text:
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. $\endgroup$

Great explanation! One question: in Albright, 2013, the section about the second order JT: 7.4.2 (page 136) they write "If the lowest lying one of
these is of the correct symmetry for $<\psi_i|d^2H/dq^2|\psi_i>$ to be non zero ($\Gamma_q=\Gamma_i \otimes \Gamma_j$)." So I guess that must be a typo since they should rather refer to the other second order term wich is the sum of the virtual excitations, right?

@R_Berger, indeed, when I was reading it I noticed a typo in that line too. I believe you are correct and that $\langle \Psi_i | \partial^2 H/\partial q^2|\Psi_i\rangle$ should be replaced with $\langle \Psi_i | \partial H/\partial q | \Psi_j\rangle$. Of course, this term can only be nonzero precisely if $\Gamma_q = \Gamma_i \otimes \Gamma_j$, which they correctly explain.

@orthokresol: How would things look like if we include SOC and go to double groups? Is there a quick answer?

@R_Berger Sorry, but the quick answer is I don't know ;) I had to do lots of research just to write this answer.

@orthokresol: Would you agree that at the end of the dashed line in the last figure there should be $^3$B${1g}$ (since A${2g}$ from $D_{4h}$ descends into B${1g}$ in $D{2h}$)? Or is there any reason why it is not displayed in the figure?

@Rudi_Birnbaum I’m a bit busy so cannot check the descent of symmetry tables carefully (these particular point groups are often a real pain), but it sounds like the correct reasoning. We can only speculate why it’s not in the figure. I guess the most likely explanation is that the authors were only interested in the singlet states.

The point is that if cyclobutadiene were symmetric the highest occupied orbital would be doubly degenerate with 2 electrons to occupy it, and so by Hund's rules it would be a diradical, just as for molecular oxygen. But the Jahn-Teller theorem says that such a state is unstable with respect to a distortion which splits the degeneracy and lowers the symmetry. This is exactly the same as $\ce{Cu^2+}$. Hence the symmetrical molecule distorts splitting the degeneracy, and both electrons occupy the lower lying state. Thus not only is the molecule not symmetric, it is also not a radical, and hence there is no ESR signal.

How would it be doubly degenerate? For example Eg represents a set of two degenerate orbitals, but that doesn't mean a complex will be doubly degenerate (only d9, low-spin d7 and high-spin d4 are).

But $\ce {Cu^{2+}}$ is $d^9$, no?

@Marko I don't really understand what you are saying. The high symmetry state leads to degeneracies, hence the theoretical square cyclobutadiene and the theoretical Cu2+ in an octahedral environment would have doubly degenerate HOMOs. But the Jahn-Teller theorem says that such geometries must distort, and further that that distortion will split the degeneracy, and so the HOMO is no longer degenerate. In the case of cyclobutadiene the splitting is so large that the lower of the states is double occupied, hence no ESR signal.

I think that doubly degenerate means that there are two energetically equivalent states, but I see only one for cyclobutadiene and that is the one where both HOMOs are filled with 1e each.

Ahhh, maybe it is the word "state". Let's avoid that word, let's use orbital instead as that's what we're really talking about here (and I see why state might be confusing). In your diagram I see two horizontal lines at the same vertical level, representing two orbitals which have the same orbital energy. This is the two-fold degeneracy being referred too, and as we have 2 electrons to occupy 2 orbitals by Hund's rule one electron goes in each. And it is this degeneracy that is split by the distortion.

I see what you mean, but as I have mentioned in my first comment, if we look at transition metal complexes the same way, then many different configurations would give the Jahn-Teller distortion, but only  d9, low-spin d7 and high-spin d4 do, because there are "two different ways" for filling the eg orbitals, not because there are "two eg orbitals" and double degeneracy should refer to the former.

I think that the confusion is that only the d9 , d7 & d4 spin systems exhibit measureable JT effect. The fact that a spitting is possible does not mean that it can be predicted per se  if this is large or small

So do molecules in triplet states with two degenerate HOMOs always exhibit JT effect?

@porphyrin I have searched and many different sources point to the same definition: "In a nonlinear molecule, if degenerate orbitals are asymmetrically occupied, a distortion will occur to remove the degeneracy." According to this, cyclobutadiene shouldn't exhibit JT distortion.

@Marko. JT is an interaction between electronic and nuclear motion. This means that in any non-linear system there exists a vibrational mode that removed the degeneracy of an orbitally degenerate state. Another definition is that a molecule possessing a degenerate state in either spin or orbit, must distort to remove the degeneracy. Which does not mean that it forms a degenerate state then distorts on discovering this, but that if the molecule is predicted to have a regular configuration this then proves to be impossible.

continued. The $t_{2g}^6e_g^3$ configuration is JT active as the 'hole' in the $d^{10}$ can be in one of two d orbitals so we expect a complex to distort. Similarly according to Levine (Quantum Chemistry), the cyclopentadienyl cation $\ce{C5H5^+}$ has a triplet ground state, ($e^2$ and electronically the same as cyclobutadiene), $\ce{C5H5\cdot}, ~e^3$  is orbitally degenerate so should distortfrom regular pentagon by JT, the anion$\ce{C5H5^-} ,~e^4$ is stable. The 'asymmetrically' occupied part of the definition seems to be unclear.

@porphyrin So cyclobutadiene and cyclopentadienyl cation shouldn't exhibit JT-distortion? But then again, cyclobutadiene is rectangular.

@Marko, I agree no, JT would seem to be logical as you suggest as one electron in each orbital does not lead to degeneracy as there is only one way of arranging them (same as for 4 electrons) but unlike one or three.

@porphyrin Yet evidence (ESR and geometry) suggests the opposite. So we are at the beginning again.

@marko  I agree, but if experiment shows that is the case then either (a) distortion has other cause than JT or (b) energy released by JT is greater than that needed to make singlet ground state from triplet and singlet then undergoes JT distortion.


text:
"Alternative" derivation of the spin-only formula
The spin-only formula
$$\mu_\mathrm{so} = \mu_\mathrm{B} \sqrt{n(n+2)} = \mu_\mathrm{B}\cdot 2\sqrt{S(S+1)}$$
is usually a good first approximation to calculate the magnetic moment of transition metal complexes. (I am aware of its limitations.) The usual derivation of this formula involves the Zeeman splitting of the ground state into its separate $M_S$ states under the influence of a magnetic field, followed by evaluating the average magnetic moment using a Boltzmann distribution.
However, if I were to write that $\vec{\mu} = \gamma_\mathrm{e} g_\mathrm{e} \vec{S}$, then I could arrive at the simpler "derivation"
$$\begin{align}
|\vec{\mu}| &= |\gamma_\mathrm{e}|\cdot g_\mathrm{e} \cdot |\vec{S}| \\
&= \frac{e}{2m_\mathrm{e}} \cdot 2 \cdot \sqrt{S(S+1)}\hbar \\
&= \mu_\mathrm{B}\cdot 2\sqrt{S(S+1)}
\end{align}$$
Is this a coincidental result? I'm assuming there is something wrong with the second "derivation", or even worse, it is just an entirely unphysical interpretation, since I am using $S$ instead of $S_z$. What is the mistake?

The first equation has units of Bohr Magneton as does your last one. As I understand it the 'derivation' applies to the spin of an isolated electron, i.e it relates its spin magnetic moment to its spin angular momentum, so I think that the similarity is accidental.

@porphyrin Which part of it would not be applicable to a multi-electron system (after coupling of $s_1$, $s_2$, .. to give $S$)?

With  more than one spin then you will have to work out how to properly add the angular momentum, so probably end up with the full derivation anyway.

source: Stack Exchange

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

sort: votes


text:
Confusion in understanding temperature change in exothermic reactions
The question due to which my doubt arose $\ce{->}$

The following system is in equilibrium:
  $\ce{SO2Cl2 + {Heat} -> SO2 +Cl2}$
What will happen to the temperature of the system initially if some $\ce{Cl2}$ is added into it at constant volume ?

My attempt at the question : Since if we add more $\ce{Cl2}$, concentration of products will increase which will in turn shift the equilibrium in backward direction, which will in turn release more heat. And since the heat is being released to the surrounding, temperature of the system should decrease.
Though according to my textbook the answer is temperature of the system will increase.
So please do explain where my understanding is wrong.

Is the reaction exothermic or endothermic, that is, what is the sign of $\Delta T$?

@zhe reaction is endothermic as heat is being supplied

The system is in thermal equilibrium with the surroundings. If the temperature of the surroundings increase, then the temperature of the system will increase...

@zhe Can you please explain your reply in detail . If the system is in thermal equilibrium then the temperature of both system and surrounding would always be same ?

And how do we know that the system is in thermal equilibrium ? Are all chemical reactions in equilibrium are also assumed to be in thermal equilibrium .

Pretty much. You can contrive the system as you'd like, but without specifying a time scale, you cannot answer this question as clearly as you like.

I think this comes down to what constitutes the system. If we think of the system as the reactants and products in a container, releasing heat will increase the temperature of the system.

Okay this do seems right to me , so if this is the case than my logic (my attempt in the question ) is correct if we consider only reactants as the system ?

I would think you would want to consider the mixture of all the gases to be the system, but in that case your logic is sound.

one more query on the same question .If we think of the system as the reactants and products in a container ,then releasing heat would actually just be transfer of heat from the system to the system .Shouldn't the temperature remain same in this case ?

