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NEET UG CHEMISTRY MCQ: d & f- BLOCK ELEMENTS

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I(aq)Θ+MnO4−(aq)→H+x+Mn(aq)+2I(aq)Θ+MnO4−(aq)→ weakly OHΘ neutral ( or )y+MnO2MnO4(aq)−+Mn(aq)+2→ZnSO4z+4H+\begin{aligned} & \mathrm{I}_{(\mathrm{a q})}^{\Theta}+\mathrm{MnO}_4^{-}(\mathrm{a q}) \xrightarrow{\mathrm{H}^{+}} \mathrm{x}+\mathrm{Mn}_{(\mathrm{a q})}^{+2} \\ & \mathrm{I}_{(\mathrm{a q})}^{\Theta}+\mathrm{MnO}_4^{-}(\mathrm{a q}) \xrightarrow[\text { weakly } \mathrm{OH}^{\Theta}]{\text { neutral }(\text { or })} \mathrm{y}+\mathrm{MnO}_2 \\ & \mathrm{MnO}_{4(\mathrm{a q})}^{-}+\mathrm{Mn}_{(\mathrm{a q})}^{+2} \xrightarrow{\mathrm{ZnSO}_4} \mathrm{z}+4 \mathrm{H}^{+} \end{aligned}
products x,y\mathrm{x}, \mathrm{y} and z\mathrm{z} are respectively
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Correct option: A
I2,IO3−,MnO2\mathrm{I}_2, \mathrm{IO}_3^{-}, \mathrm{MnO}_2
Option A (I2,IO3−,MnO2\mathrm{I}_2, \mathrm{IO}_3^{-}, \mathrm{MnO}_2) is the correct answer because iodide is oxidized to iodine in acidic medium, to iodate in neutral or weakly basic medium, and MnO4−MnO_4^- is reduced to Mn2+Mn^{2+} and MnO2MnO_2 respectively. MnO4−MnO_4^- and Mn2+Mn^{2+} react to form MnO2MnO_2.

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