How to solve this dimensional analysis/stoichometry problem This problem has been giving me a headache for the last hour on my homework. Does anyone have any pointers in the right direction?

  Finally, on January 28th, 2014, in Germany, a barn containing 90 cows exploded when a stray spark caught the methane trapped in the building on fire. Assume that the 90 cows were producing 250 L of methane per day and that there was a 15 hour collection in the barn when it ignited. What mass of water was generated in the explosion?

This problem is specifically about combustion, and I'm assuming I'd have to multiply the combustion equation for methane by 90 somehow. I'm also assuming I'd have to divide 250 / 15 to get the correct liters of methane produced at this time.

Figured it out, thanks to @M. Farooq's help.
I first did some cross multiplication of 250/24 times X/15 to find out that in 15 hours, there would be 156.25 L of methane. This is equal to 86.56 grams of methane. Using that, I found out there were 5.393 moles of methane, and multiplying that by 2 (because for every one methane it makes 2 waters) I found that there would be 10.79 mol of water when the reaction was done. This, times the molar mass of water, gives me 194.4 grams of water after the explosion.


What software is used to generate the PDB molecule of the month images? What software does the PDB use to generate these "cel-shaded" protein models with outlines around foreground atoms?

(Leptin: May 2012 Molecule of the Month by David Goodsell)

These illustrations are by David Goodsell, and carry his famous look. Whilst I was not able to determine what specific software he uses, this page suggests that he probably creates his illustrations as a composite of renders and that the outlines and depth cueing are achieved by running a 'find edges' filter over the z-buffer of the scene, which can be output from many renderers.

VMD is a fairly powerful program (that F'x has contributed patches to!) capable of achieving this cel-shaded look with minimal postprocessing by using a GLSL Fresnel shader.

As F'x mentioned, QuteMol can do this style as well, however QuteMol has not been actively developed since about 2007 and is not terribly feature-rich, which is a shame as it's one of very few molecule viewers with competent ambient occlusion.

How to do it with VMD and Blender

In the first instance, I tried using the GLSL shader Goodsell style in VMD on the PDB file 1RON, a neuropeptide. I was not satisfied with the results, but this would probably look decent on a protein SES as illustrated on the aforementioned VMD site.

Note that I have used 2 VMD representations to separate out the style of the backbone and the sidechains, minus the hydrogens. I'm not going for a 1-to-1 clone of the referred style, just something thematically similar.

I have shrunk this VMD output down because as F'x pointed out in chat, GLSL shaded spheres are not antialiased. This said, if you are very tricky, you might be able to force antialiasing using a shader method like FXAA, which can be enabled in arbitrary programs in the very latest nVidia beta drivers (I know...). Alternatively, you can do what I did and manually oversample :D.

Next step - Blender

I output the VdW spheres as two separate VRML files, one for the backbone, one for the sidechains. These can be imported into Blender. I'm going to assume proficiency with Blender, however if you are not familiar, Blender is fairly easy to learn, especially versions 2.5 and later.

I've combined the VdW sphere meshes of each component into a backbone and sidechain object, assigned materials for each (just changing the colours of the default diffuse shader) and have applied one level of Catmull-Clarke subdivision to the meshes, to make the spheres more spherical than the default.

I've also removed the light source and turned on environmental lighting, which lights the scene isotropically and will generate some nice ambient occlusion while we're at it.

I now render this at high resolution, obtaining a standard render and the Z-buffer for the scene. I've scaled them down here. Note that if you want the spheres to be a flat colour, you should set the material to 'shadeless'. I like the subtly shaded look better.

Oh yeah, that looks nice. And here's the Z-buffer. Regrettably, the Z-buffer in blender has no antialiasing which is why I've rendered things huge and scaled down in Photoshop.

I now run the Glowing Edges filter on the Z-buffer and threshold to remove faint spurious edges inside the spheres.

Invert and level, set blendmode to multiply and you're done.

Not perfect, but not bad.

Bonus

Here's the shadeless version, which is closer to Goodsell's ideal.

The Journey Continues!

Poking around with the Toon shaders in Blender, I have found a super easy way to make Goodsell-esque figures!

If you configure your material like so:

You have a shader that will rapidly shade to black at angles far from normal with a lightsource. To make this work, you should disable environment lighting and position a bright point light behind the camera.

You will note that I have adjusted the Toon size to 0.98, which will create a shadow just shy of perpendicular to the light source normal. You will also note that I have set the specularity to zero, which gives an approximately flat finish to the sphere.

Render it up and you will get something like this:


Which is more stable in gas phase, free radicals or ions? Answering some kinetics questions I came across this statement:

  Free radicals are more stable in the gas phase than ions.

Is this correct or not? Why? Is there a different trend in liquid or solid phase?

It is difficult to comment on such a broad statement because often it depends on the specifics. So instead, I will talk about a single example: Will the HCl molecule dissociate into radicals (atoms, in this case) or ions?

$$\tag{1}\ce{H-Cl -> H+ + Cl-}$$

$$\tag{2}\ce{H-Cl -> H. + Cl.}$$

Under many circumstances, neither (1) nor (2) happens, H-Cl is the major species.

In water, where the dipole of water solvates chloride ions and the hydrogen ion "reacts" with water to form hydronium ions and large structures, HCl ionizes completely (strong acid). In other solvents, it ionizes less but would also act as acid if there is an appropriate base.

$$\ce{HCl(aq) + H2O(l) -> H3O+(aq) + Cl-(aq)}$$

In the gas phase, ions of opposite charge have a strong long-range attraction, so reaction (1) is unlikely. Reaction (2) is endothermic (bond dissociation energy is defined as homolytic cleavage into radicals, and it has a positive value). If you raise the temperature, you favor the products in reaction (2), so at higher temperatures, the concentration of radicals would increase.

To directly compare ions and radicals in the gas phase, you would consider the following reaction:

$$\tag{3}\ce{H. (g) + Cl. (g) <=> H+(g) + Cl-(g)}$$

Taking away an electron from hydrogen corresponds to the ionization energy of hydrogen atoms (1313. kJ/mol, i.e. it takes energy to remove the electron from the electrostatic field of the nucleus). Adding an electron to the chlorine atom corresponds to the electron affinity of chlorine atoms (-359 kJ/mol, i.e. energy is released when an additional electron encounters a neutral chlorine). Overall, the process is endothermic. Different from reactions (1) or (2), however, the number of reactant particles is equal to the number of product particles, so in this case, it is not clear if the change in entropy is positive or negative, but it won't dominate the reaction at modest temperature. So at room temperature, this reaction will not happen, the radicals are favored over the ions.

  What is stabler in gas phase, free radicals or ions?

For the example I chose, free radicals are more stable than ions.

  Is there a different trend in liquid or solid phase?

For the example I chose, it is the opposite in aqueous solution.


Determine the bond angle in a compound How can we find the bond angle between 3 atoms in a compound?
Please specify an equation which can be used for all compounds.

In carbon compounds Coulson's Theorem can be used to relate bond angles to the hybridization indices of the bonds involved.

$$1+\lambda_{i} \lambda_{j} \cos(\theta_{ij})=0$$

where $\ce{\lambda_{i}}$ represents the hybridization index of the $\ce{C-i}$ bond (the hybridization index is the square root of the bond hybridization) and $\ce{\theta_{ij}}$ represents the $\ce{i-C-j}$ bond angle.

Let's look at some examples.

In the case of methane, each $\ce{C-H}$ bond is $\ce{sp^3}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt3$. Using Coulson's theorem we find that the $\ce{H-C-H}$ bond angle is 109.5 degrees.
In the case of the methyl carbocation ($\ce{CH3^{+}}$) the $\ce{C-H}$ bonds are $\ce{sp^2}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt2$.  In this case, using Coulson's theorem we find that the $\ce{H-C-H}$ bond angle is 120 degrees.
Finally, in the case of acetylene the $\ce{C-H}$ bond is $\ce{sp}$ hybridized and the hybridization index of each $\ce{C-H}$ bond is $\sqrt1$.  In this case, using Coulson's theorem we find that the $\ce{C-C-H}$ bond angle is 180 degrees.

Conversely, if the bond angle is known, the hybridization can be determined.


What if kinetic energy of ejected electron = 0 When hf ≥ work function, Then the electron still comes out. So, if I say kinetic energy of ejected electron = 0, it should still come out. Right ?
Then, how does the electron even move out or gets ejected if its $v=0$?

Rather than debating whether an electron with no kinetic energy is ejected consider how an experiment to measure the photoelectron effect might be conducted. A photosensitive electron emitter is irradiated and the kinetic energy of ejected electrons is measured. One way to do this experiment is described in MIT Physics Department lab course notes. The setup is described in the following schematic:

A photodiode is irradiated with the filtered output of a mercury lamp. Only one of the available narrow bands of the lamp's spectrum is selected by the filter. The light strikes the K coated surface of the cathode of the photodiode. Emitted electrons are collected at a ring-shaped anode separated from the cathode by a gap, setting up a small current. An adjustable retarding voltage can be applied between the cathode and the anode to suppress the current, and the minimal cutoff voltage V which completely suppresses the current is measured. This voltage is theoretically related to the work function $\phi$ of the material and to the energy $h\nu$ of the incident light as
$$V = \frac{h}{e}\nu - \frac{1}{e} \phi$$
Now note perhaps the most important point with regard to your question: only a limited number of light frequencies are available in the experiment, corresponding to the bands generated by the mercury lamp. In addition the motion of a collection of electrons (a current) is measured, not the velocity of a single emitted electron. While it is possible to perform single particle experiments, at no point in the MIT experiment is the kinetic energy of individual electrons directly measured.
The lab notes also provide a short history of the discovery of the photoelectric effect:

Crude though the early data were, the qualitative fact of  the  dependence  of  the  critical  cutoff  voltage  on  the wavelength of light emerged with sufficient clarity to induce the young Albert Einstein, working as a patent examiner in the Swiss Patent Office in 1905, to link the effect with the recent idea, introduced by Planck in 1900,that matter radiates its energy in quanta of energy hν.

and

It was not until 1912 that the technical problems of making  precision  measurements  of  the  photoelectric  effect were overcome [...]

In other words, 7 years passed between the time that Einstein postulated his theoretical description and a confirmation based on "clean" data.
Note also from the lab notes how inherently "messy" such an experiment can be:

It is not possible to precisely determine the work function for removing an electron, because the cathode surface interacts with the remaining gases in the photocell as a getter, so that the surface characteristics change  a  little  from  the  ideal  case.   It  is  important  to note that the electronic work function φ is a material constant which incorporates the different emission potentials of the cathode and anode.  The former is a difficult quantity to estimate due to the manufacturing process which makes  the  cathode  surface  inhomogeneous.   It  is  composed  of  a  mixture  of  potassium,  potassium  oxide  and oxidized  silver.   For  this  reason,  you  need  to  take  care that the same area is always illuminated.  The emission work function of the photoelectrons can vary locally!


