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Symmetry lost in orbitals?
I've always thought that orbitals lead to a loss of symmetry, and have never been able to give myself a satisfactory answer to this.
I'll explain via an example:
Let's take an $\ce{N^3+}$ atom. It's perfectly spherical, and has no distinguishing 'up' and 'down'. There is no set of 'preferred coordinate axes' for it since it has spherical symmetry (except the nucleus, but I doubt that matters).
Now, let's give it three electrons. They arrange themselves in the $2p$ orbitals, one in each (by Hund's rule). Now, suddenly, the atom has lost its spherical symmetry — we have a distinct triplet of orthogonal directions separate from the others.
This leads to these questions:
How can symmetry 'break' this way? Are the directions of the axes 'hidden' in the atom beforehand? Are they themselves wavefunctions (though a wavefunction of wavefunctions sound odd to me, this explanation makes sense-random events can break symmetries)
So, I'd like a clear explanation of how/why the symmetry breaks.

As long as the wavefunction share the eigenstates of L^2 and L^z, we call it spherical symmetric.

Symmetry is not actually broken that badly. If you take any one symmetry axis (or plane or...), it exists in both cases.
Lets do a thought experiment:

In the spherical case you have only one electron. Your only symmetry axis goes through the electron and the nucleus.
In the case of three electrons, the symmetry axis goes through one of the electrons and the nucleus.
Doing the above collapses the system in to one possible configuration from the many possible ones (you're measuring quantum system, Heisenberg, etc.).

In other words, before you measure the location of one of the electrons, you don't know the symmetry axis and thus the two cases are equally symmetric. The probability of finding the electron number one at the "north pole" of the atom is equally probable in both cases.
Note that defining the symmetry axis is purely theoretical (mathematical) construction. The symmetry axis exists even if you don't do the measurement. See AcidFlask's comment below.
Also, I'd like to highlight, that measuring electron states is becoming possible: http://phys.org/news177582885.html.

So the axes themselves are wavefunctions?

Wait, if you don't know the symmetry axis before measurement, then the p orbital will be spherical as well, right?

I haven't heard anyone calling the axes as wave functions, so I would not recommend it, but basically they act like vectors with spherical wave functions. Before measurement the p-orbital is still shaped like a p-orbital, but it exists in all directions (compare this to quantum states and superposition). After you measure the location of the electron (which is impossible with current knowledge), you can say in which direction the p-orbital points.

As a side note, remember also that the electrons can exist outside the orbitals. Finding electrons in the orbitals is just much more probable than finding them outside orbitals. Orbitals are just places where the electron is most likely to be found.

Apparently 2009 they have managed to do this: "For the first time, it has been possible to measure electron density in individual molecular states using what is known as the photoelectric effect." http://phys.org/news177582885.html. Very good question on hot topic this is.

Yeah, that's what I thought regarding the p orbitals. I do know that orbitals aren't "orbits", and I do admit that that makes it more complicated.. Only way to be truly sure is to observe an electron when it is exactly on an angular node (of a different p orbital). I'll be sure to check the article when I get time, thanks!

Placing a symmetry axis is not equivalent to making a measurement. Your answer conflates a purely mathematical issue of where to place axes in space with experiments that break the symmetry along one particular axis, thus singling out one spatial direction.

@AcidFlask: Yes, I agree with you, but how would you explain this otherwise. You have a system that you want to put into coordinate system, but you cannot do it until you find out which are the symmetry axis that you cannot know until you measure the system. I think the problem here is how to formulate the answer correctly. If you have a better formulation, please, edit the answer. I try to figure some better formulation.

May I ask what does "that badly" mean? Is it broken, or is it not?

Well, geometrically one electron and one nucleus have one symmetry axis. Two electrons and nucleus have zero symmetry axis... If you think at the orbitals, the probability that your higher orbitals point e.g. north is finite number. "Symmetry broken" cannot be used in this context as there is no or just a little symmetry to begin with.

@CHM Good point!  It is not broken. Please see AcidFlask's answer.

What you describe is unfortunately a very common misconception. Defining orbitals does not break spatial symmetry. It is still completely arbitrary which directions you define as $x$, $y$ and $z$, and thus it is entirely arbitrary how you orient the $2p$ orbitals in space. Therefore rotational invariance is still preserved.
The other thing to remember is that all the $p$ orbitals are (essentially) degenerate and so you can take any linear superposition of them you want. For example you can write down an orbital that looks like
$$\frac{1}{\sqrt{3}}\left(\phi_{2p_x}+\phi_{2p_y}+\phi_{2p_z}\right) $$
and placing an electron in this orbital does not break rotational symmetry.

In my understanding the superposition principle says that the system is in one of the states, but you don't know which state until you measure the system. In other words, if you take one electron and excite it to one of the p-states, it physically will be on one of the orbitals and this orbital is geometrically asymmetric. And superposition claims that because you cannot say which state it is, you say that the electron is in all the states with probability X. Now I have to get my quantum book to verify this...

The p orbitals are degenerate in energy. Unless you make a measurement that breaks the degeneracy, an electron cannot be observed to be in a px, py or pz orbital. You may applied a magnetic field or something similar to break the degeneracy, which then allows such a distinction to be made, but then this privileges a particular direction in space which is the direction of the field around the atom.

Are you saying that the electron is on all orbitals or just in one but is observed to be on all (because measrurement breaks...)?

@Juha The atom is in the state $\Psi_{\text{atom}}$, which one can choose to express as a LCAO. But the state is still $\Psi_{\text{atom}}$, that is, a single vector in Hilbert space.

I disagree, the particle is in orbital $\Psi_{p}$ with the probability $p = c (\Psi_{p})^2$. And if there is multiple orbitals, you cannot know in which orbital the atom is, until you measure it (and then the system collapses to one orbital and is destroyed etc.). In quantum system, you cannot know the location or orbitals of electrons (until recently, see the link in my answer).

AcidFlask is right. Atoms with partially filled shells do not develop a moment.  They are still spherically-symmetric, and the mathematical expression is the linear combination of the three (for p shell) Slater Determinants.

The following is to clarify the degeneracy issue debated in the comments.
Hypothesis: Electron orbitals are quantum wave functions (the truth value is left to reader).
The text below is an example adabted from Liboff: introductory quantum mechanics, pages 119 - 121.
The system is in degenerate state of
$\Psi = \frac{3\phi_2+4\phi_9}{\sqrt{25}}     (5.17)$
Probability of finding energy $E_n$ is:
$P(E_n) = \langle\phi_n|\phi_n\rangle$
$P(E_2) = \langle\phi_2|\phi_2\rangle = \frac{9}{25}$
$P(E_9) = \langle\phi_9|\phi_9\rangle = \frac{16}{25}$
$P(E_(2+9)) = \langle\phi_2|\phi_9\rangle = 0$
$P(E_(a+b)) = \langle\phi_a|\phi_b\rangle = 0$, if $a \neq b$
Following is a direct quotation: "In an ensemble of 2500 identical one-dimensional boxes, each containing an identical particle in the same state $\phi(x,0)$ given by (5.17), measurement of $E$ at $t=0$ finds about 900 particles to have energy $E_2 = 4 E_1$ and 1600 particles to have energy $E_9 = 81E_1$."
So, the above states that the particle can be only in one state, not in many states and not in any linear combination of states. The ensemble average can have a linear combination of states.
Now, adapt this to orbitals. Electron can be only in one orbital. On average, electrons of bulk material are on linear combination of orbitals.

iii) On average, electrons of bulk material can never be in linear superposition of orbitals, because a linear superposition is a coherent quantum state, and what you have is the analog of a classical probability distribution, called the density matrix.

ii) According to standard Quantum Mechanics, the particle can be in any state of the Hilbert space including, of course, superposition states. If you adopt the view that they can only be in eigenstates of the Hamiltonian, then any individual molecule can not have any dynamics other than the trivial dynamics (evolution = $exp(-iEt/\hbar)$); also then you would need to think of what happens when you measure any other observable incompatible with energy, etc.

I found several missconceptions in your last answer: i) Orbitals are not wave functions in any conceivable way (except, obviously, for the Hidrogen atom or any one-electron ion).

i) true, but this is the hypothesis you have to make to get the orbitals (and this step is made in the comments). ii) Only if you consider ensemble averages. I want to see a reference that says "one particle can exist on a linear superposition of states". iii) Again, this is the step back from wavefunctions to orbitals.

In general, not until recently it is not experimentally clear where the electrons are on an atom. You have to make a step from theory to real world and back. This question is about real electrons, but people tend to consider it about wave functions (and I disagree how they interpret the superposition principle).


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How do I determine the absolute configuration experimentally?
Let's say I synthesized or isolated a chiral molecule and want to know the absolute configuration (R or S) of that molecule. I could obviously solve the structure by X-Ray crystallography, but that's a lot of work and I might not want to do that or might not have access to the necessary equipment.
What other ways exist to determine the absolute configuration of a molecule experimentally?

As you alluded to in your post, XRC has a long history for absolute configuration assignment. The drawback is more than resources, collaborators exist for such reasons, but the hard part is actually growing a suitable crystal. You also need heavy enough atoms present, usually introduced via derivatization if not already present. It can takes years, sometimes, to obtain a suitable crystal.
OR and CD have been used in the past to predict configuration. For example, we don't teach it these days but there were a variety of chirality rules developed to do this on CD. These were mostly empirical in nature and as a result had many exceptions. Today there are ab initio time-dependent density functional theory calculations which are trying to make accurate enough calculations to be useful in this arena.
NMR obviously has some ancillary methods to address this, but I don't know anything about them.
Stereocontrolled reactions are another way to help deduce the structure, but that is time and effort consuming as a synthetic endeavor and particularly useless for anybody in the pharmaceutical industry. Syntheses are also error prone.
If you want to know what the pharmaceutical industry does when it cannot do XRC, then look no further than failed drugs. As an example, gossypol was being evaluated as male birth control. It failed due to a certain lack of reversibility, among other things. Anyway, the technique used to determine its conformation and absolute configuration is called vibrational circular dichroism (VCD). I should also mention at this point that Raman optical activity (ROA) is another method that is useful in some instances.
Modern DFT program suites allow pretty reasonable predictions for VCD spectra taking on the order of a day, these calculations are much simpler than those required for good OR or CD. These calculated spectra are compared with experimental and from here the absolute configuration can be assigned. The quality of the calculations can be easily be determined by comparing the calculated and measured IR spectra.
The really basic idea is that there are different interactions of right- and left-circularly polarized light during vibrational excitation. The difference in these responses is plotted versus wavenumber. Without the right instrumentation, this technique is really hard, but one can purchase equipment to make these pretty straightforward to measure. There is a lot more you can do with it, but Big Pharma definitely uses VCD when XRC is inaccessible.

