text:
Splitting of multiplets in ¹H NMR spectra?
I'm currently taking VCE (Victorian Certificate of Education) Chemistry classes, and we're currently studying the interpretation of spectra produced by Hydrogen NMR (Nuclear Magnetic Resonance) spectroscopy.
When studying the spectra of High Resolution 1H NMR, the peaks representing the different Hydrogen environments are split into multiplets based on the protons surrounding these environments. There has been a considerable amount of confusion in my classes over the actual principles / rules of thumb on how to calculate the multiplets for a particular environment of a known chemical (i.e.; known structue), based off the 'n + 1' rule.
(i.e.; an environment with n neighbours will be split into n+1 multiplets).
We are absolute on the principles that;

Peaks of a particular Hydrogen environment are not split by neighbouring protons in equivalent environments.
OH does not, and is not split by, it's neighbouring environments.

However, immense confusion arised over whether the following principle was correct.

Peaks of a particular Hydrogen environment will only be split by the protons in neighbouring environments once for each type of neighbouring environment.
e.g.  The middle "CH2" environment in "CH3–CH2–CH3" will only have 4 peaks;
Although it has 6 neighbouring protons, they are two lots of the same environment (CH3).

As a class, we found numerous examples from different text books and sources that provide examples of ¹H NMR spectra which did not clarify the matter; Some considered all neighbouring protons as neighbours, others discriminated on the repeated neighbouring environments. For example, the CH2 in CH3–CH2–CH3 was sometimes split into 4 peaks or 7 peaks, depending on the source.
Many Chemistry teachers contradicted each other on the matter.
There was repeated self corrections made by the teachers, such that now nobody really knows whether this principle is correct or not.
So, is there anybody that has the correct information on the matter?
Is there a reasonable explanation behind this strange lack of correlation,
or is there a common misconception about multiplet splitting?
Ultimately; How many multiplets should the CH2 in CH3–CH2–CH3 have?
(Note that if there is a complicated explanation that I am currently only at Year 12 VCE level, so links to resources I can pursue would be extremely helpful! It's been established that VCAA (an authority for the education system in Australia) ensures that the chemicals featured in the exams for NMR analysis will not be of a structure so as to allow the ambiguity above.)

Which nuclei you should consider as neighboring atoms for the multiplet splitting depends on the size of the coupling constant. The coupling constant are usually named after the number of bonds between the coupling nuclei, so a coupling between two hydrogens that are three bonds apart from each other would be a $\mathrm{^3J_{HH}}$-coupling. In an alkyl chain like CH3CH2 the coupling between the CH2 and the CH3 hydrogens would be a $\mathrm{^3J}$-coupling that is about 7 Hz large. Couplings over more than three bonds are usually not observed, because they are smaller than the linewidth of the signals in your NMR spectrum. The exception to that you're likely to encounter are couplings along C=C double bonds, where you can even see $\mathrm{^4J}$-couplings or more.
So the simplified rule would be that couplings along three bonds are visible, couplings along more than three bonds are only visible when there is at least one C=C double bond along the way.
The concepts of chemical and magnetic equivalence are essential to understanding how different multiplicities arise. Chemical equivalence means there exists a symmetry operation that exchanges those nuclei. Magnetically equivalent nuclei additionally need to have identical couplings to all other spins in the molecule. Magnetically equivalent nuclei don't couple to each other.
In propane the hydrogens of both CH3 groups are three bonds away from the hydrogens of the CH2 group, so the coupling between those is visible in the NMR spectrum. The hydrogens of each CH3 group are magnetically equivalent, due to the fast rotation along the C–C bond, and the two CH3 groups should also be magnetically equivalant. So you have 6 magnetically equivalent hydrogens that couple to your CH2 hydrogens. The result of that is a splitting into a septet (7).
The OH-group is an interesting exception, as you would expect it to lead to a visible coupling on hydrogens connected to the same carbon, but you don't observe that under most conditions. The reason is that the OH is acidic enough that the hydrogen exchanges quickly with the solvent, so the hydrogen dissasociates and associates quickly. This happens too fast for NMR, so the other nuclei only see the average OH-hydrogen. This eliminates the coupling to the OH, and it is also the reason why the OH-signal is often very broad or even completely gone in NMR spectra.

In addition to mad scientists answer I would like to explain what the quantum mechanical phenomenon is that leads to a J-Coupling. The interesting part about the J-coupling for chemists is that this coupling is a proof for the presence of an electron bond between the two coupling nuclei.
The coupling is based on two principles: a Fermi contact between the nucleus and an electron, the Pauli exclusion principle that two electrons must not have the same set of quantum numbers. In this case if one electron spin is up, then the other electron in this bonding orbital must be down. The other electron couples then to the second nucleus by another Fermi contact.
The chain of couplings is therefore:
Nucleus 1 ---Fermi contact---> electron 1 ----Pauli exclusion principle----> electron 2 ---Fermi contact----> Nucleus 2
This total chain is the effect which is called J-coupling. In a handwaving fashion one could say that through the first nucleus feels through the coupling if the second nucleus is up or down. This is shifting the energy levels a bit which can be then found in the spectrum as a shift in the resonance position. As the energy conservation principle holds one line is shifted to lower energies and the other by the same amount to higher energy leading to a doublet.
The OH proton is actually split as well by the J-coupling, however the line width of the OH resonance is bigger than the coupling constant, so the doublet is smeared out.

One could also talk about Hund's rules to further refine the description.

@CHM No that is not correct. Hund's rule gives guidance how to occupy orbitals in multi-electron atoms. The first rule is the Pauli principle. Other rules like having electron repulsion do not apply here. Furthermore the orbital I am talking about is an molecular orbital, because it describes the bond, not an atomic orbital.

While the above answers quite clearly explain the theory, I thought I would provide some clarification and citations:
If it is indeed propane you were taking the spectrum of, then you would get a heptet (7 peaks, in the standard pascals triangle pattern). Lets make another example though
R-CH2-CH2-CH2-R'
If R = R' then you get a quintet (5 peaks) in a pascal's triangle.
However, if R is not the same as R' then you will get a triplet of triplets, that is, three distinct triplets with the same J-coupling. This may or may not be easy to discern, depending on what R and R' are.
Here are some useful sources:
These two papers can give you some great examples to examine, and are useful enough I always keep them on  my USB key.
NMR Chemical Shifts of Common Laboratory Solvents as Trace Impurities; Hugo E. Gottlieb, Vadim Kotlyar, and Abraham Nudelman; J. Org. Chem. 1997, 62, 7512-7515
NMR Chemical Shifts of Trace Impurities: Common Laboratory Solvents, Organics, and Gases in Deuterated Solvents Relevant to the Organometallic Chemist; Gregory R. Fulmer,Alexander J. M. Miller, Nathaniel H. Sherden, Hugo E. Gottlieb, Abraham Nudelman, Brian M.  Stoltz, John E. Bercaw, and Karen I. Goldberg; Organometallics 2010, 29, 2176–2179

Is J-coupling equivalent or directly proportional to "chemical shift"? i.e. the x axis on a spectra?

No, shift is the position of the peak, measured in ppm, J-coupling is the distance between two coupled peaks, i.e. how far the two peaks in a doublet are from one another, measured in Hz.

Oh! So when you say "two distinct triplets with the same J coupling", there are two visible sets of triplets (obviously with the same distances between multiplets) or only 1 visible set of triplets (because the two sets have the same J coupling and so appear as 1 triplet)?
Thanks
(Sorry, still trying to articulate these ideas into the spectra)

You will have 6 peaks, organized into two sets of three, forming 1:2:1 triplets. The distance between the center peak and the two flanking ones will be the same in both cases.

@Canageek you write "...then you will get a triplet of triplets, that is, two distinct triplets..." was that a typo and should read "...you will get a doublet of triplets..."?

@Zalumon I think I meant a triplet of triplets. Each peak of the original triplet would get split into three more peaks.


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Timescale of Protonation and Deprotonation across micelle layers
I am curious about the timescales of protonation and deprotonation of solvent systems. As a followup, how is this affected when the proton source is separated by an organic phase? For instance, it is known that the pH inside a liposome can be maintained from the outside. What are the timescales of protonation when the protons need to be transported across various micelles thickness.

If anyone does weigh in, I'd be interested in some details regarding the self-ionization of water. How quickly do these proton transfers occur.  What order of magnitude are the energy barriers associated with these?  More importantly, does anyone have a good citation that deals with this, particularly for reaction mechanisms?  If not I may just make a new submission.

The timescale of protonation is very fast, to the point that is makes almost no sense to ask the question.
More importantly, you are asking about the timescale of diffusion of hydronium ion through the micelle.

@Lighthart, I agree with your insight. I am still looking for a quantifiable value for both.

Permeation of protons, potassium ions, and small polar molecules through phospholipid bilayers as a function of membrane thickness

Reported values for proton permeability coefficients vary over
  a
  wide
  range,
  from
  $10^{-2} \, \mathrm{cm/s}$ to $10^{-7} \, \mathrm{cm/s}$
  (Nichols
  and
  Deamer,
  1980;
  Nichols
  et
  al.,
  1980;
  Biegel
  and
  Gould,
  1981;
  Elamrani
  and
  Blume,
  1983;
  Grzesiek
  and
  Dencher,
  1986;
  Perkins
  and
  Cafiso,
  1986;
  Norris
  and
  Powell,
  1990).
  This
  variation
  relates
  to
  the
  use
  of
  different
  techniques,
  lipids,
  and
  bilayer
  systems
  (liposomes
  and
  planar
  bilayers)
  or
  different
  pH
  ranges
  (Perkins
  and
  Cafiso,
  1986).
  The
  proton
  data
  of
  this
  study
  agree
  with
  the
  higher
  reported
  values.
As revealed by Fig. 4, proton permeability coefficients strongly
  depend on membrane thickness, starting at a maximum value of $1.3 \cdot 10^{-2} \, \mathrm{cm/s}$ for the shortest lipid and steadily decreasing to a value of $4.9 \cdot 10^{-5} \mathrm{cm/s}$ for the longest lipid. It follows that proton permeability decreases by a factor of approximately 250 as the thickness of the
  hydrophobic region is increased from $20 \, \mathring{\mathrm{A}}$ to $37 \, \mathring{\mathrm{A}}$.

Can you summarize what's in the paper so the answer can be self-contained, please?

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text:
Determining the pH of an organic solvent
Our work is faced with the issue of determining the charge of compounds that are dissolved in chloroform. One could realistically determine the charge of the relevant moieties knowing the $\mathrm{p}K_\mathrm{a}$ of those groups. However,

How would you determine or calculate the $\mathrm{pH}$ of an organic solvent?

An alternative way of asking this question is how would one calculate the concentration of hydronium ions or free protons in a nonaqueous solution.

