Question

In acidic medium $H_2{O}_2$ changes $C{r}_2{O}_7$ to $Cr{O}_5$ which has two $-O-O-$ bonds. Oxidation state of $Cr$ in $Cr{O}_5$ is:

• A. $+6$
• B. $-10$
• C. $+5$
• D. $+3$

Answer 1

The correct option is A $+6$

1) Since, $Cr$ has only five electrons in $3d$ orbitals and one electron in $4s$ orbital.

2) This exceptional value is due to the fact that four oxygen atoms in$Cr{O}_5$are in peroxide linkage.

3)The chemical structure of $Cr{O}_5$ is $x–2–4=0$ or $x=+6.$