Yes often but most bond types cause the centroid of negative and positive charges to separate.
Partly correct - bonds between the same elements cause no seperation, hence the bond between (for example) two oxygen atoms is non-polar.
To see that the hydrogen bond does not necessarily create a polar molecule one need only consider H2 molecule. (Has hydrogen bonding and zero dipole.)
Hydrogen bonding neccessarily requires a polar bond. The bond between two hydrogen atoms in dihydrogen does not constitute a hydrogen bond. it is plainly and simply a covalent bond (to be precise, it is a 1σ bond).
I think that H2O's orbitals that force the hydrogens to be 105 degree separated (instead of HOH) are not common. For example, as I recall (from 40+ years ago) CO2 is a linear molecule OCO.
Yes $$CO_2$$ is a linear molecule. The explanation for the H-O-H bond angle has to do with the SP hybridization that occurs, and the way the non bonding electrons interact.
All such symmetric molecules are completely non polar in their ground states. They are dynamically polar in some of the "stretch vibration modes, such as O--C-O ---> O-C--O and in many of the flex modes. Stretch modes like O--C--O ---> O-C-O are not even dynamically polar as symmetry is preserved. Thus you are not always correct here:
More or less correct, yes, assymetric stretch vibration modes, and scissor modes have the ability to induce temporary molecular dipole moments, but stop and ask yourself a question.
Why?
The answer is because although the molecule is over all non polar, the bonds retain their polar nature, however, because of their symmetry the polarities (in the ground state) cancel out overall, and the molecule is non polar (as opposed to the situation with water where the bond polarities of the H-O bonds reinforce each other and strengthen the over all dipole moment of the water.
Consider that if you qould look closely enough at a $$CO_2$$ molecule, what you would actually see is this $$O^{\delta -}=C^{\delta +}=O^{\delta -}$$
H2 being one obvious exception of non-polar hydrogen bonding.
No, the covalent bonding in di-hydrogen is not the same thing as hydrogen bonding.
That is obvious. S is less electro negative than O. Both are in column 6 of the periodic table but S is below O and as you go down in that column the two "slots" than can accept two more electrons are farther from the positive nuclear charge.
Correct.
As I recall Hydrogen Sulfide is H2S and linear - I.e. really HSH with no dipole moment. If my memory is correct, that would be another example of hydrogen bonding with zero dipole.
Incorrect. The H-S-H bond angle in Hydrogen Sulphide is 92.1°
Yes I spoke of these and even described that two Os of different H2Os could, if their triangular planes were roughly parallel weakly bond to a single H2O
Mostly correct, the dimeric form of water has the planes seperate by an angle of something like 57° which probably has something to do with the anti bonding molecular orbitals of the water.
(Its H+ s each "grabbing hold" of the O-- of one of the attaching Os of the two other H2Os. I also mentioned that many of the xH2O complexes were 3D molecules, not the 2D chains I could illustrate. I did this when noting that it is often the shape of the 3D molecule, which determines it biological activity, more that the elements making it up.
Fair enough although (at least, according to the information I have) water clusters $$(H_2O)_n$$ are limited to 3≤n≤60
I have no reason to think that any complex 3D shapes of xH2O are stable at room temperature, so in all probability homeopathic medicine is ONLY placebo effect, but it is at least a possibility that some 3D configurations could be stable. If I were to guess what they might look like, I would put my money on a nearly spherical cluster that had only Hydrogens on the outside "skin" - that probably would make hydrophobic (oil like) or just try to grow by adding an O-- of H2O which would soon get thermally torn loose again. I.e. a complete spherical "skin" with protons only on the outside might be stable as to knock one off you need to break the strong H-O bond of water. Aslo such a large 3D shpere would have many internal vibration modes to absorb the energy of a collision and it would not need to absorb much as the colliding H2O would mainly just elastically bounce off the much more massive 3D complex.
I'm going to pass on commenting on this, except to say that at STP (in terms of clusters) the best explanation of water's properties is given when we have $$(H_2O)_8 > (H_2O)_5 > (H_2O)_6$$ however the triple point requires invoking $$(H_2O)_{24}$$
Again let me state that I think Homeopathic medicine is ONLY placebo effect, but a deeper than usual understanding of water does not proof that is the case. Only a shallow understanding of water as only a mix of H2O molecules leads on to that conclusion.
I strongly disagree.