Ten Chemistry questions

[Banedon]

The following are 10 questions that sprung at me while I was reading my textbook. They’re all my own (essentially), and they’re all beyond my knowledge to answer. Can anyone help? You need not answer all 10 (though if you can, that’d be very much appreciated!), and even guesses are helpful (I’ve found that ‘guesses’ often lead to new insights). My heartfelt thanks to anyone who can help me.

1. My Chemistry textbook likened entropy to shaking a box with 20 coins in it. It’s much more likely to get a random assortment of heads and tails then to get a perfectly ordered set of 20 heads (or tails). Let’s say then that the result of one such experiment is:

HHTTHTHTHTTTHTHTTTHH

What if I’d specified beforehand that the configuration I expected was exactly this – the first coin showing heads, the second heads too, the third tails and so on? Then this would be perfectly ordered, wouldn’t it? How can something that’s perfectly ordered in one view be disordered in another? Would it not then be impossible to assign entropy to anything?

In more mathematical terms, the equation to determine entropy is S = k * lnW, where k is the Blotzmann’s constant and W is the number of ways where a particular configuration can be achieved. But there’s only one way where I can achieve the above configuration, so it has zero entropy – and the same applies to every other configuration. Does anyone have an explanation?

2. Here’s a quote from my Chemistry textbook: “Actually, the MO (Molecular Orbital) diagram for Oxygen and Fluorine should have the sigma-2p orbital a bit lower in energy than the pi-2p orbitals.” It mentioned also that the sigma-2p orbital in Nitrogen is higher in energy than the pi-2p orbitals. Why? The Chemistry I know of the Organic molecules indicate that double and triple bonds are formed from sigma and pi bonds, where the sigma bond is significantly stronger than the pi bond. If that’s the case, shouldn’t the electrons in the sigma bond be significantly lower in energy than those in pi bonds?

3. In their ground states Copper has the electronic configuration [Ar]3d104s1 while Zinc has [Ar]3d104s2. In other words, both of them have full-filled inner shells with either one or two valence electrons. That’s comparable to the elements of Group IA and IIA. But elements from Group1A are more reactive than those from Group IIA. Why then is Copper less reactive than Zinc?

4. Consider the reaction:

Zn(s) + Cu2+(aq) --> Zn2+ (aq) + Cu(s)

We calculate the equilibrium constant to be 2x10^37 at 25 degrees Celsius. Therefore, when the concentration of Zinc ions is 1 mol/liter, the concentration of Copper (II) ions is less than 10^(-37) mol/liter. However, a concentration that small is ridiculously small, so small that in a solution of one liter there’d be less than 1^(-10) Copper ions in the solution. Since atoms can’t be divided like this, the true concentration would be zero! Does this mean theory has broken down, and if it does, where is the limit?

5. Here are two reactions lifted directly from my textbook.

H2O2(aq) + 2H+(aq) + 2e- --> 2 H2O(l) (Standard Reduction Potential = +1.78V at 25 Celsius)
O2(g) + 2H+(aq) + 2e- --> H2O2(aq) (Standard Reduction Potential = +0.70V at 25 Celsius)

Adding these two gives

O2(g) + 4H+(aq) + 4e- --> 2H2O(l) (Standard Reduction Potential = +2.48V at 25 Celsius)

That makes Oxygen a stronger oxidizing agent than Ozone, and it is plainly impossible (in fact, the listed value for the same reaction is +1.23V). Is there an error? If so, where is it?

6. The first acid-dissociation constant for Phosphorous Acid (H3PO3) is 1.0x10^-2 (a value lifted from my textbook). The first acid-dissociation constant for Phosphoric Acid (H3PO4) is 7.5x10^-3. Why is Phosphorous Acid a stronger acid than Phosphoric Acid? Is this explainable in terms of molecular structure? The addition of an extra Oxygen atom (which has high electronegativity) ought to stabilize the resulting anion after the release of the acidic proton.

7. Here are two more reactions coming from the textbook:

S2O82-(aq) + 2e- --> 2SO42-(aq) (Standard Reduction Potential = +2.01V at 25 Celsius)
2H2O(l) --> O2(g) + 4H+(aq) + 4e- (Standard Reduction Potential = -1.23V at 25 Celsius)

Adding gives:

2S2O82-(aq) + 2H2O(l) --> 2SO42-(aq) + O2(g) + 4H+(aq) (Standard Reduction Potential = +0.78V)

This means the persulphate ion is not stable in aqueous solutions (if I’m not mistaken). It also means that this ion cannot take part in any reactions, because it won't exist once it's prepared. This isn't the case, however. Why?

