Is there a systematic way to determine the expected cubic structure of metals? I've noticed that some groups of metals with similar properties tend to belong to the same cubic structure group, such as $\text{FCC, BCC, or HCP}$. What is a good explanation as to why these metals have similar cubic structures, and how can one determine a metal's expected cubic structure just from knowing its identity (if there is a way)?

For example, copper, silver, and gold all have face centered cubic structures.

This trend seems to break down at some point, I think, because $\frac{2}{3}^{rd}$ of the platinum group have $\text{FCC}$ structures and $\ce{Ru}$ and $\ce{Os}$ have simple cubic structures.

There is no systematic way (besides brute force - that is a system, right?) to predict the crystal structure of the metals.  Even sticking just to the transition metals, as you have noted, there are quite a variety.  The primary ones are as you have noted $fcc$, $bcc$, and $hcp$ (ignoring the rhombohedral Hg).

Several issues arise here. First, there is little difference between $fcc$ and $hcp$ - they are both close packed structures with the only the stacking order changing (ABCABC for $fcc$ and ABABAB for $hcp$). An estimate of the free energy difference between these two crystal structures can be obtained 'SGTE Data for Pure Elements', A.T. Dinsdale, CALPHAD 15(4) 317-425 (1991), the compilation of elemental thermodynamic data. Here one finds that for Ag ($fcc$), the $hcp$ phase is only 300J/mol higher free energy (at T=0K).  Similarly, for Co ($hcp$), the $fcc$ phase is only 427 J/mol higher. So, the free energy differences aren't necessarily that much, making an $a\ priori$ differtiation impossible. (I'll note here that Ru and Os are $hcp$ crystals, not simple cubic as you stated).

The second factor that is difficult to determine is the impact of magnetic free energy on the equilibrium crystal structure (as included in the above reference).  Iron is the most obvious candidate to consider this, and it does not disappoint.  As observed, the equilibrium crystal structure at room temperature is $bcc$.  Heating leads to a change to $fcc$ at $910C$, then back to $bcc$ at $1400C$.  If the magnetic contributions to the free energy are ignored, the observed crystal structures would instead be $hcp$ at low temperatures, a transition to $fcc$ at around $200C$, and a brief appearance of $bcc$ right before melting. I see no way that one could predict these particular changes in phases with temperature.


How does one apply the knowledge of the direction of a molecular dipole moment? I was watching Khan academy's video about molecular dipole moments, which explained what they are and how they can be calculated. I also read this article.

Both sources explain the concept similar to common organic chemistry textbooks, but they did not provide any information regarding the uses or applications of a molecular dipole's direction, nor did the google search I conducted later.

So far, two things are clear to me:

A dipole moment is a vector, and therefore has a quantity and a direction.
Larger dipole moment is associated with a more polar molecule. This a good enough reason to find out how large is the dipole moment.

My question is: What possible applications can knowing the direction of a molecular dipole moment have?

Take a look at chemlibretexts. I'll just write an overview.

A dipole is a vector representing the charge distribution in a molecule. It's drawn from the positive-density charge side towards the negative.

Some examples

Phase change

How do we explain $\ce{CO2}$ fusion temperature is $-78\,^\circ$C while $\ce{SO2}$ is $-10\,^\circ$C? Because of its dipole magnitude. In the first one, there's no net dipole, but there is for the second one.

The bigger molecular interactions the more they oppose to temperature-increasings.

Solubility is higher for alcohols than lipids* in water. Again dipole-dipole interaction is stronger (energetically more favourable) than dipole-induced dipole interaction.
CO$_2$ has no electric dipole in the symmetric vibration. According to quantum mechanics, if electric dipole does not change when it absorbes light of a specific energy, it will not be susceptible of a infrared measure. This particular vibrational mode will be (and it is) infrared inactive.


Deodorizing using baking powder instead of baking soda I am an engineer and I understand the difference between baking power and baking soda as baking powder is baking soda (alkali) with tartaric acid (acidic) and an inert starch (why? Is this a catalyst in the presence of humidity and temperature).

So why use baking soda to deodorize a carpet and not baking powder? Is it because baking powder has both an acid and base and thus one would neutralize the other and would render the baking powder useless for adsorbing smells?

My carpet would be at a temperature of about $25~\mathrm{^\circ C}$. I am guessing this temperature and a relative humidity of $50{-}60~\%$  will not catalyze a violent reaction with either baking powder or baking soda?

Any input from chemists would be useful!
I don't want to use either on my synthetic carpet in my rented house to experiment before I get enough input from people with know-how!

Baking soda is an effective deodorant because of its basicity. Many foul-smelling compounds are acidic. Neutralizing the compound gives it an ionic character which reduces the vapor pressure (meaning it is less available for the nose to smell it) and makes it much more water soluble (it can be washed away). Odorous compounds like fatty acids (butyric acid is BO) and thiols  have much higher $pK_a$s (~4.8 and 7-10, respectively) than tartaric acid ($pK_a$ ~3) means that they would remain un-neutralized with baking powder. It would not be an effective  deodorizer.


How does the polarity of the eluent and sample affect the Rf value in thin layer chromatography? A three-component mixture (X, Y, Z) was separated using normal phase TLC. Three eluents of varying polarity were used. Select the TLC plate that was developed using the least polar eluent (click on the corresponding TLC number 1, 2, or 3 on the bottom of the plate). On TLC plate 3, select the least polar compound (click on X, Y, or Z spot on plate 3).

Does normal phase TLC mean silica gel TLC? If this is the case then plate 3 has the least polar eluent because the compounds are at the top. For the same reason compound Z would be the least polar.

The term eluent means solvent doesn't it.

Never in my years in the lab have I seen a more polar eluent that causes a lower $R_\mathrm{f}$ value than a less polar eluent. It also does not make sense theoretically.

TLC 2 is obviously the one with the least polar eluent and compound $z$ is the least polar compound.

I fully disagree with half of MaxW’s proposed answer.


Which set of reactants is better for preparation of 1-methoxy-4-nitrobenzene? I've got two set of reactants:

4-nitrobromobenzene + sodium methoxide
sodium 4-nitrophenoxide + bromomethane

I've got to tell which reaction is more preferable among these two.

The first reaction takes place through nucleophilic addition of methoxide on the ring followed by elimination of bromine. I've been taught that such reactions take place under drastic conditions, but the presence of an electron withdrawing group at ortho and para position increases the reactivity, so the presence of $\ce{NO2}$ group should increase the reactivity .

The second reaction takes place by SN2 mechanism, where bromomethane is attacked by the nitrophenoxide ion. This reaction "seems" to be better than the first. (Is it?)

Hence, my attempt is that the second reaction should be more favourable than the first since the first reaction should take place under more drastic conditions than the second.

Am I right? Or does the presence of $\ce{NO2}$ group increase the reactivity so much that it is comparable to the second reaction or maybe better?

Drastic conditions for reaction 1? Not really. Reflux the reactants in $\ce{MeOH}$ and the reaction should go smoothly and in high yield.

Reaction 2 - two difficulties.
The anion of 4-nitrophenol is very stable and, because of the electron withdrawing power of the nitro group, it is not very nucleophilic. Second difficulty is that bromomethane has a boiling point of 3C so the reaction mixture has to be kept cold to keep it in the reaction mixture. This does not help you overcome the poor nucleophilicity of your phenoxide.

Without running the two reactions it is guesswork to predict which will give the better outcome, but I favour the SNAr.


How would one construct a 3D model of a molecule based on the crystal structure of its salts? Often crystal structures found online from past published crystallography experiments are of the salt of a molecule of biologic interest, as opposed to the freebase form of the molecule itself. For example, I have a CIF file for mescaline hydrobromide (CCDC 1211316) at the moment that I would like to create a 3D model of mescaline from. Would removing the extra hydrogen atom on the amine group (i.e., changing the $\ce{NH_{3}^+}$ to $\ce{NH_2}$) and the free bromide ion suffice or will I also have to optimize the structure's geometry? If I do need to optimize it, or otherwise modify it beyond the procedure I just pointed out, please tell me how to do it with free software (Avogadro, for instance).

In general, the crystal structure of a molecule is a highly accurate geometry. So yes, you can take a salt and modify as needed. (The caveat in the comment is useful - keep in mind that different groups may be protonated or deprotonated depending on solvent and pH.)

There are two other caveats:

1) The conformation in the crystal structure is not necessarily the lowest energy conformer in solution or the gas phase. You may wish to perform a conformer search to find a low-energy minima.

2) If you are performing further calculations (e.g., DFT for electronic structure, force fields for MD, etc.) the geometry of the crystal structure will not likely be a minima of the computational method, so geometry optimization or equilibration (for MD) is needed before further simulation.

That said, the crystal geometry is a good place to start for many uses. Certainly if a crystal structure is available, I'd suggest you use it rather than build a geometry by hand.


Why is Kw the same in acid, water, and alkali? Why is Kw the same in acid, water, and alkali? Given that Kw equals to [H+][OH-], doesn't concentration of these 2 differ in acids, water and alkali? Why is Kw only affected by temperature?

You are correct that the concentration of $\ce{[H+]}$ and $\ce{[OH-]}$ both vary based on the the composition (acid/alkaline) of the solution, but the remarkable thing is that their product does not. When $\ce{[H+]}$ goes up, $\ce{[OH-]}$ goes down, in the same proportion.

The reason for this is that the Kw represents the equilibrium constant of the following reaction, and any excess (lack) of $\ce{[H+]}$ or $\ce{[OH-]}$ will react by LeChatelier's principle to reestablish equilibrium:

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

We can write the standard equilibrium constant equation as follows

$$K_{eq} = \frac{\ce{[H+][OH-]}}{\ce{[H2O]}}$$

For aqueous solutions, the concentration of water is a fixed constant1, so we can multiply both sides by it, and fold it into the equilibrium constant2. Therefore we can write:

$$K_{w} = \ce{[H+][OH-]}$$

So just as standard equilibrium constants do not differ with the concentrations of chemical species in the solution, but can vary with temperature, so too does $K_{w}$ not vary based on the concentration of chemical species, but does depend on the temperature of the solution.

1 Technically, the values in the equilibrium constant equation are activities rather than concentrations (see here and ref 1 therein for a rather technical discussion), a distinction we normally ignore, as activities are proportional to concentration in dilute solutions. Regardless, the activity of water in most dilute aqueous solutions can be considered a constant, so it doesn't change the argument.

2 Note this is only true in a solution which is almost entirely water, with just a little bit of solute dissolved. If you had a water/ethanol mixture, or a very concentrated solution of a highly soluble salt, then the water activity can't be considered a fixed constant, and you'd have to start making some adjustments based on it. (This is where the distinction between "concentration" and "activity" come into play - these adjustments are more complicated than just dividing by the nominal concentration of water.)


What is the exact meaning of the term 'isostructural' molecules? Wikipedia:

Isostructural

  Isostructural chemical compounds have similar chemical structures.

