Oxidation number of Cr in Na2_{2}2Cr2_{2}2O7_{7}7 = 0
2(Na) + 2Cr + 7(O) = 0
Oxidation state of Na = +1, O = - 2
So, 2(+1) + 2Cr + 7(-2) = 0
+2 + 2Cr - 14 = 0
+2 - 14 + 2Cr = 0
- 12 + 2Cr = 0
2Cr = 12
Cr = 122\; \frac{12}{2}212
Cr = + 6
