Preferred IUPAC names for branched unsaturated hydrocarbons (polyenes)?

According to the general methodology described in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), a double bond is expressed by changing the ending ‘ane’ of the name of the corresponding saturated parent structure to ‘ene’.

  P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride (…) is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

  (…)

For example:

If the compound contains multiple double bonds, a multiplicative prefix is placed before the ending ‘ene’ and the letter ‘a’ is inserted between the root of the name of the parent structure and the multiplicative prefix.

  P-31.1.1.2 The multiplying prefixes ‘di’, ‘tri’, etc., are placed before endings denoting unsaturation to indicate the number of multiple bonds of each kind, as required, for example, ‘diene’ and ‘triyne’. For euphonic reasons, when the endings ‘ene’ and ‘yne’ are preceded by a multiplying prefix and a locant the letter ‘a’ is inserted. There is no elision of the final letter ‘a’ of a multiplying prefix before ‘ene’ or ‘yne’, for example, ‘tetraene’ and ‘pentayne’.

For example:

The names of substituent groups are derived from the respective parent hydrides. The presence of free valences formally derived from the loss of one, two, or three hydrogen atoms at the end of the longest chain of a parent hydride is denoted by replacing the ending ‘ane’ of the parent hydride name by ‘yl’, ‘ylidene’, and ‘ylidyne’, respectively.

  The suffixes ‘yl’, ‘ylidene’, and ‘ylidyne’ replace the ending ‘ane’ of the parent hydride name. The atom with the free valence terminates a chain and always has the locant ‘1’, which is omitted from the name. This method is recommended primarily for saturated acyclic and monocyclic hydrocarbon substituent groups (…). Substituent groups formed by this method are referred to as ‘alkyl-type substituent groups’;

For example:

methyl $(\ce{-CH3})$
methylidene $(\ce{=CH2})$
ethyl $(\ce{-CH2-CH3})$
ethenyl $(\ce{-CH=CH2})$
ethylidene $(\ce{=CH-CH3})$

Therefore, the name for the first compound given in the question is ‘3-ethenylhexa-1,5-diene’.

The name for the second compound is ‘3-methylidenehexa-1,5-diene’.

For the third compound, there is a choice among two different principal chains. In accordaqnce with the above-mentioned rules, the name could be either ‘4-ethenylhexa-1,4-diene’ or ‘3-ethylidenehexa-1,5-diene’.

The current IUPAC Recommendations stipulate various criteria for the selection of the principal chain. The first relevant criterion in this case provides that the principal chain has the greater number of skeletal atoms; i.e., the principal chain is the longest chain. If this criterion does not effect a choice of a senior parent structure, the following relevant criteria are applied successively until there are no alternatives remaining. According to Subsection P-44.4.1, the principal chain

  (a) has the greater number of multiple bonds (P-44.4.1.1);

  (b) has the greater number of double bonds (P-44.4.1.2);

  (…)

  (j) has the lower locant(s) for endings or prefixes that express changes in the level of hydrogenation, i.e., for ‘ene’ and ‘yne’ endings and ‘hydro/dehydro’ prefixes (P-44.4.1.10);

  (…)

Therefore, the correct name for the third compound is ‘4-ethenylhexa-1,4-diene’ (not ‘3-ethylidenehexa-1,5-diene’) since the locant set ‘1,4’ for the ‘ene’ ending is lower than ‘1,5’.


Why is the ionic radius of Al(3+) smaller than that of Li+?

The size of the 1s orbital in $\ce{Li+}$ and $\ce{Al^3+}$ is not necessarily the same. In fact, they are quite different because of the much larger effective nuclear charge in $\ce{Al^3+}$. One can easily look up the wavefunction for the 1s orbital and see the radial dependence on $Z_\mathrm{eff}$.

Therefore, merely looking at the electronic configuration cannot tell you anything about the size. If that was the case then one would expect $\ce{Cl-}$ and $\ce{K+}$ to have exactly the same size, which is clearly not true.

  All the answers says that $\ce{Mg^{2+}}$ is above $\ce{Ca^{2+}}$, thus it has a smaller radius. By applying  the exactly same logic i can say $\ce{Li+}$ is above $\ce{Al^{3+}}$ so it must have less ionic radius but that is not the case

You are missing the point here entirely. Mg and Ca are in the same group which is what makes them directly comparable. Li is not "above" Al any more than F is above Na.

  I think it is because of diagonal relationship. As $\ce{Mg^{2+}} = \ce{Li+}$ because of diagonal relationship and $\ce{Mg^{2+}} \gt \ce{Al^{3+}}$ because ionic radius decreases for cations along the period by going left to right.

The diagonal relationship is something that is observed, not a fundamental principle of chemistry. It still has to be rationalised. Saying that it is "because of the diagonal relationship" is akin to saying "electrons move very fast because they have a high velocity".

The diagonal relationship holds because 1) you add an extra shell of electrons going vertically from $\ce{Li+}$ to $\ce{Na+}$, which decreases $Z_\mathrm{eff}$ and increases ionic radius; and 2) you go horizontally from $\ce{Na+}$ to $\ce{Mg^2+}$, which increases $Z_\mathrm{eff}$ (same number of electrons but one more proton) and decreases ionic radius.

Going one more step horizontally to $\ce{Al^3+}$, one can see that $Z_\mathrm{eff}$ should increase again. So your thoughts on the matter are correct, but they just do not get to the crux of the matter.

The origin of the much smaller ionic radius of $\ce{Al^3+}$ is therefore directly attributable to the much larger $Z_\mathrm{eff}$ in $\ce{Al^3+}$ on the 2p valence electrons, compared to that on the 1s valence electrons in $\ce{Li+}$.

Be careful that the ionic radius is not always particularly well-defined and the values can vary from one source to another depending on how they are experimentally obtained.


Why does hydration of ions depend on their charge to area ratio?

Many people have done experimental, theoretical, statistical, molecular dynamics, etc. studies of solvation free energies of ions. I encourage you to just search solvation free energies of ions on google scholar and you'll find a bunch of very interesting and quite readable papers.

One of these papers ref. [1] (here for pdf) provides a lot of data and a very simple model which can basically be used to answer your question. In this paper, they derive a simple model which has only one fitting parameter. From this, they can determine the effective cavity size of an ion in solution. This cavity size is given by $r+\Delta r$, where $r$ is the radius of the ion and $\Delta r$ is the thickness of the shell.

Essentially, the value of $\Delta r$ is determined by the fitting parameter, and also determines the number of water molecules within the cavity of the ion. A portion of the data table is shown below with actual experimental values on the far right and parameters determined from the model in the other columns.

Just focus on the alkali metals which you list in your question. Notice as we go further down, the value of $\Delta r$ approaches zero and the actual radius of the ion is also the size of the cavity. This, however, is very much not true for the smaller ions. The Hydrogen ion being an extreme example where a solvation number of 12.0 is obviously unphysical. This, however, does provide a clue for a reasonable interpretation of $\Delta r$. $\Delta r$ is a measure of how far away a molecule must be for it to not "notice" the effect of an ion in solution. Thus, somewhat counter-intuitively, smaller ions interact effectively with a larger number of water molecules. The reason for this is that as the ions get larger, solvating them requires a greater reorganization of the hydrogen-bonding network in liquid water. This means that one must lose hydrogen bonds in order to solvate the ion.

Thus, the reason the charge to area ratio dictates the solvation energy and the solvation number is because a smaller ion will bring the same charge into solution, but due to its smaller size, will only require a small reorganization of the hydrogen bonding between waters, which minimizes losses in already stable interactions. This lack of reorganization is what leads to large values of $\Delta r$ and hence large values of $n$ in the model described above.

Let me be clear, $n$ is not simply the number of water molecules pointing towards an ion. This would not be measurable experimentally, or well-defined theoretically. Rather, $n$ will always be some kind of average value which tells you the effective number of water monomers which energetically benefit from the presence of the ion in solution. So, now the picture is clear. The smaller cations do not cause too large of a first hydration shell, so that the second hydration shell is closer to the ion than for a larger ion. Hence, the free energy and the solvation number are larger for smaller ions of the same charge.

As a bonus, ref. [2] is an interesting paper which approaches this problem from molecular dynamics. They give plots showing the relative effects which van der Waal's forces and electrostatics have. Note that the smaller ions are the only ones which are stabilized effectively by van der Waal's interactions as these scale closer to $r^{-5}$ rather than $r^{-1}$

References:

[1] Marcus, Y. (1991). Thermodynamics of solvation of ions. Part 5.—Gibbs free energy of hydration at 298.15 K. Journal of the Chemical Society, Faraday Transactions, 87(18), 2995-2999.

[2] : Grossfield, A., Ren, P., & Ponder, J. W. (2003). Ion solvation thermodynamics from simulation with a polarizable force field. Journal of the American Chemical Society, 125(50), 15671-15682.


The compound is anti-aromatic. While counting the number of π-electrons, you count the electrons which are delocalized over the ring. In this case the nitrogen lone pair is localised and does not participate in resonance. The nitrogen lone pair is in an $\mathrm{sp^2}$ orbital (red) which is orthogonal to the π system (blue):

So, the total number of π-electrons is only four: two from each double bond.

Please read this for a better understanding : http://www.chem.ucla.edu/harding/ec_tutorials/tutorial04.pdf


The post that was linked to you in comments contains  a nice, short description of the 'traditional' categories of van der Waals interactions. I suggest you take a look at that, it might just be enough to clear things up. The reason I am writing this post is that the categories introduced there are defined somewhat arbitrary, so I'll try to introduce some terminology that (as of today) is considered correct in the field.

So given a molecular system, it is quite easy to factorize the forces into intermolecular and intramolecular forces. Intramolecular forces are what hold the molecules together. These are (almost completely) covalent bonds, and these interactions would be same for one single molecule and the bulk phase of the matter. But we know, from the real life, that matter does not only exist as an ideal gas: we got liquids, solids, and even the gases are non-ideal. So evidently, there has to be forces between molecules. These forces are collectively called van der Waals forces for historical reasons.

Now there are three different types of interaction between molecules. There are electrostatic, induction and dispersion interactions.

Electrostatic effects are simple classical interaction between charged particles. Thesy simply come from the Coulomb law, can be either attractive or repulsive, and they are strictly additive (because Coulomb law is additive)

It is tricky to properly understand induction and dispersion. These two effect come from quantum mechanical perturbation theory. Without getting into details, perturbation theory means that the wavefunction of two molecules together is not simply the sum (or average) of the wavefunctions of the separate system, but there is an interaction between them, but this interaction can be expressed in terms of the interactionless wavefunction. The expression for the second-order energy correction is given by the Rayleigh-Schrodinger picture as follows (this is called the polarization approximation):

$$W^{(2)} = - \sum_{mn}' \frac{ <00|H'|mn><mn|H'00> }{W_{mn}-W_{00}}$$

In this expression, $|m>$ represents the mth excited state of the original wavefunction of molecule A and $|n>$ represents the nth excited state of the original wavefunction of molecule B. This expression could be read: the energy difference between the two separate systems and the interacting system depends on the excitations in the separate systems through an interaction operator $H'$.

