Asked by Freddie
What is the oxidation state of Manganese in KMnO4 (s) and sulfur in H2S?
Answers
Answered by
DrBob222
Rule 1 is that the oxidation states must add to zero for a compound or to the charge on an ion.
Rule 2 is representative elements have their "normal" charge; i.e., group I has +1, group II has +3, group III has +3, etc.
Rule 3 is that H is +1 (except in unusual cases of hydrides) and O has -2 (except in unusual cases of peroxides or superoxides). How does that work for KMnO4.
K is +1, O is -2 so 4*-2 = -8. So to make KMnO4 zero Mn must be +7.
Check. +1(for K) + (+7)(for Mn) + (-8) (for 4 oxygen) = 0
Now you do that for H2S. Remember H is +1 each.
Rule 2 is representative elements have their "normal" charge; i.e., group I has +1, group II has +3, group III has +3, etc.
Rule 3 is that H is +1 (except in unusual cases of hydrides) and O has -2 (except in unusual cases of peroxides or superoxides). How does that work for KMnO4.
K is +1, O is -2 so 4*-2 = -8. So to make KMnO4 zero Mn must be +7.
Check. +1(for K) + (+7)(for Mn) + (-8) (for 4 oxygen) = 0
Now you do that for H2S. Remember H is +1 each.
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