8. Is it possible to say that ionic bonds and covalent bonds are actually derivatives of the same type of bond, and the only difference between the two is their percentage ionic character? Why do we say Sodium Hydride is an ionic compound (difference in electronegativity = 0.8) whereas Boron Fluoride isn’t (difference in electronegativity = 2)?

9. I’ve a solution of Copper (II) Sulphate. It’s blue in color. Why? When I shine white light on it, does the solution reflect blue light, or does it absorb it? What happens to the other colors? From what I understand, coloring occurs because electrons can absorb photons and jump up one level. They release photons when they jump back down. If these electrons absorb blue light and jump up, then every other color must pass unhindered, and if we’re opposite a white light source with a bottle of Copper (II) Sulphate between us, we’d perceive it as every color other than blue (which is not the case). On the other hand, if these electrons absorb every other color and jump up, then when they jump down this color must be released, and we’d still see the solution as every color other than blue (which is transmitted). But both these situations are illogical. Where’s the error?

10. Another quote from my textbook: “Oxygen atoms in the upper atmosphere combine as a result of collisions involving some third molecule M:

O(g) + O(g) + M(g) --> O2(g) + M(g)

In the atmosphere, M is most likely N2, but in principle it could be any atom or molecule. The role of M is to carry away the energy that is released when the O-O bond is formed. If M were not involved in the collision, the two oxygen atoms would simply bounce off each other.”

I can’t understand the explanation. Energy is of course released when the O-O bond is formed, but can’t this energy be used to accelerate the resultant Oxygen molecule? Must the extra energy be used to accelerate the third molecule M?

*Acknowledgements: The textbook in question is CHEMISTRY, written by John McMurry and Robert C. Fay, both of Cornell University (at least, this was the case 10 years ago when the book was published), and published by Prentice Hall International Editions.

[DaemianLucifer]

I hate chemistry

But the first question is more of a mathematical one.Its only about the chance and a lucky guess.Youre saying that the entropy is zero because you guessed that it would look like this.But thats not true,since you didnt know what the result would be.It could very well be something different.If someone wins on a lottery you cannot say that his chances of winning were 100% because he won.Chances are determined before the outcome becomes known,not after.

[Pitsu]

Since Corribus, who knows most of these things probably better than me, hasn't showed up yet, i will give a try to some.

1. My Chemistry textbook likened entropy to shaking a box with 20 coins in it. It’s much more likely to get a random assortment of heads and tails then to get a perfectly ordered set of 20 heads (or tails). Let’s say then that the result of one such experiment is:

In more mathematical terms, the equation to determine entropy is S = k * lnW, where k is the Blotzmann’s constant and W is the number of ways where a particular configuration can be achieved. But there’s only one way where I can achieve the above configuration, so it has zero entropy – and the same applies to every other configuration. Does anyone have an explanation?

I do not understand the question here. Shaking a box with 20 coins can give 21 outcomes (assuming that the coins are identical). What the book was trying to say is that although it is possible for all coins (atoms) act the same (like all oxygen in a room concentrating into one corner) it is not very likely to happen. The more coins/atoms the more likely that their actions "neutralize" each other.

2. Here’s a quote from my Chemistry textbook: “Actually, the MO (Molecular Orbital) diagram for Oxygen and Fluorine should have the sigma-2p orbital a bit lower in energy than the pi-2p orbitals.” It mentioned also that the sigma-2p orbital in Nitrogen is higher in energy than the pi-2p orbitals. Why?

Maybe they meant that higher energy is needed to break the sigma bond?

Therefore, when the concentration of Zinc ions is 1 mol/liter, the concentration of Copper (II) ions is less than 10^(-37) mol/liter. However, a concentration that small is ridiculously small, so small that in a solution of one liter there’d be less than 1^(-10) Copper ions in the solution. Since atoms can’t be divided like this, the true concentration would be zero! Does this mean theory has broken down, and if it does, where is the limit?

Concentration 10^-10 ions/l means in other words that you have 1 ion per each 10^10 litres. If you take a small volume, say a liter out of that, you may not fish out that ion (conc in this 1 l volume is zero) or with some lower probability catch it (actual conc 1 ion/liter). If the "20 heads" coincidence happens, you may even have 2 or more Cu ions in this 1 liter. With so low solubility you cannot say for sure the actual concentration in small volumes, but you can give the average (even if it sounds like splitting an atom) or calculate the probability for having 0, 1, 2 or more ions.

6. The first acid-dissociation constant for Phosphorous Acid (H3PO3) is 1.0x10^-2 (a value lifted from my textbook). The first acid-dissociation constant for Phosphoric Acid (H3PO4) is 7.5x10^-3. Why is Phosphorous Acid a stronger acid than Phosphoric Acid? Is this explainable in terms of molecular structure? The addition of an extra Oxygen atom (which has high electronegativity) ought to stabilize the resulting anion after the release of the acidic proton.

