text:
N in Nernst Equation
My teacher told me that in the Nernst equation, N was the number of electrons transferred. For example, $n = 2$ in $\ce{Ni^{2+} + Zn -> Zn^{2+} + Ni}$.
However, say all of the coefficients were doubled: $\ce{2Ni^{2+} + 2Zn -> 2Zn^{2+} + 2Ni}$. Would $n = 2$ or $n = 4$ in that case?

You transfer four electrons now, hence $n=4$; it is as simple as that.

And not only does n increase, the reaction quotient changes as well. You may want to think about the properties of logarithms and work through both cases to assure yourself that it makes no difference either way.

The Nernst equation is as follows,
\begin{align*}
    E_\text{red}&=E_\text{red}^{0}-\frac{RT}{nF}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)}.
\end{align*}
where $c$ is the concentration of the species specified (usually the chemical activity, $a$, is used, but the equations follow the same principle). In the first case, $n=2$. So,
\begin{align*}
    E_\text{red}&=E_\text{red}^{0}-\frac{RT}{2F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)}.
\end{align*}
Suppose we double the coefficients in the equation. Now, $n=4$. However, the concentrations of all the species have also doubled. This requires the reaction quotient (the stuff inside the natural logarithm) to be squared. Therefore,
\begin{align*}
    E_\text{red}&=E_\text{red}^{0}-\frac{RT}{4F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)^2}.
\end{align*}
The laws of logarithms means that we can move the power of 2 to be a multiplication of the logarithm,
\begin{align*}
    E_\text{red}&=E_\text{red}^{0}-\frac{2RT}{4F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)},\\\\
    &=E_\text{red}^{0}-\frac{RT}{2F}\ln{\left(\frac{c_\text{red}}{c_\text{oxi}}\right)},
\end{align*}
which is the exact equation we had previously, thus, nothing has "changed".
So to answer your equation. Yes, $n=4$, but the equation does not change, as the reaction quotient alters accordingly, also.


text:
The effect of HCL concentration on the Solubility of Hydroxyapatite
Recently I conducted an experiment to investigate the effect of HCL concentration on the solubility of hydroxyapatite [Ca5(PO4)3(OH)]. This was done by taking a sample of the solution after hydroxyapatite was left to dissolve in HCL, and conducting an EDTA titration using this solution to find how many calcium ions were present. I have already established that hydroxyapatite is more soluble in solutions of HCL at higher concentrations (lower pH). However, I was wondering what the exact scientific theory behind this reaction was.

The $Ksp$ expression for hydroxyapatite is: $\ce{ Ca5(PO4)3(OH)_{(s)} -> 5Ca^{2+}_{(aq)} + 3PO4^{3-}_{(aq)} + OH-_{(aq)}}$.
When $\ce{HCl_{(aq)}}$ is added, it reacts with the dissolved $\ce{OH-}$ to form water, thereby decreasing the $\ce{OH-}$ concentration. Le Chatlier's principle predicts that more hydroxyapatite will dissolve (shift to the right).

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/

sort: votes


text:
How do I visualise bicyclo[4.4.1]undeca-1,3,5,7,9-pentaene?
While reading my organic chem book, I came across this compound:

Since this is an exercise question, I do not know the name of this compound.
I am unable to visualize this compound (how it looks out of the paper) as I have never come across two rings joined by a bridge-like structure, my book says it is a methylene bridge.
I appreciate your assistance in determining the structure of this compound.

Here: https://chemistry.meta.stackexchange.com/questions/3736/how-do-i-draw-chemical-compounds-and-chemical-reaction-mechanisms-and-add-it-to is something for beginners.

If you want to visualize any of the bridged polycycles, go read about norbornene.

Even adding proper formula would be good 'cause I'm not even sure about bonding in this thing, looking at the pic.

@Mithoron I am sorry for the inconvenience, but this is exactly how it has been drawn in my book. But it also says something about removing the hydrogen atoms from $[10]$ - annulene and replacing them with a methylene bridge.

Are you sure the structure is correct? It is very close to 1,6-methano(10)annulene (https://en.wikipedia.org/wiki/1,6-Methano(10)annulene). If so, I can post an answer with a picture of the crystal structure that might help.

@jerepierre I was thinking it may be this, otherwise it would be partially saturated ring. And from OP's comment, it indeed looks like it is the one.

@jerepierre You are right. It is 1-6methano(10)annulene since I verified some characteristics mentioned on the Wikipedia page with my book.

This compound is 1,6-methano[10]annulene, and a classic example of an unusual compound displaying aromaticity.
Aromaticity requires the pi-system to lie in a plane. One might speculate that [10]annulene (shown below) would be aromatic, because it contains a cyclic pi-system of 10 electrons, which appears to be planar. However, because the hydrogens indicated are trying to occupy the same space, the compound is forced out of planarity, and the compound it non-aromatic.

Bianchi and coworkers reported the crystal structure of 1,6-methano[10]annulene. In this compound, the hydrogens that clash in [10]annulene are replaced by a methylene (CH2) group. From directly above, the compound looks basically like naphthalene, with a little extra space in the middle. More interestingly, from the side, it's clear that the atoms of the 10-membered ring are nearly planar, with that methylene bridge oriented perpendicular to the plane.

Reference: Structure of 1,6-methano[10]annulene, R. Bianchi, T. Pilati and M. Simonetta, Acta Cryst. (1980). B36, 3146-3148, https://doi.org/10.1107/S0567740880011089

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Dissolving cast iron in nitric acid
Decades ago my job was to find the silicon content in cast iron gravimetrically. The first step was dissolving the sample in hot Nitric acid. Did this turn all of the constituents in the sample, except the silicon, into water soluble salts? Was the silicon then bound as silicon dioxide?

When heated, concentrated nitric acid is a strong oxidizing agent. The Fe is oxidized to Fe3+. I think the Si is oxidized to SiO2, but I'm not entirely sure about that part.

No, nitric acid is a sufficiently hard acid to dissolve silicon metal without precipitating in low concentrations. In fact, nitric acid is used to dissolve silicon in ICP Standards. The oxide reacts slowly with water and water-based solutions - nitric acid included, but the acid does not increase the kinetics appreciably, otherwise glass would be unsuitable as a vessel for nitric acid. The element in an alloy, however, can be dissolved more easily like that in cast iron.


text:
Removing water/moisture from car exhaust
I know that car exhaust is almost entirely nitrogen, water and carbon dioxide. How would I go about removing the water? Would I have to condense it off? Any devices already exist for doing so?

how much of the water do you want to remove/how dry do you want the exhaust to be?

Keeping costs reasonable, as dry as possible. Ideally no water content at all but I gather that'll be pretty difficult and expensive. Thanks in advance!

Certainly devices exist to remove water vapour from gas streams. Many labs pass their nitrogen stream through a conc. sulfuric acid bubbler to ensure it is dry. Solid absorbents such as silica gel or Drierite can also be used. The question is what volume of exhaust gas do you want to dry? These absorbents need to be changed frequently to maintain the dryness of the gas stream. If you want to keep a lot of exhaust dry you'll need a lot of drying agent.


text:
In Bohr's model what is angular momentum in the lowest orbital?
Angular momentum is given by: $n\frac{h}{2\pi}$ where $n=1,2,3...$
And angular momentum is also given by: $\sqrt{l(l+1)}\frac{h}{2\pi}$
By the first equation angular momentum should be $\frac{h}{2\pi}$ but by the second it should be $0$.
And if you say that in first equation $n$ is also $0$  then first radius of hydrogen should be $0\pu{A^\circ}$ instead of $0.53\pu{A^\circ}$.

You are trying to mix the Bohr model with quantum mechanics. The Bohr model is a semi-classical treatment of the hydrogen atom in which an electron is circling around a proton. Classically, this electron has angular momentum, even in the lowest orbital because it has a well defined position vector $\vec{r}_n$ and velocity $\vec{v}_n$. The classical angular momentum is thus given by
$$
l_n=m_e|\vec{r}_n\times\vec{v}_n|,
$$
where $m_e$ the electron mass. Bohr postulated that the angular momentum must be quantized (this is the semi-classical part) resulting in a set of orbitals.
Though Bohr's model works remarkably well it is not perfect because electrons are not really particles in the classical sense, but also behave like waves. A better treatment (but not the best) of the hydrogen atom is therefore to use quantum mechanics. In quantum mechanics the electron is described by a wavefunction that is spread out over all space and has a certain shape. For $s$ orbitals, such as the lowest orbital in the hydrogen atom, the wavefunction is spherical symmetric and has no angular momentum. The second expression you mention is the quantum mechanical operator to obtain the angular momentum.
To conclude, you cannot compare the angular momentum of the Bohr atom to the Schrodinger atom because they are based on different assumptions. Note further that for principal quantum number $n$, $\ell=0, 1, \ldots, n-1$.


text:
Why are non-aromatic rings more stable than anti-aromatic ones?
Is there any non-quantum explanation to the greater stability of non aromatic compounds as compared to  anti-aromatic compounds?

Why are you specifically looking for a "non-quantum" explanation? Perhaps I don't know what you mean by that exactly, but molecular bonds fall into the realm of quantum mechanics. Any explanation that ignores this is automatically wrong

https://chemistry.stackexchange.com/questions/45040/why-cant-antiaromatic-compounds-just-escape-planarity-and-become-non-aromatic-t


text:
What does it mean when something is said to "contain ions"?
(I'm looking for a very basic level explanation because my only chemistry experience is one fast-paced high school course.)
So, according to the professor of that course, ions are never found in nature as ions; everything would be neutrally charged. My question is, how can something have ions in it, without having them be neutralized? Doesn't nature want to lower potential energy, which would mean bonding the ions asap? Or am I misunderstanding things and said ions are already bonded to other ions, but they are still called ions even though the charge of the whole molecule should be neutral?

Could you give a specific example? What sort of compound do you imagine? Btw, I don't see the connection between the title and body of your question

what I meant was, how can ions just be floating about in some solution? wouldn't they try to bond with other ions? and an example is sodium chloride in  pure water, which "does contain ions."

I see. I don't have time to write an answer, but perhaps you would benefit from reading about solvation. Basically, yes, the ions attract each other, but also molecules with e.g. some dipole moment. In solution, their charge is then shielded by plenty of water molecules.

thanks for your time, nevertheless; I'll go look into solvation.

Your teacher is almost right but some terminology needs to be clarified.
Pure substances will almost always be electrically neutral (and the exceptions involve temporary local transfer of electrons from one substance to another as when you rub a rubber balloon on a wall surface). Beyond a certain level of charge separation you get enough electrical potential to cause sparks which allows the charges to equalise again. So persistent charge imbalance isn't easy to get. I think that is what your teacher means.
But if you look inside some pure substances, they do contain ions, just an equal number of each to avoid the bulk material having an overall charge. Common salt, for example, consists of ions (Na+ and Cl- in equal numbers). The substance is neutral, but the individual atoms making it up are all ions.
So your teacher is wrong if she meant that ions don't exist in nature but correct if she meant that bulk substances are electrically neutral.

