What are the oxidation numbers of S and O in the ion S2O3(2-)?

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What are the oxidation numbers of S and O in the ion S2O3(2-)?

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Answer 1 · 2015-10-23T17:45:51

Oxygen would have an oxidation state of $- 2$ $-2$ , therefore sulfur would have an oxidation state of $+ 2$ $+2$ .

Explanation:

Let me explain:

So you have the whole compound that has a total charge of $(2 -)$ $(2-)$ . This means everything in the compound will have to 'add' up to $- 2$ $-2$ .

Break down the elements in the compound:

Oxygen's normal oxidation number is $- 2$ $-2$ . Because you have three oxygen atoms, the oxidation number is now

$- 2 \times 3 oxygen atoms = - 6$ $-2 xx "3 oxygen atoms" = -6$

Remember, the whole compound is $- 2$ $-2$ , so we have to get the charge from $- 6$ $-6$ up to $- 2$ $-2$ .

Sulfur's normal oxidation number in this case would be $+ 2$ $+2$ . There are two sulfur atoms so the number is now

$+ 2 \times 2 sulfur atoms = + 4$ $+2 xx "2 sulfur atoms" = +4$

It's perfect.

$overbrace(-6)^(color(blue)("the oxygen atoms")) + underbrace((+4))_(color(red)("the sulfur atoms"))= -2$

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What are the oxidation numbers of S and O in the ion S2O3(2-)?

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