text:
Electronic and geometric configuration of Fe(OH)₂
I'm having serious problems in the electron configuration of iron dioxide $\ce{Fe(OH)2}$ because all I could come up with was $\ce{Fe}$ double bonded with each oxygen and single bonded with each hydrogen.
Everything seems good except for the iron that's bonded with 5 atoms (which I think is weird). Could someone you give me the right electron configuration (or geometry)?

I apologize if there's something to this that I don't know about, but my understanding of it is...
We have $\ce{Fe(OH)2}$, also called iron(II)hydroxide (iron-two-hydroxide). The reason it's not called dihydroxide, is that the number (II) describes the oxidation state of the iron, so you can easily calculate how many hydroxyl-groups (which are always -1) it needs to have an overall charge of 0.
In iron(II)hydroxide you don't have covalent bonds (double bonds, single bonds)... You have ionic bonds. This particular compound consists of a $\ce{1Fe}^{2+}\ce{+2OH-}$. They don't SHARE electrons, they are simply held together by their charges!
Ionic compounds group together in lattices. They don't have "geometric" configurations like sulfur dioxide for example. Instead, they are clustered together like in the image below:

The negative dots would be the $\ce{OH-}$ and the positive dots would be $\ce{Fe}^{2+}$. Then, of course, you would have two negatives for each positive. Ion bonds are the strongest type of bonds there is.

Thak you for your fast response ,indeed It's iron 2 hydroxide .The reason I've made the mitake is that I studied them in French .Your answer is more than clear ,It's a crystal so It has ionic bonds not covalent ones...thank you very much!


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What causes Copper Rose?
We made a solution of Tetrakis Copper(I) hexafluorophosphate and the usual color is yellowish. As it oxidizes to Cu(II), it should turn blue.
However, our current mixture has an off color that is pink. Looking up alternative copper colors, there will be copper rose. What would cause that color?

quickly looked into the Wiki if you made a solution in water, disproportionation occured, gigiving  $Cu(II)$ and $Cu(0)$. Normal color of $Cu(0)$ is red. This link http://www.webexhibits.org/causesofcolor/9.html gives good explanation, why gold is yellow. Copper gets its red color roughly in same way

@Permeakra, I have a strong inclination you're right. The Cu(I) hasn't been stored correctly so that reaction is likely to happen.

I assume you mean tetrakis(acetonitrile)copper(I) hexafluorophosphate. This compound contains a tetrahedral complex of $\ce{Cu+}$ with 4 $\ce{CH3CN}$ ligands which stabilize the copper(I) center. They are, however, easily exchanged in the presence of other potential ligands, and with water, disproportionation to $\ce{Cu^2+}$ and elemental $\ce{Cu}$ will occur.
$$\ce{2[Cu(CH3CN)4]^+ + 6H2O ->~ [Cu(H2O)6]^2+ + Cu\downarrow + 8 CH3CN }$$
The pale blue color of the hexaaquacopper(II) complex might not be visible at low concentrations, and the pink color that you have observed arises from fine particles of the elemental $\ce{Cu}$ precipitate.

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Oxidation number of nitrogen in KCN
How can one determine the oxidation number of nitrogen in $\ce{KCN}$?

Could you elaborate on what you already know?

KCN is a ionic compound consisting of $\ce{K+}$ and $\ce{CN-}$. You can know that the cyanide ion will have a negative charge, because a potassium ion always has a positive. Also the elements in the cyanide ion are more electronegative - that means they will have the negative oxidation number.
So we know that since potassium has an oxidation state of +1, the cyanide ion has an overall oxidation state of -1. As stated above, ions have oxidation states corresponding to their charge. But you want to know what the oxidation state of the NITROGEN atom has.
There are rules you can use for polyatomic ions (ions with multiple atoms in it), such as $\ce{CN-}$. Nitrogen is more electronegative than carbon. Thus it will have the negative oxidation state (think: it's better at grabbing electrons, which are negatively charged). It is 3 columns away from being a noble gas, thus its oxidation state it -3. The overall oxidation state of the ion is -1, thus carbon must have an oxidation state of +2.
To sum up: K = +1, C = +2, N = -3. If you're trying to balance an equation though, you don't need to look at C and N individually. You just look at the $\ce{CN-}$ as a whole and the $\ce{K+}$.
How did I know $\ce{CN-}$ was an ion? Since potassium is a metal, and only forms ionic bonds. C and N do not form ionic bonds, they bond covalently, thus they act as a single ion in ionic compounds. Does that make sense? Other than that it's just something you learn to recognize with time!
Read rules for determining oxidation states here:
http://www.tmcleod.org/cgi-bin/chem1/wiki.cgi?action=browse&diff=3&id=OxidationNumber
There's a comment to this answer explaining what causes electronegativity. A general rule of thumb is that electronegativity increases as you go towards higher group number and low period number. You can study it for yourself in the picture below:

(source: wikispaces.com)

Some of this is a little too general and hand waving. Potassium doesnt always have a charge. Potassium metal for example.
Also "more to the right of the periodic table" is not a sufficient description of electronegativity. The cause of electronegativity is the relatively similar shielding of the nucleus across a group whilst the number of protons (and hence charge on the nucleus) increases across the group. This increased positive charge with similar shielding causes N to be more attractive to electrons than C.

This doesn't look like Pauling scale in your picture, which one is it then?

The oxidation state of $\ce{N}$ is effectively zero, while the oxidation state of $\ce{C}$ is -1. You may rationalize this by looking at the molecular orbitals (MOs) of the cyanide ion. Carbon and nitrogen are sp-hybridized and triply bonded. As the MOs are filled, the bond MOs are filled first followed by the lone pair MOs, one for N and one for $\ce{C}$. Both lone pairs are much lower in energy than the $\sigma^\ast$, $\pi^\ast$, or $lp^\ast$ MOs.
Alternatively, you may follow the 8-electron rule and use the fact, that carbon and nitrogen are covalently bonded. This also gives a -1 oxidation state for carbon.
This behavior is also consistent with various reactions for cyanide, which acts as a nucleophile binding with the carbon instead of nitrogen.

You're talking about oxidation mumber or formal charge?

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Why does attractive forces of nucleus increase more than shielding across groups?
When you study the electronegativity of the elements, the general trend is that it rises with increasing group number, and decreasing period.
Supposedly this is because the attractive forces of the nucleus increase quicker than the shielding effect of the electrons.

Is this shielding effect simply a repellant force, or is there another phenomenon at play?
Why does the shielding effect not increase as rapidly?
Why does the difference between the attractive forces and the shielding effect become less pronounced with increasing period?

(source: wikispaces.com)

Question (1)
To at least a very good approximation, the shielding effect is merely due to a repulsive force.  Like much of chemistry at the atomic level, electrons experience a balance of counteracting forces due to the nucleus and other electrons.
Question (2)
This is usually explained in the following way.  As one goes across a period from left to right, the charge on the nucleus increases at each step, increasing the attractive potential for electrons (which you already noted).  In addition, an electron is added to the valence shell for each step.  Since the added electrons are in the same shell as the preexisting valence electrons, they do not contribute as strongly to shielding as one might initially expect.  Electrons that are in lower shells contribute to shielding more effectively since they are "more likely" to be between a valence shell electron and the nucleus.
Question (3)
One way to justify the increasing effectiveness of shielding in later periods is an extension of the answer above to question (2).  There it was stated that valence shell electrons do not effectively shield other valence shell electrons since they are at the same level.  An natural extension of this would be that the further apart two electronic shells are, the more the lower shell screens the higher shell.  Compared to the valence electrons in Pb, the 1s electrons are essentially right up against the nucleus and, from the view of Pb's 6p electrons, should almost completely neutralize the charge of two protons in the nucleus (i.e. very effective shielding).
Another way of looking at this is that as the principle quantum number (n) increases, the orbitals become less penetrating.  This means they spend less time near the nucleus (where the shielding is weak) and more time further from the nucleus (where shielding is strong).

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How can you estimate how much of a substance will dissolve in a given lipid?
How can you estimate how much of a substance with a known partition coefficient and other physical/chemical data regarding the substance will dissolve in a given lipid? In particular I would like to know the answer to this question in the context of the blood-brain barrier (BBB).

The major issue with the blood-brain barrier is that aside from small hydrophobic/lipophilic compounds (including $\ce{O_{2}}$ and $\ce{CO_{2}}$), almost everything else is prevented from passing due to cellular tight junctions and will only (potentially) pass through via active transport.
However, for small lipophilic molecules passing through a phospholipid bilayer (such as on a cell surface in the blood-brain barrier), this post provides information on equations: http://www.ncbi.nlm.nih.gov/books/NBK21626/
Where:
$\frac{dn}{dt}$ is diffusion rate (mol/s)
$K$ is the partition coefficient (the equilibrium constant for its partition between oil and water) - e.g. for Urea, K = 0.0002 while for diethylurea which is more hydrophobic, K = 0.01
$C^{m}$ is the concentration in the hydrophobic part of the phosopholipid bilayer (mol/L)
$C^{aq}$ is the concentration in aqueous solution (mol/L)
$P$ is permeability coefficient
$D$ is diffusion coefficient of the substance within the phospholipid membrane
$A$ is the area of the membrane (in $m^{2}$ I think?)
$x$ is the membrane thickness (in m I think?)
$C_{1}^{aq}$ and $C_{2}^{aq}$ are the aqueous concentrations on either side of the layer (one will be the serum or CSF concentration, the other will be the intracellular concentration)

$K = \frac{C^{m}}{C^{aq}}$
$\frac{dn}{dt}=PA(C_{1}^{aq}-C_{2}^{aq})$ and $P = \frac{KD}{x}$
$\frac{dn}{dt}=A\frac{KD}{x}(C_{1}^{aq}-C_{2}^{aq})$

You would have a similar equation to 2. on the other side of the cells lining the blood-brain barrier.
It seems from my reading that $P$ tends to be better known than $D$, which is actually usually calculated using equation 2. from $P$, $K$ and $x$.
Hope that helps!


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Why does a substance with an endothermic heat of solution dissolve?
How does a substance with an endothermic heat of solution dissolve?