Chemical Structure

  A chemical structure determination includes a chemist's specifying the molecular geometry and, when feasible and necessary, the electronic structure of the target molecule or other solid. Molecular geometry refers to the spatial arrangement of atoms in a molecule and the chemical bonds that hold the atoms together, and can be represented using structural formulae and by molecular models;[citation needed] complete electronic structure descriptions include specifying the occupation of a molecule's molecular orbitals.

My Doubt

Given that, while practicing, I encountered a question asking whether $\ce{NO3-}$ and $\ce{SO3}$ are isostructural or not. If we go as per the geometry, they are isostructural ($\ce{sp^2}$ hybridization). But if we consider the electronic structure, the former has bond order 4/3, the latter has 2 (assuming $\ce{S=O}$), and so may be, they are not isostructural. (In addition, they also have different total number of electrons.) This (that they are not isostructural) is the answer.

So, this question. Is the term defined or is it used without any strict definition?

Two molecules are isostructural if they have a similar number of atoms arranged in a similar structure (spatial coordinates). In your example, both $\ce{NO3^-}$ and $\ce{SO3}$ have same number of atoms, and they have similar spatial arrangement as well. So they are isostructural. Sometimes using hybridization of the central atom works in finding out, but not always. For example $\ce{SCl3}$ and $\ce{BF3}$ are isostructural, even though the former is $\text{sp}^3\text{d}$, while the latter is $\text{sp}^2$.


Does quantum yield of reaction factor in number of electrons needed? Can someone tell me if quantum yield of a reaction factors in the number of electrons needed to create the particular species? If not, is there a metric that does?

For example, I have a 4 electron reduction which means that even if I have 100% efficiency, my quantum yield is

$$\frac{\text{# molecules reduced}}{\text{# photons absorbed}} = 0.25$$

Is there a better metric I can use so that 1.0 will be 100% efficiency?

I'm not aware of a definition of quantum yield that factors the electrons in. The normative reference was/is Silvia Braslavsky's Glossary of terms used in photochemistry (SRC).

Unless I misunderstood, you do have an overall four electron reduction, but this happens in a sequence of four excitation events.

From a formal point of view, the quantum yield(s) should be given for each of these processes. I reckon that this isn't possible, so an overall quantum yield of 0.25 is all you can get. In the end, I'd simply explain that, rather that looking for some "tricks" to come up with a more impressive number.


Are there any other reagents except HNO3 and H2SO4 for Nitration? Is nitration of benzene possible by using any other reagents except using HNO3 and H2SO4?

It's getting on for 3 decades since since I've done organic, so a proper organic chemist can probably give a more complete answer, but yes, there are other ways to do this. In the $\ce{HNO3}$/$\ce{H2SO4}$ system the "real" nitrating agent is nitronium, $\ce{NO2+}$, so anything that can form that will nitrate benzene. One example is simply to use a nitronium salt such as nitronium tetrafluoroborate, $\ce{NO2+ BF4-}$. As for others I would be surprised to find that something like $\ce{NO2F}$/$\ce{BF3}$ doesn't work, but I don't know for certain - it's been a looooong time!


What concentration of sulfur hexafluoride is safe to inhale I want to do a density demonstration, and I want to inhale helium and sulfur hexafluoride as a fun demonstration. I am open to using any source to purchase from, and money is not an issue (but preferably as inexpensive as possible), but what concentration should I buy (what concentration is safe to inhale), and from where?

I'm afraid this answer doesn't address where you can get $\ce{SF6}$, as I don't know any suppliers myself (it would probably help to specify where you are, though). I just want to straighten out the safety details and make sure you (and anyone else who reads this in the future) know what you're doing. All chemistry experiments can take a turn for the worse if done without adequate preparation.

Pure sulfur hexafluoride is a rather chemically inert gas (unlike the related but very toxic sulfur tetrafluoride $\ce{SF4}$ and disulfur decafluoride $\ce{S2F10}$), and it can be safely handled in regular situations. Decomposition of $\ce{SF6}$ usually requires aggressive conditions, such as exposure to molten sodium metal at 200°C. There are some situations where $\ce{SF6}$ reacts at low temperatures, but that requires irradiation with deep-UV photons or direct contact with solvated electrons from alkali metals dissolved in ammonia. I don't suppose you'll have to worry about such situations. It also seems that $\ce{SF6}$ can decompose into a small amount of the other aforementioned toxic sulfur fluorides if it is exposed to electrical arcs, but since it is an electrically insulating gas, the electric fields required for arcing are even higher than in air, so this also shouldn't be a problem.

All in all, both pure $\ce{SF6}$ and helium are very unlikely to cause chemical damage to the body, even indirectly. However, that doesn't imply they are completely safe. Of course, they can still act as an asphyxiant simply by displacement of oxygen in the lungs (make sure to read this!). Be aware that the brain is incapable of directly measuring the oxygen content in your blood, and instead measures oxygenation rather indirectly, by the amount of carbon dioxide. It is possible to faint and even die from hypoxia without any warning. Additionally, since $\ce{SF6}$ is relatively heavy, it requires more effort from your diaphragm muscles to expel from your lungs, making $\ce{SF6}$ inhalation slightly more dangerous than helium.

As a safety measure, do not take consecutive breaths in the non-oxygenated gas and do not spend more than 5 seconds without inhaling air. While your lungs are devoid of oxygen, not only are they unable to transport oxygen into your blood, the lungs actually work in reverse and help expel oxygen from your blood. After performing the trick, take deep breaths in regular air for a minute or two before trying it again, to be sure all the previously inhaled $\ce{SF6}$ or helium has exited your lungs and your oxygen supply has returned to normal. Hyperventilating before or after inhaling $\ce{SF6}$/helium is likely a bad idea, as it further messes with the balance of oxygen and carbon dioxide in the blood. You should avoid any possibility of becoming light-headed and fainting, especially if you decide to do the trick multiple times. Also be mindful that the gasses can accumulate in poorly ventilated spaces, with $\ce{SF6}$ being a particular problem as it will tend to stay close to the ground and disperse with more difficulty.

If you're going through the trouble to get $\ce{SF6}$, you should do the floating aluminium boat display, too. It's pretty cool! Here's another video showing you can even pour the gas into the boat and make it sink.


Why is Iron very rare in seawater compared to many other elements? I don't know the associated technical words to research this question.

There was high presence of Iron in the oceans in anoxic precambrian times, and today copper is 5 times more abundant in the oceans than iron.

Is the solubility of $\ce{Fe^{2+}}$ and $\ce{Fe^{3+}}$ and $\ce{Cu^{2+}}$ very different?

Why is there five times more Copper in seawater than Iron? Surface minerals have 1000 times higher PPM values for the latter? It's precipitating?

Why is there 4000 times less iron in seawater than Strontium? Surface minerals have 100 times higher PPM values for the iron?

Here are some measurements stated previously:
https://web.stanford.edu/group/Urchin/mineral.html
https://en.wikipedia.org/wiki/Abundance_of_elements_in_Earth%27s_crust

I'll elaborate this answer as I learn.

It's an effect owed to the low solubility of iron in the modern oxic ocean and the tendency for iron to be ‘scavenged’ from the water column by sinking particles.

Brand, L. E., Sunda, W. G. & Guillard, R. R. L. Limitation of marine
phytoplankton reproductive rates by zinc, manganese and iron. Limnol.
Oceanogr. 28, 1182–1198 (1983).

Cooper, L. H. N. Iron in the sea and in marine plankton. Proc. R. Soc. Lond. B
118, 419–438 (1935).

Goldberg, E. D. Marine geochemistry. 1. Chemical scavengers of the sea.
J. Geol. 62, 249–265 (1954).


Grignard reagents react with epoxides but dissolve in THF, what happens in oxetane (four membered ether ring)? Ring strain causes the epoxides to react and THF does not react because the ring strain is not large enough. Does the four membered ring react or not? Does it favor reaction with the harder or softer nucleophile perhaps - thus only reacting with some Grignard reagents?

In a 1979 paper entitled Copper-catalysed reactions of Grignard reagents with epoxides and oxetane, Huynh, Derguini-Boumechal, and Linstrumelle report that the copper-catalysed ring opening of oxetane by various Grignard reagents gives the alcohols "in acceptable yields":

$$ \begin{array}{cccc}
\ce{R} & \ce{X} & \text{Solvent} & \text{Yield (%)} \\
\hline
n\text{-}\ce{C4H9} & \ce{Cl} & \text{Ether} & 75 \\
\ce{C6H5} & \ce{Br} & \text{THF} & 52 \\
\ce{CH2=CH-CH2} & \ce{Br} & \text{Ether} & 50 \\
\ce{(CH3)2C=CH-CH2} & \ce{Cl} & \text{THF} & 50
\end{array} $$


Trifluoroacetic acid, Formic acid and dichloromethane I am working in a small lab (in a small company) with very poor conditions, which I have to improve (for example: no sink , I have to wash the glassware in the bathroom). My organic chemistry is very poor, I am new to using solvents, and I have several basic doubts:

How to correctly wash the glassware after use?
How to correctly dispose of wasted solvents?
Should I leave the chemical fume hood on after using it and, if so, for how long?

I basically only do electrospinning here, using the respective solvents:
- Trifluoroacetic acid
- Dichloromethane
- Formic acid

How to correctly wash the glassware after use?

The glassware must be rinsed with a non-toxic (or less-toxic) solvent such as acetone or isopropanol before washing to carry over the bulk of hazardous materials.  You should the wash the glassware with soap and water, rinse with water, followed by a rinse with DI water, isopropanol, or ethanol to eliminate water spots.

  How to correctly dispose of wasted solvents?

Your solvents should go into an appropriate waste container and be marked as hazardous waste with the contents listed.  You also need to list the companies EPA number to dispose of the waste.  Have a reputable waste disposal service receive your waste and take it to a disposal facility.

  Should I leave the chemical fume hood on after using it and, if so, for how long?

If anything hazardous is still in the hood you must leave it on. If there is nothing in the hood there is no reason you must leave it running.


Why are water droplets shaped like that? With nothing to do, I stared at the droplets that get condensed on the glass panel of my window.

Upon examination, these droplets appear to be in some sorta pattern. There are big droplets as well as small ones that fill the space between the big ones.

The ones in the middle, between rows of small droplets are usually big one.

All these intrigue me to think about what are that factors that contribute to this pattern.

My guess: I think vapor gets condensed on dust particles thus forms those droplets in the picture. But what accounts for the different sizes of the droplets?

From the Wikipedia article on surface tension:

  Surface tension is responsible for the shape of liquid droplets. Although easily deformed, droplets of water tend to be pulled into a spherical shape by the cohesive forces of the surface layer. In the absence of other forces, including gravity, drops of virtually all liquids would be approximately spherical. The spherical shape minimizes the necessary "wall tension" of the surface layer according to Laplace's law.

In short, the more surface tension is, the rounder shapes of water you get. And the opposite goes for gravitational potential energy: The lesser gravitational acceleration results in more spherical droplets of water.

The symbol for surface tension is $\gamma$.

