What type of reaction is the reaction between an acid and a carbonate? Consider the following reaction:

$$\ce{2HCl + Na2CO3 -> 2NaCl + H2O + CO2}$$

Is this a double displacement reaction or a decomposition reaction?

The first step is a proton transfer (acid-base reaction), from HCl to carbonate, to produce hydrogen carbonate ion. The second step is also a proton transfer, from HCl to hydrogen carbonate to produce carbonic acid. The third step (or maybe better described as a two-step acid-catalyzed process) is the dehydration of carbonic acid to carbon dioxide and water.


How to judge accuracy and precision of measurement devices? I am unsure about the accuracy and/or precision of measuring instruments.
For example, if we have a graduated cylinder, a pipette and a beaker, how and why can we assert them that they can be used with accuracy, precision, or both?

Beakers, Erlenmeyer flasks, dropping funnels and the like are not volumetric instruments. They are not precisely calibrated, and their scales serve only as approximate guides. Therefore, the beaker that is mentioned in the question is right out.

Graduated cylinders, volumetric flasks, bulb pipettes, graduated pipettes, and burettes are volumetric instruments. They are calibrated; however, like any other measuring instrument, they only have a limited accuracy and precision.

The accuracy and precision do not only depend on the type of the instrument (e.g. graduated cylinder or pipette). They also depend on the dimensions, graduating divisions, and accuracy class of the instrument, which are usually internationally standardized.

In principle, volumetric instruments with a small diameter have a better precision because the same difference of the liquid level (i.e. the meniscus) corresponds to a smaller difference in volume. Therefore, typical long and thin pipettes are usually better than short and thick graduated cylinders. For example:

A standardized graduated pipette (Class A or AS) with a nominal capacity of 10 ml and an overall length of about 360 mm has a maximum permissible error of ±0.05 ml.
A standardized graduated measuring cylinder (Class A) with a nominal capacity of 10 ml and a maximum overall height of 140 mm (tall form) has a maximum permissible error of ±0.1 ml.
If you use an even larger graduated measuring cylinder (Class A) with a nominal capacity of 100 ml and fill it just to the 10 ml mark, the maximum permissible error is ±0.5 ml.

These errors represent the maximum permissible error at any point on the scale, and also the maximum permissible difference between the errors at any two points.

Since the same sources of error are, naturally, inherent both in calibration and use, a volumetric instrument should have a good precision in order to be able to be calibrated with a good accuracy.

However, note that graduated pipettes are usually calibrated ‘to deliver’ (TD, Ex), whereas graduated measuring cylinders are calibrated ‘to contain’ (TC, In). Hence, graduatted pipettes and graduated measuring cylinders may not be interchangeable at will.


Slater determinant as an unperturbed atomic wave function I've deduced following postulates from studying my chem books.

  1) Slater determinants are eigenfunctions of an unperturbed atomic Hamiltonian, which contains kinetic and central potential energy parts of each electrons only, since spin orbitals constituting the determinants are originated from one-electron Hamiltonian eigenfunctions.

Here, atomic Hamiltonian = (kinetic part of each electrons) + (central potentials between atomic nucleus and electrons) + (interelectronic potentials)

Spin orbit interaction and the other effects are neglected.

  2) Slater determinants or their linear combinations are eigenfunctions of total spin angular momentum and total orbital angular momentum operator($S^2$ and $L^2$) simultaneously.

In addition to them, I would refer to information from Quantum Chemistry, 6th Ed., written by I. N. Levine.

(p. 312) Since $S^2$ and $L^2$ commute with the atomic Hamiltonian and with the exchange operator, the zeroth order functions should be eigenfunctions of $S^2$ and $L^2$.
(The zeroth order functions Levine mentioned above is indicating the single or linear combinations of Slater determinants in a same configuration)

Well, I can accept the fact that the atomic Hamiltonian, $S^2$ and $L^2$ commute.

However if the single or linear combinations of Slater determinants in a same configuration are the zeroth order functions (eigenfunctions of the unperturbed atomic Hamiltonian), how the fact that $S^2$ and $L^2$ commute with the atomic Hamiltonian makes the Slater determinants be the eigenfunctions of the atomic Hamiltonian??

For example, a Slater determinant corresponding to the one of the Helium-first excited states, $|1s\alpha~2s\beta|$, is an eigenfunction of the unperturbed atomic Hamiltonian(this determinant is a zeroth order function), $S^2$ and $L^2$, but not of the atomic Hamiltonian.

However if the single or linear combinations of Slater determinants in
  a same configuration are the zeroth order functions (eigenfunctions of
  the unperturbed atomic Hamiltonian), how the fact that S2 and L2
  commute with the atomic Hamiltonian makes the Slater determinants be
  the eigenfunctions of the atomic Hamiltonian??

Actually, you go it right, but let us spell this out step-by-step. First, forget about angular momentum operators, since you do not need them to understand that a single Slater determinant is indeed not an eigenfunction of the atomic Hamiltonian (as you called it). That is it, you are right. A Slater determinant is an eigenfunction of the unperturbed Hamiltonian which describes a system of independent electrons, but not of the exact one.

Nevertheless, a Slater determinant can be used as a trial wave function in a variational procedure. That is the whole point of the Hartree-Fock method: a Slater determinant $\Phi$ is not an eigenfunction of the atomic Hamiltonian, but we could evaluate it energy $\left\langle \Phi \mid H \mid \Phi \right\rangle$ using Slater rules, and consequnetly, we could minimize it to find an upper bound to the ground state energy. The resulting Slater determinant obtained by minimizing the energy is indeed only an approximation to the ground state wave function.

Note also that the exact wave function (which is an eigenfunction of atomic Hamiltonian) can be expressed as a linear combination of Slater determinants for the various possible electronic configurations, not just a single Slater determinant. And this is the theoretical basis for configuration interaction (CI) method.

Now back to all this business with angular momentum operators. A single Slater determinant is not necessarily an eigenfunction of neither $\hat{L}^2$ nor $\hat{S}^2$. However, as you mentioned, the Hamiltonian indeed commutes with all these operators (in non-relativistic approximation), and thus, the exact wave function is also an eigenfunction of $\hat{L}^2$ and $\hat{S}^2$ (as well as $\hat{L}_z$ and $\hat{S}_z$). To be more precise, the exact wave function can be chosen to be a simultaneous eigenfunction of all these commuting operators $\hat{H}, \hat{L}^2, \hat{S}^2, \hat{L}_z, \hat{S}_z$. And since it is really desirable, we require the same from out trial wave function: we would like it to be an eigenfunction of these angular momentum operators.

So we want to construct our trial wave function being an eigenfunction of angular momentum operators $\hat{L}^2, \hat{S}^2, \hat{L}_z, \hat{S}_z$, but a single Slater determinant would not necessarily do it, so we form what is called a spin-adapted configuration state function (CSF) - a linear combination of Slater determinants which is an eigenfunction of angular momentum operators.

Note, however, that this is not always possible.
For instance, for the case of unrestricted determinants using a linear combination would not help in making the trial wave function an eigenfunction of $\hat{S}^{2}$.


Hydrogen Bond Without Hydrogen Surely, by the same logic, any atom with a moderate electronegativity (carbon, phosphorus, silicon) can form a polarise covalent bond with an atom with a high electronegativity (fluorine, oxygen). Giving that carbon, phosphorus or silicon a partially positive charge (just like hydrogen), allowing it then to form a 'hydrogen bond' with another atom with a high electronegativity (fluorine, oxygen). Is it called a hydrogen bond because it most commonly occurs with hydrogen? Or is it that a hydrogen bond with carbon, phosphorus or silicon is instead called a carbon bond, phosphorus bond and silicon bond? Since you don't need a hydrogen to make an atom high electronegativity partially negatively charge, surely those atoms around atom would be paritally postive to balance the charge?

Just looking at the electronegativities, there seems to be no reason why atoms such as boron, carbon or silicium — which all have similar electronegativities to hydrogen — should not undergo hydrogen-bond–like intermolecular interactions.

There is one key difference that separates hydrogen from all the other nonmetals that readily form 2-electron-2-centre (i.e. standard) bonds at ambient temperature and pressure: the lack of any core electrons.

Atoms like boron and carbon (or worse: silicon) have a low-lying 1s atomic orbital that is generally not considered to take part in bonding. However, its spherical structure surrounds the nucleus and effectively shields its positive charge from the outside world. Hydrogen, on the other hand, has a basically unshielded nucleus consisting only of a positively charged proton.

The positively charged hydrogen nucleus can now interact well with other orbitals on other atoms (typically part of other molecules). Any in-between core electrons will repulse the electrons that the electropositive element is trying to interact with. Hence, the stabilisation is greatest in hydrogen’s case.


Differences in using molten NaCl compared to aq. NaCl in a galvanic cell having copper anode and cathode Since we know that copper anode will release $\ce{Cu^2+}$ ion, so I was wondering whether the cathode will thicken as copper has a lower position on the electrochemical series.
Based on my research, I found that some said that cathode will only thicken if the solution is stirred, some said that hydrogen gas will form.
Hence, are there any difference if electrolyte used is in a molten state compared to in an aqueous state?

A galvanic cell with the same electrodes ( copper ones in this case ) is usable as a galvanic cell only if cathodic and anodic electrolyte differs in the concentration of electroactive components.
E.g. if we can use 2 different concentrations of $\ce{CuSO4}$ solution, separating them by a diaphragm or a salt bridge. The electrode with more concentrated solution around would become cathode with the more positive potential and vice versa.
Putting 2 copper electrodes into homogeneous solution of $\ce{NaCl}$ ( or molten one ) does not form a galvanic cell, unless some minor undefined differences create some residual voltage.
In case you did mean an electrolytic cell with moltan salt instead, anodic copper would be dissolving.
At the cathode, only metallic sodium would depose initially. Since some copper ions reach the cathode, copper would start priority deposition. Sodium deposition would depend on the applied electrode potential.


Why is the enthalpy of combustion higher for n-butane as compared to 2-methylpropane? N-butane has an enthalpy of formation of $\pu{-125.6 kJ/mol}$, whereas it is $\pu{-134.2kJ/mol}$ for 2-methylpropane. When comparing the heat set free at the combustion process, why is it higher for n-butane? Isn't the enthalpy of formation the negative of enthalpy of combustion and therefore 2-methylpropane should be higher?

Take 4 moles of carbon and 5 moles of H2. Synthesize a mole of n-butane and measure the heat evolved (125.6 kJ). Then repeat the process, but make 2-methyl propane (you'll measure 134.2 kJ). Then, with sufficient oxygen,  prepare 4 moles of CO2 and 5 moles of H2O from each of the previous two procedures and measure the heat of combustion (more for n-butane).

The total heat evolved to form the final oxidation products from carbon and H2 will be the same, but since the paths are different, the heats evolved in the two steps of each procedure will be different. n-Butane evolves less heat upon formation, therefore can (must) evolve more heat in the next step when it is oxidized to final products.


Why is the stoichiometric coefficient in the denominator in the rate expression? For the reaction:  $$\ce{2A + B → C}$$

The rate expression is given as:

$$\ce{-1/2 * Δ [A]/Δt}$$  = $$\ce{-1 * Δ [B]/Δt}$$ = $$\ce{Δ [C]/Δt}$$

I can't get my head around why it is $$\ce{-1/2 * Δ [A]/Δt}$$  and not $$\ce{-2 * Δ [A]/Δt}$$

If A is consumed at twice the rate that B is consumed, why is the rate for A half (and not double) the rate for B?