Dissolution happens in three steps.
Solute-solute attractions must be broken (consumes energy, endotherm), solvent-solvent attractions must be broken (also endotherm), and finally solute-solvent attractions form. This results in a lower energy state and is exotherm.
Dissolution will be endotherm if it takes more energy to break the mentioned attractions, than is released in the last step. You are asking then, why is this reaction spontaneous? In other words, why does it occur? To find out we need chemical thermodynamics. $$\Delta G=\Delta H - T\Delta S$$
$\Delta G$ = Change in Gibb's free energy for a reaction. $\Delta G<0$ = reaction is spontaneous. Otherwise it is NOT spontaneous (ie. will not occur on its own).
$\Delta H$ = Change in enthalpy. $\Delta H <0$ = reaction is exotherm. If larger than 0, the reaction is endotherm. Already now we can see that an endotherm reaction will make $\Delta G$ for the reaction more positive, and thus the reaction is less likely to be spontaneous.
$T$ = temperature in K.
$\Delta S$ = Change in entropy. $\Delta S > 0$ = reaction leads to a state of higher entropy. Entropy describes disorder in a system. Dissolution of a solute in a solvent will always lead to a state of higher disorder, since we go from having all the solute concentrated in a cluster (more ordered) to being spread evenly throughout the solution (more disordered). We can see from the equation that if the reaction leads to higher entropy, then higher temperatures increase likeliness of $\Delta G < 0$, and the reaction will be spontaneous. On the other hand, if we're going to a more ordered state, lower temperatures increase likelihood for a spontaneous reaction.
So how do we apply this to dissolution?
You know that the dissolution reaction is endotherm. That means that $\Delta H > 0$ for this particular reaction. Yet we know that the reaction is spontaneous, since you report that it does indeed occur! This means $\Delta G < 0$. The only way to make that happen is for $\Delta S$ to be positive. So the reaction must go to a more disordered state. Since it's a dissolution, we know that it does. The solute goes from being concentrated in a cluster (ordered) to being spread throughout the solvent (disordered). Pay attention to where temperature is in the equation. It is quite easy to see, that for a dissolution reaction where  $\Delta S > 0$, you can increase the chance of the reaction being spontaneous by increasing temperature!
Bottom line
Your dissolution reaction is endotherm, but because dissolution reactions lead to states of higher entropy, your particular reaction is spontaneous (it occurs) at your specific temperature nonetheless! If it wasn't, heating the system could make it so.


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White powder observed after boiling water in electric kettle for many weeks
I used my water kettle (metal ones, not the plastic ones) to boil my water. Usually, if I did not consume all the water in the kettle and the water was cold, I usually topup extra water to the kettle and turn it on to boil.
So, after many weeks of boiling in this way, I notice that there are white powder outside the kettle. (I notice that when the water is boil, some water was splashed to the outside of the kettle and I left those water splashed out to dry as time goes)
Is the white powder actually called fluoride? How do I tested it to verify if the white powder chemical is safe or not?

I've never seen a kettle where that hasn't happened :) If you're concerned, clean it off. But take notice that it's actually coming out of the water. So drinking the water pure will mean you're getting more of it than you are when you've boiled it.

As BBftw's answer says, the white powder is probably largely calcium carbonate. Chalk is calcium carbonate. So the white powder is probably no more harmful than chalk.

You can clean out the kettle by boiling with some white vinegar. // By boiling the water you're boiling out carbon dioxide which makes the solution more basic. This makes mainly calcium carbonate precipitate. $$\ce{Ca(HCO3){2(aq)} -> CaCO3{(ppt)} + CO2_{(gas)} + H2O}$$

It's likely that your home is supplied with "hard" water (that is, water with a high mineral content). These dissolved minerals typically include calcium or magnesium ions (from dissolved calcium bicarbonate and magnesium bicarbonate, respectively). Heating or boiling your "hard" water in a kettle will soften it by precipitating the calcium carbonate as its solubility decreases with increasing temperature.

This (calcium carbonate) could be the white powder you have observed after evaporation. Hard water should not pose any health risks, and in fact can add some extra calcium and magnesium to your diet. If it's still concerning to you, there are hundreds of products available that can soften your water before you use it, just google "water softener" and you'll be overwhelmed.
http://en.wikipedia.org/wiki/Hard_water

I live a place where the water is extremely hard, and I regularly remove calcium carbonate from my electric heater. The method is quite simple: fill the kettle with water, add citric acid (as solid) and heat. You will notice a lot of gas which is carbondioxide (CaCO3 + 2H+ -> Ca2+ + H2O + CO2). You can use any acid but citric acid is cheap.

You can also boil a bit of white vinegar, should dissolve it as well.


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How would you Synthesise LY-293,284?
How would I synthesise LY-293,284?

Figure 1: LY-293,284
I have satisfied the conditions at the page Should we allow Synthesis questions subject to the six constraints listed below? so please don't just write off this question and close it. The constraints are satisfied as followed:

Starter material(s): 5-acetylindole.

Figure 2: 5-acetylindole

I have completed a Google search to check for LY-293,284 synthesis pathways by typing "LY-293,284 synthesis" and there were no applicable results.
I would imagine the synthesis would occur via the following pathways:

Phenylation of the 5-acetylindole to phenyl-5-acetylindole.
Oxidation of the resulting phenyl-5-acetylindole with oxidising agents such as potassium nitrate?
Di-n-propylamination of the resulting 5-acetylindolylphenol with di-n-propylamine to LY-293,284.

The synthesis I'm hoping is simple.
LY-293,284 is a 5-HT1A receptor full agonist.

I think you'd want to start from 5-acetylindole

This is 5-acetylindole: http://i.imgur.com/Qc6W49d.gif

If you look at Eli Lilly patents and other literature, there doesn't seem to be a simple synthesis... I've put together an example of how it could be done. Although there are many ways you could go depending on starting materials, this way is based primarily on literature reactions with the exception of the last 3 steps which are Eli Lilly patents. I stopped after about 16 steps as I think that's more than enough to demonstrate how complex these designer molecule syntheses can be. I also left out reagents/reaction conditions just to show the general pathway, and to save on eye strain.

So really the simplicity (or complexity) will depend on where you are able to start on the synthetic pathway (e.g. what starting materials are available to you).


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Impact of detergent on internal living tissue?
There was some spirited debate between the members of my chemistry lab today about the theoretical consequences of ingesting a large amount of the detergent Triton X-100 at a 100% concentration. The MSDS provided by Sigma Aldrich indicates that it is "harmful if swallowed," with a LD50 of 1800mg/kg in rats, but does not provide further detail.
Given that it is capable of decellularizing animal-derived tissues, would it literally "wash" one's extracellular matrix of its cells after degrading the lipid bilayers?
If Triton X-100 is incapable of doing so, are there more powerful detergents that would pose a genuine threat if ingested?

Detergent - not even once. This is just a wild guess, but I imagine that if nothing else, drinking a cup of pure Triton would wreak havoc with the esophageal and gastric mucosal barriers, the latter of which protects your stomach from your stomach acid.

That was one of our chief concerns- that damaging the stomach's ability to contain acid would allow the acid to act upon the tissues surrounding it.

To any future readers, this is a hypothetical experiment only.  Please do not attempt this.

Do you guys want some gross pictures btw?

Caustic Ingestion Injury
Triton X-100 has a pH of 9.7 at 100% concentration (according to Wikipedia) which can cause alkaline caustic injury, however it is less likely to cause severe full thickness injury, as those injuries tend to happen with ingested products of pH <2 or >12.5 or with other corrosive properties that Triton X-100 does not have. Ingestions at pH 6.0-8.0 are unlikely to cause injury unless there is long term exposure (as in gastro-oesophageal reflux disease or bile reflux); note that saliva itself often has a pH of 7.8!
Alkaline caustic injury has a large spectrum. The alkali can cause liquefactive necrosis via saponification and protein dissolution. In the best case scenario, there is only superficial injury to mucous membranes, causing pain and exudates. In the worst case scenario there is full thickness involvement, usually involving the oesophagus, where you get a perforation. One of the worst cases I have had any involvement in was a young lady who drank a large quantity of Drano (concentrated NaOH with pH 11.5-13.5) and had oesophageal, gastric and duodenal perforation with extensive chemical damage to her pancreas, liver and fortunately to a much lesser extent, mediastinum.
For reference, Zargar's grading system for caustic ingestion (usually based on findings on gastro-oesophagoscopy +/ pharyngolaryngoscopy) is as follows:

Grade 0: Normal examination
Grade 1: Oedema and hypermia of the mucosa
Grade 2a: Superficial ulceration, erosions, friability, blisters, exudates, hemorrhages, whitish membranes
Grade 2b: Grade 2a plus deep discrete or circumferential ulcerations
Grade 3a: Small scattered areas of multiple ulceration and areas of necrosis with brown-black or greyish discoloration
Grade 3b: Extensive necrosis

Triton X-100 would most likely cause Grade 1-2b injury. Grade 3 injury is unlikely but I suppose it depends on dose as well.
Ingestion at lower concentrations and effects not related to caustic ingestion:
This link contains information on the general toxicity of non-ionic surfactants such as Triton X-100.
In summary:

They cause mild irritation to mucous membranes
This is primarily due to foaming
Foaming and irritation can cause vomiting and diarrhoea
Foam can also be aspirated causing chemical pneumonitis
Alcohol ethoxylates (as Triton X-100 appears to be) are readily absorbed into the bloodstream via mucous membranes and is likely to have toxic effects in this way

The rest of the evidence I could find also states:

They can cause haemolysis- makes sense, as surfactants in the circulation would come into contact with red cell membranes
Triton X-100 causes direct cytotoxicity due to cell membrane disruption
It also inhibits NAD(P)H, which is a coenzyme involved in cellular respiration
There is dose-dependent destruction of the cytoskeleton
There is also some associated double-stranded DNA breakage
It is possible for cells to recover from this poisoning

Thus, the likely scenario from swallowing dilute Triton X-100 is this:

First you get foaming, oral & oesophageal pain and abdominal pain
Next you get vomiting and diarrhoea, at which point you may inhale some of the foam
If you've inhaled any foam, you develop respiratory distress
You develop dose-dependent haemolysis which presents as tiredness (and/or shortness of breath, chest pain, palpitations), red or dark brown urine, jaundice
In severe cases potential multi-organ failure as a result of direct cytotoxicity, haemolysis, anaemia and/or chemical pneumonitis

It seems it'd be a huge spectrum all the way from general foaminess and some pain all the way to multi-organ failure and death. I'm not entirely sure what doses would be toxic as it appears no-one has forced animals or humans to drink the stuff!
Regarding whether there are more toxic and corrosive but chemically similar compounds out there
yes.

Phenomenally detailed response about the evidently numerous deleterious effects of Triton-X. Quite a nice description of the progression of injury as well.

@nitrl it's always exciting to see a topic that I know something about, ie medicine!


