[resonance]

If you look at a water molecule the O atom is much more electronegative. This will delocalize the electron density in the O-H bonds. The result is the formation of a dipole with the hydrogen becoming positive and the oxygen negative. The question I have is; are the positive and negative dipole ends of a water molecule at two different EM potentials. The extra electron density on the oxygen is adding stability to the oxygen, which is why it took the extra electron density from hydrogen in the first place. As such, is the hydrogen side of the dipole carrying more potential, since hydrogen gains nothing that is stabilizing, like oxygen, due to the delocation of electron density?

To make my question clearer. If you look at an orbital with two electrons the spin addition or subtraction will affect the potential of the electrons. Both situations have the same amount of negative charge. The difference has to do with the addition or subtraction of the magnetic fields created by these spining charges. With the electromagnetic force is a combination of electrostatic and magnetic, the opposite spin pair has a lower EM potential than the parallel spin pair even though both have the same charge potential. In the case of the O-H bond, the oxygen gains orbital magnetic stability,  while the hydrogen doesn't. Is the net affect is a higher EM potential on the hydrogen, even though they have the same charge potential?

[Yggdrasil]

In water, you can't talk about the electronic energies of the hydrogen and the oxygen separately.  The atomic orbitals from the hydrogen and oxygen are combined to form molecular orbitals which span all of the atoms in the molecule.  So, the overall energy of the molecular orbitals is lowest when most of the electron density resides in nearer to the oxygen.  You can't talk about the energies of oxygen's orbitals and the energies of hydrogen's orbitals because atomic orbitals do not exist in a molecule.

[resonance]

I good analogy to what I am trying to ask is connected to HCl in water. The Cl- is a weak base. As such, even though Cl- has a negative charge and is the most stable negative thing in the water (besides pH 7),  this negative charge doesn't preferentially attract all the positive charges of the H+. Something more than charge potential is at work making the negative charge of Cl- less affective at attracting positive charge. One way to explain this is chlorides's EM force potential, which is charge plus magnetic, is heavy on the magnetic side, making the EM potential low in spite of having a full negative charge, i.e., weak base.

I figured water was the same way with the highly electronegative oxygen increasing its stability with the slight negative charge lowering its overall EM potential in spite of gaining charge potential. The hydrogen, on the other hand, has less electron density, and therefore can't magnetic add the electrons as well around the hydrogen. This gives it a double potential one part associated with its induced positive charge and the other is the need for better magnetic addition. The net result is more potential on the H side. On can see this with HCl, with the strongest potential associated with the H side.

[resonance]

The idea of the hydrogen carrying the primary burden of potential going into a hydrogen bond is an important understanding that I am trying to prove. I am hoping others can help. What it will do is open another branch of hydrogen bonding investigation that has a wide range of cellular applications. My proof is currently circumstantial but my gut feeling is that this is the reality of the situation.

As another analogous example, consider one of the strongest hydrogen bonds; F-H and F-. This hydrogen bond is about 40 kcals, which is about 1/2 the strength of a weak covalent bond. If you look at the F- it is an extremely weak base. The F- is able to stabilize the extra negative charge because this electron helps it to complete its octet, thereby providing full magnetic addition from the 2s2p orbitals. As such, going into the F-H and F- hydrogen bond, the F- has little need or little potential to share. The hydrogen is induced to a very high potential by its own F-. It is able to lower potential by forcing the other F- to share, even though this F- is very weak base will little need to share. 

A more direct observation that the hydrogen of water is carrying the burden of potential is connected to clusters of hydrogen bonding. Each addition of a new hydrogen bond to the cluster will strengthen all the hydrogen bonds. It you look at first two H20, the first hydrogen bond results in the two H of the receiving water gaining potential. One way to explain this is because the first is oxygen forced to share, it loses some some stability in its octet, and pulls extra electron density from its own hydrogen to regain stability. This increases their potential to form a hydrogen bond. In a cluster, all the oxygen are hydrogen bonded and all are trying to regain some stability by taking additional electron density from their own hydrogen, making all the hydrogen increase their potential to form hydrogen bonds.

What is interesting about a hydrogen bond is that the H is already sharing its 1S orbital with the covalent bond. The partial covalent nature of a hydrogen bond implies the H will have to use level 2 orbitals to accommodate the extra electron density, since there is little room for 4 partial electrons in the 1S. What this means although this will help H lower its potential it is not as stable as H-H . The hydrogen bond requires the hydrogen share electron in higher orbitals above its own ground state. This would imply the H still has some residual potential even in a hydrogen bond. The pH affect in water appears to reflect this, with the hydrogen trying to do better than just a hydrogen bond. 

