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Ring currents of anti-aromatic and aromatic systems
Why are ring currents associated with aromatic systems diamagnetic while those of anti-aromatic systems are paramagnetic?
Does it have anything to do with the number of delocalised electrons 4n+2 and 4n , respectively?

Yes it has to do with the number of electrons. Consider a simple model of the electron on a ring, then the electron has to follow a circular path and its wavelength has to 'fit' exactly  round the ring. Thus the wavelength has to satisfy the condition $L=q\lambda$ where $q$ is a quantum number $q=0, \pm 1, \pm 2 ...$ characterising each eigenstate. The energy in this simple model comes to $E=(hq)^2/(2mL^2)$ ($h$ is the planck constant, $m$ the electron mass). Now in this model filling up the energy levels according to the aufbau principle, (Pauli principle & Hund's rules) leads to 2 electrons in $q=0$ (which is singly degenerate) and 2 in $q=+1$ , and $q=-1$ etc which are all doubly degenerate in energy.
Thus $4n+2 ~~ \pi$ electrons fill up all levels with two electrons in each level. The $4n~~ \pi$ contains only a single electron each. This is because it takes less energy to add one electron to each of the $q$ levels rather than pairing them up. The sketch shows this.

The unpaired electrons are shown in the $4n$ column,  however, the levels are also shown as being split in energy. This is due to the Jahn-Teller effect. This states that a slight interaction between degenerate levels can cause the degeneracy to be split and so lower the energy. The interaction could be due to slightly different bond lengths and the energy levels are split by only a small amount relative to the gap to other levels.
Although not shown in the diagram, (which is thus slightly inaccurate) both spins will occupy the same lower level so in this sense, [4n]annulenes do not really exhibit molecular paramagnetism. The electrons interact with the magnetic field $B_0$ and this leads to a mixing of levels and this causes a paramagnetic contribution to the shielding constant. This is larger than the diamagnetic contribution for  lower orbitals. Thus in [4n]annulenes we speak illustratively of a 'paramagnetic ring current' and molecules showing this effect are called paratropic and those with no ring current atropic.

The key to this is determining whether or not your electrons in the relevant orbitals are all paired or not.  If an orbital contains only a single electron, then we expect the molecule to be paramagnetic.  Compare, for example, the MO diagrams for benzene and cyclobutadiene (these diagrams can readily be found in any organic chemistry text or simply by Googling the name of the molecule followed by "MO").  You will notice that, for a benzene molecule, all six of the electrons in the pi system are found paired together in bonding molecular orbitals.  However, if you look at the diagram for cyclobutadiene, while the lowest two-energy electrons occupy the lone bonding orbital, the other two electrons singularly-occupy the two degenerate (i.e. equal energy) non-bonding molecular orbitals (as we expect by Hund's Rule).  According to MO theory, we would expect such a species -- be it an atom, ion, or molecule -- to be paramagnetic.


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Ligand Field Theory with non-identical ligands?
Lots of undergraduate inorganic textbooks deal with complexes with $O_h$ or $T_d$ symmetry. So, if ligands are not same(like $\ce{[Ni(NH_3)_4(H_2O)_2]^{2+}}$), what methods are used to interpret their properties like electronic spectra?

What do you mean by methods to interpret? Do you mean predicting their electronic spectra just by looking at their configuration or do you mean experimental methods?

Something like molecular orbital diagram or Tanabe-Sugano diagram... Is there any way to explain like 'transition from A state to B state corresponds to this peak in electronic spectra'?

Just use the lower symmetry of $D_\mathrm{4h}$?

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Degenerate orbitals in the Hydrogen atom
I came across the following link:
Which orbitals of the hydrogen atom are degenerate for $n=3$
And all the answers said that for the hydrogen atom, the energy of an orbital depends only on n. Is this really true? And also, isn't it true that for other atoms (at least according to those who answered the other question) the energy depends upon both $l$ and $n$?

Yes, that's true. In a hydrogen atom, orbitals with same $n$ and different $l$ are degenerate, that is, have the same energy; in multi-electron atoms, they are different. Also, welcome to Chem.SE.

Oh, I didn't know about this!(Hmph. Why can't they mention this exception on textbooks?!) Do u know why this is so? Also, thanks! :)

"Why" is a tricky question. What kind of explanation would you be satisfied with? We just solve the Schrödinger equation (exactly for H, approximately for other atoms), and that's it. Some people say the degeneracy is due to certain (and quite unobvious) higher symmetry.

Well, OK, although that's not much of an explanation as to "why", I don't think I'd be able to understand anyway, since I only just graduated high school. Thanks for ur help!

It is true that if you solve the Schroedinger equation the solutions depend only on the principal quantum number $n$ and the energies (often called eigenenergies or eigenstates) have the values predicted by the Bohr formula, $E=-hcR_\infty /n^2$ and energy differences as predicted by the Balmer formula. The calculated energy does not depend on $l$; this degeneracy of wavefunctions with different $l$ is a peculiar and special feature of the symmetry of the Coulomb potential.
However, nature cares little for our equations and experiments show that in fact the energy of the $l$ levels are different in energy to one another. This is called 'fine structure' and the energy shifts are very small as the name suggests. The first cause is explained as being due to the fact that the electron has an intrinsic quantum property called 'spin', which does not mean that it is literally spinning, but crudely speaking that it can be described by the same equations as angular momentum. As the electron is charged its spin produces a magnetic field  which interacts with the magnetic field of its orbital motion, which in turn depends on $l$. This interaction is called spin-orbit coupling, and technically it is a relativistic effect.The second effect is called the Darwin term, for electrons with $l=0$, and the third effect is the Lamb Shift, which needs explanation beyond relativistic effects and requires Quantum Electrodynamics.
The fine structure due to spin-orbit coupling can be observed easily in many atoms so the $l>0$ levels are split for each $n$ and so are no longer degenerate. The common example is the two lines of Sodium 'D' emission. The difference is small, the two transitions are 589.59 nm to 588.99 nm. This is the yellow colour associated with sodium easily seen if you sprinkle salt into a gas cooker's flame and in some types of street lights. The splitting can be measured with a simple spectrometer and is a common teaching lab experiment. The other effects mentions are far,far more difficult to measure and in a Chemistry course you are unlikely ever to meet them.

I would clarify that energy differences for "fine structure" are very small. Avoid this clarification may result in a misinterpretation of your nice details.

@ user1420303, thanks, have added a line to emphasise  this.

You are right, nature is more important than our equations (and moreover, Schrödinger equation is not even the "ultimate equation of everything", for it does not include relativistic things). I stand corrected. You might want to add Wikipedia links to Darwin term and Lamb shift, and maybe also to the thing I mentioned earlier which explains that the said degeneracy is, in fact, not accidental.

'accidental', I take the point and I have reworded this part of the text.

I took the liberty of formatting the links in a different way.

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Incomplete combustion
I cannot find a definitive equation that describes the incomplete combustion of a hydrocarbon. Equations $1$, $2$ and $3$ are some of the variations I have found.
Is it,
$(1)$ $\ce{C_xH_y + O_2 \rightarrow CO + H_2O}$,
$(2)$ $\ce{C_xH_y + O_2 \rightarrow CO2 + CO + H_2O}$ or,
$(2)$ $\ce{C_xH_y + O_2 \rightarrow CO2 + CO + C + H_2O}$?

There is no definitive equation for an incomplete combustion. Sometimes incomplete combustion of hydrocarbons may even form species such as polyaromatic compounds and other hydrocarbons. The result may differ by variation of factors such as temperature, oxygen concentration, and the hydrocarbon undergoing combustion. Basically what you're asking is like asking "is there a definitive number between 0 and 1?" You may already know that numbers between 0 and 1 is an infinite set. Incomplete combustion may not be an infinite set, because nobody proved that it is, but it's still long ways off from a definitive one equation solve it all thing.
There is only a definitive equation for complete combustion; like there is only two numbers within a set of {0, 1}.
It's easier to recognize that any combustion that is not a complete combustion as a incomplete combustion.


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How to determine electrolyte concentration in solutions?
The original question is here

We've been given the sports drink vs orange juice challenge. To find out which has more electrolytes. I could test for conductivity but that alone wouldn't tell me which has more actual electrolytes in it. So I'm wondering if there's a way I could separate each ion from each other and measure them that way?

I'm posting it here because I'm very curious as to the answer.
Several people have suggested evaporation.

You can try evaporating the liquids and seeing what's left behind. The electrolytes will be left behind as solids.

However, the issue with evaporation is that it leaves behind all solids - electrolyte or not.
Some people have suggested conductivity measurements.

Wouldn't conductivity simply be proportional to the concentration of substances that promote conductivity?

However, this objection was raised:

Well my understanding was that different ions produce different voltages, therefore, there could be a lot of one type of electrolyte producing a large voltage but there could be more electrolytes in the other solution producing a smaller voltage. I'm not sure though as I am in first year and haven't had this completely covered yet. My professor said that conductivity might not give enough information so I may need a second approach as well, I'm not sure what he means exactly.

Is this a valid point? I guess we could empirically test out this assertion that "different ions produce different voltages" by comparing the conductivities of two otherwise identical NaCl and, say, MgBr2 solutions. Wouldn't the MgBr2 have a conductivity measurement that is roughly 1.5 times that of the NaCl solution? Or is it true that the ion's charge affects the conductivity?

Yes its a valid point.
https://en.wikipedia.org/wiki/Conductivity_(electrolytic)#Strong_electrolytes
Plus orange juice and gatorades contain weak electrolytes also- which would affect conductivity.

What qualifies as an electrolyte in the context of this question? Please give your definition. It is also unclear how different electrolytes are to be weighted for comparison.

Electrolytes turn water into a conductive solution.  Conductivity is affected by how free the electrons are to move.  Conductivity (which is the reciprocal of resistivity) is affected by by concentration, temperature, and how much charge the ions carry.  Conductivity alone will give an equation with both ions and concentrations; however, this equation will have multiple solutions. For some fun understanding this phenomena look at this this link.
Conductivity is different than voltage.  Voltage in this situation is likely referring to a half-cell reaction with a standard hydrogen electrode in a galvanic cell.  Using this approach has the same issue with a multiple solution equation.
Electrolytes are typically ions.  Ion concentration is measured with ion chromatography.  Ion chromatography involves separating the ions from the solution based on various properties to perform a measurement. It is explained here, here, and here.
When electrolytes of drinks are measured it is important to include if the electrolytes are bio-available.  Bio-available electrolytes depend on the activity performed.  An activity such as sports uses a different set of electrolytes than sitting.
Testing bio-available electrolytes requires more work than analytic chemistry. Analytic results both the drink and bodily fluids of a population doing a certain activity are need to draw conclusions about the electrolyte composition of a drink.
Reading a patent for a sports drink ions include $\ce{Na+}$, $\ce{K+}$, $\ce{Mg^2+}$, $\ce{Cl-}$, $\ce{SO4^2-}$, $\ce{PO4^2-}$.  The patent also includes citrate, sucrose, dextrose, ascorbic acid and pyridoxine as these are known to affect the uptake of electrolytes.
If you have access to a university library the values of these compounds for orange juice are available.

Sucrose and dextrose are in no way electrolytes.

