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This question already has an answer here:

I understand that the hybridization in the phosphate ion is $\mathrm{sp^3}$ since phosphorous forms $4$ sigma bonds with $4$ of the oxygens and there are no lone pairs on the phosphorous atom.

The outer electronic configuration of phosphorous is $\mathrm{3s^2 3p^3}$. In the case of $\mathrm{sp^3}$ hybridisation, the $\mathrm{3s}$ and the all of the $\mathrm{3p}$ orbitals are involved. Hence, among the $4$ equivalent orbitals now obtained, $1$ of them contains a pair of electrons, while the other $3$ are singly occupied. So shouldn't phosphorous be able to form sigma bonds with only $3$ of the oxygen atoms since it has only $3$ unpaired electrons?

I am trying to compare this case with that of carbon tetrachloride. There, by $\mathrm{sp^3}$ hybridization, we were able to get $4$ equivalent orbitals each with an unpaired electron, and therefore explain its tetravalent nature. In the case of phosphate ion, one of the $4$ orbitals obtained already has a pair of electrons. So, could anyone please tell me where my understanding is flawed?

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marked as duplicate by Mithoron, Todd Minehardt, airhuff, M.A.R. ಠ_ಠ, pentavalentcarbon Feb 17 '18 at 16:30

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Electrons do not know what atom they belong to. Strip one electron from P. Now it is the same as C. Now give that electron to one of the O. $\endgroup$ – Ivan Neretin Feb 16 '18 at 6:17
  • $\begingroup$ wouldnt that affect the hybridisation of oxygen? $\endgroup$ – physics123 Feb 16 '18 at 6:20
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    $\begingroup$ Why? Which way? And besides, how is that important at all? $\endgroup$ – Ivan Neretin Feb 16 '18 at 6:21
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Disclaimer: hybridization is a mathematical concept and fails at higher levels. As per the linked answer, it is clear that the participation of $d$-orbitals in phosphate ion is minimum. But this is just a high-school level, layman's explanation for your question.


So, could anyone please tell me where my understanding is flawed?

Your reasoning is flawed in assuming that carbon atom has four unpaired electrons in its orbitals after hybridisaton, while phosphorus does not. Actually, both of them do.

In fact, carbon atom in its ground state does have a lone pair of electrons in its $\mathrm{2s}$ orbital. It is only when it gets into its excited state, that that lone pair is broken up, yielding four unpaired electrons. The excited carbon atom now undergoes hybridization as depicted in the figure below, yielding four equivalent covalent bonds as in carbon tetrachloride.

Carbon atom in its ground state and excited state, before and after hybridisation (source)

As you noted, phosphorus also has a lone pair of electrons in its $\mathrm{3s}$ orbital. But, that is its ground state electronic configuration. Once it gets excited, it will have five unpaired electrons, the fifth electron going into its $\mathrm{3d}$ orbital. Now, phosphorus will undergo $\mathrm{sp^3}$ hybridization, to form $4$ sigma bonds. The fifth electron helps form the $d\pi-p\pi$ bond with the neutral oxygen atom.

I hope this helps!

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    $\begingroup$ sigh OP needed explanation how dipolar bond works not some outdated "d-orbital" nonsense. $\endgroup$ – Mithoron Feb 16 '18 at 18:05
  • $\begingroup$ @Mithoron The way OP framed his question with "excitation of electrons in carbon" and "hybridization" it is his clear that he needed a simplistic explanation of the things going on as per high school level. The question you linked to involves Natural Bond Orbital analysis, and I am not sure how that would have helped the OP in any way. I've edited my answer to make it more clear $\endgroup$ – Gaurang Tandon Feb 17 '18 at 2:19

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