$A$ greenish-yellow gas reacts with an alkali metal hydroxide to form a halate which can be used in fireworks and safety matches. The gas and halate are,respectively:
- A$Br_{2}, KBrO_{3}$
- B$Cl_{2}, KClO_{3}$
- C$I_{2}, NaIO_{3}$
- D$Cl_{2}, NaClO_{3}$
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Similar Questions
Consider the following reactions :
$(i)$ $F_2 + 2X^{-} \to 2F^{-} + X_2$ $(X = Cl, Br, I)$
$(ii)$ $Cl_2 + 2X^{-} \to 2Cl^{-} + X_2$ $(X = Br, I)$
$(iii)$ $I_2 + 2Br^{-} \to 2I^{-} + Br_2$
Which option has only the correct set of reactions?
$(i)$ $F_2 + 2X^{-} \to 2F^{-} + X_2$ $(X = Cl, Br, I)$
$(ii)$ $Cl_2 + 2X^{-} \to 2Cl^{-} + X_2$ $(X = Br, I)$
$(iii)$ $I_2 + 2Br^{-} \to 2I^{-} + Br_2$
Which option has only the correct set of reactions?
Medium
View SolutionWhich among the following does not form polyhalide ion?
Fluorine reacts with dilute $NaOH$ to form a gaseous product $A$. The bond angle in the molecule of $A$ is
$A$ solid $A$ which has photographic effect reacts with the solution of a sodium salt $B$ to give a pale yellow precipitate $C$. The sodium salt $B$ on heating with concentrated $H_2SO_4$ gives brown vapours. Identify $A$,$B$,and $C$.
Difficult
View Solution$SO_2$ acts as a temporary bleaching agent,while $Cl_2$ acts as a permanent bleaching agent. Why?
Easy
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