As pointed out by an observant viewer in the comments, the second example for N2O is actually the MORE favorable Lewis Structure, due to formal charge arguments. Sorry for the error.
This is the best lecture on valence bond theory I've found so far on the internet! Thank you so much for breaking it down and making it super easy and simple to understand. I appreciate your simple language and step-by-step approach. Thanks for sharing!
Thank you for breaking this down well. I was very confused today in my lecture, and it took me a couple additional RUclips videos to understand this concept. I feel like I now have a solid grasp. You've made college-level Advanced General Chemistry easier for me. Thanks a ton.
@@chiamakaokafor6374 I know I'm commenting this a year later, but I just wanna say chill out its not that deep. He is right lol. I wouldn't be surviving engineering without RUclips and having these explanations online, some of these professors cant teach well, and their deep accents make it harder to understand what they say.
Thank you, this video shows crucial "derivation" type material that shows not only what VBT is but where the orbital shape comes from and what the purpose and function of VBT is. Awesome!
thanks a lot! I went through your video 2-3 times and now feel comfortable when it comes down to tackling corresponding questions. will be watching your molecular orbital theory video next.
The N2O example...the triple bond on N and single bond on O is more favorable since O is more electronegative and should receive the negative charge...at least that is my current understanding and would be glad to be corrected. Thanks for the video!
+Hegeleze You are actually correct! The second structure that i drew for N2O is actually the MORE favorable one due to formal charge arguments. Thank you for catching that! That is because you always want the negative formal charge on the more electronegative atom, which is O.
***** The question posed is only regarding the carbon-carbon bond. If you wanted to state the bonds for the whole molecule, you would include the two C-H bonds, which each are one sigma bond.
+Anam Mujeeb In N2O, the best Lewis Structure is the one that minimizes formal charges. If formal charges are equal for more than one structure, then the best structure puts the negative charge on the most electronegative atom, which is O. The best N2O structure, then, is with a triple bond between the N's and a single bond between the central N and O. Thus O gets the -1 formal charge and not N.
+Apurv Zoad The lines are to represent orbitals, as in diagrams in which you put an up or down arrow on each line to represent electrons. I just show the orbitals here, not how they are filled to indicate that you end with the same number of hybrid orbitals as the number of atomic orbitals that you start with
+Anjelica Urciel Ah, you are mixing up electron configuration, where oxygen is 2s2 2p4, with the orbitals of oxygen in the hybrid model. In water, oxygen has four electron domains about it...2 bonding regions and 2 lone pairs. To create an equal environment for those four electron domains, we imagine the oxygen taking its available valence orbitals (s, px, py, and pz in most texts) and allowing them to be rearranged a bit to fit the experimental data. The current arrangement is a low energy 2s orbital and three higher energy 2p orbitals. What we'll do is mathematically mix them all up (that's 4 orbitals and all that energy) and then reconstruct 4 new hybrid orbitals that are of the same energy as each other (that is, they are "degenerate", which means the four, new orbitals are the same energy. Energy conservation must be conserved, so you can expect the 6 electrons about the oxygen to all now be at the same energy, and the total energy of those 6 electrons is the same as before the hybridization. Also, note that since we started with 4 orbitals, we must create 4 new orbitals. This is a result of the conservation of angular momentum...not important here, but it's good to know that the laws of physics aren't being violated. So, the oxygen now sits there with 4, equivalent orbitals that are the average of an s, a p, a p, and a p. We thus call each of those new orbitals an "sp three", where sp3 is used (and the 3 is a superscript). Now the oxygen is ready to bond with two hydrogens or two halogens, or other stuff that will lead to an octet being formed via two covalent single bonds. Hope this helps!!
Yes, carbon generally shares 8 electrons, forming at total of 4 bonds. At 11:58, pi bonds are being discussed for the C2H4 molecule. Each carbon has 4 bonds. Two of the bonds are to H atoms, and two bonds to the other C atom. One of those two bonds to the C atom is the pi bond being discussed at that time. Does that answer your question?
BTW. I edited my first response to you to talk about the right molecule. In answer to your question, the pi bonds between C and O are not "partial" bonds. They are simply not along the bond axis and come from the sharing of electrons in unhybridized p orbitals.
but carbon has no free electron, since it need no more sharing of electrons ... cuz carbon has already made 2 bonds with H and 1 = bond with O .. so how can it make more bonds with Oxygen's lone pairs ... ??
The double bond with O consists of one sigma bond and one pi bond. There are no additional electrons beyond that. In other words, the two bonds between C and O are very different types of bonds.
As pointed out by an observant viewer in the comments, the second example for N2O is actually the MORE favorable Lewis Structure, due to formal charge arguments. Sorry for the error.
awesome lecture...
This is the best lecture on valence bond theory I've found so far on the internet! Thank you so much for breaking it down and making it super easy and simple to understand. I appreciate your simple language and step-by-step approach. Thanks for sharing!
I'm so glad you found it helpful!
Thank you for breaking this down well. I was very confused today in my lecture, and it took me a couple additional RUclips videos to understand this concept. I feel like I now have a solid grasp. You've made college-level Advanced General Chemistry easier for me. Thanks a ton.
thanks for your comment! I really glad that it was helpful to you!!
RUclips tutorials > my foreign chemistry professor
thats racist..even a professor with an accent can be just as good. besides how many languages can u speak??
@@chiamakaokafor6374 I know I'm commenting this a year later, but I just wanna say chill out its not that deep. He is right lol. I wouldn't be surviving engineering without RUclips and having these explanations online, some of these professors cant teach well, and their deep accents make it harder to understand what they say.