I think a good first step to answering this question would be to identify exactly what ion is formed by the addition of protons to chloroform. Hydromium ions can't exist without water (is the system open to atmospheric water vapor?). Maybe one or more of the Cl atoms on the solvent gets protonated.

How about this? http://onlinelibrary.wiley.com/doi/10.1002/anie.201000252/abstract

I'm not sure why you're wanting to exert so much effort to find the charge of compounds.
If you really want to know the charge, determine it more directly. Electrophoresis can allow you to do so. A charged species is put in a capillary tube that is in between a cathode an an anode. When a charged species is subjected to an electric field, it will experience a force proportional to its charge and the field strength.
Drag forces present in the solvent are also a consideration in this setup and they depend on the "size" of your ion as well as viscosity of your solvent. One also has to worry about Joule heating, as an increase in temperature will cause poor reproducibility. Picking a suitable detector is fun, but hopefully you can get by with the common UV-vis detector.
So if you can find a suitable setup, know the radius of your ion and know the time between injection of your sample and its arrival at a detector, you can at least determine its effective charge.

Are you suggesting that I should calculate the pI of my solvent and then backtrack that overall charge?

No. I'm suggesting that you measure its electrophoretic mobility, $\mu_{ion}= \frac{q}{f}$. Once you know the friction coefficient, $f$, then you can determine its charge, $q$.

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Molecular explanation of the Hofmeister Series
The Hofmeister Series is a useful metric in understanding how to precipate proteins. However, wikipedia and many other source indicate that the order of the series is rather empirical. I'm curious about the variety of explanations of the mechanism of the series as well as the application of the series to other non-protein macromolecules.

The molecular origin of the Hofmeister series is not well understood and still a very active area of research. For example, see the upcoming Faraday discussion conference. The syllabus for the conference says:

Although it is now clear that the Hofmeister series is intimately connected with ion hydration in homogeneous and heterogeneous environments and with ion pairing, the molecular origin of these effects has been poorly understood. Biochemists and physical chemists have been typically using the term Hofmeister series to put a label on ion specific behaviour in various environments, rather than to reach a molecular level understanding and, consequently, an ability to predict a particular effect of a specific salt ion.

(So, this is not an answer, but rather too long for a comment…)

Actually, I think this is an answer, and a good one, as it records something useful about our state of knowledge. And that is useful to do here.

A similar effect occurs in synthetic polymers. poly(N-isopropyl Acrylamide), (usually just p(NIPAM)) is a thermosensitive polymer. At room temperature, it dissolves easily in water. When you heat the solution past a certain temperature (the "lower critical solution temperature," or LCST) the polymer chains undergo a conformational change and precipitate out of solution. When you cool the solution back down below the LCST, the polmer returns to its original conformation and re-dissolves.
http://pubs.acs.org/doi/abs/10.1021/jp0690603 and http://pubs.acs.org/doi/abs/10.1021/ja0546424 are two papers from a group at Texas A&M that show that the Hofmeister series applies to these thermosensitive polymers in a similar manner. In this case, the ions lower the LCST. So even at room temperature, you could salt out the polymer with enough salt. The Hofmeister series indicates the efficacy of the ions in this salting out.
Those papers divide the ions in the Hofmeister series into "chaotropes" and "kosmotropes," depending on whether the ion tends to break apart the natural structure of the water, or tends to stabilize the natural structure of the water. They claim that the chaotropes operate by changing the surface tension of the water and by direct anion binding to the polymer, and that the kosmotropes operate through surface tension and by changing the polarization of the water molecules. They correlate the magnitude of the effect of each ion on LCST with that ion's hydration entropy and surface tension increment.
Caveats: I'm only really comfortable with p(NIPAM) itself, and with the empirical fact of the salt effect. I don't know how well-understood these entropies of hydration or surface tension increments are, or if this would apply to salting-out of proteins.


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Mechanism of arene side chain oxidation by permanganate
When treated with hot, concentrated acidic $\ce{KMnO4}$, arenes are oxidised to the corresponding carboxylic acids. For example, toluene is oxidised to benzoic acid.
I've tried to examine how this happens, using the mechanism of oxidation of double bonds via cyclic intermediate as a reference, but I can't manage to cook up a satisfactory one.
In an older book, I have read that there is no (known) mechanism for many organic oxidation reactions. I'm inclined to think that this may have changed.
So, is there a mechanism for this? If so, what is it?
If not, what are the hurdles in finding this mechanism? For example, what problems are there with other proposed mechanisms (if they exist)?

Some general information on side-chain oxidation in alkylbenzenes is available at Chemguide:

An alkylbenzene is simply a benzene ring with an alkyl group attached
  to it. Methylbenzene is the simplest alkylbenzene.
Alkyl groups are usually fairly resistant to oxidation. However, when they are attached to a benzene ring, they are easily oxidised by
  an alkaline solution of potassium manganate(VII) (potassium
  permanganate).
Methylbenzene is heated under reflux with a solution of potassium
  manganate(VII) made alkaline with sodium carbonate. The purple colour
  of the potassium manganate(VII) is eventually replaced by a dark brown
  precipitate of manganese(IV) oxide.
The mixture is finally acidified with dilute sulfuric acid.
Overall, the methylbenzene is oxidised to benzoic acid.

Interestingly, any alkyl group is oxidised back to a -COOH group on
  the ring under these conditions. So, for example, propylbenzene is
  also oxidised to benzoic acid.

Regarding the mechanism, a Ph.D. student at the University of British Columbia did his doctorate on the mechanisms of permanganate oxidation of various organic substrates.1 Quoting from the abstract:

It was found that the most vigorous oxidant was the permanganyl ion ($\ce{MnO3+}$), with some contributing oxidation by both permanganic acid ($\ce{HMnO4}$) and permanganate ion ($\ce{MnO4-}$) in the case of easily oxidized compounds such as alcohols, aldehydes, or enols.

The oxidation of toluene to benzoic acid was one of the reactions investigated, and a proposed reaction mechanism (on pp 137–8) was as follows. In the slow step, the active oxidant $\ce{MnO3+}$ abstracts a benzylic hydrogen from the organic substrate.

$$\begin{align}
\ce{2H+ + MnO4- &<=> MnO3+ + H2O} & &\text{(fast)} \\
\ce{MnO3+ + PhCR2H &-> [PhCR2^. + HMnO3+] & &\text{(slow)}} \\
\ce{[PhCR2^. + HMnO3+] &-> PhCR2OH + Mn^V} & &\text{(fast)} \\
\ce{PhCR2OH + Mn^{VII} &-> aldehyde or ketone} & &\text{(fast)} \\
\ce{aldehyde + Mn^{VII} &-> benzoic acid} & &\text{(fast)} \\
\ce{ketone + Mn^{VII} &-> benzoic acid} & &\text{(slow)} \\
\ce{5 Mn^V &-> 2Mn^{II} + 3Mn^{VII}} & &\text{(fast)}
\end{align}$$

The abstraction of a benzylic hydrogen atom is consistent with the fact that arenes with no benzylic hydrogens, such as tert-butylbenzene, do not get oxidised.

Reference

Spitzer, U. A. The Mechanism of Permanganate Oxidation of Alkanes, Arenes and Related Compounds. Ph.D. Thesis, The University of British Columbia, November 1972. DOI: 10.14288/1.0060242.


text:
How may copper acetate ligands be manipulated to change colors?
I have a solution of copper acetate and I would like to play around with the ligands to get different colors.
Background: The copper acetate was made through mixing vinegar (5% acetic acid), $\ce{NaCl}$, and $\ce{C}_{\textrm{(s)}}$. The deep blue-colored copper acetate spontaneously formed during a month in my dark storage room.
[edit] I neglected to mention that the experiment simultaneously produced a $\pu{0.5 inch}$ deposit of what appears to be Copper Carbonate or Verdigris on the bottom of the 1-Liter beaker. Also, I have let the copper acetate solution evaporate for several years now. The former $\pu{1L}$ is now $\pu{0.5 L}$ and has begun precipitating crystals (like the ones on the wiki page). Fun fact: During the Renaissance, glacial acetic acid was made by dry distilling metal acetates and primarily copper(II) acetate.[/edit]
Question(s):

What easily obtained household chemicals may be mixed with samples of the copper acetate to change the ligands attached to the copper and thereby alter the color?
Will heating or cooling the solution change the color and/or ligands?

That sounds like a cool project! Do you have enough solution for some trial and error? All manner of organic oxyanions bind to copper. You might try EDTA, which can be found in a number of beverages, although I don't know what effect that would have on color.

Common off-the-shelf eye drops are one of the best sources of EDTA if I recall correctly. Citric acid might be worth a try, instead, though, since it's cheaper and will probably behave pretty similarly.

Copper arsenate is a wonderful green… but I'm not sure you could consider arnite sources as “home chemicals” :)

@Pat I have around 1/2 Liter of deep blue copper acetate.

I have found an interesting article on making Cu2O (Cuprite crystals) from Copper acetate titled, "Seed-mediated synthesis of polyhedral 50-facet Cu2O architectures." available here: http://www.rsc.org/suppdata/ce/c1/c1ce05243h/c1ce05243h.pdf I wonder if that polymorph of Cuprite is also a superconductor.

http://chemistry.stackexchange.com/questions/574/chemical-compounds-responsible-for-the-colors-in-flowers

Copper(I) is unstable in water, disproportionating into Cu(II) plus metallic copper. You have Cu(II) acetate. This lab makes Cu(I) then stabilizes it. What happens if you added excess thiourea (photography supplies) as both reductant and stabilizer? (Thiourea is mildly toxic.  Disposable gloves, do not ingest.) http://infohost.nmt.edu/~jaltig/CuCmpds.pdf

Copper tartrate (cooking cream of tartar), copper glutamate (from Accent, monosodium glutamate), by mixing the solutions with stirring - and then with some Epsome salt (magnesium sulfate).  http://stud-fier.blogspot.com/2012/01/fehlings-test.html http://ce.sysu.edu.cn/mzw/Publications/p_before2001/18440015760.html

I may have accidentally created copper cloride in addition to the copper acetate. I decided to just let the solution (made form Copper, vinegar, and salt) evaporate with a small fan speeding the evaporation for the last few weeks. The tallest copper acetate crystals started breaking the surface and then the surface of the liquid started to form a skin (pic 1) so I poured off the now green liquid into another jar. Copper acetate is blue, which is why the green solution seems to suggest that there is copper cloride.

Evaporation! Note the blue-green layer of copper carbonate below the dark copper acetate crystals.

Top View

The green liquid on the left is what was left from evaporation (and I think is copper chloride) - next to another jar of copper acetate.

The copper acetate crystals were covered in a thin layer of copper carbonate because last week I disturbed the solution and stirred up a cloud of copper carbonate accidantally. I rinsed them with vinegar and they cleaned nicely.

