Cyclobutadiene - Jahn–Teller effect or not? In transition metal chemistry the Jahn–Teller effect arises when the configuration of the metal ion and d orbital splitting set up a doubly degenerate state, which is less stable than a state without the degeneracy but with lower symmetry.

In Ian Fleming's Molecular Orbitals it is stated that cyclobutadiene would be a diradical considering Huckel theory:

but it isn't because of Jahn–Teller distortion (it is actually rectangular and ESR shows no spin).

There isn't a doubly degenerate state in this case, yet Jahn–Teller distortion happens according to Fleming. How is this so?

(In the singlet state it would be doubly degenerate, but how to know it a priori?)

Very interesting question, and it kept me up despite daylight saving time cheating me of one hour of sleep last night... A good reference is Albright, Burdett and Whangbo, Orbital Interactions in Chemistry 2nd ed. pp 282ff. which explains this in much greater detail than I can. (In general, that book is fantastic.) I will try my best to summarise what they have written - please point out any mistakes that I may have made. I think there is also much more information out there in the literature (a search for "Jahn–Teller cyclobutadiene" should give lots of good results), although it is of course not always easy to read. $\require{begingroup}\begingroup\newcommand{\ket}[1]{|#1\rangle}$

1. What is the ground state of undistorted $\ce{C4H4}$?

Based on simple Huckel theory as you have shown above, one would expect the ground state to be a triplet state with the two highest electrons in different orbitals. This is simply because of exchange energy (Hund's first rule).

However, it turns out that the singlet state with the two electrons in different orbitals is more stable. That explains the lack of signal in ESR experiments.

(Note that this orbital picture is a simplification, as I will describe later.)

In short, this is due to dynamic spin polarisation; the idea is that there are low-lying excited singlet states which mix with the ground state to stabilise it. This is the basic idea underlying so-called configuration interaction, which mixes in excited molecular states into the wavefunction in order to obtain a better estimate of the ground-state wavefunction. If you are interested in the theoretical aspect of this, I think Szabo and Ostlund's Modern Quantum Chemistry should cover it. The topic of dynamic spin polarisation is treated in depth in pp 285ff. of Albright et al.

Your question, "how to predict the singlet state a priori"? is a good one. Albright et al. write that (p 289)

  In general, high-level configuration interaction calculations are needed to determine the relative stabilities of the singlet and triplet states of a diradical.

So, it does not seem to be something that you can figure out a priori, and it is most certainly not something you can figure out by just using Huckel theory (which is an incredibly simplistic treatment).

2. What is the degeneracy of the ground state?

You also wrote that "the singlet state will be doubly degenerate". However, in the case where both electrons are in different orbitals, there is only one possible spatial wavefunction, due to the quantum mechanical requirement of indistinguishability. (As we will see in the group theoretical treatment, there is actually no doubly degenerate case, not even when the electrons are paired in the same orbital.)

If I label the two SOMOs above as $\ket{a}$ and $\ket{b}$ then the combination $\ket{ab}$ is not a valid spatial wavefunction. Neither is $\ket{ba}$ a valid wavefunction.

You have to take linear combinations of these two states, such that your spatial wavefunction is symmetric with respect to interchange of the two electrons. In this case the appropriate combination is

$$2^{-1/2}(\ket{ab} + \ket{ba}) \qquad \mathrm{{}^1B_{2g}}$$

There are two more possible symmetric wavefunctions:

$$2^{-1/2}(\ket{aa} + \ket{bb}) \qquad \mathrm{{}^1A_{1g}}$$
$$2^{-1/2}(\ket{aa} - \ket{bb}) \qquad \mathrm{{}^1B_{1g}}$$

(Simply based on the requirement for indistinguishability, one might expect $\ket{aa}$ and $\ket{bb}$ to be admissible states. However, this is not true. If anybody is interested to know more, please feel free to ask a new question.) The orbital picture is actually woefully inadequate at dealing with quantum mechanical descriptions of bonding. However, Voter and Goddard tried to make it work out, by adding and subtracting electron configurations:1

Ignore the energy ordering in this diagram. It is based on Hartree–Fock calculations, which do not include the effects of excited state mixing (CI is a "post-Hartree–Fock method"), and therefore the triplet state is predicted to be the lowest in energy. Instead, just note the form of the three singlet states; they are the same as what I have written above, excluding the normalisation factor.

The point of this section is that, this singlet state is (spatially) singly degenerate. The orbital diagrams that organic chemists are used to are inaccurate; the actual electronic terms are in fact linear combinations of electronic configurations. So, using orbital diagrams to predict degeneracy can fail!

3. Why is there a Jahn–Teller (JT) distortion in a singly degenerate state?

You are right in that for a JT distortion - a first-order JT distortion, to be precise - to occur, the ground state must be degenerate. This criterion is often rephrased in terms of "asymmetric occupancy of degenerate orbitals", but this is just a simplification for students who have not studied molecular symmetry and electronic states yet. Jahn and Teller made the meaning of their theorem very clear in their original paper;2 already in the first paragraph they write that the stability of a polyatomic molecule is not possible when "its electronic state has orbital degeneracy, i.e. degeneracy not arising from the spin [...] unless the molecule is a linear one". There is no mention of "occupancy of orbitals".

So, this is not a first-order JT distortion, which requires degeneracy of the ground state. Instead, it is a second-order JT distortion. Therefore, most of the other discussion so far on the topic has unfortunately been a little bit off-track. (Not that I knew any better.)

Albright et al. describe this as a "pseudo-JT effect" (p 136). For a better understanding I would recommend reading the section on JT distortions (pp 134ff.), it is very thorough, but you would need to have some understanding of perturbation theory in quantum mechanics. Please note that there is a typo on p 136, as is pointed out in the comments to this answer. (Alternatively, there is also a discussion of second-order JT effects in a paper by Pearson.3) The idea is that, the distortion reduces the symmetry of the molecule and therefore allows ground and excited states to mix. This leads to a stabilisation of the ground state and hence an electronic driving force for the distortion.

In slightly more detail, we can use a group theoretical treatment to do this. The second-order correction to the energy is given by

$$\sum_{j\neq i}\frac{\langle i |(\partial H/\partial q)| j \rangle}{E_i^{(0)} - E_j^{(0)}}$$

where $\ket{i}$ is the ground state (which in this case is described by $2^{-1/2}(\ket{aa} - \ket{bb})$) and $\{\ket{j}\}$ are the excited states. $E_i^{(0)}$ is the unperturbed energy of the ground state $\ket{i}$ and likewise for $E_j^{(0)}$. $q$ is a vibrational coordinate of the molecule.

Note that the denominator in this term, $E_i^{(0)} - E_j^{(0)}$, is negative. So, this correction to the energy is always negative, i.e. stabilising. (There is another second-order term which isn't relevant to the discussion here.) The idea is that in order to have a significant stabilisation upon distortion, two criteria must be fulfilled:

The denominator must be small, i.e. the excited state must be low-lying.
The numerator must be non-zero, i.e. there must be appropriate symmetry such that $\Gamma_i \otimes \Gamma_j \otimes \Gamma_H \otimes \Gamma_q$ contains the totally symmetric irreducible representation. Since $H$ transforms as the TSIR, this is equivalent to the criterion that $\Gamma_q = \Gamma_i \otimes \Gamma_j$.

In cyclobutadiene, the two SOMOs in $D_\mathrm{4h}$ symmetry transform as $\mathrm{E_u}$. To find the irreps of the resultant states we take

$$\mathrm{E_u \otimes E_u = A_{1g} + [A_{2g}] + B_{1g} + B_{2g}}$$

(Note that there is no spatially degenerate term in this direct product, so there cannot be a first-order JT distortion!) The $\mathrm{A_{2g}}$ state in square brackets corresponds to the triplet case and we can ignore it. In our case (presented without further proof, because I'm not sure why myself) the ground state is $\mathrm{B_{1g}}$. The two low-lying singlet excited states transform as $\mathrm{A_{1g} \oplus \mathrm{B_{2g}}}$. Therefore if there exists a vibrational mode transforming as either $\mathrm{B_{1g}} \otimes \mathrm{A_{1g}} = \mathrm{B_{1g}}$ or $\mathrm{B_{1g}} \otimes \mathrm{B_{2g} = \mathrm{A_{2g}}}$, then the numerator will be nonzero and we can expect this vibrational mode to lead to a distortion with concomitant stabilisation.

In our case, it happens that there is a vibrational mode with symmetry $\mathrm{B_{1g}}$, which leads to the distortion from a square to a rectangle. This therefore allows for a mixing of the ground $\mathrm{B_{1g}}$ state with the excited $\mathrm{A_{1g}}$ state upon distortion, leading to electronic stabilisation.

Another way of looking at it is that upon lowering of the symmetry from square ($D_\mathrm{4h}$) to rectangular ($D_\mathrm{2h}$), both the $\mathrm{B_{1g}}$ ground state and $\mathrm{A_{1g}}$ excited state adopt the same symmetry $\mathrm{A_g}$. You can prove this by looking at a descent in symmetry table. Therefore, mixing between these two states will be allowed in the new geometry.

Because we are talking about the mixing of states here, the orbital picture is not quite enough to describe what is going on, sadly. The best representation we can get is probably this:

However, Nakamura et al. have produced potential energy curves for the distortion of cyclobutadiene from square to rectangular and rhomboidal geometry.4 Here is the relevant one (rectangular geometry):

Solid lines indicate singlet states, and dashed lines triplet states. Note that this diagram depicts what I have described above: the singlet ground state  ($\mathrm{{}^1B_{1g}}$), as well as the first excited singlet state ($\mathrm{{}^1A_{1g}}$), both have the same irrep upon distortion. The ground state is stabilised upon the lowering of symmetry, precisely due to this mixing.

References

(1) Voter, A. F.; Goddard, W. A. The generalized resonating valence bond description of cyclobutadiene. J. Am. Chem. Soc. 1986, 108 (11), 2830–2837. DOI: 10.1021/ja00271a008.

(2) Jahn, H. A.; Teller, E. Stability of Polyatomic Molecules in Degenerate Electronic States. I. Orbital Degeneracy. Proc. R. Soc. A 1937, 161 (905), 220–235. DOI: 10.1098/rspa.1937.0142.

(3) Pearson, R. G. The second-order Jahn–Teller effect. J. Mol. Struct.: THEOCHEM 1983, 103, 25–34. DOI: 10.1016/0166-1280(83)85006-4.

(4) Nakamura, K.; Osamura, Y.; Iwata, S. Second-order Jahn–Teller effect of cyclobutadiene in low-lying states. An MCSCF study. Chem. Phys. 1989, 136 (1), 67–77. DOI: 10.1016/0301-0104(89)80129-6. $\endgroup$


How do I interpret the results of this chromatogram and does it tell me about the compounds' polarities? I am reading this journal article which discusses using high-performance liquid chromatography (HPLC) to separate dyes from wine. Here are the relevant facts about the experimental setup:

Two mobile phases are used, a solvent A and a solvent B. Solvent A is is 40% methanol and solvent B is 70% methanol. Both solvents also have 25% 0.02M tetrabutyl-ammoniumhydroxide phosphate solution as well.
The stationary phase is a LiChrospher 100 RP-18e column.

