text:
Liquid-liquid caffeine extraction
I am performing a lab experiment in which I am extracting crude caffeine from different samples of tea. I have read a couple of similar experiments in which methylene chloride has been used to extract caffeine, owing to its volatile nature and caffeine's affinity to methylene chloride. I know chloroform can be used as a substitute for methylene chloride. I was wondering if carbon disulphide can used as a substitute in the experiment.
Thank you very much!

Please have a look at the MSDS of all three solvents before you decide that you don't want to use $\ce{CS2}$ ;)

With $\ce{CS2}$, one doesn't really need to look at the MSDS. One brief encounter with the compound itself would be quite enough.

@IvanNeretin So true :D It's really nasty! But maybe, we can keep the students from using it without even trying. Unless absolutely necessary (Chugaev elimination), there's little need to touch that stinker.

I suggest to use a 1:1 mixture of dichloromethane and 0.2 M NaOH at room temperature.
The protocol was described by Tetsuo Onami and Hitoshi Kanazawa in J. Chem. Educ, 1996, 73, 556-557.
The authors mention that only trace amounts of caffeine are extracted from tea leaves with dichloromethane alone. They reason that due to the hight amount of tannines in the tea leaves, most of the caffeine exist in protonated form and is therefore bound to the tannines.
As an experimentally proven alternative, they suggest to treat the tea leaves with a mixture of dichloromethane and 0.2 M NaOH at room temperature!
According to the article, 20-30 mg of caffeine were extracted from a 2 g tea bag after shaking the above mixture for about ten minutes.

UPDATE
If you insist on the liquid-liquid extraction of previously prepared aqueous solutions and feel the urge to set up a percolator-type apparatus for continuous extraction, the Kutscher-Steudel glassware will not work with solvents that have a higher density than water.
For continuous extractions with solvents like $\ce{CH2Cl2}$, some different glassware has been designed and was described by S. Wehrli in Ein Apparat zur Extraktion von Lösungen mit schweren Lösungsmitteln and was published in Helv. Chim. Acta, 1937, 20, 927–931 (DOI)

I personally would test some solvents. The one I have seen specifically for caffeine is warm ethanol. Better still might be to use it in conjunction with Soxhlet apparatus.  Undergrad labs usually have these (at least they did when I was a lab demonstrator). The caffeine can then be crystallised from ethanol if not too dilute, depending how you want your product.
I would not touch carbon disulfide again in my life if I could help it.

I've never seen a Soxhlet extractor in an undergrad teaching lab. I've only used one once between my time in undergrad and grad school.

Isaac Newton: "If I have seen further, it is by standing on the shoulders of giants." In other words, it's time for a good literature search, not wasting time experimenting willy-nilly.

Yeah, that's for grad school. Or, even better, post-doc.

What method are you using to compare the extraction efficiency and content of caffeine between samples? HPLC w/w assay?

I don't know what undergrads learn now in practical sessions, but Soxhlet should be nothing special in a quality teaching laboratory.

Well, then I've never seen a quality teaching laboratory. =/


text:
Is it possible for a liquid chemical to have a pH value lower than its aqueous solution?
During the QC check of one of the products that we make as a company, the product had a $\mathrm{pH}$ of $10.44$ on one $\mathrm{pH}$ meter and $10.20$ on the other. (The product is a liquid, so it's straightforward to immerse $\mathrm{pH}$ probes to take the measurements.)
For QC of the product, it is required to make a $1\%$ solution using demineralised water and to take a $\mathrm{pH}$ reading.
The first $\mathrm{pH}$ meter read $11.72$ for this $1\%$ solution, and the second meter read $11.41$.
How is it possible for the raw product's $\mathrm{pH}$ to be lower than the $1\%$ solution's $\mathrm{pH}$? The product is supposed to be basic and not acidic at all.

pH is a property of solutions. Bronsted-Lowry bases work by removing H+ from the solvent. If there is no H+ to take, the base cannot act like a base. Is your chemical a liquid when pure?

Yes it is a liquid. It is used as a high pH boiler water treatment chemical. The previous batches of this product have always had a pH value greater than 12 as a raw product. This is the first time in 11 years of making this product that this has happened.

The concept of pH tends to break down when water is not greatly in excess of the solute (e.g. if your product is a highly concentrated amine base). The apparent pH of your product, when measured using a pH meter or an indicator, may be relatively low (seemingly less basic) if the solution has a less polar environment than that of pure water--because ionization will be less favorable.
If no water is present at all, the product might still give a pH reading with a meter or indicator, but the term "pH" in this case would be irrelevant.
It always helps to talk specifics with chemistry, so I am curious as to what the ingredients of the product are.

If the product has no appreciable amount of water, then dunking the pH electrode in it could ruin the electrode. The electrode works by having a hydrated layer of silica on the glass bulb.

Worst of all, the electrode would still continue to produce some readings (seemingly sensible, but in fact random).

Unfortunately the formula is proprietary so I cannot just post it here..... If you gave me a way to contact you on a personal level then I could divulge the ingredients but not the specific recipe for the product.

@DylanLeSar Keep your company secrets...the principles apply broadly in any case.

@iad22agp thank you though for the understanding and the answer provided. I have also brought it to out technical consultant's attention and he seems to agree with you. A person never stops learning ;)

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Identifying pentane isomers

Question
Which is not considered an isomer for pentane?
A) $\ce{CH3CH(CH3)CH2CH3}$
B) $\ce{CH3CH2CH(CH3)CH3}$
C) $\ce{CH3CH2CH2CH2CH3}$
D) $\ce{CH3CH=CHCH2CH3}$

I knew A and B were obviously isomers. I knew D was not an isomer, because the definition of isomers in my book states that isomers "are two DIFFERENT compounds with the SAME molecular formula" and that structural isomers are compounds with SAME molecular formula but DIFFERENT structure.
I was very confused with the choice C since it was IDENTICAL and did not match the definition. What confused me more is that there were practice questions in the book that gave like 4 examples of compounds and asked if  they are isomers or identical compounds.
So, are identical compounds considered isomers or not?

Related: Isomer Identification Using Condensed Structural Formulae

I'll point out that (A) and (B) are the same isomer - isopentane. (c) is not just pentane, but n-pentane.

can you please clarify the confusion, the clear option for the above given example seems to be D, as C as stated by @MaxW is n-pentane, or what is generally meant when pentane is mentioned.

The question could have been worded more precisely. I think if it said "Which is not a pentane isomer?", then it would more readily fall into @Kraw's interpretation, which I think is what the question meant to be asking.

Identical compounds are not considered to be isomers. However, the following explanation should clear up any confusion from your question.
A and B are both isopentane and C is n-pentane (normal pentane). These make up two of the three isomers of pentane, the third being neopentane ($\ce{CH3CH(CH3)2CH3}$). The question is referring to pentane as $\ce{C5H12}$, a group of isomers in which n-pentane ($\ce{CH3CH2CH2CH2CH3}$) is included and not identical to.

The last sentence is wrong. A and B are identical as noted.

Ah, I see how that was poorly worded. Thanks.

Very concise - great answer.

Updated 12-03-2016
Isomer(s) of a molecule are defined as molecules that have the same molecular formula, but have a different arrangement of the atoms in space
Here we see ...the same molecular formula... but different arrangement. Now lets take your example; pentane has the chemical formula $\ce{C5H12}$
but it has a few isomers, but the molecule 2,2 dimethylpropane has the same chemical formula as pentane ($\ce{C5H12}$), the difference being in their structure. So is the formula for 2 methylbutane.
These molecules having the same formula but different structure are said to be isomers of one another, As a group of isomers of ($\ce{C5H12}$) they are reffered to as the isomers of pentane which includes n pentane So When the word isomer of something is mentioned, it will include the something and it's isomers. It's like an album with the thumbnail from the same album.
If you say isomers of 2,2 dimethylpropane it will include n pentane as well as neo pentane and iso pentane and 2,2 dimethylpropane.
. In the above case 'C' the pentane is n-pentane or normal pentane with no branches[ an isomer of pentane (or what you may refer to as the pentane itself, as per simple IUPAC nomenclature of alkanes) ].iso pentane also called as 2 methylbutane is an isomer of pentane as well. neo pentane or 2,2 dimethyl propane is also an isomer of pentane.
D in above case is identical to pentane in terms of carbon number, even in terms of general structure but has one less hydrogen, so the chemical formula is not same as it belongs to a different organic class of compounds (alkenes) and thus is not an isomer of pentane.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Nomenclature of a tertiary amine

What is the name of the amine shown above? Is it N,N dimethyl (1-methyl-1-ethyl propane amine)?

Related: Amine Nomenclature – When to use “N”

You have correctly identified the given compound as amine. Amines can be named in various ways. According to Subsection P-62.2.1.2 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC names (PINs) are formed by using systematic substitutive nomenclature. Primary amines are systematically named by adding the suffix ‘amine’ to the name of the parent hydride; secondary and tertiary amines are named by the same method with further N-substitution (see Subsection P-62.2.2.1).
When there is a choice among acyclic parent hydrides, a principal chain must be chosen. The first relevant criterion in this case provides that the principal chain has the greater number of skeletal atoms; i.e., the principal chain is the longest chain (see Subsections P-44.3 and P-44.3.2).
In this case, the longest chain has five carbon atoms; i.e. the unmodified parent hydride is ‘pentane’.
Addition of the suffix ‘amine’ and the corresponding locant to the name of the parent hydride yields the name ‘pentan-3-amine’.
Finally, the substituent prefix ‘methyl’, the multiplicative prefix ‘tri’, and the corresponding locants are added to express substitution by three methyl groups, which yields the name ‘N,N,3-trimethylpentan-3-amine’ (note that ‘N’ is lower than ‘3’).

there is an ethyl group present

@Pi5 When your structure includes “1-ethylpropane”, the longest chain actually is “pentane”.


text:
How to alkalize moist fabric
Here's some practical application of chemistry for ya.
In short:
I have a moist, hard-to-remove carpet in a car that's hard to get dry.
I treated it with an anti fungal agent containing 1/5 hydrogen peroxide supposed to kill the mold. Measured it - it's very acidic - a condition mold will like in the near future.
How do I alkalize the carpet without creating a mess? I'd like to do this w easily obtainable everyday substances.
(Explaining "mess": E.g. pouring calcium hydroxide on it would probably be a hard-to-remove mess?)
The long story for context:
I had a leak between the interior of my car and the engine compartment let rainwater moisten the carpet.
The leak was fixed, but in the past few months had really the wrong weather for getting it dry, silica absorber packs didn't help.
So, it started to mold.
To stop it, I bought an anti-mold agent fluid. I tested its pH value and it's off the scale acidic for my litmus paper.
Mold likes acidity, hence, as soon as the mold killing by oxidation wears off, the mold will like it even more, which is why I think I need to alkalize the carpet.