@aventurin I'll correct the wording.  Sugars and other carbs do not form ions.  These compounds affect the uptake of electrolytes.  Reference   https://www.google.com/patents/US4322407

OVER THE TOP, 2nd ANSWER
There are many ways to analyze things so with respect to the original challenge, to find out which has more electrolytes (orange juice or gatorade), since there is not a strict definition of what a electrolyte is, you could try to define electrolyte: Electrolytes are minerals, acids, bases, or salts that carry an electric charge... but that's debatable and still leaves room for interpretation. So rather than try to be specific, it will be easier to simply list a few analytical methods that should provide a totally comprehensive analysis.
A practical preparation and test for many elemental-ions and minerals (sodium, calcium, zinc, iron etc.) and subsequent analyses will be to sample the ash content.
Measuring the pH of the solution may help confirm further analyses, but should not be technically necessary.
Atomic absorption spectroscopy (AAS) would be the most straightforward technique is used for determining the concentrations of a elemental ions in a sample. I would use AAS to analyze both the original and reconstituted (from ash) sample for veracity.
However, AAS will not account for mixed organic molecules such as citric acid or ascorbic acid. High performance liquid chromatography (HPLC), in conjunction with IR spectroscopic analysis of the analytes would be the most straighforward method for separating organic molecules/electrolytes. Combustion analysis or Mass Spectrometry could possibly be employed for obtaining more information about the carbon, sulfur, and nitrogen, hydrogen, and oxygen content of the sample.
And to put a nail in this coffin, gas chromatography (which is often coupled with NMR these days) may also be used to separate and analyze a sample of mixed materials. And the Nuclear Magnetic Resonance (NMR) data from each analyte can be obtained (which is very useful information for both approximating concentration and identifying the analytes).
So, given HPLC + IR Spectroscopy, Gas chromatography + NMR spectra, Ash content, Mass Spec analysis, AAS, and the pH... there's really nothing more that could possibly be necessary to determine the contents of an orange or a bottle of gatorade, and then you can call an electrolyte whatever you please.

This approach is called a Texas sharpshooter.

@Agriculturist I'm afraid I don't catch what you mean. I think a better analogy for this is would be, using an uzi, grenade, torpedo, sidewinder (missile), bunker buster, napalm, and a nuke. Then you can be sure you hit the target.

The Texas sharpshooter is analogous to shooting wildly then drawing a bulls eye around the tightest cluster.      Consider modifying  the line 'and then you can call an electrolyte whatever you please' .     -- Great article on the topic http://www.nature.com/news/how-scientists-fool-themselves-and-how-they-can-stop-1.18517

There are many ways to do this and essentially you are after the mineral and ash content. So, with that in mind, phosphorus boils at 280.5 °C and here are the common methods for mineral anlaysis (like for food or gatorade).
But for about the same price of a crucible (not to mention the oven) you can have someone else analyze the ash content. This is not a service recommendation, I'm just saying it's another, more affordable option.
Whatever analysis you perform, a practical consideration is that analysis will be easier and more accurate after reducing the sample to ash.

This method would be perfect for ancient Roman Energy Drinks! -  http://www.smithsonianmag.com/smart-news/did-gladiators-drink-energy-drink-made-ash-180953213/?no-ist


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Distinguish between glucose and Fructose samples
I am looking for a simple way to distinguish between glucose and fructose samples. Currently, I am doing that by melting temperature method. Can anybody suggest me some easy way to do that?

Related: Why does fructose reduce Tollen's reagent and Fehling's solution?

One of the simplest ways would be bromine-water test. Bromine water oxidizes glucose to gluconic acid, hence decolorize the solution. Being a mild oxidizing agent, Bromine water is not capable of oxidizing fructose (ketone).

Go for Seliwanoff's Test. It is based on the simple fact that when heated, ketose sugars are more rapidly dehydrated than aldoses. After acidic hydrolysis of both, we add a pinch of resorcinol(0.5%) and concentrated HCl(3N). Fructose reacts to give a deep red cherry colour whereas Glucose reacts slightly to produce a faint pink colour.

Why are ketoses more easily dehydrated than aldoses?

Taste tests also work. Fructose is about 2.3 times sweeter tasting than glucose, and the tongue is a very good sweet-sensor.

(source: wikipedia)

While this is accurate, normally one does not recommend tasting chemicals in a laboratory.

Go for Osazone test.... add phenylhetrazine +sodium acetate + acetic acid...if yellow precipitate appears within two minutes of boiling in waterbath (needle shape crystals)then the sample is Fructose.. if yellow precipitate appears between 5 to 10 minutes of boiling in waterbath (broom shaped crystals)its glucose.

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What's the limit on energy density for chemical reactions?
Considering energy-releasing devices such as bombs or batteries which operate on chemical reactions and not nuclear (i.e. strong/weak) forces, what are the upper bounds on the amount of chemical potential energy that can be contained in 1kg or 1m3 of ordinary matter at RTP?
Here I'm not asking for the energy density of known compounds, necessarily, but rather an answer of the type "no conceivable compound could exceed X J/kg because ..."
Extra kudos if you can add insight into limits on the rate of energy release and efficiency considerations.

http://chemistry.stackexchange.com/questions/54791/which-chemical-reaction-has-the-highest-specific-energy

related http://chemistry.stackexchange.com/questions/4881/a-theoretical-molecule-that-could-match-uranium-mole-for-mole

https://chemistry.stackexchange.com/questions/112634/which-substances-have-the-highest-lowest-enthalpy-of-formation-per-gram


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How to choose proper active space in method such as CASPT2 and CASSCF?
I am confused about other research paper which use CASPT2 method. I am not clear about how to choose active space. What can I refer to? (tutorial reviews, papers, or any other things)

Related: http://chemistry.stackexchange.com/questions/54108/typical-runtime-for-casscf-calculations

Choosing the active space in any calculation with a multiconfigurational reference (CASSCF, RASSCF, GASSCF, CASPT2, NEVPT2, MRCI, the list goes on) requires more chemical intuition than more commonly-used "black box" quantum chemistry methods, such as DFT. Here, chemical intuition means preexisting knowledge about 1. the system under investigation, and 2. the properties being calculated. For example, a minimal active space may be sufficient to qualitatively capture static correlation in a system, but the active space might change when targeting excitation energies, and might be different again when calculating spin-orbit coupling constants. More specifically, it is understanding what orbitals are most important in composing your wavefunction and have the largest influence on molecular properties of interest. Unfortunately, this can also involve quite a bit of trial and error.
Some very general recommendations on the workflow for performing multiconfigurational calculations are in the numbered list in this answer (yes, a shameless plug).
Here is a more complete list of references I've found useful in my own research:

[1], [2] are slides from the group that develops GAMESS, one of the premier packages for multiconfigurational calculations. The second set of slides in particular has examples that highlight the chemical motivation for different active spaces depending on the molecule and the property (internal rotation in ethylene, internal rotation in $\ce{H2C=NH}$, dissociation of $\ce{H2C=O} \rightarrow \ce{CH2} + \ce{O}$, isomerization of bicyclobutane to 1,3-butadiene, and others).
[3] (paywalled) is a more formal overview of the above. Excellent.
[4] is a set of very detailed and theory-focused slides from the original authors of DALTON. I link to them only for completeness; there aren't many practical examples here.
[5] (paywalled) is a comprehensive review article from the authors of COLUMBUS, with many examples. Excellent.
[6] (paywalled) is an article from the authors of COLUMBUS on the application of multiconfigurational methods to geometries of small molecules. It has good explanations on the choice of active space based on electronic configurations.

In addition, there are many other online resources that are more tutorial oriented: [7], [8], [9], [10]. These may appear to be program-specific, but the chemistry is transferable between different programs; only small technical details in how to set up calculations should be different.
Finally, some programs have nice examples or helpful hints in their documentation. DALTON (PDF), Molcas, and ORCA (PDF) come to mind.
References

[3]: Schmidt, M. W.; Gordon, M. S. The construction and interpretation of MCSCF wavefunctions. Annual Review Of Physical Chemistry 1998, 49, 233-266. DOI: 10.1146/annurev.physchem.49.1.233
[5]: Szalay, P. G.; Müller, T.; Gidofalvi, G.; Lischka, H.; Shepard, R. Multiconfiguration Self-Consistent Field and Multireference Configuration Interaction Methods and Applications. Chemical Reviews 2012, 112, 108-181. DOI: 10.1021/cr200137a
[6]: Shepard, R.; Kedziora, G. S.; Lischka, H.; Shavitt, I.; Müller, T.; Szalay, P. G.; Kállay, M.; Seth, M. The accuracy of molecular bond lengths computed by multireference electronic structure methods. Chemical Physics 2008, 349, 37-57. DOI: 10.1016/j.chemphys.2008.03.009


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Which has highest solubility in water?
Among $\ce{LiF}$, $\ce{LiCl}$, $\ce{LiBr}$ and $\ce{LiI}$ is $\ce{LiI}$ the most soluble compound ? As the lattice energy depends on size and smaller the size of ions greater the lattice and as $\ce{LiI}$ has greater size therefore should it be most soluble? Since lattice energy is inversely proportional to solubility, according to me $\ce{LiI}$ is most soluble and $\ce{LiF}$ is least. Is this correct ?

Soluble in water?

$\ce{LiI}$ is less soluble than $\ce{LiCl}$ in terms of moles per liter, while $\ce{LiF}$ is the least soluble of all.


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Volatile (aroma/scent-defining) compounds from Himalayan Balsam (Impatiens glandulifera)?
Has anyone identified the compound(s) that make up the distinctive and intense scent of Himalayan Balsam (Impatiens glandulifera)?
I’ve found a number of compounds associated with Impatiens sp., but it does not look like any of them would be carriers of the scent:

2-Methoxy-1,4-naphthoquinone or "lawsone methyl ether" [an anti-inflammatory, fungicidal agent]
2-Hydroxy-1,4-naphthoquinone (Lawsone) [a dye that is also the principal component of Henna]
Methylene-3,3'-bilawsone (Diphthiocol; 3-[bis(2-hydroxy-1,4-naphthoquinonyl)]methane)
α-Parinaric acid [a polyunsaturated fatty acid]

Neither the naphthoquinones nor the fatty acid are described as having any particular scent, and they are apparently all non-volatile solids, which would rule them out.
I was unable to find any more information online, but maybe someone who is more involved with this kind of research has some pointers. Thank you!

I was just pointed to this very recent research effort by two (very) young chemists to extract an oil from flowers of I. glandulifera and identify its constituents (in German):
http://www.jugend-forscht.de/projektdatenbank/2016/1/gewinnung-eines-bluetenoels-von-impatiens-glandulifera-und-identifikation-der-inhaltsstoffe.html
I’ll see if I can contact them.

Nothing yet; but apparently they identified Lidocaine in the flower oil, which is bizarre, as it’s not known to occur naturally, but that doesn’t help with the answer, as Lidocaine is definitely not the source of the scent either :)

There's seemingly a lot published on the invasion dynamics of Impatiens glandulifera, but little on the scent-giving volatile compounds.
All I could find is a poster from a 2010 symposium on Terpenes - Application, Activity and Analysis, organized by the Phytochemical Society of Europe (PSE).
The authors presented a method to sample volatile compounds from the flowers of the neophytic Impatiens glandulifera and the native Impatiens noli-tangere without cutting them off the plant and analyzed the mixture of compounds by GC-MS.
The scents seem to stem from complex mixtures of monoterpenes.
In the flowers of both plants the following terpenes could be detected:

$\alpha$-pinene
$\beta$-pinene
camphene
1,4-cineole
d-limonene
1,8-cineole
$\alpha$-terpinolene
1-terpineol
$\alpha$-terpineol
$\beta$-terpineol

Only in Impatiens noli-tangere were found

(Z)-ocimene
fenchol
geraniol

Only in Impatiens glandulifera was found

$\beta$-phellandrene

Great info, that was what I was looking for; thank you!


