Dichromate ion is treated with base, the oxidation number of Cr in the product formed is ______

Dichromate ion ($$Cr_2O_7^{2-}$$) exists in acidic solution. When treated with a base (such as $$NaOH$$), the dichromate ion converts to chromate ion ($$CrO_4^{2-}$$):

$$Cr_2O_7^{2-} + 2OH^- \to 2CrO_4^{2-} + H_2O$$

In the chromate ion $$CrO_4^{2-}$$, we can determine the oxidation number of Cr. Let the oxidation number of Cr be $$x$$:

$$x + 4(-2) = -2$$

$$x - 8 = -2$$

$$x = +6$$

Note that in the original dichromate ion $$Cr_2O_7^{2-}$$, the oxidation number of Cr is also +6: $$2x + 7(-2) = -2 \implies 2x = 12 \implies x = +6$$.

This reaction is not a redox reaction but simply an acid-base equilibrium. The oxidation number of chromium remains unchanged at +6 in both dichromate and chromate ions.

The answer is $$\boxed{6}$$.