Thank you, this video shows crucial "derivation" type material that shows not only what VBT is but where the orbital shape comes from and what the purpose and function of VBT is. Awesome!
love you for making me understand the whole chapter ….thank you so much
thanks a lot! I went through your video 2-3 times and now feel comfortable when it comes down to tackling corresponding questions. will be watching your molecular orbital theory video next.
You should teach the teachers how to to explain this topic. The only video that cleared all my confusion.
REAL
I'm so glad you found this video helpful! Thank you.
This explanation is just so adorable.Keep up the good work.❤️
when your chem professor is now a youtuber too
thx for the lesson, my prof is really smart and talks to fast for me to follow so i appreciate the detailed guide!
The N2O example...the triple bond on N and single bond on O is more favorable since O is more electronegative and should receive the negative charge...at least that is my current understanding and would be glad to be corrected. Thanks for the video!
+Hegeleze You are actually correct! The second structure that i drew for N2O is actually the MORE favorable one due to formal charge arguments. Thank you for catching that! That is because you always want the negative formal charge on the more electronegative atom, which is O.
I am writing from Turkey. You are wonderfull
Lol when your professor is a RUclipsr too. XD
Ruth this video was extremely helpful, thank you!
Thank you so much, this makes so much more sense!
Super useful! Thanks for making this!
You are an excellent teacher
Really u r good teacher
That is awesome,,,,,thanks 4 help
Wonderful!
well for understanding 👏👏
Thank u very very much for making this video!!!
best lecture...
Thank you so so so much! You just saved from a bunch of terrifying questions that may appear on my final exam tmr. Respect!
this was so very helpful thank you so much
very helpful thank you so much !!
Good review. Thanks
I found this video so helpful!!!!! Thank you for posting it!!!
for the last triple bond example, isn't it 2 sigma bonds and 2 pi bonds?
The last example is N2O, I think and yes, there are two sigma and two pi
Ruth Robinson
If you are referring to C2H2, the carbon carbon bond is just one sigma and two pie.
I was referring to the C2H2, so do you only count the bonds between the two carbons?
*****
The question posed is only regarding the carbon-carbon bond. If you wanted to state the bonds for the whole molecule, you would include the two C-H bonds, which each are one sigma bond.
I dont understand why would nitrogen have a triple bond in the questions you have given
+Anam Mujeeb In N2O, the best Lewis Structure is the one that minimizes formal charges. If formal charges are equal for more than one structure, then the best structure puts the negative charge on the most electronegative atom, which is O. The best N2O structure, then, is with a triple bond between the N's and a single bond between the central N and O. Thus O gets the -1 formal charge and not N.
What is the underline down the s and p Orbitals ? at 3:12 !
+Apurv Zoad The lines are to represent orbitals, as in diagrams in which you put an up or down arrow on each line to represent electrons. I just show the orbitals here, not how they are filled to indicate that you end with the same number of hybrid orbitals as the number of atomic orbitals that you start with
7:03 important
thanks
why do y have H2O writen with sp3? I thought oxygen was at p4? HELP!!!!
+Anjelica Urciel Ah, you are mixing up electron configuration, where oxygen is 2s2 2p4, with the orbitals of oxygen in the hybrid model.
In water, oxygen has four electron domains about it...2 bonding regions and 2 lone pairs. To create an equal environment for those four electron domains, we imagine the oxygen taking its available valence orbitals (s, px, py, and pz in most texts) and allowing them to be rearranged a bit to fit the experimental data.
The current arrangement is a low energy 2s orbital and three higher energy 2p orbitals. What we'll do is mathematically mix them all up (that's 4 orbitals and all that energy) and then reconstruct 4 new hybrid orbitals that are of the same energy as each other (that is, they are "degenerate", which means the four, new orbitals are the same energy. Energy conservation must be conserved, so you can expect the 6 electrons about the oxygen to all now be at the same energy, and the total energy of those 6 electrons is the same as before the hybridization. Also, note that since we started with 4 orbitals, we must create 4 new orbitals. This is a result of the conservation of angular momentum...not important here, but it's good to know that the laws of physics aren't being violated.
So, the oxygen now sits there with 4, equivalent orbitals that are the average of an s, a p, a p, and a p. We thus call each of those new orbitals an "sp three", where sp3 is used (and the 3 is a superscript). Now the oxygen is ready to bond with two hydrogens or two halogens, or other stuff that will lead to an octet being formed via two covalent single bonds.
Hope this helps!!
thanks so much! ^_^
Thanx
nice
11:58 .... isn't carbon stable ... with 8 electrons shared ...
Yes, carbon generally shares 8 electrons, forming at total of 4 bonds. At 11:58, pi bonds are being discussed for the C2H4 molecule. Each carbon has 4 bonds. Two of the bonds are to H atoms, and two bonds to the other C atom. One of those two bonds to the C atom is the pi bond being discussed at that time. Does that answer your question?
no ... my question is.. if carbon is stable ... how can oxygen make partial bonds with carbon !!(in the side view)! :3
BTW. I edited my first response to you to talk about the right molecule. In answer to your question, the pi bonds between C and O are not "partial" bonds. They are simply not along the bond axis and come from the sharing of electrons in unhybridized p orbitals.
but carbon has no free electron, since it need no more sharing of electrons ... cuz carbon has already made 2 bonds with H and 1 = bond with O .. so how can it make more bonds with Oxygen's lone pairs ... ??
The double bond with O consists of one sigma bond and one pi bond. There are no additional electrons beyond that. In other words, the two bonds between C and O are very different types of bonds.
I m from India
Thanks!