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Stabilization of a conjugate base: hybridization

How increased s character in a hybrid orbital stabilizes the conjugate base.

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  • leafers tree style avatar for user Alsean Toenjes
    In an earlier lecture, we learned that larger sized atoms were more acidic, or more willing to lose an H, than smaller ones due to the fact the electrons have more freedom to move around the anion and therefore more stable. Why wouldn't this concept apply here? In this lecture we learn the exact opposite.. now the atom is more stable when it has more s character because the electrons enjoy the CLOSE proximity to the nucleus. I'm sure I'm missing something.. but what? I appreciate any feedback!
    (39 votes)
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    • leafers tree style avatar for user justjacjab
      It's been a while but I figured I'd try to answer your question for anyone else who might have noticed this apparent discrepancy.
      In the earlier lecture, a larger size helps the conjugate base become more stable because the negative charge of an electron is distributed over a larger area reducing the repelling force from other electrons and this essentially makes the arrangement more stable. In this lecture, the the proximity of the lone pair results in an increase in stability due to an increased attraction to the protons. It may help to point out that in this case we are looking at a molecule with covalent sigma bonds sharing those electrons with other atoms (which operate on different "paths" and reduces repel) and that it isn't just a single atom (a halogen in the previous lecture) where the participating electrons are in the same shell and relatively closer to one another.
      (10 votes)
  • blobby green style avatar for user mattdorman555
    If we were to flip it around and rank molecules based on basicity how would that procedure work?
    (5 votes)
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  • piceratops ultimate style avatar for user karanwattamwar
    While explaining the s-character reason in the video it is explained that smaller the distance between nucleus and electrons in the nucleus more is the attraction .........
    This the point where i am puzzled. Why isn't like -
    Sp3 is a larger orbital than sp2 and more larger than sp. so in sp3 charge is spread more than it is spread in sp......
    (3 votes)
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  • male robot donald style avatar for user esoar2000
    this idea seems to contradict the idea of delocalizing electrons to increase stability?
    (3 votes)
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    • blobby green style avatar for user Adrian mendez
      Either way, the formal charges on all of those conjugate bases is (-1).
      I believed this can be explained through Columb's law stating F = kqQ / r2
      Electron's repel one another, and constantly moving so the radius constantly fluctuates.
      Protons attract neutrons
      In order for the electrons to stay in the orbital, the force of attraction must be greater than the force of repulsion. We see the radius between the electrons and protons decrease as we become more S hybridized than P hybridized

      The 2s sphere has a smaller radius than the 2p field, depending on what point you count as the radius .Therefore it is more stable in the SP because it can hold the electrons better and thus more stable.

      However I definitely could be wrong as these are one of my 4 am thoughts, so please take what I say with a grain of salt.

      In regards to surface area, yes the more surface area/volume the Molecule has (not the volume that the electrons take up), the more easier it is to de-localize charge its charge throughout that volume. In this case, all you took away was a proton from the molecule (no neutron in the majority of Hydrogen's)

      which is why I think what this guy is saying at doesn't make sense. Hybridization doesn't change an atom's affinity to attract electrons because electronegativity is a constant we've quantified through experimentation.
      (0 votes)
  • piceratops ultimate style avatar for user Gourabo
    So consider, I take a compound having many chlorines - 2 at the middle Carbons (as they have 2 valencies each ) and three for the last Carbon - isn't this gonna exert too much Inductive effect which may possibly cause the compound to be somewhat unstable?
    (1 vote)
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  • blobby green style avatar for user Elizabeth Roy
    I am a little confused as to how greater distance of electrons from the nucleus is destabilizing in hybridization, but with regard to anion size greater distance from the nucleus is stabilizing?
    (2 votes)
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  • blobby green style avatar for user Nacho Juan Manzanero
    Does the sp3 character also confer an inductive effect? That is to say, if we have a carbon double or triple bond not necessarily on the carbon to which the hydrogen is connected, but somewhere else on the same molecule, will that also increase the acidity, and decrease with distance from the hydrogen in question?
    (1 vote)
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  • leafers ultimate style avatar for user gaurav.shinde
    The instructor said that EN of atom in conjugate base going across a period is directly related to its stability and acidity, but is inversely related when going down a group. Why exactly does he say the greater EN on the hybridized carbon with the greatest % of s character directly relates to acidity then?
    (1 vote)
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  • blobby green style avatar for user Angolberth Tawampe
    how can I work out how many milliliters in the given buffer ph
    (0 votes)
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Video transcript