Below are the chromatograms and the retention time tables that this experiment produced:

From the chromatogram, we see that amaranth has the fastest elution time of 6.60 minutes. My questions relate to what inferences can we draw from elution times:

Does the elution time correlate with the polarity of the compound itself? Here, it states that similarities in polarity correlates positively with elution time. The stationary phase used in this experiment (the LiChrospher) is a nonpolar silica column. Thus, with amaranth coming out the fastest, would it make it the most polar compound?
Further, how does the use of two solvents (A and B) affect making an inference like the one above? In the chromatogram above (specifically chromatogram "a") and the associated table, the elution time for the three compounds ranged from 6.60 minutes to 9.76 minutes. Since a solvent A and a solvent B were used, how do I know which compounds were eluted by which solvent? (If my interpretation of how the solvents A and B work is wrong, please correct me). Also, how does the use of two solvents affect understanding the relative polarity of the compounds?

You have several queries. First of let us look at the gradient method used by the authors.

The HPLC gradient for synthetic colorants was 0 min A 100 p. cent, 10
min A 0 p. cent, 14 min A 100 p. cent. The HPLC gradient for natural
carmine was 0 min A 80 p. cent, 20 min A 50 p. cent, 23 min A 30 p.
cent, 25 min A 30 p. cent

Factually, notice the trend in the mobile phase for synthetic ones. It is a linear gradient elution that they started with 40% MeOH at t=0 min, and at t=14 min, you have 70% MeOH. In the second case, it is a purely empirical "step" gradient. Try to draw a graph on a piece of paper.
Regardless, the authors use a water-rich mobile phase in the beginning and go to methanol rich mobile phase. Note I am avoiding the word polarity.

Polarity is an umbrella term in chromatography. For elementary purposes, it may be okay for serious work predicting retention on the ill-defined polarity that carries little or no meaning. You linked a GC webpage for an HPLC separation; those are different worlds.
Qualitatively, in top figure (a), it is clear that amaranth spends more time in the water-rich mobile phase, followed by azo ruby and then ponceau 4R if we use a C18 phase.

All you can say is that amaranth has a higher solubility in the water-rich mobile phase than ponceau 4R. One should not over-interpret retention behavior with this limited data.
In short, polarity is not a great terminology for describing the retention order. If you are interested, see the "Hydrophobic subtraction model" of Snyder et al. This one describes the retention in a better way of C18 molecules.
2)

In the chromatogram above (specifically chromatogram "a") and the
associated table, the elution time for the three compounds ranged from
6.60 minutes to 9.76 minutes. Since solvent A and solvent B were used, how do I know which compounds were eluted by which solvent?

This is an interesting question. As you noticed, the composition of the mobile phase changing with time, so the k, the retention factor is changing with time. People have worked out the theory of simple linear gradients. You know the retention time of elution, and you have the gradient composition chart. Correlate it, but this is a very, very rough estimate. The actual treatment is way more involved. Check "Theory of Gradient Elution Liquid Chromatography with Linear Solvent Strength: Part 1. Migration and Elution Parameters of a Solute Band" by Leonid Blumberg to see the complexity of the problem of gradient elution.
One of the best books is  "Introduction to Modern Liquid Chromatography By Lloyd R. Snyder, Joseph J. Kirkland, John W. Dolan, 2011."


Do water hybrid cars really exist? I have heard a few claims about water-fueled cars. Being a student of Engineering I totally understand that water-fueled cars (using electrolysis to generate Hydrogen) are nothing more than a hoax, unless  you are giving electricity from an outer source.

However, I have heard quite knowledgeable people that water hybrid cars are quite a reality and you can save up to 60% on your fuel costs using some kind of kits. I have difficulty agreeing with this claim as well. I think inclusion of another step (hydrolysis) in the energy cycle of car power system will only make it less efficient. Reason being you can achieve at most as much energy from making a bond (burning Hydrogen) as much you spent on breaking it (Electrolysis).

Can someone please help me understand if I am missing some point? Or I am correct?

The idea of splitting water using electrolysis and pumping it into an engine's air intake has been around for a long time. The idea is that it's supposed to increase the efficiency of burning gasoline. However, even assuming that this is true (there are many reasons why it may not be—modern gas engines are in the range of 95–98% efficient at combustion), electrolysis is very inefficient, so I'm quite dubious of any potential gains.

Hunting around on the internet, it's obvious that a lot of people seem to think this works (and that people who say it doesn't are oil company shills or some such), but I wasn't able to find any compelling evidence to support it. How this kind of system interacts with a modern computer controlled engine may perhaps even make efficiency worse. There doesn't seem to be much academic research into it, though a Google Scholar search brings up a lot of kooky patents. This kind of thing is pretty difficult to test well—tests need to be performed under very well controlled conditions to get accurate results and is thus always subject to confirmation bias and the like.

I did find a single report done by Environment Canada on one of these products for diesel engines and they found that adding the hydrogen generator device didn't make any statistically significant change to combustion efficiency, exhaust emission rates, or fuel consumption. They calculated that the device was about 60% efficient at making hydrogen, at best, and used about 8 mL of diesel fuel every 100 km, so it doesn't really make efficiency worse to any appreciable degree, but doesn't help either. (Source: ERMD Report # 2004-032 If you search, it's on scribd, but I couldn't find any permanent sources)

In short, a water-fuelled car on its own or as a hybrid with gasoline or diesel don't seem to be possible. Water is a very stable compound and it takes a great deal of energy to break it into hydrogen and oxygen gas. At most, you can get the energy you put in back, and realistically, a lot will be lost along the way. This makes a solely water fuelled car a perpetual motion machine of the first kind as it violates the law of conservation of energy—one cannot burn hydrogen and oxygen to produce energy to make the same amount of hydrogen and oxygen consumed and still have energy left over to move a vehicle. A hybrid vehicle faces a similar problem. Because the part of the system that involves water consumes energy rather than contributes it to motion, the gas or diesel part has to produce more energy to make up for the difference. Given that modern gas and diesel engines burn almost all the fuel put into them, there doesn't seem to be any mechanism by which adding a small quantity of hydrogen and oxygen could produce significantly more energy.


DFT Code for Atoms : Sources Some time ago I implemented the restricted and unrestricted Hartree-Fock methods. I had a lot of fun and I learned a lot. It goes without saying that Szabo and Ostlund's book "Modern Quantum Chemistry" has been fundamental to achieve a working code.

I would like to repeat the experience implementing a DFT code. I would like to implement it for atoms with all-electron non-relativistic calculations (scalar relativistic calculation will maybe be an interesting extension as well as the generation of norm-conserving pseudopotentials). Unfortunately I did't came across a "DFT bible" such as Szabo and Ostlund's book.

Does anybody knows good references (at the same level of Szabo and Ostlund's for in-depth discussion of the practical implementation) for the practical implementation of DFT for atoms (radial SE, logarithmic grids, different functionals, ...)?

Note that I am not searching a general DFT introduction or an explanation of real PW+NCPP implementations.

This question is similar to this one but I don't think it is a duplicate. In the previous question I was asking about a book explaining the implementation details of DFT for periodic systems. DFT applied to periodic systems usually involves a PW basis set as well as pseudopotentials and FFTs. This corresponds to a simplified version of the PWSCF code of QuantumESPRESSO.

In this question I am asking for the references that will help me to write of an atomic code, i.e. no basis set expansion and no pseudopotentials (all-electron calculation, logarithmic grid, ...). This corresponds more and less to the atomic code of QuantumESPRESSO which is usually used to generate pseudopotentials.

Marx, D.; Hutter, J. Ab initio molecular dynamics: basic theory and advanced methods; Cambridge University Press: Cambridge, 2012. This has some info on implementation of algorithm in CPMD code including flow chart of algorithm.
Thijssen, J. Computational Physics, 2nd Ed.; Cambridge University Press: Cambridge, UK ; New York, 2007. Chapter 6 comes with a short computer program downloadable from companion site written by author of the book.

Maybe these two help a bit.


Double surface in the pressure temperature concentration curve for a binary mixture In the 3 dimensional diagram given, when a constant pressure plane intersects with the double surface, it produces a looped curve without a maxima or minima. What is the double surface and how is the looped curve produced? And since the line joining $T_\ce{A}$ and $C_\ce{A}$ is the temperature-pressure curve of pure A and similarly for B, I cannot understand what the curve joining the 2 critical points (of A and B) means.

The liquid phase is above the surface. The vapor phase is below the surface.

The planar cut at a constant temperature is easiest to understand. The bottom curve will show the composition of the liquid phase as a function of molar fraction, and the upper curve will show the composition of the vapor phase.

Likewise for the curves of molar fraction versus temperature at a given pressure.


Why does thallium hydroxide increase the yield of product in a Suzuki reaction? I came across a research paper by Uenishi et al. [1] where relative rates of $\ce{KOH}$ and $\ce{TlOH}$ in Suzuki coupling reactions are compared.

Thallium hydroxide increased the yield of products, see Table I below. The reaction also occurred at room temperature as compared to $\ce{KOH}$.
What is the effect of thallium hydroxide?

References

Uenishi, J.; Beau, J. M.; Armstrong, R. W.; Kishi, Y. Dramatic Rate Enhancement of Suzuki Diene Synthesis. Its Application to Palytoxin Synthesis. J. Am. Chem. Soc. 1987, 109 (15), 4756–4758. https://doi.org/10/fhmqsr.

The Suzuki coupling reaction (also called Suzuki-Miyaura coupling reactions; Ref.1) is the coupling of an aryl or vinyl boronic acid with an aryl or vinyl halide or triflate using a palladium(0) catalyst similar to Heck reaction and Negishi reactions in mechanistic aspects. In particular, Negishi reaction uses organozinc reagents instead of organoboronic acids. Both reactions are powerful cross-coupling methods that allows the synthesis of conjugated olefins, styrenes, and biphenyls. The mechanism of Suzuki coupling reaction is depicted below:

The palladium catalyzed mechanism begins with oxidative addition of the organohalide ($\ce{R^1 -X}$) to the $\ce{Pd(0)}$ (e.g., $\ce{Pd(0)L_n}$) to form a $\ce{Pd(II)}$ complex (denoted in mechanism as $\ce{R^1 -Pd(II)L2X}$). It is worth nothing that this first step in both of the Suzuki and Negishi cross-coupling reactions is identical to that of the Heck reaction. This complex undergoes halide ligand exchange with the base present in the reaction mixture.

This base is present to serve one or two possible roles: Either to react with the organoboron reagent (denoted as $\ce{R^2 -B(OH)2}$) to form a trialkoxyboronate (denoted as $\ce{^-B(OH)3R^2}$), which then attacks the palladium(II) halide complex (denoted as $\ce{R^1 -Pd(II)L2X}$), and/or to convert of the palladium(II) halide complex to a palladium(II) oxo complex (denoted as $\ce{R^1 -Pd(II)L2OH}$) that reacts with the neutral organoboron reagent ($\ce{R^2 -B(OH)2}$) (Ref.1 & 2).

