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Are refractive or density based Brix meters better for use with alcoholic drinks?
Assume that somebody wants to measure the sugar content of a liqueur. It seems that there are two types of Brix meters for it, ones which use the solution density and ones which use refraction of light.
If we have a meter intended for measuring the content of a sugar-in-water solution (and assume that we are talking pure sucrose here, not other sugars), but are using it for measuring a solution which has, beside sugar and water, also 15 to 35 percent alcohol and also unknown organic (colored or transparent) substances dissolved:

which type will give less error, refractional or density-based?
how large will the error be? For example, if the true value is 20 g sugar per 100 ml solution, how far are the values delivered by the meter likely to stray from those 20 grams (I am looking for a back of an envelope estimation, not for an exact interval)?

This was prompted by a question on Cooking, http://cooking.stackexchange.com/questions/41757/determining-sugar-content-of-liqueurs, where we had to tell the poor OP that doing the measurement won't help him much anyway. But it sparked this interesting side discussion.

My first guess would be that both methods are likely to be heavily affected by ethanol content, which would make them useless. The simplest method I would guess to have a chance would be to measure optical rotation of polarized light, ethanol is not visible there.

@MadScientist are there commercial kits for measuring polarized light rotation which are as cheap, cabinet-space-economical and easy to use for a chemistry noob as a candymaker/brewer's Brix meter?

Once I got going, this turned out to be a lengthy answer. So here's the summary: the alcohol will introduce a significant error in both methods. The density measurement will probably be closer to the actual sugar content. Th density measurement will underestimate the sugar, and the refractive index will overestimate the sugar.

Both the refractive index and the density-based methods work for measuring sugar content because adding sugar to water raises the refractive index and the density in amounts proportional to the concentration of the added sugar. As you and others noted in that thread, ethanol interferes with that measurement because it also affects the density and refractive index of the resulting solution. So how large is that effect?
To estimate this effect, I am going to restrict myself to the 15-35% alcohol (by volume) range you suggested. The charts I found for the density and refractive index of water/ethanol mixtures gives the ratio by weight; your range converts to 12-28% by weight. Pure water has a density of 0.99823 g/mL, and a refractive index of 1.3330. 12% ethanol has a density of 0.97910 g/mL, and a refractive index of 1.3410. 28% ethanol has a density of 0.95710 and a refractive index of 1.3524. Note that adding ethanol increases the refractive index, but decreases the density. I'll come back to that at the end.
This page gives a table of the index of refraction and density of sucrose solutions in water, as a function of the sucrose weight fraction. The 20 g sucrose per 100 mL solution is approximately 20% by weight, so that will have a refractive index of 1.3639, and a density of 1.0810 g/mL.
The density and refractive index changes for both sucrose and ethanol are pretty linear over the range you're interested in, if you take them independently. Assuming that the sucrose effects and the ethanol effects are independent (NB: this may not be a good assumption), the refractive index measurement will be higher by 0.01-0.02, which corresponds to an overestimate of 7-14% sucrose. The density measurement will be lower by an amount 0.02-0.04 g/mL, which corresponds to an underestimate of 5-10% sucrose. So it looks like the hydrometer measurement will be a bit better, but the two methods will be off in different directions. It's a noticeable effect: depending on the method and the alcohol content, you might be off by somewhere between a quarter and half of your intended sugar content.
There is one neat thing you could do to take advantage of the fact that the alcohol moves the density in the opposite direction to the sugar. If you had a bottle of liquor with unknown alcohol content and unknown sugar content, you could take both the density and the refractive index measurement. Again, assuming the four effects are linear, and the sugar effects are independent of the alcohol, you could write two equations for the refractive index $RI$ and density $\rho$ as functions of the alcohol and sugar: $RI = Ax + By, \rho = Cx + Dy$. $x$ and $y$ are your alcohol and sugar content in whatever units you prefer, and $A, B, C,$ and $D$ are the empirical parameters describing each of the four effects. If you take a measurement for $RI$ and for $\rho$, you can solve for $x$ and $y$.
PS: to answer one question suggested by the comments of the [cooking.se] question you linked, density-based brix meters (I'm used to seeing them called hydrometers) are typically used for home brewing beer and wine because the only source of alcohol is the fermenting sugar. You take a density measurement before fermentation begins to get a baseline. Then, at any point after fermentation begins, the difference between the current density and the baseline density is directly proportional to the alcohol content. I'm sure the same reasoning applies to the refractive index method, but I've never used that for beer-brewing (I've only used it for my real research).  See Cutaia, A.J., Reid, A-J. and R.A. Speers,  2009. Examination of the relationships between Original, Real and Apparent Extracts, and Alcohol in Pilot Plant and Commercially Produced Beers. J. Inst. Brew. 115:318-327.

Wow, this is even more detailed than I expected. But something you didn't mention: there are also other organic substances in a liqueur, which give aroma and sometimes color, and whose effect we can't correct for. Do you think that they will generally have a higher effect on density, or on refraction, or is it impossible to tell?

@rumtscho: For clear liquors such as amaretto, Kirschwasser, absinth, pernot and the like we have water, ethanol and sugar as main constitutents (each usually > 10%) - other constituents have much lower concentrations so at the end of the day they don't introduce that much error. The flavor-carrying compounds like the extracted stuff from the herbs or the benzaldehyde are usually in the ppm to maximum 1 permil range. Things will be different e.g. with advocaat but I don't think refractive index measurements will be much fun with that, anyways...

@cbeleites so even when the additional compounds are enough to change the color of a liqueur (such as in absinth or grand marnier), you think they are not enough to change the refractive index? I am not a chemist, so I don't know if the change of one optical quality (color) is indicative of the change of another optical quality (refractive index), or if the color changes at much lower concentration which are not enough to have a practically noticeable effect on the refractive index.

@rumtscho: Of course it always depends. But we're  talking about kitchen chemistry, not about high-precision measurements. There are certain things that can disturb the measurements: turbidity is a bad sign. The advocaat also has a high protein content which makes it very different from other kinds of liqueur. But there are very colorful compounds that don't disturb. Consider that the alcohol content of both white and dark red wines can be measured without any problem with the same instrument. Caramel color (and its "natural" counterpart) should also be at most in the permil range, I guess.

@cbeleites thank you. I was aware that the content of the pigments, essential oils and waxes is in a very low range, I didn't know that this is low enough to not matter. As for white and red wine, I have only seen density meters used in wine making.

The effects aren't linear - so the suggestions aren't very practical.   Among other confounders - the densities and IRs vary with temperature.  Sucrose & other sugars are almost insoluble in ethanol.  Volume of mixtures does not conserve volume - so the negative excess volume of a water+ethanol

Consider density vs. a Pousse-café.  Refractive index might do less galloping overall - but whoa! for things like Benedictine and absinthe loaded with refractive organic essences.  You need calibration tables either way.

Brewers are interested in both the sugar content and alcohol content of their products. The way they get both is by distilling a known volume of beer and measuring the density of the distillate (made up to the original volume of the beer). As it is assumed that all the alcohol has been recovered, the alcohol content of the distillate is the alcohol content of the beer.
The same is done for the sugars with the residue. It is made up to original volume and its density measured using a digital density meter. Correction for the specific gravity of the beer and distillate give the percent sugar in the original beer.
When it comes to liquers, the concern is usually more with the alcohol that what TTB calls the 'obscuration' (sugars). If the obscuration is less than 100 mg/L (not really sure I remember that right) then the same basic method as used by brewers must be used i.e. distillation for the EtOH determination and I suppose that the same method could be used to estimate the obscuration itself but I do not recall the accepted TTB procedure for this.
In any case the only time refractometers are used in the brewing industry is where they have been calibrated against a digital density meter.
To illustrate the point about other substances interfering: a fellow once asked me to proof a limoncello. The distillate was murky with suspended droplets of lemon oil. I had one conversation with the TTB lab about what to do about this but the guy dropped the project before I ever got an answer.


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Substance whose solubility increases with decrease in temprature
Is there any chemical substance, whose solubility increases with decrease in temperature ?

I think the keyword you are looking for is lower critical solution temperature (LCST). A very prominent example of a substance having such a LCST is Poly(N-isopropylacrylamide) (maybe better known by one of its abbreviations: PNIPA, PNIPAAm, NIPA, PNIPAA or PNIPAm).

Gases are usually more soluble in water at lower temperatures.

If you have some patience, you can browse this table on Wikipedia and find substances where solubility decreases with temperature. Examples are cadmium selenate, cadmium sulphate, strontium acetate, lithium carbonate, etc. It's interesting that a good fraction of solids with lower solubility in warm water are sulphates/selenates.

Maybe you could improve a little bit the question: the body of the question in my opinion could be the question itself and you could write what you are looking for: list of substance, examples, explanations, plots (:-)) in the body instead of it!

Vague question. Solubility in what solvent?

Gases are usually more soluble at lower temperature, in fact there are some solid substances the show this behavior too.
 The classical example is $Cerium \space III sulphate$ $Ce_2(SO_4)_3$. This behavior is due to the fact that $Ce_2(SO_4)_3$ has an exothermic Heat of Solution. So for the principle of Le Chatelier increasing the temperature cause a decrease of the solubility of the salt that tend to compensate the increment of temperature absorbing energy through the formation of a precipitate, that in this case is an endothermic process.

These substance are not so common, however if you deal with sulphate, selenate or some strange metals you should take a look in literature checking if they have this behavior. As suggested Nicolau Saker Neto  I've check the wikipedia list, I've made this plot from there, in the Y axis is shown how much the substance is soluble in water as grams dissolved per 100 grams of water:

The solubility of a solid depends on the temperature in the following way: When the net dissolving heat is negative (i.e. when heat is given off by the dissolving process) the solubility is higher at lower temperatures because the energy is more readily absorbed by the dissolving medium.
This works analogously the other way around (and is usually more the case, as you seem aware of).
For gases the relationship is usually the other way around (increase in temperature results in a decrease of solubility).

The reason for this gas solubility relationship with temperature is very similar to the reason that vapor pressure increases with temperature. Increased temperature causes an increase in kinetic energy. The higher kinetic energy causes more motion in molecules which break intermolecular bonds and escape from solution.

Further explanations (and the source of the quote above) can be found here.

http://adsabs.harvard.edu/abs/1992JCrGr.116..427G
http://patents.justia.com/patent/4481069
Hydrothermal synthesis of berilinite is also retrograde (negative temperature coefficient) solubility.

Would you be willing to summarize the contents of those pages you link? We want to help combat link rot.