I think I get what you mean. It would be wrong to say that a single ion can exist completely on its own. However, anions and cations can coexist in a 'substance' without forming molecules, if their collective charge is neutral?


text:
Why is the melting point of hydrogen fluoride lower than that of ammonia?
The melting point of hydrogen fluoride is -83.6°C, as compared to that of ammonia, which is -77.73°C. How does this make any sense?
Both hydrogen fluoride and ammonia show hydrogen bonding, which "elevates" the boiling points of these two compounds when compared to the rest of the pnictogen and halogen hydrides. However, I was under the impression that hydrogen bonding increased in strength the more elctronegative atom the hydrogen atom is bound to, explaining why the boiling point of ammonia is significantly lesser than that of hydrogen fluoride (-33.34°C and 19.5°C respectively).
However, this makes no sense when looking at melting points of the two elements.

Everything you say about the boiling points is true, but only when applied to boiling points. Melting points can't be compared quite as easily.

These are two fairly different compounds. Making such comparisons is sketchy at best.

@IvanNeretin why so?

@MaxW from what I can understand, they seem like very similar compounds. Why do you say they are fairly different?

Liquids are all alike; crystal structures are all very different and unpredictable.

@IvanNeretin is there any specific reason for that?

Crystals have structures, liquids don't.

@Ivan slight correction.  Liquids do have structure that we can resolve through radial distribution functions.  What they don't have is long range or directionally specific structure.

The additional energy involved here is the stabilization energy, a form of potential energy that is stored in the more stable arrangement of a solid compared to the liquid form. This energy is determined by many factors:

Size and shape of the individual particles.
How they can 'pack' into a solid arrangement.
etc...

SO, there is clearly something different about how Ammonia is able to arrange in solid form that makes the solid stable for a greater range of energy added.
When we look at a phase change diagram the slope and horizontal component of the line determine total energy needed for that phase change. This will be different for each substance considered, and this is where your discrepancy comes from.


text:
How are imaginary frequencies and force constant calculated by DFT?
Computational calculations allow us to simulate the frequencies of molecules. They can even tell us if the optimized structure is a minimum, a saddle point or a maximum according to the number of imaginary frenquencies.
I was wondering how do quantum calculations could yield these frequencies. Does it have any link to the force constants of a molecule ?

Possible duplicate of How to calculate wavenumbers of normal modes from the eigenvalues of the Cartesian force constant matrix?

Using a harmonic oscillator model, if you know the change in energy of displacement of atoms (the energy gradient) you can calculate force constants and from that and the mass of the fragments you can calculate frequencies.

@hBy2Py I don't think this is a duplicate; maybe your answer there answers the question here, too, but from the question itself I cannot find any good reason to close it as a duplicate.

@Martin-マーチン I know, I hesitated somewhat before casting the vote. This seems much more of a high-level question, rather than a specific question of the mathematical mechanics.  Community decision.

@hBy2Py On the very contrary, this is a very low-level question (and the tl;dr is given in DSVA's comment). This question is basically How to calculate the Hessian in the DFT framework? The linked question already assumes to have that done. This is not a duplicate.

@DSVA Is that worth expanding into an answer?

@DVSA I can see it being something asked in the future, so it would probably make a good dupe target.

Within DFT the vicinity of the solution to the Kohn-Sham equations is explored treating small changes of nuclei positions as a perturbation. The restoring forces that act in the system upon such perturbations give us information about the force constant matrix, from which we can get the harmonic vibrational frequencies as explained elsewhere, e.g. in the book of Wilson and Decius. Basically, you need the second derivatives.
A key point is to do it efficiently. Doing it numerically is very costly.
Since the basis functions are known and they have some nice properties one can do it analytically. Thus, researchers have devised a set of coupled equations to compute the effect of perturbations on the density matrix, which is the core of the calculation.
You will find a detailed explanation in this article by F. Neese.


text:
Which functional groups are not reduced by DIBAL.H?
According to the table here, diisobutyl aluminium hydride can reduce amides to aldehydes but according to the data given to me by my teacher DIBAL.H cannot reduce:  amides, acids, isocyanides and nitro groups.
It's I also given in my notes that DIBAL.H is  parallel to $\ce{LiAlH4}$ as a reducing agent but it is more selective.
I would like to know if DIBAL.H is able to reduce these functional groups or not.

https://en.wikipedia.org/wiki/Diisobutylaluminium_hydride - Seems the teacher underestimated it/

Here are more references: https://www.organic-chemistry.org/chemicals/reductions/diisobutylaluminiumhydride-dibalh.shtm DIBAL is definitely not LAH.

Quite a few reagents will actually react like their more reactive "equivalent" if the conditions are right e.g. NaBH4 will  reduce groups it allegedly doesn't, that LiAlH4 is used for. Solvent and temperature play a huge part. So DIBAL will reduce functionalities that "common knowledge"  says it won't. Conditions just have to be right.

DIBAL-H does the following:

Acid chlorides to aldehydes (Fast)
$3^\circ$ amides to aldehydes
Nitriles to aldehydes
Ketones to $2^\circ$ alcohols
Aldehydes to $1^\circ$ alcohols. (Slower)

Also, it works best in these solvents: THF, toluene.
We need to avoid solvents like alcohols, amines, strong acids and halogenated compounds.


text:
Understanding two-electron integrals in Gaussian 09
I was comparing the two-electron integrals (electron repulsion integrals, ERI) printed by various well-established packages (Gaussian, GAMESS, Molpro etc.) and observed that Gaussian is the oddball in including the factors coming from permutational symmetry ("redundancy factors").
Here I use water as an example, with the Huzinaga minimal basis set taken from bse.pnl.gov for consistency (built-ins in some packages had different primitive terms or number of decimal places).
\begin{array}{|c|c|}\hline
O&0.0&0.0&0.1173\\\hline
H&0.0&0.7572&-0.4692\\\hline
H&0.0&-0.7572&-0.4692\\\hline
\end{array}
I assumed no spatial symmetry (xyz coordinates) and used cartesian functions (not that it makes a difference with Huzinaga, but for later changes).
After the runs, the easiest one to understand was the output from GAMESS (option NPRINT=4) which left the integrals uncorrected for redundancy.  Molpro (INT,SPRI=2) printed some numbers that are off by factors of 2, 4, 8 but otherwise agree with GAMESS.
But I fail to understand the output from Gaussian (SCF(Conventional) IOp(3/33=6) ExtraLinks=L316 Symmetry=None).  Some of the numbers are off by factors of 2 or 4 (but not in the same entries as in Molpro) and some other numbers seem completely unrelated, even having opposite signs.
Q: Does anyone know the reason for this and how to read it correctly (to confirm agreement with other packages)?
Here are my outputs: GAMESS (line ~292), Molpro (line ~187), Gaussian 09 (line ~4810)
The indices of $[ij|k\ell]$ are in "canonical" order, ie. $i\ge j$, $k \ge \ell$, and ${i\choose2} + j \ge {k\choose 2} + \ell$.

@Tyberius: They're still alive from where I am ... Here's the link without the # reference to the lines.

I think for some reason they didn't work right on a Mac, but they are working on mobile. I believe I may have found the issue with the Gaussian integrals. By default, I believe it prints the so called Raffenetti form of the two electron integrals, which are particular combinations of the ordinary two electron integrals. You can see an example of this on the page for Gaussian's matrix element files

This would fit well on the new Matter Modeling SE.

@DCY, did you figure it out?


text:
Acetaldehyde with few drops of concentrated sulphuric acid

Acetaldehyde on treatment with a few drops of concentrated $\ce{H2SO4}$ gives?

Attempt:
First, I formed an enol intermediate in acidic medium $\ce{CH2=CHOH}$. Then, I performed electrophilic addition of another acetaldehyde molecule on this enol intermediate to get $\ce{CH3CH(OH)CH2CHO}$.
But answer given in my book is:

Please let me know the error in my attempt.

You have it kinda backwards, the enol O acts as nucleophile on a second protonated acetaldehyde molecule.

You're not wrong.
Your mechanism describes the classic acid catalyzed synthesis of 3-hydroxybutanal by dimerisation of acetaldehyde (Borodin 1869, Wurtz 1872).

Image: Whoopie23, Wikipedia
However, under different conditions acetaldehyde forms different addion and condensation products. E.g. with concentrated sulphuric acid at low temperatures metaldehyde is formed whereas at room temperature paraldehyde (the answer given in your book) is formed.

Under what conditions is the dimerisation product formed?

Suitable conditions can be either acidic or basic. Wurtz (see link above) describes the dimerisation with hydrochloric acid at -10°C.

Under different conditions, either enol C=C or aldehyde C=O can attack the protonated C+.  In the book's case it's the latter.  An example can be like this:

I'm only giving what I believe to be a reasonable explanation here, but I think it begins with protonation of the oxygen resulting in breaking of the $\ce{C=O}$ double bond and formation of a carbocationic alcohol. The next best nucleophile in solution would be another aldehyde molecule, so you get nucleophillic attack at the $\ce{C+}$ centre resulting in another $\ce{C=O}$ cleavage and the process repeating once more resulting in the 6-membered cyclic ring above. Enol formation would be unfavourable in this case as pKa of the alpha carbon on the aldehyde is much higher than anything else in the solution.
Formation of the 6 membered ring is both kinetic and thermodynamic in nature; 6 membered rings are simply favourable due to low torsional strain (thermodynamic) and the relevant atoms being relatively close in space (kinetic).

The way I've explained is confusing as the process would not be Sn1. Cannot form carbocation at primary carbon. But also Sn2 should not be possible at sp2 centre. Thus I can only assume it is a pseudo-sn2 where build up of charge on the carbon occurs and then attack occurs causing C=O cleavage.


text:
Calculate pH of a mixture of a strong base and acid. Knowing only the pH, wt/v%, and volume of both solution.
Title explains all. I have been stuck on this for an hour and for some reason cannot understand it. I have tried to do an ICE table but get stuck halfway as I do not know whether I would use the w/v to calculate the molarity and use that to neutralize the acid with the base and find the "free" H+?
I appreciate any insight!

In general the total equilibrium concentration of any acid HA is
$$c_a=\mathrm{[HA]_e+[H^+]_e - [OH^-]_e}\tag{1a}$$
and for a base
$$c_b=\mathrm{[B]_e -[H^+]_e + [OH^-]_e\tag{1b}}$$
When there is a strong acid and base the pH is not always what is expected due to the equilibrium $\mathrm{H_2O \leftrightharpoons H^+ + OH^-}$. This is particularly the case around pH = 7. To calculate the pH of a strong base start with eqn 1(a) but for the strong base the equilibrium amount of base $\mathrm{[B]_e}=0$ and so the equation to solve is
$$c_b=\mathrm{ -[H^+]} + K_w/\mathrm{[H^+]}$$
where $K_w=\mathrm{[H^+][OH^-]} = 10^{-14}$. This equation is, letting $x =\mathrm{[H^+]}$, $x^2+c_bx-K_w=0$ and solving for a concentration of, for example, $5\cdot 10^{-8}$ M produces $\mathrm{[H^+]}=7.8\cdot 10^{-8}$ or pH = 7.11.
In the case of a strong acid [HA] = 0 and equation (1a) solved but once the $\mathrm{[OH^-]}$ is known $K_w$ can be used to find $\mathrm{[H^+]}$
The figure shows the pH, for example, of a KOH solution at a range of concentrations. Around 10-7 molar the pH is greater than that anticipated. In a strong acid the pH is smaller than anticipated (more acidic) around this concentration.

