Close-Packed Structures

ChemistrySolid StateFor JEE aspirants

Close-packed structures describe how identical spheres (atoms of a metal) stack to fill space as fully as possible. Rows become square or hexagonal layers, and layers stack as simple cubic, body-centred cubic, hexagonal close-packed (hcp) or cubic close-packed (ccp, the same as fcc). For each one this page derives the edge-radius relation, atoms per cell $Z$ , coordination number, packing efficiency and density. hcp and ccp are the true close-packed structures, filling 74% of space. Solid state is now in the JEE Advanced syllabus only.

On this page11D and 2D packing2Simple cubic3Body-centred cubic4Stacking hexagonal layers5hcp6ccp = fcc7Comparison8Density problems

Key Formulas - Quick Reference

★ Must learn Edge-radius relation: sc $a = 2 r$ ; bcc $3 a = 4 r$ ; fcc $2 a = 4 r$ ; hcp $a = 2 r$ , $c = 1.633 a$ .
★ Must learn $Z$ : sc 1, bcc 2, fcc (ccp) 4, hcp 6. Coordination number: sc 6, bcc 8, fcc and hcp 12.
★ Must learn Packing efficiency: sc $\frac{π}{6} = 52.4%$ , bcc $\frac{3 π}{8} = 68%$ , fcc and hcp $\frac{π}{3 2} = 74%$ .
★ Must learn Density $d = \frac{Z M}{N _{A} a ^{3}}$ ; for hcp, volume $= 242 r^{3}$ .
Nearest-neighbour distance $d_{nn}$ : sc $a$ , bcc $\frac{3}{2} a$ , fcc $\frac{a}{2}$ ; always $d_{nn} = 2 r$ .
2D packing: square CN 4, 78.5%; hexagonal CN 6, 90.7%.
★ Must learn Stacking: hcp = ABAB (3rd layer over 1st), ccp = ABCABC (4th over 1st); layer spacing $= 2 r \frac{2}{3}$ .

1. The Idea of Close Packing

In metals and in many ionic lattices, the particles behave like hard spheres of radius $R$ (or $r$ ). The most stable arrangement is the one in which each sphere touches as many neighbours as possible: minimum potential energy, maximum attraction, and the least empty space. This is close packing. We build it up one dimension at a time.

1.1 One dimension

Spheres in a row can be close-packed in only one way: each touches its two neighbours, so the centre-to-centre distance is $2 R$ and the coordination number is 2.

Figure 1: In one dimension there is only one way to close-pack spheres: touching in a row,

d = 2 R

, coordination number 2.

1.2 Two dimensions

A 2D layer is made by placing close-packed rows side by side. There are two ways to do it.

Figure 2: Two ways to stack 1D rows. Placing each row in the hollows of the one below (right) packs more tightly: 6 neighbours and 90.7% of the area filled, against 4 neighbours and 78.5% for square packing.

Square packing (AAA)

Each row sits directly on the row below. Cell: square of side $a = 2 R$ with $Z = 1$ . CN = 4. Area efficiency $\frac{π R ^{2}}{( 2 R ) ^{2}} = \frac{π}{4} = 78.5%$ .

Hexagonal packing (ABAB)

Each row sits in the hollows of the row below. Cell: rhombus of side $2 R$ , angle $6 0^{\circ}$ , area $23 R^{2}$ , $Z = 1$ . CN = 6. Efficiency $\frac{π}{2 3} = 90.7%$ .

Hexagonal packing leaves triangular hollows between spheres; these are the starting point for everything in 3D.

Key idea

In 2D, hexagonal packing wins: 6 neighbours and 90.7% of the area filled, against 4 and 78.5% for square packing.

2. Stacking Square Layers: Simple Cubic

Place square layers exactly on top of one another (AAA...). Each sphere touches 4 in its own layer, 1 above and 1 below. The unit cell is a simple cube with an atom at each corner.

Figure 3: Simple cubic comes from square layers stacked A over A. Corner spheres touch along the edge, so

a = 2 R

. Polonium is the only element with this structure.