The largest crystal was just over 1 cm wide. These crystals were grown from a solution that started evaporating 5 years ago.

The copper carbonate goo in the bottom needed a use so I added more vinegar to it and it formed the copper acetate solution three pictures up (next to the green copper chloride).

Copper(II) is almost always blue, so I'm not sure how much luck you'll have trying to change the color. $\ce{CuCl2}$ is one of the few solutions I can think of off the top of my head that isn't blue - it's more greenish-blue.
$\ce{CuCl2}$ can be made via electrolysis of $\ce{NaCl}$ and copper plates. You could also make $\ce{CuCl2}$ if you add hydrochloric acid and some $\ce{NaCl}$.

The wiki page has a picture of various shades of blue-green Copper(II)Chloride: http://en.wikipedia.org/wiki/Copper(II)_chloride

Here are two more blue shades:

the bright blue of the hexa-hydrated $\ce{[Cu^{II}(H2O)6]^2+}$ (dissolve $\ce{CuSO4}$, maybe not household chemical, but used e.g. as fungicide in farming incuding organic farming),
if you drop ammonia into this solution, you'll first see a white (blueish with the remaining hexahydrate) precipitate or clouds of the hydroxide
which dissolves with more ammonia into the deep blue tetra ammine complex
$\ce{[Cu^{II}(NH3)4(H2O)2]^2+}$

A deeper explanation of the ammonia ligands you mentioned is found in section 1 of: http://www.chem1.com/acad/webtext/chembond/cb09.html

Old question, but interesting topic. I don't intend to reply, but rather to add some relevant info. There is no "one" copper acetate. There are many, some neutral, some basic. They all follow the rule $$\ce{[Cu(CH3COO)2]_x·[Cu(OH)2]_y·nH2O}$$ in which $x ≥ 1, \; y ≥ 0$, and $n ≥ 0$ (i.e., with ratios $x:y:n$, being basic if $y ≠ 0$).
Neutral copper acetates exist in different hydration levels: $\ce{{[1:0:0] }Cu(CH3COO)2}$ or $\ce{{[1:0:1] }Cu(CH3COO)2·H2O}$ or $\ce{{[1:0:2] }Cu(CH3COO)2·2H2O}$, being the monohydrated one the most stable and commonly found/produced.
The known types of basic copper acetates (Kühn, 1993) are 2:1:5 (blue), 1:1:5 (blue), 1:2:0 (blue, unstable), and 1:3:2 (green), but eventually other formulas and/or structures are mentioned (e.g., $\ce{Cu4(OH)(CH3COO)7·2H2O}$, allegedly isostructural with $\ce{Co4(OH)(CH3COO)7·2H2O}$, according to López-Delgado et al. 1998 & 2001).
Scott (2002) has some neat "concentration maps" (Pourbaix diagrams) which show the products obtained depending on (reagents and respective) concentrations.

References:

Kühn, Hermann (1993): Chapter 6: Verdigris and Copper Resinate. In Ashok Roy (Ed.): Artists' Pigments: A Handbook of Their History and Characteristics, Volume 2; National Gallery of Art: Washington, DC and Archetype Publications: London, UK, 1993, pp. 131–158.
López-Delgado, Aurora; Cano, Esmeralda; Bastidas, José Maria; López, Félix A. (1998): A Laboratory Study of the Effect of Acetic Acid Vapor on Atmospheric Copper Corrosion. J. Electrochem. Soc. 1998, 145(12), 4140-4147 (DOI: 10.1149/1.1838928).
López-Delgado, Aurora; Cano, Esmeralda; Bastidas, José Maria; López, Félix A. (2001): A comparative study on copper corrosion originated by formic and acetic acid vapours. Journal of Materials Science 2001, 36(21), 5203–5211 (DOI: 10.1023/A:1012497912875).
Scott, David A. (2002): Copper and Bronze in Art: Corrosion, Colorants, Conservation; Getty Conservation Institute: Los Angeles, CA, 2002 (Partially available online at https://books.google.com/books?id=yQKuSOzkLvcC).

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text:
During the development of aspartame, why was the methyl ester chosen?
One frequent argument against the safety of the synthetic sweetener aspartame is that it can be hydrolyzed, with one of the hydrolysis products being methanol, which is known to be toxic. Of course, usually not much aspartame is added in food anyway to yield a sufficient amount of methanol or its oxidation product formaldehyde, but this does beg the question: why wasn't, say, the ethyl ester chosen to be developed as an artificial sweetener? Is there a structure-activity relationship between the alcohol used to esterify the dipeptide and the sweetness of the product?

You mention the irrelevance of the toxicity ba methanol yourself. So, the inventor then either did not think about it or the methyl is essential for taste and/or degree of sweetnes. I think it is the latter, because nobody would stay with the methyl ester today, if that was not a necessity. Aspartame use is limited by chemical stability, so eg the presumably much more stable  t-Bu ester might be commercially relevant. That this is not done, speaks for itself.

This question raises more important issues than just the technical "why methyl ester," so I'll address those too.
The easiest explanation for their focus on the methyl ester is that the ethyl ester just isn't nearly as sweet. This report says it is approximately $10\times$ less sweet (see Table VI on page 2689 and the entry '$\ce{Asp-Phe-OEt}$' with "$++$" being defined on pg. 2684). It's an easy economic choice between the two, assuming the only difference is sweetness/g, but you're also worried about the perceived safety concerns of the methyl derivative.
A toxicology class I once took can mostly be summed up by a cliche phrase, "the dose makes the poison." Even too much of a good thing is by definition a bad thing, for example too much oxygen, water, or even the sun can lead to death. We also say that too little of any compound won't cause measurable harm.
It's correct to say that aspartame will be metabolized into amino acids and methanol. Methanol itself is toxic in high doses, in the same manner ethanol is, by acting as a CNS depressant. This acute poisoning can cause respiratory failure or other medical emergencies such as kidney failure. Ethanol poisoning occurs in microorganisms too: it has been proposed that some yeast produce ethanol as a means of eliminating competition. This would have been potentially deadly for a prehistoric mammal too, if it were not for an enzyme, which metabolize ethanol. This enzyme is called alcohol dehydrogenase (ADH), which leads to biochemical pathways that allow the safe removal of ethanol. However, methanol is a different case. ADH instead make formaldehyde, which is then converted to formic acid by ALDH. Formic acid, in large quantities, is used in the animal kingdom as a weapon, most notably in bee venom and from the bites of a fire ant. In low quantities it is fairly benign. Formaldehyde is just a nasty chemical. Both are produced in the body as a natural response to methanol and it's easy to see how too much of either couldn't possibly be desirable. The body also has ways to eliminate these toxic metabolites assuming you don't overwhelm these systems. Hence why a sufficiently small amount of methanol won't make you blind.
Researchers have measured the amount of these undesirable metabolites in the blood after having people, like me, ingest nearly $\pu{1.8 g}$ of aspartame in one sitting. Then they looked at urine and blood concentrations and they couldn't detect any formic acid in the blood but they could clearly see it in their urine for, up to, $\pu{8 hours}$ after a dose. This evidence suggests that acute methanol poisoning isn't going to happen after a ~$\pu{200 lbs}$ person ingests $\pu{1.8 g}$ of aspartame. Let's stop for a second and put this in terms of something we can imagine.
I don't have any reliable numbers, but let's just say that for a $\pu{12 oz}$ diet pop there are $\pu{200 mg}$ of aspartame. That means for every liter of pop, there are approximately $\pu{564 mg}$ of aspartame. That means $\pu{3 L}$ of pop would get me to $\pu{1.8 g}$ of aspartame. That means I would need to drink more than $\pu{3 L}$ in one sitting to overwhelm the excretion systems. This suggests an acute poisoning isn't very plausible.
There is, however, long-term toxicity to consider as well. This is a much more difficult question to probe. "Sure, there is no acute toxicity, but what happens in the long-term with a low dose?" For the answer to this question, we have to turn to toxicologists who have not reported long-term issues. They are looking too. The general consensus is that there is no long-term harm known for the general population. It is hard to quantitate these things, because to perfectly do the experiments necessary would require controlling a bunch of parameters that human experimentation laws prevent, let alone funding constraints. So, researchers make the best with what they have available and using these tools they cannot find any measurable problems with this particular compound in the long-term for the general population. Their analyses are limited by uncertainties in their measurements and this is just the general nature of science.
Tomorrow the majority of scientists in this area of interest could be proven wrong by a very careful experiment, but over forty years of research hasn't shown this chemical to be dangerous at the doses we're likely to be exposed to for any duration of time. We have to make decisions off of the best evidence we have available. I think it's okay for people to apply some level of skepticism to these things, because in science we're continually correcting our accepted knowledge base. Once in a while the general community is proven wrong about something long thought to be true and everybody benefits when that is the case, but we can't say the general community is wrong without substantial evidence. It is for this reason that we have the FDA to evaluate the best available evidence and via a network of reliable strangers, we can enjoy a very cheap/sweet drink that is low in calories. Frankly, I am more worried about the amount of sugar I take in than I'll ever worry about the amount of aspartame I consume -- we have already good evidence that sugar, with a dose too small to cause acute poisoning, is probably killing hundreds of thousands of people due to long-term exposure.

Aspartame’s sweetening properties were discovered by accident. the G.D. Searle company was trying to develop drugs to treat ulcers in the 1960s when one of the scientists accidentally contaminated his finger and licked it. Knowing how many versions of molecules get tested in drug development, my guess is that an ethyl ester version was made, and probably failed due to improper taste or some other property. Aspartame is already “sparingly soluble” in water; I assume the ethyl ester would be even less soluble, making it less useful for sweetening drinks. However, the amount of aspartame that would have to be dissolved is so small I don’t know if it would make a difference. It's probably taste related.
As for toxicity concerns, it takes a lot of methanol to cause toxicity, Wikipedia claims 10 mL can cause permanent blindness, and 30 mL can be fatal, but the amount of aspartame a human would need to ingest to produce 10 mL worth of methanol would be ridiculous.

I've just done the math, and to get 10mL of MeOH strictly from Aspartame contained in food, you'd have to drink at least 405 soft drinks, or chew 10 390 sticks of gum. In a short time. :)

The most likely reason to use methanol instead of ethanol is cost. [1]  Ethanol is roughly twice as expensive as methanol.
Although both aspartame and the ethanol analog can be produced with reasonably high yield[2], the ethanol analog has not yet been produced commercially, so it would be even more expensive to develop industrial scale production of the ethanol analog.
Finally, to address the toxicity part of the question, aspartame is not at all toxic.  It is most stable at lower pH (~4.3) which is in the range of carbonated beverages.  So methanol is not generally produced while the food is in storage, and the body gets rid of methanol as it is produced.
One more note, aspartame can break down via heating.  This is why food that is intended to be baked is not usually sweetened with aspartame (or it is mixed with saccharine).  Although the fear here is not toxicity so much as cake that isn't sweet!

http://web.anl.gov/PCS/acsfuel/preprint%20archive/Files/41_3_ORLANDO_08-96_0880.pdf
http://www.patentstorm.us/patents/4730076/description.html

I didn't say aspartame itself is toxic. :) I said the concern is because one of its possible breakdown products is methanol.