Origin of the names “thionyl” and “sulfuryl” Compounds in question:

Thionyl chloride - $\ce{SOCl2}$
Sulfuroyl dichloride/Sulfuryl chloride - $\ce{SO2Cl2}$

I understand that the "thio" prefix is often used when sulphur replaces oxygen in its compounds, like "thioalcohol" or "thiosulphate". I also note that sulphur and sulfur for the same element $\ce{S}$.

However, even after countless attempts, I manage to swap the two compounds with each other, and have not been able to accurately identify their formulae when their names are given (and vice-versa). So, if someone asked me the name of $\ce{SO2Cl2}$, I would respond with thionyl chloride or sulfuryl chloride, and never be sure which one is which.

The names of other inorganic compounds are generally logical and memorable. The use of prefixes like meta, ortho, per, etc. is logical as well. However, these two sulphur compounds are very confusing. This makes me ask the question:

  What is the rule/logic/reason behind the naming of these two inorganic compounds?

I hope if I know the reason I'll be better able to recall their formulae from their names and vice-versa.

Different nomenclatures by IUPAC
For the convenience of the reader, here are the names of the compounds as mentioned in the IUPAC's 2005 Red Book[1, p 129]:

$\ce{SO2Cl2 = [SCl2O2]}$ – sulfuryl dichloride or sulfur dichloride dioxide[a], sulfuryl dichloride[b], dichloridodioxidosulfur[c];
$\ce{SOCl2 = [SCl2O]}$ – thionyl dichloride or sulfur dichloride oxide[a], sulfurous dichloride[b], dichloridooxidosulfur[c]

[a] – acceptable common name, [b] – functional replacement name, [c] – systematic (additive) name
The names which will be pertinent to this discussion are marked with italics, they should otherwise still be upright.
Also note that $\ce{H2SO4 = [SO2(OH)2]}$ is sulfuric acid[a], dihydroxidodioxidosulfur[c]. The molecular entity $\ce{H2SO3 = [SO(OH)2]}$ is sulfurous acid[a], dihydroxidooxidosulfur[c].
TL; DR: The general scheme of things is acid $\to$ radical, and the suffix used in this purpose is -yl. The difference comes from overlap as "properly" both radicals would have the same result: sulfuryl.
Etymology and history
Historically, the terms 'sulfureous' (sulphureous) and 'sulfurous' (sulphurous) date back to the 16th century. Lavosier was first to introduce the adjectives 'sulfuric' and 'sulfurous' to name acids containing sulfur in 1787[2]:

SULPHURIC, SULPHUROUS
The adjectives sulphureous and sulphurous meaning "pertaining to, or of the nature of, sulphur", date from the 16th century. The well-known oxy-acid was then known as the oil of vitriol (q.v.). In 1787, in the Méthode de Nomenclature Chimique, Lavosier Proposed the name sulphuric acid (Fr. sulfurique) for oil of vitriol and sulphurous acid (Fr. sulfureux) for the related acid containing proportionately less oxygen.[2, p 208]

The suffix -yl designates (hypothetically) as having derived from something.[2]

-YL
The suffix -yl, adapted from Fr. -yle, is derived from Gk. hylē which, in addition to meaning "wood", also means "the stuff or raw material (of any kind) from which a thing is made.[2, p 231]
[---]
(2) In the name of a radical formed by the removal of a hydroxyl group $\ce{-OH}$ from an acid.[2, p 232]
[---]
(b) Inorganic acids. A number of oxygenated inorganic radicals may be regarded as derived from acids (or from hypothetical acids) by the removal of one or more $\ce{-OH}$ groups. Such a radical is normally named by replacing the suffix -ic of the name ofthe acid by -yl. The more important of these radicals are sulphuryl $\ce{SO2\!\!:}$ (cf. sulphuric acid $\ce{SO2(OH)2}$) ... nitrosyl $\ce{NO}-$ (cf. nitrous acid $\ce{NO$.$\!OH}$)[2, p 232] [---]

As we can see, the -yl name does not care about the oxidation state of the acid. For sulfur, we see that both sulfuric + -yl and + sulfurous + -yl would give the same result: sulfuryl. One of them had to be different. For this purpose, the Greek language was used for the radical of sulfurous acid, and a Latin (possibly Sanskrit[2, p 20]) form remained for the radical of sulfuric acid.[2] This is an historic accident, it could have easily been the other way round.

THIONYL
The name given by Schiff in 1857 to the radical $\ce{SO\!\!:}$ as in thionyl chloride $\ce{SOCl2}$. It was formed from Gk. theion, sulphur, + -yl (q.v.). The radical may be regarded as the characteristic group of sulphurous acid $\ce{SO(OH2)2}$; compare sulphuryl $\ce{SO2\!\!:}$, the characteristic group of sulfuric acid.
(It is fortunate that the L. and Gk. names for sulphur (sulfur and theion) are quite different so that two different names could be formed for the two radicals.)[2, p 214]
SULPHURYL
The radical $\ce{SO2\!\!:}$, the characteristic group of sulphuric acid $\ce{SO2(OH)2}$ (see -yl), e.g. sulphuryl chloride $\ce{SO2Cl2}$ (1867)[2, p 208]

For the functional replacement name, $\ce{SOCl2}$ is named sulfurous dichloride to tie the radical more directly to its acid as well, sulfurous acid. Historically, as noted, the approach here was to use different languages of origin for the main element. This was possible because the words were very different.
If you wish to operate with common names, there is always some historic memorisation. Since this is a binary system, you could remember one and automatically derive the other. But as stated, this will not work for other similar cases: nitrosyl is for nitrous acid radical, not nitric acid.

References and bibliography
[1] Connelly, N. G.; Damhus, T.; Hartshorn, R. M.; Hutton, A. T. Nomenclature of Inorganic Chemistry; The Royal Society of Chemistry, 2005. ('IUPAC Red Book')
[2] Flood, W. E. The Origins of Chemical Names, 1st Edition.; Oldbourne, 1963.


On the origin of Gibbs Free Energy It is intuitive to me that for a spontaneous process involving any system i.e. one occurring without constant human interaction $$ \text{d}S \geq 0 $$ as I can observe increasing energy dispersal in everyday life, for example my tea turning cold.

What is not intuitive to me is that for all systems $$ \text{d}S \geq 0  ⇔\text{d}G = \text{d}H-T\text{d}S-S\text{d}T  \leq 0 $$

Why is this? I imagine that the explanation requires an understanding behind the origin of the definition of Gibbs free energy, G = H - TS.

The Gibbs energy has not been derived this way. It came from a thought process. About 100 years ago, Gibbs was puzzled by the very existence of spontaneous endothermic reactions. In mechanics, objects always fall down. They never "fall up" spontaneously. They never gets more energy spontaneously. In exothermic reaction, the matter looses spontaneously energy, like a falling object. In endothermic reactions, the atoms and molecules receive energy from the surroundings. They increase their energy. It is surprising. So Internal energy U and Enthalpy H are not equivalent to the gravitational energy. A spontaneous reaction can happen in chemistry with $\Delta$H <$0$ ou >$0$. As a consequence, H or U are not the good criteria to foresee whether a chemical reaction will spontaneously happen.

So Gibbs was thinking : What is the energy that always gets out of a system of molecules and atoms in a spontaneous reaction, even when the reaction is endothermic ? The answer is of course : the electric energy. A galvanic cell can only produce energy, whatever its heat effet. A working galvanic cell may become hotter or cooler, but it can only deliver electric energy. The amount of electric energy delivered by a cell is defined by the difference of a new potential energy, called Gibbs energy $G$, and defined so that $\Delta$$G$ is always negative in a cell. $\Delta$$G$ is obtained form the expression : $\Delta$$G$ = - $zEF$, where $E$ is the voltage of the cell, $F$ is the Faraday, and $z$ is the number of electrons exchanged in the reaction inside the cell. Furthermore, $\Delta$$G$ is defined and calculated in the same way as $\Delta$$H$. Gibbs energies of formation exist like enthalpies of formation.

For example, one may write : $\Delta$$G_{react} = \Sigma$$G_{fin}$-$\Sigma$$G_{in}$ and so on. You probably know the rest to the theory. And for example the relationship between Gibbs energy and the equilibrium constant.

Now if you plot $\Delta$$H$ and $\Delta$$G$ of a given chemical reaction versus $T$. you will notice that $\Delta$$H$ practically does not change much with $T$. But the voltage $E$ and of course $\Delta$$G$ do change much with $T$. The line $H$ vs. $T$ is nearly horizontal. But the line $G$ vs $T$ is inclined. The strange result of this drawing is the fact that both lines are joining one another at $T$ = $0$ $K$. So the difference of these energies is of course : $H-G$ = $T\Delta S$, where $\Delta S$ is the slope of the $\Delta G$ curve.


Spectrophotometry for Colourant I have been looking for a paper that by using spectrophotometry method, we can determine rate of reaction of colourant. I could not find any paper. I was wondering if any body could help me.

I'm not sure I have answer to your question maybe, if you consider natural colorant too, these are the articles that go more close to what you want:

Prediction of phenolic compounds in red wine fermentations by visible and near infrared spectroscopy

Spectroscopic, stability and radical-scavenging properties of a novel pigment from gardenia

Try to have a look a these too:

Mimani, T., and Samrat Ghosh. "Combustion synthesis of cobalt pigments: Blue and pink." Current Science 78.7 (2000): 892-896.

Bernardi, M. I. B., et al. "Comparison of blue pigments prepared by two different methods." Journal of the European Ceramic Society 22.16 (2002): 2911-2919.

Visinescu, Diana, et al. "The environmentally benign synthesis of nanosized Co< sub> x Zn< sub> 1− x Al< sub> 2 O< sub> 4 blue pigments." Dyes and Pigments 87.2 (2010): 125-131.


How to identify the concentration of nitric acid? I have quite a small quantity of concentrated nitric acid I had acquired about two years ago. This was given to me by a friend in a couple of Pyrex volumetric flasks with stopper and some in a borosilicate reagent bottles. The acid does have a whiskey brown color now.
Is there any simple method or visual cue to identify what concentration the nitric acid could be with a simple test? I want to see if the acid is 70% nitric acid. If it's the higher concentration of 90%, then I want to dilute it to 70%. I have attached a picture of one of the bottles for reference to see the color below.

It looks like you have about 40 mL of the solution. You could measure the density by using a small volumetric flask (10 or 25 mL), or even a graduated cylinder (which you could calibrate with water). Add your solution to the mark, weigh reasonably accurately, and you will have your answer (which I predict will lie between 70 and 90%).
At 20C, the density of 70% acid is 1.41; the density of 90% acid is 1.48. The difference is large enough to determine easily, and if there is any deterioration, you should be able to determine the exact concentration fairly easily from density tables, e.g.: https://www.handymath.com/cgi-bin/nitrictble2.cgi?submit=Entry


Maximum number of coplanar atoms in difluorotrimethylphosphorane In $\ce{PF2(CH3)3},$ what is the maximum number of atoms lying in one plane?

