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The question arises from trying to get where does the values of $-2$, $0$, $+2$, $+4$ and $+6$ from the electron configuration.

The electron configuration for sulphur $(Z=16)$ is $[Ne]3s^23p^4$.

It makes sense that to complete the octet sulphur needs to gain two electrons hence it will have $-2$, but it can lose all the electrons $2+4=+6$. But I have no idea what sort of explanation can be made for obtaining the other values?. Does it exist a rule or something that am I unware of?.

Chris Steinbeck Bell
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  • It is pretty rare that Sulfur looses two electrons. It may "loose" the four electrons 3p in substances like sulfites and SO2. Same thing for the six electrons 3p and 3s in substances like sulfates and SO3. Anyway the term "loose" is not ideal, because molecules have molecular orbitals, which are not always simply related to s and p atomic orbitals. It is not correct to say that Sulfur "looses" electrons. it would have been a real loss only in the case of an ionic bond. – Maurice Feb 22 '20 at 09:13
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    :/ You couldn't make it any broader? This question may count as a duplicate of https://chemistry.stackexchange.com/questions/69798/an-introduction-to-oxidation-state-oxidation-number – Mithoron Feb 22 '20 at 18:19
  • Does this answer your question? An introduction to "oxidation state"/"oxidation number" – Tyberius Feb 22 '20 at 21:50

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Wikipedia has a nice article on the topic of oxidation state. It quotes the IUPAC definition:

IUPAC: Oxidation state of an atom is the charge of this atom after ionic approximation of its heteronuclear bonds [...]

For sulfur in particular, if there are no unpaired electrons, no bonds to another sulfur (i.e. all heteronuclear bonds) and sulfur has an octet, the oxidation state of sulfur will come out as an even number. The range is -2 (all electrons in octet counted towards sulfur, i.e. $\ce{H2S}$) to +6 (no electrons in octet counted towards sulfur, i.e. $\ce{H2SO4}$). For each heteronuclear bond, the electrons associated with it are either all counted toward the sulfur atom, or none of them are.

[OP] The question arises from trying to get where does the values of −2, 0, +2, +4 and +6 from the electron configuration.

The oxidation states are listed by wikipedia as the major ones (the ones in bold, see below), but others such as +1 (e.g. $\ce{H2S2}$, dihydrogen disulfide) are also possible.

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Karsten
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