Valence Bond Theory tells us that each of the “sp3” (in reality, 44% s character) oxygen line pairs in water can act as electron donors, but observing the MO diagram for water tells us that the 2 non-bonding orbitals are of vastly different energies. (One of the “lone pairs” has bonding character too.) As such, how does water even form 2 electron-donating hydrogen bonds?

  • $\begingroup$ It is not like one orbital on the diagram corresponds to one lone pair. $\endgroup$ – Ivan Neretin Feb 15 '19 at 10:20
  • $\begingroup$ I know that. The MO diagram of water has one “strictly nonbonding” (HOMO; b1 symmetry) and one “mostly nonbonding” (a1 symmetry, formed from 2s of oxygen) that has almost all electron density on the oxygen due to the energy difference. However, hydrogen bonding is often shown as the oxygen bonding to 2 hydrogens (Google it up and you’ll know what I mean), so I was inquiring about that. $\endgroup$ – ANZGC FlyingFalcon Feb 15 '19 at 11:27
  • $\begingroup$ Then why would you say that one of the lone pairs has such-and-such character, which implies that the lone pairs are different, which in fact they aren't? $\endgroup$ – Ivan Neretin Feb 15 '19 at 11:35
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    $\begingroup$ @IvanNeretin The two lone pairs of water are of slightly different energy i.e. those two lone pairs are not exactly similar. Those two lone pairs reside in different (non-degenerate) orbitals having different symmetries. In fact, photo-electron spectrum of water also tells about the difference of the two lone pairs. They are not exactly similar. $\endgroup$ – Soumik Das Feb 15 '19 at 12:32
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