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What is a level 4 carcinogen?
My friend spilt a "level 4 carcinogen" on their lap.  What is it and how dangerous is it?
They had washed if off after.

Do you know the actual chemical? I'm not sure how many different classification schemes exist, I think the US and Europe use different ones at least.

@mad pyridine might be it.

Did they tell you any other details? For example, what was the compound, was it in a solid form or a liquid/solution, how long was it in contact before being removed?? There are many types of carcinogens which affect us through different mechanisms (e.g. inhalation, absorption through skin, radiation, etc.) so this will determine the level of risk your friend has been exposed to. Also, any idea what classification system the "level 4" is from. In all of the classification systems I have looked at, the danger decreases with level (level 1 being most carcinogenic).

There are several different ways to rate carcinogens.  According to its MSDS, Pyridine belongs to Group 3, A3:  Known to cause cancer in animals, inadequate evidence in humans.
If it was just contact with the skin, your friend should wash it off with soap and water immediately for 15 minutes (this is again from the MSDS).  Short term exposure can cause "irritation (possibly severe), allergic reactions, sensitivity to light."
So, if any irritation occurs, it is probably not serious, but I would recommend seeing a doctor anyways.  I doubt there is anything to worry about, but cancer is not something to mess around with.


text:
Why do some solutions eventually stop boiling and start "popping" as they get more concentrated?
A very long time ago I performed an experiment in my kitchen, part of which was to remove all water from a solution. Unfortunately I don't remember what was in it, but the point is that as the solution got more concentrated, it gradually stopped boiling, and would instead remain completely still for seconds, until a huge bubble would form instantly at the bottom of the test tube, making a pop sound and shooting some of the solution out of the tube. The effect got stronger and stronger until it became too strong for me to feel safe to continue.
I was unable to get an explanation of this effect out of my school teachers at the time. What I expected was that as the solution got more concentrated, the dissolved compound would start to precipitate, until eventually no water is left. Why did the popping occur instead?

What you witnessed was probably something similar to superheating/bumping. Essentially, when a solution is heated too rapidly, large pockets of vapor can form beneath the surface of the liquid that have an internal temperature higher than the expected boiling temperature of the liquid. In those situations, even small physical disturbances and/or further addition of heat can provoke the bubbles to burst explosively as the vapor is released. Usually, boiling chips (made of inert, inorganic materials like aluminum oxide, silicates, carbonates, etc.) are added to the reaction vessel in order to prevent that occurrence. The boiling chips provide an uneven and porous surface around which smaller and more numerous vapor nuclei can form, rather than the large bubbles that can form in smooth containers which have relatively few nucleation sites. Furthermore, the chips themselves provide additional surface area for the radiation of heat, allowing more uniform heat distribution. The likelihood of bumping can also be reduced by using wider reaction vessels, which allow superheated bubbles to escape without disturbing all of the surrounding liquid above them.

As the solution becomes more concentrated, the boiling point elevates, making further bumping even more likely as higher temperatures are needed to bring the solution back to a boil. Increasing concentration also increases the viscosity, making it harder for the solution to move efficiently, which leads to the superheated regions.

@BenNorris If you expand on that even by just a bit, I think that would make a great answer as well.


text:
Software/server for calculation of protonation states of small molecules depending on pH
For docking calculations at different values of pH, I am working with flavonoids and would like to know about any software/server that can prepare ligands depending on the pH. Up to now I read about Maestro (Schrodinger) but it is too expensive for us. I wonder if you could suggest other alternatives.

This is a challenging problem and there are many programs that do this.
One review was published a few years ago: "Comparison of Nine Programs Predicting pKa Values of Pharmaceutical Substances", J. Chem. Inf. Model., 2009, 49 (12), pp 2801–2812

Knowledge of the possible ionization states of a pharmaceutical substance, embodied in the pKa values [...] is vital [...] We compare nine commercially available or free programs for predicting ionization constants. [...] We compared their performances by applying them to 197 pharmaceutical substances with 261 carefully determined and highly reliable experimental pKa values from a literature source. The programs ADME Boxes 4.9, ACD/pKa DB 12.0, and SPARC 4.2 ranked as the top three with mean absolute deviations of 0.389, 0.478, and 0.651 and $r^2$ values of 0.944, 0.908, and 0.894, respectively. ACD/pKa DB 12.0 predicted all sites, whereas ADME Boxes 4.9 and SPARC 4.2 failed to predict 5 and 18 sites, respectively.

Surprisingly, the quantum program did poorly.
There's even a full book on the subject: Computational Approaches for the Prediction of pKa Values by George Shields and Paul Seybold. It covers both quantum and QSAR approaches.
Some of the best seem to be:

ACD/pKa
ADMET Predictor
Propka Also available as open source code: https://github.com/jensengroup/propka-3.1
Marvin / ChemAxon

But there are dozens out there. I'd check the review articles and try a few.

Geoff, thank you for your detailed answer. Could you please point me to some documents describing how Propka can be used for ligand protonation at a certain pH? I need to modify the charge/protonation layer of my ligand at a certain pH according with the predicted pKa value(s). I searched through the web but I find only examples where Propka is used with proteins. I also give it a try but it doesn't give me the expected result. It just gives me a table with the molecule charge as function of pH. That is certainly useful but not what me and the initial question were seeking.

I suggest contacting the authors - including @JanJensen - it's not my code.

For this, you can try to use the LigpKa database server.

Could you expand on your answer a bit to say why this would be a good choice, please?

Well, mainly, because it seems as if there is no big spectrum of available software for the task at hand. So it's a good suggestion in that it's maybe the only option. But I have no experience using it.

Understandable.  It's just better to have some substance with the answer, as I don't know if you chose it because it has the best XYZ, the most reliable W, or because you like the splash screen. ;)   Anyway, anything you know about it would be helpful to future readers of the question, is all.

you can use also H++ server just go to this link;  http://biophysics.cs.vt.edu/

Welcome to Chemistry.SE. Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$.

I don't think this is exactly what the OP wants - it seems to only handle protonation of macromolecules. The question was for small molecules.


text:
What is an oxidizer?
When talking about gaseous fuel / oxidizer mixture what does one mean by oxidizer if there is no oxygen involved?
Is it correct to think of oxidizer as the more electronegative specie, because it is pulling the electron? Or is this definition only valid for solutions?
The Wikipedia article is not clear whether the definition applies to gaseous reactants or only valid for solutions.

Oxidation: losing electron(s), Oxidizer / oxidizing agent: a chemical that can oxidize another reagent. Reduction: gaining electron(s) - think of it as reduction of charge! Reducing agent: A chemical that can reduce another reagent.
And oxidation can never occur without a reduction, meaning if you use an oxidizing agent to oxidize a substance, when the oxidizing agent is in turn being reduced itself. It gains the electrons of the oxidized substance.
To describe how well a substance functions as an oxidizer, we use reduction potentials, measured in volts! A substance with a large, positive reduction potential is easy to reduce, which in turn means it is a good oxidizing agent. Similarly, if a substance has a large, negative reduction potential, it means that it is hard to reduce this substance. This is because it's good at reducing other substances itself - it's a reducing agent.
To figure out if a redox reaction is spontaneous (if it can potentially proceed on its own), we use standard reduction potentials, $E_0$. This is for concentrations of 1 M and a temperature of 25 degrees celcius. You look these up in your textbook or online.
An example: Is the following reaction spontaneous?
$$\ce{Cu^{2+}(aq) + 2Ag(s) -> Cu(s) + 2Ag+(aq)}$$
To find out we first find the standard reduction potentials for each participating species:
$\ce{Cu^{2+}(aq) + 2e- -> Cu(s)}, E_0=+0.34V$
$\ce{Ag^{+}(aq) + e- -> Ag(s)}, E_0=+0.80V$
This describes how much they want to be reduced (GAIN electrons). The silver ion wants it the most! However, notice that in our reaction we're asking for it to LOSE electrons (be oxidized). THEREFORE: No, the reaction is not spontaneous under the aforementioned standard conditions. However, the REVERSE reaction is spontaneous. To calculate the potential for our reaction, we simply say:
$$E_0 = (+0.34V) - (+0.80V) = -0.46V$$
The reduction potential for the substance that we want to gain electrons (reduced), minus the potential for the substance we want to lose electrons (be oxidized). The fact that the result is negative, is what tells us the reaction will not proceed - and that it will in fact proceed in the OPPOSITE direction.
Take notice that these reduction potentials are not given the unit Volt for no reason! They are true electric potentials. You can think of electric potentials as "electron pressures". The pole with the highest electron pressure will be (-), because electrons are negatively charged, and the other will be (+) relative to it.
Remember that the reduction potential (the willingness to take electrons) for the silver ion is higher than that of the copper ion. Silver wants to take up electrons more than copper, meaning that electrons will travel from the copper to the silver. you could just as well say that the copper wants to get rid of its electrons more than the silver - copper has the higher electron pressure.
What about electronegativity?
Electronegativity does indeed correlate somewhat with reduction potentials, at least for simple reactions involving pure elements. And you can see that Ag does indeed have a higher electronegativity than Cu, which makes sense. Electronegativity doesn't take into account the oxidation states of different species etc., and quickly it becomes a better idea to just leave electronegativity out of the picture when dealing with redox reactions.

Brian's answer is very good and thorough, but there is one rather important empirical fact you need to consider with reduction potentials.  While the difference in $\ce{E_0}$ values can tell you if a reaction can occur, it cannot tell you if a reaction will occur.  There are other factors, like reaction rate and activation energy, that can interfere with what $\ce{E_0}$ values indicate.
There are two terrific CHEM Study videos that demonstrate these points.  The first, Bromine: Element from the Sea, shows a procedure for oxidizing bromide ion in seawater to elemental bromine. At about the 9:30 mark, they discuss looking up $\ce{E_0}$ values to find an oxidizer for bromine.  They first try oxygen, which indicates a spontaneous reaction by $\ce{E_0}$, but in fact the reaction does not occur, likely because the rate is too slow.  They then try chlorine, which works to oxidize the bromine.  In this example, the difference in $\ce{E_0}$ values is higher between $\ce{Br_2}$ and $\ce{Cl_2}$ than between $\ce{Br_2}$ and $\ce{O_2}$, so you might reasonably conclude that a higher difference in $\ce{E_0}$ values indicates a faster reaction.
Unfortunately, there is more to it, as the second video shows.  The CHEM Study video on Nitric Acid shows that nitric acid--$\ce{HNO_3}$--can be used as a strong oxidizing agent, owing to nitrogen in the $\ce{+5}$ state.  At around the 10:30 mark, they discuss the potential reduction products that can be produced from nitric acid (mostly nitrogen-oxygen gases) by consulting $\ce{E_0}$ values. The highest potential is for nitrogen gas, $\ce{N_2}$.  But when an experiment to oxidize metals is conducted, poisonous nitrogen dioxide $\ce{(NO_2)}$ is the product, not $\ce{N_2}$, even though $\ce{NO_2}$ has a smaller $\ce{E_0}$ value.  The answer is that reduction to $\ce{N_2}$ requires a higher activation energy, and at room temperature this reaction mostly does not occur, but reduction to $\ce{NO_2}$ dominates.  Not realizing this possibility could literally be the difference between life $\ce{(N_2)}$ and death $\ce{(NO_2)}$!
As always in science, experimentation is the ultimate determinant of what works. Theory such as reduction potentials helps to determine what could work, but it is only the beginning of the process.