Hydrothermal growth of quartz (positive temp coeff) has source and heat input at bottom.  Plumes rise and cool, growing seed at top.  Hydrothermal growth of berlinite  (positive temp coeff) has the seed at the hot bottom, source at the cool top.

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What's the difference between graphite, nanotubes, and fullerenes?
I understand that all of these are made up of carbon atoms but I don't understand what differentiates them. Since they are carbon they all must have covalent bonds, right?

Welcome to Chemistry.SE! Have you done any online search yourself to collect some information on the topics? For the "buckyball", you might want to search for "fullerene".

Out of the research I've done I was unable to find anything that relates to their bond types, are the differences only structural?

Please don't take this too literal, but I'd say you can think about it as similar to the relation between ice, liquid water and water steam: All 3 have the same kind of link between the water molecules (hydrogen bonds) but the properties are much different due to the structural differences within the respective hydrogen bond networks (well, for water steam you don't have much of a network).

You might want to add diamond to the list.

All three carbon modifications indeed contain covalently bond carbon atoms. The $sp^2$-hybridized C atoms are connected by $\sigma$ and delocalized $\pi$ bonds in all cases. It is the geometric arrangement of the atoms that differentiates the modifications: Stacked planes in graphite, fullerenes: ball-shaped molecules (hence the name buckyball), and for carbon nanotubes, the name is quite self-explanatory.
A single layer of the graphite structure:

A $\ce{C60}$ fullerene molecule:

And finally some carbon nanotubes:

Structure and bonding is interconnected, there is not such thing as "only structural differences".
Carbon has many allotropes, including graphite and diamond. You can say they just differ in structure, yet, this difference leads to a metallic sheet in the case of graphite, and an insulating 3D structure in the case of diamond.
You can think about carbon nanotubes, fullerenes and graphene as somewhat modified  versions  od graphite with the same delocalized / metallic sp2 carbon sheets. However that the fact that you curve / roll the carbon sheet and you reconnect certain carbons e.g. to form a tube, you do actually modify the delocalized / metallic nature of the carbon sheet. Specifically, e.g. carbon nanotubes can be semi-conducting or metallic, depending on how you roll up the carbon sheet to a tube.

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How to calculate ring strain energy?

Series A and B below can be viewed in two different ways to annulate an ethano-bridge onto four-, five-, and six-membered rings. Using the $\Delta_\mathrm{f}H^\circ$ values given (in kcal/mol), how can one calculate the strain energy for each compound?

Could anyone also explain whether the trend seen in each series is consistent with expectations based on additivity of ring strain?

Typically, ring strain energies (RSE) are calculated as the difference of the heat of formation of cyclic starting materials and less cyclic products in homodesmic reactions.
For cyclopropane, the RSE would be calculated using the following reaction:
$$\ce{(CH2)3 + 3 CH3-CH3 -> 3 CH3-CH2-CH3}$$

It's called hypohomodesmotic or HD1, and is a subset of the isodesmic reaction. It describes the same amount of carbons in the their various states of hybridization on reactant and product side, and same amount of carbon atom with 0,1,2,3 hydrogens attached. However, it is not a Homodesmotic reaction (HD2) since the carbon-carbon bond hybridization of reactant and product is not balanced. spa-sp3, sp2-sp3 etc.

I ran out of space in the comment...to the comment (user9596)...to Klaus' answer.
Klaus is right. The example with cyclopropane is certainly homodesmotic (HD2). In my opinion, this should now be called RC4 based on a new scheme in a 2009 paper by Wheeler Houk, Schleyer and Allen.  You may reference http://www.ncbi.nlm.nih.gov/pmc/articles/PMC2711007/
Criteria for HD1: (a) equal numbers of carbon atoms in their various states of hybridization in reactants and products, and (b) a matching of carbon-hydrogen bonds in terms of the number of hydrogen atoms joined to individual carbon atoms in reactants and products.
OK, so the criteria for HD1 are satisfied.
Criteria for HD2: (a) equal numbers of each type of carbon-carbon bond [Csp3–Csp3, Csp2–Csp3, Csp2–Csp2, Csp2=Csp2 etc.] in reactants and products, and (b) equal numbers of each type of carbon atom (sp3, sp2, sp) with zero, one, two, and three hydrogens attached in reactants and products.
It turns out that the criteria for HD2 are satisfied as well.
As a last bit, here are the (very similar) criteria for RC4:
(a) equal numbers of each type of carbon-carbon bond [Csp3–Csp3, Csp3–Csp2, Csp3–Csp, Csp2–Csp2, Csp2–Csp, Csp–Csp, Csp2=Csp2, Csp2=Csp, Csp=Csp, Csp≡Csp] in reactants and products, and
(b) equal numbers of each type of carbon atom (sp3, sp2, sp) with zero, one, two, and three hydrogens attached in reactants and products.

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Extraction of starch from soup
I have a solution containing many different organic compounds (originally from leek and potato soup) and I would like to extract the starch from it. However, starch is insoluble in water. How could I go about extracting the starch?

You can't simply filter it and take the feed because all the vegetables fiber remain in the filter with your starch. If you want to make a starch with a decent level of purity you should start directly with the potatoes: grating the potatoes, putting them in hot water (preferably salted: this make more effective the separation from proteins) for a bit of time. Then take the filtered solution (not the residual) make the water evaporate now what you get is the starch with the potato protein, amino acids and minerals (what is called fruit water). There is this video if you want to see.

If you want to have a more pure starch also you can try to precipitate the proteins warming the solution and adding an acid. If you are at home acetic acid could be enough.
Eventually you can cool down the solution this make starch precipitate so you can remove the water with a filter and take the precipitate.
If you really want to isolate starch there are more sophisticated methods see here and here for references.

Doesn't dissolve in water?  Filter it.  Celite is your friend.

If the soup was cooked the starch is gelatinized - hydrated goo.  Centrifuge it.

thanks! I didn't know that there was a filter-option, but i'll use it!

For me this is not the way to do it, if you filter  a soup with celite what you have are for the most part the vegetables fibers with some starch in it...

@LievenB I think this is simply a soup without water...


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Why don't B, C, Si, have ionic forms?
Also, why can some elements such as $\ce{Fe}$ have different ionic forms such as $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$? What determines each form?

Although this is not an explanation, do you know about the octet rule, ionisation energies, etc? We do btw see carbocations, but they aren't very stable.

Boron: borates, borides.  $\ce{MgB2}$ is the highest temperature BCS superconductor by a large margin.
Carbon: carbonates, carbides, hexacoordinate carbon in complexes, organolithiums, Grignard reagents, carbenium ions ($\ce{Ph3C+}$).  $\ce{Al4C3}$ hydrolyzes to methane.
Silicon: silicates, aluminosilicates, silicides.  Cook up silica gel in ethylene glycol plus potassium hydroxide (NOT in glass) to get pentacoordinate siliconate,
http://patentimages.storage.googleapis.com/WO1994020507A1/imgf000014_0001.png
You asked the second half in the Physics Stack.


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How do I calculate the order of a reaction given slope?
I'm an AP Chemistry student, and we're doing the Decomposition of $\ce{H2O2}$ lab, in which we decompose $\ce{H2O2}$ with $\ce{KI}$ as a catalyst. I've gotten this data so far:
+-------+-------------+----------------+-----------+--------------+-------------+--------------+--------------------+---------------------+-------------------+
| Trial | Volume H2O2 | Initial [H2O2] | Volume KI | Initial [KI] | Temperature | Rate (kPa/s) | Initial rate (M/s) | [H2O2] after mixing | [KI] after mixing |
+-------+-------------+----------------+-----------+--------------+-------------+--------------+--------------------+---------------------+-------------------+
|     1 |           4 | 0.882          |         1 | 0.5          | 21.2        | 0.4121       | 0.000168           | 0.706               | 0.10              |
|     2 |           4 | 0.882          |         1 | 0.25         | 21.1        | 0.2011       | 0.000082           | 0.706               | 0.05              |
|     3 |           4 | 0.441          |         1 | 0.5          | 21.3        | 0.1844       | 0.000075           | 0.353               | 0.10              |
|     4 |           4 | 0.882          |         1 | 0.5          | 31.2        | 0.8211       | 0.000334           | 0.706               | 0.10ds            |
+-------+-------------+----------------+-----------+--------------+-------------+--------------+--------------------+---------------------+-------------------+

I think that this is a first order reaction with regards to both, since when you halve the amount of each reactant, the rate of the reaction also halves.
Where I am running into trouble is calculating the rate constant (k) of the reaction. I don't really know where to start on this. I thought of dividing the initial rate by the $\ce{H2O2}$, but that wouldn't account for the amount of $\ce{KI}$. How would I go about calculating the rate constant of this reaction? I don't need an answer here, but pointing me in the right direction would be great.
Thanks!

As mentioned, you don't want a direct answer so I am just going to ask you some question to think about, that hopefully help in getting you to the answer. Please reply in the comments whenever something is unclear to you!
You're expecting first order in both reactants, so if you would write down the reaction rate equation, what does that look like?
With the equation you should have one unknown only: $k$ (or two that make up $k$: $k_0$ and one other, which?, if you are interested in the temperature effect as well).
Now calculate $k$ for each reaction separately. Are they the same? Do you expect them to be? If they're not, should you average them?

So the rate equation would be k[KI][H2O2], right?

@JohnPande Yes, that is correct. r=k[KI][H2O2] to be precise

So to find the rate of each reaction, would I use the equation k=r/([KI][H2O2])?

Doing that, I got the rates of
0.001860,
0.001739,
0.001533,
0.003761

Rate constants you mean, but yes that seems right. Now, you also report the temperatures and should work those in to find $k_0$ (also often denoted as $A$) and another parameter. This wiki will be helpful: http://en.wikipedia.org/wiki/Reaction_rate


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¹¹B NMR in a borosilicate tube
I'm just wondering, if one were to run a $\ce{^{11}B}$ NMR on a boron containing sample in a borosilicate tube, how could we separate the spectra of the tube and the sample?
Apart from using a different tube - Perhaps through a pulse sequence like BIRD for $\ce{^1H/^13C}$?
Or decoupling - which I've only just started to learn about and don't really understand?

The borosilicate tube gives a very broad signal, the signals from the boron compounds you're measuring should be much sharper. If your samples are concentrated enough and have reasonably sharp lines, that should be sufficient to distinguish them from the broad background signal from the NMR tube.
You could of course also measure an identical NMR tube with the same amount of solvent, but without your boron sample. Comparing this with your real sample should make it easy to identify which signals are background from the tube.
And of course you could use an NMR tube made from a different material.

You'd run the nmr in a quartz tube, or a tube made from some other non-boron containing material.

