If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content
Current time:0:00Total duration:6:09

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.