Relation between $a$ and $r$ : corner atoms touch along the edge, so $a = 2 r$ .
Atoms per cell: $Z = 8 \times \frac{1}{8} = 1$ .
Packing efficiency: $\frac{1 \times \frac{4}{3} π r ^{3}}{( 2 r ) ^{3}} = \frac{π}{6} = 52.4%$ (47.6% empty).
Coordination number: 6 (four in the layer, one above, one below).
Example: polonium is the only element with this structure; it is too inefficient for most metals.

2.1 Neighbours in simple cubic

Neighbour	Distance	Number	Where they are
Nearest	$a$	6	along the three axes
2nd	$2 a$	12	along face diagonals
3rd	$3 a$	8	along body diagonals
4th	$2 a$	6	two cells along an axis
5th	$5 a$	24	one cell over, two cells across

3. Square Layers in the Hollows: Body-Centred Cubic

Now place the second square layer so its spheres sit in the hollows of the first (ABAB...). The spheres of one layer no longer touch each other; they spread apart slightly so that each sits snugly on four spheres below and four above. The resulting unit cell is body-centred cubic.

Figure 4: In bcc the corner spheres do not touch each other; each touches the body-centre sphere. Along the body diagonal

3 a = 4 R

, so

R = 0.433 a

Relation: atoms touch along the body diagonal only: $3 a = 4 r$ , so $a = \frac{4 r}{3}$ .
Atoms per cell: $Z = 8 \times \frac{1}{8} + 1 = 2$ .
Packing efficiency: $\frac{2 \times \frac{4}{3} π r ^{3}}{( 4 r / 3 ) ^{3}} = \frac{3 π}{8} = 68%$ (32% empty).
Coordination number: 8. Each atom touches 4 atoms in the layer above, 4 in the layer below and none in its own layer.
Examples: alkali metals (Li, Na, K, Rb, Cs), Ba, Fe (at room temperature), Cr, Mo, W, V.

3.1 Neighbours in bcc

Neighbour	Distance	Number
Nearest	$\frac{3}{2} a = 2 r$	8
2nd	$a$	6
3rd	$2 a$	12
4th	$\frac{11}{2} a$	24
5th	$3 a$	8

Figure 5: Nearest neighbours of one atom. Coordination number rises from 6 (sc) to 8 (bcc) to 12 (fcc), and so does the packing efficiency.

In bcc the 6 second-nearest neighbours at distance

a

are only

\frac{2}{3} = 1.155

times farther than the 8 nearest. That is why bcc is sometimes described as having an effective coordination of 8 + 6 = 14 and why it still packs fairly well (68%).

Key idea

bcc: touch along the body diagonal (

3 a = 4 r

Z = 2

, CN 8, 68% filled.

Quick Recall: tap to check

Why is the CN in bcc 8 and not 12?

Each atom touches 4 atoms above and 4 below; atoms in its own square layer do not touch it.

What fraction of a bcc cell is empty?

1 - \frac{3 π}{8} = 0.32

, so 32%.

Nearest-neighbour distance of K (bcc,

a = 5.2

Å)?

\frac{3}{2} \times 5.2 = 4.5

Å.

4. Stacking Hexagonal Layers: hcp and ccp

The best 2D layer is the hexagonal layer (call it A). Its hollows come in two sets, b and c, arranged alternately. A second layer B can sit in only one set, because neighbouring hollows are too close for spheres in both. So after layer B, half the hollows of A (the c set) are still uncovered.

Figure 6: Layer B can occupy only one set of hollows of layer A (the b hollows). Where the third layer goes decides the structure: over A gives hcp, over the open c hollows gives ccp.

For the third layer there are two choices, and they give the two close-packed structures:

Third layer directly over layer A (over the hollows of B that lie above A spheres): the sequence is ABABAB... This is hexagonal close packing (hcp). The c hollows are never covered; they run straight through the crystal.
Third layer over the c hollows: the third layer, C, differs from both A and B, and the fourth layer repeats A: ABCABC... This is cubic close packing (ccp).

Figure 7: The only difference between hcp and ccp is the stacking sequence. Both have every sphere touching 12 others and 74% of space filled.

