MCQ
The oxide, which cannot act as a reducing agent, is
- A$N{O_2}$
- B$S{O_2}$
- ✓$C{O_2}$
- D$Cl{O_2}$
✓
Answer
Correct option: C.
$C{O_2}$
c
(c) Reduction is accompanied by an increase in oxidation number of the reducing agent. $C$ belong to $IVA$ so the max - $O.N.$ is $+4$. In $C{O_2}$ the oxidation number of $C$ is $+4$, which cannot be further increased. Hence, $C{O_2}$ can not act as reducing agent.
(c) Reduction is accompanied by an increase in oxidation number of the reducing agent. $C$ belong to $IVA$ so the max - $O.N.$ is $+4$. In $C{O_2}$ the oxidation number of $C$ is $+4$, which cannot be further increased. Hence, $C{O_2}$ can not act as reducing agent.
Need a full question paper?
Generate a complete, print-ready paper with questions like this in minutes — across 16+ boards, with answer keys.
Explore more
Similar questions
Solid $CuSO_4·5H_2O$ having covalent, ionic as well as co-ordinate bonds. Copper atom/ion forms ................. co-ordinate bonds with water.
→On addition of conc. $H_2SO_4$ to a chloride salt, colourless fumes are evolved but in case of iodide salt, violet fumes come out. This is because
→Only two isomeric monochloro derivatives are possible for
Given are $E^o$ values for some half reactions
→$ {I_2} + 2{e^ - } \to \,2{I^ - }\,;\,\,{E^o}\, = \,\,0.54\,\,V $
$ MnO_4^ - \, + \,8{H^ + }\, + \,5{e^ - } \to \,M{n^{2 + }}\, + \,4{H_2}O\,;\,{E^{o\,}} = 1.52\,\,V $
$ F{e^{3 + }} + {e^ - } \to \,\,F{e^{2 + }}\,;\,\,{E^o}\, = \,\,0.77\,\,V $
$ S{n^{4 + }} + 2{e^ - } \to \,\,S{n^{2 + }}\,;\,\,{E^o}\, = \,\,0.1\,\,V $
The strongest reducant and oxidant respectively are
Product $(A)$ is
→Which of the following compounds forms ortho- benzenedicarboxylic acid when oxidized by hot aqueous potassium permanganate ?
Are which type of isomers
Among the given compounds, the correct order of enol content is
Which pair show cis-trans isomerism
Which of the following is a meso compound ?