Calculating percentages of microstructures in equilibrium phase diagrams (engineering) So I have an equilibrium phase diagram of steel and I am asked to 'Calculate the proportion of pearlite in the microstructure of 0.4 wt% C steel just below the eutectoid temperature (727 °C).'

I have been using the forumlas

$Wa=\frac{Cb-Co}{Cb-Ca}$%

$Wb=100-Wa$%

$Ca=weight$% of (Carbon in this example) at the first intersection point on the eutectoid line.

$Cb=weight$% of (Carbon in this example) at the second intersection point on the eutectoid line or the bottom of the 'V'. I don't know how to explain any better sorry.

So looking at my diagram, I see that $Ca=0.022wt$% Carbon and $Cb=0.76wt$% Carbon, and $Co$ is just 0.4%.

So $Wa=\frac{0.76-0.4}{0.76-0.022}$%

$=0.487805%$%

This is where I am confused. What exactly is Wa? Is that how much ferrite/austenite/cementite that is in the pearlite then? Or how much pearlite is at that point compared to ferrite? And in any case, what does that mean Wb is?

Any help would be much appreciated. Thanks

Assuming you mean this phase diagram, it looks like your calculations are correct.

Based on my understanding of the diagram, at $727\ ^\circ \text{C}$, you have 100% pearlite at $C_b = 0.76\ \%C$ and 0% pearlite at $C_a=0.022\ \%C$. $W_a$ then appears to be how far $C_o$ is between $C_a$ and $C_b$, so $W_a$ should be the amount of pearlite.

The units of $C_a$, $C_b$, and $C_o$ are $\% C\  by\  mass$. That means when you calculate $W_a$ initially, you are calculating the fraction of pearlite and not the percent of pearlite. All of the $\% C\  by\  mass$ units cancel:

$$W_a=\frac{0.76\ \%C-0.4\ \%C}{0.76\ \%C - 0.022\ \%C}=\frac{{\%C}(0.76-0.022)}{{\%C}(0.76-0.022)}=0.487805\implies48.7805\%$$


Work done by irreversible expansion of gas against piston of fixed pressure this has been puzzling me for a while.  An ideal gas expands against a piston with mass M and area A, so that pressure external Pex= Mg/A.  The initial internal pressure of the gas is Pino > Pex.  Because the gas is expanding irreversibly, Pin > Pex, until equilibrium is reached, where Pinf  = Pex.

The physical chemistry texts I've read, including Atkins, seem to characterize the work done by the expanding gas as Pex * change in volume.  I recognize this as the equation to calculate work for lifting an object at constant velocity over a given distance.

However, since the initial internal pressure of the gas is greater than the external pressure, there is an imbalance in forces, and the mass will accelerate upwards, causing it to acquire kinetic energy that was derived from the original energy of the gas.  A rocket is an example of gas causing a mass to accelerate upwards.

My question is why the standard formula of work done by an irreversibly expanding gas is Pex * change in volume, and doesn't include a term for the kinetic energy from acceleration.

In an irreversible expansion like this, the aveage pressure of the gas within the cylinder is greater than the pressure acting at the internal piston face $P_{ext} $ (where the work is being performed).  Of course, at the internal piston face, by Newton's 3rd law, the pressure exerted by the gas on the piston face is equal to the pressure exerted by the piston face on the gas.

The reason that, in an irreversible expansion, the average pressure of the gas (averaged over the volume) is greater than the value at the piston faces is that the gas pressure is non-uniform spatially within the cylinder.  This is the result both of mass times acceleration effects within the gas itself (yes, the gas has mass) as well as viscous stresses present within the gas that are proportional, not to the gas volume, but to the rate at which the gas is deforming locally within the cylinder.  So the mechanical behavior of a gas experiencing an irreversible expansion is very different from that of a gas experiencing a reversible (quasi-static) expansion.

Now let's reconsider the force balance on the piston.  If $P_{gas}(t)$ represents the instantaneous pressure exerted by the gas on the piston at time t, then the force balance on the piston becomes:
$$M\frac{dv}{dt}=P_{gas}(t)A-Mg$$If we multiply both sides of this equation by v=dx/dt and integrate with respect to t (from time = 0 when the piston is released), we obtain:  $$\int_0^{x(t)}{P_{gas}Adx}=W(t)=\frac{1}{2}Mv^2-\frac{Mg}{A}[V(t)-V(0)]$$where W(t) is the work done by the gas on the piston up to time t, v is the velocity of the piston at time t, and V(t) is the volume of the gas at time t.

Basically what will happen here at short times is that the piston will rise and gain velocity, but it will overshoot its equilibrium position (like a mass on a spring experiencing simple harmonic motion).  It will then slow down, until it begins moving downward.  But it will again overshoot the equilibrium position.

If the piston were frictionless and the gas were inviscid, this oscillation would continue forever.  However, even if the piston were frictionless, the gas is not inviscid.  So the viscosity of the gas will act as a damping factor, and the piston, rather than oscillating forever, will experience a damped oscillation until if finally comes to a stop at the equilibrium position (i.e., when v = 0).  So the equation for the work will become:  $$W(\infty)=\frac{Mg}{A}(V_{\infty}-V(0))$$
So, even for a piston with mass, the kinetic energy effect will die out and the work will be the same as just the change in potential energy of the piston.


Why is the Maxwell–Boltzmann formula inapplicable at low temperatures? This Wikipedia article states that at sufficiently low temperatures (or high pressure), quantum effects lead to a breakdown of Maxwell–Boltzmann statistics.

I know that the spacing of energy levels becomes larger at low temperatures, but doesn't the derivation of the MB distribution, as shown in the article, already consider energy levels to be discrete?
Furthermore, it also takes the degeneracy of energy levels into account.

So why and what other quantum effects are there at low temperatures that still violate the formula?

Since the explanation was a little more complicated than I initially thought, I figured it would be worth it to combine my comments (and info from Physics SE) into an answer.
Quantum particles satisfy Fermi–Dirac or Bose–Einstein statistics depending on whether they are fermions or bosons. These distributions have the form $$\langle n_i\rangle=\frac{1}{\exp[(\epsilon_i-\mu)/k_bT]\pm1}$$ where the plus/minus is for fermions/bosons.
To consider the high temperature limit, we need to note that not only is there a direct temperature dependence, but $\mu$ is also dependent on temperature. Specifically, in the high temperature limit, $\mu<0$ and $|\mu|>k_bT$. Combining this information, we can make the (very accurate) approximation $$\exp[(\epsilon_i-\mu)/k_bT]\pm1\approx\exp[(\epsilon_i-\mu)/k_bT]$$ as the exponential function will be much larger than $1$. So at high temperatures $$\langle n_i\rangle\approx\frac{1}{\exp[(\epsilon_i-\mu)/k_bT]}$$ which matches the form of the Maxwell–Boltzmann distribution. We can see that at low temperatures these two distributions would not agree, as the additional $\pm1$ term in the denominator would become more significant as the exponential got smaller.
It's important to remember that the particles are always indistinguishable; all we have done in the high temperature limit is made an approximation that simplified the functional form. We should not take this coincidental agreement with the MB distribution (for which the particles are assumed to be distinguishable) to imply that the quantum particles have become distinguishable at high temperatures.


Why conductivity of electrolyte decreases on dilution, but molar conductivity increases? I learnt that on dilution of an electrolytic solution, it's conductivity decreases. However, there is an increase in the molar conductivity of the solution. The book explained the reason by comparing the definition of the two terms. However, it gave a hypothetical example which contradicts the statement it earlier stated. Maybe there is a printing error, but it has got me confused now.

Can someone please explain why the conductivity of an electrolytic solution decreases but molar conductivity increases on dilution, maybe with the  help of a good hypothetical example?

I've just been doing an experiment on this, and my understanding's not that great, but as I understand it the conductivity decreases because you have a lower concentration of charge carriers in the solution. The molar conductivity increases however, because as the charged ions get further apart they interact and slow each other down less. Essentially each ion becomes more able to move through the solution, but the increase in conductivity from this effect is outweighed by the decrease due to there being fewer ions in a given volume.

Personally I find it easier to think about the other way around. As you increase the concentration of the solution you get more charge carrying ions, so the conductivity increases. But because the ions get closer together they begin to interact and slow each other down so the conductivity per mole of the ions is reduced.

I hope you still find this useful and that I managed to explain what I meant


How to calculate the enthalpy of reaction using Gaussian 09? I'm having a hard time trying to understand what's the correct way to calculate the enthalpy of reaction given that I've already calculated the frequencies of my compounds in Gaussian. Below I show the reaction of interest:

As you can see in the figure, experimentally, this reaction leads to the formation of ionic species. As for my calculations, I have all four species optimized (all of them separated, I'm not working with complexes) with their respective frequencies.

As for the enthalpy, I'm using the following formula:

$$
\Delta_\mathrm{R} H=
  \left[(\text{Iodide anion}) + (\text{cationic product})\right]
  - \left[(\ce{FeCpI(CO)2}) + (\text{phosphine reactant})\right]
$$

All terms are the sum of the electronic energy and the thermal enthalpy as given by Gaussian.

Am I doing it correctly? Because I'm not sure, if I have to consider both products together or separated as I'm currently doing. Can you give me some light on this?

In principle your approach is correct. There are a few pitfalls though.

Quantum chemistry is usually not well equipped handling ions, especially anions, within reactions, hence the evaluation of these reaction energies will probably contain some errors. This is mostly due to the fact that you are technically calculating everything in the gas phase, and apply thermal corrections later. In well behaved systems, e.g. organic compounds without charges, that does not matter much. But even then we try to exploit error compensation with (for example) isodesmic reactions.

If you are treating the ions separately, you are neglecting all possible interactions between them, too. That might lead to additional errors. One way to approximate this is to calculate the ion pair together. That might not always lead to the correct result since solvents will also play a role here. Obviously you could also include these.

Another problem for your system is probably conformational space. Especially the phosphine is somewhat flexible and will adopt a few conformations (isolated as well as complexed). Here it is also important to find the global minimum, to evaluate the energies correctly.

It is advisable to look for similar systems in the literature and see what they have done. You may also find benchmark studies that will give you more information on how well your methodology performs for such systems. There is no general recipe for success; it always involves a lot of hands-on work. Eventually you will want to compare your calculated results to experimental ones to verify your findings.


How do we know whether we will get a mono substituted or tri-substituted product during electrophilic aromatic substitution? For example consider the bromination of phenol we get the tri-substituted product (even though bromine is deactivating so I suppose it should deactivate the ring with each addition) while sulphonation essentially gives the mono-substituted product (even at high temperatures).