4.1 Hexagonal close packing (hcp)

Figure 8: The hcp unit cell. Corner atoms are shared by 6 prisms, face-centre atoms by 2, and the 3 middle atoms belong wholly to the cell:

Z = 6

Relation: spheres touch within a layer, so $a = b = 2 r$ and $γ = 12 0^{\circ}$ .
Height: the cell spans two layers. From the tetrahedron below, $c = 2 \frac{2}{3} a = 1.633 a$ .

Figure 9: A B atom and the three A atoms under it form a regular tetrahedron of edge

a

. Its height is

h = 2/3 a

, and one hcp cell spans two layers, so

c = 2 h = 1.633 a

Base area of the hexagon $= 6 \times \frac{3}{4} a^{2} = \frac{3 3}{2} a^{2}$ .
Volume $= \frac{3 3}{2} a^{2} \times 2 \frac{2}{3} a = 32 a^{3} = 32 (2 r)^{3} = 242 r^{3}$ .
$Z = 12 \times \frac{1}{6} + 2 \times \frac{1}{2} + 3 = 6$ .
Packing efficiency $= \frac{6 \times \frac{4}{3} π r ^{3}}{24 2 r ^{3}} = \frac{π}{3 2} = 74%$ .
Coordination number 12: 6 in its own layer, 3 above and 3 below.

Examples: Be, Mg, Zn, Cd, Ti, Co.

4.2 Cubic close packing (ccp) = face-centred cubic

The ABC layers of ccp lie perpendicular to the body diagonal of a face-centred cube. So the unit cell of ccp is the fcc cell, and the two names describe one structure.

Figure 10: Tilt an fcc cell so its body diagonal points up and the close-packed layers appear: corner A, then plane B, plane C, then corner A again. So ccp is described by the fcc unit cell.

Figure 11: In fcc each face-centre sphere touches the four corner spheres of its face. Along the face diagonal

2 a = 4 R

, so

R = 0.354 a

Relation: atoms touch along the face diagonal: $2 a = 4 r$ , so $a = 22 r$ .
Atoms per cell: $Z = 8 \times \frac{1}{8} + 6 \times \frac{1}{2} = 4$ .
Packing efficiency: $\frac{4 \times \frac{4}{3} π r ^{3}}{( 2 2 r ) ^{3}} = \frac{π}{3 2} = 74%$ .
Coordination number: 12. A face-centre atom touches 4 corner atoms of its own face and 4 face-centre atoms in each of the two cells sharing that face: 4 + 4 + 4. Seen as layers, it is 6 in its own layer, 3 above and 3 below.
Examples: Cu, Ag, Au, Al, Ni, Pb, Pt, and the noble gases in the solid state.

Neighbour	Distance	Number
Nearest	$\frac{a}{2} = 2 r$	12
2nd	$a$	6
3rd	$\frac{3}{2} a$	24
4th	$2 a$	12
5th	$\frac{5}{2} a$	24

hcp

ABAB stacking. Hexagonal cell, $Z = 6$ , $a = 2 r$ , $c = 1.633 a$ . CN 12, 74%. The c hollows form open channels. Mg, Zn, Be, Ti, Co.

ccp (fcc)

ABCABC stacking. Cubic cell, $Z = 4$ , $a = 22 r$ . CN 12, 74%. No open channels; slip planes in four directions make fcc metals very ductile. Cu, Ag, Au, Al.

Exam Trick Same 12, same 74%, different sequence. Remember: hcp is a two-step dance (AB), ccp a three-step dance (ABC). A question asking which layers repeat in ccp wants 'first and fourth'.

Key idea

hcp and ccp are equally efficient (74%, CN 12); they differ only in whether the 3rd layer repeats the 1st (hcp) or the 4th does (ccp).

Quick Recall: tap to check

In ccp, which layers are identical?

1st, 4th, 7th ...: every third layer repeats (ABCABC).

How many atoms in an hcp unit cell, and why?

12 \times \frac{1}{6}

corners +

2 \times \frac{1}{2}

face centres + 3 inside.

Distance between two close-packed layers for spheres of radius

r

2 r \frac{2}{3} = 1.633 r

(half of

c

in hcp).

