Reaction of iron with steam While studying about reversible reactions, my professor stated that the following reaction is a reversible reaction:

$$\ce{Fe + H2O <=> Fe3O4 + H2}$$

She also stated that the product $\ce{Fe3O4}$ is formed by a mixture of $\ce{FeO}$ and $\ce{Fe2O3}$. However, there was no mention of the type of reaction or the reason due to which the mixture of $\ce{FeO}$ and $\ce{Fe2O3}$ resulted in the formation of $\ce{Fe3O4}$.

So, my questions are as follows:

What type of a reaction is it? (I think it's a displacement reaction but from what I know, there are ions present in a displacement reaction).
Why have we combined $\ce{FeO}$ and $\ce{Fe2O3}$ resulting in the formation of $\ce{Fe3O4}$? (According to me, mixtures and compounds are two different things and that we can't combine them.)

What type of a reaction is it?

I would describe it first and foremost as a redox reaction. Iron is oxidised and hydrogen is reduced; the half equations would be:
$$\ce{2Fe -> Fe^2+ + Fe^3+ + 5e-}$$ $$\ce{H2O + 2e- -> H2 + O^2-}$$

  Why have we combined $\ce{FeO}$ and $\ce{Fe2O3}$ resulting in the formation of $\ce{Fe3O4}$?

This is a good observation. $\ce{Fe3O4}$ contains both iron(II) and iron(III) ions and is sometimes written as $\ce{FeO \cdot Fe2O3}$. The dot in the middle signifies that there is some form of bonding between the two compounds (another common example of this is water of crystallisation). $\ce{Fe3O4}$ is a compound and not a mixture because it does not consist of two separate $\ce{FeO}$ and $\ce{Fe2O3}$ phases but rather it is a single crystal structure containing $\ce{Fe^2+}$, $\ce{Fe^3+}$ and $\ce{O^2-}$ ions.

  $\ce{Fe3O4}$ has a cubic inverse spinel structure which consists of a cubic close packed array of oxide ions where all of the $\ce{Fe^2+}$ ions occupy half of the octahedral sites and the $\ce{Fe^3+}$ are split evenly across the remaining octahedral sites and the tetrahedral sites.


How can I determine the structure of complexes with ethylenediamine as a ligand? I am a bit confused about how to solve this problem.  I can interpret coordination complexes like $\ce{[CoCl6]^4-}$ but I just came across a coordination complex that I am unsure of how to solve.

It is $\ce{[Mn(en)3]^3+}$ where en stands for ethylenediamine.  The problem first asks that I draw the structure (not sure how to do this as ethylenediamine is a large molecule).  I would also like to know whether this complex would be tetrahedral, octahedral, trigonal planar, etc.  I'm just kind of confused as to how to solve this problem when its a large molecule as the ligand rather than a molecule like $\ce{Cl6}$.

In principle, there is no difference between having monodentate ligands such as chlorido ligand or more complex ligands such as ethylenediamine.

The first step is always to draw out the structure of the ligand by itself. Then, identify how many coordinating entities it contains. Ethylenediamine is $\ce{H2N-CH2-CH2-NH2}$ and both nitrogens can form coordinate bonds.

In the next step, identify how many coordinating entities you have in total and thus what the maximum coordination number is. As mentioned, each $\ce{en}$ has two potentially coordinating nitrogens for a total of six dative bonds. Thus, a hexacoordinated octahedron is the maximum possible.

Continue by looking at the metal. What do you know about its properties in complexes? Does it prefer any type of complex? In this case, manganese(III) is absolutely fine with having an octahedral geometry, so that’s taken care of.

Once you’ve done all this, you will notice that the metal enjoys octahedral coordination and there are enough Lewis bases for octahedral coordination to occur — the complex will likely be octahedral. If the complex had been $\ce{[Ni(en)2]^2+}$, we would have arrived at ‘square planar’ by the same methodology.

Next, draw the complex. Start with the metal and the octahedral geometry and put a nitrogen at the end of each line. Then, try to connect the nitrogens with $\ce{CH2-CH2}$ fragments. There should be two possible, enantiomeric outcomes.


Product of the addition of 1 eq. bromine to 1-methylcyclohexa-1,4-diene When we add one equivalent of $\ce{Br2}$ then which double bond will get brominated?

The double bond in the molecule with methyl group will be more stable as it will have more hyper conjugating structures. Thus the other double bond should get brominated.
But my textbook says the double bond with methyl group will get brominated. Can someone explain why?

You misunderstood what hyperconjugation is. It occurs when the electrons in a $\ce{C-H}$ $\sigma$-bond are shared with an adjacent atom's empty or partially filled $\mathrm{p}$-orbital, giving the species an extended molecular orbital and thereby stabilizing it. This means that hyperconjugation doesn't really come into play until bromine is added across one of the double bonds. Let's consider the two possible intermediates:

$\hspace{7.cm}$

$\hspace{6.5cm}$

The first one is a tertiary carbocation, and the second is a secondary carbocation. The first intermediate is more stable due to the hyperconjugative effects of the methyl group, and will require less energy to form than the second. This means the reaction will occur much faster at the more substituted double bond, and is why that is the major product.


Is fluorine sp3 hybridized in BF3 Since BF3 has the shape of tetrahedral, and in the provided solution, Boron is sp2, why is Fluorine sp3 hybridized?

First things first: $\ce{BF3}$ is not tetrahedral. It cannot be, as a tetrahedral structure requires a total of five atoms (one in the centre and four in the corners).[1] Instead, it is trigonal planar. However, that is of minor importance.

Secondly, fluorine is most certainly not sp3 hybridised. In general, hybridisation is determined by geometry and not vice-versa, thus we only need to establish the trigonal planar geometry to arrive at boron’s sp2 hybridisation. However, this doesn’t help us for terminal atoms as there is no ‘geometry’—only a single binding partner. In some cases, constraints may exist due to multiple bonds which require π-type bonding orbitals. Again, this is not the case for single bonds which rely solely on their σ orbital.

Instead, the basic assumption should be to assume the hybridisation as low as possible. In that way, fluorine’s s type orbital can be as low as possible in energy—which means a higher energy gain. It is easy and possible to assume fluorine bonds solely with a correctly aligned p orbital and so it does.

Note:

[1]: This sentence depends on the system of classification you were taught. I was taught only to label those molecules as tetrahedral which have five atoms that correspond to the four corners plus the centre. If there are four atoms as in the case of ammonia ($\ce{NH3}$), it is possible for the overall structure to be a tetrahedron with one corner removed. We were taught not to call this tetrahedral but a trigonal pyramid. Your teacher’s mileage may vary.


How to change color of water based on its acidity? I recently saw a video on YouTube in which water, when mixed with a solution (I believe iodine was involved,) changed color based on its acidity. I am attempting to replicate this experiment, but cannot concoct the solution. Any help would be much appreciated.

EDIT:
To clarify, if I had water with a ph of 10 it would turn red, but if the water had a ph of 4 it would turn green.

EDIT 2:
I found the video in question
http://youtu.be/_yi83kOB8ug

I'm assuming that you're doing this at home. A popular (and SAFE!!) experiment at home is to use red cabbage, vinegar, and baking soda.

The indicator comes from the cabbage.

Vinegar turns the solution acidic.

Baking soda turns the solution more alkaline.


Why and how does lycopene absorbance increase with heating? Lycopene, which is a carotenoid and a phytochemical is the main reason for why tomatoes are radio-protective.

Source 1, Source 2

Let's cut to the chase: I was wondering if this claim of this source is of scientific value:

"Heating will result in an increase in amount of lycopene absorbed by the body."

It may be possible that the researcher made a more "every-day" statement for the readers to understand.

I have two questions:

Is the statement scientifically correct?
If so, why and how would such a chemical be absorbed more with the heating process?

Note that if you want to make the argument that this is a biology question rather than a chemistry one, I would bring up these reasons:

The temperature in which the heated lycopene and raw tomato are going to be absorbed in the body are the same. Thus, it might be possible that heating lycopene results in a chemical reaction.
I'm not looking for the pattern of absorption by the body. I'm looking for possible changes or specifications of its structure.

"Heating will result in an increase in amount of lycopene absorbed by
  the body."
  Is the statement scientifically correct?

It is a poorly phrased statement, but it is probably generally true.  It would have been better to say something like, "heating tomatoes will increase the amount of all-trans-lycopene available for ingestion."

Here's a link to a full paper on the subject (click the "View" button near the top-right of the page).  Basically two things are going on at the same time when lycopene containg food is heated.  It is the naturally occurring, all-trans isomer of lycopene that is thought to be beneficial to the human body. Heating lycopene causes it to isomerize to a variety of cis-isomers.  Heating in the presence of oxygen can also cause oxidation.  Both of these processes (isomerization and oxidation) reduce the potency of the lycopene in the broth.

However, in parallel with these deleterious effects, heating also makes more lycopene available in the heated broth or extract.  This is because heating disrupts cell walls and releases lycopene from the cells.  The above link shows that, at least at moderate temperatures, the latter effect predominates and so more active all-trans-lycopene is made available upon heating tomatoes, despite the loss of some lycopene to isomerization and oxidation.


Reaction of NaOH with methanol and ethanol? Methanol ($\ce{CH3OH}$) and ethanol ($\ce{CH3CH2OH}$) react with sodium metal ($\ce{Na}$) to form sodium methoxide ($\ce{CH3O^-Na+}$) and sodium ethoxide ($\ce{CH3CH2O^-Na+}$):

$$\ce{2CH3OH + 2Na -> 2CH3O^-Na+ + H2}$$
$$\ce{2CH3CH2OH + 2Na -> 2CH3CH2O^-Na+ + H2}$$

Do methanol and ethanol react with sodium hydroxide ($\ce{NaOH}$) in the same way as sodium do to form sodium methoxide and sodium ethoxide respectively?

The proposed reaction:

$$\ce{CH3OH(l) + NaOH(s) <<=> CH3O^-Na+(s) + H2O(l)}$$

According to wikipedia:

  The solid hydrolyzes in water to give methanol and sodium hydroxide

The equilibrium is biased to the left, as @vapid has mentioned in the comments. To quote him:

  In the large excess of alcohol, the reaction will be considerably shifted towards alkoxide.


Which is more acidic: p-xylene or 3-methoxytoluene? Which is more acidic: p-xylene or 3-methoxytoluene?

$\ce{OCH3}$ shows -I and +M effect, and $\ce{CH3}$ shows +I effect. Since, the former shows some -I effect, it should be more acidic. But that's not the case. Any other explanation here?

$\text{Acidic Strength}\propto \text{-I, -M, Negative Hyperconjugation effects} \propto \dfrac{1}{\text{+I, +M, +Hyperconjugation effect}}$

Recall that $\ce{-OCH3}$ is a moderately activating group as its mesomeric effect overrules its inductive electron withdrawing effect.

On the other hand, you must be aware that $\ce{-CH3}$ is only weakly activating because of hyper conjugation and inductive electron releasing effect.

Now, its easy to conclude that a strongly activating group would result in more electron density over and around the benzene ring and hence destabilise the anion formed after deprotonation. Thus, we'd expect p-methoxytoluene to be less acidic than p-xylene.


Can chemicals change crystal structure by only touching? Are there any known, or suspected compounds that take the crystal structure of what ever object they are touching, i.e. if it touches $\ce{NaCl}$ it would turn into a face centered cubic crystal structure.

No, there is no "exact mimic" crystal, but the crystal habit can alter the crystal form. Consider that the environment in the uniform inside of a substance is different from that along its boundaries; growing on a different surface can alter the shape of the crystal.