Most NMR probes have boron-containing inserts that separate the NMR tube from the coil surface, and these will also contribute to 11B background. If investing in quartz tubes, it is important to check the probe design, and select the right one for the job.


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Niobium (V) Fluoride becoming sticky
Today I was making a pastille of Niobium Pentafluoride powder. The procedure includes a 6 ton squeezer, precision wipes, silica gel, filter papers and acetone.
The first suspicious part was that on product details, $NbF_5$ was suppossed to be $5\mu m$ powder. Instead it looked like having hair(forgive my bad terminology but I am not a chemist, but a physicist and  I don't know how to describe it).
The procedure is the following.

Acetone and wipe papers$\rightarrow$clean squeezers parts(made of stainless steel)
Powder gets into the squeezer(using a funnel made of the filter paper) and gets pressed
The powder becomes a pastille.
It is stored in a plastic bottle, with Silica gel.

The filter paper is also used to fill the bottle with silica gel(as a funnel) as well as a way to store the pastille while filling the bottle with silica gel.
I have made several pastilles with the exact same recipe(i.e. $B,\;ZnF_2,\;LiF$ etc) and everything went fine. This powder however became very sticky. I have also noticed that while it's stored in the silica gel box, there are some drops inside the bottle.
So I have some questions... First of all why did it became sticky and secondly why the powder wasn't actually a powder?Is it typical for $NbF_5$?

From the properties block on the Wikipedia page for $\ce{NbF4}$:
Niobium (V) fluoride is hygroscopic - it absorbs water from the atmosphere. This hygroscopicity is your main problem. When the powder absorbs water, it probably is becoming sticky.
The added water is probably also providing the extra mobility the ions need to crystallize from the powder. Niobium (V) fluoride forms monoclinic crystals, which on a very small scale could look like hair or needles.

$\ce{NbF5}$ also seems to have a tendency to oligomerize/polymerize, as suggested by the tetrameric structure displayed on the Wiki. The only article I found in a quick search that directly mentions $\ce{NbF5}$ polymerization is this (in the abstract).

I knew it was hygroscopic, but I didn't know that it can become sticky! Is this typical for hygroscopic powders? I addition, the provider should have have it enclosed on a special container to prevent moisture, isn't that right? Can I request my money back?

Well, potassium hydroxide is so hygroscopic it will keep on absorbing water until it is a solution. It may be worth contacting the supplier, but I'm not sure how far it will go.


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Why can some ions have different valence?
Like $\ce{Fe}$, it can be $\ce{Fe^{2+}}$ or $\ce{Fe^{3+}}$, same with $\ce{Ni}$ and $\ce{Co}$, what determines whether the ionic compound will be $\ce{Fe^{2+}}$ or $\ce{Fe^{3+}}$?

An empty, half-filled or fully-filled electron shell is especially stable.  $\ce{Fe}$ metal is [Ar] 3d6 4s2.  $\ce{Fe^{2+}}$ emptiess the 4s-shell.  $\ce{Fe^{3+}}$ give that plus a half-filled 3d-shell.  Vanadium II, III, IV, V.  Now try chloride, hypochlorite, chlorite, chlorate, and perchlorate.
The most stable oxidation state depends on the medium and added ligands, if any.  Cu(I) is a Lewis soft acid. It is stabilized by soft bases like pyridine, $\ce{MeCN}$, olefins, $\ce{I^{-}}$.  $\ce{Cu^{2+}}$ is a hard Lewis acid.  It is stabilized by hard bases like ammonia, water, $\ce{Cl^{-}}$.  Dissolve $\ce{CuCl2}$ in pyridine or $\ce{MeCN}$ and you have a powerful oxidizing agent. Dissolve a $\ce{Cu^{1+}}$ salt in water and it disproportionates into $\ce{Cu}$ metal and $\ce{Cu^{2+}}$.  Put $\ce{Cu}$ metal and $\ce{Cu^{2+}}$ into $\ce{MeCN}$ and you get $\ce{Cu^{1+}}$.


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Why is absorbance spectrum continuous and not quantized?
1) In text books, the absorption event is usually described by a figure shown below1: light excites electrons to higher states in atom or molecule. The difference between the ground state and the excited one matches the energy of the absorbed light. But electron states in atoms and molecules should be quantized, so why do we observe absorbance spectra that are smooth and continuous but not spiked and quantized?

2) Where does the energy go when the excited electron returns to its normal state? In fluorescence, the electrons return to lower states and excite light at higher wavelength than the absorbed one. But if I had a molecule that did not exhibit fluorescence, just plain absorption, does the molecule emit light at the same wavelength it absorbed or does the energy turn to heat? I mean, the electron cannot be excited forever.
3) What is the physical phenomenon behind absorption coefficient? I have been thinking that it has something to do with the lifetime of the excited state – shorter lifetimes increase the probability of a photon to be absorbed, resulting in larger coefficients. Am I completely wrong?
1Image taken from http://www.photobiology.info/Visser-Rolinski.html

Actually, these are three questions ;-)

The picture is rather misleading as discreet transitions are shown as lines but smooth spectra underneath. Often spectra do show very broad but vibrational lines. The look of the spectrum has a lot to do with inhomogeneous broadening of lines.

The image shows the absorption and emission spektra of molecules. For isolated atoms, you would indeed observe sharp lines. Think in the Fraunhofer lines.
UV/VIS absorption of molecules
In the case of molecules, absorption occurs from the vibrationally relaxed electronic ground state ($S_0$) to various vibration levels of the electronically excited $S_1$ state. This results in a broad absorption band instead of a sharp line.
The manifold of $S_1$ with different vibrational levels does not live forever. Within the shortest time, relaxation to the vibrational ground state of $S_1$ occurs. The electron distribution of the excited state is typically significantly different from the $S_0$ state. This results in a completely changed reactivity.
Radiationless deactivation
If you don't observe emission from an $S_1$ or $T_1$ state, the excited state is deactivated by internal conversion, i.e. by vibrational exchange with the bulk (solvent).

Thank you for the answer! Yes, now I remember that in AAS, for example, the spectrum is spiky and not continuous, and that must be due to atoms being in atomic state and not part of a molecule. What a relief! But for molecules... when you say "vibration levels", do you mean the electrons vibrate or are you referring to the same inter-atomic vibration we see in infrared spectroscopy? So, would I see anything new if I had an aromatic molecule and measured the IR spectrum and at the same time bombarded it with UV light?

It's stretch and bend vibrations of the bonds in the molecule. In principle, you would see something new, but I've never done that.

Just throwing this here to make a technical point. If our tools were sensitive enough, a continuous spectrum would look like a group of distinct lines if we could zoom in really far.  Energy is quantized whether we look at an atom or a molecule.

@LordStryker Of course, there is a minimum linewidth for any transition, which comes form the uncertainty principle. So even the most precise measurements theoretically possible will show "bands", even if they are extremely narrow.

@Nicolau Saker Neto. A small point, but the time-energy uncertainty is not really an 'uncertainty principle' in the same sense as position  - momentum is because time is not an operator in QM. It is due to the fact that we $assume$ that an excited state decays exponentially and by Fourier transform obtain a line width. In solution inhomogeneous broadening is one contributor to broadening as is energy level shifts due to solvent -solute interactions.

@LordStryker: Only for atoms/molecules at absolute zero; at any nonzero temperature, the thermal motion of the atoms/molecules will cause Doppler broadening of the lines, smearing them out into bands.

The pictures you show are misleading as they refer only to molecules in the condensed phase. Isolated molecules in the vapour phase do have discrete  (quantised) electronic, vibrational and rotational levels that can easily be measured, and have been so for at least 50 years.
In the condensed phase the spectral lines are inhomogeneously broadened, this means that what you observe is a mixture of many superimposed transitions making a broad structureless feature. This occurs because molecules suffer random collisions/interactions with solvent molecules that shifts energy levels up and down just a little. There are so many different ways this happens that the individual (narrow) spectral line become overlapped and the measured spectrum becomes broad.
We know this because if you freeze a solution close to zero K and illuminate it with a very narrow wavelength laser it is possible to 'burn away' some of the molecules and leave a hole in the spectrum. This is seen as a narrow dip in the spectrum. The hole itself has a width due to the intrinsic lifetime of (say) the excited vibrational transition.
To answer your other points, fluorescence is emitted and either escapes completely into space or is absorbed by some other molecule. The molecule after fluorescing is often left with some vibrational energy in its ground state, this equilibrates with its surroundings as heat. If a triplet is formed, then phosphorescence can be emitted and so on as for fluorescence. If internal conversion (S1 to S0 radiationless transition), all the electronic energy appears as vibrational energy ( a very 'hot' ground state) and then equilibrates with surroundings.

Thank you all for participating. So, the smoothness arises from mainly two factors:

Molecules have a wide range of energy levels due to bond stretching / vibration etc. that partially absorb the photon energy. Atomic atoms do not have such energy levels so their AAS spectra are spiky.
The measurement technology is not precise enough to visualize individual absorptions. To my understanding, the bandwidths of filters / monochromators are too wide to cover only the individual energy levels.

However, I am still uncomfortable with the scheme I pictured below. It is easy to imagine that the transitions from various $S_1$ levels to the lower $S_1$ levels
are released to bond vibration etc. because the energy is very small. But if we look at the transition from $S_1$ to $S_0$, the energy is as large as that of the absorbed photon. So, why isn't the energy released as a photon?

To answer your last question molecules absorb into electronic levels and each of these has vibrational and rotational energy levels. There are a vast number of these,(millions) because vibrations of different frequencies but same symmetry species can combine together.  However, it turns out that only a few have a large transition probability for absorption so we often see spectra with a limited number of vibrational lines. In the gas phase we do have the laser technology to quite easily measure the width of any spectral line and this has been done for very many molecules. Look at the benzene vapour spectra for example.
In solution collisions with solvent move energy levels about a bit and this broadens the lines so that they overlap and can never be resolved no matter what the spectral resolution of the laser. The spectrum is inhomogeneously broadened. All that can be done at very low temperature, (close to zero K) is to burn a 'hole' in the spectrum (for example by destroying a group of molecules with that energy) and this hole will last for a short while until inter-molecular interactions 'fill' the hole in again. Obviously the warmer the sample the quicker this happens.
As to your diagram, deactivation the excited state occurs as a competition between photon emission and radiationless precesses. The excited singlet state decays by intersystem crossing to triplet states under the influence of spin-orbit coupling (coupling of electron spin with its orbital angular momentum) and also to the ground state by internal conversion which is primarily cause by anharmonicity. As a vibrationally hot ground state is produced this energy is rapidly lost to the solvent and takes only a few picoseconds to do so and is easily observable in femtosecond laser pump-probe experiments.


text:
What are the possible products in allylic bromination of 3-methylcyclopentene?

