text:
Reaction of Silicon monoxide with alcohols
Elements from the carbon group tend to form compounds with two oxygen's. Let's say I want to produce $\ce{SiO_2}$ out of $\ce{SiO}$ by mixing it with some alcohol, for example:
$$\ce{SiO + CH_3-CHOH-CH_3 \longrightarrow SiO_2 + CH_3-CH_2-CH_3}$$
Is this likely to occur? at least before all the alcohol evaporates...

Alcohol evaporation wouldn't be a problem - it would react but hydrogen, not carbon would be reduced.

@Mithoron I don't understand you comment. What would then be produced?

Good luck finding silicon monooxide... What goes by this name usually is a very fine mixture of silicon and and silicon dioxide. Now, lower metals are a different story.

https://chemistry.stackexchange.com/questions/43103/why-sio-compound-doesnt-exist

I doubt very much its likely to occur.  Reducing an alcohol to a alkane is not trivial. strong reducing agents such as $\ce{LiAH4}$ cannot reduce alcohols, and to my knowledge $\ce{TiO}$ cannot reduce alcohols which would be an even better canidate that $\ce{SiO}$.

Yes, but relatively weakly reducing silanes can perform this reaction in the presence of acid. A major issue with LAH reducing an alcohol is the fact that LAH deactivates the alcohol to reduction by deprotonating it.


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What is the bond energy
$\ce{Cl2}$ bond energy given is as 244.
1/2 $\ce{Cl2}$ will still be 244 or 244/2? My teacher said it is still 244.

Welcome to chemistry.SE! If you had any questions about the policies of our community, please visit the help center. I don't quite understand your question. Do you expect $1/2~\ce{Cl2}$ to be a chemical species?

What units are you using? 244 by itself doesn't mean anything.

The quantity you refer to has units of kJ/mol. Thus, for 1/2 mole of $\ce {\rm Cl}_{2}$, the quantity is halved to 122 kJ/mol. Your teacher might be referring to the quantity on a per-bond basis, but the number you report agrees with references online for a per-mole basis.

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Why does olive oil make 90 degree grids on a nonstick pan?

Whenever I add olive oil to my cold nonstick pan and try to swirl it, it ends up making these 90-degree grids. Is this something to do with the oil, the pan, or the interaction between the two? I find the regular spacing between lines rather interesting.

It doesn't do on my pan …

Are you swirling it in an orbital motion, or first shaking it in the x direction, and then shaking it in the y direction?

Must be the pan... perhaps parts of the PTFE coating was not fully applied, and oil sticks there.

I have never seen this on any of my nonstick pans before. What happens if you try it with rubbing alcohol?

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Why do 3d orbitals have lesser energy than 4s orbitals in transition metals?
This is quoted from Jim Clark's Chemguide

For reasons which are too complicated to go into at this level, once you get to scandium, the energy of the 3d orbitals becomes slightly less than that of the 4s, and that remains true across the rest of the transition series.

What is the reason stated above? Why does in only transition series, 4s has more energy than 3d & not in the preceeding elements?

There is evidence that the title statement is not true to begin with. See e.g. the links in the disclaimer in my (accepted) answer.

Disclaimer: I now believe this answer to be fully incorrect.
Please consider un-upvoting it and/or downvoting it. I do not like seeing incorrect answers at +22.
However, I will leave it up for now. It is a reflection of what is taught in many undergraduate-level textbooks or courses. However, there have been criticisms of this particular graph in Shriver & Atkins, as well as of the idea that the 3d orbitals are somehow higher in energy than the 4s orbitals. I believe it was mentioned that the energies were calculated with the outdated Thomas–Fermi–Dirac model, but cannot really remember. I will ask another question about the 3d vs 4s issue, but in the meantime I would point the reader in the direction of these articles:

Pilar, F. L. 4s is always above 3d! Or, how to tell the orbitals from the wavefunctions. J. Chem. Educ. 1978, 55 (1), 2 DOI: 10.1021/ed055p2.
Melrose, M. P.; Scerri, E. R. Why the 4s Orbital Is Occupied before the 3d. J. Chem. Educ. 1996, 73 (6), 498 DOI: 10.1021/ed073p498.
Vanquickenborne, L. G.; Pierloot, K.; Devoghel, D. Transition Metals and the Aufbau Principle. J. Chem. Educ. 1994, 71 (6), 469 DOI: 10.1021/ed071p469.
Scerri, E. R. Transition metal configurations and limitations of the orbital approximation. J. Chem. Educ. 1989, 66 (6), 481 DOI: 10.1021/ed066p481.
Some criticism of Atkins' books by Eric Scerri.

While Molly's answer does a good job of explaining why electrons preferentially occupy the 4s subshell over the 3d subshell (due to less inter-electron repulsion), it doesn't directly answer the question of why the order of the 3d/4s energies changes going from Ca to Sc. I stole this figure from Shriver & Atkins 5th ed:

The red line represents the energy of the 3d orbital, and the blue line the energy of the 4s orbital. You can see that up to Ca, 3d > 4s but for Sc onwards, 4s < 3d.

As chemguide rightly points out, up to Ca, the 4s orbital is lower in energy than the 3d. The energy of an electron in an orbital is given by
$$E = -hcR\left(\frac{Z_\text{eff}}{n}\right)^2$$
where $hcR$ is a collection of constants, $Z_\text{eff}$ is the effective nuclear charge experienced by the electron, and $n$ is the principal quantum number. Since $n = 4$ for the 4s orbital and $n = 3$ for the 3d orbital, one would initially expect the 3d orbital to be lower in energy (a more negative energy). However, the 4s orbital is more penetrating than the 3d orbital; this can be seen by comparing the radial distribution functions of the two orbitals, defined as $R(r)^2 r^2$ where $R(r)$ is the radial wavefunction obtained from the Schrodinger equation:

The 4s orbital has a small inner radial lobe (the blue bump at the left-hand side of the graph), which means that a 4s electron "tends to spend time" near the nucleus, causing it to experience the full nuclear charge to a greater extent. We say that the 4s electron penetrates the core electrons (i.e. 1s through 3p subshells) better. It is therefore shielded less than a 3d electron, which makes $Z_\text{eff}$ larger. Going from the 3d to the 4s orbital, the increase in $Z_\text{eff}$ wins ever so slightly over the increase in $n$, which makes the energy of the 4s orbital lower.

Now, going from Ca to Sc means that you are adding one more proton to the nucleus. This makes the nuclear charge larger and therefore both the 4s and 3d orbitals are stabilised (their energies decrease). The catch is that the energy of the 4s orbital decreases more slowly than that of the 3d orbital, because the 4s orbital is relatively radially diffuse (the maximum in the radial distribution function occurs at a larger value of $r$). If you have studied physics, you could think of it as the interaction between two point charges; if the distance between them is large, then increasing the magnitude of one point charge has a smaller effect on the potential energy $U = -\frac{kq_1q_2}{r}$. The faster decrease of the 3d energy also makes sense because if nuclear charge were to tend to infinity, shielding would become negligible; the orbital energies would then be entirely determined by $n$, and if this were to be the case, you would expect 3d < 4s in terms of energies, as we said at the very start.
However, in Sc, the electrons preferentially occupy the 4s subshell even though it is higher in energy, and this is also because the 4s orbital is radially diffuse - the electrons have more "personal space" and experience less repulsion. One way of putting it is that an empty 4s orbital in Sc has a higher energy than an empty 3d orbital, but a filled 4s orbital has a lower energy than a filled 3d orbital. The fact that 4s > 3d in energy also explains why, for the transition metals, the 4s electrons are removed first upon ionisation ($\ce{Sc^+}: [\ce{Ar}](3\mathrm{d})^1(4\mathrm{s})^1$.)

I just want to end off with a comment that the factors that determine the electronic configurations of d-block and f-block elements are actually very closely balanced and just a small change in one factor can lead to a completely different electronic configuration. This is why Cr and Cu have an "anomalous" configuration that maximises exchange energy, whereas we don't get carbon adopting a $(1\mathrm{s})^2(2\mathrm{s})^1(2\mathrm{p})^3$ configuration in order to have "stable half-filled shells".

In the quest of re-reading old posts, I got somewhat baffled at one point here: you said up to $\ce{Ca},$ the $E$ for an electron in $\rm{4s}$ is lower than that of $\rm{3d}$ as the increase of $\rm{Z_{eff}}$ gets somewhat nullified by the increase in $n$ in the denominator. Okay. But I'm not getting why the same thing doesn't happen in elements after $\ce{Ca}$ eg. $\ce{Sc};$ $\rm{ 4s}$ is radially diffused even in $\ce{Ca}$ but does this affect the energy of the electron? Sorry, if I'm bothering you @Ortho, but would appreciate if you tell me why can't $\rm 4s\lt 3d$  in $\ce{Sc}$  [contd.]

for the same reason as in $\ce{Ca}$ viz. the increase in $n$ in the denominator is nullified by the increase in $\rm{Z_{eff}}.$ Also, there maybe a possible typo here: 3d > 4s but for Sc onwards, 4s < 3d. Thanks.

@MAFIA36790 Sorry I didn't get back to you earlier, I was travelling on that day and forgot all about it. To be honest, after a couple more years of chemistry, I am not entirely convinced how accurate a description this is. There have been criticisms of this particular graph in Shriver & Atkins, which I read before (various authors have written on it before), but I don't have the time to do thorough research into the matter right now. I will point you in the direction of these: http://pubs.acs.org/doi/abs/10.1021/ed055p2 and http://www.chem.ucla.edu/dept/Faculty/scerri/pdf/Atkins_critique.pdf

The idea in my post was that as atomic number increases, $Z_\mathrm{eff}$ of both the 3d and 4s orbitals increase. However, the 3d orbital is more greatly affected, i.e. $Z_\mathrm{eff}(\mathrm{3d})$ increases faster than $Z_\mathrm{eff}(\mathrm{4s})$. Consequently, there will be a crossover point where $$\frac{Z_\mathrm{eff}(\mathrm{3d})}{3} = \frac{Z_\mathrm{eff}(\mathrm{4s})}{4},$$ i.e. 3d and 4s have equal energies. Before this point, 4s < 3d, and after this point, 3d < 4s. Atkins' argument is that this point lies exactly between Ca and Sc.

Why do we look at the presence of radial nodes near the nucleus when measuring the amount of penetration of the electrons in a particular orbital. Shouldn't we look at the mean distance or most probable distance of the electron from the nucleus?

@TanYongBoon neither of those criteria will give you a measure of the penetration, which most critically depends on electron density near the nucleus. The bulk of the electron density is going to be on the outside, and doesn't contribute to penetration, so it can be anywhere it likes

May I refer you to this section of the answer: "The 4s orbital has a small inner radial lobe (the blue bump at the left-hand side of the graph), which means that a 4s electron "tends to spend time" near the nucleus, causing it to experience the full nuclear charge to a greater extent. We say that the 4s electron penetrates the core electrons (i.e. 1s through 3p subshells) better."

