Looking for an old (1950s) "chemical experiments" book Not sure if this is precisely on-topic, but I'll risk some reputation and give it a try.

I got a chemistry set as a kid in the early 1970s, and the subject really captured my interest. I grabbed every chemistry book I could find from our libraries. They ranged from college textbooks to "Chemical Magic" collections of demonstrations. These latter tended toward pyrotechnics, and were pretty appalling safety-wise.

There were several favorites, but there was one book in particular that stuck with me, found in a rural Virginia county library around 1976 or so. It featured adolescents in coats and ties (and maybe one in blouse and skirt) demonstrating experiments, but the experiments were a couple of levels beyond what I found elsewhere.

The ones I definitely remember:

Removing stains by soaking clothes in potassium permanganate solution, then removing the manganese-dioxide residual stain with sodium bisulfite
Preparing white phosphorus by heating red phosphorus in a test tube, and condensing the vapors onto a cold test tube nested inside it
Preparing lithium metal by melting a lithium salt in a crucible and electrolyzing it
Synthesizing Bakelite from phenol and formaldehyde

There were probably a dozen or so others. I think one involved generating chlorine gas from an aqueous solution in the bottom of a beaker, then dropping in pellets of calcium carbide to watch the acetylene spontaneously combust.

I've searched using Google, but can't seem to get past The Golden Book of Chemistry Experiments -- yes, that's another treasured memory from the time, but it's not the one I have in mind. (For one thing, the one I'm asking about had no color plates at all.)

Another near miss that's quite familiar to me is Leonard A. Ford's Chemical Magic (I remember the 1959 original). I don't have it in hand, but Google's preview let me read far enough to see familiar material -- including pyrotechnic mixes using potassium chlorate instead of perchlorate, which got me in no small amount of trouble some years later. The book I'm searching for was more oriented toward science, rather than showmanship.

If the book I've described sounds familiar, please post an answer with any additional information you find -- title and publisher would be great, but additional experiment descriptions would also be helpful. If you have a favorite book from that period that doesn't seem to match this description, I'd be interested in hearing about it as well.

Finally, if this is a bad question, I welcome guidance on turning it into a good one, or why it's not worth pursuing. Thanks!

There are 2 books in German which inspired me to become Chemist. I had them in the late 70s.
"Chemische Experiemnte, die gelingen" (Chemical experiments which succeed) and "Organische Chemie im Probierglas" (Organic chemistry in a test tube). Both published originally in 1939 by Hermann Römpp and Hermann Raaf. Kosmos Verlag (Kosmos Publishers). No idea if there ever were English translations, but you might still find them in antiquary bookshops.


How can weaker nucleophiles kick out stronger ones in the reaction of chloroalkanes with sodium iodide in acetone? I have the following SN2 reaction, in an acetone:
1,4-dichlorohexane (1 mol) reacting with NaI (1 mol) to form 4-chloro-1-iodohexane and NaCl.

However, chloride ions are stronger nucleophiles than iodide ions in an aprotic solvent and iodide ions are weaker bases than chloride ions. Reactions generally form weaker bases, why does this reaction occur? My text shows it does, but I don't understand why. What am I missing?

Acetone is the key here. The solubility of NaI in acetone is about 50 g in 100 mL while NaCl is virtually insoluble in acetone, thus the reaction is driven towards the alkyliodide due to Le Châtelier’s principle, since NaCl will be precipitating from the reaction. This process is called a Finkestein Reaction. Also see this.


Total number of stereoisomers of 1,2-dimethyl-4-(propan-2-ylidene)cyclopentane I don't think there is any change of showing geometrical isomerism across the double bond because the two groups on the right side (two $\ce{-CH_3}$) are the same.

Also I don't think there should be optical isomerism as there is plane of symmetry. Can someone please explain the solution?

Here is a similar molecule to illustrate:

Two of the ring carbons are asymmetric (chiral). This molecule itself (as shown) is chiral. If you change one of the chiral carbons, you get a molecule that is not chiral (mirror plane along the double bond and perpendicular to the ring).


Dimethylethylamine or Ethyldimethylamine? I thought that Organic Naming was based on alphabetical for alkyl branches, so I thought it was ethyldimethylamine, however google results show me Dimethylethylamine. So which one is right? And is this an IUPAC rule?

There's a technicality in that the 2013 IUPAC recommendations (Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013 (Blue Book)) have slightly awkward preferred IUPAC names for simple amines like these. The preferred IUPAC name for your molecule is "N,N-dimethylethanamine" (as described by P-62.2.2.1 on p 671, but note that there is an erratum for the triethylamine example shown there).

Putting that aside, though, what you're conceptually missing is that the methyls and the ethyls in your proposed name are (implicitly) not on equal footing. The methyls are names of substituents, but ethyl is not a substituent - it is the parent chain (since it is longer than methyl). So, $\ce{CH3CH2NH2}$ is named "ethylamine", and $\ce{CH3CH2N(CH3)2}$ is named as a N,N-disubstituted form of "ethylamine", i.e. "N,N-dimethylethylamine". In common usage the "N,N" is often dropped, leaving "dimethylethylamine". I should emphasise that this is not in line with the IUPAC recommendations, but people can and will use it, and this is why you will see it on the Internet. Returning to the preferred IUPAC name in the previous paragraph, we can see that the use of "ethanamine" rather than "ethylamine" makes the difference between methyl and ethyl a lot clearer.

In fact, according to the recommendations, it is also permissible to name tertiary amines in the manner (alkyl1)(alkyl2)(alkyl3)amine (although it does not lead to the preferred name). If we were to adopt this method, then the name would be ethyl(dimethyl)amine, because now there is no notion of a parent chain and all three substituents are merely substituents - which should be listed in alphabetical order ignoring prefixes like "di". The parentheses around "dimethyl" are required (but not around "ethyl") because of a separate rule (P-62.2.2.2, p 672 - see also the discussion in the comments).

To sum up:

N,N-dimethylethanamine: Preferred IUPAC name
ethyl(dimethyl)amine: "Alternative" IUPAC name; permissible but not preferred
dimethylethylamine: Doesn't comply with 2013 IUPAC recommendations, but the logic behind the name is similar to that used in generating the preferred IUPAC name.


What does the subscript of atomic orbital mean? As everyone knows, the atomic orbital can be classified as $s, p_z, p_x, p_y, d_{z^2},d_{xz},d_{yz},d_{xy},d_{x^2-y^2}$ and so on. I want to know the meaning of $z^2,x^2-y^2$ and so on. Maybe this is a fundamental question, but I'm not familiar with chemistry.

I have some ideas about this:

For example, we consider the $p$ orbital: $p_z$ is symmetric about $z$ axis; $p_x$ is symmetric about $x$ axis; $p_y$ is symmetric about $y$ axis. But what's the meaning of $z^2$ or $x^2 -y^2$?
We consider the symmetry of atomic orbital with group theory. The change of orbital under the symmetry operation is the same as the orbital subscript.

For each azimutal quantum number $l$, the magnetic quantum number $m$ ranges from $-l$ to $+l$. Those give you the number of atomic orbitals "subsripts" you should obtain.
Those are then expressed using spherical harmonics. If you look at the table of spherical harmonics at the $l=1$, you will find that they contain the cartesian axes as $x-y$, $z$ and $x+y$. Therefore it is advisable to form the linear combination out of them, so you get $x$, $y$, $z$. If you go to higher angular momentum, things get more complicated, but there is some tradition what linear combinations to use.

To answer your second question, all this formulas are not some inevitable result you obtain by juggling the numbers. Contrary, it is carefully crafted in a way that the formulas are so simple.


How does the choice of metal (oxide) catalyst affect the range of unsaturated compounds that can be hydrogenated? In the hydrogenation of unsaturated compounds with hydrogen gas and a catalyst, the choice of palladium on carbon is able to hydrogenate alkenes and alkynes, but is unable to hydrogenate aromatic compounds. Using rhodium or Raney nickel, however, allows one to hydrogenate aromatics. How does the choice of catalyst affect what kind of unsaturated compounds can be hydrogenated?

The currently accepted hydrogenation mechanism for metal like palladium is the Horiuti-Polanyi mechanism. (image)  This mechanism works with good yield for alkenes and alkynes.  For an aromatic compound, the activation energy is much higher because the aromaticity of the target molecule is being broken by the first hydrogenation.  Indeed, 1,3-cyclohexadiene is very unstable and will revert to benzene and hydrogen instead of continuing to cyclohexane.

Addition of a Lewis Acid makes palladium an acceptable catalyst for the hydrogenation of benzene, however.  An intermediate compound with the benzene and the LA is first formed, and hydrogen is added to this intermediate.  This paper discusses the mechanism of hydrogenation in the presence of a Lewis Acid.

Rhodium catalysts work slightly differently.  They generally have ligands attached that help stabilize the intermediates.  You can see a simplified mechanism here and there are papers discussing specific catalysts here and here.  Basically, in the case of benzene, the relatively unstable 1,2-cyclohexadiene is held in place on the catalyst until it has been completely hydrogenated.


Using a calcium metal anode in aqueous solution? In school, I am doing an electrochemistry lab in which I need to create a galvanic cell battery with the highest voltage possible using two beakers, a salt bridge, and electrodes.

To maximize the voltage, I have decided to use calcium metal for the oxidation half-reaction and a sodium persulfate solution for the reduction half-reaction. I understand that for the reduction reaction in the cathode, I would merely dissolve the persulfate in water. However, calcium metal reacts with water, so how would I go about connecting the calcium to the salt bridge (which is a requirement in the lab)?

The salt bridge has an aqueous salt and my chemistry teacher said that the redox reaction that would result from letting the calcium touch the water would prevent it from being effective in the battery. Also, my class vaguely touched on the Nernst equation, but are there any other revisions I can make to the battery to increase the voltage, either by adding $\ce{OH-}$, $\ce{H+}$, or increasing/decreasing concentration of the persulfate solution?

Basically, the reaction of calcium with water short-circuits your battery; the electrons you wanted to go through the external circuit end up going directly into the electrolyte instead.

Magnesium will work better for your anode; it reacts little or none with water when cold and it's anodic potential still packs a pretty good punch.


What is the difference between hydrogen chloride and hydrochloric acid? My teacher insists that hydrogen chloride is a misnomer, and that the proper nomenclature demands HCl to be named hydrochloric acid. However, I saw the name hydrogen chloride used in my textbook, but it was never further explained.

Hydrogen chloride is a gas. It is the molecule made up of a hydrogen and a chlorine atom bonded by a single bond. It comes in gas cylinders for use in chemistry.

Hydrochloric acid is a solution (no, not to your problem of a stubborn teacher). It is composed of water molecules and hydrogen chloride molecules. In terms of mass percentage, concentrated hydrochloric acid is often $37~\%\ \ce{HCl}$ in water. It comes in glass bottles (often brown glass for ease of bottle use; the chemical company does not have to distinguish between compounds that can live in clear bottles and those that must live in brown ones) for use in chemistry.

Hydrogen chloride can also be dissolved in other solvents such as ether. In this case too, it may be termed hydrochloric acid, although a specification of the solvent should be added.