The temperature of a system, to describe it somewhat imprecisely, is related to the kinetic energy of the various particles in the system. The heat released when the reaction runs in reverse comes from the energy released in bond formation, which then becomes kinetic energy, thus increasing the temperature. Similarly, when the reaction goes forward, the energy needed to break bonds is extracted from the kinetic energy of the system, leading to a decrease in temperature.

Oh now I got this , thanks for explaining in detail .

This really a comment but too long to fit. The question does not specify if the container holding the gas is to be treated isothermally or adiabatically. However, by inference since the question asks about temperature changes then it cannot be isothermal, i.e. it must be that no heat is gained or lost to the 'world' outside the container otherwise the question does not make sense. It is also implicitly assumed that the added gas is at the same temperature as the gases currently there.
The fact that the volume is fixed means that there is no work done against any external pressure, so all the heat generated by the reaction moving towards reactants heats up the gases.
( Note also that in thermodynamics we always move from one equilibrium state to another equilibrium state; time is never considered.)

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


text:
Can chlorine form hydrogen bonds?
My question is simple, If chlorine ($\ce{_17Cl^35^.^5}$) and nitrogen ($\ce{_7N^14}$) have similar values of electronegativity, and chlorine has way more lone pairs than nitrogen, what prevents it from forming hydrogen bonds ?
(other than in chloral dihydrate $\ce{CCl3CH(OH)2)}$

related http://chemistry.stackexchange.com/questions/31591/why-is-hcl-not-considered-to-have-hydrogen-bonding?s=1|0.2701

Or again http://chemistry.stackexchange.com/questions/21773/why-is-hydrogen-bonding-generally-defined-to-include-only-three-period-two-eleme may be a better dupe.

i guess its more likely to be classified as a dupe of the second, although it never came up when i gave the title. P.S: How do i assign it as the second one's duplicate @Mithoron

With newest Stack Exchange™ technology it is possible to mark a post as a duplicate of both questions. You guys can close it as a dupe of one post first, and then myself or another mod can edit the duplicate list.

sure, ill do it immediately, thanks(P.S: since thanks comments arent allowed, will delete this comment too.lol) @orthocresol

Don't worry about it, thanks comments are perfectly fine and we like having them here, because it means people are being nice to each other. We just clear them as we go along (once they've served their purpose).

@Supernova https://chemistry.meta.stackexchange.com/questions/3071/searching-guidelines-for-newbie

Thanks again, will do accordingly from next time @Mithoron

You can come to chat http://chat.stackexchange.com/rooms/3229/the-periodic-table for some tips/talking/stuff

Nitrogen is equally electronegative as chlorine but also more compact.  Whether you consider hydrogen bonding as purely electrostatic or having a covalent character (The lone pair on atom X overlaps the molecular orbital in the H-Y bond of an adjacent molecule, thus partially delocalizing the bond), the more compact structure of nitrogen offers an advantage.


text:
IUPAC nomenclature of 1,2,3,4-tetramethylcyclobutane
I'm having a hard time naming these molecule myself.
For the first molecule I start with one of the stereocenters, give priority "4" hydrogen and "3" to the attached CH3. How are the other two attached nodes given priority "1" and "2" in an anticlockwise fashion, like in the graph below?
For the second one I have the analogous problem, who is "1" and "2" in each stereocenter?

Related: IUPAC name for 1,2,3-trichlorocyclopropane? (may be dupe)

The first example contains pseudoasymmetric centers which have their own notation rules (page 21 http://old.iupac.org/reports/provisional/abstract04/BB-prs310305/Chapter9.pdf). R has priority over S in assigning locants.

Related: http://chemistry.stackexchange.com/questions/70293

The Cahn-Ingold-Prelog (CIP) assignments of the configurations of the two tetramethylcyclobutanes above are accomplished as follows:
1)  Tetramethylcyclobutane 1: This isomer has all of the methyl groups on the same face of the ring. There are two mirror planes passing through C1-C3 and C2-C4. Numbering the cis-isomer is arbitrary in that there is a 4-fold axis of symmetry present. The common methods for R/S-configurations do not apply. Because the ring carbons lie in symmetry planes, they are stereogenic and achirotopic requiring r/s-assignments. When an assignment is made to one carbon, the same assignment may be applied to the remaining three.
  Temporary assignments for 1a are made to C2, C3 and C4 (r0/s0). Priorities are assigned at each of these carbons such that C1 is the highest priority group.  Returning to C1 the priorities are r0>s0>CH3>H. C1 is of the s-configuration. Digraph 1b illustrates the assignment to C1 in a different way. The horizontal line represents the plane of the ring with C2, C3 and C4 above the plane of the ring and the dotted line (C1) signifies the carbon under assignment. Since r0 precedes s0, C4 precedes C2. C1 is of the s-configuration.
2) Tetramethylcyclobutane 2: This stereoisomer has ring carbons that are both stereogenic and chirotopic. Here R/S assignments apply. There is a plane of symmetry passing through the C1-4 and C2-3 bonds while a two-fold axis of rotation exists between the C1-2 and C3-4 bonds. Applying the methodology of 1a to 2a, C1 is assigned the R-configuration. Referring to digraph 2b, R0R0 precedes R0S0 [like(lk)>unlike(ul)]. Thus, C2 precedes C4. C1 is of the R-configuration.
One may apply this method to the remaining three ring carbons but there is a shortcut. Because C1 mirrors C4, C4 is of the S-configuration. The rotational axis shows that both C1 and C2 are of the R-configuration while C3 and C4 are both of the S-configuration. The numbering system requires that the lowest numbered atoms, if possible, be labeled R.

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


text:
Pressure/ vacuum on preciptation of a borax saturated solution
Trying to go back to my collage chem/physics.  We put a supersaturated solution of borax on the stove. (heated  water, added borax till it would not take more borax).  Placed them under vacuum 15 inhg, normal pressure, 20 psi.  Initial hypothesis was crystals would grow better vacuum. They do this in space vacuum and low gravity. But we can't  get rid of gravity.
Results by weight of new jars with crystallization using the same size pipe cleaner for nucleation:

pressure: 467.4g
reg: 454.5g
vacuum:   326.3g

Definitely more crystal deposition in higher pressure, less uniform
reg  pressure most uniform deposition on pipe, less deposition
vacuum, least deposition
Did the air pressure cause the solubility to drop, resulting in more crystals?
And the vacuum less crystals because of more solubility?
I initially thought it would be the opposite.
What's the explanation?
Don't know if you would observe differences under a microscope for quality of crystals.


text:
Will OH- ions be reduced or H+ ions be reduced in KOH electrolyte
I was trying to see if I can make a battery that uses KOH as the cathode electrolyte with platinum as the cathode electrode, but I got confused as to what gets reduced in this solution (Aluminum sulfate is the anode electrolyte and aluminum is the anode electrode).
$\ce{K+}$ never will get reduced because that is thermodynamically unfavored, so will $\ce{OH-}$ (from the KOH ion) get reduced to $\ce{OH^2-}$? It makes more sense to me that $\ce{OH-}$ gets reduced because KOH is very basic and there will be a ton of $\ce{OH-}$ ions that can go to the electrode and accept the electron.
I just don’t know if this is favored thermodynamically or how I can calculate it.
Or will it be $\ce{H+}$ from water? Will water ionize in the first place?

You are right that K+ doesn't want to be reduced, nor does OH-.  So that only leaves one option and you've already hit on it...

so water is being reduced?


text:
Which is more soluble in water, HF or SO2?
I was wondering which is more soluble in water, hydrogen fluoride or sulfur dioxide? Since $\ce{HF}$ will hydrogen bond to other HF particles (as opposed to $\ce{SO2}$ dipole-dipole attraction).

Wouldn't $\ce{SO2}$ dissolve better as it would be easier for the $\ce{H2O}$ particles to dissociate?
Or would $\ce{HF}$ dissolve more readily as it will form hydrogen bonds with the $\ce{H2O}$ particles?

You're on the right track. I'm not quite sure I understand your reason for SO2 being more soluble though...could you expand or restate that?

There are a couple different forces at play here. One big difference between the compounds is that $\ce{SO2}$ is much more volatile, with a boiling point of $\pu{-10 ^\circ C}$; in other words it's a gas. It does strongly hydrogen bond with water however, and is pretty soluble at $\pu{94 g/L}$.
$\ce{HF}$ is also pretty volatile, but much less so than $\ce{SO2}$ with a boiling point of $\pu{20 ^\circ C}$. It also is a more polar molecule than $\ce{SO2}$ and hydrogen bonds even stronger with water. In fact, $\ce{HF}$ is fully miscible with water, so clearly that would be the more water soluble compound.
To summarize, the water solubility of $\ce{SO2}$ is $\pu{94 g/L}$, whereas $\ce{HF}$ is fully miscible, thus $\ce{HF}$ is the more soluble compound in water.
Reference for all solubility and melting point data:
Lide, David R., ed. (2006). CRC Handbook of Chemistry and Physics (87th ed.). Boca Raton, FL: CRC Press. ISBN 0-8493-0487-3.


text:
OH band in IR of propyl ethanoate?
In lab I combined 1-propanol with acetic acid to make propyl ethanoate. I obtained an HNMR and an IR of the product. The HNMR suggests that the product was successfully synthesized. The IR has the correct bands for the expected product, but it also has an OH peak. What would this band be?

Any extra peaks in your HNMR, maybe something near 1.5 ppm?