@TanYongBoon if you really wanted to be mathematical, I suppose it would be better to use some measure along the lines of $$\int_0^{r_0} |R(r)|^2r^2,\mathrm dr$$ where $r_0$ is some defined cutoff point which indicates the boundary between core and valence electrons. Obviously this is still quite wishy washy but I think it is more representative of the concept of penetration rather than a mean or a mode value of $r$ (which is predominantly influenced by the electron density outside the core).

That said, penetration is something that I've never seen put on a mathematical footing - it is pretty handwavy and I am not a huge fan of invoking it, which is partly why there is a disclaimer at the top of this answer, and this is my only answer on SE where I have used it. Perhaps not coincidentally, it was also one of my first answers.

This is a difficult question to answer. Following the Aufbau Principle and the n+l rule, the 4s orbital should fill before the 3d orbital. So why is 3d lower in energy? In short, the Aufbau Principle is not entirely correct. It is a guideline (like many things in chemistry.)
So, orbitals fill in order of stability. That is to say, electrons will go where they will be the most stable. It takes energy to hold electrons around the nucleus. The farther away they are, the more energy is needed to keep them. So the higher the principle quantum number, the higher the energy. I.e. 3s is higher in energy than 2s. At the same time, the principle quantum number is not the only number than needs to be considered. The quantum number l, for example, is also important. The higher value of l, the higher the energy. So 3d is higher in energy than 3p which is higher in energy than 3s. The 3d orbitals are more compactly placed around the nucleus than the 4s orbitals, so they fill first, even though this contradicts the Aufbau principle. This can be seen experimentally with the electron configurations for scandium:
Sc3+: [Ar]
Sc2+: [Ar]3d(1)
Sc+:  [Ar]3d(1)4s(1)
Sc:   [Ar]3d(1)4s(2)
Now, it is important to note that the 4s level does fill before 3d is entirely full. This is due to the 3d orbital's compactness. Electron repulsion "pushes" electrons into higher energy levels with less repulsion.
I would recommend reading this as it explains this in much more detail: http://www.rsc.org/eic/2013/11/aufbau-electron-configuration
I hope that helped!

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Why does solid phase peptide synthesis (SPPS) always proceed from the C to the N terminus?
Is it just for historical reasons? Because the first person who did it decided to attach the C-terminus to the solid support?
Or are there chemical reasons why it would be impossible to proceed in the opposite direction?

It's not impossible, it's just a lot harder and much more research has been put into the C → N version.

The reasons are

The use of an acylating agent in excess (e. g. acetic anhydride or N-acetyl imidazole) drives the reaction forward.
UNCAs (urethane-type protected  N-carboxy anhydrides (e. g. Fmoc, Z, Boc, ...)) can be used. They prevent racemization of the amino acid, that it is attached to.

But there are (despite of ribosomes themselves) also instances, where N -> C directionality was applied:

http://onlinelibrary.wiley.com/doi/10.1002/anie.199710991/abstract
http://www.ncbi.nlm.nih.gov/pubmed/10891165


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Near IR spectroscopy vs. IR spectroscopy
I am a backer of this Kickstarter: SCiO Near IR Spectroscopy
Here is a link to their technology: SCiO Technology
Question: How short of full IR spectroscopy do you think this device is? I have asked them whether we could program the SCiO to act as an IR spectrometer, and they have said this would require additional programming (which I am not qualified to do). But I am wondering if there would still be limitations. Conversely, Consumer Physics seems to imply that our taking more observations with the SCiO will make the device smarter and more useful.
This is obviously a question directed at experts in IR spectroscopy, which I am most definitely not.

This is more or less speculation, since they haven't released detailed technical specifications, but given the size of the instrument and the need to avoid moving parts to keep it durable, I would wager that it uses an IR LED for illumination, a diffraction grating for dispersion (although from the picture, it doesn't look like there's room for a conventional reflective diffraction grating setup so they might be doing something else), and probably some kind of silicon CCD or InGaAs diode array for detection. If this is the case, it's likely impossible to convert for use in conventional IR spectroscopy—not as a matter of programming, but simply that the hardware is not equipped to do it. Normal mid-IR spectroscopy (below about $4000\mathrm{\ cm^{-1}}$) involves much longer wavelengths beyond what LEDs can do, which generally means some kind of incandescent lamp, optics that don't involve glass (which blocks mid-IR), and a detector that can deal with such low energy photons (there are a few semiconductor-based sensors that looks promising, but most spectrometers use thermal detectors). There is also the matter of dealing with all the water in the air—many IR spectrometers have gas purge systems in their sample chambers.
There has been a lot of interest in standoff IR spectroscopy for things like explosives detection and advances in mid-IR optics and sensors may make this kind of small instrument possible, but we're not quite there yet. Certainly an instrument designed for NIR is not likely to have any usefulness for mid-IR, without being designed with it in mind.
A related technique that is much better suited to small handheld instruments is raman spectroscopy, since the light involved is actually usually in the NIR. I doubt this thing is sensitive enough to do it well, but it might be possible to add filters to the illumination and measurement ports to give it a shot. It's definitely possible to do raman with a handheld instrument though—I saw one of these at a conference recently. Quite expensive, but very nifty.

We had a sales rep come into the office this morning with a handheld XRF, not much bigger than a football. Ran some elemental analysis on a couple of our coatings -- impressive tech.


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Temperatures Below Absolute Zero
This seems like a farce...

Quantum gas goes below absolute zero
  Ultracold atoms pave way for negative-Kelvin materials.
  by Zeeya Merali, Nature News 03 January 2013, DOI: 10.1038/nature.2013.12146

I have seen several recent articles like this one  and I am extremely skeptical. Perhaps my skepticism is based upon an ignorance of physical principles, but I am less offended by apparent violations of the speed of light than by considering that temperatures below absolute zero could exist. I have studied under scientists who were using advanced techniques and vast amounts of energy to get small samples cooled down to a few nanokelvin. The physics seems to imply that if a sample were to reach zero, even the nuclear forces cohering the particles would cease. This was hypothesized in hushed tones, in awe and reverence. Now this article (which seems akin to scientific blasphemy) is causing an existential crisis. What is the deal with negative absolute temperature?

There's the subtle but all-important issue of defining "temperature". Under the most common (Boltzmann) definition, negative absolute temperatures are possible and are actually "above infinite temperature". However, using the Gibbs definition, only positive temperatures seem possible. There's an extensive Wikipedia article on the matter, as well as several questions in the Physics.SE site.

I wrote a blog post about it here: http://proteinsandwavefunctions.blogspot.dk/2013/01/negative-temperature-equations.html

I'm voting to close this question as off-topic because it belongs to Physics.SE.

I want to add that thermodynamics tells us that we cannot reach absolute zero. Negative temperatures doesn't violate that statement.

Negative temperatures are very much a thing, even with applications such as lasers.  Such a system is not colder than absolute zero.  Quite the opposite.  https://en.wikipedia.org/wiki/Negative_temperature

It is a quirk in how temperature would be measured using quantum mechanics. In general thermodynamics is about "bulk" properties and ignores quantum mechanics. Resolving these sort of quirks is why it would be so wonderful to have a theory of everything that would tie all the forces together.


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How do I find the empirical formula for cobalt oxalate hydrate with mass of oxalate and oxide produced?
I have the formula $\ce{Co(C2O4)*H2O}$ in a reaction that forms cobalt oxide $\ce{Co3O4}$
My data here says that my oxalate hydrate weighed $\pu{0.3283 g}$ and my product oxide weighed $\pu{0.1158 g}$ which I have calculated as:
$$0.1158 \times \frac{\pu{176.79 g}\, \mathrm{(cobalt\,oxalate)}} {\pu{240.79 g}\, \mathrm{(cobalt\,oxide)}} =\pu{0.08502 g\,\mathrm{(cobalt)}}$$
Then $\pu{0.3283 g (reactant)} − \pu{0.1158 g (product)}$ gives me $\pu{0.2125 g}$ water removed divided by $\pu{18.016 g/mol}$ to get $\pu{0.01179 mol}$ $\ce{H2O}$ in the sample.
How do I use this data to determine the empirical formula? I am not sure how to determine the molar ratios from this point.

I'm not exactly sure - but I guess you are trying to find the amount of water in your sample and to determine the ratio to cobalt oxalate.
Some parts of your calculation are correct, you do however forget that $\ce{CO2}$ is released. This would be my approach:
$$n_\ce{Co3O4} = \frac{\pu{115.8 mg}}{\pu{240.79 g/mol}} = \pu{0.4809 mmol}$$
This means that there was three times as much $\ce{CoC2O4}$ in your sample.
$$n_\ce{CoC2O4} = \pu{1.443 mmol}$$
That equals a weight of $\pu{212.1 mg}$. Your original sample weighed $\pu{328.3 mg}$, so there is $\pu{116.2 mg}$ left that seem to be $\ce{H2O}$ ($\pu{18.015 g/mol}$), giving you.
$$n_\ce{H2O} = \pu{6.45 mmol}$$
The ratio of cobalt oxalate to water then is $\left(\frac{6.450}{1.443} = 4.5\right)$ which implies the empirical formula is $\ce{2CoC2O4.9H2O}$

I am doing another approach, to get another result.
Starting from $\pu{0,1158 g}$ $\ce{Co3O4}$, whose molar mass is $\pu{241 g/mol}$, this amount corresponds to $\pu{\frac{0.1158 g}{241 g/ mol} = 0.4804 mol}$ $\ce{Co3O4}$, and also to $\pu{3*0.4804 = 1.4415 mmol }$ $\ce{Co}$. This amount of cobalt is included in the original sample of oxalate, which weighs $\pu{0.3283 g}$ So the molar mass of this original oxalate is $\pu{M = \frac{0.3283 g}{1.4415 mmol} = 227.7 g/mol}$. As the molar mass of anhydrous
$\ce{CoC2O4}$ is $\pu{147 g|mol}$,  the difference $\pu{227.7g - 147 g = 180.7 g} $ is the mass of $n$ moles water. $\pu{180.7 g}$ is exactly the mass of $10$ moles water.
As a consequence the original oxalate has the formula $\ce{CoC2O4·10 H2O}$

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Why is change in gibbs energy not zero through the gibbs free energy equation?
Change in enthalpy of the system is equal to its temperature multiplied by its change in entropy (as change in enthalpy is equal to heat absorbed /given out at constant pressure and change in entropy is heat absorbed/given out divided by its temperature). So, through the Gibbs free energy equation, the change in Gibbs free energy should be zero?