Transmetalation with the organoboron reagent then follows where the $\ce{R^2}$ group of $\ce{R^2 -B(OH)2}$ replaces the hydroxide anion on the palladium(II) complex to give $\ce{R^1R^2 -Pd(II)L2}$ and boric acid. Keep in mind that organoboron compounds are highly covalent in character, and may not undergo transmetallation readily in the absence of a base. Therefore, the reaction of the base with the organoboron reagent, $\ce{R^2 -B(OH)2}$ to form $\ce{^-B(OH)3R^2}$ is very likely. Also note that transmetalation step is highly depend on the base availability and the types of ligand on the catalyst (Ref 1 & 3).

Reductive elimination then follows to give the final coupled product regenerating the catalyst, hence the catalytic cycle can begin again.

Based on this mechanism, availability of a base is very important to reaction to proceed. It is clear that AP's question does not indicate the solvent used in the reaction. However, mentioned reference has been using THF as the solvent. Since solubility of $\ce{KOH}$ in THF is very low and hence its availability to the reaction. Thallium(I) hydroxide is strong base compatible with alkali bases such as $\ce{KOH}$ and key to its capability of enhancing the reaction here should be its readily availability in such solvents. This is evidence that in recent publication (Ref.4) where $\ce{KOH}$ and $\ce{TlOH}$ are equally effective in $\ce{Pd(0)}$  catalyzed coupling reaction, yield of which increases when solvent has been changed from THF to DMF.

References:

Norio Miyaura, Akira Suzuki, "Palladium-Catalyzed Cross-Coupling Reactions of Organoboron Compounds," Chem. Rev. 1995, 95(7), 2457-2483 (https://doi.org/10.1021/cr00039a007).
Karl Matos, John A. Soderquist, "Alkylboranes in the Suzuki−Miyaura Coupling:  Stereochemical and Mechanistic Studies," J. Org. Chem. 1998, 63(3), 461-470 (https://doi.org/10.1021/jo971681s).
Norio Miyaura, Kinji Yamada, Hiroshi Suginome, Akira Suzuki, "Novel and convenient method for the stereo- and regiospecific synthesis of conjugated alkadienes and alkenynes via the palladium-catalyzed cross-coupling reaction of 1-alkenylboranes with bromoalkenes and bromoalkynes," J. Am. Chem. Soc. 1985, 107(4), 972-980 (https://doi.org/10.1021/ja00290a037).
Hiroko Koyama, Hisashi Doi, Masaaki Suzuki, "Evaluation of $\ce{TlOH}$ Effect for $\ce{Pd^0}$-Mediated Cross-Coupling of Methyl Iodide and Excess Boronic Acid Ester toward Fabrication of $\ce{[^{11}C]CH3}$-Incorporated PET Tracer," Intl. J. Org. Chem. 2013, 3(3), 220-223 (DOI: 10.4236/ijoc.2013.33030).


Heat of reaction for the decomposition of AgCl means? Certain sunglasses have small crystals of $\ce{AgCl}$ incorporated into the lenses, on exposure to light of appropriate wavelength produces a grey colour to reduce the glare following the reaction:
$$\ce{
  \underset{(colourless)}{AgCl}
  ->[h\nu]
  \underset{(grey)}{Ag} + Cl
}$$

If the heat of reaction for the decomposition of $\ce{AgCl}$ is $\pu{248 kJ/mol}$, what maximum wavelength is needed to induce the desired process?
Does this mean that by absorbing only a certain wavelength, electrons move for a lower to higher energy state and react?
Why does it only emit grey light, but does not any colours of the spectrum in order?
The question asks about the "heat of decomposition".  Does this mean the energy required to excite an electron and cause a reaction?
When a molecule absorb a photon, it is breaking the atoms apart by  moving electrons?
Please help clear up any misunderstandings in this question.

The intent of the question is that the reaction:
$$ \ce{Ag+} + \ce{Cl-} \rightarrow \ce{Ag} + \ce{Cl}$$
requires absorbing exactly one photon. That photon must have at least the energy required to perform this reaction. The grey colour produced is caused by the metallic silver ($\ce{Ag^0}$) formed by the reaction.
Because a photon's energy is inversely proportional to its wavelength, $E_{photon}=\frac{h c}{\lambda}$, in order to have the required energy or more to cause the reaction, the photon must have a required wavelength or less.  So there is a maximum wavelength that the photon may have in order to cause this reaction.
You're told that the energy required to decompose 1 mole of $\ce{AgCl}$ is 248 kJ/mol, so you have to figure out the energy per atom/photon and convert that energy to a wavelength.
This is an example of the photoelectric effect - that light comes in quantised particles (set amounts of energy) called photons, and certain processes require absorbing a single photon at a time with enough energy, rather than absorbing more lower energy light over time, as if the light was only a continuous wave.


Which is the major product of bromination of 3-(dimethylamino)benzamide, ortho or para? I know that A will most probably not happen because of high steric hindrance. However between B and C, will B become the major product? Because $\ce{-N(CH3)2}$ is less bulky compared to $\ce{C=O-NH2}$, so there will be less steric hindrance.

Here's the 3D structure of the substrate:

As you can see, the $\ce{-N(Me)2}$ group is clearly larger than the $\ce{-CONH2}$ group. Also notice that the $\ce{-N(Me)2}$ group is planar with the benzene ring. This is because this structure is the most favourable for $\mathrm{p-\pi}$ conjugation 1.
Your guess that A would not form is correct. The large size of the $\ce{-N(Me)2}$ group would also point to the fact that C is the major product. For a better explanation of the major product, we can take a look at the 3D structures of B and C:

You can clearly see that in B, the bromine atom pushes the dimethylamine group out of it's stable configuration, where it is coplanar with the benzene ring. This requires more energy and also produces an unstable product. Therefore, C is the major product in this reaction.
References:

Novikov, V. P., et al. “Molecular Structure and Conformation of N,N-Dimethylaniline by Gas-Phase Electron Diffraction and Quantum-Chemical Calculations.” Russian Journal of General Chemistry, vol. 74, no. 8,2004, pp. 1247–53. doi:10.1007/s11176-005-0146-9.


Does bleach react with disinfectants containing quaternary ammonium salts? My mom went nuts when we pulled up the laminate to discover mold growing on the subfloor. She sprayed bleach all over it while I was at the store buying Mold Armor mold remover and disinfectant, and found out from calling poison control that this product should not be mixed with bleach. The operator was not sure what I could do to make it so it was safe to apply the mold remover.

Can anyone tell me if when bleach dries, the chemical reaction will be mitigated? I just want to clean my house without killing everyone. The surface is porous, and made of wood/particle board.

Calling poison control first was the best thing you could do! Thumbs up!

Please note that we're usually reluctant to give medical advice - we don't want to get you or anyone in your house get killed or harmed either.

You might want to call poison control again and/or send them the safety data sheet of the product.

The following is just additional information to give you some direction on why it is probably safe to use the mold remover on the floor previously treated with bleach.

The College of Veterinary Medicine at Cornell University has published Guidelines for the Use of Bleach. According to this document,

  Bleach can safely be mixed with quaternary ammonium compounds that contain detergents, such as A-33 and Roccal to create a single solution with both cleaning and disinfecting properties.

So, I'd dare to say that the combined use, as well as subsequent treatment of the floor is safe, but if I were you, I would search again confirmation with official poison control rather than trusting some guy from the internet ;-)


Can ring inversion in Cis-Decalin allow the E2 reaction? This is a tutorial question I need to do, I initially thought the question was testing whether students properly understand (1) chair conformations of decalin, and (2) that the leaving group and the breaking bond should be 'anti-periplanar'.
However, the third molecule also turned out to have undergone the elimination reaction when the first didn't, so I am very confounded. I have drawn all possible conformations of cis-Decalins, but no single conformation seems to allow $\ce{C-OTs}$ bond and $\ce{C-OH}$ bond to be anti-periplanar. Am I missing out something?

The issue in this Grob fragmentation is whether or not the $\ce{C-C}$ and $\ce{C-OTs}$ bonds are anti-periplanar. The $\ce{C-OH}$ bond is free to rotate and the unpaired electrons of the negative oxygen can align with the other two bonds. In the first example a), the red bonds are gauche to one another. Standard E2 elimination is likely the source of the mixture. Examples b), c) and d) all have red bonds anti-periplanar. The blue bonds predict the geometry of the double bonds in the product. Examples b) and c) give E-alkenes while d) affords a Z-alkene.


Entropy of solid, liquid, and gas at triple point of water At the triple point of water how do the entropies of solid,
  liquid, and gas compare?

I think that either they will be equal or it will be that gas > liquid > solid.

I don't know if entropy is influenced by the fact that it will be at the triple point. Can someone please elaborate?

Triple point defines a situation of simultaneous equilibrium between the solid, the liquid and the gas phases.

For such an equilibrium, you simply write:

$$
\Delta S= \frac{\Delta H}{T}
$$

that rises from the fact that $\Delta G=0$.

Considering water molar ($m$) enthalpies for each phase transition at $273\,K$:

$$
\Delta H_{melting,m}=6.01\,kJ\,mol^{-1}
$$
$$
\Delta H_{vaporisation,m}=45.05\,kJ\,mol^{-1}
$$
$$
\Delta H_{sublimation,m}=51.06\,kJ\,mol^{-1}
$$

one realises that

the entropy of the gas phase is higher than the entropy of the liquid phase.
the entropy of the liquid phase is higher than the entropy of the solid phase

In this respect the triple point has no peculiar behaviour compared to other points where a two-phase equilibrium is established.


Unsure whether this classification of Volatile Organic Compound and/or Particulate Matter is correct I am asking on behalf of a friend so please bear with me!

Instructions:

For each chemical, classify whether it is a VOC (Volatile Organic
Compound) and/or PM (Particulate Matter 2.5 or Particulate Matter 10).

A chemical can sometimes be both VOC and PM, they are not mutually
exclusive.

If it is not VOC or PM, then label it as Other.

Use the following definition for VOC: "Volatile organic compounds
(VOC) means any compound of carbon, excluding carbon monoxide, carbon
dioxide, carbonic acid, metallic carbides or carbonates, and ammonium
carbonate, which participates in atmospheric photochemical reactions."

Use the following definition for PM2.5 and PM10: "PM10 means
particulate matter with an aerodynamic diameter less than or equal to
a nominal 10 micrometers" "PM2.5 means particulate matter with an
aerodynamic diameter less than or equal to a nominal 2.5 micrometers"

The main trouble is that while I can classify the VOC's, I am not at all sure about the PM's and others.
I have already marked some chemicals with "1" meaning that it is a VOC, but I am unsure of whether it is also a PM, if it is not a PM then just put a "0" in that cell. For example, for "2-ACETYLAMINOFLUORENE", if this is not a PM, then it should be VOC = 1, PM = 0, Other = 0.
Some other chemicals are completely empty such as "CHLORAMBEN" and "HYDRAZINE". For these, I simply have no idea how to classify them.
Could anyone check what I have done so far, and provide some hints/guidelines on what to do with the rest, especially on how to classify a PM please?
Thank you so much!