Figure. The pH of a strong acidic and strong basic solution (e.g. KOH ) at a range of concentrations.

How did you write equation 1(a) ?

material balance on $H^+$. The conc'n of acid is amount undissociated plus $H^+$ less any hydroxyl present.

The conjugate base of a strong acid and the conjugate acid of a strong base are both neutral, and do not contribute to the pH of the mixture.
However, the $\ce{H+}$ ions in the strong acid and the $\ce{OH-}$ ions in the strong base react completely to form water when the strong acid and strong base are mixed: $\ce{H+ + OH- -> H2O}$. To perform the stoichiometry calculation, the number of moles of $\ce{H+}$ ions and the number of moles of $\ce{OH-}$ ions in the acid and base solutions, respectively, can be determined by multiplying the $wt/v\%$ by the volume of the respective individual solution, and dividing the wt by the molar mass of the acid or base.
Once the reaction goes to completion, the concentration of any excess unreacted $\ce{H+}$ or $\ce{OH-}$ can be used to calculate the resulting $\ce{pH}$.


text:
How do restricted open shell calculations mathematically proceed in the context of the self consistent field iterations?
I'm specifically concerned with HF theory. I understand restricted closed shell calculations use the Roothaan equations. If I understand correctly these equations assume each filled molecule orbital is doubly occupied. Each of the electrons in a pair is described with the same spatial wave function.
I also think I understand the Pople-Nesbet equations which allow the wavefunction for each electron to be described with their own spatial wavefunctions furthermore their spins are not constrained hence spin contamination can result.
I however, have not been able to figure out how restricted open shell calculations work. All I understand is that the spin is constrained. I do not understand if the wavefunctions of the paired electrons are described with the same wave function (similar to the Roothaan restricted closed shell approach) while the unpaired electrons are simply allowed to have their own spatial wave function. Coming back to the title question is there a bunch of equations with a famous person’s name (Roothaan, Pople-Nesbet) that are used to describe the scf procedure for restricted open shell calculations or does it use some combination of Roothaan and Pople-Nesbet?

Related: Restricted open-shell versus unrestricted methods for open shell systems

The paper by Tsuchimochi and Scuseria describes it quite well.
Briefly, Roothan-Hall equation is the same but the ROHF matrix is constructed from the $\alpha$ and $\beta$ Fock matrices from UHF.
$$\mathbf{F}_\mathrm{ROHF} =
\begin{bmatrix}
    \mathbf{R}_{cc}       & \mathbf{F}_{co}^\beta & \mathbf{F}_{cv}^{cs} \\
    \mathbf{F}_{oc}^\beta & \mathbf{R}_{oo} & \mathbf{F}_{ov}^\alpha \\
    \mathbf{F}_{vc}^{cs}  & \mathbf{F}_{vo}^\alpha & \mathbf{R}_{cc} \\
\end{bmatrix}$$
where $c$ is doubly occupied, $o$ is singly occupied, $v$ is virtual and
$$\begin{align}
\mathbf{R}_{cc} &= A_{cc}\mathbf{F}_{cc}^\alpha + B_{cc}\mathbf{F}_{cc}^\beta\\
\mathbf{R}_{oo} &= A_{oo}\mathbf{F}_{oo}^\alpha + B_{oo}\mathbf{F}_{oo}^\beta\\
\mathbf{R}_{vv} &= A_{vv}\mathbf{F}_{vv}^\alpha + B_{vv}\mathbf{F}_{vv}^\beta
\end{align}$$

Different values for these parameters have been suggested in the literature. Although they do not affect the ROHF wave function and energy, they affect orbital energies whose physical meaning is obscured because of this dependence.

Takashi Tsuchimochi and Gustavo E. Scuseria, J. Chem. Phys. 2010, 133, 141102. DOI: 10.1063/1.3503173; available as ePrint: arXiv:1008.1607 [cond-mat.str-el].


text:
Why are 4s electrons removed before 3d electrons?
According to Aufbau's Principle energy of 3d orbital is greater than 4s orbital, so which filling 4s if filled beforehand 3d. Whereas while removing electrons 4s electrons are removed first. Why?
I somewhere read that Aufbau's Principle is true only for ground state. When both 3d and 4s orbital get populated due to some interaction 3d shell is raised a to higher energy level than 4s. It it the case or it has some other explanation.

This might be interesting: https://www.chemguide.co.uk/atoms/properties/3d4sproblem.html (see also "Reference" Section of the page).

That's what I already read, lol


text:
Why is Zinc reduced instead of Hydrogen, in the electrolysis of aqueous Zinc Sulfate, using Carbon electrodes?
In the electrolysis of aqueous zinc sulfate using carbon electrodes, why are zinc ions reduced at the cathode, instead of hydrogen ions? How does this relate to over potential?
It would mean a lot to me, if someone could provide a detailed answer.

Zinc ions are reduced because there are much more of them. Should hydrogen ions be equally abundant, we'd have hard time getting any Zn at all.


text:
Why is CO ligand higher than CN- in the spectrochemical series?
I've drawn MO diagrams for both molecules and they are isoelectronic. The main difference is that $\ce{C}$ is closer in electronegativity to $\ce{N}$ than it is to $\ce{O}$ so there will be less stabilisation of the bonding orbitals for $\ce{CO}$ due to the increased energy difference in the AO's.
Does this difference lead to $\ce{CO}$ having its $\pi^*$ orbital (LUMO) lower than that of $\ce{CN^{-1}}$ so it more readily accepts $\pi$ backdonation from a metal centre? Does this alone account for why $\ce{CO}$ is higher in the spectrochemical series? What about the difference between the  $\sigma$ orbitals (HOMO) in $\ce{CO}$ and $\ce{CN^{-1}}$?

You're on the right track. It has to do with the energies of the frontier orbitals. As you rightly said, both species are isoelectronic, and the orbital energies in CO are lower than those in CN−.
The lower HOMO energy means that CO is a poorer σ donor orbital towards the metal than CN−. Likewise the lower LUMO makes it a better π acceptor.
These two factors are conflicting: stronger π acceptors are stronger-field ligands, but poorer σ donors are weaker-field ligands. Empirically, since CO is higher in the spectrochemical series, it stands to reason that the π acceptor effect is the most important factor here. This isn't really a surprise, as π-effects tend to outweigh σ-effects when it comes to the spectrochemical series.
I don't pretend that the above is a rigorous proof, but short of crunching the numbers, I don't think there's a much better explanation.


text:
How does the number of electrons increase in beta decay?
So, I am having trouble understanding beta decay. Particularly the number of electrons.
For example; Carbon-14 decaying into Nitrogen-14.
Carbon-14 has 12 neutrons, 6 protons and 6 electrons.
When it decays 1 neutron is converted into: 1 proton, 1 electron and 1 anti-neutrino.
So after the neutron breaks down, and before emission, there should now be:
11 neutrons, 7 protons, 7 electrons and 1 anti-neutrino.
But now 1 electron is emitted from the atom (as a beta ray), alongside the anti-neutrino. Which means the numbers should now be:
11 neutrons, 7 protons and...6 electrons.
Since 1 electron is created in the process and 1 is emitted out of the atom, then surely the number of electrons shouldn't change from the original 6 of carbon-14.
But Nitrogen-14 has 7 electrons?? Can someone explain?
Thanks.

The world is full of electrons. Go steal one from something else, like any other nitrogen anion.

Surely you must have heard of such thing as ions. What are they? Guess what? To put it bluntly, they are atoms with a wrong number of electrons.

Beta decay refers to nuclei. It does not take into account shell electrons.

The two commenters have pretty much answered your question: after beta emission, which is a sufficiently energetic process to tear apart a molecule, the resulting fragments/ions would likely undergo a series of additional reactions with any nearby molecules to finally produce stable species. The emitted electron also would be captured by nearby molecules, leading to a cascade of reactions, also eventually producing stable species. In any case you do indeed end up with a nitrogen atom and an extra electron.
Example: if decay occurred in a bottle of liquid methane, you would get ammonium cation (isoelectronic with methane) and an extra electron loosely attached to methane - a methane radical anion. These could recombine to give neutral ammonia and a hydrogen atom, and then further reactions are possible.
I would also point out that carbon 14 has 6 protons and 8 (not 12) neutrons, but in any case your question remains valid.


text:
What happened when the cherry juice turned blue?
When I was eating cherries today, some of its juice spilled on my shirt. Attempting to clean it, I poured a solution (baking soda, vinegar, toothpaste, and salt) onto the fabric. Though it did not remove the stain, the colour turned from a wine red to a subtle blue.
So the Question is: Why did it happen?
(This is not that technical so I'm not sure how to classify it)

After adding baking soda, I guess? Juice is acidic, but if enough base is added it may act as pH indicator and change color.

There are many naturally occurring pH indicators. See https://en.wikipedia.org/wiki/PH_indicator#Naturally_occurring_pH_indicators . BTW, red cabbage goes through the whole spectrum from very acid to alkaline!

Cherries, like red cabbage, hibiscus and many red or purple colored fruits, contain a class of compounds called anthocyanins of general formula:

where $\ce{R_x}$ is either $\ce{OH}$, $\ce{OR}$ or $\ce{H}$.
In basic medium it loses a proton and it converts to a non charged chinone like structure, which has a blue color.


text:
Octet rule violation in Period 2 elements
My textbook states,

Period 2 elements typically obey the octet rule. Period 3 and later elements can expand their valence shells. Elements in Periods 5 and 6 of the p-block show variable valence (the inert-pair effect).

I was surprised that it said period 2 elements typically obey the octet rule. What are some circumstances in which period 2 elements break the octet rule? I though they couldn't because they don't have available d orbitals.

Related: Which relatively simple molecules violate the octet rule?

If I could, I would "neutralize" a vote to close this question.  It is to me quite nontrivial.

related https://chemistry.stackexchange.com/questions/50381/why-does-boron-only-need-6-valence-electrons-unlike-the-standard-8

Typically the octet rule becomes less likely to work, regardless of the period, when we move to a group where there are too few valence electrons to go around.  Prominent examples in Period 2 include compounds of beryllium in Group 2 (such as the fluoride, especially when gaseous or liquid) and boron in Group 13 (as in alkyl boranes like this one).  Carbon can also form species that are most stable with fewer electrons than the octet rule would predict, a "violation" most prominently seen in the cyclopropenyl cation, where another "rule" that is much more reliable when properly used comes into play.


text:
Identifying principal quantum numbers in natural population analysis
Following a Hartree-Fock calculation, I would like my program to perform some kind of Natural Bond Orbital (NBO) analysis. Preceding this I would like to output a list of atomic orbitals, along with their identity (i.e. 2$p$, 3$d$) and their occupancy. The occupancy is simply gathered from a population analysis:
$$ occ_k = 2 \sum_{\mu\in k} (\bf DS \rm)_{\mu\mu} $$
Where $\bf D$ is the density matrix and $\bf S$ the overlap. Of course, the angular momentum of the relevant basis function provides us with the $s$, $p$, $d$, ... character. I understand that to find the principal quantum number $n$, and label a basis function as core, valence, or excited/Rydberg state, you can use the orbital energy (eigenvalues of the Fock matrix) and occupancy, along with some intuition.
My question is, does there exist some algorithm for this task which does not rely on such intuition? I have thought of using some method of 2-objective clustering of the occupancy and orbital energies, but this seems far to over the top for this task.


text:
Major product in Reimer-Tiemann Carbonylation

Attempt:
This is Reimer-Tiemann Carbonylation reaction in which the ortho product is major when phenol is the substrate. Hence, here I expected the ortho product to be major as well as it would also make hydrogen bonding possible, which stabilizes the compound. Besides, $\ce{-OH}$ is a better activator than $\ce{-OCH3}$.
But the answer given in the module is $C$ and not $A$. Where is my thought process wrong?