Not exactly sure what you have problems with but a common source of confusion is $\Delta G$ vs $\Delta G^\circ$. At equilbrium $\Delta G = 0$ and $\Delta G^\circ = -RT \ln(K)$

The change in Gibbs free energy for a closed system is given by:
$$\Delta G = \Delta H - T \Delta S$$
What you are describing is:
$$\Delta H = T \Delta S$$
which is only true for systems at equilibrium, or when $\Delta G = 0$
Why this is true is a little bit difficult to explain. It is essentially a form of the Clausius Inequality at constant temperature and pressure. The Clausius Inequality is really just a statement of the second law of thermodynamics, and it says that at best, for any possible process you can get a zero change in the entropy of the universe, and usually it will be a positive change. From the perspective of the system you can write that as:
$$\Delta S \ge \oint \frac{dQ}{T}$$
which is basically just saying the entropy change of the system is equal to or greater than the sum of the negative entropy changes of the rest of the universe, for each step in a closed cycle.
If you integrate that at constant temperature you get
$$\Delta S \ge \frac{\Delta Q}{T}$$
At constant pressure (and in the absence of non-expansion work) $\Delta Q = \Delta H$, so we can write:
$$\Delta S \ge \frac{\Delta H}{T}$$
...and a little rearranging gives:
$$\Delta H - T \Delta S \leq 0$$
So this says that any process which can occur on its own (without adding work or heat to it) will obey this inequality. That's a pretty useful thing to know, so we let
$$\Delta G = \Delta H - T \Delta S$$
to give us an easier way to talk about it. Processes will happen (on their own) if $\Delta G$ is negative, they won't happen if it's positive, and when they are at equilibrium, $\Delta G$ will be zero.


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Relation Between Degree of Hydration and Electropositivity
Why are salts of strongly electropositive elements less heavily hydrated in aqueous solution than the those of less electropositive elements? Shouldn't it be the other way round as the salts of electropositive elements are mostly ionic and hence strongly attracted by water molecules?

related http://chemistry.stackexchange.com/questions/31868/water-exchange-at-al-iii

What state of aggregation are you referring to? Aqueous solution or solid (crystal)?

@Martin-マーチン I'm talking about salts in aqueous medium.

The degree of hydration in solution does not directly depend on electropositivity. It depends on the ratio of charge to volume. The higher charged and the smaller a fragment is, the more water molecules hydrate it.
Note that almost every salt in solution can be considered as separated ionic pairs. For those that do not seperate, i.e. have a significantly large covalent bonding proportion, you would be looking at the solvation of the non-dissociated ion pair meaning that you can’t properly compare that hydration to simple salts’ hydration.
Metals get more electropositive the closer they are to the bottom left of the periodic table with cesium being the (non-radioactive) corner. Electropositivity tends to increase with an atom’s size. But the larger the size, the less the degree of hydration is, because the charge per volume of the metal ion will be smaller!
Also, the most electropositive metals are alkaline metals, meaning the only create $\ce{M+}$ ions. However, the stronger the charge, the smaller an ion gets: $\ce{Ga^3+}$ has an ionic radius of $76\,\mathrm{pm}$ — smaller than $\ce{Li+}$ whose ionic radius is $90\,\mathrm{pm}$ although they are two periods apart! So $\ce{Ga^3+}$ will be a lot stronger hydrated than $\ce{Li+}$ because it contains much more charge per volume.


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Why are so many ionic compounds brittle?
Even when ionic compounds are strong (e.g. rubies and sapphires), they are neither malleable nor ductile in the slightest, and if sufficient stress is put on them, they will shatter, not bend. Why is that?

I'm not sure that I would characterize sapphire as an ionic compound. But, the question is a good one that points towards material science and metallurgy so I'll contemplate it.

Fair about sapphires/rubies. But they were the hardest things with a metal and nonmetal I could come up with on short notice.

Ionic crystals are hard because of tight packing lattices, say, the positive and negative ions are strongly attached among themselves.
So, if mechanical pressure is applied to an ionic crystal then ions of similar charges may be forced to get closer to each other.
Now, by doing so, the electrostatic repulsion can be enough to split or disorient completely the lattice infrastructure. Thus imparting the brittle character.

This is not a good explanation. Why is diamond brittle? It isn't ionic.

@matt_black https://www.quora.com/Why-is-the-diamond-hard-yet-brittle

They’re brittle because it’s very difficult for dislocations to move through the crystal lattice. Dislocations are what mediate plastic deformation in crystals, so the fact that creating and moving a dislocation in an ionic crystal incurs a massive energy penalty means that an ionic crystal will undergo brittle fracture.

Ionic compounds are not unusual in being brittle
Your starting assumption that ionic compounds are frequently brittle is misleading. Many, if not most, solids are brittle, ionic or not.
The reasons why things are brittle have more to do with the bulk structure of the material and less to do with the chemical makeup of the material.
Table salt is an ionic compound and is brittle. But diamond is also brittle despite being a molecular solid where all the carbon-carbon bonds are covalent. But forged iron is strong and far from brittle. Pure copper is soft and malleable and not brittle. Nylon and Kevlar are the opposite of brittle.
Strength is more or less unrelated to being brittle or not and, defined properly, is a measure of the ability to resist deformation. But this is almost unrelated to brittleness. Glass is very strong but, like salt, very brittle which is why dropping your phone on hard surfaces is a bad idea.
Toughness is a better term for the opposite of brittleness. Tough compounds can deform without shattering. Nylon is weak but tough, kevlar is strong but tough, forged iron is also strong and tough. But cast iron is strong but brittle, providing a hint that the overall chemistry isn't everything.
What actually makes for tough compounds is the ability to mitigate external stress in the molecular structure of the material. In many polymers, the bonds in the long polymer chains can rotate and rearrange themselves to relieve that stress. In some metals (forged iron but not cast iron) the crystal structure of the metal contains defects that can move and rearrange to relieve stress concentrations. In contrast glass and table salt can't do that and even small surface scratches concentrate stress and rapidly grow causing the compound to shatter. They lack a molecular mechanism to mitigate the stress concentrations caused by small cracks. This can be partly overcome by more complex treatments of the surface of the compound. "Strong" glass (like the gorilla glass used for mobile phone screens) uses a process that treats the glass to create tension in the surface. That tension minimises the stress concentrations from small surface scratches and makes the resulting glass much stronger (this is sometimes achieved by deliberately adding ions to the surface of the glass).
Toughness is a product of the bulk material not of the bonding type of the molecules or atoms that make it up.
In summary, brittleness is not a property uniquely associated with ionic compounds. Most crystals are brittle. Those compounds that are not brittle are not distinguished by the type of bonding involved but by complex mechanisms that can alleviate the concentration of stress in the bulk compound. So many covalent crystals are brittle not just ionic ones. Some metals are brittle, though many are not.


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Fundamental equation of thermodynamics for magnetocaloric materials
The fundamental equation of thermodynamics, as us chemists (and chemical engineers!) are used to seeing it, is
$$ dG = - S~dT + V~dP + \sum_{i}\mu_i~dN_i$$
This gives the Gibbs free energy as a function of temperature, pressure, and composition, assuming there are no other relevant forces other than mechanical pressure.
The other day I watched a video on the magnetocaloric effect.  Obviously, there are non-pressure magnetic forces acting in such systems.  What's the proper form of the fundamental equation for magnetocaloric materials?
Suppose that the magnetocaloric material used is chemically pure and non-reactive during the magnetization process.  Then we could get rid of the $\sum_{i}\mu_i~dN_i$ term.  I suppose its also reasonable to assume that pressure is constant during magnetization / demagnetization process, and that the volume of the material is unchanged by magnetization so probably we could dispense with the $V~dP$ term as well (is that true?).
That leaves us with $dG = -S~dT + \rm{MAGNETIC~STUFF}$.  The $\rm{MAGNETIC ~STUFF}$ term probably has a $B$ or $H$ or something like that in it to represent the imposed magnetic field, but what else goes in there?

A quick google search led me to this paper. Maybe it has the answer you are looking for. I cannot access the paper from my home.

Thanks for the link -- it is definitely relevant.  Since I'm not a physicist, it's tough going for me, but if I am able to make enough sense of it over the next few days, I might submit an answer to my own question.  It seems like they use $H$ in their treatment, so I guess the right term in the equation would either be $x dH$ or $H dx$, but I still don't know what $x$ is or should be.

The fundamental equation for paramagnetic system is as follows
$$dU=TdS+BdM+\mu dN$$
Where B is the external magnetic field intensity and M is the magnetic moment (Here I have neglected the P-V work of the system). For your chemically pure and non-reactive system the above equation simplifies to
$$dU=TdS+BdM$$
Using Legendre Transformations:
$$y(0)=U(S,M)$$
$$dy(0)=dU=TdS+BdM$$
$$dy(1)=dA(T,M)=-SdT+BdM$$
$$dy(2)=dG(T,B)=-SdT-MdB$$
Now changing the order of the equation, we can find enthalpy,
$$y(0)=U(M,S)$$
$$dy(0)=dU=BdM+TdS$$
$$dy(1)=dH(S,B)=-MdB+TdS$$

Is M the magnetic moment or the magnetization?   If the latter, it seems like in SI units it has units of Amperes per meter, or (dividing by Faraday's constant) moles per second per meter, and B has units of kg per second per Coulomb, or kg per second per mole.  Then the magnetic force term has units of $kg/s^{2}/m$, which could also be written as Joules per cubic meter.  But don't the other terms in the equation have units of Joules, or at least Joules per mol?

M is magnetic moment while magnetization is density of magnetic moment. For dimensional consistency, B=intensity of magnetic field (Tesla or T), M=Magnetic moment (J/T). So, $B \times M = J ~or~ J/mol $ in terms of per mol basis


text:
How does Aloe Vera leave a cool feeling on your skin? Why does it evaporate fast?
If you have ever grown Aloe Vera and applied it to your skin, it feels a bit slimy. However, after it evaporates it has a cool feeling on your skin.
Also, why does the Aloe Vera evaporate so fast? What is it made of, apart from water that makes Aloe Vera do this?

I am assuming Hand Sanitizer and Aloe Vera are not directly connected.

The leaf of aloe vera is composed of three layers: 1) An inner clear gel that contains 99% water and rest is made of glucomannans, amino acids, lipids, sterols and vitamins. 2) The middle layer of latex which is the bitter yellow sap and contains anthraquinones and glycosides. 3) The outer thick layer of 15–20 cells called as rind which has protective function and synthesizes carbohydrates and proteins. So overally , it consists of about 99% water and remaining are vitamins, enzymes, minerals, sugars, lignin, saponins,etc.
So what makes inner layer gelly ?? Well, the innermost part of the leaf is a clear, soft, moist and slippery tissue that consists of large thin-walled parenchyma cells in which water is held in the form of a viscous mucilage consiting of cellulose, hemicellulos and acetylated mannans.
As for you question, as most of it consists of water(99%) the cool feeling might has been felt due to its evaporation but some of its remaining constituents are absorbed in the skin. Mucopolysaccharides present in it help in binding moisture into the skin.Its one of the important medical applications is that it could potentially be used as a
penetration enhancement agent for the transdermal delivery of drugs and intestinal absorption of coadministerd drug 1.
Saponins that are the soapy substances are found in the gel of aloe vera and have cleansing and antiseptic properties 2 so you can say that hand sanitizer and aloe vera are somewhat connected.
References

Hamman, J.H. Composition and Applications of Aloe vera Leaf Gel. Molecules 2008, 13, 1599-1616. https://doi.org/10.3390/molecules13081599

Surjushe, A., Vasani, R., and Saple, D. G. Aloe vera: a short review. Indian J Dermatol. 2008, 53(4): 163–166. doi: 10.4103/0019-5154.44785

I always assumed that the primary effect was not evaporation but something like the action of menthol which directly affects the temperature sensing nerves to create the illusion of coldness.