What is the shape of this "OH" peak? Is it the typical broad hump or is it more like a small spike with a broad shoulder? That second one is the overtone for the carbonyl peak.

How did you purify the compound and have you been careful about making sure that the sample is dry?

The Spectral Database for Organic Compounds maintained by AIST includes an IR spectrum recorded for the neat film of this compound equally known as propyl acetate (CAS registry number 109-60-4):

If you superimpose an IR spectrum of your isolated product, and see the absorption band by OH and others by your starting material, the pure 1-propanol (CAS registry number 71-23-8), than your sample likely still contains 1-propanol.  This were an indication that, for example, your purification by distillation was insufficient, and a better distillation column should be used for a subsequent second distillation.

The presence of remaining propanol may be confirmed by 1H NMR, too (remember that solvents like $\ce{CDCl3}$ more likely provide an OH signal that is broad, while d6-DMSO with slow H-D exchange rates a multiplet signal; and a droplet of $\ce{D2O}$ may mute the OH-signal completely).
For a purity check, especially if you are not allowed to alter how the H NMR data is recorded (in particular, increase of t2 time to account for different relaxation times, cf. DOI 10.1021/ed085p294), search for a complementary confirmations of purity like refractive index, and -- provided enough sample is available -- boiling temperature at known pressure, for example.


text:
rate constant via flux-flux correlation function for linear potential energy surface
How does one calculate fully quantum mechanical rate ($\kappa$) in the golden-rule approximation for two linear potential energy surfaces?
Attempt:
Miller (83) proposes $\kappa=\int{Tr[\exp{(-\beta\hat{H})}\hat{F}\exp{(-i\hat{H}t/\hbar)}\hat{F}\exp{(i\hat{H}t/\hbar)}]}dt$
Where integrand is simply the flux-flux correlation function: $C_{ff}(t)$. Which can be calculated under Feynman's path integral formalism. My attempt (which is in vain) at calculating $C_{ff}(t)$ is as follows:
$C_{ff}(t)=Tr[\exp{(-\beta\hat{H})}\hat{F}\exp{(-i\hat{H}t/\hbar)}\hat{F}\exp{(i\hat{H}t/\hbar)}]$
=$Tr[\exp{(-\beta\frac{\hat{H}}{2})}\hat{F}\exp{(-\beta\frac{\hat{H}}{2})}\exp{(-i\hat{H}t/\hbar)}\hat{F}\exp{(i\hat{H}t/\hbar)}]$
By cyclicly permuting the operators we reach at:
=$Tr[\exp{(i\hat{H}t/\hbar)}\exp{(-\beta\frac{\hat{H}}{2})}\hat{F}\exp{(-\beta\frac{\hat{H}}{2})}\exp{(-i\hat{H}t/\hbar)}\hat{F}]$
The boltzmann operator and quantum mechanical propagator can be combined as follows:
=$Tr[\exp{\hat{H}(\frac{it}{\hbar}-\frac{\beta}{2})}\hat{F}\exp{\hat{H}(\frac{-it}{\hbar}-\frac{\beta}{2})}\hat{F}]$
In the golden-rule (non-adiabatic) case, we have two electronic states 0 and 1. So F is simply a projection operator. Hence one can obtain:
=$Tr[\exp{\hat{H_0}(\frac{it}{\hbar}-\frac{\beta}{2})}\exp{\hat{H_1}(\frac{-it}{\hbar}-\frac{\beta}{2})}]$
This basically is kernel corresponding to two potential energy surfaces $V_0$ and $V_1$. For trajectory starting at $x_a$ and ending at $x_b$, we have
$C_{ff}(t)=\int{\int{K_0(x_a,x_b,\frac{it}{\hbar}-\frac{\beta}{2})K_1(x_b,x_a,\frac{-it}{\hbar}-\frac{\beta}{2})}}dx_adx_b$
For a linear potential energy surfaces (PES), where my PES looks as follows:
$V_0=k_0 x$
$V_1=k_1 x$
My kernels are:
$K_0=\sqrt{\frac{m}{2\pi t_0}}\exp{(-S_0)}$
$K_1=\sqrt{\frac{m}{2\pi t_1}}\exp{(-S_1)}$
$S's$ correspond to action which is:
$S_n(x_a,x_b,t_n)=\frac{m(x_a-x_b)^2}{2 t_n}-\frac{(x_a+x_b)k_nt_n}{2}-\frac{k_n^2t_n^3}{24m}$
The problem is the integral for flux flux correlation function doesn't seem to be converging with the imaginary argument for $t$'s. I am trying to integrate w.r.t $x_a$, $x_b$ and $t$ from -Inf to +Inf. My final answer for rate should look something like this:
$\exp{\frac{k_0^2k_1^2\hbar^2\beta^3}{24m(k_0-k_1)^2}}$
Is it a gaussian integral with respect to $x_a$ and $x_b$? One has to be careful because there is also an imaginary parts in the exponent. How does one reach the final answer for rate with those integrals? Really confused! Any help is appreciated.


text:
Detecting a gas molecule moving through a small pinhole
Suppose I have the four DNA base chemicals - thymine, adenine, cytosine, and guanine. Imagine I turn each of them into a gas, make a "cartridge" of the gas for each chemical, and then have a "pinhole" that is opened or closed by a semiconductor. When the pinhole is open, the gas will, of course, expand through the pinhole and exit the cartridge. The system must emit only one molecule at a time.
If I send a command to the semiconductor to open the pinhole, what kind of sensor can I use to detect a molecule passing through? If necessary, you can modify the base chemicals as long as it would not interfere with their chemical properties in any way other than to improve detection of those molecules, if that makes sense.
The cheaper and more compact the system, the better, of course. I was thinking about the way an inkjet printer works, with a piezoelectric crystal vibrating the ink into droplets, but even the smallest droplets (a picoliter or so) are still bigger than a single molecule, so that doesn't work. Other possible methods I've thought of include a MALDI like system, a system where a weak laser is set up with a light detector; when a molecule passes through, the beam is interrupted, using a scanning tunneling microscope (STM; no idea how that would work), or a system using fluorophores.

Vapor is technically single molecules...

A picoliter of water would still have over 30 trillion molecules of water. So, it sounds like you probably have a cool idea, but there is just no low-budget way to meter out single molecules.

@Greg, so, for example, could you turn your molecules into vapor/gas, let one pass through, and then recondense it? How would you know that only one passed through?

No simple way even if you have very deep pockets;  build a molecular beam machine, i.e. expand gas through two consecutive pinholes into ultra high vacuum, align molecules in large electric field then the use short pulse laser to excite a molecule and so deflect it out of beam as it fluoresces. Good luck :)

You can move single atoms with STM

@heather Sure you can. As porphyrin said, eg with pinholes. Such things are used eg for slit experiments on molecules - and they worked.

I don't think you have any practical option to detect the molecule flying without high risk of destroying it, ie keeping it for experiments.

The best way to do this is to detect the molecules you want in the presence of others that carry it along. You should not expect to detect every molecule of the type you want but just one of them at a time.
The most sensitive detection technique is mass spectrometry, more sensitive than fluorescence because if a molecule fluoresces in a direction away from your detector its photon is lost. All ions can in principle be collected in a mass spectrometer.
To get the molecules into the spectrometer mix them with a solvent, (if necessary) and heat this with a pulsed IR laser. This has to be sufficient to vaporise the mixture/substrate  and it has to be positioned close to the input of the spectrometer since this is done outside the spectrometer. The vapour is at atmospheric pressure and mixed with air but because of the pressure differential efficiently enters the spectrometer (it is easy to see this). Here by differential pumping and using skimmers before the entrance to the mass spectrometer part some of the target molecules will be present in the beam produced. The mass spectrometer should probably be of the time of flight/reflectron type to increase throughput and mass resolution. By pulsing the laser it is then possible to gate the detector to observe the molecules at just the right time and so reduce noise. Of course once you have detected the molecule it is destroyed on the detector.
If you do not want to destroy the molecules then fluorescence detection is the only possibility provided the molecule does fluoresce as not all do with a reasonable yield. A second laser is thus needed after the mass selection and then a photomultiplier is used to count single photons of fluorescence as they arrive (v. common technique now for 50 years or more). I'm not sure what you could do after that but again there is time as photons travel effectively instantaneously compared to the sped the molecule will be travelling at. As I said in a comment not cheap, not trivial.

What did you think about a system where a weak laser is set up with a light detector; when a molecule passes through, the beam is interrupted, therefore resulting in a different value for the light detector? I assume that has some problem as well?

The amount of light absorbed would be tiny from a single molecule, you may be looking at detecting one part in a million-million. If the light formed part of a lasing cavity into which the your molecule was added then there is more possibility of doing this. Look for Cavity Ring-Down spectroscopy.


text:
How to slow/prevent corrosion of a steel post
I have a steel post set in limestone, about 18-20 inches deep, with some concrete around it.
I just noticed that I have about 6 inches of water inside the post. The limestone had a small seam in it and I believe the water seeps through it.
I am concerned about long term corrosion.
I am about to cap the post, so I will have no access to it in the future.
What can I do to prevent/slow the corrosion?
A. Nothing. Do not worry about it.
 B. Fill the post with crushed limestone 12 inches above grade.
 C. Fill the post with concrete 12 inches above grade.
 D. Install a fitting in the post and dump some concentrated washing soda solution or powder once or twice per year to raise pH.
 E. Some type of cathodic protection.
 F. Other.
I am leaning toward B. This would be easy to accomplish and inexpensive, and limestone is supposed to raise pH level, which I think should slow corrosion.

