MHT-CET Chemistry · Solid State

Unit Cells, Edge Length and Atomic Radius

The three cubic unit cells hold 1, 2 and 4 particles, and in each the atoms touch along one line — the edge, the body diagonal or the face diagonal — which fixes the edge length in terms of the atomic radius.

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Why this matters

30 PYQs, none HARD — the most-asked calculation in the chapter. Nearly every one is 'given r find a' or 'given a find r' for a named cell, with the answer wanted in cm; the rest count particles per cell or coordination numbers. Three relations and one unit conversion (1 pm = 10⁻¹⁰ cm) cover the page.

Concept 1 of 3

Particles Per Unit Cell and Coordination Number

Intuition

A corner particle is shared by 8 cells, a face particle by 2, an edge particle by 4, a body-centre particle by none. Add the fractions and the three cubic cells hold 1, 2 and 4 particles. Coordination number is how many nearest neighbours one particle touches: 6 in simple cubic, 8 in bcc, 12 in fcc and hcp.

Definition

Simple cubic: $8 \times \tfrac{1}{8} = 1$ particle. Coordination number 6. Polonium.
Body-centred cubic (bcc): $8 \times \tfrac{1}{8} + 1 = 2$ . Coordination number 8. Na, K, Fe, Cr.
Face-centred cubic (fcc = ccp): $8 \times \tfrac{1}{8} + 6 \times \tfrac{1}{2} = 4$ . Coordination number 12. Cu, Ag, Au, Al, Ni.
Base-centred: $8 \times \tfrac{1}{8} + 2 \times \tfrac{1}{2} = 2$ .
hcp also has coordination number 12 (6 in the layer, 3 above, 3 below). Zn, Mg, Cd.

Sharing fractions

n = \tfrac{1}{8}\,n_{\text{corner}} + \tfrac{1}{2}\,n_{\text{face}} + \tfrac{1}{4}\,n_{\text{edge}} + n_{\text{body}}

Worked example

A cubic cell has particles at every corner and at the centre of every edge. How many particles does it hold?

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 1Solid StateEASY

What is the total number of particles present in bcc unit cell?

[Q52 · 2nd May Shift 1 · 2023]

Counting the eight corners as eight particles

Each corner particle belongs to eight cells, so the eight corners together contribute ONE. A simple cubic cell has 1 particle, not 8.

Drill 7 more on particles per unit cell and coordination number

Concept 2 of 3

Edge Length From Radius: Where the Atoms Touch

Intuition

In each cell the atoms are in contact along exactly one line. Simple cubic: along the edge, so

a = 2r

. bcc: along the body diagonal of length

\sqrt{3}a

, which holds four radii. fcc: along the face diagonal of length

\sqrt{2}a

, which also holds four radii.

Definition

Simple cubic: $a = 2r$ , $r = a/2$ .
bcc: $\sqrt{3}\,a = 4r$ , so $a = \dfrac{4r}{\sqrt{3}} = 2.309\,r$ and $r = \dfrac{\sqrt{3}}{4}a = 0.433\,a$ .
fcc: $\sqrt{2}\,a = 4r$ , so $a = 2\sqrt{2}\,r = 2.828\,r$ and $r = \dfrac{a}{2\sqrt{2}} = 0.3535\,a$ .
Units: $1\ \text{pm} = 10^{-10}\ \text{cm}$ , $1\ \text{Å} = 100\ \text{pm} = 10^{-8}\ \text{cm}$ . The options are usually in cm — convert at the end.

Edge–radius relations

a_{\text{sc}} = 2r,\qquad a_{\text{bcc}} = \frac{4r}{\sqrt{3}},\qquad a_{\text{fcc}} = 2\sqrt{2}\,r

Worked example

A metal with atomic radius 160 pm forms a bcc lattice. Find the edge length in cm.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 2Solid StateEASY

Calculate the radius of metal atom if it forms bcc unit cell having edge length 530 pm.

