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Halogens like $\ce{Cl}$ always exhibit $+7,+5,+3,+1,-1$ as their oxidation numbers. I found the following answer after checking several sources:

Halogen atoms have $7$ electrons in their valence shell, so $1$ of those electrons is spin unpaired, which is an unstable situation. If an even number oxidation occurs, said even number of electrons are taken away, leaving still an odd number of electrons with at least $1$ unpaired. Taking away an odd number of electrons leaves an even number of electrons with the opportunity of them to be all paired.

These oxidation numbers are observed in compounds like $\ce{HClO4}$ and $\ce{HClO3}$. We know that, before forming these compounds, the valence electrons of $\ce{Cl}$ are excited into the d orbitals. Thus now, they are not in ground state, but in the excited state instead.

The quoted explanation is giving the reason for my question on the basis of the ground state of halogens. (I especially consider $\ce{Cl}$). But we know that, before formation of compounds, the halogen atoms are in their excited state instead.

So is the quoted explanation correctly justifying why $\ce{Cl}$ exhibits odd oxidation states only?

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    $\begingroup$ Counterexample: en.wikipedia.org/wiki/Chlorine_dioxide Further, d-orbitals need not be involved in even $\ce{HClO4}$, hypervalency is the more modern approach. $\endgroup$
    – TAR86
    Commented Feb 16, 2018 at 8:22
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    $\begingroup$ "We know that, before forming these compounds, the valence electrons of Cl are excited into the d orbitals." This statement is completely incorrect (like most of sentences in questions, which start with "We know"). $\endgroup$
    – Mithoron
    Commented Feb 16, 2018 at 18:52
  • $\begingroup$ All elements exhibit odd ox. states. All molecules are in relative energy minimums. Electrons are usually in the lowest available molecular orbitals, but molecules are not in the lowest energy states. Paired electrons are of higher energy than unpaired; pairing is a marriage of convenience. Cl exhibits all ox. no. from -1 to +7. A god question is "Why are the molecules with radicals more reactive?". Why was this question resurrected? $\endgroup$
    – jimchmst
    Commented Jun 13 at 20:48

2 Answers 2

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Because of the way we construct oxidation numbers. Put an atom into a molecule with all electrons paired and all the bonds heteronuclear, and by assigning all the valence electrons to the more electronegative atom of each bond we render even number on each atom. Combine that with an odd number required for a neutral atom and you count an odd difference, thus an odd oxidation number.

Note that I have assumed all electrons to be paired and all bonds as heteronuclear. But, it is not always the case. There could be homonuclear bonds as in the halogen element itself ($\ce{X_2}$), with an oxidation number = $0$. Or there could be unpaired electrons on the halogen so we count an odd rather than even number of valence electrons per halogen atom. $\ce{ClO}$ (chlorine oxidation number $+2$) and $\ce{ClO_2}$ ($+4$) are such examples.

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  • $\begingroup$ Can you please give an example of what you wrote in the first para? $\endgroup$
    – Shub
    Commented Feb 16, 2023 at 7:18
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This is due to the excitation states that chlorine can exhibit, considering that it has an unpaired electron:

Oxidation Number +1:

3s² 3p⁵ 3d⁰

↑↓ , ↑↓ ↑↓ ↑, _ _ _ _ _

Oxidation Number +3:

3s² 3p⁴ 3d¹

↑↓, ↑↓ ↑ ↑, ↑ _ _ _ _

Oxidation Number +5:

3s² 3p³ 3d²

↑↓, ↑ ↑ ↑, ↑ ↑ _ _ _

Oxidation Number +7:

3s¹ 3p³ 3d³

↑, ↑ ↑ ↑, ↑ ↑ ↑ _ _

On the other hand, it has an oxidation state of -1 when forming ionic compounds by taking an electron from the metal.

Oxidation Number -1:

3s² 3p⁶ 3d⁰

↑↓ , ↑↓ ↑↓ ↑↓, _ _ _ _ _

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