$\ce{P}$ is the central atom with the hybridization $\mathrm{sp^3d}.$ Two $\ce{F}$ atoms will form axial bonds with it, while three $\ce{CH3}$-groups form equatorial bonds. Each of these $\ce{CH3}$-groups has tetrahedral geometry, which means that one $\ce{H}$ atom lies in the same plane as the $\ce{C}$ central atom.
Thus, two $\ce{F}$ atoms, one $\ce{P}$ atom, one $\ce{C}$ atom and one $\ce{H}$ atom with it, totalling five atoms, lie in the same plane.
Is my attempt correct?

Seven is correct (the OP found this in the comments).  The methyl carbons are in the equatorial plane and each one can rotate a hydrogen atom into this plane.  In fact the hydrogen atoms fit well with this conformation so it's actually favorable.


After completing an experiment, a student finds that the percentage yield exceeded 100%. What could have occurred to the sample to lead to this? I am looking for a general answer; nothing too detailed. And no. This is not a homework question. It happened to me in lab and I am wondering what error I could have committed.

Below is a basic example.

If percent yield is based on a mass, then a >100% yield can result from a product contains materials that should not be there.  For instance, a product that should be anhydrous may not be completely dry and water would add mass to the product increasing your yield.  Other present contaminants can also do this and can even appear from a product that reacted with the surrounding atmosphere.


Why are higher-energy bonds preferred over lower-energy ones? My chemistry book states that for example the bond S-O is preferred over the bond S-S because the first has a bond energy that is almost twice as the second one. Since the energy released by the formation of the bonds is equal to the energy needed to let the reaction occurr,why does stronger bond are preferred over lower ones?
I mean, if a S atom needs let's say 170 kJ/mol to start a reaction with another S, shouldn't this be more likely to happen respect to a bond S-O that needs 340 kJ/mol?

When a mole of $\ce{S-O}$ bond is formed, $340$ kJ of heat is released.

This $340$ is not how much energy is needed to start a reaction. That energy (activation energy) cannot be derived from the bond dissociation energy.


1,4-diphenyl-1,3-butadiene (E,E) and (E,Z) Isomer Newman Projections I was wondering if it is possible to draw the Newman projections of the (E,E) and (E,Z) isomers of 1,4-diphenyl-1,3-butadiene in the following manner that I have? I wasn't quite sure about how to present the rings in such projections.

I don't know what tempts you to use the Newman projection for these compounds. Trying to draw them, you might have missed the existence of the actual conformers. (Not speaking of the confusing 120° angle representation of the planar system.)

Thanks to the totally conjugated system, the molecules tend to maintain a planar shape. The two conformations, ‘cisoid’ and ‘transoid’ (deprecated terms; recommended: s-cis, s-trans; discussed also here) resemble double bond configurations because of the stability of the planar conformation (sterical hindrance has to be taken into account as well, it can be significant for the s-cis E,Z).

For example, the s-trans and s-cis conformations of the E,E isomer can be represented as follows.

Trying to show the conformation at all the three single bonds at once wound be difficult, but could be done. And there's more missing info, so the two Newman projections would represent the (E,Z)'s conformers as well.

I am aware that excluding the hydrogens in the Newman projection is unorthodox. With or without H's, I don't think that the Newman projection is useful here, as for someone unaware of the planarity tendency, might consider e.g. torsional angle 90° much more favorable than the 0° one.


How does salt concentration affect corrosion rate of iron? I have conducted an experiment where I test the affect of $\ce{NaCl}$ concentration on corrosion rate of iron. I did this in an environment were all of the iron was submerged in water and the water was not aerated. The results I got are shown in the graph below.

Are there any research papers that could support or neglect my data? Can anyone explain this relationship? Is 1M just an outlier?

The involvement of water accounts for the fact that rusting occurs much more rapidly in moist conditions as compared to a dry environment such as a desert. Many other factors affect the rate of corrosion. For example the presence of salt greatly enhances the rusting of metals. This is due to the fact that the dissolved salt increases the conductivity of the aqueous solution formed at the surface of the metal and enhances the rate of electrochemical corrosion. This is one reason why iron or steel tend to corrode much more quickly when exposed to salt (such as that used to melt snow or ice on roads) or moist salty air near the ocean.

Water is the enabler of fast oxidation of iron so freshwater will also cause rust. However, salt water is a very good conductor (lots of dissociated ions) and so there are a number of electrolysis reactions that tremendously accelerate corrosion in salt water.
For example if you have iron in contact with salt water and also in contact with another metal such as aluminum (also in contact with the water) you effectively get a battery which drives very fast corrosion processes.
This effect can be reversed by using a metal (like zinc) which causes the current to be reversed and in effect the zinc corrodes rapidly, protecting the iron. This is the principle of a 'Cal-rod' which is used to slow the rusting of hot water tanks


Does the size of a hole in a container affect the rate of effusion? Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of density of the gas.
If there were two containers both with the same gas but one of the containers had a larger hole than the other, would the rate of effusion be quicker for the gas in the container with the larger hole?

A small hole is used. Knudsen conducted experiments similar to  Graham's but approximately 50 years later (in 1909) and used a 0.025 mm diameter hole in a platinum strip of 0.0025 mm thickness. It was observed the size was not critical as long as the mean free path of the gas was ten times larger than the hole, thus the gas was at low pressure.

In these experiments the gas is allowed to stream out of a small hole in the vessel into vacuum. The velocity of the escaping gas is just that of the trapped gas, that would be obtained if the hole was not there, and these velocities would be the molecular velocities. Thus the size of the hole must not disturb the distribution of molecular velocities, achieved by being smaller than mean free path.


How is the amino group basic? According to the Khan Academy, the amino functional group is basic because it can remove $\ce{H+}$ from systems. Why is that?

Amino: Charged (forms $\ce{R-NH3+})$ at the pH of most biological systems. Since amino groups can remove $\ce{H+}$ from solution, they are considered basic.

My current understanding:

The $\ce{H+}$ atoms bond in a polar covalent way with $\ce{N-}$. So that means the amino acid does not release that many $\ce{H+}$ ions. I can see why this is not acidic.
$\ce{N-}$ also takes electrons from $\ce{R}$, which makes $\ce{R}$ become $\ce{R+}$. This should repel $\ce{H+}$ from elsewhere in solutions.

How does this functional group remove $\ce{H+}$ from its solutions?

Polarity and elementary electrostatics have nothing to do with acid-base definitions. Sometimes we try to find a reason, using our current understanding, and that ensues the danger of over-rationalization. At a truly fundamental level perhaps no chemist can tell you why an amine can accept another proton, may be someone can do quantum mechanical calculations and say why. So avoid over-rationalization... accept some observations as facts (temporarily until we develop a full understanding).

One can titrate an amine with an acid, by this experimentally observed fact, exactly just like we titrate NaOH with HCl. So an amine group is indeed a base or say it behaves like base. The acidic proton cannot annihilate itself during a titration, it has to go somewhere. The amine group gets protonated, in return.


Why is partial pressure proportional to mole fraction in Henry's law? According to Henry's law of partial pressure,  partial pressure of a gas above a solution is directly proportional to the mole fraction of the gas in the solution.

But my question is: If the mole fraction of a gas in solution is greater, then that should mean there are less gas molecules in the air and so the partial pressure should be less. Why is it the other way round?

1st, Henry's law of Partial Pressure states:

"At a constant temperature, the amount of a given gas that dissolves in a given type and volume of liquid is directly proportional to the partial pressure of that gas in equilibrium with that liquid."

Quoting from WikiPedia here:

Henry's law is a limiting law that only applies for 'sufficiently dilute' solutions. The range of concentrations in which it applies becomes narrower the more the system diverges from ideal behavior. Roughly speaking, that is the more chemically 'different' the solute is from the solvent.
For a dilute solution, the concentration of the solute is approximately proportional to its mole fraction $x$, and Henry's law can be written as:

$$p = K_{\rm H}\,x$$
,where $p$ is the partial pressure of the gas above and in equilibrium with the solution, and $K_{\rm H}\,$ is the proportionality constant called the Henry's Law constant.
This is a conceptual conclusion as steeper the concentration gradient, the rate of diffusion (and hence the amount of diffused substance) will be greater. Here, the concentrations of the "solute" is represented by partial pressure of the gas in the gaseous state and by it's corresponding mole fraction in the solution state.
What you have supposed stands only for a closed system. As in the second quote from WikiPedia, you understand that the law is applicable only for conditions in which the behavior/condition which you supposed is either not relevant or is negligible.
If you go through the linked article, you'll find that Raoult's Law for Vapor Pressure of Solutions appears to be a special case of Henry's Law for more concentrated solutions, and can be used as such.
Note that both Raoult's Law and Henry's Law are limiting laws and they are applicable each as when:
Henry's Law: $x$ → $0$
Raoult's Law: $x$ → $1$


Why is it that only ozone absorbs UV radiation? I know the mechanism of how ozone absorbs UV but why is it that no other molecules do this?

Is there any special reason?

PS: If I'm mistaken and there are other molecules that can absorb UV please mention them.

The ISO 21348 Definitions of Solar Irradiance Spectral Categories defines UV radiation as that portion of the electromagnetic spectrum having wavelengths from 10 to 400 nm.  Different atmospheric compounds absorb more or less strongly than others throughout the UV portion of the spectrum.

The following table was reproduced from this Wikipedia article:

Although not fully quantitative, this table indicates that nitrogen, oxygen and ozone all absorb different portions of the UV spectrum.  Water vapor is also an important UV absorber that is not discussed in the above table.

Also, regarding the UV absorption of (liquid) water and your comment that "It is widely accepted that life on Earth wouldn't be possible without ozone layer", it was in large part the protection from UV radiation afforded by liquid water that allowed aquatic organisms to survive prior to the existence of the ozone layer. If the ozone layer were to vanish today, although the nature of Earth's biosphere would change drastically, I suspect many aquatic species, particularly microbes, would adapt and continue to survive in the absence of an ozone layer. It would be bad news for you and me though ;)


Why x in equilibrium calculations can be "ignored" You don't need to work out anything for me. I am just giving an example so you can explain something to me.

For example:
$$\ce{ClF3(g) <=> ClF(g) + F2(g)}\\ \ \ K_c=8.77\times10^{-14} \ \ @\  279\ \text{K}$$

What is $[\ce{ClF}]$ at equilibrium if initially $[\ce{ClF3}]=2.50 \ M$ and $[\ce{F2}]= 1.00\ M$?

$$K_c= \dfrac{[\ce{ClF}][\ce{F2}]}{[\ce{ClF3}]} = \frac{(x)(1.00+x)}{(2.50-x)} = 8.77\times10^{-14}$$

I get everything that just happened above. Thats how I would do it but then, my lecturer said the $x$ then becomes negligible for some reason something about the CIF. and now the equation becomes:
$K_c= \dfrac{(x)(1.00)}{2.50}$ and so $x=2.2\times10^{-13}$.

What happens to the $x$ and why? Also, what exactly is $K_c$ in chemistry what does it represent? I know how to work out questions but I don't know why.

In problems like this, the maths can sometimes be simplified by making assumptions so that you do not need to do difficult algebra. You could always do the harder algebra.