Bromination of an alkene by N-bromosuccinimide (NBS) in the presence of light or peroxide is a radical reaction and produces an allylic bromide. For the following bromination of 3-methylcyclopentene, which of the following allylic bromides would be products of the reaction?

In some other reactions such as elimination, there is a tendency to form the most substituted double bond, since it is the most stable. If that were the case, then the products should be A, E, F, and G, since these are all trisubstituted alkenes (the others are disubstituted). Is that a valid consideration here, or is there a different overriding rule?

I think it should be A,C,F,D ,though I am not sure. I'll do some review and get back.

Thank you, I agree with A, C, and F
wondering why it cant be G or E :/ confusing.

Alkene substitution is only really important in a couple of cases: firstly when the reaction is entirely thermodynamically controlled (e.g. aldol reactions), and secondly if the transition state in the rate-determining step contains strong double bond character (i.e. more substituted double bond = lower-energy transition state).
Neither of that is the case here. The double bond is constrained by its initial position, and cannot jump around all that much. In theory, there are $ 2 \times 2 = 4$ possible products (A, C, D, and F) because:

the initial abstraction of a hydrogen atom can occur at two different positions, and
the allylic radical thus formed can pick up a bromine atom at two different positions.

The mechanistic possibilities are drawn out fully below:

In practice, there would likely be one product that is preferred over the others. Note the use of resonance arrows, $\ce{<->}$: these are not flipping about between the two different resonance structures. Rather, the reality is that the allylic radical is somewhere in-between the two extremes. We only draw out the two extremes here because it's a convenient way of looking at its reactivity (which is sort of a mixture of both extremes).

Your comment shows that you are on the right track.
You have chosen A and C because they are formed when the initial allylic radical instantaneously reacts with $\ce{Br2}$ in the propagation step of the reaction, isn't it? (Hint: Just say YES!)
Now give the allylic radicals corresponding to A and C a bit of time, just enough to flip the dot.
How do these other radicals look like? There you have the other two products.
For an extra point: Which isomer of methylcyclopentene could lead to G?

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text:
Why aren't atomic radii calculated using Schrödinger's equation?
The atomic radii are estimated using a variety of methods: most of these involve dividing their bond length by 2. But that is a very crude way of measuring atomic radii. I mean, atoms overlap each other when bonded, surely the actual atomic radius is much more than the one obtained by this method.
Why don't they simply use the Schrodinger's wave equation to calculate atomic radii. If it can predict atomic orbitals to such accuracy, surely it can predict something as simple as atomic radius?

What is your definition of atomic radii?

I would say that the radius equals the farthest orbital in the boundary surface diagrams of the atomic orbitals.

How do you obtain the boundary surface? Take the ground state wavefunction of the hydrogen atom for instance, $e^{-r}$, where is the boundary?

I like the question, unfortunately there is nothing simple for any atom with more than one electron. I apologize if you felt offended by the "poem".

The problem is, op does not give a clear mathematical definition of atomic radii. The question is not necessarily related to the difficulty of getting the exact solution of many-electron system.

@KlausWarzecha: There was nothing "offending" about the poem. It was quite amusing. But it wasn't relevant.

Erwin with his psi can do
Calculations quite a few.
But one thing has not been seen:
Just what does psi really mean?

(Erich Hückel, english translation by Felix Bloch)
It is not easy to perform the calculation that you called simple. The Hamilton operator $\hat{H}$ for a multi-body problem of $K$ cores and $N$ electrons
\[
\hat{H} = - \frac{1}{2}\sum_a^K\frac{1}{M_a}\Delta_a - \frac{1}{2}\sum_i^N\frac{1}{M_i}\Delta_i - \sum_i^N\sum_a^K \frac{Z_a}{r_{ai}} + \sum_{i < j}^N\frac{1}{r_{ij}}+ \sum_{a < b}^K\frac{Z_aZ_b}{R_{ab}}
\]
is still a beast for $K=1$ with all the core-electron and electron-electron interactions.
So, while the question is legitimate, the calculation simply cannot be done without a lot of approximations, such as the Hartree-Fock method.
By the way, for the determination of the atomic radii of solid metals, there's x-ray diffraction, which I wouldn't call crude.

It would be helpful if you would include some explanation.

Saw this poem in my g09 output the other day. :)

I saw this in a "related questions" link.
The answer is that yes, you can use quantum mechanics to define atomic radii (e.g., van der Waals radii). Multiple people have done so with different schemes. I believe the most recent is:
"Consistent van der Waals Radii for the Whole Main Group" J. Phys. Chem. A 2009, 113, 5806–5812
Similar efforts have been made for covalent bonding radii using quantum mechanics.


text:
Conversion of s-trans butadiene to s-cis butadiene

No nodes should be introduced in the rotational interconversion between the s-trans and s-cis forms of 1,3-butadiene.

What is the explanation for that, if one of the $\pi$-systems should rotate $180^\circ$? It seems to be that two positive and negative lobes turn upside down.

During the interconversion of s-cis and s-trans 1,3-butadiene, the molecule passes through a geometry where the two ethylenic arms are rotated 90 degrees one from the other.  In this conformation the two ethylenic arms are orthogonal to each other and there is no overlap between them.  In an MO sense they can be treated as two separate ethylenes.  Rotation beyond this conformation allows overlap to reoccur and the phase of the HOMO will be correctly reestablished with only one node occurring between C2 and C3.

Thanks Ron for the explanation. Since s-trans is lower in energy and s-cis higher, I thought of an excited state to LUMO with one node between C2 and C3 in the transition state (90°), and then upon full rotation to s-cis, the node automatically disappears and hence fully symmetrical. How do you think is that realistic?

In the 90 degree transition state, the two ethylenic branches are orthogonal to each other, there is no overlap between them at that point.  Further rotation restores overlap and the 4 orbital HOMO is restored.  The butadiene LUMO is not involved in the process.


text:
Equilibrium calculation gives negative molarity

There is a reaction: $$\ce {ClF3 <=> ClF + F2}$$ with $\ce {K_\mathrm{c} = 2.48 x 10^{-3}}$ Initially, $\ce {[ClF3] = 0.25 M} $, $\ce {[ClF] = 0.031 M} $, and $\ce {[F2] = 0.02 M} $. $\ce {F2}$ is added until the concentration of $\ce {F2} $ is $\ce {0.1 M} $.
  What is the concentration of these substances after equilibrium?

My attempt:
$\ce {ClF3 <=> ClF + F2} $
Initial concentration: $\ce {[ClF3] = 0.25 M} $, $\ce {[ClF] = 0.031 M} $, $\ce {[F2] = 0.1 M} $
Concentration that react: $\ce {[ClF3] = +x M} $, $\ce {[ClF] = -x M} $, $\ce {[F2] = -x M}$
Concentration after equilibrium: $\ce {[ClF3] = 0.25+x M} $, $\ce {[ClF] = 0.031-x M} $, $\ce {[F2] = 0.1-x M} $
So, $\ce {Kc = [ClF].[F2].[ClF3]^{-1}} $
I applied the concentration after equilibrium in $\ce {Kc} $, so I got:
$\ce {2.48.10^{-3} = [0.1-x].[0.031-x].[0.25+x]^{-1}} $
After calculate the equation above, I got $\ce {x = 0.1111 M}$.
Then my question is: if $\ce {x = 0.1111 M} $, so $\ce {[ClF] = -0.0801 M} $ and $\ce {[F2] = -0.0111 M} $. The concentration is negative. Is it possible? Are my steps incorrect or the answer is correct?

The equation $2.48\times 10^{-3} = \frac{(0.1-x)(0.031-x)}{0.25+x}$ is correct, but it simplifies to a second degree polynomial, so it has two solutions, not one. You just picked the incorrect root (negative concentrations have no physical meaning). Can you find the right answer now?


text:
Software for predicting chemical reactions
I wondered if it was possible to predict the product(s) of a reaction (or even the equilibrium) based on the reactants and the temperature/pressure.
I found an answer to a similar question here:

No, there is no such thing as chemical reaction are not that easily
  predictable. You can get result right in some cases, but it is
  impossible to count all factors.

What (significant) factors also influence a reaction? Is it the arrangement or state of the chemicals? Is it too related to other physical properties to ignore them?
Please name the most important ones.

Welcome to Chemistry.SE! These are a actually a lot of rather general questions for a first post. Would you mind to narrow the scope and ask something more specific first?

@Klaus_Warzecha Thank you for your feedback. But I don't know how to make this question more specific.

For predicting chemical reaction you should use classical mechanics to take in account the simple interactions between the molecules  and quantum mechanical for all the aspects concerning the molecule itself and complex interactions.
Knowing the principal parameter such as temperature and pressure that could describe the system with the classical mechanics (kinetic energy etc. etc.) is not enough for predicting the reaction with ab initio methods.
Adding the quantum mechanical contribute is mandatory and quite hard to do. A amateur chemist do not have neither the skills nor the computation power require to make such calculations. For the moment you can't do that with any software but the last Nobel was won by three theoretical chemists who are dealing with this topic. Have a look here for more information.
In your case where the reactions are well known you can use a chemical reactions database like the NIST one.
This is a simple query:

And this is the...

So it is not possible by only considering the valency and  electronegativity? OK, that would've been too easy. ;)

@felixd I'm afraid not... :-)

This article came recently. It shows how over dependence on computational results can be disasterous!
http://cen.acs.org/articles/93/web/2015/03/Computational-Models-Multistep-Reaction-Called.html


text:
Why do solutions become saturated?
Why can a solvent dissolve only a particular amount of solute?
If we add more solute to the solution, the number of solute particles in contact with water increases. So rate of dissolution should increase, and subsequently, the solubility product should also increase. But why do we have a constant value of solubility product?
P.s. An explanation at the particle level would be appreciated

Because there's a limit to every solvent to dissolve a solute in itself, which is called Solubility.

I know that, and that's the reason I am asking this question. But can you please explain why there is a limit to solubility ? I want an explanation at the particle level

If the solute is liquid, there could be no limitation, like ethanol. If the solute is solid and there is no limitation of solubility, I have a guess, imagine there is a big crystal like salt, and we add one droplet of water, it should become a liquid. A crystal being a crystal means there is a barrier to keep the lattice structure unless reaching the melting point etc to exceed the barrier. The attraction between a droplet of water and the big crystal of salt is much less than the total barrier energy of lattice, therefore it can not form a total liquid.

