So I am completely and utterly confused about why sulfur has $12$ valence electrons. I understand in it's configuration, it has $2$ electrons for the $3\mathrm s$ subshell, and $4$ electrons from its $3\mathrm p$ subshell, and it can use the $3\mathrm d$ subshell since its in period $3$. Therefore it is able to exceed the octet, and form more than just $4$ covalent bonds.

But how exactly do you calculate the exact number of "$12$ electrons" in the valence electron, which thus allows it to participate in $6$ covalent bonds with oxygen in the sulfate ion, $\ce{SO4^2-}$? I can't seem to understand.

Thanks so much!

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    $\begingroup$ It doesn't and it was already discussed here. $\endgroup$ – Mithoron May 10 '16 at 20:34
  • $\begingroup$ related chemistry.stackexchange.com/questions/29142/… $\endgroup$ – Mithoron May 10 '16 at 20:39
  • $\begingroup$ more precisely I recommend this answer: chemistry.stackexchange.com/a/5242/9961 $\endgroup$ – Mithoron May 10 '16 at 20:40
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    $\begingroup$ Sulphur has six valence electrons. It is in group 16. I don’t get how you could assume any other valence electron count. Also, sulphur does not exceed the octet. $\endgroup$ – Jan May 10 '16 at 21:09
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    $\begingroup$ The sulphate ion is made up of sulphur (6), four oxygens (4 · 6 = 24) and two additional electrons for the charge. Sulphur has six valence electrons, full stop. $\endgroup$ – Jan May 10 '16 at 22:12

Sulfur has 6 Valence electrons, 2 in the first shell, 8 in the second shell, and 6 in the outermost layer (third layer). They can determine the number of kernel electrons and the number of valence electrons due to the bonds they form, for example Sulfur is more likely to form ions with the Alkaline earth metals and form different covalent bonds. In the case of SO4, Sulfur is sharing electrons with 4 Oxygen atoms forming a covalent bond but the overall charge is not 0, it's -2, so it's not exceeding the octet law, it's just a polar covalent bond, this has to do with electronegativity but that's a different story.


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