The only answer I could find on the internet is "To show a standard rate of reaction in which the rates with respect to all substances are equal, the rate for each substance should be divided by its stoichiometric coefficient." Source: http://chemwiki.ucdavis.edu/Core/Physical_Chemistry/Kinetics/Rate_Laws/The_Rate_Law

But this doesn't make much sense to me. Why does the rate with respect to all substances have to be equal when one substance (A) is used up twice as fast as another substance (B)?

From the stoichiometric coefficients $\nu_{i}$ in your reaction equation

$$\ce{2A + B → C}$$

you can see (as you already correctly stated) that for each molecule of $\ce{B}$ consumed, $\nu_{B} = -1$, two molecules of $\ce{A}$ must be consumed as well, $\nu_{A} = -2$. Similarly for each molecule of $\ce{B}$ consumed one molecule of $\ce{C}$ must be formed, $\nu_{C} = +1$ (and analoguously for $\ce{A}$ and $\ce{C}$).
This will also be reflected in the changes of the amounts of those species, $\Delta n_{i}$, during the reaction, e.g. if the amount of $\ce{B}$ changes by $2 \, \mathrm{mol}$ you can expect a change in the amount of $\ce{A}$ by $4 \, \mathrm{mol}$. So, the ratios of the $\{\Delta n_{i}\}$ and the $\{\nu_{i}\}$ will be equal, i.e.

\begin{align}
  \frac{\Delta n_{\ce{A}}}{\Delta n_{\ce{B}}} = \frac{\nu_{\ce{A}}}{\nu_{\ce{B}}} \ ; && \frac{\Delta n_{\ce{A}}}{\Delta n_{\ce{C}}} = \frac{\nu_{\ce{A}}}{\nu_{\ce{C}}} \ ; && \frac{\Delta n_{\ce{B}}}{\Delta n_{\ce{C}}} = \frac{\nu_{\ce{B}}}{\nu_{\ce{C}}}
\end{align}

Rearranging these equations a bit leads you to:

\begin{align}
  \frac{\Delta n_{\ce{A}}}{\nu_{\ce{A}}} = \frac{\Delta n_{\ce{B}}}{\nu_{\ce{B}}} = \frac{\Delta n_{\ce{C}}}{\nu_{\ce{C}}} \ .
\end{align}

Now, divide these equations by the reaction volume $V$ and use $\frac{\Delta n_{i}}{V} = c_{i} \equiv [i]$ to introduce concentration changes

\begin{align}
  \frac{\Delta [\ce{A}]}{\nu_{\ce{A}}} = \frac{\Delta [\ce{B}]}{\nu_{\ce{B}}} = \frac{\Delta [\ce{C}]}{\nu_{\ce{C}}} \ .
\end{align}

Finally, if you divide now by the reaction time $\Delta t$ and use the values for the stoichiometric coefficients, you get the equations from your question:

\begin{align}
  \frac{1}{\nu_{\ce{A}}} \frac{\Delta [\ce{A}]}{\Delta t} &= \frac{1}{\nu_{\ce{B}}} \frac{\Delta [\ce{B}]}{\Delta t} = \frac{1}{\nu_{\ce{C}}} \frac{\Delta [\ce{C}]}{\Delta t} \\
  \frac{1}{-2} \frac{\Delta [\ce{A}]}{\Delta t} &= \frac{1}{-1} \frac{\Delta [\ce{B}]}{\Delta t} = \frac{1}{1} \frac{\Delta [\ce{C}]}{\Delta t} \\
-\frac{1}{2} \frac{\Delta [\ce{A}]}{\Delta t} &= -1 \frac{\Delta [\ce{B}]}{\Delta t} = 1 \frac{\Delta [\ce{C}]}{\Delta t} \ .
\end{align}


Farnesyl Diphosphate: Chemical commercial synthesis Are there any practiced commercial routes for the synthesis of "farnesyl diphosphate"? I'm especially interested in options that would scale to industrial scale processing.

To add some context, there's a bunch of secondary metabolites that branch out from FPP. Assuming one can immobilize the requisite enzyme the next question is whether the substrate FPP could be somehow manufactured at scale using a chemical synthetic route.

Farnesol is commercially available as a distillate of essential oil or synthetically derived from Nerolidol. These chemicals are commonly used in the fragrance industry, as such efficient synthetic processes should exist.

Here's a potential synthesis route for Farnesol (1) from the starting material Nerolidol (2). Farnesol is then phosphorylated into Farnesyl diphosphate.

Nerolidol (2) is treated with anhydrous acetic acid under basic conditions to produce Farnesol (1). From The Total Synthesis of Natural Products
edited by John ApSimon, page 201. Also referenced in US1663817A.

If we assume that isopentenyl pyrophosphate is similar to farnesyl diphosphate (1 isoprene unit vs. 3 isoprene units), then this technique "An Improved Synthesis Of Isopentenyl Pyrophosphate" can theoretically be used.

The double phosphorylation has stated yields from 32%-51%. Let's assume the yield of Nerolidol to Farnesol is 70%. Given the price of Neroidiol is 39 USD/kg†, the least efficient price would be about 174 USD per kilogram. The price of Farnesol is 129 USD/kg†, and Geraniol is 19 USD/kg†. (If one would like to synthesize Geraniol into Nerolidiol, steps are given on page 201 above.)

This is pure conjecture, but should most likely work given that Farnesol is similar to both Geraniol and Isopentenol. Also the reason I included the step from Nerolidol to Farnesol is because Farnesol is too expensive compared to the price of Nerolidol despite being one synthetic step away.

†Prices are quoted from Vigon International, a North American reseller.


Strongest negative inductive effect group between trimethylammonium, ammonium and dimethylsulfonium groups Which of the following group exerts the strongest -I effect?

  $\ce{-N(CH3)3+}$
  $\ce{-NH3+}$
  $\ce{-S(CH3)2+}$
  $\ce{-F}$

My idea is that positive species exerts more -I effect than neutral species. thus, 4) is neglected. Moreover, as nitrogen and sulfur have almost the same electronegativity 3) has two methyl groups and 2) none so 2) should exert more -I effect. However, the answer key says it to 1). Why is this?

The inductive effect can be quantitatively measured by the Hammett equation

$$\sigma(\ce{X}) = \mathrm{p}K_\mathrm{a}(\ce{H}) - \mathrm{p}K_\mathrm{a}(\ce{X})$$

where $\mathrm{p}K_\mathrm{a}(\ce{X})$ refers to the $\mathrm{p}K_\mathrm{a}$ of $\ce{X}$-substituted benzoic acid and $\mathrm{p}K_\mathrm{a}(\ce{H})$ to the $\mathrm{p}K_\mathrm{a}$ of unsubstituted benzoic acid. A more electron-withdrawing group will lead to a more acidic carboxylic acid group, a lower $\mathrm{p}K_\mathrm{a}(\ce{X})$, and hence a more positive $\sigma$ coefficient.

Values for a wide range of substituents in the meta and para positions have been tabulated in Table I of Ref. 1. Also included are Swain and Lupton's $F$ ($F$ for field) parameters which are designed to remove the influence of resonance effects on the original Hammett coefficients $\sigma$ (given in the same table; see Ref. 2 for original discussion).

$$\begin{array}{ccccc} \hline
\text{Entry number} & \text{Substituent} & \sigma_\mathrm{m} & \sigma_\mathrm{p} & F \\ \hline
\text{main text, p 167} & \ce{NMe3+} & 0.88 & 0.82 & 0.89 \\
58 & \ce{NH3+} & 0.86 & 0.60 & 0.92 \\
241 & \ce{SMe2+} & 1.00 & 0.90 & 0.98 \\
15 & \ce{F} & 0.34 & 0.06 & 0.45 \\ \hline
\end{array}$$

So, perhaps the answer should be $\ce{SMe2+}$. I don't have a good explanation for why, given that sulfur is less electronegative than nitrogen. I also do not think that the difference between $\ce{NMe3+}$ and $\ce{NH3+}$ can solely be attributed to the inductive effect of the methyl groups, as it is quite possible that there are solvation or hydrogen bonding effects that affect the $\mathrm{p}K_\mathrm{a}$ values. The bottom line is that I don't think this can be rationalised as simply as the question seems to want.

References

Hansch, C.; Leo, A.; Taft, R. W. A survey of Hammett substituent constants and resonance and field parameters. Chem. Rev. 1991, 91 (2), 165–195. DOI: 10.1021/cr00002a004. (there is a non-paywall copy here)
Gardner Swain, C.; Lupton, E. C. Field and resonance components of substituent effects. J. Am. Chem. Soc. 1968, 90 (16), 4328–4337. DOI: 10.1021/ja01018a024.


Why does a buffer work best at the pH closest to its pKa? I think I'm having trouble understanding exactly how the $\mathrm{p}K_\mathrm{a}$ relates here on a conceptual level. For example I know that carbonic acid works somewhat well as a buffer at $\mathrm{pH}$ near 7, since its $\mathrm{p}K_\mathrm{a}$ is relatively close, but why is that exactly?

A buffer works best when there is the same amount of weak acid/base and its conjugate. If you look at the Henderson Hasselbalch equation, and set the concentration of the weak acid/base equal to each other, $\mathrm{pH} = \mathrm{p}K_\mathrm{a}$.


Why NaHCO3 added to the water used (Ion Chromatography Suppression) I'm studying HPLC detection techniques and stuck on the conductometric suppression technique. It is describing how buffer ions (in my case $\ce{Na+}$ and $\ce{HCO3-}$) are being either adsorbed ($\ce{Na+}$) or protonated ($\ce{H2CO3}$) which makes them electrolytically inactive and leaves just water in the column.

Great! I understand that background conductivity is a problem. What I do not get is why was $\ce{NaHCO3}$ added to the water in the first place?

Ion chromatography usually means chromatography where aqueous (and usually inorganic) ions are the analytes.  They are separated from each other by the use of ion exchange resins.

If your analytes are anions, then they will be separated on an anion exchange resin.  These resins have a crosslinked organic polymer that contains permanent positive charges.  The negative charges that neutralize these resin-bound positive charges come from aqueous solution.  Different aqueous anions will have slightly different propensities to bind to and temporarily neutralize the solid-phase cations.

If you don't use an anion-containing eluent such as bicarbonate or hydroxide, the analytes that you would like to eventually detect through conductometric detection will permanently stick there!  So adding e.g. $\ce{NaHCO3}$ as an eluent will force them off of the column and ensure that your analytes eventually flow to the detector.

A decent diagram of the process is found in this PDF from Thermo (formerly Dionex). $%edit$


Is [7]сirculene aromatic or anti-aromatic? This compound is called [7]circulene and is one of the examples of non-planar aromatic systems (1). From the first overview of this compound, it is observed that there are 7 benzene rings so one would conclude that it is an aromatic compound:

There are $7 \times 4 = 28$ $\pi$-electrons, so this compound satisfies the criteria of anti-aromacity. Then how come this compound is aromatic?

(1) Smith, M. March’s advanced organic chemistry: reactions, mechanisms, and structure, 7th ed.; Wiley: Hoboken, New Jersey, 2013, p. 48, compound 40.