In this case, a valid assumption is that when the equilibrium constant is very small in comparison to the concentrations given, then the change in those concentrations (which is represent by $x$) will also be very small and can be neglected. I'll work through your example the long way to show you why this assumption is valid.

$$K_c= \dfrac{[\ce{ClF}][\ce{F2}]}{[\ce{ClF3}]} = \frac{(x)(1.00+x)}{(2.50-x)} = 8.77\times10^{-14}$$

Working through the algebra exactly:

$$(x)(1.00+x)=(8.77\times10^{-14})(2.50-x)$$
$$1.00x-x^2=2.19_{25}\times 10^{-13}-8.77\times10^{-14}x$$
$$0=x^2+(-8.77\times10^{-14}-1)x+2.19_{25}\times 10^{-13}$$

Solving the quadratic formula gives

$$x=\dfrac{(1+8.77\times10^{-14}\pm \sqrt{(-1-8.77\times10^{-14})^2 +4(2.19_{25}\times 10^{-13})}}{2}=[1,\ 2.19\times10^{-13}]$$

The first root $x = 1$ is nonsense. The second is the root you were given (when rounded to the correct number of sig figs).

The value of $x$ is so small compared to the initial concentrations of $\ce{ClF3}$ and $\ce{F2}$ that the concentrations of these species do not change within the magnitude of correct use of significant figures:

$$[\ce{ClF3}]_{eq}=2.50-2.1925\times10^{-13}=2.50 \ \text{to 2 decimal places}$$
$$[\ce{F2}]_{eq}=1.00+2.1925\times10^{-13}=1.00 \ \text{to 2 decimal places}$$

The maths above are equivalent to the assumption that since $x<<1.00$ and $x<<1.00$, $1.00+x\approx1.00$ and $2.50-x\approx2.50$. Whether you do it at the beginning or at the end, you get the same result: $x=2.19\times10^{-13}$. Making the simplification at the beginning is operationally easier and avoid potential algebra and arithmetic errors.

The assumption is not valid if the initial concentrations are close in value to $x$.


Why does no Robinson Annulation occur after the Michael addition? Under basic conditions a Michael addition reaction can be used to convert compound A into product C. Explain why no Robinson annulation cyclic product is observed?

  Compound A

  Product C

I have worked out that the other compound involved in the Michael addition reaction would be compound B (below), but I don't know why no Robinson annulation reaction could occur after the Michael addition reaction?

Compound B

As illustrated in the following figure, the Robinson annulation reaction involves two steps.  The first step involves the Michael addition of a ketone enolate (the enolate derived from cyclohexanone in the example pictured below) to an $\alpha,\beta$-unsaturated ketone (methyl vinyl ketone below).  In a second step the methyl group adjacent to the carbonyl is deprotonated and undergoes an intramolecular aldol condensation to form the final product. This last step is very favorable since the reaction occurs intramolecularly and involves a favorable 6-membered transition state to form a stable 6-membered ring.

Often in Robinson annulations methyl vinyl ketone is used. It is important that  the $\alpha,\beta$-unsaturated ketone have an alkyl group attached to the carbonyl so that the enolate required in the second step can be formed.

  Explain why no Robinson annulation cyclic product is observed?

In your example, the phenyl vinyl ketone has no alkyl group attached to the carbonyl carbon - only a phenyl group. Hence, the enolizable hydrogens present in your Product C can only react with the other carbonyl through a highly-strained 4-membered transition state preventing any ring formation through an intramolecular aldol from taking place.


GOC allylic,vinylic, benzylic positions, carbocation stability Carbo-cations may be stabilized by:
(a) π-bonds only at allylic position

(b) π-bonds only at vinylic position

(c) π-bonds at allylic and benzylic position also
(d) -I effect

While the answer is obviously not (d), I am really confused about what 'allylic', 'vinylic' and 'benzylic' positions actually mean. I have heard about allylic/vinylic/benzylic carbons, but positions..?
Please, help.
Thank-you.

The question has been taken from ADVANCED PROBLEMS IN ORGANIC CHEMISTRY by HIMANSHU PANDEY

allylic position explanation
benzylic position explanation
vinylic position explanation

For carbocation stability, think about which case has opportunity for electron resonance, i.e., which case(s) allows the positive charge to be delocalized to adjacent pi bonds. This resource might help.


How estimate E0-0 of an organic dye? How can I estimate the $E_{0-0}$ value of an organic dye? I have obtained the UV-vis absorption and fluorescence spectra of this dye in solution. This article talks about (in Table 2 notes) the intercept between the normalized absorption and emission spectra, but I don't quite understand what they mean.

Does someone here know? Attached is the (non-normalized) spectra.

You're looking for the point where the normalized absorption and emission spectra cross each other.

The UV/Vis absorption spectra is almost always higher in energy (shorter wavelengths) than the emission spectra, because the molecule goes from the electronic and vibrational ground state to the electronic excited state. Usually this also means some vibrational modes are also excited. Since the molecule is typically in either the vibrational ground state or a low-energy vibrational state, the absorption process can go up several vibrational levels (i.e, the energy absorbed is higher than just the basic electronic excitation).

For emission, the molecule often loses some vibrational energy, so the emission comes from the ground vibrational state of that electronic excited state. The molecule ends up in the ground electronic state, but a vibrationally excited state (i.e., it loses a bit less energy than the $E_{0-0}$ energy difference).

So if you normalize the absorption and emission spectra, the crossing point should roughly correspond to the $E_{0-0}$ difference.


Diluting down solutions method I'm designing an experiment and it involves testing various concentrations of solutions, and I will be testing 5 in total. I proposed to make up a stock solution, and distribute it evenly in 5 different beakers and add different amounts of water to each solution but my teacher doesn't like this method and says I should simply dilute down but I don't know what he means. So if I had a 1.00 mol/dm$^{-3}$ solution and wanted concentrations of 0.800, 0.600, 0.400 and 0.200  mol/dm$^{-3}$, how do I dilute it down? To be honest, the concentrations don't really matter, it's just the method of making them - would it be easier to dilute them by half each time? Each solution is going to be used up so I can't simply reuse them...

Ultimately I will need 200ml of each solution in the experiment design, and the precision of the concentrations is not very important. How can I dilute one stock solution to 4 other solutions?

It depends on what degree of accuracy you need and what order of magnitude the final concentrations will be.

If you need high accuracy and concentrations are in the range you are describing, I would put a class A volumetric flask on a balance, add the appropriate weight of solute, add water to not quite the full volume, dissolve the solute, and add water to the volumetric line, seperately for each solution.

If you need concentrations much lower than what you are saying, when weighing becomes your largest source of error, then it is more accurate to make a concentrated solution, and the dilute the concentrated solution.  For example, you would use a class A TD (to deliver) volumetric pipette to measure the concentrated solution into a class A volumetric flask.


Combination of polydisperse polymer mixtures Say I'm given a mixture of batches of polydisperse polymers. I know the weight $w$, the number average $ \bar M_n$, and the weight average $ \bar M_w$ of each batch that was added to the final mixture. How would I go about calculating the $\bar M_w, \bar M_n$ and PD of the final mixture?

I'm thinking that the $\bar M_w$ value would simply be a weighted average of each of the batches, but I'm not quite sure how to get the number average.

In math, I'm thinking that $\bar M_{w,final} = \frac{\Sigma_{i} w_i \bar M_{w,i}}{\Sigma_i w_i} $

I haven't been able to reason what I should do to account for the differing numbers of molecules/moles of each batch. I think it should also be a weighted average, but using the numbers of each batch for weighting instead of the weight, but how would I calculate that?

Thank you

It should not be difficult to derive either of the number-averaged molecular weight or weight-averaged molecular weight by their definitions. To avoid ambiguity, I will use $i$ as the index of each actual length of chains and $b$ as the index of batches.

First, the definitions:

$\bar{M}_\mathrm{n,final}=\frac{\Sigma_in_iw_i}{\Sigma_in_i}$

$\bar{M}_\mathrm{w,final}=\frac{\Sigma_in_iw_i^2}{\Sigma_in_iw_i}$ where $n$ and $w$ are number and molecular weight of each chain length.

The numerator of $\bar{M}_\mathrm{n,final}$ is, in fact, the total mass of the mixture. Hence, $\Sigma_in_iw_i=\Sigma_bw_b$. The denominator is the total number of chains. Hence, $\Sigma_in_i=\Sigma_bn_b=\Sigma_b\frac{w_b}{\bar{M}_\mathrm{n,b}}$.

It is slightly more complicated in the case of $\bar{M}_\mathrm{w,final}$ but you got it correct. Feel free to ignore my derivation. The numerator does not have a straightforward definition. It has to be derived from the $\bar{M}_\mathrm{w,b}$ instead. $\Sigma_in_iw_i^2=\Sigma_b\Sigma_{i,b}n_{i,b}w_{i,b}^2=\Sigma_b\bar{M}_\mathrm{w,b}w_b$. Since we have already defined the $\Sigma_in_iw_i$ above, we can have the final equation.

$\bar{M}_\mathrm{n,final}=\frac{\Sigma_bw_b}{\Sigma_b\frac{w_b}{\bar{M}_\mathrm{n,b}}}$

$\bar{M}_\mathrm{w,final}=\frac{\Sigma_b\bar{M}_\mathrm{w,b}w_b}{\Sigma_bw_b}$

$\bar{M}_\mathrm{w,final}/\bar{M}_\mathrm{n,final}=\frac{\Sigma_b\frac{w_b}{\bar{M}_\mathrm{n,b}}\Sigma_b\bar{M}_\mathrm{w,b}w_b}{(\Sigma_bw_b)^2}$


Is a compound composed out of gold, manganese, xenon, krypton, and oxalate possible? Please keep in mind while reading this that I do not intend to create this, it is part of the backstory for a video game I am producing.

I created a compound based on knowledge from my chemistry class and Wikipedia.
The formula is: $\ce{AuKr4Xe2C2O4^2-C2O4^2-MnO4-}$

The 2D structure looks like:

And the 3D Structure is:

I believe the IUPAC name would be Aurodixenichexakryptic permanganate dioxalate (please correct me if I am wrong).

I know the molar mass is $\pu{1089.7413 amu}$.

It is a thick, black liquid similar to Vantablack, as in it absorbs $99.99\%$ of the light hitting it, making it seem two-dimensional. It is odorless and tasteless.

I determined the melting point to be $0.01~\mathrm{K}$ and boiling to be $\pu{7538384392 K}$. The density under normal conditions is $\pu{89.45 g mL-1}$.

I was hoping someone could tell me is all the information thus far was valid and if someone could provide me other key facts about the compound (i.e. alternate names, $\mathrm{pH}$, $\mathrm{pOH}$, $\mathrm{p}K_\mathrm{a}$/$\mathrm{p}K_\mathrm{b}$, etc.)

No, this is not possible.

Actually, if I would have to think of the most unlikely chemical conceivable, that would be it. Let's see why:

Krypton is a noble gas that doesn't bond to anything. All of the known krypton compounds can be counted on one hand, and most of them contain fluorine. Putting krypton in a large molecule like this just can't be.
This is the same for xenon. Even though it's slightly more reactive than krypton, it still has problems forming bonds with other stuff and even when it does it's extremely unstable.
Gold is another noble element that doesn't like bonding to stuff.
Permanaganate is a very strong oxidiser. This is not something you want to have around organic stuff (such as your oxalate). They are simply not stable together.

I don't know how you derived the data for it, but it seems completely unreasonable. A boiling point of 7538384392 K? How about no way? Even if you somehow could create this compound (god knows how), I would expect it to be so unstable it would decompose completely at ambient temperature. And by "ambient" I mean ambient on a Kuiper belt object.


Which of the following has maximum nucleophilicity? Question

The answer should be (D) as the inductive effect of 3 methyl groups increases negative charge on oxygen and thus makes it a good nucleophile.But the answer is (A).

According to me, the answer should be (C) in general cases but this might get changed to A) in special cases. The best way to judge their nucleophilicity order  is to check their basicity. The weak bases are generally better nucleophiles in polar protic solvents . The acidity order of the  conjugate acids are $$\ce{CH_3OH < (CH_3)_3COH  < NH_3 < CH_4}$$So, the strongest acid is $\ce{CH_3OH}$ and weakest acid is $\ce{CH_4}$. So, weakest base is $\ce{CH_3O^-}$  and strongest is $\ce{CH_3^-}$. So, the maximum nucleophilic character should be of $\ce{CH_3O^-}$,not generally $\ce{CH_3^-}$ as basicity is inversely proportional to nucleophilicity in general cases. Coming to D), the inductive effect of three methyl groups actually decreases the acidity of tert-butyl alcohol. So, the tert-butoxide ion is more prone to take $\ce{H^+}$ rather than to nucleophilically attack.  It also creates steric crowding, and preferably takes easily ionisable hydrogen ion if possible from the compound.So,  looking only at increased $\ce{+I}$ effect doesn't tell that its nucleophilicity is increased. About A), this $\ce{CH_3^-}$ is usually seen in Grignard reagents, whose basic activity is much much greater than its nucleophilic activity. If acidic hydrogen is present in the compound, it preferably takes that. But, if a compound doesn't possess any acidic hydrogen (special case), then its nucleophilic character is more as being less electronegative than $\ce{O}$, it is very less reluctant to keep yhe lone pair on itself like the case in $\ce{CH_3O^- or (CH_3)_3CO^-}$. Thus in the absence of any acidic hydrogen $\ce{CH_3^-}$ is better nucleophile. So, from the above discussion, it can be concluded that in many of the cases, where the attacked compound doesn't possess any acidic hydrogen, A) is maximum nucleophillic. But in the opposite  cases, C) is maximum nucleophilic.


What makes an epoxide stable? Overall ring strain seems to be a big issue when it comes to organic chemistry. That is why cyclopentane may be in an "envelope form" or why cyclobutane may be in a kinked, kite form. Both of these example molecules are not planar. So, why is epoxide (oxacyclopropane) a stable molecule, if at all?

Well, you have to ask: how stable? and compared to what?

Epoxides, including ethylene oxide, are generally considered relatively unstable molecules, with a high chemical activity and involvement in numerous reactions, including polymerization and thermal decomposition. However, ethylene oxide does exist as a molecule, as does cyclopropane, which also contains a three-membered ring.

So, let's compare for example the ring strain in ethylene oxide and cyclopropane. I found some computation data on their relative stability from these lecture notes, which indicate that “ethylene oxide has less ring strain than cyclopropane”. I would attribute this to the fact that the C–O–C angle has less strain because its “relaxed” value would be 104.5°, compared to the “relaxed” C–C–C angle of 109.4°.


Implications of orthonormal wavefunctions For orthonormal functions $\psi$ and $\phi$, by definition;
$$\langle \psi|\phi\rangle=0$$
However, when a wavefunction, $f$, is expressed as a linear combination of these orthonormal wavefunctions:
$$f=c_1\psi+c_2\phi$$ operating on this function with an arbitrary operator, $\hat A$ results in:
$$\langle f|\hat A|f\rangle=\langle c_1\psi+c_2\phi|c_1\hat A\psi+c_2\hat A\phi\rangle=c_1^2\langle\psi|\hat A|\psi\rangle+c_1c_2\langle\psi|\hat A|\phi\rangle+c_2c_1\langle\phi|\hat A|\psi\rangle+c_2^2\langle\phi|\hat A|\phi\rangle$$

My question is, do the crossed terms equal zero? (i.e does $\langle\psi|\hat A|\phi\rangle$ and $\langle\phi|\hat A|\psi\rangle$ equal zero?)

I initially thought that, yes, they do equal zero because, suppose that $\psi$ and $\phi$ are eigenfunctions of the operator $\hat A$ then it follows that: $$\langle\phi|\hat A|\psi\rangle=\langle\phi|a|\psi\rangle=a\langle\phi|\psi\rangle=a(0)=0$$
However, in a problem sheet I have just done, the question relies on these terms not being zero. Is my logic wrong or does this only equal zero in the case where the functions are eigenfunctions of the operator?

If you have two orthonormal states $|\psi\rangle$ and $|\phi\rangle$,
$$
    \langle \psi|\phi\rangle = 0,
$$
there is no guarantee that the term
$$
    \langle\psi|\hat{A}|\phi\rangle
$$
vanishes. All depends on the action of $\hat{A}$ on your state.

As you already noticed, if $|\phi\rangle$ is an eigenstate of $\hat{A}$ with eigenvalue $a_\phi$, the term is obviously zero:
$$
    \langle\psi|\hat{A}|\phi\rangle = \langle\psi|a_\phi|\phi\rangle =a_\phi \langle\psi|\phi\rangle = 0.
$$
But this is only a special case, since $|\phi\rangle$ is not necessarily an eigenstate of $\hat{A}$ in general.

One of the simplest situations where this term does not vanish can occour  when $|\psi\rangle$ and $|\phi\rangle$ are eigenstates of the harmonic oscillator, that we can denote $|n\rangle$ and $|n'\rangle$. As you probably know, these states are orthonormal:
$$
    \langle n|n'\rangle = \delta_{n,n'}.
$$
If you consider the specific case $|\psi\rangle=|n\rangle$ and $|\phi\rangle=|n-1\rangle$, it immediately follows that
$$
    \langle n|n-1\rangle = 0.
$$
Now if you take $\hat{A}= \hat{a}^\dagger$, for which you have $\hat{a}^\dagger|n\rangle=\sqrt{n+1}|n+1\rangle$ ($|n\rangle$ is not an eigenstate of $\hat{a}^\dagger$), you have ($n\geq1$)
$$
    \langle n|\hat{a}^\dagger|n-1\rangle = \langle n|\sqrt{n}|n\rangle = \sqrt{n} \langle n|n\rangle = \sqrt{n}\delta_{n,n} = \sqrt{n} \neq 0.
$$

As I already stated, all depends on the action of $\hat{A}$ on your state.


For molecules to be enantiomers, do their bonds have to point the same way? Enantiomers are non superimposable mirror images. What is bothering me is:
If the bonds point different ways are they still considered superimposable and hence enantiomers?

To illustrate, comparing X and XI:

If I flip XI (180 degrees around the vertical axis that runs along the $\ce{C-C}$ bond), I get Y:

Is Y considered non-superimposable on X (and hence is Y e.g. XI an enantiomer of X)?

I would say yes they are non-superimposable, as in X the horizontal groups are pointing out whereas in Y they are pointing into the page (other than this, the groups are in "matching" positions). I don't think that is "superimposable", as the groups would not overlap.

Place your two hands face down on a table. They are non-superimposable mirror images of one another (chirality does come from the Greek kheir, meaning "hand", after all). Now flip one of your hands over so that one palm is facing up. Are your hands any different now than before you flipped one over?

No. Your hands' orientations in space have no effect on their superimposability, as is the same for molecules.


Finding stationary solutions and verifying differential equations Consider a trimolecular system where the following chemical reactions occur:
  $$\ce{A + B ->[k_1] A + C}\tag1$$
  $$\ce{A ->[k_2] B}\tag2$$
  $$\ce{C ->[k_3] C}\tag3$$
  Write the differential equations that model the evolution of the three component molar concentrations and determine the stationary solutions (there are two). Finally, if initially $[\ce{A}]=0.75$, $[\ce{B}]=0.25$ and $[\ce{C}]=0$, calculate the time when the mole fraction of $\ce{A}$ is half of its initial value.

I think that the differential equations, following the expression for the reaction velocity, are
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t}=-k_2[\ce{A}]$$
$$\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t}=-k_1[\ce{A}][\ce{B}]+k_2[\ce{A}]$$
$$\frac{\mathrm{d}[\ce{C}]}{\mathrm{d}t}=k_1[\ce{A}][\ce{B}]$$
However, I don't know how to proceed with the stationary solutions. I guess that one of them would be obtained doing:
$$-k_1[\ce{A}][\ce{B}]+k_2[\ce{A}]=0$$
but I'm not sure.

Finally, I got the relations
$$[\ce{B}]=\left(\frac{1}{4}-\frac{k_2}{k_1}\right)\exp\left(\tfrac{k_1}{k_2}(-\tfrac{3}{4}+[\ce{A}])\right)+\frac{k_2}{k_1}$$
$$[\ce{C}]=-\left(\frac{1}{4}-\frac{k_2}{k_1}\right)\exp\left(\tfrac{k_1}{k_2}(-\tfrac{3}{4}+[\ce{A}])\right)+1-\frac{k_2}{k_1}-[\ce{A}]$$
but I have no idea how to obtain the time. Any help would be appreciated.

Your differential equations are correct.

From $\frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} = -k_1[\ce{A}][\ce{B}]+k_2[\ce{A}] = 0$ follow the stationary solutions $[\ce{A}] = 0$ or $[\ce{B}] = \frac{k_2}{k_1}$.

Reaction (2) obeys a first order decay law. Its half-life is $t_{\frac{1}{2}} = \mathrm{ln}(2) / k_2$.

The image below shows how $[\ce{B}]$ reaches the stationary state soon after the reaction has started ($k_1 = 0.3, k_2 = 0.2$).


Error in calculating H+ activity I'm starting to study thermodynamics of minerals and rocks, so sorry for the basic question. I am trying to work out some exercises of a book and I found this:

Calculate the activity of $\ce{H+}$ when the activity of $\ce{Sn^4+}$ is $10^{-10}$ for the reaction $\ce{SnO2 + 4H+ <=> Sn^4+ + 2H2O}$. Standard free energies of formations $\Delta$Fof are: SnO2 = -123.20 kcal; H2O = -56.69 kcal; Sn4+ = 0.65 kcal; H+ = 0.

I used the following equations:

$\Delta$For = ($\Sigma$$\Delta$Fof,products) - ($\Sigma$$\Delta$Fof,reactants)
$\Delta$For = -RTln(K), where R = 0.001987 Kcal/molK and T = 298.15K
K = $\frac{[Sn^{4+}]}{[H^{+}]^4}$, given that the activities of SnO2 and H2O are unit.