The more stable anion, the stronger acid, right? In H3PO3 is it the phosphor bound H that dissociates first? I could imagine that the OH hydrogens are less acidic than PH hydrogen (H3PO4 has no such).

2S2O82-(aq) + 2H2O(l) --> 2SO42-(aq) + O2(g) + 4H+(aq) (Standard Reduction Potential = +0.78V)

This means the persulphate ion is not stable in aqueous solutions (if I’m not mistaken). It also means that this ion cannot take part in any reactions, because it won't exist once it's prepared. This isn't the case, however. Why?

Isn't this a question of kinetics? Many things around us, for example life as such, are very far from energetically stable state, yet exists. I would guess that S2O82- solution from AD 1876 would have more SO42- than the original anion, but a solution made yesterday might be well S2O8 2- solution.

9. I’ve a solution of Copper (II) Sulphate. It’s blue in color. Why?

Cu2+ with water "coat" has its orbitals stretched so that from one level to another is about the energy of orange light. (CuSO4*5H2o crystals are blue while CuSO4 itself is white) Since others except blue light is absorbed, blue tones are overwhelming. Some energy is used for work -electron transfer, therefore emitted light has always longer wavelength than the absorbed one was. Additionally, emission is usually not into the same direction than absorbed/passing through/reflected waves go, and is more scattered and less visible.

I will have to check others later

[DaemianLucifer]

I do not understand the question here. Shaking a box with 20 coins can give 21 outcomes (assuming that the coins are identical). What the book was trying to say is that although it is possible for all coins (atoms) act the same (like all oxygen in a room concentrating into one corner) it is not very likely to happen. The more coins/atoms the more likely that their actions "neutralize" each other.

Actually,there are 2^20 different outcomes.

[Gaidal Cain]

The following are 10 questions that sprung at me while I was reading my textbook. They’re all my own (essentially), and they’re all beyond my knowledge to answer. Can anyone help? You need not answer all 10 (though if you can, that’d be very much appreciated!), and even guesses are helpful (I’ve found that ‘guesses’ often lead to new insights). My heartfelt thanks to anyone who can help me.

I'm a physics student, but these seems to mostly be about psysical chemistry, so I can answer some. Corribus is welcome to correct me if I go wrong

What if I’d specified beforehand that the configuration I expected was exactly this – the first coin showing heads, the second heads too, the third tails and so on? Then this would be perfectly ordered, wouldn’t it? How can something that’s perfectly ordered in one view be disordered in another? Would it not then be impossible to assign entropy to anything?

You can only define one "ordered" state. The chance to achieve exactly that state would be 1 in 2^20, which is approximately 1 in a million...

In more mathematical terms, the equation to determine entropy is S = k * lnW, where k is the Blotzmann’s constant and W is the number of ways where a particular configuration can be achieved. But there’s only one way where I can achieve the above configuration, so it has zero entropy – and the same applies to every other configuration. Does anyone have an explanation?

If think you have it wrtong. If W would be the total number of different states, and each have the same probability, the above equation holds.

2. Here’s a quote from my Chemistry textbook: “Actually, the MO (Molecular Orbital) diagram for Oxygen and Fluorine should have the sigma-2p orbital a bit lower in energy than the pi-2p orbitals.” It mentioned also that the sigma-2p orbital in Nitrogen is higher in energy than the pi-2p orbitals. Why? The Chemistry I know of the Organic molecules indicate that double and triple bonds are formed from sigma and pi bonds, where the sigma bond is significantly stronger than the pi bond. If that’s the case, shouldn’t the electrons in the sigma bond be significantly lower in energy than those in pi bonds?

I believe that has something to do with spin-coupling. Basically, for some atoms, it's more favourable to place electrons with the same spin than to have electrons in the lowest possible orbital...

This means the persulphate ion is not stable in aqueous solutions (if I’m not mistaken). It also means that this ion cannot take part in any reactions, because it won't exist once it's prepared. This isn't the case, however. Why?

Probably a potential barrier. Even if it's unstable, it might require more energy to go over some kind of reaction barrier before it seize to exist. For example, this is true for diamonds, which in normal conditions should dissolve into grpahite. However, there is such a huge barrier that this doesn't actually happen.

8. Is it possible to say that ionic bonds and covalent bonds are actually derivatives of the same type of bond, and the only difference between the two is their percentage ionic character? Why do we say Sodium Hydride is an ionic compound (difference in electronegativity = 0.8) whereas Boron Fluoride isn’t (difference in electronegativity = 2)?

