Why don't electrons simply stick to the nucleus on account of being *oppositely* charged, instead of moving around it? Why don't electrons stick to the nucleus on account of being oppositely charged charged (the nucleus being positively charged)? But contrary to my intuition, electrons seem to follow a whole bunch of wacky paths (orbitals) around the nucleus (as demarcated by orbital wave-functions). Why is this?

Do note:

I'm not asking "What keeps the electron in orbit". All I want to know, is why electrons don't simply find themselves attached to the nucleus, and why do they even bother moving around the nucleus (in orbitals) at all? My textbooks, unfortunately, make no reference to this issue...they wholeheartedly adopt the "current" QM model of the (hydrogen) atom.

Richard Feynman alludes to this exact problem in the first volume of the Feynman Lectures on Physics. I'll present his argument here (based on the Uncertainty Principle), albeit, in my own words ;)

The German physicist [since he dealt with atomic chemistry/physics...by all means, go ahead and call him a "chemist" ;) ], Werner Karl Heisenberg came up with what we now know as "The Heisenberg Uncertainty Principle"... a revolutionary idea that points to an "inherent fuzziness", that exists within quantum systems (which is a fancy word that refers to any region of space that is sufficiently "small" enough to introduce wave-particle duality), and becomes apparent when we try to "measure" the various parameters that constitute them.

Basically, what (one version of) the Uncertainty Principle states is that:

  The product of the uncertainty in measurement of velocity and the uncertainty in measurement of the position of a particle can never be less than a certain constant; i.e- ħ/2

Rephrasing this

  If we were to know a particle's position to a very high precision (i.e- small uncertainty in measurement of position) then the corresponding uncertainty in the measurement of the particle's velocity (or momentum, if you know its mass) increases greatly. The same holds true vice versa.

In other words,

  You cannot know both a particle's position and momentum in a quantum system to a great precision simultaneously.

Here's a good analogy for this (Part 3: Atomic Structure - Analogy with a ballon)

From this point on, I guess the question can best be dealt with by "proof by contradiction".

  First part of the question: Why don't electrons stick to the nucleus?

We know for a fact, that Heisenberg's Uncertainty Principle holds true (yeah, I didn't prove it...but for now, just take my word for it).

So, let's assume that electrons do stick to the nuclei of their respective atoms. In that event, we would know the electrons' positions and momenta (with respect to the nucleus) with almost absolute certainty...something that Heisenberg's Uncertainty Principle does not allow.

Seeing that this situation is impossible, it is obvious that our little "assumption" must be wrong (Knowing the the Uncertainty principle is indeed, correct).

So yeah, electrons don't just go and simply "stick" to atoms.

  Second part of the question: Why do electrons follow a whole bunch of wacky paths?

The Dane, Niels Bohr postulated that electrons move in certain, fixed orbits around a nucleus...a fairly convincing (but not very accurate) model that is still often brought up in high-school and under-grad physics and chemistry courses. The Bohr Model was developed before the Uncertainty Principle.

But electrons do not move in fixed orbits ...the Uncertainty Principle triumphs again!

So if electrons don't stick to the nucleus, and if they don't move in fixed orbits around the nucleus, then where exactly are they?

We can never exactly point out an electron's position in an atom. However, we can determine the probability of finding electrons in a given region of space around the nucleus. These regions of space with a very high electron probability density are called orbitals.

This answer has been greatly over-simplified. Thought it does deal with the gist of the topic, I'm personally looking forward to seeing more detailed answers myself.


Given an unbalanced chemical equation, is it possible to determine the largest possible coefficient that could be used in the balanced equation? I am writing a program to balance chemical equations, and would love to know if there is an algebraic way to determine the greatest possible coefficient that I would need to check.

In short

Just use matrices.

In long

Consider this equation to be balanced:

$$\ce{CuSCN + KIO3 + HCl -> CuSO4 + KCl + HCN + ICl + H2O}$$

(source with spoiler)

We are trying to solve this equation:

$$\ce{aCuSCN + bKIO3 + cHCl -> dCuSO4 + eKCl + fHCN + gICl + hH2O}$$

Now, we move every term to the left hand side (warning: handwavey chemistry):

$$\ce{aCuSCN + bKIO3 + cHCl - dCuSO4 - eKCl - fHCN - gICl - hH2O = 0}$$

Transform into matrix form (the element names are just for reference):

$$
\begin{matrix}
\ce{Cu}\\
\ce{S}\\
\ce{C}\\
\ce{N}\\
\ce{K}\\
\ce{I}\\
\ce{O}\\
\ce{H}\\
\ce{Cl}
\end{matrix}
\left(
\begin{matrix}
1 &0 &0 &-1&0 &0 &0 &0 \\
1 &0 &0 &-1&0 &0 &0 &0 \\
1 &0 &0 &0 &0 &-1&0 &0 \\
1 &0 &0 &0 &0 &-1&0 &0 \\
0 &1 &0 &0 &-1&0 &0 &0 \\
0 &1 &0 &0 &0 &0 &-1&0 \\
0 &3 &0 &-4&0 &0 &0 &-1\\
0 &0 &1 &0 &0 &-1&0 &-2\\
0 &0 &1 &0 &-1&0 &-1&0
\end{matrix}
\right)
\left(
\begin{matrix}
a\\
b\\
c\\
d\\
e\\
f\\
g\\
h
\end{matrix}
\right)
=
\left(
\begin{matrix}
0\\0\\0\\0\\0\\0\\0\\0\\0\\
\end{matrix}
\right)
$$

Notice that the biggest matrix is not a square. You must transform it to a square. If the elements are in excess (just like this case), add variables; otherwise, add elements. Set the extra variables/elements to zero:

$$
\begin{matrix}
\ce{Cu}\\
\ce{S}\\
\ce{C}\\
\ce{N}\\
\ce{K}\\
\ce{I}\\
\ce{O}\\
\ce{H}\\
\ce{Cl}
\end{matrix}
\left(
\begin{matrix}
1 &0 &0 &-1&0 &0 &0 &0 &0 \\
1 &0 &0 &-1&0 &0 &0 &0 &0 \\
1 &0 &0 &0 &0 &-1&0 &0 &0 \\
1 &0 &0 &0 &0 &-1&0 &0 &0 \\
0 &1 &0 &0 &-1&0 &0 &0 &0 \\
0 &1 &0 &0 &0 &0 &-1&0 &0 \\
0 &3 &0 &-4&0 &0 &0 &-1&0 \\
0 &0 &1 &0 &0 &-1&0 &-2&0 \\
0 &0 &1 &0 &-1&0 &-1&0 &0
\end{matrix}
\right)
\left(
\begin{matrix}
a\\
b\\
c\\
d\\
e\\
f\\
g\\
h\\
i
\end{matrix}
\right)
=
\left(
\begin{matrix}
0\\0\\0\\0\\0\\0\\0\\0\\0\\
\end{matrix}
\right)
$$

Note: $i$ must be zero.

Now you can use any algorithm for solving homogeneous square matrix equations to continue.


When we cannot use Friedel-Crafts acylation to insert alkyl group? Questions:

Show a probable synthethic pathway to obtain the first molecule, I.
Will the same strategy be used to insert the alkyl group for J. Explain.

My answers (please tell me if there is any incorrect answer):

Friedel-Crafts acylation -> Clemmenson reduction -> Nitration -> Reduction of Nitro Benzene

No, to insert alkyl group for molecule J, it must undergo Friedel-Crafts alkylation. In Friedel-Crafts alkylation, there is carbocation rearrangement to ensure the carbocation is stable and tertiary. In J, the carbon that will directly attach to benzene is already stable, so the C(CH3)3 substituent will attach directly to benzene. (I'm not sure what to explain more)

Is this correct for alkylation step in J?

My questions:

For no. 2, is it true if we say that, in this case, no rearrangement will occur in Friedel-Crafts Alkylation step because the carbocation is tertiary and stable so it can proceed with this step (Friedel-Crafts Alkylation). Is it possible that rearrangement will not occur in Friedel-Crafts Alkylation?
Why we cannot use Friedel-Crafts acylation to insert alkyl group in J? I am confused, if we use (CH3)3Cl, there will be no carbocation rearrangement, but why the question asked about acylation, I think it is not possible to undergo acylation step because there is no carbocation rearrangement, so the substituent attached to J will be CH3CH2CH2???

tBu-benzene is prepared exactly as you describe, see this preparation here. Follow this by nitration and reduction.
There is no obviously comparable strategy of acylation that can compete with this. Making a tBu group from acetophenone is not straightforward and is illogical when there is a perfectly good direct route.


What is the structure of powdered sulfur? Quoting this paper:

  A variety of allotropic states of elementary sulfur have been identified.
  Solid crystalline sulfur exists either as rings of 6-12 sulfur atoms (cyclohexasulfur, $\ce{S6}$, cyclooctasulfur $\ce{S8}$, etc. or as chains of sulfur atoms (catenasulfur $\ce{S_{\infty}}$).

As said on the same paper, Alpha-sulfur (orthorombic - $\ce{S_{\alpha}}$) is the thermodynamically stable form under ordinary conditions.

What happens with powdered sulfur (also called "dusting sulfur", normally used as fungicide)? Are sulfur atoms still organized in rings or chains?

Can we associate a precise crystalline phase to powdered sulfur?

Powdered sulphur is produced by grinding raw sulphur as it is mined. According to Wikipedia, mining is done as follows:

  On Earth, elemental sulfur can be found near hot springs and volcanic regions in many parts of the world, especially along the Pacific Ring of Fire; such volcanic deposits are currently mined in Indonesia, Chile, and Japan. Such deposits are polycrystalline, with the largest documented single crystal measuring 22×16×11 cm.[21] Historically, Sicily was a large source of sulfur in the Industrial Revolution.[22]

Raw sulphur is polycrystalline (meaning many crystals grown together). So basically, powdered sulphur is the same as crystalline sulphur just as very small crystals. Consequently, what you found in that paper for crystalline sulphur also applies to powdered sulphur.


Simplest way to balance any chemical equation In your opinion, what would be the simplest way to balance any size of the chemical equation?

For example here is an equation:  $$\ce{C12H26 + O2 -> CO2 + H2O}$$

I saw some ways by putting letters in front of each molecule, but I just don't get too much how to do this with this method! My teacher is always speaking about another method, but it simply doesn't feel logic/mathematics is enough for me.

You can do this in the 'guided trial-and-error' method that LordStryker showed which is probably quickest for simple reactions, or approach it in a purely mathematical fashion which is the method I will explain. This method works well for arbitrarily difficult reactions.

Your chemical equation contains 3 atomic species: $\ce{C}$, $\ce{H}$ and $\ce{O}$. This means that you need 3 equations to balance. First I write the chemical reaction as follows:
$$\ce{a C12H26 + b O2 -> c CO2 + d H2O}$$
Now I will write the 3 equalities for the 3 atomic species that we have:
\begin{align}
12a &= c \tag{for C}\\
26a &= 2d \tag{for H}\\
2b  &= 2c + d \tag{for O}
\end{align}
The numerical constants in front of $a$, $b$, $c$ and $d$ come from the number of atoms that are in the molecule for which they are included.

As you may have noticed, this system of equations is ill-defined, because we have 4 unknowns ($a$ to $d$) and only 3 equations. The reason is that we can pick any multiple of the equation without it becoming incorrect,1 just against convention. This means that we can simply set one of the unknowns to any value. Let's say $a = 1$ for now and revisit this choice after solving the equations.