It is like others have already noted that phenol is very strongly ortho-para directing, while bromine is only slightly deactivating (or sometimes slightly activating). It is so activating that you do not even need a Lewis acidic catalyst or elevated temperature to let the reaction go fast and in good yield. Moreover, if you want the mono substituted product you will actually need to cool the reaction down to < 5 °C and use the rather dangerously flammable solvent $\ce{CS_2}$, yielding 4-bromophenol in 85% yield (source: Vogel). I assume the para product is formed due to the fact that there is no significant difference in electron density between ortho and para carbon atom while ortho is slightly sterically hindered.


Rutherford's gold foil experiment Rutherford projected $\alpha$-rays onto the gold foil. $\alpha$-rays are the same as a $\ce{He^2+}$ nucleus and they can ionize atoms. Therefore, why are gold atoms not ionized when the $\alpha$-rays collide with them?

In Rutherford's experiment to show the existence of nucleus in an atom, the alpha-particles were exposed on the surface of certain metal i.e. gold. He observed that more than 99% of these particles were able to go straight and pass through the gold atoms. Thus he was able to conclude that atoms are mainly empty space.
But why did these particles not collide with the electrons present in the gold atoms? I think that the reason behind it is that the speed of the electrons travelling in orbits did not cause any obstacle in the travelling path of the alpha particles.


How many valence electrons does the azide ion have? We had a test and and we had a question asking the number of valence electrons in an azide ion, $\ce{N3^{-}}$.  I find out that the answer is 16 but I could not understand how.
 I saw the structure. It says that the end nitrogen have a -1 charge and the middle nitrogen has a +1 charge.

From where does the one of the end nitrogen get the electron assuming the other end nitrogen gets it from the middle nitrogen?

Nitrogen (the element) has 7 electrons. But two of those electrons are in the $1s$ orbital, and so are not considered valence.  Therefore, a single elemental nitrogen has five valence electrons.

There are three nitrogens, and so $5 \times 3 = 15$ electrons for $\ce{N_3}$.

Finally, azide is $\ce{N_3^{-}}$, there is an additional electron beyond the neutral form.  So, the total number of valence electrons is $15 + 1 = 16$.


Why is fermentation of cellulose to produce biofuel and nutrients so difficult? The formula for glucose is $\ce{C6H12O6}$ and that of cellulose is very similar $\ce{C6H10O5}$.  Glucose can be readily fermented by yeast and other micro-organisms to produce carbon dioxide and ethanol. Whereas cellulose is not ferment-able by yeast and only very specific anaerobic Archae bacteria can digest it, but the end product being methane, water, sugars (primarily lactose), and volatile fatty acids (acetate, proprionate, butyrate); not ethanol.  These bacteria reside in the guts of termites that allow them to digest wood, and in the intestines of cattle and other ruminants.  The termites and ruminants use the sugars and fatty acids as food.

The enzymes used by the host anaerobic bacteria in the gut of termites are of particular interest because:

termites produce very little solid waste as feces
termites can digest wood, leaves, and broader variety of plant fodder than ruminants, which mainly feed on grasses

We are not really interested in the termites, nor the anaerobic bacteria, but rather specifically the enzymes and chemical processes used by the bacteria that break down the cellulose into the lactose and fatty acids.  Can this be accomplished synthetically in an anaerobic lab setting outside of said bacteria?

This article describes a company that tried to use cellulose raw material to produce ethanol on a commercial level, but recently went bankrupt.  This company apparently was not getting ethanol directly from fermentation of cellulose but rather the result of a multiple step process of generating syngas from the cellulose, then creating alcohol from the syngas; a much less efficient process than direct fermentation of cellulose into alcohol.

Obviously, such a process if achievable would relieve the need to use food crops to generate ethanol and put to use otherwise inedible cellulose plant waste. What is the main challenge here.  Cellulose is just one water molecule different from glucose.

Your question is a little bit all over the place, but I believe I can answer it anyway. First, though, allow me to point out that your sum formula for cellulose is wrong. While glucose is indeed a single $\ce{C6}$ compound with the exhaustive sum formula of $\ce{C6H12O6}$, cellulose is, in fact, a poorly defined polymer consisting of multiple $\ce{C6H10O5}$ subunits, linked together both linearly to form long chains but also cross-linked to form extensive networks. Therefore, the better sum formula would be $\ce{(C6H10O5)_nH2O}$ or $\ce{C_{6n}H_{10n+2}O_{5n+1}}$ for very large values of $n$.

There are two reasons why glucose, its dimer maltose and its polymers amylose, amylopectin and glycogen are much more easily digested by all organisms than cellbiose (the glucose dimer that can be found in cellulose) and cellulose itself.

The first one is the type of linkage: the first set of compounds all contain α$\ce{{1}\bond{->}4}$ connections while cellbiose and cellulose contain β$\ce{{1}\bond{->}4}$ connections. This seemingly minor difference means that while the starch subunits form spirals, the cellulose subunits actually form fibres and all the enzymes designed to break up spiral subunits cannot work with fibres.

The second is the number and type of cross-linkages. Each organism has a set of enzymes that can break up the cross-links present in amylopectin or glycogen (or both) but these again are different in shape from those in cellulose owing to the subunits’ β configuration in the latter.

Enzymes to break down amylose, amylopectin and glycogen are pretty much present in every multicellular organism since these compounds are used to store energy in an available form — which would not make sense if the organism could not break the storage form back down to monomers. However, cellulose was not ‘intended’ to be energy storage and thus there was no evolutionary pressure to evolve enzymes that could digest it. Therefore, only very few species that found and populated this ecologic niche are able to digest cellulose and even less are able to digest the additional compounds present in wood that makes it hard.

Now on to the process design part of your question. In general, if we have the enzymes we should be able to degrade wood to useable monomers much like whichever organisms do that in whichever environment they live. Not only the symbionts of termites are of interest but also wood-digesting fungi, naturally. However, the key difficulty here is acquiring the enzymes required in sufficient amounts to make the process economic. I am only aware of very few enzymes that can be fermented and isolated in sufficient quantities and purity to make for a possible economic use; among them a generic lipase. That is one significant hurdle process chemists must overcome.

However, there’s another bit. It may work if we have access to the enzyme but it doesn’t have to. A number of times the most significant issue for chemical engineers is the scale-up problem: reactions work fine on a small scale but once you try to scale them up to higher quantities for more throughput they break down in one way or another. I am sure without knowing the exact titles that there are shelves of books concerning the scale-up problem. To the best of my knowledge (caveat: I am not an industrial chemist or chemical engineer), it is often down to trial and error which processes work well and which don’t; and what works well for one type of product does not work at all for others.

Needless to say that one has to have access to a significant amount of funds to bridge the research gap until one actually is able to make money so it doesn’t surprise me (although I am unhappy) that the company you mention went bankrupt.


Color imparted to flame and color of coordination compounds When we heat a ionic substance/compound, the ionic bond gets weakened and the metal is ionized. Is the color imparted to the flame due to absorption of light? i.e. Do we see the remaining wavelengths? Or is it due to the falling back of electrons to their ground state?

Shouldn't we be seeing two colours one after other? Since the electrons take up energy of particular wavelengths (so we see the remaining colours) and then the wavelength which got used up earlier returning back when electrons return back to the ground state?

The coordination compounds should also show same dual-color property. Since when electrons absorb energy they go from t2g to eg and when they return, they should give back the wavelength.

This could also be said for halogens.

Why do we see only one colour? Which colour is it in which case? The absorption one or the emission one?

You seem to be mixing up a term. Let me get that out of the way first.

Ionisation is the name for a process creating an ion from a neutral species. However, if you heat an ionic substance and break the ionic bond, you are not ionising but deionising (although that term is usually used for a different process). The reaction equation for sodium chloride would be the following:

$$\ce{NaCl ->[\Delta] Na + Cl}$$

Now to the actual question. The perception of colour is based on one of two principles (as you noted):

out of white light, certain wavelengths are absorbed leaving behind a remaining spectrum and causing the perceived colour to be complementary to that of the absorbed light
light of a certain wavelength is emitted, directly creating the impression of said colour.

As you also noted, this could mean that we see up to two colours in the same electronic system: One due to the absorption and the second, complementary one during emission. So what is it that we see?

For coloured compounds at room temperature — that includes, but is not limited to transition metal salts and coordination complexes — we typically see absorption colour. Take for example the hexachloridoferrate(III) anion $\ce{[FeCl6]^3-}$: It has an intense yellow colour which can be explained by the absorption of very short visible wavelengths. The shortest visible wavelength is a blue one, so the perceived colour is yellow. The colouring is due to ligand-to-metal charge transfer transitions. In the corresponding fluoridocomplex $\ce{[FeF6]^3-}$, the orbital energies of the fluoride ions is predicted to be lower, giving a higher transition energy and shifting that transition just into UV wavelengths — the hexafluorideferrate(III) complex is colourless.

Also in organic chemistry, there is a number of yellow-ish compounds which occur because they just absorb at high enough wavelengths to shift out of the UV range. (UV absorption is very common for organic compounds containing a phenyl ring.)

We only see one colour, however. This is because the absorbed energy — which immediately excites an electron into a higher orbital — is liberated into the surroundings as vibrations and ultimately heat. The electron does not fall back into the ground state in a single transition but via a number of intermediates loosing energy all the way; hence we do not see a corresponding emission colour.

In flame colouring, one typically takes white (or colourless) compounds and heats them in a flame (the flame ideally being colourless, too). Only in the heat of the flame do they show their colour — and in a typical ion-lottery (qualitative inorganic analysis) lab, this is done in a dark corner of the lab, because there is actual emitted light. What is happening here?

Through the heat of the flame we are transferring energy into the compound, first causing the bonds to break (liberating atoms) and then also causing these atoms to vapourise so that we end up with gaseous metal atoms and gaseous non-metal atoms. Further heat is introduced which — if it is enough — will finally cause the transition of an electron into a higher orbital. This is, of course, unstable, but here the relaxation is immediately back down into the ground state for a significant percentage of these atoms; each of them releasing a photon with exactly the wavelength of the corresponding transition. And what we see are those photons, corresponding exactly to the wavelength of the energy difference between two orbitals.

And finally, just to confuse you, I am going to quote an experiment I saw in my school’s biology course. We had isolated chlorophyll from leaves of a plant and had it in a solution in an Erlenmeyer flask. It appeared green. Our teacher held it into the strong light beam of an overhead projector and it appeared as red. The experiment was explained as follows:

Red light is absorbed by chlorophyll, causing a green impression under normal conditions.
in the strong light beam, so much red light was absorbed and so many chlorophyll molecules were excited that the energy had ‘nowhere to go’; thus, the same red light was emitted to all directions and we were able to see that red colour.

