What is the exact calculated oxidation state of the Chromium () atom in the deep blue-coloured perox — Hydrogen Chemistry Question
Question
What is the exact calculated oxidation state of the Chromium ($Cr$) atom in the deep blue-coloured peroxide complex ($CrO_5$) that is transiently formed when $H_2O_2$ reacts with acidified potassium dichromate ($K_2Cr_2O_7$) in the presence of ether?
Answer: 6
💡 Solution & Explanation
The complex formed is chromium peroxide ($CrO_5$), which exhibits a distinctive "butterfly" structure. It contains one standard oxide double bond ($Cr=O$) where oxygen has an oxidation state of $-2$, and two peroxo linkages ($-O-O-$) encompassing four oxygen atoms, each with an oxidation state of $-1$. Let the oxidation state of Cr be $x$. The calculation is: $x + (-2) + 4(-1) = 0 \implies x = +6$.
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