If you really want to neutralize the acid of your mould removal product and don’t want to leave solid deposits, you can try a dilute aqueous solution of ammonia.
You can find typical solutions containing for example about 25 % ammonia (sometimes labelled “ammonium hydroxide”) in hardware stores. Such solutions are corrosive; they can cause severe skin burns and eye damage. You should dilute the solution to well below 5 %.
Ammonia is volatile; you will notice its characteristic pungent smell and you should thoroughly air the treated area. For the same reason, excess ammonia will evaporate and not leave solid deposits. However, it can form salts with the acid of your mould removal product and with carbon dioxide from the air. Such salts are water-soluble and, if necessary, they could be removed easily.
Ammonia is not corrosive to the metal parts of your car (ammonia is actually added to the feedwater of industrial boilers in order to keep the pH high and to reduce the risk of corrosion), except for open electrical contacts made of copper. However, you should test the ammonia solution on an unobtrusive spot to see whether it affects the colour of your carpet.


text:
What liquid other than water that is safe to put on your skin will provide the most evaporative cooling?
Suppose that you are in some extremely hot and humid place where your sweat just doesn't evaporate well enough to cool you down. You can then consider putting a liquid other than water on your skin, e.g. methanol would be an excellent liquid with a decent latent heat and high rate of evaporation at room temperature, but this is poisonous. Ethanol is better, but it will also cause problems as you need to use large quantities.
The question is then what liquid that can be used safely, will provide the best cooling.

I can't really think of any volatile liquid less harmful than ethanol. Freons, maybe, but these are unsuitable for other reasons.

Ethyl acetate - Higher vapor pressure at room temp than ethanol (73 mmHg vs 44 mmHg), higher LD50 (11.3 g/kg vs 7.1 g/kg, both rats).

Good! I didn't remember this one. But not that good, come to think of it. Its enthalpy of vaporization is lower, and with higher vapor pressure it will get into your body faster, thus compensating for the marginally lower toxicity. Well, at least it smells nice.

I just got another idea. Isotopes of H2O like e.g. with O-16 replaced by O-18 would work. Even at 100% humidity, the partial pressure of such isotopes in the atmosphere is well below saturation level. Also you can safely drink this, your sweat will evaporate if your body is saturated with this heavier variant of H2O. Unlike D2O this is safe to drink. But I think this is way too expensive to be of practical use.

Although a problem will be that as your skin cools down due to evaporating that other liquid, water vapor from the atmosphere will condense on your skin releasing latent heat in the process.

Isotopes just don't work that way. You won't see any difference from regular water.

@IvanNeretin Yes, I see now. With only normal water the reason why it evaporates slowly in a high humidity environment is because there is an almost equal flux of water condensing as the water that is evaporating. With O-18 water, this dynamics will also be at play, but it will be even worse because O-18 water will actually evaporate a bit more slowly.

How about acetone?

Ethanol, like most organic liquids, also rapidly diffuses through your skin. Good luck with the various intoxications.

There aren't many liquids other than alcohols which you can easily use. Liquid butane, for example, would be a coolant, but it would probably cause frostbite (and is flammable, not to mention from cylinders it stinks).
If you can't use ethanol for any reason, you can use isopropanol (rubbing alcohol): its cooling properties are similar. Note that you should only use isopropanol outdoors, because not only does it smell nasty in high concentrations, it is slightly toxic. Make sure to use the water solutions (~75%), as gels will leave sticky and nasty residue upon evaporation.
Even ethanol is not completely "safe", because it has the potential to cause drunkenness, even in vapor form.


text:
Why is phosphorous acid diprotic and not triprotic?
In H3PO3 only only 2 hydroxide groups exist, instead of 3. If there is 3 hydroxide group, the phosphorous will have 1 lone pair electrons, which will still satisfy the valence electron of Phosphorous.

http://chemistry.stackexchange.com/questions/44210/structure-of-h3po3

The two isomers you're referring to both exist and are in equilibrium:

When only one or two protons are removed, resonance stabilizes the anions formed, e.g. the $\ce{HPHO3^-}$ anion:

The trianion (shown below) doesn't have the benefit of resonance stabilization. This makes removing the third proton much harder.


text:
Mechanism of an epoxide ring opening with trimethylaluminium

Am I right that there is a rearrangement reaction here? After coordinating the lewis acid to the oxygen, how do I come up with the correct rearrangement step?

Looks like it is partially a rearrangement to me.


text:
How is the trend of acidity in the sequence: phosphinic, phosphonic, phosphoric acid explained?
How is the trend of increasing $\mathrm{p}K_\mathrm{a}(1)$ value in the sequence:
phosphinic $(1.24)$, phosphonic $(2.00)$, phosphoric acid $(2.21)$
explained (the image was taken from Housecroft Inorganic Chemistry)?

http://chemistry.stackexchange.com/questions/38034/why-is-h3po2-more-acidic-than-h3po4 but unanswered.

Are there any references about it?

http://www.sciencedirect.com/science/article/pii/S0277538700807512 although I'm a bit too busy to look at it again now.

The stability of conjugate bases is indeed a good way of estimating the relative acidities of different compounds.
However, in the case of the three compounds you've brought up, it is not sufficient. Taking the conjugate bases of each acid, one sees that each has a doubly bonded oxygen (oxo) and a deprotonated hydroxy. These are the only relevant groups as far as resonance is concerned.
Note that phosphorus-bound hydrogens cannot delocalize the negative charge, and neither can phosphorus-bound hydroxyl groups.
Given this, all three acids actually have the same resonance stabilization. Any resonance structure where a hydroxyl $\ce{O}$ donates an electron pair to the phosphorus is insignificant because it creates increased charge, likewise any resonance structure where the hydroxyl $\ce{O}$ receives an electron pair is insignificant because oxygen already has a complete octet in its hydroxyl form.
Since the resonance argument only tells us that their $\mathrm{p}K_{\mathrm{a}}$'s are similar (looking only at first $\mathrm{p}K_{\mathrm{a}}$) we can use the argument that the additional hydroxyl groups reduce the polarization of the $\ce{O-H}$ bond. This can be demonstrated by looking at the resonance structures of the acids (in neutral form). The hydroxyl $\ce{O}$'s in the neutral form can donate electron density to the phosphorus; however when there is only one hydroxyl group, all the positive charge is localized there, while when there are two or three, the positive charge is delocalized across three hydroxyl groups.
Note that in the case of the neutral acid, it is ok to have resonance where $\ce{O}$ donates to $\ce{P}$ because $\ce{P}$ does not gain a $2-$ charge as it would with the conjugate base.

I believe that the principal cause of confusion is that we think of the $\ce {-OH}$ substituent as electron-withdrawing via the inductive effect. However, let us consider the effect of an $\ce {-OH}$ substituent on the benzene ring. Although there is an inductive pull of electron density from the ring, the overall effect of the $\ce {-OH}$ group is actually ring-activating due to strong resonance donation. I am not sure if this comparison may be too absurd but I think it is appropriate. Contrast the acidity of para-substituted benzoic acids to the case of the phosporus oxyacids.
When the substituent is $\ce {-H}$, the substituted benzoic acid is actually more acidic than in the case when the substituent is $\ce {-OH}$. This is because the strong delocalisation of the oxygen lone pair of the $\ce {-OH}$ substituent intensifies the negative charge in the carboxylate ion rather than disperse it. If you understand this reasoning, then you should also be able to see the same effect occurring in the phosphorus oxyacid species.
Although there is some inductive withdrawal of electron density from the $\ce {P}$ atom, the net effect is donation towards the $\ce {P}$ atom to form a double bond. This actually intensifies the charge in the resultant conjugate base of the oxyacid. Thus, the more $\ce {-OH}$ substituents you have, the greater intensification effect and hence, the less stable the conjugate base of the phosphorus oxyacid.
I have seen many interesting sources proposing various explanations to this acidity order in the past but none are as convincing as what I have proposed. I hope you would agree with me on that. Some sources discuss a phenomena they call "cross-conjugation", which is similar to what I have proposed. However, the source which discussed that put it in a weird way, saying that the donation of the lone pair from the $\ce {-OH}$ groups "disrupt" the delocalisation of the negative charge on into the $\ce {P=O}$ bond.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Choosing an electrolyte for a galvanic cell
I was wondering what processes are used to select an electrolyte for a galvanic cell.
I know a zinc copper battery usually has zinc and copper sulfate electrolytes. How do these electrolytes interact wishing each half cell (i.e.: how does zinc sulfate interact with water, the anion from the salt bridge and zinc itself)?

The point of an electrolyte is to allow ion flow between the anode and the cathode in order to balance the charges on either side so the reaction can continue. What you are looking for in a good electrolyte is that it won't react with either of your metals (i.e. won't form a precipitate), and that the ions will readily move (that's what makes soluble ionic compounds a great choice).
With $\ce{ZnSO4}$, for example, you wouldn't want to have $\ce{Pb^2+}$ ions anywhere in your cell, since $\ce{PbSO4}$ would precipitate and your concentrations of your ions would change, therefore changing your overall cell potential.


text:
Prevention of rusting
I'm currently learning about corrosion prevention, and have a question regarding galvanising.
Say that you have an iron nail. You galvanise it (coat it in zinc), so that is doesn't rust by preventing water and Oxygen from reaching its surface. But then wouldn't the Zinc coating rust? What would be the point of galvanising an object when its coating is going to rust anyway?

Rusting is the reaction of iron, oxygen, and water. Zinc metal does not rust.

Galvanizing isn't just a protective coating in the sense that it avoids contact between the iron and the oxidant. Zinc oxidizes more easily than iron, and since both conduct electricity, you can have zinc in contact with the iron anywhere, and it will oxidize first. The term for zinc in this case is "sacrificial anode."
Full coating does help additionally though, by forming a protective layer of zinc oxide. Iron deteriorates when it rusts, so new iron metal is always being exposed as the process continues. However, some metals form oxides that protect the metal underneath from further oxidation. Zinc and aluminum are both examples of these metals.


text:
Solving for gas law partial derivative
I don't know where to begin with this question.

A scientist discovered that the state of the unknown gas can be well described by the equation of state below:
  $$P = \frac{RT}{\overline{V}} + \left(\frac{a + bT}{\overline{V}^2}\right)$$
  Find the partial derivative $\left(\frac{\partial \overline{V}}{\partial T}\right)_P$ which is expressed in terms of $\overline{V}, R, b$, and $P$. (Hint: If $\overline{V}$ is a function of two variables $T$ and $P$ (i.e. $\bar{V} = f(T, P)$), then the total differential can be expressed as
  $$d\bar{V} = df(T, P) = \left(\frac{\partial \overline{V}}{\partial T}\right)_PdT + \left(\frac{\partial \overline{V}}{\partial P}\right)_TdP$$
  Use a particular condition to derive the expression $\left(\frac{\partial \overline{V}}{\partial T}\right)_P$ in terms of $\left(\frac{\partial \overline{V}}{\partial P}\right)_T$ and $\left(\frac{\partial P}{\partial T}\right)_{\overline{V}}$.)

This problem is related to the physical chemistry of the gas law. What is the correct approach?