  $\gamma (\ce{H2O}) = 72.8~\mathrm{dyn~cm^{-1}}$ (at $20~\mathrm{^\circ C}$)
   $\gamma (\text{mercury}) = 465~\mathrm{dyn~cm^{-1}}$ (at $20~\mathrm{^\circ C}$)[1]

That's the reason you hardly ever see mercury drops out of their spherical shape.

In short

  The spherical shape minimizes then necessary "wall tension" of the surface layer according to Laplace's law.[2]

Oh and I almost forgot: This great article - Antonin Marchand, Joost H. Weijs, Jacco H. Snoeijer, and Bruno Andreott, Why is surface tension a force parallel to the interface?. Physique et Mecanique des Milieux Heterogenes, 2012 - is very nice in case you wanted to do additional study in the case.

References:

R Nave. Surface Tension. URL (accessed April 25th, 2018)
R Nave. Surface Tension and Droplets. URL (accessed April 25th, 2018)


Reaction of a vicinal diol with acetone I have the first part correct; however, I can't seem to understand what the 2nd part's answer is. I realize the water leaves, but then what?

There was this guy who recently protected a ketone in the presence of an ester. You might have heard about him ;)


How does molecular structure affect physical properties? The question sounds vague and non-specific because my background in chemistry is limited to undergrad college courses.  However, I am interested in learning the correlations between the atom-by-atom structure of molecules (simple, complex...) and the resultant physical properties thereof.

So far, it seems most the material online directly referencing the phrases "molecular structure" and "physical properties" is in the form of highly specialized research publications.  Surely, there is a Basics to the Relationships between Molecular Structure and Physical Properties 101 out there, somewhere.

Ultimately, my questions sum up to the following:

Where may I go to learn these things?
What material may I read?
Who might I ask?
Where do I begin?

Thank you very much.

Predicting (approximate) physical properties starting with molecules is complicated, below.  Nobody has ever produced a way to accurately predict how molecules will pack in a crystal structure.  Big Pharma is desperate to predict crystal polymorphs and their properties. Slightly vast amounts of grant funding pursued predicted super-explosive octanitrocubane.  When it was finally synthesized it crystallized in a less dense crystal structure than expected, deeply disappointing the DoD.

http://www.chemspider.com/ACDLabs.aspx
http://pubs.acs.org/doi/abs/10.1021/ed064p575
http://molsoft.com/mprop/

If you have a way to predict crystal structures given complex formula units, you're a big winner.


Can electrodes be made from mercury? 'Mercury' was included in one of the possible answers for this exam question:

Name a metal that an inert electrode could be made from.

My answer was 'platinum' which was also a possible answer, but I’m curious how mercury, which is a liquid metal, could be used as an electrode during electrolysis?

The dropping mercury electrode (DME) is a working electrode made of mercury and used in polarography. Experiments run with mercury electrodes are referred to as forms of polarography even if the experiments are identical or very similar to a corresponding voltammetry experiment which uses solid working electrodes. Like other working electrodes these electrodes are used in electrochemical studies using three electrode systems when investigating reaction mechanisms related to redox chemistry among other chemical phenomena.

Source: dropping mercury electrode
Massive thanks to Ed V for advice in the comments prompting my research


Is activation energy time-dependent? I was thinking lately about activation energy's dependency on time. I couldn't find any information on the internet. Only answer that came up to my mind is the fact that activation energy is constant, and doesn't usually change with time.

But what about auto-catalytic reactions? Maybe their activation energy changes with time after creation of small amount of product, at least theoretically. What do you think?

The activation energy of a single reaction does not change with time but there might be more than one reaction happening

An activation energy is always constant for a given single reaction. But that doesn't mean the reaction speed will be constant (it might depend on concentration or temperature).

But what about autocatalytic reactions? There are two possible sources of confusion here. A very simple autocatalytic reaction might involve a changing concentration of an intermediate. In this case the reaction might go faster because that concentration is changing as the intermediate builds up (but the activation energy stays the same). In more complex cases, the catalytic route to the product involves a different reaction. What you have is not a changing activation energy for a specific reaction but the addition of a different reaction pathway to give the same product. This reaction will have a different activation energy but the activation energy of the original pathway is not changed.


How to identify the ratio of oxidation states in terbium ions in its mixed oxide? Estimate the ratio of $\ce{Tb^3+}$ to $\ce{Tb^4+}$ in  $\ce{Tb_6O_11}$.

For my solution I knew that I had to get a sum of 22 positive charges from 6 $\ce{Tb^3+}$ and $\ce{Tb^4+}$ ions, so I just added different combinations until I found that $4(\ce{Tb^4+})+2(\ce{Tb^3+})$ gave me the desired value to have an overall neutral oxidation state on the molecule. This gives a $2:1$ ratio also. Is this the correct way to do this, or is there a more formal method?

There is certainly a formal way. Let $x$ and $y$ be the number of $\ce{Tb^3+}$ and $\ce{Tb^4+}$ ions respectively. Then:

$$\begin{align}
3x + 4y &= 22 \\
x + y &= 6
\end{align}$$

This is a simple system of two linear equations in two unknowns, and I am sure you know of many ways to solve it. Let's just solve it using matrices for the fun of it:

$$\begin{align}
\begin{pmatrix}
3 & 4 \\
1 & 1
\end{pmatrix}
\begin{pmatrix}
x \\
y
\end{pmatrix} &=
\begin{pmatrix}
22 \\
6
\end{pmatrix} \\
\begin{pmatrix}
x \\
y
\end{pmatrix} &= \frac{1}{3\cdot 1 - 1\cdot 4}
\begin{pmatrix}
1 & -4 \\
-1 & 3
\end{pmatrix}
\begin{pmatrix}
22 \\
6
\end{pmatrix} \\
&= -
\begin{pmatrix}
1\cdot 22 - 4\cdot 6 \\
-1 \cdot 22 + 3\cdot 6
\end{pmatrix} \\
&= \begin{pmatrix}
2 \\
4
\end{pmatrix}
\end{align}$$

as you found. It seems that trial and error may well be faster!


How do we indicate a volume fraction like 0.05 in procedure descriptions? I'm translating a Russian biotech text, and there's this sentence with a number without measurement units:

  The contents of the wells were discarded; the wells were washed three times with 200 µL of Buffer containing 0.05 of Tween 20.

I was told that this figure indicates a volume fraction, and is dimensionless, hence no units after it. Will this be understood by the native speaker of English as-is, without any units, as in my quote?

Or is it better to write it down as:

  The contents of the wells were discarded; the wells were washed three times with 200 µL of Buffer containing 5% (v/v) of Tween 20.

Or maybe like this:

  The contents of the wells were discarded; the wells were washed three times with 200 µL of Buffer containing 5 vol% of Tween 20.

Is there a way to retain 0.05 and somehow indicate that this is a volume fraction?

I think it would be wise to include units, because how would you distinguish between mass and volume fractions? Options 2 and 3 that you present are both seen in the scientific literature and text books. My opinion is that option 3 is the least ambiguous way to present both the numeric value and the context (i.e. the units).


What is the Solubility of Common Noble Gases in Concentrated Phosphoric Acid, 85% w/v? By Henry’s Law, the concentration of a dissolving gas in a liquid is proportional to the partial pressure of that gas in contact with the liquid, mathematically,
$ p = k_\mathrm{H} c$,
where $p$ is the partial pressure of the dissolving gas, $c$ is the concentration of the dissolving gas in the liquid (solution), and $k_\mathrm{H}$ is a constant (Henry’s constant?) which depends on temperature.

What is the value of $k_\mathrm{H}$ for common noble gases (in particular, xenon) in concentrated phosphoric acid, at, say $298~\mathrm{K}$? Can the solubility at other temperatures be calculated using an extrapolation formula?

Value seems to be $k_H^{\ominus}=4.3\times10^{-3}\;M/atm$ for Xenon in water.For it in conc. phosporic acid seems not to be available as probably no one has carried the experiment in this solvent since water is most common and used commonly.Maybe you would?

Also, A simple way to describe Henry’s law as a function of temperature is:
$$\LARGE k_H=k_H^{\ominus}\times e^{\left(\frac{-\Delta_{soln}H}R\left(\frac1T-\frac1{T^{\ominus}}\right)\right)}$$
where $\Delta_{soln}H=$enthalpy of solution. Here, the temperature dependence is:
$$\frac{-d\ln k_H}{d(1/T)}=\frac{\Delta_{soln}H}R$$

Useful Links:
Table of constants, The NIST Chemistry WebBook, The Pesticide Properties Database (PPD)


Why are beta-ketoacids better at decarboxylation than gamma or delta? My book claims that beta-ketoacids and beta-dicarboxylic acids are better at decarboxylating than their gamma and delta counterparts. Why is that?

I would assume that since gamma and delta keto-acids make a larger ring intermediate (less ring strain) and thus the energy barrier for reaching that intermediate would be lower...

It has to do with the stability of the electrons that are ejected from the bond that breaks. In B-ketoacids, the electrons can resonate onto the oxygen and form an enol/enolate intermediate, which is much more stable than forming a carbanion on a gamma or delta ketoacid (because there is nowhere for the electrons to go other than go on the carbon atom that participates in the bond-breaking event).


Why does fusing benzene rings not produce polycyclic alkynes? It would make sense that when the benzene rings are fused they keep their pi bonds and form a polycyclic alkyne. However this is not the case.

My hypothesis is that because monocyclic alkynes at 8C or above are stable polycyclic alkynes would be even more stable and more stable with smaller rings than monocyclic alkynes.

However with 7 benzene rings fused in a hexagonal tiling type fashion the center ring has all single bonds like cyclohexane.

Why is it that fusing benzene or really any aromatic compound does not form polycyclic alkynes?

A resonance structure of this would be a polycyclic alkyne along with just the regular pi bond delocalization.

Bicyclo[4.4.0]deca-1,3,6,8,tetraene-4-yne is what I am thinking about. I am moving just the pi bonds in the resonance structure to get this. I am not moving actual carbon nuclei.

I can't find a picture of this but hopefully from the name you can figure it out.

The ideal bond angles for alkynes are $180^{\circ}$, which is necessary for optimal overlap of atomic orbitals combining to produce mutually perpendicular $\pi$-type molecular orbitals. Any significant deviation from this is normally inherently destabilizing. Rings, by the very nature of their geometry, often incur additional strain due to their conformational inflexibility and possible demands for non-ideal bond angles (relative to acyclic molecules, that is). Introducing $\pi$-bonds into rings can potentially compound this problem, since they force at least local planarity and shorter bond lengths. These factors all contribute to the immense instability of cyclic alkynes in general.

When you mention stable cycloalkynes, keep in mind that stability is a relative term in this context. While those with a sufficient number of carbons can be produced and isolated under specific experimental conditions, they are not thermodynamically stable relative to their open-chain equivalents, or to most of their various possible structural isomers. In some cases, they can, however, be made kinetically stable by the addition of flanking $\alpha$-substituents which inhibit reaction by their steric bulk1.