2-Acetylaminofluorene is a solid, mp 192-196 C. It satisfies the definition of being a VOC by virtue of being a compound of carbon, not excluded by the definition. It could be fine enough to be classified as PM 2.5 or 10 if it is powdered; more information would be necessary about the source of the compound. It is also carcinogenic, but that does not appear to be involved in the questionnaire.
Chloramben can be looked up and is also a solid with a high melting point: a VOC and possibly a PM.
Hydrazine is only dangerous, but not a VOC, and a liquid with a bp of 114 C. Not a PM. I don't think a spray of droplets would qualify as a PM; they would just evaporate.
Diisocyanates: VOCs for sure. Hazardous for sure. If you allow a pot of diisocyanates to evaporate, that's all: no PM. However, I can imagine a source of sprayed diisocyanates (in a paint?) that would result in polymerized particulate matter: then it could be a PM. Thus the classification would depend on the source of the material - the environment and the application.
Polycyclic aromatic compounds: most often encountered in smoke from various sources: VOC for sure, probably also PM because they could be larger agglomerates than single molecules.
This could be a fun exercise for a chemist to look up and play with mentally, to get to know and understand a large number of chemicals. For a non-chemist, this could be very difficult.


Why is arsenic more chemically similar to phosphorus than is nitrogen? Why is arsenic more chemically similar to phosphorus than is nitrogen (to phosphorus)?

I thought that it may be because both phosphorus and arsenic have d orbitals (albeit one has one that is filled and the other has an unfilled one).

Another possibility I thought was that because nitrogen held the electrons very closely, and thus the difference between the second and third period may be more pronounced than the difference between the third and fourth periods.

I'd really appreciate it if anyone could help clarify/ explain. Thanks!

Every element which is in the first place of the group, is pretty much different from the other elements of the same group. The reason is because the first element has ALWAYS the smallest radius, which leads to some changings also in group properties. Due to the small radius they are more stable than the others and they require more energy for any specific reaction. With some research you can see that also other  first elements of other groups such as Li,O,C etc.  differ from the other elements of that specific group.


Strange valence numbers? I'm trying to understand why ChemReference sometimes lists the valence electrons as a number I don't expect. Take Oxygen, for instance. It should have six valence electrons, however the site lists two. Why?

I hear that, for transition metals, the d-orbitals count towards the valence. Is there another rule like that I need to know about that might explain Oxygen?

Look closely on that site. It's not 'valence electrons' but 'valence'. There is a difference between the valence electrons and valence.

Valence electrons are the number of free electrons in an atom that can take part in making the bond. There are 6 such electrons in the oxygen atom as you stated.

While the valence or valency is the number of bonds that can be made by the atom.

That's why valency of oxygen is 2


Borax bead test technique In the observation tables provided in books and websites two different colors are assigned to a specific borax+salt bead in both oxidising and reducing flames. Not all basic radicals show this trend but some like Cu$^+$$^/$$^+$$^2$ does. It appears green in hot condition and blue in cold condition when made using oxidising flame.

My question is what is done to actually reduce the temperature of the borax bead? Is it held under running tap of water or just outside the flame?

"Hot" refers to the bead when just taken out of the flame of the Bunsen burner, while "cold" refers to the state when the bead cooled to room temperature. Provided the small size these beeds typically have (a toothpick like rod of magnesium oxide to carry them), this is just a matter of about a minute of exposure to ambient air. Duing this process, you may spot differences, too.  Conversely, intentionally cooling the hot glass-like beads under tap water often shatters them into pieces.

Side note #1:  Testing with a bead of Borax is a one quick pretest, to be confirmed by subsequent experiments.  And as a good habit: once you found the ions, compare your bead made with the analyte with a reference, a bead made with a sample of known composition.

Side note #2: Of course I miss one, an entry for cobalt ... regardless of the flame (oxidation or reduction), hot or cold: an intense dark blue. Almost Yale blue.


Does adding water to a reaction mixture shift equilibrium? Just like decreasing volume in a gaseous reversible reaction shifts the position of equilibrium towards the side that produces more moles (increasing pressure), shouldn't adding water shift equilibrium to side that produces more moles in a liquid equilibrium mixture (increasing concentration)?

On research, I've seen an explanation that it does not because the concentration of both reactants and products decrease evenly, but I don't understand this. The side with more moles should have a greater decrease in concentration because the concentration is proportional to the number of moles. Also, by the same argument, decreasing volume in a gaseous reaction should not shift equilibrium.

My guess is that adding water does affect equilibrium for a reaction with different number of moles on each side. Because the equilibrium constant is affected. Is this right?

Yes, the equilibrium does indeed readjust after a change in concentration. This can be seen by careful examination of the formula of the equilibrium constant $K$. Consider an arbitrary reaction that produces two products from a single reactant and is in equilibrium:

$$\begin{align}\ce{A &<=> B + C}\tag{1}\\[0.5em]
K &= \frac{[\ce{B}][\ce{C}]}{[\ce{A}]}\tag{2}\end{align}$$

Let’s assume initial concentrations of $\pu{2M}$ for each species; that gives us $K = 2$ in dimensionless units (you can also use M as a unit here). Remember that $K$ is a constant so we can examine whether equilibrium has been reached by examining $K$.

Now let’s double the volume of solvent. The new concentration of each reactant is $\pu{1M}$. Therefore:

$$K' = \frac{[\ce{B}][\ce{C}]}{[\ce{A}]} = \frac{1\times 1}{1} = 1\tag{2'}$$

Since $1\ne2$, we are no longer at equilibrium. Thus the concentrations must readjust. How? Obviously, the value of the fraction is too low. To increase its value, we can increase the numerator or decrease the denominator — luckily for us, both of these mean that we must perform the reaction in a forward direction to get think right again. Since $[\ce{B}]=[\ce{C}]$, we can actually calculate the ratio of concentrations and from $2[\ce{A}]+[\ce{B}]+[\ce{C}]=\text{const.}=4$, we can calculate exact values.

$$\begin{align}&&K &= \frac{[\ce{B}]^2}{[\ce{A}]}\\[0.5em]
\Longrightarrow&&[\ce{A}] &= \frac{[\ce{B}]^2}2\\[1.3em]
&&2\times\frac{[\ce{B}]^2}{2} + [\ce{B}] + [\ce{B}]&=4\\
&&[\ce{B}]^2 + 2[\ce{B}]-4 &= 0\\
\Longrightarrow&&[\ce{B}]_{1/2} &= \frac{-2\pm\sqrt{2^2+4\times4}}{2}\\[0.5em]
&&[\ce{B}]_{1/2} &= -1\pm\sqrt{5}\\[0.5em]
&&[\ce{B}]_{1/2} &\approx 1.24\end{align}$$

Thus, the product side has increased.

A more real-world example is maybe the dimerisation of hydrogenchromate $\ce{HCrO4-}$ according to $(3)$:

$$\ce{2 HCrO4- <=> Cr2O7^2- + H2O}\tag{3}$$

This is also pH-dependent, since hydrogenchromate is a weak acid and can deprotonate to give chromate $\ce{CrO4^2-}$. However, it is fairly easy to try this experiment yourself: take an orange coloured dichromate solution and add water. A colour change to yellow should be observed indicating that the reaction has equilibrated to the reactants’ side.


Melting points of compounds? Compare the melting points of:

  1) $\ce{BeSO_4}$

  2) $\ce{CaCO_3}$

  3) $\ce{BaCO_3}$

  4) $\ce{SrCO_3}$

Okay, so there are a few statements given by my teacher:

1) Thermal stability of ionic compounds is proportional to the counterbalance effect of electropositive field of cations(magnitude of +ve charge) and electronegative field of an anion (mag. of -ve charge). (I interpret this as the lesser the difference between the magnitude of charge, the more thermally stable the compound.)

2) The real magnitude of charge is directly proportional to the number of electrons lost(basically the number we write next to an element, like $\ce{Ca^{2+}}$, right?) and inversely proportional to the size of the ion.

3) Except the second group elements and halides, all anions can be considered as "large anions" and hence their effective magnitude of charge will be same (roughly).

So since all of them are "large anions", we have to compare the magnitude of charge on the cations. Now, being large, their charge is small. So for least difference, we want the cation with the smallest charge. This means the largest size, which simply means $\ce{Ba}$. So the answer should be 3. To confirm this, I Googled the melting points of these substances.

1) $\pu{110^oC}$
2) $\pu{811^oC}$
3) $\pu{825^oC}$
4) $\pu{1494^oC}$

So according to me, the melting points in decreasing order should be:

3 > 4 > 2 > 1

When it turns out to be:-

4 > 3 > 2 > 1

Where did I go wrong?

I personally think:

1) Since the charge on anions is only roughly same, I think that when we consider the difference between the charges, that would be messed up and so we would end up with wrong answers(possibly).

2) That since this question is supposed to be for high school students, we make a lot of assumptions, which obviously don't hold true in the real world, and hence the answer is wrong? That being said,is my answer correct keeping these assumptions in mind?

3) Since chemistry has a lot of exceptionss , is this another one of those? Am I going to have to memorize this too?

First, in order to compare mp's you need to know the mp's. You have two choices: 1. run melting point measurements or 2. find literature written by someone who has run the experiments. Perhaps this is a translation problem from your native language? Perhaps you meant predict the ranking of mp's for those 4 compounds.

You state that CaCO3 melts at 811 °C. As far as I know, that is incorrect. First, the thermodynamically stable polymorph is calcite, which according to wikipedia has a mp of 1339 °C but which decomposes at temperatures above ~840 °C (also according to Wikipedia!). I hope you see this is clearly contradictory. My first hand experience is that it does indeed decompose at around 820–840 °C, so I am unable to explain how they were able to measure its melting point 500 °C above that. 500 °C is an enormous gap, but perhaps they measured it under some enormous pressure (far from STP) - I really don't know.

The other major problem I'm having is understanding what you (or your teacher) means by "counterbalance the effect of electropositive field of cations ... and electronegative field of an anion." It seems to me to be meaningless, at least I don't recognize this terminology in reference to the electrostatic forces usually invoked to account for ionic solids. BeSO4 is an obvious problem since Be is known to for its tendency to form covalent compounds. Indeed, according to Wikipedia, anhydrous Be sulfate has Be-S (covalent) bonding. While you could predict that the covalent bonding would reduce the mp, what you'd do with the polymeric nature of the solid is beyond me - I've no idea how you or your teacher can explain that, except by hand waving.

This leaves the Ba and Sr salts. Simple logic of ion size would predict the order Ba < Sr and ignoring the ridiculous assertion on Wikipedia that the Ba salt's melting point is 811 °C while at the same time (you did READ the article, right?) stating that that is the temperature at which it a "polymorphic transformation" occurs. A little further digging provides a safety data sheet which claims a mp of 1400 °C "close to its decomposition temperature", but such data should be suspect.

Your teacher's choice of picking carbonate as the anion was a poor one, in my honest opinion. Because of the tendency of these salts to decompose, rather than melt, you need to be very careful where you obtain your data from. But perhaps that is the lesson your teacher hopes some of you will learn - to not just accept a number because it is readily available (without subjecting it to critical analysis (who, what, where, why, when, how - which in this case means how reliable is the reference? does it reference a primary (academic) source? if not, an authoritative secondary source? if not, then you should seek independent confirmation. Such confirmation is hard to find on the internet, since one lazy site steals from another until the noise drowns out the signal.

