Question

How do you determine the electron configuration of `"W"`?

Answer 1

`[Xe]4f^14\5d^4\6s^2`

Explanation:

We first need to find the closest noble gas to tungsten on the periodic table.

![https://sciencenotes.org/periodic-table-pdf-2/]()

From here, tungsten is number `74` on the periodic table, so the closest noble gas to it would be xenon, which is number `54`. So, we start by writing the first part of tungsten's electron configuration which is

`[Xe]`

Next, we have `74-54=20` more electrons to fill.

We know that xenon's full electron configuration is `1s^2\2s^2\2p^6\3s^2\3p^6\3d^10\4s^2\4p^6\4d^10\5s^2\5p^6`

From this picture, we see that the next shell we choose to fill will be the `6s` orbital. But don't take too much stock in it, as it doesn't work well for many transition metals.

![https://chemistry.stackexchange.com/questions/31189/what-is-spdf-configuration]()

It can hold `2` electrons, so we will continue with tungsten's electron configuration, and now have

`[Xe]6s^2`

`20-2=18` more electrons

Next comes the `4f` subshell. It can a maximum of `14` electrons, so we can write it as `4f^14`. Continuing with our original goal, we get

`[Xe]6s^2 4f^14`

`18-14=4` more electrons

Since the `5d` subshell can hold `10` electrons, but we only need `4` electrons, it becomes `5d^4`.

So, our final electron configuration for tungsten is:

`[Xe]6s^2\4f^14\5d^4`

or we can rearrange it in order of `n` to

`[Xe]4f^14\5d^4\6s^2`