text:
Does an increase in pressure necessarily lead to an increase in temperature?
Ideal gas law says that $pV = nRT$. So with compression, one way to add pressure which makes the volume smaller, can you actually increase the temperature?
It does not make sense to me that you can increase temperature, that is, add to the kinetic energy of the molecules, simply by increasing pressure, pressure meaning the force exerted on the gas from its surroundings. Can someone please explain this contradiction?

pressure does not mean the force exerted on the gas by the surroundings, it means the force exerted by the gas on the walls of the container divided by the area of the wall.

Ok fair enough feed back about wording

Karl, are you saying that, in the adiabatic compression of an ideal gas in a closed system (e.g., an insulated cylinder with a piston), there is no change in the gas internal energy or temperature?

Really.  Then please comment on the following for the case of an adiabatic reversible volume change:  $dU=nC_vdT=-PdV=-\frac{nRT}{V}dV$.  So, $d\ln T=-\frac{R}{C_v}d\ln V$.  So, when the volume decreases, the temperature increases.

I mixed up free expansion and reversible compression/expansion.

Sorry I'm a muggle. Not sure how to work this Q and As. But in answer to the question I thought that compressed gases got colder if anything. Could be wrong of course. Thought that you slow the movements of particles when you compress things and if they are moving slower they are colder. There may come a point when they can't be compressed any further and then they'd probably explode.

I think what you are asking is "why can the temperature of the gas increase when you compress it, even if the cylinder is adiabatic so that no heat can enter the gas?"  When you move the piston to compress the gas, you are doing work on the gas at the interface with the piston.  The piston is moving toward the gas, and the molecules of gas that collide with the piston leave with a greater average velocity than when they arrived.  So their average kinetic energy is increasing.  If expansion were occurring, such that the piston were moving away from the gas, the colliding molecules would leave with lower average kinetic energy.

Who said anything about compression rate?  In the end, it is only the total amount of work that determines the change in temperature, but that's just equal to the integral of the force per unit area at the piston face times the rate of change of volume (compression rate) dt.  You know, for an adiabatic process, $\Delta U = -\int{P_{ext}dV}$, where, for an ideal gas U = U(T).

I stand by what I said.

Are you talking about an ideal gas or air?  For air at least part of the reason is that the molecules attract each other and it is not ideal.  Would an ideal gas increase in temperature?  (Either way PV=nRT, does not say wither T changes.)  A good question not answered here or anywhere else I could find.

@Tuntable I am talking about both real gases and ideal gases.  Of course, an ideal gas would also increase in temperature.  PV-nRT is not the only characteristic of an ideal gas that matters.  The first law of thermodynamics also comes into play here, and, for an ideal gas, the internal energy is a function of temperature.  Did you read my comments to the OP's post?

It is not at all clear that an ideal gas would increase in temperature, at least not significantly.  Sure, increasing the pressure increases enthalpy, but enthalpy is T+PV.  If you are sure that it increase temperature of an ideal gas, then by how much?  Do you have a formula or a reference?

@Tuntable First of all, enthalpy is not T+PV.  Secondly, the ideal gas law tells us that, for an adiabatic reversible process of any gas, $dU=-pdV$.  For the case of an ideal gas, this reduces to $$nC_vdT=-\frac{nRT}{V}dV$$So, if dV is negative, dT is positive.  This is every thermodynamics textbook.  So stop wasting my time.

The ideal gas law, PV=nRT, only tells us how to calculate the pressure given the temperature and volume.  Compressing the gas certainly increases U, which includes both temperature and pressure, not sure why you hold pressure constant.  I have never seen this explained properly, just vagueness, leaps and then differential equations.

Who said the pressure is constant?  And, for an ideal gas, U depends only on temperature, not pressure.

It’s beginning to dawn on me that you are uncomfortable because you can’t see why U would not depend on p for the case of an ideal gas.  Is that your issue?

Sorry, got my U and H mixed up.  Not sure why enthalpy is not essentially T+PV, given a non-reacting ideal gas T is essentially U + constant.  When you say dU = -pdV then you are holding p constant, changing the volume, so obviously the temperature must change.  But it says nothing about what happens to the temperature if you just compress the gas increasing its pressure, adiabaticly.  I think what I am after is a good link if you know one, rather than an explanation.

Found it at last!!  Joule-Thompson effect.
https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect
As I suspected, for Helium at normal temperatures, there is no increase in temperature at all.  The rise when one compresses air has nothing to do with PV=nRT, but only to do with the non-ideal nature of air.  Thanks for your help though.

If you had a way to increase pressure with no volume change, then yes, temperature would increase by the ideal gas law.  In reality, most compression take place by reducing volume or increasing N, so the temperature effect is hard to see directly because other things are changing too.
The pressure in PV=nRT is the force exerted by the gas on the walls of the container.  As the temperature increases, the particles move faster, and therefore have greater speeds, so greater momentum and therefore greater force when they collide with the walls, so the pressure increases.

I understand what youre saying and I agree. Yes the ideal gas law says it happens in theory but does it actually happen in reality without changing the volume or the number of atoms?

How can you compress a gas without changing its volume?  Compressing means decreasing its volume.

I was commenting on the idea of changing pressure without volume, not compression.

At constant volume, you need to add heat to raise the temperature so that the pressure can increase.  The temperature rise is the cause and the pressure rise is the effect, rather than the other way around.

Ok, yes thats the idea I was getting at in the previous comment. Thanks!

@chestermiller  I didn't mean one could compress s system without changing volume, or that one could do an experiment in a lab today to raise temperature by changing pressure (thats why I said "if" and "in reality") but what I was getting at is that there is a direct correlation between pressure and temperature that would go both ways if the experiment could be done with everything else fixed.  In real experiment, yes, it usually is temperature that is raised, but thats not what the question was asking.

@JiffFulcher you can increase the pressure by putting more gas into the containers. The container stays the same size and has the same volume

The original question was whether just decreasing the volume would increase the temperature for an IDEAL gas.  Obviously the pressure increases, per PV = nRT.  But what about the temperature?  For air, the temperature increases, but air is not ideal.  Please do not provide non-answers.

We all know, Solids have defenite size and obviously have defenite volume. Liquid has defenite volume but no shape. Gases have neither shape nor volume. Gas will occupy the available volume of the container. Molecules make use of the available free space for their movement.
Thus in gases, you can externally change the degree of freedom of its molecules. When you increase the container volume, you are increasing the degree of freedom of the gas molecules. And conversely also it is true.
Coming to the question, when you decrease the degree of freedom of molecules (by decreasing the container volume), due to the limitation in their mobility, the excess residual energy has to be given out (All system trend to minimize its energy state). Naturally the gas becomes hot in a big to exchange the excess energy to the surrounding.
(Most of the natural energy exchange is done by heat energy).

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text:
How to Find the Mass of an Individual Element in a Compound/Molecule
What is the procedure/math for finding the mass of a specific element in a substance? For instance, how many grams of Oxygen are in 6.8 grams of Epinephrine (Adrenaline)?

Do you understand molar masses an chemical formulas? If yes, then what's the problem?

The problem is I don't understand the (in the example I posted) relationship between the molar mass of Oxygen, and the entire actual mass of the adrenaline sample. How do you find the actual mass of oxygen in grams in the sample?

take one molecule of epinephrine and find the mass percent of oxygen(mass oxygen/total mass). Then multiply result with 6.8.

Step 1: Find the Molar Mass of Adrenaline
To do this, you must need to know its chemical formula which is $\ce{C9H13NO3}$. Now to calculate its molar mass, we add up all of the molar masses of each atom: $$M =9\times12~\mathrm{g~mol^{-1}} + 1\times13~\mathrm{g~mol^{-1}} + 14~\mathrm{g~mol^{-1}} + 3\times16~\mathrm{g~mol^{-1}} = 183~\mathrm{g~mol^{-1}}$$
Step 2: Find the Mass of Oxygen in 1 mol of Adrenaline
In each molecule of adrenaline, there is 3 molecules of oxygen atoms. This means that in $1\ \mathrm{mol}$ of adrenaline, there are $3\ \mathrm{mol}$ of oxygen. Therefore the mass of oxygen in $1\ \mathrm{mol}$ of adrenaline is: $\mathrm{3~mol\times16~g~mol^{-1} = 48~g}$.
Step 3: Calculate Percentage Composition of Oxygen by Mass
We now know that in every mole of adrenaline, there are $48\ \mathrm{g}$ of oxygen. We also know that $1\ \mathrm{mol}$ of adrenaline weighs $183\ \mathrm{g}$. Therefore in $183\ \mathrm{g}$ of adrenaline, there are $48\ \mathrm{g}$ of oxygen. All have to do is simply divide by them by each other and multiply by $100$ to get the percentage composition of oxygen. Therefore the percentage composition is $26.23\,\%$.
Step 4: Calculate Mass of Oxygen in Sample
Now we know the percentage composition of oxygen which is $26.23\,\%$. This means that in every gram of adrenaline, there is $0.2623\ \mathrm{g}$ of oxygen. Therefore to find the mass of oxygen in the sample: $$m_\text{oxygen} = 0.2623 \times 6.8~\mathrm{g} = 1.78~\mathrm{g}$$

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text:
Creating Pi Molecular Orbitals from P Atomic Orbitals
As shown in the image, we have two free p orbitals in ethene but we need two p orbitals each for the bonding and anti-bonding orbital. That is a total of four. What is the concept behind this?

No the antibonding and bonding does not happen at the same time.

but we need two p orbitals each for the bonding and anti-bonding
  orbital. That is a total of four.

No, that is not a total of four: if you start from 2 atomic orbitals ($p_{\mathrm{1}}$ and $p_{\mathrm{2}}$), you should get the same number of molecular orbitals at the end ($\pi$ and $\pi^{*}$). The number of orbitals is perfectly conserved here.
In general, out of $n$ atomic orbitals (at most) the same number of (orthogonal) molecular orbitals could be built. In this particular case, two molecular orbitals that are built from $p_{\mathrm{1}}$ and $p_{\mathrm{2}}$ atomic orbitals are
$$
  \pi^\phantom{*}   = 1/\sqrt{2} p_{\mathrm{1}} + 1/\sqrt{2} p_{\mathrm{2}} \, , \\
  \pi^{*} = 1/\sqrt{2} p_{\mathrm{1}} - 1/\sqrt{2} p_{\mathrm{2}} \, . \\
$$
$\pi$ and $\pi^{*}$ will be orthogonal by construction if $p_{\mathrm{1}}$ and $p_{\mathrm{2}}$ were chosen to be orthonormal in the first place.