My first thought was simply to solve the equation for T in terms of P, V, R, b, and a, then take the partial derivative.  It's a fairly straightforward quotient rule.  My solution still contains a though...

That's not universal; what if we were not able to get the explicit form for either V or T?

This is purely a question about maths. You will probably get a better answer if you ask it on maths.SE.

On math.SE it would get rejected right away for not using MathJax.

Please don't remove the question. That isn't very helpful.

Think of it this way.
$$dP = \left(\partial P\over\partial T\right)_VdT+ \left(\partial P\over\partial V\right)_TdV$$
Now, we want P to be constant (that is, $dP=0$). What does that entail? A certain relation between $dV$ and $dT$ (which, BTW, are now $\partial$ rather than $d$):
$$\left(\partial V\over\partial T\right)_P=-\left(\partial P\over\partial T\right)_V\left/\left(\partial P\over\partial V\right)_T\right.$$


text:
Ab initio effective potentials
I have been reading a journal article on ab-initio pseudo-potentials, and I need some help understanding it. The article is C. F. Melius and W. A. Goddard, III. Phys. Rev. A 1974, 10, 1528.
A summary of the article for those that do not have access follows. The authors present a way to construct a pseudo-potential. Basically, their argument goes like this (according to my understanding):

Suppose that we are using the Hartree Fock method. We can represent
  the ground-state wavefunction of some atom as a function of the valence
  orbitals, core orbitals, and spin.
$$\Psi= f (\phi_{core}, \varphi_{valence},X)$$
Note: in the article they use a weird letter instead of $f$.
  After we obtain the ground-state occupied orbitals, we solve for the valence orbitals.
  We obtain the valence orbital by solving the variational equations
  $$H^{HF}\varphi_{i}=\epsilon_i\varphi_{i}$$ Where $H^{HF}$ is the Hartree Fock Hamiltonian and is given by
  $$H^{HF}=-\frac{\nabla^2}{2}-\frac{Z}{r}+2J_{\alpha}-K_{\alpha}$$
  For the valence orbital, $J_{\alpha}$ is equal to the classical Coloumb potential due to a charge density corresponding to the $\phi_{core}$. The exchange operator $K_{alpha}$ is an integral operator resulting from the antisymmetric form of the wavefunction.
The authors then state that the valence orbital of the ground-state wavefunction is NOT determined uniquely by the variational principle. This is because the solutions of the variational equations may not yield a wavefunction that is purely a valence orbital. In other words, the mixing of core orbitals with a valence orbitals does not change its energy. Hence the solution $\varphi_i$ may be of the form $$\varphi_i=\varphi_{valence}+\sum_i c_i\phi_{core}$$
  The authors then impose the constraint that $\varphi_{i}$ be orthogonal to other occupied orbitals.

What I do not understand is this. How does orthogonality allow for a unique solution? Also later on, the authors state "Although this orthogonality restriction has the desired consequence of leading to a prescription for a unique valence orbital, there is no reason to consider $\varphi_i$ to have a special significance over any other combination of $\varphi_i$ with the various core orbitals." What do they mean by this statement?

First off, I would like to note that you wrongly interpreted the very first equation. Below I quote the relevant part of the paper:

In the Hartree-Fock (HF) approximation for, say, the $\ce{Na}$ atom,
  the ground-state wave function has the form
$$
\Psi = \mathcal{A} (\Phi_{\mathrm{core}} \phi_\nu X) \, , \tag{1}
$$
where $\Phi_{\mathrm{core}}$ is a product of (ten) spatial orbitals
  very similar to the orbitals of $\ce{Na+}$ and $\phi_\nu$ is the
  valence orbital (the one removed in ionizing to $\ce{Na+}$). ($X$ is an
  appropriate product of spin functions.)

So, $\Psi$ is not just some function of spin orbitals designated by a weird letter as you said, it is, as usual, their antisymmetrized product, where I bet the calligraphic A symbol ($\mathcal{A}$) as usual stands for the antisymmetrizer. The only small difference is that small calligraphic a is used in the text rather then much more common capital one.

Now to you actual question. Both core and valence orbitals are thought in the derivation as the solutions of the Hartree-Fock equations, which have a very well-known feature: they are defined to within a unitary transformation (see, for instance, here). So, orbitals which are solutions of the Hartree-Fock equations are not unique: there are infinite different sets of orbitals that minimize the Hartree-Fock energy and you are free to choose any of these set for any further work. This is, essentially, what is said in the paper:

It will be important in our later analysis to note that the valence
  orbital in (1) is not determined uniquely by the variational
  principle. One can modify $\phi_\nu$ by mixing in an arbitrary amount
  of any core orbital (doubly occupied. in $\Phi_{\mathrm{core}}$)
  without changing the energy.

Just note that this is true not only for the valence orbital: all orbitals (i.e. core as well) are not uniquely determined by the variational principle, as I mentioned above.
Now, if the goal is to have a unique valence orbital, then it can be achieved by restricting it to be orthogonal to core orbitals. In that case when transforming orbitals you can not mix any amount of any core orbital into the valence one since it will immediately make it non-orthogonal with this particular core orbital. And since mixing in core orbitals is the only way you can change the valence one, orthogonality of the valence orbital implies its uniqueness.

In order to obtain unique solutions for $\phi_\nu$, one generally
  requires that $\phi_\nu$ be orthogonal to the other occupied orbitals.

And it is important to note (as it is done in the next sentence of the paper) that :

Although this orthogonality restriction has the desired consequence of
  leading to a prescription for a unique valence orbital, there is no
  reason to consider the orthogonal valence orbital $\phi_\nu$, to have
  a special significance over any other combination of $\phi_\nu$, with
  the various core.

Note finally, that due to the orthogonality restriction you can not also mix any amount of the valence orbital into the core ones, but you can mix core orbitals with themselves anyway you like since it won't do any harm for their orthogonality with the valence orbital. Thus, while the valence orbital is uniquely defined by the orthogonality requirement, the core orbitals are still defined to within a unitary transformation.

Very nice answer, +1. Just want to ask for some clarification: When the authors of the article state
One can modify $\phi_\nu$ by mixing in an arbitrary amount of any core orbital (doubly occupied in $\Phi_{\mathrm{core}}$) without changing the energy.
they are talking about the total energy of the whole molecule/system not the energy eigenvalue of the valence orbital, right?

@Philipp, they are surely talking about the total energy (or just its electronic part). When orbitals are transformed so does their energies: in matrix notation when spin orbitals are transformed as follows $\mathbf{\Psi}' = \mathbf{U} \mathbf{\Psi}$, their energies become $\mathbf{E}' = \mathbf{U} \mathbf{E} \mathbf{U}^{\dagger}$. It is the Slater determinant, and thus, its energy, which is invariant to a transformation, not individual spin orbitals and their energies.

Thanks, for clearing that up. Just wanted to make sure that my basic knowledge still works. I'm getting a bit rusty :)

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Where does the charge on an electrode, responsible for the formation of a double layer once submerged in an electrolyte, come from?
I've started exploring electrical double layers and electrode-tissue interfaces whilst dealing with deep brain stimulation and stimulus artefact. I've heard many times that what causes the double layer to form is a net charge at the surface of the submerged electrode that then quickly attracts ions (cations or anions depending on the surface charge sign).
I always thought that in general, objects are neutral and don't carry charge around, where then, does the surface charge on electrodes come from?

This can be answered on more than one level. Where did the word "electrode" come from, to begin with? Supposedly it is due to the fact that you apply some electric potential to that metal needle (stuck into the brain of some creature or other electrolytic medium), and that's why the electrode gets charged.
Now what if we apply no potential? Well, for example, if two pieces of different metals are in contact, they would form a galvanic couple and develop some potential on their own, resulting in galvanic corrosion.
But wait, what if there is just one metal in the electrolyte? It turns out that some contact potential would develop even then. As the link says:

Electrode potential appears at the interface between an electrode and electrolyte due to the transfer of charged species across the interface, specific adsorption of ions at the interface, and specific adsorption/orientation of polar molecules, including those of the solvent.

Think of a metal in a distilled water (which, BTW, is a poor electrolyte). Inevitably, a tiny portion of metal would dissolve, for such is the nature of chemical equilibrium: it is almost never shifted all the way to one side. Now, that metal would be in the form of cations, and the extra electrons would stay in the bulk, giving it a negative potential (not so tiny, because the electric forces are quite powerful). And if we consider other possible effects, especially if the water is not distilled but contains some ionic species, then the resulting potential on the metal can be pretty much anything (and not necessarily negative).
All in all, a random piece of metal in a random electrolyte will most likely have some potential difference with the electrolyte and hence form a double layer.


text:
Electrolysis of brine without a membrane
In the electrolysis of brine sodium hydroxide is normally formed if there is a membrane preventing the transfer of the chlorine gas. So if the membrane is removed,  will the chlorine gas react with NaOH to form sodium hypochlorite? Because I've read that the chlorine will instead just react with the hydroxide ion to produce hypochlorite. Or will it produce both? and where does the sodium ion go if it doesn't?

So if the membrane is removed, will the chlorine gas react with NaOH to form sodium hypochlorite?

Yes.

Because I've read that the chlorine will instead just react with the hydroxide ion to produce hypochlorite.

Yes. This is just another way to say the same thing.
As for the sodium, it does not go anywhere. It simply continues to stay there in the solution, like it did before the reaction. It does not even "know" about the reaction, or that it now "belongs" to the hypochlorite; it is just a spectator ion. Since the dissolution, anions and cations go their separate ways and do not "care" much for each other. This is always the case with ionic salts.


text:
mechanism in formulation of mesylates and tosylates

in these reaction(Ts and Ms), why do they proceed in different mechanism/?
more specifically,

(Ts) e- pair from O attacks S, (Ms)but not in latter reaction. why?

it makes no sense. because TsCl is more sterically hindered by benzene ring.

in Ms, e- from hydroxyl group attack H attached to methyl group instead. I'm not famillar with H that was attached to C taken by other things.
strong base can do this?

help me~

Don't post the same question twice. Edit your previous question if you want to add more information.

okay. sorry T_T


text:
Deuterium depleted water through fractional distilation
Could a packed fractional still of the same type used to distill ethanol to 95% be used to produce deuterium depleted drinking water? The still has a 1000 watt electric heating element and has a fractioning column that is 120cm long which is packed with stainless steel wool. If it is feasible what kind of performance might I expect?

Some people hope to live longer by drinking it. Well, as to me, this is harmless, so why not. As for the technical part, we had a similar question not so long ago: http://chemistry.stackexchange.com/questions/47189/deuterium-removal-from-water. Distillation in this setup is hardly going to do the trick; after all, water and heavy water are much more similar than water and ethanol.

Consuming deuterium depleted water may have health benefits.  A company in Europe sells DDW and it is very expensive at around $40 per liter.

I very much doubt the theories that deuterium depleted water provides measurable health benefits, but it would be a very expensive idea to test directly.  Has anyone ever shown that providing mice with a diet of fivefold or even one hundred-fold more deuterium than usual results in altered health in any way?

