text:
How to obtain the Raman spectrum along every coordinate of a scan in Gaussian?
I am doing a scan calculation using Gaussian09 and adding the Freq=Raman keyword to my input file. My objective is to obtain the Raman spectrum of each conformation in my compound. However, only the Raman spectrum of the last conformation in the scan is calculated. How can I solve this problem?
I understand that I will get negative frequencies, but these will be in a certain frequency range, so I can observe the variation in the signals.
I need to do these calculations, because I want to see the type of conformation that my compound adopts under heating.

Assuming that you are changing the geometry in between calculations, you may get negative frequencies because you may well pass over saddle points. Also, Raman intensities are very expensive, IIRC, and even standard frequency calculations are way more expensive than the geometry sweep I suspect you wish to do. You are much better off picking a few structures along the way and running them separately.

Why do you want perform this kind of computation? Normally, you would only be interested in the spectroscopic properties of ground states.

No, there is no automation for this, and there should not be.
Using compound jobs is unfortunately very common among computational chemists, but they should always be viewed critically. If you are using any kind of IOp, then the probability that the second part of your job is doing what you want is low. Additionally, you do not really have any control about what happens if something happens to fail. In such cases, you also give up control over the validity of your checkpoint-file, which you might want to reuse.
Another issue is that these are basically two jobs running one after another and in most cases need different memory requirements, are more effective with another setup. It's the dream of every administrator (and co-worker) when you block resources you don't need for an extended period of time just to avoid a little more waiting.
Lastly, with the introduction of %oldchk= in in revision D.01, it is completely superfluous. Every well organised queuing system should be able to run jobs one after the other when specified (i.e. let one job wait for another to finish).
Only run compound jobs if you are 100% sure they cannot fail (and even then it's questionable).
From a theoretical standpoint, you should consider how much sense your approach makes. As TAR86 pointed out, Raman intensities are expensive. Additionally frequency calculations are only meaningful at well converged local minima structures. You are employing the harmonic oscillator approximation, which will break down if you are not using a local minimum (or at least a stationary point).
If you really need this approach, then you need to do it manually by extracting the geometries (and possibly the MO) and perform the frequency job on top of it.

Afaik it's not possible to do this directly since Gaussian does only allow geometry optimization during scans.
The definitely easiest way to do this, but not the cheapest, would be running your scan, extracting the *.xyz coordinates of every (optimized) structure and then running the freq calculations separately. Extracting the *.xyz structures can be done for example with openbabel, cclib or chemcraft. Producing Gaussian inputs can be done by a simple python script and writing link0 and link1 information followed by the coordinates from the *.xyz files.
Another way to do this, which would be cheaper but much more complicated to set up, is to produce the input files for your different scan types right away and run them separately including the freq. How to do this depends a lot on your type of scan.


text:
How are the angle bending parameters defined in the MM2 force field?
I am currently looking through the openbabel implementation of the mm2 forcefield. While I understand most of the implementation, I don't get the definition of the angle bending parameters (see https://github.com/openbabel/openbabel/blob/master/data/mm2.prm), which have the following form:
angle         1    1    1     0.450    109.470    109.510    109.500
angle         1    1    2     0.450    109.470    109.510    109.500
angle         1    1    3     0.450    107.800    109.900    110.000
angle         1    1    4     0.450    109.470    112.400    109.000
angle         1    1    5     0.360    109.390    109.410    110.000
angle         1    1    6     0.700    107.500    107.700    107.400
angle         1    1    8     0.570    109.470    108.800    109.500
angle         1    1    9     0.500    109.280    110.780    109.280
angle         1    1   11     0.650    109.500    107.500    109.500
angle         1    1   12     0.560    108.200    0.000      0.000

So they are of the form
<1.-atom-type> <2.atom-type> <3.atom-type> <force-constant> <1.angle> <2.angle> <3.angle>

However, I understand the angle bending to work like this:

There is an ideal bond angle $\theta _0$, and if the actual bond angle $\theta$ differs from $\theta _0$, this results in an increase in energy, e.g. of the form $\Delta E = (\theta _0 - \theta)^2$. Therefore, given an angle with three atom types a,b,c there should be only one angle $\theta _0^{a,b,c}$. So why are there three angles given for every combination?

After some digging around I found an explanation somewhere, though I am not sure if it is correct: The first angle seems to refer to $-XR_2-$, i.e. an atom with two additional alkyl chains, the second angle to $-XRH-$, i.e. an atom bonded to alkyl and H substituents and the third angle to $-XH_2-$.

And this of course makes sense, as hydrogen atoms are sterically less demanding, thus affecting the optimal angle.

We have decided against program specific tags.

@Martin-マーチン Good to know for the future, but IMHO we should make an exception for openbabel ; )

@deusexmachina - there's an Open Babel forum outside of Chem.SE... :-)

@GeoffHutchison Yeah, I know, now : ) Guess it is also of general interest for other MM2 implementations, though.

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sort: votes


text:
How do you tell whether fluorine goes to axial or equatorial position in iodine dioxide difluoride?
I need to find the $\ce{F-I-F}$ bond angle of $\ce{[IO2F2]^-}$. But the thing is I can put 2 fluorines in the equatorial  or axial positions so I get 3 different answers. I'm guessing they're all three different molecules with slightly different properties and I have to pick out the most stable one. Looking it up on the internet I find the two in axial position which I guess is because the two fluorines pull charge towards themselves and repel the other fluorine so they want to stay far away from each other, am it right?

Related: Why does F replace the axial bond in PCl5? (please note also Martin's comments on the question.)

First, you would need to find out what the bonding situation is at all; i.e. how many bonds of which type will be formed. You can use a scheme like the one I outlined in this answer. This should give you three bonding electron pairs and 14 lone pairs. Since we have more atom connections to make than we have bonding electron pairs, we cannot consider only 2-electron-2-centre bonds as would be the case e.g. in iodate. We must utilise one 4-electron-3-centre bond to be formed from one of the lone pairs and one of the bonding electron pairs. Read more about this type of bond for example in this well-written answer by Ron.
As we are deaing with a single 4e3c bond, iodine will need to be hybridised as $\mathrm{sp^2}$ or less. This means we have one direction (axial) with a $\mathrm p$ type bonding orbital and one (equatorial) with something like $\mathrm{sp^2}$ or similar. We can now use Bent’s rule to deduce that fluorine should occupy the axial positions as it is the more electronegative substituent.
We could also have arrived there by realising that oxygen hardly ever takes part in 4e3c bonds. These bonds are basically the incorporation of a central atom into an existing $\ce{X-X}$ bond on a formal basis. Oxygen, however, needs two electrons per atom to reach an octet meaning that there needs to  be another additional bond. This cannot be the case in simple compounds such as $\ce{IO2F2-}$.

Im sorry but I didn't quite get what 4ec3 is, I didn't find the Wikipedia link helpful and I'm not able to find anything much on google. I read somewhere that it's 4 electrons shared by 3 atoms which I don't understand how it's applied here

@Vrisk Did you also read ron’s answer linked in mine? I think he explains it rather well.

okay I read it again -  P has 5 valence electrons 2 in s and 3 in p, you form 3 sp2 orbitals and you'll have 2 electrons filling that one p orbital then what? How do two chlorines combine with 1 p orbital? How did he get that MO diagram?

@Vrisk The three atoms (in Ron’s case chlorine, phosphorus and chlorine or in this case fluorine, iodine and fluorine) align in a linear fashion. Each contributes a p-type orbital to the bonding situation. Since we have three (atomic) orbitals we are mixing, we must end up with three (molecular) at the end. It is obvious that the highest and lowest must be fully antibonding and fully bonding, respectively. The central orbital must be something different so one in which the two outer orbitals point the same phase to the central atom.

Since there is no way a p-type orbital can interact in a symmetry-allowed manner towards both sides, there is no contribution by the central atom to the orbital in the middle. However, due to the (weak) bonding interaction of the two terminal aoms, this is a (weakly) bonding orbital. Therefore, we have more bonding electrons than antibonding ones and thus bond formation is favourable.

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text:
Practical differences between storing 2-electron integrals and calculating them as needed?
Some texts that I read suggest calculating all of the 2-electron integrals at once (taking into account permutational symmetry) and some suggest calculating them on demand when building the Fock matrix.
Which of these is the most computationally efficient, and when should one pick a certain approach over another?

It depends on system size. There is a crossover point where it becomes faster to calculate the integrals as needed because as a system becomes larger and larger, the number of disk operations required to store the integrals becomes very large which greatly slows down the calculation. This is because disk operations are relatively slow compared to memory manipulation.
Thus, most software packages have an option to calculate the integrals directly and not store them. In Molpro I believe this command is gdirect.
To be more specific, there are two aspects of the storage which are slow when there are a large number of integrals. One is that disk operations themselves are slower than just doing the calculation which is only memory manipulation. The other is that the integrals are stored in some fancy way so as to minimize the amount of time it will take to find the specific integral you are looking for, but when there are a very large number of integrals, some non-negligible amount of time will go into finding the value of the integral. When these two components (plus maybe others I am missing) are larger than the time required to just do the integral calculation, then it is faster to compute the integrals directly.
I'll let someone else be more quantitative with the timing of disk operations and the lookup tables as these are more of computer science questions. But quite interesting ones.
For a useful reference on how many integrals we are talking about storing on disk, see this question.

Also as a practical guide, in my experience this crossover point was around 14 hevay atoms with a triple-zeta basis set, but this means almost nothing because this will be heavily dependent on the computing architecture you are using. I would expect the crossover point would be much higher for most supercomputing systems as I did not have much disk to work with comparatively.

Several factors go into deciding:

calculation type: for most post-SCF calculation, disk-based storage is indispensable
storage speed (solid-state disk?) vs. calculation speed (GPU?)
predictability of which integrals are needed. Screening by density matrix value makes some integrals irrelevant for SCF, but for post-SCF, this is different.
core memory vs. amount of disk memory allotted

In fact, many programs actually do not decide, but do both. ORCA, for instance, allows for a hybrid approach, trying to store integrals on disk in the order that they are eventually needed. ORCA and Turbomole have algorithms to decide on a category-basis, essentially going by the total angular momentum because the recursive calculation makes high-angular-momentum integrals expensive. Some integrals may only be needed every few iterations, when doing a full recalculation of the Fock matrix instead of $\mathbf{F}_{n+1} = \mathbf{F}_{n} + \Delta \mathbf{F}$, others every iteration, which may lead to amending the storage.
Note that calculating all integrals and storing them is sometimes called the "conventional" way, whereas calculation on-the-fly is often called "direct".


text:
Looking for a pressure phase diagram of selenium at large ranges of pressure and temperature
I'm doing undergraduate research and our group is having a lot of difficulty finding a comprehensive pressure phase diagram or phase table for selenium (including most known packing geometries that are known at extremely high pressures).
I figured I may ask here as we've spend a while searching and haven't turned up so much as a good lead for where this information may be documented. Furthermore, we'd like to know where to look from now on to find information like this (such as a similar chart for sulfur? Albeit we haven't spent as much time looking for this as it's not of our particular research interest).

