Pi bonds
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Re: Pi bonds
Hi Crystal! So I kinda answered someone's question that's similar to this, so I'm just gonna paste my answer and reword some things to make it make more sense for your question:
So I'm going to talk about the double bonds in the 2nd/bottom attached picture because those bonds are the ones that have a pi bond.
Explanation of why there is a leftover unhybridized orbital in the first place: If you look at the hybridization of C and O in the lewis structure (of the bottom attached picture), you would see that they're both sp^2, meaning that their hybridized orbital has sp^2, but because p has 3 orbitals, you would say that there is an unhybridized p orbital (and that's why you have "an electron leftover in the p-orbital" for O and C.
My guess for why the unhybridized p-orbital has to be the pi bond:
We already know that the O and C atoms can only have 3 regions of electron density (because of the hybridization of sp^2), meaning they can only have 3 hybridized orbitals (as mentioned before). In each orbital (as shown in the first/top picture I attached), there is one unpaired electron (which is what gets paired when that atom bonds with another atom), and so essentially, all of the orbitals in the atom have to be bonded with another. Thus, the three hybridized orbitals all bond with another atom, following the hybridization we created based on the Lewis structure (sp^2 = 3 regions of electron density). That means that the remaining unhybridized orbital (which is the p as explained above) has to join one of the hybridized orbitals (because the Lewis structure only shows three regions of electron density), so the unhybridized orbital has no choice but to "join" one of the other single sigma bonds, which means it becomes the pi bond of the created double bond.
^^That is honestly just how I like to think about it, but I'm unsure if it's completely right. At this point, I just basically accepted that the remaining electrons in the unhybridized p orbitals will be the pi bond in the double bond.
I really hope this helps, and let me know if you have any other questions!
So I'm going to talk about the double bonds in the 2nd/bottom attached picture because those bonds are the ones that have a pi bond.
Explanation of why there is a leftover unhybridized orbital in the first place: If you look at the hybridization of C and O in the lewis structure (of the bottom attached picture), you would see that they're both sp^2, meaning that their hybridized orbital has sp^2, but because p has 3 orbitals, you would say that there is an unhybridized p orbital (and that's why you have "an electron leftover in the p-orbital" for O and C.
My guess for why the unhybridized p-orbital has to be the pi bond:
We already know that the O and C atoms can only have 3 regions of electron density (because of the hybridization of sp^2), meaning they can only have 3 hybridized orbitals (as mentioned before). In each orbital (as shown in the first/top picture I attached), there is one unpaired electron (which is what gets paired when that atom bonds with another atom), and so essentially, all of the orbitals in the atom have to be bonded with another. Thus, the three hybridized orbitals all bond with another atom, following the hybridization we created based on the Lewis structure (sp^2 = 3 regions of electron density). That means that the remaining unhybridized orbital (which is the p as explained above) has to join one of the hybridized orbitals (because the Lewis structure only shows three regions of electron density), so the unhybridized orbital has no choice but to "join" one of the other single sigma bonds, which means it becomes the pi bond of the created double bond.
^^That is honestly just how I like to think about it, but I'm unsure if it's completely right. At this point, I just basically accepted that the remaining electrons in the unhybridized p orbitals will be the pi bond in the double bond.
I really hope this helps, and let me know if you have any other questions!
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