text:
Origin of alkenes from Wurtz Reaction
According to Wikipedia:

Also, since the reaction involves free radical species, a side reaction occurs to produce an alkene. This side-reaction becomes more significant when the alkyl halides are bulky at the halogen-attached carbon.

How would a radical reaction end up producing an alkene in the context of the Wurtz reaction?
What I do see happening is an E2 elimination creating an alkene.

Also, this raises another question - can't Grignards also create alkenes "by accident"?

There is a small, inconsequential typo in the reaction depicted here. The LHS shows the alkyl group as $\ce{R'}$, but the RHS depicts it as $\ce{R}$ only.

The best explanation I was able to find is given by Avery A. Morton , John B. Davidson , Barton L. Hakan, J. Am. Chem. Soc., 1942, 64 (10), 2242–2247. If I understand it correctly, this boils down exactly to that you already propose.

Disproportionation has been so universally accepted as the criterion of a
  free radical that there has been no attempt to show a derivation from other molecular species. Nevertheless, a mechanism for this phenomenon
  by way of an organometallic intermediate appears exceptionally reasonable. If an alkylsodium meets an alkyl halide in such manner that the two alkyl chains are adjacent to each other while sodium halide is being formed, the two alkyl residues will have unlike charges and a proton will be drawn to the alkyl radical having the two unsaturated electrons. The reaction is merely the conventional prototropic change taking place between adjacent portions of two molecules whose inorganic components are effecting, or have just
  completed, a union.

  If this view be accepted, the olefin produced must come from the alkyl halide rather than at random from both participant alkyl radicals. Though conclusions are to some extent obscured by a metal-halogen interchange (see later), all available facts unquestionably point to the correctness of this deduction.

However, there might be another reason, which I cannot back up with literature. This is probably due to the persistence of the radical species; higher concentration of radicals increases the possibility that they react with each other. In other words, one radical abstracts a hydrogen from another radical, hence creating an alkane and an alkene, in a purely radical reaction.
$$\ce{2 R-CH2-CH2. -> R-CH2-CH3 + R-CH=CH2}.$$

But then again radical mechanisms are never simple and it is quite possible that this does not happen at all, yet it would appear to be at least an explanation for the Wikipedia statement.

Imho, the author of the article you linked to has used quite a convoluted language to describe the simple E2 elimination. Is it possible that E2 elimination wasn't known in the 1940s? Or did proper terminology not exist at that time? What exactly had prevented the author from just writing "E2 elimination occurs"? Do you have any clue?

@GaurangTandon The author gave us a precise description of the mechanism. There is nothing convoluted about it. E2 is a very simple, and crude approximation, or an ideal, ideal case. If you need to break it down for a student that's fine, but for a scientifically accurate description it did not nearly contain enough information.

Yes it's possible this way:

Also, this raises another question - can't Grignards also create alkenes "by accident"?

It's very common. Grignard reagents will take away any acidic proton available. That's why we use Grignard reagents in anhydrous conditions, else it would take up water's proton.

What evidence is there for this mechanism?

@jerepierre I remeber it from my teacher's notes, and if my teacher said it, it is a universal fact. He is a very respected person in my view and he really should be. He doesn't teach us anything without concerting various books and thus we only read the relevant and don't need to read the same thing from 50 books.


text:
Mechanism for acid-catalyzed dehydration of 2,3-dimethylcyclopentanol
How do you get from 2,3-dimethylcyclopentanol to 1,2-dimethylcyclopent-1-ene?
I know the -OH group leaves as water (being protonated by acid) forming a carbocation. Then the weak nucleophile (water) attacks the beta-hydrogen to form what I think is 2,3-dimethylcyclopent-1-ene. But why is it 1,2-dimethylcyclopent-1-ene?
I know Zaitsev's rule wants the more substituted alkene, but forming the 1,2-dimethylcyclopent-1-ene would leave the carbocation hanging there, wouldn't it (the carbocation would still be unsatisfied)?

Update:

Seems like OH protonation and water loss would lead to a secondary carbocation centered at C1 (I'll use the carbon numbering of the alcohol reactant).  A hydride shift could move the C2 hydrogen to C1, thus creating a tertiary (more stable) carbocation at C2.  The C3 hydrogen is then lost to solvent, leading to 1,2-dimethylcyclopent-1-ene.  Your mechanism leading to 1,2-dimethylcyclopent-1-ene is also plausible.  The key question is, what is the rate of the C2 hydride shift relative to the rate of simple deprotonation at C2 of the initial secondary carbocation?

Oh! Maybe the hydride shift is more reasonable! It satisfies the carbocation and creates a tertiary and more stable carbocation. And then further allows for the formation of the more substituted alkene. I guess I should first consider a hydride shift then move on to a methyl shift.

@liya77 You should consider a hydride shift before a methyl shift for 2 reasons. 1. The hydride group has better migratory aptitude and 2. The carbocation formed on doing a hydride shift will be more stable than that obtained through a methyl shift in most cases.

See this:

${}{}{}{}{}{}{}{}{}{}{}{}{}{}{}{}$

This answer "looks OK". But maybe you should be working on the quality of that image. Maybe a larger resolution.

@ADG There is (free) chemical drawing software available you know (e.g. Chemdraw), both online and offline. Just use Google.

@ADG I think Jori and MAR are just trying to be helpful. There really is no need to make such passive aggressive comments. Using free chemical drawing software helps clear any ambiguities in drawings, has overall better readability and presentation, is often scalable, and is better for storage purposes.

Let's address other people's comments in a constructive manner.  You are off to a rocky start already, so really reflect on what you've written before you submit it.

@jonsca see edit.

@ADG the edit is really good -- improves readability a lot.  Thanks and nice work!

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text:
How can cooling gases do not change their physical states?
It is stated in my book "All gases condense to form liquids or solids before -273.15 degree Celsius  is reached?
could you explain this statement generally and my question in the title specifically?

both are contradicting

Well since the lowest temperature that can exist is -273.15 C.
All gases condense to liquid or solids almost all to solids because at this temperature molecular activity ceases.
And the title of your question seems contradicting with what you have written. Probably you might have mistaken interpreting "before".
also only ideal gas can remain gas at this temperature as they have no force of attraction.
http://en.wikipedia.org/wiki/Absolute_zero

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text:
Ratio of rate of diffusion of two gases under constant temperature and pressure
I'm consulting one of higher classes books to study chemistry. I had been studying Graham's Law and question related to rate of diffusion. I came across a question[1] which says:

Calculate the ratio of diffusion of $\ce{SO2}$ and $\ce{CO2}$ under constant temperature and pressure.

The question mentioned the mass of elements as following:   $\ce{S}=32$, $\ce{O}=16$, $\ce{C}=12$.
My solution of the question:
Molecular mass of $\ce{SO2}=32+32=64$
Molecular mass of $\ce{CO2}=12+32=44$
$$\frac{\textrm{Rate of diff. of}\ \ce{SO2}}{\textrm{Rate of diff. of}\  \ce{CO2}} = \sqrt{\frac{44}{64}} = \frac{\sqrt{11}}{4} $$
But the book has a different answer at the back, where it says $1:2$. I rechecked and performed all the maths again, but still cannot find this answer. Is the book incorrect, or have I made a mistake? If the latter, please help me find it.

D. N. Singh: Basic Concepts of Inorganic Chemistry, 2nd Edition.
Pearson India, December 2011. ISBN: 9788131768617
URL of publisher

If you made a mistake, I'm making the same. To me, your solution looks fine!

I think you're right here as well...

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text:
Peroxide bleaching for bast fibres?
Is peroxide bleaching ($\ce{H2O2}$) suitable for treating bast fibres? I read that to be successful, it needs to be combined with chlorine bleaching. Because of the end usage, I don't want to use this kind of bleaching.

Old patents of the 1930s, particularly from I.G. Farben often describe the use of hypochlorite/chlorine in the bleaching of bast.
However, from newer literature, such as this application manual for hydrogen peroxide bleaching from Evonik Industries, I would conclude that bleaching of bast can be performed without chlorine, supposed that the peroxide bleaching is performed

in alkaline medium
at elevated temperatures $(85-90\,°\mathrm{C})$

authors: Qûôphy Â Têñ


text:
Most efficient level of theory for sulfur-containing proline analogues (one that will work)
I'm modelling a small set of proline analogues - extra functional groups, sulfur in the ring. Using Gaussian / DMACRYS.
--functional B3LYP EmpiricalDispersion=GD3BJ --basis_set 6-311G**

or
-fun B3LYP EmpiricalDispersion=GD3BJ -bas 6-311G**

Would EmpiricalDispersion=GD3BJ go under -fun, or would it go under --additional_args? (If so, please show how, I can't get this to run correctly, unsure of the syntax)
To model this more accurately, PBEPBE has been suggested, pure DFT rather than hybrid functional, so there is increased accuracy. Is there a sufficient increase in accuracy for the increase in computational time, or is it better to apply dispersion corrections to B3LYP? (Or PBEPBE and corrections?) Is there another level of theory you would suggest trying?
The electrostatic interactions seem to be over-modelled using B3LYP / 6-311G**, is there a basis set / functional combination (without or with corrections) that would not over-estimate the energy of the + / - charge interactions? (I've tried CAM-B3LYP, but doesn't work on my system.)

It would significantly help if you could indicate how you're running Gaussian. My guess is that the empirical dispersion correction goes under "additional_args" but you seem to be using some program to create the Gaussian input, so it depends on that.

@GeoffHutchison is quite right, it would be interesting to have the Gaussian input file. Pure DFT would cut down a little on the computational cost, if it enhances accuracy is probably something only benchmarks can tell you.

You seem to be running Gaussian through another program or script. Consequently, I have no idea where the various arguments should go. Conventionally, you make a Gaussian input file with keywords, such as:

%n B3LYP/6-311G** EmpiricalDispersion=GD3BJ Opt
[... molecule goes here ..]

