When I− is oxidised by MnO−4 in alkaline medium, I− converts into :
IO−3
IO−
IO−4
I2
A
I2
B
IO−
C
IO−3
D
IO−4
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Solution
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2KMnO4+2KOH→2K2MnO4+H2O+O 2K2MnO4+2H2O→2MnO2+4KOH+2O2KMnO4+H2Oalkalme−−−−−−→2MnO2+2KOH+3[O] KI+[O]→KIO32KMnO4+KI+H2O→2KOH+2MnO2+KIO3 KI+3[O]→KIO−3 Hence option A is correct.
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