Syntheses of cycloalkynes below cyclooctyne are often elaborate, and frequently proceed through extremely reactive intermediates that decompose by specific reaction pathways so as to carefully avoid other products. Benzene (and the various polycyclic aromatic hydrocarbons, for that matter), on the other hand, is an extremely stable molecule which participates in quite a limited collection of reactions. Whatever particular reaction you're suggesting would almost certainly be hugely uphill in energy (i.e., yields a significantly less stable product), and hence very thermodynamically disfavored; it would also certainly have immense kinetic barriers as well (since any such reaction pathway implies at least temporary loss of aromaticity in the transition state[s] and intermediate[s]).

Relevant reading might include Dehydrobenzene and Cycloalkynes by Reinhard W. Hoffmann. There is extensive discussion of the reactivity of cycloalkynes, and numerous examples of spontaneous dimerizations, isomerizations, etc. Evidently, isomerizations to cycloallenes and cyclobutadiene dimers are common, which is a profound testament to the instability of cycloalkynes.

1. For example, 3,3,7,7-tetramethylcycloheptyne, a good description of which (and synthetic scheme for) is given in Strained Hydrocarbons by Dodziuk.


Naming the spin system present in ethyl butanoate I am to identify the spin system of ethyl butanoate, but I am confused as to how I should start.

I know that chemically equivalent but magnetically inequivalent nuclei should be denoted as $\ce{AA'A''}$ or $\ce{BB'B''B'''}$, etc., whereas magnetically equivalent nuclei are denoted as $\ce{A3}$, etc.

But there seem to be many chemically nonequivalent protons in this molecule.
Should I just put down this: $\ce{A3B2C2D2E3}$? How can I determine if the $\ce{CH2}$ protons are not magnetically equivalent, i.e. couple differently to neighboring protons?

By Pople Nomenclature convention, coupled nuclei for which the chemical shift difference is much greater than the coupling between them (say, greater than 3-5 times) are labelled with letters alphabetically well separated (AX, AM etc). Labelling nuclei with adjacent letters of the alphabet (AB, MN, XY etc) indicates that the chemical shifts are close compared to their coupling, and this means they are typically strongly coupled ie cannot easily be analysed using first order techniques.
As for so many NMR related questions on this site, I refer you to Hans Reich's valuable online resource for further reading.

So, what to call your system. Clearly, this is an example for teaching purposes only. You have correctly identified that you have 5 non-equivalent sites with relative populations 3:2:2:2:3. If you cannot work out, or do not know what the expected approximate chemical shifts would be for each of these guys I can tell you they are well enough separated not to worry about second order coupling.

So now you need to do is determine whether you have any magnetic inequivalence in your system. Again, have a read of this useful page to understand how to test for this, but I think you are fairly safe with this case that there are no nasty surprises.

All that is left is to assign your spin labels to your system. So for the propyl group, none of the chemical shifts are close (compared to coupling) so your labels should be well separated; so not ABC, but something like AJM. Likewise for the ethyl chain; something like RX. I'll leave you to work out how to put the two together.


Hybridisation and bond strength I'm a high school student with minimal mathematical knowledge about the theory of chemical bond. (I just have a bit of theoretical background). I came across a statement which stated that "The more is the s character in a hybrid orbital, the stronger will be the bond. "
I also came across a statement the "The extent of overlap increases with increase in directionality of bond." (Example : p-p > s-p > s-s reflects the order of the extent of overlap of orbitals keeping the internuclear distance same). What I think is that the more will be the overlap , the stronger will be the bond formed. How can I account for these contradictory statements? I'm confused. Thanks in advance.

The statements are not contradictory when read both carefully. One is about overlapping and other is about strength. Also, you might be getting confused that more overlap leads to more strong bond but the actual statement is for a particular type orbital(s, p, d, etc.) greater extent of overlapping leads to more strong bond.


How to find how much there is of a compound given the following information? The detailed version of the question:

If we burn a strip of magnesium in air, there will be two main products, magnesium oxide and magnesium
nitride.  If we then treat that mixture with water, the nitride ion is converted to ammonia.  After boiling away
the water and ammonia, we would be left with only magnesium oxide.  Knowing that air is approximately
78% nitrogen and 20% oxygen, we might expect that the mixture that forms in the burning process would be
a similar distribution of magnesium nitride and magnesium oxide.  In fact, the ratio is closer to 75% of the
nitride to 25% of the oxide.  Assuming that the reactions are 100% efficient (i.e., no loss of magnesium),
how much magnesium oxide was formed initially, if the final product contains 3.22 grams of magnesium
oxide?

After thinking about this for a while and working it out, it seems like the other information provided was irrelevant to finding the amount of MgO. I believe since the reactions are 100% efficient, that the final amount of 3.22 grams of MgO would be exactly the same for the initial amount of MgO, which was left over at the part when ammonia and water was evaporated. However, I am not sure; I would like to know, do the given percentages of nitrogen and oxygen given affect the unknown amount of MgO?

Maybe I'm just too tired right now, but didn't you mention that the absolute amount of magnesium oxide doesn't change during aqueous workup?

Isn't thus the amount of 3.22 g $\ce{MgO}$ exactly what was formed initially?

Wouldn't it be more reasonable to let the students calculate how much magnesium metal was burned initially, taking the $\ce{N2}$/$\ce{O2}$ ratio in air and the stoichiometry of the magnesium nitride into account? ;)


Unknown anion (one carbon and two nitrogen atoms, charge 2−) What's the name of this ion: $\ce{CN2^2-}$? I think that it has this structure:
$\ce{{}^{-}N=C=N-}$.

The parent molecule is cyanamide.

The ion is usually referred to as either the cyanamide dianion or the carbimide dianion


Is an antiaromatic compound more stable than a non-aromatic compound because of more conjugation? (image from a LibreTexts page (archive link))

It is shown that the antiaromatic compound is more stable than the non aromatic compounds 2 and 3 because of a more conjugated system. Here's the exact statement:

  In the first structure, delocalization of the positive charge and the π bonds occurs over the entire ring. This becomes apparent when we look at all the possible resonance structures as shown below.

  In the second structure, delocalization is only possible over three carbon atoms. This is demonstrated by writing all the possible resonance forms below, which now number only two.

  Finally, the third structure has no delocalization of charge or electrons because no resonance forms are possible. Therefore, it is the least stable of the three.

Is this correct? I thought that the conjugation in an antiaromatic compound destabilized it more than a non aromatic compound.

Sometimes, only the delocalisation of electrons is not sufficient to conclude that the compound is more stable. We should also look at which molecular orbital the electron pair is delocalising. But a clear picture of these kind of $\pi$-conjugated systems can be got when it is approached quantum-mechanically, with the help of Huckel-$\pi$ Molecular Orbital Theory.  According to this theory, to get the $\pi$-electronic energy of the system, we have to solve the Secular Determinant, which is $$\ce{det(H_{ij}- ES_{ij}) =0  ,where i,j varies from 1 to n}$$ in a $n$ Carbon System.  In Huckel Theory, for solving the determinants two approximations are taken,$1$. $\ce{H_{ii} = $\alpha$ for all i, and H_{ij} = $\beta$ for |i -j|=1.}$ This approximation tells that the effective Hamiltonian for the whole conjugation is negligible if the carbons are not close to each other(i.e. adjacent).The two different values are obvious because of the in Hamiltonian, the Columb integral, the Exchange integral interferes. $2$. $\ce{S_{ii} =1; and S_{ij}= 0 else}$. This tells the Overlap integral(Note that, this is an integral, and clear overlap in the picture doesn't mean integral should be large) are negligible for any two different carbons.  Taking these two approximations, if you solve the determinant, you will get energy levels for $5$ carbon systems,$$\ce{E = $\alpha$ + 2$\beta$;  $\alpha$; $\alpha$ - $\beta$}$$Typically, the values of $\beta$ are negative and as you increase in $\alpha$ or decrease in $\beta$, the energy is increased and system gets destabilisation. .So, the lowest energy level is $\alpha$ + $2$$\beta$, and then the second and after that the third one.Your first compound has four $\pi$ electrons. So, two electrons will be in the level $\alpha$ +$2$$\beta$, and other two is in $\alpha$ only. So,net energy, is $4$$\alpha$ + $4$$\beta$ in mostly conjugated system $1$st compound.  But in the second case there are only two $\pi$ electrons. That two will go into $\alpha$ + $2$$\beta$.So, net energy is $2$$\alpha$ + $4$$\beta$ in second one.  So, second system has lower energy than the  first as coefficient of $\alpha$ is lesser in second one, though second compound is less conjugated. So, extent of delocalisation doesn't always determine stability.


Phosphoric Acid Titration I physics graduate student, but I teach an interdisciplinary lab for the university which incorporates biology, physics, and chemistry.

We recently added a weak-acid titration lab to the course and I got a result which surprised me.

The procedure was to neutralize 100 mL of 0.010 M Phosphoric acid by adding 0.10 M NaOH.

We have 0.001 mole of phosphoric acid in solution, but this acid is triprotic which means that a maximum of 0.003 moles of hydronium can be released into the solution. My understanding of the equillibrium reactions was that as I lowered the ambient hydronium ion concentration more protons would be released by the acid.

So my expectation was that I would need to add 30 mL of the NaOH solution to reach a pH of 7. In practice it only took between 12-15 mL.

I talked this over with a Chemist I work with and she wasn't sure why this discrepancy happened either. One possiblity is that the person who prepared the acid screwed up, but we don't have more stock for me to test that theory.

So my question is do these results sound normal or did I have the right idea? If I didn't have the right idea then what am I missing theoretically speaking?

You might want to read about the pH of a neutralized solution. But in case you don't, one of my answers here on chem.se covers the calculation of titration curves with and without increasing volume. It is quite easy to create a titration curve using an adapted (see exact solution for triprotic acids here) version of equation 8.

  $$V(B)=\frac{V(H_3A)~\left(\frac{k_w}{x}-x+C(H_3A)\frac{x^2 k_1+2x k_1 k_2+3 k_1 k_2 k_3}{x^3+x^2 k_1+x k_1 k_2+k_1 k_2 k_3}\right)}{\left(x-\frac{k_w}{x}+\frac{C(B) \cdot k_b}{\frac{k_w}{x}+k_b}\right)}$$

I did that for you, so you can see, why it is not wrong, that you only needed 12-15 ml to reach pH 7 but also that that pH 7 is not what you searched for.

What you probably actually wanted, was to reach one of the equivalence points at 10 or 20 ml. You will also not have a chance to reach the third EP in an aqueous solution, which is why 30 ml is not a good goal to go for.


Conical intersections - phase Can anyone explain why at a conical intersection the wavefunction changes sign? My understanding is that it is a test to see if the crossing IS indeed a conical intersection or just a coincidence of potential energy surfaces, which do not exhibit such phase changes upon periodic cycling. If this is correct ... why? Further what can we gain from knowing that?