So to sum it up:

in coloured salts at room temperature, we see the complementary colour of an absorbed photon. We do not see the emission (and thus the actual colour) because it is too unfrequent; most energy absorbed is released as vibrations and heat.
in flame colouring, we have highly excited atoms and we do not actually see the absorption since the energy necessary is transferred by heat. However, we do see emitted photons of exactly the energy difference wavelength from a significant number of electrons falling back into a lower energy state. We see the actual colour but not the complementary one.


how to remove products from reaction space?(to apply LeChatelier's principle) I thought this question merited  a discussion of its own. In industrial applications, we talk of shifting equilibrium of reaction to the right(or desired products) by removing products quickly from reaction space? I want to know what this means? If you consider a simple fixed-bed/packed-bed reactor with gaseous reactant and product (say A-->B), does it mean I need to remove B quickly from the packed-bed reactor volume to push equilibrium to the right? My point is, isn't that already happening? By virtue of flow through the reactor isn't B already constantly being "removed" from the reactor(along with unreacted A), while the reactor keeps getting hit with fresh A? so what more effort is needed to push equilibrium to the right? are there any timescales involved? or does B have to be removed quicker than A? if so, how quicker? I need to understand this better,please help. This question can apply to other types of reactors as well like fluidized-bed or moving-bed and so on..

The scheme you ask about is effectively the same as $\ce{A <=>[k_1][k_{-1}] B ->[k_2] C}$ where $k_2$ measures the effect of removing B making C the removed product. The actual reaction is much more complicated as presumably mixing/diffusion has to be taken into account. Leaving aside these complications  as $k_2$ increases then the amount of B in the reaction reduces and the amount of C increases if measured at a fixed time. As long as $k_2$ is not zero eventually all B is removed, you can see this in the figure which shows the amount of C for different rate constants $k_2$  assuming the initial amount of A = $1$ and $k_1= 1, \, k_{-1}=3$. When $k_2=0$ (top figure) no product is formed and A and B come to equilibrium. The lower figure show the effects on forming C by removing B.


How to get the valencies of elements? How to find the valencies of elements by using its distribution of electrons?
Please explain the method in simple words. Do you have to study the valencies or is there a simple way of remembering?

PS: There is supposed to be some method of getting the valency after knowing the distribution of electrons. For example the element sodium, $\ce{Na}$ has a distribution of electrons: 2, 8, 1. What is it's valency?

Valency is a concept that describes the power of an element to bind to other elements. There is an official definition in the IUPAC goldbook:

  valence
  The maximum number of univalent atoms (originally hydrogen or chlorine atoms) that may combine with an atom of the element under consideration, or with a fragment, or for which an atom of this element can be substituted.

So in principle it is not possible to know the valency just from the electron configuration, but you have to also know the chemistry it performs. It is also noteworthy, that according to this definition all elements have only one value for it's valency.

In the first period, according to the definition, hydrogen is monovalent, i.e. $\ce{H2}$ or $\ce{HCl}$. Helium on the other hand is technically divalent, since $\ce{HHeF}$ was theoretically predicted, see wikipedia. But it is somewhat safe to assume helium is inert and therefore zerovalent.

In the second period, where the octet rule holds, it is the number of unpaired electrons in the valence shell, with the exception of carbon. Unfortunately this is also the period which behaves most different from the rest of the periodic table. So the best way here is to just learn them as exceptions: $\ce{NaCl}$ (mono-), $\ce{BeCl2}$ (di-), $\ce{BCl3}$ (tri-), $\ce{CCl4}$ (tetra-), $\ce{NCl3}$ (tri-),(1) $\ce{Cl2O}$ or $\ce{H2O}$ (di-), $\ce{FCl}$ or $\ce{F2}$, $\ce{Ne}$ is zerovalent.

Below the second row it becomes a little bit more predictable. For most main group elements it is the number of electrons in the valence shell. This obviously coincides with the highest stable oxidation state.
Some examples may illustrate that. For the first group all binary compositions of the alkaline element $\ce{A}$ of the composition $\ce{ACl}$ are known. This is also the maximum of chlorine atoms any of these elements can bear, hence they are all monovalent.
 The same applies to the earth-alkalines, $\ce{Ea}$. Here all binary chlorides of the form $\ce{EaCl2}$ are known, hence they are divalent.
For the triels $\ce{M^{(III)}}$, all of the chlorides of the form $\ce{M^{(III)}Cl3}$ are known. However not all of them are very stable, i.e. $\ce{TlCl3}$ (wikipedia). So they are trivalent.
For the tetrels, $\ce{M^{(IV)}}$, the trend continues. The binary chlorides $\ce{M^{(IV)}Cl4}$ are known, even $\ce{PbCl4}$ is quite stable below $0~^\circ\mathrm{C}$ (wikipedia).  For Flerovium only predictions have been made, but it's believed to be very unstable in its tetravalent state due to the inert pair effect.
For the pnictogens, $\ce{M^{(V)}}$, all binary chlorides up to and including antimony are known. However, also $\ce{BiF5}$ is known, hence all elements can be considered pentavalent.
For the chalcogens, $\ce{M^{(VI)}}$, all binary fluorides with coordination number six are known. Most prominent is sulfur hexafluoride, for which a chloride counterpart exists. The elements can therefore be considered hexavalent. The tetra chlorides are also stable for all elements of this group, in that case the elements can be considered tetravalent. However, more common are compounds where the chalcogens are divalent, i.e. in the respective hydrogen compounds, $\ce{H2M^{(VI)}}$.
The halogens $\ce{Hal}$ are just amazing, and once you have looked into interhalogen compounds you will most likely remember a lot of them. For chlorine, the interhalogens $\ce{ClF_{\{1,3,5\}}}$ have been reported, making it pentavalent. For bromine, the same holds true: $\ce{BrF_{\{1,3,5\}}}$. Only $\ce{BrCl}$ has been observed. However this makes bromine also pentavalent. For iodine the colourless gas $\ce{IF7}$ has been observed, making it the only heptavalent element of the series. For astatine only a few compounds are known, and the current believe is, that it is monovalent.
For the noble gases, $\ce{Ng}$, there are a few compounds known, and their valency increases throughout the group. While argon can be considered at most divalent due to the discovery of the highly unstable $\ce{HArF}$, the divalent nature of Krypton is undoubtedly proven by $\ce{KrF2}$. Xenon is the most reactive and potent noble gas and it can be considered hexavalent according to the above definition, as of $\ce{XeF6}$ does exist. However, also $\ce{XeO4}$ is known, which would make it octavalent according to a different definition.(2)

Transition metals are a whole different story and it is difficult to find a generalised rule or observation for it, since the IUPAC definition is very strict on the matter. All elements have to be looked at in detail. It will certainly also exceed the purpose of this answer.
Just to illustrate the problem have a look at manganese. There are various compounds known, such as $\ce{MnCl2}$ (divalent), $\ce{MnF3}$ (trivalent) and even $\ce{MnF4}$ (tetravalent). Binary hydrides have not yet been reported. However, there are oxides of various composition, ranging from divalent to heptavalent.

Notes

The molecule $\ce{NCl5}$, pentavalent, has been theoretically suggested, see wikipedia and references within.
A more modern interpretation of the valency concept allows elements to have more than only one value. It is therefore always depending on the compound in which it is looked at. A very brief summary can be found on wikipedia.

  The most characteristic chemical property of an element is its valency. There are numerous measures of valency each with its own area of usefulness and applicability. Simple definitions refer to the number of hydrogen atoms that can combine with an element in a binary hydride or to twice the number of oxygen atoms combining with an element in its oxide(s).

(N. N. Greenwood and A Earnshaw: Chemistry of the Elements. Second edition. Butterworth-Heinemann, 1997. Chapter 2, p. 27.)


False positive in lucas test for primary alcohol due to formation of tertiary carbocation by rearrangement? From Wikipedia:

The Lucas test in alcohols is a test to differentiate between primary, secondary, and tertiary alcohols. It is based on the differences in reactivity of the three classes of alcohols with hydrogen halides via an $\mathrm{S_N1}$ reaction:
$$\ce{ROH + HCl -> RCl + H2O}$$
The differing reactivity reflects the differing ease of formation of the corresponding carbocations. Tertiary carbocations are far more stable than secondary carbocations, and primary carbocations are the least stable.

So since alkyl halides are insoluble, they give a turbid solution upon formation, which easily happens for tertiary alcohols.
Consider a primary alcohol, like $\ce{(CH3)3-C-CH2OH}$, which forms a tertiary carbocation, $\ce{(CH3)2C+-CH2CH3}$ by an alkyl shift from primary carbocation during the substitution process. So will Lucas test give a false positive in this case, since a tertiary alkyl halide can be easily formed from a primary alcohol?

In order to understand this, you need to have this basic knowledge of the kinetics of multistep reactions

In chemical kinetics, there are two ways to deal with multiple-step mechanisms

Rate determining step method- Here one specific step is the slowest. So we consider all steps after this step to be equally fast. The rate-determining step is the slowest step of a chemical reaction that determines the speed (rate) at which the overall reaction proceeds. The rate-determining step can be compared to the neck of a funnel. (1)
Steady-state approximation method- This is used when no step is the clear slowest. The steady-state approximation is a method used to estimate the overall reaction rate of a multi-step reaction. It assumes that the rate of change of intermediate concentration in a multi-step reaction is constant. (2)

So, in the Lucas test, the formation of the carbocation is the clearly slowest first step. This governs the reaction rate and the rest of the steps don't really matter. So, although it might seem deceptive at times if the R.D.S. is slow then the whole net reaction in slow.

Hence, the rate of formation of the one-degree carbocation is so less, that the entire rection does not proceed. So what you can theoretically say will be the only difference in say a Neopentyl alcohol and a 1-alkanol would simply be that after the super slow first step, the rest of the reaction will be much faster in the case of Neopentyl alcohol. But, due to the slow first step your given example will give an absolutely accurate negative Lucas test


Is there a simple way to separate deuterium oxide from tap water? I understand that tap water contains minute amounts of deuterium oxide (heavy water). Is there a simple way to separate, concentrate the $\ce{D2O}$?

It depends how pure you want the $\ce{D2O}$ to be and what you consider simple :)

Electrolysis of water strongly favors "H" being converted to hydrogen gas rather than "D", by a factor of about 8 to 1 (depending upon the electrodes).

Inexpensive Equipment for the Preparation and Concentration of Pure $\ce{D2O}$ Ohio Journal of Science volume 41, number 5, pages 357-365 (1941) explains in great detail how to carry out the electrolysis method.  The article says the process is suitable for "those who have occasion to prepare or concentrate heavy water for research, demonstration or
advanced student laboratory use".

In a first stage, $\ce{NaOH}$ solution (initially $\pu{0.5 M}$) is subject to electrolysis, until only 1/32 of the original volume remains.  This increased the original concentration of deuterium by a factor of 12.

By then the $\ce{NaOH}$ is too concentrated, and $\ce{CO2}$ is added (as dry ice) to neutralize the solution, and the first stage solution is distilled to get rid of the sodium carbonate. About $5~\%$ of the $\ce{NaOH(D)}$ concentrate would be saved and added to the distillate to provide electrolyte for the next stage.