Correct.

9. I’ve a solution of Copper (II) Sulphate. It’s blue in color. Why? When I shine white light on it, does the solution reflect blue light, or does it absorb it?

A guess is that it's able to absorb yellow light and then go down in potential energy without emitting light, or that it can go down in another manner so quickly that it sends out far less than it sends out. It would absord yellow since that's the complementary color of blue (take out yellow of white light and it turns blue).

I can’t understand the explanation. Energy is of course released when the O-O bond is formed, but can’t this energy be used to accelerate the resultant Oxygen molecule? Must the extra energy be used to accelerate the third molecule M?

I think it's due to the fact that you need conservation of momentum as well as of energy. In fact, conservation of momentum is the tougher condition as the Oxygen molecule could convert kinetic energy to internal vibration or rotation energy...

[Corribus]

Banedon I have read your questions and can answer them, but it may take me some time as this morning is quite busy.

[Corribus]

Banedon here are your answers. Feel free to ask followups if these are not clear. I wrote them quickly. I did not read others' answers so apologize if I duplicated.

These are all good, perceptive questions. Perhaps you have a chemist hiding inside you.

My Chemistry textbook likened entropy to shaking a box with 20 coins in it. It’s much more likely to get a random assortment of heads and tails then to get a perfectly ordered set of 20 heads (or tails). Let’s say then that the result of one such experiment is:

HHTTHTHTHTTTHTHTTTHH

What if I’d specified beforehand that the configuration I expected was exactly this – the first coin showing heads, the second heads too, the third tails and so on? Then this would be perfectly ordered, wouldn’t it? How can something that’s perfectly ordered in one view be disordered in another? Would it not then be impossible to assign entropy to anything?

In more mathematical terms, the equation to determine entropy is S = k * lnW, where k is the Blotzmann’s constant and W is the number of ways where a particular configuration can be achieved. But there’s only one way where I can achieve the above configuration, so it has zero entropy – and the same applies to every other configuration. Does anyone have an explanation?

The reason for your difficulties is already given in your question. If you specify the order, then that is against the entropic law. For example, there is no difference, entropically (or “orderedly”) between the result

(1) HHHHHHHHHHHHHHHHHH

And

(2) HHTTHTHTHTTTHTHTTTHH

If you had twenty coins and you could label them somehow 1-20, then there is only ONE possible way for them to be all heads. But there’s only ONE possible way for them to show up in the second combination as well. Because you have SPECIFIED that assignment, it is perfectly ordered.

In reality, though, coins are not labeled. Consider the alternate result:

(3) THTTHTHTHTTTHTHTTHHH

Result 3 and result 2 have the same number of heads and tails. They are the same “result”, but they are different because the specific assignments are different.

If you were to work out all the possible combinations that give rise to each result, you’d find that there was a Gaussian distribution. For example, if you assigned heads a value of 0 points and tails a value of 1 point, and shook the box a zillion times and counted the total points in the box, you’d find a Gaussian distribution where the most probable result is the one where half the coins are heads and half are tails, because there are the most ways to get that result.

As for the equation – you have to remember that molecules are in this experiment indistinguishable from one another. As are the coins. Because there is no difference between case 2 and 3, that is the higher entropy state because they are the SAME state – more options for the same state = entropically favored state. When you label the coins, the results are no longer equivalent, so the entropic argument no longer holds.

2. Here’s a quote from my Chemistry textbook: “Actually, the MO (Molecular Orbital) diagram for Oxygen and Fluorine should have the sigma-2p orbital a bit lower in energy than the pi-2p orbitals.” It mentioned also that the sigma-2p orbital in Nitrogen is higher in energy than the pi-2p orbitals.

The answer to this question is actually quite complicated and is usually simply accepted as a truth in General Chemistry. The cause is two related phenomena called “penetration” and “shielding” and has to do with the shapes of p and s orbitals. The shapes of S orbitals are actually quite a bit more complicated than the simple spheres that are often shown in General Chemistry. In fact, they consist of spherical “shells” with a number of “nodes” that depends on the principle quantum number n. The same goes for p orbitals, too, but for a given n, the s orbital has better penetration near the nucleus than p orbitals, which is why p orbitals fill up after s orbitals in a multielectronic atom (since an electron in an ns orbital is on average closer to the nucleus than an np orbital, that’s why s orbitals are lower in energy, that is, more favorable). As you go across a period (or more specifically, as the nuclear core charge increases), this difference in penetration becomes more important to the relative energies of these orbitals, and (Point#1) the 2s and 2p orbitals in the multielectron atoms become further apart in energy. Point #2: because they have similar symmetries, the sigma-2S and sigma-2p orbitals can interact. The result is a lowering of the sigma-2s orbital and a raising of the sigma-2p orbital, which in many of the homonuclear diatomics reverses the expected order of the sigma-2p and pi-2p orbitals. Why only some of them? This interaction (the degree of lowering and raising) is dependent on the energy gap between them, and is thus dependent on the energy gap between the constituent 2s and 2p atomic orbitals. As mentioned above, this gap becomes larger across the period, and so the interaction is larger on the left side of the periodic table than the right side… which is why at some point (nitrogen) the ordering will flip. Now that was probably completely incomprehensible and I apologize – it’s much easier to show with graphics.

