text:
Describe the function of a salt bridge, What would happen if the half-cell compartments of a galvanic cell were not connected?
I know,  The primary function of a salt bridge is to complete the circuit, so that charge can flow from one half-cell to the other. The second function is to balance the mass by allowing the anions to move to the half-cell where addition cations are being produced.But I can't seem to figure out what would happen if the half-cell compartments of a galvanic cell weren't connected?

My electrochemical knowledge is a wee bit rusty, but I believe the salt bridge helps balance the charges and not mass.
What charge you ask? So, let imagine you start up your galvanic cell (below). Electrons from the Zn move to Cu electrode. You start to build up a negative charge on the Cu which would resist/repel further addition of electrons after a period of time, hence production of electricity ceases.
By having a salt bridge, the electrolyte ions can move from one side to the other to mitigate this issue.


text:
Why is the conductivity of non-metals not equal to zero?
My chemistry book says that delocalized electrons are the reason for electrical conductivity in a metal. However it says that nonmetals like sulfur, phosphorus and silicon have "low" conductivity. How can they conduct electricity if they have no delocalized electrons? Silicon has a giant macromolecular structure while sulfur and phosphorus have simple molecular structure. What causes them to conduct even for a little bit?

When we think of molecules we think of their molecular orbitals.  If we place these molecules into an ordered solid like a crystal, then the orbitals on adjacent molecules are close enough to interact and "extended" molecular orbitals are formed.  We call these extended molecular orbitals "bands". Typically these bands are "thicker" (spread over a wider range of energies) than molecular orbitals.  That is to say that because the energy of a molecular orbital in a molecule near an edge will be slightly different than the same molecular orbital in a molecule in the middle of the crystal, there is a spread of energies associated with a band.  Ultimately, because of this "thickening" of bands, all of the bands containing electrons overlap and form what is termed the valence band.  Similarly all of the empty bands overlap and form the conduction band.

(image source)
The valence band and the conduction band are analogous to the highest occupied molecular orbital (HOMO) and lowest unoccupied molecular orbital (LUMO) that exist in individual molecules.  The HOMO-LUMO separation is termed the "band gap" in a solid.
In metals the band gap is essentially zero with the valence and conduction bands overlapping; electrons move readily from the valence band to the conduction band. Electrons in the conduction band then move freely throughout the solid and give metals their characteristic conductivity. In insulators the band gap is very large and very few electrons have the thermal energy needed to be promoted to the  conduction band, hence there is basically no conduction.  Semiconductors lie in between these two extremes and thermal energy can promote some electrons (not as much as in a metal, but far more than in an insulator) to the conduction band, enough to have some conduction.

(image source)

For an insulator to "conduct", basically an electrochemical reaction has to take place and travel through the bulk as a "defect". Above 0 K, all solids must contain defects, defects are able to move through a lattice, because they are more mobile, so all insulating solids have a finite conductivity, that generally grows with temperature.
In a semiconductor (see rons answer), the energy necessary to reach the "conduction band" is within the range available from thermal motion at moderate temperatures. In insulators, you need an energy that is so large it is able to break bonds. This is not so easily available, so the conductivity is very low.

Even in 0K defects are present.

Defects don't magically vanish at 0 K, but they are not thermodynamically necessary.

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text:
Mechanism for formation of amide by reaction of amine with ester
This is the mechanism given in my lecture handout:

I have a problem with the third species along. Surely the negatively charged oxygen would protonate before the other oxygen would. Also, why does this reaction work at all? The amine group is more basic than the "alcohol group" that must leave to form the amide so why wouldn't that protonate first and then just leave (i.e no reaction)?

In mechanism we write equilibria which lead to observed product.

Protonation / deprotonation by solvent is fast as H+  is just a subatomic particle and is therefore far nimbler than any other ion. Bonds between larger atoms are made and broken slower, because they have more mass and the baggage of inner electron shells, which makes the transition more unwieldy. Any and all reasonably conceivable protonations will exist in fast equilibrium. The rate determining step will either be 1-->2 or 3-->4  (probably the latter, due to the low concentration of species 3 for the reasons you indicate.)

Surely the negatively charged oxygen would protonate before the other
  oxygen would.

Most of the time, but not all of the time.

The amine group is more basic than the "alcohol group" that must leave
  to form the amide so why wouldn't that protonate first and then just
  leave (i.e no reaction)?

Most of the time, but not all of the time.
Your reasoning is correct, but remember, all of these steps are equilibria.  Although what you say will happen most of the time, some of the time the reaction will take the forward route.  Presumably the reaction is run with a large excess of ammonia to "force" the reaction in the forward direction.

Wouldn't protonation of the solvent (which is probably also $\ce{NH3}$) be quicker and hence a more justified mechanism than the proposed one?

@Jori Do you mean proton transfer to the solvent by the first tetrahedral intermediate instead of proton transfer to the alkoxy leaving group?

Yes that is what I meant.

@Jori  They are all in equilibrium, so they're probably all happening.  Protonation of the alkoxy group has the advantage of 1) being intramolecular and 2) making the alkoxy a better leaving group.  But who knows what the exact pathway for the proton transfer is?

Another way to explain the forward reaction is to remember that attacking an ester is easier than attacking an amide; or that amides are thermodynamically favoured compared to esters.

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text:
Proper dash to use for negative charge
What type of dash does one use when indicating something like F$^{-}$? I assumed it was a minus, since you indicate positive charge with a +, but there is one professor I know who insists that it should be typeset with an endash. Is there a style manual on chemical typesetting I can consult.

Say for instance in Microsoft Word, if you compare the minus sign from the table of symbols (which is the correct symbol) to a hyphen, an en dash, and an em dash. You may find it quite difficult to distinguish between the legitimate minus symbol and the en dash. The hyphen and the em dash easily look like sloppy choices for an alternative.

On this matter, the Green Book (‘Quantities, Units and Symbols in Physical Chemistry’) refers to the Red Book (‘Nomenclature of Inorganic Chemistry – IUPAC recommendations 2005’), which clearly states that plus and minus signs are used to indicate the charge on an ion in a formula or name.
However, this statement does not clarify which particular character shall be used as minus sign. Nevertheless, the Red Book also explicitly mentions hyphens and ‘em’ dashes for other purposes but no ‘en’ dashes. Hence, I suppose, the Red Book would have mentioned ‘en’ dashes if this character shall be used.
Fortunately, signs and symbols for quantities and units etc are standardized in the international standard series ISO/IEC 80000, which has been adopted by many national standards. This standard is intended mainly for use in the natural sciences and technology. ISO 80000 Part 2 (‘Mathematical signs and symbols to be used in the natural sciences and technology’) clarifies precisely that the symbol that shall be used as minus sign really is the MINUS SIGN of ISO/IEC 10646 and Unicode (they are identical). The hexadecimal code assigned to the minus sign by both ISO/IEC 10646 and Unicode is 2212.

[…] one professor I know who insists that it should be typeset with an en dash.

I disagree! A standard minus sign should be

on the same height as the horizontal line of a plus sign;
as long as an equal sign.

The en dash is

lower than the horizontal bar of the plus sign;
slightly longer than the equal sign.

Unicode has dedicated characters for minus and superscript minus

standard minus: U+2212: −
superscript minus: U+207B: Cl⁻

See also:
The situation, and how to realize the correct typography using $\mathrm\LaTeX$ has been described in the German article Feinheiten bei wissenschaftlichen Publikationen – Mikrotypographie-Regeln, Teil II by Marion Neubauer in Die TeXnische Kommödie, 1997, 1, 25-44 (PDF).

Do you have an authoritative source for that I can show my group members when this comes up?

@Canageek Is the journal of DANTE, the German TeX User  Association, good enough? I've updated my answer.

It would be better if it was a chemistry style manual, something by the ACS or RSC for example.

I've been talking with some people on the TeX Stackexchange. It seems that in most fonts hyphen-minus, the keyboard symbol, is the length of a hyphen, not a minus. So the correct solution is to manually insert U+2212 (MINUS SIGN). However if that is not available, the endash is usually the same length as it and is a better replacement then hyphen-minus.

@Canageek In $\LaTeX$, things are easy :) Within equations in math mode, a - is automatically rendered in the right length and with the right vertical position. The same is true for a superscripted minus within a \ce{...} expression.

@Canageek To quote the Red Book (Nomenclature of Inorganic Chemistry) by the International Union of Applied Chemistry (IUPAC), rule IR 2.3.2 "Plus and minus signs" their statement is: "The signs $+$ and $-$ are used to indicate the charge on an ion in a formula or name.  Compare with this http://old.iupac.org/publications/books/rbook/Red_Book_2005.pdf, page 25.  Not a iota of "dash".

@Buttonwood There's not doubt about what a minus sign is used for :-) The "problem" in typography arises from the fact that there is no real minus sign on an ASCII keyboard. It actually is a hyphen.

"The ndash is - lower than the horizontal bar of the plus sign - slightly longer than the equal sign."  Actually, neither of those is always the case. It depends on the font. Yes, the en dash is usually slightly longer than a true minus sign. But using Helvetica in Word-for-Mac, for instance, it's 1% shorter.  And as for the height, there is no general rule.  In Word-for-Mac, the en dash is lower with Arial and Times, nearly dead-even with Calibri, and higher with Helvetica.

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text:
Stable conformers of piperazine
which conformer stable of piperazine, when hydrogen of N in axial or in equatorial, or lone pair? why?

If you mean piperidine, see here; if you mean piperazine, you can apply the same thinking and conclude that the axial-equatorial preference will be even smaller (~0 kcal/mol).


text:
What is the difference between tempering and hardening metals and their end products?
I think I've heard hardening is when you dunk red-hot metal into cold water, and tempering is when you take that hardened metal, heat it slightly, and then let it cool slowly. However what is the difference between their end products?

Congratulations on your Yearling badge!

As the names imply, hardening makes the metal more rigid but more brittle, and tempering (from "temperate", moderate), forgoes some hardness for increased toughness. Iron alloys are hardened by rapid quenching and toughened by annealing, but some copper alloys are made more ductile by quenching and are hardened by working.
In brief, annealing allows atoms to migrate to a less-strained position, whereas quenching lock atoms in place; this is used to make tempered glass, as well as to harden metals.
Extremely rapid quenching can produce a metallic glass, where precipitation of constituents is prevented.
Edged-tools may need the body to be ductile, so that it does not shatter, but require for the edge to be hardened so that it retains sharpness. See http://www.smt.sandvik.com/en/products/strip-steel/strip-products/knife-steel/hardening-guide/purpose-of-hardening-and-tempering/ or http://www.stormthecastle.com/blacksmithing/blacksmithing-a-knife/stock-removal-method-of-knifemaking-part-4.htm.

Though the other answer provides a good insight, I think annealing and tempering could have been confused.
I will talk with respect to iron-carbon alloys or steel here.
Tempering is usually a post-quenching or post hardening treatment.
It is done to relieve internal stresses, decrease brittleness, improve ductility and toughness. The usual heating range for tempering in steel is from $150\ \mathrm{^\circ C}$ to $600\ \mathrm{^\circ C}$ and it is below the upper critical temperature or the eutectoid line. The cooling is done in still air.
Annealing can induce ductility, soften material, relieve internal stresses, refine the structure by making it homogeneous (though grain coarsening can also happen), and improve cold working properties. Cooling in case of annealing is usually furnace cooling which gives the slowest of cooling rates.

