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What Methods Would I Use To Powder Cotton?
I’m needing a method to turn cotton into a fine powder for use as a source of cellulose. What suggestions would you give to accomplish this with materials/tools that could be obtained at local stores?

Would a blender followed by using a mortar and pestle grind it fine enough?

Though the mortar and pestle had come to mind, the blender has been less than successful in application.

Have you tried shredding it by hand before putting it in the blender? To be honest, your only way to convert fibers into powder is through some sort of mechanical action.

At least several synthesis procedures I'm aware of and requiring cellulose as starting material use cotton as it is.

I have hand shredded it, leading me to the conclusion that the blender is still not an effective method. As to the function of whole fibers, the cotton will need to be ball milled for maximum effectiveness in turning cellulose into glucose.

In a lab I would freeze it in liquid nitrogen (so it becomes brittle) and then use a pestle and mortar or a cooled grinder. Probably not too easy to get liquid nitrogen for home though.

Re: Liquid nitrogen. You can sometimes get liquid nitrogen for "home use" (e.g. liquid nitrogen ice cream) from a welding supply store. How easy that is depends on your local laws and your local welding supply stores.

The content of cellulose in paper is not as high as that of cotton, however, it may be easier to use as a source if you do require a fine powder. Why do you require cellulose?

To replicate an experiment in producing glucose. Ref: (http://pubs.acs.org/doi/abs/10.1021/cs300845f).


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Stability of beryllium carbonate
Beryllium carbonate is reported to be unstable under normal conditions and should only be kept in a atmosphere of $\ce{CO2}$. Why and how this stability is maintained? Why does this compound decompose when kept in an atmosphere of oxygen?

Why and how is this stability maintained?

Beryllium carbonate decomposes into Beryllium oxide and Carbon dioxide.
$$\ce{BeCO3 <=> BeO + CO2}$$
According to Le Chatelier's principle one simply has to store it under a $\ce{CO2}$ atmosphere to push the equilibrium to the left.
$$\ce{BeCO3 <<=> BeO + CO2}~\text{in excess}$$

Why this compound decomposes when kept in atmosphere of oxygen?

When there is no $\ce{CO2}$ atmosphere the equilibrium is pushed to the right, which would be the "normal" case under standard conditions.
$$\ce{BeCO3 <=>> BeO + CO2}$$

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Why can't LiAlH4 react with alcohols?
In the reaction of $\ce{LiAlH4}$ with carboxylic acids, deprotonation is followed by a step in which $\ce{O-AlH2-}$ acts as a leaving group. The mechanism is given here in this answer.
$\ce{LiAlH4}$ does not, however, reduce alcohols. Why doesn't a similar reaction occur with alcohols - deprotonation, followed by substitution by a hydride ion and removal of $\ce{O-AlH2-}$?

Alcohols need to be turned into much better leaving groups.

I thought so too, but then as I said above, -OAlH2 does leave. And esters are able to react too, where the leaving group is -OR, which is even poorer.

The main issue is that the Al needs to remove its hydride. With a carboxylic acid and/or an aldehyde, it can stick its hydride onto the carbonyl carbon without issue. But the carbon bonded to the alcohol cannot take on a hydride.

This also explains Charles's confusion about the $\ce{-OR}$ group leaving in ester reduction (in the above comments). The hydride from $\ce{LiAlH4}$ isn't displacing the $\ce{-OR}$ group directly in a substitution-like mechanism, instead it's "only" reducing the carbonyl to a (hemi)acetal, and it's the collapse of the hemiacetal which kicks the $\ce{-OR}$ off, rather than the hydride transfer directly.

So in general, it's easier to substitute an ester and remove the -OR than an ether - since the leaving group isn't removed directly? Nucleophilic substitution is easier at carbonyl carbons of acid derivatives than at alcohols, ethers, amines, etc.? In nucleophilic substitution at any acid derivative - which is the RDS - the nucleophile's attack step or the step in which the group leaves?

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Why do BH3 and NaBH4 have different selectivities?
I've always known borane ($\ce{BH3}$), as a reagent for alkene hydroboration. Recently in class we talked about its use in the reduction of carboxylic acids to alcohols as well.
Now obviously $\ce{BH3}$ is an electrophile and I don't think it reacts by donating hydrides like sodium borohydride ($\ce{NaBH4}$) does - in any case it's not charged like $\ce{BH4-}$. Also $\ce{BH3}$ should be a pretty weak hydride donor, considering the covalent nature of the bonds and lack of any charge. It can reduce carboxylic acids, however, which $\ce{NaBH4}$ can't do! Where's the difference and what's the reason for this contradiction?
And what prevents borane from reducing acid chlorides (my instructor told me this)? This is something $\ce{NaBH4}$ is successful at.

And what prevents borane from reducing acid chlorides (my instructor told me this) || He is incorrect, acyl chlorides ARE reduced by BH3. But this occurs much slower than for acids.

What is the typical nature of a trivalent boron atom?

related, but no explanation given http://chemistry.stackexchange.com/questions/27995/reduction-of-acids-with-borane

Your assertions are correct, for the most part. $\ce{BH3}$ is a Lewis acid, and it does not reduce carbonyl groups by directly donating a hydride group like $\ce{NaBH4}$ does. The $\ce{NaBH4}$ reduction mechanism is fairly short and involves a direct transfer of the a hydride ion to an electron deficient carbonyl carbon:
$\hspace{2.4cm}$
$\ce{NaBH4}$ is incapable of reducing carboxylic acids only because the acidic hydrogen is more reactive towards the hydride ion than is the carbonyl carbon. It is, in fact, capable of reducing carboxylate anions to primary alcohols, though the reaction is very slow. Electron-rich carbonyls favor reactions with electrophilic reducing agents. To help in understanding what makes a carbonyl electron-deficient or rich, let us consider the contributing resonance structures of amides, esters, and acid chlorides:
$\hspace{4.7cm}$
Nitrogen atoms are much better than bearing a positive charge than are oxygen than is chlorine. In this way, you can think of nitrogen donating its lone pair into the carbonyl the most, making it electron-rich, and the chloride donating its lone pairs the least, making it electron-deficient.

The $\ce{BH3}$ reduction mechanism is quite a bit longer and has some interesting intermediates, but here it is:
$\hspace{2cm}$
Note that this mechanism does in fact include hydride transfers (it has to). Though you were right in thinking that $\ce{BH3}$ wouldn't immediately donate a hydride, coordinating to a lone pair weakens the $\ce{B-H}$ bonds and makes hydride transfers possible. $\ce{BH3}$ and $\ce{NaBH4}$ really aren't as different as you might first think.
Though $\ce{BH3}$ is more reactive towards electron-rich carbonyls, it is also capable of reducing electron-deficient carbonyls like acid chlorides and esters through similar pathways. This should show you that there isn't any contradiction here, and that just because two compounds have different reactivities does not mean that they cannot perform the same type of reactions—they just won't perform it in the same way.


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What is the rule for the punctuation marks for a bicyclo compound?
What is the rule for the punctuation marks within the bracketed figures for a bicyclo compound? Is it a 'dot' or 'comma'? In some books there are commas whereas dots in some others.

Why in the world did you roll back?

Mostly dot is used, so think dot is more correct

Do you have some examples of those structures that you are talking about?

I am talking of all bicyclo compunds as a whole. I am confused because commas and dots are qually distrbuted all over the net and books !

The current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-16.2.2 Full stops
Full stops are used to separate numbers that denote lengths of bridges in polyalicyclic names constructed according to the von Baeyer system (…) and of chain lengths between spiro atoms in von Baeyer spiro names (…).

Examples:
bicyclo[3.2.1]octane
6-oxaspiro[4.5]decane


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Chemical reactions involved in staining
I'm a geology student completing a project on chemical staining and was after some chemistry help on some of the reactions taking place. I am using various different stains to identify carbonate and feldspar minerals.
One of the stains, in particular, is alizarin red S (1,2-dihydroxyanthraquinone) and is used to stain calcite ($\ce{CaCO3}$) red and Fe-calcite/Fe-dolomite blue. The stain powder is mixed with a weak (2%) HCl solution to react with the carbonate and produce the stain. However, I am just unsure about what reactions are taking place between the stain solution and the mineral that produce the precipitate on the surface of the grains.
I'm not finding literature on this reaction - probably because it has been mainly used by geologists without a solid chemistry background. Any help would be much appreciated.

If my response answered your question, please mark it as 'accepted' - thanks!

Alizarin red S can stain calcium deposits through interaction of its sulfonic acid and hydroxy groups with calcium ions.
I found a reference for you: http://jhc.sagepub.com/content/17/2/110.full.pdf+html
Interestingly, however, this research seems to suggest using a solution of pH $\approx$ 9, stating that solutions of acetic acid (pH 5) did not work that well.


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What are some good references on conductometry as an analytical technique?
Non of the standard undergraduate textbooks in analytical chemistry (such as Skoog, Harris) discusses conductometry as an analytical technique. I wonder whether this technique is obsolete. Or is it taught under some other area such as physical chemistry ?
Can someone please point me to some good references on this matter ? I am particularly interested in text books.
Thanks.

I am certainly no expert in the field (in fact, I know nothing about it) but I imagine this 225+ page text will prove to be useful for you:
Oscillometry and Conductometry: International Series of Monographs in Analytical Chemistry, Volume 21 by E. Pungor is available via Google Books. You might be able to find a used copy of this book somewhere (it was published in 1965).


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Graph of compounds/reactions used on the industrial scale
I'm looking for a machine-readable database describing industrially useful reactions in terms of inputs, outputs, and (possibly) catalyzers etc. for a large portion of, or (much more preferable) the entire organic and inorganic chemistry. A tall order, I know. Non-free, paywalled data references are acceptable, although free datasets are better.
As an example:
1-Aminoanthraquinone -> Alizarin
1-Aminoanthraquinone -> Mitoxantrone

EDIT: There's of course the Wikipedia, yet I cannot currently find out how to use Wikidata to mine the relationships between compounds. Many Wikipedia articles cite Ullmann's and other chemical cyclopaedias but I'm unaware of the latter having been converted into machine-readable form.

I don't know how machine-readable they are (and such uses might be against the EULA and T&C), but such databases exist: SciFinder Scholar and Beilstein/Gmelin available through Elsevier's Reaxys.

For example, SciFinder's Terms & Conditions warn of vague consequences in the event of "above average" usage, is limited to academic, government, and non-commercial use, and specifically mentions You may not use federated search tools, robots, spiders, crawlers, or other automated downloading programs, algorithms or devices, or any similar or equivalent manual process, to continuously and automatically search, scrape, extract, deep link, or index any content.

@BenNorris - thanks, Elsevier is quite an exemplary "money for nothing" publisher. Got no desire to scrape them, though.

If it's any use there are two refs I can dig out. One is the frequency / tally of industrial organic reactions ; the other is the same but of reactions used in medicinal chemistry.  I can't help with the inorganic.

@Beerhunter - would be very much grateful. Inorganic is of lesser practical importance.


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What is the main difference between Quantum Clusters and Nanoclusters?
I know that Quantum Dots have high quantum yield and ranges from 2nm-10nm, and Nanoclusters have high fluorescence and ranges from 1nm-10nm. But the difference between them is that Quantum Dots exhibits red shift as the size gets larger while vice versa for Nanoclusters. My guess is Quantum Clusters is the combination of both, size is less than 2nm and exhibits red shift as size gets larger. Is this correct? Are there more things that distinguish them to each other?

Since these are frequently used as terms to describe a variety of different materials, it might help if you can specify some examples of this usage.. My immediate response is "nano clusters of what exactly.."

@Geoff Hutchison What I mean by Nanoclusters is Gold Nanoclusters.

OK, but what do you mean by "quantum clusters"? I don't think I've ever heard that term.

@GeoffHutchison There are papers already published about gold quantum clusters, but my concern is that the mechanism of gold nanoclusters are, as the size gets larger it tend to have a blueshift, while quantum dots tend to have a redshift as size gets larger. There some papers that were published and titled their paper gold nanoclusters but the properties exhibited is like QDs, redshift as size gets larger. So I'm confused, why not call it quantum clusters since its size is less than 2nm and it exhibited redshift?

Calling sth "Nanoclusters" or "quantum clusters" doesn't even sound scientific - more like sth found in commmercial.

@iso345 Scientific communication is precise. This means you must pick terms that others will find understandable. Quantum dots and metal clusters are very different materials, and attempting to hybridize terms is likely to be very confusing.

Short answer: "quantum clusters" is not widespread and likely confusing. I'd suggest "small nanoparticles" or perhaps "metal nano clusters."
I can't comment on optical shifts, but many of these properties are distinct to the particular material and the onset of quantized energy levels in that material. So I doubt you can simply combine properties of quantum dots and nanoparticles.
I asked a colleague who synthesizes gold nanoparticles. She hasn’t heard anyone use the term "quantum clusters" and thinks people would find it very confusing.
Typically the usage:

Quantum Dots refer to semiconductor nanoparticles, e.g., $\ce{CdS}$ or $\ce{PbSe}$ etc.
Nanoparticles refer to nanometer-sized particles of basically anything (proteins, polymers, metals, etc.)
Clusters refer to very small particles (i.e., <1-2 nm diameter) with well-defined numbers of atoms (i.e., a few hundred at most). There is an IUPAC Gold Book definition:

A number of metal centres grouped close together which can have direct metal bonding interactions or interactions through a bridging ligand, but are not necessarily held together by these interactions.

For reference, Wikipedia suggests nanoclusters are aggregates of nanoparticles. I don't think that's in common use, but certainly it suggests your suggested term would be very confusing.


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Why are the mass ratios of O to KClO3 and O2 to KClO3 the same?
I just did a lab that involved heating KClO3 (with MnO2 as a catalyst) to decompose it into KCl and oxygen gas. That part is easy. What's throwing me is the % yield calculations.
The theoretical yield of O should just be the mass percentage of O in KClO3 (48.00/122.55 = KClO3 is 39.17% O by mass).
The actual yield of O2 (measured by subtraction) was 0.840 g out of a initial mass of 2.062 g KClO3, or 40.7%.
My question is this: Given that the theoretical yield was 39.17% based on moles of O (not O2) , why is the actual yield 40.7% (well with margins of error outlined by the professor) when it's based on O2? Shouldn't it be half as much? Or did I mess my calculations up?
Please feel free to edit tags, I'm out of my depth on this particular SE site in that area.

Your calculations are correct. The key is that you are measuring percentage by mass: the mass of oxygen in 2.062 g of $\mathrm{K}\mathrm{Cl}\mathrm{O}_3$ is the same whether it is considered the mass of individual oxygen atoms or of $\mathrm{O}_2$ (or of any other allotrope of oxygen, for that matter). You have half as many moles of $\mathrm{O}_2$, but the mass is the same (since the mass of 1 mol $\mathrm{O}_2$ is twice the mass of 1 mol O).


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sp2 hybridization Group V elements
I'm confused about $\mathrm{sp^2}$ hybridization in the formation of $\ce{NO3-}$ ion. Valence electrons of N are $\mathrm{(2s)^2}$ and $\mathrm{(2p)^3}$.
Is one of $\mathrm{2s}$ electrons kicked up to one of the $\mathrm{2p}$ orbitals, and then the remaining $\mathrm{2s}$ electron hybridized with the two remaining $\mathrm{2p}$ orbitals which have unpaired electrons? If so, that gives $\mathrm{sp^2}$ orbitals to bond with $\mathrm{p}$ orbitals of the oxygen atoms.  However, there is now that remaining unhybridized $\mathrm{p}$ orbital of nitrogen, which has 2 electrons, so how can it be involved in $\pi$ bonding?
I would appreciate someone explaining this. I'm sure I'm missing something simple, but my book doesn't have an explanation of $\mathrm{sp^2}$ hybridization for Group V elements.

So the lone pair of nitrogen provides both electrons for the delocalized pi bond, and the oxygen atoms do not contribute an electron to the delocalized pi bond?

I withdrew my earlier comment. Although I could simply share with you that number of electron regions around an atom is an indication to quickly figure out its hybridization, I am not able to explain this in detail, because I have a sub question when looking at yours: I am not sure how the sp3 hybridized oxygen that has formal charge -1 is making a bond with sp2 nitrogen which has +1 formal charge.

@bonCodigo, oxygen is thought as being unhybridized in this case. Rather, you think of the whole ion as being composed of $\mathrm{sp2}$-hybridized $\ce{N+}$, two unhybridized $\ce{O-}$, and one unhybridized $\ce{O}$.

@Wildcat this has been eating me. Almost for all ions $\ce{CO3^{-2}}$ etc

@bonCodigo, well in principle, you can invoke $\mathrm{sp^3}$ hybridization for $\ce{O-}$ ions as well as $\mathrm{sp^2}$ hybridization for $\ce{O}$ atom. But, it won't change anything, so you don't need it.

However, there is now that remaining unhybridized p orbital of
  nitrogen, which has 2 electrons, so how can it be involved in π
  bonding?

Everything is more-or-less "right" up to these point. The thing you're missing is that there is only one electron in the unhybridized $\mathrm{p}$ orbital of $\ce{N}$ atom since to get the following usual configuration of $\ce{NO3-}$ ion (image courtesy of Wikipedia)

with two negatively charged $\ce{O-}$ ions one electron has to be moved from $\ce{N}$ atom to $\ce{O}$ atom first. Now note that since all $\ce{O}$ atom are equivalent to each other we need to invoke the concept of the resonance, but for simplicity think first about any single of the resonance structure above.
So, formally we start by having:

one $\ce{N}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^3}$ electron configuration;
two $\ce{O}$ atoms with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;
and one $\ce{O-}$ ion with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;

In the first step we transfer one of the $p$-electrons from $\ce{N}$ atom to an $\ce{O}$ atom, so we end up with:

one $\ce{N+}$ ion with $\mathrm{(1s)^2 (2s)^2 (2p)^2}$ electron configuration;
one $\ce{O}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;
and two $\ce{O-}$ ions with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;

At the next step we invoke the $\mathrm{sp^2}$ hybridization of $\ce{N+}$ ion, so that we get:

one $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion with $\mathrm{(1s)^2 (sp^2)^3 (2p)^1}$ electron configuration;
one $\ce{O}$ atom with $\mathrm{(1s)^2 (2s)^2 (2p)^4}$ electron configuration;
and two $\ce{O-}$ ions with $\mathrm{(1s)^2 (2s)^2 (2p)^5}$ electron configuration;

Finally, we form bonds as follows:

One $\sigma$ bond between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion and an $\ce{O-}$ ion in forming which we use 1 of the total 3 $\mathrm{sp^2}$-electrons of $\ce{N+}$ ion and the only unpaired $\mathrm{2p}$-electron of $\ce{O-}$ ion;
Another identical $\sigma$ bond between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ and another $\ce{O-}$ ion; here we use 1 of the remaining 2 $\mathrm{sp^2}$-electrons of $\ce{N+}$ ion and the only unpaired $\mathrm{2p}$-electron of $\ce{O-}$ ion;
Yet another $\sigma$ bond, but this time between the $\mathrm{sp^2}$-hybridized $\ce{N+}$ ion and the $\ce{O}$ atom in forming which we use the last remaining $\mathrm{sp^2}$-electron of $\ce{N+}$ ion and 1 out of total 2 unpaired $\mathrm{2p}$-electrons of $\ce{O-}$ ion;
What remains now is one unhybridized $\mathrm{2p}$-electrons of $\ce{N+}$ ion and one unpaired $\mathrm{2p}$-electrons of $\ce{O-}$ ion which are used to form the $\pi$ bond.

Pictorially (image courtesy of Professor Peter Bird):

Thank you, thank, thank you!  I was so frustrated trying to figure this out, and couldn't find an explanation on the web or in other textbooks.  I wish your clear, logical explanation was in texts; they seem to have only the simplest hybridization examples.  I know this took some time to write your response; I,

(to continue) I'm very appreciative of your kindness.  Am sure others looking for an explanation of NO3- on the web will benefit as well.


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Food Grade Sodium hypochlorite decomposition/neutralization
I recently purchased some 55 gallon blue barrels for rainwater harvesting.  Upon closer inspection of the barrels, I noticed they previously housed food grade sodium hypochlorite which was used to wash produce.  I was searching for food grade barrels but wasn't expecting bleach.  Because I will be watering plants and fruit tress with the rainwater, I don't want the water chlorinated which would kill the bacteria I'm trying to promote in the soil.  These barrels are empty but I was advised to rinse them when I initially purchased them.
Can I assume that just rinsing with water won't completely get rid of the bleach  and that the residual bleach which may be on the walls of the barrel will not decompose or neutralize by itself?
If I were to wash the barrels with water and Hydrogen Peroxide, will this suffice to remove the chlorine?

Several good tap water rinses should probably get rid of any significant remaining amounts of bleach, although it's possible some bleach residues might adsorb to the wall of the barrel and resist these efforts.
Per here and here, if you want to be very sure the residues are removed, dissolve a handful of vitamin C (ascorbic acid) tablets in water and use that to rinse the barrels until all bleach odor is gone.  The ascorbic acid will act as a reducing agent and should reduce the bleach levels to next to nothing.
Dmitry is right: as an oxidizer, $\ce{H2O2}$ will probably do little to help.

Hydrogen peroxide reacts with bleach to yield oxygen gas, so it would also be useful to eliminate the bleach.  The problem isn't that it doesn't react, it's that how do you get rid of the excess hydrogen peroxide after you react the bleach away?  H2O2 is probably worse for soil than oxidized vitamin C.

Sodium thiosulfate is used in aquarium chlorine removal. It's an alternative to ascorbic acid, and cheaper. In the quantities used in aquarium maintenance, it's safe for fish, and certainly for the much hardier bacteria and protists.

What are you afraid of exactly? The quantity of bleach remaining after the rinse is certainly miniscule, and it will exponentially drop each time you refill the barrels with rain water.
If you're really concerned (which IMO you shouldn't be), you can refrain from using the water you collect the first time. Also, the bleach is designed to react with organic compounds (dirt), so rinsing the barrels with dirty water will make the remaining bleach react with it.
Hydrogen Peroxide is an oxydizer just as the bleach, so it will not help.

Indeed, when washing glassware, most chemists will rinse repeatedly as you indicate.

Hydrogen peroxide reacts with bleach to yield oxygen gas, so it would also be useful to eliminate the bleach.  The problem isn't that it doesn't react, it's that how do you get rid of the excess hydrogen peroxide after you react the bleach away?  H2O2 is probably worse for soil than oxidized vitamin C.

@CurtF. Hydrogen peroxide is a common (minor) side product in oxidation of organics by aerobic organisms, and they learned to deal with small amounts of it rather efficiently.


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Potential energy Surfaces
I am trying to calculate PES for Molecular Oxygen and Atomic Nitrogen Exchange reaction. This is also known as Zeldowich reaction. I found that for  asingle reactions, there are many PES's which I believe are for different electronic configurations of the NO2 PES. I found that the multiplicity can be 2 or 4. Also I find different surface being named as A' and A" in the literature. I would like to what exactly are these and how can I compute these in Gaussian?

Could you include the citation to the literature you are referring to, please. A scheme of the reaction and/or a reference would also be very much appreciated.


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Distance weighted interpolation in predictor steps vs corrector steps
I got stuck while reviewing the journal article "Accurate reaction paths using a Hessian based predictor–corrector integrator".
Why is the distance weighted interpolation done in the predictor steps, instead of doing it in the correcter steps?
Source:  H. P. Hratchian and H. B. Schlegel, J. Chem. Phys. 2004, 120, 9918


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Photoconductivity of Selenium
Selenium is a good conductor of electricity in presence of light. This phenomenon is called photoconductivity. How does this phenomenon occur? Are there any other elements or compounds in which this phenomenon is noted?

A very handy diagram and explanation of photoconductivity is shown below, from the American Physical Society page Viewpoint: Holey Intrinsic Photoconductivity:

The caption explains:

Excitation process leading to photoconductivity in a condensed matter system (upper panel) and its cold-atom counterpart (lower panel). In the condensed matter system, incoming laser light ($hv$) excites an electron to the valence band, leaving a positively charged hole behind. In the cold-atom analogy, the optical lattice is tuned to allow an atom to move into the excited band, also leaving a hole behind.

From the Wikipedia page Photoconductivity, states that some 'classic' examples of photoconductive materials (asides from selenium) include:

The polymer polyvinylcarbazole
Lead sulfide

Some more examples exist in the paper Organic photoconductive materials: recent trends and developments include the chemical classes:

Phthalocyanines
Squaraines
Azo pigments
Perylene pigments

User15489(Santiago)'s answer explains the phenomenon of photo-conductivity in a general way. My answer typically describes the photoconductivity of selenium.
The following is an excerpt from the book Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed:- (see link)

Selenium exhibit both photovoltaic character, where light is converted directly into  electricity, and photoconductive character. The photoconductivity of grey selenium arises from the ability of incident light to excite electron across its reasonably small band gap (2.6 eV in the crystalline form, 1.8 eV in the amorphous form). These property make selenium useful in the prdouction of photocells and exposure meters for photographic use, as well as solar cells.


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Why do transition metals tend to form positive ions instead of negative ions?
I have learnt that elements on the left-hand side of the periodic table such as sodium and magnesium prefer to lose electrons to form a cation because this requires less energy to obtain a stable octet, and vice-versa for the right-hand side of the periodic table e.g. fluorine. However, using this reasoning I am not sure why all transition metals tend to lose electrons rather than gain them.

http://chemistry.stackexchange.com/a/34134/9961 - there is also tendency for anions.

See the outer configuration of some transition elements:

Sc:      $\mathrm{[Ar]~3d^1 4s^2}$
Cr:  $\mathrm{[Ar]~3d^5 4s^\color{red}{1}}$
Co:  $\mathrm{[Ar]~3d^7 4s^2}$
Cu:  $\mathrm{[Ar]~3d^{10} 4s^\color{red}{1}}$
Zn:  $\mathrm{[Ar]~3d^{10} 4s^2}$

If you are wondering about configuration of Cu and Cr read why this happens!
As you may notice, they can form ions by either losing or gaining electron in 4s orbital. In the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. When these metals form ions, the 4s electrons are always lost first, leading to a positive charge on ion. Gaining any electron in 4s orbital would decrease the stability of anion thus formed. So, they form only cations (positive ions).
In general, the outer electronic configuration of transition elements is $n~\mathrm{s}^2~(n-1)\mathrm{d}^{1-10}$. Because of reasons above, the electron removes from  the $\ce{ns}$  orbital, where n = principal quantum number of atom.
See $\ce{Zn^{2+}}$ configuration:  $\mathrm{[Ar] (3d)^{10}}$ as an example. The electron exits from 4s orbital, leading to the formation of a positively charged ion.

Please tell me if I understand this correctly: the 4s orbital is higher in energy than the 3d orbital, so the metal will prefer to lose electrons from this orbital to lower its energy. These transition metals are not close to having a full outer shell unlike elements such as fluorine which only require 1 electron to gain a full outer shell, so fluorine gains an electron to become more stable.
I think my problem was that I got orbitals and shells confused, thinking 4s and 3d were shells rather than orbitals in a shell.

Yes you get it right. Transition elements don't form ions with complete outer shell. For ex. $Co^{2+}$ has the configuration $[Ar] 3d^7$.

In simple words, s,p,d,f are orbitals and 1,2,3,4.. are shells

I know this will probably start a confusion, just to add to your answer,  lower TM's like Pt can form anions (Pt- and Pt2-) In Pt's case the relativistic stabilisation of the 6s electrons allows the negative oxidation states

As you may notice, they can form ions by either losing or gaining electron in 4s orbital. In the chemistry of the transition elements, the 4s orbital behaves as the outermost, highest energy orbital. When these metals form ions, the 4s electrons are always lost first, leading to a positive charge on ion.


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Data repository of organic compounds
Does there exist any data repository of chemical compound that gives the structure of the compounds with the help of coordinate geometry. It would be like each atom has a coordinate and the list of bonds it has with other atoms so that the 3D structure of the compound can be created from this very theoretical description?

I'd keep in mind that the 2D line drawings that we draw for molecules are the "theoretical" models; describing a molecule with just atom coordinates is way more grounded in physical reality.

@CurtF. Except that an organic molecule may exist in a wide variety of conformations, and the preferred conformation may depend on the solvent, so a huge variety of different datasets may be necessary for a single substance. A 2D line drawing with bond lengths and angles, plus a bit of knowledge of polarity and hydrogen bonding will help to decide if the model is reasonable. For example, one of the first proposed helical models for DNA had the phosphate groups on the inside. It was later realised that in aqueous solution the phosphate groups would have to be on the outside.

Very true Steve.  I was thinking of atom coordinates derived from things like x-ray crystallography, more or less experimentally measured atom coordinates.  But in the case where those measurements aren't available then everything you say is right on the money.

I'm not entirely clear on what you want - I guess a database of a bunch of 3D coordinates for many organic compounds?
By far the largest such database is NIH PubChem with over 30 million compounds - mostly organic. You can search by structure or name or identifier. Most of these have 2D and 3D geometries you can visualize or download and many computed properties.
A similar database is the RSC ChemSpider with over 34 million compounds. Much like PubChem, names, 2D and 3D geometries and other properties are available.
I'll put in a plug for the Pitt Quantum Repository which contains ~64,000 molecules and growing, all with geometries computed using quantum chemical methods. One benefit of PQR is that you can easily access the site using smartphones and tablets and use a QR code or DOI to link to the compound.

What happened to PQR? Whenever I try to access it, I'm only getting an error, which seems to be caught in a loop.

It is https://pqr.pitt.edu/502/ if that helps any, and thank you for taking a look. And it just says 'There was an error, please try again
You will be redirected to the home page in 5 seconds'. In chrome I got a warning about unsafe scripts, but I allowed that and nothing changed. I've tried a few OS (win,xubu,and) all with chrome and different networks... all samesies. It's a great resource, it should be used! :D

A pretty good database of experimental data compiled into one place has a geometry section. Check out
http://cccbdb.nist.gov/
Go to the experimental data section, then click on summary of geometry data for one molecule. The database is fairly large, but not all molecules have a full geometric data set.

I think the NIST resource is great, but I've never understood why it's so hard to get the geometries.

I agree - I imagine that finding the geometry experimentally gets more and more difficult with increasing molecule size (due to increasing vibrational modes) I forgot to mention that the NIST site also has a calculated data section, which has a ton more geometries as calculated by different models.

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text:
Why does hybridization significantly matter for organic chemistry?
I just recalled, and still see that almost every organic chemistry book starts with bonds, empahsizing hybridization. One thing is straight forward but the rest doesn't seem to connect well.
The thing that's clear is, energy level wise $\mathrm{sp < sp^2 < sp^3}$. When $\mathrm{p}$ character is higher, the energy is larger; electrophilicity and reactivity are higher as well. This is the opposite of what electrons are constantly trying to achieve: a stable lower energy state. However, hybridization allows molecules to have an energy-minimizing shape. The (higher) energy can also be released through bonding, further stabilizing itself; so bond formation is the tendency.
Hybrids ($\mathrm{sp, sp^2, sp^3}$) form $\sigma$ bonds while pure-breeds (100% $\mathrm{s}$, 100% $\mathrm{p}$) form $\pi$ bond.
I had the impression hybridization regulates a standard bond among same species of atoms, e.g $\ce{C-H}$ bond in all alkanes are the same. But that's not the case when I read brief comments indicating that the $\ce{C-H}$ bond in $\ce{CH4}$ differ from $\ce{CH3-CH3}$.
What I am unable to understand is, why hybridization is so important for organic chemistry than to any other (general, inorganic..., physical)? Is it about bond strength (how easily one can be broken) or is it about acidity or basicity? Why should we (learners) or organic chemists care?
Please provide some applications as examples. It's also great if one could point out how each $\ce{C-H}$ bond differ from alkane to alkane or any other species.

Now that you ask, I don't really know why this is taught - it's more like problem of teaching system as differences  in C-H bond strength aren't thaught well. Maybe because it's easy to tell about it.

Are you certain that "pure breeds" which are 100% s form pi bonds?

One reason why you may think that hybridisation is less significant in inorganic chemistry may be that inorganic chemistry is also made up of the study of much heavier elements, in which orbital hybridisation takes place at a much smaller extent due to larger energy gaps and size differences between the valence orbitals of different subshells.

Orbital hybridization is mostly useful in the rationalisation of molecular geometry. In fact it was invented by Linus Pauling for this purpose: to rationalise structures of organic molecules using the language of atomic orbitals. Yes, hybridization is also used to rationalise bonding properties, as well as acidity or basicity, but its primary use is to rationalise molecular geometries.

I had the impression hybridization regulates a standard bond among
  same species of atoms, e.g $\ce{C-H}$ bond in all alkanes are the same. But
  that's not the case when I read brief comments indicating that the $\ce{C-H}$ bond in $\ce{CH4}$ differ from $\ce{CH3-CH3}$.

Well, $\ce{C-H}$ bonds, of course, differ in $\ce{CH4}$ and $\ce{CH3-CH3}$, but not too much; they are approximately the same. Are you naive enough to hope that a model as simple as the orbital hybridization one would perfectly predict the molecular geometry of each and every organic molecule? Surely, it won't. It can just give a crude (but simple) approximation.

What I am unable to reason is, why hybridization is so important for
  organic chemistry than to any other (general, inorganic..., physical)?

Where did you get that? Hybridization is a concept of general chemistry, it is used in both organic and inorganic one. Its relatively big importance for organic chemistry is due to the fact that it is the only simple model which can explain (approximately) molecular geometry of organic compounds. The original valence bond theory couldn't do it, so it was extended.

On explanation vs rationalisation
Now I would like to emphasize that hybridization provides a rationalisation of a molecular geometry (as well as some other properties), but not an explanation, so it can not be used to reliable predict the properties. What is the key difference between an explanation and a rationalisation in these context?
Well, an explanation of some phenomenon describes it in way that we can observe and test the consequences of our claims and premises. We can then tell from the observations are our claims and premises true or not, or, in other words, does our explanation work or not. The whole point of an explanation is to distinguish what is true from what is not true, in a sense, what corresponds to the actual reality from what does not do so.
In a rationalisation things work differently. We insist on some premises being true without testing them against the reality. We do observations, but we do not use them to tell are our premises true or not, rather on the basis of the desired premises and the observable evidence we start to make the observed outcome seem consistent with the premise.
In psychology rationalisation is also known as making excuses. It is a defence mechanism in which a controversial (from the society viewpoint) behaviour is justified "in a seemingly rational or logical manner to avoid the true explanation". With respect to scientific theories, it can also be thought of as a defence mechanism: when the real explanation is too complex for beginners we avoid it to defend them. Instead we first provide some simple rationalisation and later in the course of study the explanation is introduced.

The way rationalisation works for all these general chemistry models is as follows:

We do know geometry or some other property of a particular molecule, say, from an experiment, and we are seeking for some rationalisation of why the things are the way they are.
The rationalisation usually has to be as simple as possible, so that, for the case of molecular geometry, we usually start from VSEPR theory.
If the simplest possible theory can not rationalise the observed property, we choose a more sophisticated model. For molecular geometry, for instance, we could then try to use the original valence bond (VB) theory.
But even the VB theory (in its original formulation) won't work for organic compounds, so it has to be extended by two concepts: resonance and orbital hybridization. Note that this concepts are also required in inorganic chemistry as well, but not always: for some inorganic molecules the original VB theory is enough.

But however deep you go into this hierarchy of chemical models, all them are just rationalisation models: they provide simplified (and in many cases oversimplified) descriptions of the reality which is approximate (and in many cases rather crude). And besides all this process of seeking for an rationalisation is a bit like cheating:

If some model doesn't work, let us try another model.
If none of them works, let us introduce a new concept with some funky name in one of them, the purpose of which will be to rationalise this particular thing which couldn't be rationalised by the original model.

As a result, the predictive power of such models is quite small. They can be used to rationalise what is already known, but any predictions based on such models will be unreliable. And yeah, the concept of orbital hybridization is just one of these cheats mentioned above.

Is it really only about molecular geometry? Doesn't hybridization play a role in each functional groups' reactivity? How does it affect the bond cleavage? Can you please give some examples.

@bonCodigo, to be honest, it is all about nothing. Orbital hybridization is an oversimplified model which has very little to do with the reality. It can be used here and there to explain some things, but there is very little generality in such explanations.

I did read after you pointed out. Yes I understand it's a model. What's the reality? is it a too naive question to ask? Will a goo gray dark reality struck students in graduate level studies under chemistry or pharmacology?

@bonCodigo, if students realize that it is just a crude oversimplified model which can not explain everything, has a lot of exceptions and a very small predictive power, then it won't struck them. But if they naively hope that such simple model is the end of the story, then there are some bad news for them.

@bonCodigo, grad students are involved in some scientific research and if in their research project they will try to blindly use such oversimplified models to obtain some data about molecular species under the study (geometries, bond strengths, etc.) they are guaranteed to have big problems. These simple general chemistry models can be used as a part of the so-called "chemical intuition" to quickly skip a lot of the details and possibilities, and concentrate on the most chemically possible things, but that is it. Then you need something less crude to do the actual job.

I just want to be prepared. Care enough to guide me to a material where grad level (coursework not research) chemistry theory is related to these topics (the crude/skipped details)?

@bonCodigo, I have never seen any such source and I even doubt it exists for a couple of reasons. 1) Hybridization is just a piece of the VB theory, so it is not used lonely to explain properties of organic molecules, rather the whole VB theory is used. I don't know where did you get the impression on the very special role of hybridization, but I would say it is the wrong impression. The whole VB theory is what matters, not its individual parts.

@bonCodigo, 2) But even the whole VB theory is not the whole story. Many book also use another theory, MO theory, to explain some properties of organic molecules.

@bonCodigo, both these theories are integral parts of organic chemistry, you can't just use a part of one of them (hybridization) to explain a particular property or a trend in it. Organic chemistry is not that simple. For some properties (like geometry) original VB theory + hybridization is all you usually need. For some other properties and some molecules you need also to consider resonance. For third properties we use MO picture, etc. There is a little generality, as I said, just read or at least skim over any textbook on organic chemistry (the whole book, not just the first chapters).

Thanks much! My main reference is a gold Morrison and Boyd and also skimming a bit from early chapters of I. Fleming text book (which is for grad level) while facing many stumbling blocks. Anyway, I was attempting to answer this question using basics of Lewis dot, hybridization, bond order, formal charge, resonance but I was stuck. I have a sub question. If time permits you, please clarify.

it read like a charm @Wildcat

would it be correct to say that the only thing actually governing "bonding" is electromagnetism and quantum mechanics, and that pretty much all of chemistry is just an attempt to intuitively interpret the abstract physics?

@OVERWOOTCH more or less. You certainly can't say something like that about all of chemistry, but theoretical chemistry (consisted nowadays primarily of quantum chemistry) does nothing but explains chemical phenomena by methods of theoretical physics. Now, almost any such explanation necessarily involves some simplifications and model concepts, such as, for instance, the concept of chemical bonding.

@OVERWOOTCH And yes, you got the reason for all this business right - without such simple models it would not be possible to intuitively understand chemistry since purely physical descriptions of molecular systems are insanely complex. All in all only computers can deal with "real" physical descriptions of molecules, while we, humans, have to be satisfied with something simpler.

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text:
Why is tetraamminecopper(II) a square planar and not a tetrahedral species?
In $\ce{[Cu(NH3)4]^2+}$, the $\ce{Cu^2+}$ ion has 9 electrons in the $\mathrm{3d}$ orbital with only one unpaired electron. How is it a square planar geometry? Where is that unpaired electron going?

It is very convenient to use crystal field theory to discuss this.
It is usually assumed that in octahedral coordination the energy levels of the five d-orbitals are split, with two orbitals ($d_{z^2}$ and $d_{x^2-y^2}$) well above the other three. The splitting is assumed to be large enough to overcome electron pairing energy.
The first six electrons will populate three lower orbitals ($d_{xy}$, $d_{xz}$, $d_{yz}$). The next two electrons should occupy $d_{z^2}$ and $d_{x^2-y^2}$. However, if these two electrons occupy $d_{z^2}$, two corresponding opposite ligands become well shielded and leave. This is the common case of $d^8$ complexes, such as $\ce{Ni^{+2}}$, $\ce{Pd^{+2}}$, $\ce{Pt^{+2}}$, $\ce{Rh^{+1}}$, $\ce{Cu^{+3}}$ etc.
Adding one more electron makes the remaining ligand's bonding weaker. The charge of bivalent cation $\ce{Cu^{2+}}$ is large enough to bond four charged ligands if available, such as in $\ce{[CuCl4]^{2-}}$, but the ligands are bound weakly and easily dissociate.
However, $\ce{Cu^{+2}}$ ions usually adopt a distorted octahedral geometry, with two ligands having a longer bond length than the four others. These two are very weakly bound and exchange quickly. For this reason the $\ce{NH3}$ complex is written with only four molecules; the two other are so weakly bound.
This type of distortion is a case of Jahn–Teller distortion, very common in chemistry.

Why should we use octahedral splitting system if its a square planar species ?

@DhrubaBanerjee square planar species usually have enough place for two extra ligands. We do need to understand why it does not accept these ligands if available, so we have to consider octahedral splitting as well.  It helps to draw the d-orbitals in octahedral splitting to understand why $d^8$ often (but not alwasy) square planar.

Where does that unpaired lone electron get accomodated in such case ?

@DhrubaBanerjee $d_{z^2}$ orbital, which in square planar species is perpendicular to the ligand sqare. In $\ce{Cu^{+2}}$  $d_{z^2}$ contains two electrons and also is perpendicular to square of strongly bound ligands, while $d_{x^2-y^2}$ with one electrons positioned near ligands. Other three $d$-orbitals occupied by two electrons each are positioned either between ligands ($d_{xy}$) or are perpendicular to the square ($d_{xz}$, $d_{ys}$)

@permeakra Okay, but where did that unpaired electron go? Because, if it remained in the $d_{x^2-y^2}$ orbital, then the ion would be $sp^3$ hybridised, but it is not. What is happening here?

@AritraDas | the ion would be sp3 hybridised | Why so? Forget about hybridization when talking about 3d metal complexes, they are guarded by d-shells and ion radii.

Since you are describing bonding in your answer, you are actually using ligand field theory as opposed to the older and a bit incomplete crystal field theory.

@Martin-マーチン Is it true that the unpaired electron in d-orbital of Cu is promoted to 4-p orbital for greater stability by involvement of d-orbital in bonding?

The short answer is that $\ce{[Cu(NH3)4]^2+}$ does not exist; the compound you are observing is $\ce{[Cu(NH3)4(H2O)2]^2+}$. It is not square planar but a Jahn-Teller distorted octahedron. You can check out its structure in the image below.

You can fill electrons into the energy diagram in a standard fashion. You will arrive at the $\mathrm{d}_{x^2-y^2}$ orbital for the unpaired electron — the one pointing towards the four ammine ligands.
Note that other copper(II) complexes may well be tetrahedral, as is the case for $\ce{[CuCl4]^2-}$. A combination of large ligand size around a small central metal and too much negative charge for a small species causes the tetrahedric structure to be preferred over $\ce{[CuCl6]^4-}$.

This is quite interesting. Can you tell me the source by any chance? Because I need to use this piece of information for a write-up.

All Cupric complexes irrespective of the kind of ligands is always dsp2 square planar hybridization . Now the question is that why this is so?
In tetrahedral complexes when Cu is in +2 state the one electron from 4s2 and one from 3d orbital gets removed and so it a 3d9 complex .Now from 3d9 complex one electron goes to 4p orbital and in this way it now becomes 3d8 and dsp2 complex.

Promotion from $3d$ to $4d$?  Huh?

Just don’t use hybridisation for transition metal complexes; it does more harm than good.

I have edited my answer and its 4p instead of 4d.

Hybridisation theory for TM complexes is questionable at best. I will retract my downvote, but I really cannot recommend using hybridisation.


text:
Why does attraction always cause release of energy making the system more stable?
Suppose $\ce{Na^+}$ and $\ce{Cl^-}$ form an ionic bond. In doing so they will come closer by attraction. Now as they come closer their speed will increase and hence the kinetic energy increases. So basically the electrostatic potential energy will be controverted into kinetic energy.

My question is why do these processes that involve attraction cause a release of energy from the system? And after this release why is the system said to be more stable?

My understanding is there might be two concepts involved.
First, as the ions approach each other their speed increase, but as soon as they touch each other they collide and reflect back. So the ions keep on oscillating. These oscillations are what we know as heat and the temperature of $\ce{NaCl}$ will be higher than its surroundings so it will release its energy in the form of heat.
Second, it might be because any accelerating charge releases electromagnetic waves, and so as long as the ions oscillate they'll keep releasing energy in the form of EM waves which may be trapped by other ions or may escape.
After the energy will be released from the system(NaCl ion here) our system must become more stable. Stable in the sense that NaCl crystal will not convert back into ionic form separated from other ions because for coming back to that state the system must be provided the lost energy back by some external agency. This is the way how I understand the term stabilty; I don't what does this term mean in chemistry. I also don't know why " every system in nature wants to be more stable" -- This is the phrasing that I heard in school days from chemistry teacher. This fact must be related to some thermodynamic fact of kind that every system wants to release energy, but why?

I'm afraid it doesn't have much to do with chemistry.

@Mithoron This is what I was told by many chemistry teachers when I was a student. They also told be that lesser energy --> More stability. And every thing in nature wants to be more stable. Although I never quite got their 'nature' arguments, like XYZ chemical reacts in a particular  way because of its nature etc etc.

It's very basic physis problem - in chemistry it's get as given. Enthalpy should decrease entropy increase.

@Mithoron That's exactly what my question is. Why should Enthalpy decrease and entropy increase?

http://chemistry.stackexchange.com/questions/19150/conditions-for-spontaneity - your q. is more basic than that?

If your question is as incredibly basic as I think, it stems from why forces exist you can read for example https://en.wikipedia.org/wiki/Force but it's  primary propery of the universe - it's like asking why time exist or what is mass.

@Mithoron No, I am not aking why force exists.

Attraction is a shorthand way of saying "potential energy gets lower as the distance gets lower".  The opposite is true of repulsion.  The potential energy of two like-charged particles, e.g. two $\ce{Na+}$ atoms, will increase as they are brought closer together.
Attraction (and its opposite, repulsion), are forces with units (in SI) of Newtons, or $\frac{\rm kg\cdot m}{\rm s^2}= \rm N$.
Energy has units of Joules, or $\frac{\rm kg\cdot m^2}{\rm s^2} = {\rm N\cdot m} $.
The relationship between a force $F$ between two particles separated by a distance $r$ and the potential energy $E$ of those particles is $F = -\frac{dE}{dr}$.
If the energy gets lower ($dE$ is negative) when two particles get closer together ($dr$ is negative), then the force $F$ is attractive. If the energy gets higher ($dE$ is positive) as the distance gets lower ($dr$ is negative), then the force is repulsive.
Force and energy are two different mathematical ways of looking at the same thing.  That's why they are always related.$%edit$

Is $E$ total energy here or only the potential energy. Is $F$ magnitude of force? How can we prove $ F=−dE/dr$ in this case? As far as I know Work done = $\vec F.\cdot \vec{dr}$, where work is being done on a particle and $|\vec{dr}|$ is the net distance moved not the distance between two particles. And lastly in which form the energy lost will appear? K.E or heat?

$E$ is potential energy; I'll edit my post to make that clear.  The way in which energy is lost depends on the system.  For a sodium ion and a chloride ion in a vaccum, the "lost" potential energy is converted to kinetic energy as the particles accelerate towards each other.  For a sodium ion and a chloride ion in a viscous (uncharged) solvent, the potential energy is lost as heat through viscous dissipation (i.e. the friction of the ion moving through the solvent).

And yes the vector form of the relationship is $dE = \overrightarrow{\bf  F}\cdot \overrightarrow{d{\bf r}}$

I think I didn't phrase my question well. I know that P.E. will be converted into K.E. I need to edit my question.

I think your answer isn't rigorous.  You might mean to take one $\ce{Na^+}$ atom as the origin. "when two particles get closer together (dr is negative)" But $dr=r_2-r_1$ so if two particles get closer then $dr=$postive. The equation you are using, $F=−dE/dr$ seems to come from vector calculus but it is missing vector notations.

If $r=r_2-r_1$, then $dr=dr_2 - dr_1$, so if two particles get closer together, then $dr$ is negative, no matter what coordinate system you are in.  Also, often times 1D versions of vector equations are used to describe ideas from vector calculus, for simplicity.  Is your question about vector calculus or about the concepts of force and energy?  I'm confused about where this is going and am not sure I can be of much more help...

I misinterpreted because $r$ is usually the position vector. Now I get where are you coming from. You mean F=-dU/dx. But here in this expression dx=$\displaystyle \lim_{x_2 \to x_1}x_2-x_1$ not $d(x_2-x_1)$ because $x$ is the coordinate not the difference of coordinates.

And if we take NaCl in vacuum then the P.E. will be concerted just into K.E. making the total energy conserved and soon after collision K.E. will be converted back into P.E. separating Na+ and Cl- ions. Why is this system more stable?


text:
How does molecular collision excites the electron?
I have this question for a long time that in LASER when we do the pumping by heating how does molecular collision causes the atoms(molecules) to reach in excited state. E.g. in $\ce{He-Ne}$ laser when the excited $\ce{He}$ atom collides with the $\ce{Ne}$ atom the $\ce{Ne}$ atom gets excited and $\ce{He}$ atom reaches its ground state.
My question are

What happens at the molecular level that causes $\ce{Ne}$'s electron to jump into higher state? Does Ne atom absorb some kind of photon by some mechanism?
Another thing is if we take $\ce{He}$ atom as the frame of reference then the high energy $\ce{Ne}$ atom is striking $\ce{He}$ so the high enrgy $\ce{Ne}$ must give its energy to $\ce{He}$ not the other way round.Where did I go wrong here?

The general rule is that you do not need to absorb the photon to excite an electron. What you need is to provide some portion of energy.
It can be done by means of a photon which is absorbed by an atom, of course, but it is not the only way.
Alternatively, you could collide an atom with some other object and upon the collision some of the kinetic energy is converted into internal energy which is then redistributed by its components. One of the components of the internal energy is the electronic energy, so if enough kinetic energy is converted into this particular part of internal energy, an electron gets excited. This process is usually referred to as the collisional excitation. This is the process by which an electron in $\ce{He}$ atom is excited: $\ce{He}$ atom in its ground electronic state collides with an energetic electron which causes electron excitation.
A collision between $\ce{He}$ atom in an excited electronic state and ground state $\ce{Ne}$ atom also involves an energy transfer. But this time it is the excitation energy (possibly together with some amount of kinetic energy as well) which is transferred from $\ce{He}$ atom to $\ce{Ne}$ atom. Due to a near coincidence between the $\mathrm{2\,^3S_1}$ and $\mathrm{2\,^1S_0}$ energy levels of the $\ce{He}$ electronic states, and the $\mathrm{3s^2}$ and $\mathrm{2s^2}$ energy levels of $\ce{Ne}$, the transfer of excitation energy from $\ce{He}$ atom to $\ce{Ne}$ atom is quite efficient.

Image courtesy of Wikipedia.

Disclaimer: I'm a professional chemist, not a physicist, so I might be wrong in what follows...
... but as far as I understand the matter at the microscopic level the energy is transferred by means of electromagnetic radiation the quantum of which is indeed photon. But do not think about this process as a photon being somehow "produced" in a form of a "billiard ball" by one atom, then transferred to another atom and finally magically "disappearing" here. This classical way of thinking is the root of confusion.
Before the energy transfer we have two systems each described by its own wave function. In order for the energy to be transferred the systems should start to interact with each other, but once they do so they become entangled and we do not even longer have separate systems. Rather, we have a big composite system described by its own wave function. After the interaction we again can speak about the individual systems each of which is again in own quantum state, though, the final state of each system is different from the starting one due to energy transfer. And in order to predict the final state of both systems (to be more precise, the probabilities of all possible final states) one could use the rules of quantum mechanics.

Why do we need excited $\ce{He}$? Why don't  unexcited high speed $\ce{He}$ atom collides and causes $\ce{Ne}$ to excite? How does this happen technically? When $\ce{He}$ approach $\ce{Ne}$ the outer electrons must repel each other then their must be some mechanism which transfer momentum to the outermost electron of Ne. I'm sure the mechanism has something to do with the excitation of He otherwise if we look from He's frame of reference then He itself must get excited,  creating a paradox. Perhaps He's electron goes to ground state there by creating a photon which is absorbed by $\ce{Ne}$.

@user31782, now I see that you have two questions out there. The first step of the process is inelastic collision of energetic electrons with ground state helium atoms. These collisions produce excited helium atoms as described in the answer. But when excited helium atoms collide with ground state neon atoms the story is a bit different: it is still about energy transfer, but this time conceptually an electron in a helium atoms is deexcited and this energy is transferred to an electron in a neon atom which is excited.

@user31782, unexcited high speed $\ce{He}$ atom will also cause an electron in $\ce{Ne}$ atom to excite, but no so efficiently. The excitation energy transfer is much more efficient then the kinetic energy transfer in this case due to near coincidence between the energy levels of species in question.

@user31782, well, if you are interested in the actual mechanism of the energy transfer, I would say Chem.SE is the wrong place to ask such question (Physics.SE is the right place) since there is absolutely no chemistry in the mechanism of the energy transfer.

Again, I don't think that unexited $\ce{He}$ atom would be able to excite $\ce{Ne}$ atom, because if we see from $\ce{He}$'s frame then $\ce{Ne}$ is striking with a higher energy and hence must excite $\ce{He}$ atom instead -- paradox. Secondly ,yes I want to know the actual scenario of what exactly happens with the inelastic collision. How is the kinetic energy given to the outermost electron. I can't ask question on Phys.SE for some reason. I guess my question can be on topic here because I am asking about atomic chemistry. And why does the near coincidence of energy levels [cont]..

makes the collision more efficient? How is the $\ce{He}$ atom de-excited and how that energy is given to $\ce{Ne}$'s outermost electron? Why is the transfer of direct kinetic energy(unexcited collision of $\ce{He}$ with $\ce{Ne}$) less efficient as compared to 'excited-$\ce{He}$-collision'?

Sorry for being late, but can electron-atom collisions excite the atoms too?

@dark32, yes, they certainly can.


text:
Period of Vibrational Modes
I am trying to understand more about the period of vibrational nodes. Currently, I am interested in the wavelength of the $\ce{Si-O}$ bond stretch in silica. I think this value is roughly $1000\mathrm{~cm^{-1}}$.
I am somewhat confused on how to find the period of vibrational modes. I am new to this subject. When I look up the frequency, like I said, I get units of $\mathrm{cm^{-1}}$, but I want units of $\mathrm{s^{-1}}$.
Any pointers on how I can get this value?
Is this the way to calculate it:
$$T = \frac{1}{\nu}$$
where
$$\nu = c \bar{\nu}$$
with $T$, $\nu$, and $\bar{\nu}$ being the period, frequency, and wavenumber respectively?
With this I get:
$$\begin{align}
T &= \frac{1}{c\bar{\nu}} \\
&= \frac{1}{(3 \times 10^{10}\mathrm{~cm~s^{-1}})(1000\mathrm{~cm^{-1}})} \\
&= 33.33\mathrm{~fs}
\end{align}$$
Also, as temperature increases, wouldn't the period decrease? But in the literature, I find that as temperature increases, the wavenumber decreases. Thus, the period increases. Why would the period increase? I would think at a higher temperature, bond stretching would occur more quickly. Is this not the case?

$P=\lambda/c$. Assuming $\lambda=110^{-5}$ (i.e. 1000 wavelengths in 0.01 m) and $c=310^8$ m/s have $3.(3)*10^{-14}$ s perios.... yeah your calculation is indeed correct.

@permeakra OK, so here is what I am confused about: As temperature increases, wouldn't the period decrease? But in the literature, I find that as temperature increases, the wavelength decreases. Thus, the period increases. Why would the period increase? I would think at higher temperature, bond-stretching would occur more quickly. Is this not the case?

Molecular vibrations are usually approximted as harmonic oscillator. In it no matter how much energy of oscillation is, the period is the same, only aplitude differs.  (in truth the approximation breaks  at higher vibration levels, but lower ones usually are good enough, and they are the only ones active in IR spectra)

@permeakra Interesting, but I have recently read a paper where the frequency (cm^-1) decreased as a function of temperature. If energy does not matter, why would it decrease?

(un)harmonic oscillator approximation ignores other energy levels in the molecule. However, change of temperature may change relative occupation of other energy level, say, rotational levels in gas phase or electronic states in paramagnetic molecules. Change of relative occupation of rotational levels will move and change apparent IR peak position, see wikipedia://Rotational–vibrational+coupling. In solutions change of dielectric constant of solven also may have influence. Unfortunately, can't point the exact reason without the article at hand.

Just a small comment on the notation (which I have edited). The $1000\mathrm{~cm^{-1}}$ is not a wavelength ($\lambda$), it is a wavenumber ($\bar{\nu}$). The two are reciprocals: $\bar{\nu} = 1/\lambda$. The relation for an electromagnetic wave is $c = \lambda\nu \Leftrightarrow \nu = c\bar{\nu}$. Also, the period is more commonly denoted as $T$, not $P$.

I've wondered about this -- is it reasonable to just equate (a) the frequency of a photon with the energy needed to change the quantum state of a particular vibrational mode with (b) the classical-equivalent frequency of the "physical vibration" of that mode? I would think the two physics are substantially different, and such an equality would arise mostly from either coincidence or a specific choice of units.


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Is there a way to find the kinetics data for the strong acid weak base reaction?
I am layman of chemistry. Latterly, I am asked to do an mathematical optimization task for a bio-chemical process, which contains a conditioning step to neutralize the sulfuric acid using ammonia hydroxide. I am trying to use the mathematical model of the reaction (kinetic model) to optimize the reaction time/energy/efficiency (target not determined yet). However, I could not find the kinetics data. Is there a way I could find the kinetic model of the reaction?
Another question is that: Should I treated this reaction as an element reaction and use third-order reaction rate formula?
I am confused by a former post (Finding rate of acid - base reaction). In that post, the accepted answer says "Typical acid - base reactions that occur in solution are diffusion controlled reactions. Such reactions do not follow second order kinetics because at virtually every encounter of the acid and base, rapid proton transfer occurs and the reaction is complete. " Should I use the diffusion control model instead of the reaction rate model?
Any suggestion is welcomed.

"Should I use the diffusion control model instead of the reaction rate model?" Acid base reactions are among the fastest reactions in chemistry. As soon as the solutions are mixed, the reaction can be considered complete, so the rate determining step is the diffusion/mixing of the two solutions.

Thanks for your answer. I will look into the diffusion model then.


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Does evaporation reverse salt hydrolysis?
It has been a long time since I studied organic chemistry, but one thing I do remember is that when we needed cyanide salts we were told not to bother keeping any remaining solution because the salts hydrolyze very quickly.
But can't hydrolyzed salts be fully recovered by simply evaporating the water?
E.g., we don't worry about evaporating NaCl from a distilled water solution and finding anything other than NaCl.  So I'm wondering if the lab instructions to neutralize and dispose of cyanide salt solution was just based on expediency.  Or: can something unsafe happen during the evaporation of a hydrolyzed salt solution?

Do you mean "hydrolyzed", or just "aqueous" or "absorbed water"?  When I hear "hydrolyzed" in a chemistry context, I think of covalent bonds being broken, but I can't work out if that is what you mean...

@CurtF. -- Again, I'm rusty, but I'm pretty sure I mean hydrolyzed.  If it was just aqueous then you just dry it out and everything's fine.  But my recollection of the reason we don't store cyanide salts in solution is that they actually decompose via hydrolysis, e.g., KCN(s) -> K(+aq) + CN(-aq).  Which affects the solution's strength as a reagent, I guess?

That is just dissolution.  In fact if you have KCN "in solution", that means it consists of K(+)(aq) and CN(-)(aq) ions dissolved in the solution.  Maybe you are talking about hydrolysis of cyanide complexes of transition metals, such as ferrocyanide or ferricyanide?

@austinian: Oh, that might explain it.  So cyanide salts are hydrolyzed into hydrogen cyanide?  That's a good reason not to leave the solution sitting around!

No they are not hydrolyzed to hydrogen cyanide.  OH- is a very strong base and HCN is an acid, so if any such reaction formed they would immediately back convert to CN- and H2O.  Perhaps if you had cyanide salts in open containers, over long periods of time (several days or weeks) enough HCN would evaporate to significantly lower CN- concentrations in the solution.  But that won't happen if you store in sealed containers.

@CurtF. - OK, so then the answer to my questions are: Something a little unsafe could happen if you left the container uncovered, but yes, if you evaporate the water (say under a vacuum) you will (almost) fully recover the original salt?  And the solution does not lose significant potency over time as a reagent due to hydrolysis?

Salts such as potassium ferricyanide K3 Fe(CN)6 can be hydrolyzed to yield KCN and solid iron (hydr)oxide precipitate.  Is that what you are thinking of?  https://books.google.com/books?id=q3PLBQAAQBAJ&pg=PA174&lpg=PA174&dq=aqueous+cyanide+hydrolysis+by+sunlight&source=bl&ots=JvSQTB9zLr&sig=RDoGbbsvXLzMqitiHA3lqPG3Si0&hl=en&sa=X&ved=0CCsQ6AEwAmoVChMIle6H1JPyxgIVDjaICh3j4wts#v=onepage&q=aqueous%20cyanide%20hydrolysis%20by%20sunlight&f=false

Feetwet I think that is right, for simply cyanide salts like potassium cyanide and sodium cyanide, etc.

Bah, you're absolutely correct. I was thinking about dissolution in an acidic solution for some reason. Removing my suggestion now.

@CurtF. -- That book link is fascinating.  Not what I was thinking of, but now I'm wondering why I've seen K4Fe(CN)6 stored in clear containers!  Anyway, thank you, and please post your answer so I can accept it!

Simple cyanide salts
In aqueous solution, simple cyanide salts more or less completely dissociate into their constituent ions.  E.g. for potassium cyanide:
$\ce{{KCN}(s) + H2O(l) -> K+(aq) + CN- (aq)}$
Those solutions are relatively stable.  There is an acid-base equilibrium between $\ce{CN-}$ and $\ce{HCN}$, but in unbuffered or in alkaline aqueous solution this equilibrium will be almost completely on the left:
$\ce{CN- + H2O <<=> HCN + OH-}$
Over very long times, if the solution is left in an open container, some of the tiny, minute amount of $\ce{HCN}$ formed in this way could evaporate, leading to a very slow loss of cyanide from the solution.  If stored closed this will not be a problem.
(Mixing simple cyanide salts with an acid will result in an immediate and dangerous release of gaseous $\ce{HCN}$, so in this question I assume that no acids are involved.)
Thus, for simple cyanide salt solutions stored appropriately, yes you should indeed be able to recover all of the initially added the solid salt after boiling off or otherwise removing any water.
Cyanide complexes of transition metals
Salts such as potassium ferricyanide $\ce{K3Fe(CN)6}$ consist of potassium cations and complexed ferricyanide anions, $\ce{Fe(CN)6^{~3-}}$.  They are soluble in water as well, but unlike their simple cyanide anions, they can be hydrolyzed.
The hydrolysis is catalyzed by UV light and results in the formation of free $\ce{CN-}$ anions as well as iron hydr(oxide) solid precipitates.
See this book for more information.

Thanks again for an outstanding answer.  I just had to see the photolytic reaction you referenced, and that has led to some more questions.  When you have the time could you take a look at those at http://chemistry.stackexchange.com/q/34398/5354 ?


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Nitrogen and an unknown gas is leaked into a common vessel... The gaseous mixture contains 0.4 moles of nitrogen. Molecular mass of the other gas is?
For 10 minutes each, from two identical holes, nitrogen and an unknown gas is leaked into a common vessel of 3L capacity at 300K. The resulting pressure is 4.15 bar. If the gaseous mixture contains 0.4 moles of nitrogen, then molecular mass of the unknown gas is?
attempt: According to Graham's law of effusion, $$\frac{r_1}{r_2}=\frac{V_1}{V_2}=\sqrt{\frac{M_2}{M_1}}$$ where $V_i$ is the volume of gas effused. Let $V_1$ and $M_1$ be volume and molecular mass of nitrogen. $$V_1=22.4\times0.4=8.96;M_1=28$$
Substituting, $${V_2}=8.96\sqrt{\frac{28}{M_2}}$$
and $V_1+V_2=3$, so $$8.96(1+\sqrt{\frac{28}{M_2}})=3$$
This is not possible since $\sqrt{x}\ge0$.
What am I doing wrong?

Calculate total moles using the ideal gas law and solving for $n$:
$$n = {4.15\;{\rm bar}\cdot 3\;{\rm L}\over {0.0831\;{\rm L}\,{\rm bar}\,{\rm K}^{-1}\,{\rm mol}^{-1}}\cdot 300\;{\rm K}} = 0.5\;{\rm mol}$$
Given 0.4 mol $\ce{N2}$, moles of unknown gas = 0.5 mol - 0.4 mol = 0.1 mol.
Rate of effusion of $\ce{N2} = r_{\ce{N2}} = 0.4\;{\rm mol}/10\;{\rm min} = 0.04\;{\rm mol}\cdot{\rm min}^{-1}$.
Rate of effusion of unknown ${\rm X} = r_{\rm X} = 0.1\;{\rm mol}/10\;{\rm min} = 0.01\;{\rm mol}\cdot{\rm min}^{-1}$.
Using Graham's law of diffusion:
$${r_{\ce{N2}}\over r_{\rm X}} = \sqrt{M_{\rm X}\over M_{\ce{N2}}}$$
we solve for $M_{\rm X}$:
$$M_{\rm X} = {M_{\ce{N2}}\times (r_{\ce{N2}})^{2}\over (r_{\rm X})^{2}}$$
and plugging in numbers we get the molecular mass of the unknown gas:
$$M_{\rm X} = {(28\,{\rm g}\cdot{\rm mol}^{-1})\times (0.04)^{2}\over (0.01)^{2}} = 448\,{{\rm g}\over{\rm mol}}$$
where the units on the effusion rates have been omitted for clarity and because they cancel.

This is the same answer given in my book. But am I making any conceptual mistakes?

The thing that I see is you're incorrectly calculating volume (it can't be greater than 3 L, so multiplying 0.4 mol times 22.4 L/mol is not the place to start). I would say use what is given in the problem: they have explicitly given a time, a total pressure, a total volume, and the identity of one gas. From there, you should go to calculating moles of unknown. Conceptually, I would sum it up as look at what you're given and try to work the problem using those givens, and check your assumptions along the way (like the volume issue).


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Potassium bicarbonate or sodium bicarbonate
How would I go about determining whether a sample of an unknown substance is either potassium bicarbonate or sodium bicarbonate (given that the sample is one or the other)?

Related: How can I identify whether a substance is KOH or NaOH?

Easiest: flame test - yellow for Na, purple for K. However, the Na yellow lines are extremely bright, and if there's any Na contamination, the purple K lines are hard to see without a spectroscope (or diffraction grating).

@Loong  It's a dupe, same cations to discern.


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Determining number of electrons transferred
When determining the number of electrons transferred in a redox reaction is it the total in both half equations?
For example:
$$\ce{2I- + Zn^2+ -> I2 + Zn}$$
First we split it up into the two half-reactions and get the following
$$\ce{2I -> I2 + 2e-}$$
$$\ce{Zn^2+ + 2e- -> Zn}$$
So would the number of electrons transferred in the redox reaction be two or four?

possible duplicate of How is it determined how many electrons are transferred in redox reactions?

@ToddMinehardt - While I think that the answer to the question you link will answer this question, the questions represent two very different misunderstandings. The question you link to is asking why ions have the charges they do.

When determining the number of electrons transferred in a redox reaction is it the total in both half equations?

In the total redox reaction that is properly balanced, the number of electrons transferred is equal to the stoichiometric coefficient of the $\ce{e-}$ species in either balanced half-reaction.

So would the number of electrons transferred in the redox reaction be two or four?

Your half-reactions and total redox reaction happen to be balanced, to the number of electrons transferred in this case is two.


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Seeking titanium compound, melting point <1600 °C, decomposes to TiO
The title says it all.  I’m new to glass making, and I need a titanium compound that will render me $\ce{TiO}$ through decomposition, and has a melting point under 1600 °C.  Hopefully I can get one that goes under 1500 °C, because that’s the max these guys usually keep the furnace at.  The compounds I’ve found so far are typically in the 1800 °C range and we can’t do that.  Any input is appreciated.

I don't know how the decomposition reaction proceeds, but have a look at titanite (formerly known as sphene) - CaTiSiO$_{5}$. I Googled around and see a melting point at 1382$^{\circ}$ C. That said, there's silicon and calcium in the mix, so this might be a dead end for you (I know nothing about glass making).

Obviously this isn't decomposition so may be completely useless to you, but anyway; I believe TiO can be made by reacting TiO2 and Ti at ~1500c.

I believe, if you can make it and survive it, titanium oxonitride would fill the bill (shudders). Otherwise, consider mixed oxides.

The real question here is "why do you need it?" I don't know that much about glass making as well, but for me the preparation of high-temperature glass sealants usually goes like this: (a) mix the components - oxides, carbonates, whatever (b) heat the bastards in the inert crucible. The components will decompose, melt, interact and dissolve each other. Even if the melting point of any individual component is higher than the melting point of resulting glass, there shouldn't be a problem because of eutectics. My point is that I don't see why you cannot just add some $\ce{TiO2}$ to your glass.

My guess is to use $\ce{TiCl2}$ (m.p 1035 C). It disproportionate to $\ce{TiCl4}$ at high temp. which can be then hydrolyzed to $\ce{TiO2}$. This is then can be decomposed or converted to $\ce{TiO}$.

try TiO2 + Ti at higher pressure and temperature


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Belousov–Zhabotinsky Reaction
An experiment in my lab was about the Belousov–Zhabotinsky reaction and we measured 2 times - the induction period, which was the amount of time it took for the solution to change color and back for the first time, and the oscillation period, which was the period of color change cycles after that. The lab manual says to the inverse of both those times as the rate constant K for this reaction, but I am not sure why. Please help?


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Why can't hot water break disulfide bonds when the energy storage in hot water exceeds the bond energy?
I teach Biochemistry and know that heating water can break weak bonds like hydrogen bond or ionic interaction, but not disulfide bond (without reducing agent). The bond energy of hydrogen bond in water is about 12–30 kJ/mol, while the disulfide bond is 251 kJ/mol. I initially thought the heated water would have the energy much lower than 251 kJ/mol, but higher than 30 kJ/mol. But when I looked at the energy storage in 1 L of water at 90 degrees Celsius from here, it was calculated to 294 kJ. Now I wonder whether I can compare this energy to the bond energy. What do I miss here?

In thinking that the water will cool to near zero, releasing the energy so that the disulfide bond can be broken. Also, the sulfides will reform the bond since it is the most stable bond available.

could it be because 294KJ/L

Chemical reactions do not simply occur because there is enough energy around to make something happen. They also require a mechanism that enables something to occur. It isn't enough that the product will release energy on some reaction happening.
Some reactions will happen spontaneously with enough input of thermal energy to start them off (burning coal, for example). Others, like breaking disulphide bonds) will not. Even when there is a plausible mechanism for a reaction, you have to add enough energy to make the reaction happen which may be much more energy than you expect. Most reactions have an "activation energy" to get past the barriers specific mechanisms holding back the reaction.


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Thermal decomposition of mercury thiocyanate
A video on YouTube shows the thermal decomposition of mercury thiocyanate. A strange thing I've observed is that $\ce{Hg(SCN)2}$, which is the reactant, is a white powder and the product is a yellow-brown solid serpent-like structure. How is this possible? What is the chemical reaction for this decomposition?
video link - https://www.youtube.com/watch?v=FaC4MqxUnFw

https://en.wikipedia.org/wiki/Black_snake_%28firework%29

The reaction is the decomposition of mercury(II) thiocyanate:

$\ce{2Hg(SCN)2 → 2HgS + CS2 + C3N4}$

Where $\ce {C3N4} $ is the compound that makes up the "snakes," and grow due to the evolution of gases produced by the decomposition of the other products:

$\ce{CS2 + 3O2 → CO2 + 2SO2} $
$\ce {2C3N4 → 3(CN)2 + N2} $
$\ce {HgS + O2 → Hg + SO2}$

This is also known as the "Pharaoh's Serpent" in pyrotechnics. The resulting solid can range from dark graphite grey to light tan in color with the inside generally much darker than the outside.
Click here and here for reference.


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Constitutional isomers for four carbon thioesters
Question:

How many isomers are there for thioesters with the formula $\ce{C4H8OS}$?

Answer given is four.
I can see that the following two are possible constitutional isomers:

But what are the other two?

Please visit [this page](http://chemistry.stackexchange.com/help/notation " latex="" at="" chemistry.se""="), this page and this one on how to make your future posts better.

When I tell you, that dithioesters exist ... will this be a hint?

Here's my hint: Formic acid exists.

@jerepierre Didn't thought of that. But then there are more possibilities than 4.

@PH13 How you can have a dithioester with only one sulfur?

I never said that the dithioester would be the answer.