TL;DR: We don't count π electrons here, rather the π conjugated circuits [1]. [7]-сirculene has all conjugated circles of size $4n + 2,$ and none of $2n,$ what makes it an aromatic compound.
Strictly speaking, the aromaticity $4n + 2$ and antiaromaticity $4n$ rules in their original interpretation are very limited and cannot be universally applied.
According to the original Hückel $4n + 2$ rule, if a monocyclic, planar molecule has $4n + 2$ π electrons, it is considered aromatic.
But the count of π electrons is not what matters for polycyclic compounds, therefore a generalized $4n + 2$ rule exists: if a conjugated molecule has only $4n + 2$ π conjugated circuits, it is considered aromatic.
Non-benzenoid compounds can have both $4n + 2$ as well as $4n$ number of conjugated circuits counted by taking into account all Kekule structures. Among non-benzenoid non-alternant polycyclic hydrocarbons [5]-сirculene and [7]-сirculene both have only $4n + 2$ $(6, 10, 14, \dots)$ numbers of circuits.
Further reading: Is there any special rules for checking the aromaticity of polycyclic compounds?
Reference

Gutman, I.; Cyvin, S. J. Conjugated Circuits in Benzenoid Hydrocarbons. J. Mol. Struct. THEOCHEM 1989, 184 (1), 159–163. DOI: 10.1016/0166-1280(89)85141-3.


Regarding the free energy requirements for reacting gases Continued from my previous question here.

With regards to the Boltzmann factor, I am confused on what type of terms we can substitute for E. I know kinetic energy is one, free energy as well, but can we also do internal energy, enthalpy etc.?
From this link it looks like internal energy and molecular kinetic energy are related and for real gases the internal energy term contains other factors. Since U is a part of G, is it correct to say that kinetic energy in Maxwell Boltzmann Distributions make up a part of free energy?
I am still unclear as to whether I can interpret activation energy as a difference in free energy between average reactant energy and activated complex like diagram 1?
Is it correct to say that, reaction cannot occur unless the reactant have free energy above that of the activated complex, but even if this was satisfied other factors such as minimum collision kinetic energy, orientation etc. must also be met for reaction to occur?

Addendum for Q3 and Previous Post:

My confusion is with regards to the relationship between Maxwell Boltzmann Distributions like diagrams 1 and 2, if there even is one.

The M-B distribution implies to me that

activation energy $E_a$ (which I assume to be that in Arrhenius equation) is absolute given it is marked on an absolute energy scale
activation energy refers to a minimum absolute kinetic energy
activation energy is constant with temperature.

My confusion:

My impression has always been that activation energy (in the Arrhenius equation) refers to the difference in free energy between the transition state and reactant and that $E_a = \Delta G^{++}$ as noted in the latter two graphs, which appear to contradict the M-B Distribution. I'm not sure if there's a formal definition for activation energy that defines what "energy" it is.
The relationship between KE and activation energy (if any)
The effect temperature has on reactant and transition state free/kinetic energies. I believe that higher temperature lowers activation energy if it is the relative energy between the two.

Diagram 1:

Diagram 2:

I think what you are fundamentally saying is you are trying to relate classical thermodynamics to the molecular properties. It is best to have some firm familiarity with classical thermodynamics before attempting this. Classical thermodynamics does not require a molecular interpretation at all! (It helps a lot but is by no means necessary)!. I personally feel the answer I am about to write will help you more than trying to wade through terms and ideas/principles.... I hope so at least. If you don't like it then maybe someone else can provide a more quantitative solution.

1) The whole point of statistical thermodynamics is to provide a molecular interpretation. To do this we have to divide the system up into smaller sub-systems as I described earlier. The key to it is that there can be many "configurations" that give rise to the same macroscopic properties. The aim is to find the most probable one ... which is where your distributions come in.

It is the energy of the microstates that is what you put in so that you can calculate the various macroscopic properties you quote above. This is the best way to think about it. When you get to advanced theories this definition may change slightly. For now just accept that it is the energy of the micro state you are putting in.

In my honest opinion I don't know why courses teach the Maxwell-Boltzmann velocity distribution in a statistical thermodynamics course. I feel while a nice introduction to dealing with ensembles ... it leads to confusion (when I was learning this I had very similar questions with regard to what can be related to what, it only comes with time and exposure to the subject. So I know exactly where you are coming from!).

2) Whenever you see kinetic energy you should be thinking temperature rather than trying to get the individual molecular speeds... what does knowing the instantaneous position/velocity of a mole of gas particles at one instant in time really lead to? (not much)... I haven't really seen anyone combine the thermodynamic potentials with something like velocity. If anyone reading this has then I would love to hear what you have to say!! When I was first learning thermodynamics I clung on to what I was familiar with ... potential / kinetic energies ... now I don't really think about them when you say thermodynamics at all ... I don't think about $G$ as being a "part" of $U$ ... I just treat them as if they were separate (hmm how to phrase this) "properties" of a given system under certain conditions.

In fact I would go further to say that I view them as mathematical tools ... functions of variables of the system that I can manipulate to express molecular properties in terms of measurable quantities ....

3) Yes. You can indeed As is done by most theoretical chemists use Gibbs energies to calculate the thermodynamic activation barrier in the starting material and transition state.... The problem, which admittedly I may have confused you earlier about, is that this is done via the Arrhenius equation A kinetics equation. When in early undergraduate years you don't mix kinetics with thermodynamics. Later on it becomes (bar a factor of $2\pi$ or so ) acceptable to do so ... but you should not attempt this yet ....

A crucial point is that the activation energy graph's x-axis is the reaction coordinate. The reaction coordinate is a generalised coordinate which can be used (whatever it is) to describe the progression of the reaction.

Normally we deal very nicely in cartesian $x,y,z$ coordinates ... the idea of a generalised coordinate is that it can be used to reduce the dimensionality of the mathematical problem with some nifty rethinking of how we choose to describe the system .... for instance when talking about a simple harmonic diatomic molecule .. instead of choosing the location of both atoms in space (requiring 6 position coordinates) to fully describe the system as a function of time (as it oscillates back and forth ) i could use the relative separation (the internuclear distance) relate to the equilibrium bond length.... in this way it becomes much easier to describe the system ....

That was just one example of a generalised coordinate.... something that makes the description of the problem al little easier.... (especially if the system is subject to constraints, the generalised coordinates remove the constraint equation... anyway where were we.)...

Bottom line: ** there is a big difference between reaction coordinate and the Maxwell-boltzmann graph you linked.

4) This question in itself has hit on a very big topic. This is the realm of classical thermodynamics. What you need to understand is the difference between the standard Gibbs energy and the Gibbs energy .... For the following just imagine that $A$→$B$.

In a very brief overview it all boils down to this:

The Gibbs free energy is a mathematical function... don't try to pin down WHAT it is... its just another energy we can describe the system in .... It is a function of $P$, $T$ and $N$. Lets make it easier and say that the number of particles does not change. i.e $dN=0$ ...

Since the Gibbs energy is a function of $T$ and $P$ if we change either of these variables ... $G$ will change too .... The starting materials will have a certain $G$ and the products will have a certain $G$ ... The difference between these under standard conditions is fixed by definition that we are controlling the variables. This is like the thermodynamic drive for the process if it were to occur at our specified standard conditions .... Over the course of the reaction this value does not change since as I said it is the difference between the Starting Material and the Products under given conditions. .... This is also for pure values of the starting material and the product...

So far so good right?

As your reaction proceeds the composition varies i.e $dN\neq 0$. As I pointed out in another answer from the definition of the Gibbs energy this means that the value Gibbs function now changes... if you want a molecular picture it is due to the entropy of mixing adding in another factor to the "disorder" that has so far not been accounted for...

For an illustration please follow this link:

http://www.public.asu.edu/~laserweb/woodbury/classes/chm341/lecture_set5/lecture_set5.html

There are therefore two measures of the Gibbs energy. The one under standard conditions... unchanging for a given reaction... it is this one that has the $\Delta G^{\small o}$.. it is this one you get your electrode potentials off... it is this one that leads to the the equilibrium constant....

The other one is ever changing as the reaction proceeds since the composition of the reaction vessel is ever changing as $A$ goes to $B$. At equilibrium the net rate of change of A to B and B to A is now zero right... this in fact defines an equilibrium point.... shifting it slightly in one way or the other will cause the system to respond to re-establish equilibrium .... Most importantly we define equilibrium as the point where $dN=0$ .... for both types (in this case one would go up as the other went down ... ) but at equilibrium $dN=0$ .... This means that $dG=0$ too from the mathematics I showed you before ....

Therefore equilibrium is the point where the change in the Gibbs energies of two substances in a mixture is zero.... The reaction has no more drive if you like .... You can equivalently say their chemical potentials are equal too ... but leave that for now.....

So thats it ... Im sure if you ask question 4 to someone else they will give you a more complete answer.... but since its getting on for 2:00 am here I think I'm done. Hope that helps and again if I have made any errors ... please edit away :)

edit

What you are describing on the end of question 4 is know as the steric factor... you basically compare the frequency of successful collisions in the transition state geometry "linear" for example with respect to a full sphere of reactivity .... this gives you your factor :)

Edit
So the problem is your definition of the energy. I dot know if you have seen this before but perhaps this will help,...

Yes you can indeed use statistical mechanics to obtain a value of $G$ ... The best way to do it is to get an expression for the Helmholtz free energy and then just Legendre transform it to $G$ ... $G=A+pv$.

Whilst yes at a simple level the internal energy does indeed "contain" the kinetic energy ... when you learn a bit more about quantum mechanics .. we never talk about kinetic energy ... I think it would help to actually leave the classical results behind and focus on the actual version of events...

With reference to the uncertainty relation .. how do you think kinetic energy is something we can talk about at a molecular level? The answer is you don't. Once QM comes in it is not useful to talk about velocity.

That being said how do we describe motion ? after all the Schrödinger equation is just an equation of motion. All the Schrödinger equation describes is the time evolution of a vector state in a certain vector space - Hilbert space, Which is the space in which we describe quantum mechanics...

With this in mind ... all motion of the system is quantised ... can only have discrete values .... non-continuous values ... etc etc...

Translational motion is quantised... so is vibrational ... so is rotational ..... each degree of freedom has its own quantised set of discrete energy levels ....

Equipartition theorem uses classical ideas and distributions to distribute all the available energy across these states ... roughly it says that each has energy of $\frac 1 2 KT$....

To say $U$ includes kinetic energy is a primitive way of saying the internal energy is a function that describes the energy of all the modes in the system... (you would need a distribution to see how they are populated mind) ....

Equipartition theorem doesn't get it quite right when you consider the low temperature theoretical and experimental heat capacity of a substance..... Even at low temperature you are still expected to populate all modes with $\frac 12 KT$ ... but this doesn't happen since $KT$ does not have enough energy to populate said modes ... i.e it doesn't reach the quantised energy level requirement .... therefore fewer modes are available at low temperature...

Bottom line is: just try and separate the different preconceived competing ideas you have when it comes to statistical mechanics .... have you had a lecture course on this yet?

it is certainly not about having "enough Gibbs energy to react"... The Gibbs energy between a reactant and transition state $\Delta G ^{\ddagger}$ is not a variable thing as you are saying ... (obviously for a given set of chemicals),.... If it changed all the time then why would we write them down to compare them? The Gibbs energy in your picture is not the same Gibbs energy that goes to zero at equilibrium ...
\begin{equation}
\bigg(\frac{dG}{dq}\bigg)_{T,P}=\Delta _r G^{\small o}+RT\ln Q
\end{equation}
at equilibrium the left hand side .... the gradient of the graph in that link i sent you before ... is zero ...

