Lone Pairs in expanded Octet?

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Molly_McMillen_3J
Posts: 9
Joined: Wed Sep 21, 2016 2:57 pm

Lone Pairs in expanded Octet?

Postby Molly_McMillen_3J » Tue Oct 25, 2016 1:51 pm

I'm a little confused on how I am supposed to know how many lone pairs and atom will have when drawing expanded octet Lewis Structures. For example on question 3.63, Xe has 2 bound pairs and 3 lone pairs, but on 3.65, Xe can have 4 bound pairs and 2 lone pairs, I am just kind of confused on how to tell how many lone pairs an atom should have in the expanded octet rule.

AJ Garalza 2E
Posts: 24
Joined: Wed Sep 21, 2016 2:57 pm

Re: Lone Pairs in expanded Octet?

Postby AJ Garalza 2E » Tue Oct 25, 2016 4:04 pm

I believe the number of lone pairs an atom has is dependent on what the atom is bonding with. For instance, in 3.63, Xe has 2 bound pairs and 3 lone pairs because there are 22 valence electrons total. 8 electrons each are given to F, the remaining electrons just become lone pairs (2 pairs of electrons are bonding the Xe and F btw). This ends up totaling 22 valence electrons. Then for 3.65, Xe has 4 bound pairs and 2 lone pairs because there are 36 valence electrons total. 8 electrons each are given to the four Fs, the remaining electrons become lone pairs. This totals 36 valence electrons. So according to what an atom is trying to bond with will determine how many lone pairs it has.

Sangita_Sub_3H
Posts: 23
Joined: Sat Jul 09, 2016 3:00 am

Re: Lone Pairs in expanded Octet?

Postby Sangita_Sub_3H » Wed Oct 26, 2016 11:25 am

Another thing that can help you is looking at the formal charge of each of the elements in the compound. Since you want each atom to have a formal charge of zero, you can rearrange the bonds accordingly.


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