The sum can be factorized into three components:

$$W^{(2)} = - \sum_{m\neq0}' \frac{ <00|H'|m0><m0|H'00> }{W_{m0}-W_{00}} - \sum_{m\neq0}' \frac{ <00|H'|0n><0n|H'00> }{W_{0n}-W_{00}} - \sum_{m\neq0,n\neq0}' \frac{ <00|H'|mn><mn|H'00> }{W_{mn}-W_{00}}$$

Here, the first term contains the energy contributions of the interactions of the ground state of A with the excited states of B. This is called the induction energy of B. The second term is the induction energy of $A$, which again, expresses the total response of the molecule A to the ground state of B. The third term is the so-called dispersion interaction, which is then a mixing of the excited states of both molecules.

So these expressions are the defining expressions for induction and dispersion. One should know that both of these interactions are always attractive , they are non-additive, and they are always present, between charged and neutral particles too. Also don't get confused that the energies are expressed in terms of excited states: there is no excitation going on in the system, it's just the mathematical tool that we used as definitions.

The total intermolecular force, or van der Waals force is then usually the sum of electrostatic, induction and dispersion terms. For the sake of completeness we might include resonant energy couplings and magnetostatic interactions too, but these are either rare or so small that we can neglect them.


There are two things at play here and you need to mentally separate them.
When seen as molecules, all amino acids are polar to at least some degree. Most of them are very polar, especially when compared to other organic molecules. This is (obviously) due to the carboxylic acid and amino groups both of which are very polar functional groups and both of which are ionised in neutral aquaeous solution.
However, it is desirable to categorise the different amino acids into various groups, to better explain their individual contributions to an overall peptide or protein structure. As all amino acids have a carboxylic and amino group and as these groups are connected in peptide bonds to form the peptide or protein backbone, these two groups alone are not able to distinguish between the different amino acids.
For this reason, the amino acids have been classified according to their residue R. If the residue contains only carbon and hydrogen atoms (e.g. Gly, Ala, Val, Leu, Ile), it is a non-polar residue and thus these amino acids are classified as ‘non-polar amino acids’ in simplistic terms. In a peptide or protein, these amino acids’ non-polar side chains will usually act as non-polar domains within the tertiary structure
A similar case can be made for the ‘charged’ amino acids (Glu, Asp, Lys, Arg): obviously and trivially, all amino acids have charged groups when dissolved in water but only those four have a charged side chain. So only these four are able to build up salt bridges to cross-stabilise tertiary structures and are thus classed as charged.


Isotopes and their percentage existence in atmosphere

Short answer: it's the percent by number of particles.

Longer answer: It's the proportion of particles of a particular mass for that element.  To chemists, a per particle basis is also commonly referred to as a per mol basis or a molar basis.

A way that it can be measured is by mass spectrometry - you pass the atoms through the apparatus which shows the mass to charge ratio. E.g. if you could pass $\ce{Cl}$ molecules, you'd see 2 peaks in the resulting graph - one is at 35 and the other at 37 marks which shows the mass/charge. First peak would be about 3 times larger (thus there were 3 times more particles). Thus chlorine would be about 75% isotope 35 and 25% isotope 37. So while this approach measures mass of each isotope, it also determines the relative molar amounts of each isotope.


Separation of Methanol-water in a plate distillation column

Yes, you can do that.

Specific gravity is considered as unitless ratio of density related to water density.

Density as a specific mass is mass per volume, in $\pu{ g/cm^3}$ or $\pu{ kg/m^3}$

Regardless of country specific quantity naming convention:

Numerically, for 3 valid decimal digits,considering value uncertainty, the unitless quantity and quantity in $\pu{g/cm^3}$ are about interchangeable at $\pu{20 ^{°}C}$, as water density is near $\pu{0.9982 g/cm^3}$


Do crystals have stronger intermolecular forces than amorphous solids?

Diamond crystal has much stronger bonds than amorphous red phosphorus. Amorphous sodium/calcium silicate, known as glass, has much stronger bonds than crystalline metallic sodium. (Also, in all four cases the word intermolecular doesn't really apply, because these compounds are not made of molecules.)

You see that the answer depends primarily on the particular compound, rather than on its being crystalline or amorphous. In cases when something can be either crystalline or amorphous with the same chemical composition, the bonds in both are pretty much the same, as explained in the comment by Jon Custer.

Oh, and yes, chemical and physical properties of pretty much anything are explained by its structure.


http://books.google.com/books/about/Crystal_Growth_in_Gels.html?id=pQ-JuNXEL6wC
  The classic text.  Set them up, come back in a few weeks.  (Potentially any gel - polyacrylamide, agarose, agar, gelatin...)  Meanwhile,

The Prussian Blue prep works across the transition metals, with different colors.  Alas, it is not crystalline (but maybe in gel...).  Vanadium oxidation states are a spectrum of colors.  Vanadyl, V=O, is especially fine (but not as a food supplement).

Cobalt can be preparatively more fun than iron.  Co(II) is substitution labile - put anything on it you want, in real time.  Co(III) is substitution inert.  Anything you put on Co(II) then oxidize to Co(III) stays forever (absent a trace of reducing agent to reopen the door).   There is vast and colorful Co coordination chemistry.

http://onlinelibrary.wiley.com/bookseries/10.1002/SERIES2146
  Everything in there works to spec.  Fine some interesting things to do (and scale down to grams).

Cobalt sepulcrate and especially sarcophogate (Alan Sargeson)!   Tris(ethylenediamine)cobalt(III) is a propeller, left- or right handed. It can be resolved afterward (Inorg. Syn. 6 186 (1960)) or made entirely as one hand.  Net charge is 3+.  Tris(oxalato)cobaltate(III) is a propeller, left- or right handed. It can be resolved afterward (Inorg. Syn. 8 204 (1966)) or made entirely as one hand.  Net charge is 3-.  Tris(ethylenediamine)cobalt(III) tris(oxalato)cobaltate(III) as a salt with same or opposite hands could be interesting.  The crystal will certainly be tightly bound.


Based on appearance and extent of deformation, your bottle is likely made of PVC plastic which is not compatible with oils. Plastic bottles are made from blow molding. This process leaves residual stress in the polymer chains of the materials but  creates a smooth and transparent surface. Lavender oil can diffuse into PVC and make it softer. Once it is softer, the polymer chains can relax some of the residual stress which makes the surface rough and as a result appear cloudy due to increased scattering. Given Long enough more lavender oil can penetrate the plastic and soften it to the point that the plastic is allowed to flow which causes the observed deformation of the top.

Now I cannot say with certainty it was PVC that your bottle was made from, but if it was, then plasticizer would have seeped into the oil which isn't great for people. If you want to use your oil for air freshener or candles that is fine but I would not recommend it for skin contact applications.


Well, just like any other reactions, it depends upon the the heating temperature and the reaction environment. Contrary to what the reaction tells, the decomposition products of the reaction is actually iron(II,III) oxide, $\ce{Fe2O3}$ if it is heated in oxygen environment$\ce{^{[1]}}$ and $\ce{Fe3O4}$ if it is heated in $\ce{CO2}$ environment$\ce{^{[2]}}$.  $\ce{FeO}$ also formed during the reaction but for a small period of time. Hence it is an intermediate product.

  Figure 1 shows the thermal decomposition of iron carbonate at
  temperature at 600°C. It is shown from one DTA curve in Fig. 1 one
  exothermic peak at 495°C accompanied with a weight loss of about 32%
  (theoretical value 31.1%). This step is due to the decomposition of
  iron carbonate to $\ce{Fe2O3}$.

  X-ray diffraction patterns of the decomposition products of iron
  carbonate were determined for the solid calcined at 550 and 800°C
  Figure 3 (a) shows X-ray diffractogram for the sample dried at room
  temperature corresponding to iron carbonate. The diffractogram of the
  sample calcined at 550°C consists of very poorly crystallized α-Fe2O3
  phase. .Heating of ferrous carbonateat 800°C led to the formation of
  well crystalline $\ce{α-Fe2O3}$

  In air, the X-ray pattern of $\ce{FeCO3}$ was replaced by that of
  $\ce{Fe3O4}$ as the sample was heated. The change was essentially
  complete at 500°C. No further changes were noted. If, however, the
  furnace atmosphere was changed from air to $\ce{CO2}$ or helium above
  1000°C. the sample transformed slowly to $\ce{Fe3O4}$. Restoring the
  air atmosphere again resulted in the pattern of $\ce{Fe2O3}$. When
  $\ce{FeCO3}$ was decomposed in a $\ce{CO2}$ atmosphere, the first
  decomposition product observed was $\ce{Fe3O4}$.[...]the brief
  appearance of $\ce{FeO}$ was suspected in the decomposition. By
  oscillating the Geiger counter through a very small arc encompassing
  the strongest line of FeO, the presence of this oxide was indicated.
  The line appeared for only a very short time and was at no time very
  strong. In the inert atmosphere, the products of decomposition were
  $\ce{FeO}$ and $\ce{Fe3O4}$. The $\ce{Fe3O4}$ in the sample could not
  be converted to $\ce{FeO}$ even by heating to the highest temperature
  available with the furnace (about 1150°C. in helium).

So, the reaction indeed forms $\ce{FeO}$ but it gets oxidized depending on the environment the reaction is proceeded.

$$\ce{FeCO3 -> FeO + CO2 }$$

$$\ce{4FeO + O2 -> 2Fe2O3}$$

$$\ce{3FeO + CO2 -> Fe3O4 + CO}$$

References

A.A. El-Bellihi, "Kinetics of Thermal Decomposition of Iron
Carbonate", Egypt. J. Chem. 53, No. 6, pp.871- 884 (2010) DOI: https://ejchem.journals.ekb.eg/article_1268_82097e00ce1c4e475f17cfec61edb6d2.pdf
H. E. KISSINGER, H. F. McMURDIE, and B. E. SIMPSON,Thermal Decomposition of Manganous and Ferrous Carbonates,DOI: https://ceramics.onlinelibrary.wiley.com/doi/pdf/10.1111/j.1151-2916.1956.tb15639.x


Number of stereoisomers of trisubstituted cyclopentane

Almost everything that needs to be said has been said. But I do have a few comments about configurations. In enantiomeric, trans-dibromides 1 and 2, the methyl groups are chirotopic and non-stereogenic. Inversion of the stereochemistry of the methyl group in either 1 or 2 returns the same structure, respectively. Thus, there is no designation for stereochemistry at the C4 methyl center. On the other hand, inversion of the stereochemistry of the methyl group in 3 gives 4 and the same operation on 4 produces 3. Therefore, the methyl center in achiral isomers 3 and 4 is achirotopic and stereogenic. Accordingly, the C4 centers have lowercase descriptors, r and s, respectively. This site might be of help.


Catalysts to accelerate foamy polyurethane reaction

In those preparations, the foam arises from the reaction between water and isocyanate, which leads to the production of CO2 via formation of a carbamic acid, which upon decarboxylation generates the blowing agent, CO2.

