Question

Find oxidation number:-
$B{r}_3{O}_8$ [Br]

Answer 1

$B{r}_3{O}_8$ is tribromooctaoxide.

The structural configuration of $B{r}_3{O}_8$ is $-O_{3}Br-Br{O}_2-Br-O_3$.

Each of the two terminal Bromine atoms has $3$ oxygen atoms attached to it. So, the bromine atom on the terminals will have an oxidation state $+6$. The bromine atom in the middle has $2$ oxygen atoms attached to it. It will have an oxidation sate of $+4$.

Since, the three bromine atoms have different oxidation states in the same compound, the average of the oxidation states is calculated and considered as the oxidation state of the atom in the compound.

So $(+6)+(+4)+(+6)=46$. Thus the oxidation state of Bromine in $B{r}_3{O}_8$ is $-16/3$

This is called as a fractional oxidation number.