Bromine is added to cold dilute aqueous solution of NaOH. The mixture is boiled. Which of the following statements is not true?

Options

A. During the reaction bromine is present in four different oxidation states.
B. The greatest difference between the various oxidation states of bromine is 5.
C. On acidification of the final mixture bromine is formed.
D. Disproportionation of bromine occurs during the reaction.

Answer

C

Explanation

`2NaOH_(("dilute"))+Br_(2) overset("cold")to NaBrO+NaBr+H_(2)O`
`underset("Sodium hypobromide")(3NaBrO) overset(300K)to 3NaBrO`
On acidification, the final mixture gives bromine.
`5NaBrO+NaBrO_(3)+6HCl to 6NaCl+3Br_(2)+3H_2O`
Thus, during the reaction, bromine is present in four different oxidation states ie., zero in Br2, +1 in NaBrO, -1 in NaBr and +5 in `NaBrO_(3)`. The greatest difference between various oxidation states of bromine is 6 and not 5. On acidification of the final mixture, `Br_2` is formed and disproportionation of `Br_(2)` occurs during the reaction giving `BrO^(-), Br^(-) and BrO_(3)^(-)` ions.