[positiveion]

This is my understanding of oxidation states/numbers:
oxidation states essentially means: pretend like covalent bonds are ionic bonds and give them charges.

so in the formation of water

h - o - h

because o is more electronegative than h, the electrons are more concentrated in the o area. therefore, if we pretend that this polar covalent bond is an ionic bond, the charges are as follows:

H --> +
O --> 2- (as a result of the 2 hydrogens it is bonded with)

Now I don't understand how to apply this to OH (hydroxide).

I don't really understand this image.  And I don't understand why O has an oxidation number of -2 when it only takes one electron from its one hydrogen.

[Borek]

Basically oxidation numbers are numbers calculated using rules used for calculation of oxidation numbers. They have no physical meaning and there is no measurable property of the atoms that could be linked with the oxidation numbers.

Oxygen has ON of -2 because that's the way it is defined

[cth]

This is my understanding of oxidation states/numbers:
oxidation states essentially means: pretend like covalent bonds are ionic bonds and give them charges.

Yeah, you can do it that way if it helps you. Although, that definition is a bit too narrow I think. Oxidation states can be defined even if there is no covalent bond in it.
For example, one can determine the oxidation state of copper atoms in copper metal (equals to 0), even though there is no covalent bond. You have metallic bonds.

To determine oxidation states, you consider that atoms with higher electronegativity keep electrons for themselves (gain electrons) and lower electronegative atoms lose electrons. Then, you look at how many electrons have been gained, how many lost.

For example, water H₂O.
Oxygen is more electronegative than hydrogen, so oxygen "keeps" electrons all for itself: O^2- and 2 H⁺.
Oxygen has gained 2 extra electrons, so its oxidation state is -II.
Each hydrogen has lost 1 electron, so theirs oxidation states are +I.

Second example, OH^-.
Oxygen already has one electron, which explains the - charge on this anion. Then, it grabs another electron from the hydrogen  2 electron gained. Oxygen has oxidation number -II.
Similarly, hydrogen is +I.

Oxygen has ON of -2 because that's the way it is defined

Well. What about hydrogen peroxide H₂O₂? I wonder about the oxidation number of oxygen... gniak gniak

[AWK]

Well. What about hydrogen peroxide H₂O₂? I wonder about the oxidation number of oxygen... gniak gniak

see rule 5 an 6 at:
http://www.occc.edu/kmbailey/chem1115tutorials/oxidation_numbers.htm

[Borek]

Oxygen has ON of -2 because that's the way it is defined

Well. What about hydrogen peroxide H₂O₂? I wonder about the oxidation number of oxygen... gniak gniak

Hydrogen peroxide, as well as oxygen/fluorine compounds are exclusions - but listing them would only confuse things. Oxygen being almost always -2 is one of the first rules and works in 99% of the cases.

[cth]

You're right, I was just teasing you. 

And as well, this illustrates the general risk to have a rule that works in 99% of the cases, and to get it wrong 1% of the time. 

[orgoclear]

Although it was not my thread.. I hoped to continue posting some of my doubts on oxidation states. For example in CrO5. I have few more but this is the first. I know the structure is a sort of butterfly and like this
    O
O\ ll /O
l   Cr  l
O/   \O  But how do i decide the oxidation state of the Cr?

[AWK]

Use the maximum possible oxidation state for Cr atom (+6) and calculate a mean OS for O atoms

[Borek]

Oxygens with peroxide bond are -1, just like in H₂O₂.

[cth]

Although it was not my thread.. I hoped to continue posting some of my doubts on oxidation states. For example in CrO5. I have few more but this is the first. I know the structure is a sort of butterfly and like this
    O
O\ ll /O
l   Cr  l
O/   \O  But how do i decide the oxidation state of the Cr?

If in doubt, you can count the number of Cr-O bonds. For each of them, consider the oxygen atoms get to "keep" the electrons  the chromium atom receives a +1 to its oxidation number for each Cr-O bond.

You have 6 bonds. The ON for Cr is +VI.

[orgoclear]

So, for Cr-O bond we get a +1 and for Cr=O we get +2 ?

What about this?
     O O
     ll  ll
HO-S-S-OH
     ll  ll
     O O

H2S2O6?

if I count S-O as +1, then I get 5 bonds. But what about the S-S bond? (taking H as +1 and O as -2  i get oxidation number of each S to be +5.

So does that mean for S-S bond I take the contribution to oxidation state to be zero? Why?

[AWK]

So does that mean for S-S bond I take the contribution to oxidation state to be zero? Why?

In the same way as for elements, eg O₂, S₈, B₁₂, C₆₀

[cth]

What about this?
     O O
     ll  ll
HO-S-S-OH
     ll  ll
     O O

H2S2O6?

if I count S-O as +1, then I get 5 bonds. But what about the S-S bond? (taking H as +1 and O as -2  i get oxidation number of each S to be +5.

So does that mean for S-S bond I take the contribution to oxidation state to be zero? Why?

Yes, S-S counts for 0. Because you cannot say that the electrons are more on one sulphur than the other one. They are equally shared.
In your example, if your consider one of the sulphur, it has an oxidation number of: 5*1 + 1*0 = 5  (5 S-O bonds plus one S-S bond). Its ON is indeed +V.

Oxidation number is a way to count if an atom has lost or gain electrons. In a S-S bond, the sulphur has neither lost nor gain electrons (they are shared equally), so the contribution of a S-S bond to the ON is 0.