Take a slightly more general problem of miscibility, all that in thermodynamic equilibrium. In your question you mix up kinetics and equilibrium, and additionally in a wrong way. Equilibrium is the most important part.
If you put two liquids together and shake them well, than in general, they do not mix. So you will observe a boundary between, say, diethyl ether (upper layer) and water (bottom layer).
If you separate them and analyze their composition, you will find, that the ether fraction contains little bit of water (1.5g / 100 mL) and that the water fraction contains little bit of ether (6g / 100mL). Knowing this, if you now take 100 mL of water and mix it with 5g of diethyl ether, you will observe, that no boundary is formed, and all ether has dissolved in water.
As another example, if you shake ethanol with water, they mix in any ratio.
That's observation, now the rationale. The particles of pure liquid interact with each other, in attractive way (otherwise you will have no liquid, but gas). Now you mix two pure liquids together (A + B) and look, if some of the particles of A can easily sneak into the B. If their behavior is reasonably similar, they can do it and interact with particles of B also in attractive way. The attraction should be comparable or greater, than within pure A. That is the enthalpy part.
But as we are dealing with free energy, there is also the entropy, which you have to take into account. On one hand, entropy favors mixing, as you add lot of accessible states. On the other hand, if the attraction of B to A is very strong, it starts to organize the molecules of A around itself, so they are not as free to move as they were in pure A, and limits the miscibility.
For the solubility of solids, it is technically similar, the A is solid and the interaction which hold sit together is the lattice energy.
As a side note, if you ask about solubility, miscibility and such, you should refrain from ideal approximations, as they do not predict such phenomena. Their calculation from pure substance properties is therefore terribly difficult, and physical chemistry tables are the only reasonable answer.

Why does anything dissolve at all?  The binding energy (lattice energy) of the solid must be compensated by solvation energy and entropy of the solution.  Like dissolves like (then cheats, like clathrates and micelles).  As the solution concentrates, incremental driving energies diminish.
Don't think about adding water to thiotimoline.

U didnt answer my question

I agree that his answer is misleading ^^.
Water for instance, only has so much space to hold anything inside of it; ie absorbing or becoming saturated with something, say salt.
So a few things affect Saturation such as density and tempurature(which changes density of course). If the water is hot, it is able to take in more salt.  Because the energy from the heat is expanding the space between the water; more space to take in material(salt). I'm sure you have experienced this with hot water.
So Saturation; The solution becomes saturated when the Solvant no longer has enough binding energy (lattice energy) for water to keep taking in Salt for instance.
If you add enough salt to water, the water will become saturated with salt> The salt will no longer dissolve and disappear into the water but will just float to the bottom of the solution.

"heat is expanding the space between the water; more space to take in material(salt)" This is certainly an original thought worth to consider. I would have opted for the fact that the change in entropy from crystal state to soluted one is larger in higher temperatures.

But surely it has very little to do with space between molecules since you get examples where the smaller magnesium hydroxide is less soluble than the much larger potassium sulphate. I don't think anyone has really answered the question the op asked. He's not asking why things dissolve as everyone seems to be trying to answer, the OP is asking why there is a LIMIT to the amount of compound that dissolves which is a much more interesting question

Surely it has nothing to do with space. Some solutions shrink when you dissolve more stuff in them.

In dilute solutions, small particles are formed, which can suspend in the solution. As solute amount increases the particles grow bigger. At saturation, large particles are formed and precipitate from the solution.

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text:
Why does the 3rd electron shell start filling up with scandium?
The electron configuration of calcium is 2, 8, 8, 2, where up to that point each shell, asides from the first shell counts up to 8 - why then does scandium have an electron configuration of 2, 8, 9, 2?
What causes the 3rd shell to start filling up, rather than the fourth?

Some part of the answer is contained in these posts. This essay on the matter is interesting, but it feels a bit like an a posteriori qualitative argument. This article goes a bit more deeply into the matters of the essay. I haven't properly processed all this information myself, even though I've read most of the material before.

Relevent post http://chemistry.stackexchange.com/questions/6921/why-does-ruthenium-only-have-one-electron-in-its-5s-orbital-in-the-neutral-oxida/6928#6928 (first paragraph)

This post might help you: https://chemistry.stackexchange.com/questions/73325/while-filling-electrons-we-follow-aufbau-principle-but-not-while-removing-why

The relative energies of the electronic subshells have been calculated for atoms in the vicinity of $Z=20$ (J. Chem. Educ., 1994, 71 (6), 469), and the result is surprising:

Looking at this graph, by all means the electronic configuration of scandium should in fact be $\ce{1s^2 2s^2 2p^6 3s^2 3p^6}$ $\color{blue}{\ce{3d^3}}$ in order to minimize orbital energies! The graph does not show the energy of the $\ce{4p}$ subshell, but it would lie somewhat above both curves shown. However, there is an important effect not being considered here, which is the destabilizing interelectronic repulsions. Electrons in a same subshell tend to repel each other more (intrasubshell repulsion) than electrons in different subshells (intersubshell repulsion). It turns out (and no one can really qualitatively explain why, see below) the repulsion is such that in an atom with $Z=21$, the configuration $\ce{[Ar] 4s^2 3d^1}$ is lower in energy than other potential candidates, such as $\ce{[Ar] 3d^3}$ or $\ce{[Ar] 4s^1 3d^2}$. Theoretically, the odd configurations $\ce{[Ar] 4s^2 4p^1}$ or $\ce{[Ar] 4s^1 3d^1 4p^1}$ would have lower interelectron repulsion energies than the observed ground state, but in these cases the lowered interelectron repulsions do not compensate the requirement of populating the higher-energy $\ce{4p}$ subshell. The comparison of several possible electron configurations and the joint minimization of orbital energy and electron repulsion energies does not seem to be something that can be done without resorting to heavy calculations or direct experimental validation.
In addition to some material I linked in comments, this article from Eric Scerri's blog (a chemist who focuses on aspects of periodicity, including electronic distribution) states, regarding the electronic configuration of scandium compared to calcium and to the $\ce{Sc^3+}$ ion (emphasis mine):

This amounts to saying that all three of the final electrons enter $\ce{3d}$ but two of them are repelled into an energetically less favourable orbital, the $\ce{4s}$, because the overall result is more advantageous for the atom as a whole.  But this is not something that can be predicted.  Why is it 2 electrons, rather than one or even none?  In cases like chromium and copper just one electron is pushed into the $\ce{4s}$ orbital.  In an analogous case from the second transition series, the palladium atom, the competition occurs between the $\ce{5s}$ and $\ce{4d}$ orbitals. In this case none of the electrons are pushed up into the $\ce{5s}$ orbital and the resulting configuration has an outer shell of $\ce{[Kr] 4d^10}$.
None of this can be predicted in simple terms from a rule of thumb and so it seems almost worth masking this fact by claiming that the overall configuration can be predicted, at least as far as the cases in which two electrons are pushed up into the relevant s orbital.  To those who like to present a rather triumphal image of science it is too much to admit that we cannot make these predictions. The use of the sloppy Aufbau seems to avoid this problem since it gives the correct overall configuration and hardly anybody smells a rat.

So unfortunately, it seems that even though most of the effects which combine to result in the observed electronic configurations are known, there is no qualitative way to predict where the configurations are going to mismatch with the Aufbau principle or the energy levels of the orbitals. I have read the statement that the Aufbau principle is most decidedly wrong for practically every atom with respect to the placement of the orbital energy levels, but incredibly it happens to predict the configuration of the valence shell for most atoms.

The electronic configuration model is anyway a limited model, in the sense that the different ROHF orbital energies could provide the same total energy (Table 1 in http://scitation.aip.org/content/aip/journal/jcp/125/20/10.1063/1.2393223). One may fix a form of orbital energy in ROHF then discuss on top of it, nevertheleess each explanation is based on the choice of orbital energies. Historically, many plot of orbital energy are based on a rather primitive treatment http://prola.aps.org/abstract/PR/v99/i2/p510_1, for a historical survey, see Levine's quantum chemistry 5th edi, p313-315

The official electronic configuration is extrapolated from experimental spectroscopy data, but the first-principle (without recursion from experimental data) theoretical model is not unique (ROHF, UHF, simplied HF with Thomas-Fermi-Dirac potential, KS-DFT(sometimes)), which complicated the case. If one wants to explain the experimental extrapolated electronic configuration, perhaps it is a problem to go through the mental setup of the argument which is used for the extrapolation (somehow discussed in Levine's book mentioned above).

@user26143 I appreciate your contributions in discussions, as they are always very insightful and deal with a facet of chemistry I am woefully unfamiliar with. I shall be sure to read what I can from your links.

Could you correct me if I'm wrong here? In summary, the Aufbau principle predicts the correct configurations, but not always the order of filling or the energy order. In the example of Z=21, Scandium, (not considering the interelectronic repulsions) the energy levels contrast with the predicted energy levels; 3d<4s. The electronic repulsions that arise when electrons are filled change it a bit; the first electron finds 3d to be at a lower energy, but the next two prefer 4s.

(continued)But the damage is already done; the Aufbau principle is misused to predict that 4s is always at a lower energy than 3d, which is not the case here. This is the flaw. Have I understood it correctly?

@HarryHolmes I would weaken that initial statement to something like "the Aufbau principle usually predicts the correct configurations of light neutral atoms, sometimes predicts the configurations of ions and heavy neutral atoms, and rarely predicts the correct order of subshell energy levels in any case, especially the relative energies of the inner shells". For example, consider that via the Aufbau principle, 4s is always "filled before" 3d, but for every neutral element past Sc (Z=21), the 3d subshell has a lower energy than 4s.

If you restrict yourself to using the Aufbau principle solely to explain the valence shell of neutral atoms, then the principle looks nicer because you're unintentionally (or intentionally!) throwing away most of the discrepancies. However, to my understanding, restricting the Aufbau principle to describe only the valence configuration was not an original part of the procedure when it was first proposed, nor do most people obey the restriction.

The more fundamental error is to think that energy levels of electrons are set in stone (like holes in a wall) as the number of protons and electrons changes, which is what the Aufbau diagram implicitly does.

So energy's not fixed like in the arrow diagram, but multiple factors cause it to vary. Got it, thanks.

The electron configuration for scandium is:
$$\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 3d^1}$$
The $\ce{4s}$ and $\ce{3d}$ orbitals are relatively close in energy, so there are a few anomalies in the order of filling.  A fair question might be, why isn't the scandium electron configuration $\ce{[Ar] 3d^3 4s^0}$.  There is a special stability associated with completely filled orbitals, so the $\ce{3d^1 4s^2}$ configuration is of lower energy than the alternative $\ce{3d^3 4s^0}$.