The semiconductor industry, in particular, is often faced with difficulties in growing crystalline semiconductors with layers of varying composition (epitaxy). There is often strain, leading to defects, unless an intermediate layer can smooth the transition.


"Cage-like" Formazine particles, how are molecules arranged? Question: What would the  "cage-like" arrangement of Formazine molecules look like in a Formazine particle while it is still suspended in water? Would it be totally random, or have some kind of imperfect but recognizable regularity?
Formazine is used as an easily defined and prepared calibration or reference standard for turbidity measurements.

above x2: Formazine from PubChem.
From Wikipedia:

Formazine (formazin) is a heterocyclic polymer produced by reacting hexamethylenetetramine with hydrazine sulfate.
The hexamethylenetetramine tetrahedral cage-like structure, similar to adamantane, serve as molecular building block to form a tridimensional polymeric network.

From Formazine:

The readings shown by turbidimeters are not scaled in the measured light intensity, but in concentration of a reference suspension. Because the accuracy of this calibration solution determines the reliability of subsequent turbidity measurements, it is of crucial importance.
The internationally established turbidity standard for calibration is formazine. This standard can be reproduced at any time with the recipe taken from standard ISO 7027 (Water quality - Determination of turbidity).
It is extremely important to observe the prescribed preparation temperature, because it affects the particle size of the formazine particles perceptibly. The following Figure illustrates this. Errors caused by temperature variations are on the order of 1..2% per °C. Consequently, Sigrist keeps the preparation temperature constant to within ± 1°C.

above: "Particle size distributions of formazine at various preparation temperatures" from here.

above: Enlarged from TurbidityStandards.

above: Slide #10 from What is Turbidity? John Daly, ISA NorCal President, South Fork Instruments, Inc. Click for full size.

First, note that

formazine structure attached from PubChem is a different compound;
note that there are also, related, formazan $\ce{HN=N-CH=N-NH2}$ and “formaldazine” $\ce{CH2=N-N=CH2}$ (dimethylidenehydrazine);
formazin (polymer) does not necessarily has to have cage-like structure, as this characterization belongs to hexamethylenetetramine in the quoted citation.

Hexamethylenetetramine (urotropine, $\ce{(CH2)6N4}$) can be prepared from formaldehyde and ammonia. In water it also somewhat decomposes to its precursors:

Formazin (also called formalazine) structure might be still unknown, as a related publication from 1976 states:[1]

  … This substance, an insoluble condensation product of uncertain composition, is prepared by mixing solutions of hexamethylenetetramine and hydrazine sulfate. …

Sometimes empirical-like formula $\ce{(C2H4N2)_n}$ is claimed.[2]

Sometimes, claimed structure like following one is presented:[3,4]

It was (with several other structures, less nice or likely), hypothesized (well, rather more complicated, with some ether linkages $\ce{-CH2-O-CH2\bond{-}}$ and $\ce{-NH\bond{-}}$ groups present), based on spectroscopic studies.[5,6]

References:

Rice, E. W. The preparation of formazin standards for nephelometry. Analytica Chimica Acta 1976, 87 (1), 251–253.
Ziegler, A. Issues Related to Use of Turbidity Measurements as a Surrogate for Suspended Sediment. Turbidity and Other Sediment Surrogates Workshop; Reno, NV, USA, 2002.
Kaur, N.; Kishore, D. An Insight into Hexamethylenetetramine: A Versatile Reagent in Organic Synthesis. J. Iran. Chem. Soc. 2013, 10 (6), 1193–1228.
Sadar, M. J. Stabilized Formazin Composition. US Patent 5,777,011, July 1998.
Mashima, M. The Infrared Absorption Spectra of the Condensation Products of Formaldehyde with Hydrazine. Bull. Chem. Soc. Jpn. 1966, 39 (3), 504–506.
Bondybey, V. E.; Nibler, J. W. Infrared and Raman Spectra of Formaldazine. Spectrochimica Acta Part A: Molecular Spectroscopy 1973, 29 (4), 645–658.


Number of planes with maximum atoms Number of planes in $\text{SiH$_2$FCl}$ which contain maximum atoms but all must be different?

I am interested in the above problem. I predicted the answer to $10$ which indeed is $5\choose 3$ because maximum number of points in a plane will be $3$ since the structure is tetrahedral. But the book writes an answer $7$ to the above problem!! Any help will be highly appreciated.

I predicted the answer to 10 which indeed is $5\choose 3$ because maximum number of points in a plane will be 3 since the structure is tetrahedral.

Yes, 10 is the number of sets with three atoms, and each set defines a unique plane.

  But the book writes an answer 7 to the above problem

3 of those (HHSi, HHF, HHCl) don’t count, so 7.


Is there a hypervalent molecule which has no contribution of d-orbitals at all? I thought that d-orbitals were used to "extend the octet rule" in these molecules until I looked it up on Google and saw that they actually contributed only a little to the molecule. A little, but not zero.

Is there a hypervalent period-3-element molecule that has mathematically been proved not to involve d-orbitals at all?
Do they have a role, any role, in the chemical properties of the final molecules?


What is the proper laboratory technique to place a solid in a beaker that will be weighed? I was doing a lab at school yesterday which involved transferring a precise amount of solid reagent into a beaker.

The way my professor told me to do it was to tap the reagent container so that the solid would fall out into the preweighed beaker and then weigh the beaker. However, I always thought that the correct procedure was to weigh a piece of paper, then place the solid onto the paper, reweigh the paper to obtain the correct mass reading, and then use the paper to pour the solid into the beaker.

What are the advantages of using the teacher's way over the way I proposed?

There is no correct or universal weighing protocol per se. The key question is what is the desired level of accuracy? Your teacher's approach is quite cautious but this is not the universal way, as what you did is also common. He/she is apparently avoiding a spatula as well. I personally dislike paper and plastic weighing boats. If you ever had a chance to weigh dry salts or organic crystals on a dry day, these materials charge like crazy and literally all particles fly apart the moment you touch the plastic boat with gloves.

Pharma companies, where a high level of accuracy is required and legal formalities are important, chemists often use weighing funnels. They are not popular in academia (pricey!). In the picture, you see a latest type of balance which avoids static issues. On the top you see a weighing funnel. This one is plastic coated with anti-static material but glass is more common. After weighing, the funnel containing the weighed substance is placed on the mouth of a volumetric flask and washed very carefully. You can't do this with paper.

https://www.laboratorynetwork.com/doc/new-smartprep-weighing-funnel-weigh-in-samples-more-accurately-0001


reaction mechanisms of peroxides Task: Explain the reaction mechanism of 1,4-Dioxane in terms of how you get peroxides. Start here with the autoxidation with one radical X.

My explanation:

in the first step 1,4 Dioxan reacts with Dioxygen, so one gets an initiatorradical and and HOO* (* symbolises here a free electron, a radical). In the second step the initiatorradical reacts with oxygen and one receives a peroxyradical. In the end the peroxyradical reacts with an hydrogen atom and one comes to hydroperoxide. Is that right?

I could not find a definitive mechanism for 1,4-dioxane, but this is the mechanism for the formation of THF peroxide is likely very similar

diagram from here


Can -NH2 groups get nitrated? In the chapter of amines, and I see reaction occuring between amines and nitrous acid. However I am thinking why I don't see reactions of amines also react $\ce{HNO3}$ in presence of stronger acid like $\ce{H2SO4}$, that can protonate $\ce{HNO3}$, causes $\ce{H2O}$ to leave and then finally have amines to react with $\ce{NO2}$, making nitroamines?

Yes, you can form a nitroamine (nitramine) from a secondary amine and nitric/sulfuric acids. Some explosives have a nitramine structure (e.g. tetryl, RDX, HMX, HNIW).


Role of zinc chloride in reaction of ether I am trying to do a reaction between acetyl chloride $\ce{CH3COCl}$ with diethyl ether. I know that upon using anhydrous $\ce{ZnCl2}$ and heat, I'll get an ester and an alkyl chloride, but I'm not able to guess the mechanism.

Does $\ce{ZnCl2}$ break the $\ce{C-O}$ bond in ether as it would've done to alcohol? Please help me with the mechanism.

I think that it reminds Fischer esterification at the beginning. Zinc dichloride acts as a Lewis acid by coordination at the carbonyl oxygen in AcCl. It activates C=O bond because of larger contribution of carbocation form. And this cation undergoes coordination of Et2O with formation of zwitterionic tetrahedral intermediate. This particle eliminates our catalyst and cloride-anion. The product of this step is acetyl diethyl oxonium. Cloride-anion is quite nucleophilic, but we have an excellent leaving group without charge (AcOEt), so Sn2-reaction occurs and lead to the final products


Is it possible to form an 8 carbon ring using a Diels-Alder reaction? I need to convert the starting material to the product on the right but I was wondering if it was possible to use a Diels-Alder reaction and use $\ce{KMnO4}$ to cleave the ring.

You can obviously use Diels-Alder Reaction for the synthesis of the desired product but not so directly. But Diels-Alder will give an optimal yield when the ring formed is Six-Membered. After the formation of that Six-Membered product, you have to increase the no. of carbon atoms in the carbon-chain by some way, and the yield of your product will be also better if you no. of steps as less as possible.

I will not suggest you to directly add 1,3-butadiene to acetylene because there will be a chance of aromatisation, and then you cannot do anything, as to break the aromatic rings are highly unfeasible. Instead I can suggest you the following steps:

First you partially reduce the acetylene to ethene, and then perform Diels-Alder reaction with butadiene. You will get cyclohexene.
In the second step, you perform ozonolysis, and then to the resulting compound, perform  Wittig reaction with excess reagents. You will get octa-1,7-diene.
Lastly, you can perform hydroboration, oxidation to get less substituted diol, and then oxidise that to acid by any strong oxidising agent like $\ce{KMnO4}$ or $\ce{K2Cr2O7}$ in slightly acidic medium.

Thus, you will get your desired product. The path of reaction is shown below:


Why do we have glycogen in the muscles and not glucose? Why do we have glycogen in the muscles and not glucose?

I know this is in contrast to liver cells, which, on demand, readily do break down their stored glycogen into glucose and send it through the blood stream as fuel for other organs. Glycogen is also a suitable storage substance due to its insolubility in water, which means it does not affect the osmotic pressure of a cell.

I think you already answered your own question.  If the glycogen, either in liver or muscle, were all converted to glucose at once it would have a disruptive osmotic effect. But muscles also readily break down glycogen, when they need glucose to contract.  They just don't ship the glucose to other organs the way liver does.


Why is the residual proton resonance in deuterated benzene a singlet? In deuterated benzene ($99\%$) the residual protio peak is a singlet at $\pu{7.16 ppm}$. This however doesn't make sense since the protons will be randomly distributed among various benzene molecules. Therefore there won't be a mixture of $99$ $\ce{C6D6}$ molecules and $1$ $\ce{C6H6}$ but $94$ $\ce{C6D6}$ and $6$ $\ce{C6D5H}$. Considering that deuterium has a spin of $1$ the residual solvent peak in D6-benzene should be a septet. Any thoughts?

To a very good approximation, $J_\ce{HD}$ in a deuterated molecule is related to $J_\ce{HH}$ in the corresponding protonated molecule by the gyromagnetic ratios:

$$\frac{J_\ce{HD}}{J_\ce{HH}} = \frac{\gamma_\ce{D}}{\gamma_\ce{H}} = \frac{\pu{41.066 * 10^6 rad s-1 T-1}}{\pu{267.522 * 10^6 rad s-1 T-1}} = 0.15$$

In benzene where $^3J_\ce{HH} = \pu{7.6 Hz}$ and $^4J_\ce{HH} = \pu{1.4 Hz}$ (Reich NMR tables), we would thus have $^3J_\ce{HD} \approx \pu{1.2 Hz}$ and $^4J_\ce{HD} \approx \pu{0.2 Hz}$. The four-bond meta coupling would be all but invisible, and the three-bond ortho coupling if resolved would give rise to a $1:2:3:2:1$ quintet.