Then a second stage of electrolysis is performed. By the end of the second stage they reached about $10~\%$ deuterium. The process can be repeated to achieve a desired deuterium level.  They used three stages.  Yield can be improved by capturing the deutrium/hydrogen gas of the late stages, burning it to make water, and feeding this back to an earlier stage.


Method for measuring volume of gas I would like to setup something up my office to accurately measure the amount of Carbon Dioxide (CO2) being released from beans.

I looked at the site http://examples.yourdictionary.com/examples-of-ways-to-measure-volume.html which suggests several methods, the first of which, fluid displacement, I was going to employ. I know that CO2 is slightly soluble in water so I was going to use a saline solution/brine to reduce solubility. Can anyone help me here in explaining additional things that I should be doing to reduce errors in measurement with this fluid displacement method, other than of course reducing the possibility of air leaks and CO2 solubility? For example I have been reading about the risk of water evaporation in the measuring cylinder so I wonder how this can be minimised? I've also heard about the vander waals methodology for accuracy in calculation.

What other methods should I be considering? I'm not asking for better methods as I realise that's subjective, but other methods that you believe I should be investigating as possible alternatives. I have looked at a gas syringe and a balloon as alternatives.

Unless you have a way to separate CO2 from the rest of the gases, you won't be able to measure it properly. Which isn't easy. You could put a substance that absorbs carbon dioxide inside the container for the beans. However, the substance needs to be able to indicate that it has reached capacity. And also, the volume of CO2 that it can absorb must be known.

You could also set up a respirometer. You could make some assumptions about how the oxygen is consumed. Nevertheless, it should give you a good estimate. There are several videos and tutorials explaining them. And if you still have questions, you can ask Bio Stack Exchange.

The comment about the van-der-Waals equations: I think that you were probably thinking on using the ideal gas law to estimate the number of gas molecules in a container.
$$pV=nRT$$

According to the equation above, you can measure the number of particles if you know the pressure and the volume. This is true for ideal gases. For a gas molecule to behave like an ideal gas particle, it would have to have no volume and have no interactions with other gas molecules. Real gases, however, do not behave like an ideal gas.  Therefore, a better model to describe real gases is needed. The van-der-Waals equation is one of these attempts.


Orbitals SPDF. Why they named like that? I know there are orbitals named S, P, D and F.

But why they named like that. is there any significance for the letter?
Is there any order in filling the orbitals?

From the abstract of Structure of the Line Spectra of the Elements as published in the 1890 Journal of the Chemical Society.

  There are three kinds of series — principal, sharp (well-defined), and diffuse (ill- defined). The principal series form the most vivid lines in the spectra, and only occur in the first, periodic group ; next come the diffuse (really double) lines ...

See also On the Structure of the Line Spectra of the Chemical Elements, also from 1890 by Rydberg:

  Liveing and Dewar were the first to remark the existence of the series, as well as their different appearances. They are of three kinds : diffuse, sharp, and principal series. The first two are formed by the above-mentioned doublets or triplets ...

Later, "fundamental" was added.

For more information see The Origin of the s, p, d, f Orbital Labels J. Chem. Educ., 2007, 84, 757-758


Is it possible for a substance to absorb a longer wavelength of EM wave and emit a shorter wavelength? I know a fluorescent lamp works by emitting UV first and then the specific substance inside absorbs UV and finally emits visible light. An object may emit infrared under sunlight due to heating. But those examples are absorbing short wavelength of EM and then reemitting a longer one. Is there a substance that can do the opposite (e.g.: absorb visible light and then emit UV)?

It is possible, but impractical. You can "upconvert" light. This effect, as Wikipedia relays, is due to the sequential absorption of two photons per emitted photon.

This is certainly possible to make in a fluorescent / phosphorescent light scenario, but the engineering challenges (or, more specifically - the cost) involved in such a solution would be steep.

There is a chemical model of such a compound in the two-photon-absorbance wiki site, but I have not the nomenclature skills to name it for you. I even tried using online search for the structure (and it took me a while. I hate nested benzene groups - probably why I chose inorganic chemistry)


What is the structure of CF3 radical? In the class, I was told that $\ce{H3C^.}$ has a trigonal planar structure with the unpaired electron in $\mathrm{2p_z}$ orbital. But $\ce{H3C -}$ has a trigonal pyramidal structure.

But why does this happen?
Why does $\ce{F3C^.}$ also doesn't have a trigonal planar?

Edit : I have seen one possible answer in a different question.  But is there any answer that is explained with the help of VSEPR or MOT or VBT?

CH3 radical is Sp2 hybridised carrying the lone electron in a pure p orbital. This is so because Sp2 hybridised orbitals are lower in energy than Sp3 hybrid orbitals (The more the S character the lesser is the energy of the orbital). When there is a lone electron present, it is more efficient to keep the bond pair electrons in lesser energetic Sp2 hybrid orbitals and lone electron in higher energy pure P orbital than keeping all bonding pairs and the lone electron in Sp3 hybrid orbitals. So most of the radicals are Sp2 hybridised.

But in situations like •CF3 where the atoms/groups attached to C are of high electronegativity things may change. According to Bent's rule more electronegative atoms/groups will favour less electro negative orbitals i.e. orbital s with lesser s character(Sp3 has lesser S character compared to Sp2). This is quite obvious because the more the S character of the C center the more is the tendency of the C center to attract the bond pair towards itself and in cases like electronegative atom, it will also try to attract the bond pair towards itself. This makes some unhappy situation. This is reflected by the Bent's rule. That is why •CF3 radical favours Sp3 hybridisation.

It's not that only such radicals like •CF3 can acquire Sp3 hybridisation. In many other cases like bridge head of adamentum also acquire Sp3 hybridisation becauseof opposition by its rigid structure to acquire planarity on generation of radical though it's not that much easy process to form radical here.

In conclusion, a radical will favour that hybridisation which will be energetically favourable for it.


Slope ≠ 1 in Henderson-Hasselbalch graph for determination of pKa The $\mathrm{p}K_\mathrm{a}$ of indicators can be determined via plotting a graph of $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right)$  vs $\mathrm{pH}$ with the help of a spectrophotometer, from the equation $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = \mathrm{pH} - \mathrm{p}K_\mathrm{a}$.
From this, the graph should have a gradient of 1, and $x$-intercept = -($y$-intercept) = $\mathrm{p}K_\mathrm{a}$. However, my graph has a slope of ~0.8 which means my $x$-intercept ≠ -($y$-intercept) so which value do I use as my calculated $\mathrm{p}K_\mathrm{a}$?
From the papers I've read, most if not all of them use the intercept with the $\mathrm{pH}$ axis instead of the $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right)$  axis in this situation, why is this?

OP's question: From the papers I've read, most if not all of them use the intercept with the $\mathrm{pH}$ axis instead of the $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right)$ axis in this situation, why is this?

Answer for this question is easy. According to the graphs plotted using the equation $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = \mathrm{pH} - \mathrm{p}K_\mathrm{a}$ (Henderson-Hasslebalch equation), when $\log \left(\frac{\ce{In-}}{\ce{HIn}}\right) = 0$ (meaning $[\ce{In-}] = [\ce{HIn}]$), then $\mathrm{pH} = \mathrm{p}K_\mathrm{a} $. Therefore, the $x$-value where the straight-line crosses the $x$-axis is the best value for $\mathrm{p}K_\mathrm{a} $.


Why don't ligands affect the oxidation state of the central metal ion? If $\ce{Fe^3+}$ were complexed with 6 $\ce{H2O}$ ligands, the resulting complex ion would be $\ce{[Fe(H2O)6]^3+}$, and the central metal ion would still have the same oxidation state of +3 as it did before complexation. However, didn't the $\ce{Fe^3+}$ ion receive 12 electrons (2 from each water molecule) in forming the complex? Is this not a kind of reduction?

Also, is the positive charge of the complex ion still localized on the metal ion, or is it spread over the entirety of the complex?

If you take a glimpse at the molecular orbital scheme, you will notice that the electrons donated by the water molecules still reside in orbitals that are largely ligand-centred (and have a relatively low energy). On the other hand, iron(III) also has five d electron itself; these reside in the $\mathrm{t_{2g}}$ and $\mathrm{e_g}^*$ orbitals, which are largely metal-centred (and have a relatively high energy).

This is true for most complexes. The metal tends to have rather high-lying orbitals while the ligands tend to donate rather low-lying ones to the overall complex. As long as this order is retained, there is no chemical redox behaviour going on.

For some systems, however, the orbitals are either so similar or even crossed-over that suddenly a metal orbital will become lower-lying than a ligand orbital. In this case, an electron that used to reside in a ligand orbital will relax into a metal orbital — a redox process. You can for example think of the reaction of iodide with copper(II) in this manner: copper(II) will immediately get reduced to copper(I) by iodide ions. While this is a unidirectional process, some systems are also known to bind ligands in a redox fashion reversively, meaning that the electrons will be exchanged both ways. The binding mode of oxygen to haemoglobin is understood in this manner.

Finally, sometimes there is a change in oxidation state that is merely formal. Tetracarbonylhydridoiron(0) $\ce{[Fe(CO)4H]}$ is such a case. It can be protonated to formally give tetracarbonyldihydridoiron(II) $\ce{[Fe(CO)4H2]+}$ or deprotonated to give tetracarbonylferrate(–II) $\ce{[Fe(CO)4]-}$. Both changes in oxidation state are merely formal and chemically all that is happening is a protonation/deprotonation.


Cleavage of benzyl group from alkane? Is there a method to selectively cleave the C-C bond between a benzyl group (-CH2Ph) and a linear or cyclic alkane? Can this be done under standard conditions with somewhat widely available reagents? Can the benzyl group be forced out if it is under significant steric clash when it is attached?

The following is just a mind game:

Oxidize at the benzylic position to get an arylketone using oxone
Convert the ketone to an ester by Baeyer-Villiger reaction
Cleave the ester
Run a Barton-McCombie deoxygenation on the resulting alkanol

Please note that I havent't done it and I'm not planning to. I'd rather try to avoid the benzyl group in the starting material if possible.


Hill equation with inhibition and activation The Hill equation for an activating enzymatic interaction with cooperative multiplicity $n$ is

$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = V_\mathrm{max}\frac{[\ce{S}]^n}{K_\ce{S} + [\ce{S}]^n}\tag{1}$$

and for an inhibitory interaction

$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = V_\mathrm{max}\frac{K_\ce{I}}{K_\ce{I} + [\ce{I}]^m}\tag{2}$$

What would the equation be for competing activating and inhibiting compounds $\ce{S}$ and $\ce{I},$ respectively?

My sort of ansatz is that it would be of the form

$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = V_\mathrm{max} \frac{c_\ce{S}[\ce{S}]^n}{K_\ce{SI} + c_\ce{S}[\ce{S}]^n + c_\ce{I}[\ce{I}]^m}\tag{3}$$

I spent about an hour trying to derive the actual answer, but I'm no expert in physical chemistry/biochemistry and increasingly have the sense I should turn to someone who is.

