The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

Question

ExtremeBDS

11 days ago

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The unit cell for cr2o3 has hexagonal symmetry with lattice parameters a = 0.4961 nm and c = 1.360 nm. If the density of this ma

terial is 5.22 g/cm3, calculate its atomic packing factor. The atomic weights of cr

Chemistry

2 answers:

alisha [2.9K] · Answer 1 · 2026-04-10T21:28:25+03:00

Answer:

The value of the atomic packing factor is 0.76

Explanation:

The area is

A = 6r²√3 = 6(a/2)²√3 = 1.5a²√3

For HCP, it holds that:

a = 2r

Substituting

A = 1.5 * (4.961x10⁻⁸)²√3 = 6.39x10⁻¹⁵ cm²

The volume of the cell can be computed as

V = A*c = 6.39x10⁻¹⁵ * 1.36x10⁻⁷ = 8.7x10⁻²² cm³

For n results in

n = (e * N * V)/(∑Ac + ∑An) = (5.22 * 6.022x10²³ * 8.7x10⁻²²)/(2 * 52 + (3 * 16) = 18 formula unit/unit cell

There are 18 units of Cr₂O₃ or equivalently 36 ions of Cr₂O₃. The overall volume computes as

V = 36 * (4π/3) * (rCr³) + 54 * (4π/3) * (rO³) = 36 * (4π/3) * (6.2x10⁻⁹)³ + 54 * (4π/3) * (1.4x10⁻⁸)³ = 6.57x10⁻²² cm³

The atomic packing factor becomes:

APF = 6.57x10⁻²²/8.7x10⁻²² = 0.76

KiRa [2.9K] · Answer 2 · 2026-04-10T21:27:38+03:00

To determine the packing factor, begin by calculating the area and volume of the unit cell.

The area is found using:

$A=6R^{2}\sqrt{3}$

In this case, R represents the radius and is connected to a as shown:

$R=\frac{a}{2}$

Substituting the value into the area formula,[ [TAG_14]]

$A=6(\frac{a}{2})^{2}\sqrt{3}=1.5a^{2}\sqrt{3}$

The value of a is 0.4961 nm

As, $1 nm=10^{-7}cm$

Therefore, $0.4961 nm=4.961\times 10^{-8} cm$

Substituting the value,[ [TAG_27]]

$Area=1.5(4.961\times 10^{-8}cm)^{2}\sqrt{3}=6.39\times 10^{-15}cm^{2}$

Next, the volume can be calculated by the following method:

$V=Area\times c$

The value of c is 1.360 nm or $1.360\times 10^{-7} cm$

Inserting the value,[ [TAG_40]]

$V=(6.39\times 10^{-15}cm^{2})\times (1.360\times 10^{-7} cm)=8.7\times 10^{-22}cm^{3}$

Now, to determine the number of atoms in the unit cell, the following equation can be employed:

$n=\frac{\rho N_{A}V_{c}}{A}$

In this context, A represents the atomic mass of $Cr_{2}O_{3}$ which is 151.99 g/mol.

Inserting all the necessary values,[ [TAG_53]]

$n=\frac{(5.22 g/cm^{3})(6.023\times 10^{23} mol^{-1})(8.7\times 10^{-22}cm^{3})}{(151.99 g/mol)}\approx 18$

Consequently, there will be 18 $Cr_{2}O_{3}$ units within 1 unit cell.

Given, there are 2 chromium atoms and 3 oxygen atoms, thus, the total units for chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii for $Cr^{3+}$ and $O^{2-}$ measure 62 pm and 140 pm respectively.

Transforming them into centimeters:

$1 pm=10^{-10}cm$

Therefore,

$r_{Cr^{3+}}=6.2\times 10^{-9}cm$

and,

$r_{O^{2-}}=1.4\times 10^{-8}cm$

The total volume of the sphere will be the combined volume of all cations and anions, thus,

$V_{S}=V_{Cr^{3+}}+V_{O^{2-}}$

Since, the volume of a sphere is $V=\frac{4}{3}\pi r^{3}$ ,

$V_{S}=36\left ( \frac{4}{3}\pi (r_{Cr^{3+})^{3}} \right )+54\left ( \frac{4}{3}\pi (r_{O^{2-})^{3}} \right )$

Inputting the respective values,[ [TAG_93]]

$V_{S}=36\left ( \frac{4}{3}(3.14) (6.2\times 10^{-9} cm)^{3}} \right )+54\left ( \frac{4}{3}(3.14) (1.4\times 10^{-8} cm)^{3}} \right )=6.6\times 10^{-22}\times 10^{-8}cm^{3}$

The atomic packing factor reflects the ratio of the volume of spheres to the volume of the crystal, so,[ [TAG_98]]

$packing factor=\frac{V_{S}}{V_{C}}=\frac{6.6\times 10^{-22}cm^{3}}{8.7\times 10^{-22}cm^{3}}=0.758$

Thus, the atomic packing factor equals 0.758.