5. Comparing the Four Structures

Property	Simple cubic	bcc	hcp	ccp (fcc)
Stacking	square AAA	square ABAB	hexagonal ABAB	hexagonal ABCABC
$a$ in terms of $r$	$2 r$	$\frac{4 r}{3}$	$2 r$ ( $c = 1.633 a$ )	$22 r$
$Z$	1	2	6	4
CN	6	8	12	12
Packing efficiency	52.4%	68%	74%	74%
Examples	Po	Na, K, Fe, Cr, W	Mg, Zn, Be, Ti	Cu, Ag, Au, Al, Ni

Figure 12: Packing efficiency, computed from the exact formulas at right. hcp and ccp tie for the best 3D packing at 74%; diamond, with only 4 neighbours, is the emptiest at 34%.

JEE Advanced Packing efficiency does not depend on the size of the atom:

r

cancels in every ratio, so a lattice of tiny atoms and one of huge atoms are equally full. A useful shortcut: the separation of two close-packed layers is

\frac{2}{3} \times

(sphere spacing). With sphere spacing

2 r

, it is

2 \frac{2}{3} r

; in the fcc cell this equals one third of the body diagonal,

\frac{3 a}{3}

, as the three layer spacings A to B to C to A must fill the diagonal

3 a

. Check:

\frac{3}{3} (22 r) = 2 \frac{2}{3} r

Key idea

More neighbours means better packing: CN 6 → 52%, CN 8 → 68%, CN 12 → 74%.

6. Density of a Crystal

Because the crystal is only the unit cell repeated, the density of one cell is the density of the crystal:

d = \frac{mass of cell}{volume of cell} = \frac{Z \times M}{N _{A} \times a ^{3}}

With any three of $d$ , $Z$ , $M$ and $a$ you can find the fourth. Finding $Z$ from measured $d$ , $a$ and $M$ tells you the lattice type (1 sc, 2 bcc, 4 fcc).

Figure 13: One route solves every density, edge-length, radius or molar-mass problem. Most wrong answers come from skipping the unit conversion.

Exam Trick Quick conversions: 1 Å

= 1 0^{- 8}

cm, so

(1 \overset{˚}{A})^{3} = 1 0^{- 24}

^{3}

; and

N_{A} \times 1 0^{- 24} = 0.6022

. So

d = \frac{Z M}{0.6022 \times a ^{3}}

with

a

in Å gives g cm

^{- 3}

directly.

For hcp the cell is not a cube. Use volume

= 32 a^{3} = 242 r^{3}

with

Z = 6

Key idea

d a^{3} N_{A} = Z M

links structure (

Z

a

) to measurable quantities (

d

M

Quick Recall: tap to check

With

a

in Å, what is

d

d = \frac{Z M}{0.6022 a ^{3}}

g cm

^{- 3}

A metal has

d = 2.7

g cm

^{- 3}

M = 27

a = 4.05

Å. Lattice?

Z = \frac{2.7 \times 0.6022 \times 4.0 5 ^{3}}{27} = 4

: fcc.

Radius of Na (bcc,

a = 4.29

Å)?

\frac{3}{4} \times 4.29 = 1.86

Å.

6.1 The whole concept at a glance

Figure 14: Close-packed structures on one page: each branch holds the

a

R

relation,

Z

, coordination number and packing efficiency.

7. Solved Examples

Solved Example 1

How many nearest and next-nearest neighbours respectively does potassium have in a bcc lattice?
(A) 8, 8
(B) 8, 6
(C) 6, 8
(D) 8, 2

Solution:

Answer: (B). 8 nearest neighbours at $\frac{3}{2} a$ (the body-diagonal contacts) and 6 next-nearest at $a$ (the body centres of the six adjacent cells).

Solved Example 2

If a metal has a bcc structure, its coordination number is 8, because
(A) each atom touches four atoms in the layer above it, four in the layer below it and none in its own layer
(B) each atom touches four atoms in the layer above, four in the layer below and one in its own layer
(C) two atoms touch four atoms in the layer above them, four in the layer below and none in their own layer
(D) each atom touches eight atoms in the layer above, eight in the layer below and none in its own layer

Solution:

Answer: (A). In bcc the atoms of one square layer are pulled apart and do not touch; all 8 contacts are with the layers above and below.