If your compound has been synthesized before, you can measure its optical activity and compare it with the data in the literature.
If your compound hasn't been synthesized before, it can be possible to differentiate between two enantiomers with NMR by measuring the spectra of your molecule in a chiral environment. Here's a review article on how to resolve enantiomers with NMR.
Otherwise, diastereomers can be differentiated easily. By looking at your molecule, you can (sort of) predict which spectrum it should give rise to by using the concepts of enantiotopic, homotopic, heterotopic and diastereotopic protons. We had to analyse and solve 2D-NMR spectra (1H and 13C) of diastereomers last year, and to a beginner, I have to say it was a challenge.
As always, it boils down to forming diastereomers.
EDIT
Take a look at these diasteromers:

(I've got finals coming up), it's easy to see that the hydroxyl will shift the ethyl's CH2 peaks towards lower field (higher ppm) in the second isomer. The interpretation of chiral media NMR spectra would have to be similar. My impression is that the acquisition of chiral media spectra is more involved, but the interpretation isn't. I might be totally off track, though.
Someone might want to share a more thorough answer.

The NMR thingy seems to distinguish between enantiomers--but can it tell you which is which? (I'm not sure, I've only skimmed the article). If so, could you give a bit on the basic reason behind why NMR allows this? Not necessary, but it would make your post awesome.

I've never worked with such spectra. Since the principle is to generate diastereomers, I would think it's possible to determine absolute configuration of your molecule if you know the absolute configuration of your chiral media.

really? I mean I guess you can determine dextro and laevo--but you need to have a theoretical matching of d/l with R/S. I doubt there will be a direct experimental matching (ie without theoretically relating the two first), since the R/S precedence system is arbitrary.

@Manishearth: Oh wait... NMR may be able to do that. Maybe. :/

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Brittle d-block metal trend
I remember seeing a periodic table that had the top-left corner of the d-block shaded and marked as "brittle". If I recall correctly, the elements were $\ce{Sc,Ti,V,Cr,Mn,Y,Zr,Nb,La}$. I think (sorry, I can't find much on it on the net).
The triangular shaded area is a clear indicator of a groupwise and period wise trend (similar to how the non-metals form a triangle due to the electron affinity/Ionization potential trends).
So, I ask, what are the trends leading to the brittleness of a metal, and, more importantly, why are these trends there (usually trends can be explained in terms of simpler things like atomic size/shielding/etc)?

That's quite interesting! Brittleness can also be described as low malleability.. and since Gold is the most malleable (and it is at the bottom-right corner of the d-block) it could most likely be a trend. Malleability depends on the strength of the metallic bonding in the metal crystal. Since this depends on so many factors, I don't think the trend would be very regular if at all it exists.

So, I ask, what are the trends leading to the brittleness of a metal

Brittle materials absorb very little energy before fracturing. The only trend that can be linked to this is the strength of bonding in the metal (Other factors like crystal structure and purity do not have any trends and so there are a lot of exceptions). This also means that the trend in brittleness is not very regular since it is complex and depends on too many other factors.

why are these trends there? (usually trends can be explained in terms of simpler things like atomic size/sheilding/etc)

Since brittleness cannot be explained by a very simple trend, I will explain what reasons I have found.
The stronger the bonding in the metal, the more difficult it is to change its structure and thus it tends to break instead of deforming. Here's a graph (melting point vs. Group number) that gives a general measure of strength of metallic bonding in all the transition elements of the first three series.

According to the graph, we predict that Tungsten (Period 6 Group VIB) should the most brittle and this is true for impure tungsten. Again, this successfully predicts that Silver, Gold and Copper (Group IB) are least brittle (and so most malleable and ductile). Also Lanthanum (Period 6 Group IIIB) is predicted to be quite malleable and this is also true.
Note that this also explains the high malleability of alkali metals (the graph is not very clear at that point).
There may be much more exceptions to this trend when examined closely so take this only as an approximation.

Crystal structures also play an important role in brittleness but they cannot be explained by a trend.
Purity of a metal tends to increase malleability and decrease brittleness because there are less points at which a fracture can take place.

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How can we synthesize para-nitrophenol
During nitration of phenol, both para- and ortho-nitrophenols will be formed.

Is there any way in which we can synthesize para-nitrophenol  only?

You can always fractionally distillate the two. In this case, since the boiling points have a large difference, simple distillation works as well.
(I'm not too good with "collateral damage" of reactions, so if a reaction affects an extra group that it shouldn't, please let me know)
Otherwise, you can use the Claisen rearrangement to protect the ortho position.
(I'll see if I can get a chemdrawing software later)
Take phenol, add $\ce{Na+ CH2=CH-CH2-}$ (you may use lithium instead of sodium--easily prepared from $\ce{CH2=CH-CH2-X}$). You will get $\ce{Ph-O-CH2-CH=CH2}$
Heat it, the group migrates to ortho.

Repeat, and now you have a phenol with two $\ce{CH2=CH-CH2\bond{-}}$ groups at each ortho position.
Now nitrate. The nitro group will go to para, since $\ce{R\bond{-}}$ groups are weakly activating/orthopara directing, whereas $\ce{-OH}$ is strongly activating.
Now, vigorously oxidize with acidic $\ce{KMnO4}$. The $\ce{CH2=CH-CH2\bond{-}}$ groups become $\ce{-COOH}$ groups, and the other stuff stays intact.
Now, decarboxylate it. I'm not exactly sure of it, but we can do this via:

Treating with $\ce{NaOH}$ or soda lime.
(or) Treating with Vit. C

(few more from Wikipedia):

Barton decarboxylation
(check the stuff here)

The acidic and basic media may have some adverse effects on the compound, but I doubt it.

dear friend thanks for the answer, I found that ortho and para nitrophenol has two different boiling point(217 and 279 °C respectively) whether this property can be used for fractional distillation the two.

@test: For such a large difference, even simple distillation works. Fractional distillation is only necessary when the difference is less that 25° C (IIRC). And fractional distillation works for any BP difference (not zero).

You would use an allyl halide, not an allyllithium, to react with phenol.

Steam distillation removes the ortho isomer.

This can be done by first reating the phenol with $\ce{HNO2}$ Which gives us this

Then reacting with dil $\ce{HNO3}$ we get 90% yield of para nitro phenol
This happens because $\ce{N=O}$ has a very strong $- I$  effect thus it concentrates the electrons more towards the para position so on reacting with dil $\ce{HNO3}$ we get max yield of para
P.S - Another way of getting only para is using a protecting agent but  i cant recall the reaction now. I will post it later

You can use \ce{H2O} in math to auto format chemicals. \ce{N=O} also formats the bond.

Reactions work as well \ce{BF3 + H2O -> BOOM} gives $\ce{BF3 + H2O -> BOOM}$ . Full documentation.

I don't know if your answer is correct, though. He wanted "only" para, not 90% para. Of course, nothing's perfect, but 10% ortho is still a lot--so I refrain from upvoting it. Otherwise, it's a nice method--I admit I wouldn't have thought of $\ce{-N=O}$-ifying(nitriting?) it :)

The current set of solutions are rather too complicated, I think. Here's what I'd do:

Make phenyl acetate with acetic anhydride and phenol.
Nitrate your ester with the usual conditions; the bulky acetyl group makes o-nitration less likely.
Separate out your p-nitrophenyl acetate
Hydrolyze your nice ester.

There are a number of bulky protecting groups one could use for the phenolic -OH, but acetic anhydride is rather cheap in my neck of the woods, and it's what I'm accustomed to.

Uhh, 'the separate out' could have been done directly via distillation. I've included that in my answer, but I felt that it was cheating, hence the long 'complicated' procedure (also, I wanted to have fun)


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A Chemdrawing software?
I would like to know about a Chemdrawing sotware for 2-d molecular structures. it should be mainly linux based (but if it supports other OS's then that would be great)

There are tons of platform-independant HTML+JS ones like http://www.emolecules.com/ . Also: This question has no correct answer, per se, and should be CW. Also2: This is easily googleable. Also3: Maybe this should be asked on chat instead?

Community: please chime in here on your opinion on on-topicness of such questions: http://meta.chemistry.stackexchange.com/questions/73/are-software-questions-on-topic

Questions asking for a list of items with no critera for evaluating the answers are very problematic. There are many answers that are equally valid, resulting in voting purely based on popularity. Their value also tends to decay a lot when they're not curated carefully.

You have an up-to-date list here; just select your criteria and choose “2D Draw” as the category.

I'd really reccomend the free Marvin suite from ChemAxon.
MarvinSketch (the actual drawing package) hasn't disappointed me in the past and has a wide feature set.  In addition to just being able to draw compounds in Lewis format, the software includes plugins to name what you have drawn (systematic/traditional), predict properties, change atom and bond properties, generate stereoisomers and much more that I haven't got the experience to do anything more than experiment with!
You can view what you have drawn in Lewis form and also as a 3D structure should you wish to.  It supports many chemical drawing filetypes.
Here is the product page, including brochure and technical details.  It is Java based so works on Mac, Windows, Linux and is even available on the web (but is quite slow to generate properties etc. - I'd reccomend doing this locally).

thanks that is a great soft though i have to switch between sketcher and viewer

You can switch quickly between the two (and save as image etc.) using CTRL-SHIFT-V in sketch :)

Go to this link, it has links to many chemical drawing softwares
 but I will recommend Chemdraw and Chemsketch. Chemdraw is a commercial version but chemsketch is a freeware and it works on both windows and  linux operating system. I am currently using Chemsketch and it is very good.


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What factors are important for quantitative analysis of a proton 1D-NMR spectrum?
Besides elucidating or verifying a chemical structure, NMR can also be used e.g. for quantifying a mixture of different chemicals.
Depending on the quality of the spectrum and the specific substance, integrating the NMR signals for the same molecule can result in significant variations from the actual ratio of nuclei, which is known from the chemical structure.
What factors do affect the quantification of NMR signals in a proton 1D spectrum? What precautions should one take during acquisition and processing of NMR spectra to make sure that the spectra can be analyzed quantitatively?