The result I found was $3.82 \cdot 10^{-5}$, but the book says the result is $10^{-0.40}$. I assume this is a fundamental error, but given it is also a fundamental principle I need to understand it to proceed in my study.
Thanks in advance for any help.

$$ \Delta G = \Delta G^\circ + RT \ln{\frac{a_{\ce{Sn^{4+}}}a_{\ce{H2O}}^{\;\;\;\;2}}{a_{\ce{SnO2}}a_{\ce{H+}}^{\;\;4}}} $$

At equilibrium, $\Delta G = 0$.

For $\Delta G^\circ = 2\times(-56.69) + 0.65 - (-123.20 - 4\times0) = +10.47\mathrm{\frac{kcal}{mol}}$

The question says "activity", but in the absence of any other information, let's assume that "activity" is equal to "concentration".  That is, let's assume activity coefficients are 1.

Thus, we get $$ \Delta G = 0 = +10.47  + RT \ln{\frac{a_{\ce{Sn^{4+}}}\cdot1^2}{1\cdot a_{\ce{H+}}^{\;\;4}}} $$

Thus, we get $$ {\left( \frac{-10.47}{RT}\right)}  = \ln{{10^{-10}} a_{\ce{H+}}^{\;\;-4}} = -4\ln{a_{\ce{H+}}} -10 \ln{10} $$

$$\frac{\frac{-10.47}{RT} + 10 \ln{10}}{-4} = \ln{a_{\ce{H+}}}$$


When boiling water, how is evaporation rate affected by the surface area of the container? Imagine two pots boiling on a stove. One is tall, while one is wide. Both contain the same volume of water.

What will be the difference in the rate of evaporation between the two containers?

Boiling water

When you boil water on a stove, the temperature on the bottom of the pot is higher than the boiling point. Steam bubbles are generated, and they help to heat the water at the top of the pot (convection also helps, and conduction of the pot itself). Once you have a rolling boil, the rate is determined by how quickly you can supply more heat (almost all of it now goes into the phase transition). If you use the same burner, and mostly heat the pot (so no tiny pot on a big burner), at that stage the rate of boiling off should be the same.

Evaporating water

When the temperature of the water is below boiling, the mechanism is different. Rather than steam displacing the air above the water, single water molecules break away from the liquid and mix with the air. Evaporation rates depend on the partial pressure of water just above the surface (this influences condensation, i.e. the reverse reaction), the surface area, and the vapor pressure of water in the liquid state. Anything that lowers the partial pressure of water (such as "wind" bringing in dry air), increases the surface area (different pot geometry) or increases the vapor pressure (temperature) will increase net evaporation rates.

Beaker vs Erlenmeyer

Tall vs. wide pot changes the contact area with the stove top and the area of the liquid:gas interface at the same time. If you compare a beaker with an Erlenmeyer (same bottom area), it should show similar boiling rates (maybe a big mess with overboiling for the Erlenmeyer) but very different evaporation rates.


Oleum concentration description What is the meaning of "free" $\ce{SO3}$ and "combined" $\ce{SO3}$ in the oleum sample concentration description?

Also what are these maximum and minimum values of % labelling in oleum sample?

I think the term combined "$\ce{SO3}$" is nothing but $\ce{SO3}$ of $\ce{(SO3 + H2O)}$ and free $\ce{SO3}$ means that is $\ce{SO3}$ in oleum sample.

And value of % leveling of oleum is from (100% to 122.5%).


Energy levels in an atom I have been thinking about energy levels of an atom. When we study line spectra of hydrogen atom we say that when electrons jumps back from a higher shell to a lower shell it emits photons of certain frequency, but in that example we say that hydrogen has infinite energy levels and electron can jump back from any level. But when we study bonding we say that some atoms lack in d orbitals and for example say Nitrogen has no d orbitals, but when doing line spectra it has infinite energy level. How these two statements hold up with each other?

The H atom has an infinite number of energy levels spanning a finite energy range. This range is 13.6 eV, the ionisation energy, and is equal to the Rydberg R in energy. In the simplest (basic) theory the energy is $E_n=-R/n^2$, where n is the principal quantum number, $n=1..\infty$) thus as the energy increases the energy levels become closer to one another. The energy also rises (becomes less negative) as n increases.

In addition, for each level n there are other orbitals of nominally the same energy which describe the angular momentum & shape of the orbital. The orbitals are  1s, [2s, 2p], [3s, 3p ,3d], [4s, 4p, 4d, 4f] etc. where the levels in brackets are nominally of the same energy. (However,  interaction between electrons changes these energies but only slightly compared to their total energy ).

The light emission (fluorescence) you refer to comes from transitions between any two of these levels (subject to energy and angular momentum conservation). Other types of atoms behave similarly but because there are multiple electrons the equations describing the energy become far more complex.

Thus H atoms do have d orbitals just as do N atoms, but in their lowest energy state there are not enough electrons to fill any of these. The d orbitals only start to become filled as one reaches Sc in the transition metals.

In H and N atoms higher orbitals can be reached, by for example by absorbing photons or imparting energy from fast moving electrons in a discharge. However, in bonding, d orbitals will not be involved in the ground state bonding orbital of a molecule if there are not enough electrons and not enough energy to initially fill the d atomic orbitals. As soon as there are enough electrons, as in transition metal complexes, d orbitals become essential to understanding bonding.

Note: (I'm ignoring  ideas such as hybridisation between orbitals)


why does the gap between the energy levels of a hydrogen atom keep decreasing as we move further and further away from nucleus? why is the gap between the first two orbits greater than the gap between second and third orbit?
shouldn't the gap keep decreasing as the radius keeps increasing? Can someone explain this ?

Whatever the radius, the allowed energy levels are proportional to $1/n^2$.
The first gap is proportional to $1/1^2 - 1/2^2 = 0.75$.
The second gap is proportional to $1/2^2 - 1/3^2 = 1/4 - 1/9 = 5/36 = 0.139$. This is smaller than $0.75$.
The third gap is proportional to $1/9 - 1/36 = 1/12 = 0.083$. It is still smaller.


What is the process involved in performing a mass spectrometry experiment? I have heard that mass spectrometry is very time consuming: why is this?  What is the process involved in performing a mass spectrometry experiment?  What steps are the most time consuming?

This is hard to answer, because in my experience mass spectrometry isn't that time consuming at all. The experiment itself doesn't take that long. In this experiment the sample is ionized and charged fragments are accelerated in an electrical or magnetic field. Depending on their mass/charge ratio they will end up on specific locations in the detector.

The process around the experiment itself is dependent on your specific lab, but typically the mass spectrometer is operated by only a few people that do all the samples of the entire department. Because of that, there will often be a queue which could be several days. Additionally, not all mass spectra can be digitally identified as coming from a certain chemical, especially for larger molecules there are often various options that require human interference for the identification. In our lab this human interpretation is left to the person owning the sample, but if that part is also done by the mass spectrometer operators (I know in some labs this is the case) then I can imagine that this last step will be time-limiting.

To summarize: the experiment is short, the procedure around it (dependent on your specific lab) can take quite some time.


How does absorption spectroscopy work? So a beam of white-light photons passes through a gas and various photons, having exactly the same energy as the gas's molecular energy transition state deltas, get absorbed. We then get a characteristic absorption pattern. But wait a minute, don't the energised electrons fall back to their ground states within a nanosecond and create photons of precisely the same energy as those that caused the disturbance? How come we still see the absorption band?

In an absorption experiment a collimated beam of light is passed through the sample. The intensity of this is measured before and after the sample. The absorption is measured as optical density which is $D=[c]\epsilon _{\lambda} l$, where $[c]$ is the concentration, $\epsilon_\lambda$ the extinction coefficient at wavelength $\lambda$ (a measure of the strength of absorption), and l the path length through the sample. At each wavelength the measured light intensities are $I=I_0\exp(-D)$ where $I_0$ is the initial intensity. This is called the Beer–Lambert law.

Now as you mention the sample is placed into its excited state where it remains for some period of time. This depends on the molecule being studies but may be from picoseconds (relatively rare) to several nanoseconds to microseconds (also rare). In an absorption spectrophotometer the light intensity used is low, so the excited state population is always very small. As any fluorescence produced is emitted over $4\pi $ radians (in solution) the amount captured by the detector will be small, just on geometric grounds as the detector is placed some distance away from the sample and the light passes through slits before entering the detector.(The slits & grating give spectral resolution). Also note that the fluorescence is mainly at a longer wavelength than absorption but the zero zero line in absorption and emission can overlap. The geometric reason just given suffices to make absorption accurate even in this case.

In laser experiments the situation can be different, particularly when pulsed lasers are used. It is quite easy to observe 'gain' in  the sample, this means that transient signals that are more intense than the measuring (probe) beam are observed. This is due to stimulated emission caused by the probe beam. It is also quite easy to see amplified stimulated emission (ASE) from a cell containing a fluorescent dye if the laser beam is focussed.

EDIT:

An explanation as to why stimulated emission is not normally observed in an absorption experiment and why fluorescence is observed under similar conditions is given below.

In the simplest case of two levels $1$ and $2$ , where $1$ is the lowest level the rate of absorption from $1$ is $N_1B_{12}\rho(\nu)$ where $N_1$ is the population of level $1$, $B_{12}$ the Einstein coefficient for (stimulated) absorption from level $1$ to $2$ and $\rho (\nu)$ the energy density per unit frequency interval ($\pu{Jm^{-3}/s^{-1}}$). The rate of transitions from $2$ to $1$ is $N_2[B_{21}\rho(\nu) +A_{21}]$ where $A_{21}$ is the rate of spontaneous emission from level $2$ and $B_{21}$ the coefficient of induced (stimulated) emission. Note that $A$ does not depend on the number of photons present. In is convenient to assume that the molecule is bathed in thermal radiation that follows  Planck's blackbody law at temperature $T$,

$$\rho(\nu)= \frac{8\pi h\nu^3}{c^3}[e^{h\nu/(k_BT}-1]^{-1}$$

At equilibrium the rate of change $dN_1/dt=dN_2/dt=0$ then

$$ \frac{N_1}{N_2}= \frac{B_{12}\rho(\nu)}{B_{21}\rho(\nu) +A_{21}}$$

The ratio of populations is also given by the Boltzmann distribution, also at temperature $T$

$$ \frac{N_2}{N_1} = \frac{g_2}{g_1}e^{-h\nu/(k_BT)} $$

where $g_1,\,g_2$ are the degeneracies of the two levels.

Combining these last two equations and solving for $\rho$ gives

$$\rho(\nu) = \frac{A_{21}/\,B_{21} }{ \frac{g_1B_{12}}{g_2B_{21}}e^{h\nu/(k_BT)}-1}$$

For this relationship to be true $g_1B_{12}=g_2B_{21}$ and $A_{21} = n(\nu)B_{21}h\nu$ where

$$n(\nu)= 8\pi\nu^3/c^3$$

and is the number of radiation modes per unit volume per unit frequency.

This relationship means that the ratio of induced to the spontaneous emission rate (leading to fluorescence) is equal to the number of photons in this mode.  The figure shows that in the visible part of the spectrum,  $\approx 400$ to $700$ nm at at thermal emission at room temperature the number of photons mode is somewhere between  $10^{-10}$ and $10^{-100}$, the exact number does not matter much as it is so small. Thus this means that the spontaneous emission rate (i.e. fluorescence) far exceeds that of stimulated emission. Under normal circumstances where the radiation is weak, as in a spectrophometer, and $N_2 \ll N_1$, the linear absorption spectrum is obtained.  If many photons can be concentrated into a few modes,  then for these modes, stimulated emission may exceed spontaneous emission; the result is a laser amplifier.

The figure shows the average number of photons/ mode with temperature, assuming Planck black body distribution vs. the photon frequency. The vertical lines are approximately the range of visible light. The horizontal dashed line is at $300$ K.


Effect of impurities on enthalpy of fusion of ice When I freeze water with some impurities (1% NaCl by mass), the mixture seems to have a much lower enthalpy of fusion that pure water. The impurities cause the freezing point to depress to 0.6°C. I only have a simple calorimeter available but it looks like the enthalpy of fusion is dropping by 25%+ compared to pure water (tested by melting the ice into water and measuring the temperature drop).

Additionally, the ice does not have a nice crystal structure. It is "mushy" and easily deformed.

This effect is surprising to me since colligative theory tells us that most properties changes in a pretty small linear way at low concentrations (like the freezing point).

Am I observing a real phenomenon or is there some problem with my testing?

If its a real phenomenon, is there a way to calculate/estimate the effect?

According to the article An Empirical Correlation between the Enthalpy of Solution of Aqueous Salts and Their Ability to Form Hydrates (Pandelov et al. 2010), your observation of a lower enthalpy is correct:

  We have found a correlation between the enthalpy of solution of the salt and its ability to form a hydrate, namely, that the salt’s enthalpy of solution is lower than the standard enthalpy of fusion of ice (6 kJ/mol)

However, the drop in enthalpy you observed of 25%+ seems excessive, as can be seen in the graph below:

Latent heat of fusion comparison between water, $\ce{NaCl}$ solution and ethanol solution.

Image source: Colligative properties of water cited from the original article Formulation of the latent heat of fusion of ice in aqueous solution (Kumano et al. 2009)

KUmano et al. 2009's methodology centred around the use of

  Differential scanning calorimetry analysis was performed and the measurements and calculated values compared with respect to the specific enthalpy of the ice/solution mixture.

As for the 'mushy ice' that you observed, it is likely to be a mixture of ice and brine, as per the diagram below:

Image source: Wikipedia


How much CO2 gets free when combusting 6 kg n-heptane (C7H16) I am very new to chemistry and have a question to an exercise.
It's about combustion (complete oxidation) n-heptane ($\ce{C7H16}$).

There were several questions I could answer, for example the chemical equation:
$$\ce{C7H16 + 11O2 -> 7CO2 + 8H2O}$$

But what I can't answer is, how I calculate how much $\ce{CO2}$ gets free when burning 6 kg of the n-heptane.

Can someone help me with this?
Thanks a lot!

Edit:
I am not sure if this information is needed:

molar mass (g/mol): $\ce{C}=12$ $\ce{O}=16$ $\ce{H}=1$

density of n-heptane: $0.684\ \mathrm{kg/L}$

For these types of problem, concept of mole is very handy. Here, you know the reaction stoichiometry, that means you have at least three information from that reaction equation. How?

** For every mole heptane, you need 11 mole oxygen for complete combustion
** For every mole of heptane combustion, you will get 7 mol carbon di oxide and 8 mol water.

So, now if you know the mole corresponding to 6 Kg heptane you can easily find out the amount of carbon di oxide produced in mole unit.
The relation is
$$n_{CO_2}=7 \times n_{Heptane}$$
$$n_{Heptane}=\frac{m_{Heptane}}{molecular~weight~of~heptane}$$
Now if you want to also calculate amount of carbon di oxide produced in mass unit or volume unit, you need to use the second equation or ideal gas law equation.


Periodic table for printing I there a Good Version of a printable Periodic table Which is minimalistic and gives necessary information only such as -

At. Number
Element Name
At. weight
Shell configuration
color coding of various groups or Series

The IUPAC periodic table (2016-pdf) is good because it quantifies the accuracy of the natural abundance atomic weights, but it doesn't give electron configuration.

The NIST periodic table has electron configuration, and probably is closest to what you want.

The NIST table is also good from a copyright point of view, because you can reproduce it without worrying about copyright infrigment, as US government works are generally not subject to copyright protection.


How to evacuate a glass vessel containing powders? I would like to evacuate a vessel (a glass round-bottomed flask) and which contains a few grams of platinum. However, I am worried about the platinum powder being sucked up by the vacuum.

A few people told me that this is possible if I take some care, but I am still not convinced. I use a turbo pump in combination with a diaphragm pump for the rough vacuum. So in the end a very expensive vacuum cleaner! I know that the powder will destroy the turbo pump if the powder is sucked in. Do any of you have some experience with powders in vacuum?

I often high-vac powders on Schlenk. Issues only arise when going between positive and negative pressures though to minimise any risk I often lightly plug the sidearm of the glass stopper/adapter(?) with cotton wool. Glass wool would also suffice and I am certain there are already stoppers/adapters which have a porous mesh build in to prevent powder from entering the vacuum system.


What is that process called in which we add water molecules to an ion to form a complex? An example would be $\ce{[Fe(H2O)6]^2+}$ in which it has 6 water molecules associated with it. What is the process of adding these water molecules called?

I believe it is normally just called "addition of water".
There is also the term "hydration" but this can also refer to the addition of water to alkenes or alkynes which forms alcohols (in the case of alkene; I am not sure about alkynes).
For example, $$\ce{CH2=CH2 \text{(ethene/alkene)} + H2O -> CH3CH2OH \text{(ethanol/alcohol)}}$$

I personally think "addition of water" is just fine to use and people will get what you are saying.


What's wrong with using tin in medicinal chemistry? I just read this In the Pipeline post and I was slightly confused by a statement on the use of tin. Lowe reports on this paper, which describes a synthetic route to spiro heterocycles using tin compounds, and caveats the finding with the comment that

  No one's crazy about using tin, but the transformation is too useful to pass up.

Why is the use of tin undesirable in synthetic chemistry?

Organotin compounds are rather toxic.  They are also persistent in the environment and have a long biological half-life.

The problem is that trialkyltin byproducts from your reaction are difficult to separate from the product.  In the lab, this is painful and (usually) involves multiple columns, but when you want to get a drug past the FDA into the clinic you have to meet purity standards for heavy metal contamination.  To a medicinal chemist, a transformation that won't make it into pilot plant and production is useless.


Does RMS velocity depend on number of gas molecules in a gas sample? The expression of RMS velocity clearly has square root of number of gas molecules in the denominator and square root of squares of velocities of the gas molecules in the numerator. Then why RMS velocity does not depend on number of gas molecules? By looking at the expression, it seems as if it does depend on number of gas molecules.

No, it does not depend directly on the number of molecules. The expression is
$$c_{rms}=\sqrt{\frac{3RT}{M}}$$
where $T$ is the temperature in K, $M$ is the molar mass in kg/mol and $R$ is of course the gas constant in J mol-1 K-1.
However, it is possible to have a situation where the number of molecule affects the temperature of the gas, which would change the r.m.s. velocity. For example, in Joule-Thompson effect, increasing the pressure causes the gas to heat up. It might be possible to do this by pumping in more gas into an insulated constant volume vessel. (although it sounds a bit contrived to be honest!)


Does sodium produce water when it reacts with water? My 10th grade chemistry book states that:

  "Metals react with water and produce a metal oxide and hydrogen gas. Metal oxides that are soluble in water dissolve in it to further form metal hydroxide. Metals like potassium and sodium react violently with cold water. In case of sodium and potassium, the reaction is so violent and exothermic, that evolved hydrogen immediately catches fire."

So $\ce{Na}$ should react like this:

(i) $\ce{2Na + H2O -> Na2O + H2}$

(ii) $\ce{Na2O + H2O -> 2NaOH}$

When hydrogen catches fire -- that is, when hydrogen reacts with $\ce{O2}$, right? -- does it form $\ce{H2O}$?
If water is not formed, then does the water level decrease?
Is (ii) an exothermic reaction?

The reaction of sodium metal and water could (and usually does) indirectly produce more water, if enough heat is produced to spontaneously ignite the hydrogen gas which is produced by the initial reaction. This cyclic type of reaction is often found in explosive or extremely exothermic reactions, e.g. $\ce{K+ + I- + H2O2}$ will also react violently in a cyclic way.

Combustion of hydrogen in air is this:

$$\ce{2H2 + O2 -> 2H2O}$$

So, in a (well) closed system with excess oxygen, water could re-form and collect. It's worth noting that dissolution of $\ce{NaOH}$ in water ($\ce{NaOH -> Na+ + OH-}$) is also exothermic and adds even more heat to the system.


How to calculate ring strain energy? Series A and B below can be viewed in two different ways to annulate an ethano-bridge onto four-, five-, and six-membered rings. Using the $\Delta_\mathrm{f}H^\circ$ values given (in kcal/mol), how can one calculate the strain energy for each compound?

Could anyone also explain whether the trend seen in each series is consistent with expectations based on additivity of ring strain?

Typically, ring strain energies (RSE) are calculated as the difference of the heat of formation of cyclic starting materials and less cyclic products in homodesmic reactions.

For cyclopropane, the RSE would be calculated using the following reaction:

$$\ce{(CH2)3 + 3 CH3-CH3 -> 3 CH3-CH2-CH3}$$


Magnetic properties of potassium manganate(VII) According to me, $\ce{MnO4^2-}$ is paramagnetic and $\ce{MnO4^-}$ is diamagnetic.
On the basis of this, the corresponding potassium salts inherit these magnetic properties, i.e. $\ce{K2MnO4}$ is paramagnetic and $\ce{KMnO4}$ is diamagnetic.
However, according to my class teacher $\ce{KMnO4}$ is paramagnetic.
Is potassium manganate(VII) dia- or paramagnetic?

Actually it is diamagnetic but due to a phenomenon called charge transfer spectrum (CTS), oxygen transfers one electron to manganese and $\ce{KMnO4}$ as a whole becomes paramagnetic. This is also the reason why it shows a characteristic colour while diamagnetic complexes are colourless due to absence of d-d transitions.


Why is CO2 not in the C∞v point group? It seems to me that CO2, since it can be linearly rotated around the z-axis without change of shape, ought to be in the C∞v point group.  However, in all the character tables I can find, it's listed as a D∞h.  What's the difference between these two, and why does CO2 fall in the D∞h and not the C∞v?

Both are linear with a $C_{\infty}$ axis, but $D_{\infty h}$ has a center of inversion and $C_{\infty v}$ does not.


Mechanism for Bial's test In biochemistry laboratory, it's common for undergraduate students to perform identification of carbohydrates and its functional groups. Bial's test, a method to check if any pentose is present, is one of some ways for the identification. The overall reaction can be found in Wikipedia page (https://en.m.wikipedia.org/wiki/Bial's_test).

The conversion of pentose to furfural has known to me (because I've found the paper) but not the second reaction. I guess some organometallic reactions are involved but not sure to write the mechanism.