Solved Example 3

Potassium crystallises in a body-centred cubic lattice with

a = 5.2

Å. Find (a) the distance between nearest neighbours, (b) the distance between next-nearest neighbours, (c) the number of nearest neighbours, (d) the number of next-nearest neighbours, (e) the density of crystalline K (

M = 39

g mol

^{- 1}

Solution:

(a) $d_{nn} = \frac{3}{2} a = \frac{3 \times 5.2}{2} = 4.5$ Å. (b) $a = 5.2$ Å. (c) 8. (d) 6.

(e) d = \frac{2 \times 39}{( 5.2 \times 1 0 ^{- 8} ) ^{3} \times 6.022 \times 1 0 ^{23}} = 0.92 g cm^{- 3}

Solved Example 4

A metal crystallises in two cubic phases, fcc and bcc, whose unit-cell lengths are 3.5 Å and 3.0 Å respectively. The ratio of their densities (fcc : bcc) is
(A) 3.12
(B) 2.04
(C) 1.26
(D) 0.72

Solution:

Answer: (C).

\frac{d _{f cc}}{d _{b cc}} = \frac{Z _{1} / a _{1}^{3}}{Z _{2} / a _{2}^{3}} = \frac{4}{( 3.5 ) ^{3}} \times \frac{( 3.0 ) ^{3}}{2} = \frac{108}{85.75} = 1.26

Solved Example 5

In a ccp structure, the
(A) first and third layers are repeated
(B) first and fourth layers are repeated
(C) second and fourth layers are repeated
(D) first, third and sixth layers are repeated

Solution:

Answer: (B). ccp is ABCABC, so layer 4 is a copy of layer 1. 'First and third' describes hcp.

Solved Example 6

Lithium borohydride (

LiB H_{4}

) crystallises in an orthorhombic system with 4 formula units per unit cell. The cell dimensions are a = 6.8 Å, b = 4.4 Å and c = 7.2 Å. If the molar mass is 21.76 g mol

^{- 1}

, the density is
(A) 0.6708 g cm

^{- 3}

(B) 1.6708 g cm

^{- 3}

^{- 3}

(D) none of these

Solution:

Answer: (A). For a non-cubic cell use volume $= ab c$ :

d = \frac{4 \times 21.76}{( 6.8 \times 4.4 \times 7.2 ) \times 1 0 ^{- 24} \times 6.022 \times 1 0 ^{23}} = \frac{87.04}{129.7} = 0.671 g cm^{- 3}

Solved Example 7

An fcc lattice has lattice parameter a = 400 pm. The molar volume of the lattice, including all the empty space, is
(A) 10.8 mL
(B) 96 mL
(C) 8.6 mL
(D) 9.6 mL

Solution:

Answer: (D). One cell ( $a^{3}$ ) holds 4 atoms, so one mole occupies $\frac{N _{A}}{4}$ cells:

V_{m} = \frac{( 4 \times 1 0 ^{- 8} ) ^{3} \times 6.022 \times 1 0 ^{23}}{4} = 9.6 cm^{3} = 9.6 mL

Solved Example 8

(a) What is the coordination number of Cr in its bcc structure? (b) Cobalt crystallises in a hexagonal close-packed structure. What is the coordination number of Co? (c) Describe the crystal structure of Pt, which crystallises with four equivalent atoms in a cubic unit cell.

Solution:

(a) 8. (b) 12. (c) $Z = 4$ in a cubic cell means face-centred cubic, that is, cubic close packing.

Solved Example 9

Titanium has a density of 4.54 g cm

^{- 3}

and a cubic edge length of 412.6 pm (

M = 48

g mol

^{- 1}

). In which cubic unit cell does it crystallise, assuming a cubic form?

Solution:

Z = \frac{d N _{A} a ^{3}}{M} = \frac{4.54 \times 6.022 \times 1 0 ^{23} \times ( 4.126 \times 1 0 ^{- 8} ) ^{3}}{48} = 4.0

$Z = 4$ , so the cell is face-centred cubic.

Solved Example 10

Sodium crystallises in a bcc lattice with cell edge 4.29 Å. What is the radius of a sodium atom?