Yes, that was what I was referring to (methanol vs ethanol).  @J. M. - Sorry if I wasn't clear enough in my answer - the point I was trying to make is that aspartame isn't toxic in use since it rarely breaks down.  You pointed out the low quantity used in food too, which is another mitigating factor.

Speculation, not an answer. Besides cost isn't a big issue with things like aspartame.

Methyl ester was chosen by the aspartame manufacturer for two specific reasons:

The methyl ester molecule binds the Phenylalanine and Aspartic Acid molecules together forming the Aspartame molecule.
The unnatural methanol in the methyl ester provides the artificial sweetness of aspartame.

As an artificial sweetener, aspartame’s job is complete when it has tricked our brain into thinking we have consumed glucose when we have not, its toxic waste (methanol) is then swallowed for our body to get rid of.
Aspartame per se can do us no harm; it does not enter the bloodstream (EFSA 2013)
At a temperature of 86 degrees F in the GI tract, the methyl ester in aspartame converts back to pure methanol releasing its bond with the two amino acids. Aspartame’s three components are then free to enter the bloodstream to metabolise separately (EFSA 2013)
Methanol is by far the most toxic of aspartames components – the other two are amino acids.
To assess the severe metabolic toxicity of methanol as a food in humans is first necessary to establish its ADI. This can then be used to compare this toxicity to the current aspartame ADI.
As has been mentioned in your comments, the blinding dose of methanol in humans is 10 ml (one tablespoon) and the fatal dose is 30 ml (3 tablespoons) a simple calculation can express these in (mg/kg) as follows :

The blinding dose of methanol in a 70 kg adult is (114 mg/kg) and the fatal dose is (343 mg/kg). To calculate the ADI we need an NOAEL, which must be between (0.0 mg/kg to 114 mg/kg). If we use for example 10 % of the blinding dose = (11.4 mg/kg) to establish the ADI, this must be divided by a safety margin of 100; therefore the ADI of methanol in humans is (0.114 mg/kg). The ADI of aspartame (40 mg/kg) contains 10 % w/w of methanol or (4 mg/kg).

Conclusion: The current ADI of aspartame contains 35 times more methanol than is safe for humans.

Your conclusion that 40mg/kg is a bit high for an ADI may be valid, but it takes some effort to consume 40mg/kg. In any case, this is not really the place for propaganda. There are also a number of issues with your science. A. 86F?! Body temperature of humans is 97.8F. B. examination of the aspartame molecule shows a phenylalanine residue in the middle with aspartate at one end and methyl ester at the other, so if anything it's "phenylalanine" holding the molecule together, not "methyl ester" C. Methanol isn´t sweet, it's the whole molecule together which has different properties to its parts.


text:
How large is the energy difference between keto- and enol-form of guanine and uracil?
The nucleobases in DNA and RNA are generally present in the keto-form, and not in the enol-form. As an interesting historical note, James Watson and Francis Crick did initially believe them to adopt the enol-form, which isn't compatible with the correct model of the DNA they later proposed.
But there are some reports indicating that in certain RNA structures guanosine and uridine form essentially a Watson-Crick basepair where one of the two bases is present in the enol-form, instead of the usual G-U wobble base pair.
Is there anything known on how large the energy difference between the keto- and the enol-form of those bases actually is?

Do you mean in situ (i.e. in a BP) or on their own?

Both alternatives would likely answer my question, but I was mainly wondering about the nucleobases on their own.

I am not an expert in the field, but a quick literature check turns up a good amount of papers on the topic. In particular, I found this paper [1] which nicely answers the question for guanine:

So, in the gas phase, the most stable enol tautomer has a 0.9 kcal/mol free energy difference with the most stable keto tautomer. That free energy difference is very exacerbated in aqueous solution, with now a ∆G° = 8.7 kcal/mol. We also notice that the most stable keto tautomer is not the same in the gas phase and in solution, and that both keto and enol have many tautomers close in free energy, showing the limits of the simple keto vs. enol line of thought.

Regarding uracil, the first reference that comes up in a bibliographic search is this paper [2]. I only quote the abstract:

The effects of the solvent on the tautomeric equilibria of cytosine and uracil are studied using Onsager's reaction field model in the framework of density functional theory. […] Our results are in good agreement with available experimental results and confirm that the polarization of the solute by the continuum has important effects on the absolute and relative solvation energies.

pKa Values of Guanine in Water: Density Functional Theory Calculations Combined with Poisson-Boltzmann Continuum-Solvation Model, Y. H. Jang et al., J. Phys. Chem. B 2003, DOI: 10.1021/jp020774x
Solvent effects in density functional calculations of uracil and cytosine tautomerism, L. Paglieri et al, Quantum Chem., 2004, DOI: 10.1002/qua.560560517

The large difference between gas phase and aqueous solution is very interesting, as the inside of a complex RNA structure is likely closer to the former.

What I don't understand is why they did not include the tautomer, which is protonated at the external amine group. That should be prone to protonation.

The data provided by @F'x are very helpful towards answering the question. But the relative stabilization of the bases when hydrogen bonded to each other is germane to the question. Because the role of stacking is still being worked out, I don't think this question has a simple answer.
Since wobble pairs are observed, but I haven't seen them outside a context where conventional base pairs are also present in the same molecule, I'm not even sure their formation is exergonic.


text:
How to differentiate a metal from its ion?
I am working on metal and its ion. The problem is, I would like to know whether the metal has been converted to its ionic form or not.
The hypothesis for the analysis is that metals are good conductors and ions possess redox potential. Therefore, by measuring the conductance and red-ox potential is it possible to  differentiate whether I have metal or metal ion.
The questions are

Based on the hypothesis, can I expect a positive result?
Apart from this, is there any other technique where I can qualitatively / quantitatively measure  metal and its ion?

Hi Anil, this question really needs some clarification. In what context are you trying to detect whether a metal M is present in its zero-charge metallic form, or in its ionic form M$^{m+}$? Is that for an aqueuous solution? Or for some other system?

Also, some context into your own level in chemistry would help tailor a better answer, once the question will be clear…

I find your question extremely vague. "Working on a metal and its ion" is like "working on carbon and its compounds" - not very informative. What exactly is the problem you're facing?

If you're looking for an electrochemical method for identifying ionization states, I would suggest cyclic voltammetry.
I like this quote from online lectures notes:

Cyclic voltammetry has been called the "electrochemical equivalent of spectroscopy" because it can be used to determine oxidation state, E° of each redox process that the compound can undergo and can even be used to study the kinetics of the redox process.  Cyclic voltammetry is one of the standard characterization methods in most inorganic laboratories.

Other than the excellent answer by F'x, there is also actual spectroscopy, specifically UV/Vis spectroscopy. This'll only work if the metal has transitions in the right range, though -- transition metals, lanthanides and actinides are usually fine, s or p-block metals and metalloids, probably not.
Then you're looking for the absorption/transmission spectrum of the ion and the metal separately, and how much of each there is in a combined spectrum. It's not strictly quantitative, but you can get a good idea of the ratios.


text:
Why are arenes with conjoined benzene rings drawn as they are?
Whilst we are taught to represent benzene as below (A) due to its delocalised electrons, however when two benzene rings share carbons (e.g. naphtalene) it seems to be more commonly represented in textbooks as two conjugated rings (B).  Whilst I realise that the result is the same, what is the advantage gained by drawing them like this rather than (C)?  Is it simply a matter of clarity?
A:

B:

C:

Whilst draing those in Marvin I wondered if it was as the delocalisation between the shared carbons isn't shown?  I.e. it looks like the electrons are delocalised across both rings separately rather than as one big ring?

Questions about why we represent things in one way or another are hard to answer in general, because they appeal to one’s sense of taste… de gustibus non est disputandum. Here, I want to state the position and the reasoning of IUPAC on this matter. The relevant publication on this are the IUPAC 2008 Recommandations, titled Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008) [1].
Section GR-6 of the report deals with “Aromatic rings and other types of electron delocalization”. I quote part of it (emphasis is mine):

GR-6.5 Curves should only be used when delocalization is being represented
Curves represent delocalization, yet lots of structures have some delocalized elements. Although all such elements could be depicted with curves, there is little gained by doing so. On the contrary, the arbitrary use of curves can draw the viewer’s attention to insignificant portions of a structural diagram, and away from areas that are chemically more important such as an active site or a reactive group. Accordingly, curves should only be used when the delocalization is specifically being highlighted as an important feature of the structure.
When curves are not used, any alternating configuration of double bonds is acceptable within the further constraints discussed in GR-3.5.

The text then goes further to note:

It is generally not acceptable to use curves in two adjacent fused rings, since such diagrams are at best ambiguous in terms of the character of the shared fusion bond between the two rings.

Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008), Pure Appl. Chem., 2008 DOI: 10.1351/pac200880020277

PS: hmm, the leader for the IUPAC report works at CambridgeSoft…

I've added to the why question :) Turns out "it is generally not acceptable" should be "it is nonsensical".

Both ways of drawing benzene should mean the same thing to you. Use one or the other depending on what you want to express.
The Kekulé structure, B, is more useful when you're interested in a reaction mechanism, or to visualize bonding once aromaticity is lifted. Here is an intermediate way to draw it, which is aesthetically more pleasing (to me):

The A and C structures are called Clar structures, but they first appeared in reference 1. Actually, C is nonsensical. A circle represents a sextet of $\pi$ electrons, like those of benzene. Naphtalene has only 10 delocalized electrons. After all, it is aromatic, and thus possesses 4n+2 $\pi$ electrons.
Here are the rules for drawing Clar structures:

Never inscribe circles in adjacent benzenoid rings.
On ignoring rings with circles, all remaining benzenoid rings must have a Kekulé structure, that is, must have either one or two double
  bonds, as in 1; a ring can be “empty” as in 2 but there must be no
  unpaired electrons.
Subject to the above constraints, a Clar structure must have the maximum number of benzenoid rings with inscribed circles. When a
  choice is available within a linear string of benzenoid rings, the
  circle can occupy any of the available rings, and this may be
  indicated by an arrow.

Read reference 2 for more information on Clar structures and their use in theory.

