HW 1E.23

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Michelle Chan 1J
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Joined: Thu Jul 25, 2019 12:16 am

HW 1E.23

Postby Michelle Chan 1J » Thu Oct 24, 2019 4:06 pm

I am a little confused on how to figure out the number of unpaired electrons for question 1E.23. For example, how many unpaired e- would be in Ga or Ge? Can someone explain? Thanks!
Last edited by Michelle Chan 1J on Thu Oct 24, 2019 4:16 pm, edited 1 time in total.

Cooper Baddley 1F
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Joined: Wed Sep 18, 2019 12:19 am

Re: HW 1E.23

Postby Cooper Baddley 1F » Thu Oct 24, 2019 4:12 pm

Finding unpaired electrons is easiest if you draw out the electron configuration first and then if done correctly the unpaired valence electrons are just the electrons don't have a pair in the configuration.

William Francis 2E
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Joined: Wed Sep 18, 2019 12:20 am

Re: HW 1E.23

Postby William Francis 2E » Thu Oct 24, 2019 5:39 pm

As an example, Gallium's electron configuration can be written as [Ar] 3d10 4s2 4p1. The fourth shell of electrons is the outermost shell for Gallium, so we can determine that it has 3 valence electrons in total. However, the 2 electrons in the 4s subshell are paired, so Gallium has only one unpaired valence electron.

William Francis 2E
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Joined: Wed Sep 18, 2019 12:20 am

Re: HW 1E.23

Postby William Francis 2E » Thu Oct 24, 2019 5:49 pm

Also, Germanium would have two unpaired valence electrons in its expected ground state since it has one more electron than Gallium in the 4p subshell and these electrons don't begin to pair in p subshells until the fourth electron is added.

Ellis Song 4I
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Joined: Thu Jul 11, 2019 12:17 am

Re: HW 1E.23

Postby Ellis Song 4I » Fri Oct 25, 2019 7:23 pm

For me the easiest way to find unpaired electrons is to think about how they would be put into orbitals and how you have to put an electron in each orbital before you start pairing. Just figure out the number of valence electrons and put them into orbitals using the arrows.


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