JEE CHEMISTRY MCQ: REDOX REACTIONS

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Highest oxidation number that is exhibited fluorine is

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Correct option: B

Option B (0) is correct because fluorine is the most electronegative element and can only have a negative oxidation state or zero when it is in its elemental form.

More Questions from REDOX REACTIONS

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The oxidation number of

\mathrm{H}

in sodium hydride is A reagent, which can decrease the oxidation number of an element, is called Consider the following decomposition reaction

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

In the above reaction, Assertion (A):

\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

\mathrm{HNO}_3

can only decrease. Assertion :- Bromide ion is serving as a reducing agent in the reaction.

\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

\mathrm{Br}

increases from

-1

+5.

A substance, that by its sharp colour change indicates the completion of reaction is known as :