Acidified $K_{2} {C r}_{2} O_{7}$ solution turns green when ${N a}_{2} S O_{3}$ is added to it. This is due to the formation of:

C r O_{4}^{2 -}

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

{C r}_{2} {(S O_{3})}_{3}

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

C r S O_{4}

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

{C r}_{2} {(S O_{4})}_{3}

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

Open in App

Solution

The correct option is C ${C r}_{2} {(S O_{4})}_{3}$
Acidified $K_{2} C r_{2} O_{7}$ solution turns green when $N a_{2} S O_{3}$ is added to it. This is due to the formation of
${C r}_{2} {(S O_{4})}_{3}$

$K_{2} C r_{2} O_{7} + 3 N a_{2} S O_{3} + 4 H_{2} S O_{4} \to {C r}_{2} {(S O_{4})}_{3} + K_{2} S O_{4} + 3 N a_{2} S O_{4} + 4 H_{2} O$
Hence option D is correct.

Suggest Corrections

Similar questions

{S O}_{2}

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

S O_{2}

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

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

S O_{2}

H_{2} O_{2}

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

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

{S n}^{+ 2}

{S n}^{+ 2}

Join BYJU'S Learning Program