The products for annealing will depend on the carbon % and the temperature ranges from which it is cooled. But usually it is pearlite + ferrite/cementite.
The product of tempering is usually just tempered martensite or bainite.
Hope this was helpful.


text:
How to extract wavefunction from GAMESS output?
Is there a way to extract the molecular wavefunction from GAMESS output?
I know there are packages that can extract the eigenvalues or the exponents and show them as a table but that is not of much help as anyone can easily find those values in the output file by looking at the numbers.
What I am looking for is the complete wavefunction equation. Is there such a package or something available. Something like this simple extractor I have written for STO-3G (this is just a very primitive one that only works for small STO-3G outputs ) :

What kind of calculation your doing(SCF/CI/MCSCF/MPPT)?  How large?  The numbers  are in a few different places and the actual wave function can easily have millions or billions of terms and shouldn't need to be written down in this form if it is any system of real value.

@XiaoleiZhu It is SCF and the basis set is 6-311.  The system is not that large ( maybe just ~100-150 orbitals )

100 orbital can be difficult to handle in Mathematica.  Say you have 20, electrons, then it is a 20x20 determinant, each element of which is a 100 term summation of AO, and each AO is a summation of from 1 to 6 Gaussian functions multiplied by an angular part.

If you specify what you want to do with the wave function, then we can probably provide more help.  There is nothing that generate Mathematica code from GAMESS output as far as I know, but there are a number of programs that can extract the wave function and calculate properties or plot orbitals for you.  I can try to show you how to write a program to do it, but I am not sure if it will worth the effort.

@XiaoleiZhu What I want to do is related to my other question that you answered ( calculation of molecular volume based on that wavefunction ). Now I am trying to find my molecular dimensions and topography ( not just the volume ).          In Gaussian I could calculate the volume based on 0.001 e/bohr but it is just the volume. I need the dimensions for example at 0.05 e/bohr. even arbitrary dimensions can help me a lot .

In Gaussian you can generate an electron density "cube" file, which tabulates density in a specified grid.  That may be useful.  I am not sure about GAMESS

@XiaoleiZhu That is a function that exist in practically all the quantum chemistry program. I agree, going the Mathematica way is not really an option.

I'm not too familiar with the input structure of GAMESS. But, I found and input in which you can find how to call the AIMPAC keyword GAMESS input with AIMPAC keyword and this is how wfn file looks like. In principle, when you indicate the AIMPAC keyword, the GAMESS itself print the wave function in a file with wfn extension at the same folder of the input/output.
After you get the wfn file, you have to integrate it with (I recommend) 1. AIMPAC or 2. MULTIWFN (Both are free).


text:
CASTEP Geometry Optimization Movie
I am using a calculation software called CASTEP, within Materials Studio, and I want to make a "movie" of my geometry optimization, but don't know how to. CASTEP allows one to do many many different types of calculations, and one of the simpler ones it does is a geometry optimization for a unit cell. I remember hearing that it was possible to show a "movie" of the geometry optimization. This "movie" would show all of the steps from the beginning atom positions in the unit cell to the final converged one. I don't know how to make this "movie" and was wondering if someone else knows how.

Welcome to Chemistry Stack Exchange! Find out everything you need to get started by taking the tour

The geometry optimization routine does not appear to have a method to output frames to an .xtd file for CASTEP in Materials Studio. Usually a movie makes more sense for a molecular dynamics simulation.
However, if you construct a script that performs a sequence if geometry optimizations with MaxIterations = 3 (must be >2), you can output each result to .std (study table) or .xtd. You would do this by outputting a geometry optimization after three iterations, then repeat the calculation on the result, until convergence. Then you take the frames and make a video
The help file in Materials Studio is rather a good resource, be sure to dig around in there, especially for the scripting API


text:
Are all deliquescent salts highly soluble?
$\ce{Na2CO3}$ is deliquescent in nature, and $\ce{NaHCO3}$ is not, for example, so $\ce{Na2CO3}$ should have been more soluble than $\ce{NaHCO3}$ in water ($\ce{NaHCO3}$ I always thought was less soluble since it precipitated easily in the Solvay process). But my book says $\ce{Na2CO3}$ is less soluble than $\ce{NaHCO3}$ in water. Isn't deliquescence though an indicator of high solubility in water?

As already pointed out, dissolution and deliquescence are different phenomena.
For dissolution to take place, the energy produced by hydration of the salt's constituent ions should compensate for the energy required to break the crystal lattice.
Solubility then depends on temperature, pH and also the amount of salt already present in the solution. (Recall: definitions of solubility products)
Hygroscopic characteristics in general can be attributed to both absorption and adsorption (at the surface).
Particularly, deliquescence occurs when the vapour pressure of the saturated aqueous solution of a substance is less than the vapour pressure of water in the ambient air.
When water vapour is collected by the pure solid compound, a mixture of the solid and liquid or an aqueous solution of the compound forms until the substance is dissolved and is in equilibrium with its environment; when this happens, the vapour pressure of water over the aqueous solution will equal the partial pressure of water in the atmosphere in contact with it.
A crystalline salt aerosol particle will deliquesce in the atmosphere when the relative humidity surpasses a characteristic value, i.e., the deliquescence point.

Need not be for a salt to be soluble the $\Delta_{hydration}H≥\Delta_{lattice}H$ so it must break the lattice.
[see this][http://www.britannica.com/EBchecked/topic/156605/deliquescence]
Given there is that

deliquescent salts need not be soluble but all soluble salts are deliquescent

But http://www.britannica.com/science/deliquescence says that deliquecsent salts absorb moisture until they dissolve in the absorbed water to form a solution.

First of all, deliquescence is not an indicator of high solubility. Secondly, Solubility depends on factors such as temperature, pH value etc. It will be less confusing if keeping these in mind when reading the textbook. Precipitation of the bicarbonate sodium in Solvay process is due to the high pH condition. Please check wikipedia for details.

No. Very good example is Quick Lime. Buy some lumpy Quick Lime and leave it out in a humid atmosphere for some days. It all turns to powder. It will absorb water as a dessicant but is like 1% soluble in water and horribly stains your glassware, steel etc.
Fun fact, if you need to add alkalinity to anything using Calcium, I find that keeping rocks out in the open atmosphere works best. Before, I used to pay the pulverizing cost but leaving them out in the open turns them into ultrafine dust any appropriate powder handling system can accommodate. Life hack


text:
Accelerated Oxidation Of Iron When Coating Breaks
I was presented with the following information:
If iron is painted, then oxygen and water are prevented from contacting the metal, and corrosion is avoided.
Sometimes the iron is coated with a thin layer of another metal, such as Tin (the layer of Tin keeps oxygen and water away from the iron)
If the layer of Tin is worn away or scratched, the presence of Tin will actually accelerate the corrosion of the underlying iron. (http://www.mikeblaber.org/oldwine/chm1046/notes/Electro/Corrode/Corrode.htm)
I understand why Iron would be oxidized when the coating breaks, as an observation of the reduction (oxidation) potential would show that it is more favorable to oxidize iron, over tin.
However, the portion I am confused on is:
Why would corrosion be accelerated?
The passage I was reading suggested that it was because Tin acts as a cathode and thus would accelerate the corrosion, but that does not make a lot of sense to me. As I would see it, there would be little difference whether the Tin is there or not, and thus the reaction (corrosion) should not be accelerated, but simply as quick as it would have been without the coating.
Any insight would be greatly appreciated! Thanks!

The main factor for this phenomena is the the values of the standard electrode potentials for of the metal ions.
The UC Davis ChemWiki page Corrosion, uses this scenario as an example:

the presence of the second metal can actually increase the rate of corrosion. The values of the standard electrode potentials for $\ce{Sn^2+}$ (E° = −0.14 V) and $\ce{Fe^2+}$ (E° = −0.45 V) show that $\ce{Fe}$ is more easily oxidized than $\ce{Sn}$. As a result, the more corrosion-resistant metal (in this case, tin) accelerates the corrosion of iron by acting as the cathode and providing a large surface area for the reduction of oxygen

Tin is therefore more of a noble metal than iron - a galvanic corrosion chart is shown below, showing that tin is more 'noble' than iron (Image source):

The significance of tin being more noble than iron is that, according to the UC Davis ChemWiki page Electrochemical Corrosion, it can act as a depolariser, removing more electrons than the water/air alone - hence, accelerate the corrosion process.

Instead of the copper used in the example, it would be tin in your example.
A somewhat simplied illustration of the process is shown in the diagram below:


text:
Why are the axial bonds of PF5 longer than those of the equatorial bonds? (Hybridization)
The axial bonds of $\ce{PF5}$ are longer than those of the equatorial positions.
One explanation is that because the axial bonds are experiencing more repulsion
than those of the equatorial and this leads to longer bonds.
Another explanation is that the axial bonds of p/d hybrids, whereas the equatorial bonds are s/p hybrids.
I am able to comprehend and appreciate the first explanation. However, I am unable to understand how there are differences in the hybrids. I had always considered d hybrids a dubious argument at best, and I believed the hybridization would lead to degenerate orbitals.
However,  the explanation provided would suggest otherwise, that there are different hybridized orbitals of different energies at different portions of a single hybridized atom (excluding lone pairs. I understand those being non-hybridized in a hybridized atom).
Thus, my question is how the hybrid argument that was presented works and how would one be able  to realize the different hybridized states of an atom in a molecule (as an approximation).

I've been thinking and I thought that perhaps rather than hybridization, the p/d explanation can account for the Non-hybridized state as the bonding is not degenerate and could help explain why a single bond is longer. For Fluorine, instead of a hybridized state, it would bond with a p, not sp3. For P, it would bond with a d? (I would say p) instead of p (I would say s). However, this is only a guess and it would be nice to have someone give their ideas.

This earlier SE Chem post answers your question using $\ce{PCl_5}$ as the example, but the same arguments apply to $\ce{PF_5}$ as well

Older textbooks decided to use a d-orbital formalism for elements of the third and lower (in the periodic table) periods to explain hyperconjugation. Nowadays, most of these compounds are explained with multi-centre bonds.
I could go stealing pictures and explanations from ron’s great answer, but I’m not going to. Rather, I’m just going to quickly explain the gist and redirect you to that answer.
If you take away two of the fluorines, then you can imagine the $\ce{PF3}$ fragment to be sp²-hybridised with a remaining p-orbital containing two electrons. (This description is wrong, as can be seen from the pyramidial structure of $\ce{PCl3}$ and its analogues, but we’ll keep it as a working basis.) The p-orbital can now interact with two surrounding further fluorine atoms that approach it from above and below with an electron each in the way that is shown in the image in ron’s answer. The resulting three orbitals will thereby create a four-electron three-centre bond. This bond can also be described mesomerically, by assuming a $\ce{P-F}$ single bond, a positive charge on phosphorous and a negative charge on the other fluorine, with a second mesomeric structure as a mirror image of the first.
From this mesomeric consideration you can see that the bond order of the axial $\ce{P-F}$ bonds is not 1 but rather closer to 0.5. The greater the order the shorter the bond so the smaller the order the longer the bond — and there you have your result.


text:
Why does an acid salt possess acidity as well as basicity?
For example, $\ce{BaHPO_{4}}$ has acidity of 2 and basicity 1.

