Asked by anonymous
determine the oxidation number of each of the atoms of the following compounds:
1) H2S03
2) 0H^-
3) HPO4^2-
1) H2S03
2) 0H^-
3) HPO4^2-
Answers
Answered by
DrBob222
Here is a very good site on determining oxidations numbers.
https://www.chemteam.info/Redox/Redox-Rules.html
https://www.chemteam.info/Redox/Redox-Rules.html
Answered by
anonymous
can someone help me do the first one.
my answer is:
H2 = +1
S = +4
O3 = -2
Is my answer correct?
my answer is:
H2 = +1
S = +4
O3 = -2
Is my answer correct?
Answered by
DrBob222
Yes, you are correct on all. I would have liked it better if you had written EACH after H and O: like this:
H = +1 each
S = +4
O = -2 each
I shall be happy to check the others if you wish to post.Here is how I do them.
H is +1 almost always. ONly in hydride is it -1. Generally, if H is listed first, as in H2S, H2O, H2SO4, etc, it is +1. Listed second as in NaH, LiH etc it is a hydride and is -1.
Oxygen is almost always -2 except in peroxides where it is -1. But is is -2 here. Remember that the compound is zero charge so everything must add to zero.
So H is +1 each x 2 = +2.
O is -2 each x 3 = -6.
Then +2 + (S) -6 = 0
What is S? It is S = -2+6 = +4
You did great.
H = +1 each
S = +4
O = -2 each
I shall be happy to check the others if you wish to post.Here is how I do them.
H is +1 almost always. ONly in hydride is it -1. Generally, if H is listed first, as in H2S, H2O, H2SO4, etc, it is +1. Listed second as in NaH, LiH etc it is a hydride and is -1.
Oxygen is almost always -2 except in peroxides where it is -1. But is is -2 here. Remember that the compound is zero charge so everything must add to zero.
So H is +1 each x 2 = +2.
O is -2 each x 3 = -6.
Then +2 + (S) -6 = 0
What is S? It is S = -2+6 = +4
You did great.
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