I'm no expert on the chemistry stack exchange, but I think you may get a more practical answer on Home Improvement

Perhaps. I am afraid most people do not give this much thought, so the scientific approach might yield better results.

Do you know what kind of steel it is (carbon steel, stainless steel, galvanized)?

Just regular carbon steel.

Carbon steel can handle slightly basic conditions, so concentrated soda wouldn't be a go to solution. If the post isn't critical for safely supporting something or a related critical function, I think you can go with the limestone idea. If you could borrow/rent a pump and get rid of the water, that would be ideal. If you poured some concrete, could you use the ready to mix stuff that would react with the water in the post?

Unfortunately, the water infiltration will be an ongoing process, so I would have to have a permanent pump that would come on automatically as needed. Doable, but way too much hassle. After thinking some more, concrete is not a good idea, as any condensate above could pool on top of the concrete and freeze in the winter, cracking the post. This is a structural post, so it is somewhat critical, but probably not life or death critical.

Your idea of cathodic protection might work. Perhaps you could melt some zinc and pour it into the (dried out) post, effectively galvanizing at least part of it. It might not adhere well, but should provide some protection.

Could you put a cap on the end that is open to the atmosphere after emptying it? If the water has room to expand in the vertical direction, I don't think freezing will crack the pipe.


text:
A different type of photosensitive material
I'm curious as to whether there is any small, solid or gel-like material that dissolves/liquefies/evaporates under brief exposure to bright artificial light? So, some small element or agent that would form a 'wall' between two liquids but then vaporise when exposed to light (preferably artificial)
I think I found a crystal that changed to oxygen awhile ago via google, but I cannot locate it again so any help would be well appreciated.

There are many types of photoresist that might suit your purpose. Many are commercially available from electronics suppliers; even Amazon carries some. A thin film of photoresist could be used to separate two liquids, and dissolve when exposed to light, if at least one of the liquids is transparent to the wavelength needed.
You might also consider a photo-sensitive valve using a plastic that changes shape in light.


text:
Why is potassium monoxide (K2O) coloured?
Potassium monoxide is rarely formed and is a pale yellow solid. I don't understand how is this compound coloured because in oxide anion electron cannot excite in the next orbital due to huge energy difference. Also there are no empty molecular orbitals like there are in peroxide and superoxide anion. The colour cannot come from $\ce{K+}$ cation because electron cannot be excited. So how is this compound coloured?

What's potassium "monoxide", $\ce{K2O}$? (Just double-checking)

Yes it is K2O (potassium monoxide). You can also search it on the internet.

Yes, well, it's more commonly known simply as potassium oxide.

Not just potassium.  Heavier alkali metals form colored oxides too.  So how are all the heavier alkali metal oxides colored?

Probably as good an explanation as any can be found here.
https://uk.answers.yahoo.com/question/index?qid=20090729062525AA0gZlS
Basically, the electronic structure of an ionic compound like potassium oxide has "valence bands", which are based primarily on the anions (oxide ions), and empty "conduction bands" based on the cations  (potassium ions).  Typically, to excite electrons from the valence band to the conduction band you need to absorb ultraviolet photons and thus you get no color.  But potassium oxide has just a small enough energy gap between valence and conduction bands to allow some electrons to be excited with violet light.  That absorption gives the yellow color.
If we go further down the alkali metal oxides the band gap gets smaller, and more visible light with longer blue/green wavelengths may be absorbed.  Then the color becomes stronger and more reddish.

I was thinking that the answer is related to the molecular orbitals in the lattice. Thank you so much for the answer!

Hmm, but why does it decrease? May it  be due to presence of traces of suboxides?

Heavier, more electropositive alkali metals impart more negative charge to oxygen, raising the valence band energy.


text:
Algorithm used by PubChem to determine the 2D layout assignment for Compound entries?
The PubChem database provides 2D structures for entries in its Compounds database:
ftp://ftp.ncbi.nlm.nih.gov/pubchem/Compound/ (ftp link to download examples)
On the page linked below, under the "NCBI-generated records" heading, it's mentioned that the data processing of records for the Compound database "includes 2D coordinate layout assignment":
https://pubchem.ncbi.nlm.nih.gov/docs/subcmpd_summary_page_help.html#DataProcessing
What algorithm does PubChem use to generate these 2D structures?
I'm familiar with similar algorithms used in RDKit (e.g. compute2dCoords defined in rdkit/Code/GraphMol/Depictor/RDDepictor.cpp), but I'm specifically interested in a resource that describes the algorithm that PubChem uses.
Thanks very much!

According to Xemistry, PubChem uses Cactvs for 2D layout coordinate computation.
A friend pointed this out to me. If you scroll to the bottom of the CACTVS Toolkit FAQ (http://www.xemistry.com/docs.htm), it says the following:

Interesting. I did not know that PubChem is built with Cactvs. I heard it was implemented with OpenEye's OEChem library.
Well, that is not wrong, but only part of the story. Both Cactvs and OpenEye software are used, in combination with the in-house NCBI toolkit. Software built with Cactvs is in charge of structure submission pre-parsing, structure input, structure ASN.1 encoding, 2D layout coordinate computation, structure rendering, most property computations, structure unification, the complete structure search system and Web-based structure sketching for queries. The OpenEye toolkit is used for structure standardization, some property computation (names, canonic SMILES), and SD file output. There are applications at NCBI which link to both OpenEye and Xemistry libraries - they can peacefully co-exist.


text:
Why doesn't the vapor pressure of water depend on the humidity?
Correct me if I'm wrong, I'd like to think of vapor pressure of water as the number of water molecules escaping from the water surface per unit area. Greater the leaving number, greater the vapor pressure. If this is right, can I conclude that the rate of evaporation is directly proportional to the vapor pressure?
My textbook says that the vapor pressure of water depends only on the temperature of the water. It doesn't depend on external pressure or humidity. But we know that the rate of evaporation decreases with increase in humidity.(Wet clothes dry faster in less humid air.) Doesn't this imply a decrease in vapor pressure with increase in humidity?

Vapor pressure is not the number of molecules escaping. It is the force exerted by the liquid molecules per unit area (on the surface). It is an intrinsic property of the fluid, not a measure of phase-change frequency.
It is of course related to how many molecules escape, but that is not what it is.

@ khaverim - Shouldn't it be "it is the force exerted by the vapor molecules..."?

Thanks! I  see the pressure is not same as the number of molecules haha :)

One important concept you're missing here is that of dynamic equilibrium - That is, for equilibrium processes, reactions are going backwards at the same time as it's going forward.
So it's incorrect to think of a evaporation as a situation where water is going only from a puddle to the air. If there's any amount of water in the air (humidity), some of the water in the air goes "backwards" from the air into the puddle. The more water in the air, the more water in the air goes "back" into the puddle.
To a first order approximation, the forward reaction rate (evaporation) is related to the temperature and the surface area of the liquid. This isn't all there is to vapor pressure, though. The "pressure" in vapor pressure actually comes from the reverse reaction, where the rate is related to the temperature, the surface area of the liquid, and the number of water molecules per unit volume that are in the air. (The higher concentration of water in the air, the faster the reverse reaction rate.) Now, if we treat water vapor as an ideal gas, from the relation PV = nRT we can see that (at a given temperature) the number of water molecules per unit volume (n/V) is equivalent (within a scaling factor) to pressure: P = RT * n/V -- we call this the "partial pressure" of the substance.
What the vapor pressure is representing is the balance between these two effects (the forward and the reverse). If you fix the temperature and take the ratio of forward reaction rate and the reverse reaction rate (e.g. as you would to get a reaction quotient/equilibrium constant), you'll see that the effect of the surface area vanishes (as it's the same for both forward and reverse), and the only things left are temperature (which we're holding fixed) and the partial pressure of the water vapor. If we assume equilibrium, there's a particular value of the partial pressure which results in equilibrium at this temperature ... this is defined as the vapor pressure.
So the vapor pressure of a liquid is actually representing an equilibrium constant. At equilibrium, the amount of water vapor in the air will be equal to the vapor pressure. Have less water vapor in the air than the vapor pressure? Water will evaporate to bring the partial pressure of water in the air closer to the vapor pressure. Have more water vapor in the air than the vapor pressure? Water will condense out.
The other issue is that you're attempting to use an equilibrium concept (vapor pressure) in a non-equilibrium setting. While the general principles still apply - that with air less than the vapor pressure water will go from the liquid phase to the gaseous one - because you're continually replacing the air with new air you'll never actually reach equilibrium. In non-equilibrium settings you need to go beyond the equilibrium constant (the vapor pressure by itself) and look at the true reaction rates.
As we've already seen, the macroscopic reaction rate (the net rate of evaporation) is related to the balance between the two microscopic processes: the forward rate of evaporation and the reverse rate of condensation. While the rate of water molecules leaving the liquid into the air is related only to the temperature and surface area, the reverse reaction is also related to the amount of water in the air. This is why clothes dry slower in humid air at the same temperature: the (constant) forward rate of evaporation is offset by the humidity-dependent reverse rate. The more humid the air, the faster the reverse reaction rate, and the slower the clothes dry. (Though because you're below the vapor pressure, there's still a net positive reaction rate in the forward direction, and your clothes do dry eventually.)