Mechanism of cooperative enzyme activity

The easiest way to explain cooperativity is the example of hemoglobin, where four nearly identical subunits switch cooperatively between two states T and R of different binding affinity. When no ligand (L) is bound, hemoglobin is in the T state; once at least one ligand is bound, it is predominately in the R state.

$$\ce{T4 + 4 L <=> R_4L + 3 L <=> R4L + 3 L <=> R4L2 + 2 L <=> R_4L3 + L <=> R4L4}$$

The states with 1-3 ligands are less populated (because the R state has higher affinity for the ligand L thant the T state), and we will ignore them:

$$\ce{T4 + 4 L <=> R4L4}$$

From this, you can write the law of mass action and the binding isotherms.

If we apply this to an enzyme with a Hill coefficient of $n$, and call the active state of the enzyme E and the inactive state F, we would have the equilibrium:

$$\ce{F_n + nS<=> E_nS_n}$$

For sufficiently high concentration of substrate S, all enzyme would be in complex with substrate, leading to a rate of $v_\text{max}$.

Mechanism of cooperative inhibition

In the OP's question, the Hill coefficient for inhibition $m$ is different from that for enzyme activity ($n$). So we could have a regulatory protein with $m$ subunits attached to the enzyme. It binds cooperatively to inhibitor, which allosterically affects enzyme activity. The regulatory protein would also have two states, T and R, either free or bound to $m$ inhibitors:

$$\ce{ T_m + m I <=> R_mI_m}$$

When the allosteric inhibitor is bound to ligand (i.e. in the R state), the enzyme lacks activity.

Combination of the two

Only a certain fraction of enzyme is bound to substrate, and only a certain fraction of enzyme-bound substrate is not inhibited by the regulatory protein. For this scenario, the enzyme activity is $v_\text{max}$ times those two fractions.

$$\frac{\mathrm d[\ce{P}]}{\mathrm dt} = v_\mathrm{max}\frac{[\ce{S}]^n}{K_S + [\ce{S}]^n} \cdot \frac{K_I}{K_I + [\ce{I}]^m}\tag{2}$$

What happens when there is competitive inhibition?

It gets more complicated. Will the inhibitor also cause a switch in the enzymes conformation? In which direction? For the simpler (non-cooperative) Michaelis Menten model, a competitive inhibitor changes the apparent affinity of enzyme to ligand in a concentration-dependent manner (the enzyme partitions between free, ligand-bound and inhibitor-bound state). For the cooperative case, you would have to decide whether mixed states (such as $\ce{ES2I2}$) are populated; if they are, they would contribute to product formation, but less than $\ce{ES4}$ because not all active sites are turning over substrate.


Density of water and temperature Is there a formula that shows how the density of water is affected by temperature?

For example, with the speed of sound it can be shown how temperature will affect the speed:

$$V = 331\ \mathrm{m/s} + \left(0.6\ \mathrm{\frac{m/s}{°C}}\right)T$$

Is there a similar formula that relates the density of a liquid to temperature? (in this case, water)

In general, we can derive the expression for the density of any substance as a function of temperature using the definition of thermal expansion.

$$\frac{\mathrm dV}{\mathrm dT}=\alpha V\tag1$$

By differentiating $\rho=m/V$, we can show that $\mathrm d\rho/\rho = -\mathrm dV/V$. So using this and equation $\text{(1)}$,

$$\frac{\mathrm d\rho}{\mathrm dT}=-\alpha V\tag2$$

Integrating this above expression and approximating the expression,

$$\rho = \rho^\circ\cdot\left(1-\alpha\Delta T\right)\tag3$$

where $\rho^\circ$ is the initial density at the initial temperature, $\alpha$ is the volumetric thermal expansion coefficient and $\Delta T$ is the temperature.
For water, $\alpha$ is typically $207\times10^{-6}\ \mathrm{K^{-1}}$ at $20\ \mathrm{^\circ C}$. Using this, we can observe how temperature affects the density of water.


Is there a relationship between density and viscosity of a fluid? I'd like to know if there is a connection between the two physical quantities of density and viscosity.

e.g. How can you explain low density liquids such as mineral oils have a lower density than water but far greater viscosity?
e.g.2 Instead, for honey (18% of water and 82% of sugar...) a density of about 1.4 kg/litre but a greater viscosity?

They are not directly related as such. Density is defined as mass/volume units $\ce{kg/m^3}$. Water has a value 1000, (liquid) iron 7000, (liquid) lithium 516 , ethanol 789 and carbon tetrachloride 1632 all in units of $\ce{kg/m^3}$.

Viscosity is the rate of transfer of momentum exerted on a plane of fluid when a second plane of fluid a parallel distance $x$ away is moved with viscosity $v$.  The force exerted is $F=\eta \frac{v}{x}$, where $\eta$ is the viscosity.

The fundamental process that takes place when a liquid changes shape is that one layer of it has to slide past another. If we suppose that the liquid could be divided up into cells, then one cell has to 'hop' past another so that the liquid can move. The more difficult that this is the greater the viscosity. Thus to 'hop' a cell has to have an activation energy, that is it has to accumulate in itself enough energy to overcome resistance from other cells that prevent it from moving. Now defining just what a cell is is somewhat vague, but it is larger than a single molecule. But leaving that aside, the activation energy does reflect the inter-molecular interaction and shapes of individual molecules. Spheres with little inter-molecular interaction, e.g. small van-der-Waals forces, no hydrogen bonds, will have a lower viscosity than an awkward shaped molecule such as glycerol with the ability to form lots of hydrogen bonds.

Viscosity has units $\ce{Pa~s = N~m^{-2}S^{-1} = kg~m^{-1}~s^{-1}}$ but more commonly centipoise where $\ce{1~cP = 10^{-3} Pa~S}$. Some values  (in CP at 25 $\ce{^oC}$) are water 1, $\ce{CCl4}$ 0.97, iodoethane 0.56,  ethanol 1.07, cyclohexane 0.89, cyclohexanol 57.5, 1-propanol 1.945, ethylene glycol 16, glycerol 934.


What's the relationship between Van der Waals forces and hydrophobic interactions? A question on a past exam for a course I'm studying for asks:

  What's the relation between Van der Waals forces and hydrophobic
  interactions?

From what I understand, Van der Waals forces are just a name to summarize a bunch of little weak forces that only come into play when molecules are placed close to one other, based off this previous answer to my question.

I understand that hydrophobic interactions arise because non-polar components of a molecule are unable to form bonds with hydrogen in water, but disrupt the hydrogen, forcing them to make a cage-like structure (although I'm not clear on why), based off this entry in the Chemistry wiki.

Is the relation between these two forces that Van der Waals forces can be encouraged by hydrophic forces? Is there some other relation I'm missing here? Does it have something to do with the interactions denaturing at high temperatures?

Van der Waals forces are the attractions/repulsions - the forces - between molecules or atoms, other than attractions like ionic attractions, and covalent attractions. These forces are:

Keesom Effect - This is an effect caused by two polar atoms interacting with each other. Two permanent dipoles are involved, meaning the molecules/atoms involved are polar. This may be attractive or repulsive, depending on the dipoles involved.
Deybe force - This is an effect caused between a molecule with is polar, and one that is not. Because the one that is polar affects the electrons on the second (non-polar) molecule, it creates an attractive force between the two molecules.
London dispersal effect - This is a force that acts between two non-polar molecules/atoms. Because the electrons around each molecule/atom repel each other, it creates a redistribution of charge, inducing an instantaneous dipole moment. This is the force that acts in liquid noble gases, to make them liquid, or between methane molecules in liquid form.

Hydrophobic forces are caused because molecules like ethane and other hydrocarbons - petrol etc. and oils like sunflower oil etc. are non-polar, and therefore don't "like" being dissolved in a polar solvent like water. In water, there are hydrogen bonds between the individual molecules, which is why water is a liquid - the van der Waals forces are far too small in water to bind the water as a liquid at room temperature. If we introduce non-polar substances into the water - polar solvent - it upsets these hydrogen bonds, and creates an increase in enthalpy because of this. Therefore the lowest energy state is for the hydrophobic hydrocarbons to separate themselves from the water.

The relationship between van der waals forces and hydrophobic interactions is that the van der waals act to bind the hydrophobe - non-polar substance - together, to separate from the polar solvent/water, and these contribute to the energy needed to separate the two substances. The seperation causes a decrease in the entropy of the system. To counter this decrease in entropy, there must be some decrease in enthalpy. Because the hydrophobe disrupts the hydrogen bonding in water, when they separate, the hydrogen bonding then causes an decrease in enthalpy, because of the favorable interactions. This is also where the van der Waals forces come in. They are favourable interactions, so cause a decrease in enthalpy, and this helps the separation of the hydrophobe and the water into two separate phases, because it makes the separation more energetically favourable. They are not necessary, but make separation more favourable, because the enthalpy change is greater for the separation process. This makes it more favourable.


Formula of entropy change In my 12th standard book, the formula for entropy change is given as $\Delta S = \frac{q_\text{reversible}}{T}$. What is the importance of absorbing heat reversibly and not irreversibly? What does absorbing heat reversibly mean? Does it mean that the process should be reversible, and if it does, then how would we define change in entropy of a spontaneous process i.e an irreversible process?
 Would its formula also include $q_\text{reversible}$?

A process is thermodynamically reversible if it is essentially at equilibrium. Specifically, the system and its surroundings stay infinitesimally close to equilibrium with each other throughout a reversible process. Small changes in intensive variables of the system are perfectly balanced by changes in those variables in the surroundings. For example, $T_{\text{system}} = T_{\text{surroundings}}$ if heat can be exchanged reversibly; any temperature difference will result in an irreversible exchange of heat.
What's special about reversible heat transfer? When a tiny amount of heat $dq_{\text{rev}}$ is released by the system during a reversible process, the surroundings will absorb exactly that amount of heat, no matter what path the process takes.

If you have an irreversible process, the amount of heat (and work) depends on how the system and surroundings change during the process.
What's the difference between reversible and irreversible heat? The heat transferred by an irreversible process $dq_{\text{irrev}}$ is less than $dq_{\text{rev}}$, where $dq_{\text{rev}}$ is the heat that would have been transferred if the process had been performed reversibly. The difference between the two is the amount of work you'd have to do on the system to reverse the effects of the irreversible process.