For dispersion-correction (i.e., GD3BJ) you would need an appropriately recent version of Gaussian and use the separate term as indicated. It is not part of the functional designation. (Although some functionals implicitly have dispersion correction, B3LYP does not.)
I don't know a priori that PBEPBE would give more accurate results than B3LYP or any other functional. Your best bet is to either consider results from papers in the field (e.g., molecular crystal prediction or runs with DMACRYS) or testing a few functionals and finding out for yourself. Consider reading this question and answers.
Intermolecular interactions are not likely to be handled well by B3LYP. This is the whole reason for range-separated functionals and dispersion corrections. As to a specific functional suggestion, I highly recommend trying a few and looking at how to choose a functional.


text:
How to calculate the pH of an acetic acid/ calcium acetate buffer?
Problem:

A buffer was made by mixing $\pu{500 mL}$ of $\pu{1 M}$ acetic acid and $\pu{500 mL}$ of $\pu{0.5 M}$ calcium acetate.
What are the resulting concentrations of acetic acid (henceforth called $\ce{AcH}$), $\ce{Ca^2+}$ and acetate (henceforth called $\ce{Ac-}$, and what is the resulting $\mathrm{pH}$? $K_\mathrm{a}$(acetic acid) = $1.75\times10^{-5}.$

My attempted solution:
This problem gives me two equilibrium reactions to take into account.
\begin{align}
\ce{AcH + H2O &<=> Ac- + H3O+}\tag{1}\\
\ce{Ac- + H2O &<=> AcH + OH- }\tag{2}\\
\end{align}
In (2), I assume that calcium acetate completely dissociates in water. I also assume that the concentration of $\ce{OH- }$ from the auto ionization of water is negligible.
First I calculate the concentrations of $\ce{AcH}$, $\ce{Ac- }$ and $\ce{H3O+}$ in (1), ignoring what I assume to be minuscule amounts of $\ce{H+}$ from the auto ionization of water. To do this, I use the equation for the acid dissociation constant:
$$K_\mathrm{a}=\frac{[\ce{Ac- }][\ce{H3O+}]}{[\ce{AcH}]}\tag{3}$$
Initial values for $\ce{AcH}$, $\ce{Ac- }$ and $\ce{H3O+}$ are set to 1, 0 and 0 M, respectively. Equilibrium concentrations will then be $1-x$, $x$ and $x$ M.
Plugging this into (3), I get $x = 4.17\times10^{-3}.$
I have now found (or can at least easily calculate) the concentrations of the compounds in (1) at equilibrium.
On to (2), I know that $K_\mathrm{a}K_\mathrm{b}=K_\mathrm{w}$, and I have the values for $K_\mathrm{a}$ and $K_\mathrm{w}$, which gives me the value $5.71\times10^{-10}$ for $K_\mathrm{b}.$
Initial concentrations in (2) for $\ce{Ac- }$, $\ce{AcH}$ and $\ce{OH- }$ are 1 (because 0.5 moles of calcium acetate gives $2\times0.5=1$ moles of acetate), 0 and 0 M, respectively. Equilibrium concentrations will then be $1-x$, $x$ and $x$. Plugging this into (3) (though using the equation for the base dissociation constant instead), I get $x = 2.39\times10^{-5}.$
I have now found all the values I need to calculate the final concentrations and pH when adding the two solutions to each other, and up till this point I am fairly sure that all my calculations are correct.
Questions:

How do I write the chemical equation for the mixing of the two solutions? I tried the following; $\ce{AcH + Ac- <=> Ac- + AcH},$ but this doesn’t seem to help at all.
From my younger days, I seem to remember a much simpler and quicker method of doing these calculations than what I've shown here; does such a method exist? Please note that if it does exist I’d like to learn about it, but I would also like answers based on the method I've used.
Let’s say you mix one weak acid with one weak base, and they aren't conjugate pairs. How would the method differ from the one in this problem? I am specifically wondering about whether both the dissociation constants must be provided to be able to solve, and also how to write the chemical equation, but other observations are more than welcome as well.

In general: $\ce{acid + H2O <=> base + H_3O^+}$
It is called the Henderson-Hasselbalch equation
$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{\ce{[base]}}{\ce{[acid]}}$
I believe the equilibrium constant for a such a neutralisation would look like $K_n = K_\mathrm{a}K_\mathrm{b}\frac{1}{K_\mathrm{w}}$

Problem: A buffer was made by mixing 500 mL of 1 M (mol/L) acetic acid and 500 mL of 0.5 M calcium acetate.
  What are the resulting concentrations of acetic acid, Ca2+ and acetate  and what is the resulting pH? Ka(acetic acid) = 1.75×10−5

Final volume = $500+500$ mL = 1L
Concentration of Acetic Acid = $\frac{\rm moles}{\rm volume}=\frac{0.5*1}{1}=0.5$ M
Concentration of Calcium Acetate = $\frac{\rm moles}{\rm volume}=\frac{0.5*0.5}{1}=0.25$ M
So using Henderson equation:
$$pH=pKa+\log\frac{0.25*2}{0.5}=4.76$$
Rest is easy.

authors: امیر اسان ناز


text:
What is the significance of 273.16 K?
Today, at my chemistry class, we were learning how to convert Celsius to Kelvin with exact values. I said we add 273.15 K to given Celsius but my colleague said we add 273.16 K.
Yes, I did check on internet, we use 273.15 K for conversion, but I also found some relation with 273.16 K and I don't understand the concept of that. It's something to do with thermodynamics and triple point of water.

The Celcius scale was originally based on the freezing and boiling points of water, so 0 °C was chosen as the freezing point until 1954. But now, the size of one degree on both Celcius and Kelvin scales is defined as 1/273.16 of the difference between absolute zero and the triple point of VSMOW (for various reasons relating to the actual measurement), but the 0 point of the Celcius scale was left as the freezing point of water which means that even though both scales are defined at 0 K and 273.16 K (the triple point), the Celcius scale still starts at the freezing point and 0.01 °C was defined as the triple point of VSMOW, which means 0 °C = 273.15 K.
Absolute zero is the coldest possible thermodynamic temperature, 0 K. The triple point is the temperature and pressure at which solid, liquid, and gas phases can exist in a stable equilibrium. If you look at the phase diagram below, it's the point at which the solid, liquid, and gas regions all meet. For water, the temperature of the triple point (273.16 K) is remarkably close to its freezing point at atmospheric pressure (273.15 K). Because 0 °C was originally defined as the freezing point of water, instead of changing the zero point when they redefined the scale based on absolute zero and the triple point, they decided to define the triple point as 0.01 °C.
We add 273.15 to °C to get kelvins because that's the difference between 0 K and 0 °C. We don't add 273.16 because the triple point is not 0 °C.
Source

Hello, thanks for the reply. But can you simply the following for me please?
"But now, the size of one degree on both Celcius and Kelvin scales is defined as 1/273.16 of the difference between absolute zero and the triple point of VSMOW"

@Hardik It's just a fancy way of saying there are 273.16 degrees between absolute zero and the triple point of water on both scales. The difference is just where the zero point is chosen.


text:
Pseudochirality with more than two groups on one carbon atom
Can pseudochirality be considered when three or four groups on the carbon atom are constitutionally the same, but configurationally different?
For example, when the C atom is bonded to one H, and then three more structurally identical groups with 2 chiral C each - in configurations R,R and S,S and R,S respectively - then by assigning each of the 4 groups a priority value (since they are all different - though 3 have the same structure) we can assign a configuration to this C atom.

See this earlier answer and also follow the link describing "prochirality".  That web page also discusses proprochirality.  If that doesn't answer your question, post back with a bit more detail in your question (in these cases a drawing of the molecule would be very helpful)

Ah, now I think I understand what you're asking.  You're saying that we have a central carbon to which is attached a hydrogen and 3 groups with the same structure. Each of these 3 groups contains 2 chiral carbons.  One group is R,R one is R,S and one is S,S.
So yes, the chirality of the central carbon (it is chiral, it is not meso or prochiral - there is no plane of symmetry bisecting the central carbon) can be assigned using the Cahn–Ingold–Prelog priority rules where "R" has higher priority than "S".  Therefore the priority ranking around the central carbon in your molecule would be R,R > R,S > S,S > H.

Yes, that's exactly what I meant. Although I think this carbon cannot rotate light in its path, and isn't optically active - I think its configuration will be designated using symbols r and s rather than R and S.

Your molecule is chiral, there is no symmetry element present.  It will rotate the plane of polarized light..  The enantiomer would be R,R S,R S,S.  R and S will still be used to denote the configurations at the 6 chiral carbons.  r and s are only used with prochiral carbons (see my link in my comment up above). There are no prochiral carbons in your molecule.

Oh - a C which has H, OH, and a group in R, R and R, S configurations (the group has 2 chiral C) attached is optically active? A non superimposable mirror image does form about this C - the configurations of the groups attached become S,S and R,S.

"a compound in which a C has H, OH, and a group ", what is the fourth group attached to C?

Sir, what about Cis trans or E Z in case of pseudo chiral center, if one attached is Z and other is E, which is given more priority in these cases?

@ChiragLathi  "Z" has higher priority than "E"

@ron thanks for the reply,. So in case of pseudochiral center R will have more priority similarly in Z,E Z has more so in Cis trans then Cis would have more priority?

I have belatedly come across this inquiry by Charles (unseen by ChemSE in 3 years) and answered by @ron. If Ron's interpretion of what Charles intended is correct, then I presume that the hexaol (arbitrary numbering) shown below would fit into the conversation. If not, my apologies.
The central carbon $\ce{C2}$ bears, in addition to hydrogen, three groups that have the same constitution but differ configurationally. They are identified as RS ($\ce{C5-C4-C6}$), RR ($\ce{C11-C1-C16}$) and SS ($\ce{C10-C3-C19}$). The pseudo-asymmetric carbon $\ce{C4}$ is readily assigned the configuration $\ce{C4-s}$ according to CIP rule 5 (IUPAC Blue Book 2013). Carbons $\ce{C1 and C3}$ are non-stereogenic (ns) and are designated as such. The priorities are $\ce{C4-s > C1-ns = C3-ns}$ as prescribed by CIP rule 4a. The impasse between $\ce{C1 and C3}$ is resolved by CIP rule 5 where R > S. Thus, the central carbon is $\ce{C2-s}$ because the priorities are $\ce{C4-s > C1-(R) > C3-(S) > H}$. Carbons $\ce{C2 and C4}$ are also described as stereogenic and achirotopic. The hexaol has a plane of symmetry (meso) and would not rotate plane polarized light. The meso isomer $\ce{C2-s,C4-s}$ is but one of four meso stereoisomers. Again, if this structural interpretation is not what Ron gleaned from the OP's inquiry, my apologies.

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text:
In solid Iron, what is the charge of the positive atom?
I'm familiar with the sea model of metals. What I'm curious about is: exactly how many electrons are lost by the iron, to be absorbed into the sea of electrons? For a general metal, how many electrons are lost to the sea of electrons?
Thanks

I'm curious to know whether this aspect of the electron sea model can be quantified, and how well it correlates with the melting and boiling points of the pure substances.


text:
How to determine bond length and angle in a conjugated pi-system (polymethine)?
I have a polymethine molecule with let's say 9 $\ce{C}$ atoms.
$\ce{H-CH=CH-CH=CH-CH=CH-CH=CH-CH3}$
How can I calculate the length of the whole molecule?
I need this number to compute the wavelengths the molecule will absorb (model of linear potential well).
So what are the average bond lengths and bond angles in this molecule and how are they calculated?
EDIT: Another example for polymethine (just the red part of it):

So what are the average bond lengths and bond angles in this molecule
  and how are they calculated?

1,3-butadiene should serve as a reasonable model for your work.

We can find the bond lengths in the literature.  The carbon-carbon double bond length is 1.338 Å, typical for a double bond (reference).  The carbon-carbon single bond length is 1.454 Å, shorter than expected due to resonance.  All of the carbons are $\ce{sp^2}$ hybridized, which means that all of the bond angles should be ~120°.
Using this information and vector addition you can determine the length of any polymethine.

1.338Â are 0.1338nm, right? Then every double bond is 0.1338nm * sin(60°) = 0.1159nm and every single bond is 0.1454nm * sin(60°) = 0.1259nm projected to the axis through the molecule chain. That would lead me to the conclusion that my molecule above (4 C=C and 4 C-C bonds) is about 0.9672nm long. Or did I misunderstand you?

Right on the nm.  Couldn't you just add the full C=C bond length plus 0.1454 * cos(60°) ?

I don't think so. I need the length of the molecule, which is the distance between the C atom at each end, assuming the molecule has alternating bonding angle directions (looks like the example image added to the question).

Your way makes sense and I get the same final number as you.

I merged this formula into another so that I get wavelengths of a polymethine just dependant of its C number. But it gives me wrong results. Could you help me please? In chat maybe?

OK, see you in chat

Shouldn't the central C-C bond be smaller than the other two C-C bonds since if you like at the resonance structures: $H_2C=CH−CH=CH_2,
H_2C^--CH=CH-C^+H_2, H_2C^+-CH=CH-C^-H_2$, the central C-C bond becomes a double bond 1 time more than the other two C-C bonds.

@InternetGuy  No, resonance structures involving charge separation do not count nearly as much as neutral resonance structures in terms of describing the molecule.

Think in a chain of triangles.

Given that the distance between C1 and C2 (= $a$), the distance between C2 and C3 (= $b$), and the bond angle $\gamma$ are known, the distance between C1 and C3 is
$c = \sqrt{a^2 + b^2 -2ab\cos\gamma}$
(cosine rule)

If you have a site license (or a torrent) to use ChemDraw you can generate a 3D model of such a structure and optimize it to find the most stable configuration.  It will give you the bond angles on each  to a high degree of accuracy.  You can use the MM2 function to optimize and find bond lengths as well.  ChemDraw will give you a simple list.  It can also generate bond rotations, but you can bet that the conjugated pi system will remain planar, as the energy barrier to rotation about the sp2 carbon bonds is very high.
If you utilize a simple one-dimensional particle-in-a-box model, the length L of the "box" would be the length of the conjugated system.  This would be the path along which electrons are conjugated.  It's not exactly the jagged line between all the sp2 carbons, but it's pretty close, so you can say L=(number of bonds in conjugated system)x(average length of these bonds).  Note that this is ABSOLUTELY NOT the distance between the C1-C3-C5-etc. carbons the other guy mentioned--I don't think he understands exactly what you're trying to calculate.
Using this model, you can calculate the wavelength of maximum absorption from the quantum numbers of HOMO-LUMO transitions.  Look at the equation:

In this equation nf and ni are the electronic quantum numbers of the final and initial states of a transitioning electron, h is Planck's constant, m is the mass of an electron, and L is as described earlier.  The quantity in parentheses simplifies to N+1, as ni=N/2 and nf=N/2+1, where N is the number of pi electrons in the conjugated system.  To understand this conceptually, the ground state of a molecule will be populated such that the N/2 lowest energy levels will be filled (since electrons fill them in pairs), and all higher energy levels will be empty.  When it absorbs light, one of its electrons jumps from the highest filled energy level (HOMO, with ni=N/2) to the lowest unfilled level (LUMO, with nf=N/2+1).  It's important to understand that if an electron is promoted, it can't simply skip an energy level, so if you know the number of pi electrons you also know what the HOMO-LUMO transition will be.  If you can count the number of pi electrons in the conjugated system (ex. 1,6-diphenyl-1,3,5 hexatriene has 3 double bonds in its box, which means 6 pi electrons), then you can use this equation to find the desired wavelength of maximum absorption.  Surely you've seen the classical equation:

where c is the speed of light.  If you substitute this into the first equation you should be able to solve for the wavelength of maximum absorption.  Mind your units!


text:
What non-poisonous chemicals can get rid of SLS effectively?
Whenever I brush teeth, there are some kind of bubbles sticking inside the cavity and that's uncomfortable, it should be SLS as found on the internet. Does anyone know how do I get rid of those bubbles things easily?
Eat/Drink sth? Tell me if anything is wrong.
Any help would be appreciated, thanks.
Addition: SLS represents sodium lauryl sulfate, chemical exist in our cleaning product: toothpaste, shampoo,   etc. I don't know what it is for but seems it produce bubbles as long as the product is used.

Welcome to Chemistry.SE! Find out how to beautify your posts in the [help center}(/help) and take the [tour] to learn more about this site. Enjoy!

It would definitely help most readers if you explained what SLS is.

SLS is highly soluble in water (100g/L)
Therefore, in the specific case of your teeth having lots of bubbles around them after brushing, you should simply gargle your mouth with water to remove the excess SLS.


text:
Calculating reaction products programmatically?
I was wondering if there was a known algorithm which could calculate simple reaction products - not anything organic, but things like
$$\ce{NaBr + LiF -> NaF + LiBr}$$
While these are quite easily to be worked out mentally, I cannot think of an easy way in which these reaction products could be predicted. Is there some software which does this (most I have found simply uses databases of known reactions), or even better an algorithm which does it?

You are looking at the most simple form of double displacement reactions of salts with
\[\ce{M^1X^1 + M^2X^2 -> M^1X^2 + M^2X^1}\]
Dissolving these salts leads to complete dissociation $\ce{MX -> M+ + X-}$. From the solution, pairs of anions and cations recombine.

Using a lookup table with elements, you could identify the cation from each string that represents a starting material (=salt). The remaining string is the anion.
Build three lists:

a list of cations cations = ['Na', 'Li']
a list of anions   anions = ['Br', 'F']
a list of the starting materials

With Python, itertools.product(cations, anions) will give a generator object that yields tuples of all possible cation-anion combinations
These tuples can be joined to strings for a possible salts
Put these strings in a list and subtract the list of the starting materials to obtain the possible products

import itertools as it
reactants = ['NaBr', 'LiF']
# splitting the string via lookup table is something I leave up to you
cations = ['Na', 'Li']
anions = ['Br', 'F']

pr = it.product(cations, anions)
pairs = list(pr)
# yields list of tuples [('Na', 'F'), ('Na', Br'), ...]

salts = [''.join(pair) for pair in pairs]
# yields a list of strings by joining the tuples:
# ['NaBr', 'NaF', 'LiBr', 'LiF']

products = [salt for salt in salts if not salt in reactants]
# only gives those salts that are not in the list of reactants

@Mithoron Yes, it does :) And the code tends to look readable, rather than like two porcupines mating on the keyboard. At least, I hope that the snippet above did. But this is of course our very personal opinion. The same result certainly could have been achieved using Ruby or any other modern language I'm just not aware of.

Well I do not have a software but can offer some info on this.
If it is a double replacement:

A liquid, solid, or gas must form.

If it is a single replacement:

Element replacing other element must be higher on the activity series.

If you really want to make sure if the reaction happens, just
  calculate the $\ce{\Delta G}$ for it. If it is (-) then it is
  spontaneous and the reaction is thermodynamically favorable.

This is how I usually remember it. Works for me.
But otherwise Klaus got you covered :)


text:
Why are 2 ways to calculate the energy of the ground state?
Setup
I found in Szabo, Quantum chemistry page 88, these two ways to calculate the energy of the ground state:
$$E_0= \sum_a^N \langle a|h|a\rangle + \frac{1}{2} \sum_a^N \sum_b^N \langle ab||ab\rangle $$
and
$$E_0= \sum_a^N \langle a|h|a\rangle + \sum_a^N \sum_{b>a}^N \langle ab||ab\rangle $$
Question
Why?

Looks like by the basal state the OP actually meant the ground state.

Yes, thank you to enlightenme and in that moment I didn't remember the correct word in english

And a rather short parenthetical comment: try to avoid temptation to understand the meaning of an expression without following its derivation. Try to follow the whole Szabo's derivation of the HF equations and you will understand the matter way better.

Yes @wildcat that procedure is the obvious way and... also, as one day you suggest me... try to improve the functional analysis skills which I don't have well developed them. Of course, I want to improve those skills, but now I'm running out of time. Thanks again to you and all that have lending me a hand.

Both formulas have exactly the same meaning. They are just two different ways of not double counting the Coulomb and the exchange interactions. In principle, the goal can be achieved in two ways:

double counting in the first place and then dividing by two (first formula);
counting wisely so that there is no double counting in the first place (second formula).

Obviously, for any practical purposes when you one does not want to double count one would almost always choose the second procedure, but the first representation in some cases provides more apparent picture when dealing with the formulas analytically.

Where does all this come from?
The whole derivation of the expression for the ground state Hartree-Fock energy basically starts from writing down the expectation value of the ground state electronic energy for the ground state Slater determinant $\Phi_{0}$
$$
\langle \Phi_{0} \mid \hat{H}_{\mathrm{e}} \mid \Phi_{0} \rangle \, ,
$$
in terms of the individual spin-orbitals of which $\Phi_{0}$ is constructed.
And in doing so we use the so-called Slater rules and it happens so that the electronic Hamiltonian $\hat{H}_{\mathrm{e}}$ contains only two different kinds of operators:

one-electron operators of the form
$$
    \hat{F} = \sum\limits_{i=1}^{N}\ \hat{f}(i)\,,
$$
and two-electron operators of the form
$$
    \hat{G} = \frac 12 \sum\limits_{i=1}^{N} \sum\limits_{j=1\atop{j\neq i}}^{N}\ \hat{g}(i,j)\,.
$$

Already at this stage an expression for any two-electron operator $\hat{G}$ could as well be written as
$$
    \hat{G} = \sum\limits_{i=1}^{N} \sum\limits_{j>i}^{N}\ \hat{g}(i,j)\,.
$$
The mathematical meaning is exactly the same: we do not want to double count pairwise interactions, so that if, say $\hat{g}(1,2)$ is already counted, $\hat{g}(2,1)$ has to be excluded.
Note carefully that we also have to exclude physically meaningless self-interactions, such as, say, $\hat{g}(1,1)$, which is done in the first case by an additional restriction on summation $j\neq i$ and in the second case by the strict inequality in the lower limit of summation $j>i$.
With respect to the exclusion of this non-physical self-interactions note now that they it is missing from the HF energy expression being written in the first form. But that is perfectly fine since each and every Coulomb self-interaction is perfectly canceled by the corresponding exchange self-interaction. Thus, you do not necessarily need the above mentioned additional restriction of the form $j\neq i$ ($b \neq a$ this time), although you could include it if you would like to,
$$E_0= \sum_{a=1}^{N} \langle a|h|a\rangle + \frac{1}{2} \sum_{a=1}^{N} \sum_{b=1\atop{b \neq a}}^N \langle ab||ab\rangle \, ,$$
or adopting a rather usual convention for such summations, as follows,
$$E_0= \sum_{a=1}^{N} \langle a|h|a\rangle + \frac{1}{2} \sum_{a=1}^{N} \sum_{b \neq a}^N \langle ab||ab\rangle \, .$$

Thank Wildcat so much for your answer and for sharing some of your time. But, I still don't understand the details of why those expressions are different.

They are not different, they are the same. These are two different mathematical ways to express the same idea: we don't want to double count pairwise interactions.

ap.... dough ... slater determinant, only take the upper triangle of the determinant.

@beginner, not, it has nothing to do with the Slater determinant. Give me a few minutes, I'll try to elaborate a little bit more.

beginner, what @Wildcat is saying is that for a pair (i,j) interaction you only have to count it just one time. Note that (i,j) interaction is the same that (j,i) interaction. So if the interaction (i,j) is $I_{i,j}$, it is true that:   $I_{i,j} =  I_{j,i} = (1/2) ( I_{i,j} + I_{j,i}) $


text:
Nucleic Acids Bonds
In a paper I am reading regarding instability and decay of DNA, the author says:
"The chemical price paid for the greatly increased resistance of the nucleic acid phosphodiester bond (gained by removal of the sugar 2'-OH group) is the labile N-glycosyl bond"
I am trying to understand the link between the two chemical bonds. Are they dependent on each other (e.g. one causes the other?)

related: http://chemistry.stackexchange.com/questions/10991/why-does-depurination-happen-at-a-higher-ph-or-under-less-harsh-conditions-tha

I am trying to understand the link between the two chemical bonds. Are they dependent on each other (e.g. one causes the other?)

One does not "cause" the other.  Instead, the author is only trying to say that removing the 2' hydroxyl group from nucleotides does two things: (a) vastly increases stability of 3' bond to phosphate, and (b) decreases the stability of the N-glycosyl bond.  There's no way to get just one effect without the other, since you can either have a 2'-OH group or not.
A source to look for more information on N-glycosidic bond stability is the work of Richard Wolfenden.  The linked paper shows that in 2'-deoxynucleosides, the N-glycosidic bond has a half-life at 37 °C of between 12 and 43 years.  (That is the water-catalyzed half-life.  Of course hydrolysis in strong base, strong acid, or by enzymes accelerates this rate, but the "un"catalyzed or merely water-catalyzed rate is of most interest as an indicator of bond stability.)
A later paper also by Wolfenden et al. directly compares the stability of the N-glycosidic bond of adenosine to 2'-deoxyadenosine and finds large differences.  At 25 °C, the half-life of adenosine's N-glycosidic bond was 6000 years, while that for deoxyadenosine was only 180 years.  (Note the difference in temperature between this paragraph and the preceeding.)
Those papers help indicate the instability of N-glycosidic bond in 2'-deoxysugars.  Yet another Wolfenden paper examines the stability of phosphodiester bonds, although because of experimental difficulties there was no direct comparison between say A-A and dA-dA hydrolysis rates.


text:
Spectral Peak-Picking in MS Excel
I'm working often with spectral data of various gases, and the bulk of the data analysis is assigning the peaks with maximum signal (y-axis) to frequency (x-axis). Often we have several thousand data points, and it can be quite difficult to manually find peaks. I've tried getting around this in Excel with clever IF/AND statements, but the major obstacle is signal noise: I simply cannot devise a clever Excel statement to help me pick peaks.
For example, using an =IF(AND(greater signal than adjacent points),"max","") statement gives tons of extraneous hits where the noise causes a slight fluctuation in the data. Similarly, creating a column of slopes and doing an =IF(slope goes from + to -),"max","") statement also gives bad data.
The most elegant solution I have is to pick an arbitrary value # that most of my peaks are above and use an =IF(AND(slope goes from + to -,B>#),"max","") statement. That seems to work for stronger peaks (maybe 70% of the peaks on a good day), but I still have to go in and manually look at the weaker peaks (and often when doing these spectra many energetic transitions that need labeling are statistically unlikely-->low signal). I'm wondering if anyone here has a better solution able to detect local maxima not resulting from noise with a better success rate.

What signal-to-noise ratios are you trying to successfully observe peaks within?

The SNR seems pretty high, I can very easily see the peaks and just look through those regions to find local maxima.  Are you thinking about using the noise standard deviation to weed out bad hits?

Yep. Also, if you only have positive peaks and have a good horizontal baseline centered around zero, you could calculate noise from the negative noise component, and double it. This will give you low SNR peaks. Presumably you data set has a large number of points.

You're right, it's a large data set.  That actually worked pretty well for part of the data, but unfortunately our IR isn't very good, so the higher frequency region sees the baseline deviating downwards to below zero.

I would smooth your data with a rolling average and the repeat your if(and(...)), adding as one of the constraints that the absolute signal value be greater than some threshold.  If that doesn't work, it may be time to move beyond Excel.  The python/scipy code scipy.signal.find_peaks_cwt() would probably be a one-liner for you.

One thing you can do before searching for peaks is to smooth the data with a 9 point (or 15 point) moving average. I'm not sure how the ends are handled, but the average of 9 points replaces the data point so noise fluctuations are reduced.


text:
Can dimethylmercury originate from cleaning up a broken thermometer with disposable wet wipes?
I broke a CFL in my home (so elemental mercury is present in a small amount) and wiped up some of the mess with a clorox wipe.
The wipe in question contains dimethyl benzyl ammonium chloride and dimethyl ethylbenzyl ammonium chloride.
I'm not chemical-phobic, but can someone please tell me, I didn't accidentally expose myself to dimethylmercury? I realize that it is something you don't find outside of very specific lab environments.
Edit: Thanks everyone for the quick responses. It's been a long time since I took a chem class! Can you help me to better understand why this will not methylate the mercury?
For example, I had been cleaning my carpet at the same time with resolve (water, hydrogen peroxide, sodium lauryl sulfate, probably other ingredients too) and threw out paper towels soaked with it to clean the rug into the same bag as the CFL. Am I correct to assume that despite the methyl group on the SDS, it is also stable like the BZK and thus unlikely to methylate the mercury?

Don't worry, the "dimethyl" part is locked up in a stable form, and there's no pathway for it to somehow create methylmercury and other organomercury compounds in your conditions. Furthermore, the mercury content in CFLs is rather small, and if it was a well-used lamp, most of the mercury vapour will have been trapped in the glass/phosphor, reducing the already small inhalation risk. Ventilate the area well, to be sure, but it's extremely unlikely you're in trouble.

According to the US EPA, disposable wet wipes are a recommended tool for cleaning up a broken CFL. Although having a number of adverse health hazards at high concentration BZK is a common ingredient in many sanitizers, wet wipes and disinfecting wipes, where it occurs at very low concentrations; typically much less than 1%. These ingredients are not likely to react adversely with the released mercury.
For future reference to cleaning up broken CFLs, be sure to read this guide.

No, because you don't have a strong oxidizing agent to oxidize the mercury to $\ce{Hg^{2+}}$.

You generally don't use or need strong oxidizers to prepare dimethylmercury. However those methyls are definitely not able methylate the mercury, so the answer "no" is still correct.


text:
What are the resonance structures of 3-aminocyclohex-2-en-1-ylium?
How would you draw the resonance structure for the compound below:

This is one of the resonance forms I think... but I'm sure there is another one too.

The resonance structures and how to construct them for 3-aminocyclohex-2-en-1-ylium are shown below. The second structure in your question is none of them, as it is a different molecule: cyclohexa-1,3-dien-1-amine. For this molecule it is also possible to construct a resonance structure, as also shown below.

However, the two structures can be interconverted via an acid-base reaction.

And for the stability of 3-aminocyclohex-2-en-1-ylium, the last structure would be most stable (with the positive charge on the EN nitrogen)?

There is no such thing as a most stable resonance structure. Resonance is only a concept that breaks down a highly complicated electronic structure into imaginable contributions of Lewis structures. All of these contributions are present all of the time. However, there are usually a few, that would describe the overall character best. Sometimes there is even only one predominant one.
The concept of resonance is necessary to describe delocalisation effects, hence the Lewis-like representation below is probably much closer to the actual structure than any of the resonance forms depicted above.


text:
Why must the rate law be experimentally determined?
I am familiar with the idea that rate law must be determined experimentally. I am also familiar with the idea that there is no formulated way for coming up with the rate law besides just experimentation.
My question is: why isn't the rate law of:
$$ aA + bB \rightarrow \mbox{something}$$
just:
$$r = k[A]^a[B]^b$$
?
I mean, the equilibrium constant expression involves something to that effect, and the equilibrium constant expression should be precisely talking about when the rates for each direction are equal. So if the rate law is not the above, then why does the equilibrium have that expression?
EDIT: A bit off topic, but how is it possible for something to be zeroth order?

Zeroth order in this schema: If a or b = 0, the rate is no longer proportional to the concentration of that reactant as any number to the power of zero is 1. The reaction is then zeroth order with respect to that reactant.

These two past questions may be of some use.

@NicolauSakerNeto Sneaky linking it on these and two, I nearly missed that. And yes, I did also write something related to the topic, but it is more an example than an actual explanation. Shameless self promotion. !( ^_^ )!

Since rate of a reaction depends actually upon the way reactants react, there is not surety that the quoted stoichiometric coefficients may be the same for rate of a reaction, the rate depends on the way or mechanism of the reaction, and since all reactions have a bit different mechanism no strict formula could be quoted. It is only experimentally derived and can be theoretically approached if we know that it is an elementary (simple step) reaction or it falls under a already classified or studied reaction.
Zero Order, what if a reactant is in very large excess, now the rate is not limited or dependent on its concentration, also it's concentration is almost constant, so it is also merged into the constant and hence we have zero order, some adsorption reactions are maybe zero order too, it is because it actually depends on the surface area of the substance and it's constant.

The order of reaction with respect to each of the reactants, sometimes called to partial order is not equal to the stoichiometric coefficients of the reaction. They are instead dependent on how the reaction proceeds, this is why it must be determined experimentally. So basically it depends on the reaction mechanism.
The equilibrium constant has nothing to do with time as it is measured once the equilibrium has been reached using the concentrations of the reactants and products at that time.
Your rate could be:
$$rate=k[A]^x[B]^y$$
Where $x$ and $y$ are the orders with respect to A and B respectively rather than the stoichiometric coefficients.


text:
Any free tool for Calculating Gasteiger-Hückel charges for ligands?
I want to know if there is any free open source tool available for Gasteiger-Hückel charge calculation? I know it is available on Sylbyl but is there any other(free) way to calculate these charges for small molecules?
I was actually trying this paper, if any one is interested,
How to improve docking accuracy of AutoDock4.2: a case study using different electrostatic potentials.
Hou X1, Du J, Zhang J, Du L, Fang H, Li M.X et al.

-1 means there are no software or it is not the right place to ask this question?

I think question is ok,

The method is not fully published according to Gasteiger.
There is an online server for PETRA run by the Gasteiger group, and in principal, it's available from Molecular Networks. I note that PETRA is no longer listed on the products page, but it's probably part of some other product now.
The open source CDK package has an implementation of the PEPE method, although I don't know if it has been validated against Gasteiger's code.

Thanks for the answer. So, it means its not available for free as Molecular Networks is paid software. Never used CDK is it an api?

CDK is indeed an api (i.e., a toolkit in Java) but also has a web service here


text:
Measuring the % of pectin in a fibre after enzymatic treatment with pectinase
I will remove the pectin from hemp fibres with Bioprep 3000L pectinase. How can I measure the % of pectin before and after for loose fibres? I only found a way to measure it for fabric, but I will perform this test on just loose hemp fibres.

The meter for measurement for fabrics is called: ruthenium red dye method.

Note that I do not have professional experience in the textile industry!
Supposed that the stems have been (mechanically) treated, and the "bark" is removed, the ruthenium red method should work.
In Crystal-Structure of Ruthenium Red and Stereochemistry of its Pectic Stain by Clarence Sterling, Am. J. Bot. 1970, 57, 172-175, the specific binding of Ruthenium Red (RR, CAS-RN: 11103-72-3) along a pectin chain, that is, between a $\ce{COOH}$ of one galacturonic acid unit and the $\ce{OH}$ of the next, has been described.
In a more recent publication, Qualitative distinction of carboxyl group distributions in pectins with ruthenium red, published in Bot. Bull. Acad. Sin., 1999, 40, 115-119 (PDF), the method was applied to orange peel pectines of different origin. Here, the dye-pectin adducts were precipitated by addition of $\ce{CaCl2}$ and centrifugated. Then, the absorbtion of the supernatant solution and a stock solution of the dye at $\lambda = 534\,\mathrm{nm}$ were compared.
Consequently, it should be possible to take fibre samples, suspend them and detemine whether they decrease the absorption of RR (= still contain pectin).
Are you planning to bleach these fibres in a way recently discussed in another question? It might be interesting to figure out whether these conditions (alkaline hydroperoxide) are sufficent to (partly) degrade the pectin.


text:
Why is the bond order in the SO₃ molecule 1.33 and not 2?
$\ce{SO3}$ molecule has three double bonded oxygen to the central sulfur atom.
Sulfur has $\ce{sp^2}$ hybridization and it has 6 outer electrons which make the bonds with the oxygen.
So shouldn't the bond order be 2?

I just don't have time to write a complete answer, so instead I leave this link to a previous answer of mine that somewhat deals with this question. The doubly bonded picture is a very crude oversimplification, though.

Aaaah! I'm used to drawing the structures one of these three ways though.

@MARamezani This is essentially the concept of resonance, and it's intended purpose. You need all three structures though, not only one. Because one alone does not describe delocalisation.

Indeed @Martin. The resonance hybrid is the only correct structure for this.

The bonding situation in $\ce{SO3}$ is a tough nut to understand. In a historical context this molecule belongs to the species of hypervalent molecules, that disobey the octet rule. The concept of hypervalence is still very much debated. Recently there was a question raised by ron, seeking for more guidance: Hypervalency and the octet rule.
I stated in Is the bond enthalpy of S-O stronger in SO₃ or SO₃²⁻?

The bonding picture is usually trivialised as each $\ce{S−O}$ bond being a double bond. But this is actually far away from the truth, as it does not respect the charge of $q=+2$ at the sulfur atom and the charges at the oxygens with $q=−\frac23$. This is due to the fact, that the sulfur atom actually only contributes to one of the three π bonding orbitals, resulting in electron density deficiency.

There are a couple of molecules that belong roughly to the same class. I call them Y aromatic systems: $\ce{CO3^2-}$, $\ce{NO3^-}$, and $\ce{SO3}$. The description is best carried out in the framework of resonance. (Please also see Why is there a need for resonance?)

I have explained the delocalisation in the nitrate ion, but I am happy to repeat it here for $\ce{SO3}$. The molecule has an overall symmetry of $D_\mathrm{3h}$, hence the central sulfur can be regarded $sp^2$ hybridised. Since oxygen has only one bonding partner, it can sufficiently be described as $sp$ hybridised (compare here). The combination of these orbitals form a $\sigma$ bond each. The remaining $p$ orbitals perpendicular to the molecular plane form the delocalised $\pi$ system. Since delocalisation is not part of the original Lewis description it is impossible to address this with such structures, which ultimately lead to the concept of resonance. It is important, that the bonding situation can only be described by all resonance structures, as there is no predominant. The blue structure tries to emulate delocalisation, but here formal charges and charges are quite messed up.
However, we can take home an approximate bonding order for the $\ce{S-O}$ bond. Since all of the resonance structures are equally dominant, we can see that all of these bonds consist of a $\sigma$ bond. We also see, that there is a $\pi$ bond in the first three structures. Each of them contributes equally, therefore it makes a third of a full contribution per bond, hence the bond order is expected to be about $1\frac13$.
Now the molecular orbital scheme is quite similar, but it suffices with one structure. The explanation of the $\sigma$ system remains unchanged. From the four $p$ orbitals we can form four symmetry adapted linear combinations. Two of the are non-bonding degenerate, the completely symmetric one accounts for the huge stabilisation. The antibonding orbital is not occupied. See below for the schematics of $\ce{NO3^-}$, which is analogous to $\ce{SO3}$. You can find the complete MO scheme here.

Since each of the $\ce{S-O}$ bond contributes equally to the lowest lying $\pi$ orbital, the bond order is found to be about $1.33$. The calculated bond order is a little bit higher, most likely due to attractive electrostatic interactions.

Great question, I had to do a little reading to piece together an answer as I haven't worked with sulfur oxides in quite some time:
Sources:

Wikipedia: Sulfur Trioxide
Shriver & Atkins, Inorganic Chemistry, 4th ed., p388
Some Yahoo Answer
Greenwood & Earnshaw, Chemistry of the Elements, 2nd ed., p703

As a simplistic explanation, the above sources state that the lewis structure of $\ce{SO_3}$ contains a 2$^+$ charge on the central sulfur and negative charges on two of the three bonded oxygen atoms. In that case, $\ce{SO_3}$ contains one double bond and two single bonds, which is why people tend to list the overall bond-order as 1.33. The actual bonding structure of $\ce{SO_3}$ is a little more complicated than that, as J. LS points out, so you might need to brush up on molecular-orbital theory to get into the nitty-gritty of its bonding structure.

There is no need to invoke d-orbitals in bonding for early p-block elements; it's been theoretically demonstrated to be unimportant. Really you need to use MO theory to understand how the bonding in compounds like this actually works.

I was giving the basic historical interpretation of why people tended to draw it like that, followed by the straightforward explanation of why it isn't. I highly anticipate your (correct) MO-based explanation, though.

The simplest correct explanation is as you say, simply having three sigma bonds, one pi bond and two ionic interactions, giving a bonding order of 1.33. Invoking d-orbitals is unnecessary.

Fair enough, I agree, it's a bit of fluff that I carried over. I will edit the answer to reflect that!

The most important feature of this compound is, that it is an Y aromatic system (I have done calculations on it and linked the post to the question). There is one pi orbital delocalised over three bonds, hence one third per bond (the calculated bond order as a little higher though). The description with one SO-pi-bond and two ionic interactions is only half correct in the framework of the symmetry equivalent resonance structures - but they all have to be present.

@Martin-マーチン Why is the calculated bond order a little higher?

@paul 1) DFT orbitals are complete crap. While HF/post-HF orbitals are a thing within their framework, DFT orbitals are not. 2) consequently, the bonding analysis in DFT framework is by definition buggy. 3) don't forget about ionic bonding. Even though it isn't a line in the picture, it still puts some electronic density between two atoms, enough for bonding analysis to be confused.

@permeakra Kohn Sham orbitals most of the times resemble molecular orbitals quite well: http://chemistry.stackexchange.com/q/316/4945 The bond order is only well defined for localised bonds, basically it only works well for two atomic molecules. You cannot measure bond order directly.

During the reaction of $\ce{SO2}$ and $\ce{O2}$, there's a formation of nascent oxygen which oxidises sulphur. In doing this sulphur co ordinates it's lone pair to the nascent oxygen. Since only 2 electrons take part here in this co-ordinate bond formation it is just a single bond. Now the $\ce{SO3}$ thus formed has 4 bonds shared between 3 places. Hence the bond order.


text:
What causes "stringy" melted cheese (and similar behavior in other materials)?
I was wondering about this. What is the physics underlying the "stringy" consistency of melted cheese? That is, when the melted cheese is hot, and it is pulled apart, it forms strings of cheese.
This behavior also occurs with a number of other fluid substances. It seems to happen with polymer substances (like cheese which is made of proteins like casein, also plastics when melted), gums, etc. What is the technical term for this kind of behavior and what is the general mechanism by which it happens?

Unsure whether this is physics or chemistry. In any case i believe the answer is that the substance in question involves long protein molecules which somehow interact to produce longer chains.

You could've googled this actually..
https://www.cdr.wisc.edu/sites/default/files/pipelines/2000/pipeline_2000_vol12_01.pdf
http://www.finecooking.com/item/64019/the-science-of-melting-cheese
http://www.reddit.com/r/askscience/comments/sge85/what_makes_string_cheese_stringy/

Interesting question:   I found the same site as listed above.   The gist of it - end of page 1:  "Stretch is the
result of casein–casein interactions that are broken
easily but also readily reform at different locations in
the casein network. Think of holding a piece of
warm Mozzarella, take one end in each hand and
gently pull it apart. The casein molecules are
grabbing and releasing each other while sliding past
as you pull the cheese."

When cheese is heated to 140 degrees fahrenheit, the proteins line up in rows.  Soaking the cheese in a salt solution before pulling it helps to stiffen it and keep it pliable, as the salt neutralizes acids which could cause the long lines of protein to break.
Polymers, similar to cheese, can be stringy because their molecules are very large (macromolecules), and tend to entangle with one another.  @IanF1 is correct in saying this may be a chemistry question.  Here is a good explanation of polymerization:
http://courses.chem.psu.edu/chem112/materials/polymers.html


text:
Why do electronegative heteroatoms increase C=O stretching frequency (bond strength)
Why does a more electronegative heteroatom (i.e. $\ce{Cl}$ or $\ce{O}$) increase the stretching frequency of a carbonyl?
I'd suspect it's that they donate their lone pairs and we end up with a negative charge on the oxygen in one of our resonance structures which makes the electrostatic attraction between the carbon and oxygen larger?

Also, inductive effects create a larger dipole in the c=o bond which increases the electrostatic contribution to the bond strength


text:
What is the the full name of TAPD? Structure provided
I have a substituent called TAPD on a molecule I'm working with. I have a hard time locating information about its properties. The bold square in the picture below is supposed to be the rest of the system on which the TAPD is attached. I am a physicist and thus not very good at chemical naming, so I would have good use of some help.

TAPD = Tetraalkyl-p-phenylenediamine
This is because you have a phenyl (benzenoid) ring, with two amino groups (diamine) opposite (para) to each other, and each of those amino groups has two alkyl substituents, giving four (tetra) in total.


text:
Henry's constant and solubility
According to Henry's law

The solubility of gas in a liquid is directly
  proportional to the partial pressure i.e S=Kp

My book says that higher the value of K at a given pressure lower is the solubility of the gas in the liquid.
I don't understand how. I know I am overseeing something really simple over here as all the books that I have and online websites seem to conclude the same from the given equation. A little help maybe. Thanks.

Actually solubility $x$ is given as a function of pressure $p$ and henry's constant $K_H$:
$$p=K_Hx$$

My book says that higher the value of K at a given pressure lower is the solubility of the gas in the liquid. I don't understand how.

Since at a given pressure, $p$ is constant, if you increase $K_H$, logically solubility $x$ would decrease.

My book says that x is mole fraction! :(

@KaranSingh mole fraction can be treated as an indirect measure of solubility.

So would it be correct to say that mole fraction is directly proportional to solubility?

@KaranSingh approximately for dilute solutions.


text:
Saturation of multiple solutes
Can a liquid which is saturated with a single solute, dissolve a different solute, or is saturation a universal thing? I ask because I’ve seen that different solutes have different points (amounts) at which they become saturated in water, for example, which caused me to think that maybe each solution was perhaps independent, even though that would produce some interesting extreme implications. Something like, take a water & sucrose mix that is saturated; can this mixture then dissolve salt?
The specific implication I am asking about is this: If I dissolve xylitol (a sugar alcohol) in water until it is saturated (200 g xylitol to 100 g water at 25 °C / 77 °F) can I then dissolve erythritol (a different sugar alcohol) into the solution?
Obviously, we could heat the liquid and enter super-saturation, but that’s not what I’m asking. I’m asking if this can be done without super-saturating (i.e. risking precipitation/crystallization) once cooled.

Do you know about colligative properties

No, I've never heard about that. Chemistry's not my strong suit, I just figured here was the best place to ask the question.

Different solutes in the same solution indeed can affect each others' solubility. One example where this effect can be used to advantage is the "salting out" of a protein from water solution. Adding a salt (ammonium sulfate is especially good for this purpose) reduces the solubility of the protein in water, causing it to precipitate or crystallize out. Salting out can also be used to more fully separate an organic solvent from water. I imagine that in your example with the sugar alcohols, adding one component to water will reduce the solubility of the second component by a similar degree. But there may be cases where one solute increases the solubility of another.

"But there may be cases where one solute increases the solubility of another." Indeed, there's also Salting In for proteins, because the salt at small concentrations will increase the proteins solubility.

You Answered this for me 2 years ago, and I just realized that I never accepted your answer. Wow. I'm sorry man. I just did. Sorry, about that.

Hey, no problem. Glad you enjoy the forum!

The conundrum arises because dissolution is treated as a physical process when it is a chemical process. Solubility is dependent on the activities of the solvent, solute, and the precipitates [! no good word for the yet to dissolve or will not dissolve material]. The properties of the pure solids and [pure] solvents are assigned activities of one and shoved into the equilibrium constant and activity coefficients. Who really believes that the activities of water, solid NaCl, fused Al2O3, H2 gas etc. are all the same? It does make life easier. Adding something to a solution will change the activities of everything there; an application of LeChalelier's principle is a place to start. Someone working in formulation or even cooking will know that strange things happen when working near solubility limits. The timing, amounts, rate of addition order of addition can have profound effects on a product [Throw in some "normal" reactions and havoc can ensue].
The two sugar alcohols will probably compete for water lowering the total solubility. Look at the molecular structure and work on a molecular level, each alcohol molecule is probably equivalent to 6 waters. Also check on the formation of hydrates.


text:
Comparing wt% to at% which is normalized to 100%
I have measured the element composition of a sample using ICP-OES, and reported my results in wt%. A study which has analyzed the same sample types, but with an electron microprobe, has reported their element concentrations in at%, which in addition are normalized to 100 %. I would like to compare my results to those in the article, but I am not sure how.
I am able to convert the atomic percentages to weight percentages, but I am not sure how to tackle the normalization of the data. When I add the wt% of my sample, I only get to about 50%. This loss is attributed to that I have not analyzed oxygen and carbon, which I suspect would make up a significant portion of the sample weight.
However, the study I want to cite has indeed analyzed both oxygen and carbon, and all the other elements I have measured. I do not suspect that they have a very large portion of at% unaccounted for. Therefore I believe their un-normalized data is quite similar to their normalized data.
So, how should I proceed with comparing my data to theirs? Should I just argue, like I have done here, that I think their data is "comparable enough"?
I would appreciate any insight on the subject!

I think this question will be difficult to answer without knowing more about the data.  First can you elaborate on what you mean by "normalization"?  Do you mean that in the at% study, all measurements added together were only 96% or 90% or 80% or something, and so to get all measurements to add to 100%, the simply divided each measurement by 0.96 (or 0.9 or 0.8)?

Second, how to the raw at% measurements work?  Atom percent has units of atoms of X per total atom (times 100).  Are they measuring the raw number of total atoms in their OES experiment?  Or are they just comparing their results to some calibration standard for each element measured?  If the former, it should be straightforward to convert to wt%.  If the latter, you should convert their calibration standards from at% to wt% and then recalculate their results.

If you're missing at% or wt% data on more than one element, then I think you have an underdetermined system of linear equations with infinite solutions. If you don't measure every single element directly, which would be ideal, you can either only have one element missing (and assume it is entirely responsible for all the missing at% and wt%, which can be dangerous), or you need some sort of a priori knowledge of the chemical composition of the sample (e.g., if you know for sure that all nitrogen and oxygen atoms are in the form of $\ce{-NO2}$ groups, then clearly %at O = 2 * %at N)

Curt: the article (Gunst et al. 2000, J. Environ. Monit., vol. 2, pp. 65-71) does not clarify the method of normalization; they simply state that they normalize the data to 100%. The technique with which they have measured the concentrations is wavelength-dispersive X-ray analysis, using an electron microprobe. The authors say that, "The X-ray count rates for each individual particle and for standard samples were obtained from elemental distribution maps.". I remember from laboratory exercises that the software asked it we wanted the percentages to add to 100, so I think it's a common thing.

Nicolau: The chemical composition is complex, and a variety of elements are present in several different chemical forms. Manganese, for instance, would be present as different oxides, or as SiMn complexes. I cannot attribute the lacking wt% to any one element, or any one molecule, but I assume that carbon and oxygen make up a significant part of the lacking sample weight.

The terminology used in the article concerning the normalization is this: "at% normalized to a total of 100% at%)

This question is difficult to answer without knowing more about the data.
First is the issue of "normalization". Do you mean that in the at% study, all measurements added together were only 96% or 90% or 80% or something, and so to get all measurements to add to 100%, the simply divided each measurement by 0.96 (or 0.9 or 0.8)?  This is what I infer from your comments.
Second is the issue of measuring at%.  Are they measuring the raw number of total atoms in their OES experiment? Or are they just comparing their results to some calibration standard for each element measured?  Your comments make me think they are using calibration standards.  From a description of those standards, you should be able to convert the standards to wt%, and thus map from wt% to at%.
Whether you want to map to the "normalized" at% or the unnormalized at% is up to you.


text:
Is hydrolytic polymerisation the mechanism by which the precipitate forms from the neutral complex?
From basic coordination chemistry and acid-base equilibria we understand that a metal hexa-aqua ion can be deprotonated until it's positive charge is balanced by the negative charge of the hydroxides rendering a neutral complex. We then say that this complex precipitates.
Recently in my inorganic class we studied hydrolytic polymerisation where complexes with hydroxyl groups condense into $\ce{OH}$ bridged complexes, which can form polynuclear structures. These structures then release the hydroxide when the solubility limits are reached.
My question is:

Are these two ideas the same, and is hydrolytic polymerisation the mechanism by which the precipitate forms from the neutral complex? Another way of phrasing it would be, are these complex precipitates extended structures or are they bound via intermolecular forces?

Your title seemed fine, but please never hesitate to write good titles!


text:
Molecular orbital diagram of a complex including an oxido ligand
With highly oxidized metal centers, we expect that the hydroxido ligand's proton becomes acidic and will be lost to the surrounding media, creating an oxido ligand.
The oxo ligand is a $\pi$ basic ligand and we expect both $\sigma $ and $\pi$ donation into the metal center.
The metal's $\mathrm{t_{2g}}$ orbitals are lowered in energy by the presence of the oxido ligand's filled p orbital donating into the $\mathrm{t_{2g}}$. My question is regarding the symmetry of the ligand group orbitals. A p orbital does not have $\mathrm{t_{2g}}$ symmetry and as we know non-symmetry interactions are forbidden. Therefore I am struggling to understand the details of the oxido ligand.
My only explanation is that the complex's symmetry is reduced with the presence of an oxo ligand maybe from $O_\mathrm{h}$ to $C_4$ or similar. This would then change the symmetry of the $\mathrm{t_{2g}}$ orbitals on the metal center, but I really need some help explaining this!

maybe an example to convey your problem would be better.

Fact is that $\pi$ orbitals of an oxo ligand in $z$ direction can combine with $d_{xz}$ and $d_{yz}$ orbitals of the metal center. The irreducible representations for the orbitals is another topic for which we need more information from you.