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Potassium Metavanadate Synthesis
I am looking to synthesize potassium metavanadate. Possible precursors that I have are $\ce{V2O5}$ and $\ce{NH4VO3}$.
I have considered the following procedure:
Combine $\ce{V2O5}$ and $\ce{KOH}$ then adjust pH and precipitate out with $\ce{CH3OH}$. The problem is multiple species exist at various pH values, so I am looking for a method to obtain $\ce{KVO3}$. I read a while ago a source for synthesis of different alkali metal vanadates, but I cannot find it again.

I found a synthesis from Inorganic Syntheses vol 15 using potassium carbonate and vanadium oxide. I will try it this week, but if anyone else has any other references, please let me know!

Potassium Vanadate can be prepared by any of the three process:

Potassium Metavanadate, $\ce{KVO3}$, is obtained as white or colourless crystals by dissolving vanadium pentoxide in hot, strong
  caustic potash solution. By varying the concentrations several
  hydrates have also been prepared: $\ce{2KVO3.3H2O}$;
  $\ce{KVO3.2H2O}$; $\ce{2KVO3.5H2O}$; $\ce{KVO3.3H2O}$. On being
  heated, all these hydrates lose their water and leave a white,
  nacreous mass of the anhydrous salt, which melts at 495° C. Two
  other hydrates were prepared by Rammelsberg: $\ce{KVO3.H2O}$ and
  $\ce{KVO3.7H2O}$. (source)

[...] the reaction of vanadium(V)
  oxide from waste catalyst with potassium chloride in the presence of
  steam.
  $$\ce{2KCl + V2O5 + H2O(g) → 2KVO3 + 2HCl↑}$$
  (source)

Ammonium metavanadate react with potassium carbonate to produce
   potassium metavanadate, ammonia, carbon dioxide and water. Potassium
   carbonate - concentrated solution. The reaction takes place in a
   boiling solution.
$$\ce{2NH4VO3 + K2CO3 → 2KVO3 + 2NH3 + CO2 + H2O}$$ (source)

Thanks, I know the reaction schemes, but the problem is that due to vanadium speciation in water, the pH and concentrations need to be precisely controlled. If you look at the Inorganic Syntheses link I sent, you can see the care taken to control these variables. Therefore I was asking for a detailed procedure with analytical results of the product obtained.

For example, see this speciation figure:

https://www.researchgate.net/profile/Li_Zeng6/publication/257130795/figure/fig4/AS:297569017188357@1447957390467/Fig-4-The-distribution-of-vanadium-species-in-water-as-a-function-of-pH-with-a-total-VO.png

And that chart is only valid for a certain range of concentrations and doesnt include other possible ions such as K+, NH3, (CO3)2-, etc. The balanced equations are trivial to obtain and general methods are easy to find, but a detailed synthesis with analytically pure product, is what I was looking for (lab scale) and the Inorganic Syntheses was the only one I found.


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Why is osmotic pressure only for solvent
I believe that the definition of osmotic pressure means that it is only about solvent passing through a semipermeable structure.  But is there a such thing as a solute passing through a semipermeable structure because it is very small too?
This has been bothering me for a couple days.

You mean screening or filtering? Like gravel or coffee?

Osmosis depends on the separation of a solution from its pure solvent by means of a  membrane permeable to solvent but not to solute. The semipermeable membrane can take several forms, cellophane for example is impermeable to high molecular mass compounds, such as polymers, but allows water to pass though. Palladium foil is permeable to hydrogen gas but not nitrogen or oxygen so osmosis can be performed in the gas phase.
The experimental arrangement is for the pure solvent to be separated from the solution by the semipermeable membrane. The natural tendency is for the solvent to flow from the pure solvent chamber into the solution. The tendency is counteracted by applying pressure only to the solution side. The excess pressure is called the osmotic pressure.
Now if the solute as well as solvent could diffuse through the semi-permeable membrane there would soon be a solution of different concentrations on either side of the membrane. Osmosis would now work both ways across the membrane and result in equilibrium when the same solution concentration is on both sides.


text:
Why add H2O when calculating the pH of an acid?
My question is very easy.
Case 1: I have to calculate the $\mathrm{pH}$ of $0.01\ \mathrm{M}$ of $\ce{HNO3}$ solution.
This is my solution:

Case 2:I have to calculate the $\mathrm{pH}$ of a $0.01\ \mathrm{M}$ $\ce{H3PO4}$ solution.
This is the provided solution:

My question is why do we add the watermolecule in our calculations in the second case and not in the first? (or vice-verca)

Based on what I can make out, you did not use the concentration of $\ce{H2O}$ in either of your calculations.
In the second case, you wrote it as a part of your reaction equation. Now, the reason for doing this becomes clear when you realise that when you are talking about a $\ce{H+}$ ion, you are in fact referring to a "proton"-- a bare positive charge. Moreover, a proton is extremely tiny, and thus consequently the charge density is quite high.
This makes it extremely reactive (in a sense), and thus in a chemical system of any sort would immediately seek out and associate with the electron clouds of a surrounding molecule.
In an aqueous solution an obvious, readily available target is the water molecule, and you end up with hydronium ($\ce{H3O+}$) ions.
In these particular problems, the solvent i.e water doesn't have a significant effect on the equilibria, however, this may not always be the case; which is why I recommend writing your reaction equations as you did in the second case, as it allows you to identify the acid/base, and conjugate acid/conjugate base in the reaction.

The reason appears to be that in the nitric acid case you are assuming that it is a strong acid and dissociates completely. You could write $\ce{HNO3 + H2O ->H3O+ + NO3-}$ if you wanted to, as its technically more correct, and we can define pH as $\ce{pH = log10([H3O+])}$.  In the second example the phosphoric acid is not such a strong acid as nitric acid. It can also loose up to three protons, one at a time, so an equilibrium is necessary to calculate how much of the various phosphates are present. For the first step you could just write $\ce{H3PO4<=>H+ + H2PO4-}$ if you wanted. If dissociation was complete to $\ce{3H+ + PO4^3-}$ then you would treat it just as you did for a strong acid.

I don't think you did add h2o to that calculation...you just added it to the reaction equation.  The concentration of water doesn't change and so is not included in the calculation.

ok, but why is it there in the second example and not in the first?

Because acids donate H ions to water,they don't just ping them off. Strictly speaking h2o should be in the first equation also but we sometimes just leave the water out for simplicity. ..maybe we shouldn't


text:
How is latent heat of vapourisation utilized in changing state?
If latent heat of vapourisation is utilized in overcoming intermolecular forces of attraction between the participle of a material then how is it still hidden there? If latent heat is used in overcoming forces of attraction between the particles of a material then why doesn't it increase the kinetic energy of the particles of that material?

It is raising the energy level of the system by breaking the bonds that hold it together as a liquid. Once the bonds are all broken within a certain volume, the liquid will vaporize.
The energy seems to 'disappear' or 'hide' because we expect to see a change in the properties of the chemical as energy is added, but the breaking and reforming of intermolecular bonds is happening constantly within a liquid.
When the liquid has enough energy, random bond-less distributions will occur within the liquid that are large enough to overcome the pressure of the liquid and its container. This is how water boils. If enough of the water molecules become unbound, they will form a bubble. Inside the bubble, the water vapor will be exactly equal to the pressure of it's environment.
This gives us our definition of boiling point: vapor pressure equals atmospheric pressure. The water or other liquid can't stay in the liquid state because there's not enough surrounding pressure to hold it together.


text:
Carius method for quantitative analysis of halogens

Carius method: A known mass of an organic compound is heated with fuming nitric acid in the presence of silver nitrate contained in a hard glass tube known as Carius tube, (Fig.12.17) in a furnace

I want to know :

What is the significance of using $\ce{HNO3}$ ? Can we carry out oxidation with other acids like $\ce{H2SO4}$?
Why $\ce{AgNO3}$ is added to $\ce{HNO3}$ and not to the organic compound? Will it be wrong if I add $\ce{AgNO3}$ directly to the organic compound after it is subjected to fuming nitric acid?
Wouldn't precipitation of compounds other than that of halogens occur since we are carrying the reaction in a closed furnace?

Any help is greatly appreciated.

1) $\ce{HNO3}$ is a more oxidizing acid than $\ce{H2SO4}$.
2) The silver nitrate would not dissolve well in the organic solution.
3) Sulfide is an example of something you might worry about interfering as it forms a very insoluble compound with silver. But silver sulfide and most anything else that might interfere will decompose in the fuming nitric acid.


text:
Heating organic compound in a stream of nitrogen gas
This is related to estimation of percentage of oxygen in an organic compound and my textbook states:

A definite mass of an organic compound is decomposed by heating in a stream of nitrogen gas.

I would like to ask will all oxygen present in the compound evolve as molecular oxygen? Isn't there any possibility that carbon and oxygen of organic compound combine and form CO2 thereby producing error in measurement?

I'd say none of the oxygen present in the compound will evolve as molecular oxygen. Nor does the book say that it will, BTW.

@IvanNeretin    The book says o2 is evolved see the edited question.

Then I believe the book is wrong on that one.

@IvanNeretin   Could you please refer other book which has same procedure? Could you also prove why it's wrong?

The procedure will work all right. See those "other gaseous products". We don't care in which form the oxygen evolves, nor do we have to. It will all be converted to $\ce{CO}$, no matter what.

@IvanNeretin    Will red hot coke reduce all the other oxides like Nitrogen oxides, sulfur oxides, and eventually form CO?

I guess it will.

This is a strangely written text. In a practical setting, the way to get oxygen concentration is by sending an organic sample to a lab for an ultimate analysis.  The ultimate analysis chemistry is explained elsewhere.
The highlighted text seems to describe anoxic pyrolysis of organic matter well above 450°C (for coals this temperature may be higher).  Pyrolysis proceeds as follows.
$$\ce{Organics ->[heat]  syngas + tar + char  }$$
If any molecular oxygen were present in the system these two reactions would very quickly happen.
$$\ce{Organics + O_2 ->[heat]  2CO }$$
$$\ce{2H^+ + O_2 ->[heat]  H_2O }$$
The syngas contains elemental oxygen mostly in the form of $CO$ and $H_2O$, however, other gasses like $CO_2$, $H_2S$, and perhaps even other hydrocarbons may be present.  For the sygas to have most of the elemental oxygen as opposed to the char or tar phase high temperatures exceeding 450°C is important.
If the syngas containing the volatilized matter is passed over a bed of hot coal around 900°C then several things happen.