On a lower-field spectrometer this relatively small coupling is usually not properly resolved. A suitable window multiplication, though, will reveal all. I don't have a d6-benzene spectrum on hand, but I do have a $\pu{500 MHz}$ d8-toluene spectrum here. The two spectra on the left are the aromatic residual peaks, but at the bottom-left a Gaussian window function has been used to improve resolution (at the cost of signal:noise ratio). This allows us to resolve the three-bond ortho couplings of $\pu{1.1 Hz}$.

At the bottom-right is the signal corresponding to residual $\ce{C6D5CD2H}$. Since the two-bond H–D coupling has a larger magnitude ($^2J_\ce{HD} = \pu{2.2 Hz}$), it is much easier to resolve, and can be seen even with the normal exponential window function (which sacrifices resolution for signal:noise ratio improvement).

Similarly, in d6-DMSO (shown below at $\pu{500 MHz}$), the two-bond coupling is more easily resolved.


On what basis can we say that one nuclide is more stable than another one? I read in a textbook that the nuclide $^{40}_{20}\ce{Ca}$ is more stable as compared to $^{30}_{13}\ce{Al}$. I checked it online and saw that most sources describe the observation with respect to odd and even number of protons in $\ce{Al}$ and $\ce{Ca}$ respectively.

What is the reason for this? Is it due to proton-proton repulsions overcoming the attractive forces between nucleons?

In general it is very difficult to estimate the stability of a nucleus based on the number of protons and neutrons it contains, as it is an interplay of different forces (repulsion between protons and the Strong force). However, a crude estimate can be made based on these empirical rules:

There are no stable nuclei with $Z>83$ (Also no stable nuclei for Technetium ($Z=43$) and Promethium ($Z=61$)).
The stability of a nucleus depends on the ratio $$\eta=\frac{A-Z}{Z}$$
there are no stable isotopes with $\eta<1$, except for $^1$H and $^3$He. Stable nuclei with relatively small atomic number have $\eta \gtrsim 1$, while nuclei with larger $Z$ need more neutrons to stabilize the nuclear charge, for instance $^{206}_{\,\,82}$Pb has $\eta=\tfrac{124}{82}\approx1.51$.
Most stable nuclei have an even number of protons and neutrons, there are only a few stable nuclei that have both an odd number of protons and neutrons.
Nuclei that have a number of protons or neutrons that equals the so-called magic numbers 2, 8, 20, 58, 50, 82, 126 are especially stable. Nuclei that have both a magic number of neutrons and protons (such as $^4_2$He, $^{16}_{\,\,8}$O, $^{40}_{20}$Ca, $^{48}_{20}$Ca, and $^{208}_{\,\,82}$Pb) are exceptionally stable.

In your example, $^{40}_{20}$Ca belongs to this special category.


What is the bandgap value of SiO2? Using Diffuse Reflectance Spectroscopy, I have obtained the band gap for $\small\ce{SiO2}$ (purchased from a chemical retailer) in the range of 3.25 to 4.75 eV.

I would like to know if the values are in the right ballpark and, if not, would like to review the DRS procedure that I carried out.  I would also like to know if the band gap determined through DRS would have such a wide range.

For thermally grown amorphous $\ce{SiO2}$ films, a bandgap of 9.3 eV was measured by Weinberg.

Measurements on $\alpha$-quartz by Calabrese give $E_g(\Gamma)$ = 6.3 eV, compared to 9.2 eV calculated by Chelikowsky and Schlüter.

Calculations by Ramos on $\beta$-cristobalite with the space groups $Fd3m$ (face-centred cubic, fcc), $I\bar{4}2c$ (body-centred tetragonal, bt), and $P2_13$ (simple cubic, sc) gave direct bandgaps of $E_g(\Gamma)$ = 5.48, 5.79 and 5.73 eV, respectively.


Why is creating water exothermic? I have to calculate the bond dissociation energy of steam. I'm a total noob, so don't go too hard on me.
(I'm translating the exercise by myself, sorry if there is a mistake.)

Calculate the bond dissociation energy of gas-water/steam ($\ce{H2O}$). Whereas the total enthalpy of the bound is the sum of the "sub-bounds". Use the following table:

Now, we have: $\ce{2H2 + O2 -> 2H2O}$
Imagine we have a tank with one $\ce{O2}$ and two $\ce{H2}$, then, according to the table (if I read it correct), we need $2\cdot436\ \mathrm{kJ} + 497\ \mathrm{kJ}=1396\ \mathrm{kJ}$ to "break" the bonds. We then get two $\ce{O}$ and four $\ce{H}$. They react to two $\ce{H2O}$
Now, according to the table, two $\ce{H2O}$ need $2\cdot 2\cdot 463 \ \mathrm{kJ}=1852\ \mathrm{kJ}$ to bind together.
So, we start with a system of total inner energy $1396\ \rm kJ$ and end up with one of $1852\ \rm kJ$. Clearly, we have to put energy into the first system to get to the second.
Now, that's obviously false - but I don't get why. I'd expect a exothermic reaction because 1. I just know it is and 2. because the $\ce{O2}$ have a double bound which I just expect to be harder to break.
So where is my thought wrong?

You’re mistakenly conflating the bond dissociation energies between the diatomic components and the product molecule. If the bond dissociation energy of water is larger than the bond dissociation energy of the individual diatomic components, then the system is “lower energy” (preferred) in the $\ce{H2O}$ state and that “extra energy” in the atomic state will be released in the form of exothermic heating. Remember that elemental $\Delta H = 0$, and the water molecule has stronger bonds/lower enthalpy and would require more energy to dissociate than the elemental components (which is what your math shows), meaning water won’t spontaneously dissociate into the diatomic gases since that would be an endothermic process and require some external energy. As an aside, keep in mind that dissociation of bonds in a molecule like water won’t be consistent between atoms (after the first hydrogen bond has broken from an $\ce{H2O}$ molecule, the remaining hydroxyl group requires less energy to dissociate). In sum, if your product has a higher bond dissociation energy than your reactant, it’s at lower enthalpy and the net of the reaction will be a release of energy in the form of exothermic heating.


Irreversible Adiabatic Process While studying about the entropy change of surroundings in irreversible adiabatic process, I came across a sentence.

  For infinite reservoir,  all processes are considered reversible

And suddenly it resumes with the equation:

q/T = dS (surrounding)

Can anyone explain the line and how the conclusion was drawn?

You'll probably find a partial answer in this post. The reversible/irreversible tag refers to what is happening to the system, not the surroundings. Events in the surroundings are as a rule treated as reversible.

You should also keep in mind that heat is a signed quantity, and that there can be different conventions with regard to what is a positive/negative heat transfer. Usually chemists say q>0 when the system receives energy as heat, and when the surroundings loses energy as heat. Since heat is regarded as the sole source of change in the entropy of the surroundings (assumed to be an inifinitely large reservoir at constant T$^\ast$), we write $$dS_{surroundings} = -\frac{dq}{T}$$ for an infinitesimal heat transfer and $$\Delta S_{surroundings} = -\frac{q}{T}$$ for a finite heat transfer.

$^\ast$ An infinite reservoir can provide or accept energy in the form of heat without temperature change (or any other change in its state).


Computing accurate vibrational and rotational contributions to the free energy of transition states and loosely bound complexes Recently I have been dealing with a lot of transition states and relatively loosely bound ion-dipole complexes and I have some trouble figuring out how to make sure that the rotational and vibrational contributions to the Gibbs free energy are as accurate as they reasonably can. (I am using Gaussian 09 rev C)

I am using MP2/aug-cc-pVTZ, based on literature results it should be in close agreement with even better methods for the type of molecules I am working with.
First I have optimized the geometry to verytight convergence, then reoptimized it with verytight, CalcAll to make sure the geometry is as converged as possible, and get the thermochemistry output at the same time.
I have once read in a CCL post, that tightening SCF convergence can increase the accuracy of analytic derivatives, so I am performing all calculations with SCF=Conver=12.

My concerns are mostly related to having a number of low frequency modes, and the possibility of coupling between vibrations and rotations.
For example, in the case of the transition state between $\ce{MeCl + F-}$ and $\ce{MeF + Cl-}$, I have three low frequency modes: 258.4867 $cm^{-1}$, 258.4867 $cm^{-1}$ 303.2302 $cm^{-1}$ (and of course an imaginary frequency). Sure enough, Gaussian warns that:

  Warning -- explicit consideration of   3 degrees of freedom as vibrations may cause significant error

Based from what I gathered from various sources, this means that these low frequencies have significant uncertainties in them, and they may or may not be internal/hindered rotations. I have ran a calculation with Freq=hindrot but it did not identify any of the modes as rotations. However, the Gaussian manual says that the hindered rotor analysis is not always successful at finding rotations when transition states are involved.

My other concern is the validity of the rigid rotor approximation for molecules having low frequency modes (ie. low force constants). My intuitions tell me, that the lower the force constants in a molecule, the worse are the errors caused by the centrifugal distortions of the molecule.

So my actual questions are:

1.What is the proper way to accurately handle low frequency modes? How does one make sure that they are not some sort of rotation, when hindered rotor analysis does nothing? If they legit low frequency vibrations, are they reliable? If they are not reliable how can they be corrected?

2.Do I have to be concerned about low frequencies also implying that the rotational contributions are also affected due to centrifugal distortion?

Given the accuracy it sounds like you're looking for, and the systems you're studying, this indeed appears to be a Very Hard Problem. I think your intuitions are well founded.

Briefly, to your questions:

1) How to handle the low-frequency modes?

A lot of manual exploration of the PES, doing things like running normal mode scans and optimizing intermediate geometries along suspected internal rotations. This gives you an idea how the various energetic barriers to non-harmonic motions compare to a Boltzmann distribution of states.

If the barriers are high, the harmonic approximation is probably decent and you can just use Gaussian's results
If barriers are low, then you have ~free internal rotation.
If intermediate, then you're facing hindered internal rotation.

There's literature on evaluating enthalpy/entropy contributions of free and hindered internal rotations.  Some can be found in my answer to a different question, here.  Other potentially relevant resources include:

Czaszar (WIREs Comput Molec Sci 2 273, 2012)
Overview of anharmonic force fields and their computation.
Bloino (J Chem Theor Comput 8(3) 1015, 2012)
Overview of the VPT2 anharmonic computational system implemented in recent versions of Gaussian.
Carbonniere (Chem Phys Lett 392 365, 2004)
Description of the vibrational-rotational (Coriolis) coupling anharmonic correction and its potential significance for some systems. More recent references are likely available.
Meal (J Chem Phys 24(6) 1119, 1956)
Meal (J Chem Phys 24(6) 1126, 1956)
Classic discussions of Coriolis coupling.

You should be able to turn up quite a bit more information with some quick further searches, if need be.

2) Do I need to be worried about contributions from centrifugal distortion?

Very likely. Unfortunately, I don't know of very many examples where people have attempted corrections such as this. Certainly you're looking at manually developing a suitable analysis; I can't imagine any software packages out there at present are capable of automatically chugging through this for you.


Units of mass on the atomic scale what are the systems of recording atomic masses and their units?

