Oxidation Number of Common Elements

Q.

The oxidation state of Oxygen in $K{O}_2$ is

1. +1/2

2. -2

3. +2

4. -1/2

Q.

How do you find the oxidation number of nitrogen in $\mathrm{NO}$?

Q.

You are given four compounds in which an element is kept in square brackets.

$[Mn{]}_2{O}_3$, $H\left[N\right]O_3$, $K\left[Mn\right]O_4$ and $H_2\left[C\right]O_3$

Select the option that represents the correct order of the compounds on the basis of increasing oxidation number value of the element in the square bracket that is higher the value of the oxidation number of the element in the square bracket higher its placement.

1. $[Mn{]}_2{O}_3$ < $H_2\left[C\right]O_3$ < $H\left[N\right]O_3$ < $K\left[Mn\right]O_4$

2. $H\left[N\right]O_3$ < $K\left[Mn\right]O_4$ < $[Mn{]}_2{O}_3$ < $H_2\left[C\right]O_3$

3. $H\left[N\right]O_3$ < $[Mn{]}_2{O}_3$ < $K\left[Mn\right]O_4$ < $H_2\left[C\right]O_3$

4. $H_2\left[C\right]O_3$ < $[Mn{]}_2{O}_3$ < $K\left[Mn\right]O_4$ < $H\left[N\right]O_3$

Q. You are given four compounds in which an element is kept in square brackets.

$[Mn{]}_2{O}_3$, $H\left[N\right]O_3$, $K\left[Mn\right]O_4$ and $H_2\left[C\right]O_3$

Arrange them on the basis of increasing oxidation number value of the element in the square bracket that is higher the value of the oxidation number of the element in the square bracket higher its placement.

Q.

The oxidation state of O in $O{F}_2$is:

1. -2

2. +2

3. -1

4. +1