How do we know that? Energy $U$ is a state function, so it'll be the same whichever way we do the process:
$$dU = dq_{\text{irrev}} + dw_{\text{irrev}} = dq_{\text{rev}} + dw_{\text{rev}}$$
so the difference between the two heats is
$$dq_{\text{rev}} - dq_{\text{irrev}} = dw_{\text{irrev}} - dw_{\text{rev}}$$

The differences on either side are positive, so $$dq_{rev} > dq_{irrev}$$
This means that an irreversible process will always absorb less heat from the surroundings and perform less work on the surroundings than it would have if the process had been done reversibly.
Why $dS = dq_{\text{rev}}/T$? With a reversible process, $$\left(\frac{dq_{\text{rev}}}{T}\right)_{\text{system}} = -\left(\frac{dq_{\text{rev}}}{T}\right)_{\text{surroundings}}$$
so if we define $dS \equiv dq_{\text{rev}}/T$, then we have $dS_{\text{universe}} = dS_{\text{system}} + dS_{\text{surroundings}} = 0$.

Consider what would happen in an irreversible process that took the system from state A to state B. The inequality $dq_{\text{rev}} > dq_{\text{irrev}}$ together with the definition of entropy in terms of reversible heats gives
$$\Delta S = S_B - S_A = \int_A^B\frac{dq_{\text{rev}}}{T} > \int_A^B\frac{dq_{\text{irrev}}}{T}$$ for any such process. This also implies that for any cyclic process (e. g. A to B back to A), we have $$\oint\frac{dq_{\text{irrev}}}{T} \lt 0, \ \text{and}\  \Delta S = \oint\frac{dq_{\text{rev}}}{T} = 0$$
This shows that $q_{\text{rev}}/T$ is a state function (it's zero for any cyclic process) but  $q_{\text{irrev}}/T$ is not a state function. If entropy is to be a state function, then, you can't define it as simply $q/T$; you've got to use $\Delta S = q_{\text{rev}}/T$, or for the differential form $dS = dq_{\text{rev}}/T$.
How could we relate the entropy change for an irreversible process to $dq_{\text{irrev}}$? From the difference between $q_{rev}$ and $dq_{irrev}$ above and the definition of entropy, we have
$$dS = \frac{dq_{\text{rev}}}{T} = \frac{dq_{\text{irrev}}}{T} + \frac{dw_{\text{irrev}}-dw_{\text{rev}}}{T}$$

TL;DR: There IS no TL;DR; you're asking some important questions here, and the answers need to be made carefully.


Why are Ka and Kb only used for weak acids and bases, respectively? The only answer which I found to this is that it's because strong acids and bases virtually completely dissociate, so the dissociation constants would be too large and hence aren't used. This doesn't really seem reasonable to me and I have a feeling there is another answer. Is there really another answer or is this the only answer?

This is absolutely not true. $\mathrm{p}K_\mathrm{a}$ values are frequently quoted amoung organic chemists for very strong bases and sometimes even for very strong acids. For reference, check out the Evans $\mathrm{p}K_\mathrm{a}$ table and the values quoted therein.


To what degree can the Rare Earth Elements be interchanged in its workaday applications? In their humdrum pedestrian "high-tech applications due to their unique magnetic properties", how interchangeable are the Rare-Earth Elements (REEs) if, e.g., the US lacks one of them?

RRautamaa comments on Pentagon in talks with Australia on rare earths plant.

  Rare earths are not rare, but they're chemically very similar. (This is because when you go to higher atomic numbers, they start filling the $\mathrm{f}$-shell, but the $\mathrm{f}$-shell is lower in energy than the $\mathrm{d}$-shell, so it's "hidden" by the $\mathrm{d}$-shell. The $\mathrm{d}$-shell is the valence shell and is the same in all of them, so their chemistry is almost the same.)

  Separating two chemically similar substances is expensive. Separating 15 of them is much worse. This means that the mining part is easy, but the processing plant on the site costs about $500 million. This is why rare earth mines are so few in number.

This question was spurred by the Pentagon's halt on May 22 2020 on rare earths funding program pending 'further research'. "Of course, the problem with rare earth elements has never been scarcity. They're actually pretty common. The problem is just that the cost of mining them exceeded their market value everywhere but China, so China ended up with a monopoly and the ability to drive up the price. So it's not clear that this changes much.

It depends on the particular use of the "rare earth". Mischmetal, containing an assortment of mixed lanthanides, can be used as-is for lighter "flints" or fire-starters. On the other hand, ultra-high purity lanthanide elements are needed to get specific spectral lines for lasers, for example.

Other uses, such as in magnets and in high-temperature superconductors, may be somewhat less demanding of purity. Didymium glass, for example, used in glasswork to block yellow sodium light, is a mix containing primarily samarium, praseodymium and neodymium.

Another expense involved in lanthanide production is separation and disposal of the radioactive isotopes from the ore. At one time, for example, gas lantern mantles contained mildly radioactive $\ce{ThO2}$, but those are no longer available in many countries due to concern about the alpha emitting thorium and the alpha and beta emissions from daughters.


Complete Oxidation of Methanol I know that complete oxidation of primary alcohols produces acids:

$$\ce{R-CH2(OH) ->[\ce{[O]}] R-CH(OH)(OH) ->[][-\ce{H2O}] R-CHO ->[\ce{[O]}] RCOOH}$$

I wonder if that is the case of methanol. Why doesn't formic acid complete oxidation to $\ce{CO2}$?

$$\ce{HC(O)OH ->[\ce{[O]}] C(O)(OH)(OH) ->[][-\ce{H2O}] CO2}$$

Note: by [O] I mean an oxidizing agent as $\ce{KMnO4}$ or acidified potassium dichromate.

Although the direct complete oxidation of methanol (or formaldehyde or formic acid, on that matter) to carbon dioxide ($\ce{CO2}$) using  oxidizing reagents such as $\ce{KMnO4}$  or $\ce{K2Cr2O7}$  is hard to find in literature, electrochemical oxidation of methanol to $\ce{CO2}$ through formaldehyde and formic acid is one of the concepts used in fuel cell research (Ref.1). The extensive use of simple organic compounds such as methanol, formaldehyde, and formic acid as fuels is because they have several advantages, likes of being easy to store and handle, and having possess a high energy density of the order of $\pu{kWh/kg}$. Aside from that, one would also expect they should have the simplest and most straightforward reaction mechanisms of all the possible organic fuels because of their simple structure. Yet, in their review article (Ref.1), Parsons and VanderNoot revealed that the oxidation of methanol to $\ce{CO2}$, involves $\ce{6 e-}$ so the oxidation must occur in several steps with several products or intermediates. The product resulting from complete oxidation of methanol (and other individual organic fuels such as formaldehyde and formic acid) is of course $\ce{CO2}$. The reactions of all these three fuels have been studied on a variety of electrodes, and with a wide spectrum of experimental conditions to optimize the outcome (most of those references are sited in Ref.1). Keep in mind that the studies involving methanol are outweighed those of formaldehyde and formic acid, simply due to the fact that its complete oxidation to $\ce{CO2}$, generates 6 electrons compared to 4 and 2 for formaldehyde and formic acid, respectively. Electrode potentials of these reactions are still debatable, and Koppenol and Rush have collected and calculated some reduction potentials for $\ce{CO2/CO2^{.-}}$ couple and plotted them in a Frost diagram (Ref.2).
We noted above that the electrochemical oxidation of methanol to $\ce{CO2}$ involves $\ce{6 e-}$.  As expected, the oxidation occurs in several steps with several products or intermediates, and, for example, mass spectral measurements has showed that both $\ce{H2CO}$ and $\ce{HCO2H}$ were produced (Ref.3).  Their conclution was listed as:

(a) At room temperature, a stationary decomposition of methanol takes place on a platinum catalyst, with formation of formaldehyde and formic acid;
(b) $\ce{HCO2H}$ reacts with $\ce{CH3OH}$ to form $\ce{HCOOCH3}$ (methyl formate);
(c) $\ce{HCO2H}$ is also produced during electrolysis, and is an intermediate in the oxidation to $\ce{CO2}$;
(d) $\ce{H2CO}$ reacts with $\ce{CH3OH}$ to form $\ce{CH2(OCH3)2}$ (methylal), but, contrary to the results of other authors, $\ce{H2CO}$ is not an intermediate of electrolysis.

Accordingly, you’d see even under electrochemical conditions, this conversion is not straightforward. Ref.1 also proposed a step-wise mechanism to explain all those intermediate and product, involving main byproduct, carbon monoxide ($\ce{CO}$), which is believed to be poisoning the $\ce{Pt}$ electrode in most cases.
References:

R. Parsons, T. VanderNoot, “The oxidation of small organic molecules: A survey of recent fuel cell related research,” Journal of Electroanalytical Chemistry and Interfacial Electrochemistry 1988, 257(1–2), 9-45 (https://doi.org/10.1016/0022-0728(88)87028-1).
W. H. Koppenol, J. D. Rush, “Reduction Potential of the $\ce{CO2/CO2^{.-}}$ Couple. A Comparison with Other $\ce{C1}$ Radicals,” J. Phys. Chem. 1987, 91(16), 4429-4430 (DOI: 10.1021/j100300a045).
T. Iwasita, W. Vielstich, “On-line mass spectroscopy of volatile products during methanol oxidation at platinum in acid solutions,” Journal of Electroanalytical Chemistry and Interfacial Electrochemistry 1986, 201(2), 403-408 (https://doi.org/10.1016/0022-0728(86)80064-X).


Stereoselectivity of dihydroxylation with osmium tetroxide Dihydroxylation responds to steric effects and proceeds at less hindered positions. What will be the major product in the following reaction if the reagent used is $\ce{OsO4}$ followed by $\ce{NaHSO3}$?

Won't interactions between $\ce{- OH}$ bring $\ce{OsO4}$ on the same side as $\ce{- OH}$? Or will the $\ce{OH}$ sterically hinder the reagent from attacking the olefin from the same side?
Why I am I having this doubt is because I read that $\ce{m-CPBA}$ can be directed by hydroxy groups. For example, if the compound given below were to react with $\ce{m-CPBA}$, the product formed is:

The reason is because of interactions between $\ce{m-CPBA}$ and the substrate like this:

EDIT:

It's given in Classics in Stereoselective Synthesis by Erick M Carreira and Lisbet Kvaerno that $\ce{OsO4}$ with $\ce{NMO}$ does not respond to directing effects while $\ce{OsO4}$ with $\ce{TMEDA}$ responds to directing effects.

It's given in the same book that the reason why $\ce{OsO4}$ doesn't attack on the same side as $\ce{-OH}$ is because of electrostatic repulsions between oxygen atoms. These repulsions increase in E-alkenes. Fair enough. But won't the same reason apply for $\ce{m-CPBA}$? Why won't the oxygen atoms repel $\ce{m-CPBA}$ but only $\ce{OsO4}$?

Dihydroxylation responds to steric effects and proceeds at less hindered positions. What will be the major product in the following reaction if the reagent used is osmium tetroxide followed by sodium hydrogen sulfite?