[Q91 · 4th May Shift 1 · 2023]

Inverting the bcc relation

a = \sqrt{3}r/4

and

a = \sqrt{3}/4 \cdot r

are both offered beside the right

a = 4r/\sqrt{3}

. Check the size:

a

must be LARGER than

r

(about 2.3 times), so any option that makes

a

smaller than

r

is wrong.

Drill 15 more on edge length from radius: where the atoms touch

Concept 3 of 3

Unit Cell Volume and the Volume of Its Atoms

Intuition

Once the edge is known the cell volume is just

a^3

; the volume the atoms occupy is

n \times \tfrac{4}{3}\pi r^3

. Substituting the edge–radius relation turns that into a multiple of

a^3

— for bcc, one atom is

\sqrt{3}\pi a^3/16

and both atoms are

\sqrt{3}\pi a^3/8

Definition

$V_{\text{cell}} = a^3$ . Convert the edge to cm FIRST: $a = 400\ \text{pm} = 4 \times 10^{-8}\ \text{cm}$ , so $a^3 = 6.4 \times 10^{-23}\ \text{cm}^3$ .
Volume of one atom $= \tfrac{4}{3}\pi r^3$ ; volume of all atoms in the cell $= n \cdot \tfrac{4}{3}\pi r^3$ .
bcc in terms of $a$ ( $r = \sqrt{3}a/4$ ): one atom $= \dfrac{\sqrt{3}\pi a^3}{16}$ , two atoms $= \dfrac{\sqrt{3}\pi a^3}{8}$ .
fcc ( $r = a/2\sqrt{2}$ ): four atoms $= \dfrac{\pi a^3}{3\sqrt{2}}$ . Simple cubic ( $r = a/2$ ): one atom $= \dfrac{\pi a^3}{6}$ .
Handy: $141.4\ \text{pm} \approx 100\sqrt{2}$ , so an fcc metal with $r = 141.4$ pm has $a = 400$ pm exactly.

Cell volume and atom volume

V_{\text{cell}} = a^3,\qquad V_{\text{atoms}} = n \cdot \tfrac{4}{3}\pi r^3

Worked example

An fcc metal has atomic radius 106.05 pm. Find the unit cell volume in cm³.

Practice this conceptself-check · 4 quick reps

From the bank · past-year question

Example 3Solid StateMODERATE

Calculate the volume of fcc unit cell if radius of a particle in it is 106.05 pm.

[Q82 · 9th May Shift 2 · 2023]

Cubing the picometres

400^3\ \text{pm}^3

is not a number any option shows. Convert to cm before cubing:

(4 \times 10^{-8})^3 = 6.4 \times 10^{-23}

; cubing

10^{-8}

gives

10^{-24}

, and the mantissa's cube moves the exponent up.

Drill 5 more on unit cell volume and the volume of its atoms

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 (3)

Particles Per Unit Cell and Coordination Number
Sharing fractions
$n = \tfrac{1}{8}\,n_{\text{corner}} + \tfrac{1}{2}\,n_{\text{face}} + \tfrac{1}{4}\,n_{\text{edge}} + n_{\text{body}}$
Edge Length From Radius: Where the Atoms Touch
Edge–radius relations
$a_{\text{sc}} = 2r,\qquad a_{\text{bcc}} = \frac{4r}{\sqrt{3}},\qquad a_{\text{fcc}} = 2\sqrt{2}\,r$
Unit Cell Volume and the Volume of Its Atoms
Cell volume and atom volume
$V_{\text{cell}} = a^3,\qquad V_{\text{atoms}} = n \cdot \tfrac{4}{3}\pi r^3$

Watch out for (3)

Counting the eight corners as eight particles
→ Particles Per Unit Cell and Coordination Number
Inverting the bcc relation
→ Edge Length From Radius: Where the Atoms Touch
Cubing the picometres
→ Unit Cell Volume and the Volume of Its Atoms

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