With $a = 1$ it immediately follows that $c = 12$ and $d = 13$. From those two it then follows that $b = 18.5$. Our balanced reaction becomes:2
$$\ce{C12H26 + 18.5 O2 -> 12 CO2 + 13 H2O}$$

In principle this is already correct, but convention is to have only integer numbers in the equation (i.e. no decimal points) therefore we need to revisit our choice of $a = 1$ and pick it such that all numbers will become integers. In this case you can easily see that this will happen for $a = 2$ which then results in
$$\ce{2C12H26 + 37 O2 -> 24 CO2 + 26 H2O}$$

If for some reason it is not easy to see which value for $a$ you would need then you can multiply by a large power of 10 to make all the decimals disappear and then check the greatest common divisor of the 4 numbers and divide by that to obtain the same equation.

Notes

Compare $10=10$ to the multiplied version $30=30$, both are correct, but they are just different by a factor $3$.
I'm not sure whether you are familiar with linear algebra, but if you are you will probably have noticed that the set of equations is a linear set so you could solve it through matrix manipulations which makes this method applicable to arbitrarily complex chemical reactions.


Limits on contaminant size in filtration for extraction, and standard alternatives? Following up on How to perform solvent extraction of a solid I realized that I haven't quantified the size of the contaminants.  I know they aren't soluble in acetone and the target is, but I don't know how small the contaminants are.

Once the target is dissolved presumably I must assume that suspended contaminants are as small as their molecular size.  In this case is a filter-based extraction still possible?  I don't see lab filters like extraction thimbles sold by filtration size.

In a case like this is the correct approach to run the solution through a centrifuge and then suck out the middle solution with a pipette?  (Or what is the lab best practice for mechanical extraction following spinning?)

There are two approaches regarding the size selectivity of solid to be filtered. Most common is some sort of "tortuous path", that is a fine weaving of some type of fiber having some average size cutoff regarding what gets filtered out and what passes through.  Secondly there are filters that are actually grids of a specific size of hole or square to restrict the solids from passing through. These look like a screen sieve through a microscope and I've seen them as small as 1 micron, though this is extremely small and usually used for sterilization.

In reality, you are more likely to have problems with your filter becoming clogged than with too much solids getting through. Your idea of centrifugation is good, but as a pre-filter step to help prevent clogging the filter.  In this case you could either decant or suction off the bulk of the liquid before the finer filter finish. The requirement for final filtration size is application dependent.


Can a organic compounds such as hydrocarbons contain an ionic bond? Can organic compounds like hydrocarbons have types of bonds other than covalent bonds? Can they also possess ionic bonds?

Azulene has a 1.05 debye dipole moment,

A hydrocarbon as such bonded as an isolated cation-anion pair with no heteroatoms could happen in principle given sufficient steric hindrance.   Penta-tert-butylcyclopentadienide or pentakis(2,4,6-tri-tert-butylphenyl)cyclopentadienide with like fully substituted cyclpropenium or tropylium might work.

Similar substitution on fulvalenes' four inner carbons would dihedral angle rotate the rings by steric hindrance.  This prevents the double bond from forming, giving sort of a sigma covalent plus pi-dative bond.


How hot will a carbon electrode get in an electrolysis setup? For the electrolysis of sulfuric acid, I plan to use carbon graphite electrodes at voltages up to 6 V. I would like to know if my carbon electrodes would get considerably hot or warm (45 °C or higher).

Supposing there is a powerful DC source, then quite hot, unless small enough electrode current density is managed.

6 V is quite high voltage for electrolysis. But it is possible the source will not be able to provide the current for the voltage 6 V, unless the electrode area is small.

Consider also the graphite electrodes may get deteriorated by the released gas, releasing small carbon particles to the solution (remembering my own first electrolysis experiments with carbon electrodes from old Leclanché batteries, when I was 12).


Melting of Covalent Networks When you melt a covalent network, you break intramolecular covalent bonds, is that not a chemical change?

Also, is ionisation a chemical or physical change?

A chemical change is defined as when your substance reacts with another substance to form some products, or decomposition into other products. In a chemical change, there will be formation of new chemicals that are different in terms of chemical formula.

Thus, breaking of intramolecular bonds will not be a chemical change in the case of C-C bonds in diamond/graphite, because there is no formation of new chemical substances with an alternative formula. Similarly, ionization (without formation of complex ions) would be physical change


Enolate formation and tosylate displacement in cis/trans decalin What will happen in the following reactions? This is Problem 3d from Chapter 6 of Zweifel and Nantz's Modern Organic Synthesis.

I thought that after the hydroxyl group is converted into a good leaving group ($\ce{OTs}$), an enolate would be formed which would attack $\ce{CH_2OTs}$, but I guess the fused cyclobutane thus obtained would be too strained. What else could happen?

I went looking for this question again today after talking to someone about a similar reaction, then realised I never got round to trying to actually answer it.

Leading reference: Tetrahedron 1969, 25 (21), 5267–5274.

In the first case, with the trans-fused ring system, we first get tosylation as expected. Using the bulky potassium tert-butoxide, we get enolate formation at the least hindered alpha position. This enolate is then setup, as can be seen in the conformational diagram below, to displace the tosylate forming the new ring. This is confirmed in the cited literature.

In the second case, with the cis-fused ring system, things are a little more complicated. Once again we get tosylation, however the the cis-fused product can be drawn in multiple 3D conformations, only one of which is productive (shown in orange below).

In this case, the steric argument for enolate formation doesn't hold, as the thermodynamic enolate is unproductive, and as such only the enolate formed at the most hindered position goes forward to form the product.

As you mentioned in your initial post, "the obtained structure would be too strained", which is what prevented me from thinking this pathway happened in the first place. However, looking at the original Tetrahedron paper, they do have evidence to back up the formation of these ring systems. One possible conclusion is that due to the conformational preferences of the 6,6-ring systems, the bonds are closer together than you'd maybe expect, meaning there isn't such a penalty to formation of the cyclobutane.


Anthracine blue composition In 1873 a German chemist named Springmühl announced a derivative of alizarin which acted as a blue dye and he sold this dye for a high price. The dye was sometimes called "anthracine blue" and Springmuhl kept the process for producing it a secret.

Has the composition of this dye ever been revealed?

From Chemical News, April 25, 1873:

  Anthracene Blue.— A few years since aniline was the great source of new and beautiful colours. Now that every possible shade of colour, surpassing in number and beauty the hues of the rainbow, have been produced from aniline, the chemist has taken up the study of anthracene and alizarine, also coal-tar products. While preparing artificial alizarine from anthracene, Springmühl has obtained a by-product, from which he has made a beautiful blue colour, superior in some respects to any of the aniline blues. The process by which it was prepared he keeps a secret. Dried in vacuo, it is a blue powder with a few little crystals. In this it differs from the aniline dyes, which are one colour when dry, another when in solution. When pure hot water is poured over anthracene blue it mostly dissolves, but leaves a little insoluble residue. The addition of an alkali destroys its colour, which is restored, however, by an acid. The strongest mineral acids are unable to destroy its colour, but rather heightens its tone. Unlike aniline dyes, it is insoluble in alcohol and ether. Experiments show that it withstands the action of light better than aniline blue. Unfortunately, it is at present very expensive, for Springmühl obtained but 2'5 grains of anthracene blue from 25,000 grains of anthracene, which makes it cost about 3000 dols, per pound at present. A cheaper method of making it is certainly desirable.—American Artizan.

Springmühl indeed reported Anthracene Blue (orig. "Anthracenblau" (German)) from an intermediate with brutto-formula $\ce{C14H7O3}$. It seems like this method only worked well at a lab scale.
According to Springmühl's Lexicon der Farbwaaren und Chemikalienkunde [1, p. 111-112], Anthracene Blue is obtained as a byproduct of oxidation of anthracene to anthraquinone by fuming nitric acid. When gas (hydrogen in original text) evolution stops, resulting solution is neutralized with potash and heated under pressure in a sealed glass tube for 8 hours. Dark-green precipitation is then filtered off, and deep-blue solution containing Anthracene Blue is then treated with sodium chloride to precipitate the product in a form of blue water-soluble powder. But, treated with tannic acid, Anthracene Blue forms dark-blue precipitate and cannot be dissolved neither in water, nor in alcohol; but slightly soluble in acidified solutions.
Original German text:

Springmühl, der Entdecker des Anthracenblau, stellte dasselbe zuerst dar, indem er eine Verbindung von der Zusammensetzung $\ce{C14H7O3}$, welche er bei der Oxydation des Anthracens neben Anthrachinon erhielt, mit rauchender Salpetersäure behandelte und auf das erhaltene Product Wasserstoff in statu nascendi einwirken liess, die resultirende Lösung mit Kali versetzte und unter Druck erhitzte. Es bildete sich eine tiefblaue Färbung in der zugeschmolzenen Glasröhre und nach 8stündiger Erhitzung fand sich ein dunkelgrüner fast schwarzer Niederschlag in derselben, während die Lösung tiefblau gefärbt war und das Anthrancenblau enthielt, es liess der Farbstoff aus filtrirter Lösung sich mit Chlornatrium niederschlagen und mit Gerbsäure konnte der in Wasser wieder gelöste Farbstoff völlig als dunkelblaues Pulver gefällt werden, welches weder in Alkohol noch Wasser, hingegen leicht in Säuren sich löste.

Later synthetic process was changed to adapt it for industrial production. Beilstein catalogue from 1925 lists Anthracenblau-WR as 1,2,4,5,6,8-hexahydroxy-9,10-anthraquinone $\ce{C14H8O8}$ [2, p. 569], which can be obtained from 3-hydroxybenzoic acid with concentrated $\ce{H2SO4}$ in the presence of $\ce{H3BO3}$ at $\pu{260 .. 280^\circ C}$. The product is then poured into water and boiled. Alternatively, 3,5-dihydroxybenzoic acid can be used. There are also plenty other synthetic techniques slightly deviated from the mentioned above.
These days Anthracenblau is also known as commercially available dye Anthracene Blue (S)WR.

Reference

Springmühl, F. Lexicon der Farbwaaren und Chemikalienkunde: practishes Handbuch für Kenntiniss und Prüfung aller in der Industrie und im Gewerbebetrieb verwendeten Farbstoffe und Chemikalien; Leipzig: Gustav Weigel, 1877. (in German, Archive.org)
Beilsteins Handbuch der Organischen Chemie, 4th Ed.; Prager, B., Jacobson, P., Schmidt, P., Stern, D., Eds.; Springer: Berlin, 1925; Vol. 8. (in German, Archive.org)


Is the d-block ever part of the valence electrons? The d-block sub-shell is always one shell lower than the s-block and p-block sub-shells. Assuming only the electrons in the highest energy shells count toward the set of valence electrons (is that correct?), d-block is never in the highest energy shell, and so none of the electrons in the d subshells would ever count toward the valence electrons.

I suspect I'm mistaken. Am I? If so, where have gone wrong?

The answer is merely a matter of definition of "valence electrons".

Many texts define "valence electrons" in a way that explicitly includes d electrons of unfilled (but not filled) d subshells.

For example: Chemistry structure and dynamics by Spencer et al. at page 125:[1]

  We can define valence electrons as electrons on an atom that are not present in the previous rare gas, ignoring filled d or f subshells.

Many books published in the last 10 years use this definition.

That d electrons may be valence electrons is also supported by the 18-Electron rule (at least to the extent that there is such a rule).

[1] James N. Spencer, George M. Bodner, Lyman H. Rickard: Chemistry: Structure and Dynamics, 5th Edition. John Wiley & Sons: 2010. ISBN: 978-0-470-58711-9


Cyclic Voltammetry - HOMO and LUMO levels I'm a physicist, so I apologize if these are obvious questions.