I had that assumption too initially but I couldn't find any of it's  chemical constituents like that of menthol. So I stated that evaporation "might" solve the mystery to some extent. Likewise some of its constituent are absorbed but I am not sure they would play any role.

So,your edit to the answer with authentic logic is highly appreciated.

So, you are saying its just mostly evaporation?

It might be one of the logical options(considering that 99% is of water). We can feel considerable extent of cooling effect due to evaporation of water that we have practically observed as well.


text:
Bonds and pairing of spins
I was reading Peter Atkins' Elements of Physical Chemistry where I came across this line:

Bonds do not form because electrons tend to pair; bonds are allowed to form by the electrons pairing their spins.

What does this line mean? What does Atkins want to say through this line & why did he italize the words I bolded above?

I dunno if the author meant exactly this, but it is a well-known fact that pairing of electrons actually consumes energy.
For example, free nitrogen atom in ground state has three unpaired electrons. To form triple bond in dinitrogen molecule, three pairs of electrons must be, well, paired. The process of pairing does consume energy.  However, the energy freed in the bond formation is so big, that it overwhelms pairing energy by huge margin. On the other hand, in oxygen molecule two highest-energy electrons has two orbitals to occupy, and they stay unpaired, so dioxygen is paramagnetic.
So, by no means it is the fact of being unpaired that drives electrons to form bonds. In fact, it prevents formation of weak bonds. However, when two atoms come closer to each other and have two half-occupied orbitals that can form a bond, the fact that electrons may be paired allows this two orbitals to combine into occupied bonding orbital, forming a bond.

How does the spin enter the picture?

@ToddMinehardt going with huge simplification, the pauli principle states that no two electrons can have exactly same state. Electrons on same orbital must have exactly same wavefunction, however as they have spin, they still can be in states, characterised by same wavefunction, but with different (opposite) spin orientation ('up' and 'down'). Thus, they may be 'paired'

The statement "Electrons on same orbitals must have exactly same wavefunction" is patently incorrect: That's not possible. The quantum number that distinguishes them for us is spin. But back to the OP's question: Why, in your opinion, does Atkins say what he does in reference to spin?

@ToddMinehardt Orbital != Spin-orbital ; Spin-wavefuction != wavefunction. Two electrons on same orbital have same '3d-wavefunction', or, if you are annoyingly nitpicking, same 3d component of the wavefunction but different spin component.  I've no idea what Atkins said and don't care about it at all, I wrote how the things are.

@permeakra This is coming across as somewhat rude.  Please be mindful of how your comments may be perceived.  Thanks!

Some actual numbers for the cost of electron pairing can be found in this answer of mine. The magnitude of the pairing energy is usually around 2 - 3 eV.


text:
Why is cyclooctatetraene non planar but the cyclooctatetraenide anion planar?
I was reading some organometallics and I came upon the preparation of uranocene, $\ce{U(C8H8)2}$. It has two cyclooctatetraenide anions, $\ce{C8H8^2-}$, as ligands, with a uranium atom between them.
It was mentioned that the neutral cyclooctatetraene molecule $\ce{C8H8}$ is non aromatic and has a tub-shaped conformation, but the dianion $\ce{C8H8^2-}$ is planar but aromatic. Why is this so? Can fulfilling Hückel's rule force a cyclic molecule to become planar?

related http://chemistry.stackexchange.com/questions/29428/acidity-of-unsaturated-five-membered-ring-relative-to-seven-membered-analogue

Fulfilling Hückel's rule means that you get to be aromatic.
The reason cyclooctotetraene is non-planar is because its planar arrangement would be antiaromatic. It twists to avoid it, because antiaromatic compounds are unstable with respect to localisation of the π-electrons (see here: What is the justification for Hückel's Rule?)
The veritable nanosecond that a molecule has the potential to become aromatic though, it will leap at the opportunity. Any angle strain from being planar is so quickly overtaken by the huge jump in stability that aromaticity brings that the molecule will twist to form an aromatic structure in a blink.

Fulfilling Huckel's rule is just one of the conditions to become aromatic. Planarity is another one. I think there are compounds that fulfill the rule but aren't planar and hence not aromatic.

In Cyclodecapentaene angle strain overcomes stabilisation

Under most circumstances, aromaticity > angle strain.

Yeah, in smaller, more common rings rings

No, the jump in stability is not huge, except for small aromatic rings. The energetic effect of aromaticity decreases with larger ring size.  This explains why [10]annulene notoriously fails to find an aromatic conformation and larger [4n+2]annulenes, when planar, tend not to show the peculiar reaction characteristics we see in benzenoid systems.

If cyclooctatetraene were planar, it would be an antiaromatic compound according to Hückel's rule, because it has 8 π-electrons. However, since its lowest energy conformation is non-planar, cyclooctatetraene can also be considered a non-aromatic compound.

(source: Wikimedia Commons)
The dianion has 10 π-electrons. Since this fulfills the $4n+2$ criterion, it adopts a planar conformation to enjoy aromatic stabilisation.


text:
Fundamental physical forces behind osmosis
I wonder which fundamental forces act behind osmosis process? Sum of these forces oppose gravity force, like in this figure:
So I wonder what are these forces?

Short answer: Entropy.

As far as I know, entropy is not a fundamental force. According to this: https://en.wikipedia.org/?title=Force#Fundamental_forces, there are 4 fundamental forces, while entropy is not one of them. All other forces are compositions of one or more fundamental ones

It's much better answer than "Electomagnetism"

If you really want to ask about 4 "fundamental" forces than virtually everything in chemistry is about electromagnetism...

The water molecules move across the membrane due to thermal energy. If concentration of solute on both sides of the membranes is the same, the amount of water crossing the membrane in either direction is same but when the concentration is unequal, the side with more solute particles (higher concentration, that is) attracts the water molecules and thus stopping them from going back. This process goes on until the particles cannot stop more water molecules from going back and equilibrium is achieved.
Answering the question, the physical force that causes osmosis is the one that is responsible for dissolution of the solute in the given solvent which is (as others pointed out) electromagnetic in nature.

This is not correct.  Osmosis is fundamentally entropically driven (i.e., it results from the statistics of large numbers of particles), and thus does not rely upon interactions between the particles.  For instance, if you had a semi-permeable membrane that could distinguish between two different theoretically non-interacting gases, you would be able to generate an osmotic pressure.  Electromagnetic interactions can alter osmotic pressure, in the same way that electromagnetic interactions can cause activities to be different from concentrations, but this is a second-order effect.

If you dribble a basketball, the force of your hand on the ball, that keeps it bouncing, is not (1) gravity; (2) the weak force; or (3) the strong force. It must be electromagnetic force--the electrons in the molecules of your skin repelling the electrons in the molecules on the surface of the ball. Also, your skin and the ball remain (largely) intact because of the electromagnetic force that hold them together, respectively. Same for osmotic "force". The motions of the solute and solvent molecules are not governed meaningfully by anything other than electromagnetic force.


text:
C-N bond lengths in 2-bromo-1,3,5-trinitrobenzene
Are all the $\ce{C-N}$ single bonds in 2-bromo-1,3,5-trinitrobenzene the same or different? I am of the view that the $\ce{C_1-N}$ and $\ce{C_3-N}$ bonds are of the same length and they are longer than the $\ce{C_5-N}$ bond, due to steric hindrance. Am I right or am I missing out something?

(source: nist.gov)

Seems you're right.

Those pentavalent nitrogens are causing my eyes to bleed.

Subhamon, I see that you are new here and that this is the first question you've asked.  So let me mention that if an answer is helpful please consider accepting and\or up-voting it.  Accepting \ upvoting an answer is a nice way to say "thank you".

Bromine is go to out of the plane due to steric repulsion ,and C1 and C5 bond length are same but longer than the bond length C3 .

By rotating the molecule 180° about the $\mathrm{C_2}$ symmetry axis that passes through the bromine and the nitro group in the 5-position, you can interconvert $\ce{C_1-NO2}$ and $\ce{C_3-NO2}$.  Groups that can be interconverted by a rotational axis must be equivalent.  There is no symmetry element that interconcerts $\ce{C_1-NO2}$ (or $\ce{C_3-NO2}$) and $\ce{C_5-NO2}$, therefore these groups cannot be equivalent. So symmetry tells us that the  $\ce{C_1-NO2}$ and $\ce{C_3-NO2}$ bond lengths are exactly the same, and both are different from the $\ce{C_5-NO2}$ bond length.
Symmetry can only tell us if things are the same or different, it can't tell us the magnitude of any difference.  So I don't know for sure which set of carbon-nitro bonds is longer or shorter, but your argument based on steric strain seems reasonable.  It is also possible that the bulky bromine causes the adjacent nitro groups to twist out of the plane of the aromatic ring, reducing resonance overlap between nitrogen and carbon and also lengthening those adjacent nitro bonds.

I agree with much of this, but isnt there some circularity in this argument? The C2 symmetry axis only exists IF the C-N bonds are equivalent, so we shouldn't use it as proof of their equivalence?

Ha, just seen how old this is

@PCK  There is no circularity in the argument. Unless there is some outside influence that affects the two nitro groups differently (e.g. crystal packing forces in a solid), then the nitro groups must be equivalent.

As I thought this could be interesting because I would have argued the same way as both of you did I have done some geometry optimizations.
The final niveau was B3LYP-D3/def2-QZVPP in the gas phase (no frequency calculation has been done because it would have taken too long for me to wait so I have to believe that it is a ground state) and the result from this one calculation with this single functional is that all three $\ce{C-N}$-bonds have de facto the same lengths: $1.482~\mathrm{\overset{\circ}A}$ "long" and $1.480~\mathrm{\overset{\circ}A}$ "short"
The two bromine-neighboring nitro-groups do indeed turn out of plane (dihedral angle is about 61...62°) as it was suggested by ron. This seems enough to compensate the proximity to the bromine.
$\hskip2in$

What program are you using to generate that structure?

The structure itself is made with Gaussview.