J.W.Armit and R.Robinson, Polynuclear heterocyclic aromatic types. II . J. Chem. Soc. (1925):1604–1610.
Alexandru T. Balaban (2004): Clar Formulas: How to Draw and How not to Draw Formulas of Polycyclic Aromatic Hydrocarbons, Polycyclic Aromatic Compounds, 24:2, 83-89

(Too long for a comment.)
With drawing it as in diagram B, although we lose the information that the electrons are in reality delocalized, it makes for easier electron accounting when we start "pushing electrons" (that is to say, in illustrating the mechanism of a reaction).
The thing is that representing electron delocalization in Lewis structures requires crutches like resonance structures or that circle/circle arc business. At least you can say that there's an electron pair in a single bond. How many electrons are there in the circle?
If you don't deal much in the illustration of reaction mechanisms, I suppose the first and third drawings are okay. My point is that those circles don't really play nice with arrows.

The history of printing technology (price to print formulas in books) has played a big role in old days. The variants a and b are the most recent leftovers of that. BTW napthalene is neither conjugated nor conjoined, but annulated.

I would have used the term "fused" instead of "conjoined" myself, seeing as that's the standard term in discussing polycyclic aromatic hydrocarbons. @Georg: why do you not consider naphthalene to be conjugated? What definition are you using?

@J.M. While naphtalene is aromatic, it's not conjugated. Conjugation is the delocalization of charge over alternating double-single-double bonds, as is present in fatty acid side-chains. Stilbene is conjugated, but naphtalene is not.

You should draw the resonance structures of benzene and 1,3,5-hexatriene and compare them to those of benzene.

@CHM: Ah. I was accustomed to considering PAHs as conjugated, but I guess I was using the wrong definitions. Thanks,


text:
What are known examples of drugs that racemize/stereoconvert in vivo, and how are they converted?
It is known that although only the (S)-enantiomer of the infamous sedative thalidomide possesses teratogenic properties, it is not very useful to administer the pure (R)-enantiomer since it is racemized within the body. Are there other known examples of drugs that are stereoconverted in vivo? How does the body perform the conversion?

""only the (S)-enantiomer of the infamous sedative thalidomide possesses teratogenic properties"" How was this fact estabished? You always have both enantiomers in vivo, right?

@Georg probably through study of the mechanism of interaction.

Although I can't think of any drug examples other than thalidomide, here's information on thalidomide's mechanism:
The chiral carbon of thalidomide can tautomerize in basic conditions into an enol, which is achiral.  A reversal back to the ketone results in a mix of (R) and (S) enantiomers.

In the body, this tautomerization is generally catalyzed by basic amino acids.  Specifically, albumin is the main catalyst in humans.
While this is beyond the scope of your question, the reason that only (S)-thalidomide causes birth defects is that it can insert itself into DNA and suppresses certain genes necessary for embryonic development.

Reddy - Chirality in Drug Design and Development
http://www.ncbi.nlm.nih.gov/pubmed/9860497
http://www.ncbi.nlm.nih.gov/pubmed/9499573

According to the relevant article, HSA does catalyze interconversion at pH 7.4, but once other components of blood plasma are added to the mixture, catalysis is inhibited. Don't forget that amines can themselves be chiral, as in this case, but that their interconversion barrier is low. The barrier is greater when the amine is cyclic, as in thalidomide, but it still can spontaneously interconvert.

I added that article to the list.  Certain components do inhibit the reaction, although blood plasma on the whole is also a minor catalyst for the interconversion although not as good a catalyst as HSA.

Drugs racemised in-vivo
The following table shows some of the most common (not an exhaustive list) of drugs racemised in-vivo.
Since the biological targets are necessarily chiral, the molecules formed upon epimerisation (inversion at the centre being racemised) are usually inactive against the desired target (annoying, but ultimately fine), issues only arise when the epimerised molecule has its own biological activity, which is often undesirable (for instance in the case of toxicity of thalidomide).

Pathways of racemisation
The mechanisms by which drugs are racemised is not always well known/studied (i.e. it isn't always possible to pin down the exact thing responsible, however two generalised 'mechanistic pathways' have been proposed: 1
1) formation of a covalent intermediate followed by a stepwise enzyme-catalyzed transformation
2) interaction of two opposing metabolic processes, such as oxidation and reduction
One thing that is generally agreed upon, is that the racemisation is the result of more than simple deprotonation/protonation (such as the way chiral centres adjacent to aldehydes tend to epimerise), since the pH in the body is, in most places, not basic enough to do so.
The interconversion of (R) and (S) ibuprofen is one example where a mechanism has been proposed, and involves multiple intermediates:

1: Curr. Drug Metab. 2004, 5, 517–533.
Apologies for the lack of reference to the table, it was taken from a set of lecture notes, though I suspect the lecturer had previously copied it from the primary literature somewhere

Ibuprofen is another example (have a look at e.g. here).
Ibuprofen is an optically active compound with both S and R-isomers, of which the (S)-isomer is the more biologically active. Although it is possible to isolate the enantiopure compound, Ibuprofen is produced industrially as a racemate. An isomerase (alpha-methylacyl-CoA racemase) converts (R)-ibuprofen to the active (S)-enantiomer.


text:
What makes some metals melt at higher temperature?
I'm looking at the melting temperature of metallic elements, and notice that the metals with high melting temperature are all grouped in some lower-left corner of the $\mathrm{d}$-block. If I take for example the periodic table with physical state indicated at $\pu{2165 K}$:

I see that (apart from boron and carbon) the only elements still solid at that temperature form a rather well-defined block around tungsten (which melts at $\pu{3695 K}$). So what makes this group of metals melt at such high temperature?

Look at Carbon and Boron and think again.

@Georg sorry but I don't find your comment very useful… could you be any more explicit? I know about carbon and boron, and I'm asking about metals…

@F'x, I really like your question. It's hard! Georg, I'm not quite sure of your direction either, but the very tight covalent bonding of boron and carbon is so different from the metallic bonding typical of all metals that I'm pretty sure the key insight to this one lies elsewhere. I would also observe that atomic density (not quite the same as mass density) likely plays an important role, as the island of metals you have pointed out contains (I think) some of the most atoms-per-cc dense metals. Does that very cool tool you are using (?) by any chance show that also?

@TerryBollinger there is a flash tool http://www.rsc.org/periodic-table

A pure metallic bond makes rather soft and malleable things like sodium or copper. The metals in that central area (tungsten and co) do have some localized bonds mixed with metallic bond! Think of the W-Type structure! A real metal should have some closest packing, nothing else.

Some factors were hinted, but let me put them in an order of importance and mention some more:

metals generally have a high melting point, because metallic interatomic bonding by delocalized electrons ($\ce{Li}$ having only a few electrons for this "electron sea") between core atoms is pretty effective in those pure element solids compared to alternative bonding types (ionic $\pu{6-20 eV/atom}$ bond energy, covalent 1-7, metallic 1-5, van-der-Waals much lower). Also, ionic lattices like $\ce{NaCl}$ have a higher lattice and bonding energy, they have weak interatomic long-range bonding, unlike most metals. They break apart or are easily solvable, metals are malleable but don't break, the electron sea is the reason for their welding ability.
the crystal structure and mass play an inferior role among your filtered elements (just look up the crystal structure of those elements), as metallic bonding is not directional unlike covalent bonding (orbital symmetry). Metals often have half filled $\mathrm{s}$ and $\mathrm{p}$ bands (stronger delocalized than $\mathrm{d}$ and $\mathrm{f}$) at the Fermi-edge (meaning high conductivity) and therefore many delocalised electrons which can move into unoccupied energy states yielding the biggest electron sea with half or less fill bands.
noble metals like $\ce{Au,Ag}$ have a full $\mathrm{d}$ orbital, therefore low reactivity/electronegativity and are often used as contact materials (high conductivity because of "very fluid" electron sea consisting only of $\mathrm{s}$-orbital electrons. Unlike tungsten with half or less occupied $\mathrm{d}$-orbitals they show no interatomic $\mathrm{d-d}$ bonding by delocalized $\mathrm{d}$-electrons, and more importantly, a half filled $\mathrm{d}$-orbital contributes 5 electrons to the energy band, while a $\mathrm{s}$ only 1, $\mathrm{p}$ only 3, the electron sea is bigger among the $\mathrm{d}$-group.
The "packaging" of core atoms in the lattice (interatomic distance) among the high $Z$ atoms (compared to e.g. $\ce{Li}$) is denser (more protons, stronger attraction of shell electrons, smaller interatomic radius), means stronger interatomic bonding transmitted by the electron sea:

You can see here that in each series ($\ce{Li,\ Na,\ K}$) the melting points rise to a maximum and then decrease with increasing atomic number (lacking unoccupied energy states for delocalized $\mathrm{d}$-electrons), bigger electron sea being here a stronger factor than a bit more dense packaging.

Boron as a semi-metal shows metallic and covalent bonding, Carbon strong directional covalent bonding and is able to build a network of bonds unlike other non-metal elements showing covalent intramolecular bonding, e.g., in diatomic molecules but not strong intermolecular bonding in macromolecules because of lacking unpaired electrons.

So there are some bigger trends for melting points explaining the high melting points of $\mathrm{d}$-metals, but also some minor exceptions to the rule like $\ce{Mn}$.

Actually " shows metallic and covalent bonding" is rather inaccurate description to highly non-trivial bonding in elementary boron.

I think there are two properties at play here: atomic velocity and stability of the lattice structure.
Recall that temperature is a measure of the average kinetic energy of the molecules, so $v \propto \sqrt{\frac{T}{m}}$, or to achieve the same velocity, $T \propto m$. So at the same temperature, two heavier atoms will move more slowly past each other than two lighter molecules, giving them more time to interact.
Now recall that in a solid, molecules are held in a lattice by intermolecular forces, and in a liquid, atoms have enough energy that the forces between them are no longer strong enough to hold the atoms in a lattice. So the more stability a substance's atoms gain from being in their solid, crystalline arrangement, the higher its melting point will be.  Now, why would the metals in that area gain more stability in their solid forms? I'm fairly sure the answer has to do with the completeness of orbitals and half-orbitals.  I'm not sure of the particulars of the delocalized electron cloud in metals, but I think it's likely that it allows these metals to in some manner fill or empty their incomplete orbitals.

Yes, they're heavy, but that's only half of the story, right? Cause it's not all the heavy atoms (mercury is heavy, and it's a liquid at ambient conditions…)

Weight of atoms plays very little if any role in transitions from solid to liquid. Only energy is important when it comes to bonds breaking and temperature is a measure of average kinetic energy not speed.


text:
How do orbitals coexist with a nucleus?
Many diagrams of orbitals I've seen involve electrons moving through a center point—where the nucleus is. How can this be? Clearly they don't actually pass through the nucleus, so what does happen?

""moving through a center point"" that isnt true. In fact s orbitals have a certein electron density within the nucleus, but there is nothing about "moving" in orbitals.