I've carried out CV measurements on a few different types of material (with a ferrocene reference). I am interested in determining the HOMO and LUMO levels of the material.

My CV measurements result in a current-voltage curve (IV curve). It has been explained to me I just need to measure the mV difference between the peak of the ferrocene and the oxidation or reduction curve, then subtract/add that onto the reference for ferrocene (4.8eV) to get the HOMO/LUMO levels.

Sounds simple, right?

But:

Do I measure to the onset of the oxidation/reduction curve or the
peak?
What justification do I use to convert the voltage to an energy? How can I say that a difference in mV corresponds to eV? Maybe this might make sense if it was a one electron process, but how do I know that?
Are these peaks even the HOMO/LUMO levels at all? In this eminent paper it warns me

  " Often cyclic-voltammetry-based ionization potentials and electron
  affinities are inappropriately referred to as "HOMO" and "LUMO"
  energies"

Yes, you can "convert" this way, but you're correct to be skeptical.

Let's start with interpreting the cyclic voltammetry curves themselves.

(figure from Wikipedia)

Note that the "peak" actually has two sides. When you sweep the potential with CV, the cathodic peak ($E_{pc}$) and the anodic peak ($E_{ac}$) won't exactly line up. If you have a well-behaved, reversible redox species, the two peaks will be separated by $E_{pa}-E_{pc}=\frac{56.5\text{ mV}}{n}$ for $n$ electrons.

  Do I measure to the onset of the oxidation/reduction curve or the peak

You do neither. Instead, you take $E^0 = (E_{pa}-E_{pc})/2$ - the average of the anodic and cathodic peaks. This is the standard potential for that redox reaction.

You mention that you're including ferrocene as an in-situ reference potential, so you would compare the $E^0$ potential of your redox couple to the $E^0$ of ferrocene.

  What justification do I use to convert the voltage to an energy? How can I say that a difference in meV corresponds to eV?

In principal, you're monitoring an oxidation or reduction. So you measure the oxidation potential relative to ferrocene.

The correspondence to the HOMO energy is based on an interpretation of Koopmans' theorem. In short, the oxidation potential involves removing an electron from the molecule, and the lowest ionization potential should come from the HOMO.

So there should be a correlation between the oxidation potentials from electrochemistry and the orbital energies. Some people take this as fact, but it's good that you're skeptical. I see many papers that do this "conversion" without any warning that these are different things.

You asked about units - as you guessed, the conversion between potential (V) and eV (for orbitals) occurs, because in most cases, one electron is transferred, and so the charge is known. In the general case, this is certainly a cause for concern.

  Maybe this might make sense if it was a one electron process, but how do I know that?

Heh. Generally, it's one. That's a good question in electrochemistry. There are multiple ways of determination. If you integrate the peaks, the area should correspond to the number of charges transferred. But usually, you don't oxidize or reduce the whole solution.

You can estimate the number of electrons based on the peak spacing, as indicated above $\frac{56.5 mV}{n}$.

You can change the scan rate and use the Randles-Sevcik equation

$$i_p = 268,600 \ n^{\frac{3}{2}} AD^{\frac{1}{2}} Cv^{\frac{1}{2}}$$

  Are these peaks even the HOMO/LUMO levels at all?

They are not. There should be a correlation between oxidation and reduction potentials and HOMO and LUMO energies.

But step back and think - in one case, you have electrochemistry in the electrolyte solution at an electrode surface.

Can the solvent change the observed redox potential? Yes.
Can the electrode change the observed redox potential? Yes.
Could the molecules aggregate and change the observed redox potential? Yes.
Could the oxidized or reduced species undergo reactions that change the observed redox potential? Yes.
Can the sweep rate change the observed redox potential? Yes.


Reactions between aqueous or solid LiOH I've been working on a project to absorb CO$_2$ from fossil fuel power plants.  I found these reactions:

$$\ce{2Li(s) + 2H2O(l) → 2 LiOH(aq) + H2(g)} \tag{1} $$

$$\ce{2 LiOH(s) + CO_2(g) → Li2CO_3(s) + H2O(l)} \tag{2}$$

The first one returns an aqueous solution of $\ce{LiOH}$, and the second one requires solid $\ce{LiOH}$. I was wondering if it's possible to use the aqueous solution in the second equation, or does it have to be solid?  Why?

it would actually produce a faster reaction to have the $\ce{LiOH}$ in a aqueous solution rather than having it as a solid. all you would do is bubble the $\ce{CO2}$ through the solution and it will go cloudy. you've probably done a similar thing but with $\ce{Ca(OH)2}$ to form calcium carbonate as a test for $\ce{ CO2}$.


Epoxide ring opening via decarboxylation The answer for the above multiple choice problem is option (C), from which I can understand that at some point in the mechanism the molecule must've undergone decarboxylation.

Also, I think the first step should be protonation of the epoxide oxygen. (No other choice, isn't that the most nucleophilic site?)

I'm not able to understand how the ring opening takes place (guessing it is via decarboxylation, but how?)

You are correct about protonating the epoxide of the glycidic acid 1 (by some other molecule of glycidic acid). Ring opening gives stable, tertiary cation 3 and NOT cation 6, which is destabilized by the carbonyl group. Collapse of 3 gives CO2 and the enol 4 which tautomerizes to the aldehyde 5.


Whose concentration is more 1 molar or 1 molal? I was reading about concentration terms like molarity, molality, etc. and then I came across a question: which concentration is more:

1M(molar) of solute x dissolved in water vs.
1m(molal) of solute x dissolved in water

When I Google it, I did not find exact answers but did find some related information.
My approach is:

Convert 1M into molality and then compare if it is less or more than 1m. (?)
The density of water is 1kg per L.
When solute is added to it, density would definitely increase(?)
And therefore 1L of that solution will weigh more than 1 kg.
now if we calculate molality of 1M solution it would be less than 1m . (?)
am I correct ??
Density would always(?) Increase after adding solute
**Hence 1m will always be more than 1M **

To convert from molal to molar, one generally needs to know the density of solution as well as the formula weight of solute.  Suppose there is a single solute $x$ in water, and that $x$ has a formula weight of $f_x$ g/mol.  Suppose also that the solution of $x$ has a density of of $\rho$.
Let the molarity of $x$ in the solution be $C_x$ mol/L.  Then:

the mass concentration of $x$ in g/L is $C_x f_x$.
the weight fraction of $x$ in grams of $x$ per total gram of solution is thus $w_x = \frac{C_x f_x}{\rho}$.  The weight fraction $w_x$ must be a number between 0 and 1.
The weight ratio of $x$ in grams of $x$ per gram of water is $\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}} = \frac{w_x}{1-w_x}$.
The molality is simply the weight ratio divided by the formula weight, $\frac{1000}{f_x}\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}}$.  The factor of 1000 is because molality is moles per kg of solvent, not moles per g.
That last expression can be written as  $\frac{1000}{f_x}\frac{\frac{C_x f_x}{\rho}}{1-\frac{C_x f_x}{\rho}} = \frac{1000}{f_x} \frac{w_x}{1-w_x}$

This last formula shows that molality values will always be higher than molarity values.  Because $w_x$ has to be between zero and 1, then the molality formula will divide by a smaller number than then molarity formula.  One molar water "dissolved" in water (say we are dissolving distilled water in tap water, or dissolving isotopically labeled water into regular water) has a molality of around 1.02.

My approach is:

Convert 1M into molality and then compare if it is less or more than 1m. (?)

Good approach.

The density of water is 1kg per L.

True, but as noted in the comments, the density of solutions in water can be either higher or lower than the density of pure water.

When solute is added to it, density would definitely increase(?)

Not necessarily true.

And therefore 1L of that solution will weigh more than 1 kg.

Not true, and this is the point where I think your logic mislead you.  To make one liter of a 1 molar solution, less than one liter of pure water is usually required.  This is because part of the liter of solution is taken up by solute.  So not all of the liter needs to come from pure water.  And since water has a density of 1 kg/L, if we need less than one liter of pure water, we need less than 1 kg of pure water.  This then means that our one mole of solute is being dissolved by less than kg of solvent: thus the molality of a one-molar solution is greater than one.
The neat thing is that this is true regardless of the density or formula weight of the species being considered.


1,6-dimethylcyclohexene and 2,3-dimethylcyclohexene Is there a difference between these two molecules?

If so how does one tell? What is the correct nomenclature for cycloalkenes? I know that the double bond is assumed to be between C1 and C2 but why can't we number a cyclohexene to get 2,3? Is it because a lower sum is not preferred but simply lower individual numbers if possible?

With regard to numbering of locants within the same compound class, multiple bonds have seniority over simple substituent groups (e.g. methyl, ethyl, etc.) that are used for naming of branched alkanes. On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (…)

Note that Rule e takes precedence over Rule f.

For example, a preferred IUPAC name is 3-bromocyclohex-1-ene (not 1-bromocyclohex-2-ene): (This example is given in the Blue Book.)

  P-14.3.5 Lowest set of locants

  The lowest set of locants is defined as the set that, when compared term by term with
  other locant sets, each cited in order of increasing value, has the lowest term at the first
  point of difference; (…)

Therefore, the structure given in the question is named as 1,6-dimethylcyclohex-1-ene rather than 2,3-dimethylcyclohex-1-ene since the locant set ‘1,6’ is lower than ‘2,3’.


Which overlap is stronger, and why? I was just wondering which overlap is stronger, 2s-3p or 3s-2p? I believe this can probably be explained by involving some sort of mathematics (?) which I haven't used yet.

Edit: Note that 2s-3p refers to the overlap of 2s of one atom and 3p of another atom. Similarly for 3s-2p.

I haven't really come across an example for this case, guessing it's hypothetical. It'd be great if someone could give actual examples too if they exist.

An atom's wavefunctions (orbitals) are orthogonal. This follows from the fact that the Hamiltonian is Hermitian, i.e eigenfunctions (wavefunctions) of quantum mechanical operators are orthogonal.  If they are on different atoms then they are not orthogonal and the overlap clearly depends on how far apart the atoms are.

Mathematically orthogonality means that (in one dimension) two wavefunctions $ \psi_i$ and $\psi_j$ satisfy the condition
$$\int_{-\infty}^{\infty} \psi_i\psi_jdx =\delta_{ij}$$

where $\delta_{ij} = 1 \text{ if } i=j \text { else } 0$ and is the Kroncker delta. (In this case if $i=j$ then the wavefunctions are also normalised as the product is $1$, this joint condition, orthogonal & normalised  is sometimes called 'orthonormal')

In the case of atomic orbitals there are three quantum number, the principle one n and angular ones l and m. The orthogonality condition is then
$$ \int_0^{\infty}r^2\int_0^{\pi}\sin(\theta) \int_0^{2\pi}\psi^*_{n'l'm'}(r,\theta,\phi)\psi_{nlm}(r,\theta,\phi)~dr ~d\theta ~d\phi=\delta_{nn'}\delta_{ll'}\delta_{mm'}$$

In your examples $n=2$ and $n'=3$ so the orbitals are orthogonal on these grounds alone.

As a check the 2s orbital has the form
$$ \psi_{2s} = A(2-\sigma)\exp(-\sigma/2) $$
where A contains only constants and $\sigma=Zr/a_0$ where Z is the atomic number, $a_0$ the Bohr radius and r the radial distance. One 3p orbital ($l=1, m=0$) has the form
$$\psi_{3p} = A'\sigma(6-\sigma)\exp(-\sigma/3)\cos(\theta)$$

Integrating the product over the angles $\theta$ which is  the integral $\int_0^{\pi}\cos(\theta)d\theta $. We can see that this integral is zero from a plot of cosine.