Whether or not either of the two names is a misnomer depends on what you are addressing. Using hydrogen chloride for the liquid solution is as wrong as using hydrochloric acid for the gas.


Storing large quantities of KOH I need to handle large quantities of $\ce{KOH}$; close to 1000 liters of 85 % conc $\ce{KOH}$ for about an hour. Any suggestions on how I go about doing it? Also the temperature during the process might touch 500 degrees Celsius.

The first thing to clear up is – Why do you need to store 1000 L of KOH in liquid form? It is as others have pointed out, very concerning that someone who is planning to store this quantity needs to ask on the internet.

My advice is don't store 1000 L of liquid KOH, store 40 × 25 L of liquid KOH. This will reduce the risk considerably, and help ensure that a single leak isn't a catastrophic event, but is something you can deal with. This is a common philosophy, and is used, for example, in trailers of dangerous gases, which are organized into racks of 20 smaller gas bottles instead of one large tank.

Keep your 25 L containers off the floor in a bunded area for spill containment, and have appropriate means to deal with a spill. I'm not going to tell you the best way to handle a spill because I don't know. – You need to find out. Also, make sure the store is properly monitored to detect any problems.

Now, going back to my initial point: Let's accept that, for whatever reason, you need to store 1000 L of KOH (as you haven't shared with us what you are doing.) Solid KOH is much safer than liquid KOH, so why do you even need to mix up the solution all at once?

One of the causes of the Bhopal disaster (the worst industrial disaster ever, which killed many thousands of people due an uncontrolled venting event) was the unnecessary storage of a toxic and volatile compound (methyl isocyanate.) If the pesticide they were making had been synthesized via a different route, or even if the methyl isocyanate had been made only in the quantities needed, the size of the disaster could have been reduced or even eliminated (despite the extremely poor state of maintenance of the plant.)

I suggest you mix up the KOH in small batches as you need it to reduce the risk

  the temperature in the process might touch 500 degrees Celsius

By "process" I assume you mean mixing solid KOH with water, and that you do not actually need to use the KOH at this temperature. If you expect to reach these temperatures, you are doing it wrong. As can be seen from https://koh.olinchloralkali.com/TechnicalInformation/Phase%20Diagram%200-80.pdf KOH solution in all compositions from 0 to 85% is liquid at below 150 °C. Note also that concentrations above 50% are solid at room temperature – so do you really mean 85% conc KOH?

The safe way to do this is to add the dangerous material to the safe material, particularly as water has such a high heat capacity and can lose heat by boiling. You are going to need either a lot of patience, or mechanical screw feeder, to perform the addition at a safe rate. You will also need decent stirring equipment. Also, I would expect most of the heat will be released in mixing the first 50% of the KOH, because after that you are effectively mixing KOH with KOH.

In any case, if you are planning to do this, you should check the thermochemistry yourself and include it in your Risk Assessment, along with any relevant details from industry experience regarding specific hazards of mixing (spattering for example?)

I really hope you make sure what you doing before you do this. Alkalis are a lot worse on human flesh than acids, partly because they turn the oils in your skin to soap, and partly because the spills won't evaporate and therefore persist until something is done about them.

It would be a good idea to dye your KOH so you know where any spills are. I personally have spilt 50L of antifreeze solution in the lab (it was intentionally dyed beforehand in case of spills, and needless to say we used propylene glycol as it is much less toxic than the ethylene glycol used in car radiators.) Whatever you intend, something in your experiment will go wrong.

Please also check the legal aspects in your area. Also, consult with your supplier! The fact that you need to ask on the internet instead of talking to them suggests you are up to something unusual. Would you care to share with us what it is?

EDIT - Now you've clarified you're an engineer, one more very important point. Keep KOH away from aluminium. Alkali will dissolve the oxide layer that protects aluminium from corrosion, and unmask the true (very high) reactivity of the metal. A lot of heat and hydrogen gas will be generated by the reaction $\ce{3 KOH + Al -> K3AlO3 + 2/3 H2}$.

I'm guessing most stainless steels should be fine (but obviously check.) I have used equipment from the catering industry in some experiments, but stay away from stainless steel cookware that has the base thickened with aluminium. Even though it is external, it's not worth the risk.

Also, note that mixtures of acids and bases with water are often most corrosive at a medium concentration, so don't assume that just because a material is OK at 90% concentration, it's going to be OK at 50% concentration.


Is sublimation a property of all substances? I wonder if sublimation can be done only on certain substances like iodine, or it can be done on all substances.

In other words, can sublimation be done, for example, on water? If it cannot, then why?

Look up the term “phase diagram”. At different temperatures and pressures different phases are favored. Think about water. At 0 °C and 1 atm water and ice are equally favorable. There can be freezing or melting. At other temperatures and pressures there can be sublimation and deposition.

The sublimation of water is involved in freeze drying and freezer burn of ice cream.

It turns out that iodine sublimes at room temperature and atmospheric pressure but there’s nothing special about iodine compared to other substances on a fundamental level.


Determining Molecule Based on Proton NMR (HNMR) Based on the NMR data, I have determined that the molecule must be 1-phenoxy-2-propanone, however I may be wrong because given the proton NMR, there are only 4 unique carbon environments, however in 1-phenoxy-2-propanone, there are 5 carbon environments. In addition to that, on carbon environment "B" there is only one proton, which isn't indicated on the NMR, which makes me doubt my choice in molecules. I've basically tried every molecule and this was the closest I could get!Could anyone please correct me?

Thanks in advance!

Your assignment seems reasonable. One thing to keep in mind is that even though  phenyl protons are in different chemical environments, they may be too close to resolve and with a low-field instrument, some may appear as a "blob" of superimposed signals. In this case, the peak from (B) is simply unresolved from that of the (C) protons, giving a single signal that integrates to 3 protons. See here with a 90 MHz instrument. That spectrum is a bit different than the one you have, but the one you've been given may be a simulated spectrum.


Where is the mirror plane in 1-bromo-1-chloroethene? I am studying point groups in inorganic chemistry and having a very difficult time understanding some of these concepts.

In my book, Inorganic Chemistry: Fifth Edition by Miessler, Fischer, & Tarr, I am presented with a table showing that $\ce{CHFClBr}$ is in the $C_1$ point group and has no symmetry other than the identity operation.

$\ce{H2C=CClBr}$ is in the $C_\mathrm s$ point group and its symmetry is listed as having only one mirror plane.

I'm having difficulty seeing this mirror plane. I would have listed this compound as being in the $C_1$ point group. Does this have something to do with the molecule being flat?

For all symmetry operations, you should assume the atoms reduced to a point size. Then, if you have a perfectly flat molecule (like in the case of your disubstituted ethene), all atoms will be in the same mathematical plane meaning there is nothing above or below the plane.

Therefore, the plane is an element of symmetry that puts the nothing above the plane on the nothing below the plane and leaves the atoms themselves (in the plane of symmetry) untouched.


General Alkaloid Extraction Process I'm getting into alkaloid extraction from various ethnobotanicals. It is a bit frustrating trying to extract alkaloids from ethnobotanicals without good guides out there, and I wanted to make my own techniques and experiments. However, I need to know what the basic outline is for extraction. Are there any guides or basic principles I need to know?

I'm interested in extracting for example Leonurine from Wild Dagga or Mitragyna and 7-OH-Mitragyna from Kratom.

What I was looking for was Acid-Base or Liquid-Liquid extractions.

7-Hydroxymitragynine

The extraction and crystallization of the parent alkaloid, mitragynine, from dried leaves of Mitragyna speciosa (Kratom), has been described in great detail by A. H. Beckett, E. J. Shellard and A. N. Tackie in Planta Med., 1965, 13, 241-246 (DOI).

(Interestingly, the procedure also furnished speciofoline, a spiroindolo indolizidine alkaloid.)

Leonurine

A first extraction was seeming reported by S. Kubota and S. Nakajima in Nippon Yakubutsuqaku Zasshi, 1930, 153. I don't have a DOI and the source is most likely in Japanese.

I don't have personal experience with the extraction of guanidines from biological material and would probably rather try to synthesize leonurine, following the procedures described in Tetrahedron, 1969, 25, 5155-5161 (DOI).

General guidelines

I have given some hints on extraction in an answer to a question on the use of TLC in alkaloid identification.


Computer killed by insecticide; what compounds are likely? A friend of mine sprayed "Black Flag Home Defense" insecticide under her
computer. It died immediately. I'm trying to figure out why, and how
fixable it might be.

I found an MSDS, but it only says what the active ingredient is,
deltamethrin, 0.02%. Apparently everything else is proprietary and/or
non-hazardous.

Deltamethrin is not water-soluble, so I'm thinking that this probably
doesn't contain any water; more like nonpolar hydrocarbon sorts of
compounds. I'm thinking there's some particular nonpolar solvent that
makes up the bulk of it.

So, my questions:

Information or informed speculation as to the likely solvent
involved.
For those with knowledge of electronics, information or informed
speculation about the effect of such solvents on PC boards and
electronic components.

Answers to 1 alone are also useful, as I can then ask on
electronics.se what those compounds would do.

UPDATE: I don't have access to the computer myself. I passed on some of
the suggestions to my friend, and she will bring it in to a shop and see
how repairable the damage seems to be. Thanks for all your thoughts.

UPDATE 2: I spoke with the tech who looked at it. Apparently the
computer was very dirty inside; years of dust, cigarette smoke residue,
and insects. She's not sure exactly what happened, but there seems to be
a short somewhere on the motherboard, and that may have fried the power
supply as well. Given the dirt, it was probably not in good shape even
before the incident; the heat sinks were thoroughly clogged.

Overall conclusion: keep your computer clean! Keeping unknown chemicals
away from it is a good idea too.

Ingredients

So we could probably guess what's in the insecticide based on what's used in some of their other products.

Black Flag Flying Insect Killer

Tetramethrin - Sounds like the "active ingredient" here. By "active ingredient", I mean poison.

D-phenothrin - Another poison. This product sounds like it kills more than one insect.

Petroleum Distilates - Likely a solvent. Unspecified

Liquefied Petroleum Gas - Propane or butane. An aerosol propellant. Was your product from a bucket or can? If a can, then it probably has this stuff.

So what happened?

My guess is that one of the products created a short circuit

  A short circuit is an abnormal connection between two nodes of an electric circuit intended to be at different voltages. This results in an excessive electric current limited only by the Thévenin equivalent resistance of the rest of the network and potentially causes circuit damage, overheating, fire or explosion.

Typically, when something like a computer experiences a short circuit, it will stop functioning, "die", once the irreversible damage is done. As specified in the above text, circuit damage is likely done, and the circuits will need to be rewired. If there was no damage, then removing any remaining substances through a solvent, followed by thoroughly drying the computer, would remove the short circuit and allow it to function normally. It's possible that if the short circuit was created by a solvent, that the solvent has already been removed by evaporation, due to common organic solvents' melting points.


Solubility of p-nitrophenol in various substances? So I had an organic chemistry lab where we had to dissolve p-nitrophenol in water, dilute $\ce{NaHCO3}$, dilute $\ce{NaOH}$, and dilute $\ce{HCl}$ to test for solubility.

