There's a question in the chemistry textbook I use with my GCSE class which asks pupils to identify which of various compounds feature ionic bonding. One of the compounds is phosphorus(V) oxide, which I gather is an accepted name of phosphorus pentoxide (or tetraphosphorus decoxide, if you prefer). But why? One of my pupils reasoned that it must have ionic bonding, because I'd just explained to them how Roman numerals in brackets indicate the number of electrons lost. This is evidently an exception, but none of the guides to chemical naming I've found explain why!

I figure the explanation must lie in subtleties of the oxidation state that I haven't yet got my head round - Wikipedia's page on oxidation state lists phosphorus pentoxide as having oxidation state +5, and phosphorus trioxide as +3, but I'm not entirely following why.

Any clarification much appreciated!

  • $\begingroup$ $\ce{P4O10}$: assign oxidation states. Result: $\ce{O^{-II}}$ and $\ce{P^{V}}$. Anything unclear? $\endgroup$ – Jan Sep 30 '15 at 18:01
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    $\begingroup$ "Roman numerals in brackets indicate the number of electrons lost" - They indicate the oxidation state. This only truly corresponds to the number of electrons lost when one talks about pure ionic bonding. Phosphorus in $\ce{P4O10}$ is in oxidation state +5 but it does not mean it exists in the form of $\ce{P^5+}$ ions. It means that, if you cleaved all P-O bonds heterolytically and assigned all the electrons to the more electronegative element, oxygen, then the phosphorus would exist as $\ce{P^5+}$. Until you do that, P is happily sharing electrons with O in a covalent bond. $\endgroup$ – orthocresol Sep 30 '15 at 18:05
  • $\begingroup$ This previous question (and answer) about the assignment of oxidation states may help clarify: chemistry.stackexchange.com/q/10944/16683 $\endgroup$ – orthocresol Sep 30 '15 at 18:11
  • $\begingroup$ Thanks @orthocresol, that helps a lot. Having never studied (or taught) oxidation state in any detail, I've always been a little hazy on how it works in covalent compounds, but it's starting to make sense. $\endgroup$ – Oolong Sep 30 '15 at 20:39

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