Recently, I was studying about oxyacids and my teacher took example of a general compound $\ce{X-O-H}$. He told that that on dissolving it into water ,water would break apart either $\ce{O-H}$ bond or $\ce{X-O}$ bond depending upon the electronegativity difference and hence they would be called an oxyacid or oxobase.

But I want to know whether water would also separate the remaining part as the left out atoms also have electronegativity difference in $\ce{X-O}$ and $\ce{O-H}$. Is the explanation hidden in negative charge of remaining part? If not why?

  • $\begingroup$ That's a huge simplification. Continuing in the same vein, though, it would suffice to say that the difference in electronegativity is not great enough. $\endgroup$ Sep 10, 2021 at 10:05
  • $\begingroup$ Let's take the simplest case, hypochlorous acid : $\ce{H-O-Cl}$. If the bond $\ce{H-O}$ is broken, it produces the ions $\ce{H+ + OCl^-}$ like every acid in water. If the bond $\ce{Cl-O}$ is broken, it produces the ions $\ce{Cl^+ + OH^-}$. What could these ions make later on ? If they react with water, could they make new substances ? $\endgroup$
    – Maurice
    Sep 10, 2021 at 10:10
  • $\begingroup$ I don't think we're getting what you're imagining here. In first approximation nothing happens to conjugated bases. $\endgroup$
    – Mithoron
    Sep 10, 2021 at 13:02
  • $\begingroup$ Let me put it simply : Will water break XO- in a similar fashion it broke OH bond in XOH $\endgroup$
    – user265825
    Sep 10, 2021 at 13:10
  • $\begingroup$ To answer simply: Only if X-O bond visits the same fashion salon as the O-H bond, and even then, the extra electron purse makes the difference. $\endgroup$
    – Poutnik
    Sep 11, 2021 at 9:35


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