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OpenStudy (anonymous):

oxidation question

OpenStudy (anonymous):

what is the oxidation number of Cr....|dw:1429939740091:dw| so what i reasoned was that since O has a -2 (except in peroxides) and multipling it by 7 will give you -14 and what i know for Cr is that it mus be +12 BUT the oxidation # for Cr is the ionic charge "+3" since Cr is monatomic so 3 multipying two is +6 so how do i get twelve?

OpenStudy (anonymous):

@hoslos

OpenStudy (anonymous):

ok nvm i think i get it

OpenStudy (anonymous):

what is the question ?u want the oxidation no of cr

OpenStudy (anonymous):

correct, i think the problem was that i thought Cr oxidation number has to be the ionic charge +3 becuase its monatomic but its not a set number, it could be used in some case or not and in this case your trying to have a net charge of -2 becuase its a polyatomic ion therefore +6 ..... not +3

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