Can someone help assign oxidation numbers to the following compounds?

Question

Does anyone want to help assign oxidation numbers to the following?

H2O

H2O2

Na2O

NaO2

K2Cr2O7

(MnO4)-

(Fe)3

Cl2

OF2

HCN

N2H4

(S2O3)2-

Answer 1

First some general guidelines.  Oxygen always gets to be -2 (unless it's a peroxide - as in H2O2-, and then it's -1).  Hydrogen is always +1 (unless its a hydride - comes second - and then it's -1).  The oxidation number is the same as the ionic charge.  

The oxidation number of an element in the elemental state (even diatomic or triatomic molecule, as Cl2 or O3) is zero.

The sum of the oxidation numbers will always be zero for a compound or the charge on a polyatomic ion.

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Pandra got most of these, but there are a couple that need tweaking.

H2O -> H = +1, 0 = -2, OK

H2O2 -> H = +2, O = -2, NO, H=+1, O=-1 (peroxide)

Na2O -> Na = +1, O = -2; OK

NaO2 -> Na = +4, O = -2, No, Na=+1, O= 1/2 (superoxide)

K2Cr2O7 -> K = +2, Cr = +5, O = -2, NO, K=+1, Cr=+6

MnO4 -> Mn = +8, O = -2, NO this is an ion, Mn = +7

(Fe)3 -> Fe = +3 If this is just Iron (Never heard of Fe3) it is 0

Cl2 -> Cl = -1, NO, Cl=0, it's elemental

OF2 -> O = +2, F = -1, OK (this is the only compound where O has a positive oxidation number.)

HCN -> H = +1, C = -4, N = -3, NO (there's no way this adds up to zero.  H = +1, C = +2, N= -3

N2H4 -> N = -2, H = + 1, Or you could do N=+2, H=-1 if you consider this a hydride.

(S2O3)2- -> S = +2, O = -2 , plus e2-, NO - this is an ion.  S = +4, O = -2.

Answer 2

When combining elements their oxidation #'s must balance

H2O -> H = +1, 0 = -2

H2O2 -> H = +2, O = -2

Na2O -> Na = +1, O = -2

NaO2 -> Na = +4, O = -2

K2Cr2O7 -> K = +2, Cr = +5, O = -2

MnO4 -> Mn = +8, O = -2

(Fe)3 -> Fe = +3

Cl2 -> Cl = -1

OF2 -> O = +2, F = -1

HCN -> H = +1, C = -4, N = -3

N2H4 -> N = -2, H = + 1

(S2O3)2- -> S = +2, O = -2 , plus e2-