- [Voiceover] Hybridization can have a large effect on the stabilization of a conjugate base. So if we start off with ethane, here's the ethane molecule, we know the hybridization of this carbon, we know this carbon is sp3 hybridized. So let's say that ethane donates a proton and let's make it this proton right here. So the electrons in this bond, the electrons in magenta are left behind on that carbon to form the conjugate base. So here are the electrons in magenta and this carbon is sp3 hybridized which means the electrons in magenta occupy an sp3 hybrid orbital. So that's meant to represent an sp3 hybrid orbital. We know from the videos on hybridization that an sp3 hybridized orbital has 25% s character and 75% p character. So I'm just going to write down here 25% s character. Let's more on to ethene or ethylene. This carbon is sp2 hybridized so we know that this carbon in ethene is sp2 hybridized. If ethene donates this proton, the electrons in magenta are left behind. So here are the electrons in magenta. This is the conjugate base to ethene and this carbon is sp2 hybridized. So the lone pair of electrons, the electrons in magenta occupy an sp2 hybridized orbital. So that's supposed to represent an sp2 hybrid orbital. An sp2 hybridized orbital has approximately 33% s character. So, I'm gonna write down here 33% s character. Finally, we have acetylene. This carbon in acetylene is sp hybridized. So if acetylene donates a proton, if acetylene donates this proton, then these electrons are left behind. So the electrons in magenta are these electrons and this carbon is sp hybridized so the electrons in magenta occupy an sp hybrid orbital. An sp hybrid orbital is 50% s character. So this is 50% s character. Now, let's look at pKa values. So the pKa for this proton on ethane is approximately 50. The pKa value for this proton on ethene is approximately 44. And the pKa value for this proton on acetylene is about 25. We know the lower the pKa value, the stronger the acid. So as we move to the right, we see a decrease in pKa values. And therefore, that's an increase in the acidity. So we're talking about increase in the acid strength. So acetylene is the strongest acid out of these three. If acetylene is the strongest acid, that must mean it has the most stable conjugate base. So this conjugate base here to acetylene must be the most stable out of these three. So as we move to the right, we are increasing in stability. So increasing in the stability of the conjugate base. So how do we explain the increased stability of the conjugate base in terms of hybridization? Well, let's look at the hybrid orbitals that we were talking about here. For the first conjugate base, our lone pair of electrons occupy an sp3 hybridized orbital and that was 25% s character. And as we went to the right for our conjugate bases, we increased in s character to 33% to 50%. So as we move to the right, we increase in stability. We also increase in s character. So increasing in s character increases the stability of the conjugate base and we can explain that by thinking about s and p orbitals. On average, an s orbital has electron density closer to the nucleus than a p orbital. So as you increase in s character, you're increasing in electron density closest to the nucleus. So let me go ahead and point out what I mean here. So let's look at this lone pair of electrons in the conjugate base to ethane. We think about the distance of those electrons to the nucleus. An sp3 hybridized orbital has the smallest amount of s character therefore those electrons are on average further away from the nucleus. That's less stable, that's higher in energy. As we move to the right, we can see that that distance decreases. So the distance decreases. And finally, for an sp hybridized orbital, that's the shortest distance between that lone pair of electrons and the positively charged nucleus. If you decrease the distance between the positively charged nucleus and the electrons, that means you increase the force of attraction. So this conjugate base is the most stable because there's a greater attraction to the nucleus for those electrons. So the nucleus is better able to hold onto those electrons, is a greater force and that means increased stability or lower energy. So this is the most stable conjugate base. If that's the most stable conjugate base, then acetylene is the most acidic compound out of those three. So this also has an effect on electronegativity. If an sp hybridized carbon is better able to attract electrons, well think about our definition for electronegativity. It's the power of an atom to attract electrons to itself. So since the electrons are closer to the positively charged nucleus in an sp hybridized carbon so that must mean that sp hybridized carbons are more electronegative. So an sp hybridized carbon is more electronegative than an sp2 hybridized carbon and an sp2 hybridized carbon is more electronegative than an sp3 hybridized carbon. So it has to do with the amount of s character. And that might seem weird because so far we've said that carbon has a certain value for the electronegativity and we've always assumed that it's the same but now we can see that it's different. An sp hybridized carbon is actually the most electronegative.