Young, D. A. Phase Diagrams of the Elements; University of California Press, 1991; 2. Olin, A., Mompean, F. J., Eds.; Chemical Thermodynamics of Selenium; Elsevier: Amsterdam, 2005.

This paper is a computational study that computes the phase diagram of selenium and sulfur. They reference an experimental paper that looks at this, but you might not be able to access that one.


text:
Why is crystallography still used in some cases for small molecule structure determination?
I was wondering for some time why in some cases x-ray crystallography and not NMR is used for structure determination of small molecules? Does crystallization maybe require a smaller amount of the compound than NMR?

Because there are guys who sit closer to an X-ray machine than to NMR machine.

One gets different data from each of these methods.

Chirality can sometimes be determined by NMR, but more often by x-ray. Though it depends a bit on the nucleus you measure by NMR, x-ray will usually take a lot more substance.

Solid state NMR is a tricky business. If you already have a single crystal, that's an easy choice.

TL;DR: When both methods are applicable for structure determination, SC-XRD eats NMR for breakfast; collected data/refined parameters ratio in case of SC-XRD is simply way higher.

Generally nmr gives topology, H next to C next to C etc, whereas x-ray crystallography gives the structure, bond lengths & angles. However, it is generally far harder to crystallise than to dissolve. Also you may not need  bond lengths & angles but just to know what your molecule is.

Our group uses both methods a lot so here are some examples why you would use X-ray, in addition to NMR, in organic synthesis.

Your compound isn't soluble enough:
Colleagues produce very large aromatic systems for organic electronics which are basically insoluble in all solvents. With a good NMR machine (600MHz and up) it might still be enough to get spectra, but full structure elucidation might be difficult. Since the compounds crystallize well it's quite easy to get a X-ray structure.

Low number of carbon and hydrogen:
I've produced compounds with a C:H:N ratio of 2:3:5. The 1H is two singlets, the 13C is two signals. COSY, HSQC, HMBC,... won't tell you much either and it's much easier to use the nice crystals and get a structure that way than trying some correlated NMR experiments including nitrogen.

3-D orientation/stereochemistry:
Imagine this compound which is quite similar to something we produced:

Since you got two quaternary centers here it's really hard to tell the configuration on those centers relative to each other using NMR. There's no coupling to determine the stereochemistry and a NOE experiment isn't very conclusive here. With X-Ray it's no problem to find out.
Actually, these problems are definitely the ones we use X-ray the most.

Absolute Configuration:
Theoretically it's also possible to determine absolute configuration with single crystal X-ray diffraction, while it's extremely hard to do, if not impossible, with NMR.

Summary:
Single crystal X-ray diffraction gives you the configuration of your molecule, including relative configuration of stereocenters and possibly even absolute configuration, right away. If you only care about configuration (and not behaviour in solution) X-ray is definitely superior to NMR, there's basically no question on how your molecule looks.
On the other hand you need crystals and it's much more complicated to perform than NMR, so this definitely stays the number 1 method for synthetic chemists.

Coming from natural product chemistry, of course the structure elucidation by NMR is the most commonly used method, especially in isolation. NMR requires only a little substance dissolved in whichever deuterated solvent you have on your shelf and is rapidly set up. The only downside is that for very small amounts of sample you will be blocking the strongest NMR your institute has for extended amounts of time to record all the required experiments. X-ray structures on the other hand requires crystals, and those isolating natural products don’t want to trial and error forever hoping for them.
The strongest disadvantage of NMR is that there is no direct way to determine absolute stereochemistry and only a limited number of methods to determine relative stereochemistry. Often, Mosher ester experiments will be performed but they do consume a non-neglegible amount of substance that could otherwise be used for bioactivity tests etc. If the natural products contains carbohydrate side chains, figuring out which one will be immensely difficult by NMR alone.
On the (total) synthetic side of natural products chemistry, each intermediate is of course subjected to an NMR structure analysis. This is okay because stereochemistry is typically only built up in a few steps with known stereochemical outcome and an optical activity is often taken as a good enough indicator that the chirality turned out as expected. Most reactions won’t touch distal chiral centres and if they do one will arrive at a mixture of diastereomers which can easily be detected by NMR. So by taking this route, one can also confirm to a certain extent the structure of the synthesised natural product.
However, NMR is often ambiguous and sometimes error-prone. In my previous group and all through the literature I have often witnessed structure corrections following total synthesis. Maybe the original NMR analysis of the natural product gave a wrong diastereomer, maybe the enantiomer was incorrect and sometimes not even the connectivity was as proposed. Thus, one must always take proposed structures with a grain of salt until the structure is proven.
The only thing able to definitely prove a structure is an X-ray crystal structure. You get the positions of atoms and can deduce the connectivity. There is no margin for error if you performed it correctly. Most notably, you get both the absolute and relative stereochemistry for all stereocentres. The only question remaining is whether a certain atom is carbon, nitrogen or oxygen, but even that can be refined well enough. Finally, if you have good enough crystals to collect the scattering at very large angles, you can even locate the positions of the hydrogen atoms.
Typically, all structure elucidation is tentative until there has been a crystal structure. This goes for both isolation and synthesis.

The point in the fourth paragraph bears repeating -- there have been more than a few structures determined by NMR that later turned out to be completely incorrect. Whereas if you do manage to get a good X-ray structure of a small molecule, the chance you're wrong about it plunges almost to nil.

@R.M. of a small organic molecule. There have been a number of inorganic molecules with structures determined by XRD that have been proven incorrect. As I understand it, that is getting rarer these days, but could still happen.


text:
How to calculate the probability of a particular value of an observable being measured

A particle on a ring is prepared with a wave function equal to $\frac{1}{\sqrt{\pi}}$ between $\theta=0$ and $\theta=\pi$, and $0$ between $\theta=\pi$ and $\theta=2\pi$. If a measurement of the angular momentum is made, calculate the probability of finding a particular value $l\hbar$. [The angular momentum eigenfunctions are $|l>=(2\pi)^{-1/2}e^{il\theta}]$

I've been reading the Oxford Chemistry Primer on Quantum Mechanics and they don't provide solutions. I've been struggling with this question.
I think I need to calculate the coefficient of expansion for a particular basis eigenfunction, but I'm unsure how to proceed.

In the future, write out your question rather than including a picture of the text. That way, people can find this particular question from a search within the site or from Google.

Related: https://en.m.wikipedia.org/wiki/Semicircular_potential_well

$$
\newcommand{\ket}[1]{\,\lvert{#1}\rangle}
\newcommand{\bra}[1]{\langle{#1}\rvert\,}
\newcommand{\braket}[2]{\langle{#1}\vert{#2}\rangle}
\newcommand{\bracket}[3]{\langle{#1}\vert{#2}\vert{#3}\rangle}
\newcommand{\op}[1]{\hat{#1}}
$$
Solution should be a pretty straightforward application of the Born rule.

In this particular case, we have a self-adjoint operator $\op{L_z}$ with eigenfunctions and eigenvalues satisfying
$$
  \op{L_z} \psi_l(\theta) = l \hbar \psi_l(\theta) \, ,
  \quad \text{where} \quad
  \psi_l(\theta) = \frac{1}{\sqrt{2 \pi}} \mathrm{e}^{\mathrm{i} l \theta}
  \quad \text{and} \quad
  l = 0, \pm 1, \pm 2, \dotsc
$$
such that any wave function $\psi$ can be expanded over orthonormal basis $\{ \psi_l(\theta) \}_{l=-\infty}^{\infty}$ of the corresponding eigenfunctions of $\op{L_z}$ as follows,
$$
  \psi(\theta)
  =
  \sum\limits_{l=-\infty}^{\infty} c_l \psi_l(\theta) \, ,
  \quad \text{where} \quad
  c_l
  =
  \braket{\psi_l}{\psi}
  =
  \int\limits_{0}^{2 \pi} \psi_l^*(\theta) \psi(\theta) \mathrm{d} \theta \, .
$$
And a measurement on a system in arbitrary state $\psi$ can yield any of eigenvalues $l \hbar$ with the probability given as follows
$$
  \Pr(l \hbar) = |c_l|^{2} = c_l^* c_l \, .
$$

Now, since $\psi(\theta)$ is defined to be
$$
  \psi(\theta)
  =
  \begin{cases}
  \frac{1}{\sqrt{\pi}} & \text{if} \quad \theta \in [0, \pi] \\
  0            & \text{if} \quad \theta \in [\pi, 2 \pi]
  \end{cases} \, ,
$$
the coefficient $c_l$ is equal to
$$
  c_l
  =
  \int\limits_{0}^{2 \pi} \psi_l^*(\theta) \psi(\theta) \mathrm{d} \theta
  =
  \int\limits_{0}^{\pi} \psi_l^*(\theta) 1/\sqrt{\pi} \mathrm{d} \theta
  +
  \int\limits_{\pi}^{2 \pi} \psi_l^*(\theta) 0 \mathrm{d} \theta \, ,
$$
where the second term trivially vanishes leading to
$$
  c_l
  =
  \int\limits_{0}^{\pi}
  \frac{1}{\sqrt{2 \pi}} \mathrm{e}^{-\mathrm{i} l \theta} \frac{1}{\sqrt{\pi}}
  \mathrm{d} \theta
  =
  \int\limits_{0}^{\pi}
  \frac{1}{\sqrt{2} \pi} \mathrm{e}^{-\mathrm{i} l \theta}
  \mathrm{d} \theta \, .
$$
And then some simple algebra leads to the final answer, which I leave to the OP, since it is far into the night for me and I'm afraid I will make a lot of stupid mistakes.

Mister Wildcat, welcome back! We missed you. ;D


text:
How does the phosphate ion look like?
I am trying to understand how a phosphate ion looks like using the VSEPR theory.
The phosphorus element has an atomic number of 15, and therefore it has a electron configuration as follow by the Aufbau principle.
$$\pu{1s2 2s2 2p6 3s2 3p3}$$
It needs to have one π-bond with one oxygen atoms and 4 σ-bonds with all of the 4 oxygen atoms, so it is hybridizing 4 orbitals to form some hybridized orbital and leaving one p orbitals around for the pi bond? How does that work?

As a side note: there is no double bond in phosphate, only four $\ce{P-O}$ single bonds. Each oxygen carries a formal negative charge while phosphorus carries a formal positive charge. This leaves us with phosphorus being $\mathrm{sp^3}$ hybridised as (hopefully) expected.

