In chromyl chloride test for confirmation of $$Cl^-$$ ion, a yellow solution is obtained. Acidification of the solution and addition of amyl alcohol and $$10\% H_2O_2$$ turns organic layer blue indicating formation of chromium pentoxide. The oxidation state of chromium in that is

We need to find the oxidation state of chromium in chromium pentoxide ($$CrO_5$$).

Identify the structure of $$CrO_5$$.

Chromium pentoxide ($$CrO_5$$) has the structure with one oxide ion ($$O^{2-}$$) and two peroxide ions ($$O_2^{2-}$$).

$$CrO_5 = CrO(O_2)_2$$

Calculate oxidation state.

Let the oxidation state of Cr be $$x$$.

$$x + (-2) + 2(-2) = 0$$

$$x - 2 - 4 = 0$$

$$x = +6$$

Alternatively: One $$O^{2-}$$ contributes $$-2$$, two $$O_2^{2-}$$ contribute $$2 \times (-2) = -4$$. Total oxygen charge = $$-6$$. So $$x = +6$$.

Conclusion.

The oxidation state of chromium in $$CrO_5$$ is $$+6$$, which matches Option A.

Therefore, the answer is Option A.