Although chemists and chemistry students alike often indiscriminatorily use the $\mathrm{t_{2g}}$ and $\mathrm{e_g}$ descriptors for the d-orbitals of a d-block metal in coordination complexes, those designations are only truly correct in an octahedral environment. The vast majority of coordination complexes, however, are not octahedral but distorted octahedra, octrahedroids with non-symmetric ligands and others. This is also the case for oxido-ligand coordinated complexes that you are asking about.
To restrict the discussion to a single species (although it can easily be expanded to others) I will craft my answer explaining the situation in the vanadyl cation $\ce{VO^2+}$. I will discuss the pentaaquaoxidovanadium(IV) cation $\ce{[VO(H2O)5]^2+}$ as Ballhausen and Gray analysed.[1] I will label the direction of the oxido ligand $+z$ (in line with their scheme).
Constructing the molecular orbital scheme
The first important thing to note is that the complex indeed does not display octahedral $O_\mathrm{h}$ symmetry but $C_\mathrm{4v}$. That is great for us because it facilitates practically every step in the construction of molecular orbitals. A character table for the $C_\mathrm{4v}$ point group can be found in orthocresol’s collection (Note: MathJax-heavy link. Give it time to load). In the tables below I have given the different metal and ligand orbitals and their irreducible representations.
$$\textbf{Table 1:}\text{ irreducible representations of the metal orbitals in }\ce{[VO(H2O)5]^2+}\\
\begin{array}{cccc}\hline \text{irrep} & \text{s orbital} & \text{p orbital} & \text {d orbital} \\ \hline
\mathrm{a_1} & \mathrm{4s} & \mathrm{4p}_z & \mathrm{3d}_{z^2}\\
\mathrm{a_2} & & & \\
\mathrm{b_1} & & & \mathrm{3d}_{x^2 - y^2}\\
\mathrm{b_2} & & & \mathrm{3d}_{xy}\\
\mathrm{e} & & \mathrm{4p}_x, \mathrm{4p}_y & \mathrm{3d}_{xz}, \mathrm{3d}_{yz} \\ \hline \end{array}$$
$$\textbf{Table 2:}\text{ irreducible representations of the ligand (group) orbitals in }\ce{[VO(H2O)5]^2+}\\
\begin{array}{cccc}\hline \text{irrep} & \ce{O^2-} & \ce{$eq${-}H2O} & \ce{$ax${-}H2O} \\ \hline
\mathrm{a_1} & \mathrm{p}_z & {\sigma_1} \atop {\pi_1} & \mathrm{p}_z\\
\mathrm{a_2} & & \pi_2 & & \\
\mathrm{b_1} & & {\sigma_2} \atop {\pi_3} & \\
\mathrm{b_2} & & \pi_4 & \\
\mathrm{e} & \mathrm{p}_x, \mathrm{p}_y & {\sigma_3, \sigma_4}\atop{\pi_5, \pi_6, \pi_7, \pi_8} & \mathrm{p}_x, \mathrm{p}_y\\ \hline\end{array}$$
If required, I can supply images of the π and σ symmetric orbitals of the equatorial ligands once I am back at a PC that has ChemDraw (not before the 8th of January). It is not always fully obvious which orbitals interact with which ones; however, it is very obvious that the π2 ligand group orbital is nonbonding with respect to the $\ce{L\bond{->}M}$ bond.
In a second round, we need to consider the strength of interaction. The oxido ligand is much closer to the vanadium centre ($d(\ce{V=O}) \approx170~\mathrm{pm}; d(\ce{V\bond{<-}OH2}) \approx 230~\mathrm{pm}$) meaning that any interactions with that will be considerably stronger. (There are hints pointing towards an even more elongated axial aqua ligand, but that would be secondary.)
The resulting scheme you construct should show a strong interaction of the oxido ligands $\mathrm{e}$ orbitals ($\mathrm{p}_x$ and $\mathrm{p}_y$) and the metal’s $\mathrm{e}$ orbitals ($\mathrm{d}_{xz}$ and $\mathrm{d}_{yz}$). This can be interpreted as the oxido ligand donating both p-type lone pairs to vanadium; while the resulting bond is typically depicted as $\ce{V=O}$ a better depiction may be $\ce{{V^-}#O+}$. The other interactions of $\mathrm{e}$ orbitals are considerably weaker.
The orbital scheme as published
Ballhausen and Gray, who I mentioned previously, also included a quantitative scheme in their paper, albeit having omitted the ligand-metal π interactions with the exception of those of the oxido ligand.[1] Thus, their scheme is a little bit simpler than the one I would have drawn. The low-lying σ orbital is the oxido ligand’s; the orbitals labelled π are obviously the oxido ligand’s $\mathrm{p}_x$ and $\mathrm{p}_y$ orbitals. Due to the lack of aqua-π orbitals, the $\mathrm{d}_{xy}$ orbital appears unmodified. The $\mathrm{e}$ orbitals are raised notably, as is the $\mathrm{b_1}$ orbital ($\mathrm{d}_{x^2-y^2}$) due to the ligand’s σ interactions. Finally, the most destabilised metal d orbital is $\mathrm{a_1}$ corresponding to the $\mathrm{d}_{z^2}$ orbital.

The resulting scheme can be seen as an atypical Jahn-Teller distortion of an octahedral complex. In typical Jahn-Teller distortions, the $z$-ligands would be moved away resulting in a stabilisation of orbitals with $z$ contribution. Here, one $z$-ligand is moved closer resulting in a destabilisation of anything including a $z$ contribution.

Reference
[1]: C. J. Ballhausen, H. B. Gray, Inorg. Chem. 1962, 1, 111. DOI: 10.1021/ic50001a022.


text:
Why is ligand substitution only partial with copper(II) ions and ammonia?
When studying ligand substitution (at UK year 13 level), the following example has been given:
\begin{align}
\ce{[Cu(H2O)_6]^2+ + 2NH3 &<=> [Cu(OH)_2(H2O)_6] + 2NH4^+}\\
\ce{[Cu(OH)_2(H2O)_4] + 4NH3 &<=> [Cu(NH3)_4(H2O)_2]^2+ +2OH- +2H2O}\\
\end{align}
There has been no explanation given of why the substitution is only partial, whereas the substitution of, say, $\ce{[Co(H2O)_6]^2+}$ is complete.
To me, it seems that this partial substitution wouldn't be favourable, as it produces a slightly distorted octahedral shape, which is less stable than a regular octahedral shape, surely? Moreover my textbook states ammonia is a better ligand than water, so why would $\ce{NH3}$ not replace those final two aqua ligands?

Metal ion complexes have stepwise stability constants:
\begin{align}
\ce{[Cu(H2O)6]^2+ + NH3 &<=>[$K_1$] [Cu(NH3)(H2O)5]^2+ + H2O}\\
\ce{[Cu(NH3)(H2O)5]^2+ + NH3 &<=>[$K_2$] [Cu(NH3)2(H2O)4]^2+ + H2O}\\
\ce{[Cu(NH3)2(H2O)4]^2+ + NH3 &<=>[$K_3$] [Cu(NH3)3(H2O)3]^2+ + H2O}\\
\ce{[Cu(NH3)3(H2O)3]^2+ + NH3 &<=>[$K_4$] [Cu(NH3)4(H2O)2]^2+ + H2O}\\
\ce{[Cu(NH3)4(H2O)2]^2+ + NH3 &<=>[$K_5$] [Cu(NH3)5(H2O)]^2+ + H2O}\\
\ce{[Cu(NH3)5(H2O)]^2+ + NH3 &<=>[$K_6$] [Cu(NH3)6]^2+ + H2O}\\
\end{align}
with
\begin{align}
K_1 &= \frac{{\left[ \ce{[Cu(NH3)(H2O)5]^2+} \right]}}{{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_2 &= \frac{{\left[ \ce{[Cu(NH3)2(H2O)4]^2+} \right]}}{{\left[ \ce{[Cu(NH3)(H2O)5]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_3 &= \frac{{\left[ \ce{[Cu(NH3)3(H2O)3]^2+} \right]}}{{\left[ \ce{[Cu(NH3)2(H2O)4]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_4 &= \frac{{\left[ \ce{[Cu(NH3)4(H2O)2]^2+} \right]}}{{\left[ \ce{[Cu(NH3)3(H2O)3]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_5 &= \frac{{\left[ \ce{[Cu(NH3)5(H2O)]^2+} \right]}}{{\left[ \ce{[Cu(NH3)4(H2O)2]^2+} \right]\left[ \ce{NH3} \right]}}\\
K_6 &= \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}{{\left[ \ce{[Cu(NH3)5(H2O)]^2+} \right]\left[ \ce{NH3} \right]}}\\
\end{align}
The overall stability constant is:
$$K_\text{B} = \frac{{\left[ \ce{[Cu(NH3)6]^2+} \right]}}{{\left[ \ce{[Cu(H2O)6]^2+} \right]\left[ \ce{NH3} \right]^6}} = K_1 \cdot K_2 \cdot K_3 \cdot K_4 \cdot K_5 \cdot K_6$$
(In particular, note the exponent 6.)
Furthermore, usually $K_i/K_{i+1}>1$; thus, $K_1 > K_2 > K_3 > K_4 > K_5 > K_6$. Consequently, it gets more and more difficult to exchange further ligands.
Therefore, the predominating complex strongly depends on the concentration of $\ce{NH3}$.
And you probably work with dilute aqueous solutions.
Simply speaking, the missing $\ce{[Cu(NH3)6]^2+}$ in aqueous solution can be attributed to the high concentration of water which is competing with ammonia for the coordination sites.

To my understanding, $\ce{Cu(II)}$ actually prefers four-fold coordination with ligands like water and ammonia, in the square planar configuration. Gas-phase DFT / MD simulations bear this out$^1$, at least for $\ce{H2O}$, and the pioneering work$^2$ of Jannik Bjerrum indicates, apropos Loong's answer, that the first four step-wise stability constants for the substitution of ammonia ligands for water are significantly higher than the fifth (IIRC the 6th was un-measurable at that time).  I'm pretty sure this is because the $\ce{[Ar]}3\mathrm d^9$ ground state configuration of $\ce{Cu(II)}$ just doesn't have enough un-occupied d-orbitals to support 'proper' six-coordination.
You're absolutely correct that $\ce{NH3}$ is a better ligand than water -- this explains why the first four molecules substitute readily into the $\ce{Cu(II)}$ coordination environment.  However, if the remaining "axial coordination sites" are really not coordination sites at all, but instead are part of a non- or weakly-coordinating solvation shell, then it just becomes a statistical competition between excess $\ce{NH3}$ solutes and the $\ce{H2O}$ solvent, in which case the water will win out until the ammonia concentration is quite high.
On the contrary, the comparatively d-electron-poor $\ce{Co(II)}$ $(\ce{[Ar]}3\mathrm d^7)$ has plenty of room in the $3\mathrm d$ subshell to support six-coordination.