The rate of change of the Gibbs energy with respect to the extent of the reaction is by definition zero at eqm.

When equilibrium is reached $Q=K$ that is to say the reaction quotient becomes the equilibrium constant ....


When disolving a iron pill for titration, why do I use sulfuric acid instead of nitric or hydrochloric acid? In a titration to decide the iron content of a pill we dissolved the iron in sulfuric acid and then titrated it with cerium(IV) sulfate. But I just wondered what's wrong with using nitric and hydrochloric acid in this scenario?

In this case, cerium(IV) sulfate is a strong oxidising agent, hence, using $\ce{HCl}$ would likely be oxidised to form $\ce{Cl}$ of higher oxidation state. Therefore, the titre would be vastly higher than expected.

$\ce{HNO3},$ on the other hand, is an oxidising agent and would oxidise $\ce{Fe}$ to a higher oxidation state than expected (of $\ce{Fe^3+}$). Hence, not as much cerium(IV) sulfate would be needed to reach the endpoint, resulting in a titre lower than expected.

Therefore, $\ce{SO4^2-}$, being a relatively unreactive species, is the best reactant in this case.


Complex organic molecules I am studying astronomy and came across the following term in the astrochemistry course called 'complex organic molecules' or also written as COMs. My question is: What is exactly meant with these molecules? Is it just a molecule with more than one carbon atom?

tl;dr: two different definitions. Astronomy: multiple carbon atoms in molecule. Chemistry: polymer

Interestingly enough, after reading about COMs here, as well as reading the Wikipedia page and the corresponding arXiv paper, it seems like chemists and astronomers have different definitions of what a complex organic molecule should be!

As far as I knew, in chemistry complex organic molecules were long polymers, such as proteins, which were composed of thousands upon thousands of amino acid units. In the astronomy paper, however, they cite other types of molecules.

$\ce{CH3OH,
CH3CHO, HCOOCH3 and CH3OCH3}$, all cited as "complex" (haha) organic molecules in the paper, would appear to chemists as relatively simple molecules. (I read the paper, because it piqued my interest that something like a protein could be found in space). I then read the Springer article.

  The term “complex organic molecules” is used differently in astronomy and chemistry. In astronomy, complex organic molecules are molecules with multiple carbon atoms such as benzene and acetic acid. These molecules have been detected in interstellar space with radio telescopes. In chemistry, “complex organic molecules” refer to polymer-like molecules such as  proteins.


Modern open-source tools for simulation of NMR spectra I am interested in predicting the NMR spectrum of small organic compounds.  It doesn't matter to me if the prediction is very accurate.  I'll eventually be comparing the prediction to experimental results.

It looks as if several free online tools offer the ability to do such predictions:

NMR DB is a free website that offers a prediction of 1H, 13C, and supposedly even 2D NMR experiments.  Structures are entered via a GUI editor.
VeSPA and PyGamma are Python packages (or Python wrappers around C libraries) that supposedly simulate NMR spectra of molecules.  However, both packages are very old and have not been substantially updated in some time.

My question is, what is the best available (free) computational tool for predicting NMR spectra?  As I mentioned, I understand that predictions are imperfect.  What software, if any, are widely used by practicing chemists for NMR simulation?  What is the best reference to read about such software, including the heuristics and algorithms it uses to simulate spectra?  Is Vespa and/or PyGamma state of the art?  The NMR DB site is nice, but I'd very much prefer an interface that is amenable to scripting, and where structures could be supplied computationally rather than via a GUI interface.

(I'm a mass spectrometrist and biochemical engineer with little training in NMR methods.)

In Org Biomol Chem 2016, 14, 3943, Goodman reports the replacement of some "tradition" (expensive) programs with free/open-source ones, as applied to their dp4 method.

Specifically:

Molecular mechanics with tinker (I believe as a replacement for MacroModel), you may or may not want this step depending on what you're doing. The paper is looking at molecules with stereocentres and as such, they appear to have wanted to ensure the NMR values calculated matched the most likely conformation of the molecule experimentally studied.
NMR prediction using NWChem (replacing Jaguar as the gold standard) to do the DFT calculations. There is some mention that this was slower, but again, it would depend on what you're interested in studying.

I think that fulfils your requirement for being able to do the prediction using open-source programs.

As for scripting, there is a mention in the paper of wrapping the whole thing up in a python script (and to my knowledge, there isn't actually a GUI for either piece of the aforementioned programs), so it should be amenable to this.

One thing that it doesn't appear to do, is to graphically plot the resulting data generated (I believe they just compare a list of values to a list of experimental values), but you should be able to plot this in something like MatLab, or, if you play enough, it should be importable into TopSpin via something to turn it into a set of suitably written values.


Why is water more acidic than alkanols? Why is $\ce{H-OH}$ more acidic than $\ce{CH3CH2-OH}$? A common answer to this question is because of the $+I$-effect of $\ce{CH3}$ the negative charge at $\ce{O}$ increases making it less acidic. But if we compare electronegativity difference, the difference between $\ce{O}$ and $\ce{H}$ is more than that between $\ce{O}$ and $\ce{C}$ so there should be even more negative charge at $\ce{O}$ in case of $\ce{H2O}$. In this way $\ce{H2O}$ should be less acidic than $\ce{CH3CH2-OH}$. This is also the reason that $+M$ (electron donating ability) of $\ce{-OH}$ (which is having more negative charge at $\ce{O}$) is greater than the $+M$ of $\ce{-OR}$.

\begin{array} {|l|r|r|}
\hline
  &\mathrm{pK_a~in~water} &\mathrm{pK_a~in~DMSO}\\
\hline
\ce{H2O} &15.74 &31.4\\
\hline
 \ce{CH3CH2OH} &16 &29.8 \\
\hline
\end{array}

As we can see from the $pK_a$ data water is slightly more acidic than ethanol in water but in DMSO ethanol is notably more acidic. In DMSO hydrogen bonding is minimal so the acidity is primarily dependent on the acid and not the solvent it is in.

However, in water the hydroxide ion is hugely stabilised by hydrogen bonding (the ethoxide is as well but not nearly to the same extent) so this accounts for the greater acidity of water in this situation.

http://research.chem.psu.edu/brpgroup/pKa_compilation.pdf
http://www.chem.wisc.edu/areas/reich/pkatable/index.htm


Where does the rate equation come from? (r = k [A]^x [B]^y) Equations are great, but they're much better when you can understand them. I don't quite get why this works. Could someone explain it to me?r = k [A]^x [B]^y

A to the power of x or B to the power of y are telling you how the concentration of that reactant will affect the rate of reaction. For example, if you find that doubling the concentration of reactant A quadrupled the reaction rate than A could be raised to the power of 2. If doubling the concentration of B doubled the rate of reaction than you could raise B to the power of 1. K is the Arrhenius equation and its value is taking into account the frequency that the particles are colliding, the amount that are colliding with the right orientation and the energy required for the reaction to take place. K = Ae^(-Ea/RT) where A is the product of the frequency of the colliding particles and the amount that have the correct orientation. e is 2.718281828 and is raised to the power of negative the activation energy divided by RT where R is the ideal gas constant with a value of 8.314 and T is the temperature. You could use algebra to solve for A if you have the other values.


Which compounds react positively in the aniline dye test? I read that to test for the presence of phenol groups in an organic compound the aniline dye test can be performed. If the solution has phenol present it gives a red or orange precipitate to confirm the presence, however as a side note it was mentioned that this test is applicable for the detection of small quantities of phenols, which have a free para position. Why is it so?

First, the aniline is converted to a diazonium halide salt in the presence of acid and sodium nitrite, per this reference.

Then, the diazonium salt reacts with the phenol to give an azo compound, per this reference (see the section entitled Coupling reactions of diazonium ions

Note that the diazonium ion is shown reacting with the para position of the phenol. That is because the -OH group in phenol is an ortho,para-director (reference), meaning that it enhances the nucleophilicity of the ortho and para positions on the phenol ring. However, the para position is significantly more nucleophilic because of the relative lack of steric hindrance in comparison to the ortho positions, which are hindered by the proximity of the -OH group.

So, having a free para position on the phenol means that the phenol is especially reactive towards electrophiles, including the diazonium salt in this example, meaning that it will react faster and in higher yields - increasing the sensitivity of the aniline dye test.


Why does dimethyl ether have a bond angle bigger than 109.5 degrees? Why does dimethyl ether have a bond angle bigger than 109.5 degrees, in contradiction with VSEPR theory?

In case of $\ce{Me2O}$, the positive deviation of the bond angle can be explained on the basis of steric crowding of two methyl groups.

There are other contradiction too like this:

In case of $\ce{(SiH3)2O}$, the huge bond angle increase is mainly due to back bonding and partly due to steric repulsion.


FFC Cambridge Process and impure reactants The FFC Cambridge process with platinium electrodes allows the reduction of almost any metal oxide to a pure metal, with an oxygen byproduct.
So far, so good.  What I'm trying to understand is how it would handle mixtures of of materials.  For example, if it was a mixture of iron oide and titanium oxide, what would it do?  Which oxide would be reduced preferentially?  (I think in reverse order of reactivity?) If it was contaminated with calcium carbonate, would the chloride ions switch places with the carbonate?
Ultimately this is a question about how electroylsis works with complex solutions - so what would it do?

[FFC process][1]
[1]: https://www.youtube.com/watch?v=97dfa1GoF7w

next step is that the pellets are placed into a crucible or reduction vessel that's filled with very pure molten salt normally this will be calcium chloride.


Oxidation state of hydrogen in the LaNi5H intermetallic This question is from an introductory chemistry course:

  What is the oxidation state of $\ce{H}$ in $\ce{NaH}$, $\ce{HCl}$, and $\ce{LaNi5H}$?

I've looked in the textbook to no avail.

Google doesn't help me much either. It does tell me that $\ce{LaNi5H}$ is an "intermetallic" compound.

The best I've found is this:

  Allen electronegativities also yield zero for the oxidation state of all such $\eta^2$ hydrogen atoms.

What is the justification? Preferably, is there an "intro to chem" level explanation for this?

At an intro chem level, the oxidation state of all elements in an alloy or intermetallic is zero.
At a deeper level of analysis, fractional, integer, and/or decimal oxidation states may be more appropriate, depending on the system in question.

Per Wikipedia:

An intermetallic, also called an intermetallic compound, intermetallic alloy, ordered intermetallic alloy, and a long-range-ordered alloy, is a solid-state compound exhibiting metallic bonding, defined stoichiometry and ordered crystal structure. Many intermetallic compounds are often simply called alloys.

However (emphasis added):

Schulze in 1967, defined intermetallic compounds as solid phases containing two or more metallic elements, with optionally one or more non-metallic elements, whose crystal structure differs from that of the other constituents.

So, in a high-level description, an intermetallic containing hydrogen such as the $\ce{LaNi5H}$ of the question can be considered essentially to be an alloy with hydrogen as a constituent.  Based on this, assigning a $0$ oxidation state to all elements is warranted as a first approximation:

For intermetallic compounds, the ultimate choice of the oxidation state zero at all atoms is best if needed in redox chemistry.  (source | doi)

A more detailed investigation might turn up evidence supporting assignment of nonzero fractional, integer, and/or decimal oxidation states to various atoms:

Subtler estimates and round offs are required for compounds with electrons delocalized over non-equivalent atoms, as expressed by several resonance formulas with weights in arbitrarily long decimal numbers. Without round offs of bond orders in Lewis formulas, decimal values of oxidation states would be obtained for certain bonding connectivities. [...] [U]nambiguous and reasonable fractions of small integers are obtained for oxidation states in compounds such as dithiolate and catecholate (see Section 11) or in (car)boranes such as $\ce{B6H10}$ (see Section 6) and $\ce{B10C2H12}$ (p1041), or when vicinal oxidation states are indistinguishably mixed, such as in $\ce{YBaFe2O5}$ (p1058). Reasonable fractional oxidation states appear also in ions where the charge is distributed over several equivalent atoms such as $\ce{C7H7+}$, $\ce{B6H6^2−}$ (p1040), $\ce{I3−}$, and $\ce{N3−}$ (p1037). (source | doi)


How can the thermal conductivity of a binary gas mixture be estimated? Recently, I am seeking for a simple but relatively accurate method to estimate the thermal conductivity of a binary gas mixture, say ethanol vapor + nitrogen. I assume the mixture as ideal gas.

The molar fraction and thermal conductivities of the components are known. The parameters for a try are: mole fraction of ethanol vapor: $x_A=0.2$, thermal conductivity of ethanol vapor and nitrogen are $k_A=0.016\ \mathrm{W/(m\,K)}$, $k_B=0.026\ \mathrm{W/(m\,K)}$, temperature of the mixture gas $T=330\ \mathrm{K}$, and the bulk pressure is $101\ \mathrm{kPa}$.

I really hope someone could give me a hint or point me to a classical reference.
Thank you in advance!

I'm thinking you could use the Fourier equation for heat flux

$$
\phi =K \frac{S\theta}{e} \,\, ,
$$
where

$\phi$ = Heat flux
$K$ = Termal conductivity coeficient in units of  $kcal  s^{-1}\ m^{-2}\ K^{-1} /m^{-1}$
$S$ = Area of the section you're dealing $m^{2}$
$θ$ = Difference of temperature $K$
$e$ = Thickness in $m$.

And use for the resulting thermal conductivity the fraction

$$
K_{r}=\frac{\sum K_{i}J_{i}(M_{i})^{1/3}}{\sum J_{i}(M_{i})^{1/3}} \,\, ,
$$
where:

$K_{r}$ = Resulting termal conductivity of the moisture;
$K_{i}$ = Termal conductivity of each gas;
$j_{i}$ = Molar fraction of each gas;
$M_{i}$ = molar mass of each gas.

I think it'll work with some 2% of standard deviation. If you're able to read in portuguese, here is the link of my reference:

Reference

And here is the link of the Book cited by @Chester Miller.

I hope I hope I was Helpfull!


Liquid that does not spoil for a mineral aquarium I want to create a decoration - a mason jar "aquarium". I want to take a large mason jar, put some nice mineral specimens in it, fill the jar with liquid and seal it with lid.

I believe I can't use pure water because there is a chance that it will go bad after some time.

So I want to know if there is some liquid that is not too aggressive (in terms of mineral decay) or dangerous that has indefinite shelf life. I want it to last for years.

I know that vodka basically never goes bad, but it's pretty expensive and the alcohol would probably be too harsh for the minerals in the jar.

Additional information: I am considering water with salt.

There are three things to take into consideration here: volatility of the liquid you will be using (yes, the jar will be tightly closed, but some candidates might still evaporate slowly and/or have toxic or foul-smelling vapors), stability, and refraction index. Mineral oil is a great candidate, as Todd Minehardt commented. It's really stable, releases no vapors whatsoever, but its refractive index is a little higher than that of water, which might make your aquarium look funny. There are substances with similar properties that might be useful and also readily available, but with a refractive index closer to water's. Amyl alcohol (fusel oil) or ethylene glycol (car antifreeze) are good candidates and shouldn't be too hard to find. Glycerol and propylene glycol are also pretty good options, but their refractive indexes are closer to mineral oil than the other two.


How does haloform reaction take place if the reactants NaOH and X2 react with each other? If $\ce{NaOH}$ reacts with halogen to produce $\ce{NaX}$ and $\ce{NaXO3}$, then how does haloform reaction take place?

The disproportionation of halogens $(1)$ is an equilibrium reaction; depending on the concentrations of reactants and products it can go both ways. Since hydroxide (or more generally: a base) is among the reactants, increasing pH will drive the reaction towards the product side.

$$\ce{3 Br2 + 6 OH- <=>> 5 Br- + BrO3- + 3 H2O}\tag{1}$$

The first step of the haloform reaction is to enolise a ketone to the corresponding enole $(2)$. This is also an equilibrium lying on the reactants’ side this time.

$$\ce{RC(=O)-CH3 + OH- <<=> RC(-O^-)=CH2 + H2O}\tag{2}$$

So you are correct, both equilibria should be away from what is required for the haloform reaction to proceed. However, both are equilibria. For both, we can measure an equilibrium constant and use it to calculate the concentrations of both the enolate and the free dihalogen. I don’t have all the numbers here, but the result will be that both reactants are present in sufficient concentrations to drive the reaction forward. Remember that the full haloform reaction is no longer an equilibrium process; each bromination and the final separation is a non-reversible step; see $(3)$.

$$\begin{gathered}\ce{RC(-O^-)=CX2 + Br2 -> RC(=O)-CX2Br + Br-}\\[0.8em]
\ce{RC(=O)-CBr3 + OH- <<=> RC(-O^-)(-OH)-CBr3} \\[0.4em] \ce{-> RC(=O)-OH + CBr3- -> RCOO- + HCBr3}\end{gathered}\tag{3}$$

The mere presence of non-reversible steps will drive any reaction forwards.


Axial vs equatorial attack of Grignard reagent on dioxan-5-one and dithian-5-one How to account for the diastereoselectivety (axial vs equatorial attack) of these reactions?

I've found the solution in Fleming: Molecular Orbitals-Reference edition:

  The dioxan-5-one is attacked by nucleophiles in the gas
  phase from the equatorial direction, because of electrostatic repulsion from the fraction of negative charge
  carried by the two oxygen atoms in the ring, as Houk had predicted. The selectivity is not particularly high
  (67:33), because the usual torsional strain is still present, enhanced in this case by the shorter C—O bond
  lengths making the ring less puckered, and increasing the dihedral angle by which the carbonyl oxygen atom
  is pushed below the neighbouring C—H bonds. In solution, however, the ketone is attacked
  from the axial direction, just like cyclohexanones. Evidently the solvent insulates the charge on theincoming nucleophile from the electrostatic repulsion, the torsional strain remains high and there are no axial
  substituents hindering axial attack.The sulfur atoms in the dithia analogues are less electronegative than the oxygen atoms in the acetals but the axial lone pairs bulge further up, and are not heavily solvated. However, the carbonyl group is
  displaced even further down by the changes in bond length and bond angles from having two C—S bonds in
  the ring. The one effect makes axial attack less favourable, and the other makes equatorial attack less
  favourable. In the event both solution and gas phase reactions take place with a strong preference for
  equatorial attack.


What is a "hydrogen-like" or "hydrogenic" atom? I'm studying some chemistry on my own in anticipation for the new school year and in my book, I came across the Rydberg equation for the first time. I worked through some examples and everything was fine until I came across this comment on the question, "Calculate the wavelength of the radiation released when an electron moves from n=5 to n=2":

  For future reference: the Rydberg formula only works for hydrogen-like
  atoms.

What is meant by "hydrogen-like"? I've heard that solving for multi-electron systems is (near) impossible, so I understand why hydrogen is used here, but I don't understand what "hydrogen-like" is.

A hydrogen-like atom (or ion) is simply any particle with a nucleus and one electron.

That should be sufficient to answer the question at hand, but I thought I should say a bit more, as some of these answers are potentially confusing.

The historical reason why the Rydberg formula only works for hydrogen-like atoms is because it was originally formulated to explain the spectral lines of hydrogen. It was never intended to explain the spectra of multi-electron atoms.

The physical reason, though, is because the Rydberg formula uses energy levels that only depend on the principal quantum number $n$, which has to be a positive integer:

$$\bar{\nu} = Z^2\mathcal{R}\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right) \qquad n_1,n_2 \in \mathbb{Z}^+$$

and nowadays we know that this is only true for hydrogen-like atoms;$^*$ the energy levels of multi-electron atoms depend on both $n$ and $l$.$^\dagger$

The $n$-dependency was later successfully rationalised by the Bohr model, but saying that "the Rydberg formula only works for hydrogen-like atoms because the Bohr model only works for them" is misleading and misses the point, as:

This implies that the Rydberg formula was derived from the Bohr model, which is not true; it was merely empirically determined, and the formula predated the Bohr model by 25 years.
The Bohr model simply does not work for hydrogen-like atoms. The fact that it reproduces the Rydberg formula should merely be considered a serendipity; Bohr arrived at the correct result by the wrong method.
It does not lend any real insight into the proper reason why the Rydberg formula does not apply for helium, etc. (which I briefly mentioned above).

$^*$ In fact, the energy levels of hydrogen are not only dependent on $n$ (due to various small effects such as – but not limited to – spin-orbit coupling, and hyperfine splitting). Wikipedia has a good overview of the topic here and most QM textbooks have a chapter on the hydrogen atom, where they discuss these perturbations to the Hamiltonian and their effects on the energies. Not surprisingly, the inability to explain this was one of the failures of the Bohr model.

$^\dagger$ Of course, there is also a series of approximations here. The energy levels of multi-electron atoms are only approximately described by sums of orbital energies, so the transition energies are only approximately equal to a difference in energy between two orbitals.


What are glues/adhesives for cardboard boxes/crates made of? Open cardboard boxes/crates, often for foodstuffs, are often glued to hold their shape. What is the glue/adhesive made of?

I ask to know if it is biodegradable, such that it is suitable for composting/gardening.

My preliminary research seems to indicate polyvinyl acetate (PVA), which seems to biodegrade fast enough for my application.

See example box (from CTHOE at Wikimedia):

Sorry if this is the wrong Exchange, it seemed the most suitable for my question.

Because the glue is not in direct contact with food you can use basically any glue. And cardboard is easy to glue so there is no limitation. Packaging should be cheap and therefore I would not expect a biodegradable glue.


Calculation of the cell potential of a lemon battery experiment I have the following problem that I would like to get checked.

I’m told to calculate the cell potential of a simple lemon battery experiment.

After Finding out the oxidizing agents and the reducing agents I find that the cell reaction is as follows:

$$\ce{2H+(aq) + Zn(s) —> H2(g) + Zn^2+(aq)}$$

with a cell potential of $E_\mathrm{cell} = +\pu{0.76 V}$.

Are the reactions and the cell potential correct?

I am afraid that nobody can predict the true potential of Zn/ Cu lemon battery. It is not a standard system. The electrode potentials you see in the published tables are measured or calculated under unknown or controlled conditions and one of the basic requirement is that the metal must be in equilibrium with its own ions.

The lemon battery, although looks very simple, it is too complicated to deal with.

When you insert the electrodes into the lemon, there are no metallic ions to begin with, i.e., the concentration of Zn$^{2+}$ and Cu$^{2+}$ is essentially zero. Just using plug & chug approach for the Nernst equation will give you infinite potential difference. This is not what is experimentally observed.

