Change in spin angular momentum during transition in one-electron atoms I have a really basic qustion about spin and electronic transitions.

In one-electron systems (like H or $\ce{He+}$), will the electron change its spin during an electronic transition? Let's say the electron de-excites from 5p UP to 3s, will the spin be UP or DOWN in the 3s orbital?

From crystal-field theory I remember that a d-d transition where the electron had to switch to opposite spin was spin-forbidden (although it may still happen due to certain vibrational states). Is this still true for one-electron systems?

In general, transitions in which the "direction" of spin changes are forbidden. More formally, the spin conservation rule (also referred to as the Wigner rule) states that

  For both radiative and radiationless transitions,
  transitions between terms of the same multiplicity are spin-allowed,
  while transitions between terms of different multiplicity are
  spin-forbidden.

But spin-orbit coupling relaxes this rule, so in general, lines corresponding to spin-"forbidden" transition are just much weaker and not totally absent from the spectrum. And the bigger the atomic number, the stronger is the spin-orbit coupling, and consequently, the bigger are the intensities of spin-"forbidden" transitions.

$
\newcommand{\linop}[1]{\hat{#1}}
\newcommand{\ket}[1]{\left|\,{#1}\right\rangle}
\newcommand{\abs}[1]{\left|\,{#1}\right|}
$
With respect to your question on addition of spin angular momentum you raised in comment section: you formulated it incorrectly which confused me, so my comments (now deleted) were wrong. Thus, I decided to write a short answer here.

So, for angular momentum $J$ the simultaneous eigenstates of $\linop{J}^{2}$ and $\linop{J}_{z}$ operators are given as follows,
$$
\begin{align}
        \linop{J}^{2} \ket{j(j+1),m} &= j(j+1) \hbar^{2} \ket{j(j+1),m} \, , \\
        \linop{J}_{z} \ket{j(j+1),m} &= m \hbar \ket{j(j+1),m} \, ,
\end{align}
$$
where $j$ could take non-negative integer and half-integer values,
$$
    j = 0, 1/2, 1, 3/2, \cdots \, ,
$$
and $m$ ranges from $-j$ to $j$ in integer steps
$$
    m = -j, -j+1, \cdots, j-1, j \cdots \, .
$$

When there are two sources of angular momentum (like two electrons with their spins) we first we have to make a choice when specifying the state of the system: either we use the uncoupled picture $\ket{j_{1}, m_{1}, j_{2}, m_{2}}$, which leaves the total angular momentum unspecified, or we use the coupled picture $\ket{j_{1}, j_{2}, j, m}$, which leaves the individual components unspecified.

The very language OP uses (i.e. talking about multiplicities of states) suggest that we choose the second picture for further discussion. In the coupled picture the question arises as to the permissible values of total angular momentum quantum numbers $j$ and $m$. It is known from quantum theory of angular momentum that the allowed values of $m$ follow immediately from the relation $m = m_{1} + m_{2}$, while the situation with the allowed values of $j$ is more complicated: its permited values are given by
$$
    j = j_{1} + j_{2}, j_{1} + j_{2} - 1, \dotsc, \abs{j_{1} - j_{2}} \, .
$$

All this is true for any kind of angular momentum (orbital or spin) with the only one exception: with respect to half-integer values of $j$ we note that for orbital angular momentum, where the Born interpretation requires cyclic boundary conditions to be satisfied, only integer values are admissible.

Now for each and every electron we have a source of spin angular momentum characterised by quantum numbers $j = 1/2$ and $m = \pm 1/2$. Thus, in accordance with what is written above for a system of two electrons we have 4 possible state which in the coupled picture are written as follows: $\ket{1/2, 1/2, 1, -1}$, $\ket{1/2, 1/2, 1, 0}$, $\ket{1/2, 1/2, 1, 1}$ and $\ket{1/2, 1/2, 0, 0}$. The first three states are three components of a triplet, the last one is the only component of a singlet.

Note, however, that projection quantum numbers for individual components ($m_1$ and $m_2$) are not specified in the coupled picture, so while it might be tempting to think that for $\ket{1/2, 1/2, 1, -1}$ state they are $m_1 = m_2 = -1/2$ this is wrong. And that what was essentially wrong with the question raised by OP in the comments section: he used the coupled picture terminology ("multiplicities") and at the same time specified the values of projection quantum numbers for individual components ("two electrons both have spin of -1/2"). This could not be dome simultaneously.


What exactly is petuntse? (mineral) What exactly is petuntse, the "china stone" used as one of two ingredients in porcelain along with kaolin?

The Wikipedia article (and books I have read) describe petuntse as a mineral containing feldspar. Well, fine, but there are a lot of minerals that are certainly not petuntse that contain feldspar. Feldspar is one of the most common minerals, so saying it is a material with feldspar in it, is not specific.

What is the specific composition of traditional petuntses used to make porcelain?

According to the Pottery Histories web site:

  Petuntse
  Petuntse is a historical Chinese term for rocks with a high content of
  mica and feldspathic minerals that have been used as one of the
  ingredients in the manufacture of Chinese porcelain. It is generally
  similar in composition and use to China stone. Petuntse, with china
  clay, forms the base ingredients for hard-paste porcelains similar to
  those produced in China in historic times. See China stone.

and

  China stone (Cornish stone, Cornwall stone)
  China stone is a
  feldspar-rich mineral derived from partially decomposed granite. China
  stone includes feldspar, mica, fluorospar, quartz and other
  granite-derived minerals such as kaolinite, but lacks iron-bearing
  minerals.  China Stone is used worldwide as a flux in the manufacture
  of ceramic bodies and glazes. In the United Kingdom, the material is
  mined in one small area of Cornwall.

I don't know if this meets the degree of specificity that you were looking for.


How much carbonation is in one liter of commercial fountain Coca-Cola From what I understand generally Coke is carbonated to roughly 6.2 g per liter in bottles and cans, or 3.1 volumes of CO2.  I was watching this video because I was curious about how fountain soda machines worked.

https://www.youtube.com/watch?v=RRHNt67xbMs

By the end of the video they suggest that the soda is carbonated to around 5 volumes of carbonation.  Which as I understand would be somewhere in the realm of 11 g of CO2 per liter, almost twice that of the normal soda.  I imagine there are a couple other factors involved, like they are dispensing more CO2 than is actually dissolved in the soda as a consequence of trying to instantly carbonate.  According to everything I know about carbonation I know from beer making they are carbonating almost 20 times higher than a forced beer carbonation technique in home brewing.  Can anyone help me better understand this?  I would think that the carbonation level difference between bottled and fountain soda would be something highly controlled by beverage companies for the sake of product consistency.

I did some additional reading and found that the correct maintenance levels on soda fountains with chilled water to roughly 40 degrees F is 75 psi if that helps.  For 3.1 volumes of carbonation as I understand equilibrium would be just like 5 psig.  I know I missing something big.

I can only assume that the rate of carbonation loss is inversely exponential as the carbonation approaches closer to equilibrium for the environmental pressure.  Would 5 volumes rapidly decarbonate to 2-3 volumes but not get much lower very quickly?

If the soda is not distributed, the rate of loss of $\ce{CO2}$ would be an exponential decay. So changing from 5 volumes to 3 volumes is faster than from 3 to 1 volumes.

A can of soda with 3 volumes of $\ce{CO2}$ has about 15 psi of extra pressure inside the can. If it was increased to 25 psi (5 volumes, like the fountain sodas), the can would need to be thicker to ensure it doesn't rupture, which would increase the cost of the can. There is a balance of the amount of $\ce{CO2}$ and the cost for the drink.


Why isn't N2 a Lewis base? Why is $\ce{N2}$ (dinitrogen) not a Lewis base even with two lone pairs?

Actually, it is, but a very weak one.

The lone electron pairs in dinitrogen are in $\mathrm{sp}$ hybridization state, and nitrogen is quite electronegative, so they are not easily donated. Still, if a Lewis acid is strong enough, dinitrogen may form a complex.

It is known that dinitrogen adsorbs on acidic $\ce{OH}$-groups in zeolites, forming complex with IR-active frequency. The nature of the complex may be debated, as no full proton transfer occurs, but the electron pairs are definitely involved.

Diazonium salts may be considered as a lewis-acid complex of a carbocation and dinitrogen molecule. While no direct synthetic route exists from dinitrogen to diazonium salt (because typically a stronger Lewis base is available), they typically decompose with formation of dinitrogen molecule.

Significant amount of dinitrogen complexes with transition elements are known. Typically they are obtained indirectly, by forming dinitrogen ligand in-place, though.

Finally, adsorption of dinitrogen on the surface of ammonia synthesis catalysts starts as a Lewis-interaction. Similarly, nitrogen adsorption on cations in zeolites is known.


Oxygen or carbon with a positive charge? (comparison of canonical structures) Which of the contributing structures of the resonance is more stable?

I'm watching a video lecture by a professor of my college where he puts this question to the class. The class unanimously says B. At first I thought they were wrong, but then the professor agreed with them.

Is this true? The reason he gave was that oxygen's octet would be complete in B, hence it contributes more. But I don't understand this.

1) Isn't the octet complete in A too? It has 2 covalent bonds - 4 electrons (2 from sharing) and 2 lone pairs. Adding up to 8.
2) Oxygen being more electronegative than carbon, would pull the electrons towards itself, leading me to think that it would be more comfortable with a lone pair rather than another bond. Also, won't it prefer to lose out that '+' charge which the bonding's burdened onto it?

Am I wrong?

Another thing which made me believe that Prof. may have made a mistake while talking about the more relevant contributor, is that when he added Br$^-$ to the compound, he added it to the canonical structure A. And don't you explain reactions with the most relevant contributor, i.e. B?

Answer

The presence of an extra bond in the canonical structure B, along with the completion of the carbon valency makes it more stable than A. The reaction with Br$^-$ is best explained looking at A, though, because the incoming nucleophile ends up bound to carbon, not oxygen.
Thanks, everyone!

Which of the contributing structures of the resonance below is more stable?

Technically, neither. Both structures A and B are resonance contributors to the same true structure of the ion. Structures A and B, because they both represent the same species, cannot have different energies, and therefor they cannot have different stabilities.

What the question is asking is "Which contributor is more important in describing the structure and behavior of the hybrid?" Contributor B probably approximates the structure better (i.e. the positive charge is more on the oxygen than on the carbon), but structure A represents the reactivity better (i.e. nucleophiles want to attack the carbon and not the oxygen).


Do we get high on "methampetamine" naturally? First off, I'm new here and I cant tell if this is a chemistry or a biology question. It's both and if I ask on biology forums, they say to ask on chemistry forums and vice versa. So sorry if this question is more geared towards biology.

I am really amazed that our brain produces Phenethylamine any time we do something rewarding. I need to confirm this, because if this is true, I'm high right now... (wait so meth isn't bad?)

Phenethylamine gets biosynthesized in our brain to get converted into dopamine (and also L-DOPA). But the funny thing is that the molecular formula of Phenethylamine is exactly the same as methamphetamine just without the two methyl groups.

From my understanding, if you consume methamphetamine the liver metabolizes methyl groups into hydroxyl preventing it from metabolizing Phenethylamine instantly and making the hydroxyl group a free radical, so the hydroxal group is separated once it reaches the brain, therfore consuming methamphetamine would be 10x more potent then if you consume Phenethylamine. The same way the high from Dextroamphetamine (the chemical used in Adderall which is the same as methamphetamine, just with one methyl group instead of two, making it not as potent) is 5x weaker then methamphetamine.

Consuming pure Phenethylamine would make the liver metabolize 90% of it making the brain receive only around 10% of it. Phenethylamine is naturally occurring and is sold on Amazon as pure powder.

So my question: is Phenethylamine the same as meth, just with 1/10th of a high, considering that is what methamphetamine gets metabolized into?

I didn't mean get high on methamphetamine in the title, but the end product of methamphetamine (apples and oranges)

And does that mean that meth isn't that bad as the media claims it is (besides hepatoxicity)?