Important indeed :-)

Any electron-withdrawing reactant is an oxidizer, whether in gas or liquid phase (solid too). They're called this simply because they act like oxygen, oxygen being the most common oxidizing agent on earth.


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Where did the INN name of fentanyl come from?
Fentanyl is a potent anilidopiperidine opioid. Where did its INN name come from?

Figure 1: Fentanyl
I know that the INN name of paracetamol came from its chemical name para-acetylaminophenol (par-acet-am-ol). Is there a meaningful origin of the INN name of fentanyl?

My guess: the name may be based on two of the immediate precursors to the drug, N-Phenethyl-4-piperidone (FENT, from Phen + ethyl?) and aniline (ANYL):

Definitely not as straight-forward as your para-acetylaminophenol example.


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What is the mechanism of functionalisation of nanoparticle using long chain alcohols
How can long chain alcohols(not thiols) be functionalised on the surface of silver or gold nanoparticles?
I have read in an article that Polyvinyl alcohol is used as a capping agent for silver nanoparticle synthesis, but what is the mechanism for that?
What type of bonds form between nanoparticle and the long chain alcohol(covalent or ionic)?
Is there any relationship between chain length of alcohols(or thiols) to the functionalisation of nanoparticles?

The longer the alykylchains, the more likely you can form self-assembled monolayers (SAMs), which will hold themselves together by covalent interactions between chains.  The nanoparticle analog would be miceles.
However, alcohols are a fairly weak ligand for silver, so I'm not sure this is feasible.  SAMs are known to form on silver with carboxylate ligands.


text:
Reaction of haloalkane with hydroxide ions
My teacher gave me the following question:

What are the possible products of these reactions? Then, what is the mechanism of those reaction (elimination (E1 or E2) or subtitution (SN1  or SN2)? Explain in as much detail as possible.
$$\ce {CH3Cl + OH- ->} ?$$
$$\ce {(CH3)3CCl + OH- ->}?$$

I suggest that the first reaction will produce methanol and chloride:
$$\ce {CH3Cl + OH- -> CH3OH + Cl-}$$
and it is a SN2 reaction, because $\ce{OH-}$ attacking from behind and this makes $\ce{Cl}$ leave the chloromethane. It becomes $\ce{Cl-}$ because it brings one electron from the carbon atom.
Then I suggest the second reaction is a SN1 reaction because there are some barriers behind the $\ce{C-Cl}$ (the barrier is $\ce {(CH3)3}$). But I don't know the product that can be formed.

You are on the right track with the hydroxide substituting the chloride, so you can predict the alcohol product in each reaction. In order to determine whether the reactions will be SN1 or SN2, we must look at the mechanism of each reaction:
SN1

SN2

Wikipedia:Nucleophilic_Substitution
So we can see that in an SN1 reaction, a stable carbocation intermediate must be formed, while in an SN2 reaction there is no intermediate. We can determine the stability of our carbocation intermediate by looking at the following stability diagram:

Wikipedia:Carbocation
From the diagram, you can see that carbocation stability increases with the number of alkyl groups attached to the ionized carbon. If the carbocation is relatively stable, the reaction will likely be SN1, if the carbocation is relatively unstable, the reaction will likely be SN2.
Another important idea, which you have mentioned, is that a central carbon surrounded by bulky groups will be more difficult to attack due to steric hindrance (thus not likely an SN2 reaction), and becomes more easily attacked when dissociated to a carbocation.
You can use these concepts to find/explain the answers to your question.


text:
Nitrogen triodide energetics
Ammonia has a heat of formation of $-46 \:\mathrm{kJ/mol}$ (which means that it's formation from the elements releases heat). When mixed with iodine at room temperature, it makes nitrogen triodide (heat of formation +146 complexed with ammonia, +287 pure). With the slightest touch it decomposes explosively back into nitrogen and iodine. Thus the net reaction for both steps seems to be ammonia decomposing into its elements.
Iodine seems to be catalyzing its decomposition. However, such a highly endothermic reaction seems unlikely (evaporating water is only $40.65 \:\mathrm{kJ/mol}$). Also, ammonia's formation is spontaneous ($-26 \:\mathrm{kJ/mol}$) despite the entropy cost with $1 \:\mathrm{atm}$ reactants. This shouldn't go backwards.
So where do we get the energy to make this unstable compound? Iodine is a weak halogen, it barely reacts with hydrogen. Could oxygen drag this reaction forward by stealing hydrogen from ammonia?

I'm not sure what your question is. The reaction from NI3 is largely exothermic (Wikipedia gives –290 kJ/mol as the reaction enthalpy), which is consistent with its explosiveness. The reaction from NI3 · NH3 is also exothermic: you quote the heat of formation of the NI3 · NH3 as being +146 kJ/mol, and the heat of formation of product NH4I is –202 kJ/mol. Both are largely favored entropically because they release gas.

The point is how is nitrogen triodide made from ammonia and iodine if making it is so unfavored?


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How to calculate weight of CO₂ in one kg of fuel
I'm trying to do homework and I have the answers to the problems but I can't read the explanation because it's illegible and I can't find how it's done in my notes.
Here's the simplest example:
$$\ce{ C + O_2 \to CO_2}$$
It looks like the answer is 44/12 but I don't understand where the 12 comes from at all. How is a question like this approached? (Didn't know what to search in google to get an explanation).

Do you know what the concept of "molar mass" is? And "relative atomic mass" (aka "atomic weight")?

You have 1 mol $\ce{C}$ reacting with 1 mol $\ce{O_2}$ to give 1 mol of $\ce{CO_2}$
What you need now is something to relate weight and mol. This is called molar mass, notated with the letter M, and has the unit g/mol. You'd just look that up on the internet, but it's calculated from the weight of the atoms in each molecule. If a molecule weighs 32.2 units, then it'd have a molar mass of 32.2 g/mol.
The calculations are pretty straight forward. How many moles of fuel per kilo? $1000g / M(\ce{C})$, where M, as noted, indicates molar mass.
You know that 1 mol carbon turned into 1 mol carbondioxide. So the number you just calculated is thus also the moles of carbondioxide. You need to find out how many grams those x mol carbondioxide corresponds to: $n(\ce{CO_2}) * M(\ce{CO_2}) = m(\ce{CO_2})$.
If you're not confident in this, pay attention to the units. n = mol; M = g/mol; m = g. Thus multiplying M by n yields a figure in grams. Is that clear?


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Enthalpy Calculations, Heat of Formation/Combustion
I just don't understand how to do these and would really appreciate your help.
1) Ammonium dichromate decomposes in a reaction when heated. Calculate the heat transferred for the decomposition of 53.0 g of ammonium dichromate according to
$\ce{(NH4)2Cr2O7(s) -> Cr2O3(s) + N2(g) + 4H2O(g)}$ ΔH° = -315 kJ
Do I convert grams to moles and then multiply by ΔH° ?
2) The standard heat of formation $\Delta H_f$ for $\ce{C2H4}$ is +52.3 kJ/mol. If $\ce{C2H4}$ (ethene) reacts with $\ce{H2}$(g) to produce $\ce{C2H6}$ (g) (ethane) ΔH = -137 kJ. What is the heat formation of ethane?

Thanks for at least hinting at your approach for number 1, can you give a similar statement for number 2?

Exercise 1
The decomposition of ammonium dichromate has a standard enthalpy change of $-315 \:\mathrm{kJ/mol}$. So you have a figure that relates the heat to moles. You need to find out how many moles correspond to 53.0 g of the ammonium dichromate. We use molar mass for this. You can calculate molar mass by simply summing the weights of all atoms in the molecule, but I use this tool to do it. $$\ce{M}_{\ce{(NH4)2Cr2O7}}=252.06 \:\mathrm{\frac{g}{mol}}$$
$$\:\mathrm{g/\frac{g}{mol}=mol}$$
$$\Downarrow$$
$$\frac{53.0\:\mathrm{g}}{252.06 \:\mathrm{\frac{g}{mol}}}=0.21\:\mathrm{mol}$$
So far so good. Our figure of enthalpy change is per mole, thus you simply multiply the two:
$$\:\mathrm{\frac{kJ}{mol}*mol=kJ}$$
$$\Downarrow$$
$$\Delta H^\Theta-315 \:\mathrm{\frac{kJ}{mol}}*0.21\:\mathrm{mol}=-66.15\:\mathrm{kJ}$$
Exercise 2
To solve this exercise you need to apply Hess' law:
$$\Delta H^\Theta_{reaction}=\sum \Delta H^\Theta _{f(products)}-\sum \Delta H^\Theta _{f(reactants)}$$
So we actually need a figure for hydrogen. You look that up on the internet for example, and see that  $\Delta H^\Theta _f \ce{(H2)}=0 \:\mathrm{\frac{kJ}{mol}}$.
We now use Hess' law. We have the enthalpy of the reaction, and we have the formation enthalpies of the reactants. So we rearrange the equation to find the formation enthalpy for the single product. Notice we leave out hydrogen only because it's 0:
$$\Delta H^\Theta_{reaction}=\Delta H^\Theta _{f(ethane)}-\Delta H^\Theta _{f(ethene)}$$
$$\Downarrow$$
$$\Delta H^\Theta _{f(ethane)} = \Delta H^\Theta_{reaction}+\Delta H^\Theta _{f(ethene)}$$
$$\Delta H^\Theta _{f(ethane)} = -137\:\mathrm{\frac{kJ}{mol}} + 53.3\:\mathrm{\frac{kJ}{mol}}=-83.7\:\mathrm{\frac{kJ}{mol}} $$


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Why does water become less dense when it cools
Most materials become more dense as they cool, but ice floats in water, meaning that it's less dense. Why is this?

This explains it pretty well.