Edits:

As the question is now amended, the answer above holds for a single atom, but when there are two atoms then the overlap depends on their separation as the wavefunctions are no longer orthogonal because they have different coordinate origins, one on one atom the other on the other.

The overlap integral is written as
$$S= \int_{-\infty}^{\infty} \psi_i\psi_j~d\tau$$

where $\tau$ is the set of coordinates. Even in the simplest case of $\ce{H2^+}$ the coordinates needed are prolate spheroidal ones so to give an example it is simplest to use a one dimensional example since the principle is the same but technically more complex in 3D.

Starting with H atom 1s orbitals
$$ \psi_{1s} = a_0^{-1/2}\exp(-x/a_0)$$

if the atoms a and b are separated by 2R then the wavefunctions are

$$ \psi_a = a_0^{-1/2}\exp(-|x-R|/a_0) \\
\psi_b = a_0^{-1/2}\exp(-|x+R|/a_0)$$

and the overlap is the integral
$$ S=\int_{-\infty}^{\infty} \psi_a\psi_b dx =a_o^{-1}\int_{-\infty}^{\infty} \exp(-|x-R|/a_0)\exp(-|x+R|/a_0) dx$$

which evaluates to $$S=\frac{(a_0+2R)\exp(-2R/a_0)}{\pi a_0^3}$$

The wavefunctions with $a_0=1$, (and $R=1$ for the first two plots), $\psi_a$ and $ \psi_b$ vs. x  overlap $\psi_a.\psi_b$ vs. x and overlap integral S vs. the atom separation R are shown in the figures.


How exactly does electrolysis occur if the molar conductivities of cation and anion are different? I'm having trouble wrapping my head around Kohlrausch law of infinite dilution and molar conductivities in general.
For a salt $\ce{AB}$ the molar conductivity at infinite dilution can be expressed as the sum of molar conductivities at infinite dilution of the respective elements.
$$\Lambda^\circ_\mathrm{m}(\ce{AB}) = \Lambda^\circ_\mathrm{m}(\ce{A^+}) + \Lambda^\circ_\mathrm{m}(\ce{B^-})$$
But I don't get how the conductivities of $\ce{A^+}$ and $\ce{B^-}$ can be different.
I thought it means that the ions take different times to reach the electrodes so there should be, let's say, greater movement of $\ce{A^+}$ towards the electrodes so as time passes there should be more negative charge due to $\ce{B^-}$ accumulating in the solution. That's obviously incorrect.
So what exactly does conductivity signify physically if not different rates of migration?

You are completely right that ions migrate at different rates, and the relative rates can be quantified by a property of the electrolyte called the transport number, which is the fraction of the current that can be accounted by one of the ions:
$$t_\circ^+ = \frac{\Lambda_\mathrm{m}^\circ(\ce{A+})}{\Lambda_\mathrm{m}^\circ(\ce{A+})+\Lambda_\mathrm{m}^\circ(\ce{B-})}$$
$$t_\circ^- = \frac{\Lambda_\mathrm{m}^\circ(\ce{B-})}{\Lambda_\mathrm{m}^\circ(\ce{A+})+\Lambda_\mathrm{m}^\circ(\ce{B-})}$$
(here $t_\circ^i$ designates a limiting transport number ie one in the limit of dilute electrolyte solution).
In the early days of the study of electrolytes, Hittorf developed a way of determining the ratio of transport numbers by measuring the concentrations of electrolytes in different parts of an electrolytic cell. Say the cell contains a cathode, an anode, and a bridging chamber. Cations are attracted toward the cathode chamber and depleted at the cathode, and anions are similarly depleted at the anode but also migrate into the anode chamber. The change in the concentration of cations in the cathode chamber is therefore proportional to the current at the electrode minus the fraction of current carried by cations replenishing the cation chamber:
$$\begin{align} \frac{dc_+}{dt} &= \frac{I_+}{z_+F} - \frac{I}{z_+F} \\  &= \frac{(t_+-1)I}{z_+F} \\  &= \frac{-t_-I}{z_+F} \end{align}$$
A similar equation can be written for the anions. The resulting concentrations in the chambers can be related to the migration rates $v_i$ of the ions , which can in turn be related to the transport numbers as
$$\frac{t^+}{t^-} = \frac{|Z^+v^+|}{|Z^+v^-|}$$
Therefore from measurement of electrolyte concentrations in different parts of a cell it is possible to deduce the relative molar conductivities of the components of an electrolyte.
Because the counterions have different mobilities a concentration gradient will develop and this leads to a liquid junction potential. This results in a voltage drop within the cell that counters the EMF, such that the measured cell potential is not the ideal EMF. In Hittorf's experiment this is not a problem. Where it is a problem it might be avoided by using a salt bridge.
For a textbook that provides a reasonably clear introduction to electrochemistry see Atkins Physical Chemistry, from which I derived some of this answer.
Transport numbers are important in practical applications. See for instance Ref. 1 for an interesting thesis that focuses on the determination of transport numbers in cements.
References

J.A. Muka (1992). Transport Number Determination by Analysis of Transport of Aqueous Electrolytes Through Cementitious Materials Using Emf Method. University of Nairobi.


Purification of Solid Phase Synthesized Peptide I began working in a newly setup lab, and I'm having a dilysine (K2) peptide synthesized by solid phase method. I plan to cleave the peptides from the resin with a cleavage cocktail of TFA. Unfortunately, our freeze-drier is missing a part and not working then. We also don't have a rotary evaporator yet. We have our HPLC setup, but do you think it is a good idea to take the resulting solution of cleavage (after filtration to remove beads), which contains Boc, Fmoc, TFA, and possibly other impurities, directly to the HPLC?

I've read that HPLC sample should be dissolved in initial composition of my HPLC solvents, but it is in the case of having a crude dry sample.

It is my first time doing a peptide synthesis so I appreciate your ideas.

If you are a using reversed-phase (RP) column, it is okay to directly inject your crude sample in the running HPLC solvent system.  But, make sure to filter the injecting sample through a membrane filter to remove minute particles (from glass beats, etc.) from the original sample after dissolving it in suitable chosen solvent. Please also note that your chosen solvent for peptide separation usually contain small percentage of TFA. Therefore, you may need at least a rotary evaporator  for final concentration.


What kind of structure could give you a HNMR spectrum consisting of a singlet integrating to 5 protons? A clean spectrum was given to me with the integrations filled in. I just need to assign peaks.

I just don't see how it's possible to have a singlet integrate to 5 protons.
I've considered symmetry, but still, 2 or 3 way symmetry wouldn't give me anything.

The chemical formula is $\ce{C11H14O2}$, with 5 degrees of unsaturation.
The spectrum is as follows:

5H, singlet, 7.5 ppm
2H, singlet, 5.1 ppm
2H, triplet, 2.4 ppm
2H, quartet, (or sextet?) 1.8 ppm
3H, triplet, 1 ppm

I'm just looking for a hint on what a 5H singlet could be.

The 5H singlet is likely due to five protons on a monosubstituted benzene ring.

The peak comes in at 7.5 ppm which is the aromatic region, immediately telling you that they are hydrogens on an aromatic ring. Benzene rings are the most common of these.

But why are they all equivalent? They actually aren't, but their shifts are so close that they appear to be equivalent. Sometimes they will appear as different peaks (usually one peak for the 2 hydrogens close to the substituent and one peak for the 3 hydrogens away from it) if there are strong enough electron withdrawing groups on the substituent that are close to the ring. It is likely that the singlet you're looking at isn't clean cut and looks a bit like a multiplet (see this question).


Determine if two molecule structures are the same molecule (perhaps in different conformations)? I have two .pdb files describing two (small ligand) molecular structures. I need to determine if these two pdbs describe the same molecule, perhaps in different spatial conformations. Is there a program, or a combination of tools that will allow me to do this?

Moreover, if the two pdbs are determined to be from the same molecule, it could be that the atoms are in different orders. So I also need a way to identify the atoms in one PDB with the atoms in the other PDB.

I am dealing with thousands of pdbs, so I need an automatic way to do this comparison.

This is actually a very difficult problem, because a given compound may exist in multiple conformations (have multiple structures), and it might not be obvious what conformation is most relevant.

For example, a seemingly simple compound like glucose has multiple stereoisomers: if you say "glucose", you probably mean alpha-D-glucose, as opposed to beta-D-glucose, or even L-glucose. But a PDB structure file may not specify the stereochemistry to this level, or ignore it completely, and then you have no way of knowing what compound was intended. Conversely, sometimes a specific structure is given, like this open form of D-glucose; but in a water solution this structure spontaneously interconverts with the closed ring form (they are tautomers). And there are other problems with protonation (which depends on solvent and pH), charge distribution, and what not. Chemistry is a mess :)

To sum it up, the fundamental problem is that more than one completely distinct structure can actually represent the same physical compound. Therefore, comparing physical compounds is difficult; it depends on the level of detail of specification you want. I don't think there is a generic solution for  this.

Comparing structures, on the other hand, is perfectly feasible: I would recommend you take a look at the InChi format developed by IUPAC. It is an identifier string that uniquely identifies a molecular structure, so testing for structural identity is equivalent to simply comparing the corresponding InChi strings. The PDB has InChi strings associated with its entries, so this should be straightforward. (But again, this simple procedure will conclude that, for example, the open form and the closed-ring form of glucose are different things.)

I'm not sure about the atom numbering. The InChi format determines atom numbering in some canonical way according to predefined rules, so in this representation the numbering is by definition always the same (and therefore meaningless). But you would need to look into the details of the PDB--InChi conversion procedure to determine which atom gets mapped to which.


Why is the formation of slacked lime accompanied by a hissing sound and the formation of bubbles? My teacher made me write that that the formation of calcium hydroxide (slacked lime) from calcium oxide (qucklime) is described as

  $\ce{CaO + H2O -> Ca(OH)2}$

and that the process

  forms bubbles
  is an exothermic reactions
  makes a hissing sound

Why does it make a hissing sound and form bubbles?

There is no gas evolved in the reaction, then why does formation of bubbls take place?

And what is the reason of the hissing sound being produced?

Good eye for the inconsistency.  Cultivate that, it will serve you well.  But this is an apparent, rather than a real inconsistency.  The reaction is VERY exothermic.  Check out how they make quicklime to find out how much heat it takes.

Where the water touches the quicklime, enough heat is evolved to boil a little bit of the water.  Hence the bubbles and hiss.


Calculating the standard reduction potential for the oxidation of water I was working with Latimer and Frost diagrams for oxygen when I came across what seems to me a contradiction.

From the Latimer diagram for oxygen below, we know the standard reduction potential for the reduction of $\ce{O2}$ to $\ce{H2O2}$ and the standard reduction potential for the reduction of $\ce{H2O2}$ to $\ce{H2O}$, as represented in the following chemical equations:

$$\ce{O2} \overset{+0.70}{\longrightarrow} \ce{H2O2} \overset{+1.76}{\longrightarrow} \ce{H2O}$$

$$\begin{align}
\ce{2H+ (aq) + 2e- + H2O2 (aq) &-> 2H2O (l)} & E^\circ &= \pu{+1.76 V} \\
\ce{O2 (g) + 2H+ (aq) + 2e- &-> H2O2 (aq)} & E^\circ &= \pu{+0.70 V} \\
\end{align}$$

If we add these two reduction chemical equations, we can find the standard reduction potential for the reduction of $\ce{O2}$ to $\ce{H2O2}$, which would be $\pu{+2.46 V}$ according to the aformentioned reduction potentials and chemical equations.