I know that the nucleon number is the number of protons and neutrons

I also know that the mass number on the periodic table is the masses of all the isotopes and their relative abundance (I believe this is called relative atomic mass and is equal in numerical value to the molar mass of that element)

I am not sure what units relative atomic mass is measured in. some say there is no units as it is a relative scale, others say that it is measured in atomic mass units (AMU), unified mass units (u) or daltons (Da).

I believe (but would like it to be confirmed) that the atomic mass unit (AMU) is an obsolete unit based on the mass relative to oxygen.

I also believe (but would like it to be confirmed) that the unified mass units (u) and dalton (Da) are equivalent units (u=Da=1/12 mass of a carbon-12 atom)

what is the difference between isotopic mass, relative atomic mass and atomic mass and what units do they all have?

The quantity atomic mass (quantity symbol: $m_\mathrm{a}$) is defined as rest mass of a neutral atom in the ground state.
The dimension of the atomic mass is
$$\dim m_\mathrm{a} = \mathsf{M}$$
The coherent SI unit for atomic mass is ‘kilogram’ (unit symbol: $\mathrm{kg}$).

The quantity relative atomic mass (quantity symbol: $A_\mathrm{r}$) is defined as the ratio of the average mass per atom of an element to 1/12 of the mass of an atom of the nuclide $\ce{^12C}$; i.e. the relative atomic mass is the ratio of the mass of an atom to the unified atomic mass constant:
$$A_\mathrm{r} = m_\mathrm{a}/m_\mathrm{u}$$
where $m_\mathrm{a}$ is the atomic mass and $m_\mathrm{u}$ is the unified atomic mass constant.
The relative atomic mass is a quantity of dimension one (for historical reasons, a quantity of dimension one is often called dimensionless):
$$\dim A_\mathrm{r} = 1$$
The coherent SI unit for relative atomic mass is the unit one (symbol: $1$).
For historical reasons, the IUPAC accepts the use of the special name ‘atomic weight’ for the quantity relative atomic mass. The use of this traditional name is deprecated.

The value of the unified atomic mass constant (symbol: $m_\mathrm{u}$) is defined as 1/12 of the mass of a neutral atom of the nuclide $\ce{^12C}$ in the ground state at rest.
$$m_\mathrm{u} = \frac{m_\mathrm{a}(\ce{^12C})}{12}$$
The recommended value is
$$m_\mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$ (source)
Note that the value in SI units is obtained experimentally.

The unit dalton (unit symbol: $\mathrm{Da}$) and the unified atomic
mass unit (unit symbol: $\mathrm{u}$) are non-SI units that are accepted for use with the SI. Actually, ‘dalton’ and ‘unified atomic mass unit’ are alternative names for the same unit, equal to 1/12 times the mass of a free carbon-12 atom, at rest and in its ground state, i.e.

$$1\ \mathrm{Da} = 1\ \mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$

Thus, the value of the unified atomic mass constant is, by definition, equal to one dalton or one unified atomic mass unit:

$$m_\mathrm{u} = 1\ \mathrm{Da} = 1\ \mathrm{u} = 1.660\,538\,921(73) \times 10^{-27}\ \mathrm{kg}$$

The old relative atomic masses (unfortunately called ‘atomic weights’) and the corresponding atomic mass unit (amu) were originally referred to the relative atomic mass of oxygen, which was taken as 16. However, physicists used the atomic mass of the nuclide $\ce{^16O}$ whereas chemists used the average atomic mass of natural oxygen. This unit became obsolete when IUPAP (1960), IUPAC (1961), ISO, CIPM (1967) and CGPM (1971) agreed to assign the value 12 to the relative atomic mass of the nuclide $\ce{^12C}$.


How and why does this Cu change into C (graphite)? Source: gutefrage.net

At the right side of the graphic, the Cu changes while discharging into C as graphite. How and why does this happen?

I thought it happens because of the dissolving of the Cu when the electrons stream from the Cu to the Al.

The electrode in here has two parts – the Cu and the graphite (C). The Cu won't dissolve. Only the graphite (C) will react. So there is no changing from Cu to C.


Why are there distortions in electron cloud in molecules? How does increase surface area amplify these distortions? In the physical properties of hydrocarbons,it is said that branched hydrocarbons have relatively less boiling point than it's linear form. From what I have learnt is that it is due to the increase in the surface area of the molecule which increases the intermolecular force. Now, this intermolecular force is the london dispersion force which happens because of the instantaneous distortions in electron cloud. This means the more the surface area of contact the more the ease of these distortions in electron clouds. This explains the strength of intermolecular force of branched and linear form of hydrocarbons and thus the difference in boiling point. But why there are distortions in the electron clouds? Why surface area has to do with these distortions?

The distortions are a quantum effect

Yes, branched chain hydrocarbons tend to have lower boiling points and this depends on their surface area. But the "distortions" in the electron clouds don't depend on the surface area.

The ultimate reason why there are "distortions" (more accurately described as temporary dipoles caused because electrons are not evenly distributed) is because of quantum effects. The uncertainly principle says we can't know the combined location and momentum of quantum things beyond a certain limit. When you work through the mathematics of this, this means that electron clouds always contain random fluctuations that appear as short lived temporary dipoles. The fluctuations in one molecule's cloud can be felt by the electron clouds in other molecules (and can cause second order effects that alter their electron clouds as well as interacting with their random dipoles). When you work out the (fairly complicated) mathematics of this you get a weak, short range, attraction between even non-polar molecules. We call those London Dispersion forces (or Van Der Waals forces).

But the underlying electron fluctuations don't change because of the shape of a molecule. However, the total force of all the fluctuations added up depends on the surface area of the molecule (how can they interact with other molecules except at the surface?) More surface, more opportunity for the fluctuations to interact with other molecules.

So it isn't that the distortions themselves depend on the surface area, it is that in a larger surface there is more opportunity to have them interact and hence the total force caused by the focus is larger.


Why is hot concentrated sulfuric acid preferable to cold? Why is hot concentrated sulfuric acid sometimes used to react with metals instead of concentrated sulfuric acid which is at room temperature?

I have ascertained that concentrated sulfuric acid has quite strong oxidizing properties.  Still, some metals which are below the hydrogen in electrochemical series do not react with concentrated sulfuric acid which is kept at room temperature.  When the acid is warm/hot, the reaction takes place.

All I want to know is how does temperature affects its oxidizing properties?

A really short answer is kinetics. Heating up chemicals adds more energy to the system, allowing molecules to colide with more force increasing the probably of a reaction taking place.


Order of priority in hybridization of d-orbitals During hybridization of d-orbitals why the $d_{z^2}$ and $d_{x^2-y^2}$ orbitals are used up before the $d_{xy}$, $d_{yz}$ and $d_{zx}$ ?

The rotational frame of reference which defined which axis is the z-axis is changed such that $d_{z^2}$ and $d_{x^2-y^2}$ are hybridized first is a convention / practice.

In theory (meaning mathematically), you should be able to get equivalent geometry if you used the other d orbitals for hybridzation.

Basically the donut thing is weird enough we try to fold it in so we can deal with the more normal cloverleafs as leftovers.


Multiplication of the reaction and Equilibrium constant $$\begin{align}\ce{2A + 2B &-> 2C }\\ \ce{A + B &-> C}\end{align}$$

Why does the equilibrium constant change?

And why the rate of the first reaction square the rate of the second? I understand it mathematically, but logically and scientifically I can't understand it at all.

Rate will not be square in most of the cases and who said so? The equilibrium constant scientifically changes because Gibb's free energy is extensive quantity and it is related to equilibrium constant through relation that you might know $\Delta G=\Delta G^\circ +\mathcal{R}T\ln Q $. Where $Q$ is reaction quotient which at equilibrium becomes equilibrium constant and then it gives $\Delta G^\circ = -\mathcal{R}T\ln K$


Determining amounts of product and excess reagent remaining post reaction $$\ce{2FeS2 + 11/2 O2 -> Fe2O3 + 4 SO2}$$
Mass of $\ce{FeS2}$ is $\pu{600 g}$ and mass of $\ce{O2}$ is $\pu{800 g}$. Find the amounts of $\ce{Fe2O3}$ and $\ce{SO2}$ and the remaining amount of the excess reagent.

My Attempt
$$\frac{n_\ce{FeS2}}{2} = \frac{600/120}{2} = \frac{5}{2} \lt \frac{n_\ce{O2}}{11/2} = \frac{800/32}{11/2} = \frac{50}{11}\implies \ce{FeS2}~\text{is limiting reagent}$$
By Mole-Mole Analysis, we have
$$\frac{n_\ce{FeS2}}{2} = \frac{n_\ce{Fe2O3}}{1} = \frac{n_\ce{SO2}}{4}\implies \begin{bmatrix}n_\ce{Fe2O3} \\ n_\ce{SO2}\end{bmatrix} = \begin{bmatrix}5/2 ~\text{mol} \\ 10~\text{mol}\end{bmatrix}$$
$$\text{Excess reagent remaining} = \frac{800}{32} - \dfrac{5}{2}\cdot\dfrac{11}{2} = \pu{11.25 mol}$$
However, my answer does not match with my other classmates. What am I doing wrong?

Note the following relations from the equation you wrote:
1 mol $\ce{FeS2}$ = (11/4) mol $\ce{O2}$;
1 mol $\ce{FeS2}$ = (1/2) mol $\ce{Fe2O3}$;
1 mol $\ce{FeS2}$ = 2 mol $\ce{SO2}$.

You already determined that $\ce{FeS2}$ is the limiting reactant because we have 600/120 mol of iron sulfide and 800/32 mol of $\ce{O2}$;

This implies that 5 mol $\ce{FeS2}$ will require =5x (11/4) mol $\ce{O2}$.

Remaining mol oxygen = 25-13.75=11.25

Now use the relations established above: 5 mol iron sulfide would produce (5/2) or 2.5 mol iron oxide, and 5 mol iron sulfide would produce 5x2 = 10 mol sulfur dioxide.


Can redox reactions involve only non metals? When non metal react they form molecule, and whether the bond is polar or not, the elements in the molecular compound will not have charges what so ever, because they are not ionic. But can non metals be the only that is involved Ina redox reaction ?

A redox reaction is one in which oxidation state of some atom changes, and such reactions need not involve metals, nor any ionic species. Oxidation state is the idealized charge that a given atom would have if all its bonds are treated as completely ionic. Thus, strictly for the purpose of calculating oxidation state, the electrons of a covalent bond are typically assigned to the more electronegative atom. (I'd note, however, that IUPAC provides a different set of rules for calculating oxidation number.) Hence, in order for a redox reaction to occur, there does not need to be a complete transfer of electrons (in the sense of ionization). Examples of redox reactions involving molecular compounds are numerous.

For example, consider the reaction $\ce{N2 + 3H2 <=> 2NH3}$. This is the Haber process for the production of ammonia. Here nitrogen is reduced from an oxidation state of $0$ to $-3$, while hydrogen is oxidized from $0$ to $+1$. Again, nitrogen being the more electronegative atom, is assigned the electrons of the bonds. This idealization is useful in that it provides some rough idea of electron distribution, and hence has some utility in predicting reactivity.

Edit: Also, the statement "[...] the elements in the molecular compound will not have charges what so ever [...]" is inaccurate. There is a concept of formal charge that applies to neutral molecules as well as ions.


Thermodynamic vs. Kinetic products With regards to complexes, I read that thermodynamic products are favored over kinetic products, which are labile. Why wouldn't the first product formed be the most prominent?