Relying on the difference in vapor pressures of $\ce{HDO}$ and $\ce{H2O}$ is equivalent on relying on the equilibration of
$\ce{HDO(g) + H2O(l) <=> H2O(g) + HDO(l)}$
The fractionation factor $\alpha_{\ce{D}(l)/\ce{D}(g)}$ for this reaction is 1.015 at STP.
Deuterium ($\ce{^2H}$ or $\ce{D}$) is separated from protium ($\ce{^1H}$) commercially, usually via the Girdler sulfide process, which instead of relying on the above reaction, relies on
$\ce{ HDS + H2O <=> H2S + HDO}$
The fractionation factor $\alpha$ for this process is about 2.2.
Steam distillation for deuterium enrichment is certainly possible, but is not nearly as favorable as the industrial process. Given the very low concentration of deuterium in natural waters, the Rayleigh fractionation equation would be applicable:
$$R_{\ce{D}/\ce{H}} \approx R_{\ce{D}/\ce{H}_{init}} \times f^{\alpha-1}$$
In this equation:

$R_{\ce{D}/\ce{H}}$ is the ratio of deuterium to hydrogen in the liquid water
$R_{\ce{D}/\ce{H}_{init}}$ is the initial ratio of deuterium to hydrogen, or about 0.000156.
$f$ is the fraction of the liquid water remaining after distillation.
$\alpha$ is the fractionation factor mentioned above.  However the value of 1.0015 is for the condensation of vapor phase water into liquid; performing the reverse reaction we should use the inverse, i.e. for distillation of existing liquid into vapor, $\alpha = 0.985$

With this equation and with mass balances (the total amount of deuterium and of protium doesn't change in the distillation process) you can find the amount of deuterium in the distillate.
If you distill off 100 grams of an initial 1000 grams of water, then $f=0.9$, $f^{\alpha-1} \approx 1.000156$, meaning the remaining 900 grams of liquid water will have a deuterium-to-hydrogen ratio of 0.0001562 instead of 0.000156.
The mass balance for deuterium would be:
$1000 \times 0.000156 = 100 R_{\ce{D}/{\ce{H}_{distillate}}} + 900 \times 0.0001562$
If I did all my math right, that means your distillate would have a deuterium-to-hydrogen ratio of 0.0001542, or about 1.2% less than the original water.
Hopefully this exercise convinces you that you would have to do a lot of distillation and burn through a ton of water to generate reasonable amounts of deuterium depleted water through simple steam distillation.  There's a reason they don't do it that way commercially.

You mention "simple steam distillation" in your closing remarks.  I completely agree with you on that idea.  However, the method in fractional distillation is much more efficient as it performs many simple distillations (dozens or more re-distillations in a single pass) within the fractioning column apparatus.  Would this not improve upon the 1.2% given in your formula?

You are right, bu....  Roughly speaking, say you get a 1.2% decrease in relative deuterium content for each equilibrium stage, assuming each stage produces 10% vapor and 90% liquid.  Then you get an improvement factor $1.012^n$ for $n$ equilibrium stages, but the mass you obtain goes down by tenfold each time.  So if you had three stages, you would get 3.6% less deuterium than you started with, but you would have only one gram of the resultant water.  By optimizing the liquid/vapor ratio at each equilibrium stage, you could improve on this somewhat, but not by much.

Cody from Cody's Lab on YouTube used electrolysis and managed to concentrate an amount using time and solar electricity.  -  https://www.youtube.com/watch?v=VC4dk1JU4tQ


text:
Active methylene group nucleophilic substitution

The first product formed after reaction with $\ce{^-CH(COOEt)2}$ followed by hydrolysis is [2-(naphthalen-1-yl)ethyl]propanedioic acid:

But is the reaction $\mathrm{S_N1}$ or $\mathrm{S_N2}$?
On heating, decarboxylation takes place because of an $\alpha$-carboxylic acid group.

Then $\ce{SOCl2}$ followed by $\ce{AlCl3}$ is Friedel–Crafts acylation because using $\ce{SOCl2}$ gives an acid chloride. After that, an intramolecular electrophilic substitution takes place.
But where will the electrophilic attack take place?

You have given correct suggestions for the first two part of the question. Now, the question is where would be the electrophilic attack of acid chloride taken place under Friedel-Crafts acylation conditions. As Ron correctly pointed out in his comment that there are only 2 possibilities to cyclization: That would be to the 2 or 8 positions of naphthalene nucleus, and it would certainly be the 2 position, which produces a 6-membered ring. The method is a named reaction, which is called Haworth Phenanthrene Synthesis (Ref.1 & 2). According to the abstract of Ref.2:

Haworth synthesis is a multistep preparation of phenanthrenes from naphthalenes by means of the Friedel–Crafts acylation with succinic anhydride, followed by a Clemmensen reduction or Wolff–Kishner reduction, cyclization, reduction, and dehydrogenation. This synthesis is also referred to as the Haworth phenanthrene synthesis. Additionally, the same procedure used for the preparation of naphthalenes from benzenes is called the Haworth reaction. This reaction has been modified to react with 4‐pentenoic acid ester to form alkyl substituted phenanthrenes. This reaction has a wide application in preparation of fused aromatics, including naphthalenes and phenanthrenes.

The Haworth synthesis is depicted in following Scheme under the same title in top box:

You see after the reduction by $\ce{Zn/HCl}$, you get the same compound (4-(1-naphthyl)butanoic acid) you got after acid hydrolysis and heat in your scheme, as indicated in top yellow box. In Haworth reaction, $\ce{H2SO4}$ has been used instead of $\ce{AlCl3}$ for the convenience, the reaction under goes same pathway to give tetrahydrophenanthrene keto derivative by acylating 2-position of the naphthalene nucleus (see second yellow box). There was no evidence for the formation of 7-membered ring (by undergoing cyclization through 8-position).
However, if you had 4-(2-naphthyl)butanoic acid instead of 4-(1-naphthyl)butanoic acid, you certainly would get anthracene derivative (e.g., compound 4) as the major product with phenanthrene derivative (e.g., compound 5) as a trace product (See second box in the Scheme)(Ref.3).

References:

R. D. Haworth, “145. Syntheses of alkylphenanthrenes. Part I. 1-, 2-, 3-, and 4-Methylphenanthrenes,” J. Chem. Soc. 1932, 1125-1133 (DOI: 10.1039/JR9320001125).
Zerong Wang (Ed.), In Comprehensive Organic Name Reactions and Reagents – PART TWO; John Wiley & Sons, Inc.: New York, NY, 2010, “Chapter 299: Haworth Synthesis (Haworth Phenanthrene Synthesis),” (https://doi.org/10.1002/9780470638859.conrr299).
Z. Zhang, R. Sangaiah, A. Gold, L. M. Ball, “Synthesis of uniformly $\ce{^{13}C}$-labeled polycyclic aromatic hydrocarbons,” Org. Biomol. Chem. 2011, 9, 5431-5435 (DOI: 10.1039/C0OB01107J).


text:
What is the difference between pH and pKa?
Is there any difference between the two, when talking of an acid? I am a bit confused. Please help me out.

See also: http://chemistry.stackexchange.com/questions/17326/how-do-you-explain-pka-to-non-professional/17341

$\mathrm{p}K_\mathrm{a}$ is the intrinsic property of an acid. It does not depend on how much acid there is in a solution‡. For example, acetic acid has a $\mathrm{p}K_\mathrm{a}$ of $4.76$. This $\mathrm{p}x$ means $-\log x,$ we can calculate $K_\mathrm{a}$ to be $10^{-4.76}$ is $1.7\times 10^{-5}$, which means that at equilibrium, $\frac{[\ce{H+}][\ce{CH3COO-}]}{[\ce{CH3COOH}]} = 1.7\times 10^{-5}$. On the other hand, $\mathrm{pH}$ is related to the actual amount of $\ce{H+}$ ions, so if you had more acetic acid, the solution would be more acidic ($\mathrm{pH}$ is lower). This is why on Wikipedia, you see $\mathrm{p}K_\mathrm{a}$ given for acids, rather than a $\mathrm{pH}$, because $\mathrm{pH}$ is amount dependent. Similarly, in experiments, we are more interested in $\mathrm{pH}$ because that tells us the actual acidity of a solution.
Also know that though $K_\mathrm{a}$ may always be constant,‡ percent ionization isn't; as there is more weak acid, they dissociate less.

‡ Just like the ideal gas law, aqueous equilibrium constants based on concentrations are for ideal solutions, but they is very accurate for most work.

pKa is the pH for which the protonated ($\ce{HA}$)  and unprotonated species ($\ce{A^{-}}$) of the acid are equal. So $$ \ce{10^{pKa} = \dfrac{[H+][A^{-}]}{[HA]} = \dfrac{[H+]^{2}}{[HA]}} $$

The Ka, like all other equilibrium constants, gives the concentrations of the substances at equilibrium. From that, you can find the concentration of H+ at equilibrium, which in turn can be used to find the pH.
Note that Ka can only be used for equilibrium. pH can be used any time you have a concentration for H+ and just need to convert it to a scale so you can measure it/compare it to others.


text:
Will t-butoxide still form less substituted alkene when there is a possiblity for more stabilisation?
Consider this reaction:

Will a usual Hoffman elimination take place or will a carbocation form by leaving of $\ce{Br}$ and rearrange to give a conjugated alkene like this one:

I have heard that reactions with t-Butoxide are E2 and hence carbocation are not formed. So wouldnt the major product still be the least substituted alkene?

t-Butoxide is likely too strong of a base/nucleophile to be starting an E1 reaction; I don't think a carbocation will form here.
In this case, since t-Butoxide is bulky, the Hoffman product will be favored over the Zaitsev for E2 elimination.


text:
Why can a face-centered cubic lattice not be redrawn as a body-centered tetragonal lattice?
A base-centered cubic lattice can be redrawn as a primitive tetragonal lattice, therefore we do not include it in the list of Bravais lattices. Why is it then that a face-centered cubic lattice cannot be redrawn as a body-centered tetragonal lattice? If they can be redrawn as each other, why is it then that they are still listed in the Bravais lattice list? What is the ultimate condition to be in the list?

Every centered lattice can be redrawn as some other primitive lattice. This is not the reason why it is or is not included in the list.

@IvanNeretin, I do not see how that is true. How should one redraw, for example, the body-centered lattice with an another primitive lattice (triclinal excluded)?

(1) Suppose you have a lattice based on three vectors: $\vec a, \vec b, \vec c$, and it just happens to be body-centered. Then it is equivalent to a primitive lattice based on the vectors ${\vec a+\vec b-\vec c\over2},{\vec a+\vec c-\vec b\over2},{\vec b+\vec c-\vec a\over2}$.
(2) Then the answer to your question is quite straight-forward too: it surely can be redrawn this way, and they are no different (as long as we consider just the lattices).