3. In their ground states Copper has the electronic configuration [Ar]3d104s1 while Zinc has [Ar]3d104s2. In other words, both of them have full-filled inner shells with either one or two valence electrons. That’s comparable to the elements of Group IA and IIA. But elements from Group1A are more reactive than those from Group IIA. Why then is Copper less reactive than Zinc?

Answer: transition metals are funny. You can’t directly compare Group 1A and IIA with transition metals, even though they nominally have the same valence orbitals. Those s electrons often interchange with the d electrons under a lot of different circumstances. E.g. Copper could easily go to [Ar]3d94S2. It’s hard to predict trends among transition metals. Each column is very different from the other. Spin states often play a role. People can dedicate their whole careers to studying the chemistry of one metal, which may not be applicable to any other metal. Sorry that’s not much of an answer for you, but transition metals are very tricky to understand in any sort of general way. Easier to predict trends down a group than across a period.

4. Consider the reaction:

Zn(s) + Cu2+(aq) --> Zn2+ (aq) + Cu(s)

We calculate the equilibrium constant to be 2x10^37 at 25 degrees Celsius. Therefore, when the concentration of Zinc ions is 1 mol/liter, the concentration of Copper (II) ions is less than 10^(-37) mol/liter. However, a concentration that small is ridiculously small, so small that in a solution of one liter there’d be less than 1^(-10) Copper ions in the solution. Since atoms can’t be divided like this, the true concentration would be zero! Does this mean theory has broken down, and if it does, where is the limit?

Equilibrium constants (rate constants) are not meant to be used this way. They are related more to probabilities. For instance if you calculate that there should be 10^-10 Cu ions in solution, that doesn’t mean that in any given solution you’re going to find a small fraction of a copper ion. That means ON AVERAGE there are 10^-10 ions in solution at a given time. Put another way, let’s say you had 10^10 identical such solutions. On average, at any given time ONE of those solutions would have a single free copper ion; the rest of them would have zero, giving you an average of 10^-10 ions per solution. At some time later, a different solution might have the one free ion. Remember, this is a dynamic process (equilibrium), not a static one. Things are constantly in flux.

5. Here are two reactions lifted directly from my textbook.

H2O2(aq) + 2H+(aq) + 2e- --> 2 H2O(l) (Standard Reduction Potential = +1.78V at 25 Celsius)
O2(g) + 2H+(aq) + 2e- --> H2O2(aq) (Standard Reduction Potential = +0.70V at 25 Celsius)

Adding these two gives

O2(g) + 4H+(aq) + 4e- --> 2H2O(l) (Standard Reduction Potential = +2.48V at 25 Celsius)

That makes Oxygen a stronger oxidizing agent than Ozone, and it is plainly impossible (in fact, the listed value for the same reaction is +1.23V). Is there an error? If so, where is it?

I will answer this question in two parts. Part one is, why don’t your numbers add up? The easiest way to answer this is to consider that the added equation (rxn 3) involves the direct reduction of oxygen by 4 electrons simultaneously. The first two involves the reduction of oxygen, but in smaller 2 electron steps. The latter is a more favorable and it is the one that most often occurs in biology (using enzymes). In fact, there are also 1 electron pathways as well. Analogy: consider if someone gives you four burgers to eat. Now you COULD eat them all at once, or you could eat two of them, wait, and eat two more. The end result is the same, but which is easier? Obviously, it is easier to eat the meal in smaller chunks rather than cramming 4 burgers in your mouth at the same time. Thus you don’t have a numerical error – it’s simply a conceptual one. You are assuming that the two processes are the same. They aren’t, though the end results are.

