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I ask because I'm trying to write an equilibrium equation for this reaction and I'm wondering whether an extra 2 protons will be required after one is taken to attack the $\ce{OH}$ group and turn it into $\ce{H_2O}$ in order to convert the freed ammonia (with one proton missing) into a conjugate acid.

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  • $\begingroup$ You make hydroxylamine sound a lot more salacious than a molecule has any right to be. $\endgroup$ – Richard Terrett Jun 4 '13 at 8:39
  • $\begingroup$ @RichardTerrett Some people consider atoms sexy, who's to say that molecules don't feel the same way? $\endgroup$ – Josh Pinto Jun 18 '13 at 14:52
  • $\begingroup$ subatomic particles^ $\endgroup$ – Josh Pinto Jun 18 '13 at 19:43
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Well, hydroxylamine is technically amphoteric, meaning it can act as either a Bronsted acid or base. The acidic proton is presumably the one on the oxygen (although the resulting salts are apparently explosively unstable), while the nitrogen is the proton acceptor. I wouldn't expect the oxygen to be basic (which, unless I've misunderstood, seems to be what your question implies), at least not in any measurable way or for any appreciable length of time, given hydroxylamine's instability and tendency to spontaneously decompose in myriad ways. When treated with acid, it forms hydroxylammonium salts, which disproportionate upon heating.

Note: this book was my primary reference; it contains a concise but informative summary of the reactions hydroxylamine undergoes.

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