I think there is already a intramolecular hydrogen bonding between methoxy and hydroxyl in the Ortho position,it will not formed another hydrogen bonding in another Ortho position.hence forth it goes to para position giving vanillin.

You are right in that the $\ce{OH}$ is the main director, especially in alkaline media.
The para position is preferred over ortho because it has less sterical hindrance.
Therefore, the major product will de C.
In the case of phenol, there are two ortho positions and therefore there is more ortho than para substitution.
If it were a Kolbe-Schmidt reaction, instead of Riemer-Tiemann, that is, the reaction of sodium or potassium phenoxide with carbon dioxide, there is an interaction between the cation and the reagents.
In the case of sodium, although the mechanism is not clearly understood, there is an interaction between the sodium cation and the oxygen in carbon dioxide. Therefore, and because the reaction is performed without solvent, the carbon dioxide goes to the ortho position.
When potassium is used instead, there is evidence that the complex is between the aromatic compound and the carbon atom of $\ce{CO2}$.
In the case of the Riemer-Tiemann reaction, dichlorocarbene is formed in aqueous solution. In solution, the sodium and phenoxide ions are separated and hydrated, because both are perfectly stable. Therefore, the formation of a complex is very unfavorable, and its ortho directing effect negligible.
Besides, a complex between an alkaline cation and a covalent chlorine atom seems still more problematic, especially in aqueous solution.
The only complexes of alkaline cations with organics, of which I am aware, are the crown ethers, which is, of course, a completely different case.
For more details on both reactions, cfr. Jerry March (pp. 544-547 of the fourth edition).

Is it correct to conclude that the para product will always be major if there is any group ( irrespective of nature of group) at ortho?

The product of the Reimer-Tiemann reaction actually also depends on the counter cation of the base we use. It is found that if we use $\ce{NaOH}$ as the base used for formation of carbene electrophile, ortho product is major. But surprisingly, if we use the base $\ce{KOH}$, the para product is produced in major amount.  The reason behind this is the stability of the transition state via formation of an ordered cyclic arrangement.Due to the smaller size of $\ce{Na+}$, this can get fit between the Oxygen and Chlorine atom in ortho positioned electrophile via electrostatic attraction and give rise to a very ordered transition state which governs the major formation ortho product. See for details, https://chemistry.stackexchange.com/a/98197/58497
On the other hand, $\ce{K+}$ is huge in size related to $\ce{Na+}$, which obviously can't increase the stability of the ortho product by forming a cyclic orderd Transition state. That's why, the para product generally gets preference in the case of $\ce{KOH}$.

Do you have a source to support what you are saying?

You have mixed up the Reimer-Tiemann and Kolbe-Schmidt reactions. See my edited answer


text:
Regioselectivity of epoxide hydrogenation using Pt versus Ni catalyst

  What conditions are applicable for the transformations X and Y, respectively?
  (a) $\ce{LiAlH4}$, and $\ce{NaBH4}$
  (b) $\ce{NaBH4}$, and $\ce{BCl3}$ followed by $\ce{NaBH4}$
  (c) $\ce{LiAlH4}$, and $\ce{AlCl3}$ followed by $\ce{LiAlH4}$
  (d) $\ce{H2/Pt}$, and $\ce{H2/Ni}$

This question is about the hydrogenation of an epoxide (2,2,3-trimethyloxirane) to two different alcohols (2-methylbutan-2-ol and 3-methylbutan-2-ol). I put (c) as my answer, but apparently (d) is also a correct answer as well.
I have never seen the difference in reduction by change of catalyst, so had no idea about option (d). Is there any difference?

First of all, I should clarify that hydrogenolysis of epoxides does not always proceed with such predictable regioselectivity simply based on a switch in catalyst. Often it depends on the substituents present, as well as their relative configuration (for example, cis/trans for 2,3-disubstituted epoxides).[1]
In Reaxys I find no records of the epoxide in the question (2,2,3-trimethyloxirane) being hydrogenated. However, I did manage to find one study in which the authors measured the rates of hydrogenolysis of propylene oxide over Pt and Ni, finding that Pt gives more 2-propanol and Ni gives more 1-propanol.[2]

On top of that, there are also a few more examples that show similar reversals in regioselectivity for Pd (not Pt) and Ni.[3,4]

Bartók and Notheisz have put forward an explanation for the difference in regioselectivity between Ni and Pd/Pt.[2] They suggested that the key step is an oxidative addition of the metal into the C–O bond in the epoxide. This is fairly logical, considering the weakness of the C–O bond in a strained three-membered ring. For a Pt catalyst, the regioselectivity is determined by steric effects and insertion into the less sterically hindered C–O bond is preferred:

On the other hand, the authors write that for Ni

Because of the higher electron affinity of Ni, its insertion becomes ionic, formation of the surface species is controlled by electronic factors, and rupture of the C–O bond precedes formation of the Ni–C bond.

In modern parlance the way of expressing this is that the oxidative addition is asynchronous: the two new bonds are not being formed at exactly the same time. Because of the greater electron affinity of Ni compared to Pt, the Ni–O bond forms earlier than the Ni–C bond. If we were to draw a mechanism using curly arrows, it would be something like this: the flow of electrons depicted by the red arrow occurs earlier than the green arrow.

Consequently, there is a buildup of positive charge on the carbon undergoing bond breaking (hence "ionic"). This is not unlike the acid-catalysed opening of epoxides. The "electronic factors" that the authors refer to is simply the fact that the more substituted carbon is better able to carry this positive charge, which leads to the reverse in regioselectivity.

References

Murai, S.; Murai, T.; Kato, S. Reduction of Epoxides. In Comprehensive Organic Synthesis; Trost, B. M., Fleming, I., Eds.; Pergamon: Oxford, U.K., 1991; Vol. 8, pp 871–893. DOI: 10.1016/B978-0-08-052349-1.00249-3.
Bartók, M.; Notheisz, F. Stereochemistry of the hydrogenolysis of oxacycloalkanes on metal catalysts. J. Chem. Soc., Chem. Commun. 1980, 667–668. DOI: 10.1039/C39800000667.
Mitsui, S.; Kudo, Y.; Kobayashi, M. Stereospecific hydrogenolysis of benzyl-type alcohols. Tetrahedron 1969, 25 (9), 1921–1927. DOI: 10.1016/S0040-4020(01)82813-1.
Augustine, R. L. Organic Functional Group Hydrogenation. Catal. Rev.: Sci. Eng. 1976, 13 (1), 285–316. DOI: 10.1080/00087647608069940.


text:
Is there a finite number to the possible state of electron in an atom?
Okay, I know you may say, there are infinite numbers of possible state of an electron between its ground state and ionic state, since energy levels get continuous as electron gets farther away from the nuclear. But that's not what I mean.
Here is my concern: Any linear combination of eigenstates is also a state for an electron in an atom. For example, an electron in 3d orbital (n=3, l=2) is thought to have only five possible states: dxy, dxz, dyz, dx2-y2, dz2, with literally no m eigenvalue. Yet, what about there linear combinations? Aren't they also possible states of an electron in 3d?
Thanks for any help!

Well, if you have two or more degenerate eigenstates, then any their linear combination is also an eigenstate, and you have a continuum of those.

You can have a superposition of any number eigenstates, not just for same n and l. It just won't be an eigenstate anymore (except for a few special cases). What do you mean by "no m eigenvalue".?

For degenerate eigenstates any (normalized) linear combination of them will be again an eigenstate of same energy. This in fact means that there are infinitely many of them.
So where does the 5 (in case of $d$-orbitals) come from, when we actually have infinitely many? It means we can select at most 5 of them, which are orthogonal (or more generally speaking: linear independent). We are free to choose which 5, and thus there are (infinitely) many set of 5 $d$-orbitals.
Here are 2 possible sets (again for the example of $d$-orbitals) that are typically used:

The general solutions are $Y_l^m(\theta,\phi)=f_l^m(\theta) g_m(\phi)$ for $l=2$ and $m=\{-2,-1,0,1,2\}$. Where we have
\begin{align}
f_l^m(\theta) &= (-1)^m \sqrt{\frac{(2l+1)(l-m)!}{4\pi (l+m)!}} P_l^m(\cos \theta) \\
 g_m(\phi) &= e^{i m \phi}
\end{align}
and $P_l^m(\cos \theta)$ are the associated Legendre polynomials. As those are complex functions due to $g_m(\phi) = e^{i m \phi}$ (except for $m=0$), those are hard to plot or imagine.
The real linear combinations from the above:
\begin{align}
    d_{z^2} &= Y_2^0(\theta,\phi) \\
    d_{x^2-y^2} &= \frac{1}{\sqrt2}\left[Y_2^{+2}(\theta,\phi) + Y_2^{-2}(\theta,\phi)\right] \\
    d_{xy} &= \frac{1}{\sqrt2 i}\left[Y_2^{+2}(\theta,\phi) - Y_2^{-2}(\theta,\phi)\right] \\
    d_{xz} &= \frac{1}{\sqrt2}\left[Y_2^{+1}(\theta,\phi) + Y_2^{-1}(\theta,\phi)\right]\\
    d_{yz} &= \frac{1}{\sqrt2 i}\left[Y_2^{+1}(\theta,\phi) - Y_2^{-1}(\theta,\phi)\right]\\
\end{align}

would the different linear combinations  have different shape?

@santimirandarp Yes, they do.


text:
Can you convert smells to variables?
Is there a way to map smells to numerical values of different variables?
I.e. With RGB colours, each colour is made up of numerical values from 0-255. For example, red is (255,0,0), and yellow is (255,255,0). Is there a way to do this with smells. I.e, the smell of Candy is (140,70,255), and the smell of grass is (255,34,27)? How many different sets of numbers (like red, green and blue) would you need, and what would they represent?

Not really a chemistry question, your question would be more appropriate for Biology.SE.  That said, there are around 400 types of receptors for the sense of smell in humans so I'd say plausible depending on the degree of perfection you desire.

I'm voting to close this question as off-topic because it belongs on Biology.SE

I'm not familiar with Biology.SE, but if they are like us, they too will close this question as offtopic. As to the point of the question, I never heard of such a map for smells, nor do I have any idea on approaching this problem. Then again, the idea of a map for colors is also far from obvious, yet here is RGB. Maybe the problem with smells is inherently more difficult.