@PH13 I read you now. I didn't realize those isomeric possibilities were still called thioesters. Should be 8 possibilities then, yes?

Let us continue this discussion in chat.

One way to break things down in by the number of carbon atoms on either "side" of the thioester.
There are 4 carbon atoms to work with, of which one will be taken up by the thioester moiety.  Thus there are three left to work with.
On the carbonyl "side" of the thioester, between zero and three carbons are possible.  However, on the sulfur "side" of the thioester, at least one carbon atom is required; otherwise the compound is not a thioester but a thioacid.  Thus the possibilities for carbon atom distribution are:

one carbon on S side, two on CO side.
two carbons on S side, one on CO side.
three carbons on S side, zero on CO side.

You have drawn the only structures for the first two possibilities.
For the last possibility, there are two possible ways to attach a three-carbon alkyl chain to the thioester sulfur:

as a $n$-propyl group
as an isopropyl group

Thus, the four possible thioesters are:

methyl thiopropanoate
ethyl thioacetate
n-propyl thioformate
isopropyl thioformate

Actually, it looks like due to haphazard naming of thionoesters (where the O and S atoms of the thioester moiety are reversed), it is less ambiguous to say S-propyl thioformate and S-isopropyl thioformate, etc., at least for formates.  People apparently use the name O-propyl thioformate for the thionoester, confusingly.

Thanks for your answer @Curt F. I already did consider the two additions you made to my answer but here is the thing: I googled the general formula for thioester and the following image showed up:
https://upload.wikimedia.org/wikipedia/commons/thumb/9/90/Thioester-2D-A.svg/150px-Thioester-2D-A.svg.png
So the two additions (formates) you made to my answer can only be achieved by replacing R(in the image) with H. Is that allowed?

What people mean by R varies by context.  In this case, according my understanding of your question and my understanding of the definition of thioester, the R bonded to the carbonyl carbon could be H, but the R' bonded to the sulfur could not.  Unfortunately there isn't a general answer for what folks mean by R in organic chemistry.

Thanks for you input. Just to make sure, is there some link you can share where I could verify what you said and read more about thioesters? I haven't been able to find one by myself.

if you get stuck drawing more than 2 isomers what i do is take a carbon out and put it somewhere else if you need any more help lmk and on 3 i messed up on the chemical equation but the structure is correct


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How big is the gauche interaction of methyl with phenyl in a disubstituted cyclohexane?
I'd like to compare the equatorial 1-methyl-2-phenylcyclohexane conformation of with axial conformation of 1-chloro-3-methylcyclohexane conformation. In which molecule there is less steric repulsion?
In other words: Do you know a guess of energy in kcal/mol for a gauche Ph//Me interaction and for an axial Ph//Cl interaction in this cyclohexane ring?
I would prefer to have a comprehensive two-dimensional table for allylic A(1,2) strain, but I didn't find any.
It might sound crazy or even unbelievable: A gauche Ph//Me interaction in 1-phenylpropane is only about 0.07 kcal/mol higher in energy than its antiperiplanar conformation (according to: http://www.sciencedirect.com/science/article/pii/S0040402001915509) compared to 0.9 kcal/mol for a gauche Me//Me.
However a diaxial Me//H only weights about 0.9 kcal/mol compaired to 2.7 kcal/mol for a diaxial Ph//H interaction; underlining that phenyl is very often more sterically hindered than a methyl, if the phenyl group can't rotate away.

You might be interested why I'm so concernced with my question:
I read the paper http://pubs.acs.org/doi/pdf/10.1021/ja035001g and I was curious why in Scheme 5 (with 5a) the phenyl group is placed axial for R1=Me, R2=H.
(In other words I would rather take into account the allylic A13 Ph//H plus gauche Me//Ph than the diaxial Ph//Cl interaction) It's because the the isomerization of the trans hydrazone costs too much energy!

I just want to make you aware of the fact that the Tetrahedron paper is from 1983(!) and that the MM2 force field is cited from a 1977 paper. I'm not sure if those calculated values can be seen as reliable.

Does anyone knows a freeware to calculate the relative energy difference?

GAMESS.  Also, this question can't really be answered without a specification of cis vs trans.


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Why is lithium plating irreversible in Li-ion cells?
Several sources (1, 2) suggest that charging Li-ion batteries in low-temperature conditions result in lithium plating. This seems pretty logical since low temperature reduces ion mobility, and provided there exists a high enough charge current, lithium ions don't have time to intercalate into carbon layers of the anode and are deposited on top.
What I fail to understand is why this process is irreversible (Wikipedia is careful to say "may not be removable"). This is counterintuitive to me. Since the plated lithium ions don't have to overcome any van der Waals forces, they should be the first to go when the battery is supplying current. So as long as the metallic lithium does not grow dendrites, thereby damaging the internal structure of the cell, a little plating should be mostly harmless.
Does my reasoning make sense? And if not, what did I get wrong?

It's true that when the battery is re-discharged, it's thermodynamically 'easier' to oxidize lithium than lithiated graphite. However, keep in mind that the lithium will react with electrolyte to form an SEI (solid electrolyte interphase) layer. This process consumes lithium ions, so on recharge you don't get all the lithium ions back. This is why the process is considered "irreversible."

I concur with both statements above. You are both right. What needs to be considered as well is the following: given time the plated Lithium will chemically intercalate into the anode starting from the interface to the anode. This leads to increasing porosity of the plated Lithium and can lead to electrical Isolation thereof. That in turn translates into electrochemically inactive Lithium Floating around in the electrolyte.
However, those lose metallic Lithium platelets are still chemically active and will react with the elctrolyte, leading to electrolyte Degradation.
All this depends very much on how the cells are operated in detail. Under certain operating conditions, theoretically, Lithium plating can be close to completely reversible.


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Are biological macromolecules organic or inorganic compounds?
We know that organic compounds are compounds which contain carbon atoms. I want to know whether all biological macromolecules (such as proteins, lipids, carbohydrates, and nucleic acids) are organic compounds, or are only a subset of them considered organic?

Based on your definition, what do you think?

The definitions of the "biological macromolecule" and of the "organic molecule" are vague, but, I would say, yes, all biological macromolecules are organic. But what the point of such statement? What do you think you achieve with that information?

The definition of "organic" is somewhat vague: it can mean containing carbon, or it can mean coming from living things, particularly as defined by the United States Department of Agriculture.
To further confuse the issue, some carbon compounds, such as sodium carbonate or tungsten carbide, are sometimes defined as inorganic.
So state a particular definition, and examine those compounds according to that definition.

There is a field of chemistry referred to as bioinorganic chemistry.
While all the biological macromolecules are organic, if a metal-ion-containing protein (metalloprotein) is being studied, especially if the focus is upon the metal ion, such as oxidation or reduction of the metal ion, the term "bioinorganic" is often used.

There are a number of different definitions of organic compounds existing, from the oldest (anything that comes from living beings (and is assumed to have been created by a vis vitalis)) to a few contemporary ones. So the definition of organic compound is rather fuzzy, with edge cases such as trifluoroacetic acid sometimes belonging to the organic realm and sometimes not.
However, I am not aware of any definition of organic compound that would exclude the biomacromolecules and deem them ‘inorganic’. So unless your definition of organic is extremely exotic, they are included. (And I cannot imagine an exotic definition of organic compound that would not include lipids, anyway.)


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Why is glycol used as an antifreeze?
Lowering of the freezing point on adding a solute is a colligative property of solutions, it doesn't matter what the nature of the solute is (of course we have to assume ideal solutions). So when it is required to use an antifreeze - why are glycerol, glycol, methanol, etc. commonly used? For a given decrease in freezing point, a given value of molality is required - in other words a particular number of moles for the liquid's given mass. For the same number of moles, it should be simply more economical to use any substance with lower molar mass. Solubility in the solvent will obviously play a role - but there are many more solutes which are more soluble than alcohols. Why isn't $\ce{NaCl}$ used, for instance? Its molecular mass is not very high, and it is soluble to an appreciable degree. It's also more economical (moreover it will dissociate in water and generate twice the number of moles, so twice the decrease in boiling point).

You're mixing two things - alcohols lower melting point because they have low melting point themselves - not via collgative properties.

The Wikipedia article on the matter is concise and nicely informative, seems to me.

The main reason ionic substances like salts are not used is because they tend to be corrosive to metals.
Polar organic liquids tend to be the best because many are very soluble in water, and they are fairly inert under the conditions antifreeze is used.


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Where can I find CAD blocks for lab glassware?
I want to draw a sketch of my apparatus for my report. I am conversant with the AutoCAD drafting software. Where can I download CAD blocks for commonly used labware items (like round bottom flask, allihn condensor, drying tube, heating mantle, retort stand, etc.)?

Possible duplicate of http://chemistry.stackexchange.com/questions/4731/glassware-test-instruments-cad-blocks-library

Yeah, thanks for pointing out that one, but I see there are no answers to that question too.

@Todd I don't think it makes sense to close a question as a duplicate of an unanswered one.

@inɒzɘmɒЯ.A.M - Agreed, and my bad for not following up. Let's keep it open.

@Abhishek - does it have to be CAD? Found this which looks more like UML but might be of use.

@Todd, yeah it will "work" for the time being, but drawing custom made modified glassware items is very simple in CAD...anyway thanks!

ChemDraw has a couple of templates with all kinds of laboratory glasswere.
Here is a screenshot of some of the possibilities:


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Why do toothbrushes have to be dried?
I carry around a toiletries bag. I put my toothbrush there right after brushing in a restaurant or some public area.
These strangely similar articles(1, 2) suggest toothbrushes have to be aired. Why exactly? What harm does the wetness on the bristles cause?

There might be some issue with the bristles, but the much bigger problem is that the wet environment furthers the grows of bacteria and mold...

I'm voting to close this question as off-topic because it isn't very chemistry related. You might try biology.SE for it, though I doubt they'll accept it with open arms without research.

The good people at the American Academy of Pediatric Dentistry have a number of useful resources available to shed light on questions such as yours.
I direct you to a manuscript entitled Microbial contamination of toothbrushes and their decontamination in which it is written:

Storage conditions of toothbrushes are an important factor
  for bacterial survival. Dayoub et al. and Meier et al. reported
  that the number of microorganisms in the toothbrushes kept
  in aerated conditions was lower than in toothbrushes stored in plastic bags. Several authors have reported that bacterial contamination can be reduced by washing toothbrushes after use, and drying in aerated conditions. Caudry et al. reported that a wet environment increases bacterial growth and cross contamination. Therefore, as time increases between one toothbrushing and another, more microorganism development can occur in the toothbrushes a wet/moisture environment.

The references cited in the above passage are detailed in the linked PDF. Happy brushing!