$^1$ Bérces et al. J Phys Chem A 103: 9693-9701 (1999), doi: 10.1021/jp992612f
$^2$ Bjerrum, J. "Metal Ammine Formation in Aqueous Solution." Haase and son, 1941. (link)

In this complex you're going to see an arrangement with 4 equatorial and 2 axial ligand positions.  In such compounds, ligands that are stronger σ-donors and weaker π-acceptors preferentially occupy the axial positions.  On the other hand, weaker σ-donors and stronger π-acceptors preferentially occupy equatorial positions.  A quick look at the spectrochemical series (which ranks ligands by increasing field splitting effects which have been measured experimentally, which is the same as ranking them by increasing π-acceptor strength) shows that $\ce{NH3}$ is a stronger π-acceptor than both $\ce{H2O}$ and $\ce{HO}$, which means that it will preferentially occupy equatorial positions while the $\ce{HO/H2O}$ preferentially occupies the axial positions.  This is because the equatorial plane is electron-rich, which is a more stable environment for a stronger π-acceptor.  Conversely, the axial positions are electron-deficient, and thus they attract the stronger σ-donors.
The reason you only see 4 substitutions is because these substitutions are reversible.  It is energetically favorable for a stronger π-acceptor to displace equatorial ligands, but it's not energetically favorable for a weaker σ-donor to displace an axially-substituted ligand.
You can sort of see this in the structures of the ligands.  $\ce{NH3}$ has one lone pair, $\ce{H2O}$ has 2 lone pairs, and $\ce{HO-}$ has 3 lone pairs.  σ-bonding interactions involve free lone pairs on the ligands, so in this case you can see that more lone pairs means a greater ability to σ-bond.  Now look up how the spectrochemical series ranks these three ligands in terms of increasing σ-donor/decreasing π-acceptor strength.

I've never heard of such a theory before and would be very interested to see a book that explains it.

It's a direct consequence of MO theory, any graduate-level text covering coordination chemistry should do the trick.

Do you have a particular book in mind? Or a website, somewhere out there? I have genuinely never heard of such a thing and it doesn't make sense to me when considering that in an octahedral geometry, all six positions are the same, so there is no such thing as "equatorial" or "axial". If you're talking specifically about the Jahn–Teller distorted geometry of Cu(II), then that's not an issue, but the text seems to imply that it's generally true for everything. Lastly, I've never seen NH3, H2O, or OH− being referred to as π acceptors.

You could also explain the phenomenon based on σ-donor strength if that's a more comfortable way to think about it.  It's probably the better way to think about it since the three are σ-donors, perhaps I was generalizing where I shouldn't.  You'll still come to the same conclusion as to which bond where.  Another term you want to look up is Ligand Field Theory, that will give you resources.

Honestly this is like 4 years old and I have spent those years studying medicinal organic chemistry, I'm really rusty on MO theory and LF theory.  Good luck!

From the question:

To me, it seems that this partial substitution wouldn't be favourable, as it produces a slightly distorted octahedral shape, which is less stable than a regular octahedral shape, surely? (Emphasis mine)

This is a misconception. The shape of the hexacoordinated copper(II) complexes is independent of the type of ligands. Indeed, the hexaaquacomplex $\ce{[Cu(H2O)6]^2+}$ already features the same distorted shape. This distortion is termed Jahn-Teller distortion. The distortion actually results in the distorted complex being more stable than a regular octahedron. In general, you can (and should) assume that the shape a complex assumes (in solution and in crystal structures) is the most stable or at least more stable than other easily accessible variations.
Ammonia, in general, does give a stronger ligand-metal interaction with transition metals than water as a ligand. Therefore, there should be a general trend towards amination to give more stable complexes. However, in the case of axial ligands in a Jahn-Teller distorted complex it is actually slightly better to have a slightly weaker ligand there. The hexaammin complexes can be attained in liquid ammonia but when dissolved in water two ammin ligands redissociate and are replaced by aqua ligands.


text:
Placing the atomic mass and number of an element...
Why are the atomic and the mass numbers of an element flipped in the textbooks relative to the periodic table?

what do you mean by flipped here

Welcome to chemistry.SE! This is a bit unclear. Do you mean why is the info about A and Z about every element is included in "books about periodic table"? Also, it would help if you clarify what books you exactly mean.

In the periodic table, the largest number is indeed the atomic number. This is the number of protons. Since this defines the element, it is considered the most important number.
What you probably mean is
$^A_Z\mathrm{Element}$. Here, by convention the lower $Z$ (from the German word Zahl is the atomic number, while $A$ is the sum of protons and neutrons. This defines a particular isotope. This notation is often used in the context of NMR spectroscopy ($^{13}\ce{C}$ NMR), or when radioactive decays are described:
\[\ce{^{14}_6C -> ^{14}_7N} + e^- + \bar\nu_e\]
In summary, I think this is just a set of different conventions which show that in differents contexts either the element itself or a particular isotope and its properties are considered more important.


text:
How are Pauling's rules applied to Nitric acid?
The Lewis structure of nitric acid is:

So, since Pauling's rule is $\mathrm{p}K_\mathrm{a}=8-5p$, where $p$ is the number of oxo groups, how is the rule applied in this instance? $p=1.5$? $p=2$?

Have you tried looking at its $\mathrm{pK_a}$? What about comparing it with the $\mathrm{pK_a}$ of its less oxygenated congener nitrous acid, $\ce{HNO2}$?

Both oxygens count as oxo - it doesn't matter how they are bonded but that they stabilise charge of anion counts.


text:
Potentiometric titration using a pH meter
Why is the pH of a buffer solution measured before measuring the pH of an unknown solution?
I do not have any idea about it. I had read a chapter about ionic equilibrium and electrochemistry, but have not been able to find out an answer.

In principle, you should know the pH of the buffer solution. Measuring it before an unknown sample is a straightforward way of calibrating the detector (pH detectors can get a bit funny after a while).
If you put the detector into a pH 5.5 solution and it reads 7, you're going to have to tweak the detector (if possible) or find another.
Measuring unknown samples with a detector that incorrectly measures a known buffer solution is a bad idea but an even worse idea would be to measure unknown samples without knowing your detector was incorrect in the first place.


text:
Aromatic ether to phenol reagents
I have been asked to identify suitable reagents for the following transformation:

(Apologies it's back to front - the question is a map of reactions and I took a snapshot).
I really have no idea about this one. Don't even know where to begin, sorry. For the purposes of the question I only need reagents but I would appreciate a mechanism if you know it. Thanks.

The answers below are very good. If you have access to a book on protecting groups in organic synthesis (Greene and Wuts is a popular one), you'll find an extensive list of methods that have been used for demethylating a phenol.

The reaction is demethylation of methoxybenzene. You can do it by using an acid which has a nucleophilic counter ion.

It is not actually obvious why the iodine nucleophile attacks the methyl group and not inside benzene ring. The other option would be:

This reaction doesn't happen because otherwise the iodine ion would appear inside the benzene ring and resulting product would be in a very strange conformation (high energy).
More information:
 Clayden, Jonathan; Greeves, Nick; Warren, Stuart; Wothers, Peter (2001). Organic Chemistry (1st ed.). Oxford University Press. p. 434, 589.
As it was correctly stated in the other answer there are also other reagents which can be used to cleave the methyl ether. $\ce{BBr3}$ is a Lewis acid which also can be used in this reaction. A relatively new paper describes a computationally calculated bimolecular mechanism for this reaction.

I guess there are two reasons, why the iodine is not attacking from inside the ring. The first should be, because it is much too big to even fit in there. SN2 is just not possible, because that requires the donation in the antibonding orbital of the bond to be cleaved. The antibonding orbital corresponding to the phenyl oxygen bond is way higher in energy than the LUMO of the aromatic system and even higher than the CH bonds (this is an educated guess though). And also don't forget there is partial double bond character between the phenyl and the oxygen. Btw.: Excellent answer!

Boron tribromide is a standard reagent for cleaving aryl methyl ethers to the corresponding phenol. It is a general method and can be used in aromatic methyl ethers with sensitive substituents where acid or base cannot be tolerated.
The above Wiki link also shows the following mechanism
1) complex or adduct formation
$$\ce{ArOCH3 + BBr3 -> ArO+(^{−}BBr3)CH3}$$
2) bimolecular reaction of 2 adducts
$$\ce{2 ArO^{+}(^{−}BBr3)CH3 -> 2 ArO(BBr2) + 2CH3Br }$$
3) hydrolysis
$$\ce{ArOBBr2 + 3H2O → ArOH + B(OH)3 + 2HBr}$$


text:
How to calculate the pH of a buffer after dilution

You have a beaker containing $\pu{542 mL}$ of a solution of $\pu{0.48 M}$ acetate buffer ($\mathrm{p}K_\mathrm{a} = 4.76$) at $\ce{pH 4.2}$. To this solution, you add $\pu{342 mL}$ of pure water. What is the $\ce{pH}$ of the resulting solution?

The answer is $4.6$. However, I do not know how to arrive at that answer. I know the H-H equation must be used:
$$\ce{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$

Adding water means diluting the buffer.
Use the known concentration, the volume of the buffer and the volume of water added to calculate the new concentrations.

I went down that path. However, the question just gave the concentration of the buffer. That means ([A-] + [HA] = 0.48M). In order to calculate the new pH, I'd need the concentration of the weak acid and its conjugate base. I was thinking of solving for its ratio using H-H, but I am not sure if I am on the right path.

@Michael That sounds reasonable, since you have the initial pH and the pKa :-)

Calculate the number of moles in the original buffer solution - you should get around 0.26 (that is using ur n=cv equation), now add up the new volume to the old one which would give u 0.884L then use your other mole number you got earlier to use in the n=cv equation to get new concentration, it should give around 0.294. No use pH=pKa+log([A-]/[HA]) equation bearing in mind that you just calculated the new concentration of ([A-]/[HA]) so just substitute it with the 0.294 and u should get the new pH to be 4.22 or the 4.2.


text:
Osmolarity and osmotic pressure

You have discovered a new strain of bacteria called Tankeromyces subtilis that has a special outer membrane that is capable of withstanding high levels of osmotic pressure. These bacteria can survive osmotic pressure as high $\pu{1.3 kJ L-1}$, but not higher. The $(iC)_\mathrm{total}$ value for the inside of these bacteria is $\pu{630 mM}$.
What is the lowest concentration of $\ce{NaCl}$ (in otherwise pure water) in which these bacteria can survive without bursting due to osmotic pressure?
$R = \pu{8.3153 J mol-1 K-1}$ and $T = \pu{296 K}$.

The answer is $\pu{0.10 M}$, but I have no idea how to drive at that answer. However, I do know that it involves this equation:
$$p = iCRT$$

The units of osmotic pressure given may be confusing you.  kJ/L is a pressure unit becaues a Joule is equal to a Pascal-cubic meter.  $\textrm{J}=\textrm{Pa} \cdot \textrm{m}^3$  Does that give you enough to get started?

Ohhh. I didn't even think of it that way. Yes that was very helpful in getting started.

The highest pressure they can handle is $\pu{1.3 kJ L-1}$, or $\pu{13000 J L-1}$. In the formula you use the ideal gas constant $0.08206$ and the temperature given to find the concentration the maximum pressure it can handle.
This comes out at $\pu{0.53 M}$ or $\pu{530 mM}$ and that's the maximum concentration it can handle. To find the minimum, you subtract $530$ from $630$ giving you $\pu{100 mM}$ or $\pu{0.1 M}$, which is the lowest concentration it can handle.

I find this rather confusing to follow; probably you should've included the formula for osmotic pressure and why exactly you stick with these units.