Can Someone Please Explain Oxidation Numbers

[lazykid]

i was away from school when my chem class learnt about it
and i just when to get upto date so i understand everything else we learn
so yeah can someone please give me tips on it?

[jasoN-]

Rules for determining oxidation numbers
1. In the elemental form (Eg. O2, H2, Ag, Mg etc.), oxidation number is 0

2. For an ion, the charge = the oxidation number
Example for the compound CuCl2, the oxidation number of Cu is 2+ (the charge on a Cu ion)

3. Hydrogen's oxidation number is usually +1
exceptions include NaH and CaH2 where the oxidation number is -1 (Na+ and H-)

4. Oxygen's oxidation number is usually -2
exceptions include H2O2 and BaO2 where the oxidation number is -1 (H+ and O-)

5. In a neutral compound (eg. MgCl2), the sum of the oxidation numbers equals to 0
eg. MgCl2 -> overall oxidation number is 0
seperate oxidation numbers are: Mg (+2) and Cl (-1)

6. In a polyatomic ion, the overall charge =/= 0
eg. NO3-, HS-
to work out the oxidation numbers in these examples, the general rule of thumb is to use the typical oxidation numbers of H ions (+1) and O ions (-2)
NO3-: charge on O is -2, therefore overall charge of O3 is -6
As the overall charge of the compound is -1, the charge on the N ion must be +5
i.e. (+5)(-6) = -1

[lazykid]

thank you so much for that. i friend tried giving me a quick 5 minute crash course but that didn't help
this explained clearly thanks once again!

[vea]

Hi guys I have a redox reactions test tomorrow and I have a quick question.

What does it mean when the question asks you to "State the oxidation states for ______"

Thanks in advanced

[jasoN-]

Oxidation states is the same as oxidation number
so basically follow the rules for applying oxidation numbers

Eg. "State the oxidation states for Kr2Cr2O7"
First do O7:
(-2)x7=-14
Then typically the outer sides
Kr2: (+1)x2=+2
Therefore for Cr2, the oxidation number is
-14+2+oxidno.(Cr2)=0
oxidno.(Cr2)=+12
oxidno.(Cr)=+6

[|ll|lll|]

And to add a quick note, oxidation is the loss of electrons --> increase in oxidation no. Likewise, the rule is vice versa for reduction