Complex particles break down (hence the lysis in pyrolysis.)
The gas stream picks up elemental carbon from the bed of hot coal.
$$\ce{CO_2 + C ->[heat]  2CO}$$
$$\ce{H_2O + C ->[heat]  H_2 + CO}$$

By this point, if the temperature is hot enough, all but perhaps an insignificant fraction of the elemental oxygen will convert into $CO$.
Testing for $CO$ is done with the iodine pentoxide test.


text:
H2O2 (Hydrogen Peroxide) catalysis in clear plastic bottle?
I am buying some 30% H2O2 which will be in probably black HDPE bottles. If I want to make some 3% - 5% solution, can I store that in clear bottles? How much faster will it catalyse versus having it in a dark bottle?
Will shelf life be significantly shorter? How much effect does sunlight (or other light) have on the H2O2 solution? Any way to measure this?
Regards,

An easy test for this would be to react H2O2 with yeast or baking soda (sodium bicarbonate) before and after sun bathing. Or you could simply measure gas production in the sun vs in a box for say 24 hours.
An easy way to measure the O2 released is to use a closed vessel with a ballon, but the more scientific method uses an inverted, graduated column.

You can measure light intensity with a photometer.

The 3% - 5% solution is fine with baker's yeast? It won't kill the micro-organism?

Yes it will kill yeast and react quickly (like a 3rd grade valcano). You need enough yeast to completely react with all of the H2O2. I would recommend using a syringe to inject H2O2 through a rubber septum into a distillation flask with an excess amount yeast. Kind of like picture #1 above. Or you could use baking soda. Or you don't have to use anything if you just put clean H2O2 in a closed vessel (with a balloon maybe) in the sun, you will see how much O2 is released during the day. However, balloons leak slowly and water will absorb some gas... so measuring quickly is more accurate.

Ah, ok I understand now.


text:
Why is sodium sulfate neutral while its conjugate acid is weak?
It is commonly taught in chemistry class that the conjugate base of a weak acid (well, at least if it's not weaker than water) is a weak base, as shown by the relation
$$\mathrm{p}K_\mathrm{a} = 14 - \mathrm{p}K_\mathrm{b}  \iff K_\mathrm{a}=\frac{K_\mathrm{w}}{K_\mathrm{b}}$$
where the $K_\mathrm{a}$ is of the conjugate acid, and $K_\mathrm{b}$ of the base. For weak monoprotic acids this means that their conjugate bases are weak, e.g. various carboxylic acids, phenol and hydrofluoric acid. For polyprotic acids, if the conjugate base's next $K_\mathrm{a}$ is smaller than its $K_\mathrm{b}$, then it would also form an alkaline solution but not fully dissociate, e.g. bicarbonate ion (carbonic acid), bisulfide ion (hydrogen sulfide), ammonia (ammonium ion).
However, according to Wikipedia as well as my teacher, sodium sulfate $\ce{Na2SO4}$ is neutral, but Wikipedia also says the acidic hydrogen on the bisulfate ion has $\mathrm{p}K_\mathrm{a} \approx 2$, so sulfate would have $\mathrm{p}K_\mathrm{a} \approx 12$ and a $1\ \mathrm{M}$ solution of sulfate with a perfectly non-acidic counterion would have $\mathrm{pH} \approx 8$. Is sodium sulfate in fact neutral? If so, what would be the reason? Could it be that $\ce{Na+}$ is slightly acidic? Or is the formula for conjugate basicity inapplicable here?

Duplicate (though without answer): http://chemistry.stackexchange.com/questions/36913/why-is-sodium-sulfate-a-neutral-salt

@a-cyclohexane-molecule Hm... Is it then a good idea to close this?

@a-cyclohexane-molecule I mean delete. Should I delete it?

I don't know, considering that the duplicate post has no answer. I would wait for someone more experienced to reply.

Do not use mhchem in question title due to search- ability  issues.

Given $\text{pK}_{a_1} = -3$ and $\text{pK}_{a_2} = 1.9$ for sulphuric acid, the pH for a 1M solution is indeed expected to be $7.9$, $7.4$ for a 0.1M, and $8.0$ for a saturated solution. So it is slightly basic.

So the claim of neutrality is either a very rough approximation or an oversimplification of theory?

It depends on your definition of neutrality. From a theoretical point of view, if you would define neutral to be the state where you have exactly equal concentrations of hydronium and hydroxide ions, you would hardly ever get a neutral solution in practice. On the other hand, it might be OK to call a solution with 6 < pH < 8 neutral for many applications.

Wouldn't other neutral salts like $NaCl$ form pH 7.0 solutions though?

Since $\ce{HCl}$ is a very strong acid I whould expect that it would not affect the concentration of hydronium ions by reaction of water molecules with the chloride anions in a measurable way. However, the ionic strength of a higher concentrated $\ce{NaCl}$ solution would cause a slight deviation from pH 7.

When deviates from 7 it's still neutral right? Well I just thought a difference of 1 on the pH scale was too measurable to still be called neutral, but then again $ [\ce{OH-}] = 1nM$ is indeed pretty neutral...

Yes, according to the principle of electroneutrality it must be neutral.


text:
Calculate the mass of Mg needed to produce 5 kg of vanadium
I need to calculate the mass of magnesium needed to produce $\pu{5 kg}$ of vanadium from the following reaction:
$$\ce{3Mg + 2VCl3 -> 2V + 3MgCl2}$$
given relative atomic masses for $\ce{Mg}$ and $\ce{V}$ as $24$ and $51$, respectively.
I multiplied $24 × 3$ and $51 × 2$ to get
$72 : 102 $
I changed $\pu{5 kg}$ into grams, which is $5000$.
I cross-multiplied to find the answer and then divided the answer by $1000$ to turn it back to kilograms because the question stated it. My answer is $\pu{3.5 kg}$. Is this correct?

You answer is correct.
It is given that for every 3 moles of $\ce{Mg}$ that reacts, 2 moles of $\ce{V}$ are produced. Here's one possible way to proceed towards the answer.
First, find out how many moles of $\ce{V}$ are in a mass of $5000\,\mathrm{g}$:
$$5000\,\mathrm{g}\cdot\left({1\,\mathrm{mol}\,\ce{V}}\over 51\,\mathrm{g}\right) = 98\,\mathrm{mol}\,\ce{V}$$
Second, calculate how many moles of $\ce{Mg}$ will yield $98$ moles of $\ce{V}$ by multiplying the ratio of the stoichiometric coefficients (3:2) by the number from step 1:
$$\left({3\,\mathrm{mol}\,\ce{Mg}\over 2\,\mathrm{mol}\,\ce{V}}\right)\cdot 98\,\mathrm{mol}\,\ce{V} = 147\,\mathrm{mol}\,\ce{Mg}$$
Finally, calculate the mass of $\ce{Mg}$ by multiplying the number of moles from step 2 by the atomic mass:
$$147\,\mathrm{mol}\,\ce{Mg}\cdot\left({24\,\mathrm{g}\over 1\,\mathrm{mol}\,\ce{Mg}}\right) = 3528\,\mathrm{g} \approx 3.5\,\mathrm{kg}$$


text:
Is beryllium difluoride covalent or ionic?
My textbook  says that despite the large electronegativity difference $\ce{BeF2}$ is covalent since the beryllium ion will have too much charge density and it will attract the fluorine electron cloud and therefore forms polar covalent bonds.
I looked it up on Wikipedia and found out that it has a crystal lattice,  is soluble in water, and has a quite high melting point.
What else does one need to call a compound ionic? Am I or the textbook right?

I am not putting this as an answer but bond between 2 different atoms is usually not completely covalent nor completely ionic: it is a mixture of them. We prefer to talk about the "covalent" or the "ionic" character of a bond.

And BeF2 probably has more ionic then covalent character, not that it matters much.

Nearly anything has a crystal lattice. Many covalent compounds are soluble in water (and quite a few ionic ones are insoluble). As for the melting point, well, it's moderately high, so consider $\ce{BeF2}$ moderately ionic and stop pursuing this false dichotomy. Also, I suggest to close this question as a duplicate of AlCl3: ionic and covalent?, because though the title compounds are different, just about every argument applies equally well to both.

related http://chemistry.stackexchange.com/questions/17064/why-is-kf-the-most-ionic-compound

@IvanNeretin "All bonds to fluorine, with the exception N-F, O-F,
and F-F, have considerable ionic character, so the vast majority
of AFn molecules are more appropriately described in terms of
an ionic model rather than a covalent model." http://pubs.acs.org/doi/abs/10.1021/ic961315b BeF2 is much more extremely ionic than AlCl3 , with Be having a +1.8 charge.  The AlCl3 answer does not apply well to BeF2.

"Is beryllium difluoride covalent or ionic?"  Yes.

There is a very relevant article by Gillespie Covalent and Ionic Molecules: Why Are BeF2 and AlF3 High Melting Point Solids whereas BF3 and SiF4 Are Gases? J. Chem. Educ., 1998, 75 (7), p 923.
According to the article, the charge on Be is +1.81 and the charges on the Fs are -0.91. (citing to his earlier article Reinterpretation of the Lengths of Bonds to Fluorine in Terms of an Almost Ionic Model Inorg. Chem., 1997, 36 (14), pp 3022–3030)
So while nothing is completely covalent or ionic, BeF2 is extremely toward ionic.

Thing is they used specifically "Almost Ionic Model" and therefore got what wanted. This is a reinterpretation, so I'm pretty sure other calculations shown much less charge separation.

@Mithoron "We have calculated the wave function and the electron density
distribution ... The
calculations were done using the Gaussian 94 package at the HF/6-
311++G(2d,2p) with 6d functions level and at the B3LYP/6-311++G-
(2d,2p) level. ...  The analysis of the electron density
distributions was carried out using the AIMPAC suite of programs."

Well, they sure didn't pull it out of a sleeve, but different methods give different results, and this case may be an outlier. For example Geoff's answer here has lower % even for LIF. So, I wonder if other calculations also give that much charge separation...

According to my chemistry book (Pearson Baccalaureate IB) beryllium forms covalent bonds due to it being one of the expectations to the intramolecular bonding.
Furthermore, according to the book the molecule should be non-polar because all the lone pairs are bonded and thus not exerting a repulsion and altering bond angles.

I think you are confusing polar bonds with dipole moment. Also, I doubt that any element will always form a covalent bond. Maybe you meant that beryllium forms covalent bonds with itself?

I didn't say always , furthermore no, I'm not confusing those two together

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


text:
Mechanisms involved in iodination of salicylamide protocol
In the laboratory, I performed an eletrophilic aromatic subsitution by reacting $\ce{I+}$ with salicylamide. The mechanism for this reaction is pretty straightforward, but it's the mechanisms for the supplementary reactions used in this lab that I can't seem work out.

In order to prepare the electrophile  $\ce{I+}$, sodium iodide was reacted with sodium hypochlorite:
$$\ce{ ClO- (aq) + I- (aq) + 2H3O+(aq) -> I+ (s) + Cl- (aq) + 3H2O (l)}$$
The equation (I think) is right, but can anyone provide some mechanistic insight or literature on how the hypochlorite manages to oxidize the Iodide?
Does it first have to go into an iodine ($\ce{I2}$) form? The reason I propose this is because as the hypochlorite was added to the salicylamide/sodium iodide absolute ethanol solution, color change was observed from dark red-brown to increasingly lighter shades of yellow.
It seems a similar question was asked about a year ago, but I was hoping to take it a step further.
Once the reaction had finished, a 10% (w/v) sodium thiosulfate solution was added. I assume this was because thiosulfate is a well known bleach (hypochlorite) neutralization agent: $$\ce{HOCl + 2 S2O3^2- -> Cl- + S4O6^2- + OH-}$$
Assuming the above equation is correct, how exactly (mechanistically) does thiosulfate neutralize hypochlorite/hypochlorous acid? The $\ce{S-S}$ bond of the tetrathionate is whats tripping me up.