1b. Even if we assume that some metal dissolves around in a few seconds, still the metal concentration is not high enough to allow a reliable use of Nernst equation.

Citric acid is not only an acid, but it has complexing/chelating property as well.
Any voltage which comes close the calculated potential from redox tables is a mere coincidence.

I think the proper way to determine the potentials generated by contact potentials. This is the realm of physics. Two dissimilar metals, when brought in contact, will generate a potential difference. See my previous answer Electrochemical reactions in a salt water battery


Is a ring intermediate formed in hydroboration? This was the mechanism taught to us for hydro-boration

Something really strange in that mechanism is that in step 2  hydrogen has two bonds.
Also , the  the doubly bonded $\ce{H}$ , has one bond $\ce{C-H}$ which contains no electrons

Is this mechanism plausible?

In clayden they have shown the cleavge of $\ce{H-B}$ bond  and formation of the $\ce{H-C}$ bond in the same step.
But it doesn't mention the formation of a ring anywhere.

Section 1
Let's assume that the ring does form but after cleaning the $\ce{H-B}$ the electron pair goes into the bonding orbital of H not the anti bonding molecular orbital between $\ce{C-H}$ .

Primarily my doubts  are

The bond with no electron in it (does it actually form)
The reasoning in section 1 (the previous section).

That should be a transition state, not an intermediate.

Also, there should not be a positive charge on your boron in the final compound you drew.


Understanding the use of Le Chatelier's principle in the industrial production of ammonia I understand why the reaction

$$\ce{N2 + 3 H2 -> 2 NH3} \qquad \Delta H = \pu{-92.2 kJ mol-1}$$

isn't carried out in the lowest possible temperature (even though it seems that it should be as it favors the right side of the reaction). As the rate of reaction would be very slow, the amount of $\ce{NH3}$ produced would be lesser than the amount that would've been produced if the reaction was carried out at the optimum temperature in a given time span.

Why isn't then the reaction carried out at the highest possible temperature since the rate of reaction would be very high and the yield would be very high consequently as well?

Why isn't then the reaction carried out at the highest possible temperature since the rate of reaction would be very high and the yield would be very high consequently as well?

The rate constant would be high, but that does not mean that the net forward rate is high. If you start without product, the initial rate would be high, but would drop to zero very quickly as equilibrium is reached. The equilibrium concentration of product would be very low at high temperature, as you said, so the yield would be very low.

If you had a very cheap and highly efficient process to remove product directly from the reaction vessel, you could run the reaction at high temperature. However, the product removal is slow and expensive (cool down the reaction mixture, remove ammonia, and reheat the remaining reactant to feed back into the reactor). Because this is costly, the reaction is run at a compromise temperature where at equilibrium, about 15% of the reaction mixture is product.


Increase in equilibrium concentration of the reactant via decrease in container volume Consider the hypothetical reaction $\ce{3A(g) -> B(g) + C(g)}$; how to increase the equilibrium concentration of $\ce{A}$, given that the reaction is endothermic?

  increasing the volume of the container
  decreasing the temperature
  decreasing the volume of the container
  adding $\ce{B(g)}$ at equilibrium

Answer given - 2, 3, and 4. I have trouble with the option 3.

I know according to the Le Chatelier's principle, equilibrium will shift forward on reducing the container volume, but the solution says it still manages to have an increase in the overall equilibrium molar concentration of $\ce{A(g)}$; and that's what I'm not understanding.

You're right that at first glance, this is indeed self-contradictory. However, on a closer look, here's how you can realize that it isn't!

The catch here is that when the reaction shifts forward - to counter-balance the decrease in volume - the reduction in the moles of $\ce{A}$ is much lesser than the decrease in volume, hence, the overall ratio, $\frac{\text{moles}}{\text{volume}}$ actually increases!

You can demonstrate this via your own reaction. Assume $K_\mathrm{c}=\pu{1M-1}$, and all equilibrium concentrations to be $\pu{1M}$. When volume is halved, the concentrations double, so $Q=\pu{0.5M-1}$. The reaction thus shifts forward as $Q<K$.

Let the new concentrations be $\pu{2M}-3x, \pu{2M}+x, \pu{2M}+x$ for $\ce{A,B,C}$ respectively. Solving for $x$, you'll get $x\approx0.117$. Hence, the new equilibrium concentrations are $\pu{1.64M}, \pu{2.117M}, \pu{2.117M}$. In fact, not only the reactant, but all species show an increase in their concentration!

Moreover, the same result is obtained in the reverse direction. If the container volume was increased instead, all species would show a decrease in their volume. (the overall ratio $\frac{\text{moles}}{\text{volume}}$ decreases, as the moles increase far less proportionately as compared to the volume)

I must state that this isn't a rigorous proof but a layman's explanation for why this is happening, and how to understand it in simple terms. I would leave it up to a more experienced physical chemist, to demonstrate the applicability of the rules I've stated in bold above, via proofs or other methods.


Polymer Slurry used in boring foundation holes? I watched a video on the construction of Burj Khalifa, Dubai. The construction engineers answered that they're using a special type of protection known as Cathodic protection to protect from the corrosion due to local ground water. They told a compound named "Polymer Slurry" (What's that?). They're using viscous Polymer Slurry as their basic coat to dump the high density & low permeability concrete in order to create a "GOOD FOUNDATION". What is it made of? (Simply any polymer?) And what does it do? Does it actually repel water? Many organic compounds repel water... Then Why this?

After reading the link provided in the question, and doing a little googling, of "polymer slurry" and "foundation", I found something useful.

Polymer slurry seems to be made of one or more hydrophilic superabsorbent polymers, such as polyethylene glycol, polyvinyl alcohol, or carboxymethylcellulose that produce a thick viscous slurry when mixed with water. It does not repel water so much as the opposite. It is so hydrophilic that it absorbs water before the water can reach the concrete or the steel. Then this slurry becomes thicker the more water it absorbs so that the rate of diffusion of water through the slurry decreases. Ultimately, this action creates a hydrogel, which may be quite strong, despite being mostly water, due to extensive hydrogen bond crosslinks.

Other applications of polymer slurry seem to be in the drilling industry to modulate the viscosity of drilling fluids.


No. of photoelectrons vs frequency and intensity We know that when the intensity of incident light increases, number of photoelectrons ejected increases irrespective of the frequency (assuming it to be greater than threshold frequency. But does it depend on frequency too in some situations like in case of high frequency or something?

The number of photoelectrons per incident photon doesn't really vary with photon energy (frequency). Particularly a photon won't eject multiple "photoelectrons." By definition a photoelectron is a loosely bound electron (i.e. not in an atomic or molecular orbital) which is ejected in a one-to-one basis by a photon and thus satisfying the equation $$ \rm{KE}_{electron} = \rm{h}\nu - \phi$$ where:

$ \rm{KE}_{electron}$ is the kinetic energy of the electron
$\rm{h}\nu$ is the energy of the incident photon
$\phi$ is the work function for the substrate (how much energy it takes to remove electron from substrate).

In general the substrate would be a metal and the electron would be in the metallic (electrically conducting) layer.

If the electron is bound in an atomic or molecular orbital then the
technique is Electron Spectroscopy for Chemical Analysis (ESCA).
If the photon is too energetic then it penetrates too deeply to eject an photoelectron.

The gist of the above conditions is that photoelectrons are created with relatively low energy photon just above the needed threshold given by the work function, $\phi$.

Another aspect of this is that a photon can only interact with one electron at a time. However if the photon is greater than $2\phi$ in energy it is possible that the electron can receive enough energy to knock yet another electron out of the metal. These two electrons wouldn't be "photoelectrons" since they don't fit the equation. They would be noise in the experiment and the experiment would need to be designed to minimize counting such electrons. So the ejection of multiple electrons by a photon is possible, but not multiple photoelectrons by definition.


What is the Series Limit in a Spectral line What is exactly the series limit in a spectral line graph? Do we talk about series limits for each series (Eg: Lyman series in Hydrogen) or collectively for an atom?

Is series limit the transition having least energy or least wavelength?

So like for Lyman series, will series limit be that corresponding to $\ce{1 -> 2}$ or $\ce{1 -> Ionize}$?

Every series of wavelengths produced from the hydrogen atom excitations have a certain minimum limit. These minimum wavelengths are called the series-limit for that particular series. Take a look at this picture for example:

The minimum series of the Lyman series would be for the largest transition, which is from ∞$\ce{ -> }$1. Plugging the values into the Rydberg's equation:

$$\frac{1}{\lambda}=R_{∞}\left(\frac{1}{1}-\frac{1}{∞}\right)$$

That should give us a value of 91nm. Which is the series-limit for the Lyman series.

Now you can find series-limits for the other series as well.


How can a methyl halide react by intramolecular SN2? Out of the following, $\ce{CH3Br}$ doesn't react by which mechanism?

  SN2
  SN1
  E2
  Intramolecular SN2
  E1

My textbook gives 2, 3 and 5 as the correct answers, which means it does react by 1 and 4.

But how can a methyl halide react by intramolecular SN2, without a nucleophile present in the molecule? There's just a leaving group ($\ce{-Br}$).

I agree. There is no way that a one-carbon and one-heteroatom compound could react by intramolecular substitution. All intramolecular nucleophilic substitutions will require some ring-like transition state. The smallest ring you can build is a three-membered ring — but bromomethane does not have three atoms that could build up that ring (hydrogen does not participate).


Symmetry and NMR linewidths of quadrupolar nuclei The linewidth of peaks when taking NMR spectra of quadrupolar nuclei (11B or 14N for example) is often very broad, making their interpretation challenging.

Recently, when acquiring 11B-NMR spectra of some boronic esters I'd made, the NMR technician pointed out that the linewidth is much improved in highly symmetrical environments (which my boronic esters were not) – she proved this by running a sample of $\ce{LiBH4}$, but couldn't really offer an explanation of why this was the case, other than to point out that its a relaxation effect.

Claridge[1] confirms this point about symmetry:

  14N linewidth of $\ce{N(Me)4+}$ is less than 1 Hz whereas that for $\ce{NMe3}$ is nearer to 80 Hz.

Arguably, both $\ce{N(Me)4+}$ and $\ce{NMe3}$ are fairly symmetrical, being Td and pseudo-C3v respectively, which leads me to think that what the molecule needs is some kind of spherical symmetry, which only Td (and Oh, arguably less useful in synthetic organic chemistry) can provide.

Some further reading through Claridge[1] reveals that this is something known as quadrupolar relaxation, with Reich[2] confirming, though not really expanding beyond an equation describing the relaxation process and a comment that there is "friction" between the nucleus and the surrounding electron.

All of the explanations I've found have focused a lot on explaining this mathematically (Levitt describes it quite nicely, but slightly beyond my understanding). Is there an intuitive explanation for the relaxation taking place in quadrupolar nuclei, and why high symmetry environments lead to narrow peak shape?

[1]: Claridge, T. D. W. High Resolution NMR Techniques In Organic Chemistry; Elsevier:Amsterdam, 2016

[2]: https://www.chem.wisc.edu/areas/reich/nmr/08-tech-01-relax.htm, accessed 20/08/2017.