The p-nitrophenol dissolved in everything but water. I just want to make sure if these results are theoretically correct. From what I understand, the nitrophenol should have also dissolved in water because its a polar substance and it shouldn't have dissolved in $\ce{HCl}$ because I think the Hydrogen in the phenol is itself acidic due to electron withdrawing effects of the benzene.

Can someone please explain what should happen?

p-Nitrophenol is actually modestly soluble in water at neutral pH (16 g/l at 25 °C) because of its weak acidity ($p\mathrm{K_a}=7.16$ at 22 °C) and the formation of intermolecular hydrogen bonds between the phenolic OH group and water molecules. The latter would also explain why the compound dissolves in diluted $\ce{HCl_{aq}}$ despite being protonated at lower pH values. The importance of intermolecular hydrogen bonds is also demonstrated by the significantly lower solubility of the ortho isomer, which forms intramolecular hydrogen bonds and therefore appears to be less polar to the solvent (source):

In basic solutions, p-nitrophenol is deprotonated to p-nitrophenolate. The anion is stabilized by delocalization of the negative charge into the benzene ring and the electron-withdrawing nitro group (only two of the possible resonance structures are shown):


Why do strong acids dissociate (almost) fully on a molecular level? In the Brønsted-Lowry theory, strong acids $\ce{HA}$ dissociate using water and react into hydronium ions and the conjugate bases $\ce{A-}$. However I never understood in the strong acids case what stops the products of the reaction(ions) into reforming chemical bonds (why the reaction is one-way) and what enables them when the reagents aren't strong acids. How is this explained macrocosmically and microscopically?

Deprotonation is always a reversible reaction. Hence why, even strong acids, have $\mathrm{p}K_\mathrm{a}$ values (which rely on the presence of equilibria).


Are the Phosphate and Sulfate ion diagrams with double bonds an invalid picture? Whenever you see an image of phosphate, it's always shown having 5 bonds. 3 single bonds to anionic oxygen atoms and a double bond to an extra neutral oxygen.

Not only that, but they're shown being arranged in a tetrahedral manner. All of this additionally applies to sulfate, only that it has 2 double bonds instead of  just one.

Now, from what I've been able to gather from the net, including this website is that this can be explained by invoking sp3 hybridization and a d orbital. The 3 anionic oxygens who already have a charge of -1 pair up with the 3 partially filled hybrid orbitals in phosphorous, while the remaining filled sp3 orbital on phosphorous donates it's electron pair to the empty orbital on an extra oxygen.

Once again, this is exactly the same for sulfate, except that it can only bond to 2 anionic oxygens while also being able to donate 2 electron pairs to other oxygen atoms.

This leaves us with 4 single bonds to oxygen atoms. This doesn't seem unusual at all. Nitrogen can easily take on 4 bonds just like phosphorous, and seeing how row 3 elements have a looser hold on their outer electrons, it makes sense that they can donate their complete electron pairs much easier than row two elements, seeing how their electron pairs would "stick out" more, making them much more available for bonding.

However, in Phosphate when you get to the 5th bond you have to explain it by being a little more creative. You could say that a p orbital on one of the oxygen atoms donates it's electron pair to an empty d orbital on phosphorous, creating and extra bond between phosphorous and one of the oxygens. The same thing happens in sulfate only that for some reason it can form two double bonds. (which I really wonder if that would be pushing the limits of stability at all. I mean wouldn't the two d orbitals kind of overlap at an awkward angle causing some kind of repulsion?)

Now, apparently from what I've heard, experimental evidence has shown that there is very little d orbital involvement in molecules like phosphate or sulfate, and they're practically all single bonds. So my question is, is this true? And if so, then why is the double bond picture still in practically every textbook, diagram, and piece of scientific literature that you come across? How did this picture of these ions emerge? I know it had something to do with linus pauling, but was this just accepted based on theory alone and no experimental evidence at all? Was there some experimental evidence for it in the beginning?

Sorry if it sounds like I'm asking too much, but I just find this whole situation so confusing. I'd appreciate your answers.

Recent experimental evidence shows that d orbitals are involved in the extra double bond of sulfate or phosphate ions. Therefore a stable sulfate or phosphate ion has 2 double bonds instead of only one to make the formal charge on sulfur or phosphorus zero. Any Lewis structure with expanded octet (periods 3 and higher) should have zero formal charge on the central atom.


Why does moving from left to right on the periodic table result in higher conductivity? Question is in the title. I understand that the metals towards the left form stronger bonds with other chemicals compared to metals towards the right, and that the ones with weaker bonds break down easier and pass more current. (if that's wrong, please correct me) Why is this the case? Could someone please explain this concept to me more?

Actually, this only applies for metals. This trend does not apply for non-metals: non-metals form covalent bonds, which are due to the sharing of electrons - thus, there are no free electrons that can carry charge. Additionally, atoms are held tightly by covalent bonds and thus cannot dissociate into ions that carry electric charge.

With regards to metals, metallic bonding consists of metal ions surrounded by a sea of delocalized electrons - and it is these delocalized electrons that can carry charge. Each atom in a metal has the same number of electrons as the charge on the ion. For example, Na+ has one delocalized electron for one atom, Mg2+ has two, Al3+ has three, and so forth. So as the charge increases, the number of delocalized electrons per atom increases - and so the number of delocalized electrons in the metal overall also increases. This allows it to carry more charge per unit time, leading to a higher rate of conductivity.

Similarly with regards to ionic lattices, a greater charge means there will be more ions formed per molecule that is made molten. So one mol of NaCl will have 2 x 6.022 x 10^23 ions total, while 1 mole of MgCl2 will have 3 x 6.022 x 10^23 ions. So, MgCl2 has more ions to conduct electricity and thus shows greater conductivity.

I'm not sure if it has much to do with the actual 'ease' of breaking bonds. Both NaCl and MgCl2 would have to be in a molten state to be able to conduct electricity anyway, so I don't think it plays a role for ionic compounds. For metallic compounds, again, as they are simply ions surrounded by electrons there isn't any bond-breaking to be done.


Ignoring spin, consider an electron in a hydrogen 2p orbital, what is its orbital magnetic moment? I know that a magnetic dipole moment is given by $$\mu=\frac {-e}{2m}I$$ and that the z component of angular momentum is $$m_j\hbar.$$ However, I have also seen that angular momentum $I$ is given by $$I=\hbar\sqrt{l(l+1)}.$$ Is this also the component in the z direction or is it different? Are both expressions for angular momentum right? Which one do I use to answer the question?

(This question was asked in a problem sheet that I have been asked to do over the vacation. See below for context of question).

The first equation you have is for spin. Let's look at it in a operator form

To properly treat the upcoming discussion, quantum electrodynamics is introduced to describe the eigenstates of spin. The eigenstates of spin has a direct correlation with spin. Formally we depict it as the following

$$ μ_B =-g \frac {eℏ}{2m}  \frac { \boldsymbol {S}}{ℏ} $$

where μ is the magnetic moment operator, g is the g factor, ℏ is plank’s constant over 2π, m is the mass of the particle and S is the spin operator.

For orbital magnetic moment (your question) you use:

$$\mu =\hbar\sqrt{l(l+1)} \boldsymbol {\mu_B} $$

Recall:

s: l=0
p: l=1
d: l=2
f: l=3

So plugging in:

$$\mu =\hbar\sqrt{1(1+1)} \boldsymbol {\mu_B} $$
$$\mu =\hbar\sqrt{2} \boldsymbol {\mu_B}$$

(Units are $ \boldsymbol {\mu_B}$)


How is the Felkin-Anh model of stereoinduction correctly explained with MO theory? One method to predict stereoinduction in asymmetric reactions is the Felkin-Anh model (derived from the more basic Cram model) that applies to nucleophile attacks on α-chiral aldehydes. In a simple case such as (S)-2,3,3-trimethylbutanal, assignment of large, medium and small residues is trivial and we predict nucleophile attack as indicated in scheme 1 to generate (2R,3S)-2-hydroxy-3,4,4-trimethylpentanenitrile.

Scheme 1: Felkin model for all-carbon systems.

In some cases — I was taught ‘in cases of a group with high electronegativity on the α-carbon’ — we may no longer consider simply the size of the substituents but instead need to consider the electron-withdrawing substituent to be orthogonal to the carbonyl bond axis. Examples would be α-haloaldehydes, α-silyloxyaldehydes or α-aminoaldehydes as shown in scheme 2 with (S)-2-substituded butanal.

Scheme 2: Felkin-Anh model for systems containing electronegative $\ce{X} = \ce{Br, Cl, OTBS, NHR, etc}$.

How is the Felkin-Anh model explained on a molecular orbital basis? Please include diagrams showing the orbitals involved and what exactly is being stabilised.

Also, if you can confirm (or unconfirm) my list of possible electron-withdrawing substituents, please do so in comments to the question.

From the organic chemistry textbook by Clayden, Warren, Wothers and Greeves on pp. 890-891:

  The $\pi^{*}$(C=O) and $\sigma^{*}$(C–X) orbitals add together to form a new, lower-energy molecular orbital, more susceptible to nucleophilic attack. But,
  if X is not a leaving group, attack on this orbital will result not in nucleophilic substitution but in addition to the carbonyl group. Again, this effect will operate only when the C–X and C=O bonds are perpendicular so that the orbitals align correctly.

  What does this mean for stereoselectivity? Conformations of the chiral carbonyl compound that place an electronegative atom perpendicular to the C=O bond will be more reactive-size doesn't matter.

Update

In response to the requests specified in the bounty description:

  1) What about the mixing of sigma(C-X) and pi*(C=O)? 2) What about pi(C=O) and sigma*(C-X)? Shouldn't that raise the sigma*'s energy?

As you probably know, the above treatment of the effect of one fragment orbital interacting with another one is basically just a pictorial way of applying perturbation theory to the problem. From this theory you get the result that the interaction strength of two fragment orbitals gets stronger the better the overlap and the closer the orbitals are in energy - a very(!) detailed derivation of the fundamental equations involved has been done by @Martin here (an answer he truly deserves much more upvotes for :)). Overlap will not be an issue, but the energy differences will be. So, when thinking about which orbital interactions are most important for the problem at hand we have to look at their supposed energy levels. Thus, I've done a rough sketch, taking into account that:

the $\ce{C}$ atomic orbitals are lower in energy than the $\ce{O}$ or $\ce{X}$ atomic orbitals but that the $\ce{O}$ and $\ce{X}$ orbitals will be rather close in energy (similar electronegativities etc.) and
$\pi$-interactions are weaker than $\sigma$-interactions due to less overlap thus resulting in a lower level-splitting between the bonding and antibonding orbitals.

On the right side of the picture I added arrows indicating the energy differences between the most important orbitals.

As you can see the energy differences between $\sigma(\ce{C-X})$ and $\pi^{*}(\ce{C=O})$ and between $\pi(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$ are very large, much larger than the one between $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$. Thus the two interactions you ask about will only have a very minor influence on the LUMO and the single most important interaction is the one between $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$. Certainly, some mixing will occur (see below for some more on this) but when employing this method you are mostly interested in the big effects at work and can ignore the small ones.