The reduction of $\ce{O2}$ to $\ce{H2O}$ is the reverse reaction of the oxidation of water:

$$\ce{2H2O(l) -> O2(g) + 4H+(aq) + 4e-}\qquad E^\circ = \pu{-1.23 V}$$

Why does the standard reduction potential for the oxidation of water calculated from the Latimer diagram differ from the value reported in the table of standard reduction potentials? More specifically, why is the value calculated from the Latimer diagram double of that reported in the table? I checked my work multiple times and do not understand why this occurs, but I might have still made a silly mistake.

References
Shriver, D. F., Weller, M. T., Overton, T., Rourke, J., & Armstrong, F. A. (2014). Inorganic chemistry 6th Edition.

According to Hess’ law, we expect the Gibbs energy change $\Delta G$ of the two partial reactions to add up to the total Gibbs energy change:

$$\Delta G_1+\Delta G_2=\Delta G_3$$

For an electrochemical reaction, the change in Gibbs energy corresponds to an electrochemical potential $E$ according to

$$\Delta G = -zFE$$

where
$F$ is the Faraday constant and
$z$ is the number of electrons transferred.

Since for the two partial reactions the number of electrons is $z_1=z_2=2$, but for the total reaction $z_3=4$, we expect:

$$\begin{align}\Delta G_1+\Delta G_2&=\Delta G_3\\
z_1E_1+z_2E_2&=z_3E_3\\
2\times0.695\ \mathrm V+2\times1.763\ \mathrm V&=4\times1.229\ \mathrm V
\end{align}$$


What is chemical free energy? What is chemical free energy?  How can we practically explain its existence?  How can it be used to do some useful work like in an electrochemical cell?

I couldn't understand clearly about it when I referred a few books

Chemical Free Energy or Gibbs Free Energy is best understood as potential energy for reactions.

If I have a lot of Gibbs Free Energy, then I have a lot of potential energy.
Kind of like if I were on top of a hill, it's very easy for me to roll down that hill.

Often what's talked about in books is dG or the change in Gibbs Free Energy. If we think of it is simply being on top of a hill (High Gibbs Free Energy/Potential Energy) G(initial)
 going to the bottom of the hill (Low Gibbs Free Energy/Potential Energy) G(final)

Then the change in Gibbs Free Energy will be negative for something spontaneous (like the rolling down a hill example).
Change is always calculated as: change = final - initial

Gibbs Free Energy is used in the same was as potential energy. But instead of translating to kinetic energy, it translates to Enthalpy (heat) and/or Electrochemical Energy (being able to move electrons). Both Enthalpy and Electrochemical Energy can then be converted to mechanical work.

The way to think about it is not so much that you can use Gibbs Free Energy, but rather it's just the innate property of whatever situation you're working with. Molecules have a tendency to be reduced or oxidized under different conditions, the Gibbs Free Energy just tells you which way is spontaneous.

dG = dH - TdS

This gets kind of complicated, since dG is affected by many factors, but it gives you an idea, that the Change in Gibbs Free Energy can become positive or negative under different conditions.

Keep in mind that Gibbs Free Energy isn't a real thing, but rather a concept that helps explain why things happen.
 Why do gases expand?
 I don't know, but the universe has a tendency to do this. So we say that it is spontaneous, just like how we would say things roll down a hill because of gravity.
 What's gravity?
 It's our way of explaining why things roll down hills. :p

I hope that helps a little bit.


Oxalate Anion MO Diagram I figured I’d study $\ce{C2O4^2-}$ as an organometallic ligand, and found that it is weak field (pi donor). I hence decided to construct a MO diagram.

The oxalate anion is of D2h symmetry, so one can immediately use that, form irreducible representations and find the C-C orbitals:

Following which, I tried to use projection operator to look at all the group orbitals.

Unfortunately, this means that there are a lot of orbitals with Ag and B2u symmetry, and a couple with B1g and B3u symmetry. How would these orbitals interact?

Here's how I would approach it. I would start by making some (over)simplifying assumptions:

1) O $p_x$ orbitals are mostly nonbonding, so I'm going to ignore interactions of those orbitals with carbon orbitals

2) Due to $sp$ mixing, the carbon $p_y$ and $s$ orbitals can be merged into $sp$ orbitals aligned on the $y$ axis.

3) Same for the O $p_y$ and $s$ orbitals. In this case, the mixed $sp$ orbital with its large lobe oriented away from C is mostly nonbonding, so I'll ignore interactions of it with carbon orbitals

With those assumptions in place, I focus on building and ordering the occupied molecular orbitals. The highest energy occupied orbitals will be some of those oxygen-based nonbonding orbitals, so let's start by putting all eight of them at the top of the diagram. The relative energy levels among them will be affected by the number of positive and negative interactions, so the highest energy one will likely be the $b_{1g}$ where all of the $p_x$ orbitals have anti-bonding interactions with their neighbors. I mention this orbital in particular because it is the HOMO, and I think that the $\sigma$ interaction between oxalate and a metal center would come from the interaction of this orbital with the metal $d_{x^2-y^2}$ or $d_{xy}$, for example.

Mixed in with the $p_x$ and $sp_y$ nonbonding orbitals will be the two $\pi_{nb}$ orbitals, also centered on the oxygens. These are the $b_{2g}$ and $a_u$ orbitals in your set above. Since there isn't a carbon $\pi$ orbital with the right symmetry to interact with either of these, the resulting MO's have a nodal plane aligned with the C-C bond. These are the orbitals that will have $\pi$ interactions with a metal center if the oxalate is acting as bidentate ligand.

Below the nonbonding $\pi$ orbitals, we have the two $\pi$ bonding orbitals formed from the $b_{1u}$ and $b_{3g}$ central and oxygen $p_z$ orbital combinations. The $b_{3g}$ is higher energy of the two because of the C=C $\pi^*$ interaction.

At the bottom of the occupied orbitals, we have the five primarily $\sigma$ orbitals. I think the highest energy of the group is the $\sigma_{C-C}$ formed from the mixed $sp_y$ carbon orbitals oriented towards each other, ie a mix of the two $a_g$ central orbitals. Since the density is primarily between the carbons, I'm going to ignore the interactions with the O orbitals.

The four $\sigma_{C-O}$ orbitals come from the pairings of the central $b_{2u}$, $b_{1g}$, $b_{3u}$ and $a_g$ orbitals with their ligand orbital counterparts. In energy, from low to high I would guess $a_g$ ($\sigma_{C-O}$ and $\sigma_{C-C}$), $b_{3u}$ ($\sigma_{C-O}$ and $\pi_{C-C}$), $b_{1g}$ ($\sigma_{C-O}$ and $\pi^*_{C-C}$), then $b_{2u}$ ($\sigma_{C-O}$ and $\sigma^*_{C-C}$).


Relationship between the partial pressure $p_J = x_J p$ and the perfect gas law $pV = nRT$ I am currently studying Atkins' Physical Chemistry, 11th edition, by Peter Atkins, Julio de Paula, and James Keeler. Chapter 1A.2 Equations of state says the following:

When dealing with gaseous mixtures, it is often necessary to know the contribution that each component makes to the total pressure of the sample. The partial pressure, $p_J$, of a gas $J$ in a mixture (any gas, not just a perfect gas), is defined
$$p_J = x_J p \tag{1A.6}$$
where $x_J$ is the mole fraction of the component $J$, the amount of $J$ expressed as a fraction of the total amount of molecules, $n$, in the sample:
$$x_J = \dfrac{n_J}{n} \ \ \ n = n_A + n_B + \dots \tag{1A.7}$$
When no $J$ molecules are present, $x_J = 0$; when only $J$ molecules are present, $x_J = 1$. It follows from the definition of $x_J$ that, whatever the composition of the mixture, $x_A + x_B + \dots = 1$ and therefore that the sum of the partial pressures is equal to the total pressure:
$$p_A + p_B + \dots = (x_A + x_B + \dots)p = p \tag{1A.8}$$
This relation is true for both real and perfect gases.

When all the gases are perfect, the partial pressure as defined in eqn 1A.6 is also the pressure that each gas would exert if it occupied the same container alone at the same temperature.

It is this last part that I am unsure about:

When all the gases are perfect, the partial pressure as defined in eqn 1A.6 is also the pressure that each gas would exert if it occupied the same container alone at the same temperature.

I don't understand why this is true. Furthermore, is this not in reference to the perfect gas law, $pV = nRT$? Then how does this relate to $pV = nRT$?
I would greatly appreciate it if people would please take the time to clarify this.

Partial pressure is defined as the pressure a component of a gaseous mixture would exert if it alone were to occupy the container at the same temperature as that of the mixture.
If the gases in the mixture are ideal, you can apply the ideal gas equation for it $$p_A V = n_ART$$where $p_A$ and $n_A$ represents the partial pressure and number of moles of a component A. Also $$p_tV = nRT$$ where $p_t$ and $n$ represents the total pressure and number of moles of the gaseous mixture. Dividing these equations you will get $$\frac{p_A}{p_t}=x_A$$ This equation will only be true for ideal gas mixture.
Also, partial pressure is a notional concept. It is not a the actual pressure exerted by the components. It was introduced to explain and form physical properties of gaseous mixtures.


Hypervalency of orthonitrate I have always believed that it is impossible for the atoms of period 2 elements (when they are the central atoms in polyatomic ions/molecules) to accommodate more than 8 electrons in their valence shell. This is because of their small atomic size, which results in a significantly higher electron density being borne by the central atom and thus, instability can easily arise if too many electrons are held by it. Having seen many posts on this forum on the concept of hypervalency, I understand that many seemingly "hypervalent" molecules (e.g., $\ce {PCl5}$, $\ce {SF6}$, $\ce {ClO4^{-}}$) do not really have their central atoms possessing more than 8 electrons in their valence shells. This is due to the electron-withdrawing effect of the electronegative atoms that are bonded to the central atom, often resulting in there being less than 8 electrons populating the valence shell of these central atoms (Gillespie and Silvi, 2002).
However, my belief has recently been challenged by the existence of the orthonitrate ion, in which the $\ce {N}$ atom is claimed to possess $\ce {8.65}$ electrons in the valence shell (as written in the "Alternative definition" section of this article). This seems rather nonsensical to me as the $\ce {O}$ atoms surrounding it are more electronegative than the central $\ce {N}$. I would like to request for a verification of the claim regarding $\ce {N}$ possessing more than 8 electrons in the valence shell. Additionally, I would also like to ask if the ionically-bonded structure with one $\ce {N^+}$ and four $\ce {O^-}$ is an accurate representation of the bonding in the ion.
Reference
Ronald J. Gillespie, Bernard Silvi, "The octet rule and hypervalence: two misunderstood concepts," Coordination Chemistry Reviews 2002, 233-234, 53-62 (https://doi.org/10.1016/S0010-8545(02)00102-9).

