Question

What is the electron configuration (arrangement) by spdf notation of a sodium ion?

Answer 1

`1s^2 2s^2 2p^6`

Explanation:

Sodium loses 1 electron from its outermost shell to get
the configuration of neon, or form the sodium ion.

• The first shell has `1s^2`
• The second shell has `2s^2, 2p^6`

Answer 2

`1s^2\2s^2\2p^6` or just `[Ne]`

Explanation:

A neutral sodium atom has an electron configuration of `[Ne]3s^1`.

Since it has `11` electrons, it would want to lose one electron to remove its third shell, while keeping the other `2` electron shells full. So, in a sodium ion, it becomes `Na^+` and has only `10` electrons.

So, it will remove `1` electron, and lose its `3s` orbital. The electron configuration therefore just becomes `[Ne]` or if you want the full configuration, it is `1s^2\2s^2\2p^6`.