The difference between Thermodynamic and Kinetic products is the difference in activation energy and the difference in (Gibbs)-Free-Energy.
When the activation energy is low, the reaction proceeds fast, while the product might not have as low a Free-Energy as an other product (with higher activation energy). When the reaction conditions allow however, the thermodynamically more stable product (with lower Free-Energy) will be formed. Even though the kinetic product first forms, most complexes are form in equilibrium reactions. Therefore, the kinetic product reverts to starting materials and then, slowly forms the thermodynamic product. This thermo dynamic product, though it is also in equilibrium, will not revert to starting materials as easily as the kinetic product will, due to the lower Free-Energy.

To conclude: First, the kinetically favored product will form. Then, when the reaction conditions allow it, the equilibrium shifts towards the thermodynamic product.
Note that the equilibrium is essential for the thermodynamically favored product to form.


Difference between precipitation and crystallisation What is the main difference between precipitation and crystallization.
Is precipitation a chemical change and crystallization a physical change.
How can we say that the resulting solute is precipitate or a crystal?
Is it necessary that when we want precipitates to form then its ksp should be less than its Qsp or the precipitates form even after mixing two unsaturated aqueous salt solutions?

The two words are often used interchangeably (perhaps incorrectly), though there is a slight nuance.
IUPAC defines the following:

Crystallisation: The formation of a crystalline solid from a solution, melt vapour or a different solid phase, generally by the lowering of the temperature or by evaporation of a solvent.
Precipitation: The sedimentation of a solid material (a precipitate) from a liquid solution in which the material is present in amounts greater than its solubility in the liquid.

In essence, a crystallisation is the formation of a crystalline solid (i.e. not amorphous), for instance leaving a beaker of $\ce{NaCl}$ in water out for a long time will eventually grow crystals as the water evaporates.
Precipitation can be thought of a something 'crashing out', for instance if you have certain kinds of amines, you can add $\ce{HCl}$ in diethyl ether to precipitate out the $\ce{HCl}$ salt of the amine as they are not soluble.
There are of course in-between cases. Attempts at 'crystallisation' can yield amorphous solids if the conditions are not right, and, equally, precipitation can sometimes give crystalline solids.
The choice of words is therefore always ambiguous, but in general the IUPAC nomenclature should be followed where appropriate (eg. you can never crystallise to give an amorphous solid).
To answer the other part of your question about chemical vs. physical changes, you are correct that crystallisation is always a physical change (different isomorphs of the same material: amorphous, crystalline needles etc). Precipitation can be either, for instance in the $\ce{PPh3}$ case described by Jan, or in the $\ce{HCl}$ amine salt I previously described.


Cahn–Ingold–Prelog Rules, cyclic structures, and breaking ties (3S,5R)-3-ethyl-5-methylcyclohex-1-ene

Does anyone know how one reaches the conclusion that the 5C is R? I tried using the Cahn–Ingold–Prelog priority rules but kept getting confused.

From chem.libretexts.org:

  When looking for the first point of difference on similar substituent chains, one may encounter branching. If there is branching, choose the branch that is higher in priority. If the two substituents have similar branches, rank the elements within the branches until a point of difference.

From Wikipedia:

  To handle a molecule containing one or more cycles, one must first expand it into a tree (called a hierarchical digraph) by traversing bonds in all possible paths starting at the stereocenter. When the traversal encounters an atom through which the current path has already passed, a ghost atom is generated in order to keep the tree finite. A single atom of the original molecule may appear in many places (some as ghosts, some not) in the tree.

You don't need a digraph for this one. Your structure is A. Remove the double bond and add duplicate carbon atoms to each of the double bond carbons (B). These duplicate carbon atoms each bear three phantom atoms of atomic number zero. Clearly at C5 hydrogen has the lowest priority and methyl is the next lowest. The methylene groups at C4 and C6 are a tie. [There is no need to consider C2 since it comes into play for both routes. It is a wash.] The red carbon of the ethyl group bears C,H,H while the red duplicate carbon at C1 bears 0,0,0. Place your right hand on C5 and point your thumb in the direction of the hydrogen. Now your remaining fingers will point as follows: C4>C6>CH3>H.  C5 is of the R-configuration. See, Twisting stereoisomers with rings to determine R/S.  .


How does alcohol act as a dehydrant? I was doing histological slide preparation with rat's liver. There as a part of the microtechnique the fixed tissue had to be dehydrated using alcohol in the order 30%, 50%, 70%, 90% and 100% for 20-30 min each.
This had to be done to get rid of the water (possibly extracellular) before it was embedded in paraffin for microtomy because paraffin is immiscible with water.

My question is how does the alcohol extract water out of the tissue?

The alcohol displaces water gradually through the process of osmosis. To prevent rupture of cells from the osmotic pressure, samples are immersed in a series of baths of increasing alcohol concentration.

After immersion in 95% or stronger alcohol, it is sometimes the practice to displace the alcohol with toluene or other non-polar solvent so that the specimen can be permanently mounted with paraffin or balsam.


Decomposition products of sodium acetate I heated some sodium acetate to a high temperature (about 500 degrees celsius). Upon doing so, I could smell an irritating smell reminiscent of vinegar. I stopped then, assuming they were acetic acid vapors. Are they really acetic acid vapours, and if so, what is the remaining solid?

The fumes are likely to be acetic anhydride, as per this equation:
$$\ce{2CH3COONa -> Na2O + (CH3CO)2O}$$
This smells like vinegar, and reacts with moisture in the air to give acetic acid.


Is delta H a state function? My teacher says $H$ is a state function but $\Delta H$ is not. But I think both of them should be state function. Is he correct?

$\Delta H$ is a function of two states, the initial state and the final state.

For a given final state, there can be infinite $\Delta H$ values depending upon what the inital state was.

For a given inital state, there can be infinite $\Delta H$ values depending upon what the final state is.

Therefore, $\Delta H$ is not a state function.


How many p-orbitals are there in each period? How many p-orbitals are there in each period?a. 1b. 3c. 5d. 7e. 14

I take this to mean "how many p-orbitals are there in each period of the periodic table?" Looking at the p-block on the periodic table, it looks like there are 6 elements with p-orbitals per period. This isn't an answer, so I'm a little lost.

For the p subshell, the azimuthal quantum number, which describes the angular momentum and shape of the orbitals, is $l=1$. For the associated magnetic quantum number, discrete values in the range $-l, -l+1, ... 0 ..., l-1, l$ are allowed, so the number of different $m_l$ values is $2l+1$.

With $l=1$, it follows that there are $2 \cdot 1+1=3$ p orbitals with the $m_l$ values of -1, 0 and +1, respectively. Each orbital can hold 2 electrons, so the p subshell can contain a maximum of 6 electrons. This also explains why there are 6 elements in the p block of each period. With increasing atomic number $Z$, one p electron is added per element in this block.


Concentrations appearing in the Ka of buffer solutions I've just finished studying buffer solutions and I think I understood more or less how they work, but there's something I can't make sense of:

In the $K_a$ of equilibrium of a buffer solution made from Sodium acetate and acetic acid, we just have the concentrations of acid, $\ce{H+}$ and acetate ion. So, why isn't the concentration of $\ce{Na+}$ taken into consideration in the $K_a$? Why just the others?

Moreover as $[\ce{Na+}] = [\ce{CH3COO-}]$ (the acetate ion in solution is almost only the one dissociated from the salt) shouldn't in the Henderson-Hasselbach solution occur a $a^2$ factor? So in the numerator having $[\ce{CH3COO-}]^2$ and no just a single one?

Thank you

Because Na+ is a spectator ion in this situation.

$$ \ce{Na+ + CH3COOH <=> Na+ + CH3OO- + H+}$$

In other words, Na+ is on both sides of the chemical equation, so [Na+] would appear in both the numerator and denominator of the equilibrium constant if Na+ is included.

$[\ce{Na+}][\ce{CH3COO-}][\ce{H+}]/[\ce{Na+}][\ce{CH3COOH}] = [\ce{CH3COO-}][\ce{H+}]/[\ce{CH3COOH}]$


Pressure over a solution I was wondering something, if I have the same molar fraction for two different substances for a vapor on top of a solution, will this mean that the pressure of each individual substance will be the same ? Since they have the same quantity of molecules ?

This question has a link with Raoult's law : $p_{solvent}=x_{solvent}p^*_{solvent}$, where $p_{solvent}$ is the partial pressure above the solvent, $x_{solvent}$ is the mole fraction of solvent in the mixture, and $p^*_{solvent}$ is the vapor pressure of pure solvent.

If the vapor (i.e. gas) phase above the solution can be treated as an ideal gas, then yes, if two species have the same partial pressure, then they have the same mole fraction.

For an ideal gas mixture, $p_i = y_i P$, where $p_i$ is the partial pressure of species $i$ and $y_i$ is the mole fraction of $i$ in the vapor phase, and $P$ is the total pressure.

The ideal gas assumption will be very accurate in many cases commonly encountered in chemistry, but not always.  Binary mixtures of $\ce{CO2}$ and another compound at high pressure would be one example of vapor phases that are not ideal.  Concentrated vapor of aliphatic carboxylic acids is also not very ideal.


Naming alkenes with double bond branches After reading the IUPAC rules, I found the parent chain to be from the $\ce{CH2}$ numbered 1 to the one numbered 9, and I considered both methyl groups as branches. However, my confusion lies in naming the branch in the middle (depicted in red color) since it contains a double bond:

According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the traditional name ‘vinyl’ for $\ce{CH2=CH-{}}$ is no longer used for preferred IUPAC names and is only retained for general nomenclature. The preferred name is the systematic name ‘ethenyl’.

Therefore, the name for the compound that is given in the question ‘3,7-dimethyl-5-vinylnona-1,8-diene’ could be found in the literature; however, the preferred IUPAC name in accordance with current recommendations is 5-ethenyl-3,7-dimethylnona-1,8-diene.


What did Richter mean regarding an acid base reaction? Recently I've been reviewing concepts belonging to the history of chemistry. But I came stuck at trying to understand a passage which I read from wikipedia entry (and which it seems has been mentioned in different sources). This is related with the law of reciprocal proportions.

It states as this:

  It took 615 parts by weight of magnesia ($\ce{MgO}$), for example, to
  neutralize 1000 parts by weight of sulfuric acid.

There's an existing answer which explains a similar concept (law of multiple proportions) but not specifically to Richter's.

From what I understood in the law of reciprocal proportions there is a set of three elements. Two of them react with a fixed amount of a third element. The ratio of these two elements is the same when they combine between.

For example:

When $63.5~\rm{g}$ copper combines with sulphur produces copper sulphide

$\begin{array}{lllll}
\ce{Cu}&+&\ce{S}&\rightarrow&\ce{CuS}\\
63.5~\rm{g}&&32~\rm{g}&&95.5~\rm{g}\\
\end{array}$

Conversely when those $63.5~\rm{g}$ of copper combines with oxygen produces cupric oxide.

$\begin{array}{lllll}
2\ce{Cu}&+&\ce{O_2}&\rightarrow&2\ce{CuO}\\
63.5~\rm{g}&&16~\rm{g}&&79.5~\rm{g}\\
\end{array}$

Therefore the proportion between the mass of sulphur and oxygen masses is:

$\frac{m_{\ce{S}}}{m_{\ce{O_{2}}}}=\frac{32}{16}=2$

When sulphur and oxygen combine together they make sulphur dioxide.

$\begin{array}{lllll}
\ce{S}&+&\ce{O_2}&\rightarrow&\ce{SO_{2}}\\
32~\rm{g}&&32~\rm{g}&&64~\rm{g}\\
\end{array}$

Then the proportion between these two becomes:

$\frac{m_{\ce S}}{m_{\ce{O_{2}}}}=\frac{32}{32}=1$

The latter is in proportion which is half of the first ratio. Hence follows the Richter law.

But how can I use this information to understand what it was mentioned in the paragraph from above? How is exactly related with equivalent weights?