Solution:

Atoms touch along the body diagonal: $4 r = 3 a$ .

r = \frac{3 \times 4.29}{4} = 1.86 \overset{˚}{A}

Solved Example 11

Platinum (atomic radius 1.38 Å) crystallises in a cubic close-packed structure. Calculate the edge length of the fcc unit cell and the density of platinum (

M = 195

g mol

^{- 1}

Solution:

$a = 22 r = 22 \times 1.38 = 3.90$ Å.

d = \frac{4 \times 195}{( 3.90 \times 1 0 ^{- 8} ) ^{3} \times 6.022 \times 1 0 ^{23}} = 21.8 g cm^{- 3}

This matches the real density of platinum (21.4 g cm $^{- 3}$ ) closely.

Solved Example 12

A metal of molar mass 56 g mol

^{- 1}

and effective atomic radius 1.42 Å crystallises in an fcc lattice. Calculate its density.

Solution:

$a = 22 r = 22 \times 1.42 \times 1 0^{- 8}$ cm $= 4.016 \times 1 0^{- 8}$ cm.

d = \frac{4 \times 56}{6.022 \times 1 0 ^{23} \times ( 4.016 \times 1 0 ^{- 8} ) ^{3}} = 5.74 g cm^{- 3}

Solved Example 13

An element crystallises in an fcc lattice with density 5.20 g cm

^{- 3}

and edge length 300 pm. Calculate the mass of the element that contains

3.01 \times 1 0^{24}

atoms.

Solution:

M = \frac{d N _{A} a ^{3}}{Z} = \frac{5.20 \times 6.022 \times 1 0 ^{23} \times 27 \times 1 0 ^{- 24}}{4} = 21.1 g mol^{- 1}

$3.01 \times 1 0^{24}$ atoms $= 5.0$ mol, so the mass $= 5.0 \times 21.1 = 106$ g (105.7 g with unrounded values).

Solved Example 14

The fraction of empty space in a body-centred cubic lattice is closest to
(A) 0.26
(B) 0.32
(C) 0.48
(D) 0.68

Solution:

Answer: (B). Packing efficiency of bcc $= \frac{3 π}{8} = 0.680$ , so empty space $= 1 - 0.680 = 0.320$ .

Solved Example 15

Copper (fcc,

M = 63.5

g mol

^{- 1}

) has a density of 8.92 g cm

^{- 3}

. Find the edge length and the atomic radius of copper.

Solution:

a^{3} = \frac{Z M}{d N _{A}} = \frac{4 \times 63.5}{8.92 \times 6.022 \times 1 0 ^{23}} = 4.73 \times 1 0^{- 23} cm^{3}

$a = 3.62 \times 1 0^{- 8}$ cm $= 362$ pm, and $r = \frac{a}{2 2} = 128$ pm.

Solved Example 16

An element with molar mass

2.7 \times 1 0^{- 2}

kg mol

^{- 1}

forms a cubic unit cell with edge length 405 pm. If its density is

2.7 \times 1 0^{3}

kg m

^{- 3}

, what is the nature of the cubic cell?

Solution:

Work in SI units: $a = 4.05 \times 1 0^{- 10}$ m.

Z = \frac{d a ^{3} N _{A}}{M} = \frac{2.7 \times 1 0 ^{3} \times ( 4.05 \times 1 0 ^{- 10} ) ^{3} \times 6.022 \times 1 0 ^{23}}{2.7 \times 1 0 ^{- 2}} = 4.0

$Z = 4$ : face-centred cubic (this is aluminium).

Solved Example 17

Silver forms an fcc lattice with a = 408 pm. The distance between nearest neighbours is
(A) 204 pm
(B) 288.5 pm
(C) 353 pm
(D) 408 pm

Solution:

Answer: (B). In fcc the nearest neighbours lie half a face diagonal apart: $\frac{a}{2} = \frac{408}{1.414} = 288.5$ pm. Option (C), $\frac{3}{2} a$ , would be the bcc answer.