Why does this so? Its basicity is 1 because it has 1 replaceable hydrogen ion.
But why does this have acidity?

Wat do you mean by "acidity of 2"?

Acidity is defined for a base and it is the number of H+ ions, 1 molecule of the base takes up.
1 molecule of BaHPO4 of  can take up to 2 H+ ions... and thus it has acidity 2...
@Mithoron

Hmm, heard of "basicity" so "acidity" indeed should mean this but never heard this term - it's quite archaic, I'm afraid.

Lol... @mithoron

This is because Barium Hydrogen Phosphatev ($\ce{BaHPO4}$) is an amphoteric species:

In chemistry, an amphoteric species is a molecule or ion that can
  react as an acid as well as a base

(Wikipedia)
Whether it acts as a base or as an acid depends on the medium the salt is dissolved. In a pure water solution, it's an acid, since the following occurs:
$$\ce{BaHPO4 + H2O <=> Ba^2+ + PO4^3- + H3O+}$$
The phosphate ion also has its reactions with water forming other species:
$$\ce{PO4^3- + H3O+ <=> HPO4^2- + H2O}$$
$$\ce{HPO4^2- + H3O+ <=> H2PO4^- + H2O}$$
Since these are all weak bases, you won't "consume" all $\ce{H3O+}$ and the solution will be acid.
If, however, you add a strong acid, the scenario changes:
$$\ce{BaHPO4 + H3O+ <=> Ba^2+ + H2PO4^- + H2O}$$
You'll still have all phosphate species as above, but now Barium Hydrogen Phosphate has acted as a base, since it received some hydrogen atoms, which is a Brønsted-Lowry base definition.
Remember the same kind of behaviour happens with water, even if pure, because of self ionization:
$$\ce{2H2O <=> H3O+ + OH-}$$
Here, water acted as both the base and the acid, which is why it's an amphoteric species.


text:
Stereochemical aspect of heating of cyclohexane-1,2-diol
Does stereochemistry play any role in the heating of Cyclohexane-1,2-diol?
If I remember correctly, different products are formed depending on whether the diol is cis or trans. (I think the products are cyclohexanone and cyclopentanecarbaldehyde). Please clarify this along with the reason why different products are formed depending on the stereochemistry.

The reaction you are thinking of is likely the thermal decomposition of the cyclic sulfites prepared from the cis- and trans-diols.  Here is a link to the abstract.

In both reactions $\ce{SO2}$ is extruded.  Whichever group on the vicinal carbon that is lined up in an anti-periplanar fashion to the departing oxygen is the group that migrates.  This is because the antiperiplanar geometry allows for good orbital overlap during the migration; if a choice is available then hydrogen migrates in preference to an alkyl group just as with carbocations. In the cis-isomer, the vicinal hydrogen is properly aligned (I darkened the affected bonds to highlight the geometric arrangement), its migration produces cyclohexanone.  In the trans-isomer, only a $\ce{C-C}$ bond in the cyclohexane ring is properly aligned, its migration produces cyclopentanecarbaldehyde.

Is this also valid for cyclohexane-1,2-diol?

No, that's why I think this must be the reaction you are remembering.  It gives the products you mentioned and the sulfites are prepared from the two cyclohexane-1,2-diols..  Cyclohexane-1,2-diol itself does not undergo any thermal reaction.

Compounds containing a cyclohexane-1,2-diol structure can undergo pinacol-like rearrangements. Usually, the substituent that is anti to the leaving group is the one that undergoes migration. Therefore, cis-cyclohexane-1,2-diol mainly yields cyclohexanone, whereas a trans-cyclohexane-1,2-diol mainyl yields cyclopentanecarbaldehyde.

Can you please elaborate the answer with a figure?

But isn't the pinacol rearrangement an acid-catalyzed rearrangement, not a thermal rearrangement? The OP didn't mention acid, just heating.


text:
Gaussian Convergence Issue for N2+N PES
I am trying to obtain a potential energy surface (PES) for $\ce{N2 + N}$ combination using scan feature of Gaussian. I have tried different combinations of basis sets and methods but unfortunately at some point in the iteration, the convergence criteria is not met and I get an "Aborted Core dump" closure in the log file. The Gaussian code is:
# HF/6-31G test scan

NNN PES

0 2
N
N 1 R1
N 2 R2 1 120.

R1 0.5 80 0.1
R2 0.5 80 0.1

Where am I going wrong. More importantly, I would be really thankful if someone could suggest how to choose a method and basis set for a type of problem. I was thinking if I can get some answer using HF/6-31G, I could use this as an input for a higher order method like cassf. What that be a good idea?

You did not specify at which geometry you have convergence problems, so at first I thought it happens when all three $\ce{N}$ atoms are far away from each other. But then I run the calculation by myself and found out that it is not the case: the SCF convergence failure happens at R1=1.2 and R2=1.1. At this geometry SCF starts to oscillate after 20 iterations back and force as shown in a graph below, so simply increasing the number of SCF iteration would not help.

For that reason I quickly checked few other well known tricks to avoid the problems with SCF convergence (see, for instance, here) and I recommend to use a quadratically convergent SCF (QC-SCF) by adding scf=qc to the route section:
# HF/6-31G test scan scf=qc

Note though that while QC-SCF is more reliable than the default DIIS algorithm it is also slower, so expect an increase in the computation time. For that reason, there is also an interesting option scf=xqc which first tries the default algorithm and switches to a quadratically convergent one only if the default has not converged.
# HF/6-31G test scan scf=xqc

However, switching to a quadratically convergent algorithm did not help, since SCF still fails to converge but at some other geometries. So, I investigated the matter more closely, and in addition to scf=xqc I also recommend to improve both radial and angular flexibility of you basis set by adding diffuse and polarization functions into it
# HF/6-31+G(d,p) test scan scf=xqc

After running the calculation for an hour it reached 2656 points (which is about 40% of all points) without SCF convergence problems, but in case if this variant still fails:

Consider lowering down the convergence criteria by changing scf=xqc to scf=(xqc,conver=n) with n being, say, 6 (it is 8 by default).
If failures happen when at least one $\ce{N}$ is far away from two others, consider lowering down the maximum distance for your PES scan. In my opinion 8.5Å is way to big, 5Å would probably be more than enough.

I finally succeeded with the following route section
# HF/6-31+G(d,p) Scan SCF=(XQC,Conver=6)

Here are the results and below is the plot of the PES. :D

How did you make you plots? I mean which program do you use? It certainly looks very nice. Good job anyway. ;)

I see a lot of jaggedness in the results. Is this correct? Or is it because of HF method and with CASSCF, things will be much smooth. The reason I have complains about the smoothness is because I would want an analytical fit of the entire PES as the next step.

@TapanMankodi, we know a priori that HF fails to correctly describe the PES in question, but I'm not sure that the lack of smoothness is the resulf ot the HF failures. I think smoothness also depends on the step used for PES scan which was set to relatively big value of 0.1 A. I would not expect a smooth PES with such settings.

@Wildcat  Hmmmm fair enough reasoning about the step size being 0.1A. What recommendation do you have regarding better method should I use. I know that is bit vague question. In other words which should I use that will give me more reliable answers. Also how good cc-pVDZ orbitals? Are they more recommended than STO?
Being a mechanical engineering venturing into computational chemistry, your answers are being a great help!

@TapanMankodi, I'm not a big specialist in the area, but it seems like the system in question requires very advanced electronic structure methods (CASPT2/NEVPT2/MRCI) for its qualitatively correct description. These methods are not even available in Gaussian. I mean, this system is for real experts, not for beginners, you'd better start your venture from something simple.

The prblm is I am planning to get crosssection v relative energy plot for air chemistry reaction using quasiclassical trajectory which will use the PES of the said reaction. Hence solving somethng simple is not an option. On top of it being a mechanical engineering, I will have to rely on existing codes/software and not venture into coding everything in my way(although I wish I could but that would takes years). I found that CASPT2 is available in GAMESS, so have downloaded installed and started working with it but am in need of good examples for CASPT2. Suggestion welcome @Wildcat, Martin

Since this system is very small (only 3 N atoms), I would recommend using a high quality method with a large basis set, CCSD(T) with aug-cc-pVTZ may be feasible, but do not use Gaussian for CC methods, because it is extremely slow. A free and open source alternative would be the NWchem package, it has a quite reasonable CCSD(T) implementation. If this system suffers from multireference problems then you could try getting the MRCC program from M. Kállay, but if you need that, then you are in for a bumpy ride. Good luck with your PES developmet!

Apart from the technical aspects, that Wildcat already provided, I would assume, that the problem is far too complicated to be treated with Hartree-Fock correctly. It might be even not possible.
Hartree-Fock is a method that describes no coulomb correlation at all. I would assume, that this is vital to the system in any case.
At long distances you will have the problem, that you cannot describe the system in a doublet state any more. Nitrogen as an atom has five valence electrons, and according to Hund's rule it should be a quartet spin state. So if you have the separation $\ce{N + N2}$ this should also be a quartet state. In this case you will have (nearly) degenerate orbitals, that will cause the SCF to oscillate.
When you come to the situation $\ce{N + N + N}$, you are effectively treating it as a super-molecule, where you couple three quartet states. The only way of qualitatively treating this correctly would be a complete active space ansatz. Here again it is most likely, that the SCF will oscillate because of (nearly) degenerate orbitals.
Since you are freezing the bond angle at 120.0°, you should also be aware, that the equilibrium geometry will not be part of your designed potential energy surface. I would assume, that it is linear, analogous to the azide ion.
In any case I would expect the wave function to be highly spin contaminated.
Running a scan at the CASSCF level of theory is very time consuming and involves high effort computations. I would guess you need to pick at least 15 active electrons in 15 orbitals. For starters it might work only treating the π system with CAS, but that still leaves 9 electrons in 6 orbitals.

Yes, Martin, you are perfectly right. HF is hardly appropriate for the calculation in question and spin contamination is indeed quite big. For instance, at a point on PES with $R_1=6.6, R_2=3.7$, SCF converges to a state with $S=1.7416$ which is far away from a quartet state.

So, one should definitely account for electron (Coulomb) correlation, as usual, and should not use HF results by themselves. Doing CASSCF would probably be too expansive, so thinking of the HF calculations as merely an educational exercise, one could do then MP2 or DFT to take electron correlation into account.

@Wildcat To be fair to the method, you are looking for a doublet wave function (when you use the provided input). That, of course, does not make the spin contamination any better. I would not recommend MP2, since you are dealing with nearly degenerate states and the correction to the energy might simply go through the roof, hardly making anything better at that point, maybe even worse. As an educational exercise HF might work, but from the question it is not clear, what the intent behind the study is in the first place.

Sure, sure! To make it more clear: multi-configurational nature of the system in question at long distances can, of course, be dealt with multi-configurational methods only. The meaning of my previous comment was that even at short distances, where 1 CSF is presumably enough, you would better avoid HF and account for Coulomb correlation with MP2/DFT.