The signals you get in a simple 1D proton spectrum are in most cases roughly quantitative, though there are some exceptions and some aspects you need to consider if you need high accuracy.
Acquisition
There are several things you need to keep in mind while setting up the experiment to ensure that the resulting NMR spectrum will be quantitative.
The spectrum should be of high quality, any artifacts and noise will affect the quantification.
Relaxation delay
The relaxation delay is the time between experiments, if it is too short some protons won't relax completely back to the equilibrium state. This affects the intensity of the NMR signal and will cause the integrals of those signals to be off.
The relaxation delay (including the acquisition time) should be around $5 \cdot \mathrm{T_1}$. You can either estimate the $\mathrm{T_1}$ or measure it.
13C satellites
The 13C satellites are around 1% of the total signal, if you need high accuracy you should use decoupling to get rid of those.
Other aspects
You also need to ensure that the pulses used in the experiment have a reasonably flat excitation profile for the area you're looking at. This is usually no problem for standard 1D proton experiments, but might be a problem for other kinds of experiments or if you have an unusually long pulse and/or and unusually wide distribution of your chemical shifts.
Processing
Baseline
The baseline should be absolutely flat and at exactly zero, any error there will lead to large errors in the quantification. Using a digital filter during acquisition is a good idea for that (baseopt option for Bruker spectrometers), this will ensure a very flat baseline at zero, as long as no other problems distort the baseline.
Phasing
The phase-correction should be exact, it might be necessary to perform a manual phase-correction.

13C satellites are not a problem, as they are in almost all cases of a know integral (0.55% of the center line). Also the integral regions have to be optimized. Typically one chooses to take a width which is a fixed multiple of the line width of the corresponding line. By doing that one can get very accurate results.

Yeah, I wouldn't count them as a major problem, but you'd have to be consitant about them; You can't have one peak with both included, another with neither and a third with only one.

Mad Scientist provides some good advice. I have a different approach that either renders some of those actions unnecessary or cancels them out through signal averaging.
One big issue for quantification is the signal to noise ratio. Doing many of the things in Mad Scientist's answer will increase the signal to noise ratio, but the easiest two changes are increasing the concentration and taking more scans. If you can increase the concentration in your sample, do it. I often would increase the concentration to the point where I would be able to collect publication quality 1D 1H NMR spectra in a single scan. If you collect just one scan, then the problems resulting from poor phase averaging and from delay times that are shorter than the relaxation time.
If you are unable to change the concentration, then increasing the number of scans increases the signal to noise by the square root of the number of scans. 4 scans provides twice the singal to noise as 1 scan. 16 scans provides twice the signal to noise as 4 scans, and so on. When you opt to take multiple scans, then you must be worried about the things that Mad Scientist mentions. A phasing problem that aggregates over 100+ scans will not help you. Having too short a delay time such that all of your spins are not relaxed only gets worse over a large number of scans. In a recent project, I had to quantitatively integrate the 1D 1H NMR spectrum of a reaction mixture, looking for minor products and unreacted species (less than 5% of the total sample). I ended up taking a 3200 scan proton spectrum with long delay times. I was not popular with my fellow grad students, but I did get the data I needed.

I know someone who burned 2 weekends of 600 MHz NMR time trying to quantify natural abundance deuterium. Didn't work, but hey, what your prof says 'Do X'....


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What makes diazo compounds so unstable and explosive?
I once had an Orgo TA refer to a diazo compound as "diazo-boom-boom" (the technical term).  I have always been curious as to the reason behind the instability and reactivity.
According to Wikipedia

Some of the most stable diazo compounds are α-diazoketones and α-diazoesters since the negative charge is delocalized into the carbonyl.  In contrast, most alkyldiazo compounds are explosive

What is it about the alkyldiazos that makes them so much more unstable?  There doesn't appear to be any bond strain or other factors.  Naively, I can say that having a resonance structure should make it marginally more stable.

Diazo compounds can easily liberate Nitrogen gas, a pretty good leaving group.

Well, your question is equivalent to “what is it about α-diazoketones that makes them so much more stable?”, which is easier to see. Compared to an alkyldiazo, the α-diazoketone has a resonance structure in which the negative charge goes to the ketone’s oxygen (and far away from the positively-charged nitrogen atom):

Because the oxygen is a quite electronegative element, the resonance form is quite stable and explains the extra stability of α-diazoketones. It is for the same reason that the protons in position α to the ketones are always more acidic than alky chain protons.
Coming back to alkyldiazo compounds, you have to realize that merely being able to write a resonance structure does not intrinsically imply stabilization: the resonance structure has to have some intrinsic stability factor. In the alkyldiazo, the resonance form you wrote is a carbanion, which is considered quite unfavourable unless it has a further stabilizing factor. Moreover, the most common reactions gives N2, which is a very stable compound… the reaction is thermodynamically very favourable.

@jonsca: from an entropy viewpoint, the production of a gaseous product like nitrogen is very favorable indeed (what could be more "chaotic" than a gas?). That's why your diazo group does quite well as a leaving group...

@Fx Since you only mentioned it, I explained a bit more why $\ce{N2}$ is more stable at STP.

@JM Chaotic isn't the right word. 1 mole of $\ce{N2}$ in the gas phase in a 1L volume simply has more populated energy levels than the same amount of $\ce{N2}$, but frozen. In other words, there are more microstates associated with its macrostate, than when frozen, say.

Very short answer:
$\Delta G = \Delta H - T\Delta S$
As this is an org-chem question, no need to solve the equation numerically. As you might know, $\ce{N2_{(g)}}$'s standard formation enthalpy ($\Delta H_\text{f}^\circ$) is 0 – at standard $T$ and $p$, $\ce{N2_{(g)}}$ is gaseous. Therefore, its enthalpic contribution is 0. Its $S^\circ$ is $191.32\ \mathrm{\frac{J}{mol\ K}}$, which indicates that, at 273.15 K, $\Delta G^\circ_\text{f}$ is not only negative, but also large.
Thermodynamically, $\ce{N2_{(g)}}$ is favored. Take a look at nitrogen's phase diagram, generated by Wolfram Alpha.
EDIT
I've just noticed you're considering diazonium ions, not azo compounds. For the reasons F'x explained, the ions are more prompt to explosion than are the azo compounds. I think azo compounds are used in the fabrication of CDs, or so my teacher told us... They're supposedly light sensitive.

Well, I wouldn't say anything about the sign of the reaction from a single formation Gibbs free energy… you don't need to show that N2's $\Delta_f G$ is negative, but write a reaction free energy, which is much harder.

well, I think as currently written it only generates confusion… also, the basis for your calculation that N2 is stable is that N2 is the reference state of nitrogen, which it is because it is more stable… Well, I've given my opinion, I'll leave it at that.


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What's the chemical nature of a dishwasher's rinsing agent?
It seems that a rinsing agent helps the dishes in the dishwasher dry much better than they do without it (which I recently noted on my new dishwasher). What molecules are used in the composition of these rinsing agents? How do they help the plates dry?

According to the MSDS for Jet Dry (a common brand name) it contains:

Sodium 2-phenylpropane-2-sulfonate (Sodium cumene sulfonate)
Propane-1,2-diol
Citric acid

I think it may be safe to assume that the citric acid is in there for a "clean" smell and also to soften the water (thanks to the commenter).  The sulfonate is presumably a surfactant to help break the surface tension of the water, and promote evaporation, and the propylene glycol may help serve as a coolant (and is favored due to low toxicity).

Right, the first component is a surfactant. Since presumably this will be used without caring for whether the dishwater is hard or soft, you do not want to use the ester of a fatty acid here. The sulfonate, which is the conjugate base of a relatively strong acid, will tend to be more soluble, and will not form scum with hard water ions.

The citric acid is there to soften water by converting calcium, iron, magnesium (and others) salts into the more soluble citrates, which also decreases the likelihood of scum.

Citric acid does not really smell.  I also don't see how propylene glycol would serve as a coolant.


text:
How to identify zero order reactions?
I have come across many reactions that are zero order reactions but at one glance I'm unable to tell if they are zero order or not.  Is there any criteria that can be used to identify these reactions or are they purely experimentally determined?

Zero order reactions imply the reaction rate is independent of concentration. That only happens if your considering gas reactions where you use a platinum adsorbent catalyst.

@PrittBalagopal  I've studied that gas decomposition reactions (e.g. $\ce{NH3}$) on a heterogeneous catalytic surface are not 'true' zero order reactions, but are pseudo-zero order reactions. Are there any true zero order reactions?

To your question of whether zero-order reactions can be recognized “at one glance” and from an non-experimental criterion, the answer is clearly no. The only theoretical way of firmly establishing the order of a reaction (if it has one) is to fully calculate its kinetic law (or rate equation), which can only be done if you know the scheme of the reaction, i.e. the list of all elementary reactions involved.
There is one case where you can take a shortcut, namely if you know (or strongly suspect) that the reaction in question is an elementary reaction, for example, if it involves no catalyst, proceeds in a single step with only one transition state and no intermediate. If that is the case, then the order of the reaction is determined by the stoichiometry ($\ce{A –> X}$ is first order, $\ce{2 A -> X}$ or $\ce{A + B -> X}$ are second order). However, this does not apply in your case, because elementary reactions cannot have zero order!

My doubt was due to $\ce{H2 + Cl2->2HCl }$  in the presence of light .this is a zero order reaction but isnt this also a bimolecular elementary rxn?? if it is then it does not go according to your ans

$\ce{H2 + Cl2 -> 2 HCl}$ proceeds (like many photochemical reactions) by a complex mechanism… See here for an example given about bromide. Photochemical reactions are usually not elementary reactions.

oh yeah now i remember that a photon splits the halogen into two free radicals and thus the reaction proceeds....thanks

Determining the rate of the kinetics of a reaction isn't done with only one measurement, or looking at only one graph.
What zero order kinetics means, is that the rate of the reaction is independent of the concentration of chemical species. To determine the kinetics of a process, you need to take a series of measurements while altering one variable, namely the concentration of one of the chemical species under investigation.
In zero order kinetics, you should observe that as you change the concentration of reactant, for example, the rate of the reaction remains constant.

