# Homework Help: Radii of Mg2+ and O2- in MgO

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1. Apr 11, 2016

### 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 O2--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?

2. Apr 11, 2016

### Staff: Mentor

Somehow I am not convinced there is enough data for the radii determination. Sum of radii, yes, individual ones - no.

3. Apr 11, 2016

### TheSodesa

What makes you say this? Since I know that r = 0.414R, and looking at the picture the edge of the unit cube is clearly e = 2R + 2r, methinks I should be able to extract both R and r from the information given. I was able to calculate e (although my calculation could be wrong, and according to the book it is).

Last edited: Apr 11, 2016
4. Apr 12, 2016

### Staff: Mentor

You can calculate edge of the cube, no doubt about it, but I am not convinced the idea of Mg2+ being hidden in the holes between O2- is right.

Sure, it can work this way for some compounds, but without further information there is no way of telling which compounds look this way (what if the radii of the anion and cation are comparable?)

Not your fault, I just don't like the question in general. It requires you to make an assumption ($r=R(\sqrt 2 - 1)$) which I don't see as necessarily justified.

Sanity check:

Mass of the 1 cm3 of the solid can't be 3.58 g and 2.84 g at the same time, only one of these numbers is correct.

5. Apr 12, 2016

### TheSodesa

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.

Last edited: Apr 12, 2016
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