There is also a single crystal structure investigation (1) of polymorphism among picryl bromide structures. Geometry of the molecule in crystal structure is a little bit different, showing bigger difference in $\ce{C - N}$ bonds: 1.483 and 1.493 (ortho) vs 1.461 (para) in comparison with the calculated one for the gas phase:

I tried to align the molecule in the same way as it's been done by @pH13-Yet anotherPhilipp for better visual comparison. It's also worth noticing that in real crystal structure nitro groups are more tilted due to interlayer interactions:

(1) Parrish, D. A.; Deschamps, J. R.; Gilardi, R. D.; Butcher, R. J. Crystal Growth & Design 2008, 8 (1), 57–62. DOI: 10.1021/cg700727n

The bond lengths of carbon 1 & 3 are same and they are longer than that of carbon 5 and its attached nitrogen. This is due to resonance and in resonance we see 1 structure at a time.

Welcome to Chemistry.SE. There is no need to address the poster or put in a signature. Every post is signed, and every answer is directed back to the asker. See here. I have edited your post to remove these and clean up your grammar.

While we're at it, do you have any additional data that makes your answer significantly different than PH13's, which makes the same point and has a DFT gas-phase structure to back it up?


text:
How can hybrid orbitals be equivalent?
How can hybrid orbitals be equivalent? Take, for example, $\mathrm{sp^3}$ hybridization. The wave functions of each hybrid orbitals are given by:
\begin{align}
\psi_1 &= \psi_s + \psi_{p_x} + \psi_{p_y} + \psi_{p_z}\\
\psi_2 &= \psi_s - \psi_{p_x} + \psi_{p_y} - \psi_{p_z}\\
\psi_3 &= \psi_s - \psi_{p_x} - \psi_{p_y} + \psi_{p_z}\\
\psi_4 &= \psi_s + \psi_{p_x} - \psi_{p_y} - \psi_{p_z}\\
\end{align}
No one of them has same linear combination. So, how can they be equivalent? I am not getting the sense.

Since the coefficients are different, the orbitals end up pointing in different directions. As for why they happen to point in nice directions, and why their energies are exactly the same, you kind of have to trust that the maths works out. Hopefully someone else can answer because I'm not too familiar with the application of MO theory to hybridisation.

see this related question and answers

Gosh, @ortho, you really didn't know anything at the end of first year, did you! They are equivalent, only in the same sense that the x, y, and z axes are equivalent in a cube...

You are right in that the sign of each orbital, especially the p ones, is different in every hybrid orbital.
Your equations are, however, lacking the normalised contribution factors that should precede the wavefunctions. A better equation for the first orbital would be:
$$\psi_1 = a_s \psi_s + a_{p_x} \psi_{p_x} + a_{p_y} \psi_{p_y} + a_{p_z} \psi_{p_z}$$
The point is, that the contribution factors are equal for every sp³ hybrid orbital — every hybrid gets ¼ contribution from every p and the s-orbital. That way, all orbitals are rendered equal because their atomic orbital contribution factors are equal, too.
The signs before each p-orbitals contribution do not matter for the orbital’s energy value, only for its orientation. So inspite of all orbitals having different combinations of signs, their energy can be equal (but they will ‘point’ in different directions).

(Note that this is a simplified view as are hybrid orbitals in general. Spectroscopically, you can measure two different energies of the bonding orbitals in methane — the explanation cannot be given by hybridisation theory but requires proper molecular orbital theory.)

The standard set of orbitals (s, p, d, f, ...) form an orthogonal basis set, such that you can describe any continuous function as a linear combination of them.  Now, this also means that you can take some subset of them, and make a new set of orthogonal functions, by appropriate normalized linear combinations, that will contain the full symmetry of the subset chosen. The $sp^{3}$ hybridization is just such a set of combinations of $s$ and $p$ orbitals, chosen to make a set of equivalent orbitals emphasizing a mixed symmetry.
Note that the standard orbitals are not the only orthogonal basis set that exists - it just happens to be the one that nicely gives solutions to the Schrodinger equation separating out $n$ and $l$. One could use spherical Bessel functions, or any number of other orthogonal basis sets, but it would not be as clean of a solution just because of the form of the differential equation under consideration.


text:
How were old style stereographic structures produced?
Old papers on chemical structures often contained "3D" renderings of the molecule by using 2 2D images and requiring the reader to go cross-eyed to get the 3D image. A good example I recently found was figure 2 from Carter and Kraut, 1974, which I copied below:

However, this paper is from 1974, which feels like ancient history for 3D graphics. Since I have unrelated interests in molecular biology and 3D graphics, I was wondering how they produced these images so long ago, and if the code to produce them is floating around somewhere. I know there are programs to do it for me because I have seen modern images of proteins that do this, I want the code because I might try to build it into my crappy 3D rendering project that I'm wasting too much time on.

I'm going to hold out for an answer for someone that knows for sure, but much of this kind of stuff was done with https://en.wikipedia.org/wiki/Camera_lucida.  You can see that there are some inconsistencies in the bond lengths and angles, so I don't think these were done automatically (!!)

I like your project, watched the video - it's cool, keep going with it. Don't know how these images were produced, sorry. Just driving by.

UCSF Chimera is able to produce such images, though I have no idea how they do it or how it was done back then.

IBM mainframes with Tektronix graphics screens could certainly do the needed calculations and terminal output back in the 1970s. As for printed output, the first HP pen plotter came on the market in 1968. (Although, as I recall from the early 1980s the then-common giant Tektronix screens also had a thermal printer option.)

Created in FORTRAN by Carroll K. Johnson, of the Oak Ridge National Laboratory (ORNL), and first released in 1965, ORTEP (Oak Ridge Thermal-Ellipsoid Plot Program) rapidly became a favorite of crystallographers and protein crystallographers to produce illustrations of structures for conference presentations and publications. A key strength of ORTEP was its capacity to generate stereoscopic images automatically.

For source of above and more information see: History of Visualization of Biological Macromolecules
For the ORTEP program itself, check ONRL Report #3794: https://web.archive.org/web/20170210061629/http://web.ornl.gov/info/reports/1965/3445600599779.pdf (big file warning)

Wow. Looking at that FORTRAN code makes me appreciate C++. Did they sell magnetic tapes with the code, or would you have type the whole thing yourself if you wanted to install your own copy?

@Martin-マーチン This gives contact information at Oak Ridge for ORTEP's "guardian and developer" http://www.ccp14.ac.uk/ccp/ccp14/ftp-mirror/ornl-ortep/pub/ortep/aca97/aca97.pdf if anyone needs further information.

@user137, yes type the code by hand onto punched cards, starting col 7 for fortran, one card to a line of code, put into a metal box hoping no typos , hand box over counter and wait for  hours for job to be run. This was modern; the structure of insulin was determined by using programmes written on paper tape written in assembler. Editing was done by splicing tape and using sellotape to repair.


text:
Why are real orbitals used to define the orbital of hydrogenic atoms instead of complex orbitals?
On the physics site, I asked the difference between real & complex orbital. Real orbitals are the superposition of complex orbitals having definite magnetic quantum number states.
\begin{align}
p_z &= p_0 \\
p_x &= \frac{1}{\sqrt{2}} \left(p_1 + p_{-1} \right) \\
p_y &= \frac{1}{i\sqrt{2}} \left( p_1 - p_{-1} \right)\\
\end{align}
where $p_0 = R_{n1} Y_{10}$, $p_1 = R_{n1} Y_{11}$, and $p_{−1} = R_{n1} Y_{1−1}$, are the complex orbitals corresponding to $\ell = 1$.
My question is:
Why aren't the orbitals of hydrogenic atoms taken as complex orbitals instead of the real ones? They are valid states for an electron, aren't they? So, what is the problem in having complex atomic orbitals?

Quick and dirty answer: Because the complex orbitals don't have a direction, which likely helps calculation. But wait for one of the theoretical chemistry gurus to chime in here ;)

You can choose whichever you want or best suit for your purpose. For actual calculations often the complex ones used. I guess Chemistry textbooks prefer the one which is easier to draw and seems less complicated mathematically.

Indeed, one of the major pitfalls of using real orbitals is when students start asking what magnetic QNs they should assign to the p_y and p_x orbitals. I was never quite satisfied with the answer of "just pick one, it doesn't really matter." It wasn't until I took quantum that I really appreciated how ill-defined the question was. It still beats trying to explain complex orbitals at that level though.

AFAIK, when investigating a stationary state under Born-Oppenheimer, complex component may be separated into its own multiplier depending only on time and, since it has time-independent norm, it gives time-independent contribution for most stationary properties like energy. Since other multiplier depends only on space coordinates, for simplicity of math and faster computations,  almost entire computational chemistry uses real math only.

The main disadvantage of complex atomic orbitals (AOs) is that many of them are hard to visualize.  For example, try making a simple sketch explaining how $p_{-1}$ differs from $p_{1}$. Conversely the complex AOs have very few  advantages over real MOs - they are all eigenfunctions of $\hat{L}_z$ but that's not really important for molecular  and closed shell systems.
So with many key advantages and very few disadvantages real orbitals are understandably the most popular choice.

Totally agree, but I'd also add that because orbitals are not unique you can unitarily transform any real orbitals into complex orbitals. They only differ by a phase. Of course we must remember that orbitals are not physical but density is---which is always real---since the the inner product of the wave function eliminates any imaginary components.

To add my 5 cents to the discussion, note that we also usually use real basis functions ("atomic orbitals") in electronic structure calculations, although, molecular orbital expansion coefficients may be complex. So, each and every MO $\phi_{k}$ is usually expressed as follows,
$$
    \phi_{k}(\vec{r}) = \sum\limits_{q=1}^{m} c_{qk} \chi_{q}(\vec{r}) \, ,
$$
where $\chi_{q}(\vec{r})$ are real functions and $c_{qk}$ may in general be complex numbers. The point is that there is no additional variational freedom in using both complex basis functions and complex expansion coefficients. Besides, the use of real basis functions reduces (due to permutational symmetry) the number of two-electron integrals to be calculated for a basis of $m$ real functions from $m^4$ to $M(M+1)^2$, where $M=m(m+1)/2$ is the number of distinct pairs of products of two basis functions.


text:
Hydration number: Carbonic acid in water
Carbonic acid ($\ce{H2CO3}$) is known to be soluble in water, and that this particular compound is unstable, so much so that according to the 2014 article from Berkeley Lab:
New Insights on Carbonic Acid in Water, it only exists for a fraction of a second (300 nanoseconds according the article).
Newly reported experiments and observations from the Berkeley Lab found that:

“Our results support an average hydration number of 3.17 with the acid’s two protons each donating a strong hydrogen bond to solvating waters, the carbonyl oxygen accepting a strong hydrogen bond from solvating water, and the hydroxyl oxygen molecules accepting weak hydrogen bonds from the water”

What is not clear is how is the hydration number calculated and how does this relate to the short existence of carbonic acid in water?

Please note, the first two links point to the linked questions indicated for this question.