I really don't know if density functional theory (DFT) quantum mechanical modeling by supercomputers is able to accurately predict the mp's of these material. My guess is that because of the change in crystal structure(s) that it remains beyond our current capabilities (for some things, calculating the value has become reasonably accurate – as far as I know (and I'm no expert) mp isn't one of the successes. If it was accurate it would be a third way to "compare" the mp's.


Sulfuric Acid - Bond Lengths and Charge Separation So I know that the best Lewis Structure representation of sulfuric acid is one with only single bonds - i.e. sulfur isn't hypervalent.

However, why are the $\ce{S-O}$ and $\ce{S-OH}$ bonds not equivalent then? Why do their lengths differ?

I suspect this has something to do with the addition of the hydrogen on the oxygen atom.

What else could be a cause for the discrepancy?

I ask because my teacher challenged the fact that sulfur might not be hypervalent on the basis that the bond lengths differ.

In retrospect, I think that he:

1) Is placing too much weight on Lewis Structures - which are just rough renditions of what chemicals really are.

2) The $\ce{S-O}$ bonds and $\ce{S-OH}$ bonds may differ in length not because the $\ce{S-O}$ bonds are necessarily double bonded but also because the $\ce{S-OH}$ bonds are different in that S is bound to a hydroxyl group rather than a single oxygen. Also can't there be very minor resonance contributors involving 5-6 bonds on sulfur? It's not that sulfur absolutely can't access its d-orbitals but rather it doesn't to any great extent.

Good questions, let's step through them one at a time.

1) "So I know that the best Lewis Structure representation of sulfuric acid is one with only single bonds - i.e. sulfur isn't hypervalent. "

Look at the resonance structures below.  The one on the left with sulfur hypervalent is the only uncharged resonance structure.  I suspect that that counts for something.  Here is a link that says it is the most important of the structures (don't get confused when you view the link, they use equilibrium arrows (wrong) in their diagram instead of  resonance arrows (correct).

2) "why are the S-O and S-OH bonds not equivalent then? Why do their lengths differ?"

This is a very fundamental question.  My answer is, they are not equivalent because they are different.  Just as you wouldn't expect the same $\ce{C-O}$ bond length in $\ce{CH_3-O-CH_3}$ and $\ce{CH3-OH}$, you wouldn't expect the same bond lengths in an $\ce{S-O}$ bond and an $\ce{S-OH}$ bond.  In fact, since these bonds are not superimposable one on the other by a symmetry operation, we can say that they MUST be different (i.e. have different lengths).  As one of my favorite professors used to say, now if were to actually measure the $\ce{S-O}$ and $\ce{S-OH}$ bond lengths and our measurements gave rise to the same numeric bond length out to 10 decimal places, we could still step back and say that both we and God know that the lengths are not the same, we just need to measure out to more decimal places!  Let me say the converse, the only bonds that can be identical are those that can be interconverted one with another through a symmetry operation.  For example in chloroform, $\ce{CHCl3}$, there is a $\ce{C_3}$ axis containing the $\ce{C-H}$ bond.  Rotation about this axis interconverts the 3 $\ce{C-Cl}$ bonds, therefor they are equivalent bonds.  Their bond lengths MUST be the same.


Among Benzyl carbocation (Ph-CH2+) and Methyl methoxy carbocation (CH3-O-CH2+), why is the latter more stable? Benzyl Carbocation could form multiple conjugation, then why isn't that stabler? In Methyl methoxy carbocation, Oxygen donates the lone pairs but so does the double bond in the benzyl Carbocation.

let's see the resoance forms of benzyl carbocation and methoxy carbocation:
-Benzyl carbocation:

You are right, benzyl carbocation have multiple ressonance forms which means that It must be very stable. In fact, benzyl carbocation is one of the most stable carbocations.
-Methoxy carbocation:

In this case, we only observe 2 resonance forms.
so the question is, why methoxy carbocation is more stable despite of having less resonance forms??
Well, the key is the stability of this resoance forms...
In benzyl carbocation, none of its resoance forms have a complet octet, wich mean that the resonance forms are not very stable and therefore the positive charge is not very delocalized.
However, in the case of the methoxy carbocation, one of the resonance form have a complet octet, which means that the resoant form are pretty stable and therefore the positive charge is quite delocalized between the O and de C
In conclusion, number of resonance forms are important, but their stability is also a really important factor.


Why does the PH3 geometry deviate more from the trigonal planar one, than does NH3? $\ce{PH3}$ has a more bent structure than $\ce{NH3}$. The HOMO-LUMO gap for $\ce{PH3}$ is smaller than for $\ce{NH3}$, and so the distortion from the trigonal planar geometry is said to be larger. However, why is the HOMO-LUMO gap smaller for $\ce{PH3}$?

It should be possible, I suppose, to argue for why this is so by using qualitative molecular orbital theory, and from looking at the construction of the molecular orbital diagrams, but I am not sure how to construct accurate MO diagrams for $\ce{PH3}$ and $\ce{NH3}$.

In the answer to the possible duplicate question, it seems to be taken for granted that some distortion from the planar structure is stabilizing. In my question, the starting point is the planar structure, and then looking at why $\ce{PH3}$ is in a more distorted geometry than $\ce{NH3}$, relative to the planar structures.

This distortion is often explained in terms of the "HOMO-LUMO gap", but I fail to see how the HOMO-LUMO gap can be assessed qualitatively in order to predict which of $\ce{NH3}$ and $\ce{PH3}$ is in a more distorted geometry. So what factor ultimately leads to the smaller HOMO-LUMO gap in $\ce{PH3}$? $\ce{BH3}$ exists in a trigonal planar geometry - why is not this molecule pyramidalized?

Apparantly it was community consensus that this question be clear enough to stay open.

Neither $\ce{PH3}$ nor $\ce{NH3}$ have any incentive whatsoever to assume a trigonal planar geometry. The most stable ground state of both nitrogen and phosphorus atoms is $n\mathrm{s^2},\ n\mathrm{p^3}$ where $n = 2$ for nitrogen and $n=3$ for phosphorus. Since the p-orbitals are higher in energy per se, it makes sense to use them for bonding and lower their energy — which, as a side note leads to much more efficient overlap than using s-orbitals for bonding, since p-orbitals have a direction in space.

Therefore, the most stable geometry for both would be trigonal pyramidal with $90^\circ$ angles between the substituents. This would allow all-p bonding orbitals and an all-s lone pair on the central atom. If you check out $\ce{PH3}$, you will find almost exactly that geometry save a little distortion.

The distortion you see in $\ce{PH3}$ is of the same type as the one which moves the hydrogen atoms much further apart in $\ce{NH3}$: Steric stress. The hydrogens themselves repel each other; they require a certain space. Around a large phosphorus atom there is enough space for them to crowd at the ideal bonding angle while the small nitrogen does not offer that space. The final bond angle of $\approx 107^\circ$ is a fine-tuned result of trying to keep the hydrogen atoms as far away from each other as necessary and keeping the bonding angle as close to $90^\circ$ as possible.

$\ce{BH3}$ is an odd one out in this comparison. Boron does not have a lone pair and thus we expect isolated $\ce{BH3}$ to assume trigonal planar geometry. However, we also expect isolated $\ce{BH3}$ to have a low stability since boron does not have an octet. Hence it dimerises to $\ce{B2H6}$ with two $\ce{B-H-B}$ banana bonds of the 2-electron-3-centre type. The result can be described as a distorted tetrahedron around boron.


Iodine Titration with Free SO2 in wine To determine the free $\ce{SO2}$ in wine, a common method used is called the Ripper Titration Method.

The equation is as follows: $$\ce{SO2 (aq) + I2 (aq) + 2H2O (l) -> 4H+ (aq) + SO4^2- (aq) + 2I- (aq)}$$

This method involves titrating an Iodine solution into a sample of wine. I have found that triiodide is made from reacting $\ce{I-}$ with $\ce{I2}$. However, if it is the triiodide we are titrating, how is the iodine reacting with the wine?

Is it triiodide ($\ce{I3-}$) that is being titrated with the wine sample or iodine ($\ce{I2}$)?

Reaction goes along the lines $\ce{3 I2 -> 2 I3- -> 6 I-}$. The beginning of titration is $\ce{I2}$. The end of titration is $\ce{I-}$. $\ce{I3-}$ forms as an intermediate, but it can be ignored in titration.


Oxidation State of Nitrogen in N2O If we try to calculate the oxidation state of nitrogen in $\ce{N2O}$ using the familiar algebraic method, we get oxidation state $+1$ for both nitrogen atoms and that's what I found when I looked it up on the internet.
I tried to do it with the structure and that's where I got confused:
$$\ce{\overset{-}{N}=\overset{+}{N}=O}$$
The $\ce{N=N}$ bond is a coordinate bond, so that should give a $-1$ oxidation state for the left $\ce{N}$ and $+1$ for the middle one. Since the middle one has a double bond with oxygen, it gets additional $+2$ for a total of $+3.$
But if I consider this resonance structure:
$$\ce{N#\overset{+}{N}-\overset{-}{O}}$$
The left $\ce{N}$ is getting an oxidation state of $0$ and the middle one gets $+2.$ So I am confused as to what is actually correct.
Is it possible for atoms to exist in superpositions of oxidation states? Or maybe I am overlooking some basic concept?

If we try to calculate the oxidation state of nitrogen in $\ce{N2O}$ using the familiar algebraic method, we get oxidation state $+1$ for both nitrogen atoms and that's what I found when I looked it up on the internet.

Well … you get an average oxidation state. This calculation arguably implicitly assumes that all nitrogen atoms be equivalent. In some cases (e.g. hydrazine) they are and the result you get algebraicly is as you would expect from a Lewis depiction. In other cases (e.g. here) this is not the case as the nitrogens are not equivalent (only one is bound to oxygen). Thus, it is obvious that a simple algebraic approach should fail.
But what about the ‘real’ result? What about the resonance structures? Well, this is where things get really difficult. You essentially have two π systems orthogonal to each other and each occupied by four electrons which can manifest as a lone pair on either end and a multiple bond to the other atom. If you really wanted to play the game, you could include another resonance structure as shown below, where all π lone pairs are centred on the terminal nitrogen:
$$\ce{N#\overset{+}{N}-\overset{-}{O} <-> \overset{-}{N}=\overset{+}{N}=O <-> \overset{2-}{N}-\overset{+}{N}#\overset{+}{O}}$$
(It is clear that this third resonance structure contributes least to the overall picture as it has a greater charge separation and charges are separated opposite to what electronegativity would predict.)
The experimental structure shows that the $\ce{N-N}$ distance is slightly shorter than the $\ce{N-O}$ distance which one could use to assume that the $\ce{N-N}$ bond have a slightly larger bond order than the $\ce{N-O}$ bond. But ultimately, they are still very similar (the difference is merely $\pu{4pm}$) so equal bond orders might also be an option. Long story short: short of calculating the electronic distribution (i.e. solving the Schrödinger equation) you won’t be able to arrive at a definite answer for the ‘real’ oxidation states.
So what can you do on paper? Comparing the resonance structures, the leftmost as I have ordered them is slightly better than the central one as the formal charges are distributed according to the different electronegativities. Thus, I would be inclined to give it a slightly higher weight and – in a classroom setting – use it to determine oxidation states. That said, it is too much of an ambiguous example to be seriously used in any examination unless the goal of said examination is to develop the chain of arguments that formed this answer.