@Jan - this is not a duplicate. That post you mentioned is about counting electrons, mine is about molecular shape.

I am confused, Wikipedia says phosphate has a P double bond O.
https://en.wikipedia.org/wiki/Phosphate
Are you saying Wikipedia is wrong? It could be ...

Wikipedia is not wrong, it is just... a simplified representation, perhaps? It is tedious to draw the correct thing every time, so sometimes we fall back on simplified representations; as long as we know what it really means, then it is fine.

I marked as a dupe because once you get the bond counting right (you only need four single bonds) the structure explains itself and there is no need to look for a hypothetical orbital to make a hypothetical fifth bond.

It is kind of sad that the solution is removed. It was wrong, but a lot of discussion associated with it was useful.
I realize now that VSPER is an outdated theory and it cannot describe phosphate, what is a better one?


text:
Direction of flow of current in electrolytic cell
In an electrolytic cell, does current flow from

Anode to Cathode

or

Cathode to Anode

I've seen various websites, including this and this, but they seem to contradict each other.

The cathode is defined as the electrode at which reduction happens. The anode is the electrode at which you oxidise. This is always true. I remember it by saying ‘anodic oxidation is the alpha and omega.’
In a galvanic cell, the anode will produce ‘electron pressure’: the compounds being oxidised leave there electrons there until the electron density is too high. These electrons flow through the circuit from the anode to the cathode and are consumed in reductions on the other end.
In an electrolytic cell, the battery creates an ‘electron pull’ from its positive pole. This pole is connected to the anode and therefore electrons are pulled away from the anode into the battery. On the cathodic side, the battery produces an electron pressure again to drive the reduction.
So in both cases electrons flow from the anode to the cathode.
(This analysis is complicated by something I learnt in school as the technical current flow definition. According to Bavarian textbooks, technicians defined current as flowing from plus to minus which is exactly the opposite of the flow of electrons. In that case (only), the direction of current is reversed but due to a different definition.)

so in this case, the direction of current and electrons is the same ?


text:
Ionisation Energy vs First Ionisation Energy
It was stated in a textbook that the first ionisation energy is

the minimum energy needed to remove one mole of electrons from one mole of gaseous atoms in their ground state

Is there a difference between ionisation energy and first ionisation energy?

There is a first, second, third, fourth, … ionisation energy. Does that answer your question?

@Jan not really. http://www.chemguide.co.uk/atoms/properties/moreies.html explains the successive ionisation energies with the equations, but ionisation energy isn't explained.

Well, ionisation energy is the general term and first ionisation energy specifies. As you would expect from language.

https://en.wikipedia.org/wiki/Ionization_energy

There is no difference between "ionization energy" and "first ionization energy". Another definition from Wikipedia:

The ionization energy (IE) is qualitatively defined as the amount of energy required to remove the most loosely bound electron, the valence electron, of an isolated gaseous atom to form a cation.

Here it's more clear that they are the same, since "the most loosely bound electron" is a single electron, and each successive ionization only removes a single electron at a time.
Additionally, if one were to remove all electrons from an atom, it would be explicitly stated as "complete ionization".

From both your and my definitions, I can see a difference. Can I confirm that the difference is that ionisation energy removes electrons from ONE atom, while the first, second third... ionisation energy removes electrons from ONE MOLE of atoms?

No. The definition of ionization energy is $\ce{X + \text{energy} -> X+ + e-}$, with nothing about atoms or moles. It's all about the units, which work out to be per mole. Units are consistent between all successive ionization energies.


text:
Scan type of DFT calculations adding Freq=Raman keyword
I´m doing a scan calculation by adding Freq=Raman keyword. My objective is to obtain the Raman spectrum of each conformation in my compound. However, when i write the Freq=Raman Keyword, only the Raman spectrum of the last conformation in the scan is calculated . How can i solve this problem?
Thanks!


text:
Citric Acid from Aspergillus Niger
Can anyone point me towards an experiment guideline of some sort regarding fermentation of citric acid from Aspergillus Niger? I haven't been able to find one by myself. I've found plenty of research papers, but none with any sort of guideline.

Not that this isn't Chemistry...but this is (probably) better off on the Biology.SE [ https://biology.stackexchange.com/ ]. You're asking for an experimental guideline, which is within the scope of the [organic-chemistry] tag...just letting you know there's another site for Biology-ish questions ;)


text:
How should a triple bond be depicted in skeletal structures?

Draw the skeletal structure for the following compound:
  $$\ce{HOCH2CH2OCH2C#CCH2CH(CH3)2}$$

I can't figure this out.
I can draw it up until the triple bond. My professor marked off because the triple bond was not drawn correctly.
So, is the triple bond depicted any differently from the other bonds in the molecule? If so, why?

My guess, is that you forgot to take the linear disposition of the carbon-carbon triple bonds into consideration.
So the bond lines cutting through four carbon atoms will be linear, and not zig-zagged as in the rest of the molecule (the latter implies tetrahedral geometry accorded by the $sp^3$ carbon-carbon bonds)

The first diagram may feel more "graceful" (the term I used when I defended my incorrect depiction of the triple bond to my teacher when he asked me a similar question a while back), but it is incorrect.

A triple bond involving carbon should be drawn linearly. In general, if there are multiple consecutive atoms in a molecule which have a linear geometry, they should be drawn linearly. This is the case for the C–C≡C–C unit in an alkyne.
According to Graphical Representation Standards for Chemical Structure Diagrams (IUPAC Recommendations 2008):

Atoms that are isoelectronic with carbon and bearing two double bonds (or one single bond and one triple bond) should be drawn with the bonds separated by a 180° angle.

This includes carbon itself, of course. Some examples are given, including:

Note the 180° angles at the nitrile carbon (on the left), as well as both the carbon and nitrogen (on the right). Furthermore:

Large rings that contain triple bonds should depict the triple bonds in a linear fashion, just as they would be depicted in an acyclic diagram.

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text:
Looking for a Solvent for Fragrance Oils
Hi I am looking for a solvent for fragrance oils, that has low evaporation and high diffusivity (low molecular weight and hence spreads well in the air). The solvent should also be soluble in water and should have a low evaporation rate (I tried methanol, it diffuses well but it evaporates very fast, thus not serving the purpose)

Possibly related: https://chemistry.stackexchange.com/a/4927/194


text:
What gives saliva its adhesive properties?
The act of sealing off an envelope by licking the flap and "gluing" it down so that the envelope becomes closed rather tightly is probably rare nowadays, but remains well known to many people.
I was wondering which component of saliva gives it the ability to glue paper tightly enough so that the envelope is considered sealed off.
Here's the link for the Wikipedia entry of saliva, for composition.

Water-based adhesive envelopes are not completely a thing of the past - I still get them with bills. It is the water part to pay attention to - nothing required saliva, even if some of the nicer envelopes actually tasted pretty minty. A little sponge and water was fine, and actually quite useful if sealing a number of envelopes.

As @JonCuster states, the adhesive is on the lip of the envelope.


text:
Sodium metal in acetone
When I place a chunk of 99,99% pure sodium metal in a beaker of (100% chemically pure) acetone, the sodium metal starts reacting and dissolving into the acetone and it forms a yellow/brown liquid. When the acetone evaporates it leaves a dark-brown goo that is like a gel. And it smells like tar or something similar. What is this substance?

related https://chemistry.stackexchange.com/questions/6527/how-does-sodium-in-ethanol-reduce-carbonyl-compounds

Brown goo is the ultimate product of any organic reaction. Some reactions happen to produce something useful besides it, some don't.

I can suggest only two possibilities:

Your acetone or sodium are not pure. Acetone contains water, or something.
Sodium might react like magnesium does, though it required different conditions. That's possible, but maybe you just discovered a new reaction or had some weird impurities that served as a catalyst.

Self-condensation of the acetone catalysed by NaOH (formed from trace water) is also a possibility

I'm not sure if its concentration would be sufficient to cause this rapid condensation

It only requires a catalytic amount


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Is the flame temperature of a burning fuel affected by pre-heating the fuel?
This page on Adiabatic temperatures helpfully lists the theoretical maximum temperatures that flames will reach under certain conditions for a number of fuels. They assume initial atmospheric conditions of 1 bar and 20 °C.
However, I am remain confused about what would happen if the fuels (and air) were pre-heated to temperatures close to these maximums before they were mixed and ignited.
Would the flame temperature then be higher than these listed maxima?
Would the reaction be slowed down by the higher temperature? (I understand that this might happen for exothermic reaction in equilibrium with an endothermic one, but I don't think that describes burning fuel.)
Is there some other mechanism that prevents the flame temperature from exceeding these minima?

This question was inspired by discussion on a Skeptics.SE question similar to this closed Chemistry.SE question about jet fuel and steel beams. I think it is sufficiently focussed and apolitical to be on-topic here. My background level is an ancient undergraduate science degree; I have forgotten more chemistry than I remember.

I'm not really sure the concept of "pre-heating" makes sense... like if I have a block of charcoal and burn it starting at room temperature, then the claim is that it cannot go above some temperature T. But now let's say I were to cut it in half and burn each half in succession. Then technically the second half would've been in a "pre-heated" environment, and hence the max temperature would be... different? Even though it's equivalent to the first situation? Something doesn't make sense here. Either the initial temperature is irrelevant, or (probably) "adiabatic" doesn't give an upper bound.

@mehrdad Why don't you start a separate thread with your question(s)?

@ChesterMiller: I'd have to see what the answer to this question is first, it's somewhat of a duplicate...

I think that ruling out further reaction (for instance water dissociation, that has been mentioned) a simple energy balance tell me that the in-flame temperature will be higher if combustibles are pre-heated.

Yes and no: preheating a mix a little bit may increase the final temperature somewhat, but adding even more initial heat has a decreasing effect on the final temperature.
There are two competing factors that limit the final temperature in a closed system:

The total amount of energy as the sum of the initial amount of thermal energy and the chemical energy, divided by the thermal mass [mass times heat capacity] of the system, and
The tendency for a reaction to be forced backwards by applying additional energy.

For example, $\ce{2H2 + O2 -> 2H2O}$ + 286 kJ/mole, but water dissociates more as the temperature rises. At 2,200 °C water is about 3% dissociated. At 3,000 °C  it is about 50% split, so half the chemical enegy is unavailable.
BTW, thermodynamically, a heat engine produces power based on the difference in temperature, so preheating a fuel-air mixture actually decreases efficiency, though it might be needed to vaporize the fuel (e.g. in Otto-cycle engine in cold weather).

When you say "final temperature" do you mean "peak temperature"? Because I'd expect the final temperature to be room temperature...

@Mehrdad: "factors that limit the final temperature in a closed system" implies closed thermally as well as chemically. Of course, the final temperature in an open system decreases to that of the environment... in infinite time;-)

Okay, well then I'm completely confused. In a closed system, with a (presumably unlimited) supply of oxygen and fuel, why should there be a maximum flame temperature at all? The only reasons I can see are if other non-thermodynamic effects come into play (e.g. matter turns into plasma, or molecule is split before it can react, etc.). You mention something about reactions being forced backwards but I do not see how heating up an already-burning piece of wood would suddenly turn it back into unburnt wood. I must be misunderstanding a claim becuase my (layman) eyes just can't make sense of this.

A closed system has limited fuel, oxygen and no thermal flow. If any of those are not true, it is not closed.

Oops, yes, you're right. But then wouldn't it matter (...no pun intended) what the distribution and density of the fuel and oxygen are? How can they be completely irrelevant?

@Mehrdad: My understanding is that it wouldn't turn back to wood, but very hot CO2 and very hot H2O might produce combinations of H2 + O2, or C + O2 or (I'm guessing wildly) methane + O2, and maybe some longer chained hydrocarbons.

@Oddthinking: Interesting. So basically the argument is "there will always be some endothermic reaction that would occur at higher temperatures to absorb the excess heat"? That sounds like quite a theorem (or observation) -- "there is always an endothermic reaction that occurs at a temperature higher than any exothermic reaction, using the same chemicals". (a) This is completely unintuitive; why should it always be true? and (b) What if the reactants aren't mixed or distributed properly to allow this? (Which goes back to my previous comment: how can the matter distribution be irrelevant?)

@Mehrdad, That certainly isn't my theorem, but time to take this back to chat, I think.

@Oddthinking: I don't know how to take this to chat, there's no link here. Or do you mean the Skeptics chat? Either way though, the part I was asking about the distribution still needs an answer here I think...

@DrMoishePippik does the equilibrium explanation work if the hydrocarbon fire is thought of as an irreversible reaction? Is there a temperature where there is too much heat for the reaction to push more heat into?  Would the reaction stop happening then (leaving a bunch of unburnt fuel)?  For me this question was "why do fires have a maximum temperature?"

Since it's not irreversible, heat pushes the reaction backwards. Take the simple case of 2H2 + O2 <--> 2 H2O. Of course, burning wood will not return to wood, but carbon dioxide also splits as above.


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Has this way of comparing internal energy got any sense?
Can internal energy between molecules be compared?
For example, $\ce{H2}$, $\ce{O2}$ and $\ce{N2}$ have some internal energy ($U$) but taking infinite distance between atoms as zero. But each reference point is different. So, can $U$ for those molecules be compared?

Yes. This is the binding energy.
In general, the binding energy is defined as the amount of energy released upon bringing infinitely separated particles in to contact. This would then be a negative number. Another sign convention is the binding energy can be viewed as the amount of work required to separate two atoms or molecules (or whatever) which are in contact. This binding energy is the negative of the first one (is a positive number).
For molecules, you might be more familiar with the term bond energy instead of binding energy. For a diatomic, these two things are the same. Notice that when we talk about the bond energy in a molecule, we treat that bond as being a diatomic molecule.
For the diatomics you mentioned, the electronic energy for the heavier diatomics will be artificially larger than the lighter ones simply because there are more electrons. For this reason, when comparisons between systems like this are made, people usually divide by the total number of electrons. So, in this case you can visualize putting in positive work to bring the protons near to each other and then adding the electrons one at a time to the system until it has no net charge. The binding energy per electron is then a measure of the average amount of stabilization the system gained for each new electron. The same thing can be visualized for the nucleus of an atom, but in that context you will read of the binding energy per nucleon.
This same concept is also quite useful for molecular clusters and other non-covalently bound systems as a measure of the strength of intermolecular interactions. In that case it is infinitely separated molecules which define the zero of energy.
This concept of energy relative to infinitely separated components of a system is a very useful one as it gives an unambiguous definition of the strength of interactions. As a fun thing to ponder, consider how to determine the strength of an internal hydrogen bond? As it happens, to the best of my knowledge, there is no general method for determining the strength of internal hydrogen bonds as there is no natural zero of energy to choose. Some people have tried using the red shift in vibrational frequency, but this too is still a bit ambiguous and hard to determine.

An energy relative to what?

Yes, but you can only know this when you bring the atoms into contact and this is exactly why the binding energy is different for the different molecules. They have different electronic and vibrational energies as molecules (and as atoms), but by choosing the zero as the atoms (with all of their electrons), you get a measure of the energy change which comes from just bringing the atoms together. If you wanted, you could choose the zero to be the bare nuclei, but this is a measure of something else entirely, and if you do this you're probably more interested in the atom than the molecule.

What? I'm not sure what is confusing you? If you're confused by choosing the zero of energy as the infinitely separated atoms, consider choosing the zero of energy as the minimum of of the potential energy curve. This is clearly an unambiguous point, and the the potential energy of the system is easily found there. You can choose to think about the kinetic energy of the nuclei if you want.  One of these is $D_e$ and the other is $D_0$ where the two differ only by zero-point energy. These will be approximately described by the Morse Potential

Let us continue this discussion in chat.

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


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Where I went wrong and got an unwanted reaction
I want to apologize for my bad English - but there was no one to turn to.
I had begun to etch a printed circuit board and used 100ml. 3% H2O2, 2 tablespoons of citric acid, 2 tablespoons  NaCl.
After insertion of the printed circuit board - start reaction and the liquid turned blue, but in the middle saw that is less quantity and add hot water and the reaction stopped.
I noticed there was dirty on the circuit board and I removed it with a carpet knife. It immediately became red in contact with the liquid.
I poured the liquid into a plastic box and dipped a washed carpet knife and go to the pharmacy to buy H2O2
But I made a mistake because I bought 30% H2O2. I made a new liquid with 200ml. H2O2 and citric acid and NaCl. The PCB reaction was much faster. After I finished and pulled out the PCB , I wanted to see if more honey would be deposited, and I took the same knife I had dipped into the front solution and immediately got wet and the knife became black. A reddish froth floated on top of the surface. I went to wash the printed circuit board with water and heard some noise. I came back and some smell terrible and my eyes were red, I was breathing hard while I was throwing the liquid out and blowing.
The first day my eyes were watered and my sun was irritating, and in the shade everything was fine. The second day the sunlight did not irritate me, but when I looked at the light, I see something about the lamps that comes out when there is evening fog and looks at the lamp. On the third day I'm better and the vision returns.
What chemical reaction has occurred when:
H2O2 + citric acid + NaCl + Cu + Fe (I guess there are other substances in the mock knife)

There are way too many compounds with the sum formula $\ce{C6H8O7}$ for us to even attempt answering just from this.

1 - Hydrogen peroxide 30%   , 2- Citric acid , 3-sodium chloride

I doubt there was some specific reaction. The peroxide simply decomposed rapidly, the bubbling threw H2O2 + citric acid in the air, and you inhaled that. Very unpleasant. Just dilute the peroxide previously in future, and you should be fine.


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What is the product when ferric chloride reacts with propan-1-ol?
I tried to dissolve ferric chloride in propan-1-ol and I saw ferric chloride partially dissolve in it, but I couldn't identify which reaction had occurred. If any one knows about this, please give me your suggestions.

Who said there is a reaction in the first place?

There is no reaction between propan-1-ol and iron(III) chloride because propan-1-ol isn't acidic enough to react with it. Instead, phenol reacts with iron(III) chloride because it is acidic enough to form either the phenoxide or iron-phenol complex. On this basis, aliphatic and aromatic alcohol are distinguished. The mechanism of reaction of phenol and $\ce{FeCl3}$ can be found here.
According to this quora post:

Ferric chloride test is specific for determination of phenol presence.
  So as mentioned above, phenol is more acidic and can form phenoxide
  anions. This can form complex with Fe(III) in $\ce{FeCl3}$, which has a
  blue/green/red color depending on the nature of the phenol. Alcohol is
  too weakly acidic to even form the alkoxide or the complex.

So what's happening?
It actually forms a sol-gel. It is a part of iron-oxide based gels.
One noble application of this is non noble metals electrocatalysts for high temperature PEM fuel cells. For this purpose, Fe-containing carbon aerogels were prepared in resorcinol-furfuraldehyde resins. The resins are made in premixed solution of $\ce{FeCl3}$ and propanol.

This answer is thoroughly wrong. What even gave you the idea to cite this nonsense from quora?

@Mith I'm curious (since my knowledge of Org. chem. is limited to what I've learnt at school): As far as I know the ferric chloride test is one method to detect phenols (so that bit seems okay). The thing about alcohols not forming alkoxides seems off though (Maybe it's correct in the context of the answer?) Agreed, Quora answers aren't always great...but what parts are wrong in the answer here? :/

@Nilay Oh a pointer! The 3rd link (to the "iron-oxide based gels" bit) doesn't seem to work :/

I would say there is not enough information for us to think about the reasons for your outcome. I would assume you had propan-1-ol, not propen-1-ol, as the latter is not really stable and rearranges given time.
Dissolving is a physico-chemical process, which involves some particular changes on molecular level (I never state FeCl3 is a molecule). I am not sure if it matters whether you chloride was hexahydrate or anhydrous, anyway, dissolving would include destruction of ionic bonds/ionic structure of FeCl3. Iron (III) ions would attract negative sides of propanol dipoles, which are represented by oxygen atoms and form a kind of complex coordinated compound with them.
The partiality of your dissolving may just be caused by limited solubility of FeCl3 in propanol.

Sir can you suggest me any catalyst or reactant by which we get complete solubility of ferric chloride (II) in propan-1-ol.

I don't think you need really a catalyst. You might use just more propanol.

But I want to maintain the conc. of ferric chloride in propanol. If I use more propanol than conc of ferric chloride will decrease which is not helpful for me.

did you check FeCl3 solubility in propanol? Maybe you're trying to prepare an oversaturated solution... If that's fine, you may try to increase temperature while dissolving

Ok sir i will try this thing also.


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Heat and Sugar Syrup Consistency
I make my own sugar syrups and have been observing how heating the syrup to different degrees results in syrups of radically different consistencies, I wonder why. Heating up syrups (4:3 Sugar:Water) to greater temperatures causes the resulting syrup to be relatively viscous while low temperatures yield relatively thin syrups. I do really think that my observations are valid and thought this'd be the right place to get the answers. I want to know of the chemical and physical processes involved in solutions of sugar and why their peak temperatures affect their consistencies.

You're not the first one who notices. https://en.wikipedia.org/wiki/Caramelization

@Karl Um. I didn't mean it to be that. Just that my searches didn't put out anything good enough wrt my "why"s. And this link- I'm afraid it's not what I was looking for.  My question concerns the consistencies, not the process of caramelization, which is something I do possess the knowledge of.

Question: Since you didn't specify the temperatures you're dealing with, and  hence we don't know if this is due to caramelization as @Karl points out... could you mention if the syrup regains it's original consistency when cooled back to room temperature? If not, I suspect the increased viscosity could be either due to caramelization (Kudos to Karl) or perhaps it was simply the evaporation of large amounts of water (if the solution wasn't boiling) which would thereby increase the concentration, and hence the viscosity, of the syrup. Good question though, and welcome to Chemistry.SE! ;)

@paracetamol I'm dealing with the critical caramelization temperatures in most cases, and my sugar does indeed get caramelized. So the caramelized sugars are more viscous, then, I guess? Is it the polymers, carmelen and caramelin, that make the caramelized products denser?


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How can I decrease the Krafft temperature of an hexadecyl sulfate surfactant?
I need to prepare an anionic surfactant from a treatment of cetyl alcohol (hexadecanol) and $\ce{H2SO4}$, and then neutralize it using $\ce{NaOH}$ or $\ce{NH3}$; but sodium hexadecyl sulfate has a standard Krafft temperature of around 45 degrees Celsius, which is too high for our purposes.
Is it possible to reduce the Krafft temperature to room temperature by adding electrolytes, solvents or modifying counter-ions?


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Reactivity of Aluminium Oxide
I know that Aluminium Oxide is an amphoteric substance meaning that it reacts with both acids and bases. I recently learned that Aluminium has very good corrosion resistance due to the "Aluminium spontaneously forms a thin but effective oxide layer that prevents further oxidation."
However, in the test for Nitrate Ions, an Aluminium foil is added into the solution, the Nitrate Ion is reduced by the Aluminium, but this Aluminium foil surely has Aluminium Oxide which should prevent the reaction from happening, but it happens anyway. How can this phenomenon be explained?
Source: http://www.aluminiumdesign.net/design-support/aluminium-corrosion-resistance/

What other chemicals are added in the test for nitrate ions?

Aluminum oxide coating can relatively easily react with either strong base or acid, dissolving in them, exposing naked elemental aluminum to reaction medium. You can read more  here.
To be concise, there are many ways to make aluminum react: you can use a base, you can use an acid, and you can scratch a sample of aluminum and coat it with mercury. Mercury would form an alloy with Al, but that wouldn't affect its reactivity towards the solution of anything, it would just make it harder for oxygen to oxidize the fresh aluminum.
Answering you question: as far as I know, the nitrate test involves strong base, which serves as a reactant for aluminum oxide, dissolving it and letting aluminum to react further.


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Identify A, B and C and write the chemical equation

Please solve this question. I really want to know what is the answer of this question. Please help.

"Nickle", eh? :D

@paracetamol Looks like Nickie to me, must be some girl in the lab...

We are not here to do your homework.

I wonder what they mean with $\ce{C{0}_2}$ and $\ce{H2{0}}$

Second sentence just says us that 1 mole of C contains 2 moles of carbon atoms and 6 moles of H atoms, and we don't really know whether O is present. Thus the formula of C is C2H6Ox.
Though reaction of B with hydrogen in presence of Ni forming C gives a clue that B was an alkene or alkyne, and C is an alkane, and C has no oxygens.
Moreover, alcohols are known to give up water in presence of sulfuric acid to form alkenes.
In this case, C is ethane C2H6, B is ethene C2H4 and A is ethyl alcohol C2H5OH.

Avoid answering homework type question.


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What will be the colour of HCl with methyl orange?
I am new in chemistry and this question can be silly, but I am troubled with this question.
According to the pH range of methyl orange, if I add methyl orange into water, it gives a yellow color (pH 7). Then, if I add some HCl drops (a very little amount) into the water, what will be the color?
Usually, the colour of HCl is red with methyl orange (because of acid), but when HCl is added to water, will it be a colour within yellow and red, or pure red in colour?

HCl is colorless. It stays colorless when dissolved in water, whatever its concentration. If this solution n is colored, it is not due to HCl or to water. it is due to some other substance, like methyl orange, whose color changes if enough HCl is added.

Methyl orange (MO), as you know, is an acid-base indicator.
The problem with those is that they do not distinguish between acidic and basic medium, but rather between different pHs. MO has a transition interval of 3.1-4.4, which means it's red below 3.1 and yellow over 4.4. Within this interval it's all about transition colors (red through orange to yellow).
Answering your question: a few drops of HCl would affect water differently. Let's say, you had 1 L of water and you added 3 drops (0.15 mL) of concentrated saturated HCl (~40% w/w). That leads us to  ~0.002 mol of HCl added and to pH lower than 3, so the MO would turn red. If you had diluted solutions, like 0.01M or so, the calculations would tell you the pH would still be higher than 4.4, so the MO would still be yellow.


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How do Invert Sugars deter crystallization in Sucrose solutions?
Invert Sugars, by themselves, are notorious for their reputation to crystallize fast (think honey), but their addition to solutions of Sucrose (think simple syrup) deters the crystallization process. Take Caramel, Sucrose that has already decomposed into Fructose and Glucose and subsequently into the numerous products of caramelization: It crystallizes or, at least, turns grainy without the addition of Invert Sugars or a weak acid (Citric, Tartaric) (although the crystallization process is retarded, taking days even at low temperatures). Why do Invert Sugars deter crystallization when added in small quantities in solutions such as Caramel, which, I think, do already contain Invert Sugars as they are already largely decomposed? I would like to know of how Invert Sugars physically interact with the other sugar molecules and prevent them from coming together, even though they're not really resistant to crystallization (as pointed out above).
An additional (and optional) question: Why does the rate of cooling of a sugar solution affect its crystallization? So, while a slowly cooled solution crystallizes more, a faster cooled solution is more likely to crystallize less spontaneously.

I would like to know of how Invert Sugars physically interact with the other sugar molecules and prevent them from coming together

Glucose and fructose molecules (being essentially parts of sucrose) can merge into sucrose crystals during their formation, slowing or even completely stopping their growth in certain directions (if their concentration is high enough).

even though they're not really resistant to crystallization (as pointed out above)

Glucose and fructose crystallization process is much slower than that of sucrose. From confectionery process's point of view, once the fast growing sucrose crystal structure is disrupted and its growth hindered no crystallization occurs.

Why does the rate of cooling of a sugar solution affect its crystallization?

Because crystals' growth takes some time. If cooled fast, neither existing crystals will have a chance to grow, nor new seeds will appear. If cooled really fast, saturated sucrose syrup can even turn into glass with no crystals at all.
[literature]: Sgualdino, G., Aquilano, D., Vaccari, G., Mantovani, G., & Salamone, A. (1998). Growth morphology of sucrose crystals. Journal of Crystal Growth, 192(1-2), 290–299.


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Is Hydrogen Dioxide Water? If so, what is the chemical compound and differences between H2O and H2O2?
Four of us were debating what the actual compounds and terms (scientifically) for water. H2O we all agreed upon until one person said, "..yes and its Hydrogen Dioxide". Myself and 1 of the four challenged that answer by saying, "One does not drink Hydrogen Dioxide"! This turned into an half an hour debate and furious flipping through the internet, (which DID NOT provide a clear and concise answer).
Please help make this clear.

"Please help make this clear". The "di-" prefix means "two". Given that water is $H_2O$ (Two $H$s and one $O$), I think you are quite capable of clearing things up yourself ;)

The very first hit you get by typing 'hydrogen dioxide' in Google yields the answer you were looking for.

https://chemistry.stackexchange.com/questions/53533/why-is-h2o2-named-hydrogen-peroxide

$\ce{H2O}$ (Dihydrogen monoxide) is water, while $\ce{H2O2}$ (dihydrogen dioxide) is hydrogen peroxide, a strong and unstable oxidizing agent.

The terminology is such that the term "oxide" means a negative ion with a double negative charge, or oxidation state of -2, on a single oxygen atom. Thus, H2O, known as water, can be interpreted as an oxide, but H2O2 cannot, as the -2 "charge" is equally distributed between both oxygen atoms, meaning each has only -1.
Side fact: F2O is not an oxide either, as oxygen hosts positive charge and has the oxidation state of +2, so the compound should be refered to as OF2 and named oxygen difluoride.


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How to prepare 4-aminophenyl zinc iodide?
I'm trying to follow a published synthetic procedure from Direct preparation of new organozinc reagents, aminophenylzinc iodides, and their applications:[1]
$$\ce{H2N-Ph-I + \underset{(Rieke)}{Zn} -> H2N-Ph-ZnI}$$
However, I can not manage to reproduce the results.
The Rieke zinc was prepared by lithium naphthalide and $\ce{ZnCl2}$.
[1] Jung, H.-S.; Kim, S.-H.; Tetrahedron Lett. 2015, 56 (8), 1004–1006 (DOI: 10.1016/j.tetlet.2015.01.053).

Could you describe your problem in more detail?

I didn't read the Tet Lett paper, but the point is - what are you having issues with? Extraction of product? Yield? Colour change not going as reported in the paper? NMR doesn't correspond to product? No change in TLC even after 7 days? Just saying "it's not working" is sorta like... going to the doctor and saying "I'm sick". You need to describe symptoms, otherwise the doctor can't try to give you anything useful.