Under normal conditions (meaning your regular everyday freezer, or a cold day) water forms a crystal lattice below its melting point (as shown below). This structure is held together by hydrogen bonds, which are no longer overcome by the internal energy of the molecules. The hexagonal arrangement of the water molecules in this structure leads to a lower density due to the spaces left in the center of each hexagon (you can now fit fewer molecules in the same volume).

From Wikipedia
There are many other solid phases of water (at least 16 on wikipedia), some which have densities higher than normal liquid water, though most are not naturally formed (at least on Earth).

Can I ask why the specific preference to hexagonal structure? Is it the lowest energy configuration?


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Why do metals tend to lose electrons, as opposed to maintaining electric neutrality?
Metals tend to lose electrons to obtain the stable noble gas configuration of 8 valence electrons.
Why do they want to obtain this configuration, and how does the strength of their "desire" to obtain this configuration compare with the "desire" to maintain neutral charge.  If the answer depends on the chemical, I'm happy for you to provide some examples.
Thanks.

The key point here is that overall, in a compound electrical neutrality is maintained, and you have additional stability added to the mix because of the bond. I may write up an answer on this later :)

Firstly, atoms "want" to achieve the noble gas configuration of 8 valence electrons because it is the most stable form. All that means is that it doesn't tend to react under normal conditions that we experience on Earth, therefore it will stay in that configuration for quite a while and are less likely to react. There is a more complex quantum physical answer for that but you'll have to go elsewhere for than.
The main force that keeps electrons in atoms is the electrical attraction between the electrons and the protons in the nucleus and so, if it is more energetically favourable to lose that electron in order to form a bond, then that is what will happen.
Focusing on the Alkali metals as an example, as you move down the group, they get more and more reactive. This is because of two main reasons that are a result of the electrons being further away from the nucleus:

Because they're further away, the attraction between the protons and the outer most electron is less
Secondly, taking Rubidium as an example, it has 37 electrons and 37 protons. From the perspective of the outer-most electron, there are 36 electrons repelling it, and 37 protons attracting it, therefore acting as a net charge of 1. However, if you take into account the first point, the repulsion of the closer electrons is stronger than the attraction of the protons so it could even be less than one

Do you know which specific energies are involved? Bond formation energies, activation energies etc.?

I have a vague conceptual knowledge, yes, but I'm still on my second year of my GCSEs


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Chemical Equilibrium - Why do changes in pressure cause a shift in the ratio of products and reactants?
I understand Le Chatelier's principle and how every change to an equilibrium system causes an opposing reaction from the system. I also understand how, when pressure is increased, the equilibrium shifts to the side with the lowest number of moles of gas and vice versa.
So, why does this happen? How is it more energetically favourable - or is it something else?
Also, are these explanations of why the ration changes for the other factors right:

Temperature - equilibrium moves to the endothermic side of the reaction as temperature increases because the activation enthalpy of the exothermic reaction increases as the overall energy of the system is less
Concentration - ratio of products/reactants changes immediately after one part of the reaction's concentration increases, it then approaches the same ratio (assuming one whole side of the reaction is added)
Catalysts do nothing more than decrease the time it takes to reach equilibrium as they catalyse both sides of a reversible reaction and reduce the activation enthalpy

With gasses, what you're doing by changing the pressure is you change the partial pressures or the reactants. As long as there's the same moles of gas on either side, the equilibrium is unaffected, but if there's an un unequal number, the reaction quotient is changed. The same would happen if you added water to an aqueous reaction. You can play with the numbers yourself, I'll give you an example to use:
$$\ce{N2(g) + 3H2(g) <->2NH3(g)}$$
We can use the reaction quotient with partial pressures, but it's more clear if we use the one with concentrations:
$$Q_c\ce{ = \frac{[NH3]^2}{[N2][H2]^3}}$$
Using $c=\frac{n}{V}$:
$$Q_c\ce{ = \frac{\frac{n(NH3)^2}{V^2}}{\frac{n(N2)}{V}\times \frac{n(H2)^3}{V^3}}}$$
Take notice of how this fraction depends on volume! So it's really just the system reacting to attempt to reach equilibrium again (making it so that K = Q).
As for temperature. My understanding is that it's not to do with activation energy. It IS related to the enthalpy of the reaction though, and your understanding of what a temperature change means for a particular reaction is correct.
Concentration is the same deal as I've talked about in the beginning of this post, and incidentally it matches your own understanding of it.
You've got catalysts right too. Have a look at the following diagram:

I'm glad you used concentrations over partial pressures because I have actually read about concentrations and equilibrium constants

I'll just explain the effect of changing temperature, as this isn't covered so well as the others in Brian's answer.
Gibbs energy and the equilibrium constant of a reaction is given by this identity:
$$\ce{\Delta G = -RT * ln(K)}$$
The gibbs energy is also given by:
$$\ce{\Delta G = \Delta H - T\Delta S}$$
Equating the two RHS:
$$\ce{-RT*ln(K) = \Delta H - T\Delta S}$$
Divide through by $\ce{-RT}$:
$$\ce{ln(K) = \Delta H/-RT} + \Delta S/R$$
$$\ln(K) = \frac{-\Delta H}{R}*\frac{1}{T} + \frac{\Delta S}{R}$$
If $\Delta H$ is positive - an endothermic reaction, then $-\Delta H$ is negative. Increasing the temperature makes $\Delta H$ less negative, and therefore makes $ln(K)$ less negative, and thus $K$ bigger- it therefore shifts the equilibrium towards the side of the products:
$$K = \frac{Products}{Reagents}$$
Meaning more products.
If $\Delta H$ s negative - and exothermic reaction - then $-\Delta H$ is positive. Increasing the temperature makes $-\Delta H$ less positive, and therefore makes $ln(K)$ smaller, and $K$ is therefore smaller, shifting the equilibrium to the side with the reagents on it.
$\Delta H$ and $\Delta S$ also depend on temperature, but over a small temperature range this can be ignored.


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Effect of light on ionic compounds
If I had an ionic molecule which needed $\pu{4 eV}$ to break the ionic bond, a $\pu{7 eV}$ photon is shot at it. If it is absorbed by the molecule and breaks,  where does the rest of the energy go?

I'm not confident enough to post as an answer, but I'm guessing as heat. Or maybe some electrons end up in an excited state, resulting in light emission when they go back. We'll see when someone more knowledgable comes along :)

The question as posed is confusing. An ionic bond is really just electrostatic attraction between a cation and its neighboring anion. You can speak of how much energy it would take to separate the ions to infinite distance given a starting distance. But 'breaking' the bond as is done photochemically with covalent bonds is not meaningful.

Breaking an (single) ionic bond is a nonsensical concept.  Ionic bonding it the aggregation combination of many electrostatic interactions with a bulk solid.  To liberate an ion from an ionic solid you have to break many bonds, not just one.

Aside of that, a ionic crystal like common salt will vaporize as a mixture of sodium and chlorine atoms, when heated to about 1000 °C. having ions in vacuum needs still higher temperatures, like in electric arcs or in plasma generally.

@Lighthart I think "ionic" in this context is some sort of "non-covalent" bond rather than a crystal. It may be a red herring in terms of the answer to the question.

The energy can go to several different places:

The electrons may get some of the energy. In this case the dissociated atoms will be in excited states.
A photon might be emitted.
The atoms may have some kinetic energy left over after they are separated.


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How to arrange substances in order of increasing vapor pressure?
During the lecture in my chemistry class, my teacher posed the following problem:

Arrange in terms of increasing vapor pressure:
a) $\pu{0.5m}$ sugar
b) $\pu{0.5m} ~ \ce{NaCl}$
c) $\pu{2m}$ ethylene glycol
d) $ \pu{2m} ~\ce{ LiBr}$

The answer turned out to be: d, c, b, a. However, I do not understand why this is the case. I know the formula for vapor pressure is:
$$ \Delta P = -XP^\circ $$
which means that the pressure of the solvent is proportional to the mole fraction of the solute. I'm still not grasping why the aforementioned answer is correct though. Can anyone provide me with an explanation?

To solve this problem, use Raoult's law. In Raoult's law, the vapor pressure of a fluid is lowered proportionally with the mole fraction of solute particles in the fluid.
You know the formula, but you only need it if you want to calculate the actual resulting vapor pressure. We can know which is bigger like this:
A. $\pu{0.5M }$ sugar. Sugar does not ionize, thus there are $\pu{0.5 M }$ particles in the solvent.
B. $\pu{0.5M \ce{NaCl}}$. Sodium chloride dissociates to two ions. Thus there are $0.5 \times \pu{2 M}$ particles in the solvent.
C. $\pu{2M}$ ethylene glycol. Again, this doesn't dissociate, so there are $\pu{2M}$ particles.
D. $\pu{2M \ce{LiBr}}$. Dissociates, resulting in $2 \times \pu{2 M}$ particles.
Thus, listing the concentration of particles in the solvent, you see where the answer comes from:
D: $\pu{4 M}$
C: $\pu{2 M}$
B: $\pu{1 M}$
A: $\pu{0.5 M}$


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Is there any substance that's a 4-4-4 on the NFPA diamond?
A lot of the organometallics are rather... interesting compounds to work with. The most famous (among those who care, anyway) is tert-butyllithium or t-BuLi. It is the textbook example of a pyrophoric substance, demonstrated to pretty much every chemistry major as an air-sensitive chemical requiring special handling (Syringe and cannula transfers, gas-tight septa, argon/nitrogen blankets, that kind of thing). Despite the inherent hazards of the stuff, it's widely used in the industry (hence widespread notoriety) to add butyl groups to an organic molecule (most organometallics are useful for this kind of carbon-carbon bond formation), and you can buy it by the - airtight - gallon canister.
There are, surprisingly (except to most organic chemists), even more dangerous compounds in the average organic chemistry lab. Along the pyrophoric line, many of the multi-methyl-metallics are violently pyrophoric (even compared to BuLi), including trimethylaluminum, dimethylzinc, and dimethylmagnesium. All of these are also extremely poisonous (anything these compounds can do to your wonder-drug-in-progress, they can also do to various key structures in your own body), with dubious honors reserved for dimethylmercury. Because many of these, especially the organoalkaline compounds, react violently with water, they also automatically rate at least a 3 on the "reactivity" scale.
When leads me to wonder, because I wonder about these things from the safety of my office chair in a completely unrelated field; just how bad can it get? Specifically, is there any compound with an accepted use in laboratory or industrial application that is nasty enough to max out the entire NFPA-704 diamond? The closest I can find is trimethylaluminum, at a 3-4-3 (heath-fire-reactivity). T-BuLi is a 3-3-4. I can't find NFPA data on straight diazomethane (which may be because nobody in their right mind ever works with the stuff in its pure gaseous form; it's always used in a dilute diethyl ether solution, and even then is never sold or shipped that way) but it would probably be a finalist, as the gas is acutely toxic, autoignites at room temperature, and detonates on standing (something for everyone!). I'm thinking that these two inorganic families - light alkyl-metallics and organo-polyazides - would be the most likely candidates to produce a compound so toxic, so flammable, and so readily reactive, yet so interesting to chemistry, that the NFPA would see fit to rate it, and would give it highest honors.