All the drugs I know of are closely related to a molecule we produce naturally or are a molecule that gets produced by our body. This is kind of the point of drugs. But the main issue is the dose. We all have uranium in our bodies, does that mean consuming it isn't bad for you? Or to put it differently: We all have bones, don't we? But putting another bone into your body isn't really considered healthy.

First of all: Chemically, those two molecules of course are not the same. Let's get the obvious out of the way.

I think this is the wrong forum again because really what you are asking is more of a legal/ethical question than a scientific one (both questions). But I'll try to answer the 2nd question "scientifically" with a light esoteric touch: Consuming drugs means messing with a delicate balance in our brains that is so complicated we do not even begin to understand it. I'm not saying don't use drugs, I'm just saying maybe you want it to be not as bad as it really is.


Why does NaBH4 reduce double bonds conjugated to carbonyl groups, while LiAlH4 does not? I have been going through reduction of aldehydes using $\ce{LiAlH4}$ and $\ce{NaBH4}$. If there is a double bond conjugated with the carbonyl group, $\ce{LiAlH4}$ doesn't reduce it, leading to an allylic alcohol. However, using $\ce{NaBH4}$, some of the fully reduced alcohol will also be formed. Why is this so?

Neither $\ce{LiAlH_4}$ nor $\ce{NaBH4}$ are able to reduce an isolated $\ce{C=C}$ bond. But if you have an enal (a conjugated aldehyde) it can react (as an electrophile) either at the β-carbon or at the carbonyle group's carbon. According to the HSAB Principle the β-carbon is a “soft” center and would react preferably with “soft” nucleophiles while the carbonyle carbon is a “hard” center and prefers to react with “hard” nucleophiles.
Now, $\ce{NaBH4}$ is a rather soft nucleophile and thus it reacts with an enal at the β-carbon. After this reaction the $\ce{C=C}$ bond is gone. But what is left behind is still a very reactive simple aldehyde that gets reduced to the alcohol by $\ce{NaBH4}$ in a second step.
$\ce{LiAlH4}$ on the other hand is a rather hard nucleophile and thus reacts with an enal at the carbonyl carbon. After this reaction there is just an isolated double bond left that can't be reduced by $\ce{LiAlH4}$ in a second step.


What do 1T and 2H mean in front of compounds? I have seen 1T and 2H used as prefixes for metallic compounds in literature, but I am having trouble finding what they mean. For example, 1T-HfS2 or 2H-HfS2.

These are notations that are used in identifying polytypes in layered compounds.
2H indicates 2 layers per Hexagonal unit cell (H for hexagonal).
1T indicates 1 layer per Trigonal unit cell (T for trigonal).
At least, that's my understanding.


How P(C2H5)3 acts as a ligand? How does $\ce{P(C2H5)3}$ acts as a ligand and forms dπ-dπ bonds with transition elements. Why not pπ-dπ bond?

I understand that transition elements have d subshell available to accept electrons. But what about the electrons from phosphorus? From Chemistry Class 12 (NCERT), Part 1, p. 168.:

  Another factor which affects the chemistry of nitrogen is the absence of d orbitals in its valence shell. Besides restricting its covalency to four, nitrogen cannot form dπ-pπ bond as the heavier elements can e.g., $\ce{R3P=O}$ or $\ce{R3P=CH2}$ ($\ce{R}$ = alkyl group). Phosphorous and arsenic can form dπ-dπ bond also with transition metals when their compounds like $\ce{P(C2H5)3}$ and $\ce{As(C6H5)3}$ act as ligands.

As I already mentioned in my comment, phosphines such as $\ce{PEt3}$ can bind to transition metals with partially filled d orbitals as a σ-donor as well as a π-acceptor.

The σ interaction is simply a donation of electrons from the lone pair on phosphorus to empty metal d orbitals.

The π interaction is backdonation of electrons from filled metal d orbitals to an empty orbital on the phosphine. Traditionally this was thought to be the 3d orbitals on phosphorus, hence "dπ–dπ". However, as Martin pointed out in the comments this is a rather outdated mode of thinking. The current accepted view is that the acceptor orbitals are $\ce{P-R}$ σ* orbitals (to be precise, a linear combination of them that has appropriate symmetry to overlap with metal d orbitals).


Do ligands with a positive charge exist? Are there positively charged ligands which can bind to a central metal atom to form coordination compounds?

My thoughts:

I know that ligands are Lewis bases which donate a pair of electrons, and the central metal atom is usually a Lewis acid.

However, a positive charge on a species usually means it is electron-deficient. In that case, there would be no suitable electrons for it to bond to the central metal atom. So, even if positively charged ligands do exist, I think there would be very few. In such cases, it might bond with central atom which is electron rich forming the coordinate bond in the opposite way. Is that possible?

Aromatic cations like the tropylium cation ($\ce{C7H7+}$)[1] or the cyclopropenyl cation ($\ce{C3H3+}$)[2] can coordinate to metals, creating sandwich or half-sandwich compounds. The positive charge is shared by all carbon atoms of the rings.

Anderson, J. E.; Maher, E. T.; Kool, L. B. Electrochemical and spectroelectrochemical properties of the titanium sandwich complexes (Cp)Ti(L) and (Cp*)Ti(L), where L is either cyclooctatetraene (C8H8) or the tropylium ion (C7H7+). Organometallics 1991, 10 (5), 1248–1254. DOI: 10.1021/om00051a009.
Lichtenberger, D. L.; Hoppe, M. L.; Subramanian, L.; Kober, E. M.; Hughes, R. P.; Hubbard, J. L.; Tucker, D. S. Electron distribution and bonding in η3-cyclopropenyl-metal complexes. Organometallics 1993, 12 (6), 2025–2031. DOI: 10.1021/om00030a011.


How to know if acid is polyprotic? I know that some acids for sure and for certain are polyprotic. For example sulfuric acid $\ce{H2SO4}$ is diprotic giving you disulfate $\ce{HSO4-}$ and sulfate $\ce{SO4^2-}$

However not all acids with multiple hydrogens are polyprotic, even if 2 or more hydrogens are attached to very electronegative atoms. So if I am given an acid and it's chemical structure how would I know if it is polyprotic or not? Acid base reaction with something like sodium hydroxide $\ce{NaOH}$  and then distillation to separate the product/products from the water that is produced? Or is there a better way to prove whether or not an acid is polyprotic?

As discussed in your question and ensuing comments, there are ways to predict via theory whether a particular compound would be a polyprotic acid, but your last sentence asks how to prove that an acid is polyprotic. This pretty much limits us, as you proposed, to laboratory experiment, namely acid-base titration.

Although your suggestion "Acid base reaction with something like sodium hydroxide NaOH and then distillation to separate the product/products from the water that is produced" is on track, the experiment needn't be nearly so complex.

The experimental procedure is as simple as preparing a solution of the unkown acid, then titrating with a strong base solution while monitoring the pH.

The following is a titration curve for the diprotic oxalic acid, using sodium hydroxide as the strong base titrant. (Image taken from this boundless.com post):

Note that there are two equivalence points, at 25 and 50 ml of added titrant. These equivalence points are the features to look for when titrating the unknown acid. A single equivalence point indicates a monoprotic acid, two indicate a diprotic acid, etc.

There are a couple drawbacks to this experimental procedure. First, the equivalence points are not always as well defined as that shown in the figure. For instance, two acidic moieties may have similar Ka values and thus may almost appear as one.  Making a derivative plot of the titration curve can help to distinguish the equivalence points. In other words, you plot the change in pH vs volume of titrant added rather than just the pH of the titrant added.

The second drawback of this procedure is that it doesn't tell you which protons were titrated by the base. This would require a far more complex experiment, for instance by synthesizing the acid using deuterium in place of hydrogen at specific sites, then analyzing the compound after each equivalence point to see whether the deuterium was pulled off.

I hope I understood your question correctly and that my explanation of the titration procedure to determine polyprotic acids was clear. Don't hesitate to ask for clarification in the comments.


How to Make Ferrofluid in a Bottle I am trying to accomplish the very famous magnetic Ferrofluid in a bottle aka office desk toy.

Before anyone suggest to buy online, I would like to make it myself and learn the whole process VS store bought item.

here is a short video of it

Grade 11 chemistry was the last time I studied chemistry.

I would like to create the similar/same toy

My first question is: is ferrofliud thick liquid you can just buy online or do you have to make it yourself using ink toner or magnetic powder etc

I tried making it  using ink toner powder mixing it with vegetable oil. The results were not the best but worth a shot, it ended up thick blob and alot of particles it was not clear at all and very hard to see through it.

for the clear liquid I used 50% cleaning alcohol,
i tried using water before and it just ended up making the whole glass black.    So in short this is the measurements I used and once again I apologize if I forgot to mention anything.  I used black ink toner mixed it with vegetable oil, ink toner 10 grams vegetable oil 20ml, and 100ml was cleaning alcohol

Alot of people suggest use magnetic powder instead of ink toner what do you recommend? What different liquids should I be using so it does not mix and stays clear.  Some people suggest online make sure you only use water without minerals and iron and many other liquids

It is always easier to buy ferrofluid than to make it. Some of the steps don't work well unless the conditions are just right and it is hard to get them right on your first attempt. Luckily there are plenty of suppliers (this UK supplier is top of my Google search for "ferrofluid sales" and they also supply lots of cool magnets).

Ferrofluid is a suspension of small magnetic particles (often magnetite, the magnetic oxide of iron) in a fluid. To avoid the heavy and magnetic particles clumping together or precipitating, they need to be coated in a surfactant. And this only works if the particles are the right size (very small) and you have the right mix of surfactant that works in the carrier fluid (the fluid is usually mineral oil or vegetable oil; alcohols will be a poor choice).

Laserprinter toner isn't usually a mix of suspended magnetic particles: only specialty toner inks are (the ones used for machine readable printing). If you can get the right sort, it will save a lot of effort, but the wrong toner won't work at all and will just make a mess.

There are quite a few recipes on places such as Youtube, but some of them don't work well or don't explain some of the important details. So, if you don't want to spend a lot of time experimenting to get the synthesis right, just buy the ferrofluid.


Solubility of oxygen in water with temperature, why does the change in solubility decrease around 40 °C? Better set in context using a graph:

The solubility seems to fall quite persistently from 0 °C to 30 °C, then gradually slowing down in fall, then dropping faster and faster from 80 °C. Why does that change in drop happen?

The relatively abrupt changes in slope (do a derivative plot to get an even more distinct picture of what is happening here) are due to changes in the solubility function dominating the solubility behavior at a given temperature and $\ce{O2}$ concentration.  For example, Henry's Law will describe the behavior at low $\ce{O2}$ concentrations.  As Henry's Law applies to the theoretical "infinite dilution" condition, this assumption falls apart at higher concentrations.

The greater the concentration of $\ce{O2}$ the more it departs from the ideal infinite solution assumed for Henry's Law behavior.  At higher concentrations, one or more of the different gas laws then controls the solubility.  Based on your graph, it seems that there are three distinct regions, and thus three different gas laws defining the solubility of $\ce{O2}$ as a function of temperature.

One can also take an empirical approach to predicting the solubility based on experimental measurements.  This website has a live "$\ce{O2}$ solubility in water" calculator based on empirical data.  The authors use two different equations for two different temperature regimes: $\ce{0 ^oC - 30 ^oC and 30 ^oC to 50 ^oC}$.  So their $\ce{30 ^oC}$ cutoff seems to agree with the feature at that temperature in your plot.  I don't know the maximum temperatures / concentrations they used for their second equation.


Why does water boil vegetables, but oil does not? When we add vegetables to hot water (boiling), it boils the vegetables, but when we add vegetables to hot oil (boiling) it fries the vegetables. Why?

The maximum temperature for water boiling is around 100 Celsius and there is a great deal of water present.  Oil can reach greater temperature and the absence of water allows other reactions, such as the Maillard reaction, to occur.  That's why you never see browning in boiled food.


Which of the given compound has the highest rate of dehydration? I came across this question recently:

For which compound acid catalyzed dehydration rate is highest?