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Derivation of the van der Waals equation
I’m trying to rationalize the van der Waals equation for a non-ideal gas. I understand it conceptually, but I cannot finish the algebra:
Begin with the ideal gas law: P V = n R T
This is really: P(ideal) V(ideal) = n R T       (1)
Now consider the particle-particle interactions. This has the effect of decreasing the pressure, relative to the ideal condition. It also decreases the real, or effective volume occupied by the gas.
Express this as: P(real, or observed) = P(ideal) – P(due to particle-particle interactions)
Rearranging: P(ideal) = P(real) + P(interactions)        (2)
We also have: V(real) = V(ideal, or measured) - V(due to particle-particle interactions)
Rearrange this to: V(ideal) = V(real) + V(interactions)   (3)
Now consider again the volume occupied by the gas, and start with the understanding that the expression “volume of gas” is not literally correct. The “P”, “n”, and “T” are intrinsic properties of the gas particles, but “V” refers to the volume that the gas occupies, not the sum of the volume of the individual gas particles.
In the ideal condition, the measured volume of the container is equal to the volume occupied by the gas. Given that a real gas has finite volume, the volume available for the gas particles to occupy decreases by the volume of the gas particles themselves.
Express this as: V(occupied by the real gas) = V(total, of the container) – V(of the gas particles)
I consider that the measured volume of the container is the theoretical maximum volume that the gas particles may occupy (maybe here is where I’m going off the rails, but what is the correct argument?)
Rearrange this to: V(ideal) = V(occupied by the real gas) + V(of the gas particles)         (4)
Substitute equation (2) into equation (1): [P(real) + P(interactions)] V(ideal) = n R T     (5)
But now, substitute equations (3) and (4) (they express the same concept; V(real) < V(ideal)) into (5), and I get:
[P(real) + P(interactions)] [V(occupied by the real gas) + V(of the gas particles)] = n R T        (6)
The volume expression is incorrect.
I’m thinking that the logic error is somehow related to the fact that “volume” is the volume occupied by the gas, and not an intensive property of the gas particles themselves, but I can’t find how that translates to an error in my argument above.
So, let’s cheat. Logically, at any rate. Let’s assume the van der Waals equation, and see if I can go backwards, without logical error:
The volume term in the van der Waals equation is: (V – nb)
Since it arises by substitution into the ideal gas law, it follows that: V(ideal) = V – nb.
This makes the “V” on the right a real volume. If it was ideal, then either “n” is zero (no gas), or “b” is zero (no volume), or both. Logically impossible.
It must be that: V(ideal) = V(real) – nb
This implies that the volume that may be occupied by a real gas is always greater than the volume that may be occupied by an ideal gas. But the volume occupied by an ideal gas is equal to the volume of the container. Therefore, the volume of a real gas is greater than the volume of the container it occupies. This cannot be.
Where am I going wrong?
I appear to be left with the notion that deriving the van der Waals equation by substitution into the ideal gas law is not correct, but why?
And if it is correct, is there a rationale for the volume expression like that for the pressure expression? In other words, what is the logical argument, that makes sense when implemented algebraically?

This webpage may be of interest to you. It gives two derivations of the equation. The first is most similar to the method you are attempting.


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What constraints are imposed on a wavefunction by the symmetry of the system?
As a follow-up to my answer here, I'd like to ask what exactly does it mean for a wavefunction to "respect the symmetry" of the system.
The original context is: immediately after ionisation of $\ce{CH4}$, the wavefunction of $\ce{CH4+}(T_\mathrm{d})$ cannot be adequately described by removal of an electron from an $\mathrm{sp^3}$-type hybrid orbital, as the resulting Slater determinant $|\phi_1\phi_2\phi_3\phi_4\overline{\phi}_1\overline{\phi}_2\overline{\phi}_3|$ does not reflect the tetrahedral symmetry of the system immediately after ionisation ($\phi_i$ being the four hybrid orbitals).
Intuitively, it is clear that this is related to the fact that the four C–H bonds must be equivalent. Mathematically, how is this formalised, i.e. what is the criterion that the wavefunction must obey?
As a side question, how does the criterion above apply to a boron atom, for example? Is the wavefunction $\mathrm{1s}^2 \mathrm{2s}^2 \mathrm{2p}_x^1$ acceptable on symmetry grounds? My gut feeling is that it's not (because the electron density needs to be spherically symmetric?), but I might really be overthinking this.

Hmmm... Something seems weird about this. The $T_d$ symmetry which comes from the $\ce{CH4}$ will change to $C_{2v}$, which I know you are aware of. But the point is that the wavefunction is not initially in a bound state. You would need to use the time-dependent Schrodinger equation to follow the time-evolution of the wavefunction. I'm not sure what physical meaning any constraint on this initial state would mean because independent of the constraint, the wavefunction will relax to the $C_{2v}$ state. I don't know the answer, I'm just throwing out a thought which might be useful.

@jheindel I believe what he is getting at is what does it mean for a wavefunction to reflect the symmetry of the system. What types of wavefunctions are disqualified from describing tetrahedral, octahedral, etc molecules?

I think you constrain the coefficients of the AO of symmetry-equivalent atoms.

@jheindel Actually, I sort of get what you mean. I was also thinking about the particle in a 1-D box (box lies between $-a$ and $+a$); since the potential is an even function of $x$, the Hamiltonian must commute with the parity operator, and all stationary states must be eigenfunctions of the parity operator, i.e. either even or odd. This is what Ivan was describing in his answer. But that of course only applies to stationary states, and it is obviously possible to construct a wavefunction that is neither even nor odd by taking an appropriate superposition of even and odd stationary states...

and surely the same thing applies here? The wavefunction immediately after ionisation is obviously not in a stationary state, so why does it have to obey the symmetry of the system? I don't actually know, and the more I think about it the more confused I get, but I guess it is a problem for another question.

Ya that's what I was getting at @orthocresol. I think part of the problem is the answer to the question title is what Ivan said but you're asking a more complicated question than that. Namely, how does symmetry apply, if at all, to a symmetrical non-stationary state? I wonder if this places constraints on how the system relaxes.

I use quantum chemistry about as much as a lumberjack uses math (that is, just enough not to fell a tree on one's own head), so my answer will be short and hopefully useful as a tl;dr to someone else's more comprehensive take.
In quantum chemistry, if two operators commute, their values are simultaneously observable. In other words, there is a basis of functions which are eigenfunctions to both operators at once. For an opposite example, think of coordinate and momentum; they do not commute, hence the Heisenberg uncertainty and stuff.
When our problem has some symmetry (including, but not limited to, geometrical symmetry like $T_d$), this means that symmetry operators commute with its Hamiltonian. Therefore the eigenfunctions of the latter must also be eigenfunctions of the former. In other words, the application of symmetry operator to any of our states is equivalent to a multiplication by some number. That's the constraint you were after. Since a symmetry operator of a finite point group must have finite order (by applying it a few times, you get back where you started), the said number must be a root of unity.
That's about the size of it. The numbers for different symmetry operators can be chosen in a multitude of ways, each corresponding to some representation. The character tables tell you how many functions of each representation do you have. Since the functions of different representations are orthogonal to each other and don't interact, this helps us to split a huge problem into smaller chunks.

With methane, the qualitative result could have been envisioned in a simpler way, almost without any calculations. Think of a carbon atom with four $sp^3$ orbitals. They are all equivalent and orthogonal to each other. Now bring in the four H atoms. This will produce four $\sigma$ orbitals (and $\sigma^*$ too, we just don't care about those at the moment). But wait, those $\sigma$ orbitals are no longer orthogonal, so they must interact, if only a little. And because of symmetry, each two of them overlap by exactly the same amount. So it is as simple as Hückel can be: find the eigenvalues of the matrix $$\begin{pmatrix}
 0&-1&-1&-1\\
-1& 0&-1&-1\\
-1&-1& 0&-1\\
-1&-1&-1& 0
\end{pmatrix}$$
The symmetry approach I outlined above still applies; it is just that the system became so small that one can manage without it. The thing can be solved by hand. Or it can be noted that we're dealing with a complete graph, and their spectra are well known (just remember that math people have it upside down, for they use $\mathbf1$s instead of $\mathbf{-1}$s). Either way, the eigenvalues are $\{-3,1,1,1\}$, which kinda explains the 1:3 thing.
So it goes.

Obligatory reference to a lumberjack.

Yes... thank you! I thought something along these lines too (cf. my comments on the question, in response to jheindel). I think I am still a bit unsure on how to deal with degeneracy, which is what I was aiming for with the boron example; eg. the wavefunction $\mathrm{1s}^2 \mathrm{2s}^2 \mathrm{2p}_x^1$ is not an eigenfunction under, say, a 90° rotation about the y-axis (which transforms it into $\mathrm{1s}^2 \mathrm{2s}^2 \mathrm{2p}_z^1$). How can we deal with this? (I realise spherical symmetry isn't the easiest example to use, but maybe we just restrict it to 90° rotations.)

If you have degenerate orbitals, they need to have same occupation numbers. Mathematically, the wave function still musst obey the symmetry of the molecule. Sometimes you can only achieve this by mixing some determinants together.
Lets stick with the Boron example (I happend to have a calculation about that lying around). The trick here is to do a CASSCF calculation (in $D_{2h}$), with active orbitals 2px 2py 2pz. Thus the orbitals can be state-averaged and describe all required states in a balanced way. This will results in some fractional occupation numbers for the orbitals.
Due to symmetry we need the 3 p orbitals to be degenerate, and we also expect 3 degenerate electronic states (because we have a $^2P$ term).
The resulting state-averaged orbital occupations and energies are:
1s  2.00000    -7.697613
2s  2.00000    -0.495495
px  0.33333     0.025103
py  0.33333     0.025103
pz  0.33333     0.025103

And the CI vectors are (first column is orbial occupation, a for $\alpha$ spin, 0 for empty)
 CI vector for state symmetry 1
 ==============================

 a 0 0           1.0000000

 TOTAL ENERGIES                       -24.52694952

 CI vector for state symmetry 2
 ==============================

 0 a 0           1.0000000

 TOTAL ENERGIES                       -24.52694952

 CI vector for state symmetry 3
 ==============================

 0 0 a           1.0000000

 TOTAL ENERGIES                       -24.52694952

In principle, one should be able to do this within HF as well, but this requires a calculation that can deal with the full point group of the spherical atom.
So for $\ce{CH4+} (T_d)$ we can construct the WF from four configurations, one for each $sp^3$ hybrid missing one electron.

I've been wondering about the boron case for a long time, so thanks! Now, if I interpret your data correctly (not a computational chemist), the CI wavefunction is then an equal admixture of the three determinants corresponding to $\mathrm{1s}^2 \mathrm{2s}^2 \mathrm{2p}_x^1$, $\mathrm{1s}^2 \mathrm{2s}^2 \mathrm{2p}_y^1$, and $\mathrm{1s}^2 \mathrm{2s}^2 \mathrm{2p}_z^1$. I guess I am still slightly confused, because putting in three determinants should give three linear combinations. What happens to the other two - or am I on an entirely wrong track?

In this case we have 3 states with only one configuration each, due to symmetry. But the orbitals are optimized for all 3 states simultaneously (state-averaged), it's what MCSCF (CASSCF) does.


text:
Does (2-fluoroethyl)benzene undergo elimination via E1cb or E2?
I have seen in two different sources claiming that $\ce{PhCH2CH2F}$ gives styrene $\ce{PhCH=CH2}$ when treated with alcoholic $\ce{KOH}$, but one source suggests an E1cb mechanism, while the other suggests an E2 mechanism.
I think that the phenyl group should not be able to promote the formation of a carbanion in an E1cb mechanism. All the examples of E1cb mechanisms I have seen involve deprotonation of compounds which have $\mathrm{p}K_\mathrm{a}$ values of 20 or less, and toluene does not fit this.

According to the Evans $\mathrm pK_\mathrm a$ table, toluene is more acidic than propene, so the stabilisation of the phenyl group is larger than that of the allyl group. But I will acknowledge that that is a tough call for an $\mathrm{E1cb}$ mechanism, especially if the base is something weak like $\ce{KOH}$.