This is a visual analogy I came up with a few months back that sort of explains the situation in water. Picture O and 2H at a *Ignore me, I am dishonest* table playing cards. Because O is more electronegative it will win most of the hands. At the end of the night, the O gains extra chips while the two H will lose some of their chips. The next day, the room is full of *Ignore me, I am dishonest* tables with an O and 2H at each table (ice). Once again oxygen is the better player and wins more chips. The hydrogen, to stay in the game, reaches behind itself and picks the pocket of an adjacent O in another table (hydrogen bond). The O are still more electronegative and continues to win the chips while the hydrogen continue to pick the oxygen's pocket to keep the game going. Eventually the oxygen begin to notice that their pockets are being picked, since their winning do not add up to their higher electronegativity (skill). To avoid the theft, the oxygen begins to twist away requiring the H lean over even further (liquid water). This results in the unique situation of lliquid water contracting upon melting. In other words, if the O and H shared the burden of potential equally, water should expand upon melting, since the connected potential would increase due to ambient energy. But if H carries the burden of potential and the oxygen is more stable, the extra thermal energy in liquid water, will allow O a means to make it harder for the hydrogen to pick its pocket. By twisting, the hydrogen can't line up the O for the partial covalent bond (easy pick). What is left is electrostatic attraction, requiring the H get closer while still not being able to lower its potential as well. The result is, liquid water will contract upon melting. This evasiveness allows O to win more chips at the end of the game causing a slight potential to remain in the hydrogen of liquid water. This potential may be why liquid water will help catalyze oxidation (potential left in the hydrogen will excite electrons making it easier for O2).  

In liquid water all types of hydrogen bonding and nonhydrogen bonding structures occur. All this deviation from the lowest energy hydrogen bonding found in ice is an artifact of the evasive O in combination with the potential in the H, with the result liquid water is slightly electrophilic.  

[resonance]

I would like to clarify a subtle distinction. The analysis I was presenting is connected to a H, which is covalently bonded to highly electronegative atoms like O, N, Cl and F, before it forms a hydrogen bond, i.e., isolated water molecule. By the very nature of the electronegativity difference, the more electronegative atom will end up with more electron density because it is stabilized by accepting this extra electron density. Such atoms are trying to do more than share the octet, they would prefer to have most of the octet to itself, but can not easily shake the H. The net result is the hydrogen is induced to have a slight potential. This potential is lowered by forming a hydrogen bond.

Once an optimized hydrogen bond is formed, the hydrogen will lower this induced potential, such that the potential is more equally distributed between the H and the more electronnegative donor atom. If we assume an optimized hydrogen bond is 6kcal, the hydrogen starts out with at least 6kcals of potential. It is actually a little higher since the donor atom will lose some of its octet stability into the hydrogen bond. The starting potential of the H may actually be in the order of 8kcals, with the hydrogen gaining 8kcals and the donor atoms losing 2kcals for a net of 6kcals exothermic output. (as way of example).

What is significant about thinking in terms of H potential, is that if the optimized hydrogen bond does not form, hydrogen potential will remain in the H. For example, in a protein, if the H begins at 8kcals potential and the formation of the H-bond only gives off 4kcals, due to steric constraints, the H still has a residual potential. In other words, its own electronegative atom is still inducing 8kcal of potential in the H, even if the H is only able to get back 4kcals by its hydrogen bonding situation. The net result is the H still has potential to spare that it can't get rid of under the circumstances. This will add a little zone of 4kcals of electrophilic potential to the structure.

I got this off the internet; The occurrence of hydrogen bonds in protein structure has been extensively reviewed by Baker & Hubbard (1984), albeit before the pdb database was as large as it is today. They found that 90% of N-H---O bonds in proteins lie between 140 and 180°, and that they are centred around 158°C. For C=O---H, the range is more broadly distributed between 90° and 160° and centred around 129°. .

If you think in terms of H potential, knowing that the optimized H-bond is at 180 degrees (less than 180 degree  angles will leave potential) the above statement implies that the vast majority of NH--O H-bonds, and all the C=O--H H-bonds in all proteins will contain some residual H potential; i.e., electrophilic potential. If you look at enzyme catalysis, it uses a lock and key approach for the reactant. But in more general terms, all enzymes excite electrons on the reactant. It is possible the residual H-electrophilc potential, distributed throughout the enzyme, contributes to this electron exciting affect, since they collectively contain residual potential for electron density. Theoretically, the residual H potential could have allowed enzymes to nonselectively work even befoe they evolved their modern lock and key selectivity.

[Sam (NG)]

What is interesting about a hydrogen bond is that the H is already sharing its 1S orbital with the covalent bond. The partial covalent nature of a hydrogen bond implies the H will have to use level 2 orbitals to accommodate the extra electron density, since there is little room for 4 partial electrons in the 1S. What this means although this will help H lower its potential it is not as stable as H-H . The hydrogen bond requires the hydrogen share electron in higher orbitals above its own ground state. This would imply the H still has some residual potential even in a hydrogen bond.

