[TheSodesa]

1. The problem statement, all variables and given/known data
Below is an image of a unit cell for MgO. Does MgO have the lattice structure of NaCl or ZnS? If the density of MgO, ##\rho## = 3,58g/cm³, evaluate the sizes of the radii of Mg²⁺- and O^2--ions.



Answers: 149pm and 62pm

2. Relevant equations


3. The attempt at a solution

Looking at the picture, MgO seems to have the lattice structure of NaCl. Therefore as an ionic solid it has octahedral holes in the middle of the larger ##O^{2-}##-ions, where the ##Mg^{2+}##-ions are located. A relationship between the radii of ions in an octahedral arrangement can be derived to be ##r = \sqrt{2}R - R \approx 0.414R##, where r is the radius of the smaller ion and R the radius of the larger ion.

We know ##\rho## = 3,58g/cm³. We also know from looking at the picture, that the unit cell contains 4 whole ##O^{2-}##-ions and 4 whole ##Mg^{2+}##-ions, since there is 1 whole Mg-ion in the middle of the cell (the O-ions are cut either in half or into eights by the edges of the cube if it is drawn, and the Mg-ions on the edges of the cube are cut into quarters).

Therefore if we take 1cm³ block of MgO, it's mass ##m = nM = \rho V_i##, where ##V_i## is the total volume of the ions in the unit cell.

Now ##\frac{V_{ions}}{V_{unit \ cell}} = \frac{4(\frac{4}{3}\pi R^3) + 4(\frac{4}{3}\pi r^3)}{(2R+2r)^3} \stackrel{r = \sqrt{2}R-R}{=} \frac{(19\sqrt{2}-27)\pi}{6\sqrt{2}-9} \approx 0,7931##.

If we then take ##1cm^3## block of MgO, its mass ##m = 0,7931 \rho V = 0,7931(3,58g/cm^3)(1cm^3) = 2,8393g##.

Now the amount of MgO in this sample is ##n = \frac{m}{M} = \frac{2.8393g}{40,3044 g/mol} = 0,07447mol##, and since the lattice structure of MgO is stoichiometric, meaning the ratios of ions is the lattice match the chemical formula, each unit cell contains 4 MgOs.

Since there are ##N = nN_A=(0,07447mol)(6,022\cdot 10^{23} 1/mol) \approx 4,2423 \cdot 10^{22}## MgOs in the sample and each unit cell contains 4 MgOs, the amount of unit cells in the sample is ##\frac{N}{4} = 1,0606 \cdot 10^{22}##.

This number is equal to the ratio ##\frac{V_{sample}}{V_{unit \ cell}} = \frac{1cm^3}{e^3} = \frac{N}{4} \iff e = \sqrt[3]{\frac{1cm^3}{N/4}} = 4,55148 \cdot 10^{-8}##.

Since the edge of the unit cell ##e = 2R + 2r = 2R + 2(\sqrt{2}R-R) = R(2+2\sqrt(2) - 2)\\
\iff R = \frac{e}{2\sqrt{2}} = \frac{4,55148 \cdot 10^{-8}}{2\sqrt{2}} = 1,60919 \cdot 10^{-8} cm##.

Therefore ##r = 0,414R = 0,414( 1,60919 \cdot 10^{-8} cm) = 6,66548 \cdot 10^{-9} cm##.

Answer:
\begin{cases}
R \approx 161pm\\
r \approx 66,7pm
\end{cases}

This is not what the book says. What am I doing wrong?

[mjc123]

Therefore if we take 1cm3 block of MgO, it's mass m=nM=ρVi, where Vi is the total volume of the ions in the unit cell.

Why do you say this? The macroscopic density is the mass divided by the total volume occupied - not just the actual volume of the ions.
So if we take a 1 cm³ block of MgO its mass is ρV = 3.58 * 1 = 3.58 g. That is what is meant by saying the density is 3.58 g/cm³.
Hence your answer is out by a factor of cube root(0.7931).

[TheSodesa]

Therefore if we take 1cm3 block of MgO, it's mass m=nM=ρVi, where Vi is the total volume of the ions in the unit cell.

Why do you say this? The macroscopic density is the mass divided by the total volume occupied - not just the actual volume of the ions.
So if we take a 1 cm³ block of MgO its mass is ρV = 3.58 * 1 = 3.58 g. That is what is meant by saying the density is 3.58 g/cm³.
Hence your answer is out by a factor of cube root(0.7931).

Yeah, I realized my mistake a few hours ago. I think I might have been obsessed with a certain example in my book where they calculated the ratio of the volume taken up by the atoms in a unit cell compared the the unit cell itself.

A massive brain fart, in other words. I finished the question on physics forums, so I might as well post the text here as well:

You're right. How didn't I realize I had two different masses for the same volume?

Alright, so if instead we use m = 3,58g to calculate ##n=m/M = (3,58g)/(40,3044g/mol) = 0,088824mol##.

Then ##N=nN_A = (0,088824mol)(6,022 \cdot 10^{23} 1/mol) = 5,34898 \cdot 10^{22}##

Then the amount of unit cells is ##N/4 = 1,33725 \cdot 10^{22}## and ##e = \sqrt[3]{\frac{1cm^3}{N/4}} = 4,21305 \cdot 10^{-8}cm##.

Then ##R = \frac{e}{2\sqrt{2}} = 1,48954 \cdot 10^{-8}cm## and ##r = 0,414R = 6,16986 \cdot 10^{-9}cm##.

Then \begin{cases}
R \approx 149pm\\
r \approx 62pm
\end{cases}

Booyah!

I think I might have been too stuck on a certain example in the book (Zumdahl), where they estimate how much actual space is taken up by the atoms in a unit cell.

Thank you for the sanity check. I will mark this as solved.

EDIT: Zumdahl's book is very clear in saying that the ##r = (\sqrt{2}-1)R## is an approximation arrived at by assuming that all of the atoms in the ionic lattice are closest-packed hard spheres touching each other. That is how the relation was derived from simple geometry.