Consider the reaction of ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ with excess of ${\text{H}}_2{\text{O}}_2$ in acidic medium. What is the oxidation state of chromium in the resulting blue colored complex?

Solution

1. The reaction of ${\text{K}}_2{\text{Cr}}_2{\text{O}}_7$ with ${\text{H}}_2{\text{O}}_2$ in acidic medium produces ${\text{CrO}}_5$.
2. The structure of ${\text{CrO}}_5$ is a butterfly structure with four peroxide oxygen atoms and one double-bonded oxygen atom.
3. The oxidation state of $\text{Cr}$ in ${\text{CrO}}_5$ is calculated as $x+4(-1)+1(-2)=0$, which gives $x=+6$.
4. Hence the answer is (B).