On treating a chromium(III) solution with $NaOH$ and $H X_{2} O X_{2}$ and then acidifying with dilute $H X_{2} SO X_{4}$ , adding more $H X_{2} O X_{2}$ and shaking with ether, a blue colour develops in the ether layer. The blue species is $CrO X_{5}$ . The oxidation state of Cr and the number of peroxide ( $O - O$ ) groups in it are

Answer: A

CrO X_{5}

CrO (O X_{2}) X_{2}

, with one oxo oxygen (

- 2

) and two peroxo groups (4 oxygen at

- 1

). Then

x + 4 (- 1) + (- 2) = 0

gives

x = + 6

. A value of

+ 10

would arise if all five oxygens were assumed to be

- 2

, which is impossible for chromium.