I don't find this answer very satisfying. To say something is the way it is because the total energy of the system is lower that way, while entirely correct, is too shallow; you could answer hundreds of completely different questions with the same idea. What is really interesting is why should the energy be minimized in that specific configuration. Is there a reasonable theoretical justification for why the $(n-1)d$ orbitals fill after the $ns$ orbital, without necessarily having to go into lengthy and complex computer-assisted calculations?

@NicolauSakerNeto yes, what you have described is precisely what I am after - the 'why' of this phenomena intrigues me

@ron This would better be explained in terms of electron spin pairing energy.  Your answer would be very nice to have this concept incorporated.

Perhaps I am missing something.  I have read and reread the comments several times.  I didn't say "something is the way it is because the total energy of the system is lower" and just left it like that.  The points I tried to make are: 1) the 3d and 4s orbitals are close in energy, so small factors can tip the energetic balance favoring the formation of one electronic configuration over the other (3d1 4s2 vs. 3d3 4s0 example I gave), and 2) that full shells have a special stability and lower energy (inert gas configuration), hence the 3d1 4s2 configuration is energetically favored over

the 3d3 4s0 configuration.  If questions remain, please ask them in a clear, specific way and I will try to answer them.

I hope I didn't come across as insulting, as that was certainly not my intention. I'm fine with the statement that the orbital energies for 3d and 4s are close, hence more subtle effects would need to be taken into account. But then you write about the "special stability" without going into it more, so it sounded like a bit of a handwave to justify that the $3d^14s^2$ has a lower energy than $3d^34s^0$ and hence the former is observed instead of the latter. I was hoping I could stir someone into giving a more rigorous, possibly semi-quantitative answer.

No, no offense taken.  You're right about me "glossing over" the "special stability".  I did that because what more can be said than, empirical observation shows us that the inert gas configuration (e.g. completely filled orbitals) has "special stability."

What is being looked at, is how to know or predict what an electronic configuration is going to be.
When some speak of filling up shells with electrons, one has to consider what that means. Does it mean a kind of cartoonish idea of take a particular element whose electronic configuration we know eg. Carbon atomic number 6, electronic configuration 2,4.  Add a proton(making Nitrogen), and then add an electron, where will that electron go. And it goes in the last shell making Nitrogen 2,5.
Another way that might be meant by filling up is if you take the neutral atom, remove all  or some of the electrons and start adding them in then what order would they fill up.
For example neutral scandium has 21 electrons, if you remove 3 electrons, to give 18 electrons which is Argon's configuration, so that goes from neutral scandium to Sc 3+. Then if you were to add electrons, where would the 19th, 20th and 21st electron go.  Or, just for one electron, take a neutral atom, e.g. neutral Scandium,  remove an electron, so we are looking at the 1+ cation, (maybe assuming that the electronic configuration is now that of the previous element, or maybe checking to find out what is the electronic configuration of that 1+ cation), then add the electron and consider where it's going to go.
Or for a simpler example, taking Carbon, imagine it has one less electron so has configuration 2,3 (Boron's configuration), then add an electron, and it becomes 2,4.
There's the idea that each shell is identified by a number n. The second goes further out than the first. The third goes further out than the second etc.
It'd be wrong/inaccurate to say that shells are filled inward to outwards, they aren't necessary.(Though the first and second shells do follow this).  It'd be wrong/inaccurate to say that a shell fills to completion before the next shell fills, they aren't necessarily. (Though the first and second shells follow that). And it'd be wrong to think that the third shell has a max of 8, it doesn't.
Some basic material is excellent.
https://www.middleschoolchemistry.com/lessonplans/chapter4/lesson3
"When the third energy level has 8 electrons, the next 2 electrons go into the fourth energy level."
and
"The third energy level can actually hold up to 18 electrons, so it is not really filled when it has 8 electrons in it. But when the third level contains 8 electrons, the next 2 electrons go into the fourth level. Then, believe it or not, 10 more electrons continue to fill up the rest of the third level."
And you see the electronic configurations in that format here https://ptable.com/?lang=en#Properties e.g. Zinc 2,8,18,2
A view that explains a bit further than just looking at shells, is to use a model involving "subshells" rather than shells, and to say they are filled in order of energy level.  That speaks of the Aufbau principle and order of energy levels and subshells, the madelung rule  aka n+l rule.  At 16-19 you might learn about it.
The electronic configurations for neutral atoms are shown in abbreviated form here https://sciencenotes.org/list-of-electron-configurations-of-elements/
That rule, the n+l rule, or madelung rule. Some might call it the afbau rule (though calling it afbau rule might be a bit ambiguous 'cos by that some just mean the idea of filling up in order of increasing energy level, whatever that order is).
The n+1 rule gives us the following order of subshells.  "1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, and 7p". Those are all listed here https://chem.libretexts.org/Courses/Valley_City_State_University/Chem_115/Chapter_2%3A_Atomic_Structure/2.4_Electron_Configurations
n and l are quantum numbers.. n is the principal quantum number.  Each subshell s,p,d,f respectively, have l values of 0,1,2,3 respectively. For The 4s orbital ,  n=4 and l=0, thus n+l=4+0=4. The 3d orbital has n=3 and l=2, thus n+l=3+2=5. So  the n+l rule  has 4s below 3d.  4s lower energy than 3d.  The n+l rule always comes out with that order. It's a rule that helps to find the electronic configuration of neutral atoms.
When it comes to the transition metals, scandium onwards, technically, the order is a bit different. (i.e. not the n+l rule). Here, 3d is before 4s.  (which helps to explain the electronic configuration of some of the ions but doesn't help to explain scandium's electronic configuration, that would take some further explanation).
The n+1 rule gives an easy "explanation" of scandium's electronic configuration. If we assume as the rule says that 4s is below 3d.
Also an explanation behind why 4s is below 3d, is looking at a radial probability distribution graph for the orbitals, "Why 4s before 3d" by mommachem https://www.youtube.com/watch?v=5hzv5KQ4LoM     4s is said to penetrate the nucleus more than 3d.
The n+l rule also answers a more foundational question to the one you ask. Of why is Potassium [Ar]4s1
You might have thought that the third shell has a max of 8 electrons, because some basic books wrongly state that(and are wrong in doing so).  Infact a shell n, has a max of 2n^2 electrons. So the third shell has a max of 18 electrons.
But because 4s is less than 3d , (in energy level),  for (neutral)  Potassium and Calcium, the fourth shell is next to fill after 3p.
An s subshell can take a max of 2 electrons, a p subshell can take a max of 6 electrons, and a d subshell can take a max of 10 electrons.
Notice that before the third shell is maxed out. Specifically, after it has 8 electrons, (3s and 3p), then, the fourth shell fills up a bit(4s), getting two electrons, and then the third shell continues to fill(3d).
So that explains(or at least justifies!) why/how potassium, calcium and scandium have the electronic configurations they do.
And if you look at a periodic table with blocks drawn in, you see that.  (e.g. looking at the fourth row  left to right. 4s then 3d then 3p).

As you can see, an inner shell can have electrons added to it. And so that's where many books will put aside a simple view of just the shells, and from scandium onwards will tend to put away the format of electronic configurations of K,L,M,N like 2,8,9,2 And they'll go to showing electronic configurations as e.g. 1s2 2s2 2p6 3s2 4s1 e.t.c. And that model is in a way, simpler for elements from scandium onwards, and even Potassium onwards, in the sense that it's clear which subshell is going to fill up next.
When using the madelung rule(n+l rule), for getting the electronic configuration of neutral atoms, which is what it's for, then there are 21 exceptions.   No exceptions in the first three rows. Two exceptions in the fourth row - Chromium and Copper.  (Notice that Scandium is not an exception). And the rest of the exceptions are in the fifth sixth and seventh rows.  No exceptions in the s and p blocks aka the main blocks.

Also, at AP level like 16-19, or A level in UK like 16-18,  you might also be expected to know electronic configurations for ions of transition metals.  Removing electrons and knowing the electronic configurations of various cations.
And that's where another rule comes into play. For working out electronic configuration of transition metal ions.  That is that electrons come into 4s first, and go out of 4s first.
Some will explain  that "into 4s first and out of 4s first", by saying that, assuming n+l order of energy levels with 4s below 3d, that for transition metals,  electrons go in based on order of energy level, but they go out based on  which shell is further out. So the fourth shell is further out, hence they leave 4s first. And 4s is lower in energy so they enter 4s first.
There is another explanation.. Which is that according to HF calculations, when it comes to neutral potassium and calcium, 4s is below 3d(so, like the n+l rule).  But for transition metals scandium onwards, 3d is below 4s.  So that then explains very well why electrons leave 4s first, i.e. leaves 4s before leaving 3d,  it's because 4s is higher in energy than 3d.  (So we wouldn't be using the n+l rule there).
Here is a graph based on HF calculations, that shows that , for Potassium and calcium 4s is below 3d. But for Scandium onwards, 3d is below 4s. So it's not just an explanation, it's well grounded.
https://pubs.acs.org/doi/abs/10.1021/ed071p469
Transition Metals and the Aufbau Principle
L. G. Vanquickenborne, K. Pierloot, and D. Devoghel

