Question

What is the formula for the ion formed when oxygen obtains a noble gas configuration?

Answer 1

Hint: An oxygen atom can gain electrons in order to achieve a noble gas configuration which will represent the noble gas at the end of its period, which is neon. Noble gases have an electronic configuration in which all their outer shells are completely filled. This means that they have a stable configuration.

Complete answer:
The formula of the ion that oxygen forms when it obtains a noble gas configuration is \[{O^{2 - }}\]
The electronic configuration of oxygen is:
\[1{s^2}2{s^2}2{p^4}\]
So if it gains two more electrons in the 2p orbital we can say that its outer shell will be completely filled. To do so oxygen has to gain two electrons making it negatively charged. So we can say that the charge of oxygen in doing so will be -2.
Thus we can easily say that oxygen after gaining two electrons will become \[{O^{2 - }}\]
Now the electronic configuration of oxygen when it becomes \[{O^{2 - }}\] can be represented as:
\[1{s^2}2{s^2}2{p^6}\]
This is the same electronic configuration as that of neon which has an atomic number of 10.
Thus we can say that oxygen in its \[{O^{2 - }}\] ion state acquires a noble gas configuration.

Note:
Mostly oxygen shows only -2 oxidation state. But there are certain compounds in which it shows a different oxidation state. One of the prime examples of these are peroxides such as \[{H_2}{O_2}\] . In this compound the oxygen shows an oxidation state of -1. There are other compounds in which oxygen acquires a positive oxidation state. These are compounds with fluorine which is more electronegative than oxygen. In superoxide the oxidation state of oxygen is +0.5