Constraints on the polar angle (θ) in the particle in a sphere problem For constraint in $\phi$,  $\psi(\phi+2\pi$)= $\psi(\phi$) seems logical as after 2$\pi$ rotation we go back to the same point, therefore functions should be same. But how does it hold for $\theta$ as rotation of $\pi$ gives diametrically opposite point on the sphere, why should then   $\psi(\theta+\pi$) be equal to $\psi(\theta$)?

First of all, as orthocresol and porphyrin said, we should keep in mind that $\phi$ and $\theta$ must span different sets, and there are different conventions on them:

Any spherical coordinate triplet (r, θ, φ) specifies a single point of three-dimensional space. On the other hand, every point has infinitely many equivalent spherical coordinates.
[...]
If it is necessary to define a unique set of spherical coordinates for each point, one must restrict their ranges. A common choice is:
$r$ ≥ 0
0° ≤ $\theta$ ≤ 180° ($\pi$ rad)
0° ≤ $\phi$ < 360° ($2 \pi$ rad)

(Source)
In other words (using the convention above), you must have both $\psi(r, \theta, \phi) = \psi(r, \theta, \phi + 2 \pi)$ and $\psi(r, \theta, \phi) = \psi(r, \theta + 2 \pi, \phi)$ for spherical periodicity.
This is the basic symmetry on a sphere.
But the assignment $(r, \theta, \phi) \rightarrow (x, y, z)$ becomes "ambiguous" (not injective) if the function is defined as an application from $[0, \infty) \times [0, 2 \pi] \times [0, 2 \pi)$ to $\mathbb{R}^3$ and hence not invertible.
You can easily see this if you imagine mapping the surface of earth by having a very long rope with its ends fixed at both poles (half diameter, 180°) and an airplane carrying the middle of this rope around the globe (full diameter, 360° movement).
That being said, the mapping $(r, \theta, \phi) \rightarrow (x, y, z)$ must be taken from $[0, \infty) \times [0, \pi] \times [0, 2 \pi)$ to $\mathbb{R}^3$.
But that does not mean that $\psi(r, \theta, \phi) = \psi(r, \theta + \pi, \phi)$, and as higgsss observed, $\psi(r, \theta + \pi, \phi)$ is not necessarily equal to $\psi(r, \theta, \phi)$.
But higgsss also observed that, for some problems, some solutions might have symmetries allowing $\psi(r, \theta, \phi) = \pm \psi(r, \theta + \pi, \phi)$.
Solutions for the hydrogen atom are good examples of this: for the minus sign, p atomic orbitals, and, for the plus sign, d atomic orbitals.
You can see other orbital symmetries by looking at them here.


What's the energy density of liquid nitrogen? Liquid nitrogen seems to be attracting a bit of attention at the moment as a medium of energy storage, both for electricity grid applications and for transport.
For example, Highview (via the Internet Archive) are doing round-trip electricity storage via liquid nitrogen.

The Dearman Engine Company (via the Internet Archive) are developing a "liquid-air" vehicle engine.
And the UK’s Institution of Mechanical Engineers has formed a Liquid Air working group (via the Internet Archive).
What is the energy density of liquid nitrogen, per unit volume and per unit weight, assuming we’ll be bringing it up to 20 °C, 1 bar? How does this compare to the energy densities of isooctane and of Li-ion batteries?

Just skimmed over the links:

Highview:

  Using ambient heat to warm it, the process recovers around 50 per cent of the electricity that is fed in, says Highview's chief executive Gareth Brett. The efficiency rises to around 70 per cent if you harness waste heat from a nearby industrial or power plant to heat the cryogen to a higher than ambient temperature, which increases the turbine's force, he says.

  Batteries under development .. have efficiencies of around 80 to 90 per cent, but cost around \$4000 per kilowatt of generating capacity. Cryogenic storage would cost just \$1000 per kilowatt because it requires fewer expensive materials, claims Brett.

Here's some evidence that people who know about gas and liquid air are interested in the technology: strategic partnership between Messer and Highview

Actually Highview has a presentation on their web-site that contains a lot of information. Note that they propose to run a 4-stage process against heat (ca. 100 °C), not ambient temp.

Here are some values for energy storage efficiency comparison found after a very quick search:
Batteries (Wiki)
German Wikipedia has lower efficiency values
Energy conversion table (Wikipedia)

Pump-stored hydroelectricity: 70/75 - 80% (Wiki + de.Wiki)
(Note the difference between technology that is actually in use, so efficiencies that are achieved in practical use are known and prototypes/scientific studies about achievable efficiency.)

Dearman claims 70% efficiency for the engine.

Unfortunately they don't say % of what (remember the low-temp heater manufacturers claiming > 100% efficiency because they used achievable efficiency with high temp exhaust gas as their denominator?).
Neither do they say something about the power at which they have this efficiency (ideal thermodynamic efficiency is at P $\rightarrow$ 0 ($t_{cycle} \rightarrow \infty$))
Neither can we assume that they start their calculation taking into account the efficiency of producing liquid nitrogen.
Nor storage loss.

=> so far, Highview's stationary plant concept looks more sound.

Let's have a look what thermodynamics says:

ideal efficiency
boiling T of liquid air: 77 K
ambient T: assume 290 K => $\eta_{ideal} = 1 - \frac{77}{290}$ = 73%
waste heat: assume 370 K => $\eta_{ideal} = 1 - \frac{77}{370}$ = 79%
(in other words, Dearman claims close-to-ideal efficiency)
energy density of liquid $\ce{N2}$: 620 kJ/l = 712 kJ/kg (from presentation at Highview)
The Linde process is said to need approx. 500 kWh / t = 1750 kJ/kg of air
(that would be ≈40% efficiency; no real source, just something someone answered in a German-language chemistry forum)
Here's short report that gives ≈20% efficiency for producing liquid air in Egypt. They also give storage loss (ca. 0.17 \%/d)

Thus, the 50% (achieved by Highview) to 70/80% (short term goal Highview against waste heat, quote for Dearman's engine) mean that

either the liquefaction was not part of the efficiency calculation,
or they have a super efficient air liquefaction that is about 2 - 4 times as efficient as the two numbers I found.
(Anyone around here who knows? Linde, Messer, AL?)

Update: just had a closer look at the Highview presentation. Their 50 - 80% are cycle efficiency based on "Available enthalpy for cold cycle ~300kJ/kg", so less than half of the energy density they give. Calculating with 1750 kJ/kg to produce the kg liquid air, we're at 17% (close to the "Egyptian" 20% efficiency for producing liquid air), of which they recover 0.5 - 0.8.

Conclusion: electric -> liquid air -> electric cycle with an overall efficiency in the order of magnitude of 10%


Are pKa values of acid and conjugate base the same? When you look at $\mathrm{p}K_\mathrm{a}$ table, you can read $\mathrm{p}K_\mathrm{a}$ values of acids. Do conjugate bases also have $\mathrm{p}K_\mathrm{a}$ values?

If both can have $\mathrm{p}K_\mathrm{a}$ values, why $\mathrm{p}K_\mathrm{a}$ value of conjugated bases are same as acids?
Can acid have only $\mathrm{p}K_\mathrm{a}$ values, not $\mathrm{p}K_\mathrm{b}$ values?

Let consider the case of the phosphoric acid.

The phosphoric acid $\ce{H3PO4}$ has $pK_\mathrm{a1}$.
Its conjugate base $\ce{H2PO4-}$ has $pK_\mathrm{b1} = 14- pK_\mathrm{a1}$.

But as $\ce{H2PO4-}$  is at the same time a ( weaker ) acid as well,
it has its own acidity constant $pK_\mathrm{a2}$ independent  on $pK_\mathrm{a1}$ and $pK_\mathrm{b1}$.

Generally, acids like $\ce{HA-}$ have $pK_\mathrm{a2}$ and $pK_\mathrm{b1}$ values.
The former associated with $\ce{HA- <=> H+ + A^2-}$,
the latter associated with $\ce{HA- + H2O <=> H2A + OH-}$


Why is CrCl3 acidic? Why is $\ce{CrCl3}$ acidic? I know that it has a high charge density given that it is a transition metal ion, and can hence polarize neighbouring water molecules, resulting in hydrolysis and formation of $\ce{H+}$. However, I am a little confused over how this happens? Is the water molecule that it polarizes the ones that have formed a coordinate bond with it, or simply any water molecule? Also, with the high charge density, it should be attracting electrons from the water molecule. But does it mean that it attracts the 2 electrons forming the bond between $\ce{O}$ and $\ce{H}$? And if it attracts them to itself, why doesn't it form an extra coordinate bond? Why is there still only 1 coordinate bond? Another thing that I don't understand is that $\ce{OH-}$ has a lower d-orbital splitting ability than $\ce{H2O}$, doesn't that mean coordinate bonds with $\ce{H2O}$ are preferred? Why do they still polarize $\ce{H2O}$ to get $\ce{OH-}$ instead?

For starters, note that you will probably be using the hexahydrate $\ce{CrCl3 * 6 H2O}$. In fact, there is not the hexahydrate, but three distinct hexahydrate crystal structures known. These can be described as $\ce{[CrCl2(H2O)4]Cl . 2 H2O}$, $\ce{[CrCl(H2O)5]Cl2 . H2O}$ and $\ce{[Cr(H2O)6]Cl3}$. In solution, these ions will dissociate, giving you the corresponding free cation, one to three free chloride anions and potentially additional water molecules. It is safe to say that these three anions will take part in an equilibrium reaction $(1)$.

$$\ce{[CrCl2(H2O)4]+ + 2 H2O <=> [CrCl(H2O)5]^2+ + H2O + Cl- <=> [Cr(H2O)6]^3+ + 2 Cl-}\tag{1}$$

It doesn’t really matter which one is the most prevalent species, the argument is generally the same for all three. In all three cases, we have a positively charged cation to which water forms a dative bond. In neutral solution, water is known to self-ionise according to equation $(2)$

$$\ce{H2O <<=> H+ + OH-}\tag{2}$$

The displacement of a proton is facilitated by neighbouring electron withdrawing groups: They reduce the charge density making an anion more stable. And for all intents and purposes, cations are electron withdrawing, displaying a $-I$ effect (negative inductive effect). Therefore, forming a coordinate bond to a chromium(III) ion — even if the overall complex only has a $1+$ charge — will increase the tendency of a water molecule to deprotonate. Or, in equation terms, the equilibrium of equation $(3)$ is right-shifted when compared to equation $(2)$’s equilibrium. (I’ll only discuss this for the monochloridocomplex for simplicity.)
$$\ce{[CrCl(H2O)5]^2+ <=> [CrCl(OH)(H2O)4]+ + H+}\tag{3}$$

Since displacing a proton is easier, the solution is more acidic, the $\mathrm{pH}$ is lower.


What can be used as an electrolyte in water to perform electrolysis other than salt? I am looking for an electrolyte to use in water to perform electrolysis. It can't be salt, as salt ($\ce{NaCl}$) would cause the chemical reaction of
$$\ce{2H2O + 2NaCl -> Cl2 + H2 + 2NaOH}$$
or

$$\ce{H2O + NaCl -> HCl + NaOH}$$

Which both make products with poisonous chemicals (ex: $\ce{NaOH}$ = Sodium Hydroxide (aka lye) which is poisonous). I am looking for a chemical I can use to create the pure forms of hydrogen and oxygen from the electrolysis process. I understand how electrolysis works, and have tried it before, but am looking for something that could be used as a non-poisonous electrolyte. I read that baking soda could be used, would that produce what I am looking for?