What is the physical interpretation of "internal pressure"? $$ \require{begingroup}
\begingroup
\newcommand{\md}[0]{\mathrm{d}} \md U = \pi_T\,\md V + C_V \,\md T$$
  where  $\pi_T$ is the internal pressure and is given by $\displaystyle {\left(\partial U \over \partial V\right)_T}$.

What does the internal pressure actually physically mean? I can see it is the slope of the $U$-$V$ graph at constant temperature but that does not clear things up.
Also why is this quantity called the internal pressure and not something else? $\endgroup$

See orthocresol's derivation of an internal pressure $\pi_T$ relation $(1)$ here.

$$\pi_T = T\left(\frac{\partial P}{\partial T}\right)_V - P\tag{1}$$

Internal pressure $\pi_T$ measures what you would expect from its definition:

$$\pi_T \equiv \left(\frac{\partial U}{\partial V}\right)_T.$$

It shows how internal energy changes when volume is changed and temperature is constant. For ideal gases the change is zero, something that also follows from the equipartition theorem.

$$\mathrm{d}U_{\text{ideal gas}}= \frac{\nu}{2}nR\mathrm{d}T \implies \left(\frac{\partial U}{\partial V}\right)_T = 0 \ \  \text{for an ideal gas}$$

In other words, the total energy an ideal gas has is not dependent on volume. There is no repulsion nor attraction amongst ideal gas particles.

As a first approximation to model real gases, the van der Waals equation $(2)$ is used.

$$\left(P+a\frac{n^2}{V^2}\right)\left(V-nb\right)=nRT\tag{2}$$

Hence,

$$P = \frac{nRT}{V-nb} - a\frac{n^2}{V^2}\tag{3}.$$

Using relations $(1)$ and $(3)$

$$\left(\frac{\partial P}{\partial T}\right)_V = \frac{nR}{V-nb} \overset{(3)}{=} \frac{1}{T}\left(P + a\frac{n^2}{V^2}\right).$$

Again via formula $(1)$

$$\pi_T = T \cdot \overbrace{\frac{1}{T}\left(P + a\frac{n^2}{V^2}\right)}^{(\partial P/\partial T)_V} - P = a\frac{n^2}{V^2}\tag{4}.$$

Result $(4)$ $\pi_T = a\frac{n^2}{V^2}$ demonstrates that for real gases internal energy will change when volume is decreased or increased. Real particles have interparticular forces that vary with distance. A hint as to why $\pi_T$ is called internal pressure will be given when $(4)$ is substituted into $(2)$:

$$\left(P+\pi_T\right)\left(V-nb\right)=nRT,$$

or even better, $(3)$:

$$P = \frac{nRT}{V-nb} - \pi_T.$$

If $\pi_T > 0,$ then $P$ will be smaller. Therefore, attractive forces are dominant.
If $\pi_T < 0,$ then $P$ will be bigger. Therefore, repulsive forces are dominant.


How fast a surface oxide layer is formed on bulk chromium? If a chromium piece is cut in half, how long would it take for an oxide layer to form on the new surface (room temperature)? i.e. is it a matter of seconds/minutes/hours/days/weeks?

The closest study I found is by G. Thurner and P.H. Holloway (1992), but I can't really tell from their results what would happen at ambient conditions:

(I found several publications about high temperature oxidation, but none for surface oxidation at ambient conditions)

First, a definition of the Langmuir unit, used in the graph in the question, which is not often run across except by vacuum (particularly UHV) folks. One Langmuir is the exposure of a surface to a gas pressure of $10^{-6}$ Torr for one second. It roughly corresponds to one monolayer equivalent of gas hitting the surface. At normal atmospheric pressure, with 20% oxygen in air, 1 Langmuir of oxygen hits the surface in about 6.6 nanoseconds (yes, nanoseconds).

Another paper that could be of interest is C. Palacio et al., "AES, XPS and EELS study of the initial oxidation of polycrystalline chromium", Surface Science 182(1-2) 41-55 (1987). They see chemisorption of oxygen in the first 1-2 L, nucleation and lateral growth of oxide islands up to 10 L, and continued growth of an oxide film above 10 L.

Overall, there is not a huge difference between the two papers. By ~10 L of exposure to oxygen there is a continuous oxide on the surface at 300K. Following from the numbers in the first paragraph, that occurs in less than 100 nanoseconds. Basically, it oxidizes as it is cut.


Does tetrachlorodibenzo-p-dioxin (2,3,7,8) have a nonzero dipole moment? I've looked at and drawn the structure, and I originally felt that it would be nonpolar - as I believe all the dipoles would cancel out, but I'm not 100% sure. I might be too dependent on the fact that it looks symmetric.

Also, is it true that the overall molecule shape would be planar? And is the hybridization of the oxygen atoms sp3 like I think they are?

Edit: Thanks to a commenter, I have changed the question to ask about the dipole moment in hopes of being more clear. I'd also like to add that this is a homework question that I am stuck on (for general chemistry l) and the teacher hasn't been the most thorough which is why I am here trying to teach myself.

The best answer for this seems to be in Structure, Energy, Vibrational Frequencies, and Potential Energy Curve of
2,3,7,8-Tetrachlorinated Dibenzo-p-dioxin: Ab Initio MO Studies J. Phys. Chem., 1996, 100 (12), pp 4810–4814.

This research finds that though planar is the lowest energy conformation, the energy for bending with respect to the O-O axis from 180 degrees to 170 degrees is only 0.008 kcal/mol.  The molecule flaps like a butterfly.

The article says:

  At ambient temperatures, nonplanar conformations
  will be significantly populated for this potential. This prediction
  accounts for the experimental results that the electric dipole
  moment of the DD molecules is reported to be 0.64 D. [reference 24]

Where "DD" means the unsubstituted dioxin and where reference 24 is Higashi, K.; Uyeo, S. J. Chem. Soc. Jpn. 1941, 62, 396.


Aluminium foil on windows: what is the residue, and how to remove it? I put aluminium foil on some windows' glass pane wetted with plain tap water, as a typical trick to stop sunlight. After 5 days, I removed the foil to find some kind of whitish pattern on the glass. The pattern follows the imperfections in the aluminium foil: bubbles, wrinkles, even a very slight embossed pattern in the foil.

To remove the pattern, I tried window cleaner, rubbing alcohol, vinegar, washing soda. Nothing worked.

Finally I tried caustic soda, at a solution stronger than that recommended for drain opening, and it mostly worked; but still there are patches where you can see the pattern if light reflects in the glass in just the right angle.

My question is: what caused this pattern? And, how can I get rid of its remains?

My next ideas are to try bleach or a stronger solution of caustic soda, but I'm afraid of etching the glass - what would be the maximum concentration advisable?

Also - if aluminium foil reacted to tap water like this, what happens when wrapping food and even cooking with it?

(I asked basically this same question in the DIY SE, but it's not getting actual answers, so I am hoping to get better insights here)

Finally I used a glass scraper. Bleach didn't help, and I didn't dare to use even stronger caustic soda.

Fuller details at the answer I wrote at the DIY SE site: https://diy.stackexchange.com/questions/151160/how-can-i-remove-residue-left-after-removing-aluminum-foil-from-my-windows#answer-152162


Am I solving this pH problem correctly? A liter of a solution contains $\pu{0.120 mol}$ of $\ce{C6H5COOH}$ $(K_a = 6.5
×10^{-5})$ and $\pu{0.105 mol}$ of $\ce{C6H5COONa}$. I'm asked to calculate the pH of the solution after adding $\pu{100 mL}$ of a $\pu{1.20 M}$ $\ce{NaOH}$ solution, the pH of the initial solution is $4.13$.
Here's what I've done. Since $\ce{NaOH}$ is a strong base it will disassociate completely, so we will end up with $0.120$ moles of $\ce{OH^{-}}$. Calulating the $\ce{pOH}$ we get $0.92$ and with that we can calculate the $\ce{pH}$ which will be $13.07$. Is that the correct way of doing it ?

Nope.

While $\ce{NaOH}$ is a strong base, the mixture of $\ce{C6H5COOH}$ and $\ce{C6H5COONa}$ is a  buffer solution and tends to resist large changes in pH.

After adding the $\ce{NaOH}$, the extra $\ce{OH-}$ ions would react with the undissociated $\ce{C6H5COOH}$, effectively being removed from the solution. The solution would now be $1.1 \text{L}$ and contain a bunch of ions, like $\ce{Na+}$, $\ce{C6H5COO-}$, $\ce{H+}$, and $\ce{OH-}$. You can calculate the value of $\ce{[Na+]}$ now.

Use the Henderson's buffer equation:

$$\text{pH}=\text{p}K_a + \log{\left(\frac{\ce{[C6H5COO^-] }}{\ce{[C6H5COOH]}}\right)}$$

I think you can continue from here.


Reaction between carboxylic acids and amines Would some be able to explain why when I mix a carboxylic acid, like ethanoic acid, and an amine, say ethylamine, I get ethylammonium ethanoate. But when I mix a bi-functional group carboxylic acid (say ethanedioic acid) and a bi-functional group amine (say 1,2-diaminoethane) I get a condensation reaction and an amide bond is formed.

This also leads to why does ethanoic and ammonia yield ammonium ethanoate and not ethanamide. I understand that the ammonia is acting as a base in this reaction, but why doesn't it favour being a base rather than a nucleophile?

To me both of the reaction seen the same. Can someone elaborate this for me, or am I missing a key piece of knowledge.

You have a false dichotomy.

Diamines and diacids can form salts with doubly charged anions and cations. Sadly, simply searching SigmaAldrich’s catalogue for the phrase diammonium did not turn up any example; however, I assume this is because these salts are unlikely to be used and thus wouldn’t sell. Chemically, proton transfer is one of the fastest reactions possible and thus simply mixing a diacid and a diamine in a suitable solvent would result in the diammonium diate salt as it would for an ammonium monoate salt.

Amide bond formation is a different reaction that typically requires activation or harsher conditions (e.g. heating). It is a condensation that results in the release of water. All that is required is the presence of an amino group and a carboxylic acid. So aminoethane and ethanoic acid can be heated in solution to produce N-ethylacetamide.

The one difference between the mono and the di systems is that the mono system has no of polymerising or cyclising. So the amide bond will always consist of one acid fragment and one amine fragment while in the case of a diamine and a diacid polymerisation or cyclisation can occur.


Existence of polyaminomethyl compounds Amines such as triaminomethane, tetraaminomethane, and many polysubstituted amines do not exist because of their degradation into ammonia.  Is it possible for derivatives of these amines to exist at all? Take for example tetra(t-butylamino)methane. Will there be enhanced stability due to the tert-butyl groups or will degradation to t-butylamine still occur?  Is there a point at which this degradation will stop, if so where?

I wouldn't say that they don't exist, but there's very little to keep them from collapsing to an amidine, which makes isolation quite difficult.

