Chemical Bonding Result Question 24
Match the Columns
- Match the reactions in Column I with nature of the reactions/type of the products in Column II.
(2007, 6M)
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Answer:
Correct Answer: 24. (b) $\mathrm{A} \rightarrow 1,4 ; \mathrm{B} \rightarrow 3 ; \mathrm{C} \rightarrow 1,2 ; \mathrm{D} \rightarrow 1$
Solution:
- (A) In the reaction: $\mathrm{O}_2^{-} \longrightarrow \mathrm{O}_2+\mathrm{O}_2^{2-}$
Oxygen on reactant side is in $-\frac{1}{2}$ oxidation state. In product side, one of the oxygen is in zero oxidation state, i.e. oxidised while the other oxygen is in -1 oxidation state, i.e. reduced. Hence, in the above reaction, oxygen $\left(\mathrm{O}^{-1 / 2}\right)$ is simultaneously oxidised and reduced disproportionated.
(C) $\mathrm{MnO}_4^{-}+\mathrm{NO}_2^{-}+\mathrm{H}^{+} \longrightarrow \mathrm{Mn}^{2+}+\mathrm{NO}_3^{-}$
The above is a redox reaction and a product $\mathrm{NO}_3^{-}$ has trigonal planar structure.
(D) $\mathrm{NO}_3^{-}+\mathrm{H}_2 \mathrm{SO}_4+\mathrm{Fe}^{2+} \longrightarrow \mathrm{Fe}^{+}+\mathrm{NO}$
The above is a redox reaction.
BETA