There's no reason why either mechanism is wrong. It might depend on conditions. The only way to know with greater certainty is to run the mechanistic experiments.

Run Zhe's experiment in ROMetal/ROD. Check if deuterium is incorporated in unreacted starting material at the benzylic position. Incorporation of deuterium means that at least some of the elimination occurs stepwise; no incorporation means a concerted elimination.

Please mention the "sources" along with page number/ topic. That's really going to be helpful for those answering your questions. Thanks.

I think that it is easier for the molecule $\ce{Ph-CH2-CH2-F}$ to undergo E1cb mechanism than E2 mechanism. It can be also possible that the molecule undergoes elimination through a mixture of E1cb and E2, but it can be said  that majority portion of the molecules will prefer the E1cb rather than any other pathway.
First of all, an E2 mechanism goes via simultaneous cleavage of the $\ce{C_{\beta}-H}$ and $\ce{C_{\alpha}-LG}$ bonds (LG being the leaving group). So, the rate of the reaction depends on the ease of breaking these two bonds. In this case the leaving group is fluoride ($\ce{F-}$), which has a very strong bond with the $\alpha$-carbon, making it a bad leaving group. So, due to this reason, the rate of E2 should lower for this molecule.

Now, if we check the plausibility of E1cb reaction pathway, we will find that it is most suitable in this case.
Firstly, the proton abstraction by the base gives a highly resonance stabilised benzylic carbanion intermediate, and this carbanion formation is the slow step. As this intermediate's energy will decrease, the activation energy will decrease resulting in an increase in the rate constant as, $k = A\exp(-E_\mathrm a/RT)$, helping to increase the rate for E1cb.

Secondly, the breaking of $\ce{C-F}$ bond doesn't happen in the rate determining step, which also doesn't alter the rate of the reaction  in this case.

So, observing the above mentioned cases, it can be definitely concluded that  E1cb is the more preferable pathway to undergo $\beta$-elimination to form Styrene.

Subscripts should be as such: $\mathrm{S_N1}$, $\mathrm{S_N2}$, $\mathrm{E1}$, $\mathrm{E2}$, $\mathrm{E1cb}$. For the elimination ones, you don't need MathJaX anyway as there are no subscripts. On topic, calling the benzylic anion "highly" resonance stabilised is a bit disingenuous. Toluene has a $\mathrm pK_\mathrm a$ of roughly 40. That's not a very stable anion.

Okay but relaive to the other pathway, E1cb provides better feasibility by this stabilisation. And we are mainly concerned about relative rates.


text:
How to determine the highest freezing point by van't hoff factor when concentration is same?
Whch of the following will have the highest freezing point?
a) 0.1M KCl
b) 0.1M glucose
c) 0.1M BaCl2
d) 0.1M AlCl3

Raoult's law originally was meant to describe the properties of non-electrolytes soltions. When it comes to the solutions of electrolytes, such as salts, you need to introduce the van't Hoff factor, sometimes called isotonic coefficient.
To deal with it, you need to know the folmula:
i=1+α(n-1),
where
i - van't Hoff factor
α - dissociation degree
n - number of ions formed from 1 ionic formula of electrolyte
In the case of strong electrolytes, we just assume α=1, so i=n.
Thus, you just need to calculate the number of ions which are formed from 1 ionic formula (for instance, it's 5 for Al2(SO4)3, as it goes into 5 ions - 2 aluminums and 3 sulfates), and multiply the cryoscopic constant and molality by it.
So in your case the biggest number of ions comes from d)

What if  concentration and van't hoff factor is same? Can i consider molar mass then?

No, colligative properties are about the number of particles, not their nature by definition. Like, 0.01 M solutions of KCl and NaBr would have same freezing points, as both dissociate giving 2 ions (if we assume their dissociation is 100% complete)

But, depression in freezing point is a colligative property, not freezing point.    Am I right?

You are right. If depressions for both compounds are the same, aqueous solutions of these compounds would have the same FP.

Let's say the depression for both KCl and NaBr of some concentration is 0.5 degrees. Then the freezing point of both solutions is (FP of water)-(depression)=-0.5 degrees Celcius


text:
Is the triplet state of the cyclopentadienyl cation really aromatic?
While checking the ASAP articles of the Journal of the American Chemical Society I came across a very interesting contribution by Costa et al. who synthesised a $\ce{BF3}$ stabilised cyclopentadienyl cation 1 and confirmed that its singlet state was indeed antiaromatic as predicted by various definitions.[1] The authors, however, also included the following sentence:

According to Baird’s rule, “the lowest triplet state for $4n$ rings is aromatic since the bonding energy is significantly greater than for the diradical reference structure”.[1]

This statement confuses me. In my introductory organic chemistry classes, I was taught the $4n+2$ rule along with simplified Hückel analyses to predict the relative energy levels. It was always pointed out that $4n$ (if planar and a single π system) represents an unstable, antiaromatic state which the system will avoid by distortion. This was explained by the triplet, biradicallic structure of the $4n$ systems. Yet here by Costa et al. I read that $4n$ triplets are actually aromatic.
How can I reconcile this new information with what I was taught previously? Reading the article and the references has not been able to help me.

Reference:
[1]: Costa, P.; Trosien, I.; Mieres-Perez, J.; Sander, W. Isolation of an antiaromatic singlet cyclopentadienyl zwitterion. J. Am. Chem. Soc. 2017, 139 (37), 13024–13030. DOI: 10.1021/jacs.7b05807.

Well, the Wiki article for Baird's rule [ https://en.wikipedia.org/wiki/Baird%27s_rule ] does mention that the Hückel $4n+2$ rule is for ground-state (typically singlet) molecules; whereas Baird's rule (aromatic nature of triplet in $4n$ electron species) is, apparently, only when you're dealing with the excited state...

Maybe triplet state should implies a transition to a 2p orbital of the electron in the positive carbon so it favors the resonance..

Maybe it also is about the Moebius aromaticity?

@paracetamol I’m not convinced that all triplets discussed in the article are actually excited states. Case in point: the authors did have extreme difficulty actually isolating the singlet of 1.

This compound may be a zwitterion to some extent, but it's still adduct of carbene, not real cation.

@Jan that's a really good point, but also tricky to address. I think that in general the assignment of singlet/triplet as the ground state must be done by high-level calculations (for cyclobutadiene, the singlet is the ground state, even for the undistorted geometry - cf. previous answer here). For the purposes of this question, though, I suspect it doesn't actually matter, so we can circumvent the problem entirely. At least, in my answer, I (hopefully!) haven't written anything that only holds true if the singlet is the ground state.

@MELScience also a very interesting point. The topology of the pi system (Möbius or Hückel) isn't part of this question, but is definitely related to the topic. As everyone knows, ordinary ground-state Huckel-topology benzene is aromatic. As discussed in my answer, triplet Huckel-topology benzene is anti-aromatic. For Mobius-topology benzene, these are reversed; the ground state is anti-aromatic and the excited triplet state is aromatic! Ref 5 in my answer has some explanation, if anybody is interested.

TL;DR Yes. For cyclic conjugated systems in the $T_1$ state (i.e. lowest energy triplet excited state), $4n$ electrons means aromatic and $4n+2$ means anti-aromatic. The $T_1$ state of the cyclopentadienyl cation, which has 4 pi-electrons, is therefore aromatic.

Aromaticity
The (quite understandable) confusion here arises over the interpretation of the word "aromatic" or "anti-aromatic". These terms can be used differently depending on context, and there is no unified physical description of (anti)aromaticity (although progress has been made in recent years, e.g. nucleus-independent chemical shifts, etc.)
When anybody claims that something is aromatic, it is therefore imperative to check the context, and to figure out what their basis for assigning aromaticity is. In this case, the assignment of the triplet excited state of the cyclopentadienyl cation as aromatic by Costa et al.1 is based on Baird's rules. In turn, looking up Baird's article2 reveals that his assignment of (anti)aromaticity is based on the concept of Dewar resonance energies (DREs).
The concept of DREs has been explained by Baird elsewhere.3 Essentially, it is based on the comparison of idealised cyclic structures with a reference acyclic structure; so, benzene is compared with hexatriene, and cyclobutadiene ($D_\mathrm{4h}$) is compared with butadiene. For the cyclopentadienyl cation, the relevant comparison is with a penta-2,4-dien-1-yl cation.
The question thus arises as to how this comparison can be made. The way that is commonly taught, at introductory level, is to use Hückel molecular orbital (HMO) theory. So, let us use cyclobutadiene as an illustration. Note that within HMO theory the singlet and triplet configurations are of the same energy, as electron-electron repulsions are not taken into account.

The resonance integral $\beta$ is negative. So, singlet cyclobutadiene is anti-aromatic because it has less π-bonding energy than the singlet state of its acyclic counterpart butadiene. However, triplet cyclobutadiene is aromatic because it has more π-bonding energy than triplet butadiene does.
There are two things of note here: firstly, the Jahn–Teller distortion in cyclobutadiene isn't relevant for the purposes of finding whether it is aromatic or not; the aromaticity is assigned based on a hypothetical planar structure with no bond length alternation. Secondly, the aromaticity is determined with respect to the corresponding electronic state of the acyclic reference. That is to say, the triplet state of cyclobutadiene is compared with the triplet state of butadiene and not the singlet state of butadiene.

Perturbational MO theory
Qualitatively, we have reached the correct conclusion. However, this was not the approach that Baird used. We will restrict our discussion now to that of triplet states, since it is relevant to the question. Baird used a more sophisticated method involving perturbation theory. That is, Baird used MO theory to calculate the stabilisation/destabilisation that would occur upon bringing together two radicals to form either the cyclic compound or the acyclic compound.
This is confusing, so a bit more explanation is in order. The triplet state of butadiene can be described by the promotion of a π electron to a π* electron. Upon this promotion, the C=C bond twists until it has a dihedral angle of 90°, which corresponds to the breakage of the π bond.

In the triplet state of butadiene (1 ($T_1$)), the π systems of the allyl and methyl radical fragments are orthogonal to each other; they are simply joined by a C–C σ bond. Hence the total π-bonding energy of 1 ($T_1$) is simply equal to the sum of the individual π-bonding energies of 2 and 3.
On the other hand, the triplet state of cyclobutadiene (4 ($T_1$)) can still be considered to be an allyl fragment plus a methyl fragment.4 The difference is that, due to the planarity of the ring, the two radical systems are forced to overlap in a π manner (depicted by the red lines).

Now, if this overlap is stabilising in nature, then the π-system of 4 ($T_1$) is more stable than the π-system of 1 ($T_1$), and therefore we can say that 4 ($T_1$) is aromatic. Conversely, if the overlap is destabilising, then it is anti-aromatic.
Finally, we get to the crux of the answer. (I make no apology for taking so long to get to it - I thought the buildup was pertinent.) Baird first constructed the MOs of the radical fragments 2 and 3, and labelled each MO with their symmetry under a reflection. The mirror plane is drawn in light blue; antisymmetric MOs are in red and symmetric MOs in green.