That is to say, explicitly, that the order for each component in the reaction mixture is experimentally determined.

Of course, and if all are 0, then the reaction follows zero order kinetics. My bad.

To be honest, we could only use scientific data to determine the order of an reaction , Say an example of one species determining reaction. You can only plot graphs to find out the shape of the curve and thus discovering whether it is zero first or second order reaction
If you would like , it is in a chapter called determination of reaction order


text:
Why does the sulfone ring in tazobactam open when the lactam is hydrolyzed?
One of the drugs I work with is a beta-lactam (4-membered ring with an amide bond) fused to a sulfone ring, tazobactam.

It's relatively stable in water; the lactam is not significantly hydrolyzed without being catalyzed.  When the lactam is hydrolyzed, however, typically by an enzyme, and the lactam ring opened, the sulfone ring also opens cleaving the bond between the sulfur and bridgehead carbon.  A paper by Kuzin gives the following reaction scheme (this is part of figure 2, click for the whole thing)

and says of it (emphasis mine):

The Ser70-bound moiety (2) is thought to be the initial intermediate [...] Ring opening after departure of the sulfinate from C5 produces a reactive imine (3) and the more inert tautomeric enamine forms (4, 5).

But why is the sulfuryl (sulfone, sulfonyl(?)...) leaving only when the lactam is opened and not spontaneously beforehand?
Other fused bicyclic beta-lactam drugs don't open the other ring, where in lieu of the sulfone there is commonly a thioether (as in penicillin).  The sulfone group seems more electron hungry, but I don't see how the broken lactam bond affects this.
I've heard some people suggest that the lone pair on the nitrogen is able to make this happen, but why doesn't it then happen in the intact drug?

By "intact drug", you mean its crystallized form? If so, I can imagine that the proton transfer doesn't occur so well in solid form, whereas in solution it is facilitated, which could explain the compounds crystalline stability. Also, your sulphur is $sp^3d^2$ hybdridized, which leaves 3 vacant $d$ orbitals. This might explain its reactivity. I'd like to see what would happen if you swapped SO2- for PO2- (different geometry).

@CHM I mean the with the lactam bond not yet hydrolyzed.  AFAIK the drug isn't terribly reactive (from the point of view of the sulfone ring opening) in water.

The lactam will be hydrolyzed in water, under acidic/alkaline conditions. What's the optimum pH of your enzyme?

@CHM biological, so about 7.  Spontaneous (non-enzyme catalyzed) hydrolysis is negligible.

This is interesting. But I'm a bit worried about his mechanism proposal, in step 3. He shows two electrons moving to the sp2 carbon adjacent to the N, but the carbon stays sp2. There's something missing. Have you got an idea of the other AA residues near Ser70?

Here's a draft:
First, read this paper. Considering we're under alkaline conditions (blood pH between 7.35-7.45) we'll take their alkaline mechanism proposal.
Mechanism

This is obviously incomplete, with good reason. Without knowing what the active site of the enzyme looks like (except for Ser70), it's difficult to devise of a mechanism. I could imagine another alkoxy attacking the sp3 carbon between S and N, and shifting two electrons to the sulfur thus generating your SO2- in one step, but that would require a proper active site.
The geometry of the molecule inside the active site of the enzyme is obviously optimal for this reaction to proceed. I'd like to know more. For some reason my school doesn't give me access to this paper.

The intramolecular proton transfer you show in your last step is unlikely. At biological pH, that carboxylic acid is already deprotonated.


text:
What's a minimal yet chemically-meaningful kinetic system for an oscillating reaction?
Oscillating reactions are a funny aspect of chemistry. I have tried to find various simplified kinetic models of oscillating reactions such as the Belouzov-Zhabotinsky, the Briggs–Rauscher or the Bray–Liebhafsky reactions in order to be able to study them. However, the models I have found so far are either too complicated (8 or more species considered), or loose chemical significance. For an example of the second case, Ball's 1994 model[1] corresponds to:

While this is useful and introduces a nice circularity in the model, creating the possible feedback, it has lost all chemical sense — by which I mean that no correspondance can be established between the species of this model and a real oscillating system.
So, my question is: what's the simplest known chemically-meaningful model of an oscillating reaction?

[1] Ball, P. 1994 Designing the molecular world: Chemistry at the frontier. Princeton, NJ: Princeton University Press.

"Chemically meaningful" is maybe not the good expression. Also, most organic and inorganic reactions are much (much) more complicated than you see in textbooks, so a reaction system with 8 species is very simple...

After some more research, the Oregonator appears to be “the simplest realistic model of the chemical dynamics of the oscillatory Belousov-Zhabotinsky (BZ) reaction” or any other oscillatory reaction I could look up. It is described in Field and Noyes (1974) [1], and take its name from the University of Oregon where these researchers worked.
It is described by the following set of reactions:

where each entity correspond to a chemical species in the real system: X = HBrO2, Y = Br–, Z = Ce(IV), A = BrO3–, B = CH2(COOH)2, and P = HOBr or BrCH(COOH)2.

[1] R. J. Field, R. M. Noyes, “Oscillations in Chemical Systems IV. Limit cycle behavior in a model of a real chemical reaction”, J. Chem. Phys. 60 (1974) 1877-84.

I'm not sure if this is what you're looking for (it may well still be too abstract), but oscillations can occur in the Gray-Scott system, which is
$$
A + 2B \to 3B\\
B \to P,
$$
where $P$ is an inert product, and the reaction is assumed to take place in a flow reactor that provides a supply of A, giving rise to the dynamics
$$\begin{align}
\frac{\mathrm da}{\mathrm dt} &= f(1-a) - ab^2;\\
\frac{\mathrm db}{\mathrm dt} &= ab^2 - (f+k)b,\\
\end{align}
$$
where $f$ is a rate determined by the flow reactor, $k$ is the rate of the $B\to P$ reaction, and the rate constant of the autocatalytic reaction has been set to 1 without loss of generality by scaling $f$ and $k$ relative to it.
With the appropriate choice of the parameters $f$ and $k$, oscillation can happen because $B$'s concentration increases autocatalytically, but then it overshoots its food source (i.e. the concentration of A), which then builds up again, allowing the cycle to repeat.
You might not like the trimolecular step, but I've found you generally get similar behaviour if you split it up into something like
$$
A + B \to C + B\\
B + C \to 2B\\
B \to P.
$$
(Simply using $A+B\to 2B$ doesn't work, because its kinetics don't have the right sort of nonlinearity.)
I would say that this has an advantage over the Ball model that you posted in that it obeys the laws of thermodynamics. (At least from what you showed in your question, it looks to me that Ball's model only oscillates because the reverse reactions have been neglected, and if they weren't then it would have to go to equilibrium, because it's a closed system.) It makes explicit that you need a power source (the supply of $A$) for oscillation, and illustrates the connection between oscillatory behaviour and autocatalytic kinetics.

I guess you could modify it to be a closed system, with the supply of $A$ coming from the decay of some precursor species - so something like $R\to A;$ $A + 2B \to 3B;$ $A\to P;$ $B\to P$. If you start with a large initial supply of $R$ there should be a regime in which $B$ oscillates as $R$ decays.

The reaction system $A + 2B \rightleftharpoons 3B; B \rightleftharpoons P$ is also referred to as the Schlögl model and has been studied quite extensively in well stirred conditions. Small modifications to this reaction can make it oscillate.

@F'x: With the 3 reactions of Ball's system, there is no way that you can assign free energy values to species $A$, $B$, $C$ such that the combined free energy of the reactant species is always greater than the combined free energy of the product species.
e.g. Try it on Wolfram Alpha:
solve (A+B) > 2A, (B+C) > 2B, (C+A) > 2C
Therefore, from a thermodynamics point of view, I would say that the Ball reaction system is invalid.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Effect of periodic acid on cyclohexane derivatives
Periodic acid (per-iodic acid, $\ce{HIO4}$ ) cleaves vicinal diols(to carbonyls) via a cyclic intermediate.
Due to the cyclic intermediate, it needs a syn configuration for the $\ce{-OH}$ groups for the reaction.
Exactly which configurations are allowed on cyclohexane derivatives?
Let's take 1,2-dihydroxycyclohexane(boat conformer). If the two hydroxy groups are both on the axial positions (trans- ), then obviously the reaction will not take place.
What about when the two are cis- (one axial, one equatorial), and when they are both equatorial (trans- )?
(My original question had to do with diastereomers of cyclic glucose--aldohexoses--but I realised that the extra hydroxy groups make the reaction more complicated)

I think your problem with this reaction lies in the easily confused term syn and cis. Both appear to mean that two functional groups are oriented in the same direction. On a cyclic molecule, syn and cis are the same. However, cis is a stereochemical descriptor and syn is a conformational descriptor. Simplifying this reaction to an acyclic case may help. Consider the meso isomer of  2,3-butanediol shown.

This isomer is drawn in the anti conformation, which is probably the lowest energy for steric reasons. This isomer has a syn conformation, which looks like the following. Note that the two -OH groups are not both oriented forward with solid wedges. That drawing implies a high energy eclipsed conformation.

(source: nist.gov)
Now back to the reaction with HIO4. In order to form the cyclic intermediate, the -OH groups need to be close in space. On an acyclic molecule, this requires a syn conformation, which most acyclic molecules can achieve even if it is not the lowest energy conformation. In cyclic molecules, particularly cyclohexanes, the syn conformation might not have the -OH groups close together. For cis-1,2-cyclohexanediol, which is also syn respective to the "plane" of the ring, to move the two -OH groups into the correct arrangement, it would need to wiggle into the higher energy boat or twist-boat conformations. However, trans-1,2-cyclohexanediol, which is anti has the two -OH groups oriented close together. In a sense, the trans isomer has the two -OH groups oriented syn with respect to the C-C bond that they share. Thus, trans-1,2-cyclohexanediol should react more readily with HIO4 that cis-1,2-cyclohexanediol, because the lowest energy conformer of trans-1,2-cyclohexanediol has the close arrangement of two -OH groups. The cis isomer would have to change conformation, which increases the energy input required, increasing the activation energy and slowing the reaction rate.
The diaxial case (whether boat or chair), would definitely be unreactive.