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Titration curve of strong acid with strong base
I have a question regarding the titration curve for $\ce{HCl}$ titrated with $\ce{NaOH}$. Why does it look like it does, with a steep slope near the equivalence point and a flat slope at the beginning and end?


text:
Regioselectivity in nitration of a 1,2-disubstituted aromatic compound
For 1,1,4,4-tetramethyl-1,2,3,4-tetrahydronaphthalene:

What would be the products following a standard nitration with $\ce{NO2+}$?
My reasoning is thus:

Since the substituent group is alkyl then it is ortho- or para- directing. But surely there are steric hindrance factors to be considered due to the bulk of the ring which would mean para- is favoured instead of ortho. However, when I try to draw out the structure it seems the arrows are meta- instead!

The molecule is symmetrical so the two positions on the far side of the alkyl groups are exactly the same. Similarly, the two positions closest to the alkyl groups are the same. For the far positions, each one is para with respect to one alkyl group and meta with respect to the other. This is not a contradiction because ortho, meta and para are purely relative terms.
I suspect that you are right that substitution will mostly occur at the far sites. The directing effect of the alkyl groups is essentially the same for every site because they are all ortho or para to one group and meta to another. Therefore, the steric effect of the bulky methyl groups adjacent to the near positions will disfavour substitution there.

Rewording my five-year-old comments into an answer:
The idea of "meta" or "para" is a bit different here, since you have not just one, but two alkyl groups. Being meta to one means ortho/para to the other, and vice versa. Since there are two substituents to start off with, you have to consider both their directing effects separately and add them up.

The carbons labelled a are ortho to one alkyl group (stabilises the cationic intermediate a lot), but meta to another (stabilised a bit).

On the other hand, the carbons labelled b are meta to one alkyl group (stabilised a bit), but para to another (stabilised a lot).

So, purely based on electronics, both positions are almost equally favoured. Now you turn to sterics to help you decide: as you said, b is favoured because it is less hindered.

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Titration of hydrochloric acid with sodium hydroxide
I have a question regarding titration of hydrochloric acid with sodium hydroxide. Assume you have a solution of NaOH with pH = 12. With this NaOH you titrate a HCl solution. Then, will the pH of the titration product, regardless of how much NaOH you add, ever reach a pH of 12?

Theoretically, no.

Practically then, would it ever reach 12?

In practice, if you add enough, it would be so close to 12 that no instrument could tell you that it wasn't 12.

Okey, thanks! I understand mathematically why it cant equal 12 but can this be explained in terms of chemistry?

you could think of it in the reverse way - you add a fixed amount of HCl to a changing amount of NaOH. no matter how much NaOH you have, some of it will be neutralised by HCl and therefore the pH will be below 12. to be honest, no matter how you look at it (mathematical pov, or what I wrote), it's all the same thing in the end.


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Mechanism for chloromethylation of benzene with formaldehyde and HCl

What is the mechanism of the above reaction? I have thought of one possibility:

Would this work? How exactly is the chlorine installed on the alkyl chain?

The actual mechanism turns out to be in different order when you use ZnCl2 instead of AlCl3. It is probably the same case here. See: https://en.wikipedia.org/wiki/Blanc_chloromethylation

This reaction is chloromethylation, similar to Blanc chloromethylation, but using $\ce{AlCl3}$ as co-catalyst instead of $\ce{ZnCl2}$. These are reactions belonging to a group related to Friedel-Crafts reactions, but characterized by usage of protonation instead of coordination with molecular Lewis acids
Mechanism of your reaction (not accounting for influence of co-catalyst) can be written as:

(source)
$\ce{AlCl3}$ coordinates to $\ce{HCl}$ creating adduct ($\ce{HAlCl4}$) which is much stronger acid.


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What is the first organic compound to be discovered ? Urea or alloxan?
It is widely accepted that urea is the first organic compound to be synthesized from inorganic chemicals. Wikipedia states that:

In 1828, the German chemist Friedrich Wöhler obtained urea artificially by treating silver cyanate with ammonium chloride.
$$\ce{AgNCO + NH4Cl → (H2N)2CO + AgCl}$$
This was the first time an organic compound was artificially synthesized from inorganic starting materials, without the involvement of living organisms. The results of this experiment implicitly discredited vitalism — the theory that the chemicals of living organisms are fundamentally different from those of inanimate matter.

But, alloxan (a not-so popular organic compound!) was isolated in 1818 by Brugnatelli (according to Wikipedia) 10 years before the discovery of urea!
So, is urea the first ever organic compound to be discovered or is it alloxan?
I also found a link which claimed that urea is not the first organic compound to be discovered.

The first organic compound to be discovered is probably not alloxane either, since ancient Egyptians used certain esters in their perfumes. The milestone of synthesizing organic compounds from inorganic ones is what that's important.

The elders had no (sound) idea about atoms and molecules. Before defining "Organic Chemistry", it makes no sense to say anyone discovered an organic compound. If you insist on an artificially produced and/or (somewhat) pure organic substance, i'd vote for ethanol, but really imo this is pointless.

The problem herein lies in the different possible definitions of organic chemistry.

The IUPAC defines an organic compound as a structure containing at least one carbon atom that can be named using the principles of organic nomenclature.

A less concise definition is

All compounds containing carbon and at least one other element that are not carbon monoxide, carbon dioxide or the latter’s inorganic salts.

Another one:

All compounds that contain at least one $\ce{C-H}$ bond.

Originally, the definition meant compounds isolated from organic materials i.e. organisms, living beings.

You see that the exact definition of organic compounds is difficult. But it was once very clear, historically: Anything not isolated from organisms or a derivative thereof was inorganic.
Note that isolation of organic compounds can be traced extremely far back in history: The ancient Egyptians isolated indigo as a colouring agent. However, they didn’t synthesise it and most attempts pre-19th century failed. Nobody questioned the possibility of isolating pure organic compounds from living matter, it was merely questioned whether they could be synthesised without involvement of organisms.
Alloxan was synthesised by oxidising uric acid with nitric acid as Wikipedia states. Thus, it was synthesised from an organic compound and would have been considered to contain enough vis vitalis even for further synthetic steps.
The compound your link states, 5-amino-3H-1,2,4-dithiazole-5-thione, is not described in any prime source as having been isolated from organisms. In fact, it is unknown to Wikipedia, so with all due skepticism, one can assume it to be of little to no relevance in living beings.
Compounds we today would call organic have been synthesised even earlier. Ethylene was first discovered in 1669 and it was successfully chlorinated to give 1,2-dichloroethane in 1794 — way before 1818, 1821 or 1828. Due to its name Dutch oil it is possible that it was considered an inorganic chemical. If it was synthesised from ethene that was, in turn, synthesised from ethanol, that would have just meant it carried over the vis vitalis from the original ethanol.
Wöhler’s discovery was that significant, because he used compounds that were (and are) unanimously declared inorganic (silver cyanate and ammonium chloride) and arrived directly at a compound that had previously been isolated from living material namely urea.


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Why aqueous ammonia is added to glucose solution to determine optical rotation?
In order to determine the specific rotation of glucose, it was mentioned in a practical chemistry book that 10g of dehydrated glucose must be dissolved in 80ml of water, then, adding 0,2ml of aqueous ammonia. After 30 minutes, we have to complete with 100ml of water.
For me, I would prepare only a glucose solution to use in the polarimeter.
So why aqueous ammonia have to be added to the glucose solution ? and why completing with 100ml of water after exactly a half-hour? (What happens during that half-hour?)
Note : This is not a homework. I'm a pharmacy student and this is a part of the lab work I'll have next week.

Have you already heard of the different isomers of glucose in solution?

For sure! As far as I know, there are two possible isomers of glucose (Either D-glu or L-glu) in solution. They are alpha-glucose and beta-glucose.

Approximised you can work with only those four, yes. I'm assuming you know what kind of glucose you will be working with (D or L)? And you have heard of mutarotation?

For now, I don't really know whether we'll work with D or with L. Yes, mutarotation occurs when alpha-glucopyranose turns into beta-glucopyranose and vice-versa. That's it?

More or less. You might get relatively pure beta-glucose to work with, or maybe not, so using ammonia and waiting for thirty minutes should equilibrate the solution so the optical activity no longer changes. Ammonia can act as a catalyst for the transformation.

Now I pretty understood it. Can you provide me the mechanism of action of ammonia ? I mean, how it does catalyze this transformation?

I think you can manage yourself; $\ce{NH3}$ is nothing more than a base in the catalytic mechanism. ;)


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How to eliminate electrostatic charges from titanium dioxide powder
I have synthesized $\ce{TiO2}$ nanoparticles via sol-gel method and I need to calcine it at elevated temperatures for the crystal phases formation. The powder is very highly charged and not only difficult to work with but I also lose significant amounts as it gets repelled by quartz glass and other tools and fly all over the places.
Is there a way that I can alleviate the repulsive forces between the powder and the tools that I use?

Related: http://chemistry.stackexchange.com/a/29459/15065

The best way I've found is to use an anti-static gun (sold by chemical companies such as Sigma-Aldrich).
If this isn't available, taking gloves off helps enormously, as the gloves can easily become 'static' in the same way as rubbing a balloon against some fabric. Gently tapping the flask against a metal object also helps, as the metal 'takes' the static away.
I have also known chemists to wear the same anti-static/grounding bracelets used by people who work with computers so as not to damage the electrics internally, however for a bench-chemist only occasionally working with fine powders, this would probably be over kill.

I don't want to take the gloves off because my sample is doped with fluoride but the anti-static gun is what I should look for.


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Can chemicals be identified by their molar mass?
Wouldn't it be possible to identify a chemical by measuring its molar mass? How much accuracy does the measured molar mass need to be to guess what chemical it could be?
EDIT:
@ortho makes an important point about isomers, but organic chemistry is much harder ;). Let's just focus on inorganic chemistry (though there are still isomers, usually there aren't).

To some extent it is, for example you can look up high-resolution mass spectrometry. http://www.chemguide.co.uk/analysis/masspec/mplus.html However if you only have the molar mass there's no way you can differentiate between isomers (e.g. n-butane and isobutane)

It’s not like there be no isomers in inorganic chemistry, see cis-diammindichloridoplatinum and trans-diammindichloridoplatinum.

@Jan i know, but usually there are no isomers, while organic chem is much different..

Sorry, wrong...

CO and N2 has same molar mass. There is a bunch of other examples as well.

@permeakra exactly the same?

@TanMath Not exactly, but close enough.

@permeakra You make an excellent point and I would like to encourage you to write up a short answer. || Also the molar mass is not what you can measure of a compound, it is more likely you measure the actual mass of a or many compounds. The molar mass also considers natural abundance. And $\ce{^12C-^16O}$ has a mass so close to $\ce{^14N-^14N}$ that you will need to consider isotope distribution to distinguish them.

It depends.
This was pretty much the way that the elements were discovered.
For inorganic chemicals in general the problem is really one of specificity. So I want to work with a chemical sample between 1 gram and 20 grams and I have an analysis good to 1%. That only yields about 300 statistically significantly different numbers $1.01^{300} = 20$. So for various inorganic salts this wouldn't work too well.
To take this into another field think of geology. Hardness and specific gravity are simple tests used to help identify minerals. Neither is very good by itself, but combined they sort out the minerals into a lot more bins.


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Memorizing polyatomic ions? Using Periodic Table
In my Chemistry course, we must memorize a list of common polyatomic ions. Is their an easy way of memorizing ions such as Sulfate $\ce{SO4^2-}$ by looking at just the periodic table. I listed the ones we have to memorize bellow. I know that if they contain oxygen (which is most of them) they usually end in "ate" or "ite." But how do I know how many Oxygen an ion will contain and its charge. My teacher said the only reference table we can use on our test is the Periodic Table of Elements.
I have to memorize the name and the formulas of the ions. Any methods will be much appreciated. I guess the real question is their any trends in the periodic table that helps predict the charge and number of oxygens and polyatomic ions will have? So if I have Sulfate for example, Can I predict the formula by looking at Sulfur on the periodic table
Update: My question is not a duplicate of When to use -ate and -ite for naming oxyanions? because my question wants to **use the periodic table to identify and memorize polyatomic ion **. Voldemorts question had nothing to do with memorization what so ever!

But, what I don't understand is how does the book know that NO−3 is Nitrate: how does the book know that NO−3 is "the most common oxyanion for the element". How does it know that a charge of −1 and 3 oxygen atoms create "the most common" Nitrogen oxyanion?

His question had nothing do with specifically looking for patterns in the periodic table to help him or her memorize the polyatomic ions. In fact he or she wasn't even asking for a technique or method to memorize polyatomic ions, but rather figure out which form of a polyatomic ion is "most common." For Example: He wanted to find out how the author of his book knows $\ce{NO3^-}$ is more common than  $\ce{NO2^-}$

Grouping it by charge is a horribly inefficient way of looking at things. You should look at the individual groups in the Periodic Table and observe the common oxidation states. Then you will notice that they tend to be 2 apart: for example, the halogens have -1 ($\ce{Cl-}$), +1 ($\ce{ClO-}$), +3 ($\ce{ClO2-}$), +5 ($\ce{ClO3-}$), +7 ($\ce{ClO4-}$). This is also true for other elements. Also, there are some exotic species there. I don't understand why you should be forced to memorise them...

@orthocresol I didn't group it by charge. The table above was the table that my professor gave me of poly-atomic ions we are expected to know. But, I will look into individual groups in the periodic table and look for patterns in the oxidation state.

Not that exotic, also hypophosphite is (H2PO2)- 2 hydrogens don't dissotiate.

@orthocresol Also another way of looking at the samething you said is poly-atomic ions composed of the same atoms in different proportions have the same charge. And if the atom other than oxygen is in an even group the charge is an even number, otherwise it is an odd number

Possible duplicate of When to use -ate and -ite for naming oxyanions?

@Mithoron His question doesn't ask for help using the periodic table to memorize a list of polyatomic ions. My question is on how I can use the periodic table to memorize a list of poyatmoic ions.

Nowhere did I say that you were the one who grouped by charge. I just said it's inefficient. Your teacher gave you a list of ions, including the nonexistent PO3(3-) and PO2(2-), grouped by charge to study.

Number of oxygens:
If your anion is in:

the 2nd period;
the VIIth main group/the 17th group/a halogen; or
is silicon:

then the -ate anion will have three oxygens.
All other -ate elemental anions will have four oxygens.
(Take note of ‘aluminate’ which I haven’t physically seen in that form yet but your teacher seems to insist be $\ce{AlO2-}$; disregarding this general trend.)
From the -ate anion, remove one oxygen to arrive at the -ite, remove two for hypo-ite. Add one for per-ate.
Charge:
Start at an anion with the most oxygens. (Note: Usually this is the -ate but because the halogens and permanganate are special they have the per-ate anion which is more oxygen-rich and important here.) Assume the maximum possible oxidation state for the non-oxygen atom. Assume $-\mathrm{II}$ oxidation state for every oxygen. Add up and take the negative value to arrive at the charge.
A hydrogen- will add $\ce{H}$ and lower the charge by one.
A thio- means one oxygen is replaced by sulphur.
A di- means take two of the (hydrogenated) anion and subtract water. Pyro means the same thing.
Memorise all the remaining! That is most importantly:

Chromite. Similar to aluminate I’  not sure if I ever saw it out in the real world.
Cyanide/cyanate. Thiocyanate is just like thiosulphate.
Oxalate, acetate and tartrate. Organic ions that will never fit into this scheme nicely.
Peroxide and superoxide. But those are -ides anyway.
Permanganate; which is almost like a perhalogenate. Luckily, molybdate and chromate fall into a category described above.

Periodic Table

This is about the charges of polyatomic ions, determined by the number of valence electrons (i.e. group number in the periodic table) brought by all the atoms in the ion:

If the polyatomic ion contains an odd number of atoms from elements with odd atomic number the charge of the polyatomic ion is odd.

Otherwise the charge is even. The most common charges for anions are -1 and -2, and for cations +1 and +2.
https://www.papazyan.org/oddoddodd_rule_for_polyatomic.html

Are there any exceptions?

@KarstenTheis Thanks for copying the rule over here. The only exceptions should be (unstable; reactive) radical ions with an odd number of electrons.

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text:
What are the rate laws for each elementary step?
Overall reaction
$$\ce{2NO + H2 -> N2O + H2O}$$
Mechanism
\begin{align}
\ce{NO + NO &-> NO2}\\
\ce{NO2 + H2 &-> NO + H2O}\\
\ce{N + NO &-> N2O}\\
\end{align}
I am aware that this is not a valid mechanism for the overall reaction however I am still supposed to find the rate law for each elementary step. I am having trouble determining what this rate law would be in step 3 of the mechanism.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. Please do not use markup in the title field, see here for details.

For elementary reactions, the rate law can be directly read from the chemical equation. In your last equation this is a bimolecular reaction, i.e.
\begin{align}
\ce{N + NO &-> N2O} & r &= k\cdot c(\ce{N}) \cdot c(\ce{NO})
\end{align}
To be more precise here, one molecule of $\ce{NO}$ reacts with on atom $\ce{N}$, and in a wider sense, that is two (bi) molecules reacting. The order of the reaction is two.

That being said, there are a couple of things you need to pay close attention to. An elementary reaction is always a reaction between the number of molecules given as reactants, with exactly one transition state leading to the products. If this number or reactants is greater than three it is highly unlikely to happen this way, since a concerted transition state is statistically very unlikely. An elementary reaction therefore also has to be balanced, so your first reaction needs to be:
$$\ce{2NO -> NO2 + N}$$
In this case we could see $[\ce{ON\bond{~}O\bond{~}N}]^\ddagger$ as a transition state.
I very highly doubt that your second reaction as an elementary reaction, from my point of view, there are too many bond that would need to be broken. Instead, the following reactions might be possible and/or necessary decompositions:
\begin{align}
\ce{NO2 + H2 &-> NO + OHH}\\
\ce{OHH &-> OH + H}\\
\ce{NO2 + H &-> NO + OH}\\
\ce{2OH &-> H2O + O}\\
\ce{NO + O &-> NO2}\\\hline
\ce{NO2 + H2 &-> NO + H2O}
\end{align}
And there are more possibilities.
In all but one the elementary reactions are bimolecular - guess which one would be unimolecular.

As for the mechanism itself, I think these are reactions that occur, just not all of them, maybe not even the majority of them.

Remember that when a reactant is used in more than one step you have to use the rule of
rate of production = rate of consumption
This leads to $k1[NO]^2 = k2[NO_2][H_2]$
See where you can go from there remember in the last reaction its $k3[N][NO]$
Hope that helps.

Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. For more information in general have a look at the [help]. It appears that you are commenting on the mechanism, which is only part of the question for context. The actual question is: What is the rate law in the third step?

I don't see why your saying that is wrong. He posted a reaction mechanism, and the rate law for the third step assuming it's the rate determining step is what I commented. He even posted that neglect that it is an invalid mechanism and you posted why it isn't a valid mechanism. I think you misread his question.

"I am having trouble determining what this rate law would be in step 3 of the mechanism." You did not answer how to determine the rate law for that step.

I showed him that if one of the species is used in more than one step then the rate law has to be made to account for that. I gave him everything he would need to determine it....

The question is not about the rate law of the mechanism. It is only about the rate law of the last step.


text:
What is the enol form of 4-methylcyclohex-2-en-1-one?

What is the enol form of this compound?

I have made an attempt, but it doesn't look quite right.

Which hydrogen should be taken off (left or right hand side of the carbonyl)? And I don't think the ring is aromatic so I don't know how it is stabilized.

Why should it be stabilised? I doubt this enol form is more stable than the α,β-unsaturated carbonyl.

There are two possible enol forms of this compound. 2) Being an enol is not intrinsicly more stable than being a ketone (in fact, ketones are usually more stable than enols) but with a strong base you can deprotonate the system. The enolate will then be more stable. 3) Which hydrogen you take off depends on your base.

As @Jan points out, there are two possible enols. One (the one you've drawn above) has a hydroxy group located on the diene double bonds; the other (produced by removing the proton on the methylated carbon) has both a hydroxy and a methyl group attached to the diene double bonds.  One enol has one more substituent, the methyl group, located on the diene. Methyl groups stabilize double bonds.  Therefore the enol with both the hydroxy and methyl group located on the diene will be the more stable enol and will predominate at equilibrium.

More substitutes alkenes are more stable.

Also which hydrogen should be taken off (left or right hand side of the carbonyl)? And i dont think the ring is aromatic so i dont know how it is stabilized...

When creating an enol the hydrogen alpha relative to the carbonyl is generally ionized. The alpha hydrogens are somewhat more acidic than the other hydrogens because of the possibility for resonance withdrawal of electron density by the carbonyl oxygen.
The ring is conjugated, but not fully conjugated, so the ring isn't aromatic. But partial conjugation is stabilizing.
Whether the enol or the keto form of this molecule is more stable requires further consideration; we cannot simply say that one is conjugated and the other is less conjugated and the one that is more conjugated must be more stable.

It does not have to be $\alpha$ in the presence of an enone system …

@Jan edited to reflect that fact

But Dissenter, isn't the OP really asking which of the 2 possible enols is more stable ("What is the enol form of this compound")?

There are two possible enol forms of this ketone. The first one is:

This is the one you drew, which shows the α hydrogen being abstracted. It is perfectly valid.
The other one is:

Note how the alkene next to the carbonyl group "extends" the range of the carbonyl group. This allows the γ hydrogen (i.e. the one three carbons away) to be removed, and then you can "push" electrons into the carbonyl group through the alkene group. In general, this is a property of α,β-unsaturated ketones (i.e. ketones which have a double bond between the α and β carbons).
Note that in this enol form, the methyl substituent is now in the same plane as the C=C next to it. This means that, if you were to reverse the tautomerisation to get back the keto form, there's an equal chance of the hydrogen coming in from above the plane and from below the plane, and racemisation would occur. In practice, there is a rapid equilibrium between keto and enol forms, and this means that if you were to dissolve this compound in any kind of acid (or base), the racemisation would occur pretty quickly.
Here, I have drawn the mechanisms for acid-catalysed tautomerisation. I assume you are now able to draw the mechanisms for the reaction in base (which in general forms an enolate instead of an enol).

At this stage, you might wonder why we cannot abstract the other α hydrogen. There's a couple of reasons. Firstly, it's attached to a $\mathrm{sp^2}$ carbon, and the C-H bond is therefore somewhat stronger. Secondly, even if you could take off that proton and form the "enol", you would have two double bonds next to each other. That's a huge problem because those double bonds want to be linear, but don't quite have the space to do so since there's a ring:

Regarding the stability of the two legitimate enol forms, I would not say that they are any more stable than the keto form. In general, the keto form is more stable. There are cases in which the enol form is relatively stabilised, most often when there is aromaticity or extra conjugation in the enol form (other factors such as solvent will also influence the position of equilibrium):

However, that is not the case here. The starting keto form is as conjugated as any enol you can form.

I was taught that it depends on the solvent whether acetoacetate adopts a diketo or enol form: Diketo in hydrogen-bonding solvents and enol in those solvents that cannot be hydrogen bond donors.

@Jan hmm, I know it does depend on solvent, but never learnt that little detail. I'll go look it up when I have time.

Enol form of aceto acetic ester is more stable than the diketo compound because of intramolecular hydrogen bonding between OH and carbonyl Oxygen. The enol form is only stable in non polar medium. In polar solvent, the solvent interrupts the intramolecular Hydrogen bonding. I don't think it's because of conjugation

Ok, I looked it up: https://en.wikipedia.org/wiki/Dimedone has a 67:33 ratio of keto:enol in $\ce{CDCl3}$. In this compound, there's no possibility of intramolecular H-bonding, so the extra stability must come from conjugation (normally, the ratio is $10^5:1$). Presumably, the intramolecular H-bonding in ethyl acetoacetate further stabilises the enol form in a solvent that cannot form hydrogen bonds.

In basic medium, one possibility is for hydrogen four in the ring (according to iupac name) to get abstracted by the base to form a carbanion at position 4, which is in resonance with the alkene and the carbonyl group, hence in one resonating structure the carbanion will be at carbon number 2 (which was originally sp2 hybridised, the one you talk about). In this position the carbanion could abstract an hydrogen to form a second keto form (4-methylcyclohex-3-en-1-one); although this form is less stable than the other keto/enol forms, it should surely exist in equilibrium, correct?

(this can also happen in acidic medium, I just find it easier to think in basic medium)

yes your answer is right. this is more stable as it is a conjugated diene. but, initially the H+ attacks on the lone pair of =O then the oxygen takes up the pie bond to neutralize the formal + charge on it.

I wouldn't say that an electrophile (i.e. H+) attacks anything.

@Dissenter If it is acid catalysed, it totally does. If not, then there is no electrophile.

Also it is not necessarily more stable. The carbonyl is also conjugated so just saying that it is a conjugated diene is not enough.

@Dissenter yes, it will. otherwise its illogical to show that the pi- bond to shift directly to an H+.  if an electrophile hadnt attacked anything, how would any reaction proceed?

@bon yes, the initial form is also stable, but to justify that enolisation takes place you have to say that the diene is more stable.

@ritabrata I mean the lone pair attacks the proton. Which way do you draw your arrows?

@dissenter lone pairs are not capable of any movement. actually when a collision takes place , the H+ may come near the l.p. and form a bond, which on a broader view looks as if H+ goes and attacks. but still it would be wrong to say l.p. itself attacks.

Regardless of whether that's the case in reality it's convention for nucleophiles to attack electrophiles

@Dissenter lone pair are the ones which move up the to the electrophile, so the arrow is put that way. mobile nucleophiles can also attack immobile electrophiles. anyways its actually a human imagination and not the actual phenomenon, so you put in any way.

@Ritabrata I might be wrong but I believe that electrons do move in real life; electrons are the more mobile sub-atomic particles (rather than protons).

Now, @Dissenter would you say that electron jump out of their mother atom's shell and form a bond? electrons do move otherwise there wouldn't have been any photoelectric effect. but here it is completely out of context.

@Ritabrata You seem to be contradicting yourself when you say lone pairs are not capable of any movement and lone pairs are the ones which move up the to the electrophile. Your answer and comments are generally confusing and unclear. Please clarify your argument.

@bon if properly read what i said and work out a pencil and paper, you will understand what i meant to say. H+ approach but lone pair move up to form the bond.

Guys please do note again that wrong answer (if this is one) isn't equal to VLQ.


text:
Difference between calcination and pyrolysis
What is the difference between calcination and pyrolysis?
I understand they both involve heat treatment but each under a different type of atmosphere, which leads to some sort of reduction reaction? What are the differences in underlying mechanisms?

Have you read https://en.wikipedia.org/wiki/Calcination,  https://en.wikipedia.org/wiki/Pyrolysis, or other info on internet?

Yes but i was hoping to find out more about the chemical mechanisms at play

Essentially calcination is decomposition at high temperature with inorganic reactants such as calcium carbonate in making cement.
$$\ce{CaCO3 ->[\Delta ] CaO + CO2}$$
Pyrolysis is decomposition or oxidation at high temperatures of organic materials such as cellulose in making charcoal.
$$\ce{(C6H12O6)_{n} ->[\Delta] C7H4O + C_{x}H_{y} + H2O}$$
where $\ce{ C_{x}H_{y}}$ is volatilized byproducts.


text:
Six independent papers reporting epitaxial formation of silicene in 2012
According to a Nature news article cited below (info-box "The rise of silicone"), six independent papers were published reporting successful epitaxial formation of silicene.
Does anyone know which those were? I am just able to find the following two papers:

Fleurence, A., R. Friedlein, T. Ozaki, H. Kawai, Y. Wang, Y.
Yamada-Takamura (2012). "Experimental evidence for epitaxial silicene
on diboride thin films". Physical Review Letters 108(24)
Vogt, P., P. De Padova, C. Quaresima, J. Avila, E. Frantzeskakis, M.C. Asensio, A. Resta, B. Ealet, G. Le Lay (2012). "Silicene: compelling experimental evidence for graphenelike two-dimensional silicon". Physical Review Letters 108(15)
Nature (2015). "Graphene’s cousin silicene makes transistor debut". Nature news article. URL: http://www.nature.com/news/graphene-s-cousin-silicene-makes-transistor-debut-1.16839

Does anyone know where to find the other four?

Could you please expand your link into actual citations, I (and maybe a few more people) prefer not going through pubmed. What puzzles me more is, that nature does not cite these six papers. Could you also quote the section where they mention it and what references they use instead.

Since this is a Nature News article, usually it is a way of promoting a scientific article in one of the current Nature publications - that's why it can get away with stating "Six independent reports" without six citations attached.  In this case, it is citing this 2015 article:

Tao, L. et al. Nature Nanotechnol.
  http://dx.doi.org/10.1038/NNANO.2014.325 (2015)

There are many citations in that article from the 2010-2013 time period reporting silicene sheets and nanoribbons.  Keep in mind, the Fleurence et al. paper you mentioned might not be one of the six mentioned, because it uses a diboride substrate instead of silver.

Ahh, nice catch, I didn't notice that a debarred substrate was used. I will look through the references and see what I can find.

Hi and welcome to chemistry.stackexchange.com. Have an upvote ;)


text:
What accounts for the high spin state of the complex Tris(acetylacetonato)iron(III)?
I understand that there's 5 d-electrons for $\ce{Fe^3+}$ ion, but why it doesn't fill up the lower energy orbitals first to form one unpaired electron, but rather, filling up all the orbitals to form 5 unpaired electrons?

I am unsure how the ligands in this complex affects the crystal field splitting. Do we consider [acac] to be a weak field ligand?
Could it be the geometry of this complex that give rise to high spin state?

High spin complexes are rather common with $\ce{Fe^3+}$. It would need a high-field ligand to fall into a low-spin state. Compare other typical $\ce{Fe^{III}}$ high-spin complexes such as $\ce{[FeF6]^3-}$

Actually, it seems like we do not yet have a good question/answer complex covering high and low spins. I think this question could be expanded to give such a pairing.

Without going into detail about the spectrochemical series and its origin, the short answer is that the energy required to pair electrons in the $t_{2g}$ orbital is larger than $\Delta_{\text{oct}}$ (i.e. the energy gap between the $t_{2g}$ and $e_g$ orbitals). Therefore it becomes more favourable (or one could in fact say less unfavourable) to promote the electron, instead of forcing it to share an orbital with another electron.

Whether a complex is high- or low-spin therefore depends on two factors: the size of the pairing energy $P$ and the size of $\Delta_\text{oct}$. The former doesn't change that much within the 3d block, which makes the latter the more important factor to consider. I'll stop here for now but if you want to know more, ping me in a comment (@my username).

Note that your picture of the acac ligands is not perfect; a resonance structure in which we have $\ce{O=C(Me)-CH=C(Me)-O-}$ would be better.

@orthocresol How do we know if the energy required for Δoct is lower than  t2g? I'm so sorry, my knowledge of crystal field theory only limits to understanding how different ligands affects the splitting of energy field.

@Jan Thank you for pointing out. I suppose a resonance structure would be better too!

Crystal field theory is very limited in that it treats ligands as point charges - so theoretically... small, anionic ligands like F- would be a stronger than neutral ligands like CO. But this is not true. The rationalisation involves molecular orbital theory - I think I will just attach this for you to read https://drive.google.com/file/d/0B0GATVY1ly6ccHpZN0lUR1k1b2M/view?usp=sharing @Jan has kindly promised to explain more to you so I will leave it to him :)

I will go into the theoretical details first and come to the practical answer for your exact case later; separated from the theory by a horizontal rule.

Approaching the complex theoretically, our first observation is that we are going to have six atoms coordinating our $\ce{Fe^3+}$ centre. Six ligands means that we should consider an octahedric environment as initial approximation. For the next bit of the discussion, I am going to ignore the fact that two ligands each belong to the same molecule and treat them all as $\ce{O^{\frac{1}{2}-}}$ or maybe as $\ce{OH-}$ since the latter is close enough.
The crystal field theory, our first approximation, considers these ligands to be point-shaped negative charges that approach an $\ce{Fe^3+}$ ion. The lonely iron(III) in vacuum will have five d-orbitals of degenerate energy; the approaching negative charges will destabilise all those orbitals that point towards them — for simplicity, let us assume that the corners of the octahedron are the axes of a coordinate system, and thus the destabilised orbitals are $\ce{d}_{x^2 - y^2}$ and $\ce{d}_{z^2}$. (The other orbitals point between the ligands and are therefore unaffected.) This will give rise to the most basic of all transition metal complex energy diagrams as shown in figure 1.

Figure 1: splitting of d-orbitals in an octahedral complex according to crystal field theory. The energy difference always corresponds to $10~\mathrm{Dq}$
Crystal field theory, unfortunately, gives us no information about the energy difference between the two sets of d-orbitals — this is important, since we need a significant energy difference if we want to keep our electrons paired at a lower level. Figure 1 chose to keep a high-spin complex simply for spin retaining. Therefore, the again most simplistic approach to analyse different octahedral complexes by their actual energy differences (remember that $10~\mathrm{Dq}$ or $\Delta$ is only a variable) is to assume a certain contribution both from the metal’s side and from the ligands’ side:
$$10~\mathrm{Dq} = g_\mathrm{M} \times f_\mathrm{L}$$
The $g_\mathrm{M}$ contribution depends on the metal and needs to be experimentally measured, similarly for $f_\mathrm{L}$ on the ligands’ side. $f_\mathrm{L}$ is normalised so that $f_\mathrm{L}(\ce{H2O}) = 1$. $g_\mathrm{M}(\ce{Fe^{III}}) \approx 1.40 \times 10^4 ~\mathrm{cm^{-1}}$. By knowing that $f_\mathrm{L}(\ce{OH-}) \approx 0.9$ we could calculate the energy difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ to be $1.26 \times 10^4~\mathrm{cm^{-1}}$ and compare that to other iron complexes to find out that this would mean high-spin. (Using a value for $\ce{acac}$ should change the result slightly but not much.) Note that a typical low-spin iron(III) complex such as $\ce{[Fe(CN)6]^3-}$ has the following value: $10~\mathrm{Dq} = g_\mathrm{M}(\ce{Fe^{III}}) f_\mathrm{L}(\ce{CN-}) = 1.40 \times 10^4~\mathrm{cm^{-1}} \times 1.7 = 2.38 \times 10^4~\mathrm{cm^{-1}}$ — almost double the difference.
However, if you put more research into this phenomenon, it quickly becomes highly dissatisfying. The approximation for $10~\mathrm{Dq}$ up there is no real good, because the same ligand has different $f_\mathrm{L}$ values for different complexes — and even worse, the same metal-ligand combination can sometimes be found in both high- and low-spin states resulting in wildly different values for $10~\mathrm{Dq}$: For high-spin $\ce{[Fe(ptz)6]^2+}$, $10~\mathrm{Dq} = 1.23 \times 10^4~\mathrm{cm^{-1}}$ while for low-spin $\ce{[Fe(ptz)6]^2+}$, $10~\mathrm{Dq} = 2.00 \times 10^4~\mathrm{cm^{-1}}$. And whenever simple approximations don’t adequately explain the true picture, we need to resort to molecular orbital theory.
The simplest approximation in molecular orbital theory gives a similar picture as crystal field theory does: We have a set of six s-type orbitals approaching our metal ion, and these six orbitals will form σ-bonds with the metal. The main difference is that we can better explain why the metal orbitals will split: The ligand orbitals transform as $\mathrm{a_{1g} + t_{1u} + e_g}$ which means that they will interact with the metal $\mathrm{e_g}$ orbitals. Since the ligands are typically lower in energy, we realise that what we are in fact observing are antibonding $\mathrm{e_g^*}$ orbitals on the metal. Thus, electrons in these orbitals should weaken the metal-ligand bond (order); or complexes lacking electrons in $\mathrm{e_g^*}$ should have a shorter metal-ligand distance. However, this approach is also unsatisfactory: negatively charged ligands such as $\ce{OH-}$ or $\ce{acac-}$ should have donor orbitals of higher energy, meaning more similar to those of the metal, meaning a greater splitting – which we do not observe.
Therefore, in a second approximation we need to add the ligands’ p-orbitals to the picture. These transform as $\mathrm{t_{1u} + t_{2u} + t_{1g} + t_{2g}}$ — we have an additional set of orbitals that we can linear combine with the metal’s $\mathrm{t_{2g}}$ orbitals. We now have additional ligand to metal π-bonding and can assume the metal-centred $\mathrm{t_{2g}}$ orbitals again to be antibonding ($\mathrm{t_{2g}^*}$). This again raises their energy levels and thus lowers the difference between $\mathrm{t_{2g}}$ and $\mathrm{e_g}$, therefore lowering $10~\mathrm{Dq}$. We finally have an adequate approximation of $10~\mathrm{Dq}$.
Now to get back to high-spin versus low-spin. Remember that it is not favourable per se to pair electrons — when applying the aufbau principle, you must never forget Hund’s rule which tells you to fill orbitals of the same energy with parallel spins first before adding the first antiparallel spin/pairing the first electron. In certain cases such as chromium, it is even favourable to have $\mathrm{3d^5 4s^1}$ — all valence spins unpaired. This is no different when going from elements to compounds: A certain energy gain is always required to make electron pairing favourable. Oftentimes, this energy gain is simply not supplied as is the case for most $\mathrm{3d}$ transition metal complexes that are not $\ce{d^8}$. The high-spin iron(III) complexes are the rule, not the exception, for the $\mathrm{3d}$ metals.
To raise the energy difference, we need strong-field ligands. The above paragraphs tell us what they need to: σ-bonding must be strong to raise the energy of $\mathrm{e_g^*}$ while π-bonding (if present at all) must be weak to not raise the $\mathrm{t_{2g}}$ level. Cyanide is good at doing this because its σ-donor orbital has roughly the same energy as the metal, so the bonding will be largely covalent with a large difference.
If, by the way, we accidentally manage to turn a high-spin into a low-spin centre, as was highlighted above with the ptz-ligand, a number of follow-up changes will also occur: Since the $\mathrm{e_g^*}$ orbitals are no longer occupied, the ligand to metal bond order increases and the ligands come closer to the metal. Therefore again the energy difference will rise because the covalent bond will be strengthend. This is the reason why the energy difference suddenly jumps from $\approx 1.3 \times 10^4~\mathrm{cm^{-1}}$ to $\approx 2.0 \times 10^4~\mathrm{cm^{-1}}$ when transitioning between high- and low-spin states.

In the case of the $\ce{acac-}$ ligand, for starters we need to realise that oxygen (the donating atom) is rather electronegative and thus we can assume its orbitals to be rather low-lying energetically. Low σ-donor orbitals means less destabilising of $\mathrm{e_g^*}$ and therefore a lower field split.
Secondly, we always have p-type orbitals on oxygen that do not participate in the π-system so that they can π-bond to the metal, too — or it could even be the π-system itself bonding to the metal. In any case, $\mathrm{t_{2g}}$ will be split and the metal component, $\mathrm{t_{2g}^*}$ will be raised in energy. This further reduces the field splitting, altogether putting $\ce{acac-}$ into the weak field department.
Finally, even though we have iron(III) and even though higher oxidation states are generally more prone to give low-spin complexes, iron(III) is not one of them. It takes significantly stronger ligands than water or even ammonia to generate low-spin iron(III) complexes, so generally any anionic oxygen-donor will not be enough to overcome the pairing energy.

Abbreviations:
acac: acetoacetone (or -ate if charged)
L: ligand
M: metal
ptz: 1-propyl-1H-tetrazol-κN4.

Acknowledgement:
Most of the knowledge is taken from Prof. Klüfers’ lectures on coordination chemistry and refreshed by checking his course’s internet scriptum.

I think because the $\ce{Fe(acac)3}$ is in the high spin state. Why in the high spin state? Because of the weak ligand (acetylacetonate). For the high spin state, it is easy to transfer the electron from one state to other.
Briefly for weak ligand, low splitting of the orbital occurred. For low splitting high spin state occurs, in which the electrons can fill all the states sequentially starting from the ground states to excited states individually.


text:
Trying to understand the statement: 'Resonance is not a flickering between the contributing states. '
So, the famous resonance definition: it is the weighted average of different Lewis structures....
Well, this is purely wrong.
Resonance is nothing but quantum 'superposition of the wavefunctions representing different electron distributions in the same nuclear framework.'-Peter Atkins.
For-instance, to describe the wavefunction of $\ce{HCl}$, quantum superposition is applied viz.  $$\psi_{\ce{HCl}}= \psi_{\ce{H^+ Cl^-}}C_1 + \psi_{\ce{H-Cl}}C_2$$ where $C_1$ is the probability-amplitude to find the electron-pair at $|\psi_{\ce{H^+ Cl^-}}\rangle$ provided it was prepared in $|\psi_{\ce{HCl}}\rangle$ that is, $C_1 =\langle \psi_{\ce{H^+ Cl^-}}|\psi_{\ce{HCl}}\rangle;$ also $C_2 = \langle \psi_{\ce{H-Cl}}|\psi_{\ce{HCl}}\rangle$ and $C_1^2 + C_2^2= 1.$
As written by Peter Atkins in his book Elements of Physical Chemistry:

A better description of the wavefunction for the molecule is a superposition of the covalent & ionic descriptions, & we write (with a slightly simplified notation) $$\psi= \psi_{\ce{H_Cl}} + \lambda\psi_{\ce{H^+ Cl^-}}$$ with $\lambda$ some numerical coefficient. In general, we write $$\psi = \psi_\text{covalent} + \lambda \psi_\text{ionic}$$ [...]According to the general rules of quantum mechanics, in which probabilities are related to squares of wavefunctions, we interpret the square of $\lambda$  as the relative proportion of the ionic contribution. If $\lambda^2$ is very small, the covalent description is dominant. If $\lambda^2$ is very large, the ionic description is dominant.
[...] The interpretation of the wavefunction, which is called resonance hybrid, is that if we were to inspect the molecule, then the probability that it would be found with an ionic structure is proportional to $\lambda^2$. For instance, we might find that $\lambda = 0.1,$ so the best description of the bond in the molecule in terms of a wavefunction is a resonance structure described by the wavefunction $\psi= \psi_\text{covalent} + 0.1\psi_\text{ionic}.$ This wavefunction implies that the probabilities of finding the molecule in its covalent & ionic forms are in the ratio $100:1.$

This is what is known as wavefunction-collapse in Copenhagen interpretation which means $\psi$ collapses to one of its base state after the measurement; here the wavefunction is more prone to collapse to the covalent description. Hence, in this case, the state of the molecule has mostly covalent properties.

I've pretty much conceived the upper part; but speaking at the present context of the title of the question, let me start from Peter Atkins' description of benzene in his book Elements of Physical Chemistry:

One of the most famous examples of resonance is in the VB description of benzene, where the wavefunction is written as a superposition of the wavefunctions of the two covalent Kekulé structures: $$\psi = \psi_{\text{Kek}1} + \psi_{\text{Kek}2}$$ The two contributing structures have identical energies, so they contribute equally to the superposition. The effect of resonance in this case is to distribute the double-bond character around the ring and to make the lengths and strengths of all the carbon-carbon bonds identical. The wavefunction is improved by allowing resonance because it allows for a more better description of the location of the electrons, and in particular the description can adjust into a state of lower energy. ...

The statement that baffled me comes from Atkins in the last paragraph of resonance as:

Resonance is not a flickering between the contributing states: it is a blending of their characteristics, much as a mule is a blend of a horse and a donkey. It is only a mathematical device for achieving a closer approximation to the true wavefunction of the molecule than that represented by any single contributing structure alone.

The cause of my bafflement is due to the following excerpt of Feynman's lecture The benzene molecule:

There is a mystery about this benzene molecule. We can calculate what energy should be required to form this chemical compound, because the chemists have measured the energies of various compounds which involve pieces of the ring—for instance, they know the energy of a double bond by studying ethylene, and so on. We can, therefore, calculate the total energy we should expect for the benzene molecule. The actual energy of the benzene ring, however, is much lower than we get by such a calculation; it is more tightly bound than we would expect from what is called an “unsaturated double bond system.” Usually a double bond system which is not in such a ring is easily attacked chemically because it has a relatively high energy—the double bonds can be easily broken by the addition of other hydrogens. But in benzene the ring is quite permanent and hard to break up. In other words, benzene has a much lower energy than you would calculate from the bond picture.
Now we want to resolve these mysteries—and perhaps you have already guessed how: by noticing, of course, that the “ground state” of the benzene ring is really a two-state system. We could imagine that the bonds in benzene could be in either of the two arrangements.[...] You say, “But they are really the same; they should have the same energy.” Indeed, they should. And for that reason they must be analyzed as a two-state system shown in Fig.

Each state represents a different configuration of the whole set of electrons, and there is some amplitude $A$ that the whole bunch can switch from one arrangement to the other—there is a chance that the electrons can flip from one dance to the other.

In the blocked-letters Feynman told that if at one time the molecule be found in $|1\rangle$, at some time later, it may be found at $|2\rangle$ (& vice-versa) as the Hamiltonian elements $H_{12},H_{21}$ are not zero but $-A$, the probability of transition per unit time.

As we have seen, this chance of flipping makes a mixed state whose energy is lower than you would calculate by looking separately at either of the two pictures in Fig. Instead, there are two stationary states—one with an energy above and one with an energy below the expected value. So actually, the true normal state (lowest energy) of benzene is neither of the possibilities shown in Fig., but it has the amplitude $1/\sqrt2$ to be in each of the states shown. It is the only state that is involved in the chemistry of benzene at normal temperatures.

Why should one found the benzene at $|1\rangle$ at one time but at other time, at $|2\rangle?$ Because the transition-amplitudes are non-zero and 'the whole bunch can switch from one arrangement to the other—there is a chance that the electrons can flip from one dance to the other.'The electron can flip from one state to the other. But doesn't this statement is at dagger's drawn with the common statement 'Resonance is not a flickering between the contributing states.'? I am completely sure both Feynman and Atkins is telling right, I am misinterpreting both the statements but don't know how. Can anyone please, please explain me how does flipping of the electrons from one dance to the other doesn't contradict the statement Resonance is not a flickering between the contributing states?? Please help.

Discussion in chat.

Related: http://chemistry.stackexchange.com/questions/51632/what-is-resonance-are-resonance-structures-real

When it comes to resonance, you should not take the notion of superposition of different states literally. Feynman mentioned it explicitly right from the get go, just read carefully what he says:

We could imagine that [...]

See, we could imagine. The superposition of states in a description of what is called resonance in chemistry is not the superposition of real states of a molecular system, it is a superposition of imaginary states. The only real state is to the left of the equality sign, everything to the right is imaginary. Thus, your analysis of this "superpositions" in accordance with the Copenhagen interpretation is plain wrong.
Atkins seems to be going a bit too far taking resonance almost literally along the road, but at the end rightfully warns us that it is "not a flickering between the contributing states".

It is only a mathematical device [...]

Concise! And here I was yammering about in http://chemistry.stackexchange.com/a/48334/28144

Why not go back to the experimental evidence that the C-C bond lengths are all the same length so that a picture of alternating single and double bonds is not correct and so this makes any combination of resonance structures as just a way of reconciling a valence bond picture with experiment. That is, we are used to single and double bonds as its a extremely convenient (and good) way of describing lots of chemistry and are reluctant to let go. In a molecular orbital approach such a resonance does not need to be invoked. Its just a change of point of view, changing a basis set if you like.


text:
Find the order of acidic strength
Arrange the following in decreasing order of acidity

I think it's S>P>R>Q
(Q) forms an anti aromatic conjugate base. So its the least acidic compound here.
(R) forms forms a conjugate base which is only stabilized by electron withdrawing (inductive) effect of $\ce{S}$.
(P) forms an aromatic conjugate base and (S) forms am ion stabilised by resonance (two resonance structures). But negative charge in $\ce{O}$ is more stable than negative charge on $\ce{C}$. So (S)>(P).
Is this correct?

Looks right to me.

I would say P>S>R>Q as I think aromaticity of P  would take preference over non-aromatic anion of S.

Let's look at some data. pKa of cyclopentadiene is about 16, while that of acetylacetone is about 9. Now who's stronger?

The question is confusing. Normally a person would go for the P,S,R,Q because of resonance stabilization in P.

Normally a person would know that resonance stabilization is not an absolute weapon. It can be measured and thus compared to other things; some are weaker than it, and some are stronger.

@IvanNeretin would the vacant 3d-orbitals of Sulphur not delocalize the negative charge? I understand that the extent of delocalization would be less and it won't change the answer yet no mention of it is making me confused.

3d-orbitals of S are now commonly thought to be less important than we believed before. (Maybe unimportant at all.)

@IvanNeretin Thanks! I had read that 3d orbitals of $CCl_3^-$ stabilize it quite a bit so thought the same would apply here.

(Q) is the least acidic compound because the conjugate base formed is anti aromatic and hence unstable.
(R) is more acidic than (Q) because the conjugate base is stabilized by the electron withdrawing inductive effect of two sulphur atoms.
Conjugate bases of both (P) and (S) are stable due to resonance. Among the two, conjugate base of (S) is more stable because a negative charge on Oxygen atom is more stable than a negative charge on carbon atom. Here are the resonance structures:

This is verified by @Ivan Neretin's comment on the OP.

Let's look at some data. pKa of cyclopentadiene is about 16, while that of acetylacetone is about 9.


text:
pKa of Tris corrected for ionic strength
I have a question regarding the molar fraction calculation of Tris buffer at different temperatures.
Tris dissociates according to the following equation:
$$\ce{Tris + H2O -> Tris+ + OH-}$$
So, in order to calculate the amount of $c(\text{Tris}^+)$ and thus the amount of strong acid needed for titration, one needs to use derivation of Henderson-Hasselbalch eq.:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a}' - \log_{10}\left(\frac{[\text{Tris}^{+}]}{[\text{Tris}]}\right)$$
However, $\mathrm{p}K_\mathrm{a}'$ is $\mathrm{p}K_\mathrm{a}$ corrected for ionic strength ($I_c$) at a certain temperature.
The correction of $\mathrm{p}K_\mathrm{a}$ is done via the Debye-Hückel eqation:
$$\mathrm{p}K_\mathrm{a}' = \mathrm{p}K_\mathrm{a} -0.509\times\left| z-z^{+}\right| \sqrt{I_c}$$
Now, for Ic itself one needs to calculate the molar fraction of $[\ce{Tris+}]$:
$$I_c = 0.5\times \left([\ce{Tris+}] + [\ce{OH-}]\right)$$
which requires the use Henderson-Hasselbalch equation making this problem somewhat circular. or am I missing something?

If you write also mass balance and charge balance can't get [tris+]?

First decide the Ic you want to work with. Second, you should correct the pka with the Debye Huckel equation and finally you use the pka corrected with the Henderson Hasselbach equation.


text:
Why is the radius of hydrated Na+ greater than that of hydrated K+?
Although $\ce{Na+}$ has ionic radius of 95 pm and an atomic weight of 23, and $\ce{K+}$ has ionic radius of 133 pm and an atomic weight of 39, hydrated $\ce{Na+}$ is larger than hydrated $\ce{K+}$: This is evident from observing that $\ce{Na+}$ moves through a cell membrane with greater difficulty than $\ce{K+}.$
Why is hydrated $\ce{Na+}$ larger in size than hydrated $\ce{K+}?$

This post needs to be extensively edited - as it stands, it's incomprehensible (for instance, what does ppm have to do with ionic radius?). And what's your question? What are your thoughts?

@ToddMinehardt Iwould guess that pm = Picometers so .95 Ångstrom and 1.33Å for Na and K. IIRC the actual reason is that the backbone carbonyls of the strands lining the pore can't replace the solvating waters of Na as they are too far away.

@gilleain do you have a link? I would be interested to read about that

@orthocresol I don't have any specific references to hand, but 10.1021/ar5002878 looks relevant. Seems like I was remembering one possible model for the selectivity filter, and other filters might use side chains.

@gilleain. please send the link about it.

For example, this paper : http://www.ncbi.nlm.nih.gov/pmc/articles/PMC3266868/ says things like "Therefore, unlike K+-channels, the NavAb selectivity filter appears to select and conduct Na+ ions in a mostly hydrated form." So I would have to read more to really answer this...

Ion transport across cell membranes is quite complicated. If the OP is only interested in solvation of K+ and Na+ they should rephrase the question. In terms of ion channels, both K+ and Na+ have charge +1 but the K+ ion is larger so the water molecules are less tightly bound. This can partly explain why Na+ is transported in hydrated form and K+ is not.


text:
interpenetrating polymer network (IPN)
I just learend about an interpenetrating polymer network (IPN) and that this is a way of mixing polymers. But I thought it was thermodynamically impossible to mix 2 polymers because in ∆G= ∆H -T.∆S the enthropy will never be big enough to overcome ∆H. So why is it possible with an interpenetrating polymer network (IPN)??
thanks in advance

It is thermodynamically possible to mix gases, sometimes liquids, and sometimes even solids. What is so special about polymers? Why would it be impossible to mix them?

because ∆H= (δ stof 1 - δ stof 2)^2  (with δ hildebrand solubility parameter.) and because of the ^2 it's always positive. so it's the enthropy ∆S that has to overcome ∆H in ∆G= ∆H -T.∆S so that ∆G = - so that the mixing is thermodynamically possible. but if you only have some long polymers the enthropy is too low to overcome the ∆H

Now you are explaining the distinctive features of polymers in terms which only make sense for polymers. But that's not important anyway. Forget thermodynamics. Imagine two polymers which were polymerized simultaneously into an IPN, and remain that way. Probably they would like to untangle, but how? That's kinetically impossible.

ah ok, they are actually just stuck. So there are no intermolecular bonds keeping the polymers togheter? they are just tangled togheter?

I think so. More precisely, it is not important whether there is any specific bonding or not.

authors: إبراهيم حميد المسوري


text:
Why is gold so popular in nanotechnology?
Gold is a very popular metal in nanotechnology. It is often used as a substrate in electronic applications, as a core of functionalized nanoparticles, and more.
Why is gold so attractive? Why are other metals not as popular? What makes gold the go-to material for new applications?

Not only gold. Other noble metals are useful: platinum, palladium, for example.

I can say something form catalysis perspective. It's because of Sabatier principle. To make a good catalysts you must have an optimum trade off between adsorption and desorption facility. Also in most of the cases gold's antibonding orbital is slightly higher than it's fermi level. That's why it offers very low interaction between metal and adsorbate which is interesting for the reaction to occur and also low binding energy facilitates the desorption and hence increases the yield.

You should be more specific in regard to which field of nano-technology. I thought this question related to electronics and would have answered as such, but you seem interested in colloidal solutions.

@A.K. I'm not sure how you got this idea. I'm very much interested in electronics actually

Gold is easy to purify to very high degrees, and hence comparably cheap. Even if this sounds counterintuitive.

gold nanoparticles are easy to prepare and unlike most metals gold nanopowder is not readily oxidized by air.

First, let's look at gold's properties.

Gold is the most malleable of all metals. This malleability alone is very useful in aiding scientists to create such small gold nanoparticles.
Gold also reflects infrared radiation very well and is an excellent conductor of heat and electricity.
Generally, gold is not easily affected by oxygen and is generally nonreactive with most acids and bases. This property serves well especially in the human body where there are an abundant of acidic and basic fluids and the circulatory system is enveloped with oxygen. Because gold is generally unaffected by most acids, bases, and oxygen, there is more assurance for scientists that their golden nanoparticles for medicinal uses would not react with other things before reaching their specific target.
Going back to gold's ability to reflect infrared radiation, an essential reason why physicians, scientists, and researchers utilize nanotechnology is to track down certain cells (cancer, etc), molecules, or processes. Thus, it is important to use such a metal that can reflect radiation to tell them where their targeted molecules are or where their targets are going. In essence, gold nanoparticles are suitable for "staining," or showing contrasts in cells more visible, such as in tissue samples. This can provide higher spatial resolutions for labeling applications.
Gold, as a nice heat conductor, can also absorb heat very well, as I mentioned earlier, and in these biotechnologies, strong absorption of light energy is also another crucial aspect for allowing gold to then dissipate this absorbed light energy in its surroundings so that these heated areas can also be tracked and help with drug delivery purposes.

Here are some references:

Instant insight: A golden future
Nanotechnology Research into the Uses of Gold Nanoparticles

point one is not important, imo. gold nanoparticles are made by precipitation, not dispersing bulk material.

I don't know why everyone missed this point. At nanoscale materials tends to have some novel physical and chemical properties comparing to its macro scale. One such phenomenon which is seen especially in metal nanoparticles is Surface plasmon resonance (SPR)
Wikipedia:

"Surface plasmon resonance (SPR) is the resonant oscillation of
  conduction electrons at the interface between a negative and positive
  permitivity material stimulated by incident light. The resonance
  condition is established when the frequency of incident photons
  matches the natural frequency of surface electrons oscillating against
  the restoring force of positive nuclei. SPR in subwavelength scale
  nanostructures can be polaritonic or plasmonic in nature."

It is because of this phenomenon gold nanoparticles have different colours according to its shape and size.

This phenomenon thus enables gold nanoparticles to be used in wide applications such as sensors, fluorescent agents, etc...

Gold is special:

Electrodes: gold electrodes are advantageous because bulk gold is inert, so electrodes are air/water stable (do not oxidise), such that devices are more reliable in the long term.
Catalyst: gold nanoparticles can be used as a catalyst (high surface to volume ratio, unlike bulk gold), for filtering purposes for example.
Quantum dots: gold nanoparticles change color as a function of nanoparticle diameter (30-90 nm), making it useful in quantum dot applications.
Low melting temperature: melting temperature of gold nanoparticles is relatively low (~300C), making it susceptible to thermal pre/post treatment.
Easy functionalization: gold can be easily functionalized using thiol (sulfur) groups, making it possible to design adsorption behaviour and self-assembly.
Conversion of light to heat: the absorption spectrum of gold nanoparticles can be tweaked by modifying the size of the nanoparticles, which influences on gold's conversion of light to heat. This can be used to thermally kill tumours.

In my knowledge we don't use gold nanoparticles as quantum dots. Quantum dots are semi-conductor nanoparticles, whereas gold nanoparticle is a metal.


text:
What to do when there is an odd number of valence electrons left on the the central atom?
For example, I'm trying to write the formula for $\ce{NO}$, and $\ce{ClO2}$.
If you count the total number of electrons in these molecules, it's odd. This means that no matter how you arrange the electrons, some atom is not going to have a full octet, and some atom is going to have an unpaired electron somewhere - it's inevitable. So my advice is: don't worry about the central atom not having a full octet. Just try and get as many full octets as you can. Of course, don't ever stuff more than 8 valence electrons into a Period 2 element. That's a big no-no. Once you have a structure that you think is reasonable, just check Google for the structures.
The steps my teacher gave me always assume that the central atom has a full valence, or more than full, after the outer atoms have been bonded with it. I'm running into trouble here because my central atom in these cases does not have enough electrons!
I know I could Google what they look like, but it doesn't help me understand why.

$\ce{NO}$ is a radical that does not dimerize, because the free electron on the N atom is on an anti bonding molecular orbital ${\pi}$* . Try to draw the MO diagram of $\ce{NO}$ to be convinced about this situation.

@Maurice, it does dimerize, but as far as I remember mainly in condensed phase.

There are some molecules where the full octet isn't possible (like you've found with $\text{NO}$ and $\text{ClO}_2$). In these cases, the important thing to consider is trying to keep the formal charge (Wikipedia) on each of the atoms as close to zero as possible.
For example, for the $\text{NO}$, we know that $\text{N}$ contributes 5 electrons and $\text{O}$ contributes 6 electrons for a total of 11 electrons. We essentially have to decide between the following Lewis structures:
$$
(1) \qquad :\ddot{\text{N}}::\ddot{\text{O}}. \quad \text{Single electron on O} \\
\qquad \text{vs.} \\
(2) \qquad .\ddot{\text{N}}::\ddot{\text{O}}: \quad \text{Single electron on N}
$$
Here is where calculating the formal charge is important. For $(1)$ above the formal charge on the N is (5 - 4 - (4$\div$2) = -1) and for O is (6 - 3 - (4$\div$2) = +1, whereas for $(2)$ the formal charge on the N is (5 - 3 - (4$\div$2) = 0) and for O is (6 - 4 - (4$\div$2) = 0.
Therefore the second Lewis structure is more favorable.

The octet rule requires eight electrons around each atom. However, $\ce{ClO2,NO2,NO,}$ etc. are some stable molecules that disobeys this rule and are known as odd electron molecules.
If one of the atoms in a molecule has an odd number of valence electrons, the molecule will have an odd electron bond. Or we can say that the molecule has unpaired electrons. These molecules are quite stable and paramagnetic in nature. On dimerisation, the odd electron molecules are usually converted to more stable molecules with even number of electrons.

The nitrogen oxides, such as nitric oxide ($\ce{NO}$) and nitrogen dioxide ($\ce{NO2}$) are the most common examples. In spite of being an odd-electron molecule $\ce{ClO2}$ does not dimerize because the odd electron of $\ce{Cl2}$ is delocalised.

Usually, I would recommend to view molecules like $\ce{NO}$ as examples which demonstrate the limits of the valence structural formula we use today. Although the representations shown in the other answers are formally correct, they do not really help to understand the bond situation in the molecules.
$\ce{NO}$ for example has a bond order of 2.5 which you cannot deduce from the structural formula. Therefore, I would say it makes sense to avoid drawing such structural formula altogether, if possible, and rather stick to $\ce{NO}$ as a graphical representation.
In order to get an idea of the bond situation of $\ce{NO}$, it is worthwhile to study the molecular orbital diagram of the molecule.


text:
pH meter and paraformaldehyde
We have had an ongoing debate in our lab and several bad pH electrodes. I would like to ask the following question
Do paraformaldehyde soloutions (2-4% w/v) damage pH meter electrodes?
If so, why?

One would expect that the oligomers block up the membrane in the electrodes. Keeping them in an acidic solution for a while might help, polyformaldehyde degrades to formaline under acidic conditions.


text:
Is there a quantum chemistry package which calculates a second-order wavefunction?
I am trying to generate a change in density which corresponds to a second-order wave-function in standard Rayleigh-Schrodinger perturbation theory. I would like to try out a simple test on an atomic dimer. However, I am not sure if such a method is implemented in any of the standard quantum chemistry software packages. Does anyone know of such a package?
EDIT: What I am after is the following wavefunction:
$\Psi = \psi^{(0)} + \psi^{(1)} + \psi^{(2)}$

What do you mean? Second-order Møller–Plesset perturbation theory (MP2) is available in a wide variety of packages. Be careful, though, since in some packages by default the HF density is used for post-HF methods unless you explicitly tell the program to do otherwise.