Then, Baird identified two types of orbital interactions in the cyclic structures that could lead to either stabilisation or destabilisation. The first type is called a Type I interaction, and it involves the overlap of the two SOMOs on the two fragments. In this case, since the two SOMOs are of different symmetries, they cannot overlap and therefore the Type I interaction leads to neither stabilisation nor destabilisation.
The second type of interaction (I leave the reader to guess what it was called) involves the interaction of the SOMO on one fragment with doubly occupied or empty orbitals on the other fragment. In this case, the SOMO of the allyl radical does not find any symmetry match on the methyl radical, and so there is again neither stabilisation nor destabilisation. However, the SOMO of the methyl radical can overlap with both symmetric MOs of the allyl radical; this is a stabilising interaction.
Overall, therefore, the π-type overlap between the two radical fragments leads to a favourable interaction. As such, triplet cyclobutadiene is considered to be aromatic.
For the cyclopentadienyl cation, one can construct an analogous diagram, starting from an ethene radical cation and an allyl radical.

Again the SOMOs do not match in symmetry, so the Type I interaction is necessarily zero. And again, the Type II interaction is stabilising; this time, both SOMOs find symmetry matches on the partner fragment. Based on this analysis, the triplet state of the cyclopentadienyl cation is indeed aromatic.

Triplet benzene
At this stage one may wonder: when two orbitals interact, the only possibility of destabilisation is if both orbitals are fully filled. Since both interactions involve orbital overlap between not-fully-filled orbitals, how can any of them be destabilising, and how can anything be anti-aromatic? To address this issue, and for the sake of completeness, let us analyse the case of benzene. We expect the triplet state of this 6π system to be anti-aromatic. The interaction diagram may be constructed from two allyl radicals:

First, we see that by symmetry, the Type II interaction is identically zero. Now, we turn to the Type I interaction. Surely the overlap of two SOMOs must be a bonding interaction? The astute reader will notice, though, that we are talking about a triplet state of benzene. The overlap of the two SOMOs still forms a bonding and an antibonding orbital, but if we were to put both electrons into the bonding orbital, we would obtain a singlet state.
As such, the Type I interaction leads to both the bonding and antibonding MOs having one electron each. Since the antibonding MO is more antibonding than the bonding MO is bonding, this is a net destabilisation. Hence, triplet benzene may be assigned anti-aromaticity.

Further reading
The concept of excited state aromaticity has been reviewed by Rosenberg et al.5, which is a good read (I found the explanation more accessible than that in Baird's original article2).

Notes and references

Costa, P.; Trosien, I.; Mieres-Perez, J.; Sander, W. Isolation of an Antiaromatic Singlet Cyclopentadienyl Zwitterion. J. Am. Chem. Soc. [Online early access]. DOI: 10.1021/jacs.7b05807.
Baird, N. C. Quantum organic photochemistry. II. Resonance and aromaticity in the lowest 3ππ* state of cyclic hydrocarbons. J. Am. Chem. Soc. 1972, 94 (14), 4941–4948. DOI: 10.1021/ja00769a025.
Baird, N. C. Dewar resonance energy. J. Chem. Educ. 1971, 48 (8), 509–513. DOI: 10.1021/ed048p509.
Technically, there is one hydrogen less on both fragments, but the C–H bonds do not contribute to the π framework so they can be ignored - after all, we never bothered taking the C–H bonds into account, not even in the Hückel approach. The point is that whether it is a methyl fragment or a methylene fragment, the π system remains the same (likewise for the allyl fragment).
Rosenberg, M.; Dahlstrand, C.; Kilså, K.; Ottosson, H. Excited State Aromaticity and Antiaromaticity: Opportunities for Photophysical and Photochemical Rationalizations. Chem. Rev. 2014, 114 (10), 5379–5425. DOI: 10.1021/cr300471v.

Let us look at the cyclopropenyl system, which is simplest to work with and gives the most prominent effects.  To assess aromaticity, the $\pi$ electron stabilization compared with isolated $p$ orbital is modeled using Huckel MOs,  and the cyclopropenyl ring is compared with the open-chain allyl system having the same charge and spin.
Start with the cationic, singlet state.   We all know what happens:  in both the cyclopropenyl and allyl systems the two $\pi$ electrons go into a bonding orbital, but the cyclopropenyl orbital gets extra stability from having an additional bonding overlap.  So the $\pi$ electron energy is $4\beta$ in the cyclopropenyl cation versus $2.83\beta$ in the allyl cation, showing that closing the ring gives an aromatic coupling (since $\beta$ is negative, a larger positive coefficient means lower energy).
Now try the singlet anion.  In the allyl system the two additional electrons go into a nonbinding orbital so the $\pi$ electron energy remains $2.83\beta$.  But in the cyclopropenyl anion the additional $\pi$ electrons have to go into an antibinding orbital so the $\pi$ electron energy is now $2\beta$.  The allyl anion is more stable, the cyclopropenyl anion is identified as antiaromatic.
Now for the triplet anion.  In the allyl case the third and fourth $\pi$ electrons have to have the same spin so only one can be in the nonbonding orbital, the other is in an antibonding orbital and the $\pi$ electron energy ends up as $1.41\beta$.  The corresponding cyclopropenyl anion, with two $\pi$ electrons in antibonding orbitals like the singlet anion, gives $2\beta$.  The ring system is now more stable conforming with Baird's Rule for an aromatic triplet.
But there are two key differences between the Baird-aromatic triplet cyclopropenyl anion and the Huckel-aromatic singlet cation.  First, the extra stabilization obtained by closing the ring is only half as large in the Baird-aromatic case as in the Huckel-aromatic one.  Second, the Baird-aromatic ring is compared with a relatively unstable open-chain state, quite unlike its Huckel-aromatic counterpart.  For these reasons, the Baird-aromatic system will not exhibit the strong extra stabilization or overall stability offered by Huckel-aromatic systems.


text:
Why are different colors, according to sugar concentration observed in Benedict's test?
Colors range from green yellow orange to red. How does the sugar concentration and Cu2O concentration cause this?
I know Cu2O is red. Why is there a green or yellow color instead of just red or some kind of red?

Benedict's Reagent is just simply $\ce{Cu^2+}$ ions in the form of copper citrate. When Benedict's Reagent finds an aldose (a sugar with an aldehyde group), it can oxidize the aldose to a carboxylic acid. For example:

Since D-Glucose is oxidized, Cu is reduced to a red precipitate ($\ce{Cu2O}$). The colors range form green to red because the original Copper Citrate ($\ce{C6H8Cu2O7^4+}$) is blue in color.
Source: Quora

how does it become green if cu2o is red?

"The compound can appear either yellow or red, depending on the size of the particles." https://en.wikipedia.org/wiki/Copper(I)_oxide yellow + blue =...

How does that relate to concentration? And what about green?

Color theory - Yellow Copper(I) Oxide + Blue Copper Citrate = Green solution

So at higher concentrations copper oxide particles clump together?


text:
Why is chitin ordinarily insoluble in water, but soluble after deacetylation?
Chitin is usually insoluble in water. How does its chemical structure explain this?
Furthermore, when chitin undergoes deacetylation, it becomes soluble and forms hydrogen bonds with water. How is this so?

Chitin is a homopolymer of β(1,4)-linked N-acetyl-D-glucosamine residue .

The structure of chitin, an extended ribbon, is identical to that of cellulose, except that the $\ce{-OH}$ group on each C-2 is replaced by $\ce{-NHCOCH3}$, so the repeating units are N-acetyl-D-glucosamines in β (1→4) linkage.

The acetyl group can contribute to the formation of hydrogen bonds that can stabilise the crystalline structure.
The chains of chitin form extended ribbons and pack side by side in a crystalline, strongly hydrogen-bonded form. The acetyl groups contribute to the formation of stable H-bonds along the inter sheet direction, fracture along this direction requires more energy to break these high occupancy H-bonds (which prevent association with water molecules).
As a result of the hydrophobic property of acetyl group, chitin is ordinarily insoluble in water. This explains the insolubility properties of chitin.
When deacetylated, the solubility apparently increase due to increased hydrophilicity of the introduced amino groups (as well as the weakened crystalline structure allowing formation on H-bonds with water molecules).

hydrophobic/philic argument isn't good argument here

@Mithoron From most research articles I reviewed, this was often the case, at first I was tempted to conclude that there is no free -OH groups available to H-bond with water molecules, but then again looking at cellulose (which has free -OH instead of glucosamine on C2) it isn't soluble either. After a bit of digging thats when I realized that when chitin is deacetylated (with a strong base), it becomes more soluble woth the properties coming from the more hydrophilic amines hence the answer.


text:
Why is the expectation value of Hermitian conjugate operators $RR^\dagger$ always real and non-negative?
I've been reading through a derivation of the wavefunctions and energy levels for the quantum harmonic oscillator. It defines
$$\hat R^\pm=\frac{1}{\sqrt{2}}[\hat p \pm \mathrm{i}\omega \hat q]$$
in mass-weighted coordinates $q=x\sqrt{\mu},$ such that $\hat p= -\mathrm{i}\hbar(\mathrm d/\mathrm dq)$ and $\hat q$ is the position operator.
It then asserts that that $\langle \psi|\hat R^+\hat R^-|\psi\rangle$ must be real and non-negative, since $\hat R^+$ and $\hat R^-$ are hermitian conjuagtes. I know that $(\hat A\hat B)^\dagger = \hat B^\dagger \hat A^\dagger$ and so $\hat R^+\hat R^-$ is a hermitian operator meaning that the bra-ket must be real, but I can't see why it should be non-negative. Could anyone shed some light on this?

Let $\hat{R}^-|\psi\rangle = |\psi'\rangle$ (a new ket; we don't care what it is). If you take the adjoint of this equation you get
$$\langle\psi'| = \langle\psi|(\hat{R}^-)^\dagger = \langle\psi|\hat{R}^+$$
and hence
$$\langle\psi|\hat{R}^+\hat{R}^-|\psi\rangle = \langle\psi'|\psi'\rangle$$
which by definition of the inner product must be real and non-negative.

Where you've taken the adjoint of the equation, is that a valid step? It looks like you're saying that the conjugate of psi dash is equal to the conjugate of psi multiplied by the R+ operator, but what is it operating on?

Operators don't need to operate on anything. Your post has a couple examples of standalone operators. In this case you could think of it as operating on the bra to the left, though. If you think in terms of matrices, then bra times operator is no different from operator times ket.