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HW 1E.23

Posted: Thu Oct 24, 2019 4:06 pm
by Michelle Chan 1J
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!

Re: HW 1E.23

Posted: Thu Oct 24, 2019 4:12 pm
by Cooper Baddley 1F
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.

Re: HW 1E.23

Posted: Thu Oct 24, 2019 5:39 pm
by William Francis 2E
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.

Re: HW 1E.23

Posted: Thu Oct 24, 2019 5:49 pm
by William Francis 2E
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.

Re: HW 1E.23

Posted: Fri Oct 25, 2019 7:23 pm
by Ellis Song 4I
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.