The trick would be that of making water more reactive towards the isocyanate, without (as expected by a catalyst) partecipating actively in the reaction.

Several compounds of this kind have been reported and are sold, for instance: https://patents.google.com/patent/WO2009048807A2/en mentions N,N,N',N',N",N"-pentaethyldiethylenetriamine and morpholine derivatives, and commercially (I won't link any product, but googling "Blowing Catalysts" might help you) catalysts based on  N-methylmorpholine, Bis[2-(N,N-dimethylamino)ethyl] ether, DMEA and PMDTA are available.

This said, for practical applications the formulations have probably the best ratio of reactants, surfactants and catalyst for obtaining a "good" foam (with good physical properties, and possibly with negligible smell), but if your goal is that of making the reaction go faster, maybe an excess of (any?) tertiary amine, and possibly some drops of water in excess should work in making things faster (with the possible drawback of obtaining a low quality foam, maybe with "bubbles" larger than intended, or with bad "texture").

Additional note: if speed is what you are looking for, gently heating the mixture should work without adding anything else


Equilibrium reaction where the number of moles of reactants and products are the same

A simple example:
If $T > \pu{100 °C},$ and $p \leq \pu{1 atm},$ then water will be in the gas phase and:
$$\ce{CH4(g) + 2 O2(g) → CO2(g) + 2 H2O(g)}$$
Note, however, that while this reaction's equilibrium will be highly insensitive to pressure changes (within the regime that all molecules remain in the gas phase), it will not be entirely insensitive to pressure.
That's because, even though there are the same number of moles of gas on the LHS and RHS, these are real gases, and thus their molar volumes will differ slightly. This is a small effect compared to the effect of actually having different numbers of moles of gases on the LHS and RHS.


Cyclobutadiene - Jahn–Teller effect or not?

Very interesting question, and it kept me up despite daylight saving time cheating me of one hour of sleep last night... A good reference is Albright, Burdett and Whangbo, Orbital Interactions in Chemistry 2nd ed. pp 282ff. which explains this in much greater detail than I can. (In general, that book is fantastic.) I will try my best to summarise what they have written - please point out any mistakes that I may have made. I think there is also much more information out there in the literature (a search for "Jahn–Teller cyclobutadiene" should give lots of good results), although it is of course not always easy to read. $\require{begingroup}\begingroup\newcommand{\ket}[1]{|#1\rangle}$

1. What is the ground state of undistorted $\ce{C4H4}$?

Based on simple Huckel theory as you have shown above, one would expect the ground state to be a triplet state with the two highest electrons in different orbitals. This is simply because of exchange energy (Hund's first rule).

However, it turns out that the singlet state with the two electrons in different orbitals is more stable. That explains the lack of signal in ESR experiments.

(Note that this orbital picture is a simplification, as I will describe later.)

In short, this is due to dynamic spin polarisation; the idea is that there are low-lying excited singlet states which mix with the ground state to stabilise it. This is the basic idea underlying so-called configuration interaction, which mixes in excited molecular states into the wavefunction in order to obtain a better estimate of the ground-state wavefunction. If you are interested in the theoretical aspect of this, I think Szabo and Ostlund's Modern Quantum Chemistry should cover it. The topic of dynamic spin polarisation is treated in depth in pp 285ff. of Albright et al.

Your question, "how to predict the singlet state a priori"? is a good one. Albright et al. write that (p 289)

  In general, high-level configuration interaction calculations are needed to determine the relative stabilities of the singlet and triplet states of a diradical.

So, it does not seem to be something that you can figure out a priori, and it is most certainly not something you can figure out by just using Huckel theory (which is an incredibly simplistic treatment).

2. What is the degeneracy of the ground state?

You also wrote that "the singlet state will be doubly degenerate". However, in the case where both electrons are in different orbitals, there is only one possible spatial wavefunction, due to the quantum mechanical requirement of indistinguishability. (As we will see in the group theoretical treatment, there is actually no doubly degenerate case, not even when the electrons are paired in the same orbital.)

If I label the two SOMOs above as $\ket{a}$ and $\ket{b}$ then the combination $\ket{ab}$ is not a valid spatial wavefunction. Neither is $\ket{ba}$ a valid wavefunction.

You have to take linear combinations of these two states, such that your spatial wavefunction is symmetric with respect to interchange of the two electrons. In this case the appropriate combination is

$$2^{-1/2}(\ket{ab} + \ket{ba}) \qquad \mathrm{{}^1B_{2g}}$$

There are two more possible symmetric wavefunctions:

$$2^{-1/2}(\ket{aa} + \ket{bb}) \qquad \mathrm{{}^1A_{1g}}$$
$$2^{-1/2}(\ket{aa} - \ket{bb}) \qquad \mathrm{{}^1B_{1g}}$$

(Simply based on the requirement for indistinguishability, one might expect $\ket{aa}$ and $\ket{bb}$ to be admissible states. However, this is not true. If anybody is interested to know more, please feel free to ask a new question.) The orbital picture is actually woefully inadequate at dealing with quantum mechanical descriptions of bonding. However, Voter and Goddard tried to make it work out, by adding and subtracting electron configurations:1

Ignore the energy ordering in this diagram. It is based on Hartree–Fock calculations, which do not include the effects of excited state mixing (CI is a "post-Hartree–Fock method"), and therefore the triplet state is predicted to be the lowest in energy. Instead, just note the form of the three singlet states; they are the same as what I have written above, excluding the normalisation factor.

The point of this section is that, this singlet state is (spatially) singly degenerate. The orbital diagrams that organic chemists are used to are inaccurate; the actual electronic terms are in fact linear combinations of electronic configurations. So, using orbital diagrams to predict degeneracy can fail!

3. Why is there a Jahn–Teller (JT) distortion in a singly degenerate state?

You are right in that for a JT distortion - a first-order JT distortion, to be precise - to occur, the ground state must be degenerate. This criterion is often rephrased in terms of "asymmetric occupancy of degenerate orbitals", but this is just a simplification for students who have not studied molecular symmetry and electronic states yet. Jahn and Teller made the meaning of their theorem very clear in their original paper;2 already in the first paragraph they write that the stability of a polyatomic molecule is not possible when "its electronic state has orbital degeneracy, i.e. degeneracy not arising from the spin [...] unless the molecule is a linear one". There is no mention of "occupancy of orbitals".

So, this is not a first-order JT distortion, which requires degeneracy of the ground state. Instead, it is a second-order JT distortion. Therefore, most of the other discussion so far on the topic has unfortunately been a little bit off-track. (Not that I knew any better.)

Albright et al. describe this as a "pseudo-JT effect" (p 136). For a better understanding I would recommend reading the section on JT distortions (pp 134ff.), it is very thorough, but you would need to have some understanding of perturbation theory in quantum mechanics. Please note that there is a typo on p 136, as is pointed out in the comments to this answer. (Alternatively, there is also a discussion of second-order JT effects in a paper by Pearson.3) The idea is that, the distortion reduces the symmetry of the molecule and therefore allows ground and excited states to mix. This leads to a stabilisation of the ground state and hence an electronic driving force for the distortion.

In slightly more detail, we can use a group theoretical treatment to do this. The second-order correction to the energy is given by

$$\sum_{j\neq i}\frac{\langle i |(\partial H/\partial q)| j \rangle}{E_i^{(0)} - E_j^{(0)}}$$

where $\ket{i}$ is the ground state (which in this case is described by $2^{-1/2}(\ket{aa} - \ket{bb})$) and $\{\ket{j}\}$ are the excited states. $E_i^{(0)}$ is the unperturbed energy of the ground state $\ket{i}$ and likewise for $E_j^{(0)}$. $q$ is a vibrational coordinate of the molecule.

Note that the denominator in this term, $E_i^{(0)} - E_j^{(0)}$, is negative. So, this correction to the energy is always negative, i.e. stabilising. (There is another second-order term which isn't relevant to the discussion here.) The idea is that in order to have a significant stabilisation upon distortion, two criteria must be fulfilled:

The denominator must be small, i.e. the excited state must be low-lying.
The numerator must be non-zero, i.e. there must be appropriate symmetry such that $\Gamma_i \otimes \Gamma_j \otimes \Gamma_H \otimes \Gamma_q$ contains the totally symmetric irreducible representation. Since $H$ transforms as the TSIR, this is equivalent to the criterion that $\Gamma_q = \Gamma_i \otimes \Gamma_j$.

In cyclobutadiene, the two SOMOs in $D_\mathrm{4h}$ symmetry transform as $\mathrm{E_u}$. To find the irreps of the resultant states we take

$$\mathrm{E_u \otimes E_u = A_{1g} + [A_{2g}] + B_{1g} + B_{2g}}$$

(Note that there is no spatially degenerate term in this direct product, so there cannot be a first-order JT distortion!) The $\mathrm{A_{2g}}$ state in square brackets corresponds to the triplet case and we can ignore it. In our case (presented without further proof, because I'm not sure why myself) the ground state is $\mathrm{B_{1g}}$. The two low-lying singlet excited states transform as $\mathrm{A_{1g} \oplus \mathrm{B_{2g}}}$. Therefore if there exists a vibrational mode transforming as either $\mathrm{B_{1g}} \otimes \mathrm{A_{1g}} = \mathrm{B_{1g}}$ or $\mathrm{B_{1g}} \otimes \mathrm{B_{2g} = \mathrm{A_{2g}}}$, then the numerator will be nonzero and we can expect this vibrational mode to lead to a distortion with concomitant stabilisation.

In our case, it happens that there is a vibrational mode with symmetry $\mathrm{B_{1g}}$, which leads to the distortion from a square to a rectangle. This therefore allows for a mixing of the ground $\mathrm{B_{1g}}$ state with the excited $\mathrm{A_{1g}}$ state upon distortion, leading to electronic stabilisation.

Another way of looking at it is that upon lowering of the symmetry from square ($D_\mathrm{4h}$) to rectangular ($D_\mathrm{2h}$), both the $\mathrm{B_{1g}}$ ground state and $\mathrm{A_{1g}}$ excited state adopt the same symmetry $\mathrm{A_g}$. You can prove this by looking at a descent in symmetry table. Therefore, mixing between these two states will be allowed in the new geometry.

Because we are talking about the mixing of states here, the orbital picture is not quite enough to describe what is going on, sadly. The best representation we can get is probably this:

However, Nakamura et al. have produced potential energy curves for the distortion of cyclobutadiene from square to rectangular and rhomboidal geometry.4 Here is the relevant one (rectangular geometry):

Solid lines indicate singlet states, and dashed lines triplet states. Note that this diagram depicts what I have described above: the singlet ground state  ($\mathrm{{}^1B_{1g}}$), as well as the first excited singlet state ($\mathrm{{}^1A_{1g}}$), both have the same irrep upon distortion. The ground state is stabilised upon the lowering of symmetry, precisely due to this mixing.

References

(1) Voter, A. F.; Goddard, W. A. The generalized resonating valence bond description of cyclobutadiene. J. Am. Chem. Soc. 1986, 108 (11), 2830–2837. DOI: 10.1021/ja00271a008.

