Question & Answer

A) Use the ionic radii of 0.072 nm for Mg2+ and 0.14 nm for O2- to demonstrate which type of interstitial location would be most favorable for the cations in MgO "Magnesium Oxide"

In magnesium oxide (MgO), the cations (Mg^2+) occupy the interstitial sites between the anions (O^2-) in a crystal lattice structure. Given the ionic radii provided (0.072 nm for Mg^2+ and 0.14 nm for O^2-), we can determine which type of interstitial location would be most favorable for the cations.

The interstitial sites in a cubic close-packed (ccp) structure are of two types: octahedral and tetrahedral. In both cases, the cation sits within a void surrounded by anions.

1. Octahedral Interstitial Sites:
   • In an octahedral site, the cation is surrounded by six anions, forming an octahedron.
   • The effective radius of the octahedral site can be calculated as the distance from the center of one anion to the center of the opposite anion along a body diagonal of the cube.
   • For a ccp lattice, this distance is equal to the length of the cube diagonal, which can be calculated as $\sqrt{2}$ times the edge length of the unit cell.
   • The edge length of the unit cell ($𝑎$) can be determined using the formula $𝑎=2\times {𝑟}_{\text{anion}}$, where ${𝑟}_{\text{anion}}$ is the radius of the anion (O^2-).
   • Thus, $𝑎=2\times 0.14\text{nm}=0.28\text{nm}$.
   • The length of the cube diagonal (${𝑑}_{\text{cube}}$) is then ${𝑑}_{\text{cube}}=\sqrt{2}\times 𝑎=\sqrt{2}\times 0.28\text{nm}\approx 0.396\text{nm}$.
   • Therefore, the effective radius of the octahedral site (${𝑟}_{\text{oct}}$) is half the cube diagonal, minus the radius of the anion.
   • ${𝑟}_{\text{oct}}=\frac{1}{2}\times 0.396\text{nm}-0.14\text{nm}\approx 0.058\text{nm}$.
2. Tetrahedral Interstitial Sites:
   • In a tetrahedral site, the cation is surrounded by four anions, forming a tetrahedron.
   • The effective radius of the tetrahedral site can be calculated as the distance from the center of one anion to the center of the cation.
   • It's worth noting that the center of the cation is the same as its ionic radius (${𝑟}_{\text{cation}}$).
   • Therefore, the effective radius of the tetrahedral site is the sum of the radii of the cation and the anion.
   • ${𝑟}_{\text{tet}}={𝑟}_{\text{cation}}+{𝑟}_{\text{anion}}=0.072\text{nm}+0.14\text{nm}=0.212\text{nm}$.

Comparing the effective radii of the octahedral and tetrahedral sites:

• ${𝑟}_{\text{oct}}=0.058\text{nm}$
• ${𝑟}_{\text{tet}}=0.212\text{nm}$

The cation (Mg^2+) would preferentially occupy the octahedral interstitial sites because the effective radius of the octahedral site is smaller than that of the tetrahedral site, providing a more stable arrangement within the crystal lattice.

B) Demonstrate the fraction of these sites in an FCC lattice that Mg2+ ions would need to occupy in order to maintain charge neutrality with the 02- ions sitting in FCC lattice positions of a unit cell.

To determine the fraction of octahedral interstitial sites that Mg^2+ ions would need to occupy in an FCC (face-centered cubic) lattice of MgO in order to maintain charge neutrality with the O^2- ions sitting in FCC lattice positions of a unit cell, we can use the following approach:

In an FCC lattice, each unit cell contains:

• 4 lattice points at the corners (each contributes 1/8 to the unit cell)
• 1 lattice point at the center of each face (each contributes 1/2 to the unit cell)

Thus, the total number of lattice points contributed by the anions (O^2-) in the FCC lattice of MgO is:

$4\times \frac{1}{8}+6\times \frac{1}{2}=1+3=4$

The number of octahedral interstitial sites in an FCC lattice is equal to the number of lattice points, as each lattice point corresponds to one octahedral interstitial site.

So, the total number of octahedral interstitial sites in the unit cell is 4.

Since each Mg^2+ ion occupies one octahedral interstitial site, the fraction of these sites that Mg^2+ ions would need to occupy in order to maintain charge neutrality is:

$\text{Fraction of sites occupied by Mg}=\frac{{\text{Number of Mg}}^{2+}\text{ ions}}{\text{Total number of octahedral interstitial sites}}=\frac{4}{4}=1$

Thus, all the octahedral interstitial sites in the FCC lattice of MgO need to be occupied by Mg^2+ ions in order to maintain charge neutrality with the O^2- ions sitting in FCC lattice positions of a unit cell.