When chlorine reacts with a cold and dilute solution of sodium hydroxide,the products obtained are:
- A$Cl^{-} + OCl^{-}$
- B$Cl^{-} + ClO_2^{-}$
- C$Cl^{-} + ClO_3^{-}$
- D$Cl^{-} + ClO_4^{-}$
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The correct statement$(s)$ regarding $(i) HClO$,$(ii) HClO_2$,$(iii) HClO_3$,and $(iv) HClO_4$ is(are):
$(A)$ The number of $Cl=O$ bonds in $(ii)$ and $(iii)$ together is two.
$(B)$ The number of lone pairs of electrons on $Cl$ in $(ii)$ and $(iii)$ together is three.
$(C)$ The hybridization of $Cl$ in $(iv)$ is $sp^3$.
$(D)$ Amongst $(i)$ to $(iv)$,the strongest acid is $(i)$.
$(A)$ The number of $Cl=O$ bonds in $(ii)$ and $(iii)$ together is two.
$(B)$ The number of lone pairs of electrons on $Cl$ in $(ii)$ and $(iii)$ together is three.
$(C)$ The hybridization of $Cl$ in $(iv)$ is $sp^3$.
$(D)$ Amongst $(i)$ to $(iv)$,the strongest acid is $(i)$.
When chlorine is passed over dry slaked lime at room temperature,the main reaction product is
On heating $KClO_3$,we get
Medium
View SolutionThe products obtained when chlorine gas reacts with cold and dilute aqueous $NaOH$ are:
Aqua-regia reacts with $Pt$ to yield
Easy
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