5
$\begingroup$

I need help sorting out the molecular structure of tri-lead tetroxide, aka lead(II,IV) oxide $\ce{Pb3O4}$.

This data sheet shows this image:

enter image description here

The Wikipedia's lead (II,IV) oxide's link to a Jmol interactive image looks like this:

enter image description here

Even the Wikipedia article's Talk Page's discussion on the naming and structure makes for interesting reading.

The first drawing shows a $\ce{Pb-O-Pb-O-Pb-O -}$ ring, the second shows a 4-bond $\ce{Pb}$ in the middle with two 2-bond $\ce{Pb}$'s on either side.

Is there a "right way" and a "wrong way" to draw a single molecule of $\ce{Pb3O4}$ that represents how it might actually be configured? Are both, either, or neither of the above representations "right"?

| improve this question | |
$\endgroup$
  • 12
    $\begingroup$ It's not a molecule, it's an extended solid. Drawing this is as wrong as drawing a covalently bound molecule of NaCl. For the structure of Pb3O4 see e.g. Wells Structural Inorganic Chemistry 5ed, p 558 $\endgroup$ – orthocresol♦ Mar 30 '17 at 10:45
  • 3
    $\begingroup$ If you have access to scientific papers, this, Fayek, Leciejewi, Zeitsch. f. anorg. u. allg. Chemie, 2004, 336, 104-110 could be worth reading. Also, what orthocresol said. Its a crystal, painting a molecule is in no case right. $\endgroup$ – Fl.pf. Mar 30 '17 at 10:49
  • 5
    $\begingroup$ No, a crystal cannot in general be subdivided into individual molecules. I remember a question on this site about it, I will try to search. $\endgroup$ – orthocresol♦ Mar 30 '17 at 10:54
  • 1
    $\begingroup$ Yes, but ice melts to form water molecules. Pb3O4 doesn't melt to form Pb3O4 molecules. In fact it doesn't even melt, it decomposes (Wikipedia). NaCl doesn't melt to form NaCl molecules, it forms Na+ and Cl- ions. Therefore it would not be correct to think of NaCl solid as an aggregation of NaCl molecules. $\endgroup$ – orthocresol♦ Mar 30 '17 at 10:57
  • 3
    $\begingroup$ Yes, but even if it sublimates it's still water, with H-O-H covalent bonds. The truth is that Pb-O bonding is not very covalent. As far as I can tell, there is no potential for the formation of discrete Pb3O4 molecules. Maybe it would help if you added some context as to why you're curious about the structure of a single Pb3O4 molecule. $\endgroup$ – orthocresol♦ Mar 30 '17 at 11:05
5
$\begingroup$

What you call a molecular structure is in reality an asymmetric unit. These two terms are not related, and in general are not interchangeable. The same way $\ce{Pb3O4}$ is not a molecular formula, but a formula unit.

All pictures you presented are correct in terms that they do represent an asymmetric unit, but actually an IUCr-recommended asymmetric unit (can be generated with Olex2's command standardise) looks like this:

$\color{#FF0D0D}{\Large\bullet}~\ce{O}$; $\color{#575961}{\Large\bullet}~\ce{Pb}$.

enter image description here

... and with $Z = 4$ these are packed in a unit cell with $Pbam$ space group accordingly:

enter image description here enter image description here

The crystal structure itself can be represented as infinite layers of cross-linked chains made of $\ce{[PbO6]}$ octahedra, which are stacked via $\ce{- O-Pb-O -}$ bridges:

enter image description here

Crystallographic data is taken from ICSD #9753 [1].

  1. Gavarri, J. R.; Weigel, D.; Hewat, A. W. Journal of Solid State Chemistry 1978, 23 (3), 327–339. DOI 10.1016/0022-4596(78)90081-6.
| improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ This is very helpful! It's late here and I was just about to log out. I now understand the term asymmetric unit and I'm guessing after tomorrow morning's coffee I will understand how to put them together and still get the stoichiometry right, and why the Pb is sometimes drawn with thee bonds to O, and sometimes with four. $\endgroup$ – uhoh Aug 23 '17 at 16:15
  • 1
    $\begingroup$ Later than I expected it seems, sorry for the delay and thank you for the clear, concise, and well-sourced answer! $\endgroup$ – uhoh Mar 16 '19 at 9:30

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.