Smelling is not a 3D problem, but more like 400D. That's the approximate number of different receptors in the human nose.

Thanks guys. Sorry for posting to the wrong place/having such a strange question. Wasn't really sure where to place it. So far, no responses on the Biology SE. Do you think that there would be a way to graph the 400D/know where I could find any data on this?

There are three primary colors, so you can use three numbers to describe a color (the RGB and CMYK models do not use primary colors, but that is just for convenience). There are five primary tastes, so you could represent any taste with five numbers. There is not any consensus about the number of primary smells, the lowest numbers being seven or ten, depending on the source, but they seem to be very limited. Once the primary smells are fixed, you could use a vector to describe each smell, but it surely will be a huge one.

Also, the response works in essentially different ("non-linear") ways. Light blue and dark blue colors look related to us. Not so with smells. See https://chemistry.stackexchange.com/questions/62342/what-constituents-in-soft-drinks-are-the-cause-their-strong-pungent-smell/62439#62439

A very interesting read. Thanks Ivan. And thanks Raoul for explaining so clearly.

It does not work that way, or, we don't yet know, how it works. But an interesting term/keyword is often used, olfactory coding.


text:
Rietveld (FullProf) general background function
I was reading through Eric Mittemeijer's book 'Modern Diffraction Methods' when I found the following statement in the Rietveld chapter by Robert Dinnebier.

It is common practice to use orthogonal Chebyshev polynomials of higher order
  (typically 5–10) in combination with an 1/(2θi) term to describe a steep increase in the background at low scattering angle [...]. If higher-order polynomials are used, different types of correlations can occur, which require careful checking of the correlation matrix.

When I am usually describing my background in FullProf I use the '6-Coefficients polynomial function'. But this seems to cause more problems the more parameters I refine. So I switched to the 'Chebyshev polynomial' and refinded all 24 parameters (just to make sure). This caused a large jump in the X² from about 10 to 6.5 while the background was much better described.
My question is, although it is mentioned in the book, that Chebyshev is more common and I should be careful with using higher order polynomials, is this something I can use in general for my refinements? Can I just automatically assume for most x-ray determined backgrounds that using a Chebyshev polynomial with as many parameters refined as possible will give me useful results? By useful I mean good and especially 'allowed' results. Because sometimes FullProf refines parameters into regions where they are not defined like a negative η for the Pseudo-Voigt function.

If I recall correctly, Chebyshev polynomials are preferred over simple polynomials because of their termination behavior (i.e. they tend not to diverge wildly), although I don't have a reference for this handy.
Dinnebier's admonition to use higher-order polynomials with care is effectively a warning not to over parameterize your fit. If you inspect the two fits with 6- and 24-term backgrounds, my guess is that you will find the over parameterized version nudged your Bragg peaks up or down as necessary to better match your observed data. Hence, your residual has diminished but your fit is not more meaningful.
An overfitted background precludes accurate extraction of the observed Bragg intensities, leading to correlated errors in related factors (e.g. occupancy, position, etc.).

Welcome to Chemistry.SE! To help you become acquainted with the site, I would just like to suggest if you haven't already, please take a minute to look over the help center and tour page to better understand our guidelines and answer policies. To learn more visit Meta.Chemistry, Meta.SE, or chat

In my understanding, Chebyshev polynomials diverge as wildly as other polynomials, but they are orthogonal (in some sense), and as such, produce more independent refinement variables.


text:
Does the ignition temperature of a substance decrease with the concentration of oxygen?
Is it possible for a substance to undergo combustion at a temperature lower than its ignition point if the concentration of oxygen increases?

You asking a very general question for which there are likely to be exceptions. However, in general, if the oxygen concentration goes up (e.g. replacing an inert gas) then it would be easier to ignite the substance. The most notable exception is too much oxygen. Since ignition is a chain reaction, there has to be enough of the substance too.

i feel this question very interesting since i thought the same way. I also wish to test impact of air pressure on combustion if possible someday.


text:
Is *rinsing* (after washing) with cold water is as efficient as rinsing with hot water?
Let say I have, a dishwashing soap, plenty of cold water (5 °C) and only have 2 L of hot water (60 °C) to clean a t-shirt with oil stains. Should I use the 2 L to wash the t-shirt and then the cold water to rinse the shirt or should I wash the t-shirt with only 1 L of hot water and then use the second liter for a first rinse and then the cold water to finish rinsing?
If the soap has already "captured" the fat molecule (in the micelles), I shouldn't need the hot water to rinse the micelles, right?

Every sensible washing machine rinses with cold water.

Depends on carry-over & solubility
Generally, when you have two cleaning fluids, one of which is less effective than the other (the "rinse"), you need to use the good solvent to get the contaminant concentration below the solubility limit of the poor solvent, before switching solvents. Otherwise, you will need a heck of a lot of the poor solvent (albeit that in this case it is only cold water.) How do you do that with only 2 L of good solvent (soapy hot water)?
This sort of question is formally studied in both industrial cleaning plants, and decontamination processes. Determining the ideal process is not exactly rocket science but it is fairly complicated.
It depends on the solubility of the contaminant in the washing fluid -- usually, the dynamic solubility, since processes like these can't sit around waiting for a long soak. However, even if the contaminant is perfectly miscible in the cleaning fluid, it still also depends on "carry-over": the fraction of cleaning fluid that remains on or in the cleaned object between rinses.
In the case where the contaminant is miscible and the carry-over fraction is very small, then you actually get the best results (theoretically, at least) from a very large number of very small rinses. But that won't apply in your case: a wet t-shirt is capable of holding quite a large fraction of 2 litres.
For more practical size of problem like yours, the optimal solution is challenging to find analytically, but straightforward to find numerically. For example, you can set up a simple simulation in (say) MS Excel and run the Solver to get lowest residual concentration against an input parameter of "fraction of hot water used for each wash."
Oh, and it simplifies things considerably to know: you get best result when each titre is the same. That is, you don't need to consider using, say, 2/3 of the hot water on a first wash and 1/3 on a second. If you have n washes, just use 1/n each time.
Incidentally, for many practical parameters, the solution is so sensitive to carry-over fraction that often the most economical approach is to minimise carry-over; in your case, give the t-shirt a really thorough wringing out between rinses!


text:
Molecular orbital diagram for terminal metal-oxo complex with dative bonds
I am interested in terminal metal-oxo complexes and their electronic structure, specifically those with coordination numbers of 6, including the oxo ligand.
I know that Fe(IV)-oxo complexes tend to exhibit the 1+2+1+1 $d$-orbital energy scheme shown in the 3rd complex below. With this, I have two questions. Let's consider an Fe(IV) center for consistency.

Is there any way to predict, without performing calculations, what might occur if the ligands are not all identical? Does that change the likelihood of a 1+2+1+1 arrangement?
How does one account for the presence of dative bonds (e.g. if two of the equatorial ligands form dative bonds whereas the rest form traditional covalent bonds)?


text:
IUPAC Nomenclature of the organic compound(See Picture)

Question:IUPAC nomenclature of the given organic compound
Given answer is Option 4

All suggested answers (1)–(4) are wrong. But anyway, what is your question?

Question : IUPAC name of this compound

In the compound that is given in the question, only the amine can be expressed as a suffix; therefore, the principal characteristic group can only be the amine. The compound has also only one parent structure; therefore, there is no question about which chain or ring is the senior parent structure. Any further substituents are expressed as prefix. Therefore, the compound is a chlorocyclohexenamine. The only remaining question is about the correct numbering of locants.
The most important simplified criteria for the numbering in such cases are:

lower locants for the principal characteristic group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in 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:
(…)
(c) principal characteristic groups and free valences (suffixes);
(…)
(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;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The lowest locant is assigned first to the principal characteristic group according to rule (c), i.e. 1-amine.
Next, a low locant is assgined to the double bond according to Rule (e). Therefore, the name of the structure without further substituents is cyclohex-3-en-1-amine.
Now the locant for the chloro substituent can only get the number 5.
Therefore, the complete name for the compound that is given in the question is 5-chlorocyclohex-3-en-1-amine.


text:
Which one of them cannot function as a buffer solution?
A) HCN and KCN
B) NH3 and (NH4)2SO4
C) HF and NaF
D) HCI and KCI
E) HNO2 and NaNO2
Which one of the following combinations cannot function as a buffer solution and why?

Welcome to Chemistry.SE.  Homework questions are okay but some attempt to answer/understand the question must be demonstrated.

A buffer can be made either by partially titrating an acid or having a weak acid with its conjugate base.
In options A, B, C, and E, there is a weak acid (HA) with it's conjugate base (A-). However, in D, there is HCl, a strong acid, with Cl-. The HCl/KCl system is not a buffer because the $\ce{Cl-}$ anion is the conjugate base of a strong acid, and therefore remains deprotonated; the $\ce{Cl-}$ anion does not contribute to the overall $\ce{pH}$. Since $\ce{HCl}$ is the only source of protons, and there is no other species to take up the protons, the HCl/KCl system is not a buffer.


text:
Distinguish between two possible configurations for angular momentum in carbon atom
Consider the possible values of $S$ and $L$ for carbon configuration $1s^2 2s^2 2p^2$ and the corresponding rapresentations with arrows indicating the spins (consider only $S=0,L=0$ and $S=0,L=2$, as for $S=1,L=1$ some are missing).

I can't understand which one of the two representations indicated in red belong to $S=0,L=0$ state and which one to the $S=0,L=2$ state (I took a guess making the picture).
The spin wavefunction is in all cases antysymmetric ($S=0$) and equal to
$$\chi =\left(\, |\!\uparrow_1 \downarrow_2\rangle-|\!\downarrow_1 \uparrow_2\rangle\,\right)\div\sqrt 2$$
Therefore the spatial wavefunction $(L)$ must be symmetric for Pauli principle and in the second red configuration is
$$\Psi =p_0 (1) p_0 (2)$$
in the first one  is
$$\Psi= \left[p_- (1) p_+ (2)+p_+ (1) p_- (2)\right]\div\sqrt2$$
The wavefunctions are different so there must be a difference between the two red configurations, but which of them should belong to $S=0,L=2$ and the other to $S=0,L=0$ and why?

You need to make a table with the columns as $m_l$ and $m_s$  and also the totals (sum of $m_l$ which gives values -2,-1,0,1,2, and $m_s$, values -1,0,1) then remove those not possible by Pauli principle (those no good and also duplicates). There are 15 possible configurations in all. You can then identify those you need. The are 5 with L=2, 9 with L=1 and 1 with L=0.

You are separating the spatial and spin part of your two wavefunctions. I don't think that is a good idea here, because it does only work for one of the two configurations.
The wavefunction of a configuration is given by a Slater determinant build up from spin orbitals. In the following I will neglected the normalization constant of $\frac{1}{\sqrt2}$ for simplicity. The wavefunction for the first configuration ($\uparrow - \downarrow$) is
\begin{align}
   \Psi_1 &= \begin{vmatrix}
     p_-(1)\alpha(1) & p_+(1)\beta(1)\\
     p_-(2)\alpha(2) & p_+(2)\beta(2)\\
     \end{vmatrix}\\
    &= p_-(1)\alpha(1)p_+(2)\beta(2) - p_+(1)\beta(1)p_-(2)\alpha(2)
\end{align}
Note that this differs from your suggested wavefunction by the spin part of the orbitals, and there is no way to separate them out.
And for the other configuration ($-\uparrow\downarrow -$) we have
\begin{align}
   \Psi_2 &= \begin{vmatrix}
     p_0(1)\alpha(1) & p_0(1)\beta(1)\\
     p_0(2)\alpha(2) & p_0(2)\beta(2)\\
     \end{vmatrix}\\
    &= p_0(1)\alpha(1)p_0(2)\beta(2) - p_0(1)\beta(1)p_0(2)\alpha(2)
\end{align}
Both, $\Psi_1$ and $\Psi_2$ look quite similar now. However, since the spatial part of both orbitals in $\Psi_2$ is the same here, we can factor them out
\begin{equation}
  \Psi_2 = p_0(1)p_0(2)\underbrace{[\alpha(1)\beta(2)-\beta(1)\alpha(2)]}_{\chi}
\end{equation}
And we get what you have suggested for this configuration. But this does only work for $\Psi_1$!
Now, why is $\Psi_1$ assigned to $L=2$, while $\Psi_2$ is $L=0$?
I assume your confusion is because if you add up the orbital quantum numbers $m_l$ you get $0$ in both cases ($1-1=0$ and $0+0=0$). However, those sums give you $M_L$, not $L$. The many-electron quantum number $L$ has $2L+1$ components with quantum numbers $M_L=-L,-L+1,\dots 0, \dots L-1,L$. So for $L=2$ we have $M_L=-2,-1,0,1,2$, much like there are three orbitals $p_-$, $p0$ and $p+$ for $l=1$
So $\Psi_1$ would be the $M_L=0$ component of $L=2$, while $\Psi_2$ is the $M_L=0$ component of $L=0$.
But there is one issue with such assignments: You cannot really do it, as the configurations get mixed together. The point is, that the atom as a whole needs to be spherical symmetric (unless there is some external electric or magnetic field). But the individual $p$ orbitals are not. For example $p_0$ is (in the usual convention) aligned along the $z$-axis and zero in the $xy$ plane.
Mathematically speaking the configurations form a basis (Hilbert space) in which the actual electronic states are expanded. This is known as the configuration interaction method.
So for example the electronic state with $L=0$ would be
\begin{equation}
\Psi_2 = \frac{1}{\sqrt3}(|\uparrow\downarrow - -\rangle + |-\uparrow\downarrow - \rangle + |--\uparrow\downarrow \rangle)
\end{equation}
where each $|\dots\rangle$ denotes a Slater determinant.


text:
Bond Angle in Diborane, B2H6
Why are the bond angles in the diborane $\ce{B2H6}$ ring much less than $\ce{109.5^o}$? Below is the structure:

This source says that the boron atoms are approximately $\ce{sp^3}$ hybridized, so I thought that the bond angles of the ring would be approximately $\ce{109.5^o}$. However, the diagram indicates that the angles of the ring are $\ce{97^o}$ and $\ce{83^o}$. Why is this so?

It's a 4-membered ring...

Diborane is better not thought of in terms of hybridization and bond angles. Think of cyclobutane: no multiple bonds, hence $sp^3$, hence should be $109.5^\circ$, but no, it is more like $90^\circ$. How so?

So it has ~90 degree bonds to relieve some torsional strain?

Related: https://chemistry.stackexchange.com/questions/228/what-makes-banana-bonds-possible-in-diborane (full disclosure, it's my question).  @IvanNeretin hope you can forgive our hybridization slant to the question :)

Possible duplicate of What makes banana bonds possible in diborane?

It doesn’t seem to me that the proposed duplicate discusses bond angles.


text:
Rationalising diastereoselectivity of hydroboration using stereoelectronics
Using bulky reagents such as 9-BBN, hydroboration is often highly diastereoselective, giving rise to the 1,2-anti diastereomer:

This is easily explained using a steric argument, with the major product resulting from minimisation of allylic strain in the transition state. This is essentially the Houk model, though in the context of hydroboration is often described as the reactive-conformer model.
While a steric argument works in many cases, it isn't fool-proof – increasing the size of a group does not necessarily increase the selectivity as would broadly be expected by the above model.
To rationalise this, Houk has proposed a stereoelectronic component in the transition state – this is perfectly plausible as electronics evidently contribute to the conformational preference of the molecule. What Houk does not expand on is the exact nature of the stereoelectronic effects (this is not unusual, stereoelectronics are often invoked to rationalise unexpected experimental results without justification).

What is the nature of the stereoelectronic component when determining diastereoselectivity of hydroboration and in what situations should it be invoked?

This is a great question. If we were all honest, then we would call steric interactions/effects as that what they are: electronic effects. If they had a (likely) mechanism candidate, then it were probably possible to investigate this from a quantum chemical point of view; and even then it would be a snapshot. I think generalisation on this topic is difficult. It is a quite interesting to investigate though, and I'm quite certain that we'll see more of this eventually.


text:
In regard to collision theory, why do molecules have to have the correct orientation in order to react?
Hypothetically speaking, if a reagent in a bimolecular reaction has enough energy to break the bond of the other reagent but is orientated in the wrong direction, why does the bond not break? Does the reagent's orientation affect the activation energy of the reaction somehow? The point I'm trying to make is if you have enough energy to break a bond, why would molecular orientation matter when the same amount of energy is being used to distort bonds regardless of orientation?

Suppose we have a simple reaction of a diatomic and a single atom $\ce{AB + C ->A +BC}$. So the reaction is breaking the bond between $\ce{A}$ and $\ce{B}$, which then bonds to $\ce{C}$. So we might expect the transition state to look like $\ce{A--B--C}$. This suggests the reaction is highly orientation dependent.
Looking at the process as a function of the angle formed by ABC we would expect that a 0 degree angle would vary likely lead to a reaction, since everything is lined up as we would expect, while a 180 degree angle would have almost no chance of forming the desired product as it would aligned more like this $\ce{C--A--B}$. While its not a perfectly linear relationship, we should expect that the reactivity will decrease monotonically as it goes from 0 to 180 degrees.
For more complex structures, the orientation becomes even more important. A large molecule may only have a very small reactive site, and if the other reactant collides with a completely different section of the molecule, there is virtually no chance the desired reaction will occur.

I understand that if the nuclei of the reagents don't collide correctly the reaction won't occur. My gripe is when temperature remains constant why wouldn't the "AB" bond break since the same amount of energy is being transferred to the molecule.

@henrey2306 well if it was ~180 degrees it might still break, it just wouldn't form the same products. If however, the angle was closer to 90 degrees, the bond would be unlikely to break at all because while the particles may be travelling with the same amount of energy in each case, the orientation of the collision determines how much of that energy goes into breaking the bond. If the particles only skim each other or collide at an odd angle, the energy of the collision doesn't go into breaking the bond.

If you have two huge molecules, say, a sterol in you want to change an OH by some reaction, then clearly the way the reactants approach one another is important.
In the case of a reaction such as $\ce{Cl + OH \to HCl + O}$ then it does depend on the angle of approach. This is because the potential energy is not spherical around the diatomic but is different when approaching along the 'line of centres' compared to approaching at right angles to the OH bond.
Additionally the vibrational energy in a diatomic can have a vital role, not only in how many quanta are excited but at what point in the vibrational motion the collisions occurs. The effect is such that even if there is enough total energy to overcome the barrier to products, i.e. to get over the transition state, vibrational motion can easily frustrate this and the reaction heads back on itself after climbing up towards the transition state.

You don't have enough energy to break the bond. The reaction's activation energy is (in most cases) lower than the energy level needed to first break one bond and only then form the other.
If your thermal energy is high enough, the two particles form a complex transition state IF the geometry is right. This transition state decays then forwards or backwards. If the geometry (orientation) is wrong, you don't get the transition state. Orbitals with wrong orientation don't overlap, energetically, and you absolutely need the energy from the new bond that is already forming to loosen the old one.
If your thermal energy was actually high enough to just break the bond, that could happen at every collision, also e.g. with inert gas/solvent molecules, or when the molecule bumps together with its own kind. That's called pyrolysis.


text:
Lewis structure for magnesium diboride
What is the Lewis structure for $\ce{MgB2}$? I just read that it is a superconducting material (up to a relatively high critical temperature, 39 K) and am wondering if this has to do with its electronic structure.
In $\ce{MgB2}$, $\ce{Mg}$ has 2 valence electrons and each $\ce{B}$ has 3 for a total of 8. I know that $\ce{B}$ is "satisfied" with 6 valence electrons, but I don't see how to get 8 around $\ce{Mg}$ with 6 on each $\ce{B}$.

It's not really a molecule, so you don't really have a Lewis structure consisting of one magnesium and two boron atoms.  For a lattice structure see Wikipedia.  The lattice structure is hexagonal with alternating magnesium and boron layers, the boron layers having twice as many atoms per unit area as the magnesium layers.

The referenced picture is given here  and consists of green balls (magnesium) plus pink balls (boron), the latter covalently linked with each other.  If we model the magnesium as $2+$ ions (yes, the electronegativity difference is only 0.8, but de la Mora at al.[1] indicates that the magnesium-boron bonding is predominantly ionic), then each boron atom has one negative charge and so becomes isoelectronic with carbon.  Then each of the hexagonal, covalently bonded boron layers is isoelectronic and isostructural with a graphite layer.
Reference
1.
Pablo de la Mora, Miguel Castro, Gustavo Tavizon (2002),
"Comparative study of the electronic structure of alkaline-earth borides (MeB2; Me=Mg, Al, Zr, Nb, and Ta) and their normal-state conductivity",
Journal of Solid State Chemistry 169(2), 168-175, ISSN 0022-4596,
https://doi.org/10.1016/S0022-4596(02)00045-2.

Thank you, I'll review the Wikipedia page with your comments in mind. I appreciate the quick response.

@SteveSchmidt also check out the Wikipedia article on borane clusters.


text:
IUPAC Name clarification
$\ce{CH3-CH(F)-CH2-CH2-CH(C2H5)-CH2-CH3}$
Will the correct IUPAC name be 3-ethyl 6-fluoro heptane or 5-ethyl 2-fluoro heptane?

The one with the lowest set of locants is correct. I.e. the one with lower first number in the locant set; if they are same, then the second one, etc. But, it is done on the ordered sets of locants, (3,6) vs (2,5) not (5,2). The substituent prefixes are then sorted alphabetically. So, the correct name is 5-ethyl-2-fluoroheptane.


text:
Heat generation in exothermic reactions
Im thinking specifically of N2 + 3 H2 ⇌ 2 NH3. Its an exothermic reaction, therefore heat is generated. Im trying to understand how Chateliers principle interacts with this reaction. I understand that if you remove heat, Chateliers principle dictates that the equilibrium shifts to the product-side. I cant quite understand why though, do the reactants kind of know how warm it is and then decide, if its too warm "No, its too warm, lets not react and make it too warm"?
A followup on that question too, these reactions happen in enclosed places where temperature and pressure can be manipulated.Is it just efficiency of these containers that prevent the containers to not get as hot as the sun? If temperature wasnt manipulated, would the temperature keep rising?