A rather quick googling shows up that there is some Longuet-Higgins theorem, which states almost what you are looking for. For instance, the abstract of this paper says:

  It is proved that if the wave function of a given electronic state
  changes sign when transported adiabatically round a loop in nuclear
  configuration space, then the state must become degenerate with
  another one at some point within the loop.

In the paper itself it is further stated that this in turn implies that

  [the corresponding] potential energy surface intersects one of another electronic state

and that the theorem

  makes it possible to diagnose the presence of an intersection between potential energy surfaces from the behaviour of the electronic wave function as it is transported adiabatically round a closed loop remote from the intersection

which answers both your questions, if I understand them correctly.


Equivalents of Oxalic acid In this question the answer key given is A and C. But i think that D should also be wrong because equivalents = molecular mass / n factor = $\frac{90}{2} = 45 \neq 40 $

I believe I am correct but I may be wrong so can someone please help me clarify my question.

I agree with Poutnik's point that this is a poor question.

Oxalic acid has two acidic protons so its equivalent weight is $\frac{1}{2}$ its molecular weight. Thus:

Answer A is false.
Answer B is true.
Answer C is false since there are 2*100*0.2 = 40 milliequivalents of acid, but only 100*0.2 = 20 milliequivalents of base.

Answer D requires some extra work. First it is not clear if Answer D applies to both reaction (i) and reaction (ii)

For reaction (i), in 100 ml of 0.2 M oxalic acid there are 40 milliequivalents of acid
For reaction (ii), the reaction is a redox reaction, so the number of electrons exchanged by oxalic acid needs to be determined.

Here is another sloppy point about the question. Reaction (ii) is not balanced.

However to figure out the number of electrons exchanged by oxalic acid, we only need the half cell reaction which is:

$\quad\quad\ce{2CO2(g) + 2 H+ + 2e− <=> HOOCCOOH(aq)}$

$\quad\quad$ Since oxalic acid exchanges two electrons in the redox reaction,
its "redox" equivalent$\quad\quad$ weight is $\frac{1}{2}$ its molecular weight.Thus there are also there are 40 milliequivalents of$\quad\quad$ oxalic acid redox wise too.

Answer D is false, because there are 40 milliequivalents of oxalic acid, not 40 equivalents


Is ozone an electrophile in the ozonolysis reaction? From the above mechanism it is clear to see that the electrons from the double bond attack one of the terminal oxygens in ozone, thus meaning it's an electrophile. How can this be? If you draw the resonance structures, the negative charge is concentrated on the terminal oxygens too. I am OK with the rest of the mechanism but I would never think to draw oxygen as an electrophile.

Electrons are concentrated on the two terminal oxygens in ozone but remember, ozone exhibits resonance. The electrons can be delocalized across each oxygen through p-orbitals.

So yes, at first if you think about the reaction in purely Coulombic terms then it doesn't make much sense. Why would electrons go to a region that already has a relative surfeit of electrons?

But what happens is that by attacking the terminal oxygen, the double bond effectively displaces electrons to the central, electron-deficient oxygen atom.

I'm sure someone can explain this in much greater detail with reference to molecular orbital theory but I'll leave that to someone else.


Work done by an expanding gas, how to solve this particular example? I just need help with one of these types of questions and I can handle the rest.  It is from a practice quiz on Coursera, Physical Chemistry.

  How much expansion work is done on the system when exactly 1 mol of solid ammonium chloride, $\ce{NH4Cl}$, decomposes completely
  to yield gaseous ammonia, $\ce{NH3}$, and hydrogen chloride, $\ce{HCl}$, at
  a temperature of 1280 K. Treat the expansion as
  irreversible and the gases formed as perfect.

I'm not sure how to compute the two volumes required to use the formula:

$$W = n R T \int\limits_{V_\mathrm i}^{V_\mathrm f}\frac{\mathrm dV}{V}$$

Do I assume $1\text{ atm} = 101.325 \text{ kPa}$.  I did that and was able to use the ideal gas law: $V = \frac{nRT}{P}$ to get the final volume of the 1 mole of gas, but I'm unsure how to choose the initial volume (0 won't work here as it leads to $-\infty$).

Is this the correct approach?
If 1., then how do I also get the initial volume.
If not 1., then how should I approach it?

You have system where 1 mole of solid (0 moles of gas) decomposes into two moles of gas at a constant temperature (given) and likely a constant pressure (not given).

Your formula for work is not correct for this problem. You cannot have $n$ outside of the integral since $n$ is not constant.

Let's write the total differential of the ideal gas equation:

$$\begin{align}
\mathrm{d}(PV)&=\mathrm{d}(nRT) \\
P\mathrm{d}V + V\mathrm{d}P &= nR\mathrm{d}T + RT\mathrm{d}n
\end{align}$$

Since $T$ is constant and $P$ is constant, we know $\mathrm{d}T=0$ and $\mathrm{d}P=0$, so this simplifies to:

$$P\mathrm{d}V = RT\mathrm{d}n$$

Can you do something with this relationship that avoids integrating with respect to (unknown) volumes?


CuCl2 has colour but CuCl does not $\ce{CuCl2}$ is green in colour while $\ce{CuCl}$ is white. What is the reason for this? Is it because the first compound exists as a complex ion and has d-d transitions?

The reason is indeed $\text d$-to-$\text d$ transitions. The d orbitals of both $\ce{CuCl}$ and $\ce{CuCl2}$ have two possible energy levels (based on crystal field theory).

$\ce{Cu(II)}$ is a $\text{d}^{9}$ metal, and $\ce{CuCl2}$ exists as a (distorted) octahedral around the copper atom. Based on crystal field theory, the d orbitals of an octahedral metal center can be represented as shown below. In this case, because $\ce{Cu(II)}$ is $\text{d}^{9}$, the three lower energy suborbitals are fully filled in the ground state, and but the top two suborbitals only contain a combined three electrons. Therefore, the energy required to excite that electron is $\Delta _o$. When that electron returns to the ground state, it gives off a different wavelength of light than it absorbed (in this case, the light emitted is apparently not visible light).

$\ce{Cu(I)}$, on the other hand, is a $\text{d}^{10}$ metal ion, so each of the d suborbitals is fully filled. In order for an electron to be excited in this situation, it would have to leave the 3d orbital (and the $n=3$ shell altogether) and enter the $n=4$ shell. This is rather large  energy jump compared to the intraorbital jump an electron in $\ce{CuCl2}$ has to make. Therefore, $\ce{CuCl}$ does not readily absorb visible light as $\ce{CuCl2}$ does, so it appears colorless. That is, until it see moist air.


Creating a Battery using azobenzene in High School I’m doing a project on isomerization in high school and have stumbled across the studies involving azobenzene in energy storage using its photoisomeric properties and sunlight . Do you think that this is something I could show in a high school lab, if not I can create a design, but it would be cool to show this phenomenon to classmates/teachers as my project. What do you experienced chemists think about this, as I am not sure if it is feasible.

Azobenzenes are among the most widely studied molecular photoswitches, yet, I'm not sure this is a project capable of doing in a high school lab. It probably need sophisticated instruments (e.g., azobenzene needs to be covalently bound to carbon nanotubes (CNTs)) and has done previously, notably by Department of Materials Science and Engineering in Massachusetts Institute of Technology (MIT) (Ref.1). Yet, I'd encourage you to consult people who know the field (such as Prof. Timothy J. Kucharski at MIT and Harvard) and get some advice how to do it in a laboratory environment or do a oral presentation (that would be cool as well). I'll attached some figures of Ref.1 for your convenience here:

Figure 1: Synthesis and characterization of azobenzene-functionalized SWCNTs (Adapted from reference 1. Copyright @ 2014 Nature Publishing Group).

Figure 2: Photochemical charging and thermal energy release (Adapted from reference 1. Copyright @ 2014 Nature Publishing Group).

Figure 3: Photochemical and thermal cycling (Adapted from reference 1. Copyright @ 2014 Nature Publishing Group).

Figure 4: Effect of packing photochemical switches in the solid state (Adapted from reference 1. Copyright @ 2014 Nature Publishing Group).

Also read Ref.2 for more information on molecular solar thermal energy storage using chemicals other than azobenzene, which also included.

References:

T. J. Kucharski, N. Ferralis, A. M. Kolpak, J. O. Zheng, D. G. Nocera, J. C. Grossman, “Templated assembly of photoswitches significantly increases the energy-storage capacity of solar thermal fuels,” Nature Chemistry 2014, 6, 441–447 (DOI: 10.1038/NCHEM.1918).
A. Lennartson, A. Roffey, K. Moth-Poulsen, “Designing photoswitches for molecular solar thermal energy storage,” Tetrahedron Letters 2015, 56(12), 1457–1465 (https://doi.org/10.1016/j.tetlet.2015.01.187).


How do I find a concentration from a sample tested on a spectrophotometer? I have an assignment to calculate the concentration from a sample. The assignment is:

You have 100 µL of RNA. You take 10 µL from the sample and put it in 990 µL of water in the cuvette. The OD is 0.05. Calculate the concentration of RNA in the sample. Calculate how many µL you have to take to make a solution of 100 ng.

Thus far I have succeeded to calculate the concentration from the OD. It is 2 µg/µL. I used the formula that I found on the internet :
$$OD\times \text{Standard coeff.}(\pu{40 µg/ml})\times \text{Sample dilution}$$
but I cannot figure out how from that I can calculate the solution with 100 ng, because there is not mentioned the final volume for 100 ng, or how to calculate how much I have to take µL from the cuvette. I do not understand where to begin from.
I tried with the formula $C_1V_1=C_2V_2$, but I could not figure that out.

The question being asked is, how many uL of your solution do you need to withdraw in order for that volume to contain 100 ng of the RNA.  If there are 2 ug/uL in the RNA solution, that is equivalent to 2000 ngm/uL.  So if you want 100 ng of RNA, you will need to withdraw
$$\ce{\frac{2000 ngm}{1 uL}~=~ \frac{100 ngm}{x ~uL}}$$
solving for x yields an answer of 1/20 uL or 0.05 uL.  0.05 uL of the RNA solution will contain 100 ng of the RNA.


When aluminium is ignited in atmosphere, will it form only aluminium oxide or also aluminium nitride? We know when lithium is burnt in air, it forms lithium nitride and lithium oxide. I am wondering if this applies to aluminium (other metals) also?

According to the paper by Gromov and Vereshchagin [1], it is indeed formed when burning aluminium, but you could have predicted this simply by considering the standard enthalpy of formation of aluminium oxide $\Delta_\mathrm{f}H^\circ (\ce{Al2O3}) = \pu{−1675.7 kJ/mol}$ which will be liberated upon ignition and this energy helps in breaking the strong nitrogen-nitrogen triple bond (bond energy is $\pu{−945 kJ/mol})$ and finally this would form aluminium nitride, which also produces some energy (standard enthalpy of formation is $\pu{−318 kJ/mol}),$ but the combustion here is not a simple one it requires high temperature as mentioned in the paper, I think this is to sublimate $\ce{Al2O}$ which will react with nitrogen and give rise to aluminium nitride.