This is completely analogous to orthoformic acid, which, as a tetrahedral intermediate, rapidly collapses.

Substitution helps to stabilize the tetrahedral intermediates a lot. For example, both trimethyl orthoformate and tris(dimethylamino)methane are commercially available.

EDIT:
Forgot to address this in my answer, but having t-butyl groups probably makes things worse due to steric hindrance. Neopentyl is bad enough...


What are the correct resonance structures of bromoethene? Can anyone explain why my textbook only gives the following resonance structures?

  Write the important resonance structures for each of the following:

I also put two resonance case where the double bond electrons are moved on the individual carbon atoms.

Your resonance structure is perfectly alright. But the resonance structure with the negative and positive charges won't be very important and here is why: For one thing, it contains seperated charges and usually those resonance structures are most important that contain as few charges as possible. This is simply a matter of electrostatics - seperating charges and holding them apart from one another costs energy. A second reason has to do with the "electronic origin" of this resonance charge: It describes a conjugation between a free electron pair of $\ce{Br}$ and the $\pi$ system of the double bond - more accurately put the filled free-electron-pair-orbital can interact with the $\pi^{*}$ orbital of the $\ce{C=C}$ bond to form two new orbitals: an in-phase combination that is lower in energy and will contain the two electrons from the free electron pair and an out-of-phase combination that is higher in energy and will remain unoccupied. This stabilizes the system. The stronger the overlap between the free-electron-pair-orbital and the $\pi^{*}$ orbital of the $\ce{C=C}$ bond and the closer they are in energy the better the stabilization of the system through this conjugation is and the more important is the second resonance structure. But in the case at hand the $\pi$-overlap between the bromine orbital and the $\ce{C=C}$ bond is very weak (the reason for this you can find here) and so the conjugation is very weak, too. Thus, this resonance structure is not very important.


Multiple hydride shifts Can we have multiple hydride or methanide shifts? Specially in the case where after first shift it becomes less stable and after the second shift it becomes more stable.
For example, the first step of acid catalysed dehydration of 3-methylbutan-1-ol. The step in which H+ is added to the hydroxy group and it departs leaving a carbocation.

Yours is a primary alcohol and it has $\beta$ hydrogen atoms also. Thus, it will undergo $\pu E2$ mechanism. In this mechanism removal of protonated hydroxyl group and $\beta ~\pu H$ happen at once (in a single step). Therefore, there will be no scope of hydride shifting.
You may compare this with the dehydration of neo-pentyl alcohol.


Can you make a magnetic fluid that's drinkable I need to make a ferrofluid that is drinkable so I obviously can't use the traditional method of using iron as it is harmful. Are there any chemicals that are not harmful that can take on magnetic properties or and magnetic metals that are not harmful to consume?

Read Nails for Breakfast? and see if breakfast cereal manufacturers agree with your assumption "I obviously can't use the traditional method of using iron as it is harmful"

  Most people know that breakfast cereals are fortified with iron and other essential minerals and vitamins. What they don’t
  know is that in some cases the iron is actually added in the form of metallic, elemental iron—nails for breakfast!

Like permeakra commented, only a small amount iron, whether metallic or as a compound, is safe to consume.  See iron poisoning for details.


Does Nucleophilic substitution require water to happen? I am under the impression that in a mixture of hydroxide ions, haloalkanes, water, and ethanol, under low temperatures, nucleophilic substitution will occur predominantly, with small amounts of elimination.

When no water is present and the solution is heated, elimination will occur to a greater degree. Is it possible for nucleophilic substitution to occur when no water is present as all?

Thanks a lot.

Keeping other parameters fixed, an elevated temperature may enhance the rate of elimination such the product of elimination is more dominant than the one by substitution.  But there is not a "critical temperature" above which only elimination will occur every time, rather than the structure of the substrate and of the reagent equally influence this.

For the other part, transesterifications as one example work better in absence of water.  See, the oxygen present in the water molecule, with highly concentrated electron density, it may act as a nucleophile, too -- a rather hard one (HASB princible) -- leading to parasitic side reactions.


Would ionic conduction be halted by shielding at the electrodes? I've done this experiment myself and seen it front of my eyes. Set up a basic circuit with a battery, an LED, and connect the circuit with a salt-water solution and copper wire. The LED lights up.
If a potential difference is applied across two ends of the wire with the battery, positive charges move towards the end with lower potential and negative charges towards the higher potential.
Why does this conduction keep continuing? Shouldn't there eventually be a pile up of Na+ and Cl- on opposite ends to cut off the circuit? This is with a typical 9V battery. Is there a teensy bit of electrolysis going on in the wire that continuously supplies electrons to the circuit?
EDIT: In doing my own reading the next day, I seem to have forgotten the very important fact that water itself is electrolyzing at the surface of the electrodes. The salt water only helps to prevent the resulting electrolysis from quickly shielding the electrodes. The reason I was confused was because I was very skeptical of NaCl being electrolyzed and turns out this is exactly not the case

Without electrolysis, there would be just an extremely short spike of a temporary initial current, until capacitors formed by electrodes charge.
There is no significant accumulation of ions, as even a tiniest such charge separation would cause strong electrostatic forces acting against it.


Effects of a non-inert electrode in a half cell reaction involving hydrogen gas as the electrolyte ion If there was a galvanic cell that had a copper cathode electrode and with a sulfuric acid electrolyte, will the copper ions be reduced and react with the electrolyte or will the H+ ions be reduced?

Normally, with a gaseous electrolyte, an inert anode material is used. I was just curious as to what happens when copper is used as supposed to an inert electrode.

If you started with a full cell with two inert electrodes immersed in sulfuric acid and applied a large enough voltage between them, a current would flow. At one electrode, the easiest to oxidize species would be oxidized (in this case, that would probably be 2 H2O -> O2 + 4 H+ + 4e-). At the same time, the easiest to reduce species would be reduced at the other electrode (probably 2H+ + 2e- -> H2). If you switched the cathode electrode from some inert metal to copper, the easiest reduction would still be 2H+ + 2e- -> H2 (Remember, there is no Cu2+ in the system to reduce). If, on the other hand, you changed the anode from inert metal to copper, the easiest oxidation would probably be Cu -> Cu2+ + 2e-, so the net reaction would change.


How do I identify how many reactions are occurring when precipitating a mixture? An example of a question is:

  After dissolving $x$ grams of a mixture of $\ce{Na2SO4}$ and $\ce{NaCl}$ in water, you add an excess of $\ce{BaCl2}$ solution. You filter off the precipitate which forms, dry it, and find that it weighs $y$ grams. What is the percent by mass of $\ce{Na2SO4}$ in the original mixture?"

If I am correct, the reaction for this would be:
$$\ce{BaCl2 + Na2SO4 + NaCl -> BaSO4 + 3NaCl}$$

However, there are some questions which are of the same format but have two separate reactions:

  A $x$ gram mixture of $\ce{CaCl2}$ and $\ce{RbCl}$ is dissolved in water. Addition of $\ce{AgNO3}$ causes the chloride ion to precipitate as $\ce{AgCl}$. The mass of $\ce{AgCl}$ collected is $y$. What is the percent by mass of $\ce{RbCl}$ in the original mixture?

For this one, would the reaction be two equations?

\begin{align}
\ce{CaCl2 + AgNO3 &-> Ca(NO3)2 + 2AgCl}\\
\ce{RbCl + AgNO3 &-> RbNO3 + AgCl}
\end{align}

I know how to do these types of questions past the point of deciding whether or not there are two equations. I just don't understand why I couldn't write that last one as a single equation, in the same way as I did the first example. Could anybody help me? Thanks in advance.

You asked two very different questions.

  (1) How do I identify how many reactions are occurring when precipitating a mixture?

There is no information in way the reactants are shown in a chemical equation as to what the products will be. That is to say that you have to know the products of the reaction.

  (2) For this one, would the reaction be two equations?

Chemists use shorthand all the time. Reactions are typically netted out to the pertinent aspect under consideration. However if I were writing the reactions for the second problem I'd do it much differently that you have presented. I think I'd write:

\begin{align}
x\ce{CaCl2} + y\ce{RbCl &->} x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{Cl-}\\
x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{Cl-} &\ce{->[excess AgNO3]} x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{AgCl v}\\
\end{align}

From an analytical chemistry perspective there are two steps. First the chloride salts are dissolved and then the silver nitrate is added. You wouldn't want to add the salts to a silver nitrate solution since the silver chloride could coat the Ca and Rb salts preventing a complete reaction.

I even might write the second reaction as:

$$ (x+y)\ce{Cl-} \ce{->[excess AgNO3]} (x+y)\ce{AgCl v}$$

I'm not a fan of one line jobs like:

$$x\ce{CaCl2} + y\ce{RbCl ->} x\ce{Ca^2+} + y\ce{Rb+} + (x+y)\ce{Cl-} \ce{->[excess AgNO3]} (x+y)\ce{AgCl v}$$


Can anyone help me with this question on vibrational modes of free base porphyrin (C20H14N4)? 4 a. The structure of free-base porphyrin $(\ce{H2P}$, chemical formula: $\ce{C20H14N4})$ is shown below. Given that this molecule belongs to the $D_\mathrm{2h}$ point group (character table given above) identify the symmetries (irreps) for all the vibrational modes of this planar polyatomic molecule;

(i) Identify the number of vibrational modes in $\ce{H2P}$. [3 marks]
(ii) Work out the reducible representation for the vibrational modes. [6 marks]
(iii) Use the reduction formula to identify the (irreps) for all the vibrational modes. [6 marks]
(iv) Identify which of the vibrational modes are Raman-active and which are IR-active. [4 marks]

In particular I'm getting stuck with figuring out the unmoved atoms when applying symmetry operations. I know i need to calculate gamma vib in order to use the reduction formula but I cannot picture how to apply the symmetry operations to calculate unmoved atoms for each symmetry class (aside from E) because it is a more complicated structure than I'm used to dealing with (h20, benzene etc)

If you rotate the molecule by $45^\circ$, the rotation axes will fall on the cartesian axes $x, y, x$, and the mirror planes will be at $xy, yz, xz$.

Now, you can see that the NH groups are unaffected by rotation about $y$ or reflection on $yz$, the deprotonated N groups by rotation about $x$ or reflection on $xz$, and all the atoms by reflection on $xy$. You can ignore the position of the double bonds because it is an aromatic system.

Whenever an atoms lies on a mirror plane or a pure rotation axis, its position will be unaffected by the symmetry operation.


SN1, SN2 and enantiomeric excess When (R)-2-bromobutane is heated with water, the $\mathrm{S_N1}$ substitution proceeds twice as fast as $\mathrm{S_N2}$. Calculate the appropriate enantiomeric excess and the specific rotation of the product of the mixture. The specific rotation of (R)-2-butanol is +13.5.

I din't understand how to interpret '$\mathrm{S_N1}$ proceeds twice as fast as $\mathrm{S_N2}$' mathematically.