P.S. You may be wondering why, if the reverse reaction happens even in air with less than 100% humidity, why don't we see water condense out on dry surfaces in those conditions? The answer is that it does. You will get small layers of water being deposited on surfaces, even in sub-saturating humidity. It's just that the forward reaction is faster, and they evaporate again before becoming noticeable. The other issue is that nucleation of liquid is a much slower processes than adding to an existing water layer, so it's much less noticeable than the slowing of the evaporation of existing water layers.

Thank you so much for the clear explanations. May I know if I get this right :  100% humidity occurs when forward reaction rate equals the reverse reaction rate (equilibrium).  If this is the case, the max amount of water vapor that the atmosphere holds shouldn't depend on the air temperature. It should depend only on the temperature of liquid water ?

@Hiiii Implicit in the discussion above was that the temperature of the water and the air were the same. If they're different, the analysis becomes more complex, and the reaction rates incorporate both the temperature of the air and the temperature of the water. ... But at the limit, the maximum amount of water vapor that the air can hold only really depends on the temperature of the air itself, as above that temperature you can typically find some surface at that temperature where you're able to draw an equilibrium and dump excess moisture. (This is what causes rain, after all.)

The partial pressure of water vapor is equal to the equilibrium vapor pressure at the temperature of the liquid only at the surface of the water.  Above the surface of the water, the partial pressure of the water vapor decreases with distance, and far from the surface, approaches the partial pressure consistent with the relative humidity of the room air.  So there is a concentration gradient of water vapor in the air above the water surface.  Based on Fick's law of diffusion, this provides a driving force for diffusion of water away from the surface.  We call this diffusive flux of water vapor evaporation.  It is driven by the difference between the equilibrium vapor pressure of water at the surface and the partial pressure of water vapor in the bulk of the room air.


text:
Relationship between the water ionization constant and ionic strength
I read in a textbook that

the water ionization constant ($K_\mathrm{w}$) increases as the ionic strength of the solution increases

and it confused me. If the ionic strength increased, then the activity of each of the ions $\ce{H3O+}$ and $\ce{OH-}$ would decrease, leading to a decrease in the value of $K_\mathrm{w}$.
Can someone please explain the mistake in my way of thinking?

Mother Nature is not always easy to understand.
I am afraid that your textbook is telling the truth.
I have not yet understood all the details myself, but here is a link and a reference that perhaps can enlighten us both: http://www1.lsbu.ac.uk/water/water_dissociation.html#prod T. Vilarifio and M. E. Sastre de Vicente, J. Solution Chem. 26 (1997) 833-846. Theoretical calculations of the ionic strength dependence of the ionic product of water based on a mean spherical approximation.

The equilibrium constant is given though $$K_x = \prod_{\ce{B}} x_{\ce{B}}^{\nu_{\ce{B}}},$$
where $x$ is a specific quantity. We can choose that to be the activity $a$.
For the auto-ionisation
$$\ce{H2O (aq) <=>[H2O] H+ (aq) + OH- (aq)}\tag{1}\label{water}$$
we find therefore
$$K_a = \frac{a(\ce{H+})\cdot a(\ce{OH-})}{a(\ce{H2O})}.$$
Assuming the equilibrium $\eqref{water}$ is largely on the left side and therefore there is an excess of water, a reasonable simplification is to treat $a(\ce{H2O})$ as constant.
$$K_a' = K_a\cdot a(\ce{H2O}) = a(\ce{H+})\cdot a(\ce{OH-})$$
In ideally diluted solutions we also find
$$\lim_{c\to0}\gamma_i = \frac{a_i}{c_i}=1,$$
and therefore
$$K_c = c(\ce{H+})\cdot c(\ce{OH-}) \overset{\mathrm{def}}{=} K_\mathrm{w}.$$
The ionic strength is defined as
$$I_c = \frac12\sum_{\ce{B}}c_{\ce{B}}z_{\ce{B}}^2,$$
or for our case
$$I_{c,\mathrm{w}} = \frac12 \bigg[c(\ce{H+}) + c(\ce{OH-})\bigg].$$
Because of $\eqref{water}$ we can write
$$I_{c,\mathrm{w}} = c(\ce{H+})$$
and
$$K_\mathrm{w} = c(\ce{H+})^2, $$
therefore
$$I_{c,\mathrm{w}} = \sqrt{K_\mathrm{w}}.$$
From this relation you can see, that when $K_\mathrm{w}$ increases, the ionic strength also increases, vice versa.

If the ionic strength increased, then the activity of each of the ions $\ce{H3O+}$ and $\ce{OH−}$ would decrease, leading to a decrease in the value of $K_\mathrm{w}$.

The first part is correct, the activity of each ion would decrease. However, there are more ions, so the activity of the ensemble increases more (given ideal dilution), therefore also leading to an increase in $K_\mathrm{w}$.

"The first part is correct" I don't see how.


text:
Finding the formula of an alcohol with a given molecular mass

Deduce the molecular formula of the alcohol that has molecular mass 158.

I tried dividing this molecular mass (158) by the molecular mass of ethanol (46) to get a ratio, and then multiply the formula of ethanol $\ce{C2H5OH}$ by this ratio, but I didn't get a whole number.

The general formula of alcohol is CnH2n+2O. If you know the individual atomic masses of C, H and O, you can do some simple math to get your answer.

Hardly a well posed question :(

Why dividing by 46 ? Nobody says the your alcohol is a polymer of ethanol. Better follow Arishta's suggestion.

As I commented 2 years ago, there is no information to show that the alcohol is saturated, to which the formula CnH2n+2O applies.

We know that the general formula for the representation of an alcohol is $$\ce{C_nH_{2n+2}O}$$.
Where $n$ refers to the number of carbon atoms in the chain.
Lets put down the values of molecular masses of $\ce{C}$, $\ce{H}$ and $\ce{O}$.
We get,
$$12n + (2n+2) + 16$$
which can be rewritten as,
$$14n + 18$$
Thus the molecular mass given to us should satisfy a value of $n$ since it is an alcohol.
$$14n + 18=158$$
Solve and you will get $$n=10$$ and the formula is $\ce{C10H22O}$. Note that this formula does not uniquely specify one particular alcohol, as there are multiple isomers with the same molecular mass.
Let's see where you went wrong. You divided the value of molecular mass given with the mass of ethanol, which could have been correct had alcohols been made up of repeating units of the same group$$\ce{C2H5OH}$$.
But that is not the case. The molecular masses of different alcohols are not multiples of any integer. For example, methanol, ethanol, and propanol have masses of 32, 46 and 60 respectively.
Thus the only way to get your answer would be to equate the molecular mass with the general molecular mass for an alcohol having $n$ carbon atoms.

While the proof is correct, there is no statement that the alcohol is saturated. If n is not a whole number, then the alcohol must have a degree of unsaturation greater than zero.

$\ce{C_nH_{2n+1}OH}$ is the general formula.
$\ce{C}$, $\ce{H}$, and $\ce{O}$ have relative atomic masses of 12, 1, and 16.
To solve:
$$12n+2n+1+16+1=158$$
is the same as
$$\begin{align}14n+18&=158\\158-18&=14n\\(158-18)/14&=n\\10&=n\end{align}$$


text:
How accurate is this classification of matter
I've recently been starting to study chemistry.
This is how my textbook depicts classification of matter on a macroscopic level.

After seeing this, I was confused because wouldn't the mixtures, be it either homogeneous or heterogeneous, consist of either elements or compounds or both too?
Please help me clarify.

You missed the point. This "is" not "consist of" or "made from". So elements mean pure carbon, pure gold etc

Heterogeneous and homogeneous mixtures consist of many elements or compounds. For example, salt solution is a homogeneous mixture. It is made up of compounds-sodium chloride and water; the elements in this homogeneous mixture are Na, Cl, H, O.
The sub-classification of pure substances-elements or compounds is based on elements and compounds in their pure state. Na, Cl, H or O will independently fall under the category of elements. Water and sodium chloride will independently fall under the category of compounds.


text:
Positive deviation from ideal behaviour
I am confused about Positive behaviour from ideal behaviour as to why it occurs.
1) Molecular attractions plays a dominant role in positive deviation
OR
2) Molecular volume plays dominant role in positive deviation
I am confused these two facts.

Are you asking about deviations of  compressibility $pV/RT$ from unity as gas pressure increases ? If so, at low T and p molecular attractions dominate vs average thermal energy. At higher temperatures and pressures repulsive forces dominate which tend to make $V \gt V_{ideal}$.

@porphyrin I am talking with respect to vapour pressure.

If in a binary solution of two components A and B, the interaction between unlike components A-B is weaker than the interaction between the like components A-A and B-B, then it means that in such solutions molecules of A (or B) will find it easier to escape in the state A-B, than when they are in their pure states (A-A or B-B). It means that molecules of A (or B) will find it easier to escape from the A-B form.
This will result in having greater vapour pressure of each component of the solution than expected on the Basis of Raoult's law, and hence the total vapor pressure will also be higher than in the case of ideal solution. Thus the deviation will be positive.
In such solution $\Delta_{mix}H$ is positive because energy is required to break A-A and B-B bonds. For such solutions, the dissolution process is endothermic, i.e, the solubility will increase with temperature (Le Chatelier's principle) .
$\Delta_{mix}V$ is also positive because there is decrease in the magnitude of intermolecular forces in the solution, the molecules are loosely held, and therefore the volume of mixing increases.

So finally which statement(fact) is correct ?

