Question

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

Answer 1

To calculate the packing factor, first calculate the area and volume of unit cell.

Area is calculated as:

`A=6R^(2)√(3)`

Here, R is radius and is related to a as follows:

`R=(a)/(2)`

Putting the value in expression for area,

`A=6((a)/(2))^(2)√(3)=1.5a^(2)√(3)`

The value of a is 0.4961 nm

Since,
`1 nm=10^(-7)cm`

Thus,
`0.4961 nm=4.961* 10^(-8) cm`

Putting the value,

`Area=1.5(4.961* 10^(-8)cm)^(2)√(3)=6.39* 10^(-15)cm^(2)`

Now, volume can be calculated as follows:

`V=Area* c`

The value of c is 1.360 nm or
`1.360* 10^(-7) cm`

Putting the value,

`V=(6.39* 10^(-15)cm^(2))* (1.360* 10^(-7) cm)=8.7* 10^(-22)cm^(3)`

now, number of atom in unit cell can be calculated by using the following formula:

`n=(\rho N_(A)V_(c))/(A)`

Here, A is atomic mass of
`Cr_(2)O_(3)` is 151.99 g/mol.

Putting all the values,

`n=((5.22 g/cm^(3))(6.023* 10^(23) mol^(-1))(8.7* 10^(-22)cm^(3)))/((151.99 g/mol))\approx 18`

Thus, there will be 18
`Cr_(2)O_(3)` units in 1 unit cell.

Since, there are 2 Cr atoms and 3 oxygen atoms thus, units of chromium and oxygen will be 2×18=36 and 3×18=54 respectively.

The atomic radii of
`Cr^(3+)` and
`O^(2-)` is 62 pm and 140 pm respectively.

Converting them into cm:

`1 pm=10^(-10)cm`

Thus,

`r_(Cr^(3+))=6.2* 10^(-9)cm`

and,

`r_(O^(2-))=1.4* 10^(-8)cm`

Volume of sphere will be sum of volume of total number of cations and anions thus,

`V_(S)=V_(Cr^(3+))+V_(O^(2-))`

Since, volume of sphere is
`V=(4)/(3)\pi r^(3)`,

`V_(S)=36\left ( (4)/(3)\pi (r_(Cr^(3+))^(3)) \right )+54\left ( (4)/(3)\pi (r_(O^(2-))^(3)) \right )`

Putting the values,

`V_(S)=36\left ( (4)/(3)(3.14) (6.2* 10^(-9) cm)^(3)} \right )+54\left ( (4)/(3)(3.14) (1.4* 10^(-8) cm)^(3)} \right )=6.6* 10^(-22)* 10^(-8)cm^(3)`

The atomic packing factor is ratio of volume of sphere and volume of crystal, thus,

`packing factor=(V_(S))/(V_(C))=(6.6* 10^(-22)cm^(3))/(8.7* 10^(-22)cm^(3))=0.758`

Thus, atomic packing factor is 0.758.

Answer 2

The atomic packing factor is 0.76

Step-by-step explanation:

The area is

A = 6r²√3 = 6(a/2)²√3 = 1.5a²√3

For HCP, a is equal to:

a = 2r

Replacing

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

The cell volume is

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

For n yields

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 unit of Cr₂O₃ or 36 ions of Cr₂O₃ The total volume is

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 APF is equal to:

APF = 6.57x10⁻²²/8.7x10⁻²² = 0.76