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Chemistry Forums for Students => Undergraduate General Chemistry Forum => Topic started by: JEE10 on December 05, 2009, 05:09:56 AM

Title: Structur of S4O62-
Post by: JEE10 on December 05, 2009, 05:09:56 AM

What is the structure of S4O6 (2-) ?

The actual question was to find the oxidation state of each sulphur atom, but I did not know its structure.

Title: Re: Structur of S4O62-
Post by: UG on December 05, 2009, 05:15:00 AM

You don't need the structure in order to find the oxidation number/state of each sulfur atom.

Anyway, it looks like this: (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.chemthes.com%2Ficon_1%2F2463.gif&hash=f66e7b0c28d0d337970960589322fdb7269b6776)

Title: Re: Structur of S4O62-
Post by: Schrödinger on December 05, 2009, 08:50:23 AM

 

Quote from: UG on December 05, 2009, 05:15:00 AM

You don't need the structure in order to find the oxidation number/state of each sulfur atom.

Anyway, it looks like this: (https://www.chemicalforums.com/proxy.php?request=http%3A%2F%2Fwww.chemthes.com%2Ficon_1%2F2463.gif&hash=f66e7b0c28d0d337970960589322fdb7269b6776)

You do need the structure to find the oxidation number/state of each sulfur atom.
Without the structure, you would be calculating average oxidation state of all sulphur atoms put together.

For instance, in this structure, the 2 S-atoms in the center have 0 oxidation state, while those at the ends have +5 state.
But without the structure, one would say that the oxidation state on each sulphur atom is +10/4 = +5/2

Title: Re: Structur of S4O62-
Post by: UG on December 05, 2009, 05:43:24 PM

Sorry, my brain was on thrid gear  :(