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Acidified $K_2Cr_2O_7$ solution turns green when $Na_2SO_3$ is added to it. This is due to the formation of
A
$CrSO_4$
B
$Cr_2(SO_4)_3$
C
$CrO_4^{2-}$
D
$Cr_2(SO_3)_3$
Detailed Solution
Acidified potassium dichromate is a strong oxidising agent; $Na_2SO_3$ is a reducing agent.
Sulphite is oxidised to sulphate and orange $Cr_2O_7^{2-}$ (Cr in +6 state) is reduced to green $Cr^{3+}$.
Ionic equation: $Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O$
Molecular equation: $K_2Cr_2O_7 + 3Na_2SO_3 + 4H_2SO_4 \rightarrow 3Na_2SO_4 + K_2SO_4 + Cr_2(SO_4)_3 + 4H_2O$
The green colour is due to chromium(III) sulphate, $Cr_2(SO_4)_3$.
Sulphite is oxidised to sulphate and orange $Cr_2O_7^{2-}$ (Cr in +6 state) is reduced to green $Cr^{3+}$.
Ionic equation: $Cr_2O_7^{2-} + 3SO_3^{2-} + 8H^+ \rightarrow 2Cr^{3+} + 3SO_4^{2-} + 4H_2O$
Molecular equation: $K_2Cr_2O_7 + 3Na_2SO_3 + 4H_2SO_4 \rightarrow 3Na_2SO_4 + K_2SO_4 + Cr_2(SO_4)_3 + 4H_2O$
The green colour is due to chromium(III) sulphate, $Cr_2(SO_4)_3$.