You may wish to refer to the publication of Saykally et al., The hydration structure of aqueous carbonic acid from X-ray
absorption spectroscopy
Basically, to determine hydration number, X-ray spectroscopy was used. This relates to the short existence of carbonic acid as, since it is so short lived, it is necessary to analyze it in such a time frame. As such, this was "the first measurement of X-ray absorption spectra for aqueous carbonic acid, enabled by the use of rapid-flow liquid microjet technology" which allowed it to be captured.
Then, by analyzing the spectra, it is possible to calculate the approximate hydration number, as "the increasing H-bond strength is proportional to the hydration
number (measured for distances up to $2.5\mathrm{\overset{\circ}A}$): $\ce{Na2CO3}$ (5.55), $\ce{HCO3}$
(4.26), $\ce{H2CO3}$ (3.17)".

I will add that the radial distribution function ($g(r)$), which is used in this and similar studies to infer how many of (some species) are distributed around (some central point) up to (some distance), is tied directly to the hydration number. That quantity (as detailed in the reference given by Andy and in Kumar et al., referenced therein) is being interpreted to correspond to stronger hydrogen bonding if it's high, and weaker hydrogen bonding if it's low.

Please visit [this page](http://chemistry.stackexchange.com/help/notation " latex="" at="" chemistry.se""="), this page and this one  on how to make your future posts better.

Brilliant answer! Thank you Andy (and thank you @ToddMinehardt) - this is much clearer now. Essentially, the hydration number is linked to the spectra.


text:
What Causes Atomic Orbitals to Hybridise?
I'm finding it difficult to visualise the process of hybridisation. Taking the example of formation of $\ce{CH4}$, can it viewed as the collapsing of the p orbitals of the Carbon atoms as they get "heavier" on excitation even before the introduction of Hydrogen atoms or is it the collision with Hydrogen atoms that causes the fusion?

Hybridisation is purely a mathematical construct, the p orbitals do not actually collapse or anything like that. (To be honest the entire concept of an orbital is also a mathematical construct.)

Hybridization is a mathematical concept developed to cope up the flaws of Valence Bond theory. It hinges solely on the linear combination of wavefunctions of atomic orbitals having more or less same energy. Hybridized state is a purely theoretical concept & cannot be detected even spectroscopically.
There is nothing like "collapse." The wavefunction of a hybridized orbital is "mixing" of the atomic orbitals only. The number of hybrid orbitals are same as the number of atomic orbitals participating in the "mixing up". The wavefunctions interfere constructively or destructively to give rise to equivalent orbitals known as hybrid orbitals.
The Methane Case:
Carbon has valence configuration: $2s^2 2p_x ^1  2p_y^1$. So, VB theory predicts that it can form two bonds. This is one of the main defect of VB theory. Then promotion comes to the rescue! One electron from $2s$ gets promoted to unoccupied $2p_z$ orbital, so as to make the configuration: $2s^12p_x^12p_y^12p_z^1$ . So, $\ce{C}$ can form four bonds. But still there is a problem: experiments show that the four bonds in methane are exactly equivalent which is not possible by this configuration since $s$ & $p$ orbitals are not  same. To deal with the problem, concept of hybridization was developed. Hybridization gives rise to four equivalent orbitals but having different orientations (four corners of the tetrahedron.) The non-normalized wavefunctions of four orbitals are as follows: $$\psi_1 = \psi_s + \psi_{p_x} + \psi_{p_y} + \psi_{p_z}\\\\\\ \psi_2 =\psi_s - \psi_{p_x} + \psi_{p_y} - \psi_{p_z}\\\\\\ \psi_3=\psi_s - \psi_{p_x} - \psi_{p_y} + \psi_{p_z}\\\\\\ \psi_4 = \psi_s + \psi_{p_x} - \psi_{p_y} - \psi_{p_z}$$

(source: prenhall.com)

Yeah I see the pattern, virtually everything in chemistry is explained by quantum theory. Anyways, thanks @user36790 for the suggestion :)

sorry I don't get your question, do you mean class as in talking about school?

Let us continue this discussion in chat.

Is hybridization just a kind of superposition of wavefunctions of the electrons?


text:
Is there a formula to determine the total number of constitutional isomers?
Is there a trick or a formula that given a molecular formula, allows you to know exactly how many constitutional isomers can be formed with that many atoms? Or is it more of a trial-and-error technique?
I'm looking for something along the lines of $2^n$ of maximum number of possible stereo-isomers, where n is the number of stereogenic centers present in a compound.

Constitutional isomers have nothing to do with stereogenic centers?

Why does an answer that's not the answer to the question, but an answer to a totally different question get the checkmark? The gilleain answer is best by virtue of being both correct and an actual answer to the question.

@EricBlack You are welcome to upvote an answer you approve of.

In principle, the maximum number of configurational stereoisomers (not constitutional isomers) is
$$N_\text{max}=2^{\left(n+m\right)}$$
where $n$ is the number of stereocentres (R or S) and $m$ is the number of stereogenic double bonds (E or Z).
However, the actual number of different stereoisomers may be smaller than the maximum number $N_\text{max}$ if constitutional symmetry is present in the molecule (meso-compounds).
Furthermore, the actual number of practically possible stereoisomers may be reduced by ring strain or other geometrical limitations.
Nevertheless, the maximum number may also be exceeded if hindered rotation about single bonds or other steric interactions result in additional stereoisomers.

There seems to be some confusion here : in order to determine the number of stereo-isomers of a molecular formula (say C4H11N) you would need to first know the number of constitutional isomers as you need to know how many stereogenic centers across all the structures.
There are a bunch of existing questions on constitutional isomers (a, b, c, ...) but the summary is : "it is tricky without software for anything more than very small formulae".
There are mathematical formulae for determining the count of constitutional isomers, although the most powerful techniques are non-trivial (such as those based on the Pólya enumeration theorem). To efficiently list a non-redundant set of compounds from an elemental formula is possibly even harder, as it requires a mechanism to avoid duplicates in the output.


text:
What is the state of aggregation (gas, liquid) of oxygen in blood?
Atmospheric oxygen is in O2 and a gas. Then we inhale the air, our efficient lungs do the magic to filter out the oxygen and push them into the blood stream.
When we say hemo and globin transport the oxygen using the iron ions. In what state oxygen is transported in the blood? as a gas or a liquid or an ion? It is hard for me to conceive of the idea that oxygen would be in gaseous form in the blood. "GAS in blood?" e.g. Arterial Blood Gas Test
Also, how does the lungs convert the gas into something that is compatible to be in blood?
References:
Amount of Oxygen in the Blood

Very highly related, might resolve your issue: http://biology.stackexchange.com/questions/8611/why-does-oxygen-bind-to-hemoglobin-in-an-specific-angle

@M.A.Ramezani I was actually thinking where to post this question, whether in Chem or Bio. The above post definitely covers majority of the chemistry required. My main questions were, how would compounds exist in blood in gas form, if so? What sort of a conversion takes place in alveolis to capillaries. Care to post an answer? :)

unfortunately, I don't have the sufficient time right now, I might strike the problem in the future. Also, considering your comments below, I think that question on bio.SE does answer your question. Most of the oxygen in blood becomes a "part" of a complex.

The oxygen isn't actually pushed into the bloodstream, it flows naturally, via diffusion, due to the concentration gradient (concentration in alveolar air is higher then on the venous blood arriving at the lung), plus its high hemoglobin affinity.

Regarding the state of oxygen in blood: It is in solution in the blood plasma (which mostly consists of water), in the form of single molecules. Think of water which you leave exposed to air: carbon dioxide will be captured and dissolved (along with the other gases in air), but these molecules are not gaseous or liquid, but rather "in solution", which is different from the "classical" states.
Back to oxygen: As your reference already states, most of the oxygen in solution will bind to hemoglobin. The actual state of oxygen in that complex has been debated, but it is believed to be reduced by the hemoglobin iron to the superoxide anion, coordinated to Fe$^{3+}$. See Wikipedia on this.
Also, the lungs do not "convert" the atmospheric oxygen to anything, they rather allow, due to their very large surface area, the quick exchange of oxygen/carbon dioxide in solution and in the air.

So the knowledge is saying in theory (practically/reality...) there won't be oxygen as it's in blood and pure molecule or compound can't be found - for stabilising reasons and other given reasons. Acceptable.

We inhale a lot of other atmospheric gases with oxygen. The mixture of gases go to the sacs in our lungs which is crawling with lots of blood capillaries.
The blood (specifically RBCs) contains a pigmented substance called haemoglobin, each of them can bind 4 oxygen molecules (if working normally) forming a substance named Oxyhaemoglobin.
This is then carried through blood vessels to the cells in the body.
The binding phenomena of oxygen with haemoglobin is reversible in nature. That is, at high oxygen concentration oxygen binds with haemoglobin to form oxyhaemoglobin, but when the oxygen concentration decreases (in cells, which are in need of oxygen) the oxygen again dissociates from haemoglobin.
The overview thus provided here, if you wish to be more specific about chemical reactions you may find this useful.

I don't mean to be rude. Can you answer this from a chemistry perspective? e.g. what happens to Oxygen state from alveoli to Capillaries?

Simple diffusion, based on the partial pressure. It goes from high pO2 (alveoli) to low pO2(blood capilaries)  zone. @bonCodigo


text:
How to write the last cell reaction for the following cell?
Consider the following electrochemical cell:
$$\ce{Zn|Zn^{2+}}(0.1~\mathrm{M})||\ce{Zn^{2+}}(0.01~\mathrm{M})\ce{|Zn}$$ How should I write the overall cell reaction for this cell? I can identify the oxidation and reduction reactions but when they are added, all the species cancel out, leaving my with no overall equation.
Anode: $\ce{Zn -> Zn^{2+} + 2e}$
Cathode: $\ce{Zn^{2+} + 2e -> Zn}$

The species do cancel out. However, the concentrations of the species do not. It is a bit of a matter of preference as to how you express this. The way I would write it is:
$$\ce{Zn^{2+}_{(aq)} ($c = 0.01\text{ M}$) + Zn_{(s)} <=> Zn^{2+}_{(aq)} ($c = 0.1\text{ M}$) + Zn_{(s)}}$$
for which $E = E^0 - \frac{RT}{2F}\,\ln\!\left(\frac{0.1}{0.01}\right) = -0.030 \text{ V}$ at 298 K, since $E^0 = 0 \text{ V}$. A negative value of $E$ agrees with the reverse reaction above being spontaneous, which is of course what you would expect since $0.1 > 0.01$.


text:
Are PTFE stir bars resistant to all lab solvents?
It seems like glass-coated stir bars are a specialty item.  Standard stir bars are coated in PTFE.  Is PTFE resistant to all typical lab solvents?  Or is there a reference to check to ensure it is inert to the chemicals being stirred?

PTFE is resistant to all typical lab solvents and even non typical.

Perhaps it's easier to give examples of things one might use in a lab that are not compatible with PTFE?