The literature reveals that cis-1,2-cyclohexanediol 1 cleaves about 20-times faster than trans-1,2-cyclohexanediol 2 although the equilibrium between the diol and the cyclic periodate intermediate slightly favors the trans-isomer. The rates of cleavage of trans-1,2-cyclohexanediol 2, (-)-(R,R)-2,3-butanediol 3 and meso-2,3-butanediol 4 are comparable. [The data is from the reference cited.]

Buist, G. J., Bunton, C. A. and Miles J. H., J. Chem. Soc., 1959, 743.


text:
What are the major factors that influence LogP
The partition coefficient is a very useful metric in characterizing compounds particularly drugs. What chemical properties influence that partitioning? Are chemical moieties additive in their LogP? What if the compound has both hydrophobic and hydrophilic ends?

The Wikipedia page you linked to has some interesting information, that could at least get you started:

It has been shown that the log P of a compound can be determined by the sum of its non-overlapping molecular fragments (defined as one or more atoms covalently bound to each other within the molecule). Fragmentary log P values have been determined in a statistical method analogous to the atomic methods (least squares fitting to a training set). In addition, Hammett type corrections are included to account of electronic and steric effects. This method in general gives better results than atomic based methods, but cannot be used to predict partition coefficients for molecules containing unusual functional groups for which the method has not yet been parameterized (most likely because of the lack of experimental data for molecules containing such functional groups).

Partition Coefficient, represented as P and defined as: P = [Organic phase] / [Aqueous phase]
Where "[]" represents concentration in the  phase and $\mathrm{\log (P)= \log_{10}(Partition Coefficient)}$ if log P value is higher then its more lipophilic ;those drugs were more absorbed by cells and elimination of those drugs from body will be less .The partition coefficient is a ratio of concentrations of un-ionized compound between the two solutions so for a chemical which is having both hydrophobic and hydrophilic ends also have log P values.
Eg: Oleic acid: $\mathrm{\log (P)= 7.64}$
chemicals which are hydrophopic(lipophilic) tends to have more Log P values and hydrophilic(lipophobic) chemicals have less log P.
eg: pentane($\mathrm{\log (P)= 3.45}$) [lipohilic], 1-pentanol ($\mathrm{\log (P)= 1.51}$) [lipophobic]


text:
Why do thiols have such a propensity for bonding with mercury?
Historically, thiols $\ce{-SH}$ were known as "mercapto-" due to their strong propensity to bond with mercury.  Certain drugs still carry this designation, though mercaptopurine has an $\ce{=S}$ group instead of a thiol.
What makes the thiol group bond so strongly with mercury in particular? Why don't alcohols make such a strong bond with $\ce{Hg}$?
If it's known, was there any historical reason why it was important to be able to "capture" mercury?

"...though mercaptopurine has an $\ce{=S}$ group instead of a thiol." - that would be the tautomerism at work. See this article for instance.

@J.M. Ah, I see, definitely didn't think of that.  Thank you, I'll have to download that later.

Well, sulfur in general, and thiolates in particular, are known for bonding very strongly to gold, lead and mercury. In the particular case of mercury, which was known very early in the history of chemistry (indeed since the times of alchemy), this is the reason thiols are often called "mercaptans", from the Latin mercurium captans (capturing mercury).
In Pearson's HSAB theory (hard and soft acids and bases), the reason the S-Hg bond is be stronger than the O-Hg can be explained because S2- is a soft basis and Hg(II) is a soft acid, making a good fit… while O2- is harder.
Finally, regarding the historical use of thiols to bind mercury (and lead to), let's just say a chelating agent for mercury and other toxic compounds was quite useful as an antidote. You can read a very nice book on the historical use of mercury and lead in John Emsley's The Elements of Murder.

As an additional note, Hazard Information sheets for mercury spills still recommend using either sulphur or a mixture of sand and sulphur to clean up anything you can't pipette. For example.

A fair amount of the chelating agents currently used in modern medicine possess a mercapto group. Off the top of my head, there's dimercaprol (BAL, British anti-lewisite), penicillamine, succimer (DMSA, dimercaptosuccinic acid), tiopronin...


text:
Are Pulay forces expensive to compute?
In a thesis I am reading, it is said that one of the reasons for using plane-wave basis sets for first-principles molecular dynamics (aka ab initio MD) is that the Pulay forces[1,2] that arise from an MD using atomic basis sets are computationally expensive to calculate.
While I understand that having additional terms means more code to write, i.e. they make writing the software harder to write, is it true that they are CPU-intensive to compute? The criterion I would use to quantify this subjective statement is:

Given that you have already calculated the energy and forces at that point, you have already computed a large number of  integrals required for this task and involving basis functions: overlap integrals, terms of the form $\left\langle\phi_\alpha\left|\hat{A}\right|\phi_\beta\right\rangle$ where operator $\hat{A}$ is either the hamltonian, its gradient, or any other operator necessary for the calculation of energy or forces. Would the calculation of Pulay forces require any more integrals to be evaluated, or can it be straightforwardly computed from those previously-calculated integrals alone?

[1] P. Pulay, Molec. Phys. 19, 197 (1969)
[2] See also slide 6 of this

can you please clarify a little what you mean by Pulay forces? Are you referring to Pulay stress or something different?

@RichardTerrett Well, Pulay forces are similar in nature to Pulay stress… even at constant cell volume, the Hellmann-Feynman theorem doesn't hold if the basis set is not fixed, and the extra term appearing is called the Pulay forces. I added a few links, to the original paper and to the first result of a Google search for “Pulay forces”.

Stress is used in materials science (haven't really encountered it elsewhere, so correct me if this is an isolated or universal thing) to indicate a force normalized to the cross-section area, so the two are comparable I'd think, right?

Yes, you need additional quantities beyond the minimum necessary for calculating energies and the Hellmann-Feynman piece of the force when the wavefunction is not variational.
Here is a very rough sketch of why:
Puláy forces arise from applying the chain rule for calculating forces and was first discussed in the context of applying the Hellman-Feynman theorem.
Recall that the force is the change of energy with changes in (nuclear) coordinate and the energy is the expectation value of the (electronic) Hamiltonian, $E = \left<\psi\left|H\right|\psi\right>$. Applying the chain rule to this:
$$-F = \nabla E = \left<\psi \left\vert \nabla H \right\vert \psi\right> +  2\left< \nabla\psi \left\vert H \right\vert \psi\right>$$
The first term is what you get from Hellman-Feynman and is the expectation of a one-electron operator of the first you have listed. The second term goes away in Hellmann-Feynman only because it assumes the wavefunction is variational.
If you expand out the orbitals in the Slater determinant in some AO basis $\chi$:
$$ \psi(r_1, r_2, ..., r_N) = \phi(r_1) \wedge \phi(r_2) \wedge \dots \wedge \phi(r_N) $$
$$ \phi(r_1) = \sum_i c_i \chi_i(r_i) $$
then it is clear that $\nabla\psi$ generates terms which are MO coefficient derivatives (usually denoted $c_i^x$) and also AO derivatives ($\chi_i^x$). If the wavefunction were variational, then by definition all these derivatives are zero. If you work through the algebra, you will find that there are some new quantities that need to be calculated in order to work out the non-Hellmann-Feynman theorem, most notably the one-sided AO overlap derivative matrices $\left<\chi_i^x\vert\chi_j\right>$ which do not normally show up otherwise. These terms turn out to be critical for getting correct ab initio molecular dynamics since the wavefunction is rarely (if ever) variational in time.
There are tricks one can do to reduce the cost of calculating MO coefficient derivatives , but they cannot be obtained for free in general. And there's really no way around calculating the overlap derivative terms.

It turns out that the MO coefficient derivatives are usually by far the more expensive piece. Solving for these analytically is usually done with coupled-perturbed self-consistent field (CPSCF) equations: for KS DFT one has CPKS, and for HF, CPHF, etc.

In response to AcidFlask: MO coefficient derivatives, whilst non-zero, aren't required for the first derivative of the energy as a result of the energy being minimised with respect to them. See Szabo and Ostlund, p. 440:
$$\frac{d E}{d x}=\frac{\partial E}{\partial x}+\sum_{i a}\frac{\partial E}{\partial C_{i a}}\frac{\partial C_{i a}}{\partial x},$$
since $\partial E/\partial C_{i a}=0$, you don't need to compute $\partial C_{ia}/\partial x$.


text:
How does conductivity work for non-redoxed ions?

Related (very similar, but here I want a mechanism) https://physics.stackexchange.com/q/21827/7433

By the Kohlrausch law, all ions contribute to the conductivity of an electrolyte.
Now, as I understand it, the mechanism of conduction in an electrolyte is thus:

Ions migrate in solution
These ions get reduced or oxidized at the electrodes and converted to electrons
These electrons continue down the wire, leading to an increased/maintained conductivity/current

But, this mechanism doesn't work for ions which do not get redoxed--movement of ions in the solution cannot be translated to movement of electrons in the wire and thus it seems (to me) that conductivity should not increase.
But ions like $\ce{NO3-}$ have a comparable $\lambda$ (ionic molar conductivity)--so they experimentally do increase conductivity.
What is the mechanism for conduction via these ions?

Thermodynamically nitrogen dioxide is more accessible from nitrate than water is from dioxygen. I think many  conductivity meters apply a potential that would make this chemistry possible. This is a separate issue than the interdependence of ionic drifts. If you would like, I can talk about the latter.

@Chris you could, but  that would make $\lambda_{\ce{NO2}}$ dependant on the potential, wouldn't it?

Experimentally, how do you determine the the conductivity? What physically happens during this measurement?

@Chris: Run current, measure voltage/current, use $V=IR$, use the dimensions of cell to get conductivity from conductance

Right. A solution of 0.5 $\mu$ moles of acetic acid will require ~4 V potential drop across the electrodes to give 20 $\mu$A. An even smaller concentration will result in an even smaller current at that potential, requiring even greater sensitivity/more expensive hardware.

The point is that current is too low for any appreciable electrolysis. The situation for your nitrate will probably be the same, depending on the exact setup. There are situations where the species do not need to participate in any redox reaction and can still change the conductivity. That is different than your nitrate case (I'm going to bet).

@Chris: There are situations where the species do not need to participate in any redox reaction and can still change the conductivity.  How is that possible? That's exactly what I'm looking for.

