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

1 Answer

Answer:

#[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#