@Martin Thanks for the detailed reply! I will be shifting to CASSCF for calculating more accurate PES. I read that it is good to start with HF and then use that as an initial guess for CASSCF. By quartet state at $N_2$ + N configuration at far away distance, do I have to use 0 4 instead of 0 2 in the Gaussian file?

@Tapan Basically yes. In the line above the geometry you indicate charge (first number) and multiplicity (second number) as 2S+1.


text:
What radius best represents reaction distance?
Molecules must be within a certain distance for a reaction to begin as well as satisfying a myriad of other factors. Whilst the exact distance varies from reaction to reaction, is their any distance/molecular radius which provides a fairly good picture in terms of how close the centre of the molecules have to be to begin the interaction given they are in the correct orientation with enough energy etc.?

I suspect this depends on the kind of molecules involved: there's atomic, ionic, covalent, Van der Waals radius, etc.

I understand but I was wondering if there is sort of an 'effective' radius where 9 times out of 10 they will interact for all reactions that can be used for calculations.


text:
Cohesive Energy Density table request
Quick question:  Any one knows a resource for Cohesive Energy Density of some organic materials ?
I am not looking for any specific molecule, actually I want to plan my project based on available data. I have found some tables for internal cohesive pressure but could not find any thing for CED.
Anything including tables/papers and books are fine.

Some resources to get you started:

Free Volume, Cohesive Energy Density, and Internal Pressure as Predictors of Polymer Miscibility (White and Lipson, 2014) (paywalled).

Where the authors use cohesive energy density as part of their research to develop

strong correlations between bulk behavior and theoretical pure component character, and use those correlations to motivate a discussion of what drives polymer miscibility.

The Materials Design webpage Organic Materials Properties: Densities, Cohesive Energies, and Heats of Vaporization contains a couple of lists and information pertaining to a computer based modelling tool.  Also included is a bibliography of further resources.
The Tampere University of Technology in Finland's webpage Exploitation of CED provides a list of CEDs of several organic compound, as well as an explanation and further information on the previous and next pages.

Cohesive energy density (CED) of polymers and organic liquids may be calculated using group contribution method based on tabular data given in:
D. W. Van Krevelen and K. Te Nijenhuis, “Chapter 7: cohesive properties and solubility,” in Properties of Polymers: Their Correlation with Chemical Structure: Their Numerical Estimation and Prediction from Additive Group Contributions, Elsevier, Amsterdam, The Netherlands, 4th edition, 2009.
To calculate cohesive energy (in $\mathrm{J/mol}$) of some compound you have to know its chemical structure. Split this structure into groups which can be seen in tables and add together contributions from each group. To obtain CED (in $\mathrm{J/cm^3}$) you have to divide summation result by molar volume of this compound.
CED calculated in this way is not very accurate. However, it may be helpful for comparing CED of different compounds.


text:
a chemical model for pde
Consider the initial boundary value problem
$$u_t=a(t)u_{xx},\,\,\,0<x<1,\,\,\,t>0$$
$$-ku_x(0,t)=hu(0,t),\,\,\,t>0$$
$$u(1,t)=u_0,\,\,\,t>0$$
$$u(x,0)=f(x),\,\,\,t>0$$
where $a(t)>0$ is time-dependent thermal diffusivity, $k>0$ constant thermal conductivity and $h>0$ constant heat transfer coefficient.
-I wonder that is there any chemical process that can be modelled by the above equation ?
-Is it possible for a physical process that the thermal diffusivity is time dependent but thermal conductivity and heat transfer coefficient are constant ?

Cross-posted to Physics, https://physics.stackexchange.com/questions/176673/a-physical-model-for-pde


text:
Effecting Chemical Change
Apologies in advance as I am not a science person by any definition. By most, I'm probably not a person but we'll let that be.
I am studying the notion of "change" in things and focusing on chemical change. Could someone please let me know if:

the change in the color of a photogray (photosensitive, photochromic) lens is a "chemical" change?
the change in a baby's diaper when it gets wet is a chemical change (and if it is similar in any nature to the change in #1) and if that is the same as the change in a pregnancy test or a urinalysis stick
if the change in a thermochromic/temperature sensitive item (a beer can that changes color in the cold, a color changing thermometer or spoon or mood ring ) is the same type of chemical change

Sorry if this is too much for one question, or if it can't be explained to a lay person, but I appreciate any responses. I'm too newbie on this board to post all the links as links so the questioned items are connectable through the list of urls.
TIA.
Lenses
Diapers
Pregnancy test
Blood detection
Coors Light Cold-Activated Bottles
Color changing thermometer
Frozensolutions
Mood ring -wiki
-----edit---------------
The question of "chemical" change would be, in my mind different from "physical" but I don't know if that distinction comes into play here, or is an issue when the change is permanent or temporary.

Welcome to chemistry.SE! If you had any questions about the policies of our community, you can visit the help center or take a tour of the website.

For starters, I think most chemists would agree that a chemical change is different from a physical change by involving the formation or breaking of chemical bonds, or changes to the charge, coordination, or bonding state of some of the atoms involved. A chemical change can be reversible, however, so reversibility is not a determining factor. Either a physical change or a chemical change could result in a change of color, or not--as the case may be.
In your examples, the absorption of water by a disposable diaper, as well as the color changing of liquid crystal-based displays (like the beer temperature indicator, thermometer, and mood ring) are clearly physical changes. The liquid crystals change their optical properties due to the physical alignment of the molecules, and not because of making or breaking of bonds. On the other hand, the pregnancy test and blood detection test probably involve chemical changes to produce a color readout (sometimes initiated by a relatively loose binding between an antigen and an antibody, or formation of a metal complex). The photochromic lenses turn dark as a result of a chemical change induced by light (according to Wikipedia, anyway--this is not my area of expertise).


text:
What is the CO2 content in the air in a compartment of air and water, as a function of temperature?
Update 31.12.2015: In a closed system of water and air, how does the concentration of CO2 change as a function of temperature, in the gas phase?
What is the graph of carbon dioxide ($\ce{CO2}$) content in the air in a compartment consisting of 80 percent air and 20 percent ocean sea water as a function of temperature, in the temperature range 0 to 60 degree Celsius?
Example but not the complete answer:

As can be seen from the graph above, increasing the temperature of the water will cause the water to release more carbon dioxide ($\ce{CO2}$) in to the gas phase. This happens for most gases as can be seen from this page:
http://www.engineeringtoolbox.com/gases-solubility-water-d_1148.html
What I am interested in is to see a graph of the percentage, or how to put it, of carbon dioxide ($\ce{CO2}$) in air as a function of temperature.
In other words, what does the plot of:
$$\%\ce{CO_2} \text{ in air (temperature)} = \frac{\text{kilogram } \ce{CO_2} \cdot 100 %}{\text{kilogram  air}}$$
look like in the gas phase in a compartment consisting of 20 % sea water and 80 % air as a function of temperature?
Ideally the compartment should consist of a column with the height upwards of average atmosphere thickness on planet earth filled with air, and a depth downwards with the average depth of sea water also at planet earth filled with sea water. But this is probably not feasible. Therefore the simplifcation 80 percent air, 20 percent sea water. If sea water is too complicated I am also satisfied with an answer with pure water.
The definition of air without water steam in volume percentages:
Nitrogen $\ce{N_2}$ = 78.08
Oxygen $\ce{O_2}$ = 20.95
Argon $\ce{Ar}$  = 0.934
Carbon dioxide $\ce{CO_2}$ = 0.036
Neon $\ce{Ne}$ = 0.00182
Helium $\ce{He}$ = 0.00052
Krypton $\ce{Kr}$ = 0.00011
Hydrogen $\ce{H_2}$ = 0.000050
Xenon $\ce{Xe}$ = 0.000009
Ozone $\ce{O_3}$ = 0.000001
Radon $\ce{Rn}$ = 0.000 000 000 000 000 006
Edit 18.4.2015:
I will try to clarify the question:

Does an increase in temperature cause an increase or decrease in the carbon dioxide (CO2) content in the atmosphere?

Edit 22.4.2015: According to engineering toolbox dot com, the four most common gases in the atmosphere have the following solubilities in water:

Related question: https://mathematica.stackexchange.com/questions/80586/how-can-i-get-the-solubility-value-of-nitrogen-in-water-at-10-degree-celsius-wit

Simple, partial pressure of $\ce{CO2}$ is misleading.
$\ce{CO2}$ actually reacts with sea water:
$$\ce{H2O + CO2 -> H2CO3}$$
which then ionizes:
$$\ce{H2CO3 -> H+ + HCO3-}$$
which lowers the pH of the sea water (making it more acid).
In previous high $\ce{CO2}$ periods, $\ce{CaCO3}$ came out of sea water
making limestone and marble rock formations geologically. $\ce{CO2}$ at high pressure is responsible for other side effects, also killing
fish, and microbes, damaging coral reefs and generally messing up ocean life.


text:
Determining aromaticity of compounds

Which of the following planar structures would be aromatic?

Could someone help me determining the aromaticity of the compounds?
My Attempt:
(i) Not an aromatic compound since it only has 4 pi electrons that does not satisfy Hückel's (4n+2) rule.
(ii) Not aromatic because it only has 4 pi electrons, however I'm not sure if the oxygen contributes a p-orbital, but it shouldn't matter since 5 pi electrons still does not satisfy Hückel's rule.
(iii) I'm very confused because I'm not sure if the ketones are considered "part of the ring" therefore do they also contribute p orbitals? However if we ignore the ketones, then I believe the Nitrogens contribute 2 e- since they are sp2 hybridized? Therefore the 2 N atoms and 2 C atoms contribute a total of 4-not aromatic either?

http://chemistry.stackexchange.com/questions/27099/when-does-nitrogen-contribute-its-electrons-to-aromaticity

(i) that's correct
(ii) is aromatic, it fit's the 4n+2 rule with n=1.  Here is a drawing of furan showing 2 lone pair electrons in a p-orbital that completes the cyclic aromatic ring. Furan's other lone pair of electrons is perpendicular to the pi system and does not play a role in furan's aromaticity.

(image source)
(iii) Edit to incorporate @jerepierre's comment:
Just like phenol can exist in equilibrium with its non-aromatic tautomeric keto form

so can derivatives of pyrimidinedione (or uracil).  However, unlike phenol were the equilibrium lies far to the aromatic phenol side, these compounds usually (it depends on the substituents) exist in the keto form.  The strength of the keto and amide bonds outweigh the weaker aromaticity of the pyrimidine skeleton and shift the equilibrium to the keto side.

(image source)
The question remains, is the keto form aromatic?  As jerepierre points out, you can draw the following resonance structure for the keto form.  In this double-dipolar resonance structure there are 6 pi electrons 2 in the double bond and 2 [a lone pair] on each nitrogen) suggesting that this resonance structure would be aromatic.  Since this resonance structure contributes to the overall description of the molecule, it would be fair to say that this molecule has some aromatic character; certainly not as much as benzene or furan - but some.

I'm not sure about the uracil. Cyclohexadienone is clearly non-aromatic because of the sp3 center. Uracil is fully conjugated, and you can draw an aromatic resonance structure of uracil. I suspect that makes it aromatic, but one could argue either way.