  3) Shouldn't we better consider the entire thing as an allyl cation system?

The system at hand has much lower symmetry than the allyl cation because it consists of 3 different atoms/fragments. That makes drawing analogies between this and the simple pure carbon case difficult. It's certainly possible but the treatment shown in the book by Clayden is much easier. But even if you do a more rigorous perturbational treatment not only considering the first order interactions between $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$ but also the second order mixing of $\pi(\ce{C=O})$ (which in effect would be like doing something similar to the allyl cation) you get qualitatively the same picture as drawn above in the book by Clayden. Basically the only effect the mixing of $\pi(\ce{C=O})$ has is a slight down-shift in energy of the low-lying $\pi(\ce{C=O})$ orbital and a slight up-shift in energy of the new LUMO you get from Clayden's method, i.e. the bonding combination of $\pi^{*}(\ce{C=O})$ and $\sigma^{*}(\ce{C-X})$.

  4) If you can perform calculations on a simple system or know where to find one that would be greatly appreciated

You could try to do calculations for the following molecule

whereby you do one calculation with the $\ce{F}$ of the $\ce{CH2F}$ group perpendicular to the $\ce{C=O}$ plane and one with one of the $\ce{H}$ atoms of said group in that position. Then you can compare the LUMOs of both conformers and see if the position of $\ce{F}$ atom actually has an energy-lowering effect.


Does selenium sulfate corrode or otherwise react with metals? I was just watching a cartoon called Motorcity, and they featured a real compound called selenium sulfate.

While I'm sure that the compound blowing up a detached steel [car part] on contact is just cartoon wackiness, I'm curious whether it actually does anything bad to any metals, especially since a close compound, selenium sulfide, is commonly used in shampoos.

Yes, selenium sulfide is well known and it is used in anti-dandruff shampoo but other sulfur-containing selenium compounds are not really well known. Selenium sulfate, properly selenium(II) sulfate, $\ce{SeSO4}$ is unknown to the best of my knowledge. The analogous tellurium compound has been synthesized recently, but still it is not a proper sulfate compound, rather it is actually an oxide-sulfate mixture type of compound.

In case you are wondering whether selenium sulfate will be obtained by reacting selenium with sulfuric acid, then you are wrong. Selenium and sulfuric acid would react to form a mixture of acids (selenous and selenic acid) but they would all disproportionate to selenium dioxide, sulfur dioxide and water.

$$\ce{Se + 2H2SO4 ->[\Delta] SeO2 + 2SO2 + 2H2O}$$

Selenium sulfite or properly selenium sulfoxide, $\ce{SeSO3}$, is however known.

So, selenium sulfate is a fictitious compound and thus used in the cartoon.


Is sodium chloride really odourless? If yes, what do I smell? I just had to read some general descriptions of sodium chloride and it was always classified as odourless (e.g., by Wikipedia). However, large amounts of table salts (around 1 kg) have a clearly perceptible odour to me. By table salt I mean salt sold for the purpose of cooking – I have not been taking notes or making experiments, but as I had the experience on dozens of different occasions, I would guess that the salt does not need to be iodised.

If I had to describe the smell, I would first say salty. It’s somewhat similar to the smell of the seaside, but the latter is much richer and has additional components. From the ten smell categories proposed here, only sweet and chemical fit somehow. It also has something of petrichor.

Is the classification as odourless just an approximation or do I smell additional ingredients (such as iodide) or impurities that cause table salt not to be pure sodium chloride? In the latter case: What causes the odour?

Since odour is generally coming from vapour / gas form of a material, and NaCl has zero vapour pressure, it should be odourless.
However:

Moving around solid materials, dust, small particles can fly around that can be dissolved in the nose and interact with the odour sensors.
You almost always have something else together with salt. And if it is not pure, you cannot be sure what makes the odour, the salt or the impurity. Eg. sea salt often contains remains of algae.


Is there a mathematical generalization for the number of isomers (structural and configurational) of a given compound? The original question was:

  How do I find the number of structural and configurational isomers of a given compound?
  is there any formula?

For a specific example, how would I find the number of structural and configurational isomers of a bromo compound $\ce{C5H9Br}$ formed by addition of $\ce{HBr}$ to 2-pentyne

Generally speaking, graph-theoretical enumeration aims at counting chemical compounds as graphs (2D structures).
In other words, it is concerned with constitutional
(or structural) isomers.

The most famous method for combinatorial enumeration of graphs is based on Polya's theorem: G. Polya and R. C. Read,
Combinatorial Enumeration of Groups, Graphs, and Chemical Compounds (Springer, 1987).
G. Polya, R. E. Tarjan, and D. R. Woods,

Notes on Introductory Combinatorics (Birkhauser, 1983) Chapter 6.
An introduction to the practical uses of Polya's theorem from a chemical viewpoint has appeared:   O. E. Polansky,

Polya's method for the Enumeration of Isomers
(MATCH Commun. Math. Comput. Chem. 1, 11--31 (1975)). free access.
Many examples of chemical application of Polya's theorem have been described in
the following book: A. T. Balaban (Ed.), Chemical Applications of Graph Theory (Academic Press, 1976).

To accomplish enumeration of 3D structures (concerned with configurational isomers or stereoisomers), in contrast, it is necessary to take chirality/achirality of substituents into explicit consideration.

Symmetry-itemized enumeration of chemical compounds as 3D structures can be accomplished by adopting Fujita's unit-subduced-cycle-index (USCI) approach: S. Fujita,
Symmetry and Combinatorial Enumeration in Chemistry (Springer 1991).
Gross enumeration without symmetry-itemization can be accomplished by
adopting Fujita's proligand method: S. Fujita,
Combinatorial Enumeration of Graphs, Three-Dimensional Structures, and Chemical
Compounds (University of Kragujevac, Kragujevac, 2013).
For the enumeration of alkanes ($\ce{C_nH_{2n+2}}$) and
monosubstituted alkane ($\ce{C_nH_{2n+1}X}$) as 3D structures,
I would like to introduce an account article, which is freely available: S. Fujita,
Numbers of Alkanes and Monosubstituted Alkanes. A Long-Standing Interdisciplinary Problem over 130 Years (Bull. Chem. Soc. Japan, 83, 1--18 (2010)).
The following article is also available freely: S. Fujita, Numbers of Achiral and Chiral Monosubstituted Alkanes Having a Given Carbon Content and Given Numbers of Asymmetric and Pseudoasymmetric Centers (Bull. Chem. Soc. Jpn., 81, 193--219 (2008)).

To grasp perspectives of the theoretical foundations of stereochemistry and of the state-of-the-art chemical enumeration, see also the following recent book: S. Fujita,
Mathematical Stereochemistry (De Gruyter, Berlin, 2015)


Why interconversion in tautomers is so rapid? Looking at the examples in Tautomer on Wikipedia, it seems that the hydrogen needs to travel a significant distance (~20Å) for the interconversion to take place. What lets it happen so fast? Why interconversion is preferred over maintaining its position?

(source)

If you are using a protic solvent (the most common situation when you're dealing with polar functional groups), then you are "cheating" because the solvent can generate its own proton transfers, making the reaction look rapid.


is there a lead compound that crosses the blood-brain barrier when ingested? for example mercury oxide has neurotoxic properties but isn't organic so, it isn't soluble in lipids and so doesn't cross the blood brain barrier when ingested, however, methylmercury chloride is organic and soluble in lipids thus it can cross the blood brain barrier

Tetraethyl lead (bp 180 °C) and tetramethyl lead (bp 110 °C) are easily resorbed through the lungs and the intact (!) skin. They pass the blood-brain barrier without any problems. The half-life for the elimination from the brain is about 500 days.

(Source: Günter Fred Fuhrmann, Toxikologie Für Naturwissenschaftler: Einführung in Die Theoretische und Spezielle Toxikologie, Springer-Verlag, 2006, p 209)

These compounds are neurotoxic and the literature is full of reported cases of intoxications and - sometimes - successful treatment, such as here.


Selecting parent chain considering substituent (prefix) functional groups I am attempting to derive the name of the above compound. Based on the IUPAC rules, I can tell that the carboxylic acid group should be taken as parent i.e suffixed and numbering should start from there. However, since the lengths of both possible choices are equal, there is some ambiguity in my mind as to the choice of parent chain.

Further, what would happen if the -OH were shifted to the carbon adjacent to its current position? Or the amine?

According to the usual nomenclature rules, two equivalent names are possible for the compound that is given in the question:

6-amino-4-(2-hydroxyethyl)hexanoic acid
4-(2-aminoethyl)-6-hydroxyhexanoic acid

In order to identify the preferred name, we have to dig deeper in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), where we finally find two potentially relevant rules:

  P-45.2.3 The preferred IUPAC name is based on the senior parent structure that has the lower locant or set of locants for substituents cited as prefixes to the parent structure (other than ‘hydro/dehydro’ prefixes) in their order of citation in the name.

and

  P-45.5.1 The preferred IUPAC name is the name that is earlier in alphanumerical order (see P-14.5). Alphabetic letters are considered first in the order that they appear in the name; all Roman letters are considered before any italic letters, unless the latter are used as locants or are a part of a compound or composite locant, for example, ‘N’ and ‘4a’. Then, if still there is a choice, numerical locants are considered in the order of their appearance in the name.

Since P-45.2.3 comes first, the preferred name is 4-(2-aminoethyl)-6-hydroxyhexanoic acid and not 6-amino-4-(2-hydroxyethyl)hexanoic acid because the locant set ‘4,6’ is lower than ‘6,4’.

By way of comparison, a similar example given in the (corrected) Blue Book is 2-(2-bromoethyl)-4-chlorobutan-1-ol and not 4-bromo-2-(2-chloroethyl)butan-1-ol since the locant set ‘2,4’ is lower than ‘4,2’.

Remarkably, also according to the subordinate rule P-45.5.1, the preferred name would be 4-(2-aminoethyl)-6-hydroxyhexanoic acid and not 6-amino-4-(2-hydroxyethyl)hexanoic acid because ‘aminoethyl-hydroxy’ is earlier alphabetically than ‘amino-hydroxyethyl’. However, the preferred name for the example given in the Blue Book would have been 4-bromo-2-(2-chloroethyl)butan-1-ol and not 2-(2-bromoethyl)-4-chlorobutan-1-ol since ‘bromo-chloroethyl’ is earlier alphabetically than ‘bromoethyl-chloro’. This example was actually given in the Blue Book by mistake but it has been corrected as part of the official Corrections to Nomenclature of Organic Chemistry. IUPAC Recommendations and Preferred Names 2013.


Why is the photon energy of lithium flame coloring lower than that of cesium? Why does $\ce{Li}$ give a red flame color which has lower photon energy compared to $\ce{Cs}$ which gives a violet flame color according to $E=h\nu$?

This is my attempt: Since lithium has smaller size it has a high ionization energy, thus, if an electron jumps from a higher energy level to the ground state then it releases a large amount of energy compared to cesium which has a larger size and lower ionization energy. Therefore lithium's photon energy should be higher compared to cesium. But it is not. Why?

