Question

In the conversion of $\;BrO_3^ -$ to $BrO_4^ -$ ​, the oxidation state of Br changes from:
A.There is no change in the oxidation number
B.Reduction takes place in basic solution
C.The reduction also takes place by $Xe{F_2}$
D.Equivalent mass of $\;BrO_3^ -$ is one half of ionic mass

Answer 1

To answer this question, recall the concept of oxidation numbers. The oxidation state of an atom is defined as the number of electrons lost and, therefore, describes the extent of oxidation of the atom. For example, the oxidation state of carbon in${\text{C}}{{\text{O}}_{\text{2}}}$ would be $+ 4$ since the hypothetical charge held by the carbon atom if both of the carbon-oxygen double bonds were completely ionic would be equal to $+ 4$ .

Complete step by step answer:
The most frequent terms Oxidation state and oxidation number are terms frequently used interchangeably. They are defined and described as the number of electrons lost in an atom. The values can be zero, positive, or negative. When $\;BrO_3^ -$ is converted to $BrO_4^ -$ ​, the oxidation number of bromines is changed to $+ 7$ from $+ 5$ . It indicates the fact that option A is incorrect and can be eliminated.
Also, the oxidation number is increased to $+ 7$ from $+ 5$ . Statements B and C indicate the fact that there has been a decrease in oxidation state. Hence, options B and C are wrong and can be eliminated.
The equivalent mass of $\;BrO_3^ -$ is the ionic mass divided by the change in the oxidation number. As the change from oxidation state $+ 7$ to $+ 5$ is a change of 2. The equivalent mass of $\;BrO_3^ -$ ionic mass divided by 2 which is half the ionic mass. Thus, option D is correct.

Hence, the correct option is D.

Note:
In most of the compounds, the oxidation number of oxygen is $- 2$ . There are two exceptions here.
Peroxides: Each oxygen atom exhibits an oxidation number of $- 1$ . Example, $N{a_2}{O_2}$
Superoxide- Every oxygen atom is allocated an oxidation number of $- \dfrac{{{\text{ }}1}}{2}$ . Example, $K{O_2}$
Oxygen is bonded to fluorine - Example, dioxygen difluoride where the oxygen atom is allocated an oxidation number of $+ 1$ .