I've tried to search the research articles related to the mechanism in Google, Google Scholar, ... or even, Scopus but can't find any paper related to it. This is indeed surprising to me because I thought this classical test should be investigated exhaustively either by spectrometry or computational perspectives.

Any thought or idea is appreciated to present the mechanism (even the mechanism is produced by direct computation).

The acid-catalyzed condensation of furfural with orcinol to form structure 1 is explained here for Molisch's test: Clarification in the mechanism for Molisch's test for glucose. Ferric chloride is the oxidant of 1 (see 2 ----> 3). Acid catalyzes the ring closure of quinoid structure 3 to intermediate 4, which rearomatizes to 5 with loss of water. Direct cyclization of 2 with acid is unlikely because it would interrupt the aromaticity of one of the orcinol rings.


Aromatic nomenclature, when to use numbers or prefixes? When you're naming aromatic nomenclature compounds, when do use numbers, such as 2,4-diiodonitrobenzene and when do you use prefixes such as p-xylenene? Thanks! I hope that makes some sense

1,4-dimethylbenzene and p-xylene are both equally correct nomenclatures. You could also say 4-methyl toluene if you wanted. I personally prefer the first example because it allows me to visualize the molecule immediately.

As for 2,4-diiodonitrobenzene, this is indicating three substituents. You have the nitro group, and two iodo groups - one para to the nitro, and one ortho to the nitro.


How do I visualise excited state MOs? I did an excited state calculation (TD-SCF)in Gaussian and wanted to see the MOs of the first excited state. But when I opened the checkpoint file in GaussView5 and clicked edit -> MOs, I could only see the configuration and MOs of the ground state. How do I see the excited state MOs in GaussView or other visualisers that can be downloaded for free?

What you see, when you open your checkpoint file, is all you can see. I assume you use TDDFT and with this method you can't obtain other orbitals than the ones from the ground state. Why is this so?

Because in LR-TDDFT you calculate your ground state and apply a perturbation to it. (You can look up on the behind theory in good books like Introduction to Computational Chemistry from Frank Jensen.) The result from this calculation is presented from Gaussian like this:

Excited State   1:      Singlet-A      1.0662 eV 1162.87 nm  f=0.0007  <S**2>=0.000
110 ->113         0.60155  (0.72)
110 ->115        -0.23139  (0.11)
110 ->116         0.22569  (0.10)
Excited State   2:      Singlet-A      1.2549 eV  988.02 nm  f=0.0002  <S**2>=0.000
109 ->113        -0.39288  (0.31)
110 ->114         0.51726  (0.54)
Excited State   3:      Singlet-A      1.2929 eV  958.99 nm  f=0.0027  <S**2>=0.000
109 ->113         0.50172  (0.50)
110 ->114         0.29522  (0.17)
(The weights in spaces are added by me.)

These are the configurations of three excited states. The first excited state, for example, consists mainly (72%) from an excitation from MO 110 (HOMO) to MO 113 (LUMO+2). There are some contributions with weights of about 10% into the MOs 115 and 116. (On how to get the weights, have a look at an answer by me on this site here.)

You ask, how you can see the excited state in GaussView. You can't ... directly. And this question is also a little bit tricky, because what is it that you want to see? Here are some options:

You could

calculate the electron density of the excited state TD(Root=1) Density(Current) and subtract the electron density from the ground state. Then this difference shows you the "hole" where now is less density and the "hole" where there is more density.
calculate Charge Density Differences (CDD) with Multiwfn or Electronic Difference Density Maps (EDDM) with GaussSum which show you the same but a little bit easier as you do not need an additional calculation because they are based on the excitations and the CI coefficients.
calculate the transition density for the excited state TD(Root=1) Transition=1
calculate Natural Transition Orbitals with, e.g., Gaussian or Multiwfn, for very diffuse excitated states as NTOs reduces/optimizes the number of contributing excitations for each root with Density=(Transition=1) Pop=(Minimal,NTO,SaveNTO)

Just have a look on the manual for TDDFT, Density, FAQ and Pop.


Is diamond an allotrope or polymorph of carbon? Is diamond an allotrope or polymorph of carbon? When is something considered an allotrope vs a polymorph?

Allotropy is a special case of polymorphism. Polymorphism is the phenomenon of a substance exhibiting different crystal structures. Allotropism is the same phenomenon limited to the subset of all substances that contains only the chemical elements. According to the Wikipedia article on polymorphism:

  Polymorphism ... is the ability of a solid material to exist in more than one form or crystal structure. Polymorphism can potentially be found in any crystalline material ...

The allotropy article defines allotropy:

  Allotropy or allotropism (coined from Greek "other" + "form") is the property of some chemical elements to exist in two or more different forms ... The term allotropy is used for elements only, not for compounds. The more general term, used for any crystalline material, is polymorphism.

Thus, diamond and graphite represent two different allotropes of carbon, which also makes them different polymorphs of carbon.

Water is well-known for having multiple solid phases. These phases are polymorphs, but, because water is not an element, they are not allotropes.


Why does the reaction between potassium permanganate and oxalic acid result in a yellow/brown solution? The reaction between potassium permanganate and oxalic acid is a redox reaction and produces $\ce{Mn^2+}$ ions, the final solution however goes yellow/brown.
If $\ce{Mn^2+}$ ions are colourless, and its complex ions are pink, where does the yellow/brown colour come from?

The reaction between $\ce{MnO4-}$ and $\ce{C2O4^2-}$ is performed under acidic conditions.

In acidic conditions only, $\ce{MnO4-}$ gets reduced to colourless $\ce{Mn ^{2+}}$ as, $$\ce{MnO4- + 8H+  +5e- -> Mn^2+ + 4H2O}$$

But if you don't maintain proper acidic condition of the medium, it continuously absorbs $\ce{H+}$. Eventually, due to lack of protons, it gets reduced to brown coloured $\ce{MnO2}$ under neutral conditions, where it needs less no of protons to get reduced than acidic conditions, as $$\ce{MnO4- + 4H+ + 3e--> MnO2 + 2H2O}$$ That's why you see the brown colour which is due to $\ce{MnO_2}$.


How do we identify left handed and right handed crystals? I was going through this article about how Louis Pasteur studied and explained the absence of optical activity in Racemic acid. Here's an extract:

  When Pasteur next examined crystals of the same
  salt of racemic acid, he found that they, too, were chiral
  but that some of the crystals were left-handed and others
  were right-handed. The crystals were related to one
  another as an object and its non-superimposable mirror
  image, and they were present in equal amounts. With a
  magnifying lens and a pair of tweezers, Pasteur carefully
  separated these crystals into two piles: the left-handed
  ones and the right-handed ones.

My guess is that the left/right handedness refers to the molecular spatial alignment. So,

How exactly can you assign the property of individual molecules to an entire crystal?
How can you determine if a crystal is left/right handed just by looking at it?

Yes, you are right that the handedness refers to the molecular 3D arrangement. Such handed molecules are stereoisomers known as enantiomers.

I think that the first part of your question will make more sense following an answer to the second part. So with regard to the part about discerning handedness by looking at it:

(I couldn't get to the article right now due to network issues, so I hope I'm not repeating too much.)

A molecule is chiral when one of the atoms in it has four unique groups attached to it.
Chiral molecules observably rotate plane-polarized light, either clockwise or anti-clockwise (see here for a picture of what plane-polarized light is).

Pasteur observed this along with the cancelling effect of a racemic mixture. As in any scientific endeavor, Pasteur was building upon the efforts of previous work--in this case, that of Jean-Baptiste Biot in studying tartaric acid. Pasteur's careful work enabled him to deduce that the naturally occurring or biologically produced form of tartaric acid was enantiomerically pure because it rotated light and the mirror image crystals also rotated light opposite directions. This was in contrast to the racemic acid which did not rotate light since it was a 50:50 mixture that cancelled the rotation in either direction.

Bryn Mawr College has a helpful explanation of how enantiomers interact with light:

  You have probably [not] learned in lecture that chiral molecules have optical rotations and that enantiomers rotate plane polarized light in equal and opposite directions... This behavior upon exposure to plane polarized light is not some magical property that defies the general principles of enantiomers. The reason enantiomers interact differently with plane polarized light is that plane polarized light is chiral. Though polarized light is a planar phenomenon, it is actually comprised of two superimposed helical forms. You can think of the electrical component as oscillating in such a way that there are two electrical vectors rotating in a cork screw fashion in opposite directions away from the source of the propagation at the same rate. Since a helix is a chiral shape , the two versions of the light constitute an enantiomeric pair. If you resolve the vectors of the two superimposed helices, you end up with oscillation in a single plane...Enantiomers interact differently with the two forms of light. The basic idea is that when plane polarized light encounters enantiomerically pure, chiral molecules, one form of light will slow down more than the other as a consequence of the interaction. This results in one helix coming out of sync with the other. When the vectors are resolved, the resultant plane is rotated with respect to the original position. This change in position is what we mean when we say a compound has an optical rotation.

So looking at a crystal being exposed to polarized light should allow you to observe whether it rotates light to the right or left. I'm not sure how easily detectable to the eye the rotation is, but polarimeters are instruments which are used to measure the angle of rotation.


how can electrons exchange from 3d orbital to 4s if their energy levels are not equal? I was taught that electrons only exchange in orbitals of equal energy or degenerate orbitals. The energy of 3d orbitals is 5 and that of 4s is 4(by n+l rule), but then how is it that in the case of copper and chromium electrons exchange from 3d orbital to 4s?

A sort of simplified answer but here goes...

Without some other excitation, electrons only "fall in" towards the nucleus. This is sort of like the fact that rocks roll down hill not up hill. So once an electron "falls in" it has excess energy that must be shed. It can either give the energy off as a photon or transfer the energy to one or more other electron(s) which then go into some excited state (Auger effect).

The transitions occur  according to selection rules. If the selection rules don't allow a particular transition then it is forbidden. If the atom is in free space with no other perturbations, then forbidden transitions would not occur. However for molecules, in liquids and gases, because of vibrational and rotational microstates such a transition may actually occur but it will be relatively weak in intensity to the transitions allowed by the selection rules. Likewise forbidden transitions may occur in solids because of the vibrations of the atoms within the crystal which break the symmetry of an atom, or its excited state.

Now if an atom or molecule is excited by an external energy source, then an electron can move from a lower energy orbital to a empty higher energy orbital. Again these transitions obey the selection rules.


Where can I access raw values for near infra red absorbance of liquid water? I have encountered several sites where water's near-infra-red absorption spectra has been plotted as a figure (for e.g, here). Is there a site where I can access these raw absorbance/transmittance values? Eyeballing is prone to errors and inaccuracies.

What I have found: NIST has a nice plot of the IR absorption spectra of liquid water, and I can potentially access the raw data by looking at the plotting functions, but as you can see it has no near-IR data.

I found a database called HITRAN that offers what I wanted. They support a python API that can be used to pull the relevant data. The code I used is written below (you'll need the hapi.py file from here).

from hapi import *
from matplotlib import pyplot as plt
fetch('H2O',1,1,7500,11500)
nu,coef = absorptionCoefficient_Lorentz(SourceTables='H2O')
plt.plot(nu,coef)
plt.show()