(2) Jahn, H. A.; Teller, E. Stability of Polyatomic Molecules in Degenerate Electronic States. I. Orbital Degeneracy. Proc. R. Soc. A 1937, 161 (905), 220–235. DOI: 10.1098/rspa.1937.0142.

(3) Pearson, R. G. The second-order Jahn–Teller effect. J. Mol. Struct.: THEOCHEM 1983, 103, 25–34. DOI: 10.1016/0166-1280(83)85006-4.

(4) Nakamura, K.; Osamura, Y.; Iwata, S. Second-order Jahn–Teller effect of cyclobutadiene in low-lying states. An MCSCF study. Chem. Phys. 1989, 136 (1), 67–77. DOI: 10.1016/0301-0104(89)80129-6. $\endgroup$


The question is interesting, but slightly ill-defined.
If you take s and p orbitals as a solution of Schrodinger equation of hydrogen-like atom, then they are eigenfunctions of Hamiltonian, therefore are orthogonal. That means - no overlap. Do not be misled by the figure, if you do the math properly, $\langle1s|2p\rangle = 0$.

If you speak of just one electron, the solution is reasonably simple as stated in the article and the electron does not simultaneously reside in both orbitals. Even if you would think of interaction with excited states, it would not help, as single excitations do not interact with ground state. You can of course excite it, but than it is not "simultaneous".

In many-electron system, the picture gets more complicated, but this was not a question.


Usually these type of complex kinetics arise when there is an inhomogeneous distribution of properties in a system. For example reaction over barriers which have a range of sizes, this can occur in proteins or in quantum dots but not in 'normal' kinetics. Fluorescence quenching in micelles can follow non-single exponential decays because there is usually a distribution of quenchers  solubilised in micelles. The fluorescence has the form $\ln(I)\sim -At+B\exp(-Ct)+D$ where , $t$ is time and $A,B,C,D$ constants.

In diffusion controlled quenching or energy transfer quenching in solution the rate constant will depend on time as $k=k_0+k_1/\sqrt{At}$. So there are many situations in which 'strange' kinetics can be observed.

However, vey often 'power law' kinetics is just a name given to an empirical method of fitting data to a function, say of the form $\exp(-kt^n)$ where $k, n$ are variables used to fit the data. In this form the decay function has no physical meaning.

Addendum. The first equation above was calculated as shown below.

As a model we suppose that the fluorophore and quencher are both in a micelle, as example of a restricted geometry, and that effectively no fluorophore exists outside. There is always an excess of quencher over fluorophore whose concentration is small so that no more than one will exist in any micelle. It is possible that zero, one or  several quenchers can be in any micelle.

We suppose that the rate constant for quenching is $k_i=k_0+ik_q$  where $k_0$ is the fluorescence decay rate constant value without quencher and $i$ the number of quenchers in any micelle, $i=0,\,1,\,2\cdots$ and $k_q$ the constant for quenching of the fuorophore.

The measured fluorescence signal is the sum of contributions from all micelles and is therefore

$$I(t)=I_0\sum_{i=0}^\infty p_i\mathrm e^{-k_it}$$

where $p_i$ is the probability of there being $i$ quenchers in a micelle. It seems reasonable to suppose that this has a Poisson distribution, and in that case $\displaystyle p_i= \frac{m^i\mathrm e^{-m}}{i!} $ where $m$ is the mean number of quenchers in a  micelle. Substituting and simplifying the intensity then becomes

$$ I(t)=I_0\mathrm e^{-k_0t-m}\sum_{i=0}\frac{(m\mathrm e^{-k_qt})^i}{i!}$$

The summation term has the form $\sum x^i/i!$ which is the expansion of the exponential thus

$$ I(t)=I_0\mathrm e^{-k_0t-m}e^{m\mathrm e^{-k_qt}} = I_0\mathrm e^{-k_0t +m(\mathrm e^{-k_qt} -1)      }    $$

which when logs are taken is the expression above.


What makes an atom favor another one when bonding, and should there be enough atoms for both parts?

The answer to your question boils down to one thing, bond energies. When a bond is formed, energy is released. When a bond is broken, energy must be absorbed. Elements prefer to be in a low energy state, so if you just added H and O into a container, the reaction with the most negative net bond energy will take place. To calculate the overall bond energy of the reaction, the equation is:

$E_{total} $= (Energy of bonds broken) - (Energy of bonds formed).

(As a side note, the answer is in joules of energy per mole)
For naturally occurring hydrogen and oxygen the bond energy is:

$2H_2 + O_2 $ ->$2H_2O$

= $(436*2 + 493*1)-(463*4)= -487 \frac{kj}{mol}$

(2 H-H bonds broken + 1 O=O bond broken) - (4 O-H bonds formed)

This reaction is favored and would occur under typical conditions if oxygen was in the presence of hydrogen and would release -487 joules of energy per mole. The bond energies for each bond can be found by simply looking them up online. The bond energy for the formation of $H_2O_2$ I calculated turned out to be -149kj/mol. Therefore the formation of $H_2O$ would be favored over $H_2O_2$. In your question, you asked about single O and H atoms, not $H_2$ and $O_2$, which would not occur naturally, but is still very possible. This result however would be the same. The atoms would first bond to form $O_2$ and $H_2$, then would proceed to form water as noted above.

Finally, to answer your last question, if there is one hydrogen and one oxygen left, they will indeed bond to make an $OH$ molecule.


How do we figure out whether reactions are endothermic or exothermic?

If you search for "standard enthalpy of formation hydrogen iodide", you get positive value $\pu{26.5 kJ/mol}$, what means the reaction is endothermic, as by convention, the values for elements in their standard state is zero.
The other one is strongly endothermic, what you realize, if you search the same for nitric oxide.
For a general reaction, search for standard enthalpy of formation of all reactants and products, multiply the values by stoichiometric coefficients and subtract the reactant values from the product values. If the result is negative, the reaction is exothermic. For the positive value, the reaction is endothermic.
For starters: enthalpy is defined as $H = U + p . V$, where $U$ is system internal energy, $p$ is pressure, $V$ is volume. Change of system enthalpy in constant pressure condition is change of system internal energy + volume work done by system: $\Delta H = \Delta U + p . \Delta V$. By other words, $\Delta H$ is the energy you have to provide to a system at constant pressure to increase its internal energy by $\Delta U$, if the system changes its volume by $\Delta V$.
For guessing the formation enthalpy patterns, there is no easy shortcut. For simple inorganic compounds, a lot of help in noticing patterns is the periodic element table.
All comes with experience and learnt expirical knowledge and patterns. Chemistry is in large extent empirical science, where many things cannot be derived. Or can, but it is much easier to just remember, unless one wants to jump in quantum chemistry. Even then not everything can be calculated.


Why group 13 elements don't show +2 as their major oxidation state?

The one p electron is energetically much higher than the two s electrons (and any d electrons, if present). So it is rather easily removed to give a monocation or the $\mathrm{+I}$ oxidation state.

Thereafter, to remove a second electron you need to rip out one of the paired s electrons. This is not going to be easy and requires a lot of energy. But once you have arrived there, at a hypothetical $\mathrm{+II}$ cation, you again have one single lone electron in a relatively (compared to the core orbitals) orbital that should, again, be easily removed. Obviously that can happen leading you quickly to the $\mathrm{+III}$ state.

Indeed, it would be rather easy to see that two $\mathrm{+II}$ atoms might come together and transfer a single electron from one to the other—a disproportionation, giving $\mathrm{+I}$ and $\mathrm{+III}$ from two $\mathrm{+II}$. This would be predicted to be an exothermic process.

This only applies to isolated atoms; where bonds are formed other oxidation states are accessible. For example, boron exhibits the $\mathrm{+II}$ oxidation state in bis(pinacolato)diboron as can be deduced easily.


Does a stoichiometric coefficient represent molecules or moles of a substance?

Balanced equations represent ratios of reacting chemical species, not absolute quantities. What we mean when we say $$\ce{2H2 +O2->2H2O}$$ is not literally that two molecules of hydrogen react with one of oxygen, but rather that two units of hydrogen react. These units could be moles or molecules or something else entirely.


Energy absorbed by a system in isothermic process in which the phase changes from liquid to gas

As indicated by @Buck Thorn,the work done by the substance on its surroundings is $P_2(V_2-V_1)$.  So, from the first law, $$U_2-U_1=Q-P_2(V_2-V_1)$$Solving for Q then gives:
$$Q=(U_2+P_2V_2)-(U_1+P_1V_1)-(P_2-P_1)V_1=H_2-H_1-(P_2-P_1)V_1$$Here, $$H_2=H_1+(H^{sv}_1-H_1)+(H_2-H^{sv}_1)$$where $H_1$ is the enthalpy of saturated liquid in State 1, and $H_2$ is the enthalpy of the superheated vapor that exists in State 2 (i.e., at a pressure lower than the equilibrium vapor pressure at the constant temperature T).  For an ideal gas, the change in enthalpy at temperature T from the saturated vapor in State 1 to the superheated vapor in State 2 is zero.  So the previous equation reduces to $$H_2=H_1+(H^{sv}_1-H_1)=H_1+x\Delta h^{vap}$$where $\Delta h^{vap}$ is the heat of vaporization at temperature T.  Substitution of this into the equation for the heat added gives:
$$Q=x\Delta h^{vap}+(P_1-P_2)V_1$$where $V_1$ is the volume of the saturated liquid at T. The second term on the rhs will typically be negligible.


Before dissolution, the substance is usually a solid and forms crystals. Its physical state is solid. After dissolution, the substance is not visible any more : it is not a solid any more. Its physical state has changed.
Before hydration, the substance is usually solid. After hydration, the color and the volume may have changed, but the physical state has not changed : it is still a solid –


Why are bonds ionic when the electronegativity difference between bonded atoms is greater than 1.7?

The first thing to consider is the difference between covalent and ionic bonding, from the UCDavis ChemWiki site Ionic and Covalent Bonds,

  In ionic bonding, atoms transfer electrons to each other. Ionic bonds require at least one electron donor and one electron acceptor. In contrast, atoms that have the same electronegativity share electrons in covalent bonds since donating or receiving electrons is unfavorable.

The electron donor has a low electronegativity and the electron acceptor has a higher elelctronegativity - so there is a difference in electronegativity $\Delta{EN}$, effectively creating a positive and negative end, an example is below:

Image source: Japan Synchrotron Radiation Research Institute (JASRI)

When the differences in electronegativities of various compounds are graphed against % ionic character, as shown below:

Image source: University of Florida Chemical Bonding page

Values of $\Delta{EN}$ greater than 1.7 correspond to an ionic character of greater than 50%, from the University of Florida website:

  What determines how the electrons are shared is the relative electronegativity (electron greed) of the bonding atoms. The degree of polarity or degree of ionic bonding of any given bond can vary continuosly zero to nearly 100%. We normally say that bonds between atoms with electronegativity difference ($\Delta{EN}$) greater than 1.7 are ionic, although this really means only more than about half ionic in character.

