[Limewire36]

How do oxidation numbers work? If someone could better explain it to me, that would be much appreciated!

Find the Oxidation Number: S in H₂SO₃
 the H₂ is  2H⁺¹ so that means the SO₃ has to be ⁺² to equal 0.
so you take the O^-2 times it by 3 = -6
therefore S -6= -2
S= +4

Did I do that problem correctly?


Edit: I think I've figured out how to do them.

N in N₂H₄ = +4

S in S^-2= -2

K in K⁺¹= +1

C in CH₄ = -4

I in I₂ = 0

Did I get any of these wrong?

[adenine135]

They're all right, but here's a guide in case you get any more difficult ones you get stuck on:


The first step is setting the compound equal to the compound's charge.

Next you should subtract the oxidations of elements that are determined by following rules from the overall compound charge:



• The oxidation state of any element such as Fe, H₂, O₂, P₄, S₈ is zero (0).
• The oxidation state of oxygen in its compounds is -2, except for peroxides like H₂O₂, and Na₂O₂, in which the oxidation state for O is -1.
• The oxidation state of hydrogen is +1 in its compounds, except for metal hydrides, such as NaH, LiH, etc., in which the oxidation state for H is -1.
• The oxidation states of other elements are then assigned to make the algebraic sum of the oxidation states equal to the net charge on the molecule or ion.
• The following elements usually have the same oxidation states in their compounds:
            * +1 for alkali metals - Li, Na, K, Rb, Cs
             * +2 for alkaline earth metals - Be, Mg, Ca, Sr, Ba
              * -1 for halogens except when they form compounds with oxygen or one another


Then you're ideally left with one unknown, divide the remaining charge by the element in question's subscript and that's your oxidation charge of that element.


So for N in N

₂H₄:
 
The charge is 0

There are 4 hydrogen atoms with an oxidation state of -1:

• 4  (-1) = -4
• 0 - (-4) = +4

So the remaining diatomic nitrogen has an oxidation state of +4, but there are two nitrogen atoms so you must divide the oxidation state by 2:

• +4 / 2 = +2

So the oxidation state of nitrogen in N₂H₄ is +2[/list]

[JustinCh3m]

doesn't the H in N2H4 serve as a "hydride"? and thus take a -1 charge, the way it does in sodium hydride (NaH).

I thought the N had a +2 charge here and the H has a charge of -1 (which is unusual I know).

Just my 2 pennies

- CollegeChemistryNotes

[adenine135]

You're correct, hydrazine is a hydride (I'm doing way too many things at once so I apologize for my mistake).

So that would mean the N₂ would have an oxidation state of +4 and N would be +2...

[Borek]

1. It doesn't matter much whether you treat hydrazine as a hydride or not.

2. It is quite correct to treat it as an analogue of ammonia, just with N-N bond. As ammonia is usually treated as N(-3) H(+1) nitrogen in hydrazine would be (-2).

But as oxidation numbers don't reflect any real property of atoms in molecules, and there is some liberty in assigning them, whole discussion is a moot.