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Among $ Cl ^- , Br ^-, I^- $ correct order for being oxidized to dihalogen is

A. $ I^- \gt Cl ^- \gt Br ^ - $ B. $ CI ^- \gt Br \gt Cl^- $ C. $ I ^- \gt Br ^- \gt CI ^ - $ D. $ Br ^ - \gt I ^ - \gt CI ^ - $

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Explanation

The correct order for the oxidation of halide ions to their respective dihalogens is based on their oxidizing ability. Iodide ($I^-$) is the easiest to oxidize, followed by bromide ($Br^-$), and chloride ($Cl^-$) is the hardest. Therefore, the order is $I^- o Br^- o Cl^-$. Thus, the correct option is $I^- > Br^- > Cl^-$. This order reflects the decreasing ease of oxidation from iodide to chloride.

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Correct order of increasing bond angles in following species is

A. $ ClO 6 - \lt Cl_2O \lt ClO_2 $ B. $ Cl_2O \lt ClO^â€“ \lt Cl_2 O $ C. $ ClO_2 \lt Cl_2O \lt ClO ^ - $ D. None of above

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When thiosulphate ion is oxidized by iodine , which one of following ion is produced?

A. $ SO_3 ^ 2- $ B. $ SO_4 ^ 2- $ C. $ S_4 O_6 ^ {2-} $ D. $ S_2 O_6 ^ {2-} $

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Explanation

$ 2Na_2S_2O_3 + I_2 \rightarrow 2NaI + Na_2S_4O_6 $

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Which one of halogen acids is liquid ?

A. HF B. HCl C. HBr D. HI

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Explanation

HF is liq. Because of int . molec. H - Bond

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When $ Cl_2 $ is passed through hot & Conc. Solution of KOH following compound is formed.

A. KCl B. $ KClO_3 $ C. $ KClO_2 $ D. $ KClO_4 $

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Explanation

$ 6KOH + 3Cl_2 \rightarrow 5KCl + KClO_3 + 3H_2O $

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Chlorine can remove

A. Br from NaBr Sol. B. F from NaF Sol. C. Cl from NaCl Sol. D. $ F from CaF_2 Sol $ .

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Explanation

$ Cl_2 + 2NaBr \rightarrow 2NaCl + Br_2 $

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Which one of following acid is weakest?

A. HClO B. HBr C. $ HClO_3 $ D. HCl

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Explanation

$ HClO \rightleftharpoons H^+ + ClO^ - $

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Which of following will displace halogen from solution of halide?

A. $ Br_2 added to NaCl $ B. $ Cl_2 added to KCl $ C. KCl added to NaF D. $ Br_2 added to KI$

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Explanation

$ Br_2 + 2KI \rightarrow I_2 + 2KBr $

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Chlorine can be manufactured from

A. Electrolysis of NaCl B. Electrolysis of brine C. Electrolysis of bleaching powder D. All above

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Explanation

$ 2NaCl + 2H_2O \rightarrow 2 NaOH_{(aq)} + Cl_{2(g)} + H_{2(g)} $

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Nitric acid converts iodine to

A. Iodic acid B. Hydro iodic acid C. Iodine Nitrate D. Iodine pentoxide

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Explanation

$ I_2 + 10HNO_3 \rightarrow 2HIO_3 + 10NO_2 + 4H_2O $

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