The second part of the question arises from the fact, that would it be okay to state this?

In an hypothetical reaction between $A$, $B$ producing $C$ and $D$.

$\begin{array}{ccccccc}
aA&+&bB&\rightarrow&cC&+&dD\\
\textrm{1 eq gram of A}&&\textrm{1 eq gram of B}&&\textrm{1 eq gram of C}&&\textrm{1 eq gram of D}\\
\end{array}$

Becoming in

$\textrm{1 eq gram of A}=\textrm{1 eq gram of B}=\textrm{1 eq gram of C}=\textrm{1 eq gram of D}$

Would it be accurate to put it in that way?

or?

$\textrm{number of eq gram of A}=\textrm{number of eq gram of B}=\textrm{number of eq gram of C}=\textrm{number eq gram of D}$

where:

$\textrm{1 eq gram A} =\frac{\textrm{grams of A compound equal to  the equivalent weight of A}}{\frac{\textrm{formula weight of A}}{\textrm{number of electrons transfered}}}$

which is different when considering moles:

$\begin{array}{ccccccc}
aA&+&bB&\rightarrow&cC&+&dD\\
\textrm{a moles of A}&&\textrm{b moles of B}&&\textrm{c moles of C}&&\textrm{d moles of D}\\
\end{array}$

where:

$\textrm{a moles of A}=\textrm{b moles of B}=\textrm{c moles of C}=\textrm{d moles of D}$

So, is what I wrote correct? How can I relate Richter's law of reciprocal proportions to the equivalent weights? How is it related with the paragraph from above? Perhaps does it meant to say that there is a definite amount of reagent which can react with a certain amount of sulphuric acid? Then why isn't it related more with the concept of limiting reagent rather than the equivalent weight? Can someone guide me on that?

I am afraid you are mixing all modern and old concepts. Moles and the concept of limiting reagents did not exist in Richter's time.

  It took 615 parts by weight of magnesia (MgO), for example, to neutralize 1000 parts by weight of sulfuric acid

This relatively famous statement has nothing to do with law of multiple proportions but it illustrates the concept of equivalents. All this observation shows is that it always took the same amount of X to react with Y. Hence the ratio of X/Y remained constant.

If we take 1000 parts of sulfuric acid as a standard, then 615 parts by weight of MgO will neutralize it. It will take 672 parts of ammonia to neutralize the same amount of acid.

Note carefully this concept does not even need the knowledge of atomic weights.

Addendum
Also note that Richter's values are incorrect as shown in the comments i.e. if we use moles. Most likely the reason was purity of his magnesia and sulfuric acid. The inaccuracy of his values was realized in the late 1860s but we should respect his idea of constant ratio of reactants. I don't know enough German to check his entire three volumes of books nor have access to them. However a table from 1860s shows the comparison of corrected values (which are still slightly off):

An Introduction to Chemical Philosophy, according to the Modern Theories, by Adolphe Wurtz, 1865.

"It will be noticed that for many bodies Fischer's numbers are widely different from the theoretical figures, and consequently the analyses of Richter from which
they are calculated were not nearly so accurate as those of Wenzel. The inaccuracy of these analyses, and the obscurity of a perplexed explanation, have not been much noticed in giving Richter credit and some authority to
his works."


Which process is used to separate Hydrogen and Oxygen from Water? What's the process called where you can use electricity to separate the oxygen and hydrogen from  water? Is it electrolysis? Can it be applied to other liquids?

Problem in your question: should be be O and H, because $\ce{H2O}$ (water) is made of H (Hydrogen) and O (Oxygen). $\ce{CO2}$ is carbon dioxide, for reference.

And yes, electrolysis can be performed on many compounds, not just water.

An electrolytic reaction is where an electric current is used to drive an otherwise non-spontaneous chemical reaction. Splitting water is just one of such reactions.

You can read more here.


Acidity of cyclopentadiene when fused to benzene rings Why does fusion to a benzene ring reduce the acidity of a cyclopentadiene ring?

For example, acidity decreases going from cyclopenta-1,3-diene to indene and to fluorene:

Clar's rule states that (Solà, 2013):

  The Kekulé resonance structure with the largest number of disjoint aromatic π-sextets, i.e., benzene-like moieties, is the most important for characterization of properties of polycyclic aromatic hydrocarbons (PAHs).

Applying Clar's rule to the the question, we would realise that in the indenyl and fluorenyl anions, the resonance structures where there are 6 π-electrons delocalised exclusively within the fused benzene rings would contribute more to the actual structure than the resonance structures where the π-electrons are delocalised along the 5-membered ring. In the major resonance structure for the fluorenyl anion, we would have the negative charge concentrated on the carbon atom, without delocalisation.

In the cyclopentadienyl anion, the negative charge is completely delocalised, as shown below in the resonance structures. Thus, it is the most stable among them. In the indenyl and fluorenyl anions, the negative charge is not as delocalised and thus, these are less stable.

Due to this, the π-electrons in the 5-membered ring experiences the most delocalisation and achieves most aromatic stabilisation in the cyclopentadienyl anion, and this gradually decreases going to the indenyl and fluorenyl anions.

Reference

Solà, M. (2013). Forty years of Clar's aromatic π-sextet rule. Frontiers in Chemistry, 1(22). doi:10.3389/fchem.2013.00022


Deprotonation of 2- and 3-methylquinoline Why are there many examples of 2-methylquinolines being treated with a base then an electrophile but none of the corresponding reaction with 3-methylquinolines? Does this have to do with the charge not being stabilised for the 3-methylquinoline?

Yes, in contrast to 3-methylquinoline, the methyl group of 2-methylquinoline (1) is easily deprotonated by potassium ethoxide.

The anion has been examined by NMR spectroscopy (DOI), but its use in synthesis was reported long before. There's an article (in German) in Chem. Ber., 1909, 42, 1140 (DOI) where the reaction of 2-methylquinoline with diethyloxalate is described.


Is there any law for pH conservation in chemical reactions? I would like to know (just curiosity) if there is any law for pH conservation in chemical reactions. (like conservation of momentum in dynamics).

EDIT

The reaction I would like to take as example is fuel combustion. Too many sources (mostly media) stating that acid products (NOx and COx dissolved in water) are obtained.

My idea is that acids cannot be created from nothing. Either the inputs of the reaction were acid or the output is not only acid but also basic outputs.
Thus the question whether this reasoning is true.

No, there is no law requiring pH conservation. And such a law would be chemical nonsense. Why? Read on.

You need to recall what pH actually is. It is defined as the negative base-10 logarithm of the activity (similar enough to concentration for almost all practical purposes) of positively charged hydrogen ions in solution.† Or in equations:

$$\mathrm{pH} = -\lg [\ce{H+}]$$

Note that most conservation laws in physics somehow relate back to the conservation of energy or something related. However, the pH of a solution cannot be easily traced back to any state function or anything else that would warrant its conservation.

In fact, the only thing conserved is the atom itself. Hydrogen, if $\ce{H+}$ is somehow present in a reaction or solution, does not disappear, it needs to be put in on one side of the equation and turn up somewhere else. In reactions that liberate protons of any kind, usually bases are added to capture them. In those that require protons, acids are added. What (if anything) is conserved is the number of hydrogen atoms but if they are not active as $\ce{H+}$ in solution they do not count towards pH. Therefore, there is no sense in postulating pH conservation.

†: $\ce{H+}$ as is does not actually exist in solution. One can treat it as $\ce{H3O+}$ for all intents and purposes, but actually it is better to think of it as a proton being shuffled back and forth by multiple water molecules. I think I recall four being stated as the lowest count in undergrad education.


Find the number of copper spheres in a box A box contains a mixture of small copper spheres and small lead spheres. The total volume of both metals is measured by the displacement of water to be $\pu{420 cm^3}$ and the total mass is $\pu{5.4 kg}$. What percentage of the spheres are copper?

I can't get the right answer. I did this:

Given:

$m_{\text{total}} = 5.4\;\mathrm{kg} =5400\;\mathrm{g}$

$V=420\;\mathrm{cm^3}$

$\rho_{\ce{Cu}} = 8.96\;\mathrm{g} ~\mathrm{cm^-3}$

$V\times{\mathrm{\rho_{\ce{Cu}}}} = 3763.2\;\mathrm{g}$

$\frac{m_{\ce{Cu}}}{m_{\text{total}}}=69.9\%$

Can someone tell me how to approach this problem?

Here is your formulation :

$m_{\ce{Cu}} + m_{\ce{Pb}} = m_{\text{tot}}$

$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}V_{\ce{Pb}} = m_{\text{tot}}$

$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}(V_{\text{tot}}-V_{\ce{Cu}}) = m_{\text{tot}}$

$\rho_{\ce{Cu}}V_{\ce{Cu}} + \rho_{\ce{Pb}}(420-V_{\ce{Cu}}) = 5400$

$V_{\ce{Cu}}(\rho_{\ce{Cu}}-\rho_{\ce{Pb}}) + 420\rho_{\ce{Pb}}= 5400$

Put values for $\rho_{\mathrm{Cu}}$ and $\rho_{\mathrm{Pb}}$ in the equation, and rest is simple arithmetic.


Chemical Equillibria Comparisons In which of the following reactions is the equilibrium farthest to the left?

a) $\ce{P4O10 + $x$ H2O <=> 4 H3PO4(aq)}$

b) $\ce{AlCl3 + $x$ H2O <=> [Al(H2O)6]^3+ + 3 Cl- (aq)}$

c) $\ce{Li2O + $x$ H2O <=> 2 Li+(aq) + 2 OH- (aq)}$

d) $\ce{H2S + H2O <=> H3O+ + HS- (aq)}$

e) $\ce{CaC2 + $x$ H2O <=> Ca^2+(aq) + 2 OH- (aq) + C2H2(g)}$

The correct answer is d, but I'm not sure how to compare these equilibria.  I know that the answer can't be e, since the evolution and purging of gas will drive the equilibrium forward.  Also, the dissociation of ions in water is favorable due to solvation, so that rules out c and b.

But I'm not sure how to compare the remaining two reactions.

Any help would be greatly appreciated.

You have already ruled out the equations b, c, and e.

  But I'm not sure how to compare the remaining two reactions.

Have a look at equation a again.

The reaction of $\ce{P4O10}$ with water generates phosphoric acid ($\ce{H3PO4}$), which is a fairly strong three-protonic acid ($\mathrm{pK_{a1} = 2.16; pK_{a2} = 7.21}$).

As a result, subsequent deprotonation of phosphoric acid
$$\ce{H3PO4 + H2O <=> H3O+ + H2PO4-}$$

will pull the equilibrium in equation a to the right.

Hydrogen sulfide ($\ce{H2S}$), on the other hand, is a rather weak acid with $\mathrm{pK_{a} = 7.0}$. Consequently, you will find a significant amount of undissociated hydrogen sulfide in aqueous solution. Therefore, the correct answer in the comparison is indeed d.


Organic Chemistry - Reaction scheme and structures of molecules Compound P and compound Q are straight-chain constitutional (structural) isomers with the
molecular formula C5H12O. Compound P can form optical isomers, whereas compound Q
cannot.
When reacted with concentrated sulfuric acid, compound P forms two products, compounds
R and S; compound Q forms only one product, compound S.

How come compound Q is pentan-1-ol and not pentan-3-ol?

Q is pentan-1-ol and not pentan-3-ol as on dehydration, of pental-1-ol, only one product (alkene) is formed. Whereas on dehydration of pentan-3-ol, we get two alkenes, as the alkene can show geometrical isomerism.