And the last sentence is just a complete nonsense.  The wave function that corresponds to the second-order corrected energy is the second-order corrected wave function. Thus, it is even more unclear what you are asking. Please, clarify.

Wouldn't the second-order energy correspond to the first-order wavefunction in standard perturbation theory? In that sense, I need the second-order wavefunction corresponding to the third-order energy. Does that make sense? Refer to: https://en.wikipedia.org/wiki/Perturbation_theory_%28quantum_mechanics%29

No, an $n$th-order corrected energy corresponds to a wave function corrected to the same order, i.e. to an $n$th-order corrected wave function . The fact that you don't need $n$th-order corrected wave functions (rather just $(n-2)$th-order ones) to calculate $n$th-order corrected energy has nothing to do with that.

And this is why I said that you must careful. For instance, for MP2 calculations in Gaussian, the population analysis, etc. by default is done using the HF density that arises from the HF wave function which is the uncorrected wave function. If you want to use MP2 density that arises from the 2nd-order corrected wave function  for analysis, you have to tell Gaussian about it. The keywords are Density=MP2.

Fair enough. I've edited my original post to be more precise. I will try the recommended keyword in Gaussian. Thanks!

As @Wildcat already mentioned, here is a Gaussian 09 input that will generate the MP3 density, which should correspond to the full second-order wavefunction, thanks to Wigner's $2n+1$ rule:
%chk=g09_mp3.chk
%mem=2gb
#p mp3(full)/sto-3g density=current pop=full output=wfn

wavefunction

-1 1
O     -2.0481500778    0.0447984581   -0.0601937089
H     -1.1514551262    0.4243541138   -0.0806303110
H     -1.8059176361   -0.8898508193    0.0217239645
Cl     1.2314805228   -0.2273431057   -0.0015462666

g09_mp3.wfn

The MO coefficients corresponding to the fully relaxed wavefunction $\Psi$ are also dumped to the *.wfn file in addition to the normal output, which I think can be parsed with cclib.

But is not that $E^{(4)}$ that requires ${ \psi_{i}^{(2)} }$ in accordance with Wigner's rule? Since to calculate $E^{(3)}$ it should be enough to have ${ \psi_{i}^{(1)} }$ ($n=1 \to 2n+1=3$).

Sorry for such a delayed response. You're absolutely correct.

I'm actually a bit hesitant to say that this is the answer, though. If this truly gave the MPn wavefunction, the printed densities would be consistent with Wigner's rule, which they aren't.


text:
What would one have to do differently if they used NaMnO4 instead of KMnO4 for redox titrations?
I am doing a chemistry report related to redox titrations (using $\ce{KMnO4}$ as the titrant), and one section of the report asks me to explain what would occur if I were to use sodium permanganate ($\ce{NaMnO4}$) instead of potassium permanganate ($\ce{KMnO4}$), and more importantly, what I would have to do differently in my experiment if I were to use this new chemical.
What I know as of now.

Sodium permanganate is not widely used because it is more expensive and is much more difficult to create than potassium permanganate.
Molar mass is different when performing stoichiometric equations, which leads to different results when calculating certain values.

I looked online and found that the chemical is "hygroscopic;" however, I don't know if this changes anything in the experiment.
Is there anything else I can "change" in my experiment if I were to use $\ce{NaMnO4}$ instead of $\ce{KMnO4}$?

Well... I think your answer is already pretty good. For the point on molar mass, what you could maybe say is that you will need to use a different mass of NaMnO4 in order to get the same concentration of NaMnO4. It's a titration, so it really shouldn't matter what mass you use as long as you record it accurately, but maybe that's what your teacher is looking for?

I think for the most part, it being hygroscopic just means you need to keep it properly (i.e. make sure to cap the thing after using it). But someone else might know better

You are right. Preparing sodium permanganate is more expensive. The second reason is, sodium permanganate is hygroscopic and occurs as a monohydrate. This means that sodium permanganate is more soluble than potassium permanganate. (Solubility of $\ce{KMnO_4}$ is $0.065$ g/ml and that of $\ce{NaMnO_4}$ is $0.9$ g/ml)
So if you require a larger concentration of $\ce{MnO_4^-}$ in the solution, you can go for $\ce{NaMnO_4}$.

Also sodium salt solutions are less stable.


text:
How to measure the percent weight of NaCl in a potato chip
I want to determine the weight percentage of sodium chloride in a potato chip.  I can see from this answer that atomic absorption spectroscopy (AAS) can be used to quantify sodium chloride in a solution.  I propose the following measurement scheme:

Dissolve the potato chip in a sample of water with known mass
Homogenize the solution until it has a uniform composition
Draw 10 aliquots from this solution
Use AAS to measure the weight percentage of salt in each aliquot
Use the mean and standard deviation of the weight percentages from
the 10 aliquots to generate an estimate of the true weight
percentage with a confidence interval

My questions:
1) What is the best way to get the mass of the aliquot that belongs to the potato chip?  I want to minimize measurement uncertainty.
2) There could be compounds other than sodium chloride that contains sodium or chloride in a potato chip.  I fear that the AAS won't detect sodium chloride, but sodium and chloride ions separately.  What is the best way to measure the concentration or mass of sodium chloride with AAS?  If AAS is not the best way, can you name an alternative method?

I'm somewhat skeptical that atomic absorption spectroscopy (AAS) would work as you outlined the procedure. The "solution" would have starches and fats floating around in it.
(1) I'm not sure that you could get a homogeneous solution.
As far as the homogeneous solution, think of some parts of the solution settling. So what happens if the solutions sit for an hour before going into the AAS?
You could test this by homogenizing a sample and analyzing it immediately, then analyze it after an hour. Do you get the same result?
(2) Even if you did I'm not sure that all the gunk would burn correctly (uniformly) in the AA flame.
Using a plasma spectrometer would solve this problem.
(3) Depending on what sort of accuracy you want there is also the problem of moisture in the chip. How "dry" is a potato chip?
(4) There is a difference between salt added to the chip and $[\ce{Na^+}]$ and  $[\ce{Cl-}]$ in the chip. I'd guess that $[\ce{Na^+}] \approx [\ce{Cl-}]$ in a biological system, but I'm sure that it isn't exactly $[\ce{Na^+}] = [\ce{Cl-}]$.
Such are the joys of method development in analytical chemistry.

If you want to use AAS you will need to make calibration functions. Instead, you can do better and faster. I would recommend using Na-selective reagents like Crown esters for example and do the complexometric titration.

I fear that Na-selective reagents would only help to quantify sodium, not sodium chloride.  Other sodium-containing compounds could exist in the potato chip, so measuring the amount of sodium via Na-selective reagents may overestimate the quantity of sodium chloride.  Am I correct in this assessment?  If not, please tell me why.

I believe you're correct. I'd look into running cations and anions by HPLC to determine the ratio of sodium to chloride, which will tell you which one is in excess - if the slope of the plot is far from unity, that will confirm other sources of the ions in question.

Your statement about the linked answer, and AAS for this application in general, is not quite right; you are not measuring "sodium chloride", but sodium.
I would modify the preparatory part of your procedure to do a clean water extraction of the potato chip, which mainly amounts to adding a filtration step after homogenization of the chip/water mixture. This will leave you with nearly a one-phase solution of water, with some oil floating on top. As sodium chloride is far more soluble in water than in water-immiscible oils, you will want to sample only the water. This is much simpler than analyzing a 3-phase (aqueous, oil, solid) system and also adds some very small degree of selectivity (for water soluble sodium only, like sodium chloride).
Unfortunately, there is no good way to measure only the sodium that came from sodium chloride via AAS; you will simply be measuring all soluble sodium, which is the vast majority of sodium compounds. The assumption I would make in the case of a potato chip is that the vast majority of the sodium comes from the sodium chloride added, usually in good excess, for flavoring.
Regarding your question about the best way to get the mass of the aliquot that belongs to the chip: just assume that you've extracted all of the salt from the chip into the water. In this case, say you used 200 mL for the extraction and then took 10 mL aliquots for analysis; this means each aliquot represents 10/200 of a chip.

You could burn the chip with oxygen to get rid of the organic material. I don't know what is inside common chips but if you're lucky NaCl might be the only thing left. But it should be easier to analyze whatever remains. Some of the remaining salts could be insoluble allowing you to separate them from NaCl.


text:
Why don't Schools follow latest IUPAC for naming groups?
In high school I am taught that Group 7 were halogens and Group 8 were noble gases, and Group 3 the boron group, Group 4 was carbon group, etc. Then when I looked up the groups, I found that noble gases were group 18, halogens were group 17, Group 7 was the manganese group, ...
The school said that their numbering is based on counting valence electrons, and the transition metal columns have no business in their lessons.
So schools don't use IUPAC conventions.
How standard is their group name scheme?

Surely this is primarily opinion-based?

In Germany, the answer would be: "Schools are extremely slow to change their curriculae". Like in "We still only teach pre-1650 mathematics because we don't know yet how to teach children modern concepts of mathematics"-slow

The reason for this is that in 1988 (actually more than a few years back), IUPAC changed their naming system for the groups. Originally the transition metals didn't have a group number and weren't counted as a group. However now it is changed so that the transition metals do have a group number. So number instead of carbon being in group 4, it is in group 14. However despite this being changed quite a few years ago, you will still see in several websites and schools like your and mine still following the old IUPAC conventions and say that carbon is in group 4 rather than 14.
The reason that they do this I suppose is that it avoids confusion in students who are just picking up chemistry as they can see there is a clear relationship between valence electrons and group number. However besides this, schools do follow IUPAC conventions.

References:
Pure Appl. Chem. 1988, 60, 431–436, DOI: 10.1351/pac198860030431
(thanks @Loong)

So it was old IUPAC after all. Changed title to reflect this.

Expanding on the argument, the periodic tables at my school always used roman numerals for group numbers and the transition metals had group numbers with a lowercase b next to them, so halogens would be VII, boron group III and manganese in group VIIb. Now as far as I know, arabic numerals are used so no confusion should occur between 7, VII and VIIb.

@Jan They don't label the groups with Roman numerals at my school.

@Loong thanks for the correction. Yeah, I am pretty sure that 1988 wasn't a few years ago. I have edited my answer now.

@Nanoputian If you want to add a reference to your answer: Pure Appl. Chem. 1988, 60, 431–436, DOI: 10.1351/pac198860030431.

I went to school so long ago that the only elements on the periodic table were earth, air, fire and water. // I'm incensed about the IUPAC change. Why did they discriminate against the actnides?