When $H_{2} O_{2}$ is added to an acidified solution of $K_{2} C r_{2} O_{7}$ , the solution :

turns green due to the formation of

C r_{2} O_{3}

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

turns yellow due to the formation of

K_{2} C r O_{4}

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

turns deep blue-violet due to the formation of

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

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

gives green precipitate due to the formation 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 B turns deep blue-violet due to the formation of $C r O (O_{2})_{2}$
When $H_{2} O_{2}$ is added to an acidified solution of $K_{2} C r_{2} O_{7}$ , a deep blue-violet coloured compound $C r O (O_{2})_{2}$ is formed. The reaction is as follows:
$4 H_{2} O_{2} + 2 H^{+} + C r_{2} O_{7}^{2 -} \to 2 C r O (O_{2})_{2} C h r o m i c d i p e r o x i d e (b l u e) + 5 H_{2} O$
Hence, option C is correct.

Suggest Corrections

Similar questions

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

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

H_{2} O_{2}

K_{2} C r_{2} O_{7}

A g N O_{3}

F e C l_{3}

H_{2} S

H_{2} O_{2}

K_{2} C r_{2} O_{7}

Join BYJU'S Learning Program