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I tried to index powder diffractogram from the following problem:

A metal (of density $\pu{0.862 g cm^-3})$ gives the following powder X-ray diffraction data $(\lambda = \pu{1.5418 Å}).$

$$2\theta \qquad 23.6^\circ \quad 33.6^\circ \quad 41.5^\circ \quad 48.3^\circ \quad 54.4^\circ \quad 60.1^\circ \quad 65.5^\circ $$

I calculated $\displaystyle d = \frac{\lambda}{2\sin\theta}$ and compiled a table:

$$ \begin{array}{cclccc} \hline 2\theta/^\circ & d/\pu{Å} & \frac{1}{d^2}/\pu{Å^-2} & \frac{1}{d^2}/Z & h^2 + k^2 + l^2 & hkl \\ \hline 23.6 & 3.769754932 & 0.07036776606~(= Z) & 1 & 1 & 100 \\ 33.6 & 2.667180594 & 0.1405708123 & 1.997659159 & 2 & 110 \\ 41.5 & 2.175894723 & 0.2112147493 & 3.001583838 & 3 & 111 \\ 48.3 & 1.884256436 & 0.2816566342 & 4.002637145 & 4 & 200 \\ 54.4 & 1.686509508 & 0.3515785836 & 4.996301621 & 5 & 210 \\ 60.1 & 1.539473673 & 0.4219446333 & 5.996277229 & 6 & 211 \\ 65.5 & 1.425021033 & 0.4924446812 & 6.998157094 & 7 & \color{red}{?} \\ \hline \end{array} $$

I know a value of $h^2 + k^2 + l^2 = 7$ is not allowed, but I cannot understand where I am going wrong with the indexing here. Any insight would be greatly appreciated as I am wondering if this is made up data that has a typo?

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    $\begingroup$ Assuming metal = element (which may be wrong), and this table here, did you give potassium a trial (COD, a = 5.247 A), bcc? And, what is the source of this table you cite? $\endgroup$
    – Buttonwood
    Apr 4 '21 at 17:17
  • $\begingroup$ I'm not sure there is any problem as the unit cell parameters were assumed to be identical, such as in the case of a cubic space group. How about in the case of a tetragonal space group? $\endgroup$
    – z1273
    Apr 4 '21 at 17:19
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    $\begingroup$ @z1273 Three nitpicks: #1 a space group (typically one of the 230 standard settings, like $P2_1$, or $P2_1/c$) may belong to a crystal class (like for the two mentioned, both monoclinic). #2 The tetragonal class is of lower symmetry than the cubic one. It is possible to describe anything in monoclinic $P1$, the lowest symmetry. But even then, one strives to identify evidence for higher symmetry in the data (some computer programs proceed this way early / lately, e.g. Platon's ADDSYM). $\endgroup$
    – Buttonwood
    Apr 4 '21 at 17:58
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    $\begingroup$ @z1273 #3: It is possible the experimental data leading to the above mentioned COD .cif entry differ to the (maybe constructed) diffractogram of the above question differ. If however the above mentioned .cif is feed, e.g. into CCDC's Mercury, after adjusting for this approx Cu K$\alpha$ radiation, then the position of the predicted diffraction peaks along $2\theta$ are very close to the ones predicted. (Only, that the diffraction indices don't match the ones in the OP ...) By this & the proximity of the density stated in the OP and in the .cif (calculated Rx density of 0.859), K is likely. $\endgroup$
    – Buttonwood
    Apr 4 '21 at 18:04
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    $\begingroup$ @z1273 True, #1 and #2 are not required here, address a meta level. And on occasion, with the information given, there may be more than one solution. $\endgroup$
    – Buttonwood
    Apr 4 '21 at 18:36
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TL;DR: The data from the assignment is correct. You confused yourself by assigning $h_i^2 + k_i^2 + l_i^2$ to the ratio of $\sin^2\theta_i$ values, and keep in mind $(100)$ might be absent.