You can actually just use salt. It is because that the chemical reaction on the top is for one condition which is that your halide solution ($\ce{NaCl}$) has a very high concentration.

If you halide solution has lower concentration,the hydrogen ion and hydroxide ion in the water will discharged and the two half equation will be

$\ce{4H^+ + 4e^- -> 2H_2}$

$\ce{4OH^- -> 2H_2O + O_2 + 4e^-}$

So, the overall equation is

$\ce{4H^+ + 4OH^- -> 2H_2 + 2H_2O + O_2}$

Thus, is okay to use salt.

Besides Salt

You can also use hydrochloric acid besides salt. Just remember that don't use a high concentration hydrochloric acid as the anode will just discharge the chloride ion instead of hydroxide ion.


Most efficient level of theory for sulfur-containing proline analogues (one that will work) I'm modelling a small set of proline analogues - extra functional groups, sulfur in the ring. Using Gaussian / DMACRYS.

--functional B3LYP EmpiricalDispersion=GD3BJ --basis_set 6-311G**

or

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

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

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

  %n B3LYP/6-311G** EmpiricalDispersion=GD3BJ Opt

  [... molecule goes here ..]

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


Aromaticity of two rings connected by double bond I want to know how to decide aromaticity of two rings connected by a double bond.
For example consider the ring in which when double bond is broken generates tropylium cation and cyclopentadienyl anion. I know that is aromatic because the ions generated are themselves aromatic.
How will the aromaticity be affected when we get an antiaromatic one?
For example consider the ring in which when we break the double bond cyclopropenyl anion is generated.
What will be the overall aromaticity of the molecule in this case.
And in general what is the rule to decide the aromaticity of the rings connected by a double bond?

Actually, the breakage of double bond doesn't happen in the way you are thinking it does. Let me illustrate the phenomena what happens in such rings connected by double bonds.

If you have two rings connected by double bonds where the breakage of that will produce anti aromatic rings, the molecule itself becomes very inert towards that breakage. The molecule always tries to gain stability through any perturbations to it. That's why production of anti-aromatic rings becomes extremely difficult in those cases.

In your given case, the molecule becomes highly polar due to unsymmetrical distribution of electron cloud (i.e. more negative charge on cyclopentadienyl ring and more positive charge on the other ring). But in a case, where you are assuming that breakage of double bond gives anti-aromatic cyclopropenyl anion or cyclopentadienyl cation, that breakage becomes extremely impossible. Even if you would subject it to extreme conditions, the molecule will never go into that transition state as that will have immensely high energy. Instead it will always shift towards that transition state by which it can gain some stability, by forming non aromatic or aromatic rings.

If it can't stabilize by that also, it will then become almost inert towards any electrophile or any external reagent which tries to make it anti-aromatic. Even if you perform ozonolysis, it will remain as a perfect keto-group, no unsymmetrical distribution will be observed for these classes of rings.

Considering the number of π-electrons in continuous conjugation of the whole molecule, we can judge if the molecule is aromatic or anti-aromatic. But if the molecules breakage of one double bond gives it anti-aromaticity, it will try to stay in its own form, whatever it may be.


Choosing fish-oil that won't oxidize in long travel? Fish oil, is susposed   to suffer from rancidity(i.e. oxidation) when stored improperly , for example in hot temperature.

I'm interested in buying fish oil from abroad(usa) it will be shipped by air and won't be kept in a cold chain,i live in a mediternean nation(climate wise).

How should i choose my fish oil , such that when i recieve it ,it will be OK and not rancid ? is it even possible ?

First off, a "fishy" smell and rancidity are two different things.  A fishy smell is likely due to impurities, primarily small volatile amines such as trimethylamine.  These impurities are unrelated to oxidative degradation of the fish oil.

Rancidity in fish oil is the result of oxidation.  Oxidation leads to the formation of lipid hydroperoxides and short-chain aldehydes, which can have "off" tastes and give rise to unpleasant odors, but in a very different way than the classic fishy smell.

Fish oil's primary components are docosahexaenoic acid and eicosapentaenoic acid.  They are more sensitive to oxidation than many other fats because they contain many double bonds.

The best bet is to look for oil that is freshly prepared and highly purified.  Some fish oils are esterified with ethanol to form ethyl esters right "on the boat".  These ethyl esters can be distilled to remove most impurities.  Optionally, the purified esters can be converted back to triacylglycerols to give "molecularly distilled" fish oil in the triglyceride form.


Is ground-state structure of a molecule entirely determined by the positions of the nuclei? I'm a computer science with only a weak undergraduate chemistry background, but have recently become interested in simulation methods for computational chemistry. Many simulation methods, such as DFT, make the Born-Oppenheimer approximation (nuceli are static) and solve for some properties of the electrons (in the case of DFT, the density). In fact, when using Gaussian-16, I only need to specify the locations of the nuclei in space.

What then, is the role of actual bonds? Are bonds, bond order, etc. just a convenient abstraction for what the electrons are doing? If I tell you the location of some nuclei in 3D, can you then (theoretically) uniquely determine the bond structure?

Yes, these methods (both wavefunction-based and electron density-based) typically solve a purely electronic system defined by an electrostatic potential which is produced by the nuclei; the Born-Oppenheimer approximation can be understood as treating the electronic system as a function of nuclear geometry, which implies that (1) nuclear dynamics are not strongly affected by electrons and (2) electrons will almost immediately adapt to any new position of the nuclei. Born-Oppenheimer can also be understood in terms of electronic and nuclear motion (and the corresponding energies) being separable; that is, electronic structure takes shape after (and as a consequence of) nuclear dynamics, but does not directly affect nuclear dynamics themselves. Since chemistry is usually interested in stationary solutions as snapshots, these approximations are fitting for pretty much all chemical questions.

In practice, nuclei are not even particles in electronic structure calculations - in both Schrödinger equation methods (HF and post-HF) and Hohenberg-Kohn methods (DFT), the only role of nuclei is to generate an external electrostatic potential, $V_{ext}$, that will be part of the Hamiltonian/energy functional of the system. The specification of nuclear positions that you mention is just a more convenient, and chemically intuitive, way of describing that external electrostatic potential, but there is no mathematical or conceptual difference if we consider that this external potential is arbitrarily imposed from outside; it's just that it's easier to define a number of point charges (with their charge defined by their atomic number) than to provide a complex 3D potential function.

Once we assume that Born-Oppenheimer is a valid approximation and that nuclei merely exert an external electrostatic potential on electrons, the fact that electronic structure for the ground state is entirely determined by the positions of the nuclei is axiomatic: it is a consequence of the postulates of quantum mechanics (and the basis of all variational methods) for wavefunction-based calculations, and one of the Hohenberg-Kohn theorems for density functional calculations.

Geometry optimisation is also performed under the same assumptions - electronic structure is calculated under a given geometry, and then resulting forces over nuclei are estimated; a small motion is performed along these tensions, and the process is repeated iteratively until forces experienced by nuclei are small enough to fall below a certain tolerance level. Note that there is no explicit treatment of the interaction between electrons and nuclei; only two process that interact iteratively through electrostatics: electronic structure is minimised for each geometry, and then geometry is relaxed, but not minimised, according to that electronic structure. We obtain a quantum mechanics depiction of the electronic system for each geometry, but we do not perform a quantum mechanics treatment of the nuclei. Ab initio molecular dynamics is a similar approach, in which nuclei move through classical dynamics but electronic structure is calculated through QM at each step.

In all these cases, chemical bonds are only emergent interpretations of the energetics of electronic structure, but their identification is largely arbitrary, as energetic effects are not localised. While we can calculate the energy of two molecular fragments as they approach from infinity, and we'll observe interactions that we identify with bonding, the limits of what is a bond are not clear, and at a quantum chemical level, it makes more sense to talk about energies associated with different geometries than to think in terms of bond "objects".

It is in principle possible to perform QM computations which treat both nuclei and electrons as quantum particles, thus obtaining a single wavefunction describing both, but the complexity of this treatment (especially given the large difference in scale for electronic and nuclear motion) restricts calculations to a very small number of particles - the highest number I remember reading about is 8, although I might have missed recent studies. In addition, the problem of describing interactions accurately (which is behind the variety of quantum chemistry methods used just to describe electrons) grows considerably worse, to the point that quantum Monte Carlo may be a better option than quantum chemistry methods, as happens with bosons.


Atoms or molecules with spin 1 in the ground state? Is there any atom or molecule that has spin 1 in its ground state?

Do Hund's rules keep this from happening for an atom?

The reason I'm curious is that it would be nice to have a spin-1 example for use in pedagogical discussions of the Stern-Gerlach experiment.

[EDIT] Clarification: when I say "spin," I mean the total angular momentum, not just the sum of the spin-1/2's. (The total angular momentum is what you are seeing in the Stern-Gerlach experiment.) I deleted the part of the question about ions, because, as pointed out by Orthocresol, they won't be usable in a normal Stern-Gerlach spectrometer.

Atoms

Atomic carbon with its $\mathrm{1s^2 2s^2 2p^2}$ configuration has a triplet ground state ($S = 1$), precisely because of Hund's first rule.

However, in the context of the Stern–Gerlach experiment, you might run into a problem with orbital angular momentum, as carbon's ground state also has nonzero orbital angular momentum ($^3\mathrm{P}$ ground state, $L = 1$). The behaviour in a magnetic field will be rather more complex and you probably need to take into account spin-orbit coupling.

At the moment I can't think of any atoms with a ground state term symbol of $\mathrm{^3S}$. I actually suspect that it's impossible, but I'm not really up to proving it right now.

Molecules

Triplet dioxygen (as Zhe mentioned) has no orbital angular momentum ($^3\Sigma_\mathrm{g}^-$ ground state, $\Lambda = 0$), but I'm not sure if the inhomogeneity of the electron density would have any impact. (as in, I'm genuinely not sure.)

Ions

The $\ce{^2H}$ nucleus (which us chemists usually refer to as $\ce{D+}$) is lightest stable nucleus that has a spin of 1.

The only issue is that you then have a moving charge in a magnetic field. If you could ignore it, or take it out of the equation, somehow...


Colorless nature of carbonyl complexes Researching the coloration of transition metal complexes one day, I came across this reference on Wikipedia 1:

I am not sure why this is so. Take $\ce{Mo(CO)_6}$, or any other carbonyl complex of the metals in group 6. If we have a strong $\pi$ acceptor like carbonyl, and hence a low-spin $d^6$ system, would we not get an orbital splitting pattern like this:

The excitation of an electron to one of the empty $e_g$ orbitals seems feasible to me. Why can't one of those paired electrons absorb radiation to exhibit colour for the complex?

Reference

1 https://en.wikipedia.org/wiki/Metal_carbonyl

Why can't one of those paired electrons absorb radiation to exhibit colour for the complex?

