JEE MAIN - Chemistry (2018 - 16th April Morning Slot - No. 17)
When XO2 is fused with an alkali metal hydroxide in presence of an oxidizing agent such as KNO3 ; a dark green product is formed which disproportioates in acidic solution to afforda dark purple solution. X is :
Ti
V
Cr
Mn
Explanation
X is Mn.
MnO2 + 2KOH + KNO3 $$ \to $$ K2MnO4 + KNO2 + H2O
Here K2MnO4 is dark green.
3MnO4$$-$$2 + 4H+ $$ \to $$ 2MnO4$$-$$ + MnO2 + 2H2O
In acidic solution, K2MnO4 changes to dark purple solution of KMnO4.
MnO2 + 2KOH + KNO3 $$ \to $$ K2MnO4 + KNO2 + H2O
Here K2MnO4 is dark green.
3MnO4$$-$$2 + 4H+ $$ \to $$ 2MnO4$$-$$ + MnO2 + 2H2O
In acidic solution, K2MnO4 changes to dark purple solution of KMnO4.
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