As Georg said, there is no motion in orbitals. These are stationary states by definition of being eigenfunctions of some one-electron Hamiltonian. The question is not meaningful as stated.

In the definition of an s shell, you will find that its $\ell$ number is zero. In classical terms, that corresponds to an orbit with zero orbital or angular momentum — which for a large object is a clear impossibility. For an electron, it gives the peculiar result that any electron in any s shell is, classically speaking, moving back and forth through the nucleus, rather than around it. So, in one curious sense, what you just asked is exactly what happens: the classical analogy is that the electrons do go through the nucleus, which is why they have such a lovely spherical symmetry.
The second part of the answer, however, is that electrons can't go through the nucleus unless they are hugely more energetic than the ones found in a typical small-nucleus atom. Clearly there's a bit of a paradox going on there!
The resolution of the paradox is that very-low-mass charged particles must be treated by quantum rules. So, for example, rather than the electron behaving like a well-defined particle, it behaves like a standing wave. That standing wave can in turn be thought of as two simultaneous versions of the electron, one going (for example) clockwise and the other counterclockwise around. (The real situation has an infinite number of such components; I'm just picking out a single pair of them that demonstrate the principle.)
Each of these components can furthermore be thought of as being refracted by the powerful spherical charge field of the nucleus, curving around it without every striking it. This refraction is not the same as an attraction. In fact, it is this refraction effect that prevents the density of the electron cloud from reaching infinity at the nucleus — that is, from striking the nucleus. If you think of how a tank of water can cause a beam of light to bounce off the surface instead of entering the tank — and that's a terrible analogy, I know, I know — you can at least get some idea of how an increasing "optical density" towards a central point might keep light away rather than bringing it closer.
So for an electron that behaves "like" two waves going both clockwise and counterclockwise, the combined waves curve around the nucleus rather than striking it. This is a very quantum sort of event, since for a classical object such "splitting" of the object is simply not possible and the object simply dives straight into the source of attraction. But if objects are light enough, that kind of particle-like behavior simply ceases to be available to the object. Instead, you get waves that neatly and with perfect spherical symmetry curve around the nucleus, never obtaining enough energy (which makes it more particle-like) to connect directly with that nucleus.
Finally, notice that electrons in s (and other) shells necessarily combine multiple paths all at the same time. For each "image" of the electron that is traveling clockwise, there must also be an exactly balancing "image" of the same electron traveling counterclockwise, so that the two images always balance out to zero orbital momentum. What an amazing thing that is! And an important thing, too, since it's what makes chemistry possible.
So, good question, even if it really is more of a physics question per se than a chemistry question. But it's such an important chemistry question! It's like asking how the engine that powers the car works. You can accept it as a given that all cars and vehicles have engines, and that they all work in a certain way. However, sometimes it's nice to dive in a little deeper and try to understand why these peculiar things do the things that make chemistry possible — that is, how the engine really works.

I really have no idea what is going on in this answer. According to the solution to the hydrogen-like atom, there are stationary states (e.g. 1s orbitals) that have the highest likelihood of electrons amongst all possible positions existing at the nucleus.

Terry, why do you say that the electron can't go through the nucleus?  How can the observation of Fermi contact by NMR and EPR be explained unless the electrons enters the nucleus? "The purpose of this paper is to point out that the isotropic HF interaction arises from
the probability density of an ‘s’ electron inside the nucleus." http://iopscience.iop.org/article/10.1088/0143-0807/21/1/303/meta

Quick response: I will have to re-read my old answer sometime today to figure out why it came over that way to you, since in other writings I've even described the 0 orbital momentum cases as classically equivalent to the electron diving straight through the nucleus. So, there's some kind of miscommunication going on here; I'll try to figure out what it is and fix it.

Electrons can be considered both particles and waves (wiki).  Basically, considering electrons to be particles is insufficient to explain many of the observed phenomena.
In this case, a particle wouldn't really pass through the nucleus, but a wave certainly could.  Take, for example, the 2p orbital, which is centered around the nucleus with two lobes.  The wave function can let us visualize where an electron is more likely to be found: (from PSU.edu)

The probability density function is found by squaring the wave function, and the PDF shows where the electron is likely to be observed:

(sorry for the small size)
So from this diagram, there is probability 0 of actually observing an electron at the nucleus although the electron must pass through the nucleus from side to side, which gets to the heart of your question.  The PDF deals mostly with the particle nature of the electron, because it shows where it is likely to observe one.  When the electron is exhibiting wave features, however, it can pass through the nucleus without actually being found there.
The best analogy I can give is if you wave a jump rope up and down, you can't really isolate the wave in the jump rope, but it is still clearly there.  Long story short, a particle doesn't pass through the nucleus, but a wave can and does.

@Georg, I chose a p-orbital because it seems like the simplest case where the electron has to move "through" the nucleus to get from one lobe to the other.  A sphere from an s-orbital didn't seem like as good an illustration IMO.  By all means, post an answer examining an s-orbital though!

The thing is s-orbitals do have a positive electron density at the nucleus. Isn't quantum mechanics strange.

The $\mathrm{p}$ orbitals, for example, have a nodal plane where the nucleus is, which means the electron density is zero there.
An orbital doesn't represent a path the electrons take when moving. An orbital is a region of probability. To make things clear and definite, when we draw an orbital we only draw the region where 95% (for example) of the probability lies. The fact that the $\mathrm{p}$ orbitals have a nodal plane simply means that the probability of finding an electron on that plane vanishes.
A positivist would thus consider an electron's trajectory around a nucleus nonsensical, since by the uncertainty principle, we can never measure it.
The use of orbitals is for visualizing electron density - where can the electron be, most of the time? It's a very useful way of interpreting phenomena such as chemical reactivity (think of $\mathrm{S_N2}$) or stability (think of benzene's $\mathrm{p}$ orbitals).

But how can an electron get from one side to the other? Does it simply go around the nucleus?

It doesn't. Talking about the location of an electron makes sense only when taking a measurement, the act of which collapses the wavefunction. The electron can be in both sides at the same time. The crux of the problem is one's interpretation of QM. Here is a very interesting discussion on the subject.

Remember, orbitals don't show where electrons are, they show the probability density of where a measurement would find them. Likely, at a scale close enough to the nucleus (where the probability of the electron being is exceedingly small to begin with, the electron's actual wavefunction is affected by the nucleus in such a way that either they can't overlap or the chance is even more vanishingly small.
I'm a bit rusty, but I believe every orbital but the 1s has a node at or on a plane through the nucleus anyway.

The wave function is defined to be centered around the nucleus, so the electron's wavefunction is not affected by the nucleus other than to say that the origin is the nucleus.
Every orbital does have a node at the nucleus though.

@Andrew As I recall, the potential used in deriving the orbitals only uses the EM force. At the nuclear scale, wouldn't the strong and weak forces change that?

most likely, but wouldn't those attract the electron to the nucleus, opposite of the EM force?  The wave function treats electrons as waves and thus ignores nuclear forces.

@Andrew: All s orbitals have an anti-node at the origin.

A bit of a rigorous answer here. This was the explanation I received in school.
The Heisenberg Uncertainty principle states that
$$\Delta x \Delta v \ge \frac h {4\pi m}$$
Now, the nucleus has a radius in the order of $10^{-15}$ m. This means that for an electron to exist within the nucleus, $\Delta x$ has to be within $2 \times 10^{-15}$ m. Substituting values and assuming the equality as the limiting case, we obtain
$$\Delta v = \frac h {4\pi m\Delta x} = \frac{6.626\times 10^{-34}}{4\pi \times 9.1 \times 10^{-31} \times 2 \times 10^{15} } \approx 2.89 \times 10^{10}$$
Which is about 100 times the speed of light. This is impossible, hence by contradiction, we see that an electron cannot exist within the nucleus.
For a more rigorous mathematical approach, the nucleus can be assumed to be a sphere (here, I have assumed that it's more of a one-dimensional path). You can then replace $\Delta x$ with $\Delta \vec r$ and evaluate the uncertainty principle in three dimensions using either cartesian or spherical coordinates. You would obtain a similar answer in those cases as well


text:
Why is the vanadium(3+) ion paramagnetic?
I know that the electron configuration of vanadium is $[\ce{Ar}]\mathrm{4s^2 3d^3}$.
None of the electrons in the 3d subshell are paired. Once it loses these three electrons, shouldn't the remainder of the electrons be paired? How can $\ce{V^{3+}}$ be paramagnetic if it loses all its unpaired electrons?

In addition to the general rules of how electronic configurations of atoms and ions are calculated, the elements from the $\mathrm{d}$-block (a.k.a. the transition metals) obey one special rule:

In general, electrons are removed from the valence-shell $\mathrm{s}$-orbitals before they are removed from valence $\mathrm{d}$-orbitals when transition metals are ionized.

(I took this formulation from these online lecture notes, but you will find equivalent statements in your textbooks.)
So, what that does mean is that if you remove electrons from vanadium(0), you will remove the $\mathrm{4s}$ electrons before you remove the $\mathrm{3d}$-electrons. So, you have the following electronic configurations:

$\ce{V}$ is $\ce{[Ar]} \mathrm{4s^2 3d^3}$
$\ce{V^2+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^3}$
$\ce{V^3+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^2}$
$\ce{V^4+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^1}$
$\ce{V^5+}$ is $\ce{[Ar]} \mathrm{4s^0 3d^0}$

And thus, $\ce{V^3+}$ is paramagnetic, because it has two unpaired $\mathrm{3d}$-electrons. In fact, all the ions above are paramagnetic, except $\ce{V^5+}$.

is there a reason why you skipped ${V^+ }$ which I suppose is [Ar]3d3 4s1   or [Ar]4s1 3d3 ?


text:
Why do elements in columns 6 and 11 assume 'abnormal' electron configurations?
When I look around for why copper and chromium only have one electron in their outermost s orbital and 5/10 in their outermost d orbital, I'm bombarded with the fact that they are more stable with a half or completely filled d orbital, so the final electron enters that orbital instead of the 'usual' s orbital.
What I'm really looking for is why the d orbital is more stable this way. I assume it has to do with distributing the negative charge of the electrons as evenly as possible around the nucleus since each orbital of the d subshell is in a slightly different location, leading to a more positive charge in the last empty or half-filled d orbital. Putting the final electron in the s orbital would create a more negative charge around the atom as a whole, but still leave that positive spot empty.
Why does this not happen with the other columns as well? Does this extra stability work with all half or completely filled orbitals, except columns 6 and 11 are the only cases where the difference is strong enough to 'pull' an electron from the s orbital? It seems like fluorine would have a tendency to do do this as well, so I suppose the positive gap left in the unfilled p orbital isn't strong enough to remove an electron from the lower 2s orbital.

