Write Lewis structure of `O_(2)^(-)` ion and find out oxidation state of each oxygen atom? What is the average oxidation state of oxygen in this ion?

Explanation

To solve the problem, we will follow these steps:

### Step 1: Draw the Lewis Structure of the `O2^(-)` Ion
1. **Count the total number of valence electrons**: Each oxygen atom has 6 valence electrons. Since there are 2 oxygen atoms, the total is \(6 \times 2 = 12\) electrons. Additionally, the ion has a -1 charge, which means we add one more electron, giving us a total of \(12 + 1 = 13\) valence electrons.

2. **Arrange the atoms**: The two oxygen atoms will be placed next to each other since they are bonded.

3. **Distribute the electrons**: Start by forming a bond between the two oxygen atoms. This uses 2 electrons, leaving us with \(13 - 2 = 11\) electrons.
...