There's t-BuOOH but one source (cited by the Wikipedia stub article) says it's 4-4-4 while another source says it's 1-4-4...

@BoltClock and other sources have it at different values… it depends how it is conditioned (in solvent, or adsorbed on porous substrates, …). As the question say, noöne in their right mind would sell it in a form where it is 4-4-4 (and noöne would probably buy it, because it would hardly be usable!)

Diborane is a pleasant 4-4-3. Seriously, that stuff is nuts. Nice example of 3c2e bonding though. Hydrazine is also 4-4-3.

I found a MSDS for diazomethane but it is not a very good one. However, all signs point to 4-4-4.

Tert-butylhydroperoxide looks promising. One nitpick (which doesn't disqualify it as I didn't require it) is that it doesn't react strongly to water (it gets its 4 for reactivity from its strong oxidation qualities and its propensity as a peroxide to explode spontaneously). You still can't use water to fight this fire, but you can use foam.

@KeithS: If you feel that t-Bu-hydroperoxide is an answer to the question, don't be afraid of answering it yourself. You seem to have done a bit more research on it -- maybe you could expand on your comment and explain why/how it is dangerous in each one of the three aspects or something (not necessary, but more info is generally better ^.^). Linking to your sources would be great too :)

Things like ethyleneoxide (4-4-3) and propyleneoxide(4-4-2) are also pretty nasty, but I haven't come across anything like t-Bu-hydroperoxide.

As of now, it seems that diborane is a 4-4-3-W while pentaborane is slightly more reactive at 4-4-4-W.

Monomethylhydrazine is one for a 4-4-4

Diborane.  NIOSH gives NFPA 4-4-4-W:

Pentaborane(9) has the same diamond, at least on Wikipedia.

Answering my own question based on the comments, tert-butyl-hydroperoxide is at least one such chemical. As stated on this MSDS from a government website, it's a 4-4-4, with additional special warning of being a strong oxidizer. The only thing that it does not do that could make the 704 diamond any worse is react strongly with water. It is in fact water soluble, though marginally, preferring to float on top (and therefore traditional water-based fire suppression is ineffective, but foam/CO2 will work).
If anyone else can find a chemical that, in a form that is used in the lab or industrially, is a 4-4-4 that is a strong oxidizer and reacts strongly with water, that's pretty much "as bad as it gets" and they'll get the check.

There seems to be some question about the rating of tert-Butyl hydroperoxide.  Perhaps in part because in practice it is never supplied or used undiluted?  The NOAA page itself links to the ICSC which lists a 70% aqueous solution as 1-4-4-OX.

I think chlorine trifluoride deserves a mention
One of the issues with relying on published NPFA triangles to judge the answer to this question is that some of them don't seem to be very reliable given the known properties of compounds.
According to one of its suppliers ClF3 has a rating of 4-0-3. So it would not count very highly here. This seems about as inaccurate as describing dimethylmercury as only mildly toxic. I mean how in hell can it have a zero on the fire score and only a 3 on the toxicity score?
Here is how John D Clarke describes it in Ignition (his history of rocket fuels):

It is hypergolic with every known fuel, and so rapidly hypergolic that no ignition delay has ever been measured. It is also hypergolic with such things as cloth, wood, and test engineers, not to mention asbestos, sand, and water—with which it reacts explosively. It can be kept in some of the ordinary structural metals—steel, copper, aluminum, etc.—because of the formation of a thin film of insoluble metal fluoride which protects the bulk of the metal, just as the invisible coat of oxide on aluminum keeps it from burning up in the atmosphere. If, however, this coat is melted or scrubbed off, and has no chance to reform, the operator is confronted with the problem of coping with a metal–fluorine fire. For dealing with this situation, I have always recommended a good pair of running shoes.

So for example, while the compound itself is arguably not flammable, absolutely everything else is flammable in its presence. And given that those reactions tend to produce far more benign compounds like HF gas, it can hardly be considered as only a 3 on the toxicity scale either (though, perhaps, if you are dead already from the corrosive or explosive effects of its reactions, the toxicity is not a major concern).
To put it mildly, ClF3 is one of the most dangerous compounds that is commercially available on any reasonable metrics and damn those published NPFA triangles.

True, there's also dioxygen difluoride that has 4 health,0 fire and 4 reactivity. It seems pretty accurate for this one,but chlorine trifluoride seems far more dangerous than a 3 on reactivity.

A chemical that reacts explosively with water, drenches everything in unquenchable fire, and not to mention is highly toxic, should definitely be treated as if it were 4-4-4.

Ethyl nitrite is rated 4-4-4 but it does not have anything on the white square.
Pentaborane has a good 4-4-4-W. However it's not an oxidizer.

Wikipedia's article on pentaborane rates it a 3-3-2; not sure what form it's in for that rating. Anyway, source? If you can back up your pentaborane rating you'll get the check; it may not be an oxidizer, but a 4 for reactivity plus a "no water" hazard means it'll do practically everything else but.

As for ethyl nitrite, that's been used as a cold remedy for centuries (though OTC sale was banned by the FDA in 1980 as being "not safe or effective"), and is rated a 2-4-4 by NIST. Notwithstanding a few doozies from the medical community and a mentality in previous generations that "if it tastes bad/hurts/makes you puke, it must somehow be good for you", anything nasty enough to get a Health-4 rating from the chemists would be a pretty obvious poison over on the medical side.

NIOSH rates pentaborane 4-4-2, no W noted.

It seems that diborane (see answer by feetwet) is the only compound rated with H4 F4 R4 code by NIOSH, though there are two substances lacking just a single point to join a 4-4-4-club:

picric acid $\ce{(O2N)3C6H2OH}$: NFPA 704 3-4-4;
germane $\ce{GeH4}$: NFPA 704 4-4-3.

Interesting about picric acid. Back in grad school, one of my fellow grad students worked in a lab that had a big bottle of picric acid on the shelf. Next to it was a big bottle of TNT. They wanted to make explosive thin film particle detectors. I saw those bottles and never visited that lab again. Thankfully, it was in another building. Later, they had an explosion of the “nitrogen tri-iodide” they had made (for potential use with their detectors) and were drying under suction and with stirring. Luckily, no detonation of the picric acid or TNT.

I cannot seem to find any more 4-4-4 chemicals (except monomethylhydrazine),except the ones that are shown above. I have found ones that are very close:
Nickel tetracarbonyl :(4 health,3 flammability,3 reactivity)
Dichlorosilane (4 health,4 flammability,2 reactivity,water hazard)
Diethyl ether peroxide (2 health,4 flammability,4 reactivity)
Dimethylcadmium (4 health,4 flammability,2 reactivity, water hazard)
Monomethylhydrazine
(4 health,4 flammability,4 reactivity)
Tert-butyl hydroperoxide
(4 health,4 flammability,4 reactivity, oxidizer)


text:
Is there a bifurcation node in the potential energy as a function of reaction progress equation of chemical reactions?

I originally posted a similar question on math.stackexchange.com, but I believe I might have a better chance of having my question answered here after giving it some time.

I was thinking about the activation energy of chemical reactions, and I was wondering if there exists a bifurcation node somewhere in the transition state.   I'm wondering about this image:

Here's what's bugging me about it. I don't have an actual equation to the curve, so I can't necessarily do a quantitative analysis. Regardless, I can at least make conjectures about it.
Personally, my instincts are telling me, provided there truly is a bifurcation node, it would be stable at the peak, which sounds odd. From the mathematical definition, stability indicating that any small perturbations from the bifurcation point will "dampen" out in time, rather than flying off into infinity. In context though, that doesn't sound correct. The reaction has to go one way or another. The peak, representing the transition state, is said to be unstable chemically.
Another issue is with the parameter of time. It has to go forward. In order to analyze it, we would more than likely have to nondimensionalize it.
These are just some of my current thoughts on it. My question is regarding the mathematics of it. I'm just debating whether there is a bifurcation point (which I believe there is), and if so whether it is stable, unstable, or halfstable. Basically, I'm not sure if my reasoning is correct right now. Any help would be appreciated.
Edit: Let me also add that instability does not necessarily mean that a phase point will fly off into infinity. Stability and instability can be be local. In other words, an unstable point can fly into another equilibrium point.

Edit 2:
I'm going to add a lot more detail to this post as I've done some more research and thinking on this topic. First off, my assumption was correct - in fact, transition state theory is, in part, based on potential energy surface concepts. Hence, saddle-points (bifurcations), are incorporated into the theory. Second, time is not necessarily the independent variable. Typically, the graph of potential energy vs. I.V. where the I.V. is usually a coordinate of some sorts - for instance, bond lengths. That said, I believe that modifies my question.
To explain, what I am curious about is the stability of the transient point as a function of time. Right now, my graph does not reflect that, and therefore my conclusions are not necessarily correct. The slope would be, in terms of instantaneous slope, $\frac{dV}{dq}$. But that's not what I want. Rather, I'm looking for $\frac{dV}{dt}$. Without showing the derivation, one can come up with an equivalency: $\frac{dV}{dt} = -(\frac{dV}{dx})^2$. What winds up happening, now, is that the local maxima in the picture corresponds to an unstable fixed point. That makes sense. Any thoughts on this?

Let me add that this question deserves better tags...may I suggest "activated-complex," "transition-state-theory," "reaction-energy-diagram," and so forth. These are just ideas for anyone who has the rep to add tags.

I'm not really sure if this is a chemistry question. You may have to explain what you mean by "bifurcation" here, it's a bit hard reconciling what the Wikipedia page for it says with the current situation. In the future, please don't cross post. You can flag your question for migration instead.

Hi Mlagma, I'm wondering what your question actually is! Could you edit it to clarify: a. what's your level of understanding of TST and statistical physics; b. what is your question. Make it shorter, more to the point, and I'll do my best to answer!