For your second set of questions: you only have a finite amount of reactants, and hence can only generate a finite amount of heat, so the temperature would only rise until the reaction has proceeded to equilibrium.

Unlike the other factors affecting chemical equilibrium, (namely concentration, pressure), temperature follows the Le Chatelier principle via a different mechanism.
Temperature alters the equilibrium constant $K$, and this change in $K$ also changes the relative concentrations of the reactants and the products. The way in which the equilibrium constant depends on temperature is given by Van't Hoff equation:
$$\frac{d \ln K}{d T}=\frac{\Delta H}{RT^2}$$
Therefore, for exothermic reactions, the change in $K$ acts as if heat is one of the products, since increasing the temperature causes the $K$ to decrease, making the equilibrium at lower temperatures more favorable for Haber's process for making ammonia. (While this only talks of the equilibrium concentrations, lower temperatures in general result in slower rates, hence a balance is always sought)
While for your other question, as the temperature increases, the equilibrium is reached faster by both increasing ammonia (the cause of the increasing temperatures) and the decreasing $K$, until at a certain temperature, the reaction reaches equilibrium. I am not sure of the practical constraints employed in temperature regulation in the industry.


text:
Why does reduction using sodium borohydride require a protic solvent?
Recently, I was looking at the chapter on chemoselectivity in Clayden et al.'s Organic Chemistry (2nd ed.). On p. 530, it is mentioned that sodium borohydride reduction only occurs in protic solvents or in the presence of electrophilic metal ions, such as $\ce {Li+}$ and $\ce {Mg^2+}$ while the more powerful reducing agent lithium aluminium hydride does not require protic solvents. Why is that the case?
Reference
Clayden, J., Greeves, N., & Warren, S. (2012). Organic Chemistry (2nd ed.). New York : Oxford University Press Inc.

Principally solubility. Sodium borohydride is essentially insoluble in aprotic solvents.

@Waylander What about the need for the ions? These ions help to increase polarity of solvent, increasing the solubility of NaBH4?

Li and Mg cations act as Lewis acids coordinating to carbonyl oxygen, Na is much weaker in this respect.

Apparently, THF (a non-protic solvent) can also be used to conduct reductions with $\ce {NaBH_4}$, according to this: https://commonorganicchemistry.com/Common_Reagents/Sodium_Borohydride/Sodium%20Borohydride.htm#:~:text=NaBH4%20is%20usually%20used%20in,to%20give%20the%20respective%20borates.

As Mithoron states a labile Lewis acid is needed in the reaction similar to the role of Mg++ in a Grignard; Na+ is not quite up to the job in NaBH4.

The roles of a protic solvent and lithium and magnesium salts can be tied together in their role as electrophiles—and they impact both solubility and reactivity.
Protic Hydrogen's evil twin
Sodium borohydride may require (or at least favor) protic solvents for good solubility, but lithium borohydride is soluble up to several percent in diethyl ether[1]. The presence of lithium ions, as well as protic hydrogen, appears to enhance the solubility of sodium borohydride, and we can expect magnesium with its diagonal-relationship similarity to lithium to do the same.
Among ions that are reasonably stable in most solvents, borohydride is unique in having hydrogen with a significant negative partial charge. With that comes boron-hydrogen bonding orbitals that are relatively strongly concentrated on the hydrogen atoms and relatively high in energy levels. Thus by both electrostatic and molecular-orbital renderings of hydrogen bonding, the hydridic hydrogens can serve as relatively good electron donors for hydrogen bonding. This interaction between hydridic and protic hydrogen in described for a magnesium borohydride-ammonia complex in Yan et al[2]. Hence the preference of borohydride for protic solvents. Lithium and magnesium ions, whose hydride salts have greater covalent character than hydrides of heavier alkali or alkaline earth metals, can also serve as electron acceptors as well. Thereby borohydride ion is drawn into solution by protic hydrogen, lithium ions or magnesium ions.
Both ends against the middle
In this answer the role of magnesium ion is Grignard reactions is described. This ion acts as an electrophile, attaching to the oxygen and polarizing the carbonyl group and thus assisting the nucleophilic attack. Magnesium ion from a salt (e.g., $\ce{MgCl2}$}, lithium from a salt (e.g.,$\ce{LiCl}$) or organolithium compound (e.g., $\ce{R2CuLi}$), or a proton transferred from a protic solvent, can have similar effects in making a carbonyl group more reactive.
References

Mikheeva, V.I., Troyanovskaya, E.A. Solubility of lithium aluminum hydride and lithium borohydride in diethyl ether. Russ Chem Bull 20, 2497–2500 (1971). https://doi.org/10.1007/BF00853610

Yigang Yan, Jakob B. Grinderslev, Mathias Jo̷rgensen, Lasse N. Skov, Jo̷rgen Skibsted, and Torben R. Jensen. "Ammine Magnesium Borohydride Nanocomposites for All-Solid-State Magnesium Batteries". ACS Appl. Energy Mater. 2020, 3, 9, 9264–9270.https://doi.org/10.1021/acsaem.0c01599

"Thus by both electrostatic and molecular-orbital renderings of hydrogen bonding, the hydridic hydrogens can serve as relatively good electron donors for hydrogen bonding." — This is interesting: do you have a source that goes more into detail about this? [I'm sorry if it's the reference above, I don't have access any more.]

@orthocresol note my new reference.


text:
Number of spatial arrangements of tetrasubstituted biphenyl

According to my book 2-bromo-2'-(1-chloroethyl)-6'-iodo-6-(prop-1-en-1-yl)-1,1'-biphenyl has eight possible spatial arrangements. How can this be so? I can think of only four – left two groups being planar with right two groups up-down and down-up, and vice versa.
For this configuration of I and $\ce{CHClCH3}$ I can think of four, for I and $\ce{CHClCH3}$ interchanged I can think of four more, but that is not the same molecule and is just a configurational isomer.

There is a cis/trans thing in one substituent and a chiral center in another.

@IvanNeretin will that be so when the substituents are planar or up/down? I'm a bit confused about the planar part actually.

Roughly speaking, the rings are perpendicular to each other. You may think that one is horizontal (I guess that's what you mean by planar) and another is up/down, or vice versa; it is all the same.

Double bond isomerism is a type of stereoisomerism as well, and certainly has to do with the spatial arrangement.

There are three independent stereogenic centers or elements in this molecule,

chiral carbon atom of the chloroethyl group,
double bond of the propenyl group,
axis of the substituted biphenyl with bulky substituents at 2,2′,6,6′ positions.

Each of them has two configurations. It means that the total number of stereoisomers is 23 = 8. (No one of them is symmetrical and cannot be superimposed with its mirror image, no one can be superimposed with any other one.)

(1M)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
(1M)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
(1M)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl
(1M)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl
(1P)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
(1P)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1Z)-prop-1-en-1-yl]-1,1′-biphenyl
(1P)-2-bromo-2′-[(1R)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl
(1P)-2-bromo-2′-[(1S)-1-chloroethyl]-6′-iodo-6-[(1E)-prop-1-en-1-yl]-1,1′-biphenyl


text:
If a chiral molecule reacts with an achiral molecule will the product be chiral or achiral?
My book says that a pair of enantiomers will react the same way with an achiral substance.
How is this possible? Does this imply that the product is achiral, even though the enantiomers are chiral? If so, why?

Look at it the other way round: The achiral substance A doesn't care wether it reacts with one R or the other enantiomer L. The product is AR or AL. AR and AL might be identical, or they are enantiomers, and then they might or might not interconvert during the reaction, giving you a more or less racemic mixture.

The key here is to look at the transition state through which the reaction proceeds.
The short answer is that the transition states are enantiomeric, i.e., also mirror images of each other, and in an achiral environment, enantiomers have the same energy. If the transition states have the same energy, then the reactions proceed at the same rate.
The long answer is, unfortunately, convoluted by the fact that a new bond may be formed in the transition state, creating a stereogenic center. In other words, you might have something like:
$$\ce{($S$)-A + B -> [($S$,$R$)-AB]^{\ddagger} + [($S$,$S$)-AB}]^{\ddagger}$$
These are diastereomeric and the rates through these two transition states is different.
But $\ce{($R$)-A}$ does the exact same thing. So the rate of reactivity is controlled via 4 transition states:

$\ce{[($S$,$R$)-AB]^{\ddagger}}$
$\ce{[($S$,$S$)-AB]^{\ddagger}}$
$\ce{[($R$,$S$)-AB]^{\ddagger}}$
$\ce{[($R$,$R$)-AB]^{\ddagger}}$

Here, 1 and 3, and 2 and 4, are enantiomeric, respectively, and the reaction will proceed through pairs of enantiomeric transition states, but since they are enantiomeric, the two sets transition states provide matched rates from the enantiomeric reactant pair.
So, we can safely conclude that indeed enantiomeric reactants react at the same rate with achiral reagents.

Enantiomers have the same chemical and physical properties (save their optical activity) and will undergo the same reaction with an achiral substrate to form a new pair of enantiomers. To separate them, you need a chiral substrate which will form a diastereomeric product. Diastereomers have unique chemical and physical properties, facilitating their separation.
My teacher gave me the analogy of shaking someones hand. If you shake someone’s hand with a pole (an achiral object) you can’t identify which hand they used, but you will immediately be able to when you use your hand (another chiral object).

It's possible for them to react the same way because the enantiomers have the same chemical and physical properties and the achiral substance doesn't care whether it is reacting with a left handed enantiomer or a right handed enantiomer. It does not imply that the product is achiral because the reaction could disrupt what made them chiral in the first place, or it could preserve what made them chiral, it depends on the reaction.

strike out "it's possible" and replace by "will always"

No I will not strike it out because the original question was "How is this possible," and I'm using the same language as the question to explain why it is possible. Also, "will always" is false because on a molecule by molecule basis you can have side products where something random causes a different reaction, it's not always 100%. Its possible for one molecule of l-whatever to react and form a side product and for one molecule of r-whatever to form the expected product. It's up to chance, and neither is more likely than the other to form a side product but they may not always react the same.

As chemists, were not looking at individual molecules, but always at the ensemble. The composition of the product is not 50:50 by chance.

Whether or not the products are chiral depends on what you react. If you react lactic acid with, say, methanol to form an ester, then the lactate group is still chiral; if you instead burn it in oxygen, you just get carbon dioxide and water, which are both achiral.

1. The reaction products may or may not be chiral.
Consider, say, the following two enantiomers

Image source

and imagine that they react with some two achiral reactants $R_1$ and $R_2$, such that:

$R_1$ is, say, an oxygen atom which approaches the hydrogen atom and inserts itself between the carbon and the hydrogen, so that the CH bond changes to a dangling COH hydroxyl. The molecule remains chiral.
$R_2$ is some hydrogen-hungry hippo that comes along, grabs the hydrogen, and takes it away, so that you get $$\ce{CBrClFH}+R_2 \to \ce{CBrClF} +R_2\ce{H},$$ where the reactant hydride $R_2\ce{H}$ is achiral and the remaining scaffolding $\ce{CBrClF}$ is now planar and therefore achiral.