They can, they do and they color. The problem is, the corresponding transition requires so much energye, that the adsorption is well above upper range of visible spectrum in most cases. Linked article provides us with 3 optical spectra of $\ce{Cr}$, $\ce{Mo}$ and $\ce{W}$ carbonyls. For all adsoprtion mostly happens above 300 nm, with only small bits between 300 and 400 nm. Visible spectrum is between ~400 and ~700 nm. Since only bit of upper blue part of the spectrum is adsorbed, only a pale yellow coloring is observed.

This, of course, if you want so badly to apply crystal field theory to carbonyls, which does not work well in the case.


What Methods Would I Use To Powder Cotton? I’m needing a method to turn cotton into a fine powder for use as a source of cellulose. What suggestions would you give to accomplish this with materials/tools that could be obtained at local stores?

The content of cellulose in paper is not as high as that of cotton, however, it may be easier to use as a source if you do require a fine powder. Why do you require cellulose?


Determine the quantitative composition of a solid sample containing Co, Mn, Al, C, and O I have a solid sample that contains Co, Mn, Al, C, and O. The sample mass is ~200 mg. I need to determine what the composition of this sample is. I've looked at atomic adsorption spectroscopy, but that method is unable to detect C or O. I've heard mention that inductively coupled plasma mass spectrometry (ICP-MS) might work, but I am looking for additional potential solutions. I have access to a large number of instruments (university instrument sharing agreement). It is fine if the sample is destroyed as part of the analysis.

It very much depends on element concentrations and the accuracy required for the total assay, but, I would do the metals by ICP-AES or AA and C by a classical combustion technique. ICP-MS would be unnecessarily expensive and since you are looking at compositional levels rather than low trace levels, it just wouldn't buy you anything.

If you know oxygen is the only other element present at compositional (i.e. percent) levels then don't bother with the finickier oxygen analysis and just determine it by difference. If carbon is present at at least a few percent then you only need a few mg, leaving enough to do multiple digestions for the metals analysis if desired.

If you want to do direct oxygen analysis, it can be done by classical pyrolysis techniques, though this presumably high-metals matrix would require a special technique (Mertz) and this analysis could have relative uncertainties of greater than 10%.


Are all lone pairs considered as π-electrons? According to me, in methylenepyran 8 π-electrons are present as there are 3 double bonds. Therefore, 2 × 3 = 6 π-electrons and 2 π-electrons as one of the two lone pairs of oxygen will participate in resonance.

But the correct answer is 10 π-electrons. How? Are both lone pairs of oxygen are being considered? If yes, then why?

I am quoting from http://www.hutters-online.de/publikationen/dmchdo.html.

  One requirement that is basic to this concept and is agreed upon by most authors is that there have to be two double bonds that are in conjugation with a third one, but not in a linear arrangement. The p electronic system thus forms a bifurcation. So the simplest cross-conjugated hydrocarbon would be 3-methylene-1,4-pentadiene (I) that can be viewed as an ethene disubstituted geminally with two vinyl groups:

Based on the above the following resonance structures are possible. These have 8 $\pi$ electrons in conjugation. You are correct.


What is the mechanism of trialkylborane oxidation with hydroperoxide? This is the mechanism of trialkylborane. In this mechanism I can't understand why should the -OH group leave and the -R group enter in its place to form borate ester. Can someone please explain me that step?

One aspect is the release of the stable anion $\ce{OH-}$.

Another aspect is that $\ce{B-O}$ bonds are much stronger than $\ce{B-C}$ bonds because the empty p-orbital of boron overlaps with one p-orbital with a lone electron pair of oxygen.

$\ce{B-C}$ $D_{298}^ \circ  = 448 \pm 29\ \mathrm{kJ\ mol^{-1}}$

$\ce{B-O}$ $D_{298}^ \circ  = 809\ \mathrm{kJ\ mol^{-1}}$

values taken from “Bond Dissociation Energies”, in CRC Handbook of Chemistry and Physics, 90th Edition (CD-ROM Version 2010), David R. Lide, ed., CRC Press/Taylor and Francis, Boca Raton, FL.


Will entropy increase in adiabatic expansion? The formula for entropy change states that change in entropy is not possible if heat change is nil.
∆S=∆Q/T
But what if we adiabatically expand a gas? Won't the molecules get more space to vibrate in, hence increasing the randomness of the system, therefore increasing the entropy?

The answer to your question is yes and no. You are correct in supposing that as an ideal gas expands the entropy will increase as it has more space to occupy and so the number of ways the molecules can be placed in the total space has increased. (Try to not think of entropy in terms of randomness but in the number of ways molecules can be positioned in space, or occupy different energy levels). However, accompanying expansion is cooling and if the adiabatic process is carried out reversibly the decrease in entropy due to cooling exactly matches that increase due to expansion, and so the entropy change is zero.
If the expansion is irreversible and adiabatic, the gas does less work, looses less internal energy and so there is entropy increase as the cooling is now not as large as when expansion is adiabatic and reversible, and so the two process, entropy increase due to expansion and decrease due cooling no longer match.


What are the known atomic oxygen species? I'm interested in chemical reaction mechanism with more exotic particles. The Wikipedia page seems to imply that the normal atomic oxygen is $O(^3P)$, is that right?

I also came across $O(^1D),O(^1S),O(^5P)$ and some longer strings like $O(3S3s)$. What are the atomic oxygen species, or at least these which we can expect to see in the parts per millions range or a little below, in oxydation processes specifically? How to know which I left out?

In this content atom means isolated atom, which have defined quantum states.  This is to be distinguished with atom in a molecule, which is an entirely conceptual construction that have not strict definition or physical meaning.  Atoms in a molecule have no defined quantum states.

That is, only an isolated oxygen atom not in a molecule can have designations like $O(^3P)$, etc.   For molecules, term symbols belong to the entire molecule, not individual atoms.

In oxidation reactions atomic oxygen is rarely involved.  It can only be formed when very large amount of excess energy is available, such as in combustion or ignition processes, or in a radiation rich environment such as outer space.  There, the majority of atomic oxygen should be the most stable species $O(^3P)$.  However, in these high energy scenarios, all the other low lying states of oxygen will also be present in smaller quantities.


Are all 'dextrorotatory' sugars in biology actually 'd' or '+' in chirality? When papers or articles say that all proteinogenic amino acids are 'levorotatory' or 'L', they often make a point of saying that only half of them are truly, optically levorotatory.  All of them (except glycine) are called 'L' because of their 'opposing' configuration to glyceraldehyde, the original molecule deemed to be dextrorotatory...
Well, all sugars used by living things are supposedly 'R', or so I keep reading....
Are they similar in configuration to glyceraldehyde, or actually levorotatory to light?

Dextrorotatory = d = (+). Levorotatory = l = (-). These terms describe the absolute sense of rotation in an optically active substance.
D and L are older terms for the sense of chirality that can be traced back to dextro- or levorotation of glyeraldehyde. The actual compound may have the same or opposite sense of rotation, but the assignment is determined by analogy. (Cf. Fisher/Rosanoff).
R and S describe absolute stereochemistry of a center, so a specific molecule with multiple chiral centers cannot be sufficiently described with a single R or S, but require one for each stereocenter. This designation is part of the more modern (and widely accepted) Cahn-Ingold-Prelog system, which has largely been adopted by IUPAC.
Note that while D and L might be sufficient to describe a sugar, anything more complicated than a 3-carbon aldose or 4-carbon ketose is not likely to be properly described with a single R or S because there are multiple stereocenters.


Why is CuSO4 used as electrolyte while purifying copper? During the electrolytic purification of copper, the electrolyte is taken as copper(II) sulphate ($\ce{CuSO_4}$). Is there any particular reason for this? Could another salt of copper, say $\ce{CuCl_2}$ work?

I mean, I don't see any reason why it shouldn't. The only difference will be, that at the anode chlorine instead of oxygen will be evolved, which doesn't seem to be such a big thing.

Why, then, is $\ce{CuSO_4}$ preferred?

During copper electrorefining, impure $\ce{Cu}$ from the anode is oxidized and dissolved, while $\ce{Cu^2+}$ is reduced at the cathode, forming the deposit of refined copper. Because anodic impurities which are less noble than copper – most notably $\ce{Ni}$, $\ce{As}$ and $\ce{Sb}$ – also dissolve in the electrolyte, the cell voltage is kept in the range of 0.2-0.4 V to avoid co-deposition of these metals at the cathode.

Copper dissolution is the main oxidation half-reaction at the anode, and the oxidation of anions from the electrolyte needs to be minimized, since it is an unwanted side reaction which reduces current efficiency of the process. In a $\ce{CuSO4}$ electrolyte, this would be oxidation of $\ce{OH-}$ (from water dissociation, so concentration will be low) to $\ce{O2}$, while in a $\ce{CuCl2}$ solution, $\ce{Cl-}$ will be oxidized to chlorine gas, which would present a greater hazard because of its toxicity.

The electric conductivity of the electrolyte is another important factor, which varies with different copper salts and their concentrations. $\ce{CuCl2}$ and $\ce{CuSO4}$ are both readily soluble in water; however, the relative molar conductivities of these solutions will be different. In this study, a copper chloride electrolyte proved to be inferior to copper sulfate and copper nitrate in terms of deposit quality and stability of cell potential.


Is there a heat capacity of an object when it is melting? Is there a heat capacity of an object when it is melting?

I’m questioning it since the temperature doesn’t change during phase transitions and the unit of heat capacity is joule per kelvin.

Yes there is a heat capacity even at the melting point.  The reason the temperature doesn't change is that you need to add the enthalpy of melting to convert solid to liquid in equilibrium.  However, under the right conditions you can supercool the liquid under the melting point, and the temperature will behave as you would expect, needing to pull a certain amount of energy out to supercool by a degree.  But this merely delays having to deal with the enthalpy of melting until solid uncleared - it is a kinetics issue, not a thermodynamics issue.  So, it is clear that the heat capacity of a phase is continuous through the phase transition (and beyond).


Finding mass percent through molality of potassium nitrate solution Question

An aqueous $\ce{KNO3}$ solution has a molality of $\pu{4.16 m}$ and a density of $\pu{1.08 g/mL}$.  Calculate the percentage by mass $\ce{KNO3}$ of the solution.

My Approach
I know that molality ($b$) is defined as:
$$b = \frac{n_\mathrm{solute}}{m_\mathrm{solvent}}$$
In order to the mass of solution I multiplied $\pu{(1.08 g/mL)}\times\pu{1 mL}$, since, $m_\mathrm{solution} = m_\mathrm{solvent} + m_\mathrm{solute}$
However, I don't know where to use this information. I am also not sure what I to do with the $\pu{4.16 m}$ since I don't have moles of solute or mass of solvent. I am thinking that I have to assume some number here, but I might be wrong about this. Any help will be appreciated.

