So the question is which is the most common oxidation state of halogens and I have to justify my answer. Is my following thought correct? Do I have to add something?

The most common oxidation state of halogens is $-1$. Halogens are the most electronegative elements of the periodic table. Their outer electron configuration is $n\mathrm{s^2}n\mathrm{p^5}$. If chlorine, for example, gains one more electron, the outer p orbitals are completely filled (resulting in a full octet). Since halogens are electronegative they can easily remove an electron from the nearby atom. So then the halogen becomes isoelectronic with a noble gas and it is more stable in this state.

  • 1
    $\begingroup$ Whatever you have thought is correct. $\endgroup$ – Prakhar Feb 27 '17 at 12:04
  • 1
    $\begingroup$ @Prakhar No, that is wrong. Please do not confuse people by posting wrong comments. $\endgroup$ – Jan Feb 27 '17 at 13:50
  • 1
    $\begingroup$ Well ,we are talking about the most common oxidation state of Halogens, and -1 is the most common oxidation state. I know that they show oxidation states from -1 to +7 (except for fluorine), but that is not what we are being asked here. We are being asked about the oxidation state that Halogens exhibit more frequently or the oxidation state that is observed mostly. $\endgroup$ – Prakhar Feb 27 '17 at 14:28
  • $\begingroup$ @Prakhar Again, this is wrong. Iodide is not iodine’s most common oxidation state. $\endgroup$ – Jan Feb 27 '17 at 22:43
  • $\begingroup$ I suspect the "book" answer is -1, but that does not make it really right for all halogens especially iodine and heavier. "Dumbing down" science for introductory courses is something we have to put up with :-( . $\endgroup$ – Oscar Lanzi Mar 26 '17 at 2:31

There is not one single most common oxidation state across all halogens. While fluorine is the most electronegative element of which compounds are known — meaning that it can only exist in the $\mathrm{-I}$ and $\pm0$ oxidation states — the electronegativity continually reduces down the group to iodine. Iodide in the oxidation state $\mathrm{-I}$ is easily oxidised to $\ce{I2}$ or $\ce{IO3-}$ with the oxidation states $\pm0$ and $\mathrm{+V}$, respectively.

What is true, however, is that all halogens except fluorine show oxidation states from $\mathrm{-I}$ to $\mathrm{+VII}$ as can be predicted from their electronic structure $n\mathrm{s^2}\,n\mathrm{p^5}$.

  • $\begingroup$ Hmm, plus/minus zero? $\endgroup$ – Mithoron Feb 27 '17 at 18:18
  • $\begingroup$ @Mithoron Yes …? $\endgroup$ – Jan Feb 27 '17 at 22:42
  • $\begingroup$ sigh That meant that it has no sense whatsoever in chemistry to put such sign in front of zero (sort of euphemism ;) $\endgroup$ – Mithoron Feb 27 '17 at 22:48
  • $\begingroup$ @Mithoron I actually learnt to always label the oxidation state as $\pm0$. Because all other oxidation states are signed, that one should be, too. And also, all other oxidation states are written in Roman numerals but there is no Roman numeral zero — having something that stands out makes it clearer we are dealing with an oxidation state. All of my answers that talk about an oxidation state $\pm0$ should have that sign; if not it is a mistake on my behalf and I shall correct it asap ;) $\endgroup$ – Jan Feb 27 '17 at 22:51

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy