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How much CO2 could you remove from the atmosphere with 2 trillion USD?
I know it's possible to capture $\ce{CO2}$ with various chemical reactions. For example NASA's space shuttle had some kind of regenerative $\ce{CO2}$ scrubber. But how expensive is it? Could a huge number of these devices, or something like it, significantly reduce the atmosphere's $\ce{CO2}$ level?
I realize this is probably not practical, but it's an interesting thought experiment. To make my question more specific: can anyone show some rough calculations of how much $\ce{CO2}$ you could remove if you had, say, 2 trillion USD to spend? To simplify, assume that energy is at current prices, but clean. (i.e., if the device requires electricity, assume it's coming from a nuclear power plant, solar farm, etc.).

Humans are increasingly producing $\ce{CO2}$, and there's never a world-wide agreement for using cleaner methods. Even if we could, we'd end up cleaning in circles.

How about we plant a bunch of trees?

Why not use non-toxic, solar powered, environment friendly, self sustaining and duplicating CO2 scrubbers? We just need more of them.

Algae is a much more efficient CO2 scrubber than trees.  http://www.treehugger.com/urban-design/microalgae-lamp-absorbs-150-200-times-more-co2-tree-video.html

Start with reading what the IPCC has to say: 1, 2. The only scenarios seriously considered are Bioenergy with carbon dioxide capture and storage  and afforestation. However, I don't think 2 trillion USD is actually a very large amount of money in this context. You need to consider indirect (e.g., socio-economic) costs.

All other scenarios are considered to have too large associated risks and "biogeochemical, technical and societal limitations".

How about stopping CO2 release to the atmosphere with those 2 trillion in the first place? For example buy as much coal power plants as you can and let them idle.

@SpaceTrucker If you try that, expect government to dispossess you. The majority of the coal power plants run in countries that really need energy (think China). However, building infrastructure to produce cheap renewable energy might be a good idea (if you don't need a return on investment).

This is very hard to answer precisely, as there are many different carbon capture strategies, and economics at the scale required is quite different from our normal understanding. However, I'd love to see some attempts to at least get order of magnitude estimates, or sources with more in-depth analyses.
Here is an implementation of carbon capture and storage which is useful to consider for its simplicity rather than its real-world applicability, to give a sense of scale. One way to remove anthropogenic $\ce{CO2}$ is to do "inverse combustion", more specifically:
$$\ce{CO2(g) -> C(s) + O2(g)}\ \ \ \ \ \ \ \mathrm{\Delta H=+390\ kJ/mol}$$
Assume this process can be done with perfect efficiency, and that the only energy expense in the process is to drive the reaction forwards (that is, zero energy consumed in transportation, collection, construction, etc). According to this source, the amount of anthropogenic $\ce{CO2}$ emissions from 1750 to 2008 has totalled about $1250\times 10^9\ \mathrm{t_{CO_2}}$. Suppose you wish to remove all this carbon dioxide (only about half is in the atmosphere, the rest is trapped in the ocean or land) using the above process. This would require about $\mathrm{10^{22}\ J}$ of energy, which Wolfram Alpha suggests is about 50% more energy than can be retrieved from combustion of all global proven oil reserves in 2003. Put another way, this is about 20 times the world energy consumption in 2012, or 150 times world electrical energy production. This would put the cost of this process in the range of hundreds of trillions of US dollars, meaning two trillion USD barely makes a dent.
Is there some other process where two trillion USD is close to enough? I seriously doubt it.
Edit: Several comments and answers mention biological sequestration, which is a legitimate carbon capture strategy. I did not consider it, however, because its costs are far more complicated to calculate. My intention with this answer was to find a quick and comparatively simple way to attach an energy cost to carbon sequestration. Whether the monetary cost even makes sense at this scale (how do you define "monetary cost" when it's larger than world GDP?), I don't know.
But here's another amusing comparison, which should somewhat temper hopes that biological sequestration will be a magic bullet. The amount of anthropogenic carbon released between 1750 and 2008, ~350 billion metric tons of carbon, is comparable to a significant amount of the biomass on Earth (all eukaryotic life contains approximately 560 billion metric tons of carbon, multicellular life is a fraction of this). Thus, biological sequestration of the majority of anthropogenic carbon would be broadly equivalent to sacrificing all eukaryotic life on Earth (or ~10-30% of all living organisms by mass) in order to collect and bury carbon, then seeding the Earth back to its current biological state.
Over 250 years of constant stimulation to produce as much energy/goods from fossil fuels, we have released a lot of carbon.

It basically boils down to the fact that we'd have to 'unburn' all the fossil fuels we've used so far, and then some, and that energy has to come from something other than said fossil fuels. Not easy to pull off.....

This answer neglects the fact that trees exist and can fix CO2 for lower energy requirements, being solar powered.

We can cut that enthalpy cost by a factor of about two (179 kJ/mol) if we sequester as limestone instead of carbon (and reuse all your simplifying assumptions).  [ @MarchHo correctly observes that I conflated enthalpy of reaction with enthalpy of activation, although neither of us used these words.  So the catalysis bit is retracted.]

@EricTowers Catalysts don't change the enthalpy, so your catalyst argument is incorrect.

@MarchHo : Photosynthesis requires 450 kcal/mol, so is not particularly competitive.  (It's at best 26% efficient.  There are experimental solar cells over 40% efficient.  Consequently, there are better ways to capture solar energy in technologically useful form than plants of any form.)

@EricTowers but is there a better way than photosynthesis to capture solar energy AND simultaneously sequester carbon?

The energy requirement of photosynthesis are irrelevant, since they are supplied by the sun and not by humans.

"there are better ways to capture solar energy in technologically useful form than plants of any form" -- and yet we eat plants (and animals fed on plants), we don't eat sugars produced from water and atmospheric CO2 via these better technological ways. Sure, the question is founded in idle speculation, and begs for us to assume technology that doesn't actually exist, but that doesn't mean it's illegitimate to involve technologies that are inferior in every way other than their actual existence ;-)

@Timbo : That is the question that was posed.  But if you're thinking plants, you shouldn't tie your hands by having two or three times the energy requirement than the alternatives and substantially less power uptake efficiency (real plant efficiencies are 0.1-2% of incident solar is captured, so this is already an order of magnitude worse and is too optimistic because the plant natters about using this energy for other things as well).

@SteveJessop : Let's develop these technologies, then see if it takes thousands of years for us to evolve into creatures that can leverage these energy sources.  (I suspect not.)

@MarchHo We're already using that - whenever you use wood for something else than burning, you're capturing carbon. So send the message - cut down more trees! Build more wooden stuff! Of course, it's still temporary - eventually, the wood will rot and release the stored carbon. Unless you make sure it doesn't have access to oxygen... hmm, perhaps we could stuff the wood in man-made bogs and leave them to decompose to peat and coal... :P

@Luaan: that's basically the idea of carbon capture: pick your form of carbon and pick your place to bury it where it won't come back out any time soon. Charcoal in mines is probably a better bet than wood in man-made bogs, though. With the given $2 trillion budget I don't know whether you'd sink the most carbon by burying sustainably-sourced charcoal (Dewi's answer) or by developing chemical means of carbon capture (this answer).

[Stressing that I wish to mention this on an apolitical, intellectual level but there is an abstract political component] The most troublesome thing about this is the economic scale required to even make such a thing close to feasible requires central expropriation & allocation of resources to the project (unless someone has a few hundred trillion dollars around they wish to gift) & expropriation necessarily requires slavery.Removing market dynamics also occludes the inherent quality of the value of the project & is quite likely a gross misappropriation of unrecoverable resources.

@K.AlanBates: although the question is about what can be done with $2 trillion, whereas this answer is about how to sink all the carbon. $2 trillion is less than the US military budget for a single Presidential term, so I don't think it would be possible to state that such a level of state intervention is "necessarily slavery" without an argument rejecting taxation in principle, which is bound to be political. 100+ times as much, sure, you have to wonder where the resources could possibly come from under any current systems and so the argument is the same regardless of one's politics.

@SteveJessop Yes, taxation is slavery and that is -unequivocally- not a political statement (by the common usage of the term); to declare anything to the contrary is either a rejection that such a thing called "slavery" exists or is subject to the sorites paradox. Assuming slavery to be a valuable thing for purposes of discussion, there still exists the conundrum of the Socialist Calculation problem, where a cabal's assertion that a course is necessary towards an outcome they internally value is neither checked nor balanced against the value scales of individuals outside the cabal.

"Assume this process can be done with perfect efficiency" This is a totally unscientific assumption going against both basic chemistry and thermodynamics. This chemical reaction doesn't exist, and if it did, it would require much more energy than the enthalpy difference between products and starting materials.

It's... feasible. There are a number of technologies that are being considered. Costs will be high, some number of billions or trillions: you're talking planetary engineering, here.
The most obvious option is to plant trees. The obvious problem with that is that even planting them at the same speed as they are being removed is infeasible. Unfortunately trees csan store only a limited amount of carbon; once a forest's got as crowded as it can, it's maxed out its carbon sequestration capacity, and will never store another gram.
An option that addresses this is to heat crop residue, storing the "biochar", the charcoal from the burned crops, in landfill or even in farmland to enrich the soil. The output of biochar systems is typically around 20% char, 20% usable biogas, 60% usable bio-oil, making them both a net energy source, and a carbon sink.
This system is being considered as an additional revenue stream for the sugarcane growers of Brazil, where it would (if universally adopted) sequester about 330,000,000 tons of carbon annually.

Out of curiosity, I did a quick calculation of how much area would have to covered by Amazonian forest to sink the amount of $\ce{CO2}$ mentioned in my answer. Using a value of 61 petagrams of carbon per 424 million hectares for above-ground carbon density of the Amazon shown in this article, we reach an area ~40% larger than Russia, which means 16% of the land area on Earth covered in dense jungle. Doesn't sound possible.

Good point. Forestry's arguably possible in theory, but not economically sane in today's world.

You don't need to grow them all at the same time. Each crop is (ideally chared first, then) burried, say in an old coal mine, and then the land used for another crop.   A first step would be to stop burnimg fossil coal in the first place.

Yes: as an alternative to mere forestry, biochar has the advantage that every year's  crops can become part of the carbon sequestration, which makes it, to me, a very economically sensible part of the toolset to clean up atmospheric CO2 (it'll take many tools: biochar, artificial trees, scrubbing towers, ocean seeding, etc). Char needn't all be put into landfill or used-up mines either: it can be ploughed back into the earth to help drainage, used in carbon filters, etc.
It could also be used in the place of regular charcoal and burned, of course: sadly the most likely outcome, economically.

The forests couldn't do it because they'd be CO2 starved. All plants are currently CO2 starved air flow is needed to keep growing but trying to grow a vast forest or any other plant solution would leave them stunted.

"Give me a half a tanker of iron and I will give you another ice age"

That's the claim of people who believe in iron seeding the ocean. The link claims that "the addition of silicic acid or choosing the proper location could, at least theoretically, eliminate and exceed all man-made CO2", but no citation is given. As to cost?

"Current estimates of the amount of iron required to restore all the lost plankton and sequester 3 gigatons/year of CO
  2 range widely, from approximately 2 hundred thousand tons/year to over 4 million tons/year. The latter scenario involves 16 supertanker loads of iron and a projected cost of approximately €20 billion ($27 billion)."

but again, citation needed!

That's pretty much free compared to all the other attempts (including those we are actually trying, and including the huge costs sunk in designing more efficient devices etc.). If they actually had good evidence this would work (and of course, that it wouldn't cause bigger issues which is always a possibility), cost wouldn't be the thing that would stop this. And of course, eventually, the carbon will either return (as fish breed out of control with their new cheap source of food) or drop to the ocean floor (possibly creating more trouble in the future).

@Luaan Experiments have been conducted several times. Iron fertilization: Experiments - and yes, there are questions of other issues and the definition of sequestration.

My own guess is that the human population will increase so much over the next 200 years that "farming the ocean" will be necessary to prevent mass starvation. So iron seeding is going to go forward no matter what.

@user14717 That's a common theme in human history. The starvation part anyway - it hasn't come to pass quite yet, no matter how strongly the guys believed in their pet hypothesis. So far, we've always been a few steps ahead of that technologically. The fixed thermodynamic limits are probably going to make human population infeasible long before we hit limits in food production (unless we somehow lose the food sources we already have, which is of course possible - we're getting better and better in self-contained food production, though).

the $64,000 Question is ... Is it economically feasible to use a 'synthetic' process over a naturally occurring one. In this case, No. Zooplankton are the prime movers of CO2 in our atmosphere. They basically turn gaseous CO2 into Calcium Carbonate a lab phrase for sea shells. Increasing the number of Zooplankton is key to the process. UV and oil slicks and the Sargasso seas of plastic are lowering those numbers.
so the answer to the question is what process would one employ that was the most cost effective solution to the problem, clearly A. clean up the oceans and promote natural zooplankton growth or B. mimic the metabolic pathway that results in the production of calcium carbonate, as a GM/organic solution. Either A or B is a far better and sustainable solution that does not reintroduce/produce CO2 for the energy to power an artificial non-organic/biologic method.

Welcome to chem.SE! This is one borderline answer. I wouldn't know if it's VLQ or not. But it would pay if you be more elaborate.

This states part of the problem (without references) but doesn't provide any solutions so it doesn't really answer the question.


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How to convert ethanal to 2-hydroxy-3 butenoic acid?
How to convert ethanal to 2-hydroxy-3 butenoic acid?
I tried crossed aldol condensation with formaldehyde followed by acid addition.After that how should I proceed ?

Note that 2-hydroxybut-3-enoic acid has four carbon atoms and ethanal has two. What is the product of the aldol self-condensation of ethanal? Does that give you any hints?

Once you have got your acrolein from the aldol condensation of ethanal and formaldehyde you can react it with $\ce{NaCN, HCN}$ under kinetic control to get direct addition of the cyanide to the enone, forming a cyanohydrin[1]. The nitrile can then be hydrolysed in aqueous acid to yield the carboxylic acid.

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How much energy is required to make an ion?
So say that I have a normal hydrogen atom (not an isotope, just regular normal hydrogen atom) and I want to make it into an ion $\text{H}^+$. Since to make an ion of that formula I need to take away an electron, it is just the same as giving the electron energy for it to transition from quantum energy level 1 to level $\infty$, so I use the energy level formula for the hydrogen atom $E_n = -\frac{1312}{n^2}$ and substitute $n$ for $1$ and $\infty$ to get $-1312 \text{ kJ/mol}$ and $0\text{ kJ/mol}$ respectively. So if I add 1312 $\text{kJ/mol}$ of energy to the electron it should leave the atom, right? But my textbook says it is wrong.

What value does your textbook give?

This was actually a multiple choice question that gives a bunch of choices and asks which of these choices are correct. However, when I got to this choice and solved it out, I got the answer like above and the choice to me was correct. The textbook however, said a different choice was the correct choice.

Might be a sign error - meaning, you have to put in 1312 kJ/mol, so it would be a positive and not negative number. Was that a choice?

No, I got the answer that the choice gave, and it was positive.

I am baffled. Perhaps someone who is more knowledgeable than me will step in and set us both straight.

Is the problem that you need the energy for a single ion rather than a mole?

I don't think so, because 1312 kJ per mol is a lot of energy, andthe formula for an individual atom has a smaller denominator.

Not sure what you mean by smaller denominator. If the units are the same, clearly the numerator will be smaller if you're only looking at one atom rather than a mole of them. Can you calculate the energy for a single atom?

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What are all the stereoisomers of 1-ethyl-3-methylcyclohexane?
What are all the stereoisomers of 1-ethyl-3-methylcyclohexane? The following image shows my attempt to answer the question. But I am not sure if I am correct.

Note: I am treating this cyclohexane as flat since it's confomers interconvert rapidly.

For me, your answer seems right ... but you should think of other abbreviations for ethyl and methyl groups.
$\hskip2.5in$
Strictly there are eight missing hydrogens per structure that you should add for clarification: three for each methyl- and five for each ethyl-group. C or C$_2$ is not a common (and as for C$_2$ not a good*) abbreviation for such groups ... Me and Et or CH$_3$ and C$_2$H$_5$ would be more usual.

* e.g., C$_2$ could be $\ce{-C#CH}$ or $\ce{-CH=CH2}$

You forgot $\ce{-CH3}$ :P

@Martin-マーチン Wavy bonds ~~~


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Mark all stereocenters in ampicillin
Shouldn't the N pointed to by the red arrow also be a stereocenter? I am asking this because I think it's lone pair cannot take part in nitrogen inversion.

This is a very good question.  I'm not an expert at N stereochemistry but have you considered the iminol form of the amide/iminol tautomer?  It would have a double-bond to the N and would not be chiral.  I know the amide form predominates but if there is any contribution from the iminol it would "destroy" the chirality wouldn't it?

Why do you think that nitrogen inversion cannot take place there?

Probably because N is part of a ring but I am not sure if any strain is involved?

The two connected rings and the attached sterically demanding groups will certainly hinder inversion

Amide groups are flat! There's no point in talking about inversion.

After reading the inputs from @Curt F., @PH13, @Martin and @Mithoron, I infer that N won't be involved in nitrogen inversion due to hindrance caused by connected rings and attached sterically demanding groups. Hence, 'N' pointed to by the red arrow will be considered as a stereocenter in ampicillin.

You didn't get my comment - it's flat so there can't be stereocenter!

@Mithoron Correct me if I am wrong, you said that amides are flat and therefore nitrogen inversion won't be possible, right? Now, the absence of nitrogen inversion facilitates in N being a stereocenter given that all the substituents attached to it are different as well (which is also satisfied in this particular case).

All substituents of N atom are in one plane - thats why speaking of prescence of inversion or lack of inversion doesn't make sense.

@Mithoron So, are you implying that a 3-D geometry is a requirement for an atom to be a stereocenter? Because I don't see it in any of the definintion of stereocenter I found on the internet. One of the legit sources being: http://www.chem.ucla.edu/harding/tutorials/stereochem/id_stereocenter.pdf

@Mithoron I think you should write up an answer. Also everyone please avoid extended discussions in comments. Use the chatroom instead.

@theprogrammer see discussion starting here http://chat.stackexchange.com/transcript/message/23058334#23058334

I edited your answer - as you could see in discussion I linked the amide group won't be flat.

Oh yeah! I forgot to change that! Thanks bro. :)


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Is there any food-safe way to break down capsaicin (chilli) so food is less spicy?
That is, is there a chemical or enzymatic reaction with capsaicin that break it down into other food-safe chemicals? Or alternatively, bind to capsaicin such that it doesn't trigger the same spicy receptors in our mouths.
I know about how casein can be used to wash capsaicin away, but it doesn't actually change it into another form or deactivate it.

And why would you want to do that? Chillis are awesome.

I love chillies too but I can't handle too much. Also my kids can't handle any at all. It would be nice to be able to reduce the spiciness of some dishes if you added too much chillis.

It is an interesting problem, from my experience the capsaicin behaves in a way which is consistant with it being a liphilic substance. I know that if you fry chilli pueppers with some cheese that the cheese helps to bring out the flavour of chilli peppers.
I know that anacdotal evidence suggests that a dairy product such as milk or yougurt helps to stop your mouth burning. I think that this is reasonable if we assume that the chilli pepper toxin dissolves in the fatty droplets in the milk or yogurt. This might make it a better mouth decontamination liquid than water.
I think that the way to test this would be to put known amounts of capsaicin on filter paers and then soak them in milk and water and then measure the capsaicin concentration in the liquids after soaking. It could be done with HPLC or if we were able to get C14 labled compound we could use liquid scintillation counting.
From my student days I recall that a vegatarian or sasuage chilli was hotter than one made with minced beef. So it is possible that the capsaicin adsorbs onto the surface of the beef. Maybe you should consider using the langmuir isotherm to model such a thing.
As a simple alternative just put less chilli in the food.

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Why atomic masses aren't integers?
Most of the elements have isotopes, so the atomic masses are calculated depending on the percentage of the existing isotopes. That is clear. However, what about elements that have only one isotope (monoisotopic) - like fluorine? Shouldn't the atomic mass for it be a whole number and not 18.9984?

I haven't got time to write a full answer but look up binding energy. I think that answers your question.

In particular https://en.wikipedia.org/wiki/Nuclear_binding_energy#Mass_defect .

Here is a relevant question.

The short answer is nuclear binding energy, which is the energy needed to disassemble an atom into its subatomic parts (or in some cases the energy released when this happens). The binding energy is a consequence of the strong and weak nuclear forces that hold atoms together.
Where does this energy come from? It comes from the mass of the nucleons! What? It is true; most atoms have less mass than the sum of their parts, and that mass defect is converted into the energy that holds them together. If you do not believe me, let us look at one atom of carbon-12, which is used as the definition of the atomic mass unit:
$$\mathrm{1\ u}=\dfrac{1}{12}\text{ of the mass of one }\ce{^12C}\text{ atom}$$
Thus, the atomic mass of an atom of $\ce{^12C}$ is by definition $\mathrm{12\ u}$. The atom is constructed from 6 protons, 6 neutrons, and 6 electrons, and the masses of those particles are:
particle   mass                     number    total mass
p          1.00727646681290     u   6         6.0436588008774    u
n          1.0086649160043      u   6         6.0519894960258    u
e          5.485799094622×10^−4 u   6         0.0032914794567732 u
                                     TOTAL:  12.0989397763600    u

The sum of the parts of the $\ce{^12C}$ atom are more than the mass of the atom! This mass defect, $\mathrm{0.0989397763600\ u}$ in this case, is the value of the nuclear binding energy when converted to energy using $E=mc^2$.
This handy graph from the Wikipedia article linked above shows that the binding energy per nucleon is not the same for all nuclei.

If it were, then we would have integer values for monoisotopic nominal masses. However, the consequences of this change in reality could be as drastic as the heavier elements (like iron) not having enough energy to hold themselves together. An additional consequence is that there would be no productive nuclear reactions (fission, fusion, or radioactivity). All nuclear transmutations are driven by conversion of nuclei with lower binding energy per nucleon to nuclei with higher binding energy per nucleon, resulting in a net conversion of mass to energy. Without a difference in binding energy per nucleon, we would not have nuclear bombs (and probably we would be better off as a civilization), but we would also not have nuclear power and radioisotopes for compound labeling, cancer therapy, and medical diagnostics. We would also not have a ready supply of deuterium for NMR solvents, and we might not be able to use variations in isotopic distributions of $\ce{H}$, $\ce{C}$, and $\ce{O}$ to determine all sorts of things like did that "locally grown" food from that restaurant you like actually come for your geographic region?

Just curious... since most people read atomic masses from periodic table; is it not average atomic mass? Hence shall relative isotopic abundance be also not included in answer for sake of completion?

@ipcamit : The posted question explicitly excludes such considerations.

oops! missed that line!

"...the binding energy per nucleon is not the same for all nuclei.  If it were, then we would have integer values for monoisotopic nominal masses."  Because of the proton/neutron mass difference, we still wouldn't have integer values.  Carbon 12 has a 1:1 ratio of protons to neutrons.  In the absence of any difference in binding energy per nucleon, any isotope with a 1:1 proton to neutron ratio would have an integer atomic mass, but an isotope with any other ratio would not.

Calculated from the mass defect, the average binding energy per nucleon for $\ce{^{12}C}$ is $7 680 151.46~\mathrm{eV}$ which matches exactly the given $7.68~\mathrm{keV}$.

Out of curiosity, where did you find these values for the particle mass? Your values are slightly different and have more significant digits than the 2014 CODATA recommended values $m_\mathrm p=1.007,276,466,879(91)\ \mathrm u$, $m_\mathrm n=1.008,664,915,88(49)\ \mathrm u$, and $m_\mathrm e=5.485,799,090,70(16)\times10^{-4}\ \mathrm u$.

The average binding energy per nucleon calculated from the new set of data provided by @Loong is $7 680 151.43~\mathrm{eV}$, which is quite close from the value before.


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R configuration for cysteine
I think my prof made a mistake. Shouldn't the answer be 1, 4 and 5 instead of 2 and 3?

If the programmers did not make a mistake it seems that you are right - ChemDraw is on your side.
$\hskip.5in$

I got 2 and 3 as the (R) isomer. 2 and 3 are the same isomer and I got a clockwise rotation, hence R, when numbering the priority of groups where H (the lowest priority) is facing away, or dashed line. I thought the order of priority would be NH2>COOH>CH2SH. Any ideas?

@LiamH It's NH2>COOH>CH2OH but S is more important than O because of it's higher atomic number and that's why NH2>CH2SH>COOH.

You beat me to my own correction PH13 haha >> Ah looks like I was wrong. Priority is NH2>CH2SH>COOH>H. Therefore I would have to agree with the above answer and 2/3 are actually an S configuration. Got confused for a minute there!


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Determining chirality after considering conformational interconversion in a cyclohexane
Basically I want to know what does the question mean when it says "taking into consideration conformational interconversions."

I assume it means, when I draw the mirror image of the given compound, the mirror image can be same as the compound itself (the case with option (a) and option (c) ) or it (the mirror image) can be same as confomer of the compound (the case for option (a) and option (b) ), for the compound to be achiral.
Note: option (a) shows both the features because option (a) is same as it's confomer and it is same as it's mirror image as well.
Is my assumption correct?

Your analysis is correct. The stipulation "taking into consideration conformational interconversions" means that conformational interconversion is fast. That's important because compounds that we consider achiral can have chiral conformations. I used the example of butane in a previous question.
In this problem, three compounds are given that have multiple stereocenters, compounds b, c, and d. A common way to decide whether a compound with multiple stereocenters is chiral is to look for planes or points of symmetry within the compound. Compound c has a plane of symmetry, so it is clearly achiral. Neither of compounds b or d has a plane of symmetry in the drawn conformations, which means that those conformations are chiral. However, as you suggest, chair flip of b gives the mirror image of the original drawing, which means the original and mirror image are equivalent, assuming that the two conformations can interconvert.
There are many cases where conformations cannot interconvert, and that leads to very interesting compounds that don't even require a stereocenter to be chiral (again see earlier question).

I can add figures if that would be helpful.

Thanks a lot for your input @jerepierre. I believe I have it figured out but you can add figures if you like as other stackexchange users might find it helpful. Thanks again. :)

For this question, if I draw the planar ring form for each of the cyclohexane rings, only d) doesn't have a plane of symmetry. Would that be a correct way to approach the question?

@kane9530 yes, because the planar form is essentially the 'average' of the conformations.


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induction of van der Waals forces
Is there a way to maximize van der Waals forces between two materials, by electrifying one material, thus creating dipoles/instantaneous dipoles and thus creating stronger van der Waals forces? has this been documented?

I think you shoud have read more about van der Waals forces and electrifying objects. What are you really asking about?

Perhaps an example of what you mean can be seen with the noble gases. As you go from helium to neon to argon, etc. you are adding inner electron shells and making the electron cloud around each atom more polarizable. That in turn increases the van der Waals attraction between atoms, resulting in a dramatic increase in boiling points down the series.
You might be able to make an atom more polarizable (and increase van der Waals forces) by electronically exciting it to a higher energy level. Can anyone else elaborate on that possibility?


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N2O production vs air distillation
After producing nitrogen by distilling air, how is that nitrogen used with the oxygen to yield nitrous oxide ($\ce{N2O}$)? I see $\ce{N2O}$ available in different purities from various suppliers, but how is that different from just distilling air? When distilling nitrogen from air, once both gasses have reached the right temp/pressure to become a liquid, what is preventing it from becoming $\ce{N2O}$ in a liquid form instead of one floating on top of the other? These questions are in reference to an automotive application.

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Just so we're clear here, liquid air is a physical mixture, a solution composed mostly of dinitrogen ($\ce{N2}$), dioxygen ($\ce{O2}$) and argon ($\ce{Ar}$). There are almost no nitrous oxide molecules. In principle you could take $\ce{N2}$ and $\ce{O2}$ and react them (combustion) to produce $\ce{N2O}$, but this process would be quite difficult to control, as there are several more stable nitrogen oxides. This probably precludes usage of nitrogen combustion in oxygen as a source of large amounts of nitrous oxide.

After producing nitrogen by distilling air, how is that nitrogen used with the oxygen to yield nitrous oxide ($\ce{N2O}$)?

It's not clear to me that $\ce{N2O}$ is produced from $\ce{N2}$: Rather, it appears that commercial quantities of $\ce{N2O}$ are produced by thermally decomposing ammonium nitrate:
$$\ce{NH4NO3(s) -> N2O(g) + 2H2O(g)}$$
as described (at least) by this manufacturer and supplier of nitrous oxide gas plants. See the comprehensive Wikipedia entry for more information.

I see $\ce{N2O}$ available in different purities from various suppliers, but how is that different from just distilling air?

I'm not sure what you mean here.

When distilling nitrogen from air, once both gasses have reached the right temp/pressure to become a liquid, what is preventing it from becoming $\ce{N2O}$ in a liquid form instead of one floating on top of the other?

Nitrogen is liquid in the range 63 to 77 K; for oxygen, it's 54 to 90 K. Thus, the temperature range for which both are liquids is bounded by the range on nitrogen, or roughly between 63 and 77 K.
The temperature at which nitrous oxide solidifies (182 K per Wikipedia) is much higher than the temperature range at which oxygen and nitrogen are in the liquid phase, and thus it would have come out of solution as a solid long before the requisite temperature was reached. Thus, we would not expect to see liquid $\ce{N2O}$ co-existing with liquid oxygen and nitrogen.
Note bene: I am using the pressure range of 5 to 10 bars as cited here in assuming that liquid and solid phases of the above moieties do not coexist during industrial gas production at these relatively low pressures.


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percent ionization explanation
Using a weak acid as an example:
$$\ce{CH_3COOH <=> CH_3COO^- + H^+}$$
The position of this equilibrium shifts to the right with increasing dilution.
I have found textbooks that say this, but do not provide an explanation. Why is there increasing ionization with increasing dilution?
I was wondering if it has to do with the free energy available; since ethanoic acid is a weak acid, energy has to be provided to promote the proton to a higher free energy state (transfer from $\ce{CH_3COOH}$ to $\ce{H_2O}$), therefore, "more" free energy for promotion and less contact with other $\ce{CH_3COOH}$ molecules?

Maybe related to http://chemistry.stackexchange.com/questions/7648/does-the-number-of-h-ions-in-solution-go-up-on-dilution-of-a-weak-acid ?

For $\ce{HA <=> A- + H+}$  with a particular $K_a$ value at a particular temperature (Also, remember it usually applies to weak acid as $K_a$ of strong acid is rather meaningless),
Consider its equilibrium expression: $$K_a=\frac{\ce{[A^{-}][H+]}}{\ce{[HA]}}$$ when water (of the same temperature) is added, all three component concentration will decrease with the same extent.
Think about whether the numerator and denominator has a bigger decrease and you will find the answer (temperature is relatively constant). And for your free energy concern, being out of equilibrium necessarily means there is a favorable driving force pushing the reaction in one direction to reach back eqm.

Acquire a physical picture about the initial and equilibrium state and the dynamic state of a reaction in terms of collision theory. Think equilibrium in the form of competing forward and backward reaction and equilibrium constant is actually derived from rate constant.

Another reminder: Temperature and concentration/pressure all affect free energy. At the initial state (when you just add water and no reaction has taken place yet), there is only change in concentration but not temperature. So even if the temperature is changed upon reaction, the percentage of ionization will also be higher in all circumstances.

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What is the difference between infrared action spectroscopy and conventional IR spectroscopy?
I'm familiar with infrared spectroscopy generally, which probes mainly the vibrational behavior of molecules under study.  However, I'm seeing references in various literature to 'infrared action spectroscopy' (e.g., here), which I'm having a hard time finding information about.
So, the title question: What is the difference between infrared action spectroscopy and conventional IR spectroscopy?

After 30 minutes of Googling, I'm thinking that we both (EDIT: nope, it was just me) misparsed the name.
When reading the name "infrared action spectroscopy," my first instinct was to mentally split it up like
$$\textrm{infrared action} \qquad\qquad \textrm{spectroscopy}$$
that is, spectroscopy of the "infrared action" flavor.
However, after running into more dead ends than a drunk rat in a maze with no exits, I realized that the correct way to split it might be
$$\textrm{infrared} \qquad\qquad \textrm{action spectroscopy}$$
or "action spectroscopy" that happens to use infrared light.
Lo and behold, the first hit on google was a wikipedia page on action spectroscopy. That link seems to have more of a biological tint to it (as do most results), but there are several physical science papers out there as well. (See Edit)
It seems that action spectroscopy in general refers to hitting a system with light and watching some kinetic property of the system (rate of photosynthesis, conversion rate of isomers, etc.) In the case of the paper you cited, they report the relative intensity, so my guess would be that they did something like a scan over frequencies and watched how the ions fragmented at each frequency, which is supported by a cursory glance over the text surrounding Equation (1). Unfortunately, I don't have time to read the paper in full, so let me know if I'm totally wrong.
The moral of the story is that English is hard.
=============
EDIT: It turns out I'm the dolt and everyone else processed this correctly. As Brian mentions in the comments, the action spectroscopy page on Wikipedia is pretty awful, so I'll try to expand on action spectroscopy below, given what little I know about the subject.
In your usual spectroscopic setup, you hit the sample with some form of EM radiation, then look at what's reflected or absorbed to learn something about the system (e.g. UV-vis, IR, NMR, rotational)
It seems that in action spectroscopy, your goal is to use the light to cause some sort of a change in the system, and measure that instead. In the example given on Wikipedia, you measure the rate of photosynthesis as you change wavelengths--this tells you at which wavelengths photosynthesis is most effective.
In the paper I linked above, they seem to mediate the conversion of cis-butene to trans-butene using light. Presumably, you could start off with low-energy photons, then gradually ramp the energy up until you see the isomerization happen. The wavelength at which the transformations first start happening would tell you the activation energy barrier for the cis-trans isomerization. (Note that I can't actually read that paper because of the paywall, so I may be wrong about this.)
In the paper Brian linked, they seemed to use IR to study the fragmentation of an ion sample. They mention "loss paths" as observed through an MS, so it seems like they hit a sample with IR light, then figure out what it breaks up into.
As Brian said in a comment, it seems that action spectroscopy is where light is used "to alter the system under study, not to just probe it passively." I'd like to add that, in general, it seems that the quantity of interest is not the amount of light that comes back out, but something else that happens in the system as a result of the exposure to light (e.g. isomerization, ion fragmentation, glucose production).

If this is right, then the question title might be misleading, since we'd really be considering the difference between "visible action spectroscopy" and "infrared action spectroscopy," not "infrared action spectroscopy" and "infrared spectroscopy"

No, I had indeed read it as "IR ... action spectroscopy." I also found the numerous references through Google to 'action spectroscopy' as pertains to biological growth responses to light, but those seemed sufficiently unrelated that I disregarded them. Now that I reread the (pitifully short) Wik article, though...  "[Action spectroscopy] shows which wavelength of light is most effectively used in a specific chemical reaction." It must be high-intensity spectroscopy, where the intent is to alter the system under study, not just probe it ~passively.

Or, perhaps the question would be better written without any flavor of EM radiation specified... "What is the difference between action spectroscopy and 'conventional' spectroscopy?"  Regardless, thank you for your persistence in pounding on Google until something coherent fell out.

I'm not entirely sure. The major block for me was getting around the issue of splitting the name correctly, but I think it's an interesting example of how a little bit of a thinking change can really help things out. I'll edit in a bit more detail for anyone else who comes along though, since you're right about the wikipedia article being pretty piss-poor.

@Brian do you think the new edited section would be sufficient for someone to get a rough idea of what action spectroscopy is?

Two thumbs up. I like.


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Relationship between the first and the second quantum number
Does the secondary quantum number tell how many subshells a specific principal quantum number shell has?
E.g., if the principal quantum number is $n$, there are ($n-1$) subshells.

Does the secondary quantum number tell how many subshells a specific
  principal quantum number shell has?

No, it does not. It is the principal quantum number $n$ which does it (as you actually mentioned yourself later in the question). The second quantum number $l$ describes the shape of the orbitals in a particular subshell. Say, $l=0$ for $\mathrm{s}$-subshell of any shell and all $\mathrm{s}$-orbitals have a spherical shape.

If the principal quantum number is $n$, there are $(n−1)$ subshells.

Why $(n−1)$? Think about the most trivial example: how many subshells the first shell ($n=1$) have? One, not zero, so $(n−1)$ is obviously wrong formula. The situation is in fact simpler: $n$-th shell has $n$ subshells.

A small detail to add to Wildcat's excellent answer: the $\ell$ quantum number does not tell you the total number of subshells, but you may infer the minimum value of the principal quantum number $n$ and therefore the minimum number of subshells. For example, if $\ell=3$ then $n\ge 4$.


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Dry Hydrogen Production
I am building a set-up to produce a controlled flow of dry hydrogen. I'm aiming for around 99.5 percent purity.
Am I right in thinking a PEM cell and Deterite drying tube should be fine for this? Or will I need a condenser of some kind?
Any tips?

Some links that might be useful for people reading this https://en.wikipedia.org/wiki/Proton_exchange_membrane_fuel_cell and https://en.wikipedia.org/wiki/Drying_tube

What you should be really careful about is storing it$\ldots$Hydrogen gas is highly explosive (even though its potential energy is well-hidden behind a kinetic barrier) but I guess you already know that.

If you might ultimately need ultra-pure hydrogen, palladium membrane hydrogen purifiers are a simple way to eliminate trace gases.

For what setting up a fuel cell would costs you can buy high purity hydrogen in a tank. It is a cheap gas to make in production quantities. The expensive part will be the regulator.

Using molecular sieves comes to my mind, they're easy to use and efficient. You can find some practical information about how this is done for example here http://www.klmtechgroup.com/PDF/Articles/articles/Molecular-Sieve-Applications.pdf

Hydrogen gas is inert towards most common drying agents, so you could try using a drying tube filled with anhydrous salts like CaCl2, Na2SO4, MgSO4 or you can bubble it through cc.H2SO4. But molecular sieves are also a good bet if you have them.


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Generating Precipitation Diagrams with CompuTherm Pandat
I am trying to develop precipitation diagrams for an Ag-Cu-Sn alloy to figure out the kinetics involved using the CompuTherm Pandat Software (Demo Version).
The precipitation diagrams need the thermodynamics/mobility database and it further needs the kinetics database as well. The software I have installed comes only with Al-Ni Kinetic database.
Does anyone know how to get Pandat to generate diagrams for other alloys?
Their website is http://www.computherm.com/


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How can I keep milk from spoiling and/or turning yellow?
I am trying to find a way to keep milk from spoiling and/or turning yellow.
Apparently there is a process for achieving this, and I am aware that there is a lady who creates milk beads for resin jewelry and she claims that she is able to prevent the milk from spoiling/yellowing over time.
I have heard that you can add baking soda to milk to preserve it, but this lady also claims that she adds nothing to the milk and that it simply undergoes a "process".
Of course, she won't share her secrets, so my question is:
What are the different ways in which milk can be preserved while preserving the white color?

Pasteurization is probably what is meant.


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Do melt flow rate and density tell you everything about pure HDPE?
I understand that melt flow rate and density can be used to classify HDPE. Does this mean that if I have the melt flow rate and density of an HDPE (with no additives), then I can know its other material properties?

Yes, as a rule the melt flow rate and density do tell you what the other material properties will be. This is why HDPE is often specified by those two properties.

authors: زهره البنفسج


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At what frequency does a non-polar molecule acquire a dipole able to participate in London Dispersion forces?
London Dispersion forces are forces between non-polar non-ionic molecules that the random fluctuation of their electrons makes them temporarily dipoles.

What is the frequency that this occurs?
Every how many milliseconds does a molecule like this becomes a dipole?
Are there such numbers for some specific molecules out there?

Good question! I don't have an answer offhand, but I suspect you're off by several orders of magnitude in your time estimate -- I would think the electronic fluctuations occur on a femtosecond or picosecond time scale.

In my point of view oscillations in time are an artifact of the orbital picture. The time independent Schrödinger equation and its exact solutions account completely for London dispersion interaction. The problem occurs when you build wave functions and interaction energies for multi-electron systems using one-electron wave functions and Fock space and so on. This picture can be patched by London dispersion. But the oscillations are pure mathematical model without physical counterpart.

According to Prof. Loren Williams of Georgia Tech, the fluctuations occur on the femtosecond $\left(10^{-15}\ \mathrm{s}\right)$ time scale:

Figure 13 shows how fluctuating dipoles of liquid Xenon (or Helium or Neon, etc) are coupled. Darker blue indicates higher electron density. The fluctuations are correlated and are very fast, on the femtosecond ($10^{-15}$ second) timescale. Adjacent Xenon atoms experience electrostatic attraction from the transient dipoles. Two different representations of fluctuating dipoles are shown.

This corresponds to a frequency of $1\ \mathrm{PHz}$ (petahertz), which is about five hundred thousand times faster than most modern computer processors.
Another estimate of the time scale (albeit one devised by me and thus of potentially suspect validity), is the SI value of the atomic unit for time, derived by dividing the reduced Planck constant by the atomic unit for energy, the Hartree (values are rounded a bit):
$$
\left\{atomic\ time\right\} = \frac{\hbar}{E_\mathrm{h}} = \frac{1.05457\times 10^{-34}\ \mathrm{J\cdot s}}{4.3597\times 10^{-18}\ \mathrm{J}} = 2.41888\times 10^{-17}\ \mathrm{s} \approx 0.024\ \mathrm{fs} = 24\ \mathrm{as}
$$
That last value is $24$ attoseconds, which is a well and truly short span of time. For comparison, this corresponds to a frequency of $41\ \mathrm{PHz}$.
Whatever the exact value, I have to believe the fluctuations are fast enough so as to be impossible to measure directly. (I would be fascinated to learn otherwise, though!)  Thus, even if different atoms/molecules do actually have different frequencies of London dipole oscillation, we have no way of finding out these values.

This is a nice description of dipole oscillation, but doesn't answer the posted question. They are asking how often formally nonpolar particles form an instantaneous dipole. The particle is polar for nearly the entire femtosecond it takes to flip polarity.

@DanBurden Uh-oh, looks like this is going to get out of hand quickly.  :-)  My answer attempts to address the question on its own premise, that the dipoles do disappear. Yours rejects the question's premise, arguing that the dipoles are never (or nearly never) absent. I leave it to the community to vote their preferences.

No worries! I think your answer is much more educational than mine, which is why I upvoted it. Just wanted to throw my 2 cents in. Plus, I learned something new!

This is speculation on my part, but perhaps the polarization time scale depends on the dimensions of the system. I wonder if sufficiently large electron clouds can be polarized in a measurable amount of time. Maybe this could be done with fullerenes, or Rydberg atoms?

@NicolauSakerNeto Not even quantum electrons can move faster than the speed of light, so there has to be some limit to the response time scale as the spatial extent increases. However, if the magnitude of the fluctuations doesn't increase proportionally to the system size, the transient dipole may not be measurable? Regardless, an interesting (thought) experiment!

Is there a kind of spectroscopy based on this oscillation? I.e., would liquid xenon absorb electromagnetic radiation at 41 PHz, thereby driving the oscillation? Just curious.

@iad22agp Great question! I think you should ask it as a new, separate Chem.SE question, though.

There is something screwy about this number 41 petahertz. That corresponds to x radiation at a wavelength of about 7.3 nanometers - hence ionizing radiation. That is a lot more energy than I would expect to be required to cause oscillations of electron density.

@iad22agp X-rays are scattered by the electrons, i.e. by elastically deforming the charge distribution, in matter. Matches perfectly.

While dispersion forces dominate in non-polar, non-ionic systems, they are actually felt by all chemical species that contain electrons. But, for simplicity's sake, let's use the example of a non-polar, non-ionic atom or molecule to describe these forces in this answer.
Surprisingly, these molecules are almost never completely "non-polar". There is almost always a dipole moment, but the strength and direction are random. The electron densities around these atoms and molecules are always shifting, which almost always leaves an uneven distribution of electron density around the nucleus.
The London dispersion forces are simply the sum of interactions between the constantly fluctuating dipoles of one particle, and the induced dipoles within the neighboring particles. The influence of the electrostatic force is directly proportional to the product of polarizabilities for each participating particle.
To summarize: Molecules or atoms with no permanent dipole are almost never in a completely "non-polar" state, due to the constantly fluctuating electron cloud. This negates the question of instantaneous dipole frequency, because it is never not a dipole.

While this is a good description of London forces, it doesn't address the actual question of the frequency of appearance of the dipoles.

Thanks! I was trying to answer the question by saying that the molecules are never not dipoles. I apologize if this is unclear.

Yep, that point was unclear, at least for me. Edit made it much better! :-)

Hmm... so, perhaps a better question is something like: "Below what time scale is time-averaging of electron positions insufficient to be able to treat a nominally non-polar atom/molecule as lacking a dipole?"

Maybe - I think it would help to define a nonpolar molecule as having a polarity below some threshold.  For example, very very small electron density asymmetries can probably be considered nonpolar. However, I think the threshold would have to be dependent on polarizability, temperature, particle size, etc. It's a tricky problem.

The electron rich/ electron poor regions tend to oscillate at a period of about 1-10 femtoseconds.
Logothetidis, Stergios. Nanostructured Materials and Their Applications. N.p.: Springer, 2012..


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What is the lethal dose ( LD50 ) of coffee?
Of course, there are many types of coffee, so I would like a range of lethal dose amounts for the average human, from the mildest of lattes to the most hardcore of espressos.
Pure caffeine's lethal dose is easily calculable, but what about coffee?

According to one reference:
LD$_{50}$ for coffee = 118 servings at 240 ml each; and
LD$_{50}$ for espresso = 175 shots (volume not specified),
for a 75 kg person.
A much more comprehensive quantitative description is given here for the LD$_{50}$ of coffee for an 80 kg person (113 cups at 250 mL each, 135 mg caffeine per serving). They account for half-life of elimination of caffeine, and arrive a the conclusion that one can't actually reach the LD$_{50}$ (15,360 mg) for caffeine due to metabolism, elimination, and the limit on how much water can be processed by the kidneys.
Finally, as noted in the last reference and hinted at above, the amount of water a person would imbibe on the way to reaching the 113 cup total (about 30 L) would likely kill the person (water intoxication).

That said individual weaknesses of cardiovascular system of otherwise pretty healthy human may lower the LD dose, and even non-letal but large dose may put one down for a time.

@permeakra Indeed. Biochemistry and pharmacokinetics aside, I suspect the motor control side effects from this rate of coffee consumption would result in a self-limitation on intake well before reaching a fatal dosage.


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What phase of matter is salt when dissolved in water?
Being an ionic solid, salt is very susceptible to being torn apart by the polar molecules of water. Inside the water, salt exists as sodium and chlorine anions; but what state of matter is it in? My instinct tells me it is in a solid state, as salt's melting point is 800°C, but it sure doesn't really act like a solid.
EDIT: I think I ought to rephrase, I mean what state of matter are the individual sodium and chlorine ions inside the water.

It's simply liquid solution! I have no idea why tou'd think otherwise

The salt is liquid? Really? At room temperature?

Phase is macroscopic - "state of matter are the individual sodium and chlorine ions inside the water" has no sense when salt is dissolving it simply becomes part of liquid mixture - that's it.

Could you turn that into an answer and provide references? @Mithoron

http://chemistry.stackexchange.com/questions/308/what-is-the-difference-between-melting-and-dissolving may answer it even better


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protecting contact between aluminium and copper wires
The quarters in which I live has both aluminum and copper household wiring. This results in corrosion of the contacts and can lead to sparking, overheating, and even fire.
Is there any way to inhibit the corrosion (if not stopping it completely), except for changing the whole wiring?

This may depend on your location and local electrical codes.  I have the same issue in my house.  In the USA, there are NEC (National Electrical Code) compliant wire nuts rated for Al to Cu connections.  These come pre-packed with a protective goop. The goop can also be purchased separately in a tube for use with ordinary wire nuts. You may have to ask at your local electrical supply store.  In my area the pre-packed wire nuts are a distinctive purple color not found on regular wire nuts (for example: Home Depot Al/Cu nuts).
For code compliance, use approved solutions only, even if you think you know more chemistry than the electricians. Having your house burn down from a wiring issue is no fun.

I'd like to add to this, that interNACHI states many of the qualities regarding aluminum for wiring use.

You won't be able to stop the corrosion unless the two pieces of metal are galvanically separated, that is, in this case, there is no ion-containg water around (I assume it's inside walls/floors/ceilings). I don't think it's really practical in a house where the wires are already mounted inside structures very apt to contain moisture.

Wire nuts are a considerably unfavorable means of connecting the house wiring to stubs for what I presume are your receptacles. It would be far safer for you to purchase the Aluminum compatible receptacles and use the binding screws for the connections. Wire nuts provide too little contact area and offer no heat dissipation ability to deal with the heat generated by the bimetallic interface of the aluminum/copper joint.

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What bond\crystal has the most regular pattern (i.e. least irregularities)?
Salt is a famous regularly repeating cubic crystal, but there are always natural deformities and flaws present. I have heard that graphene is an extremely uniform substance. With these in mind, what is the most regular crystal pattern, natural or man-made?

Silicon boules for silicon wafers.  Grown by the ton, all single crystal (with, of course, some defects since thermodynamics demands them).

In nature you need to look for precious gamestones and then do optical selection (diamante and ruby would be good candidates). The same would be true for synthetics but of course as mentioned by Jon you can also take elements (Si, Ge). General rule: they should be hard (smaller chance that pressure will affect structure, by pressure I mean also your tools) and you need to grow them slowly.
btw it's hard to call graphen crystal pattern -- it's 2D.

The electronics industry requires extremely pure crystalline silicon to make the silicon wafers on which most electronics products are based. Too many impurities ruin the chips made with the wafers. Therefore, the industry has developed good techniques to make very pure, defect free (or, more accurately, with very few defects) crystals.
Zone refining is a key technique used to achieve large, pure silicon crystals in the semiconductor industry. Impurities can be reduced to parts per trillion and defects in the resulting crystal structures are also very low.
Currently scientists are trying to redefine the kilogram by using extremely round spheres of ultra-pure silicon. The technique relies on counting the silicon atoms in the spheres and needs the absolute minimal level of crystal defects for the counts to work reliably (they measure the crystal parameters very accurately and thereby derive the numbers of atoms in the spheres: defects upset the precision of this calculation). These are probably the most regular actual crystals in the world right now.


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How can a ketone be enantioselectively reduced, in the presence of an ester?
I'd like to reduce enantioselectively a ketone in presence of an ester (which should NOT be reduced to an aldehyde and further to an alcohol).

Adding $\ce{LiAlH4}$ would also reduce the ester and gives no enantioselective reduction of the ketone.
In the following reaction (Frater-Seebach Alkylation) they used Bakers yeast:

Probably S-BINAP also give 2. I think there is a $\ce{RuCl2BINAP}$ transition metal complex in combination with $\ce{H2}$ used.
I would first use KOH, then adding Evans auxiliary, adding LiCl for chelation of acid carbonyl to Evans auxiliar carbonyl, then adding $\ce{NaBH4}$ and to a standard Felkin Anh reduction. And as a last step I would use $\ce{LiOEt}$. Does this work or will the chelating ketone also be reduced?

May be my own suggestion is easier to reproduce in lab than the use of asymmetric reduction using $\ce{H2}$ under high pressure.
I'm looking for further ways to get the (S)-hydroxy compound 2. Are there easier ways?

"Probably S-BINAP also give 2." Under what conditions? BINAP is just a ligand.

You're right. I get this reaction condition from a slide from Google. I've edited my question.

I have had great results for enantioselective ketone reduction using the Noyori Ru-TsDPEN system.
This has advantages through being catalytic with low stoichiometry and doesn't require high pressure, or molecular hydrogen for that matter. The downside might be the initial cost of the catalyst / chiral ligand but you don't need very much. The typical reduction reagents are triethylamine and formic acid, making this a very mild system to work with in terms of conditions.
It might be cheaper to purchase the RuCl2(p-cymene) and chiral TsDPEN separately. That way you can also play with the ligand-metal ratio if you need to.
This saves a lot of hassle rather than adding on and removing chiral auxiliaries

+1 to Beerhunter's answer. The Noyori $\ce{Ru-TsDPEN}$ transfer hydrogenation can work really well. As alluded to,the catalyst can be made fairly easily, just involving stirring several components together, drying, filtration and concentration. The whole procedure is relatively insensitive and making it yourself makes the whole process enormously cheaper


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Can a benzyl ether be removed in the presence of an ester?
I'd like to deprotect a benzyl group for a Fráter–Seebach alkylation. Would it be possible to do this with a Birch reduction using Li/NH3 (or Na/NH3), without also reducing the ester carbonyl group?

If not, hydrogenation over Pd/C seems to be a plausible alternative. How else can I remove the benzyl group on oxygen?

Hydrogenolysis (H2, Pd/C) is by far the most common method. For a more comprehensive list, see the books by Greene and Wuts or Kocienski. http://onlinelibrary.wiley.com/book/10.1002/0470053488

Related: Can an ether be selectively hydrolysed in the presence of an ester?

In Wutz's book on protecting groups,[1] I found an example of almost instantaneous reductive debenzylation which leaves an ester untouched:[2]

The use of BCl3 is also possible:[3]

References:

Wuts, P. G. M. Greene's Protective Groups in Organic Synthesis, 5th ed.; Wiley: Hoboken, NJ, 2014, p 131.
Sulikowski, M. M.; Davies, G. E. R. E.; Smith, A. B., III. Avermectin–milbemycin synthetic studies. Part 7. An approach to the southern hemisphere of milbemycin α1. J. Chem. Soc., Perkin Trans. 1 1992, 979–989. DOI: 10.1039/P19920000979.
Williams, D. R.; Brown, D. L.; Benbow, J. W. Studies of Stemona alkaloids. Total synthesis of (+)-croomine. J. Am. Chem. Soc. 1989, 111 (5), 1923–1925. DOI: 10.1021/ja00187a081.


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Grignard reaction - exchange of groups
Why is this reaction successful:

Why doesn't metal-halide exchange occur easily to get Ar-MgCl and R-Br?

Why do you think it should? Could you elaborate?

A similar reaction is used in the synthesis of  sibutramine, only the aldehyde is replaced with a cyanide group, and R=isobutyl.

I guess the reason is that metal halide exchange is very fast only in the case of lithium organyls while it proceeds slower for grignard reagents.

And I guess that chlorine isn't that susceptible, while Br and I would be.


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Are the tars in cannabis considerably more dangerous than those in tobacco?
The claim that "the tars in cannabis are considerably more dangerous than those in tobacco" is made in Patricks An Introduction to Medicinal Chemistry (4th edition, p. 169). It's just a simple one line statement and not much attention is given to it, but from what I already knew this is to say the least a very controversial statement (see for example the independent website about drugs Erowid) and I expect better for a frequently used University level book.
The question is to what degree (from a chemical point of view) there is truth (or falsehood) to this statement given current more or less objective (not like the Heath monkey studies) knowledge? Is the author in his right to boldly state this?

I think this is a good question.  My worry is that some will feel it approaches "personal medical advice" and is thus off-topic, but I don't believe this is true.  It is asking about the truth value of claims advanced by referenced documents in the chemical literature.  On-topic.

@CurtF. Thank-you. No it's not at all personal medical advice, I haven't smoked a joint in 4 years haha! I just noticed many many times conflicting statements regarding this, all from more or less reputable sources. This was kind of the last "drop" that made me decide,  I want to know what is actually true and I want others to know what is actually true. For example on a Dutch (I'm Dutch) health website affiliated with the government I saw the claim that 1 joint equals 20 cigarettes, but of course no source.

I seems that people have forgotten the scientific part of this claim (from a medical chemistry point of view) and only focus on the political impact. This is IMHO an excellent place to sort out the scientific part!

I think this is fair game here and on topic, but it might be a bit broad for a good answer. If I see that correctly, an answer would have to first figure out what compounds will be inhaled in both cases and then rank them according to the physiological effect. Splitting it up in smaller pieces might be worth thinking about. Also you could probably get some more detail - at least for part of the problem (e.g. side effects of medical marijuana) - on health.se. But I don't know about their scope, so better check that first...

@Martin-マーチン For the sake of research I would absolutely agree with you of course. But splitting this question would be disastrous for it being useful as a point of reference. I think it's most on-topic here. I'm specifically looking for a chemical explanation of the claim rather than empiric  experience. I'm going to two the following two things: 1) raise a 50 point bounty for attention and 2) contact the author of the mentioned book to see if he wishes to elaborate on the claim (after all, it is his book!). Thanks again Martin, good to see you on the mod team :)

From my toxicology lecture (which I remember vaguely) I think I can recall that the danger of smoke tar is, among other things, determined by genetics. The body cannot excrete tar as it is hydrophobic, so it runs it through a series of reactions, including epoxidizing, which makes the resulting compound watersoluble. There's a widespread (I think around 20% population) mutation which changes this chain, making one of the intermediaries a carcinogen, which, as I remember, is the main cancer causing compound. So assessing tar risks is pretty complicated.

This is not a complete answer (merely a step in a particular direction, namely that of the falsehood of the claim), but I contacted the author of the book (Introduction to Medicinal Chemistry) Graham Patrick and he wrote the following:

Dear Jori,
Many thanks for your e-mail and observations. Unfortunately, that
  statement was made in a very early edition of the textbook published
  over ten years ago and so I doubt if I will be able to track down the
  original reference on which it was based. As you say there is
  controversy and the statement is too sweeping. Work on the 6th edition
  is starting this year and so it will be possible to modify that
  particular passage. Many thanks for your input.
Best wishes Graham


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Making Liquid CO2
I recently saw a video on making liquid $\ce{CO2}$:
https://www.youtube.com/watch?v=8AN_XMcD3yI
Basically we seal some dry ice in a container. As the dry ice sublimates the pressure inside the container increases and so at about 5 atm we reach the triple point of $\ce{CO2}$, at which point the solid $\ce{CO2}$ melts instead of sublimating.
At some point in the video, the teacher cautions that the same experiment should not be done with a Gatorade bottle (similar to a water bottle). However, I'm not sure why this is the case.
Certainly the usual "dry ice bomb", which also has warm water poured into the sealed bottle, is not a good idea. In the present scenario, once all of the solid $\ce{CO2}$ had melted and the liquid $\ce{CO2}$ then goes into the gas phase, the increase in pressure could lead to an explosion: isn't this the same principle as in the case of the "dry ice bomb"? Is the warning because a Gatorade bottle (or water bottle, etc.) cannot withstand 5 atm of pressure?

I'd say the plastic bottles go below their glass temperature and have very low  strength in their glassy phase whereas a pyrex bottle would have thermal stability to maintain strength but I don't know the temperature involved and it would depend on which plastic a bit. It may even be chemical damage of the plastic.

@user2617804 you should add that as an answer. It didn't even occur to me to consider the effects of temperature on the plastic.

I think these three links nicely summarize what can potentially go wrong when a plastic bottle explodes.
As for your question about how much pressure the Gatorade container can hold, that's a little trickier to figure out. It looks like the Mythbusters found that soda bottles can hold up to 150psi (~11 atm), but gatorade and water bottles aren't designed to hold carbonation, so I'm guessing they'd be closer to the 60psi (~4 atm) that they found for water coolers.
I think ultimately the answer is that the teacher is playing on the safe side: sure, you might be able to hold 10 atm of pressure in a soda bottle at home, and get some cool-looking CO2 out of it, but seeing as you could lose an eye if your bottle was even slightly damaged (and seeing that this has happened before!), is it something you'd want to let high schoolers do on their own?

Thanks for that! I don't suppose you have any suggestions of common containers that would be suitable for such an experiment and/or simple methods of determining if a container could withstand that much pressure safely?

Most likely the plastic would suffers thermal and physical shock so it would shatter- that is when it suffers rapid temperature changes and rapid physical stress- plastics have very poor thermal shock resistance due to poor crack resistance and low thermal conductivity- conductivity that would stop the heat transfer occurring in one place.
If the conditions aren't rapid enough to cause shock then it probably could still fail though its hard to tell with the CO2 phase diagram (quick google search finds it) so the plastic bottles go below their glass temperature and have very low strength in their glassy phase whereas a pyrex bottle would have thermal stability to maintain strength but I don't know the temperature involved and it would depend on which plastic a bit. It may even be chemical damage of the plastic.


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How does Le Chatelier's principle work in a gaseous system?
My textbook said Le Chatelier's principle normally applies to gases but there is no explanation.
I did some Googling and it said if you increase the pressure, it will shift to the side with less moles of gas.
What does this really mean? Could someone clarify Le Chatelier's principle for me?

In the balanced equation for the reaction, the equilibrium will shift to the side with fewer moles of gas when pressure is increased. What exactly are you confused about?

Le Chatalier's principle states that if a stress is applied to a system at equilibrium, then  the system, will shift its equilibrium position in order to bring the system back to its equilibrium.
So the stress applied to the system could be temperature, pressure and the concentration of one of the components in the system.
I will choose to explain about pressure because apparently you have the problem understanding that.
Pressure.
If you add pressure to a system then the stress will be to decrease the pressure by shifting the equilibrium position towards the side with less number of particles.
If you  remove pressure from a system, then the stress will be to increase the pressure of the system by shifting the equilibrium position towards the side with the higher number of particles.
So it is true that when you increase the increase the pressure in a system, then the system is stressed to reduce the stress. In order to do that, the system's equilibrium will shift to the side with less number of particles.
For example:
2Fe$_2$O$_3$ + 3C -----> 4Fe + 3CO$_2$
So if you add pressure to this above system, then the stress is to reduce the pressure. On your reactants side the sum of the coefficients is 5 and on your products side, the sum is 7. Since you need to bring it down then the shift will be to the side with the least sum which is to the left-hand side. Therefore, the shift will be to the left-hand side.
Hope this helped you with the problem.

Just as if it was a chemical reaction of aqueous solutions, a gaseous chemical reaction has its reagents and products and these exist in an equilibrium kind of state that can be shifted either way depending on the conditons the reaction is in and the amount of reagents and products you have (in moles). Adding pressure to the reaction is like compressing the gasses into a smaller volume and this causes more molecule collisions. Since the space between the molecules involved in the chemical reaction is decreased, you have more molecule collitions among them, this makes the molecules to react with each other more rapidly and speeds up the reaction. The reaction will go the way the equilibrium constant tells it to go, but with this addition of pressure it will do it faster. Molecule collitions also creates and destroys molecule's bonds, thus this can create or absorb heat depending on the reaction that is taking place.

I'm not really sure this answers the question about Le Chatelier's principle.

What is the question exactly then? Is it what is the Le Chatelier's principle? or is it what happens when you add pressure to a chemical reaction taking into consideration the Le Chatelier's principle? The question is not even stated correctly, how can you say that my answer didn't answer the question.

@EduardoZurita - The increased pressure will increase collisions, which will increase rate. However the definition of equilibrium is the state at which the forward and reverse reactions are both occurring with equal rates. Thus, any discussion of reaction rates changing in an equilibrium situation is not helpful. The system will shift under stimuli to achieve a state of equal rates. The OP is probably stuck at the part you brush off: The reaction will go the way the equilibrium constant tells it to go. How do you know which way that is?

If you are unsure what the question is you should ask for clarification from the OP as I have done. I also agree with @BenNorris.

Delete my answer if that makes you all happy. I got on this website becuase I needed help with an issue I had at work. I received a message from this website saying that I am supposed to help out other ppl answer their questions. I did my best trying to answer a question and I only got critized for not answer it correctly. I don't have a college education like most of you all do, man, I dont even speak English well. So could any of you who are just picking on me help me out answer the question I posted instead of just being all ruth to me.  Please!


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How does the Kucherov reaction work if the alkyne is aromatic?
I know how the Kucherov reaction works for aliphatic alkynes, but what if the alkyne has a phenyl attached? How will the aromatic ring affect the reaction of alkynes with water?

Presumably you're most interested in disubstituted alkynes. Is that correct?

No, monosubstituted alkynes, I'm interested about the case of an symmetrical alkine monosubstituted with an aromatic ring for the begining, then asymmetrical alkine

A symmetrical alkyne cannot be monosubstituted. It sounds to me like you want to know about this reaction: $$\ce{PhC#CH->[\ce{Hg^2+}][\ce{H2O}]?}$$ or do you want to know about this reaction: $$\ce{PhC#CR->[\ce{Hg^2+}][\ce{H2O}]?}\ \text{R=alkyl}$$

I want both reactions


text:
Why don't periodic acids convert alkenes to 1,2 diols?
Periodic acids cleave 1,2 diols by forming an periodate ester as an intermediate. Osmium tetroxide also forms an osmate ester with alkenes and yields syn diols after a hydrolysis step. Given that their structures are so similar, why doesn't periodic acid also succeed in forming diols with alkenes? (These in another step will be further cleaved to ketones). Instead in the Lemieux-Johnson oxidation, osmium tetroxide is taken in the first step - followed by periodic acid.
Osmium tetroxide does not cleave 1,2 diols - though I thought it might be in its being not such a strong oxidising agent as periodate - it may not suffice to cleave a C-C bond. Why doesn't though periodic acid carry out diol formation as well?

Sodium periodate can be used indirectly to form diols from alkene, see:
Emmanuvel, T. M. A. Shaikh, A. Sudalai, Org. Lett., 2005, 7, 5071-5074.
The mechanism for periodate cleavage of diols also strongly suggests the periodate is an electrophile.  This may explain why in general, alkenes are not nucleophilic enough to react under normal conditions without a co-reagent. The osmium tetroxide reaction is favoured as a [3+2] cycloaddition mechanism.

Is osmium tetroxide more electrophilic than periodic acid?


text:
NaOH Normality Slowly Changing
Our lab uses a sodium hydroxide solution of a (theoretically) known normality (0.8000 - 1.0000) to titrate product samples for quality control.  However, we suspect that the normality of the $\ce{NaOH}$ left in the burette/tubing connected to our container is changing if we let it sit (our tests come out wrong, and the answers seem to get better if we run a couple of burettes worth of $\ce{NaOH}$ out first).  I expect it's absorbing moisture from the air, but I don't know for sure.
We've tried different container types and we've put a desiccant in line with the tubing to absorb stray moisture (which it seems to be doing, there's been an observable color change), but the situation remains that we need to just throw out up to 200mL of $\ce{NaOH}$ any time we need to run a test.  Does anyone on here have any recommendations for countering this normality drift?

As @iad22agp said, it's most definitely absorption of CO2; NaOH is known to do this. Maybe instead of throwing it out, you could titrate the NaOH before using it?

I would be more worried about carbon dioxide (rather than water) being absorbed into the NaOH solution. A 1 molar NaOH solution is not going to be appreciably hygroscopic. But it will absorb CO2: two moles of hydroxide yield one mole of carbonate ion, which, depending on what you are titrating, may act as only deprotonate one mole of acid per mole base - reducing the effective normality of the NaOH solution. You could try protecting any air vents with a drying tube filled with sodium or calcium hydroxide, or use a regulated supply of chemically pure nitrogen as a blanket.

Thanks!  We're going to try switching out the desiccant in our drying tube for sodium hydroxide pellets next week, and I'll let you know how it goes.  We're also going to place a small watch glass or something over the open top of the burette, and hopefully stop some absorption from there too.
Do you have any advice regarding using sodium hydroxide in the drying tube, or can you think of anything we should be aware of?

The NaOH seems to be okay in the burette now, and the stuff in the container is fine.  We're still having trouble with the caustic that's left in the tubing though (so the first run is fine, but doing the test again with the caustic that was in the tubing gives bad results).  We're going to try changing the tubing material.  I'm accepting your answer because it was spot on, and the tubing wasn't something I asked about in my original question.


text:
Produce CuCl2 from CuSO4
I have a large amount of copper sulfate available and I am interested in obtaining copper chloride from it. What is the easiest way to do so?
I have considered mixing an aqueous solution of copper sulfate with sodium chloride, but I think that it would be difficult to separate the resulting sodium sulfate from the copper chloride. My other idea is to first react the copper sulfate with sodium carbonate and then react the resulting copper carbonate with hydrochloric acid, which would produce copper chloride (the other products would be $\ce{CO2}$ gas and water).  I am trying to find a cost effective method for performing this reaction.

You cannot obtain $\ce{CuCl2}$ mixing $\ce{CuSO4}$ and $\ce{NaCl}$ because both are very soluble, the second strategy seems good because $\ce{CuCO3}$ is insoluble in water, and you can react the new salt with $\ce{HCl}$.
Good luck.

Here's a much more direct method: just add a stoichiometric amount of calcium chloride to a concentrated aqueous solution of copper sulfate:
$$\ce{CuSO4(aq) + CaCl2(aq) -> CaSO4(s) + CuCl2(aq)}$$
If the copper sulfate you have is in the anhydrous form (white powder), then add 695.3 mg of pure $\ce{CaCl2}$ per 1000 mg of pure $\ce{CuSO4}$. If your copper sulfate is in the form of the pentahydrate ($\ce{CuSO4.5H2O}$, striking blue crystals), then add 444.5 mg of pure $\ce{CaCl2}$ per 1000 mg of pure $\ce{CuSO4.5H2O}$.
The low solubility of calcium sulfate (~0.2 g per 100 g of water at 20°C) is what drives the reaction forwards. This double displacement reaction does not happen if you use sodium chloride, because sodium sulfate is highly soluble. Simply filter the solid out and obtain a relatively pure solution of copper(II) chloride, with small amounts of calcium and sulfate contamination. Drying the solution completely will net $\ce{CuCl2}$ crystals.

Try adding pure salt to aqueous CuSO4 and freeze out Glauber's salt (Na2SO4.10H2O):
2 NaCl + CuSO4 (aq) --Freeze--> Na2SO4.10H2O + CuCl2 (aq)
This path avoids the formation of problematic CaSO4 to quote a source (http://homepages.see.leeds.ac.uk/~earlgb/Publications/Vam%20Driessche%20et%20al%20New%20Perspectives%20on%20Mineral%20Nucleation%20and%20Growth%20Springer%202016_Chapter12.pdf):
"Induction time measurements performed by Liu and Nancollas (1973) suggested that the nucleation of gypsum is associated with an appreciable activation barrier, involving a critical nucleus composed of approximately six ions, as predicted based on a “nonclassical nucleation model” introduced by Christiansen and Nielsen (1952)."
The formed CuCl2 can be further purified by concentrating the solution and cooling to collect CuCl2.2H2O.

I would guess you can directly mix copper sulfate solution with concentrated $\ce{HCl}$ to yield dissolved $\ce{CuCl2}$ and $\ce{H2SO4}$. This is of course in competition with (re-)formation of $\ce{CuSO4}$ but I would be interested in seeing how excess $\ce{HCl}$ would affect yield.
Also -- I made $\ce{CuCl2}$ by dissolving copper metal in concentrated $\ce{HCl}$, adding concentrated $\ce{H2O2}$ (slowly) and then evaporating off the water solvent. The dehydrated form is a brown powder and moderately irratiting. Easily soluble in water and has a blue/green solution color. -- So some method of reducing the copper sulfate to copper-metal (e.g. electrolysis), and then this dissolution in $\ce{HCl/H2O2}$ should do it.
Acid fumes deem a fume hood necessary, but decomposition of $\ce{H2O2}$ just gives off water and oxygen gas.
Along the lines of your precipitating idea, using barium chloride or lead chloride to form the corresponding insoluble sulfate would potentially be a faster, one step process, although perhaps more expensive. After, you decant the remaining $\ce{CuCl2}$ and evaporate solvent.

Adding conc. $\ce{HCl}$ to $\ce{CuSO4}$ solution will give a yellow solution containing $\ce{[CuCl4]^{2-}}$ ions. It definitely won't produce $\ce{CuCl2}$ as all the ions will stay in solution.

I thought it may form a more complex ion, thanks. By the way --how does one write subscripts without italicizing (with LaTex)? I'd like to contribute here more often. @bon

@bon Nevermind, I can see it from your edits, ty

See here for information on formatting.

Displacement reaction of copper by sodium will work well if ethanol is used to precipitate the sodium sulfate. Copper chloride is soluble in ethanol (>50 g/100 mL) while sodium sulfate is not. A mixture of concentrated salts in water, followed by dilution with fair amounts of ethanol will allow recovery of high-purity copper chloride dihydrate after proper washing and drying. However, you must check for ethanol contamination in the crystals.


text:
Confusion with the Periodic Table
The periodic table has 7 periods and they have 2,8,8,18,18... elements respectively from 1 to 7. But from what I understand, the periods each state the number of electron shells that the elements in that period has. So if that is the case, shouldn't period 3 have more elements, since it can hold up to 18 electrons, and therefore it can have up to 18 more protons from the largest atomic number element in period 2? (Since period 3 has a M electron shell, it can also have a p orbital, and in total therefore can have 18 elements.)
The same could be said for elements in group 4, group 5, group 6 and so on since they can have f,g, and h orbitals as well. Am I missing something here?

There are a couple assumptions in your question that are really confusing me: (1) "shouldn't period 3...hold up to 18 electrons?" Why should you expect that? (2) "Since period 3 has a M electron shell" I haven't really been in a chemistry classroom for a while--what's an M electron shell? (3) " group 4, group 5, group 6...can have f,g, and h orbitals as well" No naturally-occurring element has electrons in anything higher than an f-shell. If you could explain these three points, I might be able to figure out where the confusion is coming from.

@chipbuster - M corresponds to principle quantum number 3 (K is 1, L is 2, and so on). It's old-style (or X-ray) notation. I would argue that f, g, h (and higher) atomic orbitals exist, but aren't populated for the elements in question.

The standard periodic table isn't arranged according to electronic configuration; it's arranged so that elements with similar properties appear in the same group. There are alternative versions of the periodic table arranged in different ways, including by electronic configuration.

@ToddMinehardt thanks, I had never heard x-ray notation used before. And agreed about the higher atomic orbitals.

So if that is the case, shouldn't period 3 have more elements, since
  it can hold up to 18 electrons, and therefore it can have up to 18
  more protons from the largest atomic number element in period 2?

Indeed, elements of the same period have the same number of electron shells, but the "problem" is that in accordance with the Madelung/Janet/Klechkowski  rule, the $\mathrm{4s}$ orbital is occupied before the $\mathrm{3d}$ orbital. As a result, the $\mathrm{3d}$ orbital could be filled only when the $\mathrm{4s}$ orbital is already filled, i.e. only for elements of the 4th period.
Similarly, the $\mathrm{5s}$, the $\mathrm{5p}$, and the $\mathrm{6s}$ orbitals are occupied before the $\mathrm{4f}$ orbital. As a result, the $\mathrm{4f}$ orbital could be filled only when the $\mathrm{5s}$, the $\mathrm{5p}$, and the $\mathrm{6s}$ orbitals are already filled, i.e. only for elements of the 6th period.


text:
Why are covalent bonds directional?
It is said that covalent bonds are directional, while ionic bonds are not. Why? Is it because of the orientation/directional properties of the overlapping orbitals?

The meaning of covalent bonds being directional is that atoms bonded covalently prefer specific orientations in space relative to one another. As a result, molecules in which atoms are bonded covalently have definite shapes.
The reason for this directionality is that covalent bonds are formed by sharing electrons between atoms, or, in other words, as you said, by overlapping the atomic orbitals of participant atoms. And usually only few patterns of overlap are possible, consequently, only few spatial arrangements of atoms are possible.
Ionic bonds are different: there is no electron sharing (or atomic orbital overlaps) and the number of anions surrounding a cation is limited by the charges of the ions, their sizes, and the efficiency of the lattice packing.

Very well put. Another explanation is that as covalent bonds are made by 'sharing' of electron, the electron will 'spend' more time with the element which is more electronegative. e.g in HCl, H has the electronegativity of 2.1 while Cl has 3, so the 'pull' of Cl on the shared electron is greater than H, hence the direction. Hence covalent bonds show directionality because of the shape of their electron wavefunction. Its the probability of finding an electron in a certain direction which is in turn effected by the partial charges. Even Ionic bonds show directionality (because there is no 'true' ionic' bond) but its very slight.

But there are non-polar covalent bonds too. Those are just as directional as the polar ones are.

Following on from orthocresol's point, I think you are confusing directionality with polarity.

Everyone! Technical inaccuracy != VLQ answer. You're trying to get this deleted, but in this case, as it isn't spam or link-only or comment-y, it doesn't need moderation, just vote about the quality, and the OP can decide whether they wanna delete it or not. || @Muzammil your comment makes better sense, but it's hardly simpler and easier. You might edit it into the answer though.

In covalent bond mutually sharing present between more or less electronegative element.
Eg; if we consider two elements A and B.electronegative of A is more than B.then electron density more on A and less on B.
As a result B easily lose electron means that their is direction of losing electron and it's direction depends upon electronegative of element hence covalent bond is directional bond

-1 Look at the comments on the other answers... "But there are non-polar covalent bonds too. Those are just as directional as the polar ones are." Example: C-C bond in ethane.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. At the moment this reads more like a comment than an actual answer - could you elaborate a little more. With a bit more rep, you will be able to post comments on any question/answer.


text:
Why do chloride ions not bond to water molecules/hydronium ions when hydrochloric acid dissociates?
I am a high school student and I was reading a textbook and it gave me the equation of this hydrochloric acid which was put into water and the products formed were hydronium and chlorine.
$\ce{HCl + H2O -> H3O+ + Cl-}$
I understand that $\ce{HCl}$ (Hydrochloric acid) breaks up in the water to become 2 free ions (Hydrogen & Chlorine) and the Hydrogen bonds to the water because of electronegativity that causes the oxygen to be partially charged and so the $\ce{H+}$ will bond to water giving us $\ce{H3O+}$. This makes sense to me however I do not quite understand why $\ce{Cl-}$ is left "alone".
If there is a lot of water with only little amount of acid added surely it should lead to the $\ce{Cl-}$ also bonding to water or hydronium? Why does it not bond to them?
My Hypothesis
I have attempted to use my knowledge of physics and chemistry to tackle this problem, I am not as well-versed in chemistry so please bare with my "beginners" attempt at understanding this.
Water ($\ce{H2O}$) is a polar molecule due to electronegativity so perhaps as the chlorine is trying to attach to the hydrogen part of the molecule the oxygen's negative field is slightly stronger than the attractive forces of the hydrogen in the molecule of water thus the chlorine cannot bond with the water.
Next hydrogen is covalently bonded to water so their outer orbitals are filled up so there is no need for the chlorine to even bond as the hydrogen has filled its orbitals.
Please criticize my thinking as it will only improve my scientific curiosity and intuition.

Welcome to Chemistry.SE! I see you have figured out some of our MathJax formatting, but if you wrap your chemistry in a \ce{...}, you get even prettier formatting! For example, $\ce{H3O+}$ comes out as $\ce{H3O+}$ and not $H_3O^+$. Here is a link to our [help] page on MathJax (which links to a meta post with more information): http://chemistry.stackexchange.com/help/notation.

We also discourage using Mathjax in question titles because it makes them hard to search, both here and using external search engines. I edited your question to clean up the Mathjax this time, and to choose perhaps a better tag (acid-base).

Actually @ron beat me to it!

Looks like just another dupe of this and this.

@Wildcat It's about chloride not hydrogen ions.