@jerepierre  I agree, it could be argued either way, not a good question.

i also thought that the ketone substituents didn't influence aromaticity either

@jerepierre I think those structures are rather too minor contributors, but one would need experimental data to determine this for certain.

@Nerdingout check this: http://chemistry.stackexchange.com/questions/22100/what-is-the-resonance-structure-of-nitro-aminobenzene-with-a-broken-aromaticity

but wait wouldnt (iii) be aromatic due to the 2 pi electrons from the double bond + 2 electrons from each nitrogen (4) = 6?

Yes, isn't that what I wrote?

you said (iii) is non-aromatic

I said that the keto form of phenol isnon-aromatic, I said (iii) is a romantic

Okay that makes more sense!! Also last question-so the ketone substituent does not influence aromaticity?

It does influence aromaticity.  It is only because we can draw the dipolar resonance form of the ketone that the molecule can be considered aromatic

I would again like to point out, that aromaticity is a feature, that correctly can only be obtained from experiments. Lewis structures, even resonance structures can only explain, why there is a ring current, not predict if there is. Hückel's rule is a nice guideline for an ad hoc chemical assumption, but it can hardly be treated as a law. For what it's worth, aromaticity is simply defined, as "having a chemistry typified by benzene". http://goldbook.iupac.org/A00441.html


text:
Non-radiative transitions or what features of rhodamine result in it being so highly fluorescent?

Rhodamine is a fluorescent dye. What features of the molecule result in it being so highly fluorescent ?
I assume the extended conjugated system with a stabilized first excited state is necessary, to absorb light in the visible region of the spectrum, and I would also think a rigid, planar structure is preferable for good Franck-Condon factors for absorption. Are there other factors involving the energy/stability of the $S_{1}$ state I should be aware of ?

Rigidity is definitely an important factor. At $295\ \mathrm{K}$, indeno[2,1-a] has a fluorescence quantum yield $\Phi_F$ of 0.94, (DOI). In an organic glass at $77\ \mathrm{K}$, even (E)-stilbene, which hardly shows any fluorescence at room temperature, exhibits  a $\Phi_F$ of 0.75.

Vibrational stretching of $\ce{C-C}$ bonds in substituents can significantly contribute to radiationless deactivation of an $S_1$ state and therefore reduce the fluorescence quantum yield $\Phi_F$. Admittedly, toluene isn't famous for its fluorescence ($\Phi_F = 0.15$), but the fluorescence of tert-butyl benzene is even lower ($\Phi_F = 0.03$).
Apart from radiationless deactivation through internal conversion (IC), an excited singlet state can be depopulated by intersystem crossing (ISC) to a triplet state. Under certain conditions, this is even possible for compounds that usually have very low triplet quantum yields.
Halobenzenes added to a solution of fluorescing anthracenes in dioxane (DOI) or ethyl iodide added to solutions of pyrene in alkanes (DOI) have shown to quench the fluorescence normally observed in the absence of the halogenated compounds. The latter exhibit a external heavy atom effect.
The lack of $S_1 \rightarrow T_1$ intersystem crossing is probably the most important factor that makes rigid compound show large $\Phi_F$ values.
It is striking that strongly fluorescent compounds typically have an extended $\pi$-system. Here, the $S_0\rightarrow S_1$ excitation is a $(\pi,\pi^*)$ transition.
On the other side of the fence, ketones, such as acetophenone or benzophenone, hardly show fluorescence. For these compounds, $S_0\rightarrow S_1$ excitation of the $\ce{C=O}$ group is a $(n,\pi^*)$ transition. From a $^1(n,\pi^*)$ state, spin-orbit coupling can both flip the spin and move one electron to a more distant and geometrically different orbital. $S_1 \rightarrow T_1$ thus is in fact a $^1(n,\pi^*) \rightarrow {}^3(\pi,\pi^*)$ transition.
For compounds like rhodamin B in their $^1(\pi,\pi^*)$ states, the  corresponding transition to a $^3(\pi,\pi^*)$ state is largely spin-forbidden ans does not occur. Provided that no external effects come into play, these molecules exhibit strong fluorescence.

Regarding lack of intersystem crossing, don't you typically get E-type delayed fluoresence from ISC/reverse ISC for this sort of dye ? I remember eosin itself having a similar structure.

@J.LS Fluoresceine, Rhodamine B, Rhodamine 6G, Eosin B and Eosin Y do have pretty much the same xanthene-type core, but the eosines are two- and fourfold bromo-substituted. Here (only), an internal heavy atom effect might come into play.

rigidity is a factor but 9, 10-diphenyl anthracene has almost unity fluorescence yield but still has rotatable groups. Eosin has a particularly small singlet-triplet energy gap hence reverse intersystem crossing. In the absence of heavy atoms and flexible rings etc I expect a large singlet -triplet energy gap to contribute to  a large fluorescence yield.

The trite answer to your question is that the rate constant for non-radiative transitions,internal conversion($k_{ic}$, S$_1$ -S$_0$) and intersystem crossing ($k_{isc}$, S$_1$-T) are smaller than that for fluorescence. This is seen from the definition of fluorescence yield $$\phi _f = \frac{k_f}{k_f + k_{ic} + k_{isc}}$$ Clearly if $k_{ic} + k_{isc}$ is small relative to $k_f$ then the yield is large, tending to unity in the limit. The measured fluorescence lifetime is also reduced, $\frac{1}{\tau_f} = k_f + k_{ic} + k_{isc}$. A similar equation applies to phosphorescence yield and lifetime (except that there is no $k_{ic}$).
The singlet to triplet intersystem crossing can only occur if there is an interaction between the two states. As these have different spin this interaction must be one that couples states and has to be coupling between spin and angular momentum, called spin-orbit coupling. This interaction is greater if heavy atoms are present in the molecule, such as iodine, or in the solvent, as in CCl$_4$.
Although the effect is greater for atoms with large atomic number it occurs in all atoms. The coupling is approx 21 times larger in chlorine than carbon and 180 times larger with Iodine. Thus with heavy atoms the fluorescence yield will be reduced compared to the same molecule with, say, H substituted for I or not dissolved in a heavy atom solvent. So we exclude these special cases and then ask what determines the fluorescence yield.
Another special case is chemical reaction, for example cis-trans isomerisation. In these cases the excited state has a small barrier on the path to products after which a conical-intersection is met and either the reactant is reformed or the product produced. We exclude such reactions in trying to explain why fluorescence yields are large or small. This still leaves molecule with no heavy atoms and no excited state chemistry and which have different fluorescence yields.
It used to be thought that flexible groups in a molecule cause low fluorescence yield. Some how the molecule shook itself into a triplet state, the ‘loose bolt’ idea. But this is not the case, for example rhodamine has at least three flexible groups attached as does 9,10-diphenylanthracene with a yield as close to 1 as its possible to get.The phenyl groups on positions 9 and 10 have torsional motion at some large angle to the three anthracene rings, thus motion here does not reduce the yield. In fact all molecules are flexible to the extent that they have vibrational normal modes which constantly distort the molecule from what we draw as, say, perfect hexagons in benzene.
The theoretical work in understanding radiationless transitions was done in the 1960’3 and 70’s (several papers by Robinson, Frosch, Siebrand, Lin, Bixon, Jortner). Understanding non-radiative processes was a major achievement in photophysics/ photochemistry but is not well known.
The rate of the non-radiative process is determined by the Fermi Golden Rule. In this method a single state is coupled to a continuum of final states. In its simplest form rate constant is given by
$$W_E=\frac{2\pi}{\hbar}|V|^2F_c\rho_E $$
where $V$ is the electronic coupling energy between the two states and is proportional to spin orbit coupling in S-T transitions and to vibrational coupling in internal conversion  S$_1$ to S$_0$. $F_c$ is the average Franck-Condon factor and $\rho _E$ is the number of vibrational energy levels / unit energy, or the density of states at energy $E$.
In solution it is necessary to average over the Boltzmann distribution of energy levels and in this case the rate constant becomes
$$W=|V|^2e^{-X}e^{-\gamma \Delta E /\hbar \omega} $$
and X is the displacement of the vibration with the largest frequency $\omega$ , usually C-C and C-H vibrations, $\Delta E $ is the S-T energy gap and where $\gamma = ln(\frac{\Delta E}{\hbar \omega X} -1)$ and this typically has a value in the range 3 to 5 so can be treated as constant. This equation means that the non-radiative rate constant decays exponentially on the energy gap.

This is what is observed experimentally by choosing classes of molecules that are otherwise similar, e.g. benzene , naphthalene, anthracene, tetracene etc. Similarly the deuterated versions of these molecules have been studied and the effect of the lower frequency  of CD compared to CH vibrations is clearly seen, as the deuterated species have a lower rate constant as predicted.
The same relationship has been observed in S$_2$-S$_1$ energy gap in azulene derivatives and in the energy gap in different rare earth chlorides MCl$_3$. (Azulene is an isomer of naphthalene, is blue, and has a 5 and 7 membered ring rather than two six membered rings)
Other points to note are that only some vibrational modes are important in radiationless transitions. These are called promoting modes, at least one is present and can be identified by experiment. The promoting modes are the only one with non-zero electronic coupling terms and they have zero displacement of their internuclear position between the two states, thus are asymmetrical stretch modes. There are also accepting modes in the final state. These receive the vibrational energy and are modes with a large vibrational frequency so as few of them as possible are excited to make up the energy gap. These are usually CC or CH modes.
The properties of aromatic molecules isolated from one another by low pressure in the vapour phase have also been studied. The same basic theory applies, but without having to apply the Boltzmann distribution. In this case the non-radiative rate constant increases with excitation energy in the singlet state and thus with increased energy gap. This is, however, entirely what is predicted by theory.

Hmmm, I can't think of much else. You have a conjugated, planar and rigid structure that ensures your structure doesn't change/expand much upon promotion of an electron into the antibonding orbital; hence, ensuring a good FC factor.
Possibly solvent related effects that might encourage ISC over fluoresence? I know polar solvents can encourage ISC but not sure in what chemical context and it's applicability to the particular situation.

Hadn't though about solvent issues, thanks ! I guess I was thinking about what factors might result in a more accesible S1 state beyond just "extended pi system".


text:
How to calculate the concentration of conjugate acid from given pH and pKa values?
Preface: Buffer solution (acid-base buffer). I am provided with a weak base, which I will designate B. $\mathrm{p}K_\mathrm{a}$ for $\ce{B}$'s conjugate acid, which I will designate $\ce{BH}$, is $8.1$, and its mole weight (sic) is $121.1$. I'm assuming the latter is the molar mass, though I don't know how that helps me solve this problem.

Problem: $\pu{0.1 M}$ ($\pu{mol/L}$) of $\ce{B}$ was dissolved in $\pu{1.0 L}$ water, then titrated to $\mathrm{pH}$ $8.1$ and $7.5$. I am to assume no change in volume.

