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I'm looking for an explanation of the bonding in the phosphate (PO43−) ion:

enter image description here

(Image courtesy of Wikipedia)

Phosphorus (15P) - being the fifteenth element - has fifteen electrons, five valence electrons and the following electron configuration:

1s2
2s2 2p6
3s2 3p3

This being so, by what means does it form its bonds? Naively, one might expect it to form 3 single covalent bonds and be happy. But it doesn't, it forms five bonds, presumably using each of its five electrons. My first question is - why would it do this? It seems to me that this would result in the Phosphorus atom having ten electrons in its outer shell (four from the double bond with the oxygen and two each from the single bonds with the other three oxygens).

Apart from the oxygen with a double-bond, we have three oxygens forming single bonds. Does this not leave them one electron short?

Finally, why are the oxygens shown as carrying a negative charge?

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First of all, let me state the obvious: Phosphorus is awesome. After we got that out of the way we can focus on why.


There are many different modifications of phosphorus in nature. With increasing thermodynamic stability they are $$\ce{P_{white} -> P_{red} -> P_{violet} -> P_{black}}.$$

Apart from this there are many low molecular different allotropes, like $\ce{P4 (white)}$, $\ce{P6}$, $\ce{P8}$, $\ce{P10}$, $\ce{P12}$. And because that is not enough, there are chain like polymeric allotropes, too.

Apart from this it is possible to formulate many different cations and anions, that are derived from the above molecular structures. Just to name a few, there are $\ce{P3+, P5+, P7+, P9+}$ mainly observed in gaseous phase an $\ce{P^{3-},P2^{4-}, P3^{5-}, P4^{2-}, P7-,...,}$ usually in combination with alkali metals. Most amazingly it may form anionic polymer strains of the general form $\ce{[P7-]_{\infty}, [P15-]_{\infty}}$.

But now for the most important part, phosphorus has stable oxidation states in compounds, ranging from $\mathrm{-III}$ to $\mathrm{+V}$. Here are a few examples: $$\ce{\overset{-III}{P}H3,\ \overset{-II}{P_2}H4,\ [\overset{-I}{P}H]_{n},\ \overset{\mathrm{\pm0}}{P4},\ H3\overset{\mathrm{+I}}{P}O2,\ H4\overset{\mathrm{+II}}{P2}O4,\ H3\overset{\mathrm{+III}}{P}O3,\ H2\overset{\mathrm{+IV}}{P2}O6,\ H3\overset{\mathrm{+V}}{P}O4}$$ While dealing with these compounds it is usually completely unnecessary to describe bonding with hybrid orbitals.

In case of phosphane $\ce{PH3}$ it would be wrong. Assuming $\ce{{}^{sp^3}P}$ one would expect $\angle(\ce{H-P-H})\approx109^\circ$, while it is found to be $\angle(\ce{H-P-H})=93.5^\circ$, which is almost the same angle as the $\ce{p}$ orbitals are having towards each other.

In general your assumption is correct, that it is possible to form only three covalent bonds to reach a stable configuration. And that will most likely be the case when phosphorus forms compounds with more electropositive elements.

Now dealing with oxygen, means dealing with a much more electronegative element, i.e. $\ce{En(O)}\approx3.4$, $\ce{En(P)}\approx2.2$. That also means that bonds are much more polarised towards the oxygen.

Analysing the phosphate anion $\ce{PO4^{3-}}$ it is crucial to recognise its symmetry, which is tetrahedral $T_\mathrm{d}$. In this arrangement it is perfectly safe (but not at all necessary) to describe phosphorus as $\ce{sp^3}$ hybridised.

A Natural Bond Orbital analysis (BP86/def2-TZVPP) reveals that there are 4 equal $\ce{P-O}~\sigma$ single bonds and each oxygen has three lone pairs. The contribution of the $\ce{d}$ orbitals to bonding is well below $1\%$ and can be interpreted as numerical noise (use as polarisation functions) of the DFT method. \begin{array}{rlrr}\hline & &\mathrm{\%P (hyb)} &\mathrm{\%O (hyb)}\\\hline 3\times&\ce{Bd(O-P)} & 24 (\ce{sp^3}) & 76 (\mathrm{sp^{2.3}})\\ &\ce{Lp(O)} & & 100 (\mathrm{sp^{0.4}})\\ 2\times&\ce{Lp(O)} & & 100 (\ce{p})\\\hline \end{array}

This is consistent with the partial charges, i.e. $q(\ce{P})=2.2$, $q(\ce{O})=-1.3$. Therefore a more accurate Lewis formula is with charge separation.
phosphate lewis structure

