Question

The oxidation state of oxygen is maximum in:
A.) bleaching powder $(CaOC{l_2})$
B.) oxygen difluoride $(O{F_2})$
C.) dioxygen difluoride $({O_2}{F_2})$
D.) hydrogen peroxide $({H_2}{O_2})$

Answer 1

To find the oxidation state of oxygen in all the given elements we will add the oxidation numbers of elements other than oxygen and unknown oxidation number of oxygen atom which will be equal to the total charge on the compound given.

Complete step by step answer:
Let us assume the oxidation number of the oxygen be $x$ for all given compounds.
For option A.), in the bleaching powder $(CaOC{l_2})$, the oxidation number of calcium ($Ca$) is $+ 2$ and oxidation number of chlorine is $- 1$ and for two chlorine it will be $- 2$. Also, we know that the total charge on the compound is zero. Therefore, the oxidation number of oxygen atom can be given as:
$( + 2) + x + ( - 2) = 0 \\\ x = 0 \\\$
For option B.), in the oxygen difluoride $(O{F_2})$, the oxidation number of fluorine is $- 1$ and for two fluorine oxidation numbers it will be $- 2$. Also, the total charge on the given compound is zero. Therefore, the oxidation number of oxygen atom can be given as:
$x - 2 = 0 \\\ x = + 2 \\\$
For option C.), in the dioxygen difluoride $({O_2}{F_2})$, the oxidation number of fluorine is $- 1$ and for two fluorine oxidation number will be $- 2$ and there are two oxygen atoms so there oxidation number will be taken as $2x$. Also, the total charge on the given compound is zero. Therefore, the oxidation number of oxygen atom can be given as:
$2x - 2 = 0 \\\ x = + 1 \\\$
For option D.), in the hydrogen peroxide $({H_2}{O_2})$, the oxidation number of the hydrogen is $+ 1$ and for two hydrogen it will be $+ 2$ and there are two oxygen atoms so there oxidation number will be taken as $2x$. Also, the total charge on the given compound is zero. Therefore, the oxidation number of oxygen atom can be given as:
$2x + 2 = 0 \\\ x = - 1 \\\$
As we can see that in oxygen difluoride $(O{F_2})$, we have a maximum oxidation state that is $+ 2$.

Hence, option B.) is the correct answer.

Note:
Always remember that both dioxygen difluoride $({O_2}{F_2})$ and hydrogen peroxide $({H_2}{O_2})$ have similar chair like structure but in dioxygen difluoride $({O_2}{F_2})$, the oxidation state of oxygen is $+ 1$ and in hydrogen peroxide $({H_2}{O_2})$, the oxidation state of oxygen is $- 1$.