When you learn molecular orbital theory, you learn to throw the octet rule out the window in favor of identifying the molecular orbitals that form the actual bonds.  Take a look at the seemingly innocuous case of carbon dioxide discussed in Chemistry LibreTexts (illustration from this reference):

There are really two related ways in which the molecular orbital reality blows the octet rule away even though you thought the familiar Lewis structure $\ce{O=C=O}$ obeyed this rule.
First:  The octet rule assumes that electrons in covalent bonds are shared between pairs of atoms at the ends of the respective bonds.  In the above diagram they're not.  Most of the occupied orbitals share electron density between all three atoms and those that don't are concentrated at the two oxygen atoms, not at the ends of either carbon-oxygen bond.  This three-way sharing is especially notable with the pi bonds; the structure I drew above is really not an accurate way to describe the pi bonds.  You really need a combination of pairwise valence bond structures even to begin to approach the molecular orbitals:
$\ce{\overset{-}{O}-C#\overset{+}{O}}$
$\ce{O=C=O}$
$\ce{\overset{+}{O}#C-\overset{-}{O}}$
In effect, the carbon-oxygen bonds with this multiple-atom sharing are not just double bonds.  They are a mixture of single, double and triple bonds providing more overall bonding than just plain pairwise double bonds could have.
Second:  Given this multi-way electron sharing the number of occupied orbitals that are at least partly shared by any atom can go well beyond four.  In the case of carbon dioxide this is especially true of the oxygen atoms, each of which has a finger in the pie of all sixteen valence electrons.  In the case of carbon dioxide it is true that each oxygen atom cannot have more than half of that electron density close to it, but that is a property of the specific compound carbon dioxide (which has only sixteen valence electrons to work with among all three atoms) and not a universal law of nature.
This should give you some idea of what goes on with a more complex structure such as your orthonitrate ion.  Electrons are not shared pairwise but among multiple atoms in molecular orbitals, and with this multi-way electron sharing the bonds are not really confined to specific pairs of atoms.  What looks like single, sigma-only bonds between the nitrogen and each oxygen atom are really a combination of sigma and pi bonding interactions, delocalized through a complex molecular structure; and each $\ce{N-O}$ interaction is thereby a combination of single bonds, multiple bonds and even purely ionic interactions.  Assigning a meaningful atom-by-atom electron count to such a structure is, as Mithoron suggests, fuzzy math.


Why are bridge bonds weaker than terminal bonds and, H-B-H terminal bond angles larger than H-B-H bridging bond angles in diborane? In Concise Inorganic Chemistry by J.D.Lee (Adapted by Sudarsan Guha, Fourth Edition), on page 83, under the topic "Bridge Bonding" it is given for diborane $(\ce{B2H6}):$

The $d_\ce{B-H}(\text{terminal bonds}) < d_\ce{B-H}(\text{bridge bonds})$
The energy required to replace hydrogen atoms from the bridged position is more than that needed for the terminal position. This is supported by the fact that on reaction with methyl chloride, terminal hydrogen atoms are replaced by methyl group in preference to the bridged hydrogen atoms.
The reason for this could be the fact that to break the bridge bond, the overlapping zone should be broken from both sides and the overall energy required is more as compared to do so for terminal bonds.

Structure of Diborane:

Image Credit: Wikipedia
Based on the second paragraph it can be concluded that terminal bonds are weaker than bridge bonds. But some sources on the internet including Wikipedia state the bridge bonds are weaker than the terminal bonds without any satisfactory reason. I wish to know the reason for this.
Further, kindly explain why the terminal H-B-H bond angle is larger than bridging H-B-H bond angle as evident from the molecular structure as given above. I have searched my book as well as google to my best, but I didn't get any reason for this.
Kindly note, I read the unaccepted answer to this question. But still my doubt persists and makes me type this question by adding relevant details from my book and internet.
Kindly clarify this doubt.

It depends on what you define as a bond.  A three-center two electron bond is in general stronger than an ordinary two-center bond because delocalizing the bond over three atoms instead of two makes the bonding MO more stable.  But, the bridge bond is also shared between two linkages so each individual linkage has less bonding than a single two-center bond linkage.
The question of bond angles is a bit more complex because it hides a couple other issues.  Superficially, the bond angles in the bridge region are forcibly reduced by forming a four-membered ring.  But that begs the following questions:

Why is the dimer formed instead of a trimer or tetramer, which would enable a larger ring and more natural bond anges?

Why is the bond angle through boron actually greater than the bond angle through hydrogen?  Since hydrogen forms smaller atoms steric factors would favor the reverse.

An answer to both questions may lie in the tendency of 3c-2e bonds to form closed structures.  The bonding orbital described above gets an extra bonding linkage by bringing the end atoms of a chain together.  Such a tendency is seen in the formation of a triangular trihydrogen cation in interstellar space, and in the intermediate stage of a carbocation rearrangement.
In the case of diborane this tendency shows up as boron-boron bonding. The dimer has a favorable geometry for forming such a boron-boron bond, whereas an oligomer with larger bond angles in the center would tend to hold the boron atoms too far apart.
The nature of this boron-boron bonding is revealed in molecular-orbital calculations reported by Prof. Bruce Tattershall at Newcastle Universitywinnoplus=2&outputype=HTML5.0_JSmol&basis=MO&orbital=B(p_x)_H_bridge*&plane=xz&orbno=10).  The diagram taken from this source is shown below to reveal six occupied valence MOs and defines the $x$ and $z$ axes in the molecule for later reference.

Below are the quantitative data behind this diagram for the occupied molecular orbitals.  Entries marked with a blue dot indicate boron-boron bonding, while those with a gold dot are boron-boron antibonding.

The boron $s$ and $p_x$ interactions along the sigma-bonding axes show a mixture of bonding and antibonding overlaps; but the $p_z$ interaction, using orbitals that overlap the hydrogen atoms, shows a single bonding interaction with large boron components. This indicates that the closure of the 3c-2e bonds in the dimer occurs through the $p_z$ lobes rather than along the actual boron-boron axis, forming a "$\pi$ bond over a gap" analogous to what an organic chemist might find in a homoaromatic molecule.
Note also that the 3c-2e bond closure occurs around a hydrogen atom. A larger group in the bridging position could leave the boron atoms too far apart to overlap well and thus inhibit formation of the b9ron-boron bond in a dimer. This would explain the greater tendency of trialkylboranes to exist as monomers compared with the parent (di)borane.


will the autoionization of water reduce due to common ion effect? When a weak acid is added to water , will the dissociation of water reduce due to common ion effect caused by increased H+ concentration due to acid , my text book said common ion effect is caused by a strong electrolyte
Moreover this doubt is a part of a larger doubt - if multiple weak acids are added in water will they reduce each others and water's dissociation as compared to their dissociation in their individual solutions ( due to common ion effect caused by H+ ions ) They should ? Having this doubt because the textbook I'm using just adds up their normal dissociations irrespective of common ion.

Adding acid to the water will not stop the auto-ionization of the water, so: the auto-ionization will continue to give equal concentrations of hydroxide and hydronium. But the addition of acid to the water increases the concentration of the hydronium ions in the solution, pushing water balance with its ions to the left to resist the increase in the concentration of hydronium and reach a new equilibrium. The concentrations of hydronium and hydroxide produced by auto-ionization of water are reduced, where the concentrations of $\ce{H_3  O^+}$ or $\ce{OH^-}$ from water ionization is less than$\pu{ 10^{-7}}$.

$$\ce{{[H_3O^+]}_\text{(after acid  addition)}=\ce{[H_3O^+]}_\text{(produced by acid dissociation)} +\ce{[OH^-]}_\text{(produced by water auto-ionization )}}$$


Incomplete Combustion products In class we performed an experiment which I am unable to understand the results for
Why does the the amount of soot produced vary? What is the amount of soot  produced  dependent on?

Experiment:

  We had to have four test tubes upside down in a beaker of water. The test tubes were filled with water in them as well. We had to insert pieces of calcium carbide under the test tubes and have them as follows: one full of acetylene, one 3/4 full, one 1/2 full, one 1/10 full. Then remove the rubber stopper and bring a burning splint to the mouth of the test tube to ignite the acetylene gas.

Results:

The full one: burned slow and produced a little bit of soot
3/4: burned fast and lots of soot
1/2: burned fast and lots of soot
1/10: burned fast and no soot

The reaction for complete combustion of acetylene is
$$\rm 2C_2H_2 + 5O_2 \rightarrow 4 CO_2 + 2H_2O$$

...you'll get complete combustion when acetylene and oxygen are mixed in a 2:5 mole (or volume) ratio. If you have a higher ratio than that, combustion will be incomplete. The incomplete combustion products will appear as soot and carbon monoxide.

Air is 1/5th O$_2$, so you need 5 times as much air as you'd need pure O$_2$. The optimal acetylene/air ratio should be about 2:25.

Look at the acetylene/air ratios for each tube. The gases are both at the same temperature and pressure, so the volume ratio is equal to the mole ratio. Any ratio that's greater than 2/25 will probably produce soot, and the higher the ratio, the more soot you'd expect. Any tube that contains an appreciable amount of oxygen mixed in with the acetylene will burn quickly.

The full tube burned slow because oxygen needed to mix with acetylene at the mouth of the tube. The little bit of acetylene that burned there produced soot, because the acetylene/air ratio was probably much higher than 2/25.

The tubes that were 3/4 and 1/2 full had 3:1 and 1:1 acetylene/air ratios. Both ratios are fuel-rich, so soot was produced (you'd expect more soot with the 3/4 full tube than the 1/2 full tube). Both tubes burned quickly because there was a fair amount of O$_2$ mixed with the acetylene.

The tube that was 1/10 full had a 1:9 acetylene/air ratio. 1/9 is just a bit higher than 2/25, so this tube was fuel-rich, too, but much less so than any of the other tubes. Very little soot would be expected; maybe none, if more air was drawn into the tube as the acetylene burned.


Beer's law with a polychromatic light source As is well known, Beer’s law (aka the Beer-Lambert law, among others) is predicated on a number of conditions, one of which is that the light source be monochromatic or, as a practical approximation, that the spectral bandwidth of the light source be much smaller than the spectral bandwidths of whatever absorbers are being investigated. For present purposes, a polychromatic light source may be considered to have spectral bandwidth, $\ce S_{source}$, that is not negligible relative to that of the relevant absorbers. The question posed below will deal with what happens when the light source is decisively polychromatic.

Assume a cell path length, b, of 1 cm, and assume a single absorber, having triangular molar absorptivity spectral profile (Fig. 1):

The peak molar absorptivity, $\ce \epsilon_{450 nm}$, is 100 L/mole cm at 450 nm. Assume the absorber concentration, c, ranges from 0 to 0.01 M. In the figure, the absorber’s molar absorptivity spectral profile is modeled as an isosceles triangle. In reality, absorber profiles are often modeled [1, p. 164] as Gaussian, Lorentzian, Voigt, or whatever the underlying physics requires. Here, however, an isosceles triangle is used for mathematical simplicity.

The figure also shows a narrower red isosceles triangle, centered (and having unity transmittance) at 425 nm and having full width at half maximum height (FWHM) equal to 20 nm. This red triangle is the transmittance spectral profile [2, equation 3-60, p. 70] of the light that exits a monochromator, assuming white light of uniform (unity) intensity is incident upon the monochromator and also assuming both 10 nm/mm reciprocal linear dispersion (RLD) and 2 mm slits ($\ce W_{slits}$). With white light incident on the entrance slits of the monochromator, the exiting light has $\ce S_{source}$ = RLD x $\ce W_{slits}$ = 20 nm. Note that $\ce S_{source}$
 is defined here as the FWHM of the spectral profile: this is not the only possible such definition, but it is satisfactory for present purposes. Before proceeding, note that a 0.1 M monochromator typically has something like RLD = 4 nm/mm and 2 mm slits are relatively huge, as slit widths go, so the light source’s profile is polychromatic (and not good at all). Indeed, the FWHM is 20 nm, while that of the absorber is 50 nm.

