JEE CHEMISTRY MCQ: REDOX REACTIONS

1 Easy

1 of 1

Understanding +4 | -1

An element with electronic arrangement at 2, 8, 18, 1 will exhibit the following oxidation states

View answer and explanation

Correct option: B

Option A is correct because an element with the electronic configuration 2, 8, 18, 1 (Copper) can exhibit multiple oxidation states due to the involvement of d-electrons.

More Questions from REDOX REACTIONS

Practice All Chapter MCQs →

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 :