Part 2 is about the oxidizing capacity of dioxygen is indeed a very strong oxidant, which is why it is also (although many people don’t realize it) a toxic chemical. It is dangerous to breathe pure oxygen for extended periods of time as it can damage your mucous membranes, among other things. Additionally, it’s also, as you show, very reactive. So why is it that if you mix dioxygen with a fuel (say, octane), a source of ignition is required, even though the reaction is thermodynamically favored? Again, the reason is more complicated than what you’d find explained in a General Chemistry textbook. Dioxygen is a ground-state triplet (i.e., it has unpaired electron spins). Most organic molecules are ground-state singlets (they have paired electron spins). For any chemical or physical process, total spin needs to be conserved (which is really the same as saying momentum needs to be conserved). Thus the reactions above, though they are thermodynamically allowed, have a very large kinetic barrier to reaction. For example, you can write your reaction as:

Triplet -> Singlet.

A typical combustion reaction such as O2 + CH4  CO2 + H2O, which is obviously exothermic and SHOULD be spontaneous (although in practice we know it isn’t – it requires heat to go) can be rewritten as

Triplet + Singlet -> Singlet + Singlet

These reactions violate conservation of momentum, and so they are slow, despite the large thermodynamical driving force. This is actually problematic for many biological reactions, and nature has developed enzymes which actually speed up these types of reactions by forcibly removing an electron from dioxygen prior to allowing it to react with the substrate. Another interesting tidbit is that some cancer therapeuticals operate by a technique called photodynamic therapy (PDT). They use light to generate excited species called “triplet sensitizers”. These species are singlets in the ground-state but when absorbing light photons, they generate excited states that are triplets. These triplets interact with ground-state dioxygen (also a triplet) to form excited-state oxygen, which is a singlet, via an equation like:

A(singlet) + light -> A*(triplet) (actually, it’s more like: A(singlet) + light -> A*(singlet) -> (slow) A*(triplet)

A*(triplet) + O2(triplet) -> A(singlet) + O2(singlet).

The latter reaction does not violate conservation of momentum because two triplets are allowed to react with one another to give two singlets by a process called triplet-triplet annihilation. Anyway, “singlet oxygen” is an extremely reactive species that easily tears apart just about any organic substrate that it comes in contact with, because the spin kinetic barrier is no longer present. So what you do is localize the triplet-sensitizer in the area of a tumor, shine light on it, and zap – no more tumor cells. It’s usually only really useful for surface tumors, though, because light doesn’t penetrate very far into live tissue.

The slow triplet-singlet process is also demonstrated by the phenomenon known as phosphorescence. Glow-in-the-dark toys use this principle. You generate a photoactivated triplet-state using a light source, and the triplet-state returns to the singlet ground-state by emitting photons (that’s the glow part), but since the process violates momentum conservation, the process is slow, which is why they continue to glow for minutes or even hours after you turn the lights off.

If you can track down a copy of Bioinorganic Chemistry by Bertini, Lippard, Gray Valentin, there is a lengthy section about this in there that you may find interesting.

6. The first acid-dissociation constant for Phosphorous Acid (H3PO3) is 1.0x10^-2 (a value lifted from my textbook). The first acid-dissociation constant for Phosphoric Acid (H3PO4) is 7.5x10^-3. Why is Phosphorous Acid a stronger acid than Phosphoric Acid? Is this explainable in terms of molecular structure? The addition of an extra Oxygen atom (which has high electronegativity) ought to stabilize the resulting anion after the release of the acidic proton.

My guess would be because phosphorous acid has some tautomeric forms which phosphoric acid does not. H3PO3 is actually best written as HP(O)(OH)2. The tautomeric form is P(OH)3. Usually the more tautomeric forms in the conjugate base, the stronger the acid. That’s just a guess though. The difference in pKA isn’t all that great, though, so it’s not really that big of ann effect.

7. Here are two more reactions coming from the textbook:

S2O82-(aq) + 2e- --> 2SO42-(aq) (Standard Reduction Potential = +2.01V at 25 Celsius)
2H2O(l) --> O2(g) + 4H+(aq) + 4e- (Standard Reduction Potential = -1.23V at 25 Celsius)

Adding gives:

2S2O82-(aq) + 2H2O(l) --> 2SO42-(aq) + O2(g) + 4H+(aq) (Standard Reduction Potential = +0.78V)

This means the persulphate ion is not stable in aqueous solutions (if I’m not mistaken). It also means that this ion cannot take part in any reactions, because it won't exist once it's prepared. This isn't the case, however. Why?

I’m not sure about the answer here, other than to say that you should be careful when trying to add electrochemical reactions together and then apply the results to real situations, as in the above example. These are not half-reactions. They are full reactions. The thermodynamics change when more electrons are involved.

8. Is it possible to say that ionic bonds and covalent bonds are actually derivatives of the same type of bond, and the only difference between the two is their percentage ionic character? Why do we say Sodium Hydride is an ionic compound (difference in electronegativity = 0.8) whereas Boron Fluoride isn’t (difference in electronegativity = 2)?

