[KAJLogic]

When phophorus is in phosphate it makes four bonds.

I get the 3px,py, and pz bonds but where are the other two bonds coming from? The benefit of no one ansewering right away is I keep hacking at it. I see that the oxygen bonding to phosphurus is negative so it would make sense that it is making an ionic bond. However to satisfy phosphurus' d orbital it would atleast need xy1 xz 1 yz 1  x²-y² 1  z² 1 so why is it interested in an un satisfying amount of electrons from oxygen? To my knowledge no electrons is a far superior equalibrium than an un-even volume distribution.

[antimatter101]

Good question - this effect has been seen throughout heavier elements down the table. Its real name is hypervalence, and numerous theories have been put forward. It is still ambiguous in nature, though.

Firstly, the number of electron pairs each shell can contain increases down the periodic table. Take argon, for example. In favour of a stable octet, it does not try to incorporate electrons to its empty p-orbitals - though it can theoretically. Lighter elements, such as fluorine, cannot incorporate such p-orbitals, and thus cannot experience hypervalence.

Basically, the theory is stating that the elements with unfilled p-orbitals in favour of a stable octet can fill their p-orbitals if attacked by sufficiently strong electrophiles. In this case, the electrophile's enthalpy is negative. Also, due to a decrease in effective nuclear charge, the heavier elements' valence electrons are more susceptible to electrophilic attack. For example, we see sulfur trioxide, sulphuric acid, and other sulfur hypervalent molecules bonded to electronegative atoms, but not stuff such as SH₄, SH₆ etc. Because they are simply not electrophilic/electronegative enough!

Also, scientists have also realised that the central atoms in phosphoric and sulphuric acids have formal charges of 1 and 2 + respectively - meaning that they would require two extra electrons to obtain a stable octet.

Also, bonding via p-orbitals may strengthen other bonds formed and decrease bond length, leading to stability.

And that is all there is.

[magician4]

And that is all there is.

with all due respect: no, it isn't

you didn't even beginn to discuss modern multicenter  bonds, let alone ab initio QM calculations that show you that the situation is much more complex than a simple yes or no for hypervalancy and / or formal charges, and would instead return something like "a 17% part of  one d-orbital is involved here, somehow" to you if you'd ask what a propper understanding of reality should be like

maybe that would be more than the what KAJlogic really wanted to know for starters, yes, I'll give you that ,

but :

you didn't even begin to scratch the surface of the problem with your explanation

no, it really isn't all there is, no, not by a very very long way


regards

Ingo

[KAJLogic]

Okay let me try to suss all this out.. From my understanding what is occuring is oxygen^-

- MID SENTENCE DIGRESSION -
which I assume is the electrophile- this is weird to me cause it shouldn't be due to it already over extending its columbic energy- because you said; "In this case, the electrophile's enthalpy is negative," and; "Also, due to a decrease in effective nuclear charge, the heavier elements' valence electrons are more susceptible to electrophilic attack," again to my understanding the effective nuclear charge is talking about the columbic attraction on an electron taking into account variables such as distance and other electrons.
- BACK TO WHERE I LEFT OFF -

has an electron piggybacking that it doesn't nessasarly have the energy to hold but it is a low enough energy electron that it doesn't have the energy to leave. So, oxygen bonds with phosphurus' d orbital which has an equally indifferent desire to have an additional electron (or in this case two).  Is this accurate?

[magician4]

Is this accurate?

no, it is not


regards

Ingo

[KAJLogic]

where do the extra electrons end up then? 3rd energy shell which orbital?

[magician4]

look up "3 center two electron bonds", i.e. modern bonds theory (it's the same scheme for 3c4e, except that the additional electrons would be located in the non-bonding orbital formed by, in this case, overlapping oxygen orbitals and the central s at phosphorus

general idea you can seen in this post:
http://www.chemicalforums.com/index.php?topic=70829.0  , picture  in 3. article in thread
(except that the picture shows [F-H-F]- , but the situation at phosphate insofar is identical with respect to  multicenter bonds )

looking at a more traditional description of the situation, i.e. MO -theory, situation would be described like this:


(picture taken from: link)

the "central" part is the same for the isoelectronic molecules of SiO₄^4- (shown in pic.), PO₄^3-, SO₄^2- and ClO₄^-
[just replace lefthand "Si" with "P", and shift one electron from righthand t₁ at oxygen to lefthand p at (then) P : there you are]

as you can see, the highest "bonding" orbital in the MO would be t₂ (blue), whereas the HOMO t₁ (lilac) is non - bonding (as it's at the same level than before at the oxygens).


"below" QM - results, this is the closest thing to reality we've got these days to describe the situation

regards

Ingo

[KAJLogic]

Yes thank you. I have to start utilizing the search on this forum more. Please let the owner of this site know that if he is willing I will pay for a copy of his server files. It does look like he is using some given format, but if he has access to any ftp I will help him do it (if necessary). I say this here because I noticed some of you are administrators.


P.S: You do not by any means need to ansewer this I can probably look it up somewhere, but if you have some free time. The lines in that diagram are refrencing orbitals the arrows electrons. But what do the "t"'s represent? Also (assuming the colored thick lines represent orbitals) why does P have more than 3 orbitals?

[Borek]

if he is willing

He is not.

[magician4]

(...)But what do the "t"'s represent?

those are Mulliken symbols , denoting the symmetry of those orbitals (amongst other properties like sign affection and thatlike)

Also (assuming the colored thick lines represent orbitals) why does P have more than 3 orbitals?

what you see (except of 1s, 2s, 2p  which aren't shown) are: 3s (one, red) 3p (three, blue) , 3d ( 2 +3 = 5 total , 2*e - type [green ] and 3*t₂ type [blue ])

... i.e. all orbitals that are relevant to explain the behaviour of P in MO-theory, no matter whether they might be occupied with electrons in its elementary groundstate  or not: even empty orbitals will influence chemistry


regards

Ingo

[KAJLogic]

So would I be accurate in saying that phosphorus in reaction to the situation it is put in combines orbitals? 

[antimatter101]

Yes, you can say that, but the d orbitals are so much higher in energy than the s and p ones that it is nearly impractical to hybridise or bond with it. That is why the atoms have formal charges instead.

[KAJLogic]

I was wondering about this too.

   So phosphurs is less electronegative than oxygen so it makes sense that it looses an electron to three of the oxygens it comes in contact with. Now the reason the fourth oxygen gets double bonded is because after loosing its electrons to the other two oxygens it is now more electronegative, and sense it is an sp³ hybridization it should only be able to sustain 4 bonds, however, the extra bond is simply implying that oxygen is providing 2 electrons to the now empty orbital due to the other three oxygens. Is this all correct? ( thanks in advanced you are all invaluable)

[KAJLogic]

Revitalizing this thread so perhaps I can get an answer.

[magician4]

analyzing the P-O bonds in PO₄^3- as being ionic ("So phosphurs is less electronegative than oxygen so it makes sense that it looses an electron to three of the oxygens"),  in the light of the title question will lead to meaningless results

phosphate and thatlike need modern understanding of bonds: at least MO theory, better still multicenter bonds, and QM calculations at best.

regards

Ingo