The corresponding NBO reflect the bonding picture one would expect given all details from above. It should be noted, that natural bond orbitals are a linear combination of the canonical orbitals and do not have a physical meaningful energy eigenvalue. The two top rows represent the $\ce{p}$ lone pair orbitals of oxygen, the third row represents the $\ce{sp^{0\!.4}}$ lone pair orbitals, the fourth row gives the $\sigma$ bonding orbitals. (The last row is the orientation of the molecule, core orbitals are not displayed.)

phosphate nbo

The corresponding canonical orbitals which have a physical meaningful eigenvalue are delocalised over the whole molecule, hence they are not providing a simple understandable bonding picture. While NBO fail to respect the symmetry point group, canonical orbitals are constructed to obey this principle. (Here shown from highest energy, top, to lowest energy, bottom, core orbitals not displayed.)

canonical orbitals of phosphane

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    $\begingroup$ But since each oxygen has a negative formal charge, wouldn't phosphoric acid instead be H4PO4, rather than H3PO4? Since each negatively-charged oxygen can bond with one hydrogen $\endgroup$ – Tan Yong Boon Sep 14 '17 at 23:45
  • $\begingroup$ @TanYongBoon then you would have charge imbalance and it would be $\ce{H4PO4+}$, but in principle I see no reason why it couldn't be protonated there, too. $\endgroup$ – Martin - マーチン Sep 15 '17 at 0:09
  • $\begingroup$ Ok. But which would be the dominant species in aqueous solution? It would still be H3PO4 as H4PO4+ would be a strong acid and fully dissociate in water to give H3PO4? Not sure of what I'm saying... $\endgroup$ – Tan Yong Boon Sep 15 '17 at 2:16
  • $\begingroup$ @TanYongBoon In aqueous solution of $\ce{H3PO4}$ there will be multiple equilibria with all protonation stages present. The dominant species will probably be $\ce{H2PO4-}$ or $\ce{H3PO4}$. Very unlikely without another acid is $\ce{H4PO4+}$. Without another base, $\ce{HPO4^2-}$ and $\ce{PO4^3-}$ are also quite unlikely. The formal charges are just a book-keeping tool, in the phosphate ion they are delocalised. But when you add protons, then they will be counteracted. $\endgroup$ – Martin - マーチン Sep 15 '17 at 2:23
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Phosphorus ($\ce{_{15}P}$) - being the fifteenth element - has fifteen electrons, five valence electrons and the following electron configuration:

$\ce{1s^2,~2s^2,~2p^6,~3s^2,~3p^3}$

This being so, by what means does it form its bonds? Naively, one might expect it to form 3 single covalent bonds and be happy. But it doesn't, it forms five bonds, presumably using each of its five electrons. My first question is - why would it do this? It seems to me that this would result in the phosphorus atom having ten electrons in its outer shell (four from the double bond with the oxygen and two each from the single bonds with the other three oxygens).

Very observant! What happens is that phosphorus is hybridized in the phosphate ion. Hybridization allows the central phosphate atom to form more bonds than you might expect from its ground state electron configuration.

Within the framework of hybridization, one may look at the bonds in the phosphate anion as having a great deal of ionic character.

The electronegativity of phosphorus is 2.19; the electronegativity of oxygen is 3.44. The difference is 1.25 - or rather sizable. As a result, electron density is largely withdrawn from the central phosphorus. Hence the negative charges you see on the three oxygen molecules in your picture.

Other depictions, which reject a large degree of d-orbital utilization in hybridization, might give a negative formal charge to all the oxygens and a positive formal charge to the central phosphorus by eschewing the $\ce{P=O}$ double bond.

Apart from the oxygen with a double-bond, we have three oxygens forming single bonds. Does this not leave them one electron short?

Finally, why are the oxygens shown as carrying a negative charge?

You could draw double bonds for all the oxygens but there aren't that many valence electrons to be distributed. The structure shown above is a compromise between number of valence electrons and electronegativity considerations - we wouldn't want a surfeit of electrons around the central phosphorus since that's not as thermodynamically favorable as having the more electronegative oxygen support these electrons.

In addition, the sum of the formal charges on an atom must add up to the overall charge on the atom. Phosphate anion is triply negative charged; note how there are three oxygens bearing a single negative charge.

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    $\begingroup$ The oxygen's are not 'short' of electrons if you're referring to the filling of their shells. The sigma bonded oxygens have 1 bonding pair and 3 lone pairs which gives a full valence shell. The negative formal charge occurs because oxygen has 6 valence electrons in the ground state and 2/2+6=7 electrons in this structure. 6-7=-1 $\endgroup$ – canadianer Jun 17 '14 at 7:03
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Thank you very much, both of you, for some very detailed replies. I'm going to try to synthesise them into a single answer, which I hope will be accurate. Here is how I understand it, armed with the new knowledge contained within the posts above:

In the phosphate ion we have a central Phosphorus atom, with five valence electrons.

This is bonded to four oxygen atoms, which have six valence electrons.