It’s something of a spectrum. There is always SOME sharing going on, although often the difference in electronegativity (which is something of an arbitrary scale) is so large that it is OK to assume that the electrons exist solely on one species or another. Be careful with sodium hydride. Hydrogen is a weird guy and chemists have been arguing ever since mendeleev invented the periodic table whether hydrogen belongs above lithium or above fluorine. The fact that it’s a hydride (H-) should be a clue that something funky is going on. In general, electronegative differences is a good guide to whether a bond is “ionic” or not, but not always.. because some species can exist in multiple oxidation states.

9. I’ve a solution of Copper (II) Sulphate. It’s blue in color. Why? When I shine white light on it, does the solution reflect blue light, or does it absorb it? What happens to the other colors? From what I understand, coloring occurs because electrons can absorb photons and jump up one level. They release photons when they jump back down. If these electrons absorb blue light and jump up, then every other color must pass unhindered, and if we’re opposite a white light source with a bottle of Copper (II) Sulphate between us, we’d perceive it as every color other than blue (which is not the case). On the other hand, if these electrons absorb every other color and jump up, then when they jump down this color must be released, and we’d still see the solution as every color other than blue (which is transmitted). But both these situations are illogical. Where’s the error?

Metal compounds are colored because of absorption. When white light (light that contains light of all frequencies) passes through a solution, light of a certain frequency is absorbed, which means that light that we observe includes all frequencies EXCEPT the frequencies that were absorbed. We interpret this as color. (Some compounds are also colored because of light scattering and because of emission or fluorescence, but this is usually not the case in transition metal complexes). In this case, you are seeing absorption that arises because of electrons in lower-energy metal d-orbitals being transferred to higher-energy metal d-orbitals after absorbing energy from light photons, and thus these transitions are usually called d-d transitions. They are usually weak because they violate momentum conservation, and so these compounds, unless they are very concentrated, usually are weakly colored. You can determine what color light is being absorbed by using a color wheel. If the solution LOOKs blue, then actually yellow (I believe) light is being absorbed, because when you observe every color at once except yellow, it appears blue to your eye. It’s a tricky mind-puzzle thing that I still have to stop and think about every now and then. You can convince yourself of this by taking an absorption spectrum and seeing exactly what wavelengths are being absorbed. Blue = high energy and Red = low energy. But compound which absorb red wavelengths of light do not LOOK red. It’s a confusing distinction.

Also note that often times solutions of metal salts are not always the same color for identical metals. This is because the d-orbitals are spaced differently depending on the nature of the other “ligands” involved. Thus the exact frequencies of light that are absorbed can be different.

10. Another quote from my textbook: “Oxygen atoms in the upper atmosphere combine as a result of collisions involving some third molecule M:

O(g) + O(g) + M(g) --> O2(g) + M(g)

In the atmosphere, M is most likely N2, but in principle it could be any atom or molecule. The role of M is to carry away the energy that is released when the O-O bond is formed. If M were not involved in the collision, the two oxygen atoms would simply bounce off each other.”

I can’t understand the explanation. Energy is of course released when the O-O bond is formed, but can’t this energy be used to accelerate the resultant Oxygen molecule? Must the extra energy be used to accelerate the third molecule M?

Right. Another question that is really (way) beyond the scope of a general chemistry course. You have to get into potential energy surfaces to really explain it well. Say you have two oxygen atoms and they approach one another. Depending on their distance apart, there is a characteristic electrostatic (potential) energy that depends on the nuclear and electronic charges. The point at which they form “a molecule” is not easily defined. However, in general, as you bring the oxygens close together, the potential energy gradually drops until it reaches a minimum value (the equilibrium bond distance), at which point the potential energy rises steeply to infinity (because the two atoms cannot lie directly on top of each other). If you plot this potential energy as a function of R (the distance between the two), this is called a potential energy surface. Such a plot is often (to a first approx) modeled by a mathematical function called the Morse Function. You can find a graphical depiction here: http://en.wikipedia.org/wiki/Morse_potential

Now, the total energy of the system is equal to the kinetic energy plus the potential energy. The potential energy we have already given by the Morse function. The two atoms which have a combined kinetic energy KE start off at long distances from each other. As they approach each other, the potential energy gradually drops, and so (because energy is conserved) they begin to speed up. They will not loose energy unless something happens to make them loose energy. Even when they are at a distance that they can be considered to be “a molecule”, the overall KE and PE must add up to the same value. So they speed up and speed up and even once the PE starts to rise again and they slow down, they have so much KE that they bounce off the potential barrier and continue to their trajectory in the opposite direction. I.e., they do not get trapped in the “well”. The only way they can become trapped is by crashing into something else, which can take a chunk of that energy away, so that when the return away from the short-R potential barrier, they no longer have enough energy to go on to infinite separation again. At this point, the two atoms are in a “bound vibrational state” and they begin oscillating back and forth with a frequency that depends on how much total kinetic energy they have remaining. Gradually, more collisions will generally sap even this excess KE and the molecule will end up at the bottom of the potential well, at which point they are a stable dioxygen molecule.