The order of events is probably:

Protonation of the hypochlorite ion

$$\ce{ClO- + H3O+ <=> ClOH + H2O}$$

Nucleophilic displacement of chloride by iodide

$$\ce{I- + ClOH \phantom{0}<=> Cl- + IOH}$$

Protonation of hypoiodous acid

$$\phantom{000}\ce{IOH + H3O+ <=> IOH2+ + H2O}$$

Fragmentation of the $\ce{IOH2+}$ cation

$$\phantom{00000}\ce{IOH2+ <=> I+ + H2O}$$
Alternatively, the second step could have been protonation of hyochlorous acid, followed by displacement of water by iodide, and then fragmentation of iodine monochloride:
$$
\begin{align}
\ce{ClOH + H3O+ &<=> ClOH2+ + H2O}\\
\ce{I- + ClOH2+ &<=> ICl + H2O}\\
\ce{ICl &<=> I+ + Cl-}
\end{align}$$

Great answer, but I'm not sure if it successfully accounts for the brown and purple color observed which would mean there was triiodide ion or elemental iodide present and not just I- (which is colorless).

Reaction of $\ce{I-}$ with $\ce{ClO-}$ is a redox reaction.
$$
\begin{align}
\ce{ClO-(aq) + H2O(l) + 2e- &<=> Cl-(aq) + 2OH-(aq)} &&E^\circ = \pu{0.90 V}\\
\ce{2I-(aq) &<=> I2 (s) + 2e-} &&E^\circ = \pu{-0.54 V}\\
\ce{ClO-(aq) + H2O(l) + 2I- (aq) &-> Cl-(aq) + 2OH-(aq) + I2(s)}&&E^\circ_{\text{rxn}} = \pu{0.36 V}
\end{align}$$
The reaction is spontaneous because $E^\circ_{\text{rxn}}$ is positive value. This produces $\ce{I2}$ in situ in amounts large enough to show its brown color in solutions ($\ce{I2}$ is purplish in solid state). In low concentrations it is yellowish. It is important to note that, this reaction did not use acid, thus acid is not used here for the mechanism.
The alicylamide nucleus is already activated by phenolic $\ce{-OH}$ group so that it does not need catalyst to iodonize in the presence of $\ce{I2}$. This reaction leaves $\ce{HI}$ as a byproduct, which react with $\ce{NaOH}$ in the solution. Once iodination reaction is completed the solution is still basic.
The final solution should contain both $\ce{I2}$ and $\ce{ClO-}$, if alicylamide is the limiting reagent. Thus, following redox reactions would happen when aqueous $\ce{S2O3^2-}$ is added to the reaction mixture.
$$
\begin{align}
\ce{ClO-(aq) + H2O(l) + 2e- &<=> Cl-(aq) + 2OH-(aq)} && E^\circ = \pu{0.90 V}\\
\ce{2S2O3^2-(aq) &<=> S4O6^2-(aq) + 2e-} && E^\circ = \pu{-0.09 V}\\
\ce{ClO-(aq) + H2O(l) + 2S2O3^2-(aq) &-> Cl-(aq) + 2OH-(aq) + S4O6^2-(aq)} &&E^\circ_{\text{rxn}} = \pu{0.81 V}
\end{align}
$$
and,
$$
\begin{align}
\ce{I2 (s) + 2e- &<=> 2I-(aq)} &&E^\circ = \pu{0.54 V}\\
\ce{2S2O3^2-(aq) &<=> S4O6^2-(aq) + 2e-}&&E^\circ = \pu{-0.09 V}\\
\ce{I2(s) + 2S2O3^2-(aq) &-> 2I-(aq) + S4O6^2-(aq)}&&E^\circ_{\text{rxn}} = \pu{0.45 V}
\end{align}$$
Both reactions are spontaneous because $E^\circ_{\text{rxn}}$ values for both are positive. Thus, medium is still basic and need acid ($10\%$ $\ce {HCl}$) to neutralize.


text:
Weird C-N bond in caffeine
I saw this picture on a teacher's wall and I noticed the C single bonded to the N on the right side. How is it possible that carbon has four bonding sites but is only using one in the picture? Is this stable?

That's impossible, of course. It is just another $\ce{CH3}$ that did not fit.

Compare to Wikipedia: it should be a CH₃. Also, the five-member (imidazole) ring is missing a double bond. In addition to being badly drawn.

Fun fact: The pentagon that is drawn (regardless of whether we consider it poorly drawn) tessellates unlike a 'well-drawn' imidazole ring.


text:
Why isn't the melting point of H2O very high, if the intermolecular forces are strong?
H2O has very strong intermolecular forces due to the hydrogen bonds that a formed within the compound. Usually this would mean the compound has a very high melting point as a large amount of heat energy is required to overcome the forces, however H2O has a melting point of only O degrees. How come it is not a lot higher?

Because intermolecular forces in general are weak... Water has very high melting point for three atom molecule, with only light atoms.

0 degrees is already very high for such a small molecule. Most molecules of similar mass are gases at 0 degrees.

Melting points of the group 16 hydrides:

Water: 0 °C
Hydrogen sulfide: −82 °C
Hydrogen selenide: −65.73 °C
Hydrogen telluride: −49 °C
Polonium hydride: −35.3 °C

In theory, as the molecules grow larger (moving down in the group), their melting poits become higher too. Water is a clear outlier due to hydrogen bonding. Its melting point is already way higher than what should be expected looking at similar molecules.


text:
What was wrong with Schroedinger's interpretation of the wavefunction?
I've read in a couple of different places, and was told when first learning about quantum mechanics, that Schroedinger interpreted the wavefunction as being representative of the charge distribution of the electrons in a system. This idea was later replaced by Max Born's interpretation of the wavefunction as the probability density of the electrons in a system which is the common view now.
It's quite apparent that quantum mechanics is built on this idea of probabilistic interpretations, and so without it QM would have problems. It's not obvious to me, however, how one makes the jump from interpreting the wavefunction as a charge density to a probability density. Or maybe a better question to ask is this: why did Schroedinger interpret the wavefunction as a charge density? Specifically, what physical motivation is there for this interpretation? Furthermore, in what situations does this interpretation fail? I could imagine some problems with Pauli Exclusion, but perhaps not. In that case, one would still give as an axiom that the wavefunctions of fermions are anti-symmetric upon particle exchange and then one would have that there could be no charge density in that region of space, and everything might still be fine...
So, to narrow things, in what situations does the interpretation of the wavefunction as a charge density fail?

Could you quote the mentioned "couple of different places"? I simply don't believe that Schrödinger interpreted $|\Psi|^2$ as the charge density. I mean, it is quite apparent from the very Schrödinger's work that, in general, for a many-variable wavefunction $\Psi$, $|\Psi|^2$ could not be the charge density. Such interpretation will work only for $\Psi$ that describes a single charged particle.

Found this similar question on Physics.SE, the first answer to which expresses the same concern.

I'll provide links later but at the very least there's a Wikipedia page on interpretations of quantum mechanics which says this under the history section. I've heard it a number of times by word of mouth and there's even a simple dimensional argument which refutes that interpretation. It seems Schrödinger later changed his interpretation of the wavefunction, but if you know of a paper when he did this I'd be interested to read that.


text:
Where does the extra electron come from in the Lewis dot structure of OH-?
Hydrogen has one valence electron and oxygen has 6 valence electrons. However, the dot structure shows 3 lone pairs of oxygen and 1 shares with hydrogen.How is that possible?

The extra electron comes from that "-" in "OH-". If we just took two neutral atoms, O and H, they wouldn't form $\ce{OH-}$, but a neutral particle instead.

@Mithoron However, the answer to the question you refer does not answer my question nor the question you refer. The answer just mentions the algorithm of Lewis dot structure. And that is not my doubt. Reliable theory needs reliable observations.

I think OH- should not be seen independently nor does it exist independently. The extra electron should come from some other atom, which becomes cation. However, I'm not sure whether my understanding is correct.


text:
What type of isomers are the following two compounds?
What type of isomers are the following two compounds?They do not seem to be geometrical isomers as the methyl groups at terminal position are not in same plane as such.Are they conformers?

Believe it or not, they are enantiomers. Oh, and you don't normally draw them like that, because like you said, those methyls are out of plane.

Could you explain a bit more elaborately if you have time?Thanks. @IvanNeretin

@IvanNeretin I did'nt draw them :-P!

Then these pictures are meaningless and don't represent anything. Or, if we put it another way, they are poorly drawn pictures of the same compound (like in @julien's answer), which, BTW, exists as a pair of enantiomers, but the pictures don't tell you which is which.

@IvanNeretin Oh well I think I understood why they are enantiomers!

Related: Why are allenes chiral? and How are the Cahn-Ingold-Prelog rules applied to chiral molecules without a stereogenic carbon center? i.e biphenyls and allenes

It is easier to visualise if you draw, for example,  the methyl and the hydrogen on the left  in the plane of the paper, then the methyl and hydrogen on the right, are at right angles to this and so in and out of the plane.

@Sanc Are you sure, though, that those questions are unrelated? Perhaps they're very relevant and you just stopped someone from helping you.

@TIPS I'm sorry.


text:
Number of geometrical isomers when a cyclic ring is involved along with axis of symmetry

This is an question from my Chemistry study material and it asks to find the number of Geometrical Isomers of the following compound in the picture.

Basically the method/formula given in my book to find number of Geometrical
  Isomers of a compound is:

Using this formula I put $n=3$.Morever since both ends are same I get $p=(3+1)/2=2$.So final answer according to me is $2^{3-1}+2^{2-1}=4+2=6$.
But the answer given is $2^3=8$.
Where did I go wrong?
Related: How to derive these general formulae for number of stereoisomers of a compound with a possible plane of symmetry?

Didn't read the formula, but 6 looks right to me. See, the ends are either both trans, or both cis, or cis and trans (no matter which is which, since they are otherwise identical), which makes 3 options, and on top of that they are either on the same side of the cycle or not, which multiplies it by 2.

Yeah right.Even I thought the same.In all probability the given answer is wrong.BTW i wonder why don't you write your comments as answers.They are elaborate enough to be an answer! :-) @IvanNeretin

Besides, I was wrong on this one, as explained by vapid. Indeed, if we consider enantiomers, we have 8 all right.

@IvanNeretin I think he is wrong.He considered enantiomers which are optical isomers and not geometrical.

Are they? I never was able to remember correctly. What are geometrical isomers, then? Everything with similar connectivity, yet different, excluding optical isomers?

Yes you are right.See this http://www.avogadro.co.uk/organic/isomer.gif @IvanNeretin

Wait, I got lost in this thread, so deleting my previous comment. The issue is that this molecule is symmetric, so you lose a factor of two. Consider the case of 2,4,6-octatetraene. There's 3 double bonds, but there are only 6 isomers because the two ends are the same. If you consider 2,4,6-nonatetraene, then you'll have the full 8 isomers.