As HYDR0GEN stated, this combustion isn't simple like throwing a match on aluminium sheet, it needs high temperature. According to the work by Trunov et al. [2, p. 38], as the size of aluminium particles decrease the needed temperature for combustion decreases and that's why researchers who wrote the paper used superfine aluminium powder:

  Figure 5. Experimentally determined temperatures of aluminum
  ignition as a function of the used sample size.

References

Gromov, A.; Vereshchagin, V. Study of Aluminum Nitride Formation by Superfine Aluminum Powder Combustion in Air. Journal of the European Ceramic Society 2004, 24 (9), 2879–2884. DOI: 10.1016/j.jeurceramsoc.2003.09.020. (PDF)
Trunov, M. A.; Schoenitz, M.; Dreizin, E. L. Ignition of Aluminum Powders Under Different Experimental Conditions. Propellants, Explosives, Pyrotechnics 2005, 30 (1), 36–43. DOI: 10.1002/prep.200400083.


What is a chemical daily life experience that can be modelled by superior mathematics? I'm searching for a chemical daily life experience that can be modelled by superior mathematics so that the origin of the equations involved can be easily explained to a beginner in chemistry.

By superior mathematics I mean things like partial and ordinary differential equation, Integral Calculus, group theory, differential geometry. But I want to understand how are the equations derived.

For example, in physics, I could model the trajectory of the rear wheel of a bicycle (a daily life experience) and I fully understand all the equations involved, without being a physicist. I want something similar to find in chemistry.

Diffusion is one example from the daily life. For example, the absorption of ethanol from the small intestine to blood can be modeled using differential equations. The extraction of caffeine from grounded coffee is also a diffusion phenomenon.


How does the polarizability of an apolar compound influence properties such as its boiling temperature? I have understood that the more polar is a compound, the higher is its boiling temperature. But I would like to know what is the justification for this answer.
I am new to chemistry so I would appreciate a simple answer.

In non-polar liquids, the intermolecular forces are known as London dispersion forces (or dipole induced dipole interaction). This is the weakest interaction that exists between molecules.
The strength of this interaction depends on polarisability of non-polar molecule. As an example, let's take the following examples:
$$
\begin{array}{lr|lr|lr|lr}
\hline
\text{Compound} & \mathrm{bp}/\pu{°C} & \text{Compound} & \mathrm{bp}/\pu{°C} & \text{Compound} & \mathrm{bp}/\pu{°C} & \text{Compound} & \mathrm{bp}/\pu{°C} \\
\hline
\ce{CO2}  & -78.5 & \ce{CCl4} & 76.7  & \ce{CBr4}  & 189.5 & \ce{CI4}   & 329.2 \\
\ce{CS2}  & 46.3  & \ce{CBr4} & 189.5 & \ce{SiCl4} & 57.7  & \ce{SiBr4} & 153.0\\
\ce{CSe2} & 125.5 & \ce{CI4}  & 329.2 &            &       &                 \\
\hline
\end{array}
$$
In these molecules, polarisability depends on polarisability of anions $(\alpha),$ which depends as
$$\alpha \propto q \times r,$$
where $q$ is the charge and $r$ is the ionic radius.
As charge on each anion is same, hence we can conclude that, greater the radius, greater is the polarisability, and thus greater is London dispersion forces.
Therefore, greater boiling point.


Why is the IUPAC name of 1H-benzo[d]naphthalene not 1H-benzo[i]naphthalene? In the nomenclature of fused arenes you need to add a 'fusion locant'. The first face should correspond to positions 1 and 2 and is lettered ‘a’, continuing alphabetically clockwise around the exterior of the parent structure.
In this case, the parent is naphthalene and using numbering of naphthalene the benzo is fused on position d so the name would be 1H-benzo[d]naphthalene

But IUPAC also has some rules about the orientation of such structures:
(a) maximum number of rings in a horizontal row (b) the maximum number of rings in the upper right quadrant. (See the drawing on the right.) Subsequent the numbering of peripheral atoms in the preferred orientation starts from the uppermost ring. (If there is more than one uppermost ring, the ring furthest to the right is chosen.)
 Using that numbering scheme the face of fusion should be i.

Apparently, you are mixing two different numbering schemes.

For the construction of a fusion name, the two components are numbered separately (if necessary), using letters for the parent component and numbers for the attached component. The corresponding rule P-25.3.1.3 in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows.

  (…)
  Isomers are distinguished by lettering, continuously, each peripheral side of the parent component (including sides whose locants are distinguished by letters, for example, 2a,3a) using the italic letters a, b, c, etc., beginning with a for the side numbered ‘1 ,2’, b for ‘2,3’ etc. To the letter as early in the alphabet as possible that denotes the side where the fusion occurs are prefixed, if necessary, the numbers of the positions of attachment of the other component. These numbers are chosen to be as low as is consistent with the numbering of the compound and their order conforms to the direction of lettering of the parent component. (…)

The parent component for the example that is given in the question is naphthalene (not the complete benzonaphthalene). The letter as early in the alphabet as possible that denotes the side where the fusion occurs is d. Therefore, the correct name of the complete compound is
1H-benzo[d]naphthalene.

The other numbering scheme described in P-25.3.3.1.1 applies to the complete fused ring system.

  P-25.3.3.1.1 The numbering of peripheral atoms in the preferred orientation starts from the uppermost ring. If there is more than one uppermost ring, the ring furthest to the right is chosen. Numbering starts from the nonfused atom most counterclockwise in the ring selected and proceeds in a clockwise direction around the system, including fusion heteroatoms but not fusion carbon atoms. (…)

Thus, for the example that is given in the question, this numbering applies to the complete
1H-benzo[d]naphthalene. In particular, this numbering is used for the locant of the indicated hydrogen. It would also be used for any substituents of the compound.


Changes over time in sodium hydroxide concentration in a hot water bath used for pretzel manufacture I make soft pretzels using a hot (190 °F) and dilute (2-4% w/v) sodium hydroxide solution. The bath is kept hot in a open-top stainless steel vessel like a stockpot. I make my pretzel dough and dip the dough in the hot solution for about 20 s prior to baking. As you can imagine, there is a fair amount of evaporation, as well as absorption by the dough of the solution.

Can someone help me theorize what happens to the sodium hydroxide concentration over time? pH is already 14, so pH is not a good measure of concentration. I don’t need to measure precisely, but I would like to understand if concentration is increasing as water evaporates, or if concentration stays the same because the solution is evaporating, or if there is a decrease in concentration by absorption of the pretzel dough.

This will help me understand what I need to do to keep my production going without running out of solution. In other words, do I just add water back to the bath as it evaporates, do I add water and the same concentration of sodium hydroxide I did in the beginning? Or other?

You need both.

Adding water, as water evaporates, but hydroxide does not.
Adding the hydroxide solution, as it is taken away by pretzels

But I suppose the evaporation is the major effect.

Probably the most convenient way is measuring density, with areometer $\pu{1.0-1.2 g / ml}$

Additionally, hydroxide gradually absorbs carbon dioxide, becoming sodium carbonate. pH paper strips may do good here, if solution looses it's alkality.


How can I make homemade ink? I want to make some homemade ink suitable for amateurs to practice calligraphy. I have access to many household chemicals, including acetic acid, sodium hydrogen carbonate, sugar, sodium chloride, and more.

Can someone give me an easy recipe for a colored (any color will be fine) ink that I can make at home? (preferably basic, but slightly acidic is fine too I guess)

Here are some ideas:

Caramelized sugar --- nope, too expensive and smelly.
Something with rust? (iron)
Something like india ink. (pigments are acceptable too)
Cornstarch and iodine.

I don't have a full recipe, but if you're fine with black ink, try to boil some crushed or grounded organic material containing large amounts of gallic acid (or other hydrolyzable tannins) with rusty iron and some vinegar (acetic acid).

Suitable organic materials that come to my mind are oak bark or green tea.

You should end up with an ink that darkens over time. Note that this ink will probably clog fountain pens but should be fine for calligraphic pens.


A 4.00 L sample of gas is cooled from 67 °C to a temperature at which its volume is 2.60 L A $4~\mathrm{L}$ sample of gas is cooled from $67~\mathrm{°C}$ to a temperature at which its volume is $2.6~\mathrm{L}$. What is this new temperature? Assume no change in pressure of the gas.

Hint: Use Ideal gas equation:

  $$\large PV=nRT$$

Given pressure is constant, can you tell the final temperature from this relation?


Valence bond theory and molecular orbital theory — how do they complement one another? Based on my readings, valence bond theory (VBT) and molecular orbital theory (MOT) tend to complement one another in explaining a molecule, but I don’t understand how VBT helps us explain the properties of a compound (other than bond angles). How does VBT complement MOT in aiding our understanding of molecules?

You are correct that Valence Bond Theory (VB) is still very active in the chemical bonding community along with MOs. I recommend this article/conversation by Roald Hoffmann about this topic (Acc. Chem. Res. 2003, 36 (10), 750–756). Succinctly, they both are capable of providing a description of bonding in molecules; however, VB theory is better at demonstrating the existence of localized bonds while MO theory is more generalizable to delocalized systems (and ultimately the foundation of predictive computational chemistry). If you have an organic molecule, for example, the VB picture is more helpful for understanding reactivity because these molecules (for some reason) usually behave according to the intuition accessible within such a local bonding framework. MO theory could give you a numerically exact answer, but the wavefunction is in general hard to interpret since it's delocalized across the entire molecule.

One topic that might be of interest to you in this field is MO Localization. This is the field of study which aims to take the MOs and the high accuracy wavefunction and transform them into orbitals that reflect chemistry as intuitively as VB theory. (One can take arbitrary rotations of MOs without changing the observables of the wavefunction). This problem is technically ill-defined, however, there are many approaches that have done this to great success. For example: Natural Bond Orbitals (NBOs), Adaptive Natural Density Partitioning (AdNDP), Intrinsic Bond Orbitals (IAOs), and many others. As the titles suggest, there is a belief that localized (VB-like) bonds are real things hidden in the true wavefunction.


Methods of splitting carbon dioxide into elements Is it possible to split $\ce{CO2}$ into C and O? If so, what are some methods and their efficacies?

This question originated because I was wondering if it was possible, and if so, how, to convert ice from the ice caps on Mars (made up of frozen $\ce{H2O}$ and $\ce{CO2}$) into oxygen.

I assume you meant convert $\ce{CO2}$ into $\ce{O2}$.

Lithium peroxide ($\ce{Li2O2}$) is a substance that does this. It absorbs $\ce{CO2}$ and converts itself into $\ce{Li2CO3}$ and $\ce{O2}$:

$$\ce{2Li2O2 + 2CO2 -> 2Li2CO3 + O2}$$

It was used as an air purifier in the spacecraft to scrub carbon dioxide. It offers it's bonus of releasing oxygen as well as being non-hygroscopic, so it won't remove the moisture content in the air.