Another resource is from the University of Washington Lecture 23: Ionic to Covalent Bonds.


Cheap and not messy solvent for motor oil

In terms of inexpensive, safe, and a little polar to move oxidation products, I'd try soaking in and flushing with hot soybean oil (replace elastomer hoses and seals).  Then flush with kerosene (odorless white spirits, hardware store  paint section), then blow out with compressed air.

Carburetor cleaner was mixed cresols - dangerousy on skin (heavy gloves), but great against deposits.  If it is really nasty in there, I'd go for gun bore cleaner soaking, specifically Ed's Red without the lanolin and substituting MEK (methyl ethyl ketone) for the acetone to lower flammability. At the end, blow it out, flush with kerosene, blow it out.

Contain all wastes.  Beware of skin contact (gloves) .  Do not breath the fumes (ventilation, and maybe a charcoal latex paint mask, then discard).  Eye protection (goggles).  VAPORS ARE FLAMMABLE AND HEAVIER THAN AIR.

http://www.youtube.com/watch?v=2OfaCCJjBEU
Ed's Red Gun Bore Cleaner  (wear gloves!)

1 part Dexron II, IIe or III ATF, GM Spec. D-20265 or later.
1 part Kerosene - deodorized, K1 (can be Kroil or WD-40)
1 part Aliphatic Mineral Spirits, Fed. Spec. TT-T-2981F, CAS #64741-49-9, or may substitute "Stoddard Solvent", CAS #8052-41-3, or equivalent, (aka "Varsol")
1 part Acetone, CAS #67-64-1.
1 lb anhydrous lanolin or  Lanolin, Modified, Topical Lubricant.

Note that Ed's Red is rapid death to plastics, polymers, varnishes, paints, and coatings.


Thermodynamic stability of meta-xylene over ortho- and para-isomers

The thermodynamic stability of m-xylene over o-xylene or p-xylene can be deduced by considering the hyperconjugative effect. More specifically, resonance forms with a positive charge on methyl-substituted carbons are more important than resonance forms with a negative charge on them, as the methyl group stabilises the positively charged carbon by hyperconjugation.

In the o- or p- isomers, the hyperconjugation of one methyl group produces negative charge at ortho and para position where another methyl group resides. This is highly destabilising. But, in the m-isomer, hyperconjugation does not destabilise the structure and both methyl groups' hyperconjugation effects act in a cooperative way with each other.


Hydrogen bonding in staggered vs eclipsed conformations

Disclaimer: I am not a computational chemist, I am completely new to this stuff. Just a humble student of @pentavalentcarbon

  However, my thought is that if you have interacting OH groups, the hydrogen bonding would make the eclipsed conformations more stable. Is this be the case? Why or why not? For example, a small molecule containing two carbons, each with an alcohol group (or more attached).

I looked at two model compounds, ethane, and ethane-1,2-diol. Also, I did preliminary molecular mechanics based geometry optimisations using the MMFF94s force field in Avogadro.

I performed a CCSD(T) calculation in psi4 using the aug-cc-pVDZ basis. For ethane, obviously two conformers were identified:

$$\begin{array}{cc}
\hline
\text{conformer} & \text{Energy (in hartrees)} \\ \hline
\text{eclipsed} & -79.5971\\
\text{staggered} & -79.6027\\
 \hline
\end{array}$$

The difference in single point energies between the staggered and eclipsed forms $\Delta E_{\text{staggered-eclipsed}} = -0.0056$ which means the staggered form is lower in energy by about $14.7\ \mathrm{kJ\ mol^{-1}}$ (or $3.5\ \mathrm{kcal\ mol^{-1}}) $

Now, for ethane-1,2-diol. A first preliminary calculation, and random rotor search in Avogadro gave me three distinct conformers (although, two were very close in energy), however, every trial after that just produced two. They are given below:

$$\begin{array}{cc}
\hline
\text{conformer} & \text{Energy}\ (\mathrm{kJ\ mol^{-1} }) & \ce{OCCO}\ \text{bond angle} \\ \hline
\text{conformer 1} & 72.0914 & 59.0266\\
\text{conformer 2} & 85.534 & -179.967\\
 \hline
\end{array}$$

I have attached some screen shots below:

Now, hydrogen bonding is not explicitly treated in this calculation, but based on my discussion with @pentavalentcarbon the Leonard-Jones term in the forcefield (and to a lesser extent the electrostatic term) does capture these interactions.

So clearly, some stabilisng influence is evident in the structure, because if I were to argue on sterics alone, conformer 2 should have had lower energy than conformer 1. As an additional test, I measured the distance between $\ce{H}5$ and $\ce{O}2$ and found it to be 2.62 Angstroms. This is indeed suggestive

Now, using the geometries of conformer 2 and conformer 1, and that of an additional conformer with completely elcipsed oxygens that I obtained from Avogadro, I performed another CCSD(T) calculation is psi4 (again in the aug-cc-pVDZ basis). The table given below summarises my results:

$$\begin{array}{cc}
\hline
\text{conformer} & \text{Energy (in hartrees)} \\ \hline
\text{conformer 1} & -229.744410904091524
\\
\text{conformer 2} & -229.740781143788695\\
\text{completely eclipsed} & -229.725483329882337
 \\
 \hline
\end{array}$$

Again, the difference in single point energies between conformer 1 and conformer 2, $\Delta E_{\text{1-2}} = -0.0036297603$, means that conformer 1 is more stable by about $9.5399\ \mathrm{kJ\ mol^{-1}}$ (or about $2.277\ \mathrm{kcal\ mol^{-1}})$

This is probably due to an intramolecular hydrogen bond, as suggested, which has an overall stabilising effect on the gauche form of ethylene glycol (conformer 1). This is thus one of the instances in which the anti conformation (conformer 2) is higher in energy relative to the gauche conformation.

Finally, the difference in single point energies between conformer 1 and the eclipsed conformer, $\Delta E_{\text{1-eclipsed}} = -0.0189275742$, this implies that conformer one is more stable by about $49.694\ \mathrm{kJ\ mol^{-1}}$ (or $11.877\ \mathrm{kcal\ mol^{-1}}$)

Thus, I would say the energy penalty of eclipsing is still much higher than any stabilisation that can be gained from hydrogen bonding in this conformation. Shifting the angle between the two $\ce{OH}$ groups by about 60 degrees relieves this eclipsing strain while still allowing for the formation of a hydrogen bond, and is by far the best compromise.

To be honest, I wanted to scan over several dihedrals and make a potential energy plot, and calculate an exact barrier to rotation but faced two hurdles: a) couldn't write down the z-matrix properly, and couldn't write the appropriate loop in the program. b) the calculations that I did perform and have quoted here, took a while to run and based on that, I am guessing what I had in mind would have been a little too ambitious for my poor,  little laptop.

Addendum @Jan pointed out in the comments that the gauche effect is responsible for the stability of the gauche conformer, which to be fair, I did not think about at all because I was so focused on the "hydrogen bond", however, it is worth looking into. Here's a post that discusses it (answered by @Jan himself ;) )


I am not sure of putting copper nitrate to lead. But the common way to produce lead nitrate is to put lead into nitric acid. It will produce lead nitrate and hydrogen gas.

$\ce{Pb + 2HNO_3 -> Pb(NO_3)2 + H_2}$

If you have lead oxide instead of lead it also can priduce lead nitrate by adding to nitric acid

$\ce{PbO + 2HNO_3 -> Pb(NO_3)2 + H2O}$


How is it that fructose has a different metabolic pathway than glucose but yet glucose is converted to fructose?

The metabolic pathway you are talking about is how fructose is converted into energy and how its concentration in the blood is regulated. It is indeed true that blood fructose level does not affect blood insulin level, and this is why it is more "fat-inducing", as it cannot be effectively regulated like glucose.

It is also true that in glycolysis, glucose is converted to fructose-6-phosphate. However, this should not be confused with fructose itself. The fructose phosphate that fructolysis produced is fructose-1-phosphate, not fructose-6-phosphate.


Can a strong nucleophile carry out an SN1 reaction?

You've pretty much said it all already. SN2 is ruled out at the tertiary centre. Since the attack of the nucleophile in the SN1 mechanism is not rate-determining, it doesn't matter whether the nucleophile is weak or strong; iodide works fine.

In this case, it's more likely to get intercepted by the solvent itself, which is present in much larger quantities and is also nucleophilic, but that's another story.


Reference for UV/Vis Spectroscopic Data request

To know the chemical name, molecular structure, or best: the unique registry number provided by the American Chemical Society (CAS-number) is beneficial in your search as trade names may vary depending on the supplier of the material you are interested in.  The index number of the Colour Index (like C.I. 73000 for indigo) frequently used in the business of dyes potentially is of lesser use here.  In decrease of data coverage:

There are commercial sources like Reaxys (by Elsevier) allowing to search for absorption data.  As this (in part) emerged from a much senior database about chemical reactions, Beilstein (back to 1771) this is a typical database used in Departments of Chemistry at a University or Industry. SciFinder, based on the data indexed by the American Chemical Society equally allows you to find such data -- for both it might be helpful to request assistance / initial training to fully benefit from these sources.  Note: there is so much published data, none of both huge data bases may index / point back to all of the original journal articles, patents, conference proceedings, etc. pp.  Depending on your search strategy, typically you find the data you look for in both data bases--at different places.  Sometimes "The Two" are complementary to each other, sometimes an other (third) search engine finds the data on the net that neither ACS nor Reaxys indexed (yet).

Personally, because of the design "how the numerical data are managed" I would suggest Beilstein as the better choice here: if you find your compound, the data base will list in a table wavelength and molecular absorption coefficient altogther with the bibliographic reference -- often including the solvent of the analysis.  And the thought occurs to me, it is easier to query with Boolean logic, like "find a compound of molecular mass in the range of xyz to x'y'z with an $\epsilon$ greater than 20000", for example.

Less in coverage -- yet worth to point to are the compilations Sadtlers Handbook of ultraviolet spectra (today part in Bio-Rad's "KnowItAll" database) and the handbook of stains, dyes, and indicators by Sigma-Aldrich, too mention some of the printed references usually found in the section of spectroscopy/chemistry in university libraries.  Here you find not only one $\epsilon$ at $\lambda_{\mbox{max}}$, but a complete print of the spectrum.  (Be aware, the elder the sources sometimes use wavenumbers or kilo-Kaysers instead of nanometers to reference the wavelength.)

Taking into account your user profile related to quantum chemistry, perhaps you want to train your quantum chemical model with known experimental data, to evolve from a descriptive to a predictive -- and small molecules may serve you well, too.  If so, take a look at NIST's webbook based on these references.

Last but not least: Take these data with a grain of salt, often there is some limitation: the same compound / material often exhibits a slightly different UV-Vis absorption spectrum if analyzed once in solvent A, and subsequently in solvent B.  Bathochromic / hypsochromic shifts (solvatochromism) may still occur.  Solvent polarity is among the  parameters in this play.


What is this SMILES notation: C[Si](C)C |^1:1|

The SMILES syntax in question is neither mentioned in the Daylight Theory Manual, nor in the OpenSMILES specification.