In my paper https://www.academia.edu/18391675/About_the_distribution_of_electrons_magnetic_dipole_moments_in_atoms_Part_1 I have had some doubts about the role of the electrons dipole moment distribution in chromium atom which is quite symmetrical if I use my octahedral model instead of s- and p-orbitals. ...

...Now a have read on Wikipedia about the magnetic characteristics of pure chromium: "Chromium is remarkable for its magnetic properties: it is the only elemental solid which shows antiferromagnetic ordering at room temperature (and below). Above 38 °C, it transforms into a paramagnetic state." This matches perfect into my model.

Sometimes the "why" questions are really hard to answer. The universe was built in a specific way and we may have to accept it as it is. It's really annoying that humanity cannot explain everything about the universe.

As I understand this, there are basically two effects at work here.
When you populate an $\mathrm{s}$-orbital, you add a significant amount of electron density close to the nucleus. This screens the attractive charge of the nucleus from the $\mathrm{d}$-orbitals, making them higher in energy (and more radially diffuse). The difference in energy between putting all the electrons in $\mathrm{d}$-orbitals and putting one in an $\mathrm{s}$-orbital increases as you fill the $\mathrm{d}$-orbitals.
Additionally, pairing electrons in one orbital (so, adding the second $\mathrm{s}$ electron) carries a significant energy cost in terms of Coulombic repulsion because you're adding an electron essentially in exactly the same space as there's already an electron.
I'm assuming that the effect isn't strong enough to avert fluorine having a $\mathrm{2s^2}$ occupation, and if you look at gadolinium, the effect there isn't strong enough to stop the $\mathrm{s}$ from filling (large nuclear charge and orbital extent at the nucleus is a good combination energy-wise), it does manage to make it more favourable to add the electron into the $\mathrm{5d}$ instead of the $\mathrm{4f}$ orbitals.
Also, if you take a look at tungsten vs gold, there the effect isn't strong enough for tungsten to avoid a $\mathrm{6s^2}$ occupation, but is for gold - more $\mathrm{d}$ electrons making the screening effect overcome the strong nuclear charge and enhanced nuclear penetration of an $\mathrm{s}$-orbital.

Isn't there also the fact that for large atoms, relativistic effects pull the outer shells closer in which could be a factor for more shielding?

@Alizter: Yes, relativistic effects especially shrink the s orbitals, leading to more shielding for the higher angular momentum orbitals which thus become more diffuse and higher in energy. I can't remember exactly how the magnitudes for the different contributions shake out.

This is just a confirmation to Aesin's answer...
Say, we take copper. The expected electronic configuration (as we blindly fill the $\mathrm{d}$-orbitals along the period) is $\ce{[Ar]}\mathrm{3d^9 4s^2}$, whereas the real configuration is $\ce{[Ar]}\mathrm{3d^{10} 4s^1}$. There is a famous interpretation for this, that $\mathrm{d}$-orbitals are more stable when half-filled and completely-filled. That's a complete myth. There are very few pages explaining this myth, like chemguide.co.uk.
As we fill the electrons starting from $\mathrm{3d^1}$, we'd be stuck at chromium and also at copper. While filling chromium and copper, it has been observed that the energies of $\mathrm{4s}$ and $\mathrm{3d}$ orbitals are fairly close to each other. The increasing nuclear charge (as we go along the period) and the size and shape of $\mathrm{d}$-orbital should be a reason. This similarity makes the energy for pairing up electrons in d-orbital very less than that of pairing up in $\mathrm{s}$-orbital (i.e.) the energy difference between these orbitals is much less than the pairing energy required to fill the electrons in $\mathrm{4s}$ orbital. Moreover, the energy for the configuration $\mathrm{3d^5 4s^1}$ is much less than that of $\mathrm{3d^4 4s^2}$. Since we usually fill electrons in the order of increasing energy, the next electron (in case of manganes) goes into the $\mathrm{4s}$-orbital.
The same reason for effective nuclear charge makes the $\mathrm{3d}$-orbitals somewhat lower in energy than $\mathrm{4s}$-orbitals and hence, the unusual configuration of $\ce{Cr}$ and $\ce{Cu}$.

From a paper of Richard Hartshorn  and  Richard Rendle of the University of Canterbury (*.doc file), which supports that this is quite untrue:

In the case of chromium, this means that $\mathrm{4s^1 3d^5}$ will be lower in energy than $\mathrm{4s^2 3d^4}$, because in the second case you have to "pay" the electron pairing energy. Since this pairing energy is larger than any difference in the energies of the $\mathrm{4s}$ and $\mathrm{3d}$ orbitals, the lowest energy electron configuration will be the one which has one electron in each of the six orbitals that are available. Effectively this is Hund's rule applying not just to strictly degenerate orbitals (orbitals with the same energy), but to all orbitals that are (significantly) closer in energy than the electron pairing energy.
In the case of copper, the $\mathrm{3d}$-orbital has dropped in energy below that of the $\mathrm{4s}$, so that it is better to have the paired electrons in the $\mathrm{d}$ and the unpaired one in the $\mathrm{s}$. The reason why the $\mathrm{3d}$ is lower than $\mathrm{4s}$ is tied to the high effective nuclear charge. The high effective nuclear charge gives rise to the small size of $\ce{Cu}$ compared with the earlier transition metals, and also means that orbitals in inner shells are more stabilised with respect to those further out for copper than for earlier elements.

Does that paper mean that $4s$ orbitals have less energy than $3d$ orbitals? That would be in conflict with your chemguide link, which i had incidentally read before and i am convinced of its correctness...

Perhaps this shouldn't be counted as an answer, but since this topic has been resurrected, I'd like to point to Cann.[1] He explains the apparent stability of half-filled and filled subshells by invoking exchange energy (actually more of a decrease in destabilization due to smaller-than-expected electron-electron repulsions).
According to him, there is a purely quantum mechanical energy term proportional to $$\frac{n_{↑}(n_{↑}-1)}{2} + \frac{n_{↓}(n_{↓}-1)}{2}, $$
 where $n_{↑}$ and $n_{↓}$ represent the number of spin-up and spin-down electrons in a subshell. This term decreases the potential energy of the atom, and it can be shown that the difference in exchange energy between two consecutive subshell populations (such as $\mathrm{p}^{3}/\mathrm{p}^{4}$, $\mathrm{f}^{9}/\mathrm{f}^{10}$, etc) has local maxima at half-filled and filled subshell configurations. This can be used to defend the favourability of $\mathrm{s}^{1}\mathrm{d}^{5}$ and $\mathrm{s}^{1}\mathrm{d}^{10}$ configurations relative to $\mathrm{s}^{2}\mathrm{d}^{4}$ and $\mathrm{s}^{2}\mathrm{d}^{9}$, even though there would be a slight increase of electron density in the more compact $\mathrm{d}$ subshells.
However, this clashes with Crazy Buddy's reference, which seems to deny any stabilization effect. So, which is (more) true? Or is neither?

Cann, P. Ionization Energies, Parallel Spins, and the Stability of Half-Filled Shells. J. Chem. Educ. 2000, 77 (8), 1056 DOI: 10.1021/ed077p1056.

We know that the electronic configuration of chromium is $\ce{[Ar] 3d^5 4s^1}$. It is because promotion occurs in case of $\ce{3d}$ and $\ce{4s}$ orbitals — in other words, the electron is shifted from a lower energy level to a higher one (also known as excitation). Promotional energy and pairing energy both are endergonic so the process which requires less energy would preferably take place.

Could you explain a bit more? For example, why does that process have less energy? Elaborate a bit and your post will be improved greatly :)


text:
Why is the 2s orbital lower in energy than the 2p orbital when the electrons in 2s are usually farther from the nucleus?

My chemistry book explains that even though electrons in the $\mathrm{2p}$ orbital are closer to the nucleus on average, electrons from the $\mathrm{2s}$ orbital spend a very short time very close to the nucleus (penetration), so it has a lower energy. Why does this tiny amount of time spent close to the nucleus make such a big difference? It seems like it should be the average distance that matters, not the smallest distance achieved at any one point, in determining stability. What makes that momentary drop in energy so important that it is outweighs all the time spent farther away from the nucleus with a higher energy?

All, this is for CrazyEtc to say for sure, but I think his main question was a bit simpler: Why do orbitals that have less energy sometimes extend farther out into space than ones with higher energy? That's an interesting question, and one that's worth trying to answer as directly as possible.

Due to shielding.

I think your question implicates another question (which is also mentioned in some comments here), namely: Why are all energy eigenvalues of states with a different angular momentum quantum number $\ell$ but with the same principal quantum number $n$ (e.g., $\mathrm{3s}$, $\mathrm{3p}$, $\mathrm{3d}$) degenerate in the hydrogen atom but non-degenerate in multi-electron atoms?
Although AcidFlask already gave a good answer (mostly on the non-degeneracy part) I will try to eleborate on it from my point of view and give some additional information.
I will split my answer in three parts: The first will address the $\ell$-degeneracy in the hydrogen atom, in the second I will try to explain why this degeneracy is lifted, and in the third I will try to reason why $\mathrm{3s}$ states are lower in energy than $\mathrm{3p}$ states (which are in turn lower in energy than $\mathrm{3d}$ states).
$\ell$-degeneracy of the hydrogen atoms energy eigenvalues
The non-relativistic electron in a hydrogen atom experiences a potential that is analogous to the Kepler problem known from classical mechanics.
This potential (aka Kepler potential) has the form $\frac{\kappa}{r}$, where $r$ is the distance between the nucleus and the electron, and $\kappa$ is a proportionality constant.
Now, it is known from physics that symmetries of a system lead to conserved quantities (Noether Theorem).
For example from the rotational symmetry of the Kepler potential follows the conservation of the angular momentum, which is characterized by $\ell$. But while the length of the angular momentum vector is fixed by $\ell$ there are still different possibilities for the orientation of its $z$-component, characterized by the magnetic quantum number $m$, which are all energetically equivalent as long as the system maintains its rotational symmetry.
So, the rotational symmetry leads to the $m$-degeneracy of the energy eigenvalues for the hydrogen atom.
Analogously, the $\ell$-degeneracy of the hydrogen atoms energy eigenvalues can also be traced back to a symmetry, the $SO(4)$ symmetry.
The system's $SO(4)$ symmetry is not a geometric symmetry like the one explored before but a so called dynamical symmetry which follows from the form of the Schroedinger equation for the Kepler potential.
(It corresponds to rotations in a four-dimensional cartesian space. Note that these rotations do not operate in some physical space.)
This dynamical symmetry conserves the Laplace-Runge-Lenz vector $\hat{\vec{M}}$ and it can be shown that this conserved quantity leads to the $\ell$-independent energy spectrum with $E \propto \frac{1}{n^2}$. (A detailed derivation, though in German, can be found here.)
Why is the $\ell$-degeneracy of the energy eigenvalues lifted in multi-electron atoms?
As the $m$-degeneracy of the hydrogen atom's energy eigenvalues can be broken by destroying the system's spherical symmetry, e.g., by applying a magnetic field, the $\ell$ degeneracy is lifted as soon as the potential appearing in the Hamilton operator deviates from the pure $\frac{\kappa}{r}$ form.
This is certainly the case for multielectron atoms since the outer electrons are screened from the nuclear Coulomb attraction by the inner electrons and the strength of the screening depends on their distance from the nucleus.
(Other factors, like spin and relativistic effects, also lead to a lifting of the $\ell$-degeneracy even in the hydrogen atom.)
Why do states with the same $n$ but lower $\ell$ values have lower energy eigenvalues?
Two effects are important here:

The centrifugal force puts an "energy penalty" onto states with higher angular momentum.${}^{1}$ So, a higher $\ell$ value implies a stronger centrifugal force, that pushes electrons away from the nucleus.