1st statement is certainly correct.

http://chat.stackexchange.com/rooms/54160/jee-preparation


text:
Why is this carbocation more stable?
For comparing the stability of different compounds that undergo resonance, I was taught to first check for equivalent resonance, then the number of atoms on which the delocalization was taking place and so on(e.g phenolate ion vs acetate ion)
But comparing benzyl carbocation with $\ce{CH2=CH-CH2+}$, even though $\ce{CH2=CH-CH2+}$ has equivalent resonating structures(and hence a greater extent of delocalization of electrons), it is less stable than benzyl carbocation. Why?

Benzyl carbocation has more number of resonating structures than this carbocation.

@cotton phenolate ion has more resonating structures than acetate ion too but because of equivalent resonating structures, the acetate ion is more stable.

@xasthor, see when a negative charge is delocalised on a set of electronegative elements, it can be considered stable, and similarly for a positive charge on electropositive ones, but comparing the resonance of acetate ion,with that of an allyl carbocation, doesn't seem fit. I agree with what cotton said, the no of resonating structures determines the resonance energy, hence stability.

I'm not comparing the stability of acetate ion with this allyl carbocation. I was simply giving an example of how equivalent resonance, in the case of acetate ion vs phenolate ion is more important than the number of resonating structures. Here I am comparing $\ce{CH2=CH-CH2+}$ with benzyl carbocation.

Equivalent resonance structures was not the only reason acetate ion was more stable than phenolate ion. The negative charge was delocalized on two oxygen atoms(which are very electronegative); this also gave the negative charge great stability. Added to it, negative charge was delocalized on carbon(electropositive atom) in the resonating structures of phenolate ion, which decreased their stability.

@Cotton (very first comment) Not to forget that beside "simple resonance" structures drawn on paper, benzyl cations rearrange into Hueckel aromatic tropylium cations, as seen in mass spectroscopy at $m/z = 92$, too.

More resonance structures does not necessarily mean more stable. Maybe here it does, but not in general.

Orthocresol, that may mislead new students, for whom it may be a good rule of thumb to believe that more resonance structures do indeed suggest greater stability. Can you elaborate on why this trend may not be true in general?

What all the comments boil down to is we are comparing different things:
Allyl vs benzyl cations:  Both contain onlybhydrogen and carbon, with the positive charge formally distributed over two carbon atoms in allyl versus four carbon atoms.  Putting the positive charge on more carbon atoms makes the benzyl cation more stable.
Acetate vs phenolate anions:  Here, unlike the hydrocarbon actions mentioned above, oxygen enters the picture.  The negative charge in the acetate ion is distributed over two oxygen atoms, versus phenolate having its negative charge on one oxygen plus multiple carbons.  Having the second oxygen atom wins out because of the superior electronegativity of oxygen, which is a nonfactor for the cation case mentioned above.
Note that the "rules" above are not absolute laws.  You have to consider other factors such as aromatic or antiaromatic coupling that involves the charge; e.g. cyclopropenyl cation is much more stable than benzyl cation because of the positive charge being inherent to an aromatic ring, even though cyclopropenyl has only three instead of four  carbons to take up the charge.


text:
polar covalent bond
Please take a look at the image of polar covalent bond in the $\ce{H2O}$ molecule below. My question is given in bold type in the following discussion.

Here is my flow of thought that confused me so much :

$\ce{H}$ atom is slightly positive and $\ce{O}$ atom is slightly negative.
Positive and negative charges attract one another, no matter how small their charge is.
Ionic bonds happen between atoms that have different charges. for example, $\ce{NaCl}$. $\ce{Na}$ atoms have positive charges and $\ce{Cl}$ atoms have negative charges.
So, in the image above, hydrogen and oxygen in $\ce{H2O}$ should behave just like $\ce{NaCl}$. What I mean by "behave" is that they should form ionic bonds just like $\ce{NaCl}$. $\ce{H}$ and $\ce{O}$ have different charges. $\ce{Na}$ and $\ce{Cl}$ also have different charges. Based on this argument, why does $\ce{NaCl}$ form ionic bonds but $\ce{H2O}$ does not?

I understand the basic difference between covalent and ionic bonds & polar and nonpolar covalent bonds. But based on my line of reasoning above, what prevent molecules with polar covalent bonds from turning into ionic bonds? They have different charges, just like molecules in ionic bonds?
Do I have some basic misconception about polar covalent bonds and ionic bonds? Thank you for answering. I really appreciate it, this question has been bugging me for a while.

What does 'it' refers to in your question ?

Remember that all bonds have some of both ionic and covalent character to them. I suspect that the H-O bond in water is more ionic in character than say the C-H bond in methane for example due to the delta +/- charges in the case of water.

Reopen and immediately reclose with the homework close reason imho. Since that is not an option, I’m voting to leave closed.


text:
electrochemical (Galvanic) cell solution purpose
I will try to rephrase my confusion:
I am not sure if the redox reactions occur in Galvanic cell because of the wire connecting the 2 metals or because of the solution causing electrons to leave the zinc atoms
I have 2 questions:

What is the purpose of the solution in a Galvanic cell? Why in both halves of the cell, the solution contains positively charged ions?
Do electrons leave a Zinc atom and a positively Zinc ion is added to the solution even without connecting both halves of the cell with a wire, or this process (the oxidation) occurs only when both electrodes are being connected with a wire

Thanks

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

I recently started to research this subject. About the first question I have no thoughts as I have no idea what is the purpose of it. About the second question, I understand that electrons leave the zinc and want to 'join' the copper, because zinc is more active metal than the copper but I am missing the understanding if the electrons leave zinc atom and a positive zinc ion is added to the solution because of the solution or because of the two metals being connected with a wire

Lets start with the answer to the first question:-
(1)The Purpose of the solution in the galvanic cell is to facilitate the conversion of $M^{2+}$/$M$ or $M$/$M^{2+}$ as the case may be. Basically the solution contains Metal Ions which when required may convert to the Neutral Metal Atom. Similarly the electrode contains Neutral Metal Atoms which when required may convert into ions.
For Example, in the $Cu$ Cathode:-
$$\ce{Cu_{(aq)}^{2+} + 2e  ->Cu_{(s)}  }$$
Over here an atom from the aqueous phase loses two electrons and moves onto the cathode. BUT THIS WOULD NOT HAVE BEEN POSSIBLE HAD THERE NOT BEEN A SOLUTION AROUND THE CATHODE CONTAINING $Cu^{2+}$ IONS AROUND IT.
Thus we need a soultion around the electrodes.
Moreover the solution in both the halves contains both positive cations(like $Zn^{2+}$ and $Cu^{2+}$) as well as negative anions like $Cl^-$ or $SO_4^{2-}$ if the solutions are made up of $\ce{ZnSO4}$ and $\ce{CuCl2}$ .
(2) Electrons do not leave the $Zn$ atom unless and until the two electrodes are connected by a wire, since when electrons leave the metal atom they need to get used by another metal ion, which is what happens in the $Cu$ cathode
$$\ce{Cu_{(aq)}^{2+} + 2e  ->Cu_{(s)}  }$$.
Moreover current(electrons) does not flow in a broken circuit thus the two electrodes need to be connected by a wire, and the two solutions need to be in contact too, possibly through a salt bridge or a permeable membrane.

What a detailed answer. Did the solution contained Metal Ions before being in contact with the cathode (the metal) or when they first mixed and touched each other then the solution started to "absorb" Metal Ions from the cathode?

A solution of an ionic metal always exists in the form of ions, so yeah the solution contained ions even before it touched the electrodes. Any solution of an ionic metal like Zn Cu Na or Li exists in the form of ions. However, once it comes in contact with the electrode the conversion of ions->metal atoms or metal atoms->ions depends upon the nature of the metal and the nature of the metal of the other electrode.

I get it. So if I understand well, without connecting the wire between the two halves of the cell no reaction will happen between the solution and the metal of the electrode right? No oxidation nor reduction will happen without connecting the wire, is that right?

Yes that's correct. The wire is the reason the electrons flow and facilitate the reactions. However, there's a catch, I don't know whether you need to know this right now or not but when you dip an electrode into a solution some reduction/oxidation takes place which leads to the formation of a potential on the electrode which stops further oxidation/reduction from happening. And this happens without the connecting wire. However, DO NOT research about this if you don't need to know this RIGHT NOW. Learn only what you have to right now, Don't get confused.


text:
How do you make sense of acid base reactions in organic chemistry?
My question is in regards to acid base reaction that take place in larger chemical reaction in orgo. Like, I understand acid base reaction, how to qualitatively check which is stronger acid or base or where equilibrium lies, very well. But i feel that when I am trying to study the mechanism of series of other reaction like grignard mechanism, oxidation of alcohol, reaction with aldehydes etc… I see many intermediate acid base reaction in these mechanism that don’t make sense in terms of what I learned earlier when I was exclusively studying acid base reaction.
For instance, I have seen many intermediate acid base reaction in larger reaction where an oxygen atom of one molecule attack H atom attached to oxygen atom in another molecule. And then I am thinking what the point of this reaction because equilibrium is gonna be in middle and not favor products since the conjugate base create will be acid right away anyways.
Other times I have seen H20 act as an acid while molecule with sulfur atom act as a base. And again I am thinking how this is possible because isn’t sulfur a stronger acid then water. When I  asked two of my teacher about this. One of them said she actually made a mistake. But the main professor actually said that she is correct and to explain this, she said that water would even be hydronium ion here and when there is lot of water around, this water can serve as acid. However I couldn't fully understand what she meant by that? Can anyone please explain what she meant by this or how these acid base reaction possible. I have attached picture of what I am mean below to explain it my problem more clearly.