Now, they all are secondary alcohols so their dehydration should follow E1 mechanism and the stability of the carbocation formed in the RDS should be considered while comparing their rates.
Following this rule, I think, the answer should be (B) as ketone group shows (-)R effect in (C) and (-)I effect in (A) and (B), all which destabilize the carbocation formed.
But the actual answer is (A). How can that be? Is there some other mechanism to follow? I can’t understand what am I doing wrong.

Notice, that the first compound (A)can show tautomerism, after drawing the enol form we find that the carbocation lies at the allylic position, which means that it is stabilised through resonance forms, hence the compound A has highest rate of reaction.
Hope it helps.


Why is the compound 3,4-dihydroxy-3-cyclobutene-1,2-dione more acidic than acetic acid? Why is the compound 3,4-dihydroxy-3-cyclobutene-1,2-dione more acidic than acetic acid?

I don't get how its resonance stabilization is far greater than that of acetic acid despite the fact that the conjugate base has high angular strain. How is it more stable than acetate ion? In the consideration of acidity, why is only the resonance dispersion of charge taken into account and not the angular strain?

For stability comparisons of the anions so formed I am stuck at the resonance structure too:

Finally, I want to know whether it would be more acidic that formic acid?

This molecule is commonly known as Squaric acid. The Wikipedia entry gives the possible resonance structures of the conjugate base (added here for convenience):

You can see that there are four possible resonance structures for the conjugate base, resulting from the proximity of two ketone groups and two hydroxyl groups around the ring. This proximity allows movement of electrons around the square ring and between the ring and each oxygen atom.

On the other hand, acetic acid (or rather, the conjugate acetate anion) has only two resonance structures, as seen here:

Since the principle is that a molecule with more resonance structures is more stable than a molecule with less resonance structures, squaric acid anion is more stable than acetate anion. Since a more stable conjugate base makes an acid stronger, squaric acid is considered to be stronger than acetic acid.

This conclusion is also evident by the Pka values. This table provides Pka values for many acids, including squaric, acetic and formic acid. You can see that squaric acid has two values for two acidic protons - 1.5 for the first proton and 3.5 for the second proton. acetic acid has a Pka value of 4.75 and formic acid has a Pka value of 3.75.
This means that the strongest acid of the three is squaric acid, followed by formic acid, and the least acidic of them is acetic acid.

Squaric acid > Formic acid > Acetic acid


Writing atoms of functional group backwards Can the $\ce{HO}$ bond be written as $\ce{OH}$ and still be correct?

They mean the same thing. If the OH group on the left were written as OH, it would look like there was a carbon-hydrogen bond. By inverting the order, it makes it clear that the oxygen is bonded to the carbon and the hydrogen is bonded to the oxygen. $\ce{NH2}$ might also be written as $\ce{H2N}$ in a structure drawing if a bond was coming from the right, but the functional group is the same in both cases—the group is connected to the rest of the molecule by the nitrogen atom.


Why doesn't pure water conduct when it dissociates? Sorry if this is an obvious question.
I am told water dissociates, one product of which is negatively charged hydroxyl ions. This makes sense, but then why doesn't pure water conduct electricity? Surely the hydroxyl ions could act as charge carriers to allow current to pass through the solution?
Thanks

Pure water does conduct as pointed out by the above comments. But it has very less dissociation hence there are very few ions to conduct electricity.
you may have heard that pH of pure water is 7. This statement implies that the concentration of H+ ion is 10^-7. Since it is a neutral solution concentration of OH- ion is also 10^-7. As such the conducting property of water is very less.


Why is the formation of a peptide bond not spontaneous? Why is the formation of a peptide bond from a carboxylic acid and amine not spontaneous?
I'm not looking for the free energy calculation; I have it from several sources.  I'm just asking for an intuitive explanation.  Is it because the peptide bond is planar so there's a loss of free rotation and the delta S is unfavorable?  Is the resonance stabilization in the peptide bond less desirable because the other atom (N) is less electronegative?  Do two equally sized molecules somehow have more entropy than a big molecule and a small one (water)?

Based on a quite old reference 1 (which I'm using because it's available free by open access), peptide bond formation at 25 C is unfavorable only because of a large enthalpy change, on the order of 1.5 kcal/mol (6.3 kJ/mol). The entropy change is actually favorable, with $T\Delta S$ being about 1 kcal/mol (4kJ/mol), so the net free energy change is ~ + 500 kcal/mol (approx. +2 kJ/mol).
The reason for both values being what they are is because the free acid and amine groups are ionized at neutral pH. Thus, the reaction can be represented as
$\ce{R-CO2- + ^+H3N-R -> R-CONH-R + H2O}$
This can be broken into two steps conceptually:

$\ce{R-CO2- + ^+H3N-R -> R-CO2H + H2N-R }$

$\ce{R-CO2H + H2N-R -> R-CONH-R + H2O}$

As we know, proton transfer from an acid to a base typically involves a large negative enthalpy change, so reversing that to neutralize both has a positive enthalpy change. Since the products remain neutral, we never get that energy back. And the bond energies of a neutral carboxylic acid and amine are not much different from an amide and water, so the neutralization dominates.
For the entropy change, we start with two molecules and end with two, and the degrees of freedom are similar. The authors of the cited paper propose that the positive entropy change is also related to the loss of ionization. They believe that the more structured hydration shell of the ionic species vs neutral is sufficient to make the free amino acids lower in entropy than the neutral products, making bond formation entropically favorable, but not enough to overcome the unfavorable enthalpy change.
Later paper have further shown that the amide formation reaction becomes favorable around 60 C because the favorable entropy change finally overcomes the enthalpy change at that temp.
1 Dobry, A., Fruton, J.S. and Sturtevant, J.M. (1952)  J. Biol. Chem. 195, 149–154.
(Available by open access here)


A compound that absorbs all visible light Is there a compound that absorbs all visible light?

Substances which absorb almost all the light falling on them appear black. Therefore you are looking for the blackest known compound.

The record is currently held by Vantablack[1], a substance composed of vertically aligned carbon nanotubes, which absorbs up to 99.965% of visible light incident upon it. As you can see it is really black.

Previously, the record was held by Superblack[2,3], a nickel-phosphorus alloy with an absorbance of up to 99.65%.

For comparison, regular black paint generally absorbs around 97.5% of visible light.

Sources

[1] https://en.wikipedia.org/wiki/Vantablack

[2] https://www.newscientist.com/article/dn3356-mini-craters-key-to-blackest-ever-black

[3] http://pubs.rsc.org/en/Content/ArticleLanding/2002/JM/b204483h#!divAbstract


Is the reaction between thiosulfate and hydronium an acid-base reaction? Is the following reaction an acid-base reaction:

$$\ce{2 H3O+(aq) + S2O3^2-(aq) -> 3H2O(l) + SO2(g) + 1/8 S8(s)}$$

And if so, which is the acid and which is the base?

Two of the three acid-base theories (Brønsted and Lewis) also define an acid-base reaction while the third theory (Arrhenius) does not — it merely defines what acids and bases are.

In the Lewis theory, an acid-base reaction occurs when a lone pair donor interacts with a lone pair acceptor to form a new bond without a simultaneous redox reaction happening. The less complicated Brønsted-Lowry theory describes an acid-base reaction as a reaction in which a proton $\ce{H+}$ is transferred from an acid to a base, giving us a conjugate base and a conjugate acid, respectively.

As has been pointed out, your reaction is not a one-step reaction. Rather, we have to break it down into its constituent steps. The first step, a reaction of hydronium $\ce{H3O+}$ with thiosulfate $\ce{S2O3^2-}$ has to be an acid-base reaction because that is practically the only reaction hydronium undergoes.

$$\ce{2H3O+ + S2O3^2- -> H2S2O3 + 2 H2O}\tag{1}$$

The acid is the species that donates the transferred proton, the base is the species that accepts it. Therefore, we can assign the acid and base roles as shown in equation $(1')$.

$$\ce{2 \underset{acid}{H3O+} + \underset{base}{S2O3^2-} -> 2 \underset{conjugate base}{H2O} + \underset{conjugate acid}{H2S2O3}}\tag{$1'$}$$

The second reaction is a redox reaction. Thiosulfate (and therefore also its conjugate acid) has two sulfur atoms in the oxidation states of $\mathrm{+V}$ and $\mathrm{-I}$, respectively. These undergo a redox reaction leading to compounds of the oxidation state $\mathrm{+IV}$ (sulfurous acid $\ce{H2SO3}$) and $\pm0$ (elemental sulfur). We can use the full redox reaction determination algorithm, but I’m skipping that since it is a simple one-electron transfer.

$$\ce{H2S2O3 -> H2SO3 + 1/8 S8}\tag{2}$$

And finally, we have another decomposition namely that of sulfurous acid to sulfur dioxide $\ce{SO2}$ and water. This can be seen as the back-reaction of a Lewis acid-base reaction wherein water is the liberated Lewis base and sulfur dioxide the liberated Lewis acid. Indeed, the forward Lewis acid-base reaction (the reverse direction of this reaction) occurs, too. This reaction is in equilibrium, driven to the product side because of the volatility of $\ce{SO2}$.

$$\ce{H2SO3 (aq) <=>> H2O (l)  + SO2 (g) ^}\tag{3}$$

Therefore, the overall reaction is the sum of one Brønsted acid-base reaction, one redox reaction, and one Lewis acid-base reverse reaction.

$$\ce{2 H3O+ (aq) + S2O3^2- (aq) -> 3 H2O (l) + 1/8 S8 (s) + SO2 (g) ^}\tag{4}$$


Second Ionisation Energy Confusion It is intuitive that the second ionisation will require more energy than the first. However, I'm having trouble explaining exactly why. I came across this other question on a similar topic: Why second ionisation energy greater than the first?

The answer made sense until the answer said:

  Technically, there would be an increased attractive force ($\propto kq_1q_2/r^2$) from the nucleus, because the electrons on average will now be closer to the nucleus.

Considering, say, chlorine, when a valence electron is removed, it makes sense that the electrostatic repulsion between valence electrons would decrease. However, by removing 1 electron, why does this increase the attractive force? I don't see how would increase the coulombic force between the nucleus and a valence electron, since the magnitude of both charges are constant.

If it was the radius changing, why should the electrons be closer to the nucleus just because 1 electron is removed?

Beside that the proportional relationship proposed could be discussed further, as r rearrange, all this statement are related to the sums of forces. They stated that there would be a greater attractive force because they discuss ionization and this is the case. But the attractive force is the result of all attractions and repulsions. Any distribution of electric charges would lead to this situation and is generally an exercise treated after Coulomb law. Any reader can do that providing the dynamics is not involved.


Does salinity affect seawater's pH? If so, how? It seems like excessive salinity within seawater has the same destructive effect on marine organisms just like excessive acidity. But in real terms, is there any relation between salinity and the acidity of seawater?

P.S. I don't take chemistry. This is for my biology depth study. Please try to respond in a way that is eay to understand. :)

Yes, salinity affects the pH of seawater.

See RECOMMENDATONS FOR THE
DETERMINATION OF pH IN SEA WATER
AND ESTUARINE WATERS Pure & Appl. Chem., Vol. 60, No. 6, pp. 865-870, 1988.

An approximation of pH as a function of temperature and salinity is given:

pH = (2559.7 + 4.5 S)/T - 0.5523 - 0.01391 S

Where "T" is temperature in kelvins and "S" is salinity in units of parts per thousand.

As seen above, in the temperature range of Earth's oceans, increasing salinity increases pH.

See also pH of seawater Marine Chemistry Volume 126, September 2011, pages 89-96.

Fig. 2 plots pH versus salinity at 25 degrees C and 0.000334 atmospheres of CO2 and shows pH increasing significantly with salinity.


Rate of dehydration for butan-2-ol and butane-2,3-diol A question taken from a test series (Vibrant, India):

Amongst butan-2-ol and butane-2,3-diol, which has higher rate of dehydration?

The carbocation formed in butane-2,3-diol is more stable as it will rearrange to get +M stability from alcohol group. However, the product will be aldehyde (due to tautomerism) for butane-2,3-diol and alkene for butan-2-ol.
As rate is determined by stability of carbocation, butane-2,3-diol should have faster rate. Also, the products don't give any extra hint. Right? But the answer given for higher rate is butan-2-ol. Is there any point that I have missed out?

For this question, given the lack of initial information, I'm making the following assumptions:

Butane-2,3-diol dehydrates to give butan-2-one
2-butanol dehydrates to give but-2-ene

Butane-2,3-diol dehydration is nicely described in this paper from Emerson1:

Aqueous 2,3-butanediol was shown to react in a pseudo-first-order reaction in the presence of sulfuric acid to form methyl ethyl ketone. The reaction could be described by a pinacol rearrangement mechanism and had an activation energy of 36 kcal/mol.

Finding the data for dehydration of 2-butanol proved more elusive. One study2 conducted dehydration rate experiments for secondary alcohols at 250 °C and 40 bar pressure and obtained a rate constant for 3-hexanol to be $k = \pu{(0.067 \pm 0.021) h^{-1}}$. Assuming this rate would be similar to 2-butanol, we can find the activation energy via the arrhenius equation, which comes out to be around 48 kJ/mol, which is around 11.52 kcal/mol. Another source3 conducted the same experiment over a copper-chromite catalyst, and obtained an activation energy of 13 kcal/mol. This means that 2-butanol will dehydrate faster than butane-2,3-diol.
From a conceptual standpoint, this makes sense: The pinacol rearrangement mechanism would require a hydride shift, which would be the RDS and would take more time and energy than a simple carbocation formation. However, if you're comparing absolute reaction rates, a large number of factors come into play, and given the right conditions, butane-2,3-diol can dehydrate faster than butan-2-ol.
References:

Emerson, Richard R., et al. “Kinetics of Dehydration of Aqueous 2,3-Butanediol to Methyl Ethyl Ketone.” Industrial & Engineering Chemistry Product Research and Development, vol. 21, no. 3, 1982, pp. 473–77. doi:10.1021/i300007a025
Bockisch, Christiana, et al. “Kinetics and Mechanisms of Dehydration of Secondary Alcohols Under Hydrothermal Conditions.” ACS Earth and Space Chemistry, vol. 2, no. 8, 2018, pp. 821–32. doi:10.1021/acsearthspacechem.8b00030.
Joseph Sires Cantrell, "Rate and activation energy of the first order dehydration of 2-butanol over a copper chromite catalyst", 1954, source


Making metal powders using electrolysis - chosing the right electrolyte Really fine metal powders are good for two things: making paints and pyrotechnics. For this purposes, I can see people usually use thing called ball mill.

It's basically a barell stuffed with heavy balls and the thing to be powdered. You leave it for a while, rotating all the time, and the balls slowly crush your substance into smaller and smaller pieces.
There are many disadvantages to this:

I'd need to construct it
It would be running and making loud mess all the time
soft metals, like aluminium and copper don't even crush into pieces

So I have figured out a different way. More chemical one I'd say. I can dissolve metals in acid by forming salts out of them and then electrolyse those salts. Not so long ago, I complained that instead of copper plating I got copper powder. That's just what I want now. I'm really sorry that powder I created the other day is already flushed down the drain. It could make beautiful green flames.

Now while the general idea is easy, I'm not so sure about choosing the right electrolytes. My theoretical knowledge of electro-chemistry unfortunately sucks. And I'm afraid that every metal will need different electrolyte.

For example I think using any acid for Aluminium could cause something nasty since it reacts with them little bit too quickly. For Iron, I tried hydrochloric acid and then I found out thet $\ce{HCl}$ dissolves iron to form iron chloride.

Are there general rules to figure this out? If not, what sould I consider to find electrolyte individually for these metals:

Aluminium
Iron
Copper
Magnesium

Powdered metals can be bought of course, but I don't want to waste good money for something, that can be easily made from trash (old nails, aluminium foils from chocolate...).

Fine copper powder can be prepared using thermal decomposition of copper citrate (can be prepared by mixing dirt cheap citric acid, three equivalents of a base like potassium hydroxide and copper sulphate). I once made pyrophoric (!) copper powder using this technique. There are mentions of preparation of iron powder same way. Anyway, it is much more practical than electrolysis for big batches.

Frankly speaking, fine aluminium flakes are widely employed in various paints and are dirt cheap. They should be available in shops for artists. There is no reason to make them at home (and aluminium reacts with water once cleaned of oxide film)

Magnesium is probably the worst. It is quite costly and active and making fine powder is dangerous. It also reacts with water and alcohols, so making it employing electrolysis is out of question. There are mentions of home smelting of aluminium and magnesium, producing magnalium. Magnalium is brittle and can be easily milled, and it is active enough so fine milling is usually not needed. While fancy, proper pyrotechnics can be made without magnesium with easy, aluminium is usually much more preferable. Magnesium often produce huge volumes of white smoke, a habit aluminium doesn't have.

Regardless of above, I strongly encourage to be cautions when working with metal powders, especially magnesium.


How do point defects in a crystal relate to Entropy? I am interested in Non-Stoichiometry and how it relates to crystal structures; In particular I wish to learn about surfaces of metal oxides. I am aware that point defects in such crystals are a result of an entropy driven reactions. So my question is this:

Are there any good introductory resources on this subject? I have basic undergrad chemistry and physics background.

I would strongly recommend this resource:
http://www.tf.uni-kiel.de/matwis/amat/def_en/
It is a good complete on line introduction to defects theory in crystals.
Some key points listed below.

The concentration of point defects $c_{PD}$ at equilibrium can be expressed by an Arrhenius formula:
$$c_{PD}=A \cdot e^{-\frac{G_F}{kT}}=A \cdot e^{-\frac{S_F}{kT}}\cdot  e^{-\frac{H_F}{kT}}$$
where $G_F$ is the free energy of formation, $H_F$ the enthalpy of formation and $S_F$ the entropy of formation. In solids  $H_F$ can be confused with the energy of formation, $S_F$ is associated the variation of the vibration entropy of the crystal due to the introduction of a point defect.
Formation enthalpies, and thus formation energies can be obtained by:
$$ H_F(D) = E_{tot}(D) - E_{tot}(X)+\sum_i n_i \mu_i +q (E_F -E_V) $$

$E_{tot}(D)$  and $E_{tot}(X)$: total energies of the system with and
without the defect D.
$n_i$: number of atom removed from the system,
a negative value denote the addition of one atom.
$\mu_i$: chemical potential of the atom $i$. This should consider the chemical environment at the equilibrium condition (which reservoir is to take into account).
$E_F -E_V$: Fermi energy relative to the valence band maximum.
$q (E_F -E_V)$: energy change due to the exchange of electrons and holes with the carrier reservoirs.

For the equilibrium concentration in the multiple defects case an additional entropy term comes from the standard configuration entropy which can be calculated from the Boltzmann formula
$$S=k_B \cdot \ln P$$
where P is the number of configurations for the same macrostate.


Reaction mechanism of aqueous ammonia with 3,4-dibromopyridine I am struggling on how this mechanism would occur. I am aware that the bromine will be removed by for some reason I cannot figure out the $\ce{NH3}$ attaches onto the ring, can anyone prove me with any help?

I have also had a feeling that deprotonation occurs and a benzyne is formed, in which a bromine gets kicked off and one of the $\ce{:NH3}$ attaches onto the ring. Is that correct?

The reaction between ammonia and 3,4-dibromopyridine is a nucleophilic aromatic substition reaction (SNAr). This is possible because the ring is rather deactivated due to the presence of the electron withdrawing nitrogen atom of the pyridine drawing electron density towards it. The ammonia attacks carbon at the 4-position preferably, rather than the carbon at the 3-position as this position is more deactivated than the latter. This attack results in the breakage of the double bond, resulting in a loss of aromaticity. This results in the formation of a carbanionic Meisenheimer intermediate. The bromine at the 4-position then leaves and the double bond reforms, resulting in the restoration of aromaticity in the ring. This results in a net substitution of $\ce {-Br}$ with an $\ce {-NH2}$ group.


Lewis structure of benzenediazonium In the benzenediazonium $\ce{C6H5N2+}$ the nitrogen contributes its lone pair to the carbon in benzene. So where is one electron of that carbon, since it is not used for bonding with nitrogen where has one of the electron of carbon gone?

If the diazonium ion were formed in the way you described, it would be formed by N2 acting as a nucleophile and attacking the ring with a lone pair, displacing a leaving group - a nucleophilic aromatic substituion. In this case, the "one electron of carbon" would have left with the leaving group that was on the ring before. The carbon would now be neutral as it had lost an electron and essentially gained one from the dative bond from the N.

However, this isn't the usual way to form a diazonium ion. In fact, we normally start with a primary aromatic amine, and react it with a diazotizing mixture of NaNO2 and HCl. In this case, the positive charge on the N ends up, not from the N forming a dative bond whilst acting as a nucleophile towards the ring but in the final elimination of water to form the triple bond in the diazonium ion:

Image: Wikipedia


Formula of the compound based on occupancy of lattice voids Atom of element $\ce{B}$ forms HCP lattice and those of the element $\ce{A}$ occupy 2/3rd of tetrahedral voids. What is the formula of the compound formed by the elements $\ce{A}$ and $\ce{B}$?

My doubt is that since the element $\ce{B}$ is in HCP, the number of octahedral and tetrahedral voids will be 6 and 12, respectively, which means that $2/3 \times 12 = 8$ tetrahedral voids occupied by $\ce{A}$.

Since $\ce{B}$ is in HCP lattice, there must be 12 voids, i.e. the ratio will be $8 : 12 \implies  2 : 3$, so $\ce{A2B3}$ must be the answer.

But according to NCERT textbook (example 1.2) the answer is $\ce{A4B3}$.

Since B forms an HCP lattice, it's z=6 (z is the number of atoms power unit cell of the lattice) then the ratio comes out to be 8:6 that gives us A4B3.


How do atomic orbitals interfere both constructively AND destructively to give two molecular orbitals? I am an organic chemistry student learning how atomic orbitals interfere to give rise to molecular obitals. The image below suggests that each of the hydrogens' atomic orbitals interfere both constructively AND destructively. How is this possible? More fundamentally, how does the 1s atomic orbital of a hydrogen atom have both positive and negative phase in the first place? I realize that there is a similar question on this topic but I wasn't able to understand the answer to that post due to my limited background in quantum mechanics and was hence was looking for someone who could explain in simple terms.

Part of the issue is that you're treating orbitals as objects. They aren't objects, they are quantum mechanical wavefunctions. The electrons which occupy the orbitals are objects, but the orbitals themselves are decidedly not objects.
When we talk about quantum mechanical states, the overall phase of an individual state has no meaning. It is a fundamental postulate of quantum mechanics that the wavefunction $\psi(x)$ and the wavefunction $k\cdot\psi(x)$ represent exactly the same state, so the phase $k$ is completely meaningless. 1s orbitals are states as well, so it is meaningless to talk about the phase of a 1s orbital.
However, if you have a state that is made up of two different states (like a molecular orbital), then the relative phases of the two constituent orbitals matter. That's why the bonding and antibonding MOs are different: it's because the two constituent parts don't have the same relative phase in the two MOs (in the bonding it's the same, in the antibonding they're opposite).

