STD 11 - 7. Redox reactions — NEET STD 11 Science — Question

The oxidation state of $Cr$ in $K _2 Cr _2 O _7$ is calculated as :

$2+2 x +(-2) \times 7=0$

$\Rightarrow 2+2 x -14=0$

$\Rightarrow 2 x -12=0$

$\Rightarrow 2 x =+12$

$\Rightarrow x =+6$

$\therefore$ Oxidations state of $Cr =+3$

Potassium dichromate reacts with hydrogen fluoride to produce potassium fluorotrioxochromate $(VI)$ and water. The reaction involved is :

$K _2 Cr _2 O _7+2 HF \rightarrow 2 K \left[ CrO _3 F \right]+ H _2 O$

The oxidation state of $Cr$ in $K \left[ CrO _3 F \right]$ can be calculated as follows:

$+1+ x +(-2) \times 3+(-1)=0$

$\Rightarrow+1+ x -6-1=0$

$\Rightarrow x -6=0$

$\Rightarrow x =+6$

$\therefore$ Oxidations state of $Cr =+3$

Since, there is no change in the oxidation state of $Cr$ in the reactant and in the product, the color of the solution doesn't change. Therefore, the color of acidified $K _2 Cr _2 O _7$ is not changed by $HF$.