@F'x I know it's a bit convoluted. Anyway, I'm familiar with statistical physics. As for TST, that is newer to me, but I know enough to understand the main concepts. This is my question: Referencing aspects of nonlinear dynamics, are there bifurcation points that exist in the equation (and qualitatively the graph) relating potential energy to the progression of the reaction w/r/t time? And if so, are these points stable or unstable? Essentially, is it possible to analyze aspects highlighted by TST with the concepts found in the mathematical field of nonlinear dynamics?

@Manishearth I apologize for cross-posting (I didn't realize I could flag it for migration; I'll remember that). Perhaps a better word in this context would be a "fixed point." So in this case, are there stable or unstable fixed points?


text:
Basic Understanding of Relative Atomic Mass
I’m currently trying to learn some chemistry and have run into some trouble understanding some of the basics. The confusion stems from a passage in my textbook, so I will list the brief passage and then ask my question.

In addition to predicting molecular formulas, Avogadro’s hypothesis gives correct results for the relative atomic masses of the elements. Analysis by chemists during the 18th century had revealed that 1g of hydrogen combines fully with 8g of oxygen to make 9g of water. If Dalton’s formula for water, HO, were correct, then an atom of oxygen would have to weigh eight times as much as an atom of hydrogen; that is, Dalton’s assumption requires the relative atomic mass of oxygen to be 8 on a scale where the relative atomic mass of hydrogen is set at 1.

So far so good, I believe that oxygen has a relative mass of 8 because there are 8g of oxygen in the original reaction and likewise with the 1g of hydrogen, but this next part is where it starts to get confusing.

From Avogadro’s hypothesis, however, each water molecule contains twice as many atoms of hydrogen as oxygen, so to achieve the experimental mass relationship, each oxygen atom must have twice as large a relative atomic mass. This gives a relative atomic mass of 16 for oxygen, a result consistent with modern measurements.

Alright, so basically what I’m confused about is how they arrived at a relative atomic mass of 16 given that there is 8g of oxygen and 1g of H in the original experiment. What exactly are they observing in order to establish the relative atomic mass of oxygen as 16. Any help understanding this and relative mass in general would be greatly appreciated.

It's because there's two hydrogen atoms in water, not one. In order to show you, let's assume we're making HO instead of H2O. This would clearly only require half the amount of hydrogen, but still the same amount of oxygen. Thus we have .5 grams of hydrogen combining with 8 grams of oxygen in a 1:1 relationship. Thus the oxygen weighs 16 times more than the hydrogen. Do you understand?
In H2O they combine in a 2:1 relationship. We have 1 gram of hydrogen combining with 8 grams of oxygen, but because there's 2 hydrogen for every oxygen, we actually need TWICE as many oxygen atoms to have the same amount of atoms. In other words, there will be the same number of atoms in 1g of hydrogen as there is in 16g of oxygen.

Basically, the error here is Dalton's in assuming that each mole of water contains an equal number of oxygen and hydrogen atoms. It doesn't; the molecular formula of water is H2O - each mole of water contains 2 moles of hydrogen (2*1g = 2g) and one mole of oxygen (16g). Water has a relative molecular mass of 18g. For every mole of oxygen there are two moles of hydrogen:
 $$\ce{2H2 + O2 -> 2H2O}$$
However, in this particular case, one mole of hydrogen (1g) and half a mole of oxygen (0.5*16 = 8g) are reacting to produce half a mole of water(9g). To get the original relative molar mass of oxygen they would have to double the mass of oxygen used - 8g - which would mean the result of 16g. This then gives a relative atomic mass of 16. Oxygen contains more isotopes (variants of an element with different relative atomic masses) than just oxygen-16 (such as oxygen-17 or oxygen-18), but to the level of accuracy we're working with it doesn't matter too much.
Basically, oxygen does have a relative atomic mass of 16. Ignoring isotopes, it could be worth looking at the (probably simplified) structure of the nuclei of oxygen and hydrogen . Certainly at this level, the relative mass of protons and neutrons is taken to be "1" (the two particles do have very slightly different masses, but this makes things easier). A 'typical' hydrogen nucleus just contains a proton with around 0 neutrons, so it has a relative atomic mass of 1 whereas a 'typical' oxygen nucleus contains 8 protons  and around 8 neutrons; it has a relative atomic mass of 16.
However, as mentioned before, different isotopes containing different numbers of neutrons do exist. Relative atomic mass is just a weighted mean of these (which is why copper can have a relative atomic mass of 63.5 or chlorine can have one of 35.5), but the model above, whilst being a massive simplification and better for relative isotopic mass, might help you visualise the difference better.

Why wouldn't it be $2H_{2} + O_{2} \rightarrow 2H_{2}O$ to get everything to balance?

I made a mistake. Sorry, should have spotted that. Have now corrected it, so thanks for pointing that out. (actually, I'm kind of worried that this answer got upvoted, but it is the sort of mistake that is easy to miss)


text:
Why are these molecular orbitals invalid for hexatriene?
Here are the MOs (more accurately, the breakdown of the MOs) for conjugated hexatriene:

Now, when first asked to draw the MOs myself, I drew this one for $\psi_3$:

It still has two nodes, and is still symmetric. I see nothing wrong with it, except that it doesn't match the "correct" MOs.
Similarly, for $\psi^*_4$, I can draw

Now, as I understand it, MOs are formed from various combinations of atomic orbitals, subject to a few rules regarding symmetry. From what I know, the two MOs I've given above should be valid MOs for hexatriene, or at least components of an MO of hexatriene.
Why aren't these two MOs valid? Also, is there a methodical way to obtain the sequence of MOs in the general case(of a conjugated polyene)?

The general method to find molecular orbitals of π electrons in conjugated hydrocarbons in first approximation is called Hückel’s method. It is an undergrad-level topic, and you should find lots of explanations on the internet or in undergrad physical chemistry.
The wave function you propose is actually equal to $(-\psi_1 +\psi_3 + \psi_5^*)$   (with normalization factor not indicated). Thus, it is not an eigenstate of the system, and not a molecular orbital. But… unless you know Hückel's method, I am not sure there is an intuitive reason to discard this possibility: it has the right number of nodes (i.e. +/– inversions) for $\psi_3$.

Ah, makes more sense now :)
I'll read up on Huckel's method then. Thanks!

I read up a bit about Huckel theory. There's a pdf that explains this particularly well.
At one point, the following formula is derived at one point:
$$c_s\propto\sin\left(\frac{\pi k s}{n+1}\right)$$
which gives the coefficient (sans a positive normalisation constant) of the atomic orbital of the $s$th atom in the $k$th MO of a conjugated linear or monocyclic system on $n$ atoms.
A quick and dirty method has also been provided.

Using this, one can get a general idea of how the MO should look, and then one can get the exact +/- sequence by considering the number of nodes. Note that this is really the same thing as applying the $c_s$ formula given above, just in a graphical manner.

I'm no expert, but looking at this reminds me of a problem I had with benzene. There's a totally unscientific pattern here that might help you remember the correct molecular orbitals. I only learned this from looking at your molecular system here, so I don't know how well it extends to other systems (it might!)
First, examine the chain as a ring (this is what I did for Benzene). If you imagine your proposed molecular orbital arrays as wrapping around like in a cyclic molecule, you'd see they're actually equivalent to the correct molecular orbital representations. At that point, it's a question of determining which equivalent representation is the correct one. The rule of thumb here is that the correct arrangement:

Maintains appropriate symmetry.

Appropriate symmetry refers to odd-numbered molecular orbitals having a vertical axis of reflection down the center of the array, while even-numbered molecular orbitals possess a center of inversion.

Maximizes the number of pairs of one type, a type being either a "same-sign" pairing of molecular orbitals or an "opposite-sign" pairing.

Using your example:
Psi-3: Your proposed array, from left to right, follows a pattern of "opposite-sign" pair, "same-sign" pair, "opposite-sign" pair, whereas the correct arrangement features 3 "same-sign" pairs. The correct arrangement takes precedence because it maximizes the number of pairs of a single type, having 3 pairs of one type as opposed to yours only having 2 pairs of one type.
Psi-4: Similar to the previous example, from left to right, the correct arrangement has 3 "opposite-sign" pairs. Yours has at most two pairs of one type (2 "same-sign" pairs in your case)
Psi-5: In the psi-5 case, an alternative array with superior pairing doesn't exist, but an arrangement with equivalent pairing does; however, this latter array would sacrifice the appropriate symmetry and is therefore not correct.
I just came up with this trick here on the spot, so I don't know how consistently this it works (and physically I'm sure it's terribly wrong). It does seem to work with smaller even-number systems, but it has trouble dealing with nodes in larger even-systems like octatetraene. The pattern may otherwise be generalizable to smaller even-number-electron conjugated pi systems.
Where this pattern fails
The pattern starts to break down when you examine odd-number-center systems and large even-center systems with nodes (like psi 3 and 6 in octatetraene). With some modifications, the pattern can still work, albeit perhaps becoming too complicated for practical use. Also none of it applies to conjugated cyclic systems which have degenerate orbitals.
For odd-number systems and even-system with nodes (psi 3/6 in octatetraene), counting the interactions between adjacent orbitals becomes necessary. For example, in pentadienyl radical/ion, my pattern would suggest the psi-3 orbital as down/up/up/up/down, but the correct answer is down/node/up/node/down. Counting adjacent like-sign orbitals as +1 and nodes as zero, this only makes sense if adjacent opposite-sign orbitals are valued at slightly less than -1. The sum of the orbital interaction then favors the system with nodes. This applies in even-systems as well, such as comparing the presence of nodes in psi-3 (up/up/node/down/down/node/up/up) of octatetraene to an alternative up/up/down/down/down/down/up/up array (which is a rearrangement of psi-2, similar to how butadiene's psi-3 compares to its psi-2 -- my pattern works there). Kind of complicated then I guess.


text:
Reaction mechanism for oxidation of primary alcohol to carboxylic acid
Can someone help me identify the steps for the following reaction? I'd like to keep track of everything that happens. I can't figure out what happens with oxidation in this context, and thus can't get started.
It's 2-methylpropan-1-ol oxidized with potassium dichromate(VI) in aqueous sulfuric acid to yield 2-methylpropanoic acid (also known as isobutyric acid).

