Localized vs Delocalized

[@biscuitstix_]

it's always these old videos from nearly a decade ago or more that come in clutch

[@RyanJeskeOrganicChem]

I'm glad its still helpful!

[@rahulahlawat255]

Thanks for the perfect way you explained everything!

[@RyanJeskeOrganicChem]

You're welcome!

[@valerieyoung708]

9 minutes went by so quick! thanks, great explanation.

[@RyanJeskeOrganicChem]

You are very welcome. I'm glad I could help you.

[@kyuhnfukaikage1283]

Thank you! Really helpful 👍

[@maxplank3782]

Please make the line drawings visible on a smartphone they need to be scaled up

[@narutorandomness457]

THANK YOU

[@maximus7043]

Thank You. Clear explaination

[@emkay7233]

how are you doing your electron configuration? isnt N 2s2,2p3

[@RyanJeskeOrganicChem]

Hi em kay, I think you are referring to the electron configuration in a Nitrogen atom, which is 1s2, 2s2, 2p3. The concept of localized vs delocalized is used in molecules, so we are mainly concerned with just the valence electrons, and any hybrid orbitals that are formed when the atoms form the molecule. In this case, with the first molecule (tri ethyle amine, the nitrogen connected to the three 2 carbon chains) there are 4 electron groups, so the nitrogen needs to be hybridized with 4 equivalent molecular orbitals. This case is sp3. Thus, the lone pair is in an sp3 orbital, there is no opportunity for resonance because there are no adjacent pi bonds or empty p orbitals. The lone pair in the first example is localized. Does that clear it up for you? Let me know if it does not.

[@aditi.chaudhary]

Thank u so much..u cleared my confusion.. 👍👍👍

[@rafaelapluguez-lopez6264]

Thx a lot!

[@nwakolpo]

Hi there. Thank you for the video. My question is for the last example. So after the movement of the pair of electrons we end up with the formation of +N and - C. Does that mean the last electron shell of N will be incomplete with 1e and the negative C atom will have an electron on its 3s shell ? Regards

[@RyanJeskeOrganicChem]

Hi Ven, I assume you are talking about the structure that I draw at 7:04 in the video? In that case, the N will be a cation because it has 4 bonds to it and no lone pairs. To calculate formal charge, we take half of each bond and all lone pairs to get a count of 4 electrons in the "valence" shell of the N. Normally, N has 5 electrons, so we have one less negative charge than normal, which makes this a cation. The N still has 8 electrons in its valence shell, all 8 come from the 4 bonds. The electron configuration for this N would still be 1s2, 2s2, 2p6 for a full valence shell. As for the C, we use the same method to calculate formal charge. Half of each bond (3) and one lone pair give 5 electrons around the C. Normally C has 4 valence, so we have one more negative charge than normal, giving a negative formal charge. The C still has 8 electrons in its valence shell, 6 from the three bonds and 2 from the lone pair. The electron configuration for this C would still be 1s2, 2s2, 2p6 for a full valence shell.

[@Vwgti2011]

Amazing! Thank you so much. I was struggling with this concept for a while. I really enjoy your explanation regarding the electrons in the P orbital and the general rule of if there is a double bond to an atom then chances are that atom is localized.

[@RyanJeskeOrganicChem]

Glad it helped you. Just remember that if there is a double bond to an atom, chances are it is the lone pair that is localized, not the atom.

[@glendasworld4125]

that makes no sense

[@RyanJeskeOrganicChem]

I'm not sure what you mean. Can you be more specific? Have you learned about resonance, yet? I can see how delocalization might be a bit confusing without first learning about resonance.