InterviewSolution
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InterviewSolution
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Justify giving reaction that among halogens, fluorine is the best oxidant and among hydrohalic compounds, hydroiodic acid is the best reductant. |
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Answer» `F_(2)` can oxidize `Cl^(-)` to `Cl_(2) , Br^(-)` to `Br_(2)` and `l^(-)` to `I_(2)` as : `F_(2 (aq)) + 2 Cl_((s))^(-) to 2 F_((aq)) + Cl_(g)` `F_(2 (ag)) + 2Br_(aq)^(-) to 2 F_(aq)^(-) + Br_(2 (l))` `F_(2 (aq)) + 2I_(aq)^(-) to 2 F_(aq)^(-) + I_(2 (s))` On the other hand , `Cl_(2) , Br_(2)` and `I_(2)` cannot oxidize `F^(-)` to `F_(2)` . The oxidizing power of halogens increases in the order of `l_(2) lt Br_(2) lt Cl_(2) lt F_(2)` . Hence , fluorine is the best oxidant among halogens . Hl and HBr can reduce `H_(2)SO_(4)` to `SO_(2)` , but HCl and HF cannot . Therefore , Hl and HBr are stronger than HCl and HF. `2HI + H_(2)SO_(4) to I_(2) + SO_(2) + 2H_(2)O` `2HBr + H_(2)SO_(4) to Br_(2) + SO_(2) + 2H_(2)O` Again `l^(-)` can reduce `Cu^(2+)` to `Cu^(+)` , but `Br^(-)` cannot . `4I_(aq)^(-) + 2Cu_(aq)^(2+) to Cu_(2)I_(2 (s))_(2 (aq))` Hence , hydroiodic acid is the best reductant among hydrohalic compounds . Thus , the reducing power of hydrohalic acids increases in the order of `HF lt HCl lt HBr lt Hl`. |
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