The crux of the original question seems to be why standard Upjohn dihydroxylation conditions (cat. osmium tetroxide w/ NMO as the terminal oxidant) doesn't respond to hydrogen bonding based directing effects, whilst stoichiometric osmium tetroxide in combination with TMEDA does.

One of the earliest reports of diamine ligands on osmium (of which TMEDA is one) was by EJ Corey's group in 1996 (, who obtained a crystal structure of osmium tetroxide bound to, in this case, a chiral diamine in a bidentate fashion. (If you're familiar with the Sharpless Asymmetric Dihydroxylation you'll see why this was an important first step).

With the coordination of diamine ligands to the osmium tetroxide understood, several groups began to think about why this allowed for hydrogen bond based direction of the dihydroxylation. The most comprehensive reference I've found is a chapter in Organic Reactions by Donohoe et al. (DOI: 10.1002/0471264180.or076.01 sadly not widely available, but a .doc file missing the figures is available from the publishers website which may be somewhat helpful).

Donohoe's group (in collaboration with several others) also published a detailed study of the transition states of these directed reactions in 2002 (DOI: 10.1021/jo026161y hopefully more widely available).

The essential argument being made is that the TMEDA (or other chelating diamine) is acting as a Lewis-base, providing electron density to the osmium tetroxide, which in turn makes it more willing to hydrogen bond (more electron dense carbonyls are more willing to H-bond). This is perhaps a sensible conclusion given how electrophilic osmium tetroxide is, with 4 highly electronegative atoms all pulling for electron density.

In the paper Donohoe proposes two possible TS structures for the directing effect, with either the oxygen or the nitrogen being involved in coordination to the hydroxyl of the substrate. Through NMR and IR work it was eventually determined that TS-A was the true pathway in which the TMEDA is donating to the osmium tetroxide, with the carbonyl on osmium using its lone pair to coordinate to the hydroxyl of the substate.

To conclude. Standard Upjohn conditions give the product expected by a steric argument alone (opposite to the hydroxyl) whilst modified TMEDA conditions give the product formed through H-bond mediated direction.

  But won't the same reason apply for m-CPBA

This part of the question should maybe be another question entirely, but I think its fair to say that osmium tetroxide bears zero resemblance to mCPBA, the electronics of a metal centre vs the electronics of a peracid are completely different and as such their relative ability to hydrogen bond and hence be directed will be significantly different.


Confusion on when to use mass of entire solution when performing q=mc∆T I hope you are having a good day.

I am studying for the AP Chemistry exam in May 2020. While studying, I encountered a doubt.

In Problem 1 (the question on the top), the educator saw that the problem was referring to the solution as a whole (the problem states ∆Hsoln), and used 100 g + 10 g = 110g grams for the mass in Q = mc∆T and then, on the next part, did ∆H = -Q/n to find the enthalpy of solution.

However, in Problem 2 (the question on the bottom), the educator uses the total mass of the solution (200 grams) even if the problem refers to the reaction (the problem states ∆Hrxn) and not the solution. When I saw the question, I thought I had to convert the 1 mol Ag+ and 1 mol Na+ to grams, and add those masses for (m) in Q=mc∆T since the problem referred to the reaction and not the solution.

As a result, I am confused as to when exactly I should use the mass of the entire solution instead of the mass of the reactants in my Q = mc∆T and ∆H = ±Q/n calculations.

Thanks so much!!! :)

In mcΔT, you must use the total mass of the system that is changing temperature. In this case, that is the solution.

The same thing is occurring in the second problem. The temperature of the solution changes, and therefore the solution mass is the mass you should use.

It wouldn't make sense if the solute changes temperature but the solution it is dissolved in doesn't change temperature, does it? I find it always helps me to think about the physical process that is happening for questions like this. Always identify what your system of interest is, and the initial and final states of the system. Doing so can clear up a lot of these problems, and will also help you get through more difficult problems if you ever take a thermodynamics course.


Determination of products in reactions involving carbocation rearrangement? How do you determine the migratory "aptitude" of various groups during carbocation rearrangements? Is there a experimentally determined order?
 For example, what will be the product in case of the following pinacol-pinacolone type rearrangement:
$$\ce{(cyclo-pentyl)CH3C(OH)-C(OH)(CH3)(cyclo-pentyl) ->products}$$

Will the methyl group be transferred or the cyclopentyl moiety? The same question arises if the cyclopentyl is replaced with phenyl group? In general, which shifts, methyl, hydrogen, phenyl or any other group, will take place? Or equivalently, what determines which group has a greater tendency to undergo shifts during carbocation rearrangements?

All else equal, the smaller group migrates preferientially. The energy required to more the group is related to the mass of the group:

$$W=\int_0^\vec{x}\vec{F}\cdot d\vec{x}$$
$$\frac{d\vec{x}}{dt}=\vec{v}   \ \ \ \ \ \therefore \ \ \ \ \ d\vec{x}=\vec{v}dt$$
$$W=\int_0^t\left(\vec{F}\cdot \vec{v}\right)dt
$$$$\vec{F}=m\vec{a}$$
$$W=m\int_o^t \left(\vec{a}\cdot\vec{v}\right)dt=m\int\left(\vec{v}\cdot\vec{a}dt\right)$$
$$\vec{a}=\frac{d\vec{v}}{dt}$$
$$\vec{a}dt=d\vec{v}$$
$$W=m\int_0^\vec{v}\vec{v}\cdot d\vec{v}=\frac{1}{2}m\vec{v}^2\vert_0^\vec{v}=\frac{1}{2}m\vec{v}^2$$


pH of an amino acid solution I'm trying to help a friend of mine with this chemistry question:

  What is the pH of a $1.0M$ solution of glycine?

They're given that the pKa of COOH is $2.4$ and the pKa of the amine group is 9.6. The solution they were given says the pH is just $\frac{2.4+9.6}{2} = 6.1$, but I don't see any reason why this should be true especially since this answer initially appears independent of the given concentration of glycine.

Now if I try to solve this the long way: If I initially just have a solution of glycine, I believe it would function like a weak acid, and hence the equation is:

$$Ka = \frac{[H^+][A-]}{[HA]}$$

or

$$10^{-9.6} = \frac{x^2}{1-x} \approx x^2$$

Thus pH = $4.8$

At the very least, I found this document: https://www.ch.ntu.edu.tw/~ccchan/course/genchem2003/Notes/final_ans.pdf

Which showed that the pH of a $0.05 M$ solution of glycine has a pH of $5.54$, so with a much higher concentration of glycine functioning like a weak acid, my pH should be less than $5.54$ (and hence definitely less than the $6.1$ the professor put as his solution). Am I missing something?

I think the professor is right. I would solve it like any other acid-base problem. Glycine (G, neutral as zwitterion) can either protonate to HG+; or deprototonate to G-. The mass balance and charge balance equations are:

C$_G$=1.0=[HG+]+[G]+[G-] (Mass balance for glycine)

[H+]+[HG+]=[OH-]+[G-]    (Charge balance)

Given these two equations, and the three for $K_{a1}$, $K_{a2}$ and $K_{w}$, I could solve exactly the system of five equations with five unknowns. But that would be overkill, since we can do some approximations:

Since G is both a (not so) weak acid and a weak base, we can assume that it doesn't dissociate that much to HG+ or G-. So, the predominant form would be G, which (given the $pK_a$'s) is predominant in the 3.5 to 8.5 pH range (aprox, it depends on the errors that we tolerate).
In this pH range, and given the high glycine concentration and $pK_a$ values, [HG+]>>[H+] and [G-]>>[OH-]; so that, from the charge balance:

[HG+]=[G-]

This equality solves the problem since, combining the equations for $K_{a1}$, $K_{a2}$, we get:

$K_{a1}K_{a2}=\frac{[G-][H+]^2}{[HG+]}=[H+]^2$ so $[H+]=\sqrt{K_{a1}K_{a2}}$

$pH=\frac{1}{2}log(pK_{a1}+pK_{a2})=6.0$

You can check that at this pH, [G]=1,[HG+]=$10^{-3.2}$,[G-]=$10^{-3.6}$, so that the approximations [HG+]>>[H+] and [G-]>>[OH-] are valid.

(For more details see 6.7.5 at https://chem.libretexts.org/Textbook_Maps/Analytical_Chemistry_Textbook_Maps/Map%3A_Analytical_Chemistry_2.0_(Harvey)/06_Equilibrium_Chemistry/6.07%3A_Solving_Equilibrium_Problems)


Redox reactions redefined I saw in my textbook (NCERT class 11 part2 chemistry) that oxidation is defined as

  gain of oxygen or more electronegative element by a substance.

After it was redefined and said as

  oxidation is loss of electron(s) by any species.

Again it is being redefined and stated as

  oxidation process is visualised as a decrease in electron density.

I can't understand why they are changing the definitions again and again. I mean what things they are taking into consideration, each time they define the term

IUPAC definition

The IUPAC definition is the following:

  The complete, net removal of one or more electrons from a molecular entity (also called 'de-electronation').
  An increase in the oxidation number of any atom within any substrate.
  Gain of oxygen and/or loss of hydrogen of an organic substrate.

  All oxidations meet criteria 1 and 2, and many meet criterion 3, but this is not always easy to demonstrate. Alternatively, an oxidation can be described as a transformation
  of an organic substrate that can be rationally dissected into steps or primitive changes
  . The latter consist in removal of one or several electrons from the substrate followed or preceded by gain or loss of water and/or hydrons
  or hydroxide ions, or by nucleophilic
  substitution by water or its reverse and/or by an intramolecular molecular
  rearrangement
  . This formal definition allows the original idea of oxidation (combination with oxygen), together with its extension to removal of hydrogen, as well as processes closely akin to this type of transformation
  transformation (and generally regarded in current usage of the term in organic chemistry to be oxidations and to be effected by 'oxidizing agents') to be descriptively related to definition 1.

So that is the official definition, and it also gives multiple rules (and admits that the third rule does not always work). In order to use the second definition, you also need to look up their definition of oxidation state.

Definitions in your textbook

  I can't understand why they are changing the definitions again and again.

They are changing the definition to apply the concept to more and more reactions.

In the explanation of the IUPAC definition I posted above, they mention "the original idea of oxidation (combination with oxygen)". That's where your first textbook definition comes from. So the following reaction is an oxidation because there is a reaction with dioxygen:

$$\ce{4Fe^2+(aq) + O2(g) + 4H+(aq) -> 4Fe^3+(aq) + 2 H2O}$$

If I want to call the similar reaction with fluorine an oxidation as well, I have to generalize the definition to loss of electrons (your second definition). So this would be an oxidation of iron(II) as well:

$$\ce{2Fe^2+(aq) + F2(g)  -> 2Fe^3+(aq) + 2 F-}$$

Finally, if I want to generalize further to reactions where I want to discuss loss of electrons in atoms that are part of a covalent molecule, I need some accounting scheme to assign valence electrons to individual atoms (this is called oxidation number or oxidation state). Once that is in place, I can discuss reactions such as the following in terms of oxidation (and reduction):

$$\ce{H2C=CH2 + H2 -> H3C-CH3}$$

Here, your first definition does not apply (no oxygen) and your second definition is difficult to apply (how do we know which species lost electrons - they are all neutral, and they all have different composition?). So that is where you need some rule similar to your third rule ("decrease in electron density"). Usually, though, this is stated using oxidation states so that you are back to gaining or losing an integral number of electrons (i.e. oxidation state changed by +1 rather than "the electron density decreased slightly).