As always, $\ce T(\lambda) = I_{transmitted}(\lambda)/I_{incident}(\lambda)$ and $\ce A(\lambda) = -log[T(\lambda)]$. If $\ce c = c_{max} = 0.01 M$ and b = 1 cm, then the next figure shows the absorbance (black) and transmittance (red) spectra from 380 to 520 nm (Fig. 2):

If the light source was monochromatic, at 425 nm, a Beer’s law plot would be a straight line with slope = $\ce \epsilon_{425 nm}$ x b = 50 L/mole cm x 1 cm = $50 M^{-1}$.
For absorber concentrations from 0 to 0.01 M, $\ce A_{max} (425 nm)$ = 0.5 at $\ce c_{max} = 0.01 M$.

However, for the light source profile in Fig. 1, the Beer’s law plot will curve downward, as undergrad textbooks like to show in cartoon form, and the absorbance at $\ce c_{max}$ will be less than 0.5.

Question: For the above scenario, with the specified polychromatic source profile in Fig. 1, what is the absorbance at $\ce c_{max} = 0.01 M$?

Hints: 1) Forget about undergrad textbooks for this. 2) There is no added noise or error. 3) The light source cannot be modeled as a pair of monochromatic lines.

Assume the photodetector responsivity is flat from 200 to 700 nm and dark response is zero. There is no detector or preamplifier noise, no digitization error, and no source shot noise. This is idealized in every reasonable way except for the polychromatic source, which, of course, no one would deliberately use.

References:

T.C. O’Haver, “Effect of the Source/Absorber Width Ratio on the Signal-to-Noise Ratio of Dispersive Absorption Spectrometry”, Analytical Chemistry, 63 (1991) 164-169.
J. D. Ingle, Jr. and S. R. Crouch, Spectrochemical Analysis, Prentice Hall, Englewood Cliffs, NJ, ©1988, pp. 70-74.

There are several ways to answer the question, so here are my three. The first answer involves using a simple integer wavelengths approximation, i.e., noticing that Figs. 1 and 2 have 1 nm point spacing, and then using a spreadsheet to perform a summation. The answer explained in this part also serves to define the terms used in the second answer, which involves analytic calculation of the relevant overlap integrals. Then the third answer, not discussed in detail, uses a Mueller calculus-based method. Before proceeding to the explanations, the numerical answer to the question is $A_{maximum} = 0.46951302113867$ at $C_{max} = 0.01 M$ and Fig. 4 shows the absorbance as a function of concentration:

The absorbance is 93.9% of the ideal value (i.e., 0.5) at $C_{max}$, despite the source being significantly polychromatic, and the Beer’s law plot is convex, but perhaps less than might have been expected intuitively.

All three of the answers and explanations that follow are based on the model shown in Fig. 4 below:

As per the question’s assumed conditions, the light intensity incident at the monochromator is unity, i.e., 1, with fictive units of Watts. Note that units may be ignored since they will divide out when transmittance is computed later. The 50:50 beam splitter sends half of the light it receives to the sample (upper) beam path and the other half of the light it receives goes to the reference (lower) beam path. Let $I_{incident}(\lambda )$ denote the light intensity incident upon the absorber and $I_{sample}(\lambda )$ denote the attenuated light intensity that gets through the absorber. Then $I_{sample}(\lambda )$ is the incident light intensity on the upper photodetector and transimpedance amplifier (PD-TIA).

From Fig. 4, it is clear that $I_{reference}(\lambda )$ = $I_{incident}(\lambda )$, assuming the mirror is ideal. This is simply the unattenuated incident light intensity on the lower PD-TIA. The photodiodes were stated to have flat responsivities, i.e., spectrally constant responsivities, so their photocurrent responses are independent of wavelength. Hence, the responsivities, in this idealized scenario, cancel out when transmittance is computed later. Likewise for the constant transimpedances, which are assumed equal. Hence, it suffices to simply ‘add up’ the light intensities, for each color of light, incident at each PD-TIA. Now to the answers and explanations.

Answer 1: As per Figs. 1 and 2, assume a wavelength spacing of 1 nm. There is no light outside the region 405 to 445 nm and wavelengths are effectively “discretized”, i.e., are treated as discrete integer values. Therefore, the number of discrete wavelengths, where light intensity is non-zero, is just 39, i.e., from 406 to 444 nm, in 1 nm increments. At 405 nm, $I_{incident}(\lambda )$ is 0 and it increases linearly to 0.5 (at 425 nm). It then decreases linearly to 0 (at 445 nm). The absorbance, $A_{absorber}(\lambda)$, increases linearly from 0.1 (at 405 nm) to 0.9 (at 445 nm). Hence, the transmittance through the absorber, $T_{absorber}(\lambda)$, decreases non-linearly from about 0.7943 (at 405 nm) to about 0.1259 (at 445 nm). As always, $T = 10^{-A}$, so there is no difficulty computing $T_{absorber}(\lambda )$ at any of the relevant wavelengths.

Accordingly, $I_{sample}(\lambda )$ = $I_{incident}(\lambda )$ x $T_{absorber}(\lambda )$ and $I_{reference}(\lambda )$ = $I_{incident}(\lambda )$ x $T_{reference}(\lambda )$ = $I_{incident}(\lambda )$, since $T_{reference}(\lambda )$ = 1. All of the necessary calculations are shown in the spreadsheet screenshot in Fig. 5 below:

In Fig. 5, column A contains the wavelengths and column B is a unitless “x” variable that is relative to 405 nm, i.e., at 405 nm, x = 0 and so forth. Column C contains the incident light intensities, column D has the associated absorbances and column E has the corresponding transmittances through the absorber. Column F has the unity transmittances for the reference beam path, column G has the $I_{sample}(\lambda )$ values and column H has the $I_{reference}(\lambda )$ values. Adding up the 41 values in columns G and H yields total intensities: 3.39164 and 10.000, respectively. Then T is simply 0.339164 and $A_{maximum} = 0.469590$. This is close to the true value, as computed in the second answer, despite the rather coarse “discretization” of the wavelengths.

Answer 2: From the first answer, it is clear how the relevant overlap integrals should be calculated. First, x is the independent (unitless) wavelength variable and it is continuous from 0 to 40. As x increases from 0 to 20, $I_{incident}(x)$ increases from 0 to 0.5, so $I_{incident}(x)$ = x/40. As x increases from 20 to 40, $I_{incident}(x)$ decreases from 0.5 to 0, so $I_{incident}(x)$ = 1- (x/40). For the absorbances, $A_{absorber}(x)$ = 0.1 + 0.02x for x = 0 to 40. Hence, $T_{absorber}(x)$ = $10^{0.1 + 0.02x}$ and $I_{sample}(x)$ = $I_{incident}(x)$ x $10^{0.1 + 0.02x}$.

For the sample beam path, the total intensity, $I_{sample}^{total}$, is the following overlap integral: $$I_{sample}^{total} = \int\limits_0^{40} {I_{sample}(x)\ \mathrm dx} $$

But this is simply the sum of two definite integrals: $$\int\limits_0^{20} {(x/40)\ 10^{0.1 + 0.02x}\ \mathrm dx} + \int\limits_{20}^{40} {[1-(x/40)] 10^{0.1 + 0.02x}\ \mathrm dx}$$

Using the fact that $10^{-x} = e^{-xln(10)}$, and defining $a \equiv -0.02 ln(10)$, then results in
 $$ \frac{e^{5a}}{40} \int\limits_{0}^{20} {xe^{ax}  \mathrm dx} + e^{5a} \int\limits_{20}^{40} {e^{ax}  \mathrm dx} - \frac{e^{5a}}{40} \int\limits_{20}^{40} {xe^{ax}  \mathrm dx}$$

The integrals are easily evaluated, resulting in $$ \frac{e^{5a}}{40a^2} {[(20a - 1) e^{20a} + 1]} + \frac{ e^{5a}}{a}{[e^{40a} – e^{20a}]} - \frac{e^{5a}}{40a^2} {[(40a - 1) e^{40a}] - [(20a - 1) e^{20a}]} $$

With $a \equiv -0.02 ln(10)$, $I_{sample}^{total}$ = 3.39224318670537.

For the reference beam path, the integrals are easier because $T_{reference}(x)$ = 1. Hence $$ I_{reference}^{total} = \int\limits_0^{20} {(x/40) \ \mathrm dx} + \int\limits_{20}^{40} {[1-(x/40)] \ \mathrm dx} = 5 + 5 = 10$$ Alternatively, this is simply the area of the triangle: 0.5 x 40 x 0.5 = 10. Therefore, $T =  I_{sample}^{total}\ / \ I_{reference}^{total} = 3.39224318670537\ /\ 10 = 0.339224318670537$. Then $A_{maximum} = 0.46951302113867$.

The analytical method worked, as expected, but the system was so heavily idealized that it was essentially a “spherical cow” and they don’t milk. Any realistic complications, e.g., even simply using a tungsten bulb light source, make the analytic method harder and it does not take much to make the equations all but analytically intractable. Then numerical methods may be employed, but the difficulty then mostly lies in getting the right model equations: correctly evaluating equations is not necessarily the same as evaluating correct equations.

What about convolution? Convolution, per se, is unnecessary in this heavily idealized problem, for the reasons discussed below. First, note that the spectral profile of $I_{incident}(\lambda )$ is the same as the monochromator’s transmittance profile (shown in Fig. 1), except for having a maximum value of 0.5 rather than unity. The spectral profile is symmetric. Second, the absorber’s transmittance profile is as shown in Fig. 2 and it is also symmetric. Third, the convolution of two real functions involves reversing one of the functions and then computing the overlap integrals as a function of a shift parameter, i.e., a relative displacement of the two functions. Since the two functions to be convolved are both symmetric, nothing needs to be reversed, so the convolution reduces to computing the overlap integrals as a function of a shift parameter, $\tau $. Zero displacement, i.e., $\tau = 0$, can be conveniently defined as when the two functions are perfectly aligned.

The next step is to compute the convolution function either analytically, if possible or desired, or numerically, if necessary or convenient. Then the convolution function would be evaluated at $\tau = 25$ (nm), since this corresponds to the 25 nm shift between the two spectral profiles and $\tau $ is integer valued, i.e., data values are available every nm, as shown in Fig. 5. With finer wavelength spacing, the shift would scale up accordingly. If the convolution is computed analytically, then $\tau $ is a real number.

For the data in Fig. 5, the numerically computed convolution is shown in Fig. 6:

From Fig. 6, the value of the convolution at 425 nm is 3.39164, as in Fig. 5. If the two spectral profiles do not overlap, then the value of the convolution is 10, as expected. But note that all that was needed was the single overlap integral at the known $\tau = 25$ (nm), and that value is 3.39164, as shown above. For the reference beam path, the overlap integral was 10, as shown in the trivially simple calculation above.

This problem is so heavily idealized that there is no necessary reason to compute the convolution function. In a more realistic situation, the computation would be harder and even numerical evaluation of the convolution function might be fruitless.

Answer 3: Figure 4 appears to just be a schematic block diagram, but it is actually a screenshot of a Mueller optical calculus-based computer simulation program. Although discussion of the fundamental principles of the several optical calculuses is beyond the scope of this answer, both Mueller calculus and Jones calculus were invented in the early 1940’s and there is a considerable body of information about them in the literature, e.g., [1-4].

In the simulation depicted in Fig. 4, stacks of Stokes vectors (one per discrete color) are passed from one optical component to the next optical component(s), e.g., from the lamp block to the monochromator block. Each component has a resident stack of matrices, with components as appropriate. When an optical component receives an incoming stack of optical vectors, each vector is multiplied by the corresponding matrix and the resulting stack of processed optical vectors is sent out to the next optical component (or components) “downstream”. All the stacks of optical vectors are received, processed and then sent on in a single computer simulation step.

