how phosphorus forms 2 types of bonds with oxygen i.e. 5 in $\ce{P4O10}$ and 3 in $\ce{P4O6}$ when only three are needed to complete its octet?

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    $\begingroup$ Did you consider the different oxidation states of phosphorus in both oxides? $\endgroup$ – Klaus-Dieter Warzecha Apr 3 '15 at 10:22
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    $\begingroup$ Welcome to chemistry.SE! I wouldn't rely too much on octet if I were you, especially when it's about phosphorus. Your question won't make sense if we not rely on octet, which is the case about group 15. $\endgroup$ – M.A.R. Apr 3 '15 at 10:40
  • $\begingroup$ @Ϻ.Λ.Ʀ. -- too bad there's no such simple reply for Cu(OH)2. I mean, as simple as Chinmay's. $\endgroup$ – CowperKettle Jan 28 '16 at 17:54

Phosphorus has electronic configuration of [Ne]3S2 3P3 As it is a 3rd period element it also have a vacant 3D orbital.

In case of P4O6 only the 3P3 electrons take part. Resulting in formation of 3 bonds with O. But in case of P4O10 one of the electron of 3S is excited to vacant 3D orbital. Now P has 5 unpaired electron and it can form 5 bonds. Therefore P forms 5 in P4O10.

Now, about the octet. Not every compound follow the octet rule. Just to understand The whole chemistry depends upon STABILITY i.e lesser energy of molecule. Therefore, if by forming more bonds (and does not obeying Octet rule) the molecule is gaining more stability, it will do so.

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  • $\begingroup$ The shape of the 4 Os around P is tetrahedral . Also, according to JD Lee's Concise Inorganic Chemistry, the double bond P=O in P4O10 contains p(pi)-d(pi) backbonding, where a full p-orbital on oxygen overlaps with an empty d-orbital on P ... this seems to point towards an sp3 arrangement with one electron pair from P being "given" to an empty p orbital on O, otherwise how will O attain it's octet ..? $\endgroup$ – vishesh jain Jun 15 '19 at 8:34

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