$KMnO_4$ acts as an oxidising agent in acidic medium. $X$ is the difference between the oxidation states of $Mn$ in reactant and product. $Y$ is the number of $d$ electrons present in the brown red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of $X+Y$ is $.........$
- A$5$
- B$10$
- C$15$
- D$20$
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Purple solution of $KMnO_4 \xrightarrow{Na_2SO_4, BaCl_2}$ Pink ppt. Select the correct statement about pink ppt.
Difficult
View SolutionThe product $D$ in the following reaction sequence is:
The oxidation state of copper changes when aqueous copper $(II)$ ions react with
$(I)$ $NaOH_{(aq)}$
$(II)$ $Fe_{(s)}$
$(III)$ $KI_{(aq)}$
$(I)$ $NaOH_{(aq)}$
$(II)$ $Fe_{(s)}$
$(III)$ $KI_{(aq)}$
Medium
View SolutionWhich of the following reactants can be reduced by $SO_2$ in the given reactions?
$(I)$ $O_3 \to$ product
$(II)$ $Cr_2O_7^{2-} \xrightarrow{H^{+}} $ product
$(III)$ $MnO_4^{-} \xrightarrow{H^{+}} $ product
$(IV)$ $H_2S \to$ product
$(I)$ $O_3 \to$ product
$(II)$ $Cr_2O_7^{2-} \xrightarrow{H^{+}} $ product
$(III)$ $MnO_4^{-} \xrightarrow{H^{+}} $ product
$(IV)$ $H_2S \to$ product
Medium
View Solution$X$ and $Y$ are the number of electrons involved,respectively,during the oxidation of $I^{-}$ to $I_{2}$ and $S^{2-}$ to $S$ by acidified $K_{2}Cr_{2}O_{7}$. The value of $X + Y$ is . . . . . . .
DifficultJEE MAIN 2026
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