The problem is the photon energy of flame coloration should gradually decrease down the group of alkali metals but rather it's increasing.

The colours you get in flame tests / burning metals, comes from the electrons in one shell being 'knocked' (by the heat), into a higher shell; then shortly after the electron 'drops' back (in one or more steps). The electron which dropped down a shell has a lower energy state, so to conserve energy, a photon is emitted. The further it 'drops' the higher energy/frequency/ colour it has. (Lower energy Infrared -> ROYGBV -> UltraViolet -> X-rays Higher Energy)

An element with more protons has a 'deeper' bottom shell. More protons => more attraction to electrons => more energy required to 'knock' up to a higher shell => higher energy/frequency photons.

So in general, a larger element would give higher energy photons off; in a flame test.
However it is a little more complex than that.

Because elements have many shells, an electron may jump back in more than one hop; giving off several different colours. For example Calcium has an orange flame colour, but heaver Strontium has a Scarlet red colour. If you look at a Strontium flame through a prism, you get Red, Green, and Violet.

Note: You may get infrared and ultraviolet in the mix too.

The ionization energy (IE) is qualitatively defined as the amount of energy required to remove the most loosely bound electron. Normally an electron in its outermost (filled) electron shell.
Lithium only has electrons in its bottom 2 shells. Cesium has electrons in its bottom 6 shells. Its outer most one has a single electron and is relatively "further away" from the protons in the nucleus. So it is easier to liberate Cesium's outer most electron than Lithium's outer most. In a flame test, if the flame is hot enough, electrons from deeper shells may be 'knocked' out of place and so you get higher energy photons.


Have the STRUCTURES of synthesized anthropomorphic molecules (nanoputians) been verified experimentally? Have the molecular structures of any synthesized anthropomorphic molecules (nanoputians) been verified experimentally?

This is a different question than this question as I am asking here about the structure, not the actual shape of the molecule. The importance of the distinction was pointed out in this helpful comment. Was there some property that was measured that confirmed that the correct, desired molecular structure had been in fact correctly synthesized?

The first article wikipedia's entry refers to (here) is not only a publication behind a paywall, but with freely available supporting information, too.  This in principle allows to replicate the work.

Browsing through the latter provides all / most / many details of how building blocks were prepared, how they were joined; including both numerical and graphical display of 1H and (broadband decoupled) 13C NMR spectral data.  Like the following, found on page 22 of the SI:

and 13C NMR (same page, below)

Perhaps (i.e. the following is a speculation) the "main" article in the journal includes additional spectroscopic data (UV, IR; MS) or characterisation (melting point, combustion analysis, etc.) as the article stretches from page 8750 to 8766.

To have a more rounded perception of this work done, it would be interesting to look at the echo by others in the community.  Which was the context this publication was cited?  Did the original researchers continue to work on these compounds, and which context?  In which ways other researchers than the authors echoed this work?  Because ACS's page indicates the publication was noticed in freely accessible media, it were worth to look up about this article in SciFinder, too.


Would then base react with water? Imagine you fill a tank with undecayed neutrons and water .Now neutrons will decay into proton and electrons. Assume that the wall of the container is so thick that it doesn't allow the protons to pass while allow beta rays to pass. The water will now have h + ions and they will make h3o + ions in water that is water will become acidic. Will now
1. A metal having normal reactivity will give H2 gas
2. NAOH react with solution to give water

Your proton-charged water will not last long enough to react selectivy with any electropositive metal or a base that you add.  Electrostatic forces are powerful.  The excess protons will grab electrons, generating ions that themselves grab other electrons, until eventually the beta particle electrons that were emitted to begin with get consumed to rebalance the charges.  You might get some unpredicted chemical reactions but no lingering excess charge.


Which unit cell do we obtain if spheres are arranged in ABABAB type in 2D and then that 2D layer is arranged in AAAAAA type in 3D? Arranging spheres ABABAB type in 2D gives us close packing. Now, this layer if arranged in ABABAB (in 3D) gives us HCP and if arranged in ABCABC (in 3D) gives us FCC.

What do we get if we arrange it in AAAAA type (in 3D)? (we are talking about one type of spheres only)

The question is a little confusing, so I will interpret your 'ABABAB' type in 2D to mean a simple hexagonal layer (so, the same layer as used for your hcp and fcc packing). So, a stacking of a simple hexagonal layer, each layer directly above the other, will give a structure looking like right hexagonal prisms - or right triangular prisms if you look a bit more carefully.

Anyway, the resulting structure is the simple hexagonal Bravais lattice with a single atom basis. Hcp is also a simple hexagonal Bravais lattice, but with a two atom basis, one each from plane A and plane B. I do not know of any elements that crystallize in this crystal structure (but am open to corrections!).


Why is hydrogen evolved at the copper side of a lemon battery? I have 2 quick questions about the principle of a lemon battery.
What I understand is that:

Zinc is oxidized, freeing up electrons that travel through an external wire to the copper electrode (to which they are pulled)
Once the electrons reach the copper electrode, they react with $\ce{H+}$ ions in the lemon, resulting in $\ce{H2}$ (gas).

Since the zinc electrode that generates electrons is already immersed in the lemon (containing $\ce{H+}$), why wouldn't the electrons just react at the zinc electrode with $\ce{H+}$ ions in the lemon? Why would they have to travel to the copper electrode before doing so?

Once the electrons reach the copper electrode, why wouldn't they react with the $\ce{Cu^2+}$ ions that have been eventually formed at the copper electrode (even though I understand there wouldn't be a lot of them since Copper is not very reactive)?

So you have a zinc electrode and a copper electrode and a lemon. The lemon is not a reagent grade chemical, so let's say it's the citric acid in the lemon that provides the corrosive action.

The copper essentially just sits there; as you indicated, the low reactivity of copper in acid means it doesn't dissolve readily. On the other hand, the zinc is much more reactive, and as you suggested, should just react all by itself. If it did, it would make a terrible battery; it would have a very short shelf life because the zinc would dissolve without providing an exterior current. But zinc has a high hydrogen overvoltage; that is, hydrogen does not readily evolve from zinc. The reasons are complex, but you may imagine that hydrogen ions or atoms are adhered to the zinc surface in such a way that they do not easily combine to form molecules which can form bubbles and go away. So the hydrogen evolution reaction is stifled.

However, the copper surface is different, and does permit hydrogen ions to react with an electron to form an atom which can then find another atom to form a molecule, and then more, to form bubbles. Alas! Copper has no extra electrons to spare; after you take off a few, the copper becomes positively charged and repels hydrogen atoms.

But the zinc electrode has gained a few electrons from the few zinc atoms that have dissolved in the citric acid. Let's connect the zinc electrode to the copper electrode: electrons flow from the zinc to the copper where they attach to hydrogen ions. The copper provides a surface for hydrogen ions to grab an electron and eventually become a bubble. The zinc provides electrons by dissolving as ions and leaving the electrons behind. And the battery could be useful for a long time if you disconnect the wire, because the zinc stops dissolving.

If this is still not clear, imagine a "battery" with a copper electrode and a sodium electrode. When you put the sodium into the lemon, it will react immediately without needing the copper to distribute its electrons. This "battery" might last about 2 seconds and would not be very useful.

And, as it turns out, impure zinc may have inclusions of different materials that provide a cathode surface right in the midst of the anode,  so the zinc could dissolve rather rapidly and make a poor anode. In the distant past, a tiny bit of mercury would be amalgamated onto the zinc, covering up the impurities. Mercury has an even higher hydrogen evolution overvoltage, so the zinc was protected. The electrochemical reaction then was the zinc dissolves in the mercury and travels to the mercury-solution interface before it ionizes and dissolves, but since the mercury layer was so thin, this was no problem. Eventually mercury was eliminated for potential human health reasons.


Eletrolytic rust removal I've seen quite a few youtube videos about electrolysis for rust removal. The process seems to be effective. My question is: how does it work?

I get that the water is ionised so we have O- and OH+ ions. I get that O2 is liberated at the anode. But what is going on at the cathode to release the rust? Is the rust chemically changed or is the rust somehow chemically released from the parent metal?

I'm a (rusty - graduated 1981) physicist, so don't hold back on the answer.

Electrolytic rust removal appears more like a semi-hoax to me. I think it is more like a mechanical removal of rust layer. As we dip a rusty object in sodium bicarb solution and you pass cathodic currents, we would see bubbling of hydrogen gas from the rusty object. These bubbles apparently help to loosen up the tight rust layer (bottom up). The cathodic current may also reduce some iron oxide.

The anode, which most videos suggest should be made of iron, is also useless in a sodium bicarbonate solution. Any inert conductor would have worked like a graphite anode. The reason is that the iron from the anode must go into the solution, if it were a true electroplating. However, in most videos iron oxides from the anod float as a precipitate of iron hydroxides. The whole solution turns rusty. Finally, people suggest a mechanical brushing or polishing. I am sure if a rusted object is polished with an abrasive, the results would be the same as electrochemical "rust" removal.

Now some over-the-counter rust stain removers are quite effective. They contain powerful reducing and perhaps some chelating agents. No, it is not oxalic acid, if that is coming to mind. These reducing agents immediately remove the rust by chemical means, but the iron surface is no longer shiny (greyish black). One has to polish it again to look shiny.


Why does technetium not exist in nature? Why does technetium not occur in nature? It is the only element before bismuth to net exist in nature. What's the reason for that?

Technetium does occur in nature. From the Los Alamos page on technetium:

Technetium was the first element to be produced artificially. Since its discovery, searches for the element in terrestrial material have been made. Finally in 1962, technetium-99 was isolated and identified in African pitchblende (a uranium rich ore) in extremely minute quantities as a spontaneous fission product of uranium-238 by B.T. Kenna and P.K. Kuroda.
Technetium has been found in the spectrum of S-, M-, and N-type stars, and its presence in stellar matter is leading to new theories of the production of heavy elements in the stars.

The journal article describing the 1962 discovery is Technetium in nature. B. T. Kenna and P. K. Kuroda.  Journal of Inorganic and Nuclear Chemistry 26(4) 493-499 (1964).


How many watts would I need to get CO2 from 30 °C to 4000 °C in 1 hour Scenario: After reading some articles on new sorts of ways to deal with $\ce{CO2}$ in the atmosphere, a question came to mind: If I had 100 kg of $\ce{CO2}$, and I would like to heat it up from 30 °C to 4000 °C in a given period (lets say 1 hour), how much energy would I need? What would that be in watts? And from there, assuming a solar panel with 20 % efficiency for 1 m² (therefore 200 watts/hour), what would be the area of solar panels needed to reach heat the $\ce{CO2}$?

Obs: From my very forgotten knowledge of Chemistry from high school, I found the specific heat for $\ce{CO2}$ should be 1000 to 1800 kJ/(kg K), and from this page http://www.heatweb.com/techtips/useful.html I got a formula and tried calculating it, but my units are probably all wrong, because the value was considerably low. I then multiplied 100 kg * 1500000 * 3970 °C. With the energy value, I divided by time (1 hour) to have the output in watts.

