MCQ
Which compound acts as an oxidising as well as reducing agent
- ✓$S{O_2}$
- B$KMn{O_4}$
- C$A{l_2}{O_3}$
- D$Cr{O_3}$
✓
Answer
Correct option: A.
$S{O_2}$
a
(a) The minimum and maximum oxidation number of $S$ are $-2$ and $+6$ respectively. Since the oxidation number of $S$ in $S{O_2}$ is $+4$, therefore it can be either increased or decreased.
(a) The minimum and maximum oxidation number of $S$ are $-2$ and $+6$ respectively. Since the oxidation number of $S$ in $S{O_2}$ is $+4$, therefore it can be either increased or decreased.
Therefore $S{O_2}$ behaves both as an oxidising as well 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
The standard reduction potentials at $298 \mathrm{~K}$ for the following half cells are given below :
→$\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O}, \mathrm{E}^{\circ}=1.33 \mathrm{~V}$
$\mathrm{Fe}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Fe} \mathrm{E}^{\circ}=-0.04 \mathrm{~V}$
$\mathrm{Ni}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Ni} \mathrm{E}^{\circ}=-0.25 \mathrm{~V}$
$\mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{e}^{-} \rightarrow \mathrm{Ag} \mathrm{E}^{\circ}=0.80 \mathrm{~V}$
$\mathrm{Au}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Au} \mathrm{E}^{\circ}=1.40 \mathrm{~V}$
Consider the given electrochemical reactions, The number of metal$(s)$ which will be oxidized be $\mathrm{Cr}_2 \mathrm{O}_7{ }^{2-}$, in aqueous solution is. . . . . .
Consider the following conversions
$(i)\, O(g) + e^-\to O^-(g) ; \Delta H_1$
$(ii)\, F(g) + e^-\to F^-(g) ; \Delta H_2$
$(iii)\, Cl(g) + e^-\to Cl^-(g) ; \Delta H_3$
$(iv)\, Na(g) \to Na^+(g) ; \Delta H_4$
incorrect statement is :-
→$(i)\, O(g) + e^-\to O^-(g) ; \Delta H_1$
$(ii)\, F(g) + e^-\to F^-(g) ; \Delta H_2$
$(iii)\, Cl(g) + e^-\to Cl^-(g) ; \Delta H_3$
$(iv)\, Na(g) \to Na^+(g) ; \Delta H_4$
incorrect statement is :-
$12\,g$ of $Mg$ (at. mass $24$) will react completely with acid to give
→For the reaction at $298 \mathrm{~K}, 2 \mathrm{~A}+\mathrm{B} \rightarrow \mathrm{C} . \Delta \mathrm{H}$ $=400 \mathrm{~kJ} \mathrm{~mol}^{-1}$ and $\Delta \mathrm{S}=0.2 \mathrm{~kJ} \mathrm{~mol}^{-1} \mathrm{~K}^{-1}$. The reaction will become spontaneous above. . . . . . .$\mathrm{K}$
→Values of the four quantum numbers for the last electron in the atom are $n = 4,\,l = 1,\,m = + 1$ and $s = - 1/2$. Atomic number of the atom will be
→The major product of the given reaction is
→ $[Figure]$ $\xrightarrow[{(ii)\,{H_2}S{O_4}\,,\,heat}]{{(i)\,OHCC{H_2}COCl}}$
Which one of the following is an example of disproportionation reaction?
Three isomers $A, B$ and $C$ (mol. formula $\left. C _{8} H _{11} N \right)$ give the following results
→$A\,and\,C\,\xrightarrow{{{\text{Diazotization}}}}\,P + Q\,\xrightarrow[{(ii)\,oxidation\,\left( {KMn{O_4} + {H^ + }} \right)}]{{{\text{(i) Hydrolysis}}}}$$\begin{array}{*{20}{c}} {R\left( {product\,of\,A} \right)} \\ { + \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,} \\ {s\,\left( {prosuct\,of\,C} \right)} \end{array}$
Ethylene difluoride on hydrolysis gives
Platinum, palladium, iridium etc., are called noble metals because