{oct}} \cdot N,\qquad \text{formula} = \text{A}_{n_{\text{A}}}\text{B}_{N}

Worked example

Anions Y form an hcp lattice; cations X fill 23\tfrac{2}{3} of the octahedral voids. Find the formula.
Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 15th May Shift 1 · Q97Hard

Example 4 · Solid State · Packing Efficiency and Voids

A compound is formed by two elements A and B. The atoms of element B form ccp structure. The atoms of A occupy 13\frac{1}{3} of tetrahedral voids. What is the formula of the compound?

Forgetting the factor 2 on tetrahedral voids

13\tfrac{1}{3} of the tetrahedral voids is 13×2N=23N\tfrac{1}{3} \times 2N = \tfrac{2}{3}N, giving A2B3\text{A}_2\text{B}_3. Using 13N\tfrac{1}{3}N gives AB3\text{AB}_3, which is offered as an option every time.
Drill 4 more on formula of a compound from the voids the cations fill

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (4)

  • Packing Efficiency: 52.4%, 68%, 74%

    Packing efficiency

    PE=n⋅43πr3a3×100:π6, 3π8, π32\text{PE} = \frac{n \cdot \tfrac{4}{3}\pi r^3}{a^3} \times 100:\quad \tfrac{\pi}{6},\ \tfrac{\sqrt{3}\pi}{8},\ \tfrac{\pi}{3\sqrt{2}}
  • Occupied Volume, Void Volume and the Volume Per Particle

    Occupied and void volume

    Vocc=PE⋅V,Vvoid=(1−PE) V,Vparticle=PE⋅VnV_{\text{occ}} = \text{PE}\cdot V,\qquad V_{\text{void}} = (1-\text{PE})\,V,\qquad V_{\text{particle}} = \frac{\text{PE}\cdot V}{n}
  • Tetrahedral and Octahedral Voids: 2N and N

    Void counts

    Ntet=2N,Noct=N,N=nmol×6.022×1023N_{\text{tet}} = 2N,\qquad N_{\text{oct}} = N,\qquad N = n_{\text{mol}} \times 6.022 \times 10^{23}
  • Formula of a Compound From the Voids the Cations Fill

    Cations from void fraction

    nA=ftet⋅2N+foct⋅N,formula=AnABNn_{\text{A}} = f_{\text{tet}} \cdot 2N + f_{\text{oct}} \cdot N,\qquad \text{formula} = \text{A}_{n_{\text{A}}}\text{B}_{N}

Watch out for (4)

Test yourself on Solid State

20 past MHT-CET questions from this chapter, timed at 18 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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