Correct Answer: Option B

Explanation

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

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