Practice Questions

Find the ratio of packing efficiencies of sc, bcc and fcc.Answer: $0.524 : 0.680 : 0.740$ .
A bcc metal has a = 330 pm. What is the nearest-neighbour distance?Answer: $\frac{3}{2} \times 330 = 285.8$ pm.
How many second-nearest neighbours does an atom have in fcc, and at what distance?Answer: 6, at distance $a$ .
Tungsten is bcc with a = 316.5 pm and M = 183.8 g mol $^{- 1}$ . Find its density.Answer: $d = 19.3$ g cm $^{- 3}$ .
Mg (hcp) has atomic radius 160 pm. Find the ideal height c of its unit cell.Answer: $c = 1.633 \times 320 = 523$ pm.
Silver (fcc, M = 108) has density 10.5 g cm $^{- 3}$ . Find its atomic radius.Answer: $a = 409$ pm, $r = \frac{a}{2 2} = 145$ pm.
In which cubic lattice are the second-nearest neighbours only about 15% farther than the nearest?Answer: bcc: $\frac{a}{3 a /2} = 1.155$ .

Common Mistakes to Avoid

Watch out

Using $a = 2 r$ for bcc or fcc. Atoms touch along the edge only in simple cubic; bcc touches along the body diagonal, fcc along the face diagonal.
Writing the bcc nearest-neighbour distance as $a$ . It is $\frac{3}{2} a$ ; the 6 atoms at $a$ are second neighbours.
Saying hcp packs better than ccp (or the reverse). Both fill 74% of space with CN 12.
Saying the 3rd layer repeats the 1st in ccp. That is hcp; in ccp the 4th layer repeats the 1st.
Taking $Z = 4$ for hcp. The hexagonal prism holds 6 atoms; only its one-third primitive cell holds 2.
Using $a^{3}$ as the volume of a non-cubic cell. Use $ab c$ for orthorhombic and $242 r^{3}$ for the hcp prism.
Forgetting to cube the powers of ten: $(1 0^{- 8})^{3} = 1 0^{- 24}$ , not $1 0^{- 8}$ .
Treating ccp and fcc as different lattices. ccp is described by the fcc unit cell.

Frequently Asked Questions

What is close packing in solids?

Close packing is the arrangement of identical spheres that leaves the least empty space. Hexagonal layers stacked in the hollows of each other give hexagonal close packing (ABAB) and cubic close packing (ABCABC). Both have coordination number 12 and fill 74 percent of space.

What is the difference between hcp and ccp?

Both are built from hexagonal close-packed layers and both have coordination number 12 and 74 percent packing efficiency. In hcp the third layer lies directly over the first (ABAB). In ccp the third layer covers the remaining hollows and the fourth repeats the first (ABCABC). ccp has an fcc unit cell with Z = 4; hcp has a hexagonal cell with Z = 6.

Why is ccp the same as fcc?

If you look along the body diagonal of a face-centred cubic cell, the atoms fall on close-packed planes stacked in the order A, B, C, A. So the ABCABC stacking of cubic close packing is described exactly by the fcc unit cell.

How do you calculate packing efficiency of bcc?

A bcc cell contains 2 atoms and its atoms touch along the body diagonal, so root 3 times a equals 4r. Packing efficiency equals the volume of 2 spheres divided by a cubed, which simplifies to root 3 times pi divided by 8, about 68 percent.

What is the coordination number in bcc, fcc and hcp?

The coordination number is 8 in body-centred cubic, and 12 in both face-centred cubic and hexagonal close-packed structures. Simple cubic has coordination number 6.

How do you find the type of cubic lattice from density?

Rearrange the density formula to Z = d times N_A times a cubed divided by M, with a in centimetres. If Z comes out as 1 the lattice is simple cubic, 2 means body-centred cubic and 4 means face-centred cubic.

Is the solid state chapter important for NEET?

No. Solid State was removed from the NEET syllabus from 2024 along with its deletion from the rationalised NCERT textbook, so NEET 2026 will not ask it. It still matters for JEE Advanced and for some state board and university exams.

What types of questions on close packing come in JEE Advanced?

JEE Advanced asks for packing efficiency and density of fcc, bcc and hcp, edge-radius relations, nearest-neighbour distances and counts, stacking sequences, and ratios such as density of fcc to bcc phases. Numerical and integer-type answers are common.

Previous year questions on Close-Packed Structures

2 questions from past papers, each with a step-by-step solution.

Ready to master Solid State?

Take a full mock test, practice concept-by-concept, and get an AI-powered rank prediction — all on Fundamenthol.

Practice questions →Take a mock test