Probably only strongest redox stuff would be real problem.

PTFE is attacked by alkali metals, most notably by the liquid sodium/potassium alloy $\ce{NaK}$.

Labware manufacturers will often put out chemical compatibility charts for their products. A search on Google for the phrase ptfe chemical resistance yields a large number of hits, of varying thoroughness.
One fairly comprehensive one is this one from ThermoFisher Scientific.
Scanning it, PTFE has "Excellent" resistance listed for most solvents, except for pure 3-phenyl-2-propenal, which is "merely" "Good" ("Little or no damage after 30 days of constant exposure.") Lists from other sources have a different sets of compounds, some of which caution against conditions like "dry fluorine" or some rarer compounds. This one lists fluorine, gold monocyanide, and diethylamine as not recommended. (Though a number of other sources - including ThermoFisher - say diethylamine is perfectly fine.)
So PTFE is resistant to pretty much all lab solvents which a major manufacturer of lab equipment thinks it's relevant to mention - but if you have something that you think may be atypical or which you have particular concerns about, do a search for that particular compound or contact the manufacturer for their recommendations.

This may not be the answer that you are looking for, but I remember a project, for which we had to replace PTFE stirring bars with their glass-coated counterparts.
It is true that PTFE is resistant to most of the solvents used, but it is not really hard and resistant to abrasion.
In a study where aqueous suspensions of metal silicides were stirred, we recognized very soon that the PTFE covers didn't last very long. The stirring bars were polished down to the metal core, which spoiled the experiment. Glass-covered stirring bars turned out to be a viable alternative here.


text:
Cis-trans isomers internal energy
Whilst reading about stereochemistry, I came to know that the stability and melting point of cis isomers is lower than trans isomers. By searching on net I've found the reason behind that. But one question arises - why is the internal energy of cis isomers bigger than of trans isomers?

Do you refer to cyclic or acyclic compounds? In acyclic compounds the trans isomer is more stable while in (small) cyclic the cis is more stable. Thus the internal energy of cis is bigger in acyclic while the internal energy of trans is bigger in cyclic.

I can think of two reasons:
1) Steric repulsion
2) Electrostatic interaction
Steric repulsion: cis means in same side. So, sterically it is suitable to be on opposite side. It is a fundamental nature of both human and chemical entity to define a personal boundary (My personal opinion :) )
In the image you can see the personal boundary of $\ce{-CH_3}$ is the furthest in the trans configuration and that's wh the trans configuration is more stable than the cis configuration.
Electrostatic interaction: Due to electrostatic interaction, molecules or functional group try to distance themselve as far as possible. This also results in lower energy in trans configuration.

Actually steric hindrance is not the right word to describe it, it should be termed "steric clashing" or "steric repulsion". Steric hindrance is what happens when the approach of a molecule is prevented by large, bulky groups (e.g. SN2 at t-butyl bromide, or base hydrolysis of t-butyl esters). Lastly, (someone please correct me if I'm wrong) I think steric repulsion is fundamentally electrostatic in nature - it arises from repulsions between the negatively charged "electron clouds" of atoms.

@orthocresol This is correct, but also bear in mind that there are nuclei, which also cause repulsion. But yes, the biggest effect can be assumed to come from Pauli repulsion since you are trying to force electrons of the same spin in the same space.


text:
How to calculate mass of nanomaterial grown on a substrate
I'm growing graphene on copper substrates and need to know the mass. The graphene is grown via chemical vapor deposition; amounts of reactants are not completely known. The best approach I've thought of is recording the mass of the substrate before and after using an analytical balance, but with the nature of the material being nano, this method may be elusive. What other method could I use to find the mass?

Is it a graphene monolayer? If so, then weighing will likely not work; monolayer graphene weighs $\mathrm{0.8\ mg\ m^{-2}}$. Also, does the graphene have to stay adhered to the copper? And to what precision do you need the mass to be determined? Do you know the shape/area the graphene layer has, or could you somehow mask it (i.e. measure the area covered in graphene and use the areal density to obtain the mass)?

It's vertically grown graphene. We're calculating specific capacitance using cyclic voltammetry. It doesn't have to be extremely precise; just enough to do basic calculations.

Oh, not at all the type of layer I had in mind. Quite a pretty image! Depending how tall and dense that graphene forest (coral?) is, weight may work. Unfortunately it's a rather irregular layer of material, so now I'm having a hard time thinking of alternatives. If a destructive method is allowed, maybe you can oxidize the carbon to $\ce{CO2}$ and measure that with, say, gas chromatography?

Ah, I don't think destruction of it is something we would be trying to do. And yes, I think its a great image. Wish it were mine.

One possible option for thin film thickness measurement is by Energy Dispersive Spectroscopy (EDS) in a scanning electron microscope (SEM).  There exist commercial packages for this such as LayerProbe by Oxford and STRATAGem (I have no financial interest in Oxford or SAMx).  If your campus/research facility has an SEM with an EDS (many do), but not a commercial package, you can still work it out yourself.  If I were tackling this in my own lab, I would probably do something along the following lines:
Option #1:
Obtain a spectrum from a known thickness of carbon on your Cu substrate.  This can be any method such as sputtering or evaporation so long as the thickness is known by other means.  Most geology labs have the ability to do this kind of deposition since they need reproducible carbon thicknesses on their samples.
For example, geo labs will typically place a polished brass object next to the sample to be coated in an evaporative coater.  The change in the color of the brass corresponds to the thickness of the carbon: e.g. blue is 250 Å but bluish green is 300 Å (Kerrick, D. M., et al., 1973).  So something like 25 % accuracy is possible this way.
Then you measure that thickness in the SEM using an exact beam current, size, distance from the objective lens.  Then you measure your graphene in the SEM using the exact same conditions.  If your sample is less than about 200 nm thick, then the number of counts/second is proportional to the number of carbon atoms present in the layer.  Using the mass for your standard layer, (assume a density of graphite), then you can compute the mass of your graphene.
If your thickness is more than about 200 nm, then you may need to do an absorption correction – i.e. the photons are being absorbed on their way out and your count rate is no longer proportional to the number of atoms.  You can compute attenuation coefficients using the CXRO calculator (and a couple other places if you google).  Then you can do a thin film correction.  At 200 nm, your error from ignoring the thin film correction is probably about 5–10 %, which is likely a bit less than the error in the determination of the standard film using the color method.  So you can think with that for your sample.
Option #2:
You can measure the mass directly using an oscillating crystal.  Commercial systems comprise a quartz crystal with electronics set to find the resonant frequency.  You place the quartz crystal in your chamber alongside your Cu substrate.  As your reaction happens, it grows both onto your Cu and onto the quartz crystal.  The resonant frequency of the crystal changes as mass is added.  These are sold commercially, though a bit pricey.  For example, at TedPella (again, no financial interest).  The resolution is typically billed as 1 Å thickness resolution (assuming you know the density of your material.)
For your application, you would have to be sure that the graphite grows on your monitor crystal the same as on your Cu substrate.  That may require checking in an SEM, and it may require prepping the crystal somehow.  Hence, I offer this as suggestion #2.
Finally, you could also figure this out using a focused ion beam and sectioning the sample from the side.  If this is too complex, there also exist labs where one can hire them to do this kind of measurement.
I hope one of these solutions works for you!
Ref:
Kerrick, D. M., Eminhizer, L. B., & Villaume, J. F. (1973). The Role of Carbon Film Thickness in Electron Microprobe Analysis. American Mineralogist, 58, 920–925.


text:
Why is copper(II) oxide green?
Copper(II) oxide is black, but when we find the percentage of oxygen in air by volume using ammonia, copper, and ammonium chloride, the color of $\ce{CuO}$ turns green. What is the reason for this?

These two questions seem kinda separate. I think the community will be able to cover them both, provided that you ask them separately.

Just hazarding a guess, it might be due to formation of $\ce{CuCO3}$ (by reaction with $\ce{CO2}$ in the air).

As I said in an earlier comment, you need to separate the two questions.

Cupric oxide is not green, nor does it turn green. What you see is another compound, probably $\ce{CuCl2}$. BTW, the question would benefit from a concise description of the process you are using, so that we wouldn't have to resort to guesswork.

@orthocresol that's a good guess. CuCO3.Cu(OH)2 is green


text:
Confusion about direction of dipole arrow in alpha-helices and other molecules
I understand that molecular dipoles are electric dipoles. And electric dipole moment vectors point from the negative to the positive charge.
In class we learned to draw these special molecular dipole arrows (with a "plus" at the beginning) that point from the positive to the negative partial charge. I also see these arrows all over the internet, e.g.:

But for alpha-helices, the dipole moment points from the C to the N terminus. And the detail always shows the arrow pointing from the oxygen (negative) to the hydrogen (positive):

Should I just accept that even though an alpha-helix is a molecule like water, the "standard" way to draw its molecular dipole arrow is in the opposite direction than what people normally draw for water and other small molecules?
Edit: Another image, from a recent publication, but without macro-dipole arrow...:

Are you talking about alpha helices in membranes or in protein cores? Either way it has been a while since the polarity was described as going from C to N.

Any alpha helix I guess.

The O to H in the backbone will be massively irrelevant compared to polar or charged amino acids, or non-polar (hydrophobic) residue r groups.  In structural biology you don't worry about the backbone charge direction as it will not affect the iso/leucine and polar groups that form the "surface" of the helix. I guess this question is confusing without any context. Could you cite the origin of the images?

The images were just the first that came up when googling for "molecular dipole" and "alpha helix dipole". We have this exercise where something about alpha helix stability is argued with the helix macro dipole.

The more "alpha helix dipole" google images I look at, the more confused I get. Everybody seems to have a different idea. And wikipedia says that there is some controversy about the alpha helix macro dipole, but doesn't give a direction.

The direction of the arrow is just a convention. Apparently no hard rule exists for denoting the dipole moment of alpha helices. What is important is that you know that the C-terminus is partially negative and the N-terminus is partially positive. I guess it's also important to know which way your teacher wants the arrow drawn.

Dipole moment is a vector. The molecular dipole moment that you are talking about is the net dipole moment i.e. sum (vector addition) of all bond dipole moments. Like all vectors, dipole moment has direction. When you say there is a dipole moment from C-terminus to N-terminus you are thinking of protein as a linear molecule, ignoring the tertiary structure. The net dipole moment of the protein will depend on the tertiary structure and the side chains. If this comment necessarily answers your question then I'll convert it as answer.

@WYSIWYG thanks. my question is: why do people draw the macro/net/sum dipole of a linear alpha helix from negative to positive, but for a water molecule they draw it from positive to negative.

