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The oxidation number of `Cr` in `K_(2)Cr_(2)O_(7)` is |
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Answer» Let the Ox.no. of Cr in `K_(2)Cr_(2)O_(7)` be x. We know that, Ox. No of K=+1 Ox. No.of O=-2 `2("Ox.no.K")+2("Ox.no.Cr")+7("Ox.no.O")=0` `{:(,2(+1),+,2(x),+,7(-2),=0),(,+2,+,2x,-,14,=0):}` or 2x=+14-2=+12 or `x=+(12)/(2)=+6` Hence oxidation number of Cr in `K_(2)Cr_(2)O_(7)` is +6. |
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