I am not sure you even need the density in this case.
Say you have $\ce{1kg}$ or $\ce{1000g}$ water (the solvent). That would mean that there is $\ce{4.16 mol}$ of $\ce{KNO_3}$.
Since we can figure out the molar mass of $\ce{KNO_3}$ = $\pu{101.11g mol-1}$, the mass of $\ce{KNO_3}$ in the sample must be $(4.16)(101.11)= \pu{420.62 g}$.
We already assumed that we had $\ce{1000g}$ of water, the mass percent of $\ce{KNO_3}$ is then equal to: $$\frac{420.62}{420.62+1000
} \approx 29.6\%$$


Reversablity of dehydration Aqueous magnesium bicarbonate when evaporated decomposes into magnesium carbonate. Is that reaction reversible by dissolving the carbonate in soda water?

Yes, in fact this process is similar to the processes in lime stone that form cave stalagmite and stalactites but with calcium carbonates.
$$\ce{M(HCO3)2 <-> MCO3 + H2O + CO2 }$$
$$\ce{M} = (\ce{Mg},\ce{Ca})$$
Though I should note the process in not called dehydration.


Why doesn't the carbocation form on the carbon that is attached to the ethyl group in this reaction? Why doesn't the carbocation form on the carbon that is attached to the ethyl group in this reaction?

Subsequently, why doesn't the $\ce{Cl-}$ attack the carbon that is attached to the ethyl group? Is a tertiary carbocation less stable than a secondary allylic carbocation?

Reaction:

1,2 -hydride shifting takes place after the formation of the tertiary carbocation. After the hydride shifting, the carbocation becomes resonance stabilised. And the chlorine adds up there(the first product that you have shown). Also, as I said the carbocation was resonance stabilised. So you can draw resonance structure of the carbocation intermediate and the positive charge goes to 4 position and then chlorine adds up there (the second product that you have shown) and thus we get the second product. Hope this helps.


Energy required for obtaining required mass of aluminium from alumina via electrolysis How do I know how much energy I need to obtain 20 g alumunium from alumina with 5 V?

I calculated the quantity of electricity is 2.22 F.

Any help appreciated!

Required energy $E$ can be found as a product between voltage $U,$ current $I$ and time $t$:

$$E = UIt\label{eqn:1}\tag{1}$$

Unknown multiplier $It$ can be found using united Faraday law:

$$m = \frac{MIt}{zF}\quad\implies\quad It = \frac{mzF}{M}\label{eqn:2}\tag{2}$$

where $m$ is the mass of the substance liberated at the electrode; $M$ is the molar mass; $z$ is the number of electrons transferred per ion; $F$ is Faraday constant.
Considering aluminium is produced via Hall–Héroult process or similar, at the cathode the following reaction takes place:

$$\ce{Al^3+ + 3 e− → Al}$$

and $z = 3.$
Plugging \eqref{eqn:2} into \eqref{eqn:1} allows to finally find the energy:

$$
\begin{align}
E &= \frac{mzUF}{M}\\
  &= \frac{\pu{20 g} × 3 × \pu{5 V} × \pu{96485.33 C mol-1}}{\pu{26.98 g mol-1}}\\
  &= \pu{1.07 MJ}\tag{3}
\end{align}
$$


Are there colorful metals besides gold and copper? Are there yellow metals besides gold? Are there metals with color different than yellow and grey/silver or reddish?

Osmium has a bluish-gray tint.
Well; slightly.
Cesium is silvery-golden!,
But don't wear it.


How to RAISE the melting point of water? It is common that adding common salt into water will lower its melting point. But, is there any way to RAISE the melting point of water? Does it lower the boiling point of water as well? Why?

source

One way to raise the melting point of water is to increase pressure beyond about 635 MPa.  By raising pressure you could get the melting point to be even greater than the normal boiling point.

A second way is to lower the pressure, but this can only increase the melting point by 0.01 K.

A third way would be to added enough of a high melting point solute that you pass the eutectic point. In other words, generally when you add a solute to water, the melting point decreases, but only until you add enough solute that the eutectic point is reached, beyond which adding more solute increases the melting point.

source: http://www.phasediagram.dk/binary/CaCl2.PNG

For example, in the diagram, above, A is the eutectic point.  Beyond this point, adding more CaCl2 increases freezing point.  Then beyond 38% CaCl2, melting point exceeds ) degrees C.


Pseudochirality with more than two groups on one carbon atom Can pseudochirality be considered when three or four groups on the carbon atom are constitutionally the same, but configurationally different?

For example, when the C atom is bonded to one H, and then three more structurally identical groups with 2 chiral C each - in configurations R,R and S,S and R,S respectively - then by assigning each of the 4 groups a priority value (since they are all different - though 3 have the same structure) we can assign a configuration to this C atom.

Ah, now I think I understand what you're asking.  You're saying that we have a central carbon to which is attached a hydrogen and 3 groups with the same structure. Each of these 3 groups contains 2 chiral carbons.  One group is R,R one is R,S and one is S,S.

So yes, the chirality of the central carbon (it is chiral, it is not meso or prochiral - there is no plane of symmetry bisecting the central carbon) can be assigned using the Cahn–Ingold–Prelog priority rules where "R" has higher priority than "S".  Therefore the priority ranking around the central carbon in your molecule would be R,R > R,S > S,S > H.


Does a salt always have a neutral charge? I know that a salt is the product of a reaction between a base and an acid, but does this mean that there will always be one cation and one anion and therefore a neutral charge?

In relation, are acids and bases charged?

A salt is an ionic solid compound. As such, it needs to have a zero net charge.

However, that doesn't mean it should contain only one cation and one anion. The stoichiometry of the salt depends on the charges of the ions. For instance, iron (III) sulfate is $\ce{Fe2(SO4)3}$; this stoichiometry is required to balance the charges from $\ce{Fe^3+}$ and $\ce{SO4^2-}$.

Also, other types of salt exist, called mixed salts. They contain several different cations and/or anions (e. g. $\ce{KAl(SO4)2}$).

Compounds with a net charge exist, they are just not called salts (look up Complexes).

  In relation, are acids and bases charged?

Acids may (e. g. $\ce{NH4+}$) or may not (e. g. $\ce{H2SO4}$) be charged. Similarly, bases may (e. g. $\ce{SO4^2-}$) or may not (e. g. $\ce{NH3}$) be charged.


Behaviour of water at exactly 0 °C If I had a beaker of water (i.e. many molecules), at exactly 0 °C and at atmospheric pressure, what phase would be encountered?
And would the answer to this also apply to other substances at their freezing/melting point, and out of interest, would the same effect be observed at their other "phase intercepts" too?

It depends on how you achieved water at 0 °C. If you cooled water down to 0 °C, then it will remain water, unless you remove more heat, at which point it will start to freeze. Water needs to be at 0 °C to freeze.

If you warm ice up from below freezing to 0 °C, then it will remain ice, unless you add more heat to melt the ice. Ice needs to be at 0 °C to melt.

It will be the same for all substances.


Does acetic acid give a positive result with the iodoform test? Acetic acid contains a $\ce{-COMe}$ group, so theoretically, it should give a positive result in the iodoform test (i.e. a yellow precipitate of $\ce{CHI3}$ being formed).

But apparently the iodoform test is used to distinguish between acetic acid and other compounds with a $\ce{-COMe}$ group. So, logically, acetic acid must not give a positive result. Why is this so?

The iodoform reaction proceeds by the mechanism shown below. (The final proton transfer need not occur between the two partners, any other hydroxide could abstract the acid’s proton and any water molecule can protonate iodoform.)

This can fail for carboxylic acids or esters in one of two ways:

The first step. The $\mathrm{p}K_\mathrm{a}$ value of the proton being abstracted is $26.5$ for acetone and $29.5$ for ethyl acetate according to the Bordwell $\mathrm{p}K_\mathrm{a}$ table. This can be attributed to the second oxygen of an ester donating electron density into the carbonyl group and thus lowering the tendency to form the enol(ate). But we can still argue that enough ester will be deprotonated for the iodination to occur. (Note that this fails entirely for a free acid, since the molecule is already charged. A second deprotonation to produce the endioldiate is very unfavourable.)

The final fragmentation step. In esters, $\ce{R}$ also contains an oxygen atom, so there are a total of three pathways by which the tetrahedral intermediate can fragment (one being redissociation of the hydroxide). In the ketone’s case, only one is productive, namely the formation of iodoform which precipitates and drives the equilibrium towards its formation. The ester can fragment under displacement of the alcoholate (productive, gives an acid) which is much more favourable than displacing a $\ce{CI3-}$ group due to the $\mathrm{p}K_\mathrm{a}$ value of both’s conjugate acids.
The resulting acid is merely deprotonated. The attack of a nucleophilic (hydroxide) species to the resulting anion is strongly disfavoured.

It is also to be noted that an ester can always be hydrolysed to the corresponding acid under basic aquaeous conditions. Thus, at any intermediate an acetate anion could generate inhibiting any further reactions.


What is meant by fractional oxidation state? What is the real concept behind oxidation state?

Nothing too much, really. When you have a compound like sodium tetrathionate $\ce{Na_2S_4O_6}$, if you calculate the oxidation state of sulfur, you get $\frac{5}{2}$. This means that the AVERAGE oxigation state of sulfur is $\frac{5}{2}$. In fact, the sulfurs in the compound are not equal, you have two of them with an oxidation state of $+5$, and two of them with $0$. But when you balance redox equations, you do not really care about this, and you can use the average value of $2.5$ to find the coefficients.


What is the difference between a phosphoryl group and a phosphate group? I am looking at covalent modification of enzymes and I found out that only one of these types is phosphorylation. But I do not understand what the difference between the a phosphate and a phosphoryl group is in terms of nomenclature.

The difference is basically one oxygen atom. If you consider monoethyl phosphate as an example, the ethyl alpha-carbon bears a phosphate group, whereas the ethoxy oxygen bears a phosphoryl group.


Stability of furanose vs. pyranose form of glucose? I am trying to imagine the cyclization of the Fischer projection of glucose as it forms an $\unicode[Times]{x3B1}$-D-glucofuranose and an $\unicode[Times]{x3B1}$-D-glucopyranose (right now I am trying to understand the Haworth projection, and as my 3D visualization gets better, I will extend to the more thermodynamically stable chair representations of these monosaccharides). Now, if I understand correctly, in the pyranose form, the alcohol on $\mathrm{C5}$ attacks the carbonyl carbon on $\mathrm{C1}$. In the furanose form, the alcohol on $\mathrm{C4}$ attacks the carbonyl carbon on $\mathrm{C1}$. These results in 6-membered and 5-membered rings, respectively.

I understand in nature, these are the most stable rings. However, what thermodynamic incentive does the furanose form have? That is, why would $\mathrm{C4}$'s alcohol attack and form a 5-membered ring at the expense of the $\ce{CH2OH}$ group and the $\ce{OH}$ group being near each other as a side chain on the 5-membered ring? Doesn't this introduce considerable strain? My idea is that the 6-memebered ring is more thermodynamically stable and it doesn't have the problem of a hydroxymethyl and a hydroxyl group on a single carbon of the 5-membered ring.

If you look up the statistical distribution of glucose isomers in aquaeous solution, you will find $36\,\%$ $\unicode[Times]{x3B1}$-pyranose and $64\,\%$ $\unicode[Times]{x3B2}$-pyranose. All the remaining forms lumped together only account for less than $1\,\%$ of glucose in solution; that includes both furanoses, the open-chain form and the (possible, but unlikely) seven- and four-membered rings.