Obs2: I know that I am forgetting or do not know about many other limitations (such as the dissociation above a certain temperate). I also know that it would be inefficient and very costly, but would nonetheless like to find out the values, to try and figure what would be necessary for this procedure.

Question: Is this a viable way to do this calculation? What else should I consider?

I approached this in a similar way utilizing the heat capacity of $\ce{CO2}$ over a temperature range.

An estimation can be made by approximating the change in enthalpy of $\ce{CO2(g)}$ over a given temperature range. This can be accomplished through the integral form for changes in Enthalpy:

$$\Delta{H} = \bar{H_1} - \bar{H_2} = \int_{T_1}^{T_2}C_p(T)\,\mathrm dT$$

Which has units of [energy/mass].

Often times, the function $C_p(T)$ for certain gases and liquids will be tabulated in textbooks or chemical databases (such as NIST). In this case, I will utilize a function for the $C_p(T)$ of $\ce{CO2}$ in two temperature ranges (300–1200 K, 1200–6000 K)$^1$:

$$C_p(T)=A+BT+CT^2+DT^3+\frac{E}{T^2}$$

Range 1 (300–1200 K) coefficients:

$A=24.997,$ $B=55.187,$ $C=-33.691,$ $D=7.948,$ $E=-0.137$

Range 2 (1200–6000 K) coefficients:

$A=58.166,$ $B=2.720,$ $C=-0.492,$ $D=0.039,$ $E=-6.447$

Next, we can obtain the energy required for this temperature change by multiplying $\Delta{H}$ by the mass of $\ce{CO2}$ we're interested in, such that:

$$Q = m_\ce{CO2}\,\Delta{H} =  m_\ce{CO2}\int_{T_\text{initial}}^{T_\text{final}}C_p(T)\,\mathrm dT = m_\ce{CO2}\left[\int_{T_1}^{T_2}C_p(T)\,\mathrm dT + \int_{T_2}^{T_3}C_p(T)\,\mathrm dT\right]$$

Evaluating the integrals (keeping in mind that A, B, C, and D will change based on the temperature range in this case) yields:

$$m_\mathrm{CO2}\times7.266\times10^{12}\ \mathrm{\frac{J}{mol}}$$

$$\frac{100}{0.04401}\ \mathrm{\frac{kg}{\frac{kg}{mol}}}\times7.266\times10^{12}\ \mathrm{\frac{J}{mol}} = 1.651\times10^{16}\ \mathrm J$$

In watts, this value becomes:

$$\left(1.651\times10^{16}\ \mathrm J\right) \left(\frac{1}{3600}\ \mathrm{\frac{h}{s}}\right) = 4.560\times10^{12}\ \mathrm W$$

Calculating the solar-panned area required:

$$A_\text{panels} = \frac{4.560\times10^{12}}{200}\ \mathrm{\frac{W}{\frac{W}{m^2}}} = 2\times10^{10}\ \mathrm{m^2}$$

Source [1]: “Carbon Dioxide.” Carbon Dioxide, National Institute of Standards and Technology, webbook.nist.gov/cgi/cbook.cgi?ID=C124389&Mask=1#Thermo-Gas.


What organic solvents are suitable to use with potassium permanganate? I'm attempting to oxidize a substance with potassium permanganate and I was wondering what organic solvents I have at my disposal. The compound is soluble in chloroform and ethanol, but obviously those would not do well with a powerful oxidizing agent. The phase-transfer catalyst I'm going to employ is tetra-n-butylammonium bromide.

It sounds like you need a solvent that 1) does not react with KMnO4 and 2) is immiscible with water.

As @Janice DelMar said in her comment, some of the lower-chlorinated hydrocarbons (methylene chloride, 1,2-dichloroethane, 1,1,2-trichloroethane) should work. If you can live with a high boiling point solvent, you might consider chlorobenzene.

Other solvents that fit these criteria include most alkanes (pentane, hexane, cyclohexane, petroleum ether, etc.), benzene (but not toluene), and diethyl ether.

A quick test to determine compatibility of your solvent with KMnO4 involves a TLC stain. Make (or borrow) a batch of KMnO4 stain as described here. Dip a TLC plate into your solvent and then dip it into the stain with minimal drying between. If the plate turns brown (MnO2), then your solvent was oxidized and is no good. If the plate stays purple, then your solvent is resistant.


What happens if at equilibrium pressure is increased by decreasing the volume? I think, according to Le Châtelier's principle both the conditions i.e. increasing the pressure and decreasing the volume must shift reaction towards the side on which there are less amounts of gaseous molecules. And if there are the same stoichiometric coefficients for the gaseous substances (i.e. $∆n_g = 0$) then it must not have any effect on the reaction.
But increasing the pressure and decreasing the volume increases the number of collisions between the molecules of reacting substances and hence the concentration of products must increase.

Both these thoughts confuse me.

Questions I encountered:
$$ \ce{H2O(g) + CO(g) <=> H2(g) + CO2(g) + Heat }$$
In this question $∆n_g = 0$ so there must be no effect of increasing pressure and decreasing the volume. But the answer mentions that the concentration of both $\ce{CO}$ and $\ce{CO2}$ will increase.

There was also another question of same type:
$$ \ce{SO2Cl2(g) <=> SO2(g) + Cl2(g) }$$
Here the reaction must shift backward as pressure is increased and volume is decreased.
But again the answer mentions that concentration of all the gasses increases.

When a reaction is at equilibrium, the forward and reverse rates are equal. Decreasing the volume increases the concentration of all species (both reactants and products). This will result in a higher forward rate (because the concentration of reactants increased) and in a higher reverse rate (because the concentration of products increased). If forward and reverse rate increase by the same factor, the reaction remains at equilibrium.


Why is it possible, that an equilibrium shifts? So in class, I've learned about Equilibrium. So it is basically when the rate of the reverse and the forward reaction are the same. So I was like that sounds pretty easy, I can understand this. And then suddenly there was a twist (there always is in class). The teacher was like: so if you increase or add a stressor, the equilibrium will shift to stop it. How can equilibrium shift, I thought it was a concept not a "thing" to move up or down. Is it because it shifts to maintain the Equilibrium Constant?

I do not understand this. If this question has already been asked then please point me to that. I have even searched online and could not find an explanation in layman's terms.

Let's do a simple generic example. We have a reaction:

$$\ce{2A <=>[k_1][k_{-1}] B}$$

Experimentally we determine that the rates of the forward and reverse reactions are:

$$\text{Rate}_1=k_1[\ce{A}]^2$$
$$\text{Rate}_{-1}=k_{-1}[\ce{B}]$$

At equilibrium:

$$\text{Rate}_1=\text{Rate}_{-1}$$
$$k_1[\ce{A}]^2=k_{-1}[\ce{B}]$$

Since the rate constants are constants we can combine them into the equilibrium constant to generate the law of mass action:

$$K_{eq}=\frac{k_1}{k_{-1}}=\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}$$

Now, if $k_1=2k_{-1}$, we have a value of the equilibrium constant, which gives us the relationship between $[\ce{A}]_{eq}$ and $[\ce{B}]_{eq}$ that defines equilibrium:

$$K_{eq}=\frac{k_1}{k_{-1}}=\frac{2k_{-1}}{k_{-1}}=2$$
$$K_{eq}=\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}=2$$
$$2{[\ce{A}]_{eq}^2}=[\ce{B}]$$

In this case, only concentrations of $\ce{A}$ and $\ce{B}$ that satisfy $2{[\ce{A}]_{eq}^2}=[\ce{B}]$ constitute equilibrium. Other combinations of concentrations are not at equilibrium and therefore the forward and reverse rates are not equal.

Let's see how this works. An equilibrium situation is $[\ce{A}]=1\text{ M}$ and $[\ce{B}]=2\text{ M}$ because:

$$\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}=\frac{2}{1^2}=2$$

The rates are:

$$\text{Rate}_1=k_1[\ce{A}]^2=2k_{-1}[\ce{A}]^2=2k_{-1}(1^2)=2k_{-1}$$
$$\text{Rate}_{-1}=k_{-1}[\ce{B}]=k_{-1}(2)=2k_{-1}$$

However, what if $[\ce{A}]=[\ce{B}]=2\text{ M}$?

$$\frac{[\ce{B}]_{eq}}{[\ce{A}]_{eq}^2}=\frac{2}{2^2}=\frac{1}{2}\ne K_{eq}$$
$$\text{Rate}_1=k_1[\ce{A}]^2=2k_{-1}[\ce{A}]^2=2k_{-1}(2^2)=8k_{-1}$$
$$\text{Rate}_{-1}=k_{-1}[\ce{B}]=k_{-1}(2)=2k_{-1}$$

The concentrations, when plugged into the mass action expression, do not match the relationship defined by the equilibrium constant. More importantly, the rates are no longer equal. In this example, the rate of the forward reaction is now 4 times the rate of the reverse reaction. This is how equilibrium is reestablished. The forward reaction is going faster, but as it does so, $\ce{A}$ is consumed (slowing down the forward reaction) and $\ce{B}$ is produced (speeding up the reverse reaction) until the rates are equal again.

We can even do a bit of algebra to figure out what the concentrations of $\ce{A}$ and $\ce{B}$ will be once equilibrium is reestablished. Let's have $x$ represent the change in concentration of $\ce{B}$, so the change in the concentration of $\ce{A}$ is $-2x$.

$$K_{eq}=2=\frac{[\ce{B}]}{[\ce{A}]^2}=\frac{2+x}{(2-2x)^2}=\frac{2+x}{4-8x+4x^2}$$
$$2(4-8x+4x^2)=2+x$$
$$8-16x+8x^2=2+x$$
$$6-17x+8x^2=0$$
$$x=0.44695..., 1.6781...$$

In our case, we must use $x=0.44695$ since the other value would lead to a negative concentration for $\ce{A}$. Thus:

$$[\ce{A}]_{eq}=2.00-2(0.44695)=1.11$$
$$[\ce{B}]_{eq}=2.00+0.44695=2.45$$


Does the mechanism of AChE inhibition by Isoparathion depend on chirality? It is known that the inhibition of acetylcholinesterase by isomalathion can proceed either with diethyl succinate as the leaving group or thiomethyl, depending on the specific stereoisomer of isomalathion. Does the mechanism of inhibition of AChE by the Parathion isomer O,S-Diethyl O-4-Nitrophenyl thiophosphate (Or its dimethyl analog) likewise depend on the chirality of the molecule, or is 4-nitrophenyl the leaving group for both stereoisomers?
Reference:
Doorn, J. A., Thompson, C. M., Christner, R. B., & Richardson, R. J. (2003). Stereoselective Inactivation ofTorpedo californicaAcetylcholinesterase by Isomalathion:  Inhibitory Reactions with (1R)- and (1S)-Isomers Proceed by Different Mechanisms. Chemical Research in Toxicology, 16(8), 958–965. doi:10.1021/tx030026e

It would appear that the mechanism of inhibition of AChE by both stereoisomers of isoparathion (Or, at least, its methyl analog) proceeds through the expulsion of the 4-nitrophenoxy leaving group. Several papers have studied the inhibition kinetics of isomalathion and compared the results to those of isoparathion. All of them seem to agree that, at least for isoparathion methyl, AChE inhibition proceeds through the expulsion of the same leaving group, but the O,S-dimethyl phosphorylated AChE would be chiral, as is expected. Furthermore, the inhibition, reactivation, and aging reaction rate constants would also differ depending on the stereoisomer of isoparathion.
References:
Berkman, C. E., Quinn, D. A., & Thompson, C. M. (1993). Interaction of acetylcholinesterase with the enantiomers of malaoxon and isomalathion. Chemical Research in Toxicology, 6(5), 724–730. doi:10.1021/tx00035a019
https://pubs.acs.org/doi/pdf/10.1021/tx00023a004
Jianmongkol, S., Marable, B. R., Berkman, C. E., Talley, T. T., Thompson, C. M., & Richardson, R. J. (1999). Kinetic Evidence for Different Mechanisms of Acetylcholinesterase Inhibition by (1R)- and (1S)-Stereoisomers of Isomalathion. Toxicology and Applied Pharmacology, 155(1), 43–53. doi:10.1006/taap.1998.8608


How the Azimuthal quantum number describes 3d shape of the orbital? I read yesterday that Azimuthal Quantum number describes the shape of orbital....

what they call 'orbital' of an electron in chemistry is called wave function in physics. It determines how the electron moves and how it behaves. In fact it determines everything of the electrons. The azimuthal quantum number is just the quantum number used to describe angular part of the wave function, so it determines the shape of the orbital. This is a very rough explanation. You need to learn something like quantum chemistry or quantum mechanics in order to get a better understanding on that.