The answer in your textbook is correct. Here are all possible isomers:

I added a symmetry axis to the right, so you can see that these compounds are mirror images and are not superimposable. The formula you used for calculations is too complicated. The way to calculate it is simple: there are four stereogenic centers so: $$2^4=16$$There is one axis of symmetry so:$$16/2=8$$Have more faith in your books:)
EDIT:
Because my answer has raised an interesting discussion, I will try to present my point of view. First of all, the cis/trans isomerism is a descriptive way to represent a stereochemistry of a molecule, but its scope is limited. It is useful for teaching organic chemistry, but not for a systematic nomenclature (especially for a ring chirality). I think that the main source of confusion here is understanding the difference between a local and global (absolute) stereochemistry. In the above picture the controversy is whether the four enantiomers in the two rightmost columns can be treated as geometrical isomers. Let's simplify the problem and consider 1,2-dimethylcyclopentane:

The situation is clear for the first two isomers (a and b) - these are geometrical isomers, no doubts about it. What about b and c? We see that in both isomers the methyl groups are in the trans orientation, so one may conclude that they are not geometrical isomers. Period. However, cis/trans notation describes only the local geometry (E/Z notation obviously cannot be used in this example). Isomers b and c have the same relative orientation of methyl groups, but in a global scope these are two different molecules, and thus two different geometrical isomers. Therefore, 1,2-dimethylcyclopentane has one cis and two trans isomers. This may be counterintuitive at first, because there are no such situations in case of double bonds, where the cis/trans notation is more frequently used, but in rings we have an additional layer of complexity. I am aware that some people may disagree with this explanation, but as far as I know there are no official (i.e. IUPAC) directives to resolve this controversy.

You considered optical isomers(enantiomers) but the question asks for geometrical isomers only.

@Sanchayan Dutta So are you saying that the two compounds in the first column are not geometrical isomers? That makes no sense.

Tell me you definition of geometrical isomers and optical isomers first.

This is not a case of different definitions of geometrical and optical isomerism - we all agree in this matter. So if you exclude optical isomerism there are only 3 geometrical isomers - don't you agree? Still, makes no sense.

No there are 6(exclude the enantiomers on the right) as the ring too shows Geometrical isomerism.

@Sanchayan Dutta Now explain me what is the difference between geometrical isomerism in the ring and optical isomerism.

No difference.However the rightmost two cannot be considered GI in any way as they have been taken before.

@SanchayanDutta I disagree. The author of your textbook also seems to disagree. I am stopping this discussion and letting others voice their opinion.

You are both right and both wrong in the sense that there is no accepted definition of geometric isomers. It seems that IUPAC discourages its usage . @vapid maybe you can edit this in your answer or make more research on it first

@K_P I agree with you that the problem is controversial. I will edit the question in the near future to show my point of view, today I am a bit busy.

@K_P The link is broken


text:
What's the difference between pipettes calibrated "TD", "TC", "Blow-out" and related terms?
I've never blown-out the last drop of any glass pipette I've ever used during my undergraduate studies. Recently I re-read some notes I had lying around in order to prepare a class I had to give to a group of first year undergraduates and found out that there is a class of glass pipettes which need to be "blown-out" in order to deliver the nominal volume correctly.
I researched a little bit and found what seems to me to be completely inconsistent terminology and conventions to refer to this class of pipettes.
In https://en.wikipedia.org/wiki/Graduated_pipette it is clearly stated that there are "two kinds" of graduated pipettes. "serologic/blow-out" and "Mohr/drain-out". It is also stated that serologic pipettes have calibration marks all the way to the tip, whereas Mohr pipettes do not. The same page states that pipettes are either calibrated "to deliver" (TD) or "to contain" (TC). It states that "TD" is a far more common calibration than "TC" and explains that pipettes calibrated TC need to be blown-out in order to deliver the expected volume correctly, one would assume from this that TD is a synonym fro "blow-out". Judging from this there would seem to be two kinds of pipettes, "serologic/blow-out/TD" and "Mohr/drain-out/TC". This really contradicts my experience, I've been using pipettes with marks going all the way to the tip and never been told to blow out the last drop, however, there is always the chance I've been doing it wrong this whole time.
Digging a little bit deeper I found that this document http://www.brand.de/fileadmin/user/pdf/Information_Vol/Brochuere_Volumenmessung_EN.pdf (link downloads pdf) that wikipedia lists as a source doesn't mention "blow-out" pipettes at all, but does mention pipettes calibrated "TC". In page 13 there are some pictures and descriptions of pipettes resembling what one would think are "Mohr" and "Serologic" types, and both are said to be calibrated "to deliver" actually, reading this text one would think that ALL graduated and volumetric pipettes are calibrated to deliver (however, this document is from a lab equipment manufacturer, perhaps they only manufacture TD pipettes...?).
Digging yet deeper http://faculty.clinton.edu/faculty/Mike.Lawliss/My_webpage/Sci110/Other%20documents/Pipetting%20A%20Practical%20Guide_Guzman2001.pdf is another article cited as a source by wikipedia. Here the author pretty much identifies serologic pipettes with blow-out ones, but also states that blow-out pipettes "usually" have double rings (non-colored) near the top. There is, however, a drawing of a "serologic" pipette WITHOUT rings near the end of the article, close to an "Ostwald-Folin" pipette WITH rings which the author specifically identifies as blow-out. The definition of the term "serologic" pipette is, according to the author, "a type of measuring pipette used to measure variable volumes" which in my view would make ALL graduated pipettes serologic.
To add to all this confusion Sigma-Aldrich seems to manufacture "TD, blow-out, serologic pippetes" http://www.sigmaaldrich.com/catalog/product/sigma/z325422?lang=es&region=AR and also "TD/TC volumetric pippetes", which makes me lose all hope of making sense of any of the terms listed in the question at all.
In short, is there a STANDARD way in which these terms (TD, TC, IN, EX, Blow-out, Serologic, etc) are used? Is there a standard definition regulated by some sort of organization, or at least generally agreed upon? If you think that there is, could you please provide trustworthy sources? Thanks a lot.
This probably seems like a silly question (it is =D) but it kind of bugs me not to know something so basic as "what are the different kinds of graduated pipettes?"... It seemed like something that would be quite easy to settle when I started looking into it... guess I was wrong.

It's not a silly question if you need very precise measurements. Use the equipment the wrong way, and you've just introduced a (comparatively) massive systematic error into your procedure.

Since you are explicitly asking for standards, you might want to start with ISO 4787 Laboratory glassware – Volumetric instruments – Methods for testing of capacity and for use. This international standard provides methods for testing as well as for use of various volumetric instruments made from glass. Further requirements for individual volumetric instruments can be found in

ISO 648 Laboratory glassware – Single-volume pipettes
ISO 835 Laboratory glassware – Graduated pipettes
ISO 385 Laboratory glassware – Burettes
ISO 1042 Laboratory glassware – One-mark volumetric flasks
ISO 4788:2005 Laboratory glassware – Graduated measuring cylinders

Most commercially available laboratory glassware is in accordance with these standards. However, some nonconforming volumetric instruments are also available, for example in non-standard sizes.
Due to retention of liquid on the inner surface of the volumetric instrument, the volume of liquid delivered is not identical with the volume of liquid contained by the volumetric instrument. Volumetric instruments are either adjusted ‘to contain’ (i.e. the capacity printed on the instrument corresponds to the contained quantity of liquid) or ‘to deliver’ (i.e. the capacity printed on the instrument corresponds to the delivered quantity of liquid; the wetting residue remaining in the instrument has already been taken into account in the calibration). According to international standards, the marking ‘Ex’ is used to indicate that the volumetric instrument has been adjusted to deliver; the marking ‘In’ is used to indicate that the volumetric instrument has been adjusted to contain. Most typical graduated pipettes, bulb pipettes, and burettes are usually adjusted to deliver, whereas typical graduated cylinders, volumetric flasks, and capillary pipettes (i.e. very small measuring pipettes with a capacity of up to 0.2 ml) are usually adjusted to contain.
ISO 648 and ISO 835 specify three classes of accuracy for pipettes:

Class A (higher grade, without waiting time)
Class AS (higher grade, with a specified waiting time)
Class B (lower grade, without waiting time)

‘Waiting time’ is the period to be observed in order to ensure complete delivery after apparent completion of the liquid delivery of the pipette (i.e. after the meniscus appeared to come to rest) and before the final reading of the delivered volume is taken (i.e. before the tip of the jet is removed from the inner surface of the receiving vessel).
Furthermore, four types of graduated measuring pipettes are standardized in ISO 835:

Type 1: Partial delivery (adjusted to deliver from zero line at the top to any graduation line; nominal capacity is represented by the lowest graduation line)
Type 2: Total delivery (adjusted to deliver from any graduation line down to the jet; nominal capacity is represented by the highest graduation line)
Type 3: Total and partial delivery (adjusted to deliver from zero line at the top to any graduation line; nominal capacity is obtained by delivery down to the jet)
Type 4: Blow-out (adjusted to deliver from zero line at the top to any graduation line; nominal capacity is obtained by delivery down to the jet where the last drop of liquid in the jet is expelled by blowing. Blow-out pipettes are only adjusted to accuracy Class B.)

Typical graduated measuring pipettes with a capacity from 0.5 ml to 25 ml are commercially available in accordance with ISO 835 as Type 1, 2, or 3 and in Class A, AS, or B. Bulb pipettes with a capacity from 0.5 ml to 100 ml are available in accordance with ISO 648 in Class A, AS, or B. They are all adjusted to deliver (‘Ex’); i.e. the residual liquid still left in the tip has already been taken into account during calibration and must not be expelled into the vessel by blowing out.
Very small measuring pipettes with a capacity of up to 0.2 ml (capillary pipettes) are designed as Type 4 (blow-out) in Class B; however, most available capillary pipettes are actually adjusted to contain (‘In’), which is strictly not conforming with ISO 835. Though, the formal difference between blow-out pipettes adjusted to deliver and pipettes adjusted to contain may be small for total delivery since the last drop of retained liquid in the jet can be expelled by blowing in both cases. Nevertheless, when the greatest possible accuracy is desired, it is not sufficient to simply empty pipettes that are adjusted to contain by blowing out the liquid with a pipetting aid; such pipettes should also be rinsed two to three times with the diluting medium.
The general procedure for the use of pipettes that are adjusted to deliver (‘Ex’) is described in ISO 4787 as follows. However, when the greatest possible accuracy is desired, pipettes should be used as closely as possible to the manner in which they have been calibrated (which is also specified in ISO 4787, ISO 648, and ISO 648).

10.5.1 Pipettes adjusted to deliver (see ISO 648 and ISO 835)
After rinsing with the liquid or reagent to be used, fill the pipette by suction to a few millimetres above the selected graduation line. Remove any liquid remaining on the outside of the jet. The final setting of the meniscus shall then be made by dispensing the surplus liquid through the jet. Remove any drops of liquid adhering to the jet by bringing an inclined ground glass vessel into contact with the tip of the jet. Delivery shall then be made with the tip of the jet in contact with the inner surface of the inclined receiving vessel.
If the setting after delivery is done at a lower graduation line, the liquid flow has to be nearly stopped a few millimetres above the graduation line. After observing a waiting time, if specified, complete the final setting quickly.
A waiting time, if specified, shall be observed before making the final setting for delivery of a given volume.