The concept of centrifugal force can be seen in the radial Schroedinger equation for the radial part $R(r)$ of the wave function $\Psi(r, \theta, \varphi) = R(r) Y_{\ell,m} (\theta, \varphi )$
\begin{equation}
\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \underbrace{ \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} } } - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R(r) = 0
\end{equation}
\begin{equation}
{}^{= ~ V^{\ell}_{\mathrm{cf}} (r)} \qquad \qquad
\end{equation}
The radial part experiences an additional $\ell$-dependent potential $V^{\ell}_{\mathrm{cf}} (r)$ that pushes the electrons away from the nucleus.

Core repulsion (Pauli repulsion), on the other hand, puts an "energy penalty" on states with a lower angular momentum. That is because the core repulsion acts only between electrons with the same angular momentum${}^{1}$. So it acts stronger on the low-angular momentum states since there are more core shells with lower angular momentum.

Core repulsion is due to the condition that the wave functions must be orthogonal which in turn is a consequence of the Pauli principle. Because states with different $\ell$ values are already orthogonal by their angular motion, there is no Pauli repulsion between those states. However, states with the same $\ell$ value feel an additional effect from core orthogonalization.

The "accidental" $\ell$-degeneracy of the hydrogen atom can be described as a balance
between centrifugal force and core repulsion, that both act against the nuclear Coulomb
attraction.
In the real atom the balance between centrifugal force and core repulsion is broken,
The core electrons are contracted compared to the outer electrons because there are less inner electron-shells screening the nuclear attraction from the core shells than from the valence electrons.
Since the inner electron shells are more contracted than the outer ones, the core repulsion is weakened whereas the effects due to the centrifugal force remain unchanged. The reduced core repulsion in turn stabilizes the states with lower angular momenta, i.e. lower $\ell$ values. So, $\mathrm{3s}$ states are lower in energy than $\mathrm{3p}$ states which are in turn lower in energy than $\mathrm{3d}$ states.
Of course, one has to be careful when using results of the hydrogen atom to describe effects in multielectron atoms as AcidFlask mentioned. But since only a qualitative description is needed this might be justifiable.
I hope this somewhat lengthy answer is helpful. If something is wrong with my arguments I'm happy to discuss those points.

Why do you say that the "l-degeneracy" of the hydrogen atom is accidental?

@TanYongBoon I mean "accidental" in the sense that it just so happens to mark a balance between centrifugal force and core repulsion. Of course, this is the result of an underlying symmetry and in that sense it is not accidental.

General chemistry textbooks tend to explain atomic structure exceedingly poorly using a hodgepodge of obsolete concepts. Your chemistry book provides such a typical example - the notion of penetration only makes sense in the ancient Bohr-Sommerfeld model that has been obsolete since the discovery of quantum mechanics! The idea was that orbits of electrons in $\mathrm{p}$ states were more elliptical than those in $\mathrm{s}$ states because they had more angular momentum, and so on average there will be some fraction of the time where $\mathrm{p}$ electrons would be closer to the nucleus than those in the $\mathrm{s}$ orbits, i.e. they would penetrate the $\mathrm{s}$ orbit.
We now know that electrons do not actually have classical orbits and that the Bohr-Sommerfeld model is wrong. However there remains one correct statement of fact, which is that electrons in $\mathrm{p}$ orbitals by definition have more angular momentum than those in $\mathrm{s}$ orbitals (The orbital angular momentum is $\ell=1$ in $\mathrm{p}$ and $\ell=0$ in $\mathrm{s}$). Nonetheless it is completely misleading, since $\mathrm{2s}$ and $\mathrm{2p}$ orbitals are only well-defined in hydrogen and other ions with only one electron, and furthermore that all orbitals of the same principal quantum number are degenerate in energy in hydrogenic systems. Thus angular momentum alone cannot be the answer.
It is only due to electron-electron interactions that the orbitals that correspond to $\mathrm{2s}$ and $\mathrm{2p}$ in multi-electron atoms split and become nondegenerate. The effect of having more angular momentum therefore turns out to be only indirectly responsible via the different many-electron interactions occurring in the analogs of $\mathrm{2p}$ and $\mathrm{2s}$ orbitals in many-electron atoms. The simplest explanation that is even remotely correct is the screening effect, where the $\mathrm{2s}$ electrons by their very presence make the $\mathrm{2p}$ electrons feel a lower effective nuclear charge than if they were absent. A more precise quantum mechanical statement is that Slater's variational treatment of the nuclear charge in hydrogenic orbitals results in a lower effective nuclear charge for $\mathrm{2p}$ electrons than $\mathrm{2s}$ orbitals. There is unfortunately no good a priori physical principle for this; it is merely a statement of a result from a quantum mechanical calculation.
Summary: $\mathrm{2p}$ electrons are higher in energy than $\mathrm{2s}$ ones due to screening effects that result from electron-electron interactions. Explanations involving penetration of classical orbits or purely angular momentum considerations are wrong, even though they tend to show up in general chemistry textbooks.

General chemistry textbooks tend to explain EVERYTHING using over simplified concepts. It always reminds me of Jack Nicholson's line in "A few Good Men" - You can't handle the truth!! The truth in this case being too complicated for a "simple" explanation.

It seems like it should be the average distance that matters

No. It is the average energy that matters. Note that this stuff about "spends so much time here and so much there..." is really just a (not particularly good) way of describing a quantum-mechanical wave function's absolute square. The electron is actually never at any particular place in the orbital, rather it's everywhere at the same time – until you decide to measure its precise position (thereby destroying the orbital, that has to do with the Heisenberg uncertainty principle), in which case there would be a certain probability of finding it here or there. That probability can be calculated from the wave function $\psi(\mathbf{x})$ by taking its absolute square, which gives the radial distribution function in your plots.
The energy of any quantum state is determined by the expectation value of the Hamiltonian operator, which very roughly means "measure the energy at all places in space, multiply with the probability the electron is found at that place, and add all those values up". That should sound quite intuitive. In actuality it's more complicated: the electron also has some kinetic energy at every place in space, which is itself encoded in the wave function – but you need to consider its complex phase, which is a whole extra dimension that you simply don't see in your plots and certainly not in the average distance.
Calculating the exact energies for atomic orbitals is a lot of maths that probably doesn't belong here, but it's possible to carry it out*, and it should be obvious now that the result doesn't necessarily have much to do with simple "energy at the average distance". The quantities such as angular momentum mentioned in CHM's answer can be used to make it a bit simpler again (the Hamiltonian separates into a radial and spherical part), but that alone doesn't really explain much I suppose.

*As a matter of fact, it's not possible to calculate it exactly for anything more complicated than a hydrogen atom! You have to do it with certain approximations.

For small systems, it actually is possible to compute it exactly.  You just have to do it numerically.

Because of the penetration, that is, because the $\mathrm{2s}$ electron can spend time near the nucleus, it is less effectively shielded by the core electrons than a $\mathrm{2p}$ electron. Because it is less effectively shielded, a $\mathrm{2s}$ electron experiences a higher effective nuclear charge and is held closer to the nucleus than a $\mathrm{2p}$ electron which gives the $\mathrm{2s}$ orbital a lower energy.

I think the simplest way to explain what is going on is by way of a classical analogy.
Picture an elastically bound ball that whizzes back-and-forth along the $x$ axis. Next, whack the moving ball hard with a $y$-aligned impulse. The result is an elliptical or figure-8 path, one for which the maximum radius of motion along the $x$ axis will, in a perfect world, remain unchanged.
The single-axis ball is the classical analog to an $s$ orbital, and the elliptical or figure-8 ball corresponds to a $p$ orbital. Notice that for a given maximum radius, the $p$ orbital will include two orthogonal oscillation components, $x$ and $y$, and so will always have more energy than an $s$ (or $x$-only) oscillation with the same maximum radius. That also means that you can construct an $s$ orbital that extends beyond the maximum range of the $p$, yet will still have less energy than the $p$.
Or even more succinctly, think of a rocket with a fixed energy budget. The $s$ orbital is using all of its energy to achieve the maximum possible "launch altitude" from the nucleus. It can get away with that because in quantum land a light object can head straight back towards the nucleus without ever hitting it. In contrast, the $p$ orbital diverts some of its available energy into achieving a more conventional concept of an orbit, which makes its maximum radius lower than that of the do-or-die $s$ electron. I suspect that the textbook was trying to convey this "short cut" aspect of the $s$ orbital, rather than a special attraction per se near the nucleus.
P.S. - "Conventional orbit" is a loaded phrase in this context. The peculiarities of the quantum world require orbital solutions that look like figure eights and other loopy forms, rather than conventional elliptical orbits. These loopy forms are all spherically distorted standing-wave solutions, which can in turn be thought of as akin to distorted multiple loops on a skip rope. These skip ropes rotate in the complex plane, though, not in ordinary space.

Answers involving purely angular momentum considerations  cannot be correct, as they cannot explain why the $2s$ and $2p$ orbitals are degenerate in energy in the hydrogen atom.

This is not a thorough answer. It holds when comparing orbital energies of the same shell number $n$ (in your case 2) in atoms containing a low number of electrons. This question contains more information on this.
The letter of the orbital ($\mathrm{s}$, $\mathrm{p}$, $\mathrm{d}$, etc.) is a nameplace for the azimuthal quantum number (orbital angular momentum, $\ell$). The $\mathrm{s}$ orbital has $\ell=0$, $\mathrm{p}$ has $\ell=1$. The angular momentum being higher, the energy also is.

Could not a simple way of rationalising this result be to use the formula for a classical Coulombic attraction between point charges?
Given the radial distribution as shown, and the scaling of electrostatic force with r² rather than r, any time spent closer to the nucleus is vastly more important than time spent further away in terms of energy minimisation. This means that as radius increases (going left to right on the graph) what the lines do gets increasingly irrelevant unless they have exhibited very similar behaviour previously.