(The actual mechanisms may not be too realistic - I'm a physicist and not a chemist ;-). But the principles hold.)
As you can see, the reaction products can remain chiral (which is the nontrivial property). But there is also the possibility that the chirality can be destroyed in the reaction.
If the reaction products do come out chiral, then they're obligated to produce opposite enantiomers for the products given opposite enantiomers for the starting chiral reactant.
2. The reaction rate itself is the same.
This is likely what your book means when it says 'a pair of enantiomers will react the same way with an achiral substance'.
If you go back to the two examples above, the reactant itself doesn't care which enantiomer it's interacting with: it just sees a CH bond and does stuff with it, and it doesn't much care about which way the halogens are arranged on the other side of that carbon.
In a more complicated molecule, the reaction may indeed 'bounce off' of the other parts of the chiral center, or have an even more complex interaction with the rest of the molecule. However, for every reaction pathway in the S enantiomer, there will be an equal-but-mirror-image reaction pathway with the R enantiomer, and both will be dynamically equivalent, so the reaction rates must be the same. (If the other reactant is itself chiral, on the other hand, that's no longer true, and the reaction rates and products can come out completely different.)


text:
Geometric isomerism in substituted cycloalkenes?

My book says the first and third molecules only will show geometrical isomerism. Why is this so? I can understand how the third will show (both the methyl groups can be into the screen or one can be into and one can emerge from the screen) but what about the others? And if the first shows geometrical isomerism why then doesn't the second? They are fairly similar except for the -OH substituent.

The magic prefixes to consider here are cis- and trans-

@Zhe if the first shows cis/trans isomerism why wouldn't the second?

What groups are going to be cis- or trans-? There's only one group to move up and down on the ring...

@Zhe got it!You mean the hydrogens don't you?And in the second there are two hydrogens attahed to a single C therefore there is no geometrical isomerism?

In the second and in the fifth molecules, the lowest $\ce{CH3}$ group and the $\ce{OH}$ can be above or under the plane. This makes two R/S isomers in these two molecules..

Only compounds I and III will show geometrical isomerism.
Explanation -
Compound 1 - the double bond is not causing the isomerism. In fact this double bond cant show geometrical isomerism because of twisting of the ring structure for the second pair of isomer. So, how is this showing isomerism then ?
The answer lies with the two sp3 hybridized carbons one with OH and the other with CH3 attached. Here the other atoms attached are Hydrogen so hydrogen atoms can be syn or anti with respect to each other and thus show geometrical isomerism.
Compound 2 - the double bond is not causing the isomerism. (Discussed in compound 1) Here there is only one sp3 hybridized carbon(not attached with double bond) so it is lacking one group with which it could've made isomerism.
Compound 3- Here you are correct. the CH3 groups and H attached can show syn and anti so it will show geometrical isomerism
Compound 4 - Just like compound 2 here is only one sp3 hybridized carbon and it cant show geometrical isomerism on its own. The other groups can't show geometrical isomerism due to twisting of ring structure.


text:
How is the covalency of terminal carbons of a diamond satisfied?
Are the terminal carbons of the diamond lattice remain bonded to some element like hydrogen?
Or, is it that they don't achieve octate in their outer shell?

They bond to whatever they can grab.

Probably just hydrogen atoms.


text:
What is the difference between non-adiabatic coupling and diabatic coupling?
The diabatic couplings are the off-diagonal terms of wave functions written in a diabatic basis. There is a need to use the diabatic basis when there is a breakdown of the Born-Oppenheimer approximation. Then a transformation is done from the adiabatic to diabatic basis. But what does non-adiabatic coupling mean?
The reason I am asking is that most quantum chemistry softwares (I use Molpro) calculate the non-adiabatic coupling. Then some procedure has to be used to convert them to diabatic coupling. But what is the difference between the two and why do I need to use the diabatic coupling?

Have a look at Landau - Zener coupling and Fermi Golden rule for different approaches to coupling.

May find this article useful.

The difference is in the representation: adiabatic representation, being exact, allows you to make adiabatic approximation. If you do not want to have adiabatic approximation, there will be off-diagonal matrix elements in kinetic energy, so the coupling goes through kinetic energy. In diabatic representation, the kinetic energy is diagonal, and the couplings come from potential.
As was pointed out above, diabatic coupling allows you to apply Landau-Zener rule and estimate probabilities of transition.


text:
Hückel method for biphenyl
I am trying to calculate the energy levels of biphenyl (see picture below) using Hückel method. Since this is my first time using Hückel I am a little confused.

Biphenyl has 12 carbon atoms, therefore I should get a 12x12 matrix like the one below.
\begin{array}{cccccccccccc}
\alpha-E & \beta & 0 & 0 & 0 & \beta & \beta & 0 & 0 & 0 & 0 & 0 \\
\beta & \alpha-E & \beta & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & \beta & \alpha-E & \beta & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & \beta & \alpha-E & \beta & 0 & 0 & 0 & 0 & 0 & 0 & 0 \\
0 & 0 & 0 & \beta & \alpha-E & \beta & 0 & 0 & 0 & 0 & 0 & 0 \\
\beta & 0 & 0 & 0 & \beta & \alpha-E & \beta & 0 & 0 & 0 & 0 & 0 \\
\beta & 0 & 0 & 0 & 0 & \beta & \alpha-E & \beta & 0 & 0 & 0 & \beta \\
0 & 0 & 0 & 0 & 0 & 0 & \beta & \alpha-E & \beta & 0 & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & \beta & \alpha-E & \beta & 0 & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \beta & \alpha-E & \beta & 0 \\
0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & 0 & \beta & \alpha-E & \beta \\
0 & 0 & 0 & 0 & 0 & 0 & \beta & 0 & 0 & 0 & \beta & \alpha-E
\end{array}
Is this matrix correct for biphenyl, if I neglect dihedral angles etc.?
If I take det(Matrix)=0 and solve this for the Energy term I should get several solutions. How can I assign these solutions to each energy level correctly?

The eigenvalue of the solution is the energy of the orbital and the eigenvector itself tells you the components in that solution. There's no work to do for assignment.

For a planar molecule this looks ok. If the molecule is twisted then remove the $\beta$ connecting atom 6 and 7 as now conjugation will be removed. The parameter $\beta$ is a negative number so the lowest energy is will be of the form $\alpha + C\beta$ where coefficient $C$ has the largest positive value. Note that some levels may be degenerate. The wavefunction with the lowest eigenvalue has the fewest nodes, the highest energy one has the most nodes between atoms.

@porphyrin Thanks for your comment! I have another question: If I want to calculate the excitation energies for 1->2 then I have to fill all energy levels with the 12 electrons and then I build the difference between HOMO and LUMO right?

Yes add all the electrons them move one from homo to lumo for the lowest possible excitation.

@porphyrin If I want to calculate the second lowest transition it would be HOMO->LUMO+1, right?


text:
Graphite: Thermal and electrical conductivity
I recently encounter this question:
Compare, while providing an explanation, the thermal and electrical conductivity of graphite to that of diamond.
Provided answer:
Graphite conducts electricity(only across a layer) where as diamond does not.
Diamond is better at transferring heat than graphite.
Explanation:
Diamond is covalent network solid of carbon atoms which are linked to each other through sigma bonds.
Graphite has a layered structure where each layer is bonded to the other later through vanderwall forces only. Each layer consists of hexagonal rings which have a delocalized pi electron network just above and below the plane.
These delocalized pi electrons are the reason why graphite can conduct electricity along the planes, where as diamond cannot due to the lack of delocalized electrons.
Diamond is a better conductor of heat because the transfer of heat takes place through adjacent atoms transferring their vibrational energy and diamond has a very close knit network of atoms where as graphite does not.
My doubt:
In metals it is accepted that both electrical and thermal conductance is due the the free electrons present in the metal body.
Graphite too has delocalized free electrons so according the same logic graphite should be both a good conductor of heat and electricity.
So where am I or the answer provided going wrong?
P. S. I'm writing from a mobile device due to the lack of access computers so there might be some errors.
Edit:
J. D. Lee: Concise inorganic chemistry, fifth edition :

This would mean that the electrons are truley free since they can interact with other species nearby not preventing them from being bound to a system.

Electrons are the main thermal conduction mechanism in metals only at quite low temperatures. Phonons dominate by room temperature.

@JonCuster could you please elaborate it in an answer?

The conductivity of graphite is too low to provide much to heat conductivity.

@Karl Electrical  conductivity of graphite is two-dimensional; lateral  heat conductivity in those two dimensions is very good in pyrolytic carbon (which is oriented graphite crystal).

@Whit3rd Right, but has nothing to do with electrical conductivity. Heat is transported much more easily along covalent bonds than through the vdW interaction that holds the graphite sheets together. Conincidentally, also electrons cannot move in that direction.

You are wrong to say that graphite has free electrons.  The electrons are bound to sheets of aromatic rings--delocalized but not free. In electrical conduction, the electron pairs are able to jump from bond to bond on the sheet and between sheets. It is worth noting that graphite has anisotropic conduction where between sheets conduction is lower than conduction along a sheet.
For thermal conduction, energizing an atom does not transmit the energy via free electrons like a metal would.  instead, you are limited to a predominately phonon driven heat transfer again, but now the structure is less dense and the carbons are less rigidly bonded.
Diamond as your correctly identified has a rigid structure that can easily conduct phonons but not electrons which gives good thermal, but not electrical conduction.

Can you please explain: electrons are delocalized but not free. And why graphite cannot transfer heat but can transfer electricity along the planes with ease?

@Debaditya delocalized means that the electrons are not bound to a local area (delocalized).  Free means that the electrons are unconstrained. The electrons on graphite and other aromatic rings may travel around the rings as they are delocalized but must be located on a bond/atom not in the matrix as they are not free. Thus the conduction mechanism is different for graphite than for metals.


text:
Classification of the Inside of a Grapefruit
In the text Chemistry: The Central Science, the inside of a grapefruit is classified as a homogeneous mixture. Unless I’m missing something, to me it would be a heterogenous mixture owing to pulp, seeds, and ultimately the fact that cells are not homogenous. Is this a mistake in the text, and, if not, what am I missing?

Your interpretation sounds reasonable to me, but I’m no grapefruit expert. :-) If it’s physically separable, it’s definitely not homogenous.

Yes, it is heterogeneous (unless it's put into an ultrasonic cell disruptor... and even them I'd quibble).

Which edition are you working out of? Searching through the text of the 12th ed only shows one mention of grapefruit in the acid-base intro. If it's newer than 12th, then they added that part quite recently and hence wouldn't have had as much proofreading on it.

According to Wikipedia:

In chemistry, a heterogeneous mixture consists of either or both of a)
  multiple states of matter or b) hydrophilic and hydrophobic substances
  in one mixture

The inside of a grapefruit is clearly made up of multiple states of matter (solid and liquid), and is thus heterogeneous. Text books sometimes contain errors, and that appears to be the case here.