Neither five-membered rings nor six-membered rings contain any relevant amount of angle strain. You can easily verify this buy getting yourself a molecular model kit and attempting to bild these rings from four-bond sp³ carbons: both can be built smoothly, like a charm. So angular strain is not a reason why the pyranose forms are preferred.

Considering reaction mechanisms, the formation of five-membered rings is usually kinetically preferred due to easier reachable transition states. So this also cannot be the reason for preferring six-membered rings.

The actual reason is not so much angular strain, but rather dihedral angle strain: the same strain that prevents ecliptic conformations in alkanes. The pyranose form has the perfect $60^\circ$ dihedral angle between two non-ring atoms which minimises this type of strain. Furanoses must choose between envelope or twist conformations to at least relieve some of this strain.

All this is, however, happening at an equilibrium, which is semi-rapid in adjusting itself. So there will always be a certain percentage of furanose form in solution being formed and breaking apart. If any kind of reaction selectively only happens with furanose anomers, then this will cause the formation of new furanose which again will be captured away. As I wrote before, the attack which creates a furanose is preferred over that which creates a pyranose. (But the breaking apart of pyranose is slower than that of furanose.)


Ionic bonds- are electrons transferred? In an ionic bond, are electrons actually transferred from one atom to the other to form a positive and a negative ion OR are ionic bonds just extremely polar covalent bonds? Sorry if this seems like a simple question but I've heard this somewhere so just wanted to make sure.

This is not a simple question at all, in fact, it is an interesting insight in chemical bonding.  You make a valid assertion with your question:

  are ionic bonds just extremely polar covalent bonds?

This answer is a simplistic way of looking at it.

According to ChemGuide page Electronegativity, there is no real clear-cut 'boundary' between covalent and ionic bonding. The website describes that there is a 'spectrum of bonding' based on the electronegativity (chart shown below) of the constituent elements.

Image source: UC Davis ChemWiki

So, the greater the electronegativity, the greater the polarity or "pulling power" on the bonding electrons towards elements with a higher electrongativity.

In the following diagrams (from the ChemGuide site linked before), A and B represent arbitrary elements, and the green and red dots represent bonding electrons. Starting from a situation where the polarity is zero as the electronegativity of the constituent elements are the same (e.g. diatomic molecules $\ce{O2}$, $\ce{Cl2}$ etc), as in diagram I.

As the polarity increases, the bonding electrons move towards the more electronegative member, causing the bond to become increasingly polar (II), then eventually to the point where the difference is enough (III) for:

  To all intents and purposes, A has lost control of its electron, and B has complete control over both electrons. Ions have been formed.

I bolded the 'to all intents and purposes' to emphasise that this is a simplistic generalisation

(I)

(II)

(III)

As the lowest electronegativity is not zero (it is Francium at 0.7), there will always be an influence on the bonding electrons. So, the reality is that A does not completely lose its influence on the bonding electrons.

For example, if we were to look at a compound combining the most and least* electrongative elements (as appear in the diagram above): francium fluoride $\ce{FrF}$, even though fluorine has a far stronger electronegativity than francium, the electronegativity of the latter is not zero, so, the influence of the francium is very small, but not zero - indicating that purely ionic bonds don't exist.

As noted in the comments, cesium is likely to be less electronegative than francium, however the point made above still stands.

This is nicely summarised by an article from the University of Malta:

  Pure ionic bonding is not known to exist. All ionic compounds have a degree of
  covalent bonding. The larger the difference in electronegativity between two
  atoms, the more ionic the bond.


How does O-benzoyl-N-methylhydroxylamine hydrochloride oxidise this ketone? How does the mechanism of this reaction with BzONHMe work? Can part of it act as a base and abstract the alpha proton? Would be great if someone could help me!

The first step is the formation of the Me,O-Bz imine of the ketone. This tautomerises to the enamine which then does a pericyclic rearrangement with the carbonyl of the benzoyl group to install O at the alpha position. Finally  the imine hydrolyses to reveal the ketone. See Aldrich Chemfiles (2007) 7.1, page 7.


Why does KCl come in amber coloured bottles? I remember from school that photosensitive chemicals are kept in amber coloured bottles. Is KCl photosensitive? I read up the impurities list on the bottle and it says that it may have bromides too? So is it because of the KCl or the bromides? If it is due to the bromide, what sense does it make since it would react just to form another impurity?

The potassium chloride solutions in my lab are in transparent glass bottles, as was the potassium chloride solution in my inorganic students’ lab at the LMU in Munich. If I recall it correctly, the solid potassium chloride we had there was in a transparent bottle, too, but don’t quote me on that.

There should be nothing in potassium chloride that will react under irradiation. And these types of impurities are generally not a reason to use such a bottle. But maybe the supplier of your choice just happens to have too many amber bottles?


Specifically how dangerous is a 10g chemical sample of pure arsenic and how do you properly dispose of it? I'm not a chemist.  I'm an electrical engineering student and I recently purchased an element collection.  The collection came with a 10g arsenic sample.  I assumed it wasn't much of a thing because there were no warnings or anything but afterward when I started reading more I became concerned.  How dangerous is the really, what precautions must I take in handling the sample, and most importantly how do I properly dispose of this sample?

Thank you

Treat this with very great care; don't let it come into contact with you or anyone else as it is readily absorbed through the skin, lungs and gastrointestinal tract.

Arsenic is the classic poison known also to the ancients. Ten grammes is a huge quantity as it is deadly in quantities of approximately  0.1 g in the blood stream. The oxide and compounds such as the  arsenate, are far more poisonous than the metal as they replace phosphate in the body (which is harder for the pure metal to do) so someone can be poisoned directly by taking a large dose or by gradual exposure over a period of time.

For a fascinating read on this and other poisons look up 'A is for Arsenic: The Poisons of Agatha Christie' by K. Harkup, publ. Bloomsbury. '


Contact explosion of nitrogen triiodide Nitrogen triiodide, on slightest physical contact, explodes to evolve dense purple fumes of iodine. Why does this phenomenon occur?

video link - https://www.youtube.com/watch?v=2KlAf936E90

Wikipedia states

  The instability of $\ce{NI3}$ and $\ce{NI3 · NH3}$ can be attributed to the large steric strain caused by the three large iodine atoms being held in close proximity to each other around the relatively tiny nitrogen atom. This results in a very low activation energy for its decomposition, a reaction made even more favorable due to the great stability of $\ce{N2}$.

The steric strain between the atoms can be better visualized by the representations here.

The decomposition reaction is
$$\ce{2NI3\to N2 + 3I2}$$
Note that there is an increase in the number of moles (4 versus 2), which means that the reaction is favored by entropy, because there are more possible microstates.

More specifically, the enthalpy of formation is 154.4 kJ/mol, which is incredibly favorable for a reaction.

Reference: UC Davis ChemWiki


Will ferrocene undergo electrophilic aromatic substitution? Ferrocene has two structure 1 and 2.It is bonded to an aromatic ion cyclopentadienyl anion.Aromatic species such as benzene , cyclopentadienyl anion undergo electrophilic aromtic substitutions.

Then,will Ferrocene undergo  electrophilic aromtic substitution ?

Which conformer between 1 and 2, would preferentially undergo friedel crafts acylation reaction?

Due to the aromatic character of the cyclopentadienyl ligands, ferrocene can undergo electrophilic aromatic substitution reactions typical of aromatic compounds such as benzene. Ferrocene shows very high reactivity towards electrophiles, making it more closely comparable to phenol than to benzene, references and examples here and here.


ice tables equilibirium q I was given initial concentrations for the following reaction:

$$\ce{2SO2 + O2 <=> 2SO3}\ \ \ K_c =7.5\times 10^{-2}\\
[\ce{SO3}]= 0.05\ \mathrm{M}\\
[\ce{SO2}]= 0.125\ \mathrm{M}\\
[\ce{O2}]= 0.02\ \mathrm{M}$$

What are the final concentrations?

First I calculated $Q$, and found that $Q>K_c$ so the reverse reaction is favoured.

I created an ICE table and wrote the expression for $K_c$ to solve for $x$, but the calculator says there is no solution.

Here is the ice table:

$$\begin{array}{|c|c|c|c|} \hline
\ & [\ce{SO2}] & [\ce{O2}] & [\ce{SO3}]\\ \hline
I & 0.125 & 0.02 & 0.05 \\
C & +2x & +x & -2x \\
E & 0.125+2x & 0.02+x & 0.05-2x\\ \hline \end{array}\\$$

$$7.5 \times 10^{-2} = \dfrac{(0.05-2x)^2}{(0.02+x)(0.125+2x)^2}$$

You have done everything correctly so far. Your calculator might not be able to factor the expression as it is written. It is a cubic equation with three roots and two discontinuous points. You may be used to ICE-type equilibrium problems where you can simplify parts of the expression because $x$ is small compared to the initial concentrations. This is not the case in this problem. Factoring the cubic polynomial is not always easy.

Your mass action expression can be rearranged to the following cubic equation so long as $x \ne -0.02$ and $x \ne -0.0625$, both of which are nonsense anyway. If your calculator cannot factor this expression, try using Wolfram Alpha.

$$0.3 x^3-3.9565 x^2+0.201922 x-0.00247656 = 0$$

Trying solving this equation. It has three roots, two of which are nonsense since they yield negative final concentrations. The remaining root is the answer.


Tetrahedral or Square Planar How can one predict whether a given complex ion will be square planar or tetrahedral when its coordination number is 4 using crystal field theory ?

Is it possible to theoretically predict this?

This is a really quick and dirty method since I don't know much about the calculations involved in crystal field theory. This method is based on some general observations I've made.

First of all, if your complex metal ion is not in group 8 ($\ce{Ni}$, $\ce{Pd}$, $\ce{Pt}$) or it doesn't have a $d^8$ configuration (like $\ce{Rh(I)}$ or $\ce{Au(III)}$), it is usually tetrahedral (Assuming you already know that the coordination number is 4).

If your metal ion is in group 8 or has a $d^8$ configuration, look at the crystal field splitting diagram. Square planar complexes have a four tiered diagram (i.e. four different sets of orbitals with different energies).

If it has a two tiered crystal field splitting diagram then it is tetrahedral.

But this assumes you have the crystal field splitting diagram of the complex.

As to how you obtain these diagrams (the calculations involved), I don't know exactly how it's done for specific molecules.

Like I mentioned before, this is just a very basic way to distinguish between the two geometries. I'm sure the calculations involved to obtain these diagrams would be really complicated but would make more sense in some cases if you're looking for the actual reasons behind why some complexes are square planar while others are tetrahedral.

EDIT:

Predicting the structure of a complex depends on how much info you're given. If you only have the number of ligands in the complex or the molecular formula of the complex, then you can predict the structure based on this list here.

Then again, steric factors and electronic factors are also important in determining the structure and your prediction would be more accurate if you know what ligand(s) you're dealing with (for steric effects) and the electronic configuration and oxidation state of the central metal ion.

I don't know much about how exactly these factors influence the structure but in general, you only have to be worried about these if:

the ligands are bulky or have some other special features
certain special electronic configurations (I've read something about certain electronic configurations causing distortions in the structure especially in copper(II) complexes)


How does a molecule search work by drawing in visual editor? There are a lot of chemical search engines (such as ChemSpider's
Structure search) that allow users to draw and search for a molecule. How does it work? Is the comparison of images used? Could you give me an algorithm or pseudo code?

