Q.The density of iridium is 22.4 g/cm³. The unit cell of iridium is fcc. Calculate the radius of iridium atom. Molar mass of iridium is 192.2 g/mol. (136 pm)

A bcc cell packs 2 atoms with a packing fraction of about 0.68 (68%).

For a bcc cell the body diagonal = 4r, and body diagonal = a√3, so a = 4r/√3.

Atoms per cell = 2 (1 body centre + 8 corners x 1/8).

Packing fraction = (volume of 2 atoms)/(volume of cell) = [2 x (4/3)πr³] / a³.

Substituting a = 4r/√3 gives packing fraction = √3π/8 ≈ 0.68.

Compare: simple cubic = 0.52, bcc = 0.68, fcc/hcp = 0.74.

bcc packing efficiency = 68%, hence free space = 100 − 68 = 32%.

For a body-centred cubic lattice, atoms touch along the body diagonal, giving the relation 4r = √3 a (r = radius, a = edge). The unit cell has 2 atoms.

Packing efficiency = (volume occupied by 2 atoms) / (volume of unit cell) × 100 = 68% (approximately).

Therefore the percentage of free (unoccupied) space = 100 − 68 = 32%.

A bcc unit cell has 2 atoms per cell and packing efficiency 68% (=√3π/8), so the remaining 32% of the cell's volume is empty space.

In a body-centred cubic (bcc) arrangement, atoms touch along the body diagonal, giving edge length a related to atomic radius r by a = 4r/√3. There are 2 atoms per unit cell (1 corner-contribution + 1 body-centre atom). Packing efficiency = (volume occupied by atoms)/(volume of unit cell) × 100 = (√3π/8) × 100 ≈ 68%. The unoccupied (void) space is therefore 100% − 68% = 32%.

In simple cubic (SC) packing the atoms touch along the cell edge and occupy only about half the available volume, giving 52.4% packing efficiency - the least efficient of the common cubic packings.

In a simple cubic lattice, atoms are present only at the corners of the cube, and each corner atom touches the atoms in the adjacent cells along the edge. So the edge length a = 2r, where r is the radius of the atom.

Each unit cell contains 8 corner atoms x 1/8 = 1 atom.

Volume of the atom = (4/3)pi r^3

Volume of the cube = a^3 = (2r)^3 = 8r^3

Packing efficiency = [volume of 1 atom / volume of cube] x 100 = [(4/3)pi r^3 / 8r^3] x 100 = (pi/6) x 100 ~ 52.4%

This is the lowest packing efficiency among the cubic lattices (bcc = 68%, fcc/hcp = 74%).

In a bcc lattice atoms touch along the body diagonal, and working through the geometry gives a packing efficiency of (√3π)/8 ≈ 68%.

In a body-centred cubic unit cell, atoms touch each other along the body diagonal (not along the edge).

Step 1 — relate radius r to edge length a:

Body diagonal = √3·a = 4r ⟹ r = (√3/4)·a

Step 2 — atoms per unit cell (Z):

Corner atoms: 8 × 1/8 = 1

Body-centre atom: 1 × 1 = 1

Z = 2

Step 3 — packing efficiency:

Packing efficiency = (Z × 4/3 π r³ / a³) × 100

Substituting r = (√3/4)a:

= (√3 π / 8) × 100 ≈ 68%

Face centred cubic (fcc) has the highest packing efficiency (74%) of the three cubic lattices.

Packing efficiency is the fraction of total volume of a unit cell occupied by the constituent spheres (atoms), assuming they touch each other along the relevant direction.

• Simple cubic (sc): atoms touch along the cell edge; packing efficiency = 52.4%
• Body centred cubic (bcc): atoms touch along the body diagonal; packing efficiency = 68%
• Face centred cubic (fcc): atoms touch along the face diagonal, and this is the most efficient way of packing spheres (cubic close packing); packing efficiency = 74%

Since 74% > 68% > 52.4%, the fcc lattice packs atoms most efficiently, leaving the least empty (void) space.

In a bcc unit cell, atoms occupy 68% of the total volume.

In a body-centred cubic (bcc) arrangement, atoms touch each other along the body diagonal of the cube (not along the edge).

Step 1 — Relation between radius and edge length.

For a cube of edge $a$, the body diagonal length is $\sqrt{3}\,a$. Along this diagonal, one corner atom, the body-centre atom, and the opposite corner atom touch, so the diagonal spans $4r$ (radius $r$ of two half-atoms at the ends plus the full diameter of the centre atom):

$\sqrt{3}\,a = 4r \implies r = \frac{\sqrt{3}}{4}a$

Step 2 — Number of atoms per unit cell.

Z = 8 corners × 1/8 + 1 body-centre × 1 = 1 + 1 = 2 atoms.

Step 3 — Packing fraction.

$\text{Packing fraction} = \frac{Z \times \frac{4}{3}\pi r^3}{a^3} = \frac{2 \times \frac{4}{3}\pi\left(\frac{\sqrt3}{4}a\right)^3}{a^3}$

Simplifying gives $\dfrac{\sqrt3\,\pi}{8} \approx 0.68$, i.e. 68% of the cell volume is occupied by atoms (32% is void space).

bcc packing efficiency = 68% → free space = 32%.

In a bcc unit cell there are 2 atoms and the body diagonal = 4r = √3·a. Packing efficiency = (2 × ⁴⁄₃πr³)/a³ = 0.68 = 68%.

Free space = 100% − 68% = 32%.