I went to title this post “Trends in Lewis Acid/Base Strength,” but I got the weirdest case of déjà vu when I tried. Turns out, I’ve been there and done that, which makes me feel weirdly nostalgic.
Although that post has already been written (for a year, actually), I’m back and ready to party with some new information that they kept from us in Gen Chem and Organic. That’s right, in this installment of Gen Chem III, we’ll be talking about More Lies We Were Told In Freshman Year!
Aren’t you psyched??
If you’ve taken organic chemistry, you heard a lot about nucleophiles and electrophiles. Like, a lot. Liek, approximately 5much 7u. If you haven’t, you can probably figure out what they are by their names: nucleophiles chase positive charge, and electrophiles chase negative charge.
In Inorganic, though, we’re going to use much more posh terms. Rather than all of that self-explanatory funny business, we’re just going to call them Lewis Bases and Lewis Acids.
Now, if you’ve read my posts on acid-base chem, or if you’ve studied it at some point, you’ll remember that there are multiple definitions of acidity and basicity. You’re probably most familiar with the idea that acids donate protons, and bases accept them. However, there’s a more general definition than this, and it encompasses things that this doesn’t. The Lewis definition says that Lewis acids accept electrons and Lewis bases donate them.
I won’t get into too much detail about that (if you want, you can read this post), because today, we’re not going to talk about what these are, but rather, what makes them tick. Doesn’t that sound like a joy?
First, for our Lewis acids, let’s think of what we want in a good electron-receptor. Well, we want them to be able to, you know, accept electrons. This means one of two things, from a chemical standpoint: either you’re going to want a molecule with an electron-deficient atom (BH3 could use two electrons’ worth of love) or with one that can expand its octet ([looks shiftily at silicon, phosphorus, etc.]).
For our Lewis bases, we want good electron donors. That means that they’ll need to have the electrons to donate, and that they’ll have to be willing to let go of them. Practically, this means we (almost) always need a nonbonding pair of electrons on our central atom that can reach out and share itself with the electron-seeking acid. Negative charges are good, too, but entirely unnecessary.
Okay, so I basically summed all of Lewis acid-base chemistry up in those two paragraphs. Can you go home now?
Of course not! Aren’t you eager to learn how you can manipulate these things into wanting electrons more or wanting to be more generous…?
Too bad, I’m going to tell you anyway. There are two primary categories of effects that will alter the strength of Lewis acids and bases: electronic effects and steric effects. Electronic effects deal with, you guessed it, the electronic properties of the acids and bases. Steric effects deal with their physical shapes and how they bump into each other.
First off, on the electronic side of things, the strength of a base decreases across a row of the periodic table. You’ll probably be able to guess why from your periodic trends: your atoms get more electronegative as you move from left to right. By the time you get all the way to the halides, you’ve got atoms that really just don’t want to let go of their electrons. That means coaxing them into donating them will be problematic. (By the same token, acid strength increases across a row. Because, you know. Acids are supposed to want electrons.)
Now, let’s say you take any arbitrary acid that comes out of the first two rows of the periodic table. I’ll pick H+. If you take this acid and react it with bases that only change by a shift in position in a column (let’s say you use NH3, PH3 and AsH3), it will form progressively weaker adduct bonds (acid-base bonds) with the bases as they go down the column. This is because, the farther apart the acid and base are on the periodic table by column position, the larger the energy difference is between their valence orbitals, and the harder it is to get effective overlap. (The same is true for bases.)
That leads into a concept that made me feel really weird when I first learned it, since I’d never heard of it before: turns out, there are such things as hard and soft acids and bases.
Well, this is just a really weird way of saying that certain kinds of acids (hard acids) form the best adducts with 2nd row bases, and that other kinds of acids (soft acids) form the strongest adducts with 3rd row bases and below. (You can also say the same for bases.) Hard acids include H+, BF3, cations of alkali and alkali earth metals (i.e., Li+, Mg2+, etc.), and first row transition metals in high oxidation states. Soft acids, on the other hand, include BH3 (USUURPERRR), first row transition metals in lower oxidation states, and second and third row transition metals.
As my professor and my book both neatly pointed out, the primary difference between the two is this: hard acids and bases interact primarily through electrostatic (charge) interactions, whereas soft acids and bases interact through covalent (bond) interactions. What’s very important to note is this: hard acids and bases prefer each other, and soft acids and bases prefer each other.
Now, let’s take a detour and ask an important question: if you’re looking at a jumbled mess of acids and bases, how can you tell which is which? Well, let’s look at an example, and then I’ll explain you the thing:
MeHgF + HSO3– ⇌ MeHgSO3– + HF
Keq > 1
In this case, we know from the equilibrium constant that the products are favored over the reactants (Keq = [products]/[reactants] at equilibrium, remember?). Using that information, we can figure out which acids/bases are hard and which are soft, easy-peasy.
First off, note on the left that we have an association of a scary-looking MeHg+ (acid) and F–(base), as well as an association of H+ (acid) with SO3– (base). We know that these associations aren’t favored, since equilibrium prefers the products of this reaction to the reactants. Since we know that, we can guess that these pairs are mismatched (hard/soft and soft/hard). Even if we’re feeling really conservative and decide to only be sure that H+ is hard (it’s as high up in the periodic table as you can get, son), we can figure out the whole puzzle.