There are very few, and possibly no, quick and dirty ways to explain MO theory, in my opinion. However, if you have studied linear algebra before,* then the transition from AOs and MOs can be viewed as a change of basis. A basis refers to a minimal set of vectors, from which any arbitrary vector may be constructed through linear combination. If you have a vector $(a, b)$, you can express this as: $$\begin{pmatrix}a\\b\end{pmatrix} = a\begin{pmatrix}1\\0\end{pmatrix} + b\begin{pmatrix}0\\1\end{pmatrix}.$$
It doesn't matter what $a$ and $b$ are, you can always express $(a, b)$ as a linear combination of the two basis vectors. You can think of the basis vectors $(1, 0)$ and $(0,1)$ as being the 1s orbitals on both individual atoms.
The MOs correspond to using the basis $(0.5, 0.5)$ and $(0.5, -0.5)$.† Notice that this is still a valid basis, because we can still express any arbitrary vector $(a, b)$ as a linear combination of these:
$$\begin{pmatrix}a\\b\end{pmatrix} = (a+b)\begin{pmatrix}0.5\\0.5\end{pmatrix} + (a-b)\begin{pmatrix}0.5\\-0.5\end{pmatrix}.$$
If we had rejected the "antibonding" combination $(0.5, -0.5)$, and only taken the "bonding" combination $(0.5, 0.5)$, then we would not be able to form a valid basis. For example, we could not express the vector $(1, 2)$ as just a linear combination of $(0.5, 0.5)$.
From this point of view, AOs and MOs are simply different bases.‡ The existence of both the bonding and antibonding MO is not only logical, but also mandatory in order to form a complete basis.
The familiar AOs that you have studied (1s, 2s, 2p, ...) are simply a series of basis states for the hydrogen atom. In the $\ce{H2}$ molecule, there are two sets of basis states, one for each atom. The transition from "2 sets of AOs" to "1 set of MOs" is just a change of basis: we go from two sets of atomic bases to one single molecular basis.

The proper answer is, of course, to pick up a QM textbook and study it. By QM I do indeed mean quantum mechanics, not just MO theory. It is a long journey, and you may not necessarily be fully prepared for it right now, but it will reward you with a better understanding of how this works. MO theory fundamentally relies on quantum mechanics, and to learn MO theory without understanding quantum mechanics is what leads to confusion and questions like yours.

* The analogy I use above is slightly simplified, and might look slightly contrived, but the link between QM and linear algebra is actually very deep. QM is somewhat analogous to linear algebra but on an infinite-dimensional vector space (technically, a Hilbert space).
† Technically this should be $1/\sqrt{2}$, not $0.5$. In QM we are nearly always concerned with orthonormal bases, i.e. ones where each basis vector has a length of $1$ and forms a scalar product of $0$ with every other basis vector. (Mathematically, we have a basis $\{|i\rangle\}$ for which $\langle i | j \rangle = \delta_{ij}$.)
‡ "Bases" being the plural of "basis".


How to recognize Barium sulphite? Yes it's a white ppt. which is soluble in dil. $\ce{HCl}$, but I'm not sure which of the following happens:
$$\ce{BaSO3 +$dil.$HCl->Ba(HSO)3 +Cl-}\tag1$$
$$\ce{BaSO3 +2H+->Ba^2+ +SO2 ^ +H2O}\tag2$$
Questions:

Which of the above actually happens? I have these two from different sources.
Does it becomes $\ce{BaSO4}$ on standing? Is it afterwards soluble in dil. $\ce{HCl}$, does this change require an oxidising agent, if yes then which one is more suitable - bromine water or $\ce{HNO3}$?

Sulphite$(\ce{SO3^2-})$ and Barium Choride$(\ce{ BaCl2})$ or Strontium Chloride$(\ce{ SrCl2})$:
$$\ce{SO3^2- +Ba^2+ ->BaSO3 v}$$
the precipate disoolves in dilute hydrochloric acid, when sulphur dioxide evolves:
$$\ce{BaSO3 v +2H+ ->Ba^2+ +SO2 ^ +H2O}$$
On standing, the precipitate is slowly oxidises to the sulphate and is then insoluble in dilute mineral acids; this change is rapidly effected by warming with bromine water or a little concentrated nitric acid or with hydrogen peroxide:
$$\ce{2BaSO3 v +O2 ->2BaSO4 v }\\
\ce{2BaSO3 v +Br2 + H2O->BaSO4 v +2Br- +2H+}\\
\ce{2BaSO3 v +2HNO3 ->3BaSO4 v +2NO ^ +H2O}\\
\ce{2BaSO3 v +H2O2 ->BaSO4 v +H2O}$$
The solubilites at $\pu{18 ^\circ C}$ of the sulphites of $\ce{Ca,Sr,Ba}$ are respectively $1.25, 0.033, 0.022$ all in $\pu{g/l}$
Source:

Vogel’s Qualitative Inorganic Analysis for the JEE, by Svehla/Sivasankar,  7/e Paperback, 1 January 2013


PBE vs. PBEPBE functional I've found good explanation what the PBE functional is. However, sometimes I see PBEPBE and I cannot find any good information about what this means and how it's different from PBE.

The Gaussian manual explains PBE and also provides the reference but at other points they talk about PBEPBE without any reference.

Can someone explain the difference? Does the second PBE just describe some kind of modification?

The designation PBEPBE is an actual terrible artefact from researchers using Gaussian. The actual publication[1, 2] only refers to it as PBE, and most other program packages (I know) implement it as such. Often they make the correlation and exchange parts separately available via PBEC and PBEX, or similar. Unfortunately this designation also made it into the literature.
The confusion further continues since Gaussian uses the PBE1PBE designation for the (standalone) hybrid functional, which is in most other packages known as PBE0.

But it doesn't stop there, for TPSS you may find similar naming/ reference schemes. The BP86 functional keyword requests the VWN(III) version, while many other programs use VWN(V) for this. The latter is available via BVP86, but misses any implementation of dispersion. We could go on with this...

In the end what is important is the correct referencing of the methodology so that the results are reproducible, i.e. look at the literature cited. Gaussian actually does a good job referencing those sources. And when you publish, you'd do us all a favour and not use the keyword designations of Gaussian.

[1] John P. Perdew, Kieron Burke, and Matthias Ernzerhof, Phys. Rev. Lett. 1996, 77, 3865. DOI: 10.1103/PhysRevLett.77.3865. Mirrored at Burke's page: dft.uci.edu/publications.php

[2] (Erratum) John P. Perdew, Kieron Burke, Matthias Ernzerhof, Phys. Rev. Lett. 1997, 78, 1396. DOI: 10.1103/PhysRevLett.78.1396


How to use a spectrometer to determine NPK I'm trying to determine if it's possible to use a low-grade spectrometer to read NPK values of a solution.

I believe I should be able to take readings of the individual compounds (e.g., calcium nitrate) and then use that to "subtract" from a reading of the mixed/final solution and basically deduce how much of each compound is present - is that correct?

Your spectrometer detection range is 340-780 nm. That is to say mostly visible, and a bit of UV. If your compounds are colorless, it's a first indication that you won't see anything on your spectra. These compounds seem to be inorganic, no transition metal or anything exotic, so it's not a good method.

Apart from this, yes, you would need to calibrate with solutions of known concentration (unless you know the absorption coefficient, which would give you the concentration with Beer-Lambert law).


How long would it take for sucrose to undergo hydrolysis in boiling water? I was reading the book Cocktail Codex and there was a snippet on avoiding boiling sugar and water to make simple syrup that made me a bit skeptical. Here it is:

  Heat also affects the molecular structure of sugar. For example, if sucrose, the disaccharide commonly known as white table sugar, undergoes an extended period of boiling, it will eventually convert to the monosacchariedes glucose and fructose, which taste sweeter (and cause nasty hangovers).

Does 100 °C water hasten hydrolysis, and if so, how long would you need to boil for the breakdown to occur?

(the hangover bit is a bit weird too considering your body breaks down sucrose into glucose+fructose?)

This is a nice well-defined question, and luckily there is excellent data for which we can provide a quantitative answer.
Richard Wolfenden's research group has sought for many years to characterize the spontaneous (i.e. not enzyme catalyzed) rate of many enzymatic reactions.  In general this is so that the spontaneous rate can be compared to the enzyme-catalyzed rate, so that the catalytic proficiency of enzymes can be calculated.
All that's just prefatory remarks to introduce this paper, "Rates of Spontaneous Cleavage of Glucose, Fructose, Sucrose, and Trehalose in Water, and the Catalytic Proficiencies of Invertase and Trehalase".
The abstract's first sentence is:
The half-lives for spontaneous hydrolysis of trehalose and sucrose at 25 °C are 6.6 × 106 years and 440 years.
Thus, a solution of sucrose in water should be 50% decomposed into monomers after 440 years -- at 25 °C.  That value comes from thermodynamic extrapolation of experiments done at higher temperatures.  (I don't think they wanted to wait hundreds of years to do an experiment at 25 °C.)
But higher temperatures is exactly what you're after.  Figure 1 from the paper has the data.

The lower-right datapoint is very close to 100 °C, and shows the first-order rate constant for sucrose hydrolysis is about $10^{-6}$ per second.  That means that 63% degradation would take $10^{6}$ seconds, or about 11.5 days.  Getting 99% degradation would take about 53 days.
edits
The above has been edited to fix sloppy math errors, as noted by @gremin and @John Bentin in the comments.  Thank you both.


How to define custom force in quantum chemistry packages? I don't know if this is really applicable but I want to add a custom force (say, $F(x,y,z)$) to some atoms in my geometry optimization. Is there any way to add such forces to the DFT or molecular mechanics calculations? (preferably GAMESS or Gaussian for QC or Avogadro for MM).

It can be also done as Molecular Mechanics (MM) as I just need an approximate geometry after applying the force.

Let me give an example. Say I have molecule A and molecule B and a force that changes in space as $F(x,y,z)$ but is only applied to molecule B. Now I want to calculate the new geometry of A and B.

I can think this is like adding a $V_F$ to the external energy in the Schrödinger equation, but I don't know how to do that in QC packages:

$$\left[-\frac{1}{2}∇^2 + V(r) +\color{red}{V_F(r)}\right]ψ(r) = Eψ(r)$$

Update:  The idea of making a unit cell and making shrinkage on it looks great (suggested by Geoff Hutchinson here), but the problem is I couldn't figure out how to install CP2K on MacOS and also I am looking for something with a simple interface like Avogadro. I even tried Maestro by Schrödinger software, but couldn't make unit cell shrinkage for that.

The short answer is not. You cannot define a constant force for an atom or set of atoms in a system.

If you still want to modify force constants, this is the way to do it:

Gaussian, you can modify through the keyword GEOM(IHarmonic=n), where n is part of: $\frac{n}{10^6}$ which indicates the respective force applied to the system. Its units are $\frac{Hartree}{Bohr^2}$. To more info, please refer to this link.
If you want to introduce a specific force to an atom or set of atoms in MM, the best way to do it is by parameterizing your own system into your desire environment. The most important part is that you have to have in mind that the right part of the principal ec. of any MM describes the parameters of the specific force field of your desire system or set of systems under specific conditions (i.e. aminoacids in water like OPLS)

$$\vec{F}=\frac{\partial V}{\partial x}$$

Moreover, when you talk about forces in QM you are talking about the forces acting on electrons. Not on atoms.
But, if you still want to know a way to know how the forces of electrons (From DFT) and, at the same time, nuclei are, I recommend you Car-Parrinello. Please, refer to this link
If you have a particularly interest on simulating systems in QM with specific forces acting on electrons, the best way to do it (As far as I know) is to develop your own density functional.

Update

I Gaussian input file was requested to figure out how the keyword and input is used: Geom(iharmonic=n). I did not find the respective file, instead, I give a try twice by crating two input files with different values of n. The used Gaussian input files were:

chk=Test.chk
#p opt b3lyp/6-31g geom(iharmonic=n)

Test

0 1

H
H 1 0.9
O 2 1.2 1 120.

and the used values of n were 2 (On the left) and 20 (On the right). The respective Energy and RMS profile of both simulations are:

Also, in Gamess, you can use these keywords:  Statpt, Force and its related keywords like Zmt. More info, Please see this link at page 2-121 and 2-131.


Ramachandran plot Phi(ϕ) Psi(ψ) dihedral angle Convention for Zero, Positive and Negative value- old and new Update: possible duplicate: What is the precise definition of Ramachandran angles?. Question modified.
G.N. Ramachandran et al, in their own work (PMC) (DOI), did not used phi(ϕ) and psi(ψ) as we use today. They used phi(ϕ) and phi-dash(ϕ') defined as follows:

The conventions we have adopted for the standard configuration ϕ = ϕ' = 0°) and for the positive sense of rotation for the two angular parameters ϕ and ϕ' are indicated in Fig. 1.