The last statement sounds like an odd assumption to me, how do you titrate something without [significantly] altering the volume? I always assumed titration is only done by adding a liquid to another liquid, is this not correct?
I am to find the concentration of $\ce{BH}$ at $\mathrm{pH}$ $8.1$ and $7.5$.
This question looked really easy to me at the first glance, but after a few hours I realized I had no idea on how to go about this.
I've tried calculating the equilibrium concentrations and going from there. This just ends up in $x \neq x$ in my calculations (where $x$ is the concentration of some substance), which obviously doesn't hold.
I've used the acid dissociation-, base dissociation- and Henderson–Hasselbalch equations, but nothing seems to do the trick. I might not be the brightest person to have grazed the surface of the planet, but I'm confident that I've got a firm grasp on elementary arithmetic.
The answer is provided, and should be $\pu{0.05 M}$ at $\mathrm{pH} = 8.1$.
I would highly appreciate somebody showing or explaining the thought process to me.

To answer your question about titrating without adding volume—you don’t. This problem is strange in many ways, but I will attempt to help you to understand.
Since we are working with dissolved $\ce{B}$, I feel as though it makes the most sense to express the reaction with the given by the equation:
$$\ce{B + H2O<=>BH +OH-}$$
So the equilibrium expression of the reaction is given by the equation:
$$K_\mathrm{b}=\frac{[\ce{BH}]\cdot[\ce{OH-}]}{\ce{[B]}}$$
If the $\mathrm{p}K_\mathrm{a}$ of the conjugate acid is 8.1, then by mathematical manipulation:
$$\mathrm{p}K_\mathrm{a}=-\log K_\mathrm{a}$$
$$-\log K_\mathrm{a}=8.1$$
$$\log K_\mathrm{a}=-8.1$$
$$K_\mathrm{a}=10^{-8.1}\approx7.94\times10^{-9}$$
Since $\mathrm{p}K_\mathrm{w}=\mathrm{p}K_\mathrm{a}\cdot\mathrm{p}K_\mathrm{b}$, the $\mathrm{p}K_\mathrm{b}$ of the base should be, and since $\mathrm{p}K_\mathrm{w}=1\times10^{-14}$:
$$1\times10^{-14}=7.94\times10^{-9}\cdot K_\mathrm{b}$$
$$K_\mathrm{b}=\frac{1\times10^{-14}}{7.94\times10^{-9}}\approx1.26\times10^{-6}$$
For the same reason that $1\times10^{-14}=\mathrm{p}K_\mathrm{a}\cdot\mathrm{p}K_\mathrm{b}$, $\mathrm{14=pH+pOH}$. Given that the pH of the water is 8.1 for our first calculation:
$$14=8.1+\mathrm{p}\ce{OH}$$
$$\mathrm{pOH}=5.9$$
and
$$-\log{[\ce{OH-}]}=5.9$$
$$\ce{[OH^{-}]}=\mathrm{10^{-5.9}}\approx1.26\times10^{-6}$$
Since we know that any $\ce{BH}$ that is made must come from the reaction of 1 molecule of $\ce{B}$, then:
$$[\ce{B}]=0.1-x~~~~[\ce{BH}]=x$$
Using the knowledge we have gained from the problem, we can now solve (finally!):
$$K_\mathrm{b}=\frac{[\ce{BH}]\cdot[\ce{OH-}]}{\ce{[B]}}$$
$$1.26\times10^{-6}=\frac{[x]\cdot[1.26\times10^{-6}]}{[0.1-x]}$$
$$x=0.1-x$$
$$2x=0.1$$
$$x=0.05$$
So:
$$[\ce{B}]=0.1-0.05=0.05~\mathrm{M}$$
Try using this process to solve for the $\mathrm{pH}$ of 7.5.

Thank you very much for your response, it was very helpful to me. I tried a myriad of approaches, including, from the Henderson–Hasselbalch equation, $pH = pK_a + log_{10} \frac{[Base]}{[Acid]} => 8.1 = 8.1 + log_{10}(\frac {0.1}{x}) => x = 0.1$, and I also tried calculating the equilibrium concentrations of acid and base to get $8.1 = 8.1 + log_{10}(\frac {a-x}{b+x})$, where a and b are the equilibrium concentrations before adding the acid. Do we just ignore the equilibrium concentrations because they are relatively small as compared to after having added the acid?

Well, the Henderson-Hasselbalch equation would indeed work for this problem, and all things considered, this was probably what you should have used. Your problem in your first equation was assuming that at equilibrium, $[Base]=0.1$. You had a better idea with the second equation second equation the $\log{\frac{[Base]}{[Acid]}}$, but you should have used $[Base]=0.1-x$ and $[Acid]=x$. This would give you the same result as in my method.

I have a question related to a subproblem of this problem. If it isn't too much to ask, would you please check out my work and see where my error lies?
Problem: Add 0.01 mol of base to the solution. Assume that the pH before this addition was 8.1 (i.e. [B] = [BH] = 0.05 M).
I assumed that when adding base, the equilibrium would be displaced, causing more BH to form. From the base dissociation equation I get:
$$1.26\times10^{-6} = \frac {(0.05+x)(1.26\times10^{-6}+x)}{0.06-x}$$
As far as I can tell, this should be correct, but when solving the quadratic equation (see next post),

adding the solution to the existing $OH^-$ concentration, and calculating the pH from there, my answer is off by about .2 pH no matter how many digits I include or exclude from my calculations. I also don't understand why I can't just add my solution for x to [BH] and find the pH directly from $-log_{10}(0.05+x)$, but instead have to add x to $[OH^-]$, calculate the pOH and arrive at the pH from there, since BH is basically the equivalent of $H_3O^+$ in this reaction, no?

Do you mind me asking what you are getting as your pH calculation?

Unfortunately I have a seemingly endless stream of scribbles; I have at least two values: 8.179547779 and 8.3 (didn't write more decimals in my notes).
I might not have the chance to reply again tonight, but I will definitely read your reply when I come back.

Ah, the full value of the 8.3 one is 8.382365987, but for some reason I used $3.16\times10^{-7}$ for the initial $OH^-$ concentration (final concentration will be the initial $OH^-$ concentration + reacted base), probably just some careless error.

Ringo has given an excellent answer for your questions and continued to direct you during comment session. I just want to tell you is, this question can also be answered by using Henderson–Hasselbalch equation.
Suppose after titrating, the equilibrium concentrations of BH is $\alpha$ and $\beta$ at pH 8.1 and 7.5, respectively. Thus, all [BH] comes from the original $\pu{0.1 M}$ solution of B, the equilibrium concentrations of B is $(0.1 - \alpha)$ and $(0.1 - \beta)$ at pH 8.1 and 7.5, respectively.
Apply Henderson–Hasselbalch equation separately for pH 8.1 and 7.5. For pH 8.1:
$$8.1 = 8.1 + \log \frac{0.1 - \alpha}{\alpha}$$
For pH 7.5:
$$7.5 = 8.1 + \log \frac{0.1 - \beta}{\beta}$$
I let you solve these two equations. Nevertheless, the the answers I got are: $\alpha = \pu{0.05 M}$ and $\beta = \pu{0.08 M}$.
Note: The question states that "$\pu{0.1 M}$ (moles/L) of B was dissolved in $\pu{1.0 L}$ water...," which is one of many mistakes in this question. It should have been "$\pu{0.1 mol}$ of B was dissolved in $\pu{1.0 L}$ of water...," instead.

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


text:
Why is glycine hydrochloride more acidic than chloroacetic acid?
I cannot figure out why glycine hydrochloride is more acidic than chloroacetic acid. Could I have figured this out from a $\mathrm pK_\mathrm a$ table and maybe am just not seeing it? On the table I have in my book (Organic Chemistry, Seyhan Ege, 3rd edition), neither acid nor its conjugate base are there.

Glycine hydrochloride is a salt of the amino acid glycine. This means that HCl (hydrochloric acid) has been used to protonate the molecule.
To understand better which part of the molecule is protonated we need to think about chemical structure of glycine and it's acidity in water.

In water glycine is amphoteric which means that it can react acid as well as base. The hydrochloride salt can be easily prepared by adding HCl to solution.

Now if we evaporate solvent we will get glycine hydrochloride salt, which has a pKa is 2.3 for first deprotonation and a pKa of 9.6 for the second deprotonation. Chloroacetic acid's pKa is 2.87, which means that glycine hydrochloride is more acidic.
Why? It's conjugate base is more stable.
In common language conjugate base is happy with its charge and doesn't want to get protonated. Let's take look at conjugate bases.

In chloroacetic acid's conjugate base, the nearby electron withdrawing chlorine group stabilizes the negative charge in the conjugate base, which lowers the pKa of chloroacetic acid compared to acetic acid (pKa = 4.75) ~ 2 units (remember pKa is on a logarithmic scale). In glycine hydrochloride's conjugate base, the nearby positive charge stabilizes the negative charge of conjugate base, which lowers the pKa even more to 2.3.

Glycine hydrochloride is the amino acid glycine adsorving a molecule of Hydrochloric acid, wich is a strong acid. I think this is the main reason of the acidicity difference, since the pKa's are similar (comparing glycine and chloroacetic acid).
Update: glycine is amphoteric: it can either act as a base or an acid.
In water the hydrochloric acid molecule of the glycine hydrochloride will dissociate: $$\ce{H2O +HCl -> H3O+ +Cl-}$$
Then:
$$\ce{Gly +H+ -> GlyH+}$$
$$\ce{GlyH+ +Cl- <-> Gly + H+ + Cl-}$$
You can conclude, therefore, that Glycine hydrochloride is more acid than glycine (and also the chloroacetic acid). Hope it helps, and sorry for the english mistakes!


text:
How can I easily explain to a layperson the difference between types of "contains"
I had a conversation with somebody who had somewhat confused ideas about food. He thought that "all food contains sugar" and when I gave a bottle of oil as the counterexample, he said the glycerol is the sugar there.
Now let's forget for a minute that glycerol is not a sugar. The point is, glycerol is indeed a "building block" of a triglyceride, and glucose is a "building block" of glucose. But still, if you are talking about a process which needs sugar, starch won't work unless you cleave it off first. Just like starch won't spontaneously conduct electricity or emit beta radiation even though it's chock full of electrons.
Given that we have somebody with a very basic level of knowledge, and who is already trying to wrap his head around several confusing concepts at once, what would be an intuitive way to get him to really grok that, just because a given atom or group is present in the structure of a molecule, it does not mean that a substance "contains" this atom or group in the same way that a wet sponge contains water? And that without going off on a long tangent of theory?

"Sugar is Carbohydrate, right? Is everything you eat Carbohydrate?!" And then a link to Wikipedia's Carbohydrate page. Yep. That's what I leave them to study. :)

@MARamezani it was more complicated. He thought that everything including proteins and fats gets converted into sugars before the body can use it for energy. But I'm not asking about that part, I'm asking how to explain that even though starch does get converted into a sugar after being eaten, you still cannot expect starch slurry to behave like glucose syrup during cooking - and how to tick that box off quickly, before moving on to digestive pathways.

Well, behavior on the macroscopic scale is dependent on many factors. Usually, the interactions between molecules are way more important roles in play than what molecules are. Oh, Curt just answered this, and I perfectly agree with his answer.

@Klaus that seems to me like retelling a bad memory/experience.