You seem to be considering atomic orbitals, the orbitals in water look like this:


For information about water and hydrogen bonding, and which orbitals are involved see http://lsbu.ac.uk/water/h2oorb.html and associated pages.

[edit] I believe there is a mistake on that website, it says that the 1b₂ and 3a₁ orbitals are those mainly responsible for the donation in hydrogen bonding, but this should be the 1b₁ and 3a₁ donating with the 4a₁ and 2b₂ antibonding orbitals receiving electron density.  You can see this in relation to the position of the lobes on the orbitals with respect to the expected positions of the "lone pairs" on the molecule.

[resonance]

Thanks Sam. That orbital stuff used to make a lot a sense to me many moons ago. I am trying to show that, within an isolated water molecule, the H carries the burden of potential. This makes intuitive sense to me, but I am not sure how to formally prove it. I work under this premise and spent most of my time on the applications end, i.e., the cell. I now am trying to start at the beginning to see if I can pass on my intuitive understanding.

For example, if you look at the pH affect in neutral water, this is dependant on the ability of the oxygen of water to stabilize extra electron density so it can become OH-. If the oxygen can do that, the slight dipole in molecular water should be fairly easy to accommodate. That fact that 10-7 moles of water tries to get to -OH may imply that water may not fully meet all its electron needs. The pH affect is also dependant on the enough potential in the H. This will cause H to accept anything including the less than optimum situation that occirs in H3O+.   

[resonance]

The idea of the hydrogen of water containing the burden of potential seems logical to me. With this assumption a new branch of chemistry emerges, with profound implications. What I need is a solid basis for why this is so. So I will try again. Electrons in orbitals are more than just charge, they are charge in motion. A charge in motion will give off a magnetic field. This is why electrons in orbitals need to have opposite spin. This will allow the magnetics fields to attract so the electrons can overcome some of the charge repulsion. If you alter the magnetic field direction (change the electron spin of one of the electrons) the charges and the two magnetic field will both repel, kicking one of the electrons into a higher energy level.

If you look at an atom, not only are the magnetic fields of electron pairs attracting in each orbital (opposite spin),  but all the electron magnetic fields in all the orbitals are also attracting with respect to each other. If we could just shut off the magnetic attraction between the electrons, the atom would puff out. It is very likely the orbitals, as we know them would also change, more in line with S type orbitals (Bohr atom) with all the electrons trying to get as far away from each other as possible due to mutual repulsion. It is the magnetic fields, all attracting together, which gives up the characteristic shapes of the orbitals, which helps confine the electrons in much smaller space.

When a chemist thinks in terms of a dipole, he or she often thinks in terms of charge, with opposite charges attracting. They are essentially treating the molecular dipole like the magnetic attraction has no importance. But the reality is, a molecular dipole is not just a charge potential but it also has a connection to magnetic addition and subtraction. This magnetic connection is the fundamental basis for electronegativity with more electronegative atoms better able to take advantage of better magnetic attraction. Less electronegative atoms show less magnetic attraction and some even show magnetic repulsion. For example, if you look at Cl-, if only charge was important, this charge imbalance should not occur, since there are too many negative charges. But if you add magnetic attraction, the Cl- overcomes the repulsion created by the extra negative charge, with a little extra to spare.

In the case of the dipole in a molecule of water there is indeed a charge dipole. But that is not the whole story. To get the dipole potential distribution, one needs to add the magnetic potentials within each side of the dipole. The oxygen is able to generate greater magnetic addition by accepting the extra electron density, i.e., magnetic power of the octet. If only charge was important, it would have too much negative charge and would be unstable. But this extra charge potential gives it a magnetic advantage that allows it to overcome the extra charge potential. Whether this is a wash or not, the net affect is the overall potential due to the charge plus the charge in motion, counterbalances. The hydrogen loses some magnetic advantage, so it has a full charge potential with less magnetic stabilization. The net affect is the hydrogen is at higher overalll potential.

P orbitals in atoms create the most magnetic advantage. This is why the most electronegative elements are connected to outer P orbitals. The reason this is so is connected to their 3-D shape in x,y,and z. If you apply the right hand rule for magnetism; current direction, magnetic field direction and attractive force direction, also occur in x,y,z, for each of the x,y,z p-orbitals.The result is a 3-D attraction of all the magnetic fields; perfect. The S, D and F don't have perfect magnetic addition, with the result such atoms are less electronegative. The hybrid orbitals distort magnetic perfection but still allow the oxygen to gain more from magnetic addition than hydrogen. But this imperfection is what makes O susceptable to the H in its attempt to lower its potential.