K2 Cr2 O7 find out the oxidation state of Cr2
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Let the Ox. no. of Cr in be x.
We know that, Ox. no. of K=+1
Ox. no. of O=−2
So, 2(Ox.no.K)+2(Ox.no.Cr)+7(Ox.no.O)=0
2(+1)+2(x)+7(−2)=0
or +2+2x−14=0
or 2x=+14−2=+12
Hence, oxidation number of Cr in +6.
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