Usually, structure search is implemented using a graph model. For example, let's take cyclohexane. On screen this is a hexagon - but this is converted on the server to a simple atom/bond graph model of six carbons in a ring.

This graph can then be compared to all the molecules in the database by various means. For example, subgraph isomorphism or fingerprints or even SMILES string comparison.

As far as I know, ChemSpider allows a variety of searches - with/without hydrogens, substructure search, element fingerprints, and so on. Each of these is a whole question by themselves, so I'm not sure how useful an algorithm that covers them all would be.


[Co(H2O)6]3+ vs [Co(NH3)6]3+ oxidising water Cobalt(III) complexes cannot be prepared by addition of the ligand to a solution of the metal cation because the aqua cation $\ce{[Co(H2O)6]^3+}$ has no stable existence in solution. It is strongly oxidising with an E° = +1.84 volts for $\ce{[Co(H2O)6]3+ -> e— [Co(H2O)6]^2+}$. It oxidises water according to the equation:
  $$\ce{2 [Co(H2O)6]^3+ -> H2O + 2 [Co(H2O)6]^2+ + 2H+ + 1/2O2} (2)$$

  Complex formation, however, can modify the standard electrode potentials considerably (for example, E° = +0.10
  volts for $\ce{[Co(NH3)6]^3+ -> e— [Co(NH3)6]^2+}$ and therefore, oxidation from Co(II) to Co(III) is easily
  accomplished in the presence of coordinating ligands".

Why is it that $\ce{[Co(H2O)6]^3+}$ oxidise water, but $\ce{[Co(NH3)6]^3+}$ does not? I understand how values of reduction energy potential effect the proceeding of a reaction, but what is it about $\ce{[Co(H2O)6]^3+}$ that makes it able to oxidise water more?

There are 2 competing principles.

$\ce{[Co(H2O)6]^3+}$ has  strong oxidation effect, but...

$\ce{[Co(NH3)6]^3+}$ is much more stable than $\ce{[Co(NH3)6]^2+}$.

As the consequence, the former from the next 2 equations is significantly shifted to the right.

$$\begin{align}
\ce{  [Co(H2O)6]^3+ + 6 NH3 &<=>> [Co(NH3)]^3+ + 6 H2O} \\ \ce{[Co(H2O)6]^2+ + 6 NH3 &<=> [Co(NH3)]^2+ + 6 H2O}
\end{align}$$

Therefore

$$\frac{c_{\ce{[Co(H2O)]^3+}}}{ c_{\ce{[Co(H2O)]^2+}}}\lt\lt     \frac{c_{\ce{[Co(NH3)]^3+}}}{ c_{\ce{[Co(NH3)]^2+}}}$$

As both complex pairs keep the same redox potential, their standard redox potentials are therefore very different.


Gas Pressure and Molecular Mass The ideal gas equation (or the van der Waals equation, for this matter) states that holding volume and temperature constant, increasing the number of moles of particles in the container will increase the total pressure exerted by the gas mixture.

That’s all well and good, but when it comes to looking at chemical reactions and their equilibria, Le Châtelier’s Principle states that when a gas mixture is pressurized, the equilibrium shifts to the side of the reaction that produces a lesser total number of moles. This would mean less particles, and therefore less pressure, as I see from the ideal gas equation.

What I do not understand is why the mass of each molecule does not affect the pressure exerted by the gas. If you have heavier molecules, they would each have more momentum, and would exert more force on the container’s walls—i.e. more pressure. So is this not a factor when considering Le Châtelier’s Principle? The equilibrium shifts toward the formation of more particles, but they are each lighter and exert less force. Why is this favorable over having less but heavier particles?

The Maxwell–Boltzmann distribution is based on the principle that the distribution of kinetic energies of gas molecules is the same. So in a mixture of $\ce{N2}$, $\ce{N2}$, and $\ce{He}$, the molecules of each gas would all have the same KE distribution. However since He is much lighter, its molecules would have higher velocities.

Normally we'd think of the gas mixture as being homogeneous in a liter bottle in the laboratory. However if we consider the atmosphere then gravity has an effect. So helium will drift upwards in the atmosphere since it is less dense. In fact at the edges of the outer atmosphere a helium atom can get enough velocity to escape earth's gravity. So all helium that leaks into earth's atmosphere is ultimately lost into space.


Aldolisation reaction and chalcone formation I realized a reaction of mixed aldolisation between acetone and 2-nitrobenzaldehyde, using a solution of 10% NaOH in ethanol.

I obtained a mixture of products at the end of the reaction. The problem is when I added the ethyl acetate I saw a blue liquid mixed with brown solid.  After a night of this reaction I found a lot of white crystals mixed with the blue liquid.

Can someone explain the presence of this blue liquid with the brown solid, knowing that the crystals are soluble in ethanol?

Well, actually you've made a reaction called the "Baeyer–Drewson indigo synthesis" (original article here), which was route developed in the nineteen century to obtain the indigo dye, nowadays replaced by other methods. I'm not writing the mechanism for this reaction because it is quite long and well-documented elsewhere.

Your blue compound is therefore the indigo dye represented here. As for the brown compound, it's difficult to say without more data (NMR spectra for instance), but the synthesis can give many colored intermediates or by-products such as isatin (bright red).

I hope this helps.


Are there chemicals that absorb and release upon heating oxygen more efficiently than potassium oxide? $\ce{K2O}$ absorbs oxygen forming $\ce{K2O2},$ which decomposes to $\ce{K2O}$ and $\ce{O2}$ at 500 °C. This means that $\ce{K2O}$ can absorb up to 17% its mass of oxygen. Are there other substances that exceed this ratio and release oxygen below 1000 °C?

If you are looking for something that can be made using elemental  oxygen as an oxidizer and reversibly release the excess oxygen below 1000 °C, sodium peroxide beats potassium peroxide by using lighter metal atoms.  It can be prepared by reacting sodium with oxygen in two stages (forming sodium oxide in the intermediate stage) at 130–200 °C and releases the excess oxygen at 657 °C (Wikipedia).

Lithium, magnesium and of course hydrogen form peroxides with even higher excess oxygen percentages, but these are not made via reversible reactions with elemental oxygen.


Would a melted crystal recrystallize back into the same lattice structure when frozen? Two things that I'm not sure about melted crystals:

For a material that has only one possible crystal structure, would freezing cause it to recrystallize, or could it become amorphous upon freezing?
What would happen upon freezing to a material with more than one possible crystal structure?

If cooling is rapid enough, then crystals don't have time to form, so the substance may be amorphous or glassy (though there may be some localized structure). For example, there are metallic glasses and amorphous water ice. Metallic glasses may have improved wear and corrosion resistance over a crystalline alloy. One interesting intermediate form is a quasicrysal, which can have 5-fold symmetry. Roger Penrose proposed this geometrical tiling before it was found in metallic glasses.
It depends on the conditions in which the material freezes. For example, graphite can be forced into a diamond structure by heat and pressure. A pseudomorph could form if the substrate initializes crystallization in a different form.


Why are the ozonides of heavier elements more stable? The ozonides of $\ce{Cs}$, $\ce{Rb}$, $\ce{K}$ are well known and relatively stable, but there is little mention of the ozonides of $\ce{Na}$, $\ce{Li}$. Why is this?

Potassium and heavier alkali metals ozonides are formed by treating with ozone, or by treating the alkali metal hydroxide with ozone. They are very sensitive explosives that have to be handled at low temperatures in an atmosphere consisting of an inert gas.

Lithium and sodium ozonide are extremely unstable and must be prepared by low-temperature ion exchange starting from $\ce{CsO3}$. Sodium ozonide, $\ce{NaO3}$, which is prone to decomposition into $\ce{NaOH}$ and $\ce{NaO2}$, was previously thought to be impossible to obtain in pure form. However, with the help of cryptands and methylamine, pure $\ce{NaO3}$ may be obtained as red crystals isostructural to $\ce{NaNO2}$.

As pointed out by user2617804, lithium and sodium ozonide are extremely unstable because the ozonide ion is big for the tiny sodium and lithium ions so it is stretched around it due the distributed charge on the ozonide ion- puts it under large internal stresses.

See wikipedia for more details.


How can I calculate the percentage of dissociated dinitrogen tetroxide? When $\pu{36.8 g}$ $\ce{N2O4(g)}$ is introduced into a 1.0-litre flask at $\pu{27 °C}$ the following equilibrium reaction occurs: $\ce{N2O4(g) <=> 2NO2(g)}$; $K_p = 0.1642~\mathrm{atm}$.
Find the percentage dissociation.

I tried the following:
Let $x$ be the degree of dissociation and $P$ be the initial pressure of $\ce{N2O4}$  that is $\approx 9\ \mathrm{atm}$.
Then, $$\frac{9Px^2}{1-x} = 0.1642$$
Neglecting $x$ in $(1-x)$, I got $x$ as $\displaystyle \frac{0.4}{3}$ [approx]
But the answer given as $6.44~\%$.

$\ce{N2O4 (g)}$ say initially is $P\ \mathrm{atm}$.

$$PV = nRT$$

$$P(1)= (36.8/92)( 0.082 )(300 K)$$

$$P= 9.84\ \mathrm{atm}$$

Now we have:

$$\ce{N2O4(g) <=>2NO2(g)}$$

Considering initial pressure of N2O4 as p, at equilibrium, assuming 'x' as the degree of dissociation, we will have pressure of N2O4 to be p(1-x) and pressure of NO2 as 2px.

So, $$K_p = \frac{[\ce{NO2}]^2}{[\ce{N2O4}]}$$

$$K_p= \frac{(2px)^2}{p(1-x)}$$

$$K_p= \frac{4px^2}{(1-x)}$$

assuming $1-x$ approximately as $1$,

$$K_p= 4px^2$$

substituting $p$ as $9.84\ \mathrm{atm}$ and $K_p$ as $0.1642$:

$$0.1642= 4(9.84)x^2$$

$$x^2= \frac{0.1642}{4(9.84)}$$

$$x^2= 0.1642/39.36$$

$$x^2= 0.00417$$

$$x= 0.0645$$

degree of dissociation is 0.0645

Hence % of dissociation is $0.0645 \times (100) = 6.45\ \%$ (near to your answer)


Does the addition of water to N2H4 produce N2H4 I am currently calculating the $\ce{pH}$ of a species.

The reaction is as follows,

$$\ce{N2H4 + H2O <=> N2H_{$A$}+ + OH-}$$

However, I'm not sure what the value of $A$ will be. In my textbook it says $4$, which is stumping me as in every other example with a weak base the subscript increases.

What is the value of $A$ and why?

Yes, equations must be balanced, thus the only correct way to write this is:

$$\ce{N2H4 + H2O <=> N2H5+ + OH-}$$

Easy to remember: Nitrogen typically has a free electron pair which can act as a Lewis base for a proton.

If your textbook says $4$, it is an error in your book. Good job on spotting it and realising it.