Five P electrons plus 4 times 6 O electrons gives 29 electrons. We also have a charge of 3- indicating that we need to take into account 3 additional electrons, giving us 32 electrons overall.

We could model this situation by proposing that the P atom exists at the centre, with four O atoms bonded to it. One O is double bonded to the P and the other three Os are single bonded, giving 10 electrons to the P (which it is happy with).

In the case of the double-bonded O we would have 2 lone pairs of electrons (makes 4) and a further 4 electrons from the double bond. Oxygen has 6 valence electrons and so the 2 lone pairs are its own electrons and its other 2 valence electrons form the double bond with two of the P electrons. The double-bonded O has eight electrons and is happy.

In the case of each single-bonded O we would have 3 lone pairs (presumably the oxygen's own six valence electrons?) plus an extra electron (presumably from elsewhere? I don't really understand this) which participates in a single bond with a P electron. This gives us eight electrons, which the oxygen is happy with.

This is the model shown in the wikipedia drawing:

enter image description here

Now, if we draw the Lewis structure:

enter image description here

We can see all of this put together and we can also see that, while the double-bonded O is neutral, having six electrons assigned to it, the three single bonded O atoms have seven electrons assigned and therefore carry a -1 charge.

There is no reason, however, why we should select the top O atom as the one to have a double-bond. This is not at all what we find in reality.

We can build a simplified model of reality, where the double-bond (and its electrons) is exchanged between each O atom:

enter image description here

And where there is no double-bond at all:

enter image description here

All of this, however, is really nothing more than an extreme simplification. The reality is a hybrid. It is possible to propose that we have four sigma bonds with four sp3 hybridised orbitals from the phosphorus and a fifth delocalised pi bond with the d orbital of the phosphorus. This gives our P 10 electrons and each O 8 electrons.

But the reality seems to be that, in fact, the d orbital is not involved at all and instead we have four sigma bonds formed by sp3 hybridised orbitals from the phosphorus.

And here's what I just don't understand. When I drew the Lewis structure, I added three extra electrons from somewhere. This would make sense if, as martin says, each oxygen has three lone pairs, we can imagine a sea of the five P valence electrons and these extra three electrons spread evenly with two forming each sigma bond. This gives each oxygen eight electrons and it would give the P eight electrons too. The higher electronegativity of Oxygen means that the O atoms attract the electrons more strongly and each atom receives a negative charge, whereas the P attracts them less strongly and receives a positive charge. I'm really not sure I've understood the arrangement of electrons though, I would like to know how the electrons are apportioned in the ion, where they came from and how the overall charge comes about.

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    $\begingroup$ You should ask another question. In short: It does not matter at all where the electrons came from. For each anion you will most likely have an cation and that can be the source for the extra electrons. $\endgroup$ – Martin - マーチン Jun 19 '14 at 8:17
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    $\begingroup$ I would like to clarify, that in the phosphate anion all oxygens are equal, no double bonds are formed. Phosphorus and oxygen both fulfill the octet rule. The bonding picture I provided in my answer in no simplification, it is the best representation for this molecule. A Lewis structure, however, is a very crude simplification. $\endgroup$ – Martin - マーチン Jun 25 '14 at 6:56
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Simply put: To make PO4 you must start with a neutral P and three O with a -1 charge and one O with a neutral charge.

The phosphorus can make 4 bonds by hybridizing

The phosphorus hybridizes to sp3 by loosing one electron to the neutral O. This creates four oxygen atoms that have a -1 charge and one Phosphorus atom with a +1 charge total charge -3

Once the bonds are formed they are all equal and the molecule is stable. It concerns the valance electron availability the neutral O has 6 and gains one while the other three O -1 have 7 each and the P has 5. total of 32 When the P hybridizes it then has 4 valence electrons while the neutral O gains one valence electron, thus achieving 7. the total remains the same. It doesn't mater what position the neutral O attaches to. The neutral O will have transferred the electron before attaching.
In phosphoric acid H3PO4 phosphorus undergoes hybridization to produce 5 valence electrons by transferring an electron to a higher hybrid level. Phosphorus gains stability by rearranging its valence electron shells.

Phosphoric acid striped of the hydrogen protons will leave you with phosphate.

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I happen to look at this site just now. The reason is all a matter of hydridization. Look at the electron configuration of Phosphate and that of the next element that use next orbital i.e. K. The hybridization would bring in 3s, 3p, and 4s together. To complete fill them, number of electrons or bonds required would therefore be 5. I guess that there are 5 sp3 hybrid orbitals. Because of electron negativity of Phosphorus and oxygen, there are negative charge on the oxygen atoms that are singly bonded, while the doubly bonded oxygen, being closer could have insignificantly charged negatively.

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protected by orthocresol Apr 23 '17 at 20:52

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