@KlausWarzecha there is a very big difference between somebody who learned something wrong, and somebody who'll defend his ignorance with his last breath. Both exist, but luckily, this guy wasn't aggressively arguing, he was genuinely curious.

@rumtscho Then the matter deserves another answer. Btw, Curt F. already gave a very good one!

@rumtscho: I had to star this entire question for that last quote about ignorant people. I will quote that!

I would say something like this:
All matter -- that is, all substances -- are made of atoms.  The properties of a given substance are determined by (a) the type and (b) the arrangement of the atoms in that material.
Chemists refer to the "types" of atoms as "elements".  For example, vegetable oil and table sugar (sucrose) are both composed of three types of atoms only: carbon, hydrogen, and oxygen.
The arrangement of the atoms in the two materials differs drastically.  (You could refer to structure diagrams at this point if your friend is interested.)  First, oil has much less oxygen and slightly more hydrogen per carbon atom.  Second, the way the atoms are connected -- their arrangement -- is very different.
Chemists sometimes give a name to a group of similar arrangements of atoms.  For example, "sugars" are compounds that (a) have carbon, hydrogen, and oxygen atoms in ratios that are approximately $\ce{CH2O}$, and (b) often form ring structures, as seen in table sugar, glucose, and corn starch.  But the key words are "approximately" and "often" because some molecules that chemists call sugars don't satisfy these rules.  Biochemists interested in metabolism sometimes refer to molecules that can "easily" be converted into "true" sugars as sugars.
Naming in chemistry is complicated by the fact that chemists use words for different purposes: to indicate metabolic similarity, to indicate structural similarity, or to indicate similarity in reactivity.  Those different purposes sometimes lead situation where chemist B says that what chemist A calls a "sugar"  isn't really a sugar.  But that's just an argument about humans' imperfect way of naming things, not about chemistry.  There is a lot of wiggle room in the words "approxiately", "often", "similar", and "easily".  So if we focus on metabolism only, the glycerol part of the vegetable oil arrangement (i.e. structure) is a bit like a sugar.  But it is dissimilar from sugars in terms of both structure and reactivity.
Again, what ultimately matters is (a) the type and (b) the arrangement of the atoms in a substance.  Those two things are together necessary and sufficient for a complete description of all substances.  Whether to call particular arrangements of particular atom types "sugars" or not is something people can argue about.

As a side note, my pencil is made of "carbon". Diamond is also made of "carbon". That doesn't mean I can write with diamonds.

(However, I do know that graphite and diamond aren't molecules)

I avoided even mentioning "molecules" because I think it is a bit jargon-y.  For example we could argue over whether a single crystal of diamond is a "molecule" or not.  Diamond and graphite certainly have very different arrangements of atoms.  "Arrangement" is a term that in my mind at least includes both connectivity as well as spatial geometry.

I will upvote your answer, but it's not what I was looking for, sorry. Of course this explanation is correct from a chemistry point of view, but all my intuition of explaining things to people tells me that I'll lose the audience along the way :( I hope somebody will have some more succinct idea. But thank you for this answer, it contains interesting points about sugars I hadn't realized myself.

Thanks for the upvote and sorry I wasn't able to help!  I hope folks chime in with more answers; this is a great question (+1) and deserves several answers from different perspectives.  My perspective is that lay people often have problems distinguishing between arguments about names and arguments about reality, but I can certainly see how a discourse on this subject could bore some folks.

In his excellent answer, Curt F. wrote:

The properties of a given substance are determined by (a) the type and (b) the arrangement of the atoms in that material.

Let me add some examples to illustrate how subtle differences lead to substantial property changes any layman can experience.
Containing the same number of atoms means nothing

$\alpha$-pinene (1) and camphene (2) have the same sum formula ($\ce{C10H16}$), but a very different constitution and a very different smell, just as their names suggest.
Containing the same number of atoms and having almost the same constitution

Thymene (3) and carvacrol (4) are regioisomers and only differ in the position of the $\ce{OH}$ group. However, thymene has a thyme odour, while carvacrol smells like oregano.
Mirror, mirror

Compounds 5 and 6 have the same constitution and identical boiling points. They only differ in that one group is above/below the ring plane (the world isn't flat). These are (just) mirror images: enantiomers of limonene. However the interested layman will recognize they are not identical: 5 smells like lemon, while 6 smells like orange.

Containing isn't being
A layman might know that chlorine is a toxic gas and was used in the chemical warfare of WWI. Some might even know that fluorine isn't any better.
These unhealthy properties are not inherited by compounds that "contain" these elements!

Table salt ($\ce{NaCl}$) "contains chlorine" (admittedly, writing it that way makes me shudder) but is isn't as toxic as the element. Neither is fluoridated tooth-paste.

Each molecule of diazepam (7) contains one chlorine atom, each molecule of fluoxetine (8) contains three (!) fluorine atoms. Even so, these compounds are not part of modern chemical warfare, but on the WHO List of Essential Medicines.

This is a great answer that I suspect you will be able to point to (or at least copy and paste) in response to many future chem.se questions.

For a layman you could simply mention that airplanes contain aluminium, as do rollerskate rails.
The arrangement of the aluminium is a critical factor determining which vehicle could be used for skating around the neighbourhood and which vehicle could be used for intercontinental transport.


text:
Why is the concentration halved at the intersection of Hg²⁺/Hg₂²⁺?

For the couple $\ce{Hg^2+}/\ce{Hg2^2+}$: $E^\circ = \pu{0.91V}$.
  At the border the concentration is $c = \pu{0.10 mol \cdot L-1}$ for all ions. Hence build the stability diagram of mercury.

I have:
$$\ce{2Hg^2+ +2e- <=> Hg2^2+}$$
Then by the Nernst relation I have:
$$E = E^\circ + \pu{0.03 V} \times \log\left( \frac{\left[\ce{Hg^2+}\right]^2}{\left[\ce{Hg2^2+}\right]} \right)$$
In the solution of the exercise they write:

At the border $\left[\ce{Hg^2+}\right] = \left[\ce{Hg2^2+}\right] = \frac{c}{2}$.

I don't understand why it is $\frac{c}{2}$.

I'm not sure about what's going on in this example, but apparently it was considered that the state is at the equilibrium at standard conditions, which makes $E = Eº$. Considering that, the log term has to equals zero, and that happens for log(1). This then means that the ions concentrations are equal, and therefore they are C/2. I won't post this as an answer because I'm not sure about how to justify this chemically correctly.

I don't undersand can you explain a bit more ^^ write the egality with the log please like you think may be i will understand it too

The total concentration of all ions is $c = \pu{0.10 mol L-1}$. The conservation equation of element mercury requires that
$$\left[\ce{Hg^2+}\right] = \left[\ce{Hg2^2+}\right] = \frac{c}{2}.$$
Generally, it's a convention to make the concentration of ions at the border equal. On the other hand, matter conservation implies that the concentration of each ion is $\frac{c}{2}$.


text:
What is a catemer?
What is a catemer and how does it differ from a dimer? What type of compounds could form catemers and how does their formation influence its crystal structure?

In the crystal, carboxylic acids mostly form dimers through pairs of $\ce{OH\bond{~}O=C}$ hydrogen bonds.
In some cases, infinite chains of hydrogen bonds are formed instead: here, each carboxylic acid forms hydrogen bonds to two different neighbours. These structures are known as catemers.

(Please note that the depiction of the catemer is very much simplified! In reality, the structure could also be a helix, stabilized by additional sets of interactions)
Actually, I have never used the term catemer in a crystallographic publication. It is very imprecise, almost meaningless. Instead, I have mostly used the a notation introduced by Margret C. Etter to describe hydrogen bonding pattern (dimers, stands, rings) in a crystal.
Two relevant articles in this context are:

Encoding and decoding hydrogen-bond patterns of organic compounds,
Acc. Chem. Res., 1990, 23, 120-126 (DOI)
Graph-set analysis of hydrogen-bond patterns in organic crystals, Acta Cryst. B, 1990, 46, 256-262 (DOI)

A good explanation of the various structures can be found in this paper: http://www.academia.edu/7596281/Crystal_Packing_Hydrogen_Bonding_and_the_Effect_of_Crystal_Forces_on_Molecular_Conformation


text:
Given a forward reaction with positive enthalpy, is its reverse reaction be endothermic or exothermic?
I'm pretty sure the sign changes.
But, I want to be sure if it becomes endothermic or exothermic.

Welcome to chemistry S.E.! Please tell us what you feel, we will correct you if you are wrong :)

Yes, in a reverse reaction the sign changes. That is because Enthalpy is a state function, one that depends on only the current state of the function. That means the energy required to push a reaction to its products would be directly reversed to push it back to its reactants.
A key use of this knowledge is in Hess's law, and Born Haber Cycle.

If a reaction is endothermic for some conditions would be endothermic for all conditions? In other words, could a reaction be endothermic for X conditions and exothermic for Y conditions? Thanks in advance.

The enthalpy of a reaction can be found by calculating the amount of energy required to break the bonds of the reactants, and then subtracting from this value the amount of energy required to form the bonds of the products. You can find a table of these values here.
The reverse reaction would simply be doing the opposite, meaning that the magnitude would be the same, but the sign would be opposite, as you stated.


text:
What type of bonding occurs in isolated silicon?
Would isolated silicon engage in network covalent bonding as quartz does or would it engage in a different kind of bonding? Would germanium display similar qualities?

I am asking what is the bonding of silicon, not why silicon is different from carbon.
I am interested in isolated silicon, not silicon as a "hydrosilicon".

I believe it is network covalent, but I am not sure.

possible duplicate of How come there aren't any silicon analogs of alkenes or alkynes?

Thanks for the link. It helps provide insight into the silicon silicon bond, but does not describe it as how silicon would bond with itself, but rather why it is ineffective. Even if it is ineffective, silicon does engage in bonding as an isolated compound (I would believe). However, I am curious as to what would be that bonding

You mean elementary silicon? - https://en.wikipedia.org/wiki/Silicon

@Mithoron Yes, that is exactly what I mean. I did read the wikipedia page but I saw mainly about its bonding with other elements.

Its structure is drawn in section Characteristics - it's indeed covalent, diamond type.

Oh Thanks! I guess I was so focused on the text I missed the convenient picture!

In addition to @Mithoron's comments and the Wikipedia he linked:
This is an important consideration in photovoltaics, as silicon forms the basis of the internal structure of solar technology (and image sensors). According to the PVEducation.org page Semiconductor Structure,

Each silicon atom has four valence electrons which are shared, forming covalent bonds with the four surrounding Si atoms.

Diagrammatically, the website includes the following schematic:

The caption reads:

Schematic representation of covalent bonds in a silicon crystal lattice. Each line connecting the atoms represents an electron being shared between the two. Two electrons being shared are what form the covalent bond.

In terms of germanium, according to the WebElements page Group 14 Elements,

Tin and lead are very metallic although one modification of tin known as grey tin has the same diamond structure [as carbon] as does germanium and silicon.


text:
What is this reference material in spectrophotometry?
First, I should say I never have done any spectroscopies and I am just reading about it.
As far as I know, UV-Vis spectrophotometer has 2 beam samples: sample and reference. The sample has very diluted test material in a solvent and the reference is only the solvent.
I was reading a book about some experimental values for UV-Vis spectrophotometry and I noticed this page.  My question is about that sentence saying "The data were obtained .. against water reference" .   As the reference is the solvent alone, does it mean the samples are dilute solutions in water ?  if so, how they solved benzene in water?
Here is a snapshot of the text:

Our UV-Vis is a single beam instrument. It does not require a pair of matched cuvettes since it records the reference spectrum to subtract from the sample spectrum.