That then brings up another question, which is if 3d is below 4s in neutral transition metal atoms, scandium onwards, as HF calculations show, then why is  Scandium's electronic configuration [Ar]4s2 3d1. Why isn't Scandium's electronic configuration [Ar] 3d3. And the explanation given, is that if we take Sc 3+  So scandium but with all fourth row electrons removed, so 3d and 4s electrons removed, to give Sc3+. Scandium has 21 electrons. Sc3+ has 18 electrons.   And then we add  three electrons, then the 19th electron goes into 3d. And the next two electrons initially go into 3d but then jump into 4s, because it's more stable so more preferable.  This is explained nicely on Dr Eric Scerri's blog in these two articles     here and here and on RSC here.  And also explained by Jim Clark, an educator who has written some articles explaining that   here  and  here
There is an article by a chemist, Geoffrey Neuss, here that tries to differ with Dr Scerri, and while Neuss makes some points that I think are fair and could be incorporated into Dr Scerri's explanation, such as that an ion doesn't necessarily have the same order of energy levels as a neutral atom (and perhaps shouldn't be assumed to be).. and that an ion's energy levels would be a bit different to a neutral atom. e.g. even if the order is the same, the distance between energy levels could differ. And Neuss also makes the good point, that when we see an electronic configuration lists 3d before 4s, that's just because it's listed in order of principal quantum number, it's spectroscopist notation and one shouldn't conclude from that, as some students might, that it is making a point about the order of 3d and 4s.
Neuss might not have been aware that the basis for Dr Scerri saying that 3d is below 4s, is HF calculations, and very strong.  Neuss uses NIST data that looks at electron excitations and tries to conclude that neutral Scandium has 4s < 3d in energy level, like the n+1 rule has it. But that's flawed for Neuss to do that, because when a photon is fired in to cause an excitation, the energy levels are changed quite a bit.. And it's a different set of rules to work out what transition is going to happen, called "selection rules". So e.g. even looking at something simple like Hydrogen and the excitations of that, you can't get any list of orbital energy order.  So i'd definitely go with Dr Scerri's explanation.
Orbitals fill up low energy first, then to higher energy.  But since an electron could go into one orbital then jump to another, (as explained by Dr Scerri), we can't really always predict electronic configurations.
Jim's chemguide article mentions Vanadium. We can use NIST data to see the ground configurations of neutral atoms and of ions.     For Vanadium  V+ is [Ar]3d4  and V is  [Ar]3d3 4s2
So we see that to get from V+ to V, when an electron is added to V+, one of the electrons jumps out of 3d into 4s. Or to look at it the other way, when an electron is removed from V to make it V+, an electron leaves 4s but then another electron jumps out of 4s and into 3d.  The rule that electrons leave 4s first, would not predict V+ 'cos it didn't account for an electron jumping from one orbital into another!
There might still be some further advanced issues re the discussion of ordering of 3d and 4s,  mentioned in the accepted answer here  Why do 3d orbitals have lesser energy than 4s orbitals in transition metals?
At a very high level (and not relevant to the energy of orbitals subject), they'd say electrons are not particles, some say they are waves, or something with properties of each. And also, at a high level  see this ACS paper, https://pubs.acs.org/doi/pdf/10.1021/ed200673w  they say electrons don't "occupy" orbitals, they might even be spread out. And orbitals are just visual representations mapping to some Mathematics, not direct representations showing shapes of regions of space. But for the purposes of considering electronic configurations, people are fine with the concept *as a model*, of orbitals as regions and of electrons in orbitals.

This is Dr. Eric Scerri. I would be interested in hearing why Neuss' version is considered to be superior to mine, as claimed above. Eric Scerri
www.ericscerri.com


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What is the correct product of the Diels-Alder reaction between 2-ethoxybuta-1,3-diene and (Z)-acrylaldehyde-3-d?
What is the correct product of the Diels-Alder reaction between 2-ethoxybuta-1,3-diene and (Z)-acrylaldehyde-3-d?
It keeps mentioning that my regio-selectivity is incorrect. And that I should consider the charge distributions of the diene and dienophile to figure it out correctly, but I honestly have no idea.

You're also likely to get the endo product over the exo product.

These resonance structures and the polarization shown within explain the regioselectivity (1,4 isomer instead of 1,3) in the product.


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Mono-addition of HCl to unsymmetrical dienes?

I completed 3/4 of the answers, by choosing the correct reaction pathways leading to each of the 3 different products. But I don't know how to draw the reactant diene, I really thought what I initially had was correct, should I make a trans reactant or is there something else missing?

Do we agree that the first step ist the addition of a proton to the diene? This is bit different from the addition of HCl to an alkene, where you would rely on the Markovnikov rule.
Here, stabilization of the cation through resonance comes into play.

I don't know how to draw it [...]

Draw stepwise!

In principle, a proton could add to one of the positions a to d.
How do the resulting cations look like and which of them are stabilized by resonance?

Is it now a bit more clear?

I still don't quite understand, how should I alter my structure?

@HarryJohnson Now it's me who doesn't understand ;) Do you mean the R as an abbreviation for the alkyl chain and that i didn't write out the terminal CH3?

AH, I understand it now! Basically my initial answer was very close, but in order for it to go form the second product, both double bonds need to be shifted over one carbon toward the right. :) Thanks!


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Determining standard Gibbs free energy for water?
I am dealing with the dissociation of water into its components hydrogen and oxygen and I am trying to determine the standard Gibbs free energies of each component. My professor said if the substance is an ideal gas in the standard state then
$$g_{i}^{0}=g_{i}^{\text{pure IG}}(T, P_\text{ref})$$
Now this means I can determine it for hydrogen and oxygen because they can be treated as ideal gases, but how would I consider the standard Gibbs free energy for water?

Let's look at the reaction of the formation of water:
$\ce{2H2(g) + O2(g) -> 2H2O (l) }$
The Gibb's Free Energy relates the spontaneity of various reactions by looking at the change in enthalpy, temperature, and entropy.
We can relate those by:
$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$
If we look at our reaction, it's clear that this is an exothermic reaction. Bonds are being formed as hydrogen gas and oxygen gas form water. Moles of gas are decreased and a liquid is formed so entropy is decreasing as well. We know that the formation of water is a spontaneous reaction, so $\Delta G^\circ$ must be negative. Therefore, this reaction must be run at temperatures where the difference of the enthalpy and the product of the entropy and temperature must be < 0.
The Gibbs free energy is also a state function, so looking at $\Delta G^\circ_f$, we can find $\Delta G^\circ_{rxn}$.
$$\Delta G^\circ_{rxn} = \Sigma \Delta G^\circ_{products} - \Sigma \Delta G^\circ_{reactants} $$
Both hydrogen gas and oxygen gas are in their standard states and have a Gibbs Free energy of 0. If you look at tabulated standard thermodynamic data at $25^\circ\ \text{C}$, $\ce{H2O (l)}$ has a Gibbs free energy of$ -237.13 \text{kJ}\cdot\text{mol}^{-1}$.
$$\Delta G^\circ_{rxn} = -237.13 \text{ kJ}\cdot\text{mol}^{-1} - 0 \text{ kJ}\cdot\text{mol}^{-1} = -237.13 \text{ kJ}\cdot\text{mol}^{-1} $$

@BenNorris Thank you very much. I forgot how to do the dots and was having trouble with the exponents.


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Variable oxidation state of transition metal ions
Transition metals can form stable ions with different oxidation states. But I am confused why doesn't only the most stable state exist. Let me clarify my question more:
$$\ce{Ti^{2+} -> Ti^{3+} + e^-} $$
With $E^0 = +0.5~\mathrm{V}$.
This just shows that $\ce{Ti^{3+}}$ is more stable. So my question is that why doesn't $\ce{Ti^{3+}}$ only exist in its most stable form?

Understanding Latimer, Pourbaix and Frost diagrams might help you understand the relative interconversion of different oxidation states. here and here is a good link. I will post an answer if you still want any clarifications.

if you look at the electron configurations of the Titanium cations you can see what is stable (titanium 4+ is most stable because it has the same electron configuration as Argon) and Ti(3+) has a half filled 4s-orbital and is relatively stable...

Thermodynamics proposes, kinetics disposes.  The most stable form can have a high activation energy barrier surrounding it.  The medium matters.  Hard Lewis bases (water) stabilize high oxidation states (Cu(II), tungstate).  Soft Lewis bases (acetonitrile) stabilize low oxidation states (Cu(II); $\ce{W(CO)6}$ and http://www.google.com/patents/EP1995347A1?cl=en).
$\ce{Ti^{+4}}$ is the basement, especially with oxygen.  Now rationlalize titanium suboxide Magnéli phases.  They are spectacular as inert electrodes under extreme conditions, re Ebonex.
DOI:10.1016/j.electacta.2010.05.011
http://eprints.soton.ac.uk/cgi/request_doc?eprintid=152857&_action_null=Request+a+copy


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Delocalised electrons and stability of benzene
In most high school chemistry books, it is said that the delocalised electrons in benzene makes it particularly stable. However, how does it contribute to stability? Is there any simple way of understanding this?

Related, but not a duplicate: Why are aromatic compounds so stable?

In physics (or you can google it) you will learn about the particle in a box experiment.  At the atomic level, if you confine a particle to a box, it will have some energy.  It turns out, if you make the box longer, the energy of the particle will be lowered, if you make the box smaller, the particle will have a higher energy.  The same concept applies to benzene.  Basically you start with 3 separate acetylene molecules in which the pi electrons are confined to a 2 carbon segment.  Once you trimerize the acetylene and form benzene, the pi electrons can now roam over 6 carbon atoms, a much longer "box" if you will.  So the pi electrons in benzene will be stabilized (lower energy, longer box) compared to the pi electrons in the starting acetylene (higher energy, shorter box).  Molecular Orbital theory tells us the same thing in a slightly different manner.  In each acetylene (let's just focus on the 2 pi electrons in the acetylene that will be the pi electrons in benzene) we have 2 pi electrons in the HOMO and none in the LUMO.  As we bring these 3 acetylenes closer together to form benzene, the orbitals interact.  The 3 initially degenerate HOMOs in the 3 acetylene molecules split (they are no longer degenerate) as they move closer to becoming benzene and a net energetic stabilization results.

Could you explain why, then, delocalization only increases stability in some molecules (those with multiple equally-stable lewis structures)?

For high schoolers, this is how I explain it.  Remind them, that stability = less reactive or more set in place.... i.e. the electrons aren't going to go wondering off to another atom.
In our scenario... the carbons are moms/dads and the extra electrons are excitable little children who are prone to run off (think of little preschool brothers/sisters/cousins at a family reunion in a park).  Each individual mom/dad would be spending all their energy keeping their energized little kid from running off.  In the same way each carbon has trouble holding on to the extra energy electrons.... again, just as parents have a hard time holding their excitable children in place.
However, these carbons have linked up and are supporting each other and their extra energy babies (i.e. delocalized electrons) who have to move so they zip around in this ring of carbons.  Just like energized children zip around but stay in the safe part of the park when there are extra adults looking out for them and giving them physical safety boundaries.  By linking up, each of the carbons share the extra load of holding on to the extra electrons.  This stabilizes the whole group meaning the extra energy electrons won't go running off i.e. try to react with another molecule.  The electrons are happy and content staying within the ring of carbons.  AND (for my AP students, this keeps the carbons happy having the 4 bonds it needs without having to have a higher electronegativity number to keep them bonded; and how the stability of benzene makes it hydrophobic because it prevents the hydrogen bonding in water...)
I also remind them that if the double bonds in benzene were just double bonds, there would be a complete pi bond between two of the carbon atoms - i.e. these two carbons would own/hold the extra bonding energy and covalently share it.  But with a benzene ring, it is more like a co-op of carbons sharing the extra electron load and that is why we say that in reality it isn't 3 single and 3 double bonds in the ring, but more like (6) one and a half bonds.  Then we talk about bond lengths a little bit - comparing bond lengths of a single, double, and then a benzene ring bond.
AND (for my AP students, we talk about how conjugation stabilizes other organic molecules like antioxidants, etc.)