The answer to both questions lies in the fact that a reaction can be directed to a certain extent by controlling the concentration of the reagents.  The first reaction is perfectly fine since an alkoxide ion is more basic than a hydroxide ion. As for the second reaction, a carbon attached to an electronegative element is quite acidic. Hence is capable of possessing a negative charge. This along with the excess amount of water we supply for this reaction and the fact that water is neutral in nature adds to the reaction progressing in the forward direction.
Hope this helps

Thanks for the reply. Can u explain more about the relative concentration of reagents? How does concentration of reagents help in favoring one product over another?

@TLo Please don't post comments asking for more information on a new topic. Ask a new question instead, as specific of one as possible.


text:
Titanium in HF vs HCl
So we tried the reaction between Titanium metal with hydrochloric acid and hydrofluoric acid at a similar concentration. While boiling temperatures and constant heat over a long period of time are required to etch $\ce{Ti}$ using $\ce{HCl}$ even cold $\ce{HF}$ is enough to immediately break the protective oxide layer and the metal vigorousely reacts in the solution afterwards.
My question is now why $\ce{HF}$ is so much better at etching and dissolving $\ce{Ti}$ compared to aqua regia or $\ce{HCl}$?
Of course there is a protective oxide layer on the metal but boiling it in $\ce{HCl}$ will also form a purple solution after a while and the color intensifies if heat is applied over a long period of time. So even after the layer is destroyed the metal itself continues to react with the $\ce{HCl}$. For $\ce{HF}$ it reacts way faster so I assume it's not an issue with the protective layer itself but rather the reaction with the pure metal.
I know for glass and $\ce{HF}$ it's a combination where $\ce{H2F+}$ reacts with $\ce{Si-OH}$ endgroups and increases the electron density on the bridging oxygen which makes them more basic and the $\ce{Si-O-Si}$ bond breaks to form $\ce{Si-OH}$ bonds. But this is all for the oxides and I assume the titanium dioxide will be dissolved in a similar manner.
For metals I couldn't find much information on the topic. I know there are some similar elements, Niobium, Silicon, etc. they all require $\ce{HF}$ to be etched or dissolved.
Maybe it's the case because a $\ce{Ti-F}$ bond, much like a $\ce{Ti-O}$ "bond" is way stronger than a $\ce{Ti-Cl}$ bond? Which would explain why $\ce{HF}$ starts the reaction faster as this unique $\ce{HF/H2F+}$ mechanism seems to work pretty well for tough oxides.
Maybe it's also just the fact that more energy is produced when a $\ce{Ti-F}$ bond forms but the fast reaction starts immediately so it shouldn't just be the heat that is produced enhancing the reaction.
The process itself is a redox-reaction but the essential part here is the $\ce{H+}$ not the counter-ion. So from a redox perspective it should be pretty much the same, unless the redox potential and the overall redox reaction is influenced by the $\ce{Ti-F}$ bond formation, like the hydration enthalphy adds to the oxidative potential of oxygen and suddenly $\ce{Cu(I)}$ turns into $\ce{Cu(II)}$ once the insoluble $\ce{CuCl}$ is added to water.
(Also I'm sorry, but I cannot create a suitable tag with my score at the moment)


text:
Why do we observe signal coalescence during dynamic NMR?
At low temperature, the rotation around the N-C amide bond of dimethylformamide is slow, so the two methyl groups are in different chemical environments. The H NMR thus shows a peak for each of the methyl groups.
As the T is increased, we see that the peaks begin to coalesce, until eventually they coalesce into a single average peak. The basic explanation is that, since the methyl groups are exchanging their chemical environment at a rate faster than the NMR time scale, they are considered to be within the same average chemical environment.
I am not very comfortable with this explanation, because it could also be argued that 50% of the time the methyl will be emitting signal A, and 50% of the time it will be emitting signal B - and the FID signal should still contain two signals. Alternatively, if the rotation is continuous, then I could expect a very broad peak extending from A to B.
My guess is that there is something about the structure of the signal and the Fourrier transform that I do not understand. Could someone help me out understanding why this coalescence occurs?
Thank you!

related Why does DMF (dimethylformamide) have 3 peaks in its 1-H spectrum? and NMR: Estimate rate of coalescence when coalescing signals are multiplets

Thank you Mithoron. The answer to that question provides the basic answer, and I am hoping I can understand the why a bit deeper :-)

Yeah, this could be treated as sort of follow-up to DMF question.

Do you want a mathematical answer, or would just a qualitative one suffice? The books listed in this thread (see Physical chemistry > NMR) cover the topic of chemical exchange very comprehensively. Keeler is more readable, Levitt and Gunther have the maths to back it up (QM treatment and Bloch equations respectively).

I would like to eventually understand the mathematical answer, but for now a qualitative answer that points me in the right direction to learn more about the topic would suffice!

Levitt (Spin Dynamics) p 488-495 explains very clearly how when the molecule jumps between two sites the effect this has on the FID is to cause the two signals become  broadened , merged & then narrowed to a single peak as the exchange rate constant is increased. There is also a mathematical treatment if you wish to follow that.

The basic explanation is quite simple: a spectrum reports the rate of precession of a collection (ensemble) of nuclear spins. Precession can be pictured as orbital motion about a common axis, like earth rotating about the sun (merely for purposes of illustration). Now assume you have two groups of spins (equal number in each), spins in each group precessing at a fixed rate, that is, spins in each group precess coherently. Since the NMR signal is an average over all of the spins, and you have two populations with distinct precession rates, you see two independent peaks in the spectrum.
If on the other hand there is fast exchange (fast to be defined later) between the spins in the two populations, each spin ends up with an average precession rate, the mean of the starting values of the populations. In that case, the NMR spectrum contains only one peak. The spins have been mixed and each population is no longer pure, the original populations have lost coherence.
In the intermediate regime, when exchange rate is not so fast that the populations are thoroughly averaged, the spins will inhabit mostly one population but their rates will deviate from the pure initial rate due to sampling of the rate of the other population. This jumping between populations leads to a loss of coherence, and the NMR spectrum will show broadened peaks. These eventually coalesce into a broad central peak with a precession rate equal to the average over the two starting populations. The condition for coalescence (fast exchange) is that the exchange between populations happen rapidly relative to the precession rate about the axis, thereby blurring the precession rate of individual spins.


text:
Oxidative cleavage of 2-methylcyclohexa-1,3-diene
What are the products of oxidation of 2-methylcyclohexa-1,3-diene using hot concentrated acidified $\ce{KMnO4}$ solution?
Do we approach this problem the same way we would approach oxidation of alkenes?
I'm unsure how this mechanism would work.

Yes, pretty much. After all, you have two alkenes in your starting material.

Thank you for your answer. Could you draw out the products of oxidation of 2-methyl 1,3 cyclohexene? I'm unsure of how to start with this question.


text:
How is the acetoxy group ortho/para-directing in electrophilic aromatic substitution?
Why is the $\ce{-OAc}$ group o,p-directing in electrophilic aromatic substitution?
How can I compare the activating effect of $\ce{-OAc}$ with $\ce{-OH}$ and $\ce{O-}$? Is it to do with the number of lone pairs on oxygen?

Phenoxide is more reactive than phenol towards electrophilic substitution. However, the number of lone pairs isn't the factor that controls it, because it is not possible for both lone pairs to be simultaneously in conjugation with the benzene ring. If one lone pair is in conjugation with the benzene ring, the other lone pairs will not be oriented correctly to overlap with the pi system of the ring. So, you can have one, two, or three lone pairs - it does not matter because any lone pairs beyond the first are useless.
Only the availability of the lone pairs matters. That has to be discussed with inductive or resonance effects. With an $\ce{-OH}$ group you have a hydrogen atom fighting for some of the electrons (inductive effect), and with a $\ce{-O^-}$ group there is nothing else.
Moving on to the acetoxy group $\ce{-OC(O)CH3}$, the lone pair on oxygen is delocalised into the electron-withdrawing carbonyl group, via resonance. You can think of it as the $\ce{C=O}$ and the benzene ring both fighting over the oxygen lone pair, which makes it less available for donation to the pi system. Consequently, the acetoxy group is still o/p-directing (because of the presence of at least one lone pair), but is not as strongly activating as the hydroxy group (because the lone pair is less available).


text:
Why is arsenic trichloride more likely to be polar than aluminum trichloride?
According to my chemistry teacher, $\ce{AsCl3}$ is more likely to be polar than $\ce{AlCl3}$. Why is this? As I understand it, the difference in electronegativity is what makes a compound more or less likely to be polar, and $\ce{AlCl3}$ has a much greater difference between the two elements' electronegativities. So why isn't it more likely to be polar of the two?

By the way, welcome to Chemistry.SE! Take the tour to get familiar with this site. Mathematical expressions and equations can be formatted using LaTeX syntax. If you receive useful answers, consider accepting one.

It is because $\ce{AlCl3}$ has a trigonal planar geometry (leading to no permanent dipole moment) and $\ce{AsCl3}$ has a pyramidal geometry (leading to a permanent dipole moment).

Why do their geometries differ? Both are, according to VSEPR, AB3 - type molecules without any lone pairs (unless my lewis structures and Google's are way off), so why is AsCl3's pyramidal?

