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.