Question

Bromine is added to cold dilute aqueous solution of$NaOH$ . The mixture is boiled. Which of the following statements is not true?
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 1

Hint:All halogens react with sodium to produce sodium halides. Bromine reacts with sodium hydroxide to produce sodium bromide, sodium bromate and water

Complete answer:
For bromine, the formation of the sodium bromate(V) happens at around room temperature. Sodium bromate (I) must be formed at about 0°C.
Halogens compounds have a tendency that when reacted with a cold dilute solution of sodium hydroxide their oxidation state gets changed in order to balance the reaction.
The reaction of Bromine with dilute aqueous solution of takes place as:
 \[2NaOH(dilute) + B{r_2} \to (cold)NaBr{O} + NaB{r} + NaBr{O_3}\]
This reaction contains both oxidation and reduction reactions.
When acidification of the mixture takes place bromine is obtained as a product:
 $5NaBrO + NaBr{O_3} + 6HCl \to 6Nacl + 3B{r_2} + 3{H_2}O$ .
During the reaction, bromine has four different oxidation states like zero in $Br_2$ ,$+1$ in $NaBrO$ , $-1$ in $NaBr$ and $+5$ in $NaBrO_3$ .During the acidification reaction $Br_2$ is formed and disproportionation of bromine takes place giving $Br^-$,$BrO^-$ and ${BrO_3}^-$ ions. It means that the greatest difference between the various oxidation states of bromine is six not 5.
Therefore, the wrong statement in the aforesaid question is The greatest difference between the various oxidation states of bromine is 5.

Hence the correct option is option (B).

Note:
The halogens have a tendency to get oxidized or reduced in a reaction in order to form the salt and balance the oxidation state of an equation. The halogen which behaves in same order like chlorine, fluorine and bromine