What an aqueous solution really contains? I want to know what an aqueous solution is. For example, let's consider a $1~\mathrm{M}$ aqueous solution of $\ce{MgCl2}$. Does it also contain $\ce{MgOH2}$? Though it is a simple question, I feel it is important to understand something as vital to chemistry as this.

Uh... I think the insight you need is about chemical equilibrium.
An aqueous solution is exactly the mixture formed when something dissolved in water. This is easy to understand, but here is a more difficult question: what's the component of this mixure?
To answer this is not easy.
In a glass of pure water, there are still a little $\ce {H+}$ (to say presicely, $\ce {H3O+}$) and $\ce {OH-}$. This is because sometimes proton ($\ce {H+}$) will "jump" from one $\ce {H2O}$ molecular to another one.
$\ce {2H2O <=> H3O+ + OH-}$.
(Since we usually do not use the expression $\ce {H3O+}$, this process can be rewrited as $\ce {H2O <=> H+ + OH-}$ to avoid disruption.)
This phenomenon (called the autoionization of the water) is very weak. The concentrations of $\ce{H+}$ and $\ce{OH-}$ in "pure water" is $\ce{[H+]=[OH^{-}]=}1.0 \times 10^{-7} M$ when the temperature is 25°C (298K).
Then you put something in this glass of water. As other people mentioned, most of covalent compounds, such as glucose, will not break apart in water since the intermolecular force (covalent bond) do not break. Water moleculars may surround those covalent compounds like a cage to make them dissolved. So there won't be other things.
But sometimes it breaks: when acid compounds are dissolved into water, like $\ce {HCl}$ and $\ce {H2SO4}$, they will give their hydrogen atoms as proton to water:
$\ce {HCl + H2O -> H3O+ + Cl-}$
$\ce {H2SO4 + H2O -> H3O+ + HSO4-}$
This is because they are more acidic than $\ce {H3O+}$, here "acid" means "who gives proton(s)", while "base" means "who accepts proton(s)" (This kind of definion is called as Brønsted-Lowry theory, and as you see, in autoionization of water, one of the $\ce {H2O}$ is acid and another is base).
Noticed that acid like $\ce {HCl}$ and $\ce {H2SO4}$ will nearly definitely give their $\ce {H+}$ to water compounds (they are too “willing” to do it), we call those acid as strong acid. But some acid are not: protons sometimes jump back.
For example, $\ce{HSO4-}$ will meet an equilibrium:
$\ce{HSO4- + H2O <=> H3O+ + SO4^2-}$
This kind of acid will be called as weak acid. In an aqueous solution of a weak acid, most of the weak acid compounds keep unchanged.
Yet we haven't mentioned what you concerned: how about aqueous solution of ionic compounds? They usually break their intermolecular forces (ionic bond) via hydration, i.e. combination with water compound. Water compound have some charges on them, so they will attract ions and break them apart. Especially, we will see bases we are familiar with -- those who will ionized as cations and $\ce {OH-}$ -- first.
In fact, the structure of hydrated cation will be complex ("$\ce {H3O+}$" is also not the actual situation), so we usually just write them as unhydrated form, such as $\ce {Fe^2+}$ or $\ce {Na+}$. As you know, there are weak bases and strong bases.
So bases like $\ce {NaOH}$ are quite "willing" to accept protons:
$\ce {NaOH -> Na+ + OH-}$
But $\ce {Fe(OH)3}$ may not, and it also meet an equilibrium:
$\ce {Fe(OH)3 <=> Fe^3+ + 3OH-}$
Again, in an aqueous solution of a weak acid, most of the weak base sturcture keep unchanged. And for this reason, weak bases are usually insoluable in water, since water "have no good idea" to deal with the whole relatively large ion sturcture. But in fact, the word "insoluable" means that there is another equilibrium like this:
$\ce {Fe(OH)3_{(aq)} <=> Fe(OH)3_{(s)}}$
and the solid form dominates.
So here we meet the hardest part: what about adding some salts into water, like $\ce {MgCl2}$ you mentioned?
So far you have learned about ionization of the salt, and you know most of salt will completely ionized. But how about those ions?
A balance like $\ce {Mg(OH)2 <=> Mg^2+ + 2OH-}$ shows that $\ce {Mg(OH)2}$ is a  weak base. But if we see it as $\ce {Mg^2+ + 2OH- <=> Mg(OH)2}$ it means $\ce {Mg^2+}$ is a weak acid. And if you remember I have said in autoionization of water, the water compound can be both base and acid (we say this kind of substances are amphoteric). Then we can know that there will be a new equilibrium like this:
$\ce {Mg^2+ + 2H2O <=> Mg(OH)2 + 2H+}$
(We usually call this as hydrolyzation of salt.)
And because $\ce {Mg^2+}$ is a weak base, you may expect that there will be not a lot $\ce {Mg(OH)2}$ and it's still soluable. Also there will be not a lot $\ce {H+}$ so the solution is just a bit more acidic then water.

Addition: You can also think this comes from these two balances:

$\ce{H2O <=> H+ + OH-}$ (autoionization).
$\ce{Mg(OH)2 <=> Mg^2+ + 2OH-}$. in further calculation you may meet in future,  this kind of idea can make you avoid some mistakes.

So you may expect in a $1M$ aqueous solution of $\ce {MgCl2}$, there will be a lot of $\ce {Cl-}$ ($2M$) and $\ce {Mg^2+}$ (a bit less than $2M$), a little $\ce {H+}$ and a little $\ce {Mg(OH)2}$, and there will be $\ce {OH-}$  which is negligible in the most case. (Of course there is very much water!)
While you can see in the case of $\ce {NaCl}$ there won't be $\ce {NaOH}$ and $\ce {HCl}$ compounds in the solution: because $\ce {NaOH}$ and $\ce {HCl}$ are "strong", there won't be a hydrolyzation equilibrium. But yes, you can say there are $\ce {H+}$ and $\ce {OH-}$, as you have known, there are $1.0 \times 10^{-7} M$ of them. However, we usually not concern about this. And as you see, all those $\ce {H+}$ and $\ce {OH-}$ in an aqueous solution, in effect, come from water compounds. (This is why there will be "strong" acid and base: because they are all giving/receiving protons from water, even if they actually have different ability, compared to water, they are behaving undistinguishably. This is called as leveling effect of solvent.)
There are a lot of things I haven't mentioned here, because I think you are not able to calculate something about equilibrium, and maybe you are not familiar with thermodynamic idea yet. But here I just hope what I said is interesting enough for you.
Well, a simple conclusion:

To find out what will be there in an aqueous solution, you may consider:

If the solute ionize: Is it a covalent compound? Is it acid?
If yes, whether the ions are weak acid or base (in Brønsted-Lowry theory)? Remember, a cation which can be formed from a weak base will be weak acid; while an anion which can be formed from a weak acid will be weak base. (Or you may use an awful word: they are conjugate acid base pair!)

I think these two step are enough for you for the time being.


How do the atomic orbitals look like in reality? We all know about atomic orbitals and their 3D structure. But the structure in theoretical. What does these orbital really looks like

Atomic orbitals are just the solution of Schrodinger's wave equation. They help us to understand better the behaviour of electrons. Electrons behave more like waves and atomic orbitals can be thought of as places where the probability of finding those electrons is maximum. In reality, there is no such place as an atomic orbital.


GAMESS crash course? I'm still learning about MO theory – and I thought that I would do some calculations with GAMESS to become more familiar with the concept. Even though I look forward to delving into the realm of quantum math in the not-so-distant future – it's very difficult to get a grip on all of the terminology. I'm using the GAMESS Avogadro extension to write some input files, but I don't know which options to use. Which options should I use if I want to study the orbitals of, say, shikimic acid? RHF, MP2, AM1 / MINI 6-31G(d)? Computer specs: 3.6GHz i5 with 16GB of RAM

First off, welcome to Chemistry.SE! For formatting help visit the help center and for more information about this page, take the tour.

While I do think you should learn some theory, it's definitely possible to learn some practical computational chemistry through experimentation. And yes, there are a lot of abbreviations to follow.

One thing you should track while you explore is how long different calculations take. In general, more accurate quantum chemical methods are also more computationally costly. This means that they take longer, but also as systems get bigger, the cost rises faster. You might hear some people talk about "linear scaling" (i.e., double the size of the system, twice as much time) or $N^2$, $N^3$, etc.

A good resource, IMHO, is Molecular Modeling Basics by Jensen. It's available through CRC Press but Jan also blogs extensively. He uses GAMESS and Avogadro but also Jmol and other programs.

While the Avogadro project is working on some more detailed tutorials, I'd suggest starting some calculations with HF/6-31G* or B3LYP/6-31G* and see how things go for different molecules. I strongly suggest running some calculations and then consider how the results compare to experiment. Benchmarking is important. Computational models all have systematic and other errors, and for real work it's critical to run a set of molecules and carefully choose methods that work well.

Here are a few key terms to consider:

Method: There are several types of methods.

Semiempirical methods: These include AM1, PM3, RM1, (and PM6, PM7, and others that I don't think are in GAMESS-US). Such methods are fast, since they approximate some integrals and use empirical parameters. They also have an implied basis set which is usually very small. (Avogadro doesn't support viewing semiempirical calculations with GAMESS yet.)
Wavefunction methods: These include HF and relatives. These are similar to typical quantum chemistry methods taught in school.

MP2, MP3, MP4: These are perturbation methods that improve on Hartree Fock and introduce some level of electron correlation.
CCSD, CCSD(T), etc.: These are "coupled cluster" methods that are highly accurate, but often slow, and do not handle large molecules.

