Question

When $KMn{O_4}$ reacts with $KBr$ in alkaline medium gives bromated ions. Then oxidation state of $Mn$ changes from $ + 7$ to:
A.$ + 6$
B. $ + 4$
C. $ + 3$
D. $ + 2$

Answer 1

Hint: The oxidation number is known as oxidation state, in which the atoms either gains electrons or loses its electrons to form a chemical bond with another atom. When permanganate ions react with halogen in an alkaline medium the oxidation state of manganese changes due to the formation of bond pairs between the outermost electrons of the metal ion.

Complete step-by-step answer:
Bromine is very reactive and at high temperature and readily dissociates themselves to yield free bromine atoms. When permanganate ion reacts with bromine ion in an alkaline medium to give manganese dioxide and bromate ion as a product. The bromate anion is a bromine-based oxoanion which is formed when a bromine ion reacts with ozone.
The reaction is :
$2KMn{O_4}\, + \,Br\, + \,{H_2}O\, \to \,2Mn{O_2}\, + \,KBr{O_3}\, + \,2KO{H^{}}$
Let’s write in simplified manner;
So, the reaction becomes below the following;
$2MnO_4^ - \, + B{r^ - } + {H_2}O \to 2Mn{O_2} + Br{O_3}^ - + 2O{H^ - }$
Now, let’s calculate the charge from left side of reaction;
Let’s start with $MnO_4^ - \,$
Let’s consider $Mn\, = \,x$ Here the outer charge is \[ - 1\] and each oxygen is having \[ - 2\] , we have \[4\] oxygens;
$x\, + \,4( - 2)\, = \, - 1$
$x\, - \,8\, = \, - 1$
$x\, = \, + 7$
On left side of equation we have $M{n^{ + 7}}$
Now, let’s observe the right side of equation;
We have $Mn{O_2}\,$ ;
Let’s consider $Mn\, = \,x$ and each oxygen is having \[ - 2\] , we have \[2\] oxygens;
$x\, + \,2( - 2)\, = \,0$
$x\, - \,4\, = \,0$
$x\, = \, + 4$
On right side of equation we have $M{n^{ + 4}}$
So, the overall reaction will be;
$2M{n^{ + 7}}O_4^ - \, + B{r^ - } + {H_2}O \to 2M{n^{ + 4}}{O_2} + Br{O_3}^ - + 2O{H^ - }$
When $KMn{O_4}$ reacts with $KBr$ in an alkaline medium, it forms bromate ion. In other words three bromate ions are formed and a hydroxyl group in which sharing of electrons takes place. In which the manganese gets reduced from $M{n^{ + 7}}$ to $M{n^{ + 4}}$. Therefore, the change in oxidation state of manganese changes from $M{n^{ + 7}}$ to $M{n^{ + 4}}$ .

Hence the correct answer is an option (B).

Note: Bromine is a strong oxidising agent, which easily reacts with many transitional elements to complete its outer shell. It is a weaker reducing agent which lose electrons readily rather than accept to complete its outermost shell configuration. Bromine as a halogen shares its maximum physical and chemical properties with other halogens.