I think because it makes a tetrahedral which is favorable when you have oxygen there. With an NBO analysis in Gaussian you can get the number of H-bonds in your molecule.
Regarding your first question> For the water molecule, the positive charges are localized in the hydrogen atoms, therefore you can have a localized hydrogen bond acceptor. However, the negative charges are smeared out in the lone pairs, and the distribution of hydrogen bond donors is more disordered. The next articles discuss about that:
*Acc. Chem. Res., 2012, 45 (1), pp 63–73 ,Liquid Water: From Symmetry Distortions to Diffusive Motion.
*Nature Communications volume 4,1450 (2013), Electronic signature of the instantaneous asymmetry in the first coordination shell of liquid water
* Phys. Chem. Chem. Phys., 2014,16, 24479-24483, A continuous mixture of two different dimers in liquid water .
Regarding your second question> You can have geometric or energetic criteria to define when a molecule is accepting hydrogen bonds. The image attached shows an example, you can read more in this paper:
Article Hydrogen Bonding Definitions and Dynamics in Liquid Water
The Journal of Chemical Physics 126, 204107 (2007),Hydrogen bonding definitions and dynamics in liquid water
Since H-bonding is nothing then electrostatic attraction between H and more electronegative elements so for sure there will be H-bonding in all negatively charged oxygen in carboxylate or sulfonate or phosphonate but the strength may vary.
Thanks Mr. Jhaa for your answer.. So I think that O with double bonds are more electronegative than negatively charged O therefore =O are stronger H bond acceptor than _O- . Am I right?