Density Functional: This includes another large set of methods, including BLYP, B3LYP, and newer methods like M06-X. Rather than approach things through the Schrödinger equation,

Basis: Most methods (except AM1, RM1, PM3, etc.) also require a "basis set," a set of functions to describe the atomic orbitals. Wikipedia article There are a few families, including:

Pople Type: 3-21G(d), 6-31G(d), 6-311+G(d), etc. These are older, more traditional basis sets. Generally 6-31G(d) or 6-31G* is considered minimally acceptable today. The "*" or (p,d,f) included at the end indicate that higher polarization is included on some or all atoms.
Correlation Consistent: cc-pVDZ, aug-cc-pVTZ, etc. These are newer basis sets, designed to function better with methods that treat electron correlation. The "aug-" includes various diffuse functions, which are useful particularly for anions and electron density that is particularly, well, diffuse. (In Pople basis sets, these are indicated with "+".


How to convert the D'Arcy and Watt model from activity to concentration in mol/m^3 The D'Arcy and Watt 1970 model in "Analysis of sorption isotherms of non-homogeneous sorbents" for water adsorption is as follows:

$$W = \frac{KK'a}{1+Ka} + ca + \frac{kk'a}{1-ka},$$

where:

$W$ is weight of vapour adsorbed per $1~\pu{g}$ of substrate per $1~\pu{g}$ sorbent ($\pu{g/g}$)
$a$ is the water activity or relative humidity (no units, number between $0$ and $1$ or if relative humidity is used - is percentage)
rest are constants with no units

I want to use this model to consider water concentration in $\pu{mol m^-3}$ instead of water activity.

How do I convert this?

The general relation between activity (effective concentration) and concentration is as follows (see Wikipedia for more detail):
$$a_i = \gamma_i \frac{c_i}{c^\circ}$$
The standard concentration is defined as $c^\circ=1~\pu{mol dm^-3}$. For extremely dilute systems it is possible to just use the concentration instead of the activity (often used as a first order approximation, too):
$$\lim_{c_i\to0}\gamma_i\approx1 \implies a_i\to c_i$$
Depending on the accuracy necessary for your model, you might have to look up the specific values.

In any case you should again look up the definition of activity (or relative humidity) in D'Arcy and Watt's paper, because the activity defined for the above relation could be larger than 1.


Why is quinone mono oxime more stable than 4-nitrosophenol? When we do nitrosation of phenols, we get para-nitrosophenol as the major product, which tautomerises to quinone mono oxime. This article states that it largely exists in the quinonoid form.
My question is, why? I know that nitroso form is less stable than the oxime form, and enol is less stable than keto, but on the other hand we are losing the aromaticity also. Why is being non aromatic getting more favourable than being aromatic in this case?
I read the post shared by @Safdar in the comments, but this answer there says that both forms are aromatic. Is this true?

Two factors come into play: (1) the inherent difference in resonance energy is not as great as you might think, and (2) the greater tendency for a dissociable proton to bind more tightly to an oxime function rather than a phenol function appears to counterbalance the reduced $\pi$-electron bonding difference.
Less than meets the eye
Suppose you were to use the Hückel model to estimate the pi-electron binding energy for several eight-carbon conjugated systems.  By that method I find the following results:

For the conjugated straight chain $\ce{CH2=(CH=)_6CH2}$ the total $\pi$-electron binding energy works out to $9.52\beta$, where in this theory $2\beta$ represents one isolated $\pi$ bond.

For styrene, with a $\pi$-electron pattern matching the nitroso-phenol, the result is $10.53\beta$.  A difference of about one $\beta$ unit is typical when an aromatic ring is compared with the opened ring.

For para-xylylene whose quinonoid $\pi$-electron structure matches the oxime tautomer described here, we get $9.92\beta$.

Of course the aromatic structure is most stable, but not all non-aromatic structures are created equal.  The Hückel-model calculations indicate that even though the quinonoid structure is typically called non-aromatic, in fact it captures a significant part (about $40$% based on the numbers) of the extra $\pi$-electron stabilization we usually associate with an aromatic ring.  The energetic difference between the two competing $\pi$-electron configurations is thereby less than the "aromatic" and "non-aromatic" labels would seem to imply.
Swimming upstream
Now let's see where the dissociable proton is more stable.  To do that we look up the $pK_a$ values typically associated with each prospective acidic functiin:

Phenol gives $pK_a=9.95$.

Acetoxime, $\ce{(CH3)_2C=NOH}$, gives $pK_a=12.2$.

The oxime function is clearly holding the proton more tightly; placing the acidic proton there instead of on the phenol function saves $2.2$ powers of ten in the equilibrium constant.
We cannot directly compare this with the Hückel-model calculations from the previous section, because that model is only semi quantitative and a more complex calculation would be needed to capture the effects of the heteroatoms in the conjugated structure.  But the results do suggest that the combination of the quinonoid structure capturing part of what is usually aromatic $\pi$-electron stabilization with the preference of a dissociable proton to bind to the oxime function leads the oxime tautomer to win out over the nitroso-phenol.


How to calculate what mass of halothane is inspired in one minute given the partial pressure and temperature? If the anesthetic mixture is inspired at the rate of $\pu{100 mL/min}$, what mass of halothane, $\ce{CHCIBrCF3}$, molecular weight $M = \pu{197.4 g/mol}$, is inspired in one minute if the partial pressure of halothane is $\pu{7.6 torr}$ and the temperature is $\pu{21 ^\circ C}$.

  $\pu{0.08g}$
  $\pu{0.80g}$
  $\pu{1.80g}$
  $\pu{3.36g}$

My working:
\begin{align}
P &= \pu{0.01 atm} &&(\pu{1 atm} = \pu{760 torr})\\
T &= \pu{294 K}    &&(273 + 21) \\
V &= \pu{1 L}      &&(\pu{1000 mL} = \pu{1 L})\\
R &= 0.08          &&(\text{L-atm})\\
\end{align}
Equation:
\begin{align}
  0.01 \times 1 &= n \times 0.08 \times 294\\
  0.01 &= n \times 23.52\\
  n &= 4.3
\end{align}

Which suggest $\pu{4.3 mol}$ of $\ce{CHCIBrCF3}$ which would give the mass $\approx\pu{849g}$.

That is way off any of the answers. Can someone see what I have done wrong?

Volumetric Flow Rate (VFR) is given to you.
By applying ideal gas equation to the gas, in terms of volumetric flow rate,
$$\frac{PV}{t} = \frac{nRT}{t}$$

And in an ideal gas mixture (the assumption),
the partial pressure of a gas is what is exerts alone in same volume in same temperature as of the mixture.
$$\frac{P_\mathrm{gas}\cdot V_\mathrm{tot}}{t} =
\frac{n_\mathrm{gas}\cdot R\cdot T_\mathrm{total}}{t}$$

amount of substance = mass of the compound/molar mass (or molecular weight of the compound)
$$n = \frac{m}{M}$$

$$\frac{P_\mathrm{gas}\cdot V_\mathrm{total}}{t} =
 \frac{(\frac{m_\mathrm{gas}}{M})\cdot R\cdot T_\mathrm{total}}{t}$$

Apply the values, mind the units:
\begin{align}
V &=\pu{0.1 L}\\
T &= (273+21)\pu{ K}\\
P_\mathrm{gas} &= \pu{0.01 atm}\\
R &=\pu{0.082057 L atm mol^-1K^-1}\\
M_\mathrm{gas} &=\pu{197.4 g mol^-1}\\
t &= \pu{1 min}\\
\end{align}

$$\frac{\pu{0.01 atm}\cdot\pu{0.1 L}}{\pu{1 min}} =
\frac{
  \frac{m_\mathrm{gas}}{\pu{197.4 g mol^-1}}
  \cdot\pu{0.082057 L atm mol^-1 K^-1}\cdot\pu{294 K}}{\pu{1 min}}$$

For $m_\mathrm{gas}$ I can see the value in your multiple choice.


Why does changing the temperature shift the equilibrium? I know that raising the temperature in a reversible chemical reaction causes the equilibrium to shift to the endothermic side.
I know that $\Delta G = \Delta H − T\Delta S$, but I don't know how to put them together to show why the equilibrium would shift.
To summarise, I already know what happens when, and I'm interested in why.

Suppose you have some reaction
$$\ce{A <=> B}.$$
The equilibrium constant for the reaction is $K$ and the Gibbs free energy change is $\Delta G$. The equilibrium constant is
$$K = \frac{[\ce{B}]}{[\ce{A}]}.$$
Increasing value of $K$ shifts the equilibrium towards the right, i.e. more $\ce{B}$, and reducing the value shifts it to the left, i.e. more $\ce{A}$. The equilibrium constant and the Gibbs free energy are related by
$$ K = \exp\left(-\frac{\Delta G}{RT}\right)$$
Putting in the expression you give for $\Delta G$ gives us
\begin{align}
 K &= \exp\left(-\frac{\Delta H - T\Delta S}{RT}\right) \\
   &= \exp\left(-\frac{\Delta H}{RT}\right)\,
      \exp\left(\frac{\Delta S}{R}\right).
\end{align}
If we assume the variation of $\Delta H$ and $\Delta S$ is small we can ignore the entropy term because it's a constant, and we get
$$K \propto \exp\left(-\frac{\Delta H}{RT}\right).$$
If the reaction is endothermic $\Delta H$ is positive, so we have the exponential of a negative number and this is less than one. If we increase the temperature we decrease $\Delta H/(RT)$ and the exponential increases. So increasing the temperature makes the equilibrium coefficient bigger, i.e. it drives the endothermic reaction.
To understand why this happens consider what happens when we make the temperature very high.  If you do this the value of $\Delta H/(RT)$ approaches zero and the exponential approaches unity, i.e. the concentrations of $\ce{A}$ and $\ce{B}$ are the same. What we've done is make the thermal energy much greater than the activation energy in either direction, so the rates of the forward and backward reactions are the same.


What are possible hazards of fog machines? From a chemistry point of view I would like to learn about possible hazards of fog machines.

Such devices, also called “foggers”, create vapor which appears like fog or smoke. It is used in discos, parties, etc.
It works by heating “fog juices” which are made of glycerine and distilled/de-ionized water mix or propylene glycol and distilled water mix.
I don't know the temperature these mixes are heated by the machines.

I know glycerine and propylene glycol are not hazardous, but I would like to know if heating them can pose any risk. I've talked with someone who works in this industry and he couldn't tell if there has been any study to show if the fog produced by these foggers is harmful or not.
So even though these devices have existed for years I would still like to know if they can be harmful and if they should be avoided when possible.

How you can help as chemist is by telling me if vaporizing glycerine and distilled water mix at certain temperature can cause any change to the supposedly harmful substance and make it dangerous or not.

Given that these machines have been used extensively in the entertainment industry for decades, and one (Rosco Fogger) even has an Academy Award, common sense says that they aren't dangerous.

Breathing in copious quantities of the fog causes irritation, but that applies to a lot of things. Dry ice fog[1] is basically water vapour - fill the room with it and you die from asphyxiation from the CO2

looks better in many cases as it's "fog" and not "smoke"