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

42 PYQs, none HARD — one of the most-asked calculations 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

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?

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 2nd May Shift 1 · Q52Easy

Example 1 · Solid State · Unit Cells, Edge Length and Atomic Radius

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

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

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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 4th May Shift 1 · Q91Easy

Example 2 · Solid State · Unit Cells, Edge Length and Atomic Radius

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

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

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.

Practise this conceptself-check · 4 quick reps

The same idea in a real exam question:

MHT-CET · 2023 · 9th May Shift 2 · Q91Moderate

Example 3 · Solid State · Unit Cells, Edge Length and Atomic Radius

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

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

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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Related notes

Density — the formula that uses n and a³ together →