Question

What is the ground state electron configuration for `"O"^(2-)` ion?

Answer 1

Based on the periodic table, `"O"` is atomic number `8`, which means it has `8` electrons. The first few atomic orbitals are `1s`, `2s`, and `2p`.

Each orbital can hold `2` electrons maximum, and there are `2l+1` of each type of orbital (`s,p,d,f,g,...`), where `l = 0` corresponds to an `s` orbital, `l = 1` means `p` orbital, and so on.

So, the configuration for neutral `"O"` atom is:

`1s^2 2s^2 2p^4`

where the `2p` orbitals hold the `4` electrons like this:

`ul(uarr darr) " " ul(uarr color(white)(darr)) " " ul(uarr color(white)(darr))`

When oxygen gains two electrons, it acquires a charge of `2-`. Therefore, the two electrons, which go into the highest-energy atomic orbitals, give you a new configuration of

`color(blue)(1s^2 2s^2 2p^6)`

for `"O"^(2-)`, which is the same electron configuration as `"Ne"`.

That is known as a noble-gas-like configuration, and in general it is stable if oxygen is like that in a compound (having `8` electrons in its `2s` and `2p` orbitals combined).