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JEE CHEMISTRY MCQ: REDOX REACTIONS

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Oxidation number of iodine varies from
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Correct option: B
−1-1 to +7+7
Option B (−1-1 to +7+7) is correct because iodine, having 7 valence electrons, can exhibit oxidation states ranging from -1 to +7.

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The oxidation number of H\mathrm{H} in sodium hydride is A reagent, which can decrease the oxidation number of an element, is called Consider the following decomposition reaction

2KClO3⟶2KCl+3O22 \mathrm{KClO}_3 \longrightarrow 2 \mathrm{KCl}+3 \mathrm{O}_2

In the above reaction, Assertion (A): HNO3\mathrm{HNO}_3 acts as oxidizing agent.

Reason (R): Oxidation no. of nitrogen is +5, no increase in oxidation no. beyond +5 can occur. The oxidation no. of HNO3\mathrm{HNO}_3 can only decrease. Assertion :- Bromide ion is serving as a reducing agent in the reaction.

MnO4−(aq)+Br−(aq)+H2O⟶2MnO2(aq)+BrO3−(aq)+2OH−(aq)\mathrm{MnO}_4^{-}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})+\mathrm{H}_2 \mathrm{O} \longrightarrow 2 \mathrm{MnO}_2(\mathrm{aq})+\mathrm{BrO}_3^{-}(\mathrm{aq})+2 \mathrm{OH}^{-}(\mathrm{aq})

Reason :- Oxidation number of Br\mathrm{Br} increases from −1-1 to +5.+5. A substance, that by its sharp colour change indicates the completion of reaction is known as :
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