@Zalumon Electrical dipole moment is always negative to positive. See here. I am not sure why in chemistry a reversed direction is used for molecules. Perhaps because they want to denote the direction of electron cloud density. This seems to be a case of conflict of semantics. It is better to ask this question in Chemistry-SE.

http://www.av8n.com/physics/electric-dipole.htm#sec-mutations

@Loong You can add that as an answer and you can claim the bounty too (a petty 50 points, but I don't have much to spare here)

http://scienceline.ucsb.edu/getkey.php?key=2477

I think your problem is the difference between the chemist and physicist notation of dipoles / polarization: they use opposite definition, and most of them not aware. So there might be big mix-ups in the references, if you are not careful.

I (accidentally) stumbled upon the following statement in Atkins' "Elements of Physical Chemistry" (p378):

We represent dipole moments by an arrow with a length proportional to $\pmb{\mu}$ and pointing from the negative charge to the positive charge (1). (Be careful with this convention: for historical reasons the opposite convention is still widely used.)

Unfortunately he does not go into more detail. And I know this does not really answer your question.
The definition from the IUPAC is the same as the one used by Atkins:

electric dipole moment, $\mathbf{p}$
  Vector quantity, the vector product of which with the electric field strength, $\mathbf{E}$, of a homogeneous field is equal to the torque.  $\mathbf{T} = \mathbf{p} \times \mathbf{E}$. The direction of the dipole moment is from the negative to the positive charge.

The source quoted there is from 1993, so you can probably understand my surprise, when I did a little more searching and found in C. Párkányi's "Theoretical Organic Chemistry" (1997, p239):

[...] in organic chemistry the positive direction of the dipole moment is normally defined as the direction from the center of the positive charge towards the center of the negative charge. This convention prevails in physical organic chemistry and in inorganic chemistry. However, while the dipole still points from the positive charge to the negative charge, in physical chemistry and in chemical physics the positive direction of the dipole moment is defined in the opposite way, i.e., from the negative charge to the positive charge.

There are also a few sources given, but I currently have not enough time to look them up. I find this statement (definition) highly confusing and it also does not give another reason why this is the direction it uses. Please forgive me for not going into more detail here, I really don't want to add any more to the confusion.
Conclusion
Don't add to the confusion. Use
$$\huge\ominus \overset{\mathbf{p}}{\longrightarrow}\oplus$$
as your definition from now on. Popular use does not make it right. Help using it consistently in the correct way and flush out the historic remnants that are still being taught. However, keep Atkins' warning in mind when you read books, paper, etc.. In the literature you will find both versions.

If anyone wants to argue with you about this, give the following derivation. The dipole moment operator $\mathbf{P}$ is a vector operator  that is the sum of the position vectors $\mathbf{r}$ of all $N$ charged particles weighted with their charge $q$.[goldbook]
$$\mathbf{P} = \sum_i^N q_i \mathbf{r}_i$$
For a molecule (neutral by definition)
we find
$$\begin{align}
  0& =\sum_i^N q_i &\Leftrightarrow&&
  0&=\sum_{i_+}^{N_+}q_{i_+} + \sum_{i_-}^{N_-}q_{i_-} &\Leftrightarrow&&
  \sum_{i_+}^{N_+}q_{i_+} &= - \sum_{i_-}^{N_-}q_{i_-} &.
\end{align}$$
This can be transformed into
$$\sum_{i_+}^{N_+}|q_{i_+}|\mathrm{e} =
- \sum_{i_-}^{N_-}|q_{i_-}|(\mathrm{-e}).$$
Therefore we can write
$$\begin{align}
  \mathbf{P} &= \sum_{i_+}^{N_+} q_{i_+} \mathbf{r}_{i_+}
             + \sum_{i_-}^{N_-} q_{i_-} \mathbf{r}_{i_-}
             &\Leftrightarrow&&
  \mathbf{P} &= \mathrm{e}\left(
                \sum_{i_+}^{N_+} |q_{i_+}| \mathbf{r}_{i_+}
              - \sum_{i_-}^{N_-} |q_{i_-}| \mathbf{r}_{i_-}
                \right)&.
\end{align}$$
In the parenthesis the first term is a linear combination of vectors for all positive charges with the resulting vector $\mathbf{r}_+$ and the second term is a linear combination for all negative charges with the resulting vector $\mathbf{r}_-$. The dipole operator is therefore equivalent to
$$\begin{align}
  \mathbf{P} &= \mathrm{e}\left(\mathbf{r}_{i_+} - \mathbf{r}_{i_-}\right),
\end{align}$$
which is a linear combination of vectors that points from negative to positive.

It is true, but not true in practice. Atkins and IUPAC may decide something, but majority of chemists use the chemist notations (including several theoreticians and many textbooks).

Let's take a simple case of a positive charge $q_+$ and a negative charge $q_-$ with $q_+ + q_- = 0$ placed along the $x$-axis at $x_+$ and $x_-$, respectively. The dipole at the $x=0$ is
$$\mu=q_+x_+ + q_-x_- $$
which can be rewritten as
$$\mu=q_+(x_+ -x_-) $$
which is a vector pointing from $x_-$ towards $x_+$. So this is the "natural" direction based on the definition of a dipole.
However, chemists a very concerned with the "flow" of electrons and electronegativity. So in the case of water we like to use the bond dipole to depict the "flow" of electrons from the $\ce{H}$ to the $\ce{O}$ due to the higher electronegativity of $\ce{O}$.  So we tend to invert the direction of the dipole vector.
For this reason it is important to always define the direction of the dipole "arrow" in drawings to there is no confusion.

Is the convention of reversed direction of dipole moment universally true for all cases. Is it defined like that somewhere (IUPAC) ?

Yes http://goldbook.iupac.org/E01929.html

It says that "The direction of the dipole moment is from the negative to the positive charge." Clearly contradicts the popular chemistry convention of positive to negative.

This does not really answer the question. Can you add some details on why chemists use a reversed direction of dipole? Please add a canonical reference for that if possible.


text:
Is 3,5-diethyltoluene an accepted IUPAC name?
My book says that

Monocyclic substituted hydrocarbons are named as derivatives of benzene or compounds like cumene, toluene, xylene, etc. However, if the substituent introduced into such a compound is identical with the substituent already present in the compound, then the substituted compound is named as tge derivatives of benzene.

But I guess that names like toluene have stayed due to historic reasons and are not accepted in IUPAC nomenclature. Is my book correct?

Names like toluene are "retained names".

In chemistry, a retained name is a name for a chemical compound that is recommended for use by a system of chemical nomenclature (for example, IUPAC nomenclature), but that is not fully systematic. - Wikipedia
‘Retained names’ are traditional or common well-established names that may be used either as preferred IUPAC name or as an allowed alternatives in general nomenclature, for example, naphthalene, benzoic acid and pyridine. IUPAC's Blue book,  P-12.3

Later in the same draft, we read

Toluene, xylene and mesitylene are specific parent hydrides that are composed of two components, one cyclic and the other acyclic and saturated. These names are retained due to a long and well established tradition. Toluene and xylene are preferred IUPAC names, but are not substitutable; toluene is substitutable under certain conditions for general nomenclature. P-22.1.3
Furthermore, anisole and toluene cannot be substituted by other
methoxy or methyl groups, respectively. P-46.2.3.2

Conclusion:

Names like toluene, anisole and xylene are retained names. They are preserved for historical reasons; however, their unsubstituted forms are PINs (Preferred IUPAC names).
The name 3,5-diethyl toluene isn't acceptable according to the systematic IUPAC nomenclature.
If you want to name the compound below, it's called 1,3-diethyl-5-methylbenzene.

Source


text:
Why is temperature constant when an ideal gas expands into a vacuum?
An ideal gas in a box has pressure $p$ and temperature $T$. This box is kept in a vacuum, within a large container.
When the box is punctured, what happens to the temperature of the gas as it expands to fill the entire container?
I was taught that the temperature will remain constant. How is this possible? I understand that the pressure $p$ will decrease and the volume $V$ will increase, but for the temperature $T$ to remain constant, the decrease in $p$ must exactly counterbalance the increase in $V$. Is this the case?

This process is called an expansion into a vacuum. As you described, the volume of the gas increases and the pressure drops, such that $pV$ does not change at all. Since $n$ is constant, $T = \frac{pV}{nR}$ is also constant.
I think the fact that $pV$ is constant is not very intuitive if one only uses the ideal gas law. A more formal way of explaining that $T$ is constant would use thermodynamics. If you assume that the large container does not allow any transfer of heat (i.e. it is thermally insulating), then the large container is an isolated system: $\mathrm{d}q = 0$. Since the gas is expanding into a vacuum, $p_{\text{ext}} = 0$ and $\mathrm{d}w = -p_{\mathrm{ext}}\,\mathrm{d}V = 0$. By the First Law, $\mathrm{d}U = \mathrm{d}q + \mathrm{d}w = 0$ and temperature is therefore constant (since the temperature of an ideal gas depends fully on $U$).

As gas particles expand into a vacuum there's no transfer of their kinetic energy into anything else, they do not collide with anything but with themselves. Since the temperature of a gas is a measure of the average kinetic energy of its particles, the temperature remains constant. (This is different than in the case when the expansion would involve for example pushing a piston, frictionless or not.)


text:
How to calculate the mass of an element in a sample of a compound?
What calculations are needed to express the grams of an element from a compound dissolved in aqueous solution? For example, I have a $0.5\,\mathrm{M}$ glucose aqueous solution and I want to know how many grams of $\ce{O}$ are in there.
I have tried to calculate in this way:

Molecular weight of glucose = $180.159 \mathrm{\frac{g}{mol}}$
Atomic weight $\ce{C} = 72$; $\ce{H} = 12.09$; $\ce{O} = 96$;
I am considering $0.5\,\mathrm{M}$, so in $1\,\mathrm{L}$ I have $90.08\,\mathrm{g}$ of glucose
At this point I divide the molecular weight of glucose by the atomic weight of oxygen $\frac{90.08}{96}$, but I don't think it is correct.

As suggested in the comments, keep track of the units!  If your answer doesn’t make physical sense, you made a mistake.
A thorough approach to this problem is to:

calculate how much glucose you have.
calculate the fraction of oxygen in glucose.
multiply the (unitless) fraction found in (3) by the total glucose calculated in (1).

Using the values (and associated units) given in your problem:

$0.5 \pu{\frac{mol}{L}} \text{ glucose}\times \pu{1 L} \text{ glucose}\times \pu{180.16 \frac{g}{mol}}$ glucose = $\pu{90.08 g}$ glucose.
$\frac{\pu{96 g mol^-1} \ce{O}}{\pu{180.16 g mol^-1} \text{glucose}} = 0.533$
  mole fraction of $\ce{O}$ / glucose.
$0.533 \times \pu{90.08 g}$ glucose = $\pu{48 g} ~\ce{O}$

Of course there are many shortcuts you could take or even different approaches altogether, but I like to parse a problem like this down stepwise in a way that shows what you are doing and facilitates keeping track of units.  This makes troubleshooting much easier and systematic if you find you've made a mistake or want to walk through it a second time to verify what you've done (as you definitely should!).