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When a lone proton bonds onto an H2O molecule in an acid, why doesn't the proton bond with a pair of the lone electrons in H2O and satisfy its electrons needs by having 3 single bonds? Why is it still charged?

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marked as duplicate by Mithoron, andselisk, Todd Minehardt, a-cyclohexane-molecule, Nilay Ghosh Jan 25 '18 at 3:01

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Okay let’s look at the H2O molecule first. The oxygen atom has 6 protons and each of the hydrogen atoms have a proton. Hence the H2O molecule has a total of 8 positively charged protons. Oxygen has 6 valence electrons and each of the hydrogen’s have 1 valence electron, directly involved in covalent bonding with the oxygen atom. Hence the H2O molecule has a total of 8 negatively charged electrons. As the number of protons = number of electrons in the molecule, the molecule has no overall charge. When a lone proton forms a dative covalent bond with a lone pair on the oxygen atom of the H2O molecule, there are now overall 9 protons and still the same 8 valence electrons in the molecule. Hence due to the imbalance between protons and valence electrons in H3O, there is an overall positive charge on the molecule, hence why it exists as H3O+ when a lone proton (no electrons) bonds with H2O.

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There aren't any free protons floating around it water. They are attached to water molecules or acid molecules. If a positively charged proton is transferred to a neutral water molecule (from another water molecule or from an acid molecule) then clearly that molecule becomes positively charged.

Now how is the new proton bonded to the water molecule? If you consider it to be ionically bonded then you are assuming the oxygen is hanging onto its electrons (at which it is pretty good) in which case the proton retains its positive charge. But if you assume, as you seem to be suggesting, that the bond is purely covalent, then the bond would be formed by sharing the electrons equally between the proton and the oxygen. In this case you would indeed write $\ce{H_3O^\oplus}$ i.e. the positive charge becomes a formal charge on the oxygen.

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