But at any day, you would need a lot of lithium peroxide if you want to make Mars habitable.

Edit: The partial pressure of carbon dioxide in the Martian atmosphere is $\pu{575.8Pa}$. Using the ideal gas law, and temperature of Mars is $218K$, we get amount of substance of $\ce{CO2}$ as $\pu{0.32mol/m^3}$.

Such an amount would require $\pu{14.65g}$ of lithium peroxide per cubic meter of martian atmosphere. According to SigmaAldrich, that costs about ₹21.22 (33 US cents) per cubic meter.


Books for a very advanced 3rd grader (elementary school) My daughter is a 3rd grader, but she reads at 8th grade level, and she is 2 grade levels ahead in math.

She reads everything she gets her hands on.

She took some chemistry activities, and she loved them. ("Mad Science", "Ooo, goo and stinky too",...)

She knows that matter is made of atoms, and that atoms are made of protons and electrons.

I'd like to find some chemistry books that would be appropriate for her. A good book would be one that starts from beginning, help her connect together the stuff she already knows, and then take her further.

My reason is that I don't want school to be too easy for her. I want to nurture her curiosity, love of challenge, and motivation to study.

On the other hand, I don't want to push her even further ahead in the subjects that she has to take in emementary or middle school anyway -- because that would make our dealing with the school system even more complicated... So, I think Chemistry and Physics are good subjects.

One of the best is The Golden Book of Chemistry Experiments.  By today's standards, this book is considered "dangerous", but keep in mind that it was written for teenagers in the 1960s.  In any case, parental supervision should always be required in chemistry, and if you carefully research the experiments before doing them you should be fine.  It takes you all the way through technique, building apparatus, and safety.  The only problem is obtaining chemicals: the sources cited therein are hopelessly outdated (i.e. your will NOT find most of this stuff in drugstores anymore), and you will need to use educational supply companies to obtain small quantities of chemicals.  Have fun!


Is there a hierarchy to functional groups? I think I am getting slightly confused about the names we give to functional groups. I am aware that there are different types of functional groups such as alkyl, aryl and acyl groups.

I also know that there exist functional groups such as those of esters, ketones and amides which come under the acyl group. However, thinking about this recently I also realised that there existed a carbonyl group, which is also present in a ketone?

I am also a bit confused as to whether the acyl group's general formula is RCO or RCOX where X could be a variety of groups. Also, is R a part of the functional group or is it there to show what the functional group is attached to.

Could someone please make this hierarchy of functional groups a bit clearer?

What you may be confusing is the difference between a functional group and a moiety. A moiety is a part of the molecule that could be either the whole functional group or just a part of the structure of a functional group.

Wikipedia does a decent job of explaining this.

In this illustration there is an ester functional group and an acetyl moeity.

As for your second question about the acyl group, my understanding is that the formula is RCO- and that the X is not necessary to define the acyl group. The R represents an alkyl group.


Why is pyridine used when making tosyl esters from alcohols? Tosyl chloride is used to make a hydroxyl group into a better leaving group. However, when the reaction of tosyl chloride and an alcohol occurs, a weak base such as pyridine should be used. Why?

The function of pyridine is actually not so simple and not so easy to notice at first glance. There is a fundamental reason why pyridine is used to promote the acylation reaction, which is that it can act as a catalyst.

Despite its basicity and the subsequent formation of its chlorhydrate salt after the tosylation reaction, pyridine is also an excellent nucleophile. Pyridine is, in fact, more nucleophilic than the alcohol, and it attacks the acyl chloride rapidly, forming a highly electrophilic (because of the positive charge) intermediate: N-tosylpyridinium chloride. It is indeed this intermediate the actual tosylating agent which reacts with the alcohol to give the ester. Because pyridine is acting as a nucleophile to speed up the reaction, yet is unchanged by the reaction, it is a "nucleophilic" catalyst.


Does gravity affect the trajectory of gas particles? We're studying the kinetic theory of gases in school, and one of the points that was brought up was that: "Gases consist of particles in constant, random motion." How is it possible for gas particles to move in straight lines, entirely randomly? Doesn't gravity affect them, or is their mass so small that its effect on their movement is negligible?

Yes gravity pulls on gas molecules. That is why the atmosphere doesn't just float off into space.

The gist is that the time between collisions is very short in the lower atmosphere, and the distances very short. The mean free path at atmospheric pressure is only about 70 nanometers. So the assumption is that gas particles travel in a straight line between collisions.


Derivation of the Orgel diagram for octahedral d2 complexes The Orgel diagram for a typical octahedral $\mathrm{d^2}$ complex is shown below:

(source: Kettle, Physical Inorganic Chemistry, p 146)

I understand that the right-hand half of the graph shows how the energies of electronic states of a $\mathrm{d^2}$ complex vary with the ligand field splitting $\Delta$.

However, neither the Wikipedia page nor the LibreTexts (formerly UC Davis ChemWiki) page are helpful in describing how this diagram is constructed. In particular, I have a few questions:

What are the "weak-field" and "strong-field" limits?
How are the symmetry labels for the electronic states constructed?
How are their relative energies determined?
Why are some of the lines straight, but others curved?
Why is the diagram also used for $\mathrm{d^3}$, $\mathrm{d^7}$, and $\mathrm{d^8}$ complexes (both octahedral and tetrahedral)?

1. Weak-field and strong-field limits

I will adopt the description used in Figgis and Hitchman's Ligand Field Theory and Its Applications (p 5), because I cannot really phrase it better:

  It is useful to recognise the relative importance of the terms in the Hamiltonian for different systems:

  First- and second-transition series: $\mathbf{H}_\mathrm{LF} \approx \mathbf{H}_\mathrm{ER} > \mathbf{H}_\mathrm{LS}$

  $\mathbf{H}_\mathrm{LF}$ (ligand field) takes into account the presence of the surrounding ligand molecules [...]. $\mathbf{H}_\mathrm{ER}$ (electron repulsion) takes into account the repulsions within the d or f electron set [...]. $\mathbf{H}_\mathrm{LS}$ (L-S coupling, also known as Russell-Saunders coupling) takes into account the formal coupling between the spin and the [orbital] angular momenta of the d or f electrons.

Spin-orbit coupling (which scales as $Z^4$, $Z$ being the atomic number) plays a nonzero, but relatively small, role in the properties of the 3d or 4d transition metals (compared to the 5d, 4f, or 5f metals). Therefore, it is not reflected in the Orgel diagrams above.

The weak-field limit simply refers to the case where the ligand field is weak, i.e. $\mathbf{H}_\mathrm{LF} < \mathbf{H}_\mathrm{ER}$. In this regime, the spherical symmetry of the free ion (e.g. $\ce{V^3+}$) is almost unperturbed by the six ligands (assuming octahedral symmetry). Therefore, the electronic states are similar to those of the free ion.

The strong-field limit simply refers to the opposite case, i.e. $\mathbf{H}_\mathrm{LF} > \mathbf{H}_\mathrm{ER}$. In this regime, the electron-electron repulsions are relatively small compared to the ligand-field splitting $\Delta$. As such, the electronic states are entirely determined by the electronic configuration $(\mathrm{t_{2g}})^m(\mathrm{e_g})^n$ (for a $\mathrm{d^2}$ complex ion, clearly $m + n = 2$).

As $\Delta$ increases (i.e. going from left to right), we move from the free ion limit where there is no ligand field and $\Delta = 0$, through the weak-field regime where $\Delta$ is small, into the strong-field regime where $\Delta$ is large.

2. The electronic states: their symmetries and their energies

2.1. Free-ion terms

For a $\mathrm{d^2}$ configuration, one can either enumerate microstates (a nasty, tedious method that I do not recommend), or one can use group theory (wait for the MathJax to load, then scroll to direct product tables: $R_3$). The wavefunction of a d-orbital, in the spherical symmetry of the free ion, transforms as $D \equiv \Gamma^{(2)}$. The electronic wavefunction of two d-electrons transforms as the direct product of $D$ with itself:

$$\begin{align}
\Gamma^{(2)} \otimes \Gamma^{(2)} &= \Gamma^{(4)} \oplus \Gamma^{(2)} \oplus \Gamma^{(0)} + [\Gamma^{(3)} \oplus \Gamma^{(1)}] \\
\mathrm{D \otimes D} &= \mathrm{G \oplus D \oplus S \oplus [F \oplus P]}
\end{align}$$

Square brackets indicate spatial wavefunctions that are antisymmetric with respect to interchange of the two electrons; to satisfy the Pauli exclusion principle, these have to be paired with symmetric spin wavefunctions. Adding up the spins of two electrons ($s_1 = s_2 = 1/2$), we can either get $S = 1$ ("triplet") or $S = 0$ ("singlet"). The triplet spin wavefunctions are symmetric, so these go with the $\mathrm{F}$ and $\mathrm{P}$ terms. Likewise, the singlet spin wavefunction is antisymmetric, and goes with the $\mathrm{G}$, $\mathrm{D}$, and $\mathrm{S}$ terms. This yields the term symbols $\mathrm{^1G}$, $\mathrm{^1D}$, $\mathrm{^1S}$, $\mathrm{^3F}$, and $\mathrm{^3P}$ for a $\mathrm{d^2}$ configuration, where the superscript indicates the spin multiplicity $M = 2S + 1$.

The singlet states are higher in energy due to exchange energy (Hund's first rule). As such, they get ignored in our treatment here. This is not always wise, and the more sophisticated Tanabe-Sugano diagrams do plot electronic states of different multiplicity, but as far as we are concerned, we can neglect them here.

Between the two triplet states, the state with the larger value of $L$ (total angular momentum quantum number) is the ground state (Hund's second rule). Therefore, the $\mathrm{^3F}$ state with $L = 3$ is the ground state, and the $\mathrm{^3P}$ state with $L = 1$ is the excited state.

The difference in energy between these two states is solely attributable to electron-electron repulsions.[1] The two states are separated by an energy difference $15B$, where $B$ is a Racah parameter that acts as a measure of electron-electron repulsions. I believe there is a quantum mechanical expression for it, but in practice the value of $B$ is empirically obtained (via Tanabe-Sugano diagrams). It is easier to treat it as an empirical parameter that is experimentally obtained, similar to how one would treat the ligand-field splitting $\Delta$.

The free-ion terms and the difference in energy $15B$ are shown in the middle of the diagram, where $\Delta = 0$.

2.2. Strong-field limit terms: no electron repulsions

In an octahedral $\mathrm{d^2}$ complex, there are three ways of arranging the two d electrons: $\mathrm{(t_{2g})^2}$; $\mathrm{(t_{2g})^1(e_g)^1}$; and $\mathrm{(e_g)^2}$. These are the three electronic states under consideration and are what you see at the right-hand side of the diagram. Hopefully the energy ordering is straightforward; the energy gap between each state is equal to $\Delta$ since it requires promotion of one electron from $\mathrm{t_{2g}}$ to $
\mathrm{e_g}$.