Why isn't sodium hydrogen phthalate used instead of KHP? Why is potassium hydrogen phthalate (KHP) ubiquitously synthesized and used as a primary standard, while sodium hydrogen phthalate isn't even available to buy?

For classical analytical problems you have to check at least 100 year old literature. It is easy in Google Scholar. Set the date limits to 1920s or 1950s whatever you wish to try.  In older times, the conventional wisdom, which still persists, was that one should use a higher molecular weight standard. The rationale was that the relative error will be small on a balance. KHP has a higher molecular weight than NaHP. This is one advantage. Second, advantage is that it available in extremely high purity. It can be dried as well.
Both KHP and NaHP work perfectly. There is a caveat with NaHP, that it crystallizes as a hemihydrate i.e., for two NaHP units, there is a water molecule. KHP crystallizes as a normal anhydrous salt. Analytical chemists run away from hydrated salts because you would never know what is the hydration level, what was the humidity level at which they were stored. We would never know the "true" molecular weight of the substance which is being weighed out. There are salts such as ferrous ammonium sulfate with water crystallization, but it is very stable with respect to storage. NaHP, who knows how it loses its water of hydration.
Search "Acid potassium and acid sodium phthalates as standards in acidimetry and alkalimetry" on Google Scholar. The reference is Journal of the American Chemical Society 37, 10 (1915): 2352-2359.


Why is casein hydrophobic? According to this source hydrophobic behaviour occurs when a molecule does not have a charge or is neutral and therefore cannot be attracted to the negative nor positive parts of water. Casein has an isoelectric point of 4.6 and milk has a pH of about 6.7. As a result, the casein proteins would have a negative charge, hence wouldn't the casein be attracted to the positive part of water?

In milk, casein exists as a colloidal aggregate (casein micelle) with the properties you state, but purified casein is different. It is rich in proline (shown below), has no disulfide bridges, and hence little tertiary structure to hide the hydrophobic residues in the core of the protein with the hydrophilic residues on the surface to bind to water. In addition to having a hydrophobic side chain, proline stiffens the aminoacid backbone, hindering tertiary structure. The net effect is a molecule with too much exposed hydrophobicity to be water soluble. Casein micelles are complex structures and quite different from lipid micelles.

PROLINE


Dangerous exothermic reactions at home Exothermic reactions are essentially reactions that release energy as heat (in part, at least) to the surroundings, and may cause an explosion or combustion, depending on the amount of heat released and the specific conditions (see this story as an example of an unexpected exothermic reaction at home).

We mix up all kinds of stuff at home, mostly in the kitchen, and each "ingredient" may be composed of numerous distinct chemical compounds. This huge number of possible combinations might make it probable that a redox reaction, which may be exothermic and there's no telling how much exothermic, will happen from time to time when mixing two (or more) foodstuffs together. However, it is pretty rare to hear about significant exothermic reactions that may be called "a household accident", such as the one I brought earlier.

My question is: how is this so? Is there any energy threshold that we do not normally cross in the kitchen? Or are the concentrations not large enough? How come kitchen chemistry is not more violent than it actually is? What makes "experimental" food chemistry safe for billions of people without chemistry knowledge?

Life on Earth happens in pretty tight chemical limits, except for some exotic organisms we rarely encounter.

So, on the human side, the stuff we want to eat is not really very reactive. Carbohydrates, fats and proteins will give you no great fireworks no matter how you mix the three of them. The trace elements we need would be capable of some spectacular reactions, but we don't eat them in their pure form.

Which brings me to the "what is available" side. for almost all of human history, humans ate what they found in nature. And the stuff that reacts violently when mixed with organic matter is not really lying around to be used as an ingredient. Sure, some natural events are capable of producing pure oxygen, fluorine, or whatever, but they don't lie around for a long time waiting to be harvested, they react away. And plants and animals (=our food) tend to evolve in a way such that their bodies use the nitrogen and sulfur to build amino acids, not TNT and sulfuric acid.

There are a few substances in the kitchen which can give you a bit of a bang when mixed. If you want to impress a second grader, such classics as baking soda + vinegar, or mentos + cola will do nicely. But stuff which tends to pack more energy, and likes to free it in a flash, is not really food. You will notice that you need advanced processing methods to get all these ingredients - even though some of them (vinegar fermentation) were known in antique times, they are not something that grows on trees.

You can certainly find some edge cases - I keep pure NaOH on hand for pretzels, it is also used for some styles of noodle - but they are rare and concern ingredients which are not willy-nilly mixed into random dishes.


Is there a quantitative test/analysis technique that can determine if some chloroform has degraded into phosgene? Before I go to use any chemical that I've had in storage for a while, I try to check if it's degraded into something else (e.g.: checking the $\ce{H2O}$ content of $\ce{H2O2}$, checking for $\ce{H2O2}$ in $\ce{Et2O}$, etc.). This can usually be done using the appropriate test strip.

I have a 3/4 full bottle of chloroform that's been sitting on a shelf for about a year (probably less, actually), in an amber bottle to restrict the amount of light that can get to the chloroform. I'm aware that chloroform can break down into phosgene, and I know there are some ways to test for this, but I was hoping for a quick quantitative test that can be done without any hassle, but I can't seem to find anything out there. The Wikipedia page for chloroform states:

  Suspected samples can be tested for phosgene using filter paper (treated with 5% diphenylamine, 5% dimethylaminobenzaldehyde in ethanol, and then dried), which turns yellow in phosgene vapor.

I don't have either of the diphenylamine or the dimethylaminobenzaldehyde, and would prefer not to have to purchase them just for this simple test (I don't have any use for either chemical currently).
Does anyone know if there is a simple quantitative test that accomplishes the same thing? Preferably a test strip.

Not a quantitative test but a good proof of phosgene. You could use a pH paper strip. Chloroform should show a neutral pH, but if there is phosgene/HCl the paper will show acidity.


How is graphite a covalent solid? I know it is covalent because there are covalent bonds between atoms of carbon in the same plane. But there are Van der Waals interactions between different planes. Isn't this a feature of molecular solids?

Graphite is odd in its categorisation and does by no means rigorously follow the conventions of either a covalent solid, or a molecular solid (for example it can conduct electricity). Graphite has properties of both molecular solids (it is soft), and covalent solids (it has a very high melting point. In fact it will break down before it melts.).

For this reason, the categorisation of some materials, such as graphite, can be somewhat complicated, and the idea of graphite being a covalent solid really just comes from its similarity with other covalent solids.


Dependence of change in entropy on temperature The thermodynamic definition of entropy is expressed as
$$dS = \frac{dq_{rev}}{T}$$
I understand that it depends on amount of heat transferred because heat transfer can be understood as the degree of dispersion of energy but what does the temperature dependence actually mean?

An analogy that I like is one paraphrased from Peter Atkins' great book "The Laws of Thermodynamics - a VSI":

  Imagine two rather different places in terms of noise - a silent
  library and a busy train station. Imagine that the temperature is the amount of noise at the given place - a measure of the "state of noise". Let's say that you laugh out loud,
  corresponding to supplying energy as heat.

  In the library, where there is little background noise (low $T$) this would cause a large amount of disturbance and
  disorder - corresponding to a large increase in entropy. On the busy
  train station, on the other hand, you laughing would go largely
  unnoticed - there is lots of background noise already (high $T$). The (relative) amount of additional disturbance and
  disorder would be much smaller, corresponding to a smaller increase of
  entropy.

In a very cold system, even a tiny amount of supplied energy in terms of heat would cause a large change in entropy. On the other hand, supplying the same amount of energy to a really hot system - with lots of molecular motion going on - would cause a minuscule change in entropy.

Disclaimer: Note that the analogy and explanation above are two ways of convincing yourself that the entropy change due to added heat is inversely proportional to the system temperature. They do not claim to be correct or physical in any other sense, so read them with a (large) pinch of salt.


Comparing the Lewis acidity of carbon tetrafluoride and silicon tetrafluoride I've been asked to compare the relative acidity of $\ce{CF4}$ and $\ce{SiF4}$. As both atoms have noble gas configurations, neither can act as a lewis acid and so I assume the question is inferring that they are in solution.

My guess is that under aqueous conditions they could react via some sort of hydrolysis to give $\ce{H+}$ and $\ce{F-}$ alongside an alcohol in the case of carbon or $\ce{Si(OH)_{(4-x)}F_{x}}$ with silicon.

As the energy match between the $\ce{Si}$ and $\ce{F}$ HOMO and LUMO is worse this means that this is likely to be a weaker bond and so $\ce{SiF4}$ is more acidic? Is there an argument that could be made about the relative stability of the hydrolysis products?

Both molecules can theoretically act as a Lewis acid. This would involve the approach of a Lewis base (i.e. a lone pair-carrying group), its interaction with the central atom and the formation of a four-electron-three-centre bond. These bonds with a bond order of $0.5$ can be understood by invoking the following two resonance structures:

$$\ce{\overset{-}{X}\bond{...}AF3-F <-> X-AF3\bond{...}\overset{-}{F}}$$

Where $\ce{A}$ is carbon or silicon in the context of this question. The structure that this arrangement would require is a trigonal bipyramid; the central atom is $\mathrm{sp^2}$ hybridised with its unhybridised p orbital forming the 4e3c bond to the electronegative groups.

You may immediately recognise this as the transition state of an $\mathrm{S_N2}$ reaction which is possible both with carbon and silicon centres.

These four-electron-three-centre bonds are typically more stable if the central atom has a lower and the terminal atoms have a higher electronegativity. If you compare interhalogen compounds, only iodine is able to form such bonds with chlorine ($\ce{(ICl3)2}$) while all other halogens can only form these bonds with fluorine (e.g. $\ce{ClF3}$). This already should allow us to conclude that $\ce{SiF4}$ is the stronger Lewis acid. But the discussion is not finished here.

We also need to discuss steric effects. Carbon tetrahalides are very crowded molecules — even with fluorine even though it is the smallest halogen. Giving up the tetrahedral structure of $\ce{CF4}$ to arrive at trigonal bipyramidal $\ce{CF4X-}$ would require the three fluorines that do not take part in the 4e3c bond to move closer together, increasing steric crowing even more. Indeed, the required interaction would be the Lewis base’s lone pair with $\sigma_{\ce{C-F}}^*$ — an orbital which is already well hidden by the three other fluorines. Thus, steric effects further reduce $\ce{CF4}$’s ability to be a Lewis acid to the point where we shouldn’t even be saying it be a weak Lewis acid: we should just call it Lewis inactive.

Silicon is larger than carbon and better able to stabilise 4e3c bonds due to its lower electronegativity. Thus, 4e3c bonds are sufficiently accessable and stable for silicon in $\ce{SiF4}$ to act as a Lewis acid. Depending on the nature of $\ce{X-}$, the pentacoordinated $\ce{SiF4X-}$ species may either break down under liberation of either $\ce{F-}$ ($\mathrm{S_NSi}$ substitution) or $\ce{X-}$ (reverse reaction) or — especially if $\ce{X} = \ce{F}$ — it may add another to give $\ce{[SiF4X2]^2-}$ like the well-known hexafluoridosilicate anion $\ce{[SiF6]^2-}$.