Stability constant of complexes I know that this constant tells us the strength of the interaction between the metal and the ligand,  so should it also mean that the higher the constant, the stable the complex ?
I need to make sure of it because in photometric titration, all that matters is the stability of complex, and if the constant is given, one should know which complex is formed earlier.

As $K$ is a thermodynamic constant, and if you define your reaction as

$A + B \rightarrow{} C + D$

leading to

$K = \frac{c_C \cdot{} c_D}{c_A \cdot{}c_B}$

then -- thermodynamically -- yes.  Especially for chelate complexes (like with EDTA), entropically favoured. This answer excludes (macroscopic) kinetics


Flammability (NFPA) - how is it defined? I was reading up on a wonderful little chemical compound known as chlorine trifluoride ($\ce{ClF3}$). For a primer, check out Dr. Derek Lowe's blog post here: Sand Won't Save You This Time. The title of the post is quite telling; this compound is so reactive with other compounds - any other compound, really - that it will set sand on fire, making this old standby for extinguishing lab fires useless and putting the chemical in very rare company indeed. Obviously, water is a huge no-no for chlorine trifluoride; it will exothermically react with the water to produce clouds of steam containing two nasty acids (hydrochloric and hydrofluoric) and a host of chlorate and fluoride compounds. Somewhere in the middle of all the heat, fluorine and organic-ish compounds, you could probably even get some fluorine oxides, which would just LOVE to rapidly reduce their oxidation state (an event normally accompanied by shrapnel).

I was intrigued; I went and looked up the MSDS for this chemical to see what the instructions were for dealing with industrial spills of this stuff (apparently, there's a decent market for it in the semiconductor industry, and probably also for plastics to produce highly fluorinated polymers like Teflon). "A good pair of running shoes" can't be the only answer to a $\ce{ClF3}$ fire. (well, actually, it pretty much is; get everyone in a one mile radius and two miles downwind the hell away from the area)

Now, here's the question. While this stuff has some serious "synergy" when it comes to being an oxidizing agent (it's a better oxidizer than pure oxygen, loads better than pure chlorine, better even than pure fluorine gas), the NFPA flammability rating of the chemical is 0. This, despite reports from those who originally studied it that the chemical is hypergolic with every known fuel substance, and most things you wouldn't consider fuel (like asbestos, sand, concrete, brick, earth, test engineers). The resulting reaction certainly looks like fire; intense heat, bright light, sparks, smoke, the works. So, if chlorine trifluoride doesn't "burn" by the NFPA's definition, what is the NFPA's definition of "burning"?

You have found one of biggest flaws in the  NFPA "Fire Diamond". The NFPA rates the flammability of the compound as a fuel. ClF3 is not a fuel. It is an oxidant. ClF3 is playing the role that oxygen normally does. It causes fuels (reductants) to burn by oxidizing them rapidly and exothermically. Oxidants are incapable of creating fire in absence of fuel regardless of temperature. Fires started by ClF3 continue to burn after being smothered because oxygen isn't necessary to continue the oxidation process. And since ClF3 is denser than air, it gets trapped under the sand. Many other powerful oxidants, like fluorine and perchloric acid, have low flammability ratings. Oxygen also has a flammability rating of zero. They do not burn; they cause other compounds to burn.

The scale for the flammability rating has to do with flash point, which is the temperature at which a fuel forms an ignitable mixture with air. Since ClF3 doesn't burn (isn't a fuel), and doesn't react with air (N2, O2, CO2, Ar, etc, which are all inert towards oxidation), it does not have a measurable flash point.

The danger posed by ClF3 is summed in the yellow (reactivity or stability) diamond and the white (physical and chemical hazards) diamond. Your msds rates ClF3 as a 3 in reactivity/stability, which is appropriate. Wikipedia's NPFA rating article, even mentions chlorine trifluoride:

  3 -  Capable of detonation or explosive decomposition but requires a strong initiating source, must be heated under confinement before initiation, reacts explosively with water, or will detonate if severely shocked (e.g. ammonium nitrate, chlorine trifluoride).

The white triangle will have appropriate hazard information. For ClF3, it has the symbols OXY, which means it is a strong oxidizer, and W, which means it reacts with water.

I have not found the specific reaction that ClF3 undergoes with sand (mostly SiO2). I expect that it involves ${\ce{SiO2 -> SiF}}_6^{2-}$, except that this half reaction is not an oxidation.


Difference between single point energy and dimerization energy I am investigating potential energy surface characteristics of some electrostatic monomeric interactions e.g. $\ce{F-H}$...$\ce{F-H}$.

I'm using DFT and SCS-MP2 to perform the necessary calculations for single point energies. Is it possible to compare single point energy with the electrostatic dimerization energy of the monomers?

Concepts or references will be highly appreciated.

Definitions

Rather than use the phrase "dimerization energy", which is the energy difference specific to two chemically identical species coming together, such as $\ce{2 BH3 -> B2H6}$, it is better to use the phrase interaction energy. See here and here for a detailed distinction between interaction energies and binding energies. I will simplify a bit some of the text there and in this answer.

The single point energy (SP energy, SPE) is the total energy of a system at one and only one configuration of the nuclear positions. In the case of two monomers (sometimes called fragments), the interaction energy will be the energy difference between the complex (sometimes called the supersystem) and the sum of the individual fragments or monomers at the same nuclear positions as in the complex. This gives rise to the equation

$$
\Delta E_{\text{int}}(\text{AB}) = E(\text{AB}) - \left[ E(\text{A}) + E(\text{B}) \right]
$$

where there are two molecules, A and B, leading to three calculations, one for each of the monomers and one for the AB complex. In the dimerization of borane from above, or your specific example of the $\ce{HF}$ dimer, assuming that the internal coordinates (bond lengths, bond angles) of each monomer are identical, then the equation above can be simplified as A and B are identical:

$$
\Delta E_{\text{int}} = E_{\text{dimer}} - 2 E_{\text{monomer}}
$$

To take your specific example of the $\ce{HF}$ dimer, you mention the role of the "orientation of atoms" in both the single point and dimerization energies. For a single monomer, assuming again that internal coordinates are not changing, the relative positions of the nuclei are not changing, and the total energy is invariant to translations and rotations. However, in the dimer or complex geometry, the relative orientation of one monomer to the other will change the total energy of the dimer, leading to a change in the interaction energy. For example, in addition to the linear form $\ce{F-H}\cdots\ce{F-H}$, you might be interested in the "flipped" versions $\ce{F-H}\cdots\ce{H-F}$ or $\ce{H-F}\cdots\ce{F-H}$, or forming a T-shape, etc. The interaction energy of each will tell you about the relative stability of that nuclear configuration relative to the other configurations.

The most important variables then in an interaction energy calculation are the relative orientation of the interaction species and the distance between each of the fragments. How do you decide this? Usually some intermolecular coordinate is scanned, such as the distance between closest atoms on the monomers, still while leaving the monomer internal coordinates fixed to give an unrelaxed potential energy surface scan.

Calculation Examples

For example, here is the interaction energy for the $\ce{F-H}\cdots\ce{F-H}$ configuration with a variety of methods using the cc-pVTZ basis set and an intraatomic distance of 0.927 angstroms, scanning the $\ce{H}\cdots\ce{F}$ distance:

From a plot like this, we see that MP2 overestimates the interaction and CCSD underestimates it compared to CCSD(T); the shape of the B97-D potential is probably incorrect. HF is missing some repulsive interaction at long-range that the other methods include. It isn't dispersion, as the total dispersion interaction is always attractive and the minimum of the potential is qualitatively correct. You could certainly use any density functional or other method of your choice here (SCS-MP2, SOS-MP2, CISD, ...).

You could similar plots for other configurations, or choose a single method and place multiple configurations on the same plot (here, CCSD(T)/cc-pVTZ):

From a plot like this, it's pretty clear that only the $\ce{F-H}\cdots\ce{F-H}$ configuration is ever bound; it would be interesting to compare the T-shaped configurations to the linear ones.

Interaction Energy Decomposition

You mention "electrostatic monomeric interactions", and we also see some interesting long-range behavior in plot 1. One of the most powerful techniques for understanding intermolecular interactions is symmetry-adapted perturbation theory (SAPT), which gives a decomposition of interaction energies in terms of electrostatics, Pauli exchange, polarization (called induction), dispersion, and charge transfer (though the definition of the latter is often ambiguous).

Here are each of the interaction energy components (SAPT0/jun-cc-pVTZ, monomer-centered basis):

Combining all the electrostatic (1), exchange (1), induction(3), and dispersion (2) terms gives

This makes it pretty clear that all points, $\ce{F-H}\cdots\ce{F-H}$ is bound mostly by electrostatic attraction, and dispersion is negligible. Maybe you knew that, but I didn't. My guess is that the presence of dynamic correlation in all the wavefunction methods damps the long-range electrostatics compared to Hartree-Fock.

References

Interaction energies

J. Řezáč and P. Hobza, Chem. Rev., 2016, 116 (9), pp 5038–5071

SAPT

B. Jeziorski, R. Moszynski, and K. Szalewicz, Chem. Rev. 94, 1887 (1994)
E. G. Hohenstein and C. D. Sherrill, J. Chem. Phys. 132, 184111 (2010)
E. G. Hohenstein, R. M. Parrish, C. D. Sherrill, J. M. Turney, and H. F. Schaefer, J. Chem. Phys. 135, 174107 (2011)
T. M. Parker, L. A. Burns, R. M. Parrish, A. G. Ryno, and C. D. Sherrill, J. Chem. Phys. 140, 094106 (2014)

Technical notes

All of these calculations were run with Psi4. None of the interaction or SAPT energies are CP-corrected.


Is the absolute temperature concept 100% viable? Maybe I'm wrong about it but I wanted to know whether the concept of absolute temperature is real or not.

As every element behaves differently at a particular temperatures how could we assign a common temperature to all these elements where they possess zero point energy? If they will do so then what will happen if we find an new element which ground state energy is at $\mathrm{-10\ K}$ theoretically?

The concept of absolute temperature is real. Temperature can be thought of as thee measure of the concentration of heat energy in a substance and isn't effected by different elements or compounds. Absolute temperature which is −273.15 degrees Celsius (0 kelvin) is the temperature where there is no heat energy in the substance at all. That means that all the molecules are existing in their lowest possible energy state. So conventionally is thought that it is impossible to get lower than absolute zero as to reduce the temperature, you have to remove heat energy but you can't remove heat energy in a substance which doesn't have any heat energy in the first place.

However theoretically, the Boltzmann equation does allow for substance to go minus Kelvin and in fact it actually lies above infinity Kelvin. Recently there have been several claims that this has been achieved. If you search this up on Google I am sure that you will be able to find more information on this.


Iron/Iron(III) Concentration Cell From what I understand about concentration cells - which I never learned about when I took general chemistry - they rely on the fact that if we have differing concentrations of some solute separated by a permeable membrane, they will try to exchange solute and/or solvent to achieve the same concentration or as close to it as possible. Correct?

Now, in a concentration cell, there is no physical exchange of ions. Instead, a salt bridge facilitates the flow of electrons, which either:

a) helps convert solids into ions

or

b) helps convert ions into solids

So in the above problem, we obviously have unequal concentrations of iron(III) ions. We can describe the problem using cell notation:

${|Fe|Fe^{3+} (4.8*10^-7 M)||Fe^3+ (1.3*10^-3 M)|Fe|}$

The flow of electrons will be from half-cell A to half-cell B. These are the reactions:

Half-cell A: $\ce{Fe -> Fe^3+ + 3e^-}$

Half-cell B: $\ce{Fe^3+ + 3e^- ->Fe}$

This increases the $\ce{Fe^3+}$ concentration in half-cell A while decreasing the  $\ce{Fe^3+}$ concentration in half-cell B. This will continue until both half-cells have the same $\ce{[Fe^3+]}$.