Not easily explained, but the simple answer is that the face-centered cubic lattice has greater symmetry than body-centered tetragonal lattice so it is the choice to represent the redundant types. You could think of combinations of the 7 lattice systems (Triclinic, Monoclinic, Orthorhombic, Tetragonal, Rhombohedral, Hexagonal, and Cubic ) and 7 types of lattice points (primitive, body-centered, face-centered, 3 base-centered, and rhombohedral) for a total of 49 combinations. Some aren't allowed, but over the many allowed combinations there are only 14 unique Bravais lattices.

I'd put it this way: a crystal is a fact of nature. A centered crystal is not a fact of nature, but merely a human convention. It is just our way to look at some particular crystals.

To which Bravais lattice does the above lattice belong to then?

OK, let's get it straight. A lattice is just a periodic set of points. For the sake of brevity, I'll talk about 2D lattices; trust me, with 3D it is pretty much the same, only more complicated.

Initially the lattices were just like that: periodic patterns of one infinitely repeating parallelogram, called the unit cell. Other than that, they were without form, and void; and darkness was upon the face of the deep. The very word "centered" was not invented yet.
Then people started to categorize lattices by symmetry.

First of all, there are those with $90^\circ$ rotational, or 4-fold symmetry (we may call them "tetragonal", or maybe "square"), and those with 6-fold symmetry (we may call them "hexagonal"). We put them aside and look deeper at the rest.

You see that the first one of these has no symmetry to speak about (other than the $180^\circ$ rotational symmetry, which all 2D lattices have by birthright). The link above calls such lattices "oblique"; I don't like the word, but whatever. The second one has mirror symmetry with respect to the lines running along the sides of the cell; this is an example of so-called "rectangular" lattice. And the third one has mirror lines along the diagonals of the cell. This just does not sit well with us; we'd rather have them along the sides. (We will meet these diagonal mirror lines later anyway, when we'll deal with the tetragonal and hexagonal lattices; but that's another story.) So we select another unit cell, twice bigger in area than the original, and call it centered.

See, the lattice (i.e., the arrangement of blue dots) is still the same; we couldn't and didn't make it more symmetric, or less symmetric, or in any slightest way different, for that matter. We just started to think of it in a different way.
And so the centered rectangular cell entered the list of 2D Bravais lattices. See, it did not get there because it can't be decomposed into some primitive cell (it surely can, as any centered cell can), or because other cells can't be centered (they surely can, only it would make no sense). It is there because we wanted our symmetry elements to be aligned in a certain way.
Like I said, in 3D we have the same story, only somewhat more complicated. Indeed, the face-centered cubic cell can be thought of as body-centered tetragonal (and also as some primitive non-centered cell, too); it is just that the cubic cell has 4-fold rotation axes running along all sides, and in a tetragonal cell some of them would run diagonally, which is not considered comme il faut.
My apologies to those who think they'd be better off with the diagonal mirrors and without centering. Anyway, the convention has been in use for well over a century now, so maybe it's not that bad after all.

This is exactly what I needed to read, exactly at this moment; will be posting a follow-up question (somewhat) shortly...

After going off and thinking then coming back and re-reading, it seems that all 2D Bravais lattices are subsets of o̶b̶l̶i̶q̶u̶e̶  I mean, the lattice who shall not be named, and there are no Bravais lattices which can't be defined by $(\vec a, \vec b)$  or $(a, b, \theta)$ depending on if your use involves orientation. While there are non-Bravais lattices (e.g. honeycomb) which can't, that's different. Your answer was really helpful. Thanks!

"Bravais" is not quite the word I'd use here. All lattices can be thought of as some primitive lattices. A honeycomb can't, hence it is not a lattice at all.

Crystallographic terminology associated with the honeycomb structure

In one crystallography lecture, the professor mentioned that in some areas of physics all lattices are redrawn to a smaller primitive lattice and none of the centred lattices are allowed. This is obviously mathematically possible.
However, it is often beneficial to have a lattice as symmetric as possible. While obviously the symmetry elements in both descriptions (body-centred tetragonal and face-centred cubic) must be the same by definition — we are dealing with the same arrangements of spheres after all — the cubic lattice immediately gives us a lot more symmetry that we can work with. So the choice is actually to use the smallest lattice with the highest symmetry for easier working.
Now the actual reason that base-centred cubic is not permitted is not that it can be redrawn as tetragonal per se — as you showed, the face-centred cubic can also be redrawn as tetragonal. The high symmetry of the cubic lattice not only requires $\alpha = \beta = \gamma = 90^\circ$ as with the tetragonal lattice but also $a=b=c$ while tetragonal has $a=b\ne c$. This means that in cubic we cannot have any preferential direction; all directions must be equal or else it would not be cubic.
By introducing a centred atom on only one of the faces (without loss of generation I choose the $ab$ face), we have immediately created a preferential direction: $c$ is now no longer equal to $a$ and $b$ even if their values are equal. Thus, we have already lost cubic symmetry and are dealing with a tetragonal base-centred lattice. What we are actually transforming is one tetragonal lattice into another, smaller one. We lose no symmetry.
In face-centred cubic, all directions are still equivalent, since all contain a sphere in the centre of the face. Thus, by turning it into a smaller body-centred tetragonal lattice we would be losing symmetry by making $c$ unequal from $a$ and $b$ — this is not desirable.

Thank you Mr Jan,  I'm not sure I've understood all of it but could you please tell me to which Bravais lattice does the above fcc cubic fall then? In  Fcc cubic or in body centered tetragonal ?

The first problem is with your starting argument : The problem with the "base-centered cubic lattice" is that it is not a cubic lattice! Cubic symmetry means the presence of 4-fold and 3-fold symmetry at the same time (3-fold in the space diagonal of the cube). You cannot have only centering on one face and 3-fold symmetry. Thus a hypothetical "base-centered cubic lattice" has only 4-fold symmetry and is thus a tetragonal lattice (and can be re-drawn as such without loss of symmetry). Please remember, that the crystal lattice is defined by the symmetry present, not by the random metrics of the axes!
To answer your question: all centered cells can be - trivially - redrawn as primitive cells without loss of symmetry. However, there are always additional conditions to these cells to keep the higher symmetry available in the centered cell.
For example, a face-centered cubic cell allows 4-fold and 3-fold symmetry. The body-centered tetragonal cell you propose has no restriction on c, and thus allows only 4-fold symmetry. It does allow 3-fold symmetry in the special case where c = SQRT(2)*a (in which case it can be re-drawn as face-centered cubic).
Similarly, monoclinic C can be expressed by triclinic P with a=b. Orthorhombic A can be redrawn as monoclinic P with a=b.
We can thus replace all centered cells with primitive cells, but these will not be the usual unit cells, but have to be new ones. Thus you can replace the 14 Bravais lattices (7 crystal systems plus centering) by 14 crystal systems with primitive cells, each with their own conditions.
We go for the centering approach for two reasons : (a) The macroscopic crystal shape is determined by the available symmetry elements. Thus is does not matter for the shape whether it is cubic primitive, body- or face-centered... it matters that there are 4-fold and 3-fold axes present. Since crystal classes came originally from crystal shapes, the nomenclature evolved this way.
(b) The seven crystal systems align closely with the symmetry elements present. Do you really find it easier to replace : "a two-fold axis or a mirror plane requires a monoclinic crystal system" with "a two-fold axis or a mirror plane requires a monoclinic crystal system or a triclinic system with a=b" ?

a tetragonal body centered lattice can be dimensionally equivalent to a face centered cubic lattice but under cubic symmetry a=b=c while under tetragonal symmetry a=b but c is unconstrained. so there is only one value of the a/c ratio for the body centered tetragonal lattice that makes the two dimentionally equivalent, just as two crystals could be dimensionally cubes but if one has a fcc lattice it would truly have cubic symmetry(min of 3 four fold axes) while if the other had a body centered lattice it would not truly have cubic symmetry (with only 1 four fold axis) and thus the faces perpendicular to the c-axis could have different properties to the four side faces.


text:
Hydrogen donor without hydrogen acceptor?
Is there a molecule that can donate a hydrogen to a hydrogen bond but not accept one? It would seem that since the donated hydrogen needs to be bonded to an electronegative atom, that electronegative atom will always have lone pairs available to accept a hydrogen bond.
For example, in ethanol, the H in the OH can participate in hydrogen bonding since the the O is electronegative. Then the oxygen with its two lone pairs can also accept hydrogen bonds.
In reverse, things like diethyl ether can accept but not donate hydrogens because the oxygen is surrounded by carbons, not hydrogen.
Are there any exceptions to this apparent rule?

Ammonium ions should be able to act as hydrogen donors yet lack a lone pair and are thus no hydrogen bond acceptors.

As mentioned in the comments by Philipp, ammonium ions $\ce{R3NH+}$ are a good direct answer to what you are looking for.
The nitrogen in an amide $\ce{RCONHR'}$ is similar - the N lone pair is conjugated into the C=O bond and that makes it a very poor hydrogen bond acceptor, but the N-H lone pair can be a hydrogen bond donor. For more details, see this previous question with an excellent writeup by Jan. Now an amide can of course function as a hydrogen bond acceptor via the lone pairs on oxygen, so it's not strictly what you're looking for, but it was just something related. And it is of course directly relevant to the secondary structure of proteins.

Good approximations to such a molecule are provided by chloroform and dichloromethane. In these molecules the polarization of multiple carbon-chlorine bonds renders the hydrogen sufficiently protic to serve as a good donor. In molecular-orbital terms the bonding orbitals in chloroform and dichloromethane are concentrated mostly on the carbon and chlorine atoms and an antibonding orbital is correspondingly enhanced on the hydrogen where it can have good overlap with a neighboring-molecule electron pair.
In principle, most nonmetal atoms can accept a hydrogen bond (even in some cases where a nonbonding pair is not avaialble), so the neat compounds must have some hydrogen bonding. But this interaction is weak; in mixtures containing stronger hydrogen-bond accepting atoms such as oxygen or nitrogen the chlorinated hydrocarbons hydrogen-bond almost exclusively through their donor hydrogens to the acceptors in the other species1.
Reference

Frank H Allen, Peter A Wood, Peter T A Galek (August 2013). "Role of chloroform and dichloromethane solvent molecules in crystal packing: an interaction propensity study". Acta Crystallogr B Struct Sci Cryst Eng Mater. 69(Pt 4):379-88.
doi: 10.1107/S2052519213015078.


text:
Nomenclature and chemical formulae for bridged dinuclear chromium complexes
In theory, $\ce{[Cr(H2O)6]^3+}$ ions have a bluish violet colour. However, many times in the lab we end up with a dark green solution. This is due to hydrolysis, coordination with sulphate (or other) ions, and polymerisation, most often dimerisation.
Hydrolysis$$\ce{[Cr(H2O)6]^3+(aq) +H2O(l) <=> [Cr(OH)(H2O)5]^2+(aq) + H3O+(aq)}$$
Coordination and dimerisation
The degree of coordination depends on conditions, e.g., temperature. It is common to find the following equilibria.
$$\ce{2[Cr(OH)(H2O)5)]^2+(aq) + $\mathrm{SO}_4^{2-}$(aq) <=> Picture1(aq) + 4H2O(l)}$$
$\ce{Picture1}$:

$$\ce{2[Cr(OH)(H2O)5)]^2+(aq) + 3\ $\mathrm{SO}_4^{2-}$(aq) <=> Picture2(aq) + 6H2O(l)}$$
$\ce{Picture2}$:

Question

How can one write these ions in condensed structural formulae? I wish not to lose any information.
How can one unambiguously name these two ions?

I am looking for official IUPAC names and formulae using latest recommendations for complexes. For instance, $\ce{[Pt(NH3)4][Pt(Cl)4]}$ is named tetraammineplatinum(II) tetrachloroplatinate(II); $\ce{[Cr(OH)(H2O)5]^2+}$ is pentaaquahydroxochromium(III) ion.
Thank you in advance!
P.S. Pictures were drawn using the free online software eMolecules. Also, feel free to correct any formatting and/or grammar errors. As for chemistry-related mistakes, please comment rather than edit.

It's simply dissociation not hydrolysis.

Also https://en.wikipedia.org/wiki/Olation

Thank you for the links. As for hydrolysis vs dissociation, either can be used here. Both Atkins and Wikipedia, for example, give this process as hydrolysis.

$\require{unicode}$The rules for formulating and naming coordination compounds are presented in Chapter IR-9 of the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book). An abridged version is included in the IUPAC Technical Report Brief guide to the nomenclature of inorganic chemistry. Pure Appl. Chem. 2015, 87(9–10), 1039–1049 as well as in the corresponding four-sided lift-out document, which is available as supplementary material.
The line formula for a simple coordination entity is constructed within square brackets. The symbol for the central atom is placed first and is then followed by the symbols or abbreviations for the ligands (in alphabetical order according to the way they are presented in the formula).

IR-9.2.3.1 Sequence of symbols within the coordination formula
(i) The central atom symbol(s) is (are) listed first.
(ii) The ligand symbols (line formulae, abbreviations or acronyms) are then listed in alphabetical order (…). Thus, $\ce{CH3CN}$, $\ce{MeCN}$ and $\ce{NCMe}$ would be ordered under C, M and N respectively, and $\ce{CO}$ precedes $\ce{Cl}$ because single letter symbols precede two letter symbols. The placement of the ligand in the list does not depend on the charge of the ligand.
(iii) More information is conveyed by formulae that show ligands with the donor atom nearest the central atom; this procedure is recommended wherever possible, even for coordinated water.

Note that according to the recommendations in Item (iii), where possible, the coordinating (ligating) atom should be placed nearer the central atom in order to provide more information about the structure of the complex. Therefore, the line formula of the hexaaquachromium(3+) complex ion should be written as $\ce{[Cr(OH2)6]^3+}$ rather than $\ce{[Cr(H2O)6]^3+}$.
For the same reason, if possible, bridging ligands should be placed between the central atom symbols. Bridging ligands are those bound to more than one central atom. They are differentiated in names by the addition of the prefix ‘μ’.

IR-9.2.5.2 Bridging ligands
Bridging ligands, as far as they can be specified, are indicated by the Greek letter μ appearing before the ligand symbol or name and separated from it by a hyphen; (…)
(…) In formulae, bridging ligands are placed after terminal ligands of the same kind. (…)
The bridging index $n$, the number of coordination centres connected by a bridging ligand, is placed as a right subscript. The bridging index $2$ is not normally indicated. (…)

Polynuclear complexes may be named differently according to whether only stoichiometry is to be specified or partial or complete structural information is to be included.

IR-9.2.3 Formulae of coordination compounds
(…) Different applications may require flexibility in the writing of formulae. Thus, on occasion it may be desirable to violate the (…) guidelines in order to provide more information about the structure of the compound that the formula represents. In particular, this is the case for dinuclear compounds where a great deal of structural information can be provided by relaxing the ordering principles outlined in Section IR-9.2.3.1. (…)

IR-9.2.5.1 General
(…)
As a general principle, as much structural information as possible should be presented when writing the formula or name of a polynuclear complex. However, polynuclear complexes may have structures so large and extended as to make a rational structure-based nomenclature impractical. Furthermore, their structures may be undefined or not suitably elucidated. In such cases, the principal function of the name or formula is to convey the stoichiometric proportions of the various moieties present.
(…)
Note, however, that the rules for formula writing may be relaxed in various ways in order better to display particular features of the structures in question. (…)

As a first approximation, the systematic line formula for the complex ion shown in Picture 1 is $\ce{[(H2O)3Cr($\unicode[Times]{x3BC}$-OH)2($\unicode[Times]{x3BC}$-SO4)Cr(OH2)3]^2+}$. Note: pay attention to the Greek letter; it is to be written upright.
However, since this complex ion is symmetric, its formula may be further simplified.

IR-9.2.5.4 Symmetrical dinuclear entities
For symmetrical dinuclear entities, the name may be simplified by employing multiplicative
  prefixes.
(…)

The resulting formula is: $\ce{[\{Cr(OH2)3\}2($\unicode[Times]{x3BC}$-OH)2($\unicode[Times]{x3BC}$-SO4)]^2+}$. The corresponding name is di-μ-hydroxido-μ-sulfato-bis(triaquachromium)(2+).

The corresponding formula for the complex ion shown in Picture 2 is $\ce{[\{Cr(OH2)2(SO4)\}2($\unicode[Times]{x3BC}$-OH)2($\unicode[Times]{x3BC}$-SO4)]^2-}$. The corresponding name is
di-μ-hydroxido-μ-sulfato-bis(diaquasulfatochromate)(2−).
Furthermore, if necessary, you can indicate the connectivity of the sulfato ligand using the kappa convention (sulfato-2κO).

(a) Recommendation $(i)$ states that central atom(s) should be listed first. Isn't then $\ce{[(H2O)3Cr(\mu-OH)2(\mu-SO4)Cr(OH2)3]^2+}$ in violation of this rule? Or does $(iii)$ supersede $(i)$? I also have another question. (b) IR-9.2.3.1 states that "placement of the ligand does not depend on charge". This seems to contradict what I have learned; specifically that anionic ligands are to be placed before neutral ligands. (c) Can one also use chloro- instead of chlorido- and cyano instead of cyanido (which is also what I have been taught). Thanks for a brilliant answer!

@LinearChristmas Yes, the formula $\ce{[(H2O)3Cr(\mu-OH)2(\mu-SO4)Cr(OH2)3]^2+}$ violates the ordering principles outlined in Subsection IR-9.2.3.1. This is allowed according to Subsections IR-9.2.3 and IR-9.2.5.1. I made an edit to my answer to clarify this.

@LinearChristmas The order of ligand citation was changed in the IUPAC Recommendations 2005. Now, it does not depend on the charge of the ligand. Also, the historical ligand names chloro and cyano are obsolete now. They were changed to chlorido and cyanido in 2005.

How would this generalise to the case where there is also a metal-metal bond present? More specifically, is the proper formula of chromium(II) acetate dihydrate given as $$\ce{[Cr(OH2)($\unicode[Times]{x3BC}$-O2CCH3)2]2}?$$
How would one name this compound? (If you're available, feel free to add this as an addendum to your already wonderful answer to receive a +100 bounty.)


text:
Why is ICl5 unstable?
For the question

Why is $\ce{ICl5}$ unstable?

my textbook gave an answer of

$\ce{ICl5}$ has too many highly electronegative chlorine atoms present on the central iodine.

But how does having electronegative atoms around a central atom make a molecule unstable? For example, $\ce{SF6}$ is very stable.

Steric hindrance, maybe?  Fluorine atoms are relatively small.  Steric hindrance is less in $SF_6$ than in $ICl_5$.

Could symmetry  play a role in the relative stability of these molecules?

Hypervalent iodine is a common oxidant. Iodine(V) and iodine(III) are both reactive because iodine is most satisfied with a -1 oxidation state. Here, it has an oxidation number of +5.

Actually, iodine is not most satisfied in a $-\mathrm{I}$ oxidation state, especially not in water. Iodinated table salt contains iodate, not iodide, so that it will not get reduced to iodine and cause ugly colouring.

The textbook answer must be wrong, since $\ce{IF5}$ is a sufficiently stable compound and fluorine is more electronegative than chlorine.
Rather, I think that the reason for $\ce{ICl5}$’s lesser stability is the lower electronegativity of chlorine. Creating this type of hypercoordinate compounds requires two sets of four-electron three-centre bonds of the $\ce{X- \bond{...}{I+}-X <-> X-{I+}\bond{...}X-}$ type. This type of bond requires a sufficient electronegativity on the outer atoms to stabilise the partial negative charge better — something fluorine is more capable of than chlorine. $\ce{ICl5}$ should only be possible because of iodine’s low electronegativity and $\ce{BrCl5}$ should be orders of magnitude less stable again.
A second aspect could be the larger size of chlorine atoms compared to fluorine atoms causing greater steric stress, longer bonds and therefore weaker interactions.


text:
How many atoms of Lithium are there in 14.3 grams of Lithium Fluoride?
How many atoms of Lithium are there in 14.3 grams of Lithium Fluoride?
I can't seem to single out how many atoms of Lithium are in lithium fluoride.
I know that the first step is to convert 14.3 grams to moles, so 14.3 multiplied by the molar mass of LiF divided by 1 is 0.55 moles. But where do I go after this? Do I just multiply 0.55 by Avogadro's number?

Convert 14.3 grams of Lithium Fluoride to moles which is the same for lithium and lithium fluoride since there is one atom of lithium per molecule of lithium fluoride (carry an extra significant digit to round later to 3 significant figures).

$\mathrm{moles(Li) = \dfrac{g(LiF)}{mw(LiF)} = \dfrac{14.3}{25.939} = 0.5513\text{ }moles(Li)}$

Convert moles to atoms by multiplying with Avogadro's number, and round to 3 significant figures.

$\ce{atoms(Li) = moles(Li)\text{ }(N_{a})} \mathrm{ = 0.5513(6.022E23) =  3.32E23}$

no of moles in 14.3 g of LiF is
14.3*(1/formula unit mass)=no: of moles
no:of Li atoms=avogadro number * no:of moles (as in one LiF molecule there is only one Li ion)
the answer is 3.312*(10^(23)) (approximate value)

Your answer has too many significant figures and it is a bit off. What did you use for the molecular weight of lithium fluoride and  Avogadro's number?!?

@MaxW i never intended to do the calculation.I just took the rounded of values for atomic mass.like a 19 for Fluoride and 7 fo lithium.(after all there is no point in doing the pin point calculation like even you can add the mass of electron too (which i don't think is needed when one explains the theory part).

I tried a number of variations in the values and couldn't get your answer. I just don't think you should say "the answer is" if you haven't done a proper calculation.

@MaxW thats a valid point which i missed out(didn't add "approx").well i will edit.Thanks


text:
What does addition of iodine and base do this case?

I figured that the first reaction just adds a bromine to the meta position but I don't know what step ii) does. The only thing I can think of is the base gets rid of the alpha hydrogen next to the carbonyl. Then perhaps the negative carbon now attacks iodine? But that seems a bit off to me.

$\ce{I2}/\ce{OH-}$ will perform the iodoform reaction on the the product of previous reaction. The base deprotonates an alpha hydrogen, and the resulting enolate nucleophile attacks iodine. After three substitutions, the $\ce{CI3}$ group is a passable leaving group, and a nucleophilic acyl substitution occurs. The reaction is done in basic conditions, so the carboxylic acid is deprotonated. The leaving group is a strong base, so it is protonated. $\ce{CHI3}$ will be formed along with the conjugate base of m-bromo-benzoic acid. Here is a generic version:
$$\ce{RC(O)CH3 + 3I2 + 4OH- -> RCO2- + 3I- + 3H2O + CHI3}$$


text:
Output of the Sugar & Sulfuric Acid Reaction?
When a reaction of sugar and sulfuric acid occurs, some form of carbon is produced. Is this carbon bonded as graphite? If not, than what is it bonded as?
$\ce{C6H12O6 ->[H2SO4] 6C + 6 H2O }$

@A.K. Wow, that was really dumb of me, I didn't double check my chemical reaction and somehow convinced myself that carbon was the output.

@A.K. Revised question to valid chemical equation.

Related: http://chemistry.stackexchange.com/questions/15442/how-does-sulfuric-acid-dehydrate-sugars

The product is not pure elemental carbon, but instead an amorphous solid that consists of mostly carbon atoms, with at least 5-10% of the atoms being other elements.  For example, one academic paper from 2001 found that the "carbon" formed by sulfuric acid catalyzed dehydration of sugar had a $\ce{C}:\ce{H}$ molar ratio of 14:1.
Such materials are sometimes called hydrogenated amorphous carbon.


text:
Thermodynamic versus kinetic reactions
In a lecture, teacher said
"Nucleophilicity is about 'kinetic', and basicity is about 'thermodynamic'"
but I saw the picture below from another source.

Here, both pathways start with an acid/base reaction (it's about basicity) and then nucleophilic addition happens (it's about nucleophilicity). as a result, 'kinetic' and 'thermodynamic' product come up.
So here's my question.
Are the terms 'thermodynamic' and 'kinetic' that are used above in two cases different?

The terms are used in a different context to mean slightly different things.
The kinetic product of a reaction is the one which is formed the fastest. In the case shown above this is the 1,2 product which results from formation of the most stable cation, which has a lower activation energy for formation.
The thermodynamic product is the most stable product. In this case it is the 1,4 product because it has a more substituted double bond. However, it forms from a less stable carbocation and so the activation energy is higher and it is formed slower than the kinetic product. Raising the temperature favours the thermodynamic product because the activation energy becomes less important. Another way to get the thermodynamic product is to run the reaction under equilibrating conditions, allowing the product to convert back to reactants. This will eventually form more of the more stable product.
Nucleophilicity is often referred to as a kinetic property because it has to do with how well a species can attack an electrophile and is affected by factors such as steric hindrance.
Basicity is often referred to as a thermodynamic property because it is to do with the position of the acid-base equilibrium: $$\ce{B + H+ <=> BH+}$$

I want to point out that the two cations you show in the diagram are not two separate species, but rather two "resonance structures" of the same cation. Kinetically, it appears that chloride attacks preferentially at the more highly charged carbon atom, to give the so-called kinetic product.

Oops. I wasn't paying attention to that.

but they become different after they formed product, right?

Right-- the one allylic cation potentially gives 2 different products in combining with chloride ion.


text:
What is the correct order of chemical reactivity in terms of oxidising property for the following elements: F, Cl, O, and N
This is a textbook question. I am confused as to which factors i should consider while sorting out these elements. If I take electron gain enthalpy, then Cl should be greater than F, but if i consider electronegativity F is greater than Cl . I'd be grateful to anyone who provides  some idea..

Fluoride ion has very high hydrtation enthalpy and low F-F bond enthalpy.These are the two main reasons for fluorine being a better oxidising agent than chlorine (actually the best).

See, F is the strongest oxidising agent in the entire periodic table. Also, according to periodic table trends, oxidising character decreases down the group and increases across a period. You should check the reactivity of any element according to the question given, here, according to the oxidising character. If nothing is given, usually IE and Electronegativity is considered.
Therefore, I think the order should be F>O>Cl>N. Cl>N because Cl is more oxidising than N.
I must add that a simple Google search would have yielded the same result.


text:
Rotation around CC bond in 1,2-dioxetane: transition states and local minimum structures
TLDR:
What: Potential energy surface associated with rotation around CC bond in the ring-opened version of 1,2-dioxetane.
How: State average(4)-CASSCF(12,10)/VTZP constrained geometry optimizations keeping OCC'O' and one HCC'O' dihedral angle fixed. Active space is bonding and antibonding CC', OO', CO, and C'O' orbitals in the closed ring structure, plus the two nonbonding orbitals on the oxygens perpendicular to the plane of the ring.
Question: How explain the profiles in Figure 2.

I want to look at the transition states associated with the OCC'O' dihedral torsion in the biradical structure of 1,2-dioxetane (after ring opening). This is similar to the exercises in basic organic chemistry courses, where we look at the anti and gauche conformations of isomers. In Figure 1 you can see one biradical structure with an OCC'O' dihedral of about 70 degrees.

Figure 1. 70 degree dihedral OCC'O' angle minimum structure.
I decided to "map" the PES associated with CC rotation, by doing constrained geometry optimisations. I kept the OCC'O' and one HCC'O' dihedral simultaneously fixed at 10 degree intervals from 0 to 180 degrees, and optimizing the other internal coordinates. I did this with the state-average (4 lowest singlets) complete active space self-consistent field (CASSCF) method, using the atomic natural orbital type basis sets with polarisation at the triple zeta level (ANO-RCC-VTZP), using the Molcas suite. The energies of the singlet states can bee seen in Figure 2.

Figure 2. Mapping the $S_0$, $S_1$, $_2$, and $S_3$ potential energy surfaces, obtained as described above.
Qualitatively, the PES's mainly match what I expected, but with one major change: instead of a transition state close to 120 degrees, I see a minimum.
I am trying to understand why this is what I obtained, but I cannot really explain it. The change is so major, that any errors with my constrained optimization is likely not the cause, I reckon.
Figure 3 shows the C-C bond distance at the various optimized geometries. Here, the main features are present: minimum at 60 degrees and maximum state at 120. This makes sense as in a minimum the structure is more stable, which means shorter bond lengths. However, there is some weird behavior at the 90 degree data point, which probably is due to the same effect as the energy minima around 100 degrees.

Figure 3. C-C bond lengths in Å at the various optimized geometries.
So, my question, can anyone explain the anomalies presented in Figures 2 and 3?

Updates and answers to questions:
Yes, I changed the OCC'O' and noe of the HCCH dihedral angle simultaneously. I only chose one HCCH dihedral as I thought that was enough to keep the structure rigid during the geometry optimization.
The active space is the bonding and antibonding $\sigma$ orbitals of the closed ring structure (CC', OO', CO, and C'O'), plus the two nonbonding p orbitals on the oxygens (perpendicular to the plane of the ring).
I did not use symmetry to speed up the calculations.
For each dihedral angle used, I also measured the CC bond distance. I plotted this as an overlay on the energy profile, to see how the CC bond changed relative to the energy change.
As requested by Brian, the energies of all calculated states are included in the question.

I don't really understand figure 2. Why do you mix two graphs in one? I assume the bold line is the energy profile? What is the ground state? Could you explain the active space you have chosen for this? How does the active space change? When you say you increase OCCO and HCCO in 10 deg intervals, do you say you change them at the same time? Why are you just fixing one of the protons? Are you using symmetry? I assume SA means state averaged and you use 4 singlets for this? How do they look?

I have mislabeled Figure 2. I will fix it. And answering your questions will take some time, but I will try and get it all into the question soon.

@Martin-マーチン How much detail do you need me to provide concerning the change of the active space? Not much happens during the rotation. About the singlet states, what do you need to know regarding "how they look"? I have not optimised other states than the ground state.

You really need to plot the energies of all of the states you're averaging over.  I'm no expert with CASSCF, but this looks to me like you may have some crossover between various states, and plotting their energies simultaneously may provide significant insight.   That dramatic plunge in the $\ce{C-C}$ distance at $90^\circ$ is particularly suspicious. Inclusion of some triplet and/or quintet states may also be informative.

You should be able to tell the occupations of the orbitals in the involved states. At one point I would expect, that the anti binding and the bonding orbitals between the oxygen are degenerated, the molecule should be symmetric after all. And from that, there should be one non bonding orbitals per oxygen left, why did you not include these? The natural orbitals would be great to see, to. I'm not sure what to expect, but a detailed analysis of everything in context might give valuable insight.

I have updated the question quite a bit! I don't have the relevant orbital files for all states, only the ground state. I could re-run the calculation tonight to get the files, though. Would it be helpful if I added molden images of the active space orbitals of the ground state at some angle intervals (like, 0, 30, 60, 90, 120, 150, 180)?

I kept the OCC'O' and one HCC'O' dihedral simultaneously fixed at 10 degree  || I'm pretty sure you should keep only one of theese constraints.

@permeakra I was thinking this too -- unless some sort of dissociation or other divergent behavior occurs, best to minimize the number of applied constraints.

I chose two constraints as dissociative behavior occurred in transition state searches (the TS search algorithm works like this: constrained optimization until certain gradient criteria are fulfilled, indicating that the geometry is close to a stationary state, at which point the constraints are released). However, such dissociative behavior may not happen here, as I perform regular geometry optimizations. However, the OCCO and HCCH dihedrals are very close (within 5 degrees) in the 6+ and 180 deg minima, so I don't think the effect is that large.


text:
Entropy and Heat Capacity Trend in water
It is my understanding that entropy and heat capacity essentially measure the same thing. Since entropy is the ratio of heat (translational, rotational, vibrational movement) to temperature (translational movement), a substance with higher entropy (heat/temperature ratio) would require more heat to raise a certain temperature, which is the definition of heat capacity. Why then, is the heat capacity of liquid water around 4.2 J/g-K while for water vapor, it's only 2.0 J/g-K when vapor has more entropy?

It would be perhaps clearer to state that the heat capacity is directly related to the second derivative of the free energy. Since the free energy has both an enthalpic and entropic contribution, the rest is left as an exercise...

Well, you see that entropy and heat capacity are not always directly related. That being said, water is quite unusual in this respect.


text:
Photoionisation microscopy of hydrogen - where are the p orbitals?
In this Physics Review Letters article, which was published in 2013, but I found recently, the authors report photoionisation microscopy images of hydrogen atoms in various electronic states.

A. S. Stodolna, A. Rouzée, F. Lépine, S. Cohen, F. Robicheaux, A. Gijsbertsen, J. H. Jungmann, C. Bordas, and M. J. J. Vrakking, "Hydrogen Atoms under Magnification: Direct Observation of the Nodal Structure of Stark States,"
  Phys. Rev. Lett., 2013, 110, 213001; url: http://journals.aps.org/prl/abstract/10.1103/PhysRevLett.110.213001

Here are some non-paywall summaries of this work:

physicsworld.com http://physicsworld.com/cws/article/news/2013/may/23/quantum-microscope-peers-into-the-hydrogen-atom
APS Physics http://physics.aps.org/articles/v6/58
io9 http://io9.gizmodo.com/the-first-image-ever-of-a-hydrogen-atoms-orbital-struc-509684901

The imagery is compelling:

I see images that look like $1s,\ 2s,\ 3s,\ \&\ 4s$ orbitals with appropriate number of nodes.
The solutions to the Schrodinger equation for hydrogen-like orbitals include wavefunctions corresponding to $p$ oribtals. Why did the authors not observe images that appear to look more like the $p$-orbitals?

These states aren't 1s, 2s, 3s, 4s, like you suggest. These are actually highly excited states of hydrogen. Note the negative energies in wavenumbers (cm$^{-1}$) in the rightmost column of the figure you provide. This is how much lower in energy the states are relative to the ionization limit, 13.6 eV, so we are talking about excitations ~13.57 eV in energy above the ground state. In fact, these energies are so high they refer to them as "quasibound" states.
The states the authors are looking at Stark states (the experiment is done in a field of 808 V/cm). So the analytical results they compare to are the solution of the hydrogen atom in parabolic, rather than spherical polar coordinates. This is why the quantum numbers given in the middle column are a bit funky. So the (0,29,0) does not mean $n=0$, $l=29$, $m_l=0$ (as I'm sure you can see why!) The quantum numbers define the state in parabolic coordinates, so the numbers are ($n_1$, $n_2$, $m$).
The reason they do this is that parabolic coordinates are still separable (so you can solve it analytically) when an external electric field is applied along some axis. Spherical coordinates, on the other hand, are not. So you don't see pure s, p, d orbitals (as nice as that would be!). My math gets a bit rusty here, so maybe some one else can add to this, but I believe the images you see on the left and the middle column are actually linear combinations of the spherical solutions we are used to as chemists. In fact, there is a line in the paper that in the experiment they prepare their states by exciting to a mixture of $s$ and $p$ states.
I believe you can connect the parabolic and spherical solutions in an electric field by a unitary transformation if the field is weak enough, since you can get the Stark effect in spherical coordinates by perturbation theory.
Hopefully that helps clear things up a little bit. I'm not an atomic physicist, so I'm often confused by some of the language and tools they use. There is a nice Wolfram Demo on the hydrogen atom in parabolic coordinates with and without an external electric field you may find interesting here:
http://demonstrations.wolfram.com/TheHydrogenAtomInParabolicCoordinates/

Thank for this answer. The article is behind a paywall for me and the specific details that you highlight are not in the publicly available summaries and news coverage.

@BenNorris Of course. I think this is a (not uncommon) case of over eager  scientific journalism that tends to be misleading. If you want a reprint of the paper, feel free to drop me a line and I'll gladly get you a copy.


text:
2nd Electron Affinity of halogens
I have seen questions regarding EA but they were all about the 1st EA so this question is not duplicate.
I came across a statement which goes like "2nd EA of halogens is almost zero". How is this possible? How can energy gained/released on adding an electron be almost 0, particularly when we are adding an electron to a noble gas, because, according to me, a lot of energy is required to do this.

Chart of Electron Affinities
The Electron Affinity can be thought of as the "electrical advantage" given by adding an electron to an atom.  So, if you have a halogen that gains an electron, it becomes more stable because now it has its octet, orbitals filled, etc.  This is why the 1st EA is large.
But if you want to add an electron, it needs to go to a whole new energy level, which will not be very desirable in terms of the electronic structure of the atom.  So the difference in electric stability or potential formed is nearly 0.  In other words, the halide ion has "0" want for another electron, since that would require adding another energy shell.
Another way to think about it is like this: the 2nd EA of a halide is basically the same as the 1st Ionization energy of the alkali metal to which it is isoelectronic (since adding an electron releases as much energy as removing that electron will take).  Since the 1st Ionization energy of a Group 1 metal is normally very small because the electron is so far away and shielded from the nucleus, so is the 1st electron affinity of a noble gas, and thus, the 2nd electron affinity of a halide.
Furthermore, now that the halogen has a negative charge, it experiences repulsion with the electron soon to be added. (Why does oxygen not like to be doubly anionized?)

The $\Delta E$ is nearly zero for addition of a free electron. It will be strongly positive if one is attempting to transfer an electron to $\ce{X^-}$ from some other stable species.


text:
The temperature rise after neutralisation

The answer to this question is B, however, I don't quite understand.
If considering Q= mc delta T, the heat released is to some extent being influenced by the no. of moles. If using that line of thought, wouldn't we want C where it has also the same no. of moles?
If we really have to arrive at B, my other proposition would be that the increase in heat is dissipated in a larger volume, hence giving a temperature rise of 6 degrees C.
I am a bit confused by this question. Any comments would be very helpful. Thank you in advance

Your other proposition is correct. We have twice the amount of reagents, hence twice the heat, but also twice the volume of water.

@IvanNeretin, but why is my other proposition incorrect?

Same amount of reagents means same amount of heat. But heat and temperature are not quite the same. Would that amount of heat warm any amount of water by $6^\circ$C? Would it, say, warm an ocean by $6^\circ$C?

One calorie is the amount of energy (heat) needed to raise 1 gram of water 1 degree Celsius at a pressure of one atmosphere. (It is almost constant to 3 significant figures, but there is some temperature dependence...) Also assume that 1 cm^3 of water is 1 gram which is more than good enough for the calculations needed for the problem.
So for the reaction in the problem given, 50 mL of water by 6 degrees is 300 calories.
For the other reactions I'll give calories and temperature rise for product solution.
(a) 600 calories, 12 degrees C
(b) 600 calories, 6 degrees C
(c) 300 calories, 3 degrees C
(d) 300 calories, 1.5 degrees C


text:
Ranking order of nucleophilicity

According to me the answer should be (c),(b),(a),(d). As we move left to right in a period the electronegativity increases and hence the tendency to donate electrons will decrease so nitrogen is better electron donor than oxygen. Now to select between three oxygen donating species we can use the steric hindrance concept and hence my answer.

But the correct answer is (b),(c),(a),(d).

Please tell me what is my mistake.

The first thing to note is that nucleophilicity is a kinetic parameter that cannot be addressed easily with thermodynamic parameters such as electronegativity. Instead, you need to inspect all potential nucleophiles by their kinetic merits.
One easy thing that we can use, even though it is thermodynamic, is the effective charge. The centre of negative charge of the acetate molecule is somewhere in the middle between the two oxygens and that of tosylate is somewhere in the middle between three equivalent oxygens. Also, in both systems the oxygens are bound to very highly oxidised atoms ($\ce{C^{III}}$ and $\ce{S^{IV}}$, respectively), which greatly reduces their electron density and thus their desire to attack nucleophilicly. Finally, in acetate’s case the nucleophilic attack would reduce the effectiveness of the carboxylate resonance as one of the two mesomeric structures becomes disfavoured (charge separation) which adds another grain of salt against nucleophilic attacks.
Tosylate is less nucleophilic than acetate because there are three oxygens than two but they only bear a single negative charge between them.
Now onto cyanide and methanolate. For cyanide, first note that it is both $\ce{C}$-nucleophilic and $\ce{N}$-nucleophilic. Since the HOMO, the highest occupied molecular orbital, has a much larger contribution on the carbon side, the $\ce{C}$-attack is strongly favoured over $\ce{N}$-attack and we can ignore the latter.
We can again use the bond-partners argument here. The oxygen of methanolate is bonded to a $\ce{CH3}$ group which, in general, is electron donating, thus increasing its electron density. The cyanide carbon, however, is bonded to the more electronegative nitrogen which slightly reduces its electron density. Thus, we can infer that methanolate should be a better nucleophile than cyanide. (Note, however, that both are very good nucleophiles, especially when compared to the other two.)

The conjugate base of a weak acid is going to be a strong base/a good nucleophile, so when looking at the two nucleophiles above $\ce{CH3O-}$ and $\ce{CN-}$  we should consider the pKa values for methanol and cyanide. We can see that the pKa of methanol ($\ce{CH3OH}$) is 16 and the pKa of cyanide ($\ce{CN}$) is 9.1 therefore the methanol is going to be a better nucleophile since it is the weaker acid and it will want to regain its proton after donation or stabilise its charge by attacking an electrophile. I would agree with you on bulky steric effects with (a) and (d) but you should also consider the fact that resonance stabilisation makes for a better acid so that would also decrease their nucleophilic properties with (d) having the most resonance structures therefore having the lowest nucleophilicity, due to strong acids having weak conjugate bases.
References:
http://www.masterorganicchemistry.com/2012/06/18/what-makes-a-good-nucleophile/
http://www.masterorganicchemistry.com/2010/04/19/the-conjugate-base-is-always-a-stronger-nucleophile/
http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf

I disagree that the conjugate base of a weak acid = good nucleophile. A sulfide is a stronger acid than an alcohol but a better nucleophile. Also, sulfonates apparently don't really partake in resonance like carboxylates do.

@DGS sulfides are stronger acids than alcohols, but the conjugate base of a weak acid = good nucleophile is trend that one can follow to help figure out these questions and as 99% of the time with trends there are exceptions to them which in this case RSH vs. ROH is.

I still think your trend is too general. It works for all bases of the same element, and maybe within row, but it ignores thiols, phosphines, selenides, etc.

Then why doesn't one confine the trend to the element in question in the case of the ones you mentioned only S with S etc... Then use other properties to determine the degree of nucleophilicity between different elements


text:
Creating Molecular Orbital and Walsh Diagrams
Does anyone know of a good website to create your own MO and Walsh diagrams? Could you potentially make them on a software package such as excel?
Any advice on this would be greatly appreciated.

Personally, I use Inkscape to draw MO diagram by hand.


text:
Nomenclature of saturated aliphatic hetrocyclic amines with an alkyl group on the nitrogen
For example, how would I name the following amine?

Since two of the alkyl groups on nitrogen are linked to each other (in a ring), I can't figure out what they should be called, as they don't fit into the usual alkyl (methyl/ethyl/etc.) mould.

According to Subsection P-22.2.1 of the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the preferred IUPAC name (PIN) of the functional parent compound is the retained name ‘pyrrolidine’, the nitrogen atom being numbered ‘1’.

Therefore, the name of the compound given in the question is ‘1-propylpyrrolidine’.

Without knowing what you have tried and where your specific trouble is, my guess is that you are having trouble identifying the parent chain.
This molecule is a heterocyclic amine, and the entire heterocycle is the parent chain. In this case, the parent chain is called pyrrolidine.

Now - how would you name a pyrrolidine with an alkyl substituent on the nitrogen atom?