At this point, it becomes a bit fuzzy and there will be situations where the oxidation state is ambiguous. You gain clarity in the context of electrochemistry when you separate reactions into half reactions in electrochemical cells, and can actually measure the transfer of electrons that pass through the wire connecting the two cells.


Why do n AOs only form n MOs? The textbooks I have read introduce LCAO by considering the H2 molecule. In this example, there are only two possible combinations of the two 1s orbitals (in phase and out of phase).

When there are more than two valence orbitals, however, a greater amount of combinations arise. A common description of the pi-MOs in butadiene is shown below:
There are two (and I think only these two, assuming all combining p-orbitals are identical) possible combinations of the p-orbitals which (apparently) do not contribute to the bonding in butadiene. These are shown at the bottom of the illustration above.

Why is it that n AOs only form n MOs, even when more than n LCAO are possible? You would think that all possible combinations would contribute a possible electron energy state to the molecule.

If you have $n$ functions (e.g. AOs) you can make a maximum of $n$ new linearly independent functions (e.g. MOs).  If you try to make $n+1$ MOs, then any one of them can be expressed as a linear combination of the other $n$ MOs.

The most usual way to enforce linear independence is to enforce orthogonality, i.e. all your MOs have to have zero net overlap with each other. All the "correct" MOs you drew are orthogonal to each other, but the "incorrect" MOs are not orthogonal to any of the "correct" MOs.


Spectroscopy and Excitation of Electrons As metals are heated, they glow white and then blue. How might you explain this change in color as the temperature of the metal is increased?

My answer consist of the idea that with increased thermal energy more electrons are excited and as they decay they emit wavelengths of the color blue. I am not sure how to connect the idea of white light to this answer because white light consist of all the wavelengths of the visible spectrum.

I assume it is a homework question...
You may want to into the concept of blackbody radiation, as an explanation. It is discussed in most PChem / Quantum Chem books in the first chapters. Feel free to update your question according to.

Generally, we do not say that photons are excited. Photons do not have ground and excited states in the same chemical sense as electrons. Excited states (of e.g. electron) may emit photons during de-excitation, but those are the excited state of the electron (before the photon even existed) not the photon.

Also, we generally do not say photon decay: technically you can argue that absorption is something like that, but it is definitely not a common way to phrase it, and obviously absorption is absorption, so it will not emit any radiation.

Note: After this answer, the question was changed to talk about electronic excitation instead of "photonic excitations", therefore a large part of my answer is irrelevant.


Correct definition of terminal alkynes? My textbook, Organic Chemistry (6th edition) by Brown, Foote, Iverson, and Anslyn, states:

  Alkynes in which the triple bond is between carbons 1 and 2 are commonly referred to as terminal alkynes.

Is not this definition incomplete? I can imagine scenarios where we have a terminal triple bond that does not reside between carbons 1 and 2 according to IUPAC rules. E.g., $\ce{CH2=CH-C#CH}$. Here, the terminal triple bond is between carbons 3 and 4 (according to IUPAC rules, doubles bonds take priority when numbering).

You are correct about the absolute numbering of $\ce{CH2=CH-C#CH}$. By the IUPAC priority rules, the alkene has the lower absolute numbering.
However, numbering can also be relative. In your case, the relative numbering 1,2 refer to either of the following:

relative position from the end of the chain, where the chain end is #1
relative position from the functional group you care about (where that group is #1)
suggesting that a reaction occurs on neighboring positions in a molecule.

That being said, the definition given by Brown et al. is imprecise. A better definition is one of the following:

A terminal alkyne is an alkyne in which one (or both) of the two alkyne carbons is at the end of a chain.
A terminal alkyne is an alkyne $\ce{RC#CR}$ in which one or both $\ce{R}=\ce{H}$


Why does the reaction quotient use the products (multiplications) of reactants and products, rather than their respective sums? As you may know, the reaction quotient $Q_c$ is defined by the equation
$$
Q_c = \frac{[C]^\gamma [D]^\delta}{[A]^\alpha [B]^\beta}
$$
for the chemical reaction
$$
\alpha A + \beta B \rightarrow \gamma C + \delta D.
$$

This is something I've been struggling to understand at an intuitive level. Why is $Q_c$ the ratio of the product of the reaction's product concentrations and the product of the reaction's reactant concentrations (sorry for the confusing terminology; I can't think of a more precise way to phrase that), rather than the ratio of the sums of concentrations, like this:
$$
Q_c' = \frac{\gamma[C]+\delta[D]}{\alpha[A]+\beta[B]}
$$

This second, incorrect equation makes more sense to me because concentrations of solutes add up. Is the ratio defined as such so that it is directly proportional to the concentrations of each product and inversely proportional to the concentrations of each reactant? Is there some relation to kinetics that might give me a deeper insight into the meaning of this quotient?

The derivation cited can help you understand it mathematically. But the answer also turns out to be satisfactory intuitively. Let me try to explain how.

The term $RT\ln Q_c$ is derived form the entropy part or $\Delta G=\Delta H-T\Delta S$, and in a slightly abstract way, represents entropy of mixing of the products and the reactants. This is why reactions which are spontaneous initially, as concentrations change, reach to the equilibrium state before complete conversion of reactants into products. If there was no entropy of mixing, a reaction which had negative free energy change(under any set of conditions) would continue to have a negative change (hence, spontaneous) until all of the reactants is converted to products, which will be the state of maximum entropy. But since there is entropy of mixing, the state of maximum entropy is reached before all the reactants are converted into products (because somewhere in between the entropy of mixing due to having a mixture of reactants and products outweighs the negative $\Delta G$ for the reactants to convert into the products) , and the concentrations at which this maximum entropy state can be reached is precisely what the equilibrium constant tells us.

To calculate the entropy, we need to realize the fact that certain components of the reaction may have a greater contribution in deciding the equilibrium constant than others because, at the maximum entropy state, the components are present according to their stoichiometric relations and hence one might be in a well-defined excess over other. The relative contributions of each species to the entropy is logarithmic, that is, the increase in entropy is a logarithmic function of concentration (related slightly in a abstract manner to $S=k_B\ln \Omega$). Therefore, the contribution to the free energy and consequently in deciding the $K_c$ is logarithmic in the concentrations. The stoichiometric coefficientts are multiplied to the logarithmic contributions of each(which appear as the power to which the concentrations are raised when brought inside the logarithm) and the relative contributions are simply added, but since the addition of logarithms means multiplications of its arguments, we get the familiar $RT\ln Q_c$ term which gives the mentioned definitions of reaction quotient and the equilibrium constant.

I have used reaction quotient and equilibrium constant interchangeably but the RQ represents an unfulfilled equilibrium constant and hence represents the same drive towards maximum entropy as $K_c$ (which is the state of maximum entropy itself).


Entropy change in an adiabatic expansion Question
  One mole of ideal gas initially at a pressure of 1 atmosphere and $T = 298\ \mathrm K$, is
  expanded into a volume 50 % larger adiabatically.

In this adiabatic process, no work is done on the surroundings and $\mathrm dQ=0$. And entropy is defined as $\mathrm dS=\frac{\mathrm dQ_\text{rev}}{\mathrm dT}$. However, there must be a change in entropy surely?

Is the definition $\mathrm dS=\frac{\mathrm dQ_\text{rev}}{\mathrm dT}$ only at a constant volume? Because I have also seen entropy defined as $\mathrm dS=\frac{C_V}{T}\,\mathrm dT+\frac{R}{V}\,\mathrm dV$ and there would be a change in both temperature and volume. However, how do I find what these changes are? Would it make a difference if the expansion was reversible or irreversible (the question didn't make it explicitly clear)

"One mole of ideal gas initially at a pressure of 1 atmosphere and T = 298 K, is expanded into a volume 50% larger adiabatically."

The question doesn't have sufficient information for a solution, because we don't know if the adiabatic expansion is reversible or irreversible.
If reversible:
If the expansion is done reversibly, then we know entropy of the universe can't be changing (because reversible), and that heat flow from the gas to the surroundings is zero (because adiabatic).  Without heat flow, the only way the entropy of the universe can stay constant is if both the entropy of the universe and of the gas stay constant.  In this case you can apply both of your equations.  But keep in mind that your equation for $dS$ is only true for ideal gases.
$dS = \frac{\delta q_{rev}}{dT} = 0$
$dS = 0 = \frac{C_V}{T}dT + \frac{R}{V}dV$
$\frac{dT}{T}=(-\frac{R}{C_V})\frac{dV}{V}$
This is now a differential equation you can solve for temperature if you know volume.
As the other answers have indicated, a reversible expansion will involve the gas doing work on the surroundings.  One way to think about why this is so is that for the process to be reversible, the potential of the gas to do work must not be lost, and so if this potential for doing work doesn't remain in the gas, it must go to the surroundings.  A loss in the potential to do work (of the universe) is an increase in entropy.  And in a reversible process that can't happen.
Think of gas in a piston, where the piston is held down by a pile of sand.  The reversible expansion involves slowly removing sand, grain by grain, from the piston.  As the grains are removed, the piston rises and the gas lifts the remaining sand, doing work against gravity.
If irreversible:
...then we still don't have enough information because there are many ways the process can be irreversible.  But one particular, extreme case of irreversibility is if the adiabatic expansion occurs without transfer of any energy, then we can solve the problem.  In this extreme scenario, neither heat nor work are not transferred.  You can thus infer that $dU=0$ for the gas, so
$dU=-TdS + p dV=0$
$TdS = p dV$
Using your definition for $dS$ for an ideal gas,
$T \frac{C_V}{T}dT + T \frac{R}{V}dV = p dV$
Keep in mind that $\frac{RT}{V}$ for an ideal gas is $p$, so you should get that
$C_V dT = 0$
Which is also a differential equation that you can integrate to get the temperature change.  (The obvious solution is that the temperature does not change in this case.)
In this second case, now that we know that $T$ is constant, we know $dT$ is zero, and thus using your equation for $dS$, we get that $dS = \frac{R}{V}dV > 0$, i.e., the entropy of the gas increases, which makes sense because the process was irreversible.  The potential to do work has been lost.
Think of gas confined to one half of a cylinder, and the other half of the cylinder is empty (i.e. vacuum).  If the divider between the filled and unfilled halves is suddenly removed, that would be a case of adiabatic, isoergic expansion.