In transition elements the oxidation states differ by one (+2 to +7 in Mn). However, in the p-block elements, the oxidation states differ by two (-1,+1,+3,+5 in the halogen group). Why is this so?

  • $\begingroup$ This is not true of the whole p-block. For example, carbon -1 (ethyne) -2 (ethene) -3 (ethane). It is also not true of the halogens as they can be 0 ($\ce{Cl2}$). $\endgroup$
    – bon
    Jan 25 '16 at 18:31
  • 1
    $\begingroup$ it's not entirely correct, for example almost all oxidation states from -1 to +7 are known for chlorine. HCl (-1), Cl2 (0), ClF (+1), ClF3 (+3), ClO2 (+4), ClO2F (+5), ClO3 (+6), HClO4 (+7). All oxidation states from -2 to +6 are known for sulfur. It is more complicated than that. $\endgroup$
    – permeakra
    Jan 25 '16 at 18:32
  • $\begingroup$ But in general isn't it this way? @permeakra $\endgroup$
    – Indo Ubt
    Jan 25 '16 at 18:48
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    $\begingroup$ @IndoUbt Nope. Actually, s/p block typically demonstrates that (covalency+formal charge) usually differs by two except some special cases. This goes to stability of free radicals, which are more stable when the orbital holding unpaired electron is either hidden behind other orbitals (case of 1st d-row), or is very large (case of neutral nitrogen and chlorine oxides, superoxide anion and so on). This doesn't restrict oxidation state all that much. $\endgroup$
    – permeakra
    Jan 25 '16 at 20:43

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