The corresponding procedure for the use of pipettes that are adjusted to contain (‘In’) is described in ISO 4787 as follows. Note that, when the greatest possible accuracy is desired, it is not sufficient to simply empty the pipette by blowing out the liquid with a pipetting aid; the pipette should also be rinsed two to three times with the diluting medium.

10.5.2 Pipettes adjusted to contain
Rinse the pipette with the reagent to be used to a few millimetres below the desired graduation line. Fill the pipette by suction to as close as possible above the selected graduation line. Remove any liquid remaining on the outside of the jet. Make the final setting of the meniscus to the line by withdrawing the surplus liquid by means of filter paper. For the discharge, rinse the pipette several times with the diluting medium.

As you noted, "pipettes are either calibrated 'to deliver' (TD) or 'to contain' (TC)."
Because a bit of solution typically remains in pipettes, TC pipettes must be blown out.
Serological pipettes can be both TC and TD because they are graduated. So for a 10 ml pipette graduate to 0.1 ml, then the pipette will deliver TD from say between the 9.80 ml and 5.30 marks. That would be of course 4.50 ml. However to get all of the last 0.10 ml of liquid, then you'd have to blow out the pipette.

What is your definition of a serological pipette? Can you provide a source that clearly defines what a "serological pipette" means? http://www.brand.de/fileadmin/user/pdf/Information_Vol/Brochuere_Volumenmessung_EN.pdf  (link downloads a pdf) that source, for instance, has, in page 13, pictures of what I would consider serological pipettes and explicitly states that these should NOT be blown out.

Ignacio - I doubt there is a IUPAC definition for a "serological pipette." To me such a pipette is simply one that is graduated rather than one have a single fill mark. Such graduations would also require a constant diameter. On the other hand a "volumetric pipette" would typical have a large bulb. Assuming a 0.1 ml graduation, then it would seem that the last 0.1 ml of a serological pipette could either be TC or TD.

In olden days, (I was taught to both suck up and blow out a pipet by MOUTH), TD and TC about covered it. Now days, we're talking about micro-liters and automatic pipets and use for GC/MS calibration, and all sorts of more technical more SPECIALIZED uses. Wouldn't you expect that more specialized uses require more specific equipment? Temperature, gravitational field, density, surface tension, humidity, viscosity...those are the factors that can lead to variations in the amount of liquid left inside the pipet at the end of the delivery process. (I'm not saying all of them are important at sea level on Earth). FWIW, different needs will need different things. I think a good overview would be to discuss TD/TC and then explain one specific high accuracy micropipette. FWIW.

I certainly agree with you that specific, niche needs require highly specialized equipment, but I don't think that was what I was getting at with the question. Could you provide a definition of what it means for a pipette to be "TD" or "TC"? Do you know if any international or national standards organization regulates how these terms are used?


text:
Total number of carbon atoms in buckminsterfullerene
From what I read, buckminsterfullerene is composed of 60 carbon atoms, $\ce{C_{60}}$ and it is made up of 20 hexagons and 12 pentagons. So, I decided to do a back of the hand calculation with the knowledge that each of these were tri-atomic junctions. Since one hexagonal tri-atomic site would comprise a $\ce{C-C}$ bond angle of $\frac{2\pi}{3}$, thus just $\frac{1}{3}$rd of an atom and hence two atoms in a hexagon. Since there are 20 hexagons, this gives me 40 atoms. Now the $\ce{C-C}$ bond angle at the pentagon site is $\frac{3\pi}{5}$ which makes it $\frac{3}{10}$ atoms per site and hence $\frac{3}{2}$ atoms per pentagon. With 12 pentagons, it results in 18 atoms. So the total number of atoms according to this calculation amounts to 40+18=58. Where did I go wrong?

Now that's a nice trick indeed. Think of it this way: forget the angles, just count the atoms, as if there were six in each hexagon and five in each pentagon. That would make 180. But wait, we've just counted each atom more than once. Like you said, each atom is on the junction of three polygons, hence it must have been counted three times, so ${180\over3}=60$.

@IvanNeretin but, the atoms for the corners of the pentagons cannot be counted as $\frac{1}{3}$rd since those atoms are just $\frac{3}{10}$ths. If we do that then we get $206\frac{1}{3} + 125\frac{3}{10}$ equalling 58. What you're telling me to do is $206\frac{1}{3} + 125\frac{1}{3}$ equalling 60, but I can't assume a $\frac{1}{3}$rd for the atoms within the pentagon or maybe I think I can't and I should where in the problem becomes, why should I have to consider a fraction of $\frac{1}{3}$ for the atoms in the pentagons as well and not $\frac{3}{10}$ ?

Because if you count them as $3\over10$, then each atom (which, BTW, is a junction of two hexagons and a pentagon) is counted as ${1\over3}+{1\over3}+{3\over10}={29\over30}$, rather than a full solid $\mathbf1$.

Yes, but because the fullerenes are spherical balls and not graphene sheets, hence there needs to be some pyramidization, which is why the sums of the fractions need not necessarily be equal to one. Think of it as the apex angles of every triangle of each of the faces of the pyramid of Giza, the sum of the apex angles need not add up to $2\pi$ but if I flatten the pyramid out into a mesh of 4 2D triangles, then the apex angles would need to add up to $2\pi$, similarly the fractions of the out of plane pyramidized figure need not add up to 1.

Surely the angles need not add up to $2\pi$. The fractions, however, do need to add up to 1. This is a certain place in the molecule; it is occupied by one atom. You can't split atoms.

Let's try counting the vertices of a cube this way. Each face has 4 corners with an angle of $\frac\pi2$, which is $\frac14$ of a full angle, so 1 vertex per face. There are 6 faces so a total of 6 vertices, but obviously a cube has 8 vertices... clearly this method doesn't work correctly.

@IvanNeretin   Yes the problem is that I was treating these atoms as if they were planar and was hence landing up with a situation where I was defining the fraction as $\frac{\alpha}{2\pi}$ and at every pyramidized junction $\frac{\alpha}{2\pi}+\frac{\beta}{2\pi}+\frac{\gamma}{2\pi} < 1$ and hence I would never have a full atom counted. The sum of errors I guess added up to 2 atoms. Still, thank you very much for being patient with me and for clearing my doubt about the subject.

@f'' Yes, I realized a second folly in my approach. There will be parts of an atom not within any closed structure and I will miss out on them too. The approach only works if all my atoms are enclosed in a circuital structure and if all the circuital structures are planar. Thank you very much for your answer. It was a great help.

Fun fact, Descartes' theorem of angular defect states that if you do this on any polyhedron without holes, you'll always be off by 2.


text:
Hydrogen bonding relating to molecular orbital theory
In [1, p. 232] here is an explanation of how alcohols hydrogen bond to each other and I do understand how the alcohols form hydrogen bonds to each other:

FIGURE 6.15 Water and alcohols are both Brønsted acids and Brønsted
bases. This figure shows hydrogen bonding between molecules of water $(\ce{R=H})$ and alcohols.

Is it possible to think of hydrogen bonding as a Lewis Acid-Base reaction rather than just a Brønsted Acid-Base reaction, and, by extension, think of this whole scheme as a HOMO-LUMO interaction?
Like, can I think of hydrogen bonding as a HOMO-LUMO interaction in general, arising from the idea that it can be classified as a Lewis Acid-Base reaction?
References

Jones, M.; Gingrich, H. L.; Fleming, S. A. Organic Chemistry, 4th ed.; W.W. Norton: New York, 2010. ISBN 978-0-393-93149-5.

Yes you can think of hydrogen bonding in terms of MO theory. Consider a hydrogen bond between two water molecules as a simple example.

The donor hydrogen donates electron density into the $\pi^*$ orbital of oxygen which is the LUMO of the acceptor oxygen. This thus weakens the donor $\ce{O-H}$ bond, but shortens the distance between the donor hydrogen and acceptor oxygen.
Thus, there are two components to be considered: the pure electrostatic attraction and the stabilization due to covalent interactions.
Because all you're asking about is the MO side of things, as explained above, hydrogen bonding can be viewed as donation of electrons into the $\pi*$ orbital of some acceptor atom. This could indeed be called a Lewis acid-base interaction, but I don't hear this very often.
I'm not sure what your textbook is talking about with hydrogen bonding being a Brønsted-Lowry reaction. Hydrogen bonding is not a reaction (in the traditional sense) so I would disagree with your textbook. The acceptor oxygens are accepting electrons much more than they are accepting the whole proton which would make them Lewis acids. It's certainly possible for water/alcohols to behave as either Lewis acids or bases and Brønsted-Lowry acids or bases, but the former is much more common based on the fact that auto-ionization is quite a small effect but hydrogen-bonding is a very large (and common) one.
Regarding the MO view of things, the exact same logic can be applied in inorganic chemistry when considering carbonyl compounds or dihydrogen/dihydride compounds. In that case the donation of electrons into a $\pi*$ orbital (or $\sigma*$ orbital in the case of dihydrogen compounds) is internal to the molecule however and we call it backbonding.
Here's an MO and orbital diagram showing that interaction.

Oh but my textbook isn't exactly saying that it's a reaction. I think it's just saying that hydrogen bonding can be thought of as a Bronsted Lowry interaction, though that might be misleading to a certain extent.

That sounds rather unlikely. The hydrogen atom is the electron deficient atom. How could it be the electron donor? I would think that it is more likely that the oxygen lone pair orbital is the HOMO which interacts with the O-H sigma* LUMO

If you want to think of the hydrogen bond in valence bond terms then it is possible to write down five terms that contribute to the structure$^*$.  $\ce{A-H..A}$ , covalent A-H bond  $\ce{A^- -H^+ ..A}$ and $\ce{A^+ -H^- ..A}$ ionic A-H bond  $\ce{A^- -H ..A^+}$ and $\ce{A^+ -H ..A^-}$ charge transfer A..A bond.
The ionic (electrostatic) contribution accounts for approx 65% of the bonding, with charge transfer more important for shorter bonds (<0.28nm) and ionic contribution for longer bonds.
$^*$ Jeffrey, 'An introduction to hydrogen bonding'

How does that answer my question? I'm asking for how i could think of it in terms of MO theory and whether the thinking above is legitimate i.e. Whether thinking of hydrogen bonding as a HOMO-LUMO interaction is correct

Rather than consider the dots in your picture, (which actually don't mean much) using the terms above means that you can say what electronic structures you can expect to contribute to  a hydrogen bond.  The alternative is to use a molecular orbital approach then there are terms such as electron repulsion, charge transfer, dispersion & polarisation contributing. Take you pick :)

I should add that using mo theory, as used to build up, say, $\ce{O2}$, or benzene is difficult to apply when there is a 3 centre bond such as O..H-O.

In fact, yes.
It can be explained by a HOMO-LUMO interaction.
I am currently running NBO (natural bond orbital) computations for a theoretical study of hydrogen bonds in some models. And NBO interprets hydrogen bond as the donation of electrons from the lone pair (non-bonding electrons) of the hydrogen acceptor to the (LUMO) antibonding orbital of the hydrogen donor.
For example LP1 O--- BD* (O--H)
This means that there is a donation of electrons from one lone pair (LP) of oxygen (which is in a non-bonding orbital) to the antibonding (LUMO) orbital of the O-H bond.
Interestingly, I just found out that there can be a type of hydrogen bonding between a lone pair of oxygen and C-H as the hydrogen donor. In some old papers in the literature, it is referred to as weak CH/O hydrogen bonds. (10.1021/ma0485585)
I hope that this is enlightening.


text:
Bleach in drain cleaners
I have a clogged pipe in my bathroom so I went to the store to get something to unclog it. Basically there are two types that you can find in my area:

Sodium hydroxide pellets
"Liquid plumber" mixtures. The components for those are mainly surfactants, sodium hydroxide and sodium hypochlorite.