It is not part of the canonical SMILES syntax, but an extension, developed  by the ChemAxon company.

According to their documentation, extensions are tagged by leading and trailing | (pipe) symbols. ^1:, followed by an atom index, indicates a

  monovalent radical center:


Compare stability of transition metal cations in aqueous solution

It is written in NCERT Chemistry class 12 "In a medium like water, d3 is more stable than d5 (see CFSE).

CFSE stands for Crystal Field Stabilisation Energy. Basically in water there are h2o molecules which complexate the ions. Assuming octahedral geometry the d orbitals split into d3 (lower in energy compared to barycentre) and d2 (higher in energy compared to barycentre). If you only fill the d3 then you would actually have a more stable configuration as compared to d5 situation where the e.g. levels would also be occupied.

This is a more prominent factor than half filled orbital stability. In $\ce{Co^3+}$ there are no stabilizations involved and hence it is least stable.

$\ce{Sc^3+}$ has a noble gas configuration and this dominates all effects in terms of stabilizing the ion.


Is it possible to store moist air at high pressure?

There is one importing principle to consider.
Air does not store vapor. Space does. Less space means less vapor capacity, no matter how much of air is there. (*)
If air is being compressed, the partial vapor pressure increases. It it reaches the saturated vapor pressure at given temperature, water vapor starts to condensate. When air is then expanded, its  absolute and relative humidity decreases, according to the expansion ratio.
E.g. Lets have 1 cubic metre of compressed air with 100% relative humidity. If it expands to 2 cubic metres and temperature raises back to the initial temperature, its relative humidity will be 50 %.
So compression provides liquid water and expansion needs liquid water to restore humidity.
If there is compressed air with cold water and if pressure drop e.g. 20 to 1 atm) then its relative humidity suddenly drops from 100% to 5% (when air warms up back after adiabatic cooling down.). Water exposed to low humidity air will start slow evaporation.
It will not help much in humidity restoration, unless some active evaporation technique is involved a/o the air release is slow, passing evaporation unit.

(*) Exactly speaking, saturated vapor pressure and therefore vapor capacity of space very slighly increases with total pressure. But it is negligible in context of your question.

Note that absolute humidity is in g/m3 - it is intensive, not extensive variable. Rather, the vapour content  = absolute humidity * volume.


Extracting metallic magnesium from cholorphyll

You can make the magnesium salt, but electrolysis of an aqueous solution of this salt will reduce the water, evolving hydrogen, not reduce the magnesium.  You can get metallic magnesium via electrolysis of the molten chloride or high temperature reduction of the oxide with carbon,  neither of these cheap or easy.  See here.


Does the arrangement of atoms around a chiral carbon matter?


The influence of diffuse functions on electron density calculations

The colours you see in the first two pictures are not electron densities, but phases of the orbitals.
The colours of a single molecule could matter if you were calculating the interactions between different molecules (in which in-phase or out-of-phase interactions could matter) but they make no difference between the molecule(s) within the same calculation.
The electron density arises from the square power of a function which is, say, positive in the red lobe and negative in the blue one.
The square power means that there is no difference between blue and red, or + and - (See Born's law). They are just with inverse phase in respect to each other, but they have no other influence on the system otherwise.
You might find in some software packages a "switch phases" button, which inverts the colours.


Rate of electrophilic bromination of toluene and ethylbenzene

First, I think it's probably important to note that the difference in $\text{+I}$ effect of a methyl vs. ethyl is very small. In principle, the ethyl is more electron-rich, however the strength of the inductive effect decays very rapidly with increasing distance. Ultimately, I suspect variance in steric factors, solvation, and electron-donating ability to the $\pi$-system of the ring by hyperconjugative overlap are the decisive factors, not the difference in induction.

If we consider sterics, the ethyl is obviously the bulkier group, and that may partially explain the difference in reactivity. The ethyl, moreso than the methyl, will hinder the approach of the electrophile, and may potentially raise the energy of the transition state and $\sigma$-complex intermediate slightly by:

(a) electrostatic repulsion toward the electrophile, particularly in the case of ortho- attack, and

(b) inhibit potential stabilization of the $\sigma$-complex intermediate by solvent molecules.

From the standpoint of electronic effects, I suspect the differences are marginal. To the extent of my understanding, the energies of $\ce{C-H}$ and $\ce{C-C}$ bonding $\sigma$ orbitals are very similar, and the efficiency of hyperconjugative overlap should consequently be similar as well. I've occasionally encountered claims that $\ce{C-H}$ bonds are slightly better $\pi$-donors, but I'm not aware of any conclusive evidence to that effect. If that is true, however, it might contribute to the observed disparities in reactivity.

For a quantitative approach, you may want to take a look at the substituent constants derived for various alkyl groups from the Hammett equation. The equation was originally developed with the aim of analyzing substituent effects on the thermodynamic equilibria of benzoic acid derivatives. However, since the reactivity of substituted benzoic acid is also affected by the mesomeric and inductive properties of ring substituents (albeit in the opposite sense from a benzene ring in an EAS reaction), I think the Hammett equation furnishes a reasonable proxy measure of at least the relative strengths of substituents as either donors or acceptors of electron density. I found one table of said substituent constants. (Keep in mind that those table values are for ionization of benzoic acid, so a negative value is actually indicative of an electron-donating effect, which is favorable for an EAS reaction but destabilizing to an acid.)

Finally, as to the reaction mechanism, I'm not going to reproduce it here. Every EAS reaction follows a very similar, schematic mechanism. For the mechanisms, take a look, for example, at the Wikipedia articles on electrophilic halogenation and electrophilic aromatic substitution more generally.


Calculating volumes of acid/base required to make buffer using pH and pKa

First of all, note that:

$$\log_{10}\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)=4.5- 3.74$$
$$\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)\neq\dfrac{e^{0.76}}{10}$$

but

$$\left(\dfrac{\ce{[A^{-}]}}{[\ce{HA}]}\right)= 10^{0.76}$$

This said, the volume will depend on the concentration of the acid and of the salt (if it's a solution) that you are provided with. This ratio will give you the relative concentrations required to get a buffer of the pH 4.5.


Entropy as an intrinsic property of matter

You can use the tools of statistical mechanics to compute the entropy of molecules in thermal equilibrium.

For molecules in an ideal gas the molar entropy is
$$S=R\ln z +RT\left(\frac{\partial z}{\partial T} \right) _V - R \ln N_A + R$$
Here $z$ is the molecular partition function
$$z=\sum_i e^{\epsilon_i/kT}$$
where the summation extends over all states available to a molecule and $\epsilon _i$ is the energy of state $i$ (states degenerate in energy are counted separately).

If you can identify all the possible conformations ("geometries") available to the molecule and their respective possible energy states, you can use the formula to compute the entropy.

Qualitatively speaking, if you increase the number of conformations you increase the entropy, since the summation in the partition function now extends over more states.

You might invoke Boltzmann's equation
$$p_i=\frac{e^{-\epsilon_i/kT}}{z}$$
to compute the probability of finding a particular molecule in state i.

Reference

McQuarrie and Simon, Statistical Thermodynamics, University Science Books (1999).


Heating CrCl3(bpy)3 — isomerism and ligand exchange

If you are given only the options $\ce{Cr(bpy)3Cl3}$, then it is mandatory for the bipyridine ligand to actually be attached to the chromium centre. If it were not, as in your suggestion $\ce{[Cr(bpy)2Cl2]Cl . bpy}$, this additional bipyridine can dissociate away as there is nothing keeping it close to the complex. On the other hand, isolateable substances are always neutral, thus if your complex cation is $\ce{[Cr(bpy)3]^3+}$, then the three chloride anions will remain in the same crystal structure and you can isolate a complex salt of the type $\ce{[Cr(bpy)3]Cl3}$. This molecular structure can be proven by being able to precipitate three moles of chloride per mole of complex, e.g. with $\ce{AgNO3}$.

  Why would heating that cause a $\mathrm{bpy}$ to leave the complex and two chlorides to replace it? Or is it $\ce{[CrCl2(bpy)2]Cl(bpy)}$ in which case, why does heating it not form the more stable complex I've stated above?

This somehow assumes that $\ce{[Cr(bpy)3]Cl3}$ is the most stable of these complexes. However, the chemical literature thinks different. Burstall and Nyholm first reported the synthesis of $\ce{[Cr(bpy)2Cl2]Cl}$ from anhydrous chromium(III) chloride using three (!) equivalents of $\ce{bpy}$ and heating to reflux.[1]

On the other hand, traditional methods to generate $\ce{[Cr(bpy)3]^3+}$ salts — most commonly perchlorates $\ce{[Cr(bpy)3](ClO4)3}$ — typically started from a corresponding $\ce{[Cr(bpy)3]^2+}$ and employed oxidation.[1] (This was apparantly first published in 1932 by Barbieri and Tettamanzi in an obscure journal that I can’t find.)[2] Relatively recently, a new method to synthesise the $\ce{[Cr(bpy)3](ClO4)3}$ complexes from $\ce{CrCl3 . 6 H2O}$ was published — which, however, builds on completely removing any chloride from the solution by using silver(I) salts as in equation (1).[3]

$$\begin{multline}\ce{CrCl3 . 6 H2O + 3 [Ag(bpy)2](ClO4) ->\\ [Cr(bpy)3](ClO4)3 + 3 AgCl v + 3 bpy}\tag{1}\end{multline}$$

Seemingly, having chloride ions in solution around the $\ce{[Cr(bpy)3]^3+}$ ion can’t be trusted.

Partially due to SciFinder being down for maintenance as of now, I was unable to find any references for the synthesis or properties of $\ce{[Cr(bpy)3]Cl3}$, but given the Burstall-Nyholm method I would not assume the cation to be more stable than $\ce{[Cr(bpy)2Cl2]+}$.

A likely explanation for the lower stability could be the charge balance. The dichloridobis(bipyridine)chromium(III) has an overall charge of $1-$, which is generally okay. In tris(bipyridine)chromium(III), there is a very small $\ce{Cr^3+}$ ion in the centre of a complex surrounded by otherwise neutral ligands. Any method to relieve the charge could well be favourable; amoung them reduction of the metal or ligand exchange with anionic ligands. Remember that charge balance is a thing: $\ce{[CoCl4]-}$ forms when cobalt(III) reacts with chloride ions mainly due to charge balance reasons.

References:

[1]: F. H. Burstall, R. S. Nyholm, J. Chem. Soc. 1952, 3570. DOI: 10.1039/JR9520003570.

[2]: Barbieri, Tettamanzi, Atti R. Accad. Lincei. 1932, 15 877. Reference as cited in [1].

[3]: W. Kharmawphlang, S. Choudhury, A. K. Deb, S. Goswami, Inorg. Chem. 1995, 34, 3826. DOI: 10.1021/ic00118a035.


Skin permeability, for all chemicals or just some?

"Skin's outermost layer, the stratum corneum, is an effective barrier to most inorganic nanosized particles."

But skin is quite permeable to many organic compounds. DMSO is one example, and it is warned that DMSO is a good solvent for other compounds that could be more toxic than DMSO.

