Oxidation number of Cr in Na\(_2\)Cr\(_2\)O\(_7\) = 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 = \(\frac{12}{2}\)

 Cr = + 6