A red-orange vapour obtained from a salt, K2Cr2O7 and concentrated H2SO4 is passed into NaOH solution to give a yellow solution Y. Acidification of Y with dilute H2SO4 gives an orange solution Z. The oxidation states of Cr in the vapour, Y and Z, and the nature of the change Y -> Z, are respectively

JEE Advanced levelSingle correct optionDifficulty 3 of 3qa087

A red-orange vapour obtained from a salt, $K X_{2} Cr X_{2} O X_{7}$ and concentrated $H X_{2} SO X_{4}$ is passed into NaOH solution to give a yellow solution Y. Acidification of Y with dilute $H X_{2} SO X_{4}$ gives an orange solution Z. The oxidation states of Cr in the vapour, Y and Z, and the nature of the change Y $\to$ Z, are respectively

The vapour is

CrO X_{2} Cl X_{2}

(Cr

+ 6

), Y is

CrO X_{4} X^{2 -}

(Cr

+ 6

) and Z is

Cr X_{2} O X_{7} X^{2 -}

(Cr

+ 6

). The change

2 CrO X_{4} X^{2 -} + 2 H X^{+} Cr X_{2} O X_{7} X^{2 -} + H X_{2} O

is a condensation equilibrium with no change in oxidation state. So

+ 6, + 6, + 6

and non-redox.