Where Fig. 1 is:

Also they had adapted a very different values, 0 to 360 degrees for ϕ and ϕ'. They provided the Ramachandran plot like this:

This looks very different from the Ramachandran plot we see nowadays because:

This (Ramachandran's original) graph looks flipped from todays.

The zero degrees value of present day is perhaps different from Ramachandran's Zero degrees. Because the current day zero degree correspond to Ramachandran's 180 degree (of original graph).

So my question is basically:
On todays convention; how we define the zero value, positive value and negative values of phi and psi dihedral angle? and how does the new and old convention relate?
Prior research:
I did not find anything on intensive internet search. I got one graphic without any reference

I am doubtful about this diagram because its Zero degrees seem to be the zero degree of Ramachandran's original plot.
Many thanks in advance
Update: As the user @Buck Thom said

The meaning of angles with value of zero is the same.

But sill I looked both old and new graph and found what is labelled as 0 degrees in the new graph, has been labelled as 180 degrees in the old graph as follows

Evidently the old $\phi$ is measured from an angle oriented $180^\circ$ relative to the new convention. This implies that the amide H ($H_N$) rather than carbonyl carbon was used as the reference fourth atom used to define the dihedral, since these atoms are related by a $180^\circ$ rotation. The old choice of atom is opposite on the chain to the modern convention of using the heavy backbone atom (carbonyl carbon). You can obtain angles in the new convention by subtracting $180^\circ$ ie $\phi \rightarrow \phi - 180^\circ$.

The meaning of angles with value of zero for $\phi$' aka $\psi$ is the same. However the values between $180^\circ$ and $360^\circ$ in the original Ramachandran plot can be shifted below zero, as $180^\circ\rightarrow -180^\circ$ and $360^\circ\rightarrow 0^\circ$, or more generally $\psi \rightarrow \psi - 360^\circ$.
By the way inspection shows that this is the correct interpretation. See for instance the following diagram. The extended (beta/turn/pII) region is usually larger than the alpha region. The beta region sits at large values (below  $180^\circ$) of $\psi$, whereas the narrower alpha region is close  to  $0^\circ$ (at negative angles):

In Ref. 1 Ramakrishnan and Ramachandran explain the differences in the  conventions:

At a recent conference of some of the representative workers in this field held in Bethesda, it was decided to denote the two dihedral angles about the bonds N-Ca and C.-C' by $\phi$ and $\psi$ respectively, the sense of rotation being the same as that adopted here. The fully extended chain, with N-H and C'=O trans with respect to one another, is to be taken as the standard conformation with $\phi$=$\psi$=0.
It is readily seen that the new $\phi$ is the same as the old $\phi$, but that $\psi$=$180^\circ$+$\phi'$ (and $\phi'$=$180^\circ$+$\psi$). All the data reported here are thus readily converted into the ($\phi$,$\psi$) coordinates. The diagrams in Figs. 2, 3, and 6 have to be shifted up by $180^\circ$, or half the total length, along the vertical direction. In particular, the right-and left-handed alpha helices will have ($\phi$,$\psi$) equal to ($133^\circ$,$123^\circ$) and ($227^\circ$,$123^\circ$). As before, a helix with ($-\phi$,$-\psi$) will be inverse to one with ($\phi$,$\psi$); i.e., it will  be of opposite sense, but having the same number of turns per unit.
This paper was finalized well before this meeting, and so the older conventions are adopted here. However, it is proposed to use the new notation in the following papers in this series.

References
1.Ramakrishnan C, Ramachandran GN. Stereochemical criteria for polypeptide and protein chain conformations II. Allowed conformations for a pair of peptide units.  Biophysical Journal, 01 Nov 1965, 5(6):909-933. DOI: 10.1016/s0006-3495(65)86759-5.


Why does water attack the more substituted carbon in oxymercuration? In hydration of an alkene by oxymercuration-demercuration, for an asymetrical mercunium ion, the more substituted carbon atom is attacked by water.

Why is this the case? Since R groups are electron donating, it should make the more substituted carbon less electrophilic, also water would like to attack the carbon having less steric hindrance but the reverse is true.

Using 1-methyl-1-cyclohexene as an example, the mercurinium ion is not symmetrical. If it were, the positive charge would all be on mercury. The ion is unsymmetrical with part of the positive charge on the more substituted carbon. Why this carbon? Follow Markovnikov's Rule: tertiary carbocations are more stable than secondary carbocations. This bias is manifested in the transition state permitting water to attack at the more substituted carbon. The addition is anti. I chose 1-methyl-1-cyclohexene as an example because adding a tertiary butyl group to the 4-position produces more subtle aspects of the reaction mechanism.

While structures 1 and 2 above are both tertiary alcohols, only alcohol 1 is formed. This selectivity is controlled not only by the anti-addition but also by the more favorable chair-like transition states (red arrows) over the boat-like transition states (blue arrows). These issues are provided in greater detail at
http://ursula.chem.yale.edu/~chem220/chem220js/StudyAids.html#Mechanisms as PowerPoint and pdf files.


ChemDraw: how to change the default aromatic ring style for drawing from SMILES I'm a beginner at ChemDraw.
I'd like to draw a molecule from SMILES and the ring bond should be drawn like the left image:

In my case, my image is drawn like the right image.
Is there any method to change the default style setting?

From pp. 91–92 ChemDraw 20.0 User Guide, PDF version shipped with the ChemOffice suite:

Toggle Aromatic Display
An aromatic structure can be displayed either as an alternating double and single bond or as an aromatic circle. Toggle Aromatic Display option lets you change an aromatic structure from Alternating double/single bond to Aromatic circle
and vice versa. To do so:

Select the structure.
Go to Structure → Toggle Aromatic Display.

When opening a file of non native chemical format or pasting as SMILES, you can now define the way aromatic structures are displayed.
To do so:

Go to File → Preferences. The Preferences Dialog appears.

Click Open/Save tab.
Click the Normalize Aromatic Display checkbox and select either Alternating Double/Single Bond or Aromatic Cirlces as desired.
Click Apply and then click OK. All aromatic structures in the file loaded or the SMILES string pasted will be of the defined representation.


Should moderately soluble compounds like calcium hydroxide be denoted (aq) or (s)? I'm learning to write balanced molecular and ionic equations. For "moderately soluble" compounds such as calcium hydroxide, how do I know whether to write the state symbol as (aq) or (s)? This is the equation I'm supposed to work with:

$$\ce{Al2(SO4)3 + 3 Ca(OH)2 -> 2 Al(OH)3 + 3 Ca(SO4)}$$

In principle, you cannot know without conducting the experiment. It depends on the concentration of a moderately soluble compound such as $\ce{Ca(OH)2}$ whether it is dissolved or whether there is stuff still sitting at the bottom of the vessel undissolved.

However, you are performing a reaction which is in aqueous solution. This reaction will typically only work if all the reactants are already in solution. Moreover, this is a precipitation reaction. Thus, it wouldn’t make sense to have a non-dissolved reactant to react it with a dissolved one. Given this context, you should assume both reactants are in solution.

$$\ce{Al2(SO4)3 (aq) + 3Ca(OH)2 (aq) -> 3CaSO4 v (s) + 2 Al(OH)3}$$

Note that whether or not aluminium hydroxide precipitates depends on the resulting $\mathrm{pH}$ value. If you started off with excess hydroxide, it may well stay in solution.

The exact answer for a given compound can only be known if you know both concentration and solubility product.


Can aqua regia/royal water dissolve lead? As known aqua regia can't dissolve $\ce{Ag}$ because formation of $\ce{AgCl}$ precipitate. Similarly will aqua regia precipitate $\ce{Pb}$ as $\ce{PbCl2}$? OR It is not possible to observe this because aqua regia produce heat keeping solution warm ($\ce{PbCl2}$ is soluble in hot water).

Aqua regia doesn't actually dissolve lead metal. It converts to insoluble lead salt. From here:

When we are working with an acidic chloride solution, like HCl (with
an oxidizer) or aqua regia, most of the lead will become an insoluble
lead chloride which if it is left to settle can be removed by careful
decanting of the solution and filtering, what lead does carry through
the process with this solution can then be removed with a few drops of
sulfuric acid, and precipitated out of solution as a lead sulfate.

Here is a YouTube video for your reference.


How could I create a machine that as rapidly as possible creates greenhouse gases from air? For a very large-scale project, I am trying to create a greenhouse gas demonstration device, which rapidly produces greenhouse gases from atmospheric air using only electricity, or any chemical reactions or chains of chemical reactions that can attain their input energy from electricity. At the intake would be a very powerful fan which rapidly forces air through the system, to speed up the process. How would it be done?

UPDATE:

I was thinking nitrogen oxides would be best, considering the atmosphere is approximately 78% nitrogen and 21% oxygen. But what would be the fastest way to produce them, assuming price is not really an issue?

Current concentrations of $\ce{CO2}$ and $\ce{CH4}$ in the atmosphere are around 390 ppmv and 1.75 ppmv, resp. These concentrations and the wish to obtain large quantities of both gases rapidly seem mutually exclusive, unless you use (at least one) additional carbon source.

Taking the options for interconversion into account, $\ce{CH4}$ in pressurized bottles would be an obvious choice ;).


Dry Hydrogen Production I am building a set-up to produce a controlled flow of dry hydrogen. I'm aiming for around 99.5 percent purity.

Am I right in thinking a PEM cell and Deterite drying tube should be fine for this? Or will I need a condenser of some kind?

Any tips?

Using molecular sieves comes to my mind, they're easy to use and efficient. You can find some practical information about how this is done for example here http://www.klmtechgroup.com/PDF/Articles/articles/Molecular-Sieve-Applications.pdf


Why do reactions happen at all? What makes some atoms/compounds reactive and some noble? What is the fundamental truth behind a reaction? It seems that two species can undergo number of combination to give a compound. Say Sodium and Chlorine, it seems intutive to have number of combinations from this pair such as $\ce{Na2Cl}$, $\ce{NaCl2}$, $\ce{Na3Cl2}$, and even do not react at all. But why $\ce{NaCl}$?

I had been carrying the concept of octet rule, which I now find to be exception in some cases like $\ce{AlCl3}$. So, what determines the products of a reaction? Or is it that all possible combination are formed but only few are major products, in this case $\ce{Na2Cl}$, $\ce{NaCl2}$ etc. are formed but $\ce{NaCl}$ is major product.

I think it has to do something with energy but cannot figure out what.

What makes some atoms/compounds reactive and some noble?

Unpaired electrons and how tightly these electrons are bound to their respective nuclei.

  Why do reactions happen at all? What is the fundamental truth behind a reaction?

Because atoms are more stable together than they are apart, and systems will tend towards their lowest energy state.

  But why NaCl?

This is really a matter of ratios, which is the basis of chemistry. Different elements share their electrons in different ways under different circumstances. Some can donate their electrons ($\ce{Na}$, $\ce{Al}$, $\ce{Au}$), some accept electrons ($\ce{Cl}$, $\ce{O}$, $\ce{N}$), and others are somewhere in between ($\ce{C}$, etc.). Sodium and chlorine are very consistent with how they share electrons. Sodium has 1 valence electron, chlorine lacks 1 valence electron, and they combine in a one-to-one ratio. The formula unit of their salt is $\ce{NaCl}$.

When considering the salt of aluminum and chlorine, ask how many valence electrons does each atom have/need. I think you might begin to see the pattern.

$[1]$ Bond energy diagram


Synthesis of indigo from anthranilic acid - reaction not taking place I was trying to synthesise indigo via the Heumann indigo synthesis (II.), which involves heating antrhanilic acid and chloroacetic acid, both dissolved in isopropanol. After heating up, a hydroxide or another strong base is added to form indigo

I have slightly altered the synthesis process, added trichloroacetic acid instead of chloroacetic, and also added more reactants to tilt the equilibrium towards the product.

The problem is, that any signs of reaction have taken place after more than 3 hours

I have discussed the issue with my chemistry teacher, and he said that the trichloroacetic acid had been of poor quality (made 1990, majority of the original crystals dissolved, turned to brownish liquid). I also think could be simply stopped by the forming of salt (sodium trichloroacetate) and that being unable to interact with anthranilic acid.

