The oxidation number of chromium in Na2Cr2O7 can be determined using the overall charge of the compound and the oxidation numbers of the other elements.

- The overall charge of Na2Cr2O7 is 0.

- The oxidation number of sodium (Na) is +1.

- The oxidation number of oxygen (O) is -2.

Let x be the oxidation number of chromium (Cr).

In Na2Cr2O7, there are two chromium atoms (Cr), seven oxygen atoms (O), and two sodium atoms (Na).

Setting up the equation based on the overall charge:

2(+1) + 2x + 7(-2) = 0

Simplifying the equation:

2 + 2x - 14 = 0

Combining like terms:

2x = 12

Dividing both sides by 2:

x = 6

Therefore, the oxidation number of chromium in Na2Cr2O7 is +6.