Conductivity meters work on AC! Frequencies at about 40 to 100 kHz are used. All electrode reactions stop at about  20 to 40 Hz. .

This can arise due to ion-ion interactions brought on by Coulombic fields.
A proper explanation is rooted in transport phenomena. Unfortunately chemists are largely not taught this, but chemical engineers do get the opportunity.
Under standard conditions, ions in a solution will take a random walk during diffusion. This results in no net movement of our ions over some span of time. Such a system perturbed by an applied electric field will cause ions to preferentially move in one direction, we say the ions will drift. We say they have a drift velocity. These velocities are affected by a number of parameters, as there is still continual collision, but we can say that the mobilities will be unequal for ions of unequal sizes.
Different mobilities mean concentration gradients form and charge separation too. Typically we use an electroneutrality field as a reasonable approximation to allow analytic solutions for various equations. This field depends on all the ionic fluxes present and as such will affect the total measurable current.
The result is simple: the introduction of any ion will cause a perturbation in the field that all the ions encounter in solution.

But how does this translate to electrons  in the wire? Or does the ion "pull and hold" the electron in place (basically the Coulomb force bring some extra electrons to the electrode as a once-per-ion effect?) Transfer phenomena sound interesting, thanks for the rest!

Oh I see.. You're saying that the nonredox ions affect the field and thus improve mobilities-->conductance.

This is a good answer (upvoted!) but is very "atom-focused" instead of "field-focused".  Maybe it's a chemist vs. physicist thing or something.  Anyway, ions diffusing while under the influence of electric fields are governed by the Nernst-Planck equation.  This is true of any ion, whether or not they participate in chemical reactions at electrode surfaces.  Electric fields that emanate from the electrodes of an electrochemical cell affect the behavior of every charged particle (i.e. ion) that's nearby.


text:
Experimental calculation of solubility product constant
In order to calculate the $K_\mathrm{sp}$ of lead(II) iodide, I mixed $\pu{6.0 mL}$ of $\pu{0.01 M}$ $\ce{Pb(NO3)2}$, $\pu{6.0 mL}$ of $\pu{0.02 M}$ $\ce{KI}$, and $\pu{8.0 mL}$ of water, and a precipitate (barely) formed.
From this, I calculated $[\ce{Pb^2+}]$ to be $\pu{0.0036 M}$ and $[\ce{I-}]$ to be $\pu{0.0060 M}$, giving
$$K_\mathrm{sp} = [\ce{Pb^2+}][\ce{I-}]^2 = 1.1 \times 10^{-7}.$$
However, the actual value (at $\pu{25 ^\circ C}$) is $7.9\times 10^{-9}$ (Skoog et al. Fundamentals of Analytical Chemistry), so my result is off by a factor of $\sim 10$.
I'm pretty sure I had clean equipment and measured carefully and accurately. Why is there a discrepancy? Is my maths wrong, or are there some inaccuracies in my experiment?

I don't understand your issue… “barely” is not helpful: your product of concentration is higher than the solubility product, so it precipitates, which is what you expected. I suppose “barely” means you expected a larger mass of precipitate, but unless you actually weighted it and found it to be inferior to the expected calculated mass, I don't see the issue…

@F'x I actually did a variety of concentrations to narrow it down to the point at which precipitation could be observed, varying my solutions by 1 mL. So 'barely' here means 'as close to equilibrium as I could repeat with the solutions on hand'.

It's probably an experimental error. The 'barely' may be important here.
For one, you may not be sure how accurate your initial molarities are. A mistake in one of them ruins it all. Titrate them properly and check. Titration usually gives two significant digits.
Also, since its a cube overall, a small mistake can get much larger. You may want more accuracy in your readings (one significant digit isn't good enough), and you need to improve on the 'barely'. Though a factor of 11 needs a variation by a factor of 2.3 in both readings. So I don't know.
Have you redone the experiment? Doing an experiment more than once is integral to getting good values.

I don't have time to do the calculations manually, but inputing your problem data into a chemical equilibrium software (here Dozzaqueux) reveals that the precipitate you observe is Pb(OH)2, while no PbI2 should form. The former is white, while the later is yellow; can you tell us what color was your precipitate?

It was white, and now I'm definitely confused! :) Thanks though, I will likely learn from this too

How long you observed your mixture? It is possible that you had not reached the equilibrium required for the solubility product constant to be valid. Allowing time for the system to reach equilibrium might have resulted in the appearance of more precipitate that would indicate that you had added too much material.


text:
What is natural bond orbital theory used for?
My understanding is that the NBO methodology transforms the molecular orbital picture of a molecule into a collection of pairwise interactions (bond orbitals). This is obviously handy for didactic and comprehension purposes as it distils what might be a very complex and non-local MO picture into manageable lobes conforming to the introductory theories of chemical bonding (VB, orbital hybridisation) taught in high-school and early university courses.
Does natural bond orbital theory have utility outside of this role?

The 2pz (2p0) state for hydrogen atom is a true stationary state and can remain forever. There is absolutely no constrain in quantum mechanics which states hydrogen atom must be symmetric. If QTAIM is not able to properly describe it, it is the shortcoming of quantum mechanics?

Yes, along with a family of related methods. I honestly don't fully understand the mechanics behind it, but the orbitals obtained with the NBO method can be used for a treatment called Natural Resonance Theory, which gives you an idea of what sort of Lewis-structure resonance components there are and in what weightings, as well as bond indices, and a measure of covalency.
There are three papers from the developers of the technique, covering theory, usage, and some more examples. (I'd probably start with the usage paper.)
Also, along with NBO, there are the Natural Atomic Orbitals which are often lumped in because they're also computed by the NBO program, and Natural Population Analysis which you can use the Natural Atomic Orbitals to do. These give you an idea of what sort of atomic orbital character is contributing to your molecular orbitals, and also atomic charges and net spins.

The entire premise of NBO theory is to rotate canonical molecular orbitals into orbitals that have maximal chemical meaning, in the sense that is defined in the NBO papers. I am not aware of any purpose of NBOs and their corresponding natural charges beyond interpretation.

(Fairness in reporting: I am a hard-core Quantum Theory of Atoms in Molecules partisan)
NBO is one of many attempts the apply rotations among molecular orbitals, each of which can spread over the entire molecule, (i.e. the wave function) so that they maximally-resemble localized atomic orbitals or "hybrid atomic orbitals". Then one may be able to talk about "pieces" of a overall state wave function, and draw connections to the historical descriptions of bonding.
But, the whole process of wave function localization/transformation is puzzling, because it leaves the observable--the electron density--just the way that it was, at least it ought to.
I believe that its motivation is good-intentioned.  However, Schroedinger himself cautioned strongly against attempts to analyze the wave function directly. For example, since the wave function may be complex, what does one do with a localized wave function that has real and imaginary components?  Why not just analyze the electron density (which is real and observable) directly?

In reply to the partisan of QTAIM: QTAIM is a well-defined theory in the context of quantum physics. However, it does not meet essential axioms of chemistry (see below). So why one wishes to impose it to chemical interpretations? - Atomic (and molecular) volumes can be obtained in arbitrary ways in QTAIM. For example, the volume of hydrogen atom can range from nearly zero to infinite, depending on what we like to see. - There is no connection between chemical bond and QTAIM, as Bader define a new concept (bonded atoms) which is not chemical bond. So, it is of no worth for a chemist. - QTAIM de

<con't>     delocalization indices employ orbitals as wavefunction in contrast to its claim, electron density is the origin of world!
QTAIM is absolutely inconsistent with excited states. Just analyze a hydrogen atom in 2p0 excited state. Three critical points appear, one of which is of rank two. What a funny chemistry is hidden behind!

AND too many other flaws...

Your wavefunction is wrong, not being spherically-symmetric.  Garbage in, garbage out.

What is a "chemical" bond?

Atomic Volume is precisely defined in QTAIM.  It's the area contained in the atomic basin, the union of all steepest ascent paths with their nuclear attractor


text:
Is a 2-D periodic structure isomorphic with the surface of a torus, a sphere, neither or both?
When I was reading through the ADF-BAND tutorials, one of the toy systems presented was a 1-D periodic structure involving 3 collinear hydrogen atoms. The tutorial pointed out that, topologically speaking, this is cylindrically symmetrical (more specifically, it's ring symmetrical).
In the instance of a 2-D structure, can the calculation be considered a model for the surface of a torus (this seems logical), a sphere (I doubt this because if you fit a rectilinear grid to a sphere you end up with two poles and dissimilar meridians and parallels), or other?
Bonus questions: Has anyone used periodic calculations to model electronic structure/chemistry on the surface of a sphere or torus? Can you introduce a curvature term to account for these structures having a finite size?

Yes, it's the same thing, at least according to http://physics.stackexchange.com/questions/21882/gravitation-in-a-space-that-is-topologically-toroidal

@Manishearth - the same as which?

I was saying that a repeating 2D space is the same as a torus (topologically)

Yes, a 2D-periodic space can be mapped to a torus, but that's more a question for the math.SE…
Regarding your bonus question, why would there be? What would you do with it? Molecular structures are intrinsically 3D, so I don't see what you would do in a 2D (periodic or not) space? Even when we talk about planar or pseudo-2D structures (buckyball, nanotube, etc.) they are 3D objects with 3D electronic densities and wavefunctions.

Edit: 3D structures that are periodic in two dimensions and finite in the other one can be studied by many computational chemistry codes. They are often referred to as slab calculations or surface calculations. The most common issue is that of Coulombic interaction (or Poisson equation solver), which typically requires special treatment in the 2D case.

By 2-D I mean structures which are periodic in 2 dimensions but finite in a third. A potential motivation is to model tubular structures that are too large to feasibly solve aperiodically.

@RichardTerrett OK, I edited my answer accordingly… but I don't understand what you mean by “curvature term”, then.

By this I mean an element of the calculation that corrects for the distortion(s) of the plane arising from being mapped to a torus with non-zero local curvature.

@RichardTerrett then there is no need too… while the “mathematical” view of a 2D periodic space is akin to a 3D torus, I don't believe any technique out there would really physical cast the 2D structure onto a 3D torus to perform any simulation.