I'm very desperate about this. Do you have any thoughts on how to get that reaction going?

Trichloroacetic acid was definitely the wrong way to go.

The reason is the chloroacetic acid is alkylating your anthranilic acid, adding the missing carbons Ben Norris comments on.  Ttrichloroacetic acid is not a great alyklating agent, owing to steric bulk.

If you are dead set against working it up and separating your anthranilic acid, you might try just adding the chloroacetic acid.

Otherwise clean it up and start over.


Why does copper precipitate out as a pink solid? When copper is precipitated out of a solution e.g from CuI, why does it appear pink instead of copper-coloured? Are there other situations where copper appears pink in solid form?

What you describe as pink is the actual colour of copper. The colour of a clean, solid surface of high-purity copper is typically salmon red. In your case, the appearance may be further affected by the particle size of the precipitate.

The familiar reddish-brown colour of copper is caused by copper(I) oxide, which is slowly formed on the metal surface.


Reaction of CpFe(CO)2Cl with phenyllithium The reaction of $\ce{(C5H5)Fe(CO)2Cl}$ with an equimolar amount of $\ce{Li+[C6H5]-}$ gives a product. Draw its structure and apply the 18 electron rule.

In a question I was given the molecular weight and percentages of each element in the compound. I worked out that there is $\ce{1 Fe}$, $\ce{13 C}$, $\ce{10 H}$ and $\ce{2 O}$.

The product seems to be an iron atom surrounded with ligands, $\ce{2 CO}$ ligands, $\ce{1 C5H5}$ and $\ce{1 [C6H5]-}$ ion.

The 18 electron rule expands to 8 valence electrons from $\ce{Fe}$, 2 electrons from each $\ce{CO}$ ligand, 5 electrons from $\ce{C5H5}$ and 1 electron from the $\ce{[C6H5]-}$ anion: $8+4+5+1 = 18$. But what would the structure look like?  Especially how is the $\ce{[C6H5]-}$ anion connected?

http://voh.chem.ucla.edu/vohtar/fall06/classes/277/pdf/18-Electron%20Rule.pdf
Summing ligand electron contribution.
DOI:10.1006/jmsp.2001.8470
http://www.ccp14.ac.uk/ccp/web-mirrors/farrugia/~louis/chem2x/Organomet-2.pdf
 Document p. 34.  Note error in drawing the graphic.

Direct displacement gives $\ce{CpFe(CO)2Ph}$, with the phenyl being a simple anion sigma-bonded to the metal. If there are two carbons plus two oxygens in the product, implying the CO ligands are intact, your are done.  If some fancy stiff happened with those ligands and the phenyl anion, you must work it out or look it up.


Boil to dry 50 liters water I would like to find out how long it would take to completely turn 50 litres of water to steam. Let's assume power of 2500 W and the water is already at boil. I am not very scientific so please forgive any omissions. Thanks

What actually happens in real life depends on a lot of things. Factors like the shape of the pot can make a big difference to how much of that 2500 W actually goes into vaporizing the water and how much of it is lost to the surroundings. If we assume that all 2500 W is going into the water, it makes things a lot simpler. If the water is already at the boiling point, then all we have to do is calculate the total amount of energy needed to vaporize all the water:

The enthalpy of vaporization of water is 40.65 $\mathrm{kJ/mol}$, which means 40.65 kJ of heat needs to be added to turn one mole of liquid water to steam. 50 L of water is 2775 moles:
$$\mathrm{\frac{(50\ L)\left(1000\frac{g}{L}\right)}{\left(18.02\frac{g}{mol}\right)}=2775\ mol}$$

So the total heat we need is:
$$\mathrm{(2775\ mol)\left(40.65\frac{kJ}{mol}\right)=112.8\ MJ}$$

1 watt is one joule per second, so for a 2500 W heat source:
$$\mathrm{\frac{112.8\ MJ}{2500\ J/s}=45\ ks}$$
Or, approximately 12.5 hours. In real life, it may be significantly longer for various reasons, but this is the fastest it could go.


Why is ICl5 unstable? For the question

  Why is $\ce{ICl5}$ unstable?

my textbook gave an answer of

  $\ce{ICl5}$ has too many highly electronegative chlorine atoms present on the central iodine.

But how does having electronegative atoms around a central atom make a molecule unstable? For example, $\ce{SF6}$ is very stable.

The textbook answer must be wrong, since $\ce{IF5}$ is a sufficiently stable compound and fluorine is more electronegative than chlorine.

Rather, I think that the reason for $\ce{ICl5}$’s lesser stability is the lower electronegativity of chlorine. Creating this type of hypercoordinate compounds requires two sets of four-electron three-centre bonds of the $\ce{X- \bond{...}{I+}-X <-> X-{I+}\bond{...}X-}$ type. This type of bond requires a sufficient electronegativity on the outer atoms to stabilise the partial negative charge better — something fluorine is more capable of than chlorine. $\ce{ICl5}$ should only be possible because of iodine’s low electronegativity and $\ce{BrCl5}$ should be orders of magnitude less stable again.

A second aspect could be the larger size of chlorine atoms compared to fluorine atoms causing greater steric stress, longer bonds and therefore weaker interactions.


Szabo and Ostlund's calculation of repulsion integral on page 172 via A.41 Returning to Szabo & Ostlund after a hiatus I tried to recreate his STO-3G calculation for HeH+ which begins on p. 168. It all went fine except for the two-electron, two-center integrals. I employed Szabo's values for these on p. 172 and was able to get his result for the energy and so on.
What I was not able to do is duplicate his p. 172 values for the 2-e integrals from his expression on page 416 (A.41). For example, on page 172, he gives $(\phi_{1}\phi_{2}|\phi_{1}\phi_{2}) = .1773.$
There are other approaches to this problem and I have looked at a few, but I would like to understand it as Szabo presents it. Because I was able to get the nuclear repulsion terms and so on I think my expressions for the contracted gaussians are fine but am misinterpreting some aspect of (AB|CD) in A.41.
If anyone is familiar with this particular calculation, and could show the proper calculation of (say) (12|12)...This is a very picayune question in the sense that I am interested in the particular calculation in Szabo: how to get 0.1773 on p. 172? Thanks for any illumination.
Edit
The Fortran program alerted me to a confusion. The "contracted" gaussians, using (11/11) as an example, look like:
$$G_a= .444635(2a_1/\pi)^{3/4}e^{-a_1|r-R_A|^2}+.535328(2a_2/\pi)^{3/4}e^{a_2|r-R_A|^2}+.154239(2a_3/\pi)^{3/4}e^{a_3|r-R_A|^2}$$
Regardless of how we do the integral it is:
$$\int G_a^2 \frac{1}{r_{1,2}}G_a^2$$ and we are raising the coefficients associated with $G_a$ to the 4th power and they come out of the integral. If we could do this integral in the usual naive way using linearity we would have 81 terms (some repeated) involving constants like:
$$(d_id_jd_kd_l)(2a_i/\pi)^{3/4}(2a_j/\pi)^{3/4}(2a_k/\pi)^{3/4}(2a_l/\pi)^{3/4} $$ where $i,j,k,l$ range from 1 to 3 respectively, where the $d_i$ are the coefficients $0.444365,$ etc.
That is why on p. 420 in the fortran program the quantities D1 include weights as factors, and a product of four D1's are pre-multiplied on p. 421 by the constants calculated in "TWOE" on p. 423. Using Fourier transforms to do the integral does not change the constants.
In short, the program is doing a lot of work here. It should still be possible to exhibit (11/11) easily but no wonder A.41 is not used in practice--it represents no economy over the fortran.

Restricting attention to the integrals representing "self-interaction " of helium in HeH+ we have to find coefficients of the integral,
$$\int A\cdot A\frac{1}{r_{12}}\cdot A  \cdot A$$
in which for an STO-3G calculation $A=a\cdot g_1+b\cdot g_2+c\cdot g_3$ with $g_1$ a gaussian and $a$ a coefficient formed as on page 153 sec. 3.203. The coefficients will fall out as a sum of 81 terms, some of which are identical:
$$(a+b+c)^4 = a^4+4a^3b+6a^2b^2+4ab^3+b^4+4a^3c+12a^2bc+12ab^2c+4b^3c+6a^2c^2+12abc^2+6b^2c^2+4ac^3+4bc^3+c^4.$$
That there are 81 terms may be checked by viewing the fortran at page 420 and 421. The expression for V1111 is formed by pre-multiplying by D(i)D(j)D(k)D(l) in which i,j,k,l range from 1 to 3. The coefficients DDDD do not appear explicitly in A.41.
The 12 terms in $ab^2c$ and so on may represent different quantities, and care must be taken to keep the accounting correct because (using notation defined below)
$$12ab^2c\cdot 2\pi^{5/2}/((a_1+a_2)(a_2+a_3)\sqrt{a_1+a_2+a_2+a_3})$$
$$\neq 12ab^2c\cdot 2\pi^{5/2}/((a_2+a_2)(a_1+a_3)\sqrt{a_1+a_2+a_2+a_3}) $$
Letting $\zeta_1 = 2.0925$ (page 170) we have:
$a_1=0.109818\cdot\zeta_1^2,~~ a_2=0.405771\cdot\zeta_1^2,~~ a_3=2.22766\cdot\zeta_1^2; $
$a =0.444635\cdot (2a_1/\pi)^{3/4},~b=0.535328\cdot(2a_2/\pi)^{3/4},~ c= 0.154329\cdot(2a_3/\pi)^{3/4}$
The sum represented in the appendix A.41 for the 2-center 2-electron integral
$$ \int He \cdot He \frac{1}{r_{12}}\cdot He\cdot He$$ is given below, and in Mathematica I get 1.307238. Substituting in the constants for hydrogen I get about 0.77466. These are Szabo and Ostlund's values on page 172. They are a good test because they avoid calculation of the error function used in cross terms and they are close to results for the integration of Slater orbitals using spherical coordinates, which are $(5/8)\cdot \zeta. $
The calculation exhibits the coefficients in A.41 explicitly and is in essence the fortran code for V1111.
$ a^4\cdot 2 \pi^{5/2}/((a_1 + a_1)\cdot(a_1 + a_1)\cdot\sqrt{4\cdot a_1})+$
$4 a^3b\cdot 2 \pi^{5/2}/((a_1 + a_1) \cdot(a_1 + a_2)\cdot\sqrt{3 a_1 + a_2})+ $
$ 2 a^2 b^2\cdot 2 \pi^{5/2}/((a_1 + a_1) (a_2 + a_2)\cdot\sqrt{2 a_1 + 2 a_2}) +$
$ 4 a^2 b^2\cdot2 \pi^{5/2}/((a_1 + a_2)\cdot(a_1 + a_2)\cdot \sqrt{2 a_1 + 2 a_2}) +$
$4 a b^3\cdot 2 \pi^{5/2}/((a_1 + a_2) (a_2 + a_2)\cdot \sqrt{a_1 + 3 a_2}) +$
$ b^4\cdot 2 \pi^{5/2}/((a_2 + a_2) (a_2 + a_2)\cdot \sqrt{2 a_2 + 2 a_2}) +$
$ 4 a^3 c\cdot 2 \pi^{5/2}/((a_1 + a_1)*(a_1 + a_3)\cdot \sqrt{3 a_1 + a_3}) +$
$ 4 a^2 b c\cdot 2 \pi^{5/2}/((a_1 + a_1)\cdot(a_2 + a_3)\cdot \sqrt{2 a_1 + a_2 + a_3}) +$
$ 8 a^2 b c\cdot 2 \pi^{5/2}/((a_1 + a_2)*(a_1 + a_3)\cdot \sqrt{2 a_1 + a_2 + a_3}) +$
$  4 a b^2 c\cdot 2 \pi^{5/2}/((a_1 + a_2) (a_2 + a_3)\cdot \sqrt{a_1 + 2 a_2 + a_3}) +$
$  8 a b^2 c\cdot 2 \pi^{5/2}/((a_2 + a_2) (a_1 + a_3)\cdot \sqrt{a_1 + 2 a_2 + a_3}) +$
$ 4 b^3 c\cdot 2 \pi^{5/2}/((a_2 + a_2)\cdot(a_2 + a_3)\cdot\sqrt{3 a_2 + a_3}) +$
$ 2 a^2 c^2\cdot 2 \pi^{5/2}/((a_1 + a_1)\cdot(a_3 + a_3)\cdot \sqrt{2 a_1 + 2 a_3}) +$
$ 4 a^2 c^2\cdot 2 \pi^{5/2}/((a_1 + a_3)\cdot(a_1 + a_3)\cdot\sqrt{2 a_1 + 2 a_3}) +$
$ 8 a b c^2\cdot 2 \pi^{5/2}/((a_1 + a_2)\cdot(a_3 + a_3)\cdot \sqrt{a_1 + a2 + 2 a3}) +$
$ 4 a b c^2\cdot 2 \pi^{5/2}/((a_1 + a_3)\cdot(a_2 + a_3)\cdot \sqrt{a_1 + a_2 + 2 a_3}) +$
$ 2 b^2 c^2\cdot 2 \pi^{5/2}/((a_2 + a_2)\cdot(a_3 + a_3)\cdot \sqrt{2 a_2 + 2 a_3}) +$
$ 4 b^2 c^2\cdot 2 \pi^{5/2}/((a_2 + a_3)\cdot(a_2 + a_3)\cdot \sqrt{2 a_2 + 2 a_3}) +$
$  4 a c^3\cdot 2 \pi^{5/2}/((a_1 + a_3)\cdot(a_3 + a_3)\cdot \sqrt{a_1 + 3 a_3}) +$
$ 4 b c^3\cdot 2 \pi^{5/2}/((a_2 + a_3)\cdot(a_3 + a_3)\cdot \sqrt{a_2 + 3 a_3}) +$
$ c^4\cdot2 \pi^{5/2}/((a_3 + a_3)\cdot(a_3 + a_3)\cdot \sqrt{4 a_3})$
Some copy-and-paste Mathematica code --don't know if it works in Wolfram Alpha.
z11 = 2.0925; z21 = 1.24;

