The color of acidified K_2Cr_2O_7 is not chan | vedclass

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(C) The color of acidified $K_2Cr_2O_7$ remains unchanged if the oxidation state of the transition metal,$Cr$,remains constant after the reaction.The

Vedclass · Accepted Answer

$HF$

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(C) The color of acidified $K_2Cr_2O_7$ remains unchanged if the oxidation state of the transition metal,$Cr$,remains constant after the reaction.The oxidation state of $Cr$ in $K_2Cr_2O_7$ is calculated as:$2(+1) + 2x + 7(-2) = 0$$2 + 2x - 14 = 0$$2x = 12$$x = +6$Potassium dichromate reacts with hydrogen fluoride $(HF)$ to form potassium fluorotrioxochromate$(VI)$ and water:$K_2Cr_2O_7 + 2HF \rightarrow 2K[CrO_3F] + H_2O$In $K[CrO_3F]$,the oxidation state of $Cr$ is:$(+1) + x + 3(-2) + (-1) = 0$$1 + x - 6 - 1 = 0$$x - 6 = 0$$x = +6$Since the oxidation state of $Cr$ remains $+6$ in both the reactant and the product,the color of the solution does not change. Other reagents like $H_2O_2$,$Sn^{2+}$,and $HBr$ act as reducing agents and reduce $Cr(VI)$ to $Cr(III)$,causing a color change.

The color of acidified $K_2Cr_2O_7$ is not changed by:

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