The plot shown in Fig. 3 was obtained by using 1,001 concentrations: 0 to 0.01 M, in increments of 10 µM. The wavelength range was from 405 to 445 nm, in 1 pm increments. Thus, with every one of the 1,001 simulation steps, stacks of 40,001 Stokes vectors (used in Mueller calculus) were passed, starting with the lamp block and ending with the sample and reference PD-TIA blocks. Both of those blocks, for every step, computed 40,001 intensities, multiplied them by responsivities, summed the resulting photocurrents and then multiplied by the transimpedances.

In the present case, the responsivities were simply constant default values (0.1 A/W) and the transimpedances were $1 k\Omega$. For every step of the simulation, T was computed as the quotient of the real number outputs of the two PD-TIA blocks and $A_{maximum} = -log(T) = 0.46951302122$. Compared with the exact answer, this differs in the ninth decimal place. As well, there was no need to solve any equations or use a spreadsheet: no wheel had to be re-invented.

Another of the advantages of using optical calculus-based spectrometric simulations is that they can easily accommodate non-idealities, such as realistic lamp spectral profiles and photodetector responsivities, imperfect beam splitters and mirrors, polarized light, optical activity, circular dichroism, stray light, light source modulation and so on. In fact, any linear optical effect can be modeled with an optical calculus. In addition, noise effects are easily included, e.g., detector/preamplifier noises and light source noises, including low frequency noises.

Summary: Answer 2 provided an exact solution to the posed question, but it was a bit tedious to obtain and would not be useful if the system was altered in any significant way, e.g., if the photodiode responsivities were those of a real red-sensitive photodiode. Answer 1 was adequately accurate, despite its simplicity, but not particularly amenable to generalizations. Answer 3 gave an accurate answer and made it simple to get the plot shown in Fig. 3. There is no significant difficulty in generalizing to more realistic scenarios.

Lastly, it is clear that using the decisively polychromatic source in Fig. 1 caused a relatively small reduction in $A_{maximum}$ and the Beer’s law plot is convex, but not badly so. The deleterious effect on spectral resolution was not investigated here, but could easily be done, perhaps as another question.

References:

W.A. Shurcliff, Polarized Light, Harvard U. Press, Cambridge, MA ©1962 & 1966.
D.S. Kliger, J.W. Lewis, C.E. Randall, Polarized Light in Optics and Spectroscopy, Academic Press, Boston ©1990.
Mueller calculus article in wikipedia and references therein.
Jones calculus article in wikipedia and references therein.


Why does permanent marker writing come off a white board when coated with white board marker? Why is it that dry erase markers allow permanent markings to come off?

Here's a demonstration.

The solvent of the marker dissolves the permanent ink and mixes it with the whiteboard ink, leaving a dark dust that doesn't stick to the board.

The link in Wikipedia doesn't name common ethyl alcohol (denatured or not) as being a solvent for old marks. That is what I have used.


Calculating heat of combustion via calorimetry Question:

  When $\pu{60 g}$ of $\ce{C60}$ is combusted in a bomb calorimeter that has a water jacket containing $\pu{300.0 g}$ of water, the temperature of the water increases by $\pu{10 ^\circ C}$. Assuming that the specific heat of water is $\pu{4.18 J g^{-1} K^{-1}}$, estimate $\Delta E$ of combustion per mole of $\ce{C60}$.

Answer: $\pu{-150.5 kJ/mol}$

What I have tried:
$$
\begin{align}
Q &= m \times c \times \Delta T \\
Q &= 300 \times 4.18 \times 10
\end{align}$$

That's the only equation I know and it doesn't seem to work for this problem.

You have 60 g of $\ce{C60}$ (which has a molecular weight of 720.6), so that means you'll need to multiply your answer by 720.6/60 which (cleverly) turns out to be 12.01. Another way of saying it is that you have calculated the heat of combustion for 1/12.01 moles of $\ce{C60}$, so you'll need to multiply your answer by the inverse in order to get the desired (per mole) quantity.

Using your notation, numbers, and noting that a change in one degree Celsius is equivalent to a change in one kelvin:

$$\Delta E_{\rm molar} = -12.01\;\mathrm{mol} \cdot 300\;{\mathrm g} \cdot 4.18\;{\mathrm J} \cdot {\mathrm g}^{-1}\, \cdot {\mathrm K}^{-1}\, \cdot 10\;{\mathrm K} = -150.6054\;\mathrm{kJ}\cdot\mathrm{mol}^{-1}$$


Determining standard Gibbs free energy for water? I am dealing with the dissociation of water into its components hydrogen and oxygen and I am trying to determine the standard Gibbs free energies of each component. My professor said if the substance is an ideal gas in the standard state then

$$g_{i}^{0}=g_{i}^{\text{pure IG}}(T, P_\text{ref})$$

Now this means I can determine it for hydrogen and oxygen because they can be treated as ideal gases, but how would I consider the standard Gibbs free energy for water?

Let's look at the reaction of the formation of water:

$\ce{2H2(g) + O2(g) -> 2H2O (l) }$

The Gibb's Free Energy relates the spontaneity of various reactions by looking at the change in enthalpy, temperature, and entropy.

We can relate those by:

$$ \Delta G^\circ  = \Delta H^\circ - T\Delta S^\circ$$

If we look at our reaction, it's clear that this is an exothermic reaction. Bonds are being formed as hydrogen gas and oxygen gas form water. Moles of gas are decreased and a liquid is formed so entropy is decreasing as well. We know that the formation of water is a spontaneous reaction, so $\Delta G^\circ$ must be negative. Therefore, this reaction must be run at temperatures where the difference of the enthalpy and the product of the entropy and temperature must be < 0.

The Gibbs free energy is also a state function, so looking at $\Delta G^\circ_f$, we can find $\Delta G^\circ_{rxn}$.

$$\Delta G^\circ_{rxn} = \Sigma \Delta G^\circ_{products} - \Sigma \Delta G^\circ_{reactants} $$

Both hydrogen gas and oxygen gas are in their standard states and have a Gibbs Free energy of 0. If you look at tabulated standard thermodynamic data at $25^\circ\ \text{C}$, $\ce{H2O (l)}$ has a Gibbs free energy of$ -237.13 \text{kJ}\cdot\text{mol}^{-1}$.

$$\Delta G^\circ_{rxn} = -237.13 \text{ kJ}\cdot\text{mol}^{-1} - 0 \text{ kJ}\cdot\text{mol}^{-1} = -237.13 \text{ kJ}\cdot\text{mol}^{-1} $$


Interpreting Infrared Spectroscopy (IR) Spectra I have already attempted to interpret some of the peaks. Please tell me if they are correct/incorrect and anything that I have missed.

In your first spectrum, the band at 3517 $\mathrm{cm^{-1}}$ indeed fits best to the $\ce{N-H}$ stretch of a primary amine (source). However, I would not assign the shoulder of this band to a $\ce{#C-H}$ stretch. Such a band would be sharper and perhaps more intense, and as Greg E. already pointed out in his comment to your question, there is also no corresponding $\ce{C#C}$ stretch band present (reference). The absence of bands in the range between 2500-2000 $\mathrm{cm^{-1}}$ allows you to rule out some other functional groups with double or triple bonds, like nitriles, thiocyanates, isocyanates, carbodiimides or azides. The band at 1715 $\mathrm{cm^{-1}}$ is a typical $\ce{C=O}$ stretch, and while it could be a ketone, the presence of the broad band at 2932 $\mathrm{cm^{-1}}$ (which could actually be a double band) makes it likely to be an aldehyde, since the double band can then be assigned to the $\ce{=C-H}$ stretch of the CHO group (it could also include other $\ce{C-H}$ stretches due to its width).

Your second spectrum appears a bit more complex. The broad, intense band at 3298 $\mathrm{cm^{-1}}$ in combination with the band at approximately 1680 $\mathrm{cm^{-1}}$ could be assigned to an amide, but I would expect a stronger carbonyl band in this case. Nevertheless, the intensity and width of the 3298 $\mathrm{cm^{-1}}$ band suggest the presence of a functional group capable of H-bonding. The bands between 2900-2800 $\mathrm{cm^{-1}}$ could be assigned to alkyl $\ce{C-H}$ stretches (source). For the compound to be aromatic, there should be aromatic $\ce{C-H}$ stretch bands present, slightly above 3000 $\mathrm{cm^{-1}}$ (source).


Thermodynamics in Solution (solvation) from quantum chemistry The gist of the question is "how do I get a heat of formation in solution" - but let me give it some context to be more specific.

I am interested in calculating the thermodynamic values of molecules in solution. Due to the number of jobs, time etc. I have to use DFT methods. These are not as accurate as for example CCSD related methods but still provide good answers in reasonable timeframes (plus I'm checking up on estimated values, so I don't immediately need the highest acutacy).

As atomisations are hard to calculate with DFT methods (see a book by Cramer & comment on the ORCA forum), isodesmic reactions offer a better option.
So far so good. This allows me to use literature values for reference compounds to calculate the heat of formation of a species.

Now I need the heat of formation of the species in the liquid phase.
I can repeat my quantum chemistry calculations with a solvation model and again obtain values - however I now lack the reference compounds.
Based on my understanding of chemistry (see textbook by Cramer) I can relate the Gibbs free energy of the gas phase molecule with the solvated molecule.
So using the following:

$G_{liq} = G_{gas} + G_{sol}$
$\delta G = \delta H_f - \delta S ~T_{LO}$

I can work out the liquid phase heat of formation of the molecule.
(Please correct me if I am wrong.)

Now here is a little problem:
For organic molecules, there can be multiple local minima - but only one absolute total minimum.

The total minimum can correspond to a different structure in the gas and liquid phase, so do I compare structures and consider them or do I use the total minimum in both cases, I.e the lowest energy structure to calculate my thermodynamic data?

Reference:
Essentials of Computational Chemistry: Theories and Models, 2nd
Edition Christopher J. Cramer
http://eu.wiley.com/WileyCDA/WileyTitle/productCd-0470091827.html

The best way is to take all the structures into account and use Maxwell-Boltzmann distribution to weigh them accordingly. Here is what I would do
For each chemical structures/isomers (here I used F to denote Gibbs free energy),
$ F_{liq} =  F_{gas} +  F_{solvation}$
Now, $ F_{gas} =  E_{scf} - k_BTlnQ$
Where, $Q=q_{translational} \times q_{vibrational} \times q_{rotational} $
Using appropriate method (implicit solvation or QM/MM) you can find $F_{solvation}$. Now you have $F_{liq}$ for each isomers. Then use the following method to weigh them according to Maxwell-Boltzmann distribution.

In the above image, I use $\epsilon_i$ to denote $F_i$.


What is the correct reaction mechanism? I came across this question:

I propose two things to happen:

It can undergo Aldol condensation

or

it can undergo SN by replacing $\ce{-Cl}$ with $\ce{-OH}$.

But apparently, both these mechanism give the wrong product. I’m unable to find out what is going on.

I see what went wrong. Such questions are often confused with aldol or cannizarro as they have a carbonyl group and a base is given.
Actually this is a completely different name reaction called Favorskii rearrangement. The key to identify it is that its applicable to α-halo ketones.
In Favorskii rearrangement, the base takes α-hydrogen from that side of ketone which is opposite to the side having the halogen. It then forms a 3 member cyclic ring as an intermediate product. The product formed is a carboxylic acid derivative, mostly ester.
You can learn more at organic-chemistry.org or wikipedia