The hydrophobicity of benzene doesn't have anything to do with delocalization. It's just do to with the fact that the bonds are essentially non-polar.

But you go from two parents with two children to six parents with six children. The number of children per parents has not changed. How does this make it more stable?

Non-polarity alone does not account for the hydrophobicity of benzene; it is markedly more hydrophobic than cyclohexane, for instance.

A chemical reaction is a kidnapping then? The analogy is terrible, unscientific and probably much harder to understand than introducing a little bit of molecular orbital theory. It's explanations like this, that lead to an underdeveloped abstract thinking and to serious misunderstandings in more advanced courses.


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What is the smell of 'burning' metal?
When a pass a construction site where someone is welding two pieces of metal together there is a very distinct smell that I associate with 'burning' metal, although I am not quite sure it is burning in the traditional sense. Also, when someone ignites a sparkler this same (or at least similar) smell emerges.
I would like to know what is causing the smell? Is it the smell of metaloxide particles that are dispersed from the weld site? Or is it something in the air that reacts and causes the smell?

Would you happen to know if all metals have the smell when being welded, or is it just steel? I suspect the smell comes from the release of trace amounts of carbon/phosphorous/sulphur present in alloys, especially steel, which immediately react with air and each other forming odorous compounds. Very pure metals may be odorless when welded

@NicolauSakerNeto I'm not sure whether it is only steel or more (all) metals, although I suspect that the thing I smell near construction sites will usually be from welding steel

I always suspected that it is a burning iron itself, but without proof and further thinking. As noted, similar smell comes from welding and disc grinding.

According to the Canadian Centre for Occupational Health and Safety page Welding - Fumes And Gases, the welding fumes are

a complex mixture of metallic oxides, silicates and fluorides. Fumes are formed when a metal is heated above its boiling point and its vapours condense into very fine, particles (solid particulates).

This composition varies dependent on the composition of metals being welded and fluxes.  The website lists several main types and their associated health risks.  Often associated with metallic oxides are some pungent/noticeable gases:

Gases produced from the thermal breakdown of coatings, from welding and cutting processes, stated in the article linked as being examples of "welding gases" (examples from the linked website include ozone)
And depending on the coating that may be on the metal, organic vapours (including aldehydes, diisocyanates, phosgene and phosphine)

The smell is often noticed at construction sites, when reinforcing bars are welded or cut. Constructional steel contains about 0.1% carbon and phosphorus.
Phosphorus normally isn't welcome since it renders steels more brittle. On the other hand, it increases the corrosion stability. This is is a desired property for rebars embedded in concrete.
Is it conceivable that noticable (alkyl)phosphanes are formed when this steel is welded or cut with a disc grinder?

My first impression was also to say phosphines from the P in the metal. Often, just by holding steel up to your nose you can detect a faint aroma


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How does CrF₆³⁻ hybridize?
My textbook says that $\ce{[CrF6]^{3-}}$ is an example of $sp^3d^2$ hybridization. But I don't really see how. Chromium has an electronic configuration of:
$$3d: \boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \ 4s: \boxed{\uparrow}$$
Give three more electrons and it might look something like this:
$$3d: \boxed{\uparrow \downarrow}\boxed{\uparrow \downarrow}\boxed{\uparrow}\boxed{\uparrow}\boxed{\uparrow} \ 4s: \boxed{\uparrow \downarrow}$$
This is the ground state of the electrons. I see no way for electrons to be promoted for the formation of six orbitals. And the electrons have to be promoted all the way up to the $4d$ orbital, which would require a significant amount of energy. What is the exact mechanism of bonding in this ion? Amazingly, I couldn't find anything on the Web about this.
EDIT: My textbook also says that $\ce{[Co(NH3)6]^{3+}}$ is an example of $d^2sp^3$ quite similar to $\ce{[CrF6]^{3-}}$. But when I try to understand the exact mechanism, I encounter a similar problem as a I did above.

In $\ce{[CrF6]^{3-}}$ the central atom is $\ce{Cr(III)}$ or $\ce{Cr^{3+}}$. Chromium is usually $d^6$, so now (take away 3 electrons) $d^3$.
When we start filling in the electron pairs from the ligands, we get the following:

The electrons from the metal are marked in red.
The electrons from the ligands (blue) are filled into two $d$, one $s$ and three $p$ orbitals: $d^2sp^3$ hybridization as per solution!
Now try to do the same with $\ce{[Co(NH3)6]^{3+}}$!

This is a very tempting solution, but I'd like to issue the warning, that an $d^2sp^3$ hybrid orbital would mean a sixfold degeneracy, which is clearly impossible in an octahedral (ish) field. This application stretches the concept a little bit too far.

@Martin ,So isn't it d^2sp^3 hybridized?then what should be the answer?please explain.

@Akash If you have a new question you're welcome to ask it, but I won't be explaining this sort of thing in the comments. There are also multiple posts on this site about the involvement of d orbitals in hybridisation. Please search for yourself.


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Extract acid from a solution
I have a solution containing glucose and some sulfuric acid. I want to distill out the sulfuric acid to then reuse it for other purposes. The problem is the presence of glucose in the solution. I'm afraid that, when increasing the temperature, the sulfuric acid will dehydrate glucose to carbon and other compounds - something that I don't want to happen. Would distillation cause such a reaction?
And in case it would - are there any other methods to extract the sulfuric acid?

I don't know your equipment, but the boiling point of sulphuric acid is > 300 °C.

Is there a specific reason you want to recover that sulphuric acid? Ultimately it'd be much simpler, cheaper, and less energy intensive to simply discard it (after proper neutralization) and use new sulphuric acid.

I used the sulfuric acid for hydrolysis of starch and as it is a catalyst, it would be better to recover it? Or is that more expensive than using new sulfuric acid?

Sulphuric acid is one of the cheapest chemicals of the world. Don't recycle.

As above, your labor all by itself (even if you are grad student) is worth more than the sulfuric acid.  As a pedagogic exercise, run the solution down a strong anion exchange resin washed OH-.  Only glucose comes out.  Recover the sulfuric acid by displacing it with phosphoric acid.


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Products of Reimer–Tiemann reaction of 4-methylphenol

p-cresol (4-methylphenol), undergoes the Reimer–Tiemann reaction upon treatment with chloroform ($\ce{CHCl3}$) in alkaline medium. Which of the following is the major and minor product?
A. 2-hydroxy-5-methylbenzene-1,3-dicarbaldehyde
  B. 2-hydroxy-5-methylbenzaldehyde
  C. 4-(dichloromethyl)-4-methylcyclohexa-2,5-dien-1-one
  D. 4-(dichloromethyl)-4-methylcyclohexa-2,5-dien-1-ol

The answer given is that B is the major product, and C the minor product.
I have two questions about this:

Why is the mono-aldehyde B the major product, and not the dialdehyde A?
I think both electron-donating groups should greatly increase the reactivity of benzene, allowing the reaction to take place twice. On the other hand, steric factors would oppose the second formylation. I do not know which one predominates.
How can one derive the formation of the minor dearomatised product C?

The Reimer-Tiemann reaction is known as the ortho-formylation of phenols under strongly alkaline conditions.
It's not an addition of a carbene to a double bond, like the Simmons-Smith reaction, the pyrrole to pyridine ring enlargement or the Buchner expansion, which is catalyzed by transition metals.
As ron already pointed out, pH is one crucial requirement, having a phenol (and not just any other arene) is the other.

What we're actually facing here is the addition of a carbanion to a strongly electron-deficient carbene.
As again already described by ron, there is significant charge density at the 4-position of the phenolate, and the initial addition step might indeed happen here too.

The resulting intermediate can't rearomatize other than in a back reaction and that's what it will mainly do! Only a minor part will grab a proton and yield C.
That was close, but no cigar! Let's have a look at the addition at the 2-position.

Through the interplay of dienone, phenolate and phenol we get rid of one chloride and all further steps to the final benzaldehyde are likewise assisted by this tautomerism.
We still haven't answered the question why no 1,3-dialdehyde is formed. Let's have a look at the major product B under alkaline conditions and its resonance structures.

One of them looks like the enolate of a 1,3-dicarbonyl conpound, with additional charge distribution over the $\pi$ system.
This species has a significantly lower localized electron density than the anion of the starting material. So, the party stops here.

Edit In a comment, it was asked whether the geminal dichloride will convert to the corresponding aldehyde under alkaline conditions.
Such a transformation could involve a nucleophilic substitution, leading to a geminal halohydrine which then eliminates HCl to furnish the aldehyde.
In the case of benzylic dihalides, this is possible even without the intramolecular assistance of a phenolate (cf. this example) using secondary amines as a base and in the case of diphenyldichloromethane, the transformation to benzophenone is achieved by heating in the presence of water.
In the case of C, the simple deprotonation of the dihalide might be a dead-end alternative pathway to substitution.

Seconding Klaus, this reaction is selective for the ortho position on phenols.

+1 for a nice answer. Just a couple of clarifications. Does the alkaline conditions not replace the two chlorine atoms in the minor product via SN to dive an aldehyde? Also, can you expand a bit on how do we deduce the low electrophilicty of the monocarbaldehyde anion?

@SatwikPasani You paid attention while reading - I was distracted while writing! Actually, I never intended to argue about the electrophilicity of  the anion :D

The reaction is run under basic conditions.  Early in the reaction, the phenolic proton is removed to generate the phenoxide anion.  If you draw resonance structures, you'll see that negative charge is delocalized to the ortho and para positions.  Dichlorocarbene, an electrophilic species is also generated from the chloroform and the base in this reaction.  The dichlorocarbene will preferentially attack the electron rich ortho and para positions in the phenoxide anion.  The para intermediate is often isolated as a cyclohexadienone (your product C), whereas the ortho intermediate usually reacts further due to the adjacent phenoxide oxygen playing a role, going on to the aldehyde.  Here is an old, but good reference, showing the scope and mechanism of the reaction.  It is generally not a clean or high-yielding reaction.
http://alexandria.tue.nl/repository/freearticles/587973.pdf

"The dichlorocarbene will preferentially attack the electron rich ortho and para positions in the phenoxide anion" But shouldn't the nucleophile instead attack the electrophilic species?

@gaurang tandon its the same thing. A nucleophile attacking electrophile = electrophile attacking nucleophile. Just semantics of your frame of reference.