MCQ
Colour of acidified $K _2Cr_2O_7$ is not changed by
- A$H_2O_2$
- B$Sn^{2+}(aq.)$
- ✓$HF$
- D$HBr$
✓
Answer
Correct option: C.
$HF$
c
The color of acidified $K _2 Cr _2 O _7$ is not changed if after reacting with a substance, the oxidation state of the transition metal, $Cr$, in the compound remains the same.
The color of acidified $K _2 Cr _2 O _7$ is not changed if after reacting with a substance, the oxidation state of the transition metal, $Cr$, in the compound remains the same.
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 $HF$.
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