Fragmentation involving formation of enamine I have a problem on fragmentation which requires mechanisms involving formation of an enamine from a cyclic ketone and cyclic secondary amine. This then reacts with a cyclic aldehyde via conjugate addition, but the intermediate cannot be reached.

I've been told that I have to exchange the enamine of the ketone with the enamine of the aldehyde. What does this mean?

  PROBLEM 17

  Give mechanisms for these reactions, commenting on the fragmentation.

  Purpose of the problem
  Revision of conjugate addition of enols, another ring expansion with an enolate as leaving group and an interesting piece of stereochemistry.

  Suggested solution
  The first step is enamine formation and the second is conjugate addition. This appears to lead to a dead end as we cannot find a way to make the intermediate from the product.

@Laura: Both Waylander and Zhe are correct in their respective comments. Your example is from the work of Hendrickson and Boeckman.1 Enamine 1 and conjugated aldehyde 2 were mixed neat forming bridged aminoketone 3. At some point, as Zhe has intimated, exchange of pyrrolidine from the ketone to the aldehyde must occur before cyclization to 3 can be accomplished. The usual explanation is that traces of water are present to catalyze the reaction. Quaternization of the amino group in 3 affords the salt 4 which cleaves with hydroxide via a Grob-like fragmentation to 5. Note the important anti arrangement of the dotted bond and the proximate vicinal bold bond. The product contains a double bond as Waylander has stated. The first example of this reaction, 6 --> 9, was reported by Stork and Landesman2 in 1956.

1) J. B. Hendrickson and R. K. Boeckman, Jr., J. Am. Chem. Soc., 1971, 93, 1307.
2) G. Stork and H. K. Landesman, J. Am. Chem. Soc., 1956, 78, 5129.


Basis for the hydrophobic effect? I'm confused about why hydrophobic molecules, which do not have high polarity, would have a tendency to attract and cluster with themselves. It is easier to understand the hydrophilic as long as one believes that positive charges attract negative charges. Anyone have any insights?

The driving force behind the hydrophobic effect is entropy. Suppose I put several hydrophobic molecules in water. These molecules have two choices. (1) The molecules can stay separate from each other and randomly move around in solution. (2) They can group together and form a cluster. It turns out that it is entropically favorable to cluster. The reason is that it takes more water to individually solvate the separate molecules. It takes less water to solvate a cluster of the molecules. When water solvates these molecules, it is ordering itself. In fact, it is believed that water forms cages around hydrophobic molecules in solution. Therefore, if the molecules cluster, less water will have to order itself. Less ordering means greater entropy (randomness).

It also true that London dispersion forces will occur in these clusters. However, this is not the driving force for the hydrophobic effect. The driving force is entropy.


Does it make sense to differentiate the Arrhenius equation with respect to temperature? I am trying to answer the question: "between concentration and temperature, which has a greater significance on the rate of reaction"
Hence, I'm trying to find a method to systematically compare the impact/significance of concentration and temperature on rate. This is where I ended up trying to differentiate the Arrhenius equation:
$$k = A\mathrm e^{-E_\mathrm a/(RT)}$$
I thought that deriving rate constant $k$ with respect to temperature is finding the rate of change of $k$ with respect to temperature. Is it conventionally and mathematically correct to do this?
If it is, then continuing the thought process:
I differentiated the equation assuming that the Arrhenius constant...is...constant... But is that true? Does $A$ change when temperature change (since my variable here is temperature)?
If everything was not non-sense until this point, I ended up with the derivative:
$$\frac{\mathrm dk}{\mathrm dT}=A\mathrm e^{-E_\mathrm a/(RT)}\times E_\mathrm a/(RT^2)$$
Is this correct chemistry/maths?
EDIT
Is there a better way to compare the significance of concentration with that of temperature on the rate of reaction? Maybe using percentage increase of each factor with the percentage increase of rate?

Note that $A$ is a kind of a collision frequency factor, which is temperature dependent via the mean molecule speed, even if far less then the Boltzmann exponential factor.
$$\mathrm{d}k/\mathrm{d}T = \frac {\mathrm{d}A}{\mathrm{d}T} \cdot \exp{(\frac {-E_\mathrm{a}}{RT})} + A \cdot \exp{(\frac{-E_\mathrm{a}}{RT})} \cdot \frac {E_\mathrm{a}}{RT^2} \tag{1}$$
$$A = A_0 \cdot {({\frac {T}{T_0})}}^{1/2} \tag{2}$$
$$\frac {\mathrm{d}A}{\mathrm{d}T} = A_0 \cdot \frac 12 \cdot {({\frac {T}{T_0})}}^{-1/2} \cdot \frac {1}{T_0}=\frac {A_0}2 \cdot {({\frac {T_0}{T})}}^{1/2} \cdot \frac {1}{T_0} =\frac {A_0}2 \cdot \frac 1{\sqrt{T \cdot T_0}}  \tag{3}$$
By subtitution in (1), using  (2) and (3):
$$\mathrm{d}k/\mathrm{d}T = \left( \frac 12 \cdot \frac 1{\sqrt{T \cdot T_0}}   +  {({\frac {T}{T_0})}}^{1/2}  \cdot \frac {E_\mathrm{a}}{RT^2} \right) \cdot A_0 \cdot \exp{(\frac{-E_\mathrm{a}}{RT})}\tag{4}$$
As AJKOER properly noted, temperature dependance  $A=A(T)$ can be usually neglected and we can use the simplified (1), as $\frac {\mathrm{d}A}{\mathrm{d}T} = 0$:
$$\mathrm{d}k/\mathrm{d}T = A \cdot \exp{(\frac{-E_\mathrm{a}}{RT})} \cdot \frac {E_\mathrm{a}}{RT^2} \tag{1a}$$
unless the activation energy is extremely low and reaction rate is driven by diffusion ( typical one is $\ce{H3O+(aq) + OH-(aq) -> 2 H2O(l)}$  ).
In solvents, we cannot usually ignore intermolecular interactions, so temperature dependence of $A$ may get quite complicated.


Write a Net Ionic Equation to Represent the Reaction of Aqueous Ammonia with Propionic acid I would like to know how do I begin to solve a problem like this?

That is finding the formula equation?

I would assume that it would be
$$ \ce{NH3 + CH3CH2COOH -> NH3 + CH3CH2COOH + H2O} $$

I am really confused as to how I can balance this equation. Do I give the ammonia a hydrogen? Do I break up the acid into smaller units?

From there I would like to know the best way for me to switch the equation into an ionic equation and then finally a net ionic equation.

By the way the net ionic equation is
$$ \ce{CaCO3 (s) + 2HC2H3O2+ (aq) -> CO2 (g) + H2O (l) + Ca^2+ (aq) + 2C2H3O2^- (aq) } $$

I am just clueless as to how they got there.

We know that ammonia is $\ce{NH3}$ and propanoic acid is $\ce{CH3CH2COOH}$.

Ammonia is a basic substance, and propanoic acid is... an acidic substance. More specifically, ammonia fits the definition of a Bronsted-Lowry base, so it will act as an $\ce{H+}$ accepter in solution.

Arrhenius acids are expected to increase the concentration of $\ce{H+}$ in solution, and propanoic acid is no exception. In water as propanoic acid dissociates...

$\ce{CH3CH2COOH ->[H2O] CH3CH2COO- + H+}$

This is where the $\ce{H+}$ comes (and is later bonded to the ammonia via)...

$\ce{NH3 + H+ -> NH4+}$

After those two steps occur, the products $\ce{NH4+ + CH3CH2COO-}$ are formed. Your ionic equation is as follows...

$\ce{NH3 + CH3CH2COOH -> NH4+ + CH3CH2COO-}$

There appear to be no spectator ions (ions that appear on both sides of the equation), so there is nothing to cancel out. As such, this is ALSO the net ionic equation. They both appear to be balanced.

So, yes, you had to do both. You were supposed to add an $\ce{H+}$ to the ammonia, but that happens if you dissociate the acid.

As for the equation with the calcium, carbonate ions sometimes create carbon dioxide and water (via carbonic acid readily decomposing, like in sodas).


What are the triple point pressure and temperature for pure cyclohexane? Here is a video that demonstrates the interesting nature of the triple point of cyclohexane, so I tried researching on the internet for a phase diagram, but only found data from a manufacture where the triple point was specified as 279.48 degK at 5.388 kPa. But in the video there doesn't seem to be a cooling plate. The flask appears to be at room temperature.

What are the triple point specs for pure cyclohexane? If they match the data above, how were they able to achieve the phase transitions shown in the video at room temperature?

Also where can I find a phase diagram for pure cyclohexane on the internet?

The NIST REFPROP (Reference Fluid Thermodynamic and Transport Properties) program reports the triple-point temperature of cyclohexane as 279.47 K.
The program refers to: S. G. Penoncello, R. T. Jacobsen, and A. R. H. Goodwin: “A Thermodynamic Property Formulation for Cyclohexane”, International Journal of Thermophysics, Vol. 16., No. 2, 1995, 515–531.
This publication mentions the triple-point temperature as T = 279.47 K and the triple-point pressure as p = 5.388 kPa.
The given reference is: J. G. Aston, G. J. Szasz, and H. L. Fink, J. Am. Chem. Soc. 65  1135 (1943).

The following p-T diagram (made using REFPROP) shows the melting line (solid-liquid equilibrium, blue) and the vaporization line (liquid-gas equilibrium, red), which meet at the solid-liquid-gas triple point. Unfortunately, REFPROP does not include the sublimation line for cyclohexane.


Why does the d orbital size decreases on addition of electrons? In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), on page 80 under section 3.7 "The Extent of d-orbital Participation in Molecular Bonding" it is given:

  A second factor affecting the size of d orbitals is the number of d orbitals occupied by electrons. If only one 3d orbital is occupied on an S atom, the average radial distance is $2.46$Å, but when two 3d orbitals are occupied the distance drops to $1.60$Å.

I wish to know the reason why the size of the d orbital drops when additional electrons are added. I think this is counter-intuitive since on addition of an extra electron the interelectronic repulsions increases and confining them to a small volume increases overall energy of the system thereby making it unstable. So, I think the orbital size must increase, but it is happening the opposite way.

Kindly explain the above-mentioned concept.

There is not much to explain because the concept is wrong. There is no sulphur compounds in which the 3d orbitals (remember: they have an energy level similar to 4s) are involved in bonding in any noteworthy way, shape or form.

From what I believe I may remember about the outdated concept, ‘one d orbital’ refers to cases like $\ce{SO3^2-}$ with sulphur in the $\mathrm{+IV}$ oxidation state while ‘two d orbitals’ would refer to e.g. $\ce{SO4^2-}$ or an oxidation state of $\mathrm{+VI}$. In the Lewis formalism, the answer is easily deduced if you draw octet-abiding structures: $\ce{SO3^2-}$ will include a formal single positive charge on sulphur while in $\ce{SO4^2-}$ sulphur has a formal double positive charge. The higher an oxidation state, the more contacted an atom becomes.

A more correct answer would elaborate on ($\mathrm{sp}^n$-type, formed only by sulphur’s s and p atomic orbitals) molecular orbitals and the increasing number of electronegative substituents that need to bond to the central sulphur, reducing its charge density and leading to the same effect.