If H+ is hard, it must be matched with a soft base. That means that the other base, F–, must be hard, and its complement, MeHg+, must be a soft acid.
Like I said, easy-peasy lemon-squeezy!
Okay, but can you go backwards?
Well, let’s look at this:
MeHgOH + HSO3– ⇌ MeHgSO3– + H2O
Okay, well, we know that H+ is a hard acid, and we know that MeHg+ is a soft acid. We also know SO3– is a soft base, which means that HO– must be a hard base. Since we know that, we would expect that products would be preferred over reactants, and that Keq would be greater than one.
Yes, June, that’s nice, but what if there’s only one acid or base in a reaction? Can you predict an equilibrium then?
Well, uh… not really.
Let me show you what I mean. Say, we come up with a reaction like this:
HF + SO32- ⇌ HSO3– + F–
Here, we have a proton, a hard acid, sulfate, a soft base, and fluoride, a hard base. Just based on that, we’d expect Keq to be less than one, right?
Well, turns out, that’s not really what’s happening. Why the heck??
Because sulfate is a better base, plain and simple. Sure, it’s a soft base, which means that it shouldn’t want to be with the proton as badly as fluoride does, but because sulfate’s base game is so dang strong, it really doesn’t matter. It’ll react with anything, regardless.
(Giiiirl, u basic.)
(I’m sorry, I’m very sleep deprived.)
All right, now that we’ve spent a sufficiently long time diverted from our actual subject, let’s get back to talking about electronic effects. All this has related to changing around the central atom. What happens if you keep that constant?
Turns out, you can’t dismiss things when your center stays the same. Nope, you’ve still got to worry about substituent effects.
Substituents tend to have opposite effects on acid/base strength depending on whether you’re talking about an acid or a base. For example, withdrawing groups (electronegative atoms such chlorine) decrease the strength of a base, since they pull electron density away from the central atom. However, for the same reason, they increase the effectiveness of acids.
The same works in the other direction: if you put carbon substituents on a base, it makes it work better, since carbons “donate” electron density to the central atom by induction. However, that would make an acid weaker, since it would want another molecule’s electrons less.
Resonance structures can also contribute to strength. Our professor gave us the example of the boron trihalides. Just based on electronegativity, you would expect BF3 to be stronger than BCl3, which should in turn be stronger than BBr3. Turns out, the exact opposite is true. Why?
Because the halogens can donate another pair of electrons to the boron by resonance. It’s really electron deficient, remember? All of the halides bonded to the boron could, theoretically, form a resonance structure with the boron in which the halide is double bonded to the boron (which would have a full octet). However, as we learned, as you move down columns, elements like making double bonds less and less. This means that BBr3 has the least double-bond character out of all of them, and, consequently, that the boron in this acid wants electrons more than in the others.
Okay! We’re done now right? (Yes, I’m exhausted, too.) Nope, we’ve still got to talk about steric effects, remember?
(I’m sure you didn’t forget. I kind of wish I had, ehehe…)
Adding increasingly bulky substituents to an acid or a base always decreases its strength, regardless of the acid or base you’re talking about. This plays havoc with actual acid-base reactions, making them a little more complicated than we sleep-deprived, chocolate-fueled students would like.
I’ve used my professor’s examples so far, so why stop now?
Let’s say you have two bases, pyridine and 2-methylpyridine. For reference, here are pictures, in that order:
Now, let’s say you try to react these with both B(CH3)3 and H+. Would there be a difference in reactivity? Well, of course, because otherwise I wouldn’t have posed the question.
Yup. Turns out that the 2-methylpyridine much prefers to react with the proton. Why is that? Well, because the proton is small, and it won’t bump into that methyl group the same way that the B(CH3)3 would.
This can also manifest itself in “central atom effects” of a sort. (I’m definitely not coming up with my own example here, ’cause I don’t feel that confident about this.) Let’s say you had 2,6-dimethylpyridine. Now, what if I told you that it doesn’t form an adduct with B(CH3)3, but it does with Al(CH3)3. How would you explain that?
The reason is that, while B(CH3)3 is a stronger acid than Al(CH3)3, Al(CH3)3 forms longer bonds, since Al’s orbitals are larger than B’s. This means that, even though the base is really sterically hindered, it can react with Al(CH3)3, since the bond distance is long enough that it doesn’t have to get so close that it starts bumping into things
Finally, a little thought to keep in mind is that, if you tie back your substituents (by putting them in rings, for example), you’ll eliminate the strain problem, just like, if you tie your little brothers to their bedposts, they won’t be able to follow you into the computer room anymore. 🙂
(I really am very tired. So very tired.)
Okay, that wasn’t too hard, was it? In fact, one could even say that it was pretty… (No. I can’t do this. I’m sorry. I’m not that tired.)
All right! Now, while we’re walking down memory lane, it’s time to look at another aspect of Gen Chem: solutions! First, though, we might take a detour to figure out just how semiconductors tick. Because, you know. Apparently computers are important.
I can’t think straight enough to do a plug, so, instead, here’s a video of a rap song that teaches you the days of the week, courtesy of Scout Van Fleit.