So, we can also describe half-cell A as the anode, because it is where oxidation occurs, and half-cell B as the cathode, as it is where reduction occurs.

So far so good?

And if we want to calculate the electrochemical potential of this particular half-cell (wired up to make a spontaneous reaction), we can apply the Nernst equation as follows:

This all looks correct.  It's worth noting that the voltage generated by this cell occurs through the flow of electrons from half cell A to half cell B as the system reaches equilibrium.  It's also worth noting that the final concentrations of ions should eventually become nearly identical given enough time and free interaction between the two half cells (at which point the voltage becomes 0).

For further reading, this site has some good examples:
http://chemwiki.ucdavis.edu/Core/Analytical_Chemistry/Electrochemistry/Voltaic_Cells/Electrochemical_Cells_under_Nonstandard_Conditions/Concentration_Cell

While wikipedia has a (brief) overview of the theory:
https://en.wikipedia.org/wiki/Concentration_cell


Chlorination of Water using sodium hypochlorite $\ce{NaClO_{(s)} + H2O <=> NaOH_{(aq)} + HClO_{(aq)}}$

This reaction is often used in swimming pools as an alternative to directly chlorinating the water.

I read that the water must be kept slightly acidic. Otherwise, $\ce{OH-}$ ions in the alkaline solution would react with the $\ce{H+}$ ions in the chloric (I) acid to form water. Consequently, $\ce{ClO-}$ ions are removed and the concentration of $\ce{H+}$ ions decreases.

So far this all seems fairly reasonable.

However, my textbook then said this:

  In alkaline solution, the equilibrium moves to the left.

This contradicts what I have read online and elsewhere. Surely if the concentration of hydrogen ions decreases, the equilibrium would shift to the right to increase the concentration of the products and thus the hydrogen ions?

Thanks in advance.

The reaction as shown goes only a little to the right. The $\ce{HClO}$ shown is poorly ionized; on the other hand, the $\ce{NaOH}$ is completely ionized. How far the reaction goes to the right will be indicated by the pH, which is probably about 10, considering that sodium carbonate is added to commercial bleach to raise the pH to 11 for greater shelf life.

When chlorinating a swimming pool, long shelf life (in the pool) is not desired. You want some more activity, even if at the expense of having some of the $\ce{NaClO}$ divert to other products.

There are two variables in the swimming pool:

the chlorination amount
the activity, which will be controlled by the pH.

So you put in the chlorine, then dial up the best pH. As far as moving the equilibrium, perhaps it is better to show the reaction as,
$$\ce{ClO- + H2O -> HClO + HO-}$$

The original sodium hypochlorite will be totally ionized; the hydrolysis reaction is the $\ce{ClO-}$ ion attracting a proton from water to form somewhat nonionized HClO and some $\ce{OH-}$ ions. Then the prospect of adding more $\ce{OH-}$ ions is easily seen to move the equilibrium back toward the left, generating more $\ce{ClO-}$ ions. Likewise, adding acid generates more $\ce{HClO}$ in the solution.


How to know the ionization power of an unknown metal? - difficult Imagine having the 4 following beakers:

where P, Q, R and S are metallic plates. In beaker 1 and beaker 2 nothing happens. However in beaker 3 and 4 the metalic pieces become thinner.

How would you orden the ions $P^{p+}, Q^{q+}, s^{s+}, R^{r+}$ the following ions by oxidizing power?

I think it is one of the following possibilities:

$P^{p+} < Q^{q+} < s^{s+} < R^{r+}$

I think that P and Q clearly have the highest oxidizing power as in beaker 3 and 4 the metallic plates become thinner. But I have no idea whether R has a higher oxidizing power than S. Or is it the other way around?

Could someone help me and explain how I can order these ions by oxidizing power?

I am guessing you know when a metal bar is becoming thinner, the metal is being oxidised, i.e $$ \ce{M_{(s)}} \rightarrow \ce{M^{n+}_{(aq)}}$$

$P^{p+}, Q^{q+}, s^{s+}, R^{r+}$
Let's start with beaker 3 and beaker 4

In beaker 3, you have $\ce{S_{(s)}}$ in $\ce{R^{r+}_{(aq)}}$, and you observer the metal bar becomes thinner. Which means, $\ce{S_{(s)}}$ is being oxidised by $\ce{R^{r+}_{(aq)}}$, or in other words, the reduction potential of $\ce{R^{r+}_{(aq)}}$ is greater than the reduction potential (oxidising power) of $\ce{S^{s+}_{(aq)}}$

$$\ce{R^{r+}_{(aq)}} > \ce{S^{s+}_{(aq)}}$$

In beaker 4, you have $\ce{Q_{(s)}}$ in $\ce{S^{s+}_{(aq)}}$, and again the same reasoning applies.

$\ce{Q_{(s)}}$ is being oxidised by $\ce{S^{s+}_{(aq)}}$, or in other words, the reduction potential of $\ce{S^{s+}_{(aq)}}$ is greater than the reduction potential (oxidising power) of $\ce{Q^{q+}_{(aq)}}$

$$\ce{S^{s+}_{(aq)}} > \ce{Q^{q+}_{(aq)}}$$

Thus, so far we have:

$$\ce{R^{r+}_{(aq)}} > \ce{S^{s+}_{(aq)}} > \ce{Q^{q+}_{(aq)}}$$

Let's look at beaker 1 and 2

In beaker 1 you have $\ce{P_{(s)}}$ in $\ce{Q^{q+}_{(aq)}}$, and oxidation doesn't take place. Which means the reduction potential of of $\ce{Q^{q+}_{(aq)}}$ is less than that of $\ce{P^{p+}_{(aq)}}$

$$\ce{P^{p+}_{(aq)}} > \ce{Q^{q+}_{(aq)}}$$

And similarly, in beaker 2 you have $\ce{P_{(s)}}$ in $\ce{R^{r+}_{(aq)}}$, and nothing happens. Which means that the Which means the reduction potential of of $\ce{R^{r+}_{(aq)}}$ is less than that of $\ce{P^{p+}_{(aq)}}$

$$\ce{P^{p+}_{(aq)}} > \ce{R^{r+}_{(aq)}}$$

So the final order becomes (in terms of reduction potential, or oxidising ability):

$$ \ce{P^{p+}_{(aq)}} >\ce{R^{r+}_{(aq)}} > \ce{S^{s+}_{(aq)}} > \ce{Q^{q+}_{(aq)}}$$


Reactivity of Ammonium I was going through some displacement reactions when this struck me:

What is the position of the ammonium cation in the metal reactivity series?

I can find other positive ions like carbon and hydrogen in my textbook but not ammonium. I tried to look it up in google but the search results were not up to the mark.
I could get some comparison at this site. From there, it seems that ammonium will be above zinc, lead and iron.
This post also claims its position but does not refer it Do copper and ammonium hydroxide react?.
I know 'ammonium in the metal reactivity series' sounds a bit strange but what I basically want is its position among the metals and the positive ions. Can anyone here help me with this?

Let's compare ammonium cation NH4+ and the neutral molecule, NH4 with other metals.

As Ed V pointed out, with a mercury cathode and low concentration of H+, NH4+ can be reduced to NH4 electrolytically (also with sodium amalgam), which forms an amalgam that is stable for a short period of time: the mercury bubbles up, then collapses, evolving NH3 and H2. Ammonium amalgam was discovered in 1808 by Humphry Davy and Jöns Jakob Berzelius.
https://en.wikipedia.org/wiki/Amalgam_(chemistry)

A recent (2016) paper http://scripts.iucr.org/cgi-bin/paper?xk5030 gives the crystal ionic radius of ammonium as falling between 140 and 154 pm, depending on the anion; this compares nicely with the atomic radius of mercury which is 150 - 155 pm, so an alloy is not inconceivable.

Metallic ammonium (NH4) has been considered for some off-world chemistry in a 1954 paper. The phase transition of mixed crystals of hydrogen and ammonia to metallic ammonium was found (by calculation) to take place at pressures almost certainly less than 250 kilobars, suggesting that Neptune and Uranus contain large deposits of metallic ammonium.

http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?db_key=AST&bibcode=1954MNRAS.114..172B&letter=0&classic=YES&defaultprint=YES&whole_paper=YES&page=172&epage=172&send=Send+PDF&filetype=.pdf

But on our world, the reaction of a metal like iron with a high concentration of NH4+ cation would produce H2, not NH4. The reaction could be examined on the basis of the pH, or concentration of H+ ions. As it turns out, the potentials are simply related to the concentration of H+ ion, which depends on the anion.

A 0.1M ammonia solution (anion = OH-) has a pH of 11.122 (http://www.chem.science.unideb.hu/Solutionsmonobasic.pdf). A solution of NH4Cl (anion = Cl-) gives a pH of 5.13 (https://socratic.org/questions/calculate-the-ph-of-the-following-aqueous-solutions).

When we displace from a solution of ammonium salts, we wind up displacing hydrogen. You could imagine that adding iron or zinc to a solution of NH4+Cl- could evolve neutral NH4 which immediately decomposes to NH3 plus a hydrogen radical H., which then dimerizes to H2, and the NH3 immediately redissolves in the water. Far better to think of the NH4+ dissociating in the aqueous solution before reacting with the iron or zinc, in other words, the reaction with the metal is a reaction of H+ (or, actually, hydrated H+, or hydronium ion) after dissociation of NH4+ in the presence of H2O.


A Question on the Role of Triplet Oxygen in the Formation of Artemisinin It is observed that whereas singlet oxygen is required for an ene reaction with an alkene to give an allyic hydroperoxide, triplet oxygen will react similarly with an enol to give an alpha-hydroperoxy aldehyde. (Example: see Chen HJ et al., Tetrahedron 69 (2013), 1112-1114.)

The drawing below shows two steps in the final sequence of reactions leading to the antimalaria drug artemisinin, as described in the cited article. The enol is formed by opening of the hemiacetal, and before it can tautomerize to the aldehyde, it reacts with atmospheric oxygen (in the dark, so presumably triplet oxygen) to give the alpha-hydroperoxy aldehyde shown. That in turn spontaneously folds up to give artemisinin (not shown). A concerted reaction of the enol with triplet oxygen, if it is a variant of the ene reaction, would be spin-forbidden.

Can anyone explain why this otherwise "forbidden" reaction works?

I offer the following thoughts on the paper you cite regarding the formation of artemisinin 9 from carboxylic acid 1. You are correct that the allylic hydroperoxide 2 is formed by an ene reaction with singlet oxygen (1O2). The stereochemistry of the hydroperoxide group is the result of the ene reaction occurring on the convex face of the unsaturated cis-decalin 1. The crude hydroperoxide 2 was exposed to an atmosphere of triplet oxygen (3O2) in a vigorously stirred mixture of petroleum ether and catalytic trifluoroacetic acid. The initial steps are 2 --> 6 via radicals 3, 4 and 5. Triplet oxygen forms hydroperoxide radical 3 which undergoes cyclization to 1,2-dioxetanyl radical 4. It is likely that radical 4 reacts with triplet oxygen prior to dioxetane fragmentation owing to the stereochemistry of the hydroperoxy radical in 5. The chain reaction is completed by the reaction of radical 5 with hydroperoxide 2. In the transformation 2 --> 3 the radical HOO was formed and can also intercede as a source of a hydrogen atom regenerating oxygen. Finally, acid-catalyzed cyclization of 6 leads to artemisinin (qinghaosu) 9. There is no issue of "forbidden" reactions.