As far as I recall from the water treatment courses I took in college, hypochlorite's oxidizing activity is heavily dependent on pH, as the active form is hypochlorous acid, not the hypochlorite ion. At high pH values the oxidizing activity becomes really low whereas low pH will make it stronger.
Does it make sense to buy a product that mixes a strong base, such as sodium hydroxide, with sodium hypochlorite? Wouldn't it be a waste of money?

Lye is very corrosive, just like acid, except opposite (hope that makes sense). Although lye is a very powerful chemical, and it is sufficient for many blockages, oxidizing agents (like bleach) may be helpful for breaking down paper/cellulose blockages.
Mixing bleach with acid will cause chlorine gas evolution... bad, bad, bad.
EDIT- The oxidative powers of hypochlorous acid and hypochlorite ion(s) compared:

Source: Descriptive Inorganic Chemistry, Third Edition (By Geoff Rayner-Canham, Tina Overton)
So in terms of power, yes hypochlorous acid is a stronger oxidizer, but it's not wildly different.
On a related note (see comments below) hypochlorous acid is a much better microbicide than the hypochlorite ion, because it penetrates cell walls more easily. However, regarding a drain cleaner, the oxidative difference is much less pronounced. And it's worth mentioning that at above pH 12, a cell will lyse (burst open) and become saponified, so that obviates whatever microbial effect hypochlorite or hypochlorous acid might have on bacteria... like being thrown into a poisonous wood chipper.

All your facts are right, but I don't think this answers my question. I am aware of the effects of strong acids over hypochlorite, what I wanted to know is if bleach would mantain any oxidizing power in a very basic solution (I'm guessing pH 12 or above) as to be worth using it, or I might as well get enough lye to last a lifetime of clogged toilets for the same price.

@Variax I would expect pH of no less than 14 from a drain cleaner.  Houshold bleach typically has a pH of 11. Although I've never tested the function of oxidizing power of bleach with respect to pH, I do understand the question and will say that at above pH11 all of the bleach exists as  (-OCl); none of it exists at HOCl and straight houshold bleach (pH 11) is a plenty "powerful" oxidizer (eats holes in tee shirts). As an interesting experiment, you could test the amount of time for say a 99:1 dilution of bleach to turn a piece of black craft paper white at various pH levels, for example.

I found some sources for the effect of pH on bleach. They refer to water treatment, but the science is there: http://www.cdc.gov/safewater/chlorination-faq.html (Point 6), http://oregonstate.edu/dept/larc/sites/default/files/pdf/chlorine-fact-sheet.pdf. This second one stated that ClO- is two orders of magnitude less oxidizing than its acid counterpart. So if 100 is the maximum and 1 the minimum, the oxidizing power of hypochlorite is:
pH 5: 99.68
pH 6: 96.89
pH 7: 75.79
pH 8: 24.37
pH 9: 3.96
pH 10: 1.30
pH 11: 1.03
pH 12: ~1.00
pH 13: ~1.00
pH 14: ~1.00

@Variax Your links are not working

I fixed them, they work now

There's table here showing oxidation power: Hypochlorous acid mechanism Also, the second link still doesn't work. Ok here's the PDF.

That curve with the % and the fact that ClO- is 80 times less effective is exactly why I asked this in the first place. Seems to me that I'd get the same result if I add bleach to a pH14 solution than if I added table salt unless there's something we're missing here.

@Variax I think you are seeing it wrong... the curve is simply the transition from HOCl to -OCl amount of hypochlorite- not efficacy. It is not 80 times less effective. NO! 80x efficacy for killing microorganisms because it can migrate through cell walls (it's bad for people too)... but the oxidative power is not that much different.


text:
SMARTS for cis amide
Is there a SMARTS pattern for cis (and trans) amide bonds?
This seems to be the SMARTS for any amide:
[NX3][CX3](=[OX1])[#6]

But I am not sure whether or not the SMARTS format has cis/trans definitions for “single” bonds such as amides.

No, not really.
For a double bond, you'd use the same '\' and '/' characters to indicate cis- and trans- stereochemistry that you use in SMILES. But as you indicate, there isn't a double bond for such stereochemistry.
My personal preference in scripts, code, etc. is to match the double bond, and then use my own code to check the relative positions of substituents around the central double bond. In this case, you could look at a dihedral angle around the C-C-N plane.
For future reference, an excellent resource for testing SMARTS patterns is SMARTSView.


text:
Relationship of Dielectric constant with ionic bond
How is dielectric constant related to formation of ionic bond?

@Mithoron Thanks. But chemistry and biology you cant just separate out.

You are probably referring to Coulomb's Law:
$${F\propto \frac{q_1 q_2}{d^2 \epsilon }}$$
This law, that you probably already met in physics, applies to chemistry and ionic bonds, since they are all characterized by electrostatic interactions.
You see that the strenght of a ionic bond is directly proportional to the product of the charges on the ions, inversely proportional to the square of their distance, and inversely proportional to the dielectric constant of the solvent. In vacuum, the dielectric constant is $\epsilon = 1$.
Otherwise, when the ions are in a solvent, the dielectric constant is associated to the solvent in use (you can find appropriate charts online).


text:
Is there anything that can react with molecular gaseous nitrogen?
I am looking for something to react with nitrogen in the air. Is there any substance that can react fast with molecular nitrogen at room temperature?

From Wikipedia

Elemental lithium reacts ("burns") in an atmosphere of $\ce{N2}$ to give lithium nitride:
  $$\ce{6Li + N2 → 2 Li3N}$$
Magnesium also burns in nitrogen, forming magnesium nitride.
  $$\ce{3 Mg + N2 → Mg3N2}$$

Those reactions will only take place in "pure nitrogen" not air. The OP asked about reactions in air that react with nitrogen.

While lithium reacts with nitrogen at room temperature,  reaction between magnesium and nitrogen require large amount of heat.


text:
Can pure elements have different colors?
This question was inspired by this: Are there colorful metals besides gold and copper?
I was wondering how the color of a pure element could be changed. I know that it is possible for color to change when electrons to get excited, but I'm curious about how that works. Can you change the color of any element if you heat it up enough (I'm talking about the pure elements here, not compounds)? If not, then what limitations are there about changing color? And to ask something a little more specific, could the color of pure gold be changed without adding it to an alloy or compound?
Note: These questions Why does iodine have different colors in different media? What does a molecules color have to do with its bond/orbital energies?
do talk about this topic, but I felt like they didn't answer my query satisfactorily.

You should read about allotropy I guess.

Particle size can make a difference too.

"Pure element" means that the only element present is the one considered, but this says nothing about the bonds between the element itself. As Mithoron said, allotropes are a nice example of this: just look, for instance, at the allotropes of phosphorus: just by binding phopsphorus in different ways you can get a black, white, red or violet compound.
This is due to the different wavelenghts of light interacting with different bonds.

And consider black, conductive graphite and clear, insulating diamond.


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Confusion with usage of heat capacity in Adiabatic expansion
My book says, that if a system is expanding adiabatically then $dq=0$ and $U= q+w$, so $U=w$.
It also states that $$\delta U=(C_v)\cdot\delta T$$
Here $C_v$, is the molar heat capacity at constant volume. How can they just $C_v$, which is defined for a specific volume, in a system that is expanding, i.e. its volume is changing?

For an ideal gas, U is a function only of T (and not V).  So the derivative of U with respect to T is an ordinary derivative, and is independent of V.  So why do we call $C_v$ the heat capacity at constant volume, if, for an ideal gas, it doesn't even depend on volume?  It has to do with how we measure $C_v$ experimentally. If the volume is held constant, then the amount of work is zero and $$Q=\Delta U=C_v\Delta T$$  So, in the case of constant volume, we can measure $C_v$ by measuring the amount of heat added Q.  If the volume is not held constant, so that some work is done, we can't determine $C_v$ from the amount of heat added.


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Molecular (infrared) vibrations: General relationship between molecular mechanics force constants and harmonic frequencies (in internal coordinates)?
In molecular mechanics method simple molecules, say symmetrical tri-atomics (H$_2$O), have potential energy surface governed by distances $r_1,r_2$ and angle $a$ like
$V (r_1,r_2,a_{121}) = K (r_1 - r_0)^2 + K (r_2 - r_0)^2 + K_a (a - a_0)^2$
The non-linear tri-atomic will have 3 harmonic frequencies $f_1,f_2,f_3$.
I understand that Hessian matrix seems to be important. The Hessian elements is $\frac{\partial^2 V}{\partial r_1^2}=2K$ and $\frac{\partial^2 V}{\partial r_2^2}=2K$ and $\frac{\partial^2 V}{\partial a_{121}^2}=2K_a$.
Is there a general relationship how harmonic frequencies are related to force constants? Does this relationship involve further approximations (besides the simple molecular mechanics PES)? The variables are internal coordinates and not the normal coordinates, right?
The water case I have just taken as illustrative example...
Thank you in advancement!
Miranda Christina

Several textbooks describe how to do this, so rather than repeat pages of this stuff have a look at the GF method (or FG method) on wikipedia, but more detailed examples are found in 'Atkins & Friedman 'Molecular Quantum Mechanics'; Graybeal, ' Molecular Spectroscopy'; Beddard 'Applying Maths in Chemical & Bimolecular Sciences'; Herzberg 'Infra Red & Raman Spectroscopy'.


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Estimation of nickel ions
I was doing experiment of estimation of copper and nickel ions by means of volumetric and gravimetric analysis respectively.
Suppose we analysed $\ce{Cu^2+}$ ions by precipitating with $\ce{H2S}$
But still there are some copper ions remaining in solution.
Is there possibility that they ($\ce{Cu^2+}$) can form complex with DMG
and interfere with Ni-DMG complex.

In this paper such complex is mentioned, so I guess that it's possible
http://www.scielo.br/scielo.php?script=sci_arttext&pid=S0103-50532011000500023

Copper can form a stable DMG complex (if it is there in the mixture in high concentration), it may also effect the color of the resultant Ni-DMG complex. Just go through Vogel's quantitative inorganic analysis book once.


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Vapour Pressure in Closed Containers
When the vapor pressure exerted by the gas equals that exerted by the liquid, do their densities equal?

What does it mean for a gas to have a vapor pressure?

V, you are mixing terms a little.
Vapor pressure is a quantity indirectly describing the amount of a substance must be in the gas phase for the substance to be in equilibrium between the liquid and gaseous states.  If the partial pressure of the gas (say water vapor) is equal to the vapor pressure of the liquid (must be the same substance, so water) than the liquid will have no net flow either from the gas phase to the liquid phase or vice versa.
There is no need for the substances to have the same density. This is only true for super-critical fluids.  Super-critical fluids exist once you have raised the temperature above a critical point after which there is no phase transition between liquid or gas.