Again, this is much easier to show with diagrams, in person. But it’s a good question. The reason that the excess energy can’t just go to the oxygen molecule itself, is because to become a bound molecule, it needs to loose enough excess energy so that the coulombic attraction between them is stronger than their independent translational motions. That’s basically the definition of a stable molecule. This question pretty much defines the basis of quantum chemistry of molecules.

[Corribus]

p.s. These are such good questions that I will be saving them for use with my future students. Thanks.

[Gaidal Cain]

Hmm. My Chemistry is obviously not what it should be... I was just reminded again why I'm not studying it- to many details here and there and not many general ideas that hold almost everywhere...

[ThunderTitan]

Hmm. My Chemistry is obviously not what it should be... I was just reminded again why I'm not studying it- to many details here and there and not many general ideas that hold almost everywhere...

In other words it's too much information for you...

[Banedon]

Thanks everyone (especially you, Corribus! I can't thank you enough!), I'll save this webpage and absorb it all, then see if there's any follow up questions.

Thanks again - it'll definitely help my understanding a whole lot.

[Kalah]

OK, next question should be something in the region of the evolutionary significance of ribosomes, the underlying causes of the first world war or the uses of mediaeval written vs. other sources. Then I might stand a chance.

[Caradoc]

Or maybe antique computer languages, baseball, or people I went to school with...

[Gaidal Cain]

I'm happy if it's about something that makes sense (i.e. physics or maths)...

[theLuckyDragon]

Okay, then give me the answer to the following question. I think I know the concept inolved and the things that are happening, but I never quite found a way of putting it into words. It's a highschool-level question, so it ought to be simple.

"Why is it that when we exhale with the lips close together, cold air emerges, but when we exhale with the mouth wide open, warm air emerges?"

[Gaidal Cain]

The temperature of the breath is the same, but if you test it with your skin, holding it close as you breathe with just a slightly open mouth will blow away a thin layer of hot air around your skin. Breathing with your mouth open will have the air leaving much slower (OK, there's some physics involved here). It's the same principle as for why wind feels chilly or a fan helps against the heat.

[theLuckyDragon]

Thought so! Thanks, but can you give an explanation in more scientific terms? Detail that "physics involved here", if you don't mind?

[Gaidal Cain]

Well, this is mainly fluid dynamixs, which I'm not particularly good at, and which gets terribly complicated if you try to solve it in any form beyond very low speeds and very simple geometry.

Anyway, it should be clear that the air that comes from inside you should be at about body temperature, yes? That measn it's somewhat warmer than the temperature of your skin, especially around your hands (which you'd normally use to test this effect). That explains why it's "hot" in itself.
It's also easy to understand that the air near your skin should be somewhat warmer than the room (in most cases). I'm not sure why the faster air would not simply replace this, but I have an idea: When you increase the velocity of the stream of air from your mouth, it will "suck in" air from around and become colder, perhaps even to the point where it gets colder than the air around your skin (there do exist such mechanisms, though I'm not sure if they are large enough at these speeds).

[Corribus]

Okay, then give me the answer to the following question. I think I know the concept inolved and the things that are happening, but I never quite found a way of putting it into words. It's a highschool-level question, so it ought to be simple.

"Why is it that when we exhale with the lips close together, cold air emerges, but when we exhale with the mouth wide open, warm air emerges?"

The answer to your question is quite simple, and is the same as "why does a stiff breeze cool you, even on a hot day?"

Briefly: the stream of air reduces the local pressure above your skin, particularly the vapor pressure of water. This sudden depression in water vapor pressure makes evaporation of water molecules on the surface of your skin (sweat) favorable. Evaporation requires heat energy to happen (endothermic) which is taken from your skin, which you perceive as cooling. The drop in vapor pressure is proportional to the velocity of the stream of air.

The same principle is used every day by organic chemists with a device called a rotoevaporator.

You can see the effect quite well at home if you use a low-boiling point solvent like rubbing alcohol. Get a small vial and fill it with rubbing alcohol (a few mL), and then blow a stream of air constantly over it for a few minutes (breathing probably won't do it - you need an air can or something) and the vial will get extremely cold.

[theLuckyDragon]

Okay, so that explains the sensation of cool air. But can the same explanation be applied for the sensation of warmth when exhaling with the mouth wide open?