I know that you asked for an intuitive explanation, but I'm afraid that if one wants to go beyond ron's comment, the technical aspects are somewhat necessary. It isn't incredibly important to understand the maths, but I have included it as I want to enable interested readers to go further, and it provides a basis for the last section, so I make no apology for it. I have also assumed some basic knowledge of NMR theory.1

Linewidths and relaxation

Following a pulse sequence, the spectrometer detects transverse magnetisation (i.e. magnetisation in the xy-plane). The signal that is detected is generally of the form $s(t) \propto \exp(\mathrm i \Omega t)\exp(t/T_2)$, where $T_2$ is the transverse or spin-spin relaxation time.

If one takes the Fourier transform of $s(t)$, one gets the spectrum $S(\Omega)$, and it can be shown that the widths of the Lorentzian peaks thus obtained are proportional to $1/T_2$. To be precise, the linewidth at half-height $\Delta \nu_{1/2}$ is given by

$$\Delta \nu_{1/2} = \frac{1}{\pi T_2}$$

Therefore, fast transverse relaxation (which corresponds to small $T_2$) leads to broad lines, and slow relaxation leads to narrow lines.

Quadrupoles and the electric field gradient

For quadrupolar nuclei ($I > 1/2$), a major source of relaxation arises due to interaction of the nuclear electric quadrupole moment with the electric field gradient at the nucleus.

An introduction to the above two concepts is as follows: a quadrupole moment reflects an asymmetry in the charge distribution of the nucleus, and can be thought of as two opposing dipoles placed next to each other. The electric field gradient refers to the derivative of the electric field $E$, which is itself the derivative of the electric potential $V$ (so, the field gradient is the second derivative of the potential). These derivatives are evaluated at the spatial position of the nucleus.

The electric field gradient therefore most generally has nine components, since two derivatives, each with respect to 3 axes, are taken. These are represented by $q_{ij}$ (where $i,j$ are one of $x,y,z$):

$$q_{ij} \equiv \left.\frac{\partial^2V}{\partial i\,\partial j}\right|_\text{at nucleus}$$

However, by a suitable transformation of coordinates, the cross derivatives $q_{ij}$ ($i \neq j$) can be set to zero (this essentially involves diagonalising the matrix $\mathbf{q}$). The remaining quantities $q_{xx}$, $q_{yy}$, $q_{zz}$ are referred to as the principal values, and without loss of generality we can adopt the convention $|q_{zz}| \geq |q_{yy}| \geq |q_{xx}|$. Since the electric potential obeys Laplace's equation $\nabla^2 V = 0$, this also means that $q_{xx} + q_{yy} + q_{zz}$ is necessarily equal to 0 (this is often referred to as "traceless", since the trace of the matrix $\mathbf{q}$ is 0). Therefore, there are only two independent parameters. These parameters are chosen to be $q_{zz}$, and $\eta$, a quantity called the biaxiality and defined as

$$\eta = \frac{q_{yy} - q_{xx}}{q_{zz}}.$$

Quadrupolar relaxation2

The interaction of the quadrupole moment with the electric field gradient contributes to the energy of the nucleus (exactly analogous to how an electric dipole moment interacts with an electric field).3 This interaction is modulated by molecular tumbling. That is to say, as the molecule rotates in solution, the magnitude of this interaction changes as well.4

If this interaction varies at the frequency corresponding to the transition between two spin states, then it is possible to induce this transition, thereby leading to spin relaxation. (More strictly speaking, it has to contain a component that oscillates at the transition frequency.) In this aspect, it is analogous to the more well-known case of how magnetic dipolar interactions between two nuclei can lead to relaxation, which is described quite thoroughly in Keeler's book.1 For spin-1/2 nuclei in solution-state NMR, dipolar interactions are the primary method of relaxation since these nuclei do not have quadrupole moments. For quadrupolar nuclei, though, quadrupolar relaxation is the primary method of relaxation, simply because this interaction term tends to have a large magnitude.

For our purposes, it is sufficient to use a derived result, which you seem to have already come across. I know that I am glossing over a lot, but as far as I can tell, the derivation of this is very involved:5

$$\frac{1}{T_1} = \frac{1}{T_2} = \frac{3\pi^2}{10}\frac{(2I+3)}{I^2(2I-1)} \left(\frac{e^2q_{zz}Q}{h}\right)^2 \left(1 + \frac{1}{3}\eta^2 \right) \tau_\mathrm{c}$$

Definitions:

$T_1$ and $T_2$ are the spin-lattice (longitudinal) and spin-spin (transverse) relaxation times respectively; these only account for relaxation via the above mechanism and not for other possible mechanisms
$I$ is the spin of the nucleus
$e$ is the elementary charge, $\pu{1.602 \times 10^-19 C}$
$q_{zz}$ is the main component of the electric field gradient, as explained above
$Q$ is the quadrupole moment of the nucleus
$h$ is Planck's constant, $\pu{6.626 \times 10^-34 J s}$
$\eta$ is the biaxiality, defined above
$\tau_\mathrm{c}$ is the correlation time, which is a constant that essentially measures the rate of molecular tumbling (see books in ref 1 for more information).

Please, no more maths

OK - the only remaining maths is to analyse the equation above and identify the factors which contribute to relaxation, and hence, linewidths. For the actual question, the relevant portion is $e^2q_{zz}Q/h$, which is often written as $\chi$ and termed the nuclear quadrupole coupling constant. Some representative values of $\chi$ are given in ref 2.

In particular, the value of $q_{zz}$ is mainly determined by the distribution of electrons close to the nucleus, and this provides an obvious link to the molecular geometry of the quadrupolar nucleus. For nuclei in tetrahedral, octahedral, cubic, or spherical environments, $q_{zz}$ is, by symmetry, equal to zero. Therefore, quadrupolar relaxation by the above mechanism does not take place, $T_2$ is long, and linewidths are small. One sees very narrow linewidths for the central atoms in, for example, $\ce{Cl- (aq)}$ (neglecting solvation effects, spherical symmetry); $\ce{NMe4+}$ and $\ce{SO4^2-}$ (tetrahedral), and $\ce{[Co(CN)6]^3-}$ (octahedral). On the other hand, if such symmetry is lacking, then $q_{zz} \neq 0$, there is usually fast quadrupolar relaxation by the above mechanism, and linewidths are large.

$q_{zz}$ is directly tied to the distribution of electrons and not the distribution of bonds. Although there is nearly always a correlation, there are some interesting cases where this is not true; in $\ce{Et2N-NO2}$, for example, the $\ce{NO2}$ nitrogen has an extremely sharp peak, despite the local molecular symmetry not intuitively suggesting that it would be the case.2 (The $\ce{Et2N}$ nitrogen, as expected, has a very broad peak.)

Since we went the full distance to obtain the above expression, we can also extract a bit more information from it. The rate of relaxation is also proportional to the nuclear quadrupole moment $Q$; some quadrupole moments are tabulated in ref 2. Some nuclei which have small quadrupole moments (namely deuterium, lithium-6, and cesium-133) have smaller linewidths and are much more amenable to NMR studies. Also, the factor of $I^2$ in the denominator also means that higher spins tend to give sharper lines.

A bonus: coupling to quadrupolar nuclei

$\ce{^13C}$ NMR experiments are often run with proton decoupling - this means that when the carbon signal is collected, the protons are irradiated with a radiofrequency field. This has the effect of inducing rapid transitions between the two spin states of the protons, and each carbon-13 nucleus effectively "sees" an average of the two spin states, causing the C–H couplings to be lost.

Since quadrupolar relaxation also induces rapid transitions between spin states of quadrupolar nuclei, it leads to what is effectively a "self-decoupling" of all quadrupolar nuclei. However, the couplings can still be seen when the quadrupolar nuclei are in highly symmetrical environments: for example, the $\ce{^19F}$ spectrum of $\ce{[BrF6]+}$ is shown here.6 Both $\ce{^79Br}$ and $\ce{^81Br}$ have roughly equal abundance (~50%) and a nuclear spin of $3/2$, so we expect to see two 1:1:1:1 quartets of equal intensity, one arising from $\ce{[^79BrF6]+}$ and one from $\ce{[^81BrF6]+}$.

A few other examples are presented on Hans Reich's page. Of course, these can be understood from the discussion above about how molecular symmetry affects quadrupolar relaxation.

In fact, because deuterium has a small quadrupole moment and undergoes slow quadrupolar relaxation, coupling to deuterium can often be seen. All organic chemists should recall that $\ce{CDCl3}$ appears as a 1:1:1 triplet - this is why!

Notes and references

For an exposition of basic NMR theory see either Keeler, Understanding NMR Spectroscopy, 2nd ed. (Wiley) or Hore, Nuclear Magnetic Resonance, 2nd ed. (OUP).
A necessarily mathematical, but still fairly accessible, introduction to the topic is provided in: Gerothanassis, I. P.; Tsanaktsidis, C. G. Nuclear electric quadrupole relaxation. Concepts Magn. Reson. 1996, 8 (1), 63–74. DOI: 10.1002/(SICI)1099-0534(1996)8:1<63::AID-CMR5>3.0.CO;2-N.
The full Hamiltonian is provided in Appendix A.7 of Levitt, Spin Dynamics, 2nd ed. (Wiley).
Abragam (ref 5) writes that this is best understood in terms of "a fluctuating electric field gradient acting on the quadrupole moment of the nucleus", and refers to: Bloembergen, N.; Purcell, E. M.; Pound, R. V. Relaxation Effects in Nuclear Magnetic Resonance Absorption. Phys. Rev. 1948, 73 (7), 679–712. DOI: 10.1103/PhysRev.73.679. (Non-paywall version from Harvard available here.)

However, as far as I can tell, Bloembergen et al. simply describe the case for dipolar relaxation of spin-1/2 nuclei, and then write that "the interaction of the electric quadrupole moment of the deuteron with a fluctuating inhomogeneous electric field can bring about thermal relaxation. We omit the analysis of this process, which parallels closely the treatment of dipole-dipole interaction".
This equation is given without explanation in Gunther, NMR Spectroscopy, 3rd ed. (Wiley). An explanation can be found in Abragam, The Principles of Nuclear Magnetism (OUP), including the conditions under which this equation holds true. Quadrupolar relaxation is dealt with on p 313, but going through the earlier sections is necessary to understand it. (I didn't do that.)
Gillespie, R. J.; Schrobilgen, G. J. Hexafluorobromine(VII) cation, $\ce{BrF6+}$. Preparation of hexafluorobromine(1+) hexafluoroarsenate(1-) and hexafluorobromine(1+)-fluorodecafluorodinabinonate(1-) and characterization by fluorine-19 nuclear magnetic resonance and Raman spectroscopy. Inorg. Chem. 1974, 13 (5), 1230–1235. DOI: 10.1021/ic50135a043.


What is the voltage of an electrochemical cell at equilibrium? I am having a hard time understanding what happens at equilibrium in an electrochemical cell (a full cell with two electrodes). At equilibrium, according to Nernst equation, the cell potential is $E_{cell}=E^0_{cell}-\frac{RT}{zF}lnQ_r$. Is this cell potential equal to the voltage of the cell ? At equilibrium, the voltage is zero, so does it mean that $E_{cell}=0$ and that both half cells have the same potential ? I am confused about this, as I also understood that $E_{cell}$ at equilibrium is the electromotive force of the cell, which means that it isn't zero.
Thank you for your help !

There is a cell equilibrium and 2 independent half-cell equilibrii.
There are 2 half-cell equilibrium potentials and a cell open voltage as EMF as the difference of 2 half-cell equilibrium potentials, because the cell in the whole is not in equlibrium.
When a cell gets into equlibrium as the whole system, the cell voltage is zero.