Another example: we all (should) know that mercury is toxic. However, it's not going to hurt you much if you roll it around in your hand for a brief period (don't breathe in, though!). But organic mercury compounds are terribly toxic and skin permeable, including the infamous dimethyl mercury which killed a famous researcher of toxins.


Wave and particle properties of protons and electrons

Actually, everything has wave-particle properties and is given by the deBroglie equation: $${\mathrm p}\lambda = h$$ where $\mathrm{p}$ is momentum (mv), $\lambda$ is wavelength and $h$ is known as the Plank constant ($6.626\cdot 10^{-34}\;\mathrm{J\cdot s}$).

As you can see as the mass of an object increases, the wavelength decreases. So in macroscopic objects that we an see, such as ball, its mass (therefore momentum) is so big compared to $h$ that the wavelength is really, really small, so we can't observe it. For light waves, it is the opposite. Due to their relatively large wavelength, they have a really small mass. However we can do experiments that show the particle nature of light. Such as the photoelectric effect which can only be explained by thinking of light as a particle rather than a wave.

Electrons which have a mass of $9.011\cdot 10^{-31}\;\mathrm{kg}$ are in between the two examples that I have given you. It turns out that their mass is perfect for their wave and particle nature to be observed. Protons are roughly 1840 times heavier than electrons, however we are also able to observe their dual nature.

The answer your second question is that they do exhibit these wave and particle properties at the same time. An electron can't simply decide to be a particle for a certain instant than decide to become a wave in the next moment. All objects exhibit their wave-particle properties at the same time.


Which one of the two is the stronger base: methoxide ion or hydroxide ion?

We have $\ce{MeO-}$ and $\ce{HO-}$
$\ce{-CH3}$ has a +I effect. Therefore it destabilizes the methoxide ion. $\ce{-H}$ has a comparatively lesser +I
This leads to the conclusion that $\ce{OH-}$ is more acidic than $\ce{CH3O-}$.
This seems like a valid argument, but the order varies in gaseous/aqueous phases, because we consider polarizabilty and solvation effect, rather than inductive effect there


The role of halogens in electrophilic aromatic substitution

The bromine is also not resonance donating/activating

You can't have it both ways.  If the bromine can resonance interact with the intermediate, then it can also resonance interact in the starting bromobenzene.

Let's back up to the starting bromobenzene.

  Bromobenzene will likely react, but at a slower rate than plain
  benzene, because the bromine is inductively withdrawing

That's correct, but let's compare further.  The oxygen in phenol is electron withdrawing (oxygen's electronegativity is even greater than bromine's), yet phenol is a very reactive aromatic compound towards electrophilic aromatic substitution (it will decolorize a solution of bromine, it doesn't even need a Lewis acid catalyst).  The difference between bromobenzene and phenol is that the phenolic oxygen can interact strongly via resonance and donate electrons into the aromatic ring, while the bromine in bromobenzene also resonance interacts with the aromatic ring, but much less efficiently (the larger size of bromine's $\ce{4p}$ orbitals makes overlap much less efficient with carbon's 2p orbitals).  Although the interaction is much less efficient, there still is some resonance interaction between the bromine substituent and the ring.  Look at the analogous resonance structures for bromobenzene that you drew for the intermediate, in the contest between inductive deactivation and weak resonance activation, the ortho and para positions are the least deactivated.

When bromobenzene undergoes electrophilic aromatic substitution, the whole ring is deactivated because (unlike phenol, where the resonance effect overwhelms the inductive effect) the weak resonance effect doesn't offset the strong inductive effect.  However, since the bromine substituent does interact with the ortho and para positions by resonance, these two positions are less deactivated that the meta position which cannot interact via resonance with the bromine substituent.


Which dihydroindenyl cation is the most stabilised?

The carbocation a) is mostly stabilised because it has maximum resonance structures which leads to maximum delocalisation of the charge.

None of the others has so many resonance structures.Therefore a) is most stable.


Let's look at the „[1,4]diazepine“ part more closely. (Standalone, it's correctly named 1,4-diazepine.) The „-epine“ in the Hantszch-Widman system stands for unsaturated seven-membered ring. More precisely, one with the maximum number of noncumulative double bonds. (IUPAC uses that term instead of „conjugated“ probably to avoid confusion with physical electronic conjugation, which might not be the case, as we're at the level of topology, not even geometry.)

The exact structure is specified with indicated hydrogen. There are four 1,4-diazepine structures, 1H-1,4-diazepine, 2H-1,4-diazepine, 5H-1,4-diazepine, and 6H-1,4-diazepine (not considering their stability/existence in tautomeric pair/multiple):

In the world of abstract organic chemistry nomenclature, 1,4-diazepine without indicated hydrogen can be represented as following structure (dotted bonds for unknown bond order, any arrangement of noncumulative dbl. bonds assumed; no explicit hydrogens):

When fusing with benzene, 1,4-diazepine part has to be numbered as follows (not the other way), to get lowest numbers/letters when starting with heteroatom:

… obtaining benzo[e][1,4]diazepine (new numbering of the resulting structure marked, partially; coincidentally we get the same set of numbers for nitrogens):

Indicate the hydrogen to get our 1H-benzo[e][1,4]diazepine:

(But note that fused ring nomenclature is quite long and complicated, generally the final numbering could also depend on the indicated hydrogen position.)

Finally, note that the new nomenclature has simplified the names of benzo-fused rings, the current preferred name is 1H-1,4-benzodiazepine.


Spontaneity of an exothermic chemical reaction

Not all reaction is like this but, exothermic reactions are usually spontaneous because molecules always want to be more stable. So by reacting to another molecule and releasing energy, they can lower their own energy thus making them more stable and unreactive.


Is there a name for this algorithm to calculate the concentration of a mixture of two solutions containing the same solute?

This is a so-called "Pearson's square" or "Box method" of balancing ratios, originally used extensively in dairy industry (at least since 1920s judging from Google Books search).
Earlier the similar approach has been used in sugar industry by using "Cobenz diagrams" aka spider diagrams.

Widely popularized in Soviet books for analytical chemistry at least since 1940s (probably adapted from the German literature as many other tech novelties of that time were), also used in current Russian literature by the names "Метод креста" ("Cross method"); "Конверт Пирсона" ("Pearson's envelope") or "Диагональная схема правила смешения" ("Diagonal mixing rule scheme").


Stereochemistry & mechanism of β-cyano enone methylation

Mechanism I developed from the paper @user55119 graciously provided. Appreciate it! Any refinements I appreciate. :)


Molecular orbital (MO) diagram for N2 and N2^-

I have been taught that the MO diagram is different for molecules with
  14 or less electrons than the one used for molecules with 15 or more
  electrons.

This is (partly) wrong because the change in the order of $\mathrm{\sigma_{2p_{z}}}$ and $\mathrm{\pi_{2p_{xy}}}$ MOs to the left of $\ce{N2}$ is not directly related to the number of electrons. Rather, it is rationalized by a successive decrease of the $\mathrm{s}$-$\mathrm{p}$ interaction moving from $\ce{Li2}$ to $\ce{F2}$. The $\mathrm{s}$-$\mathrm{p}$ interaction is the bonding interaction between the $\mathrm{2s}$ orbital of one atom and the $\mathrm{2p_{z}}$ orbital of another atom which (among other things) increases the energy of the $\mathrm{\sigma_{2p_{z}}}$ MO relative to the hypothetical case without $\mathrm{s}$-$\mathrm{p}$ interaction.

Now the difference in energy between the $\mathrm{2s}$ and $\mathrm{2p_{z}}$ AOs increases from $\ce{Li2}$ to $\ce{F2}$ due to increasing nuclear charge and poor screening of the $\mathrm{2s}$ electrons by electrons in the $\mathrm{2p}$ subshell. As a result, the rising effect of $\mathrm{s}$-$\mathrm{p}$ interaction on $\mathrm{\sigma_{2p_{z}}}$ MO is getting less and less prominent, so that eventually to the right of $\ce{N2}$ $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\pi_{2p_{x,y}}}$ MO.

Image courtesy of the UC Davis ChemWiki.

Now, back to the question.

  For $\ce{N2-}$ there are 15 electrons. Will the MO diagram be the same as
  that of $\ce{N2}$ (because it is actually an ionized molecule of $\ce{N2}$) or not?
  Why?

The short answer is: we could not tell it using the primitive molecular orbital theory introduced in the general chemistry courses. In exact same way we could not tell why $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\sigma_{2p_{z}}}$ MO to the left of $\ce{N2}$ and not to the left of, say, $\ce{C2}$. All this is simply because the primitive molecular orbital theory does not explain the things, rather it rationalizes them (for differences between explanation and rationalization, please, read here).

But then how do we know in the first place that $\mathrm{\sigma_{2p_{z}}}$ MO becomes lower in energy than $\mathrm{\sigma_{2p_{z}}}$ MO to the left of $\ce{N2}$ so that we can start to rationalize it? Well, we know that from a more advanced molecular orbital theory which is based on the laws of quantum mechanics. And the same advanced theory should be applied for the case of $\ce{N2-}$ to get the MO picture of this chemical species.

Now note that even in this advanced molecular orbital theory a bunch of approximations is introduced, and the answer in general depends on at which level of theory calculations are done. The simplest picture would be the so-called frozen-orbital approximation (in the Hartree-Fock picture) in which MOs of an anion (as well as that of a cation) are assumed to be the same as for the corresponding neutral molecule. But this approximate picture might be far away from the reality.

Update

The question in fact does not make a lot of sense since $\ce{N2-}$ is known to be electronically unstable, i.e. its energetically less favourable then the system of $\ce{N2}$ plus a free electron. There exist few metastable resonance states of $\ce{N2-}$ that correspond to attachment of the scattered electron to excited valence states1, but that's a different story which is beyond the scope of the question.

1) Krauss, M., Neumann, D., Valence Resonance States of N2, JResNatlStandSecA, 77A, 4, 1973. PDF.


Can the following molecules be called meso compounds (cyclic structures which usually exist in non-planar forms)?

These molecules can never be planar. There are no internal symmetry planes and therefore they are not meso.


How could you distinguish between an irreversible electron transfer couple and a chemically irreversible couple using cyclic voltammetry?

To this question there are black-and-white as well as grey answer parts.

Chemical reversibility refers to the generation of a species that cannot be reduced or oxidised furthermore. In case you scan at moderate speed (scan rate), you may reduce or oxidise a compound and this intermediate can further react to an electrochemically inert species. If the follow-up reaction occurs fast, this couple is chemically irreversible.

Electrochemical reversibility refers to the rate constant of electron transfer. High rate constants (fast electron transfer) are found in electrochemically reversible couples whereas low rate constants are found in electrochemically irreversible cases.

Okay, you can trick chemical reversibility by scanning so fast, that you oxidise/reduce the intermediate of a redox-reaction at the electrode so fast that it has no time to further react. In this case, you should have only a forward-peak using slow cyclic voltammetry without a backward-peak; but you should start to see a backward-peak as you increase the scan rate.

