Question #24a27
1 Answer
Here's how you can find the oxidation numbers for those atoms.
Explanation:
I'll show you how to do parts (a), (c), and (d), and leave the other two parts to you as practice.
I assume that you're familiar with the rules used when assigning oxidation numbers, so I won't go over that here to avoid making the answer unnecessarily long.
If you want to refresh your memory, you can read more about what oxidation numbers are and how they are assigned here:
http://www.chemguide.co.uk/inorganic/redox/oxidnstates.html
So, let's start with the first one.
- phosphorous acid,
#"H"_3"PO"_3#
Start with what you know. Hydrogen has an oxidation number of +1, while oxygen has an oxidation number of -2.
#stackrel(color(blue)(+1))("H"_3) stackrel(color(red)(?))"P" stackrel(color(blue)(-2))("O"_3)#
This means that, in order for the compound to be neutral, you need the oxidation number of the phosphorus atom must be
#3 * (+1) + 1 * color(red)(?) + 3 * (-2) = 0#
#3 + color(red)(?) -6 = 0 implies color(red)(?) = +3#
The oxidation number of the phosphorus atom is +3.
- dichromate anion,
#"Cr"_2"O"_7^(2-)#
Once again, work with what you know. The oxidation number of oxygen is -2, which means that you have
#stackrel(color(red)(?))("Cr"_2) stackrel(color(blue)(-2))("O"_7^(2-)#
This time, the oxidation numbers of all the atoms must amount ot he overall charge of the ion, which is equal to 2-.
#color(red)(?) * 2 + 7 * (-2) = -2#
#2 * color(red)(?) = -2 + 14 implies color(red)(?) = 12/2 = +6#
The chromium atoms are in a +6 oxidation state.
- thiosulfate anion,
#"S"_2"O"_3^(2-)#
Once again, use oxygen's known oxidation number and the overall charge of the anion to find the oxidation number of the sulfur atoms
#stackrel(color(red)(?))("S"_2) stackrel(color(blue)(-2))("O"_3^(2-))#
This time, you have
#2 * color(red)(?) + 3 * (-2) = -2#
#color(red)(?) = (-2 + 6)/2 = +2#
The sulfur atoms are in a +2 oxidation state.