2D periodic systems can be mapped to toruses, but not spheres. This is easy to see because in a sphere, parallel lines always intersect. In the periodic system parallel lines never intersect.
Regarding your bonus question: I don't know of anyone who has tried to use a periodic model to study a sphere or torus. But people have kind of gone the other way around, and replaced a 3D periodic model with the surface of a 4D sphere. This allows you to avoid complications associated with long-range Coulomb interactions.


text:
Gas chromatography/mass spectrometry: Can I use peak ratios and molecular mass to derive quantity?
Can derive the quantity of a substance from a GC/MS report if I know:

the ratio to another substance in the data,
the quantity of that second substance, and
the molecular mass of both substances.

The site was very dense and vague regarding how GC/MS test work, but fairly upfront regarding how they derive their ratios.
Here are the numbers (slightly tweaked, but more or less in proportion):
Substance A:

Quantity: $300 \:\pu{mg}$ (Know already, not part of GC/MS report)
Mol. Mass: $500 \:\pu{g/mol}$ (found in external source)
Peak Proportion: 3

Substance X:

Quantity: Unknown
Mol. Mass: $425 \:\pu{g/mol}$ (again, external source)
Peak Proportion: 1

To be clear, the ratio of 1:3 means that substance X has a horizontal peak $\frac13$ of substance A.  I'm not clear on what the x-axis is actually a measurement of, which is part of why I'm at a loss. The site does make it very explicit that the ratios are not directly proportional  to mass ratio (so if it found a ratio of 3:1 of glucose to arsenic, this doesn't mean the substance is 75% glucose and 25% arsenic, only that the glucose "peaks" 3 times higher, which I've taken to mean "3 times the oomph", but that may be incorrect, as well).
So, if I know that the molecular mass($\rm M$) of substance X is 0.85 of substance Y, can I derive the actual mass of substance X using the formula:
\begin{equation}\mathrm{qty}_X=\frac{\mathrm{qty}_A\times(M_X/M_A)}{\mathrm{peak}_A/\mathrm{peak}_X}\end{equation}
with data being:
\begin{equation}\rm qty_X=\frac{300\:\pu{mg}\times(425/500)}{3/1}\end{equation}
Which simplifies to:
\begin{equation}\rm qty_X=(300\:\pu{mg}\times 0.85)/3\end{equation}
and finally the result of $85 \:\mathrm{mg}$.
So I guess in the end there are 3 questions:

Is this even how GC/MS results work?
If so, is my assumption to derive the mystery quantity using the molecular mass correct?
Is the math itself in order? (I'm specifically worried that I should invert either the mass ratio or the peak ratio or both).

Of course, if the answer to the first questions is no, then my true question is: can I derive the quantity of substance X with the given data, and if so, what would be the right approach?
If anyone is curious for some context, I need to know the actual quantity of substance X as I know that, by mass, it has a threshold between harmless and toxic, so just knowing substance X is $\frac13$ "peak" of substance A doesn't let me know if I should let my dog/child/self ingest it.

The way you have presented this does not sound like you are really using GC/MS, but just gas chromatography. The height of these peaks depends on the detector response and it seems silly if you're using a MS detector in the way you imply here.

Is this even how GC/MS results work?

As cbeleites said the method you described is a proper technique but not likely to be appropriate given the information you cited.
In GC/MS you should have two sets of information. The first is the GC Total Ion Chromatograph (TIC) which will have time as the x-axis and response (abundance) as the y-axis. For each retention time on the TIC there will be a corresponding Mass Spectra (MS). In the MS the x-axis is m/z (ion) and y-axis is also response (abundance).
Different compounds have different responses so if you inject the exact same amount of two different compounds you could get a much larger response from one compared to the other. For example tramadol gives much higher response compared to hydrocodone. This is why you want your internal standard to be structurally similar to your analyte.
Quantification of a compound is often done by running 3 - 5 calibrators at known concentrations in order to make a calibration curve. Once an acceptable calibration curve is made the sample along with controls can be fit onto the curve to calculate the amount of the compound of interest in the sample and controls. If the controls are correct then the sample value can be used.
To make a calibration curve you need an internal standard in each calibrator, control and sample. You can then make the curve either with ion ratios between paired internal standard and analyte ions OR if your internal standard response is consistent in all the calibrators, controls and sample you can use the GC peak response.
In conclusion if you only have one GC/MS run with the two dissimilar compounds in it you will not be able to easily calculate the quantity of the second compound based on the GC response of the first unless you have additional information that was not listed in this question.

You're right. The link to the German chemgapedia shows the calibration. I somehow assumed that the OP tried to understand the internal standard part of an analysis procedure someone else runs - re-reading the question, I may have misunderstood what really the question is.

And certainly it is questionable whether a calibration with slope only and intercept forced to zero (which is how I read that mysterious factor) is appropriate.

This sounds like A being an internal standard for the determination of B.
Internal standards are used to get rid of (small) multiplicative variation due to certain influences, e.g.

in GC, the injected volume
in optical spectroscopy, the optical path length or the illuminated volume.

Sometimes the calculation part of the idea is called normalization.
The English Wiki page for the Response factor explains the internal standard (and calculations).
The important part is:

One of the main reasons to use response factors is to compensate for the irreproducibility of manual injections into a gas chromatograph (GC). Injection volumes for GC's can be 1 microliter (µL) or less and are difficult to reproduce. Differences in the volume of injected analyte leads to differences in the areas of the peaks in the chromatogram and any quantitative results are suspect.

So you switch your whole calibration to response normalized to the response of the internal standard (= signal analyte / signal internal standard) and the result is concentration analyte / concentration of internal standard = molar amount analyte / molar amount internal standard. With the known amount/concentration of internal standard you can calculate back to the analyte amount/concentration.
If you can read German, here's  another explanation and HPLC example calculation.
Skoog & Leary's Instrumental Analysis says that with a suitable internal standard, accuracy of < 1 % is possible.
So, if your A is an internal standard, then, yes, that is a proper technique.
Note, however, that arsenic does not sound like an internal standard for glucose (nor vice versa): the internal standard should be

as similar as possible to the analyte
but the signal should not overlap with other substances
for GC/MS that could even be an isotope labeled version of the analyte for maximum chemical similarity.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Definitions of autocatalysis
Is there a formal definition or algorithm that can take a list of reactions and tell me whether it contains a set of species that can produce itself autocatalytically?
It is clear that the following reaction is autocatalytic:
$$\ce{
A + B \to 2 B
}$$
B appears on both sides of the equation, so it's a catalysis reaction, and there is an extra B produced on the right, so it catalyses its own production from A.
But the following reaction scheme can also be considered to contain autocatalysis:
$$
\ce{B + C \to D + E\\
A + D \to 2C}
$$
If this system starts out with plenty of A and B, together with one molecule of C, then the C molecule can react with a B to produce a D, which can then react with an A to produce two C's — whereas, without the initial "seed" molecule, no reactions will take place. (I'm neglecting reverse reactions here of course, but hopefully it gets the point across.)  The autocataltic cycle can easily be seen in the following diagram, where circles represent species and boxes represent reactions:

Another non-obvious example of autocatalysis is a reaction scheme of this form:
$$\ce{
A + C \to C + D\\
B + D \to D + C + E}
$$
Here, C and D each catalyse the production of one another from the precursors A and B. I'm aware of Kauffman's definition of reflexively autocatalytic sets, which applies to schemes of this form, but would miss an autocatalytic cycle of the kind shown above, since it doesn't consist of single-step catalysis reactions. So I'm looking for a definition of autocatalysis that would include both cases, as well as (hopefully) other, more complex schemes that I haven't yet thought of.

Likely not a direct answer to your question but I'll leave it in a comment for the sake of awareness and perhaps discussion (and just popped up on my feed this morning).  This paper looks interesting with regards to autocatalysis. http://arxiv.org/pdf/1205.0584v2.pdf

@LordStryker thanks! It's based on Kauffman's RAF sets though (Stuart Kauffman is the third author), whose definition relies on all the catalytic reactions being single-step ones.  So it doesn't provide what I'm looking for, but it certainly looks interesting.

This is a very interesting question. I am nowhere near being up to the challenge of answering it, though I would love to see someone tackle it.

Is there a formal distinction between a chain reaction, and an autocatalysed reaction? Both involve the creation of temporary species which are accumulate before being used to create product.

I don't know if there's a formal difference, but the two concepts are clearly very closely related. I tend to think autotcatalysis is a more technically well-defined thing, since "chain reaction" can also refer to less interesting "chains" of reactions such as $A \to B$ together with $B  \to C$. But if there's a good formal definition of a chain reaction in the self-amplifying sense, I'd be very interested to see it.

@Nathaniel The definition of a chain reaction I saw as an undergrad (which excludes A->B, B->C) Is that a chain reaction has to include chain initiating steps, chain propagation steps, and chain termination steps. You can then define your elemental reactions in terms of these three categories, possibly overlapping if you have more than one chain carrier. This excludes intermediates such as A->B, B->C but includes A->B, B+C->D+B, B->A. Chain carriers are produced in initation/termination steps and bystanders in propagation steps.

The way this question is put is very curious, especially when you say "formal definition." You make it sound like you want something more mathematical. This is too obscure of an interest to have much luck in chemistry, so you'd probably have more luck on math.stackexchange.com

@Chris I am indeed looking for a formal, mathematical definition, but autocatalysis is a chemistry concept, not a mathematical one, and therefore if a formal definition does exist it will be in the chemistry literature.

I can't tell if the modifier "formal" applies only to "definition" or to "algorithm" also, in the first sentence of your question's body. In case any algorithm would do: this looks like a pretty standard instance of the cycle-detection problem that one learns in undergraduate computer science.
If you only want to detect the presence of cycles, then the algorithm is essentially just to attempt a topological sorting of the graph; if the attempt fails, you have a cycle, and I guess that would mean that you have some sort of catalysis going on. (I didn't major in chemistry; please yell at me if that's non sequitur.)
Okay, so mere cycles are not enough - it seems you want cycles that have "gain". How about a tortoise and hare and have these multiply a state variable by the ratio of outgoing to incoming edges to reaction nodes (denoted by squares in your diagram)? If the hare catches up to the tortoise, that implies a cycle, hence catalysis; then if you compare their state, if the hare has accumulated a larger "gain" than the tortoise, I would say that that shows the presence of autocatalysis?
I'm not sure how this works if there are multiple, linked cycles. You could have one cycle in your graph that's autocatalytic, and another cycle linked to it that "uses up" its catalyst (something of a contradiction). For example:
B + C -> D + E
A + D + H -> 2C + F
2F + G -> H

Is that still an autocatalytic system?