For electrochemical reversibility, you can use Butler-Volmer kinetics. The main message of Butler-Volmer is that you increase the rate of electron transfer by increasing the potential. Electrochemical reversibility is seen on the peak separation between oxidation and reduction peak. Reversible systems show a peak separation at standard conditions with one electron transferred of 59 mV.

So, how would I dinstinguish them using cyclic voltammetry? By the peak shape and the change of CV by varying the potentials and scan rate.


How many tetrahedral and octahedral holes are there in a bcc unit cell?

BCC doesn't have any tetrahedral voids, however we have octahedral holes at face centres. Four adjacent corners and two closest body centres. In a homogeneous lattice the octahedral is compressed along the body-centre-atom axis. However I believe we can form a hetrogenous lattice and form a close approximation to octahedral.

We have thus $6\times1/2 = 3$ distorted octahedral voids.


Is the equation pH + pOH = 14 valid at different temperatures?

$\mathrm{pH + pOH = 14}$

This equation only holds true at around $25~^\circ\mathrm{C}$, where the water autodissociation constant $K_\mathrm{w} = 10^{-14}$.

Mathematically, you can write

$$\begin{align}
K_\mathrm{w} = [\ce{H+}][\ce{OH-}] &= 10^{-14} & &\\
\log([\ce{H+}][\ce{OH-}]) &= \log(10^{-14}) & &\text{(logarithms of both sides)} \\
\log[\ce{H+}] + \log[\ce{OH-}] &= -14 & &(\log(ab) = \log a + \log b) \\
-\log[\ce{H+}] - \log[\ce{OH-}] &= 14 & & \\
\mathrm{pH + pOH} &= 14 & & \text{(definitions of pH and pOH)}
\end{align}$$

It's fairly clear from this that if $K_\mathrm{w}$ is $10^{-x}$ at some other temperature, then you will obtain $\mathrm{pH + pOH} = x$.


The name ‘3,4-diethyl-6-isobutyldecane’ that is proposed in the question is unambiguous and describes the correct structure. However, it corresponds to obsolete IUPAC recommendations.

The retained prefix ‘isobutyl’ was still contained in the 1993 recommendations but is explicitly no longer recommended in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book).

According to current IUPAC recommendations, the preferred prefix for the group $\ce{(CH3)2CH-CH2-{}}$ is ‘2-methylpropyl’.

Therefore, the preferred IUPAC name (PIN) for the compound given in the question is 3,4-diethyl-6-(2-methylpropyl)decane.


What is the origin of the differences between the MO schemes of O₂ and N₂?

This phenomenon is explained by s-p mixing. All the elements in the second period before oxygen have the difference in energy between the 2s and 2p orbital small enough, so that s-p mixing (combination) can occur lowering the energy of the σ(2s) and σ*(2s) and increasing the energy of the σ(2p) and σ*(2p) molecular orbitals. By moving towards right in a period, the s orbital gets more stabilized than the p orbital and the difference in their energies increases, making the s-p mixing for oxygen much smaller.


Why is my solution for the partial pressures of two different gases incorrect?

The answer by camd92 is correct, but there is no reason to calculate amounts. According to the chemical equation, one third of the particles in the gas phase will be dinitrogen oxide, and two thirds will be water. So you know the mole fractions, and you can directly calculate the partial pressures. (And yes, camd92 is also correct in pointing out the mistakes in the parts of your calculation that weren't necessary in the first place).

$$P_{\ce{N2O}} = \frac{1}{3} \times 3.70\ \mathrm{atm} = 1.23\ \mathrm{atm}$$
$$P_{\ce{H2O}} = \frac{2}{3} \times 3.70\ \mathrm{atm} = 2.47\ \mathrm{atm}$$

The bottom line is that chemical equations (i.e. stoichiometry) describe the ratios of amounts, and that is exactly what you needed here to determine the partial pressure from the total pressure.


What happens when a symmetry element is added or removed from a point group?

Inversion $i$ is equivalent to a two-fold improper rotation $S_2$. Introducing a $2n$-fold improper rotation axis to the existing $C_n$ point group will result in a new $S_{2n}$ point group, e.g. in your case $C_3 ⊗ i = S_6$.


Can =NH2+ in Arginine's side chain form hydrogen bonding at physiological pH?

Yes, the arginine side chain is an excellent hydrogen bond donor. Charged N-H groups are even better hydrogen bond donors than the corresponding neutral N-H groups. Here is an article that gives examples of arginine side chains in hydrogen bonds. It showcases bidendate hydrogen bonds (i.e. two hydrogen bonds from the arginine side chain) with carboxylates and with nitrogenous bases of nucleic acids.

Specifically for proteins, if the hydrogen bond donor is positively charged (e.g. Arg+) and the hydrogen bond acceptor is negatively charged (e.g. Glu-), you have a combination of ionic and hydrogen bond interaction, which is particularly strong. It is often referred to as a salt bridge, such as in this example. If you take a close look at the arginine side chain, you will see that it is protonated (i.e. positively charged) and interacts with two negatively charged carboxylate groups.


Heating of mixture of gases with one gas barely above its boiling point

You could certainly have such a situation; that is, heating two compounds in their gaseous state at constant volume such that that the total pressure of the system exceeds the vapor pressure of the lower boiling compound causing it to condense.

As the gas condenses, it will of course give off heat as you said, but it would require continued heat input to sustain the increasing pressure and condensation. I believe that is what you meant in your second question anyway.


This is a pretty straight forward equilibrium problem in disguise. If we assume the initial state represents equilibrium, we can determine the equilibrium constant by plugging in the concentrations to the law of mass action. We can get the concentrations because we have the number of moles and the volume of the container.

$$\ce{N2O4 <=> 2NO2}$$

$$K_c=Q_{eq}=\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}$$

Okay, now we let the gas expand. Being in thermal contact with the surroundings is important, since it means the temperature will not change (which means the equilibrium constants does not change). Now we will find that we are not at equilibrium any more. We will probably find that $Q_c < K_c$, so equilibrium needs to shift toward $\ce{NO2}$.

Now, we can set up the ICE (initial, change, equilibrium) table:

          [N2O4]           [NO2]
I         5.71 M           5.71 M
C         -x               +2x
E         5.71 -x          5.71 + 2x

Substitute into the law of mass action at equilibrium:

$$K_c = \dfrac{(5.71+2x)^2}{5.71-x}=13.33$$

Rearrange and solve for $X$. Being a quadratic equation, you should get two roots (two solutions for $x$), and as is always the case for an equilibrium problem, one of them will be nonsense. For example, you could get one value of $x$ to be larger than the initial concentrations!


Why do some solvents bubble more than others when boiling?

Boiling depends on certain properties of a solvent:

Tendency to form an emulsion of bubbles due to low surface tension and separation of solvent phases, as in soap bubbles. Try boiling some dish-washing detergent in water.
Tendency to superheat, i.e. to go well above the boiling point and then to suddenly form bubbles, which is more pronounced at lower pressure. This can cause violent bumping, in addition to foaming. The cleaner the apparatus, the more likely, so boiling stones (chips) are added to nucleate more gentle boiling. A glass mug of water heated in a microwave can bump violently enough to blow open the door (and blow the fuse).


The answer is yes. Aldehydes will decolourise bromine water.

The oxidation of aldehydes to acids is discussed in this answer Does Br2/H2O oxidize all aldehydes?

In addition alpha-bromination of the aldehyde via the acid catalysed bromination of the enol is a well-documented process. https://chem.libretexts.org/LibreTexts/Athabasca_University/Chemistry_360%3A_Organic_Chemistry_II/Chapter_22%3A_Carbonyl_Alpha-Substitution_Reactions/22.03_Alpha_Halogenation_of_Aldehydes_and_Ketones

http://www.chem.ucalgary.ca/courses/350/Carey5th/Ch18/ch18-3-1.html


What are better acids for removing rust from an iron nail than phosphoric, or oxalic acid?

In industry, rust is cleaned using hydrochloric acid, because besides being an acid, it is also a reducing acid.


Calcium carbonate and hydrogen phosphate producing calcium phosphate, carbon dioxide, and water

Most likely that your teacher meant to say phosphoric acid ($\ce{H3PO4}$) rather than the hydrogen phosphate anion ($\ce{HPO4^{2-}}$) so lets go with that.

$$\ce{ 3CaCO3 + 2H3PO4 -> Ca3(PO4)2 + 3CO2 + 3H2O}$$

In reality the carbon dioxide and water would exist in an equilibrium state with carbonic acid, $\ce{H2CO3}$:

$$\ce{ H2CO3 (aq) <=> H2O (l) + CO2(aq)}$$


It is common practice to define the standard state of a gas when describing its thermodynamic properties. For instance, the free energy of an ideal gas can be defined relative to the free energy in its standard state (the gas at a partial pressure of 1 bar and at the same temperature), as

$$G = G^\circ + RT\log\left( \frac{p}{p^\circ} \right)$$

where $p^\circ= \pu{1 bar}$.

In practice it may be impossible to achieve the gaseous state described in the OP, not least because absolute zero is not experimentally accessible. However, there are workarounds that might allow one to extrapolate to $\pu{0 K}$ and to standard conditions (a pressure of 1 bar). First the easy part, the pressure: if you have a very dilute gas, then intermolecular interactions become unimportant in describing its behavior, it becomes practically ideal. As such, its behavior at any other pressure can be described by using standard equations applicable to ideal gases. It does not matter that the state described is not real, it can be used as a hypothetical reference point. For instance, if we know the free energy under ideal gas conditions at a particular pressure, we can compute the free energy in the standard state as follows:

$$G^\circ = G - RT\log\left( \frac{p}{p^\circ} \right)$$

For an ideal gas, the internal energy $U$ is independent of pressure, so in the given example there is no need to perform any computation at all, provided we are at the right T. The problem is of course that we are not at the right T. In practice it may be possible to either assume that the experiment was conducted sufficiently close to $T=\pu{0 K}$, so that we can assume the equality holds, or to extrapolate to that temperature by performing experiments over a range of T.

As for the approximation $\Delta U (\pu{0 K})\approx \Delta H (\pu{298 K})$, I'll give a roundabout justification, since I am not an expert on the subject. There are two points that can be made. First, it is probably ok to assume $\Delta U (\pu{0 K})\approx \Delta U (\pu{298 K})$ if it can be assumed that the electronic transition is not sensitive to differences in the distribution of (mainly vibrational) thermal motions at $\pu{0 K}$ and $\pu{298 K}$.

Second, $$\Delta H = \Delta U + \Delta (pV) = \Delta U + \Delta n RT $$ However here we have a conundrum. The reported IE ignores the kinetic energy of the electron (it assumes that the emitted electron has no kinetic energy). But the thermodynamic expression assumes thermal equilibrium, so which is it? Well, at zero Kelvin there is no translational energy, so when considering $\Delta U(\pu{0 K})$ we can safely ignore any thermal contribution from the electron. So we are ignoring the thermal properties of the released electron, and then since $\Delta n =0$ (ignoring the electron) we have that $$\Delta H = \Delta U$$