A) \[CrO_{4}^{2-}\xrightarrow[{}]{{}}C{{r}_{2}}O_{7}^{2-}\]
B) \[BrO_{3}^{-}\xrightarrow[{}]{{}}Br{{O}^{-}}\]
C) \[{{H}_{3}}As{{O}_{3}}\xrightarrow[{}]{{}}HAsO_{4}^{2-}\]
D) \[Al{{(OH)}_{3}}\xrightarrow{{}}Al(OH)_{4}^{-}\]
Correct Answer: B
Solution :
(a) \[\overset{+6}{\mathop{Cr}}\,O_{4}^{2-}\xrightarrow[{}]{{}}\overset{+6}{\mathop{C{{r}_{2}}}}\,O_{7}^{2-}\] No change in oxidation number of Cr (b)\[\overset{+5}{\mathop{Br}}\,O_{3}^{-}\xrightarrow[{}]{{}}\overset{+1}{\mathop{Br}}\,{{O}^{-}}\] Oxidation number decreases from + 5 to +1 (reduction) (c)\[{{H}_{3}}\overset{+3}{\mathop{As}}\,{{O}_{3}}\xrightarrow[{}]{{}}\overset{+5}{\mathop{HAs}}\,O_{4}^{-}\] Oxidation number increases from + 3 to +5 (oxidation) (d) \[\overset{+3}{\mathop{Al}}\,{{(OH)}_{3}}\xrightarrow[{}]{{}}\overset{+3}{\mathop{Al}}\,(OH)_{4}^{-}\] No change in oxidation number Therefore,\[BrO_{3}^{-}\xrightarrow[{}]{{}}Br{{O}^{-}}\]requires a reducing agent.
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