In this case, the absolute configuration remains same (as R) for $\mathrm{S_N1}$ and changes (into S) for $\mathrm{S_N2}.$ This is because the priority order of non substituted groups (wrt $\ce{C2})$ remains the same before and after substitution of $\ce{Br}$ by $\ce{OH}.$

Let the initial mole fraction of (R)-2-bromobutane be 1. After the reaction, let the mole fractions of R- and S-2-bromobutane be $x$ and $1-x$, respectively.

According to question condition, as $\mathrm{S_N1}$ proceeds twice as fast as $\mathrm{S_N2},$ we get

$$x = 2(1 - x) \quad\implies\quad x = 2/3$$

So we have excess of R-enantiomer, and the specific rotation is

$$\frac{2}{3}\cdot (+13.5) + \frac{1}{3}\cdot (-13.5) = +4.5$$


Dissociation constants of leucine Leucine is a type of amino acid. Let me write its chemical formula as $\ce{NH2-C5H10-COOH}$. On the wikipedia page, https://en.wikipedia.org/wiki/Leucine, there are two dissociation constants, one for $\ce{-COOH}\,$ given by $\mathrm{pK}_{a1}=2.36$, and one for $\ce{-NH3+}$ given by $\mathrm{pK}_{a2}=9.60$. But I also know that once $\ce{-COOH}\,$ loses its proton $\ce{H+}$ into the solution, the proton may combine with the $\ce{-NH2}\,$ group of the same molecule to form the zwitterion $\ce{NH3+-C5H10-COO-}$ so that the solution now has $4$ species with $4$ dissociation equilibria

$$\begin{array}{rclll}
\ce{NH3+-C5H10-COOH}\!\!&\!\!\ce{<=>}\!\!&\!\!\ce{NH2-C5H10-COOH +H+},& \ &(1)\\
\ce{NH3+-C5H10-COOH}\!\!&\!\!\ce{<=>}\!\!&\!\!\ce{NH3+-C5H10-COO- +H+},& \ &(2)\\
\ce{NH2-C5H10-COOH}\!\!&\!\!\ce{<=>}\!\!&\!\!\ce{NH2-C5H10-COO- +H+},& \ &(3)\\
\ce{NH3+-C5H10-COO-}\!\!&\!\!\ce{<=>}\!\!&\!\!\ce{NH2-C5H10-COO- +H+}.& \ &(4)
\end{array}$$

So there should be $4$ equilibrium constants. Which two do $K_{a1}$ and $K_{a2}$ refer to and how do I know the other two? Or is it that we can assume that $\ce{-COOH}\,$ and $\ce{-NH3+}$ dissociate independently so that the equilibrium constants of the $4$ equations are $K_{a2},K_{a1},K_{a1},K_{a2}$ respectively? I've also seen in some calculation that the zwitterion $\ce{NH3+-C5H10-COO-}$ and the original molecule $\ce{NH2-C5H10-COOH}$ are treated as the same species. But I don't quite understand why this is allowed. They clearly have different charge distributions and should be treated as distinguishable particles in entropy (or Gibbs free energy) calculations.

Among the $4$ equilibria, only $3$ are linearly independent, because we have $(1)+(3)=(2)+(4)$. Let's then choose only three independent reactions

\begin{align}
\ce{NH3+-C5H10-COOH}&\ce{<=> NH3+-C5H10-COO- + H+},\\
\ce{NH3+-C5H10-COO-}&\ce{<=> NH2-C5H10-COO- + H+},\\
\ce{NH3+-C5H10-COO-}&\ce{<=> NH2-C5H10-COOH},
\end{align}

as the basis equations, which are obtained from $(2)$, $(4)$, and $(1)-(2)$. The equilibrium constants of these equations are $K_{a1}$, $K_{a2}$, and $K_{a2}/K_{a1}=10^{-7.24}\ll 1$. Therefore most of the neutral amino acid molecules would exist in the zwitterionic form $\,\ce{NH3+-C5H10-COO-}$. Neglecting the last equilibrium, the equilibrium constants $K_{a1}$ and $K_{a2}$ are those of the $\ce{-COOH}\,$ and $\ce{-NH3+}$ groups in the zwitterions $\,\ce{NH3+-C5H10-COO-}$, which is much more abundant than the original molecules $\,\ce{NH2-C5H10-COOH}\,$ making the latter negligible.


Can a metallic bond with a sea of electrons form inside the human body? I’m interested if atoms or ions can for a metallic bond inside the human body?
I’m talking about a metallic bond as described here: https://www.britannica.com/science/metallic-bond
So I’m thinking of a metallic bond where the valence electrons of the atoms or ions are not associated with specific atoms or ions of the metallic bond. My understanding is that this is described as a sea of electrons.
If such a metallic bond can not form inside the human body, I would be very interested in the reason why. Thank you very much.

We may not produce nails from our bodies, but certain enzymes and proteins do involve metal clusters, especially clusters involving iron [1].  In the natural biomolecules iron-iron and iron-nickel compounds serve as functional groups for electron transfer.  Also discussed are iron-sulfur clusters, which have characteristics similar to all-metal clusters despite sulfur usually being considered a nonmetal.  Chemists have sought to mimic these natural cluster compounds with artificial molecules, not only using iron and nickel but also moving on to less common metals such as molybdenum and ruthenium.
Reference
1.
C. Fehl C and B.G. Davis, "Proteins as templates for complex synthetic metalloclusters: towards biologically
programmed heterogeneous catalysis.", Proc. R.
Soc. A 472: 20160078. (2016)
Link


Supramolecules in Spartan Maybe someone here is familiar with supramolecules calculations using Spartan? I am new in this area and trying to figure out some things:
First - what's the best way to draw them? I tried to draw one molecule and then using copy-paste append the others but then they overlap and it takes forever to minimize their energy so they make hydrogen bonds in the places I need. Maybe there is any way to 'freeze' hydrogen bonds? Because I now where they shoud be (or at least want to calculate different structures with hb in different places of molecules). If I try to put this molecules near when sketching still when I try to minimize the energy hb forms not in the exact places I need them to be. And I need four molecules to form kinda cyclic-structure.
Thank you in advance!

hb I need between two molecules are:

You can create distance constraints between two atoms (set them to the distance you expect for the hydrogen bonds), and then do a force field minimisation.

It's in the menu called Geometry, and then "Constrain distance". You pick the atoms, and click the little lock in the lower right-hand corner, and then enter the distance you want next to the lock.


Number of isomers in a 3-sugar oligosaccharide I recently read that 12288 different oligosaccharides can be formed from the D-isomers of glucose, mannose and galactose. I wanted to try to calculate this, but I keep falling very short of this number.

Each molecule has the potential to form 6 unique bonds:

The disaccharide can form 13 unique bonds:

(Note, bonds formed by anomeric carbons shown with an arrow)

So, if a monosaccharide can form 6 different bonds, this means that:

6 possible bonds x 5 bonds on 2nd sugar x 2 anomers for seconds sugar = 60 possible disaccharides

Each of these disaccharides has 13 possible bonds it can make to a third saccharide:

13 bonds x 5 bonds of 3rd molecule x 2 anomers = 130 combinations for a single dissacharide.

130 combinations per disaccharide x 60 possible disaccharides =7800 isomers.

This is probably a counting error. But maybe I am missing a stereocenter?

This question is interesting and neglected. Bob made a sensible attempt at solving the problem. I took a different approach. [The number of isomers 12,288 was found in Stryer’s Biochemistry (Google: stryer 12288 isomers)]. Of course, oligo- means a few. I presume this means three in this case. But can D-glucose, D-galactose and D-mannose be used as many times as you want? That would mean 27 (33 = 27) permutations which would eventually make the total much greater than 12,288. However, if each hexose is to be used once, then six (6! = 6 factorial) permutations are possible for the location of the hexoses in the oligosaccharide. Division of 12,288 by 6 removes this issue from the question and leaves 2,048 as the number to justify. Now 2048 = 211. The problem reduces to finding 11 stereocenters, if possible!
Let us assume as a point of departure that diagramatic structure ABC (top of the illustration; The connectivity C1-C4 is arbitrary.) shows where bonds are formed between the C1 anomeric carbons and the hydroxyls at C2, C3, C4 and C6 of another hexose. Another pair of assumptions is that all rings are 6-membered (pyranose) and that all hexoses, as defined, are in the D-series. Finally, assume for the time being that there are no C1-O-C1 connections as typified  in figs. D and E.
Rather than define C1 of hexose A as having two stereocenters (α and β anomers), this assignment will be considered in a different way. Hexose A is given a value of zero because no attachments may be made at its hydroxyl groups at C2, C3, C4 and C6. Hexose B has four hydroxyl groups (blue dots) that can form two anomeric, glycosidic bonds at C1 with hexose A. The AB connection creates 24 (16) stereoisomers. In a similar fashion the hydroxyls at C2, C3, C4 and C6 of hexose C (red dots) form another 16 stereoisomers with C1 of hexose B. The total count thus far is 3! x 28 = 6 x 256 = 1536. This result is only one-eightth of the goal of 12,288 or short by a factor of 23. To this point the eight hydroxyl groups considered can only form two configurations each.
Now the anomeric hydroxyl of hexose C when bonded to the anomeric carbon of hexose B will give not two stereoisomers but rather four (αα, αβ, βα, ββ)(see fig. E). Similarly, the same relationship exists between hexoses B and A (see fig. D). These configurations add a factor of 16 (24) to the total. This reasoning leads to a value of (3!) x (212) = 6 x 4096 = 24,576.

If the original analysis considered a structure such as ABC having a mutarotating C1-OH, then the total might be 12,288. However, ABC is a connectivity map, not a chemical structure.

What may be concluded from all of this? The premises set down here may not be those applied by the proposers of 12,288. There may be a flaw in this, or their, analysis. The answer may be 12,288, 24,576 or something else.


References for learning about acid and bases in organic chemistry I was wondering if someone could help me find references available that will help me solve the following problems. These are not homework problems, but are suggested problems that were given to me by my instructor.

I'm specifically interested in learning how to solve question A. (b). How can I know whether a reaction will occur between the two products and further, how can I predict what the product will be if a reaction does occur (what atoms bond together, and how can I know which atoms bond together in any organic reaction)?

Concerning your first question: I found that the  book "Organic Chemistry" by Clayden, Warren, Wothers and Greeves and here especially the 8th chapter contains a very good and readable introduction to acidity and basicity. After you have worked through this chapter you should be able to explain why certain molecules are as basic/acidic as they are and you see a lot of typical $\mathrm{p}K_{\mathrm{a}}$ values for a lot of substance classes so that you get some knowledge about what $\mathrm{p}K_{\mathrm{a}}$ values you can expect for other substances. As for your second question: If you want to know why reactions occur in the way they do you'll have to take courses in Organic Chemistry. Apart from that, I wholeheartedly recommend the aforementioned book again. If you work through this book (this will take you a while: it has 1200 pages... but then again, that is 1200 very readable pages :)) you'll have a very solid basis in organic reaction mechanisms and organic chemistry in general. If you want to dig a little deeper this book by Bruckner is very thorough but a little dry.

One particular perspective that I found very helpful in explaining why reactions happen in the way they do was the frontier orbital theory. A very good and readable book about its use in organic chemistry is this one by Fleming. The book by Clayden et al. also uses this theory a lot in explaining reaction mechanisms.