a1 = .109818*z11^2; a2 = .405771*z11^2; a3 = 2.22766*z11^2;

a = .444365 (2 a1/Pi)^(3/4); b = .535328 (2 a2/Pi)^(3/
     4); c = .154329 (2 a3/Pi)^(3/4);

a^4*2 Pi^(5/2)/((a1 + a1)*(a1 + a1)*Sqrt[4 a1]) +
 4 a^3 b*2 Pi^(5/2)/((a1 + a1) (a1 + a2)*Sqrt[3 a1 + a2]) +
 2 a^2 b^2*2 Pi^(5/2)/((a1 + a1) (a2 + a2)*Sqrt[2 a1 + 2 a2]) +
 4 a^2 b^2*2 Pi^(5/2)/((a1 + a2)*(a1 + a2)*Sqrt[2 a1 + 2 a2]) +
 4 a b^3*2 Pi^(5/2)/((a1 + a2) (a2 + a2)*Sqrt[a1 + 3 a2]) +
 b^4*2 Pi^(5/2)/((a2 + a2) (a2 + a2)*Sqrt[2 a2 + 2 a2]) +
 4 a^3 c*2 Pi^(5/2)/((a1 + a1)*(a1 + a3)*Sqrt[3 a1 + a3]) +
 4 a^2 b c*2 Pi^(5/2)/((a1 + a1)*(a2 + a3)*Sqrt[2 a1 + a2 + a3]) +
 8 a^2 b c*2 Pi^(5/2)/((a1 + a2)*(a1 + a3)*Sqrt[2 a1 + a2 + a3]) +
  4 a b^2 c*2 Pi^(5/2)/((a1 + a2) (a2 + a3)*Sqrt[a1 + 2 a2 + a3]) +
  8 a b^2 c*2 Pi^(5/2)/((a2 + a2) (a1 + a3)*Sqrt[a1 + 2 a2 + a3]) +
 4 b^3 c*2 Pi^(5/2)/((a2 + a2)*(a2 + a3)*Sqrt[3 a2 + a3]) +
 2 a^2 c^2*2 Pi^(5/2)/((a1 + a1)*(a3 + a3)*Sqrt[2 a1 + 2 a3]) +
 4 a^2 c^2*2 Pi^(5/2)/((a1 + a3)*(a1 + a3)*Sqrt[2 a1 + 2 a3]) +
 8 a b c^2*2 Pi^(5/2)/((a1 + a2)*(a3 + a3)*Sqrt[a1 + a2 + 2 a3]) +
 4 a b c^2*2 Pi^(5/2)/((a1 + a3)*(a2 + a3)*Sqrt[a1 + a2 + 2 a3]) +
 2 b^2 c^2*2 Pi^(5/2)/((a2 + a2)*(a3 + a3)*Sqrt[2 a2 + 2 a3]) +
 4 b^2 c^2*2 Pi^(5/2)/((a2 + a3)*(a2 + a3)*Sqrt[2 a2 + 2 a3]) +
  4 a c^3*2 Pi^(5/2)/((a1 + a3)*(a3 + a3)*Sqrt[a1 + 3 a3]) +
 4 b c^3*2 Pi^(5/2)/((a2 + a3)*(a3 + a3)*Sqrt[a2 + 3 a3]) +
 c^4*2 Pi^(5/2)/((a3 + a3)*(a3 + a3)*Sqrt[4 a3])


Evaporating the water of a solution of only carbonate ions and water gives? I have a solution with only $\ce{CO_3^{2-}}$ ions and water.

What would happen with the carbonate ions if I were to evaporate the water of that solution (in a vacuum)?

I would love to add some quantum chemical bone to the stew we are trying to cook here.

Let us assume that it is possible to have a solution of carbonate ions, like already discussed by Nicolau in the comments. Then Rauru is also right that you will have to consider several equilibria:
\begin{aligned}\ce{
(1) && 2H2O &<=> H3+O + {}^{-}OH \\
(2) && CO3^{2-} + H2O &<=> HCO3- + {}^{-}OH\\
(3) && HCO3- + H2O &<=> H2CO3 + {}^{-}OH
}\end{aligned}

Now $(3)$ seems a little far fetched, so we will ignore it.

When you do now remove water from the equilibrium through (gentle) evaporation, you will have a mixture of only $\ce{CO3^{2-} + HCO3- + {}^{-}OH}$. Further assuming this is possible you will always have some collisions within this ensemble. Statistical it is therefore always possible to also remove $\ce{HCO3- + {}^{-}OH}$ via $(2)$ from the system. Also single entities of carbonate ions will repel each other as states by G M. So basically it all boils down to on question: Is carbonate stable in vacuum? What happens to carbonate if you put it into vacuum?

Now we have some quite good approximations to these particular conditions (gas phase and 0K), but I do not want to go into detail. I did a calculation on the carbonate with Gaussian09 and as a cheap shot BP86/cc-pVTZ level of theory. The only thing I want to show is the highest occupied molecular orbital.

What can clearly see is that this is no bonding orbital. There are lone pair contributions from all Oxygen atoms (due to imposed symmetry constraints). In addition to the high negative charge, electrons will not be stabilized by the field of the nuclei. Also, Orbitals only vanish at infinity, giving electrons a non-zero probability to actually leave the compound. Furthermore, there are two electrons in this orbital, so the spins of these are paired, which also costs energy.

My suspicion therefore is that the carbonate at least loses one electron. To where? To anywhere. (Also mentioned by Nicolau in the comments.) I do not have any actual proof for this, as it quite hard to find people actually doing experiments and/ or calculations on that topic. I remember that one question of a Professor of mine in the first semester was: "What happens to sulphate if you shoot it into space?" The answer was given without any further explanation and I never bothered to ask: "It looses an electron."

Note:

Another point that should always be considered when dealing with gas phase reactions is, that when ions have enough kinetic energy they might react in the strangest ways. One of them might be
$$\ce{2CO3^{2-} -> 2CO2- + O2^{2-}}.$$

Also collision with any particle may cause electrons to be transferred.


Is it possible to calculate energy of electron in any orbital and atom in Schrödinger wave model theory? Is it possible to calculate the energy of electron in any orbital and atom in the Schrödinger wave model theory? If so, how? E.g. energy of the $3s^2$ electron of the $\ce{Na-}$ ion.

It is surely possible. Note though, that for many-electron systems orbital picture is an approximate description, but that is an approximation on which the whole general chemistry is based on, so that is fine from that perspective. The approximation is known as the Hartree-Fock (HF) model, and thus, to get the energies of electrons occupying some orbitals all you need to do is to solve the Schrödinger equation in this approximation.

As a quick exercise one can do HF calculation in, say, Gaussian, with the following input file:

#P HF/aug-cc-pVTZ Pop=Full

Na-

-1 1
 Na                 0.00000000    0.00000000    0.00000000

to get this picture of occupied orbitals:

The HOMO with the energy of $-0.01288 \, \mathrm{Hartree} = -0.3505 \, \mathrm{eV}$ is basically the $\mathrm{3s}$ orbital you're looking for (trust me). Note, however, that this energy is approximate since the basis set (aug-cc-pVTZ) is finite. We could do better than that, but that is a different story.

In response to permeakra (since he does not trust his colleagues) I quote an authoritative reference in the field which explicitly talks about individual electrons occupying individual spin-orbitals and orbital energies. As a reference I choose the infamous book entitled "Modern Quantum Chemistry" written by Attila Szabo and Neil S. Ostlund.

Quote #1 (Szabo & Ostlund, p. 50)

  This Slater determinant has $N$ electrons occupying $N$ spin orbitals
  $(\chi_i, \chi_j, \dotsc, \chi_k)$ without specifying which electron
  is in which orbital.

Quote #2 (Szabo & Ostlund, p. 54)

  $$   f(i) \chi(x_i) = \varepsilon \chi(x_i) \tag{2.52} $$ [...]

  The
  solution of the Hartree-Fock eigenvalue problem (2.52) yields a set
  $\{\chi_k\}$ of orthonormal Hartree-Fock spin orbitals with orbital
  energies $\{\varepsilon_k\}$.

Below I also quote what Szabo & Ostlund have to say about Koopmans' theorem, because I already know that won't trust me.

  Quote #3 (Szabo & Ostlund, p. 110)

  The first theorem [Koopmans' theorem] constitutes an interpretation of
  the Hartree-Fock orbital energies as ionization potentials and
  electron affinities.

See, orbitals with their energies & occupancies do exist in the Hartree-Fock theory. It is the interpretation of orbital energies as ionization potentials which requires an additional theorem.


IUPAC naming of isocyanides What is the IUPAC name of isocyanide : isonitrile or carbylamine?

Are "methyl isocyanide" and "carbylamine" the same thing?

The most recent IUPAC recommendations[1] states that:

  P-61.9: Preferred IUPAC names are formed substitutively using the prefix ‘isocyano’ attached directly to a parent hydride.

This is a change from previous recommendations, and differs from some of your examples above.

To give a concrete example, $\ce{C6H5NC}$ should be named isocyanobenzene - this is known as a PIN by IUPAC (preferred IUPAC name). Phenylisocyanide is also acknowledged as an acceptable trivial name (a name that does not conform to the recommendations but is still commonly used).

[1]: IUPAC Recommendations for Preferred Names 2013, Chapter P-6. DOI:10.1039/9781849733069-00648


Enantiomers or Identical Compounds? I don't know why my solutions manual says these are enantiomers:

The Fisher projections I drew suggest that these are identical compounds.

The solutions manual is correct, they are enantiomers.  The compound on the left is (R)-2-chloro-(S)-3-butanol and the compound on the right is (S)-2-chloro-(R)-3-butanol.  Try to use the "R\S" nomenclature system on the 2 chiral carbons, that should help figuring out whether they are identical [(R,R and R,R), (S,S and S,S), (R,S and R,S), (S,R and S,R)] enantiomers [(R,R and S,S), (R,S and S,R), (S,R and R,S)] or diastereomers [(R,R and R,S), (R,R and S,R), (S,S and S,R), (S,S and R,S)].