2.3. Weak-field limit terms

As can be seen from the Orgel diagram, there are four states: two $\mathrm{T_{1g}}$ states, one $\mathrm{T_{2g}}$ state, and one $\mathrm{A_{2g}}$ state. (The subscript g, as well as the spin multiplicity, are omitted in the diagram. This is to allow it to be generalised to tetrahedral complexes, where the g/u labels are not applicable, as well as complexes with other d-electron counts, where the spin multiplicity will be different but the symmetries the same.) The question we will have to look at is how these states are obtained, and how they correlate with the two extreme limits described earlier.

Going from the free ion limit, we need to find out what happens to $\mathrm{F}$ and $\mathrm{P}$ terms when the symmetry of the compound is reduced from spherical to octahedral. Luckily, there are group theory tables that do just that, called tables of descent in symmetry. These tell us that:

$$\begin{align}
\mathrm{F} &\to \mathrm{A_2 \oplus T_1 \oplus T_2} \\
\mathrm{P} &\to \mathrm{T_1}
\end{align}$$

and you can see this correlation in the lines spreading out from the centre. Conventionally, the $\mathrm{T_1}$ state arising from the $\mathrm{P}$ state is labelled as $\mathrm{T_1(P)}$ and likewise the other $\mathrm{T_1}$ state is labelled $\mathrm{T_1(F)}$. This isn't shown on this particular diagram, but may or may not appear in other texts (and it is very helpful because it eliminates any ambiguity about which $\mathrm{T_1}$ state you are talking about!).

Now, we need to figure out which final configurations to join them to. This is done by taking more direct products (yes, group theory tables again). For the $\mathrm{(t_{2g})^2}$ configuration we have

$$\mathrm{T_2 \otimes T_2 = A_1 \oplus E \oplus [T_1] \oplus T_2}$$

however since we are only interested in the triplet states, we are only going to take the $\mathrm{T_1}$ term, which has an antisymmetric spatial component to go with the symmetric triplet spin component. Likewise

$$\begin{align}
\mathrm{T_2 \otimes E} &= \mathrm{T_1 \oplus T_2} \\
\mathrm{E \otimes E} &= \mathrm{A_1 \oplus [A_2] \oplus E}
\end{align}$$

So, the $\mathrm{(t_{2g})^1(e_g)^1}$ configuration gives rise to $\mathrm{T_1}$ and $\mathrm{T_2}$ terms,[2] whereas the $\mathrm{(e_g)^2}$ configuration gives rise to an $\mathrm{A_2}$ term. The only remaining question is, which of the $\mathrm{T_1}$ states correlates with which of the strong-field configurations? The answer is given by the non-crossing rule: two states of the same symmetry cannot cross each other.[3] Therefore, we can confidently join the $\mathrm{T_1(F)}$ state to the $\mathrm{(t_{2g})}$ configuration, knowing that the alternative would lead to a forbidden crossing.

Incidentally, this is also why the two $\mathrm{T_1}$ electronic states have curved lines: quantum mechanical mixing introduces an extra perturbation to the energy. The upper energy level is always pushed upwards, and the lower energy level pushed downwards, which leads to the curvature shown in the diagram. The other two states do not have such effects and are therefore depicted as straight lines.

3. Other geometries and configurations

Since a $\mathrm{(t_{2g})^3(e_g)^2}$ configuration is spherically symmetric, the electronic states arising from an octahedral (high-spin) $\mathrm{d^7}$ configuration are exactly the same as those arising from an octahedral $\mathrm{d^2}$ configuration. (The spin multiplicity is different: $S = 3/2$ gives $2S + 1 = 4$.)

Furthermore, the octahedral $\mathrm{d^3}$ configuration can be thought of as having two holes in a spherically symmetric $\mathrm{(t_{2g})^3(e_g)^2}$ arrangement. These holes give rise to exactly the same symmetry states as the $\mathrm{d^2}$ configuration; however, since holes are essentially "negative electrons", the relative energies are reversed. For example, the $\mathrm{A_2}$ state, which arises from placing the two holes in the $\mathrm{e_g}$ orbitals (i.e. a $\mathrm{(t_{2g})^3}$ electronic configuration), is actually the ground state instead of the highest excited state. This is reflected in the left half of the diagram.

(If the above two paragraphs were confusing, they were covered in more detail in another answer of mine: How do I find the ground state term symbol for transition metal complexes?)

As for tetrahedral complexes, note that the orbital splitting in a tetrahedral complex is exactly the same as in an octahedral complex, except that the relative energies of the $\mathrm{e}$ and $\mathrm{t_2}$ sets are reversed (and the g subscripts are gone).

Therefore, if we consider (for example) the $\mathrm{(e)^2(t_2)^1}$ ground state of a tetrahedral $\mathrm{d^3}$ configuration, this can be considered to have two holes in the $\mathrm{t_2}$ orbitals. These two holes give rise to exactly the same states as the octahedral $\mathrm{d^2}$ case, where in the ground state there are two electrons in the $\mathrm{t_{2g}}$ orbitals. The tetrahedral $\mathrm{d^3}$ ground state therefore has a $\mathrm{^4T_1}$ label, exactly analogous to the octahedral $\mathrm{d^2}$ ground state which is a $\mathrm{^3T_{1g}}$ term. Of course the spin multiplicity is different and the g label is gone, but we've already established that the diagram specifically says nothing on these two points for a reason.

Consequently, tetrahedral $\mathrm{d^3}$ and $\mathrm{d^8}$ complexes fit into the right half of the diagram. Using similar arguments, tetrahedral $\mathrm{d^2}$ and $\mathrm{d^7}$ complexes fit into the left half of the diagram.

Notes

[1] Quoting myself from another answer

  The classical argument behind Hund's second rule is that electrons that have the same sign of angular momentum "collide" into each other less and as a consequence repel each other less. It is like two people running around a track; if they ran in the same direction, they would bump into each other less than if they ran in opposite directions.

[2] As an aside: why is the $\mathrm{(t_{2g})^1(e_g)^1}$ configuration split into two terms? The answer lies in, unsurprisingly, electron-electron repulsions (which were neglected in the strong-field limit). Consider the electronic configuration $(\mathrm{d}_{xy})^1(\mathrm{d}_{x^2-y^2})^1$, which has electron density concentrated in the $xy$-plane. We might expect this configuration to experience more electron-electron repulsion than, for example, the $(\mathrm{d}_{xy})^1(\mathrm{d}_{z^2})^1$ configuration. While molecular terms do not necessarily have a one-to-one correspondence with orbital microstates, one can see from this discussion how it is plausible for the degeneracy to be lifted when electron repulsions are taken into account.

[3] The Hamiltonian $\hat{H}$ transforms as the totally symmetric irreducible representation of the relevant point group (which makes sense, since the energy of a system has to be constant under reflection, inversion, rotation, etc.). Consequently, the matrix element $\mathbf{H}_{mn} = \langle m | \hat{H} | n \rangle \neq 0$ if $m$ and $n$ have the same symmetry. This directly leads to the non-crossing rule, that energy levels corresponding to wavefunctions with the same symmetry do not cross under any strength of a perturbation (in this case, the ligand field).


What element has good conductivity but is also brittle? I'm confused by what element(s) can be identified to have these properties.

The following is text from my 2019-2020 Kaplan MCAT General Chemistry book under the metalloids section

  Silicon (Si) has a metallic luster but is brittle and a poor conductor

So onto the MCAT Concept Check 2.2:

  Identify a few elements that  likely possess the following  properties

  Good conductivity but brittle

I flip to the answer key and it says:

  Metalloids exhibit brittleness but good conductivity

So I'm confused how this makes sense because Si was the main example they provided in that passage, and it seems to contrast the two points they make.

So what is an actual example of good conductivity but brittle??

Whether a "metalloid" is both brittle and a good conductor depends on what you mean by "metalloid", which varies in literature, what is a "good" level of (electrical) conductivity, and even what exactly you mean by "brittle" versus "ductile".  Silicon, for instance, is only a semiconductor, with much less electrical conductivity than metals, when pure.

Wikipedia identifies beryllium and magnesium as brittle metals, although the latter can be made ductile with minor alloying.  Zinc is called "slightly brittle".


Can I make a pH 7 buffer from phosphoric acid and sodium hydroxide? I am an electronics guy with not so much knowledge of chemistry.
I would like to prepare a pH 7 buffer to test pH meters.

All web references I have found so far describing how to make a phosphate buffer use
disodium phosphate (Na2HPO4) and Monosodium phosphate (NaH2PO4) as ingredients.

I have a good supply of 85% phosphoric acid and solid sodium hydroxide. Can I make a pH 7 buffer from these ingredients instead or is there a catch?

You should be able to do this with little trouble.  The only caveats:

Make a stock solution of phosphoric acid and also of NaOH.  When you add them to each other to make your buffer, heating could be a problem, much moreso than if you were using the mono- and di-basic sodium phosphate salts.  So you may want to cool on ice when mixing these compounds.
You can calculate approximately how much NaOH and phosphoric acid to mix in order to get a pH 7 result, but it's best to use a pH meter to verify that you've reached pH 7.


Why is peanut butter viscous? Why is it that when you chop peanuts, a highly viscous paste is created?

Im assuming that the reason that chopped peanuts dont become just dust is because we cannot chop finely enough, but what causes the chopped particles to adhere to each other?

Peanuts contain peanut oil. When you crush the peanut you also compress some of the peanut oil out out of the nut particles.


Corey-Schmidt oxidation of primary alcohols Why does pyridinium dichromate (PDC) in dimethylformamide (DMF) oxidize alcohols to acids while in methylene chloride the oxidation stops at the aldehyde level?

As stated in the previously posted answer:

  This oxidation into
  carboxylic acids succeeds when the intermediate aldehyde equilibrates with a liberal proportion of hydrate; that is, when the intermediate aldehyde belongs to the aliphatic kind and is not highly hindered. The water necessary for the formation of the intermediate aldehyde hydrate may proceed from the decomposition of PDC.$^{[1]}$

Though the text does state that water may form from decomposition of the dichromate anion, since DMF is so hygroscopic as a result of its high polarity ($\epsilon_\mathrm{r}=38^{[2]}$), I would not be surprised to learn that little to no over-oxidation occurs in completely dry DMF under an inert atmosphere.

This is only speculation though, and the dichromate anion may very well be decomposing to produce water in the solution. As such, I will propose a reaction mechanism for this decomposition. The first step can only occur when pyridinium dichromate is well solvated and protons are free to be transferred between the species, which occurs in DMF but not in DCM:

$\hspace{.6cm}$

This produces chromic acid, another chromium(VI) oxidizing agent. Unlike dichromate oxidation, however, chromic acid oxidation actually produces water as a byproduct, which is then free to hydrate the resulting aldehydes:

$\hspace{.8cm}$

$[1]$ Tojo, G.; Fernández Marcos. Oxidation of Alcohols to Aldehydes and Ketones: A Guide to Current Common Practice; Springer: New York, NY, 2006; p 28.
$[2]$ Wikipedia: Solvent