A dichromate salt and aqueous sulfuric acid are used to prepare chromic acid in situ, which is among the most powerful oxidizing agents commonly available. A primary alcohol is first oxidized to an aldehyde (via a chromate ester intermediate). Under acidic conditions, the aldehyde's oxygen acts as a base and abstracts a proton, the carbonyl carbon is attacked by water (whose oxygen acts as a nucleophile) and simultaneously the carbonyl π-bond breaks, and then finally the resulting molecule is deprotonated by another water molecule, yielding a diol. That diol is then subject to further oxidation by the remaining chromic acid, which occurs by the same mechanism as the initial oxidation to an aldehyde. The ultimate result is a carboxylic acid.


text:
How much CO₂ should one expect in an IR spectrum?
I ran an IR spectrum ($\ce{KBr}$ pellet) and found a sharp absorbance at $2360 \text{cm}^{-1}$, which was unexpected. As far as I can tell, this indicates either a nitrile or $\ce{CO2}$.
I know that there is no nitrile in my product. Would you expect atmospheric $\ce{CO2}$ to show up to a noticeable degree in a typical IR spectrum?

From what I've read (and personal experience with IRs I've run), a relatively intense $\ce{CO2}$ asymmetric bond stretching signal at approximately $\ce{2350cm^{-1}}$ is an extremely common background artifact in IR spectra (along with the broad, weak signals downfield of $\ce{3000cm^{-1}}$ attributable to $\ce{O-H}$ bonds from ambient moisture). Depending on your software, it's often possible to run a "dry" IR reading with no sample present in order to establish a baseline spectrum of the surrounding air, which can then be checked against by your software (so that overlapping signals from the baseline air spectrum are deleted from the spectrum of your sample). Alternatively, you can take measures to dry the optics (I've once used $\ce{N2}$ for that purpose; alternatively, air directed through $\ce{CaCl2}$ drying tubes could possibly be used, but I can't attest to the efficacy of that). That manual drying procedure takes some effort, and might be futile in any case depending on the conditions of the lab. My one experience with that approach was disappointing.

The main source of carbon dioxide in an IR spectrum is you. Be careful not to exhale into the sample area. Or, for fun, do this and see what happens!

It's rather common in my experience. If you background substraction doesn't work perfectly, you'll get a noticeable $\ce{CO2}$ signal. It can be either negative or positive due to the incomplete substraction of the background signal.
If you measure the spectrum multiple times, this signal should vary in intensity as the $\ce{CO2}$ concentration varies slightly. Though this is probably exaggerated in the situation where I observed it, as we have an open IR spectrometer and often multiple students exhaling $\ce{CO2}$ around it.


text:
Intrinsic Reaction Coordinate - What does 'Connectivity' really mean?
I am currently in the process of connecting transition state (TS) structures to other minima in order to determine a reaction pathway.  Now I understand the basic concepts of this method.  You start with your TS, run an IRC in both the Forward and Reverse directions, then you optimize the final structure in the successful IRC run.
My question is, what constitutes connectivity between a TS and a minimum energy structure?
Examples
1.) I run an IRC and I take the last resulting structure and optimize it. The optimization takes 4 steps and the geometry barely changed.  Is this structure connected to the TS?
2.) I run an IRC and I take the last resulting structure and optimize it.  The optimization takes 20 steps and the geometry changed fairly appreciably, however, connectivity of the atoms remains the same.  Is this structure connected to the TS?
3.) I have a TS where a water is being formed next to some other molecule.  I run the IRC and following the pathway shows you that the water forms and is being moved away from the molecule.  The IRC successfully finishes so I take the last structure and optimize it.  The optimization brings the water back towards the molecule to some particular spot where H-bonding occurs with the molecule.  Is this structure connected to the TS?
Note
These examples were purposely laid out in order to determine at what point TS->Minima are or are not connected.

Various criteria may be applied in order to ascertain the connectivity of structures. A common solution is a path generated by requiring it to be a Newton path on the potential energy surface (E):
$$
(\nabla\nabla^\ast E(x))^{-1}\nabla E(x) = -\frac{dx}{dt}
$$
Two points $x_0$ and $x_1$ are connected only if $\exists x:\mathbb{R}\to\Omega,\tau_0, \tau_1\in \mathbb{R}$ such that $x(t)$ is a Newton path and $x(\tau_i)=x_i$, i.e., starting and ending structures, where $\Omega$ is your space of structures (usually centered 3D coordinates). $x$ is then a parameterization of that path on $[\tau_0,\tau_1]\subset\mathbb{R}$.
Sometimes it is additionally required that there is at most 1 maximum on $E(x)$ between $\tau_0$ and $\tau_1$.
You may be interested in reading this page. In a literature search, you may want to read up on the work by Vanden-Eijnden, Bofill, Quapp, and Henkelman. This topic is too rich to discuss in this forum.
To address the specific examples, assuming the starting point is in the convex attraction basin of the nearest optimum and the used optimizer converges in such instances, the optimum is connected in the above sense. Most optimizers used in geometry optimizations are some variant of (quasi-) Newton-Raphson optimizer, so this criterion is given. There are several reasons why the IRC may not converge to the same optimum as the subsequently used geometry optimizer, including changes of basis sets, convergence tolerances, or in case of quasi-Newton methods, the Hessian guess.
The fact that the optimizer brings the water closer to the other molecule is no indication of lack of connectedness. Instead the IRC run did not end at an optimum. It is actually expected that the energy of two (neutral) molecules decreases as they approach each other from infinity even in the complete basis set limit (van der Waals forces). You just found the vdW/BSSE-bound pre/post-complex.

Great answer, but your second link seems broken, could you fix that?

It seems the domain was sold. Sadly I don't remember whose publications I was thinking at the time (It's been almost 4 years :-). Probably Bofill, Quapp, van den Eijnden, Henkelman are good starting points in a literature search.

Thanks for taking the time to revise your answer. Unfortunately I still don't understand the first part, specifically your equation. Could you tell me what $dt$ refers to? Can I assume, that $\tau$ refers to a structure? What does it mean that the starting point is in the convex attraction basin?

@Martin-マーチン I've clarified above. Does this make sense to you now? Newton methods converge on convex functions, hence any convex region that includes an optimum is an attractive basin.

Yes, I think that is quite fine now. I'll let the bounty run for a few more days to give this a little more exposure, but I think it's yours. Thank you for your effort.


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Why does carbocation stability increase in the order 1°, 2°, 3°?
I'm reading about $\:\mathrm{S_N1}$ and $\:\mathrm{S_N2}$ reaction mechanisms. 1° carbocations are unstable to the point of not having been observed in solution, ever. 2° are more stable, and 3° carbocations are the most stable.
I know that higher stability carbocations do also require less activation energy in their formation. Why is that? Why exactly are they more stable? And is this another way of saying they are also less reactive? Does this in turn mean they require a better nucleophile compared to less stable carbocations?
A small side question. How should I think about resonance structures? You don't view it as changing between the forms, spending more time being the more stable form, but rather as a seamless blend of the forms, with more characteristics of the more stable form, right?

It may be better if you post your question about resonance in a separate post. Short answer: yep, it's a blend. More specifically a quantum superposition, but no need to get into that. Basically, the electrons of all the resonating bonds are "shared" over the given area. The probability of finding an electron is higher in the places where stability is more, but that doesn't mean that the electron is zipping back and forth in the area. You need to understand a bit of quantum mechanics to understand this.

One word: hyperconjugation! The more carbons bound to the carbocation, the more bonds can take part in hyperconjugation. This effect happens when a bond between two atoms donates electron density to the electron-deficient carbocation, making it more stable.

Why doesn't this work with hydrogen, considering hydrogen is less electronegative already to begin with?

@Brian: It is not the carbons attached to the carbocation center that donates electron density via hyperconjugation but the bonds attached to it (free electron pairs would also do). If you consider the $\cẹ{(CH_3)_3C^{+}}$ cation the carbocation center contains an empty p orbital. This can interact with vicinal $\ce{C-H}$ $\sigma$ bonds, that happen to be parallel to it, thus lowering its energy.

From what I have read, the increasing stability of carbocations is attributed to the combined inductive and hyperconjugative effects, at least in case of alkanes. In other organic molecules, the mesomeric effect can also come into play. These effects are easy to understand, so i suggest you read up on them.
Subsequently, your next question deals with why more stable carbocations have a lower activation energy in their formation. In my opinion, you're getting the cause-effect mixed up. The carbocation is stable BECAUSE the energy required to form it is lower, not the other way around. And why is the energy lower? Because of the aforementioned stabilization effects!
This is from the point of view of a 17 year old kid who's been studying organic chemistry for about 3 years, and faced the same problems as you're facing. I won't be able to explain it as ManishEarth did though, using quantum mechanics.


text:
How can a group be both a good nucleophile and a good leaving group?
I'm reading about $\:\mathrm{S_N1}$ and $\:\mathrm{S_N2}$ reaction mechanisms, and I have a few questions.
My book has a couple of tables, one lists a bunch of substances grouped as good, moderate and poor nucleophiles. Among the good are $\ce{Br-}$ and $\ce{I-}$. So a couple of halogens. I'm figuring chlorine is in that group too.
Then it has a table showing ability to function as leaving group. Iodine, bromine and chlorine anions are at the top of that list too.
My question is: How can these be quite attracted to the positive charge AND quite easy to remove at the same time?

 I'm not entirely sure of this, comments appreciated
Leaving group tendency is a thermodynamic variable. So, it depends more on the equilibrium and energies of the reaction. We calculate it from the basicity of the group (weak base -> more stable while solvated -> better leaving group).
Now, iodide is a weaker base than $\ce{OH-}$, since its conjugate acid $\ce{HI}$ is a stronger acid than $\ce{H2O}$. So, iodine is a better leaving group.
On the other hand, nucleophilic tendency is basically "how willing to sacrifice a lone pair is the group?".  It's more dependent on forces, and less on energy, as it is a kinetic phenomenon. The outer shell electrons in $\ce{I-}$ are more loosely bonded to it than those in oxygen in $\ce{OH-}$. So, it is a better nucleophile.
Another way to look at it would be via the "accessibility" of the group in solution. This basically attempts to explain the difference between a nucleophile and a leaving group. In a leaving group, you have to "give up" the ion to the solution. This means that solvation doesn't really matter here -- once you've given it up, it's gone. The process will have a similar rate for different ions. However, when looking at the reverse process, you need to check how solvated the ion is. If the ion is surrounded by water molecules (like $\ce{F-}$, it will be much harder for the electrophile to "extract" it. On the other hand, an ion like $\ce{I-}$ is less solvated, and thus more "accessible".

To further contrast with the "leaving group tendency", nucleophilicity is a kinetic phenomenon.

@BenNorris: ? That's already there, see the third paragraph :)

Doesnt the size of the nucleophile play a vital role?  Because iodide is huge wouldnt it have more of a basic nature than nucleophlic nature.

Why is a weak base more stable when solvated? Shouldn't it be the other way round?