Molality of solution having more than one solute? So here is my question,

  Question: A water sample contains 9.5% $\mathrm{MgCl_2}$ and 11.7% $\mathrm{NaCl}$ (by weight). Assuming 80% ionization of each salt. Boiling point of water will be ________. ($\mathrm{K_b}$ for water $\mathrm{=0.52}$)

You know, this question isn't that tough to bother anyone (definitely not!). What's all confusing here is the molality of two solutes.

I know that,

$$\Delta \mathrm{T_b=K_b\cdot m}$$

Degree of dissosiation $(\mathrm{\alpha}) = 0.8$

So I know very well that molality of a solution having only one solute is,

$$\mathrm{m=\frac{n\ (no.\ of\ moles\ of\ solute)}{y\ (mass\ of\ solvent\ in\ kg)}}$$

But what about the solution having more than one solute? Do they have any molality (I am sure that there will surely be, because it is there in the question). If yes then what's the method of calculating?

Also if you are solving this, here is the answer to the question (as per my answer key),

  377 K

The boiling point elevation fomula assumes that all particles are equal, only the number of particles in the solution matters.

So I would use:

m = (0.2 + (3)(0.8))(molality of $\ce{MgCl_2}$) + (0.2 + (2)(0.8))(molality of $\ce{NaCl}$)

To find molality of $\ce{MgCl_2}$ and molality of $\ce{NaCl}$, consider a 1kg sample of the solution.  There are 95g  $\ce{MgCl_2}$, 117g $\ce{NaCl}$ and 788g water.

Calculate the number of moles $\ce{MgCl_2}$ in 95g $\ce{MgCl_2}$, and divide this number by 0.788kg.  This is the molality of $\ce{MgCl_2}$ in the solution.

Repeat for $\ce{NaCl}$


Hydrogen bonding of chlorine Even though nitrogen is less electronegative than chlorine according to Pauling scale bit why don't it form hydrogen bonding as effective as of nitrogen ? ( Electronegativity valve of  N =3.0,Cl=3.2)

There is more to hydrogen bonding than electronegativity.  To get the strongest hydrogen bonds you need a compact atom to concentrate the negative charge for electrostatic attraction and to overlap well with the hydrogen 1s orbital for the molecular orbital/delocalized covalent bond contribution (yes, the latter is a thing.)  Nitrogen has that, chlorine not so much.  So, you get stronger hydrogen bonds with nitrogen than with chlorine.

Did you catch the wording?  Though chlorine has disadvantages it can form (weaker) hydrogen bonds under some conditions.  See here; the abstract indicates that the interaction can exist in some chlorinated organic compounds.


The number of geometrical isomers of complex of type [Ma₃b₂c] Find out the number of geometrical isomers of the complex $\ce{[Ma_3b_2c]}.$

I applied simple combinations to get the number of isomers as
$$\frac{6!}{3!\,2!} = 60.$$
However, my book states that only three geometrical isomers are possible. I could not see why simple combination is not giving the correct answer.

First, let's establish that we're all talking about octahedral complexes to eliminate any possible confusion.
Fix $\ce{c}$. You can have either $\ce{b}$ trans or $\ce{a}$ trans.
If $\ce{b}$ is trans to $\ce{c}$, there's only one distinct way to arrange the remaining $\ce{b}$ and three $\ce{a}$'s. So that's one isomer.
If $\ce{a}$ is trans to $\ce{c}$, then the remaining two $\ce{a}$'s can be cis or trans to each other, i.e., all three $\ce{a}$'s can be facial or meridional, respectively. That's two more isomers. That's it.
I think your confusion comes about from symmetry because some of the combinations you calculated can be rotated into others.


Why is the maximal percent concentration of acids similar to their molar mass Why is the usual percent concentration of a concentrated acid's solution similar to its molar mass?

For example:

                FW (amu)     mass %
sulfuric        98.08        98%
nitric          63.00        68%
hydrochloric    36.46        36%

Is there some reason for this similarity or is it coincidence?

Because of its solubility.
More concentrated solutions of, for example, nitric acid or hydrochloric acid, will result in fuming solutions where acid concentration evaporate out until the solution reaches an equilibrium.
Do not get confused. You can actually dissolve more than 370g of hydrochloirc acid in 630 mL of water.
In effect:

Solubility in water (Wikipedia)

823 g/L (0 °C)

720 g/L (20 °C)

561 g/L (60 °C)

But if we talk about a commercial solution, which should stay at the same concentration in time, you need to find a compromise between most quantity of acid possible in solution and less lose of acid due to vaporization.

Remember that pure hydrogen chloride is a gas at room temperature. If you continue increasing its concentration, you end up with man equilibrium shifted towards gas phase.

Also, as you can see, if you increase the temperature, solubility decreases, so if you heat a 50% aqueous solution of hydrochloric acid, you would end up not knowing what quantity of the reactant are you actually using.

Nitric acid is less volatile, and sulphuric acid is much less volatile, so you can have the latter almost in a pure form (98% is not an aqueous solution anymore! is just sulphuric acid with some water and other trace compounds)


How is hyperconjugation into antibonding orbitals stabilizing? So I learned that hyperconjugation is electron delocalization from a $\ce{C-H}$ $\unicode[Times]{x3C3}$ bond into an empty p orbital (on a carbocation for example) or an antibonding orbital. I get how it's stabilizing in the case of an empty p orbital, but how is it stabilizing in terms of delocalization into an antibonding orbital, since by definition electron density in antibonding orbitals is destabilizing?

This is nothing specific to antibonding or nonbonding orbitals. There is a very general principle behind it:

  Mix an occupied and an empty orbital with each other, and the occupied orbital will be stabilised.

You’re not actually ‘transferring electron density into an antibonding orbital’. You are simply linearcombining two orbitals: one occupied one with a rather low energy and one unoccupied one with a rather high energy. With this linear combination, you receive two new orbitals, where the unoccupied one is further destabilised while the occupied one is further stabilised.

Your electron density is still in the (now more stable) occupied orbital and the antibonding/nonbonding/otherwise empty orbital (now less stable) is still empty. Only their shapes have changed a bit, because we mixed them.


Doesn't the fact that the lone pair on the nitrogen in amides occupy a p orbital contradict Hund's rules? Since the $\ce{sp^2}$ hybridized orbitals are lower in energy than the p orbital shouldn't the $\ce{sp^2}$ orbitals fill first? Why is this not the case - two electrons fill the p orbitals while all three $\ce{sp^2}$ orbitals are singly occupied? If you look at Carbene (CH2) it's $\ce{sp^2}$ hybridized with a lone pair in one of these orbitals. However, the higher energy p orbital is empty - as would be predicted is Hund's rules are followed. Is it something to do with the conjugation in amides?

two electrons fill the p orbitals while all three $\ce{sp^2}$ orbitals are
  singly occupied

Not really, hybridization occurs in response to bonding, so we can only talk about an atom's hybridization when it is in a molecule.  In amides the nitrogen $\ce{sp^2}$ orbitals form molecular orbitals (MO's) with whatever else they are bound to.  In the case of a simple amide with the nitrogen bound to hydrogen, the MO would involve a nitrogen $\ce{sp^2}$ orbital and a hydrogen 1s orbital and 2 electrons would occupy this orbital (one from the nitrogen $\ce{sp^2}$ orbital and one from the hydrogen 1s orbital)  - so what was the $\ce{sp^2}$ orbital will be filled with 2 shared electrons when a bond is formed.  Therefore the 3 $\ce{sp^2}$ orbitals, now MO's, and the p orbital on nitrogen will each be filled with 2 electrons in the molecule.

  Carbene (CH2) it's $\ce{sp^2}$ hybridized with a lone pair in one of these
  orbitals

Carbenes can exist in either a singlet or triplet state, as shown below.  There are 2 possible triplets depending upon how the carbene is hybridized, a bent $\ce{sp^2}$ hybridized carbene, or a linear $\ce{sp}$ hybridized carbene. In the case of methylene ($\ce{:CH2}$), and actually for most carbenes, the triplet states are lower in energy due to Hund's rule (don't pair electrons unless you must); and the triplet has one electron in the p orbital and one in the $\ce{sp^2}$ orbital.


What question could be answered by "because the metabolites in the cell are in ionized form" in the topic of anaerobic metabolism? I know it might sound a bit absurd, but my teacher gave us some "answers to question", that is, given one answer, write at least one question the answer to which is the given one.

I've solved most of the answers, but there's one I just can't figure out: "because the metabolites in the cell are in ionized form", and the topic is anaerobic respiration and fermentation. No additional details from her.

I'm in high school (a "scientific" school in Italy), if that can help for the knowledge level.

Any idea?

Interesting reverse question. I could figure out three questions:

Why milk changes from a liquid to a gel when it turns into yoghurt?
Why venous blood has lower pH than arterial blood?
Why do we get tired on intense physical exertion?

The answer to the first one would be because the metabolite of lactose fermentation in the lactobacilli cells is lactic acid, that in ionized form releases $\ce{H+}$, lowering the medium pH until the isoelectric point of the milk proteines is reached, causing their agglutination. The answer to the second one is that because the venous blood carries away carbonic acid produced as a metabolite in cellular respiration, that has two ionized forms, $\ce{HCO3-}$(bicarbonate ion) and $\ce{CO3^2-}$(carbonate ion). The answer to the third one would be that, as in lactobacilli case, our body resorts to anaerobic lactic fermentation in activities that require a large power output, beyond the oxygen delivery capacity of the circulatory system. So lactic acid builds up in the muscles, lowering the pH and causing the fatigue sensation. This last explanation was broadly accepted, but seems to be controversial in light of recent studies, as we can see in the abstract for this article:

  In addition, many textbooks report that muscle fatigue is mainly the
  result of a decrease in pH within the muscle cell due to a rise in
  hydrogen ion concentration ([H+]) resulting from anaerobic metabolism
  and the accumulation of lactic acid.6–8 Recent literature, however,
  contradicts this assertion.


Geometry of AsF5 molecule Which of the following is the best description of the arragement of fluorine atoms around the arsenic atom in a molecule of $\ce{AsF5}$?
  (a) trigonal bipyramid
  (b) octahedron
  (c) tetrahedron
  (d) square pyramid
  (e) planar pentagon

The correct answer is (a).

I know that the answer can't be (b), (c), or (e) since octahedron requires 6 ligands and tetrahedron requires 4 ligand, and planar pentagon is only observed in $\ce{[IF5]-}$ and $\ce{[XeF5]-}$. I recall that first row transition metals prefer trigonal bipyramidal configurations, thus trigonal bipyramidal is the more likely geometry, but I'm not sure why this is true?

Any help would be greatly appreciated.

There are no lone pairs in $\ce{As}$ after it forms a bond with each of the five $\ce{F}$.

Hence, the number of lone pairs of electron around the central atom ($\ce{As}$) is zero, and the number of sigma bonds the central atom forms is five (a single bond has one $\sigma$-bond only).

Hence the hybridization number of $\ce{As}$ in $\ce{AsF5}$ is $0+5=5$ and it follows $\ce{sp^3d}$ hybridization geometry with no change in actual shape because of absence of lone pairs.

So, it has a trigonal bipyramidal structure: two $\ce{F}$ ions in the axial positions and three others in equitorial positions.