Arsenic centers will look like nitrogen centers, which do have a lone pair.


text:
Betaine HCl stomach pH
It seems betaine HCl is often recommended for those suffering from "low stomach acid" -- which, as I understand, is having too high stomach pH for proper digestion (especially for proteolysis via pepsin). However, I have a few questions then -- though I'm not sure if my reasoning is correct here:
Understandably, one wouldn't want to drink pure or highly concentrated HCl by itself to increase stomach acid! But then, what role does the betaine play?
I'm guessing betaine HCl probably does not dissolve in water to give the same pH as just straight (or concentrated) hydrochloric acid -- since then it would seem just as dangerous as drinking plain HCl!
In that case, if it doesn't decrease the pH as much, making it safe for oral consumption, what value does it bring for "lowering stomach pH" anyway? I mean, one could just drink a little vinegar or citric acid for the same effect?
Or, perhaps, is betaine-HCl just a means of some "delayed release" of HCl to lower stomach pH without hurting the mouth and esophagus during its initial consumption? Betaine may just be a useful amphoteric here, given it's quarternary ammonium and carboxylic acid groups (although not just any amphoteric, such as water, may suffice here!)
Alternatively, could the value of betaine HCl simply be in providing a source of $\ce{Cl-}$ anions, possibly for increased pepsin activity? (I'm not sure if pepsin requires merely low pH or specifically also needs $\ce{Cl-}$ anions as well) Betaine-HCl seems to be often formulated with additional pepsin enzyme(s) as well...

One factor could be that betaine has a pKa of 1.8, considerably higher than HCl and somewhat close to stomach pH.  Pure HCl could have a pH of 0 or lower.  Betaine is also a biocompatible osmolyte.

I had trouble with food not digesting whenever I ate specific foods. Now, I pop a tablet or two of betaine HCl and the problem is over. Before this Zantac 75 was the only thing that stopped me from puking. And believe me, I tried all of the products. Anyway, betaine HCl actually does work if somebody has indigestion. I don't know about all that other stuff quoted. Most of the stuff you hear is nonsense anyway. Anyway. That's my two cents.

It's often formulated with additional pepsin (and possibly other enzymes too). That could be what's helping you, especially in the quantities (ie milligrams) of betaine-HCl you'd have to take!

Looks like I was right to be skeptical, according to WebMD:

Betaine hydrochloride is a chemical substance made in a laboratory. It
  is used as medicine.
Betaine hydrochloride has an interesting history. Betaine
  hydrochloride used to be included in over-the-counter (OTC) products
  as a “stomach acidifier and digestive aid.” But a federal law that
  went into effect in 1993 banned betaine hydrochloride from use in OTC
  products because there wasn’t enough evidence to classify it
  “generally recognized as safe and effective.” Betaine hydrochloride is
  now available only as a dietary supplement whose purity and strength
  can vary. Promoters still claim that some health conditions are due to
  inadequate stomach acid, but this claim has not been proven. Even if
  it were true, betaine hydrochloride wouldn’t help. It only delivers
  hydrochloric acid but does not itself alter stomach acidity.
Betaine hydrochloride is also used to treat abnormally low levels of
  potassium (hypokalemia), hay fever, “tired blood” (anemia), asthma,
  “hardening of the arteries” (atherosclerosis), yeast infections,
  diarrhea, food allergies, gallstones, inner ear infections, rheumatoid
  arthritis (RA), and thyroid disorders. It is also used to protect the
  liver.
Don’t confuse betaine hydrochloride with betaine anhydrous. Use only
  the FDA-approved betaine anhydrous product for the treatment of high
  levels of homocysteine in the urine (homocystinuria). This is a
  symptom of some rare genetic diseases.
How does it work? It isn't known how betaine hydrochloride might work.

from http://www.webmd.com/vitamins-supplements/ingredientmono-312-betaine%20hydrochloride.aspx?activeingredientid=312&activeingredientname=betaine%20hydrochloride

Betaine HCl is usually delivered in capsules which pass through the mouth and esophagus intact.
Regarding the webmd article that says " Even if it were true, betaine hydrochloride wouldn’t help. It only delivers hydrochloric acid but does not itself alter stomach acidity."
This statement seems contradictory. Indeed, I found article the tests stomach ph after betaine hcl usage and confirms that indeed it reduces it:
https://www.ncbi.nlm.nih.gov/pubmed/23980906

But in that study it looks like the pH measurements were taken on rather fasted patients, not after having a moderate amount of solid food intake. I'd imagine a much greater amount of betaine HCl would be required in that case to significantly raise stomach acidity -- which, im guessing, is another reason why the FDA banned it in the first place.

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


text:
How do I increase the amount of electrons transferred in a battery?
I made a simple battery with $\ce{Al}$ as anode electrode with $\pu{0.7 M}$ aluminum sulfate electrolyte. Platinum is my cathode electrode with $\pu{2 x 10^-4 M}$ $\ce{KOH}$ solution. I am generating $\ce{85 mA}$ (my cell produced $\pu{0.04 V}$ under a $\pu{47 \Omega}$ load). How can I increase amount of electrons to increase the current produced by my cell?
My half cell reactions are
\begin{align}
\ce{2Al &-> 2Al^3+ + 6e-}\\
\ce{3H2O + 6e- &-> 3H2 + 6OH-}
\end{align}

Can you provide a picture of your setup, and some more details of the shape/configuration/etc. of your electrodes? As the question sits, there's not enough information to help you debug your apparatus.

simply increasing your electrode areas by either physically using large electrodes, or by sandpapering your electrode surfaces to make rougher surfaces should help. I do agree with hby2py though, more details would help in debugging.

Please note that the proper term for "number of moles" is amount of substance. The former would be the same as referring to the mass as "number of kilograms".

Have you acidified your aluminum sulfate electrolyte at all? Are you stirring either electrolyte chamber?

Also, why are you chilling your KOH?

yes, I stirred both the solutions. I did not acidify the aluminum sulfate solution. The addition of aluminum sulfate to water made the pH decrease. I just wanted to know how the KOH will react under a lower temperature


text:
Aromatic Overtones
For IR Spectroscopy, I noticed that there are aromatic overtones or "benzene fingers" a bit below the $\mathrm{2200~cm^{-1}}$ region. Why do these occur in IR spectroscopy? And is there any way to prevent them?

In general there are 'fundamental' lines at frequency $v_1, v_2$ etc. Weak overtone transitions can occur for any fundamental vibration at frequency $nv_1$, $nv_2$ etc. where $n=2, 3,..$ and so on for each vibration $v_i$. Combination bands can also be present and involve exciting two (or more) vibrations and have frequency $nv_1+mv_2$, where n and m are integers and $v_1,v_2$ represent any two different vibrations. Relative to the fundamental frequencies the overtone and combination transitions are inherently weak. There is no way to prevent them unless you are able to align molecules in your sample and do polarisation spectroscopy and even then many lines will still be present depending on dipole direction and selection rules.
Transitions from the ground vibrational state are the most intense as these have the greatest population, but 'hot' bands starting from $v=1$ can also occur if the temperature is high enough. Greatly lowering the temperature will remove these.


text:
proteolysis - how to determine the partial or completeness of the cleavage reaction?
Though the question is about biomolecules, we agree the underlying functions, formation, degradation etc come down to chemistry, (then physics and maths). I am posting it here to receive an understanding from chemistry perspective. I am referring this info here to grasp the concept.
e.g. Trypsin is a protease that regulates specific hydrolysis/proteolysis by cleaving at the c terminus side of Arg and Lys.
Similarly, given two catalysts: (proteases) papain and bromelain. Both are said to regulate selective peptide cleavage reactions (specificity of proteolysis) - breaking peptide (amide) bond from C-terminal analysis (enzymatic hydrolysis). This is clear to me.

These enzymes only break c-terminus peptide bonds of specific amino acids. e.g. Papain for Arg and Leu. Bromelain for Ala, Leu, Ser, Gly, Tyr.
How to determine if papain or bromelain hydrolyse a given polypeptide chain completely or partially? And what's the resulting residue and chain? I am not expecting an one-liner, I want to understand the reaction mechanism and concept clearly for further application.
e.g. what's the resulting fragments of this after Papain act on it?
Arg-Leu-Ser-Ala-Tyr

@Martin-マーチン the poor question is lying here for weeks. Some of your expert advice and insight would be very helpful.

Are you asking experimentally how to work out if the peptide has been cleaved? Or asking if the enzymes given will cleave the example pentapeptide sequence ?

@Baran I am asking for the latter. I have started to read on Edman degradation. But not sure if that's the theory I need to read about. Is there resource/articles at y1 Uni level, explaining the concepts how certain enzymes do selective cleavage, the resulting amino acids sequences, and whether it's partial or complete. I do feel that my question is not clear enough as I am a bit lost in terms of the context of this whole scenario.

A biochemistry textbook would probably be a good place to start. Lehninger and Berg are both great books and cover proteins

@NotBaran do you know which part of Berg I should look at? Do you know what concept is this question trying to test?

A possibly helpful paper (I'd try to read it and answer your question, but I can't access the paper from where I am....): http://rstb.royalsocietypublishing.org/content/257/813/237

@ChildishJack I updated my question, the main issue remains the same. I am not able to find a proper reference within Yr 1,2 Uni levels to sort an answer in determining partial or completeness of the cleaving process/reaction.