As the reference is the solvent alone, does it mean the samples are dilute solutions in water ? if so, how they solved benzene in water?

I come to a different conclusion.
To me it reads as follows:

sample: solution of benzene, acetone, acetonitrile, etc. in water
reference: water

Benzene is soluble in water: about $1.8\,\mathrm{g\cdot L^{-1}}$ at 25 °C.
Edit
As a reference (blank), a solution of everything except the sample is used. In organic (photo)chemistry this usually means: just the solvent. The situation is different in biochemistry, medicinal chemistry, etc. when complex aqueous buffer systems are used. Here, everything except the probe really means that the whole buffer is used as a blank.

Thanks.  Can we say that reference and solvent are always the same ?

@Aug Mostly, but not always. I've updated my answer.


text:
Is ln(K2/K1) = (∆h°rxn/R)(1/T1 - 1/T2) even when ∆h°rxn varies strongly with temperature?
In all of my physical chemistry books I find the same expression for estimating the equilibrium constant of a reaction at a non-standard temperature. The derivation starts with $R \ln K_\text{eq} = -\Delta g^\circ/T = -(\Delta h^\circ/T - \Delta s^\circ)$. Then one takes the temperature derivative of each side assuming that $\Delta h^\circ$ and $\Delta s^\circ$ can both be considered constants, which eliminates the entropy term. The equation
$\mathrm{d}\ln K = (1/R)(\Delta h^\circ/T^2)\mathrm{d}T$ is then integrated between $T_1$ and $T_2$ to give
$\ln(K_2/K_1) = (\Delta h^\circ/R)(1/T_1 - 1/T_2)$.
Here's the question: $\Delta h_\text{r}^\circ$ is really NOT a constant, but a function of temperature. Assuming constant heat capacities, $\Delta h_\text{r}^\circ$ at $T = \Delta h_\text{r}^\circ + \Delta c_p(T - T_0)$. Likewise $\Delta s_\text{r}^\circ$ at $T = \Delta s_\text{r}^\circ + \Delta c_p \ln(T/T_0)$. Where $\Delta c_p$ is the difference in the heat capacity of the products and the reactants.
I have tried to insert these expressions into the typical derivation but am getting wonky results. Can someone help me with the derivation or point me to a source where $\ln K$ vs. $T$ is derived with temperature-sensitive reaction enthalpies and entropies.
Edit 4/19/2015: Would it be more effective to invoke $\mathrm{d}\Delta g = \Delta v\mathrm{d}P - \Delta s\mathrm{d}T$ which at constant pressure, and in the case of free energy of reaction, gives $\mathrm{d}\Delta g_\text{r}^\circ/\mathrm{d}T = \Delta s_\text{r}^\circ + \Delta c_p \ln(T/T_0)$ ?
Thanks in advance for any help.

Why wonky results? What do you obtain?

Relevant: http://en.wikipedia.org/wiki/Van_%27t_Hoff_equation

For each species, the quantities $g^0$, $s^0$, and $h^0$ are defined as the free energy, the entropy, and the enthalpy of the pure species, respectively, at 1 atm. pressure and temperature T. So, for constant pressure of 1 atm., $d\Delta g^0=\Delta s^0dT$.  But, $\Delta s^0=\frac{\Delta g^0-\Delta h^0}{T}$. So $\frac{d\Delta g^0}{dT}=\frac{\Delta g^0-\Delta h^0}{T}$
So, $\frac{d(\Delta g^0/T)}{d{(1/T)}}=\Delta h^0$

I'm getting that the expression comes out as
$\frac{d(\Delta g^o/T)}{d{(1/T)}}=\Delta h^o$  without the $ /T $.

Found an answer to my original question in Castellan's 1971 book Physical Chemistry (equation 11-59). Starting with  $d(ln K)/dT = \Delta h_r^o/RT^2$ expand $\Delta h_r^o$ as a polynomial series $\Delta h_o^o + AT + BT^2 + ... $ then integrate.  That was a lot simpler than I expected. Thanks everyone for the ideas that put me on the right track.

Thanks for spotting my math error.  I corrected it.  In my equations, the superscript 0 signifies at 1 atm pressure.  However, the standard heat of reaction $\Delta h^0$ is a function of temperature (as your latter post indicates).  The terms in your polyn0mial are obtained from the standard heat of reaction at 298 K and the temperature-dependent heat capacity variations of the reactants and products (by integration).


text:
When does hydrogen bonding happen?
What kinds of molecules will experience hydrogen bonding (inter and intra)? Why does it occur for some H atoms but not others?

What kinds of molecules will experience hydrogen bonding (inter and intra)?

Intramolecular hydrogen bonds:

The hydrogen bond which is within the same molecules is called intramolecular hydrogen bond.
It will form when 2 functional group of a molecule can form hydrogen bonds with each other.
In order for this to happen, both a hydrogen donor an acceptor must be present within one molecule, and they must be within close proximity of each other in the molecule.
EXAMPLE:

 Source
Intermolecular hydrogen bonds:

The hydrogen bond which occurs between two separate molecules is caleed intermolecular hydrogen bond.
They can occur between any number of like or unlike molecules as long as hydrogen donors and acceptors are present an in positions in which they can interact.
EXAMPLE:

 Source

Why does it occur for some H atoms but not others?

Hydrogen bond depends on certain factors like electronegativity, viscosity, atomic size.
Viscosity:
Those substances which are capable of forming hydrogen bonds tend to have a higher viscosity than those that do not. Substances which have the possibility for multiple hydrogen bonds exhibit even higher viscosities.
electronegativity:
Hydrogen bonding cannot occur without significant electronegativity differences between hydrogen and the atom it is bonded to.
Thus, we see molecules such as PH3, which no not partake in hydrogen bonding.
Atomic size:
The size of donors and acceptors can also effect the ability to hydrogen bond. This can account for the relatively low ability of Cl to form hydrogen bonds. When the radii of two atoms differ greatly or are large, their nuclei cannot achieve close proximity when they interact, resulting in a weak interaction.

Source for detailed explanation

Are lone pairs a necessity for an H bond to be formed? Check this link they advertise lone pairs as the means of salvation (for an H bond to form). I'm commenting because you haven't mentioned the role of lone pairs in your answer even though the question is "When does hydrogen bonding happen?"

Most people seems to think that hydrogen bonds are limited to $\ce{X-H\bond{~}Y}$ interactions, where $\ce{X}$ and $\ce{Y}$ are heteroatoms.
In the crystal, a $\ce{C-H}$ of an aromatic system or a $\ce{C=C}$ double bond can serve as a donor as well and interact with heteroatoms, such as oxygen atoms in $\ce{C=O}$ or $\ce{C_{arene}-O-C_{alkyl}}$.
These interactions can form complex networks of rings and strands.
We've examined and published that a while ago for a couple of different phthalimides, such as N-(4-methoxybenzyl)phthalimide or N-allylphthalimide, and some more.
While the articles are behind a paywall (they were published before Acta Cryst. E went open source), the CIF files can be downloaded and played with/examined.
A crucial parameter for hydrogen bonds is distance, if the acceptor is too far away, there is (obviously) no hydrogen bond.
As a rule of thumb, $2.3\,\mathrm{\mathring{A}}$ is a distance where hydrogen bonds can be expected.


text:
Does intramolecular hydrogen bonding occur in dicarboxylic acids?
Consider ethanedioic acid:

Could it form intramolecular hydrogen bonds (between the C=O and HO)?

It's theoretically possible, but I'm not sure if it's observed in practice

To add to @Mithoron 's comment, here is a link to a paper where calculations suggest that intramolecular hydrogen bonding (like what you've drawn) occurs in oxalic acid.  Here is a link to another reference that discusses this result.

No, this would not happen the way your second drawing depicts because the hydrogens do not orient themselves that way. Intramolecular hydrogen bonding would be no more feasible in ethandioic acid than any other monocarboxylic acid. See picture below:
Source: http://commons.wikimedia.org/wiki/File:Oxalic-acid-3D-balls-B.png

I'm afraid It's not a proof.

Why do you believe that the hydrogens "do not orient themselves that way"?  See my comment above, calculations suggest that a conformation with intramolecular hydrogen bonds is the most stable conformation.

It would be a large concentration of charge in a relatively small region of space. It is my understanding that the electron clouds and atoms orient themselves in a molecule so as to reduce this repulsion between the substituent atoms.


text:
Why does mercury not form amalgam with iron?
I am learning about solutions and while giving an example of a liquid (solute) and a solid (solvent), my teacher told us that mercury forms an amalgam with all the metals except iron.
I want to know why this is so?

Iron does form an amalgam with mercury: see for example this paper http://iopscience.iop.org/0953-8984/4/44/024.

Pt, W and Ta also resist dissolution with Hg https://en.wikipedia.org/wiki/Amalgam_(chemistry)

Iron does not form an amalgam readily, hence the use of iron electrodes in mercury switches.  Apparently, iron does form alloys with mercury under specific conditions; see http://www.madsci.org/posts/archives/2011-04/1304143502.Ch.r.html.
By contrast, mercury spilled on gold (e.g. gold-alloy jewelry) soaks in immediately, discoloring the gold, and migrates further into the bulk metal in a few hours, thus restoring the color.

Mercury also forms an amalgam with aluminium, one of th reasons why it is banned on planes.

As does gallium, both of which creep along the grain boundaries, destroying the integrity of the alumin(i)um in short order. https://www.youtube.com/watch?v=IgXNwLoS-Hw BTW, the (very toxic) mercury or (safer) gallium can be recovered by dissolving aluminum in the alloy in NaOH solution.

The solubility of iron in mercury is less than 5x10^-7 wt%. This means a litre of mercury will dissolve less than 70 micrograms of iron (or 17 grams of gold), which is negligible. However, mercury still wets the metal and adheres to it, so iron and copper amalgams are possible by mixing mercury with a very fine metallic powder. Copper lays somewhere between gold and iron in solubility and it is still possible to make amalgam that hardens, though not with solid metal, only powder.

@Libor, that might also be expanded to an answer, with citations for the solubility. Thanks for the info. BTW, the solubility of iron seems to be quite temperature-dependent, decreasing above 87°C, in the answer.

In Fe,Co,Ni The Metallic Bonds Are Very Strong Hence Don't Allow Mercury To Diffuse Into It Hence No Amalgam With Them While In Other Metals Like Ag,Au,Na etc(Soft Metals) Mercury Can Easily Diffuse And Form Metallic Bonds With Them(Ag,Au,Na,etc) Hence Amalgam Formation In Them.
Note:-All Metal Mentioned Above Are Of Comparable Size Of Hg i.e. Fulfilling Amalgam formation Condition.

This is thoroughly incorrect.

@Vishal Gupta  Is there a reason why you capitalize the first letter of every word ?