I'm explaining for HI03
First we assign oxidation numbers to Oxygen and Hydrogen then to any other atom.
So oxygen is -2*3 and H is +1.
So now let I's oxidation number be x.
+1+x+(3*-2)=0
(0 because the overall compound has 0 charge)
+1+x-6=0
x-5=0
x=5
Determining oxidation numbers for remaining atoms in HIO3, H3PO4, Ba(OH)2, and BaPO4
I have no idea if I am doing this right I don't even know how to do this
For HIO3 I added the charges as +1 and 3(-2) which i got -5 then I did it the similar way for the others and the answers I got is for H3PO4 is -3
Then for Ba(OH)2 I got 0 then for BaPO4 I got 22.
Please show me how to do this correctly....
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