First, let's get bold and assume that we are indexing a cubic system. This is usually the first step in analytical indexing, plus it's an extremely common system for metals. Combining Wulff–Bragg's condition

$$2d_i\sin\theta_i = n\lambda\tag{1}$$

with the relationship between the quadratic form of the relation between interplanar spacing $d_{hkl}$ and the cell parameter $a$

$$\frac{1}{d_i^2} = \frac{h_i^2 + k_i^2 + l_i^2}{a^2},\tag{2}$$

the following relation between the scattering angles and Miller indices is evident:

$$\sin^2\theta_i = \frac{\lambda^2}{4a^2}\left(h_i^2 + k_i^2 + l_i^2\right)\tag{3}$$

Since $h_i, k_i, l_i \in \mathbb{Z}$ and $\lambda^2/(4a^2)$ is a constant, normalized series of $\sin^2\theta_i$ ratios corresponds to the $h_i^2 + k_i^2 + l_i^2$ sums. Your mistake was that you took normalized $d_i^2/Z$ values (by the way, $Z$ is the number of formula units in crystallography) as $h_i^2 + k_i^2 + l_i^2.$

$$ \begin{array}{cccccccc} \hline i & 2\theta/^\circ & d_i/Å & \sin^2\theta_i & \displaystyle\frac{\sin^2\theta_i}{\sin^2\theta_1} & h_i^2 + k_i^2 + l_i^2 & hkl & a_i/Å \\ \hline 1 & 23.6 & 3.77 & 0.042 & 1 & 2 & 110 & 5.33 \\ 2 & 33.6 & 2.67 & 0.084 & 2 & 4 & 200 & 5.33 \\ 3 & 41.5 & 2.18 & 0.126 & 3 & 6 & 211 & 5.33 \\ 4 & 48.3 & 1.88 & 0.167 & 4 & 8 & 220 & 5.33 \\ 5 & 54.4 & 1.69 & 0.209 & 5 & 10 & 310 & 5.33 \\ 6 & 60.1 & 1.54 & 0.251 & 6 & 12 & 222 & 5.33 \\ 7 & 65.5 & 1.43 & 0.293 & \color{red}{7} & 14 & 321 & 5.33 \\ \hline \end{array} $$

As you can see, we have a number seven again which is … good news! Not only our initial suggestion to go with the cubic system was right (integer relations), but we also stumbled upon $\sin^2\theta_7/\sin^2\theta_1 = 7$ thanks to the author of the assignment who was kind enough to give us at least seven peaks. As you have rightfully suggested, it is impossible to find the sum of the squares of three integers for the number seven, but we don't have to — these are ratios, and can be multiplied by two.

At this point I pre-calculated cell parameter

$$a_i = d_i\sqrt{h_i^2 + k_i^2 + l_i^2}\tag{4}$$

to find the average from each signal, but thanks to the rounding error that wasn't necessary, and $a = \pu{5.33 Å}.$

Multiplying by two is equivalent to assuming that the first reflection in the diffractogram is obtained as a result of x-ray scattering from the atomic plane with indices $(110).$ At this point any number can be represented as the sum of the squares of three integers and we populate the table to the end with Miller indices.

Now thanks to that seven we also know what out of three cubic systems we have exactly: body-centered cubic (bcc) with $Z = 1 + 8 \cdot 1/8 = 2:$

Body-centered cubic cell

This means we can check ourselves and deduce what metal has been analyzed by finding its molar mass $M$:

$$ \begin{align} M &= \frac{\rho a^3N_\mathrm{A}}{Z} \\ &= \frac{\left(\pu{0.862 g cm^-3}\right)\left(\pu{5.33E-8 cm}\right)^3\left(\pu{6.02E23 mol^-1}\right)}{2} \\ &\approx \pu{39.3 g mol^-1},\tag{5} \end{align} $$

which leads us to believe the metal we analyzed was potassium with the density $\pu{0.89 g cm^-3}$ as reported by Wikipedia, which in my opinion is close enough to the value of $\pu{0.862 g cm^-3}$ from the assignment.

Final note: keep track of significant figures. You are not allowed to present values with ten sig figs calculated from the scattering angles given with three sig figs accuracy.

Indexing is more an art than a science” —A. Le Bail [1], aka the quote that crystallographers love to insert into presentations and emails for no reason:)

Reference

  1. Bail, A. L. Monte Carlo Indexing with McMaille. Powder Diffraction 2004, 19 (3), 249–254. DOI: 10.1154/1.1763152.
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