When $H_{2} O_{2}$ is added to an acidified solution of $K_{2} {C r}_{2} O_{7}$ :

solution turns green due to formation of

{C r}_{2} O_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

solution turns yellow due to formation of

K_{2} C r O_{4}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

a deep blue-violet colored compound

C r O {(O_{2})}_{2}

is formed.

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

solution gives green ppt. of

C r {(O H)}_{3}

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is C a deep blue-violet colored compound $C r O {(O_{2})}_{2}$ is formed.
A deep blue layer is formed due to $C r O (O_{2})_{2}$ is present in product.

$C r_{2} O_{7}^{2 -} + 2 H^{+} + 4 H_{2} O_{2} \to 2 C r O (O_{2})_{2} + 5 H_{2} O$

Suggest Corrections

Similar questions

K_{2} {C r}_{2} O_{7}

{N a}_{2} S O_{3}

{S O}_{2}

K_{2} {C r}_{2} O_{7}

K_{2} {C r}_{2} O_{7}

K_{2} {C r}_{2} O_{7}

S O_{2}

S O_{2}

K_{2} {C r}_{2} O_{7}

Join BYJU'S Learning Program