It was intended to be parsed as "(formal definition) or (algorithm)" - if an algorithm exists I won't quibble if it's stated informally :) You're right that cycles imply catalysis, but it's distinguishing this from autocatalysis that's the difficult bit. If you remove the double edge in the figure (i.e. change the reaction scheme to $\ce{B + C -> D + E}$ and $\ce{A + D -> C}$) then there's still a cycle, but only catalysis, not autocatalysis. If there's only one cycle in the graph it's easy to distinguish the two, but if there are many then I don't know how to do it.

I've added some more paragraphs; I think maybe you should pose this as a computer science / programming question on one of the other SO sites.

But the tortoise and hare algorithm is for finding cycles in sequences, not graphs, right? The problem here isn't so much to come up with an algorithm to detect autocatalysis - it's to find a suitably general definition. Once I have that I can probably come up with an algorithm myself (and I'd ask on a CS site if I couldn't). "A single cycle with some 'gain'" is only a special case of what I'm looking for - in general there will be many interlinked cycles - see the 3rd reaction scheme in my OP for an example.

The example you post contains an autocatalytic cycle but also contains some other stuff - that's OK. I'm looking for a way to find just the bits of the graph (subgraphs) that are autocatalytic.


text:
Is there a theory behind selecting elements that may be successful in potential superconductors?
Looking at something like $\ce{YBa2Cu3O7}$ which was one of the first cuprate superconductors to be discovered, I'm always curious how the selection of these substances as likely superconductors comes about?
Does it have to do with optimal crystalline structure, unpaired electrons, or is it similar to the way semiconductors are designed with doping to create "holes"?  Or are these superconductors simply determined experimentally based on small changes of what has worked in the past?
As a side question: I've always been curious as to how they stumbled upon Yttrium for this particular case.  It's not something I think of as being carried in ready supply.  Would this have been after YAG lasers were already around?

Suffice to say, nobody really knows, and a Nobel prize surely awaits the first people who can explain the relationship between chemical composition and high-temperature superconductivity.

That's what my feeling was, but surely there must be some rule of thumb, right? Otherwise what would dictate experimental protocols and the like?

Someone more qualified than I am will probably have more intelligent things to say here, but there is a lot that can be done with sheer brute force trial and error, looking at empirical trends and trying to extrapolate them (which is really another kind of educated trial and error). That's how a lot of science is done in the absence of guiding theoretical principles.


text:
Z-matrix of a series of 5 colinear atoms
How can I write a Z-matrix for the following complex?
  O==C==O  H--O
               \
                H

I have managed to write:
O
C   1  a
O   2  a     1  180
H   3  d     2  180    1  ?
O   4  b     3  180    2  ?
H   5  b     4    t    3  ?

but I don't know how to specify the dihedral angles indicated with ?. Are they all zero? Or completely unspecified? If so, will I be able to do a coordinate optimization with these variables or will it be a problem?

You cannot have bond angles of 0 or 180 degrees in a Z-matrix. This is because the dihedral becomes degenerate. To solve this, you can add 'dummy atoms' which provide an auxiliary point of reference to remove the 180 degree angle. These atoms are usually denoted $\ce{X}$ or $\ce{Xx}$ in computational chemistry suites. Usually, dummy atoms are placed at 90 degrees to a pair of atoms and the third collinear atom is then defined with a bond angle of 90 degrees and a dihedral of 0 or 180 degrees with respect to the previous 3 atoms.
Example
To illustrate what I'm talking about, here's a schematic of your pair of molecules, with three dummy atoms:

This could have a Z-matrix like:
O   1
C   1    a
Xx  2  1.0  1  90
O   2    b  3  90     1  180
Xx  4  1.0  2  90     3  0
H   4    c  5  90     2  180
Xx  6  1.0  4  90     5  0
O   6    d  7  90     4  180
H   8    e  6  104.5  7  180

Notes
There are no hard and fast rules about how many you should use (obviously we could simply define subsequent dihedrals in terms of the leftmost dummy atom) however it's best to strive for locality so that small changes in some variable don't correspond to large displacements elsewhere, in which case rounding errors or large gradients can ruin your day. You could also have defined all of the dihedrals in your molecule with respect to that convenient non-collinear hydrogen, however this is the Z-matrix equivalent of spaghetti code.
You will note that I use different letters for each variable. This is because forcing the $\ce{H-O}$ and $\ce{C=O}$ bonds to be of equal length is probably an unrealistic constraint for a geometry optimisation of this system.
Finally, note that dummy atoms do not take part in the electronic structure of a molecule, so feel free to put them in anywhere and set the dummy atom 'bond' length to whatever you want. 1.0 Ångström is traditional as far as I know.
P.S.
Molden has a complete, if idiosyncratic graphical Z-matrix editor.

Nice answer (beat me to it to boot!).  To others tuning in, stay away from using Molden's auto-Z-matrix generator.  It will likely give you junk.

@LordStryker - this is a fair observation. Molden is best used for constructing Z-matrices manually or through the 'reorder Z-matrix' command, which lets one rebuild an iffy Z-matrix by simply clicking on the atoms in the desired order.

You can use SMILES in Avogadro or openbabel and it produces 3D coordinates that
you can edit. And you can create Gaussian, NWChem, gamess... inputs.

How does that relate to Z-matrices? Can Avogadro or openbabel produce Z-matrices from 3D coordinates? Are they then in “reasonable” physical order?

@F'x Yes, both Open Babel and Avogadro automatically create Z-matrices with "reasonable" physical order. For some definition of reasonable. For your question, I think the answer above is much better - it can be very hard to optimize so many collinear atoms.


text:
Where do I find datasheets of zeolites?
I'm looking into uses of zeolites as water absorbers for heat storage and absorbtion cooling. However, I don't find hard numbers on the likes of absorption enthalpies at different temperatures, saturation moistures, etc.
I'm not an institution, chemical suppliers are traditionally reluctant to sell to private individuals. I found zeolitic absorbers in few catalogues, but only with a descriptor of the type, no further data. One company I asked (that sell zeolites in amounts of kg to private customers) was unwilling to share data on their product (They're an engineering company doing product developement with zeolitic absorbers for heating/cooling etc.).
I only found (sketchy) data on single zeolites in working papers, but no comprehensive source.
So: seeing that vendors don't share the info, where do I find datasheets of zeolites?

These data are time-consuming to compile together and expensive to produce. You might get lucky and find a Handbook having what you want, perhaps "Handbook of Zeolite Science and Technology" ?

found the TOC, http://www.crcnetbase.com/isbn/9780824740207 doesn't suggest datasheets. Still, answering as an answer and not as a comment would be more in the spirit of this site

I've searched for this same data.  My interest regards electronic proporties of zeolite-class media.  I've written this same kind of question on Quora.
I think the reason this is rarely answered, and why there are no 'catalogs', is several-fold:

There exist a wide range of applications, based on the composition
of the water / etc.
The treatment media usually varies in
composition, meaning that no directive for the use of a zeolite
product can be given by a sales rep (only on-staff engineers, who
know how to use it).
Some uses create by-products that are regulated, creating possible liabilities for the seller by indicating a fitness-for-use.
Regulations change; per year, per locale, etc.

I think the groups that know about these are environmental engineering firms, whom have a stable of engineers that provide end-to-end services for their clientele.  Many of of these firms have Chemists on-staff, or a consultancy relationship that suits the same.  They know the chemical properties of their media, and need a zeolite / catalysts that can effect something very specific.  As a result, they're out their hunting the journals for something that targets their needs.
Hard to out-sell that without already having a dog in the game.


text:
What dictates the lifetime of a solvated electron in a given solvent?
Solvated electrons have a long lifetime in ammonia solutions, but their counterparts in water (called hydrated electrons) have a much smaller lifetime, of the order of microseconds in very pure water.

(image from: Boero et al, Phys Rev Lett 2003, 90, 226403)
What properties of the solvent account for these very different lifetimes? Both are polar protic solvents, so what other(s) factor(s) could be involved here?

For those who have never had the pleasure of personally doing this, see this video. It has been known since 1807 that dissolving sodium in liquid ammonia results in a beautiful color. It was originally thought the color was due to some familiar complex instead of a solvated electron. A similar phenomenon happens with other alkali metals in ammonia.
Research in the field says that it takes, at least, forty some ammonia molecules to solvate a given electron. Metstable cavities form and their stability could depend highly on electrostatic interactions to solvate our electron. Ammonia will slowly react by evolving hydrogen gas,
$$\ce{2 NH3 + 2 e- -> H2 + 2 NH2-}$$
Differences in solvent can play a huge role in the stability of our solvated electron. An analogous decomposition occurs in water,
$$\ce{2 H2O + 2 e- -> H2 + 2 HO-}$$
Famously the latter occurs faster than the former, see here in comparison to my earlier link. We could say that these differences in rate reflect different stabilities of our solvated electron. Although, the addition of an appropriate catalyst to our ammonia will result in rapid evolution of hydrogen.
A more quantitative description for the difference in energies is obtained by measuring the UV-vis spectrum for a solvated electron in both water and ammonia. One will quickly notice that the band appears at higher energies in water than it does in ammonia and there are considerable differences in the band shapes (the band is wider in ammonia).
So Why Do They Differ
So here I have to make the disclaimer that no current theories quantitatively reproduce the observed phenomena, such as the absorption spectrum for a solvated electron. This question has no established/accepted answer as it is an ongoing area of research.
The self-ionization of water has an equilibrium constant on the order of $K = 10^{-14}$ and that of ammonia is on the order of $10^{-30}$. Hydronium formation occurs in the case of water and ammonium in the case of ammonia. Hydronium has a considerably lower $\mathrm{p}K_\text{a}$ than ammonium and so it’s reasonable to see why a reaction in water would be more likely with an electron. (Ammonia rarely produces a weakly acidic species, but water often produces a strongly acidic complex.)
A slightly deeper reason may be that water forms more ordered local domains/structures in solution than ammonia and this influences the rate by resulting in a larger entropy of activation when the hydrolyzed electron breaks up these larger structures, recall that $k \propto \exp (\Delta S ^\ddagger /R) $.
Just speculation though.