Skip to main content
Back to REDOX REACTIONS

JEE CHEMISTRY MCQ: REDOX REACTIONS

1 Easy
1 of 1
Application +4 | -1
Maximum oxidation state is present in :
View answer and explanation
Correct option: A
CrO2Cl2andMnO4−\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}\mathrm{ and MnO}_4^ -
Option A (CrO2Cl2andMnO4−\mathrm{Cr}{\mathrm{O}_2}\mathrm{C}{\mathrm{l}_2}\mathrm{ and MnO}_4^ -) is correct because Cr in CrO2Cl2 has +6 and Mn in MnO4- has +7 oxidation states, which are the maximum for these elements.

More Questions from REDOX REACTIONS

Practice All Chapter MCQs →
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 :
Type at least 2 characters or jump directly:
🩺 NEET MCQs 🔬 JEE MCQs 🎯 JEE College Predictor Study guides 📊 Dashboard
🔓

Save your learning progress

Sign in with a free account to save your practice answers and quiz results, then return to them when you revise.

Continue with Free Account →