If you add AlCl3 to water, why does the pH decrease? You are not adding any hydrogen ions, so why does the pH of the solution decrease?
Thank you in advance.

EDIT: I will add the context of my asking this question to clarify some details. This is not a homework question. I am preparing for an upcoming examination in Chemistry, and I found this statement in the mark scheme of a practice exam I was doing, and yet I don't understand why it is true.

I have learnt pH as simply being the negative log of the concentration of H+ ions. I fail to see how adding the above salt changes the pH, especially since HCl is a strong acid (hence it completely dissociates in solution).

Thought process: Is the pH change a result of Aluminium forming Aluminium Oxide?

A more clarified reaction in addition to Berry Holmes' answer:

$$\ce{AlCl3 + 4H2O <=> [Al(OH)4]- + 3Cl- + 4H+}$$

The reason this happens is the Lewis-Acidity of $\ce{Al^3+}$ as well as the higher affinity for $\ce{OH-}$ ions when compared with $\ce{Cl-}$.


How does increasing the ionic strength of an aqueous solution decrease the activity coefficient? Activity is a new concept for me, and I am having a little bit of trouble understanding it.

For the equation:

$$\ce{aA + bB<=>cC + dD}$$

$$K_c= \frac{[\ce{A}]^a[\ce{B}]^b}{[\ce{C}]^c[\ce{D}]^d}$$

As I understand it, this is not the true equilibrium constant expression for the chemical equation above. My textbook calls $K_c$ in this instance the "concentration quotient." The technically correct equilibrium constant expression is:

$$K=\frac{(a_\ce{A})^a(a_\ce{B})^b}{(a_\ce{C})^c(a_\ce{D})^d}$$

Where $K$, in this equation, is the "true" equilibrium constant determined by the activity, $a$, of each species in equilibrium with each other. The activity is represented by the equation: $a_\ce{A}=\gamma_\ce{A} [\ce{A}]$, where $\gamma$ is the activity coefficient.

My textbook Quantitative Chemical Analysis (Ninth Edition, by Daniel C. Harris), states that "the equilibrium 'constant' is not really constant." And then it goes into how increasing ionic strength increases the solubility of ionic compounds in aqueous solution. It then later states "the activity coefficients must decrease with increasing ionic strength."

This would seem to contradict the idea that increasing ionic strength increases the solubility of an ionic compound, would it not? If the ionic strength increases the solubility of an ionic compound, then wouldn't that mean that the concentrations of the ions that make up that ionic compound would increase, and thus the activity coefficient, therefore, must increase to show an increase in activity, otherwise known as the "effective concentration?"

Or is the activity independent of the ionic strength such that in order to keep the activity the same under higher concentrations, the activity coefficient must be smaller? This would then bring up the question: is the equilibrium constant, $K$, (determined by the activities of the components in the equilibrium expression) constant even under different ionic strengths, while the "concentration coefficient" (determined merely by the concentrations of the components in equilibrium with one another) does vary under different ionic strengths? I'm not sure I'm getting this very well.

Thermodynamic activity is essentially the same the chemical potential.  At the solubility limit of a compound, the dissolved species is in equilibrium with the solid species.  For example, a saturated solution of sodium chloride is in equilibrium with solid sodium chloride crystals.  In such equilibria, the chemical potential of the dissolved sodium chloride is equal to the chemical potential of the solid sodium chloride crystals.

  If the ionic strength increases the solubility of an ionic compound, then wouldn't that mean that the concentrations of the ions that make up that ionic compound would increase, and thus the activity coefficient, therefore, must increase to show an increase in activity, otherwise known as the "effective concentration?"

Let's assume that increasing the ionic strength doesn't alter the crystals of solid sodium chloride that dissolved sodium chloride is in equilibrium with.  That means the chemical potential of the crystal is unchanged.  At the solubility limit, the dissolved sodium chloride is still in equilibrium with the crystals, and as before, this equilibrium requires that the crystals have the same chemical potential (aka thermodynamic activity) of the dissolved species.

If there is more of the dissolved species (a higher concentation), the only way the activity can stay the same is if the activity coefficient goes down.

So you were right when you wrote:

  ...the activity [is] independent of the ionic strength such that in order to keep the activity the same under higher concentrations, the activity coefficient must be smaller...

  This would then bring up the question: is the equilibrium constant, $K$, (determined by the activities of the components in the equilibrium expression) constant even under different ionic strengths, while the "concentration coefficient" (determined merely by the concentrations of the components in equilibrium with one another) does vary under different ionic strengths?

The equilibrium constant is a ratio of thermodynamic activities.  The "concentration coefficient" can vary and will not be constant any time that activity coefficients are not 1.0.  Ionic strength is one of many variables that can affect activity coefficients.  So to conclude:

Yes, the equilibrium constant $K$ is determined by the activities of the components in the equilibrium expression.
The "concentration coefficient" $K_c$ can vary as a function of ionic strength (or other variables).


Why is the radial distribution at the nucleus 0 but not the radial wave function? The modulus squared of the radial wave function gives the probability of finding an electron in an infinitesimal volume dv. On the other hand, the radial distribution gives the probability of finding an electron in a shell of radius r and thickness dr, or in other words, the probability of finding an electron at a radius r.
I know how the radial distribution is derived and it being 0 at the nucleus makes sense mathematically but not physically. The radial distribution being 0 at the nucleus means that an electron cannot be found in a shell of r=0 & thickness dr ( a sphere of radius dr), but we already know from the radial wave function that there is a non-zero probability of finding an electron in an infinitesimal volume at the nucleus. These results appear to contradict each other.
Is there something wrong in my physical interpretation of the 2 functions?

You state 'but we already know from the radial wave function that there is a non-zero probability of finding an electron in an infinitesimal volume at the nucleus.'
To add to Poutnik's answer, I think that this sentence states the misunderstanding. The radial part of the wavefunction is $R_{n\ell}(r)$ and when normalised
$$\int_0^\infty R_{n\ell}^*(r)R_{n\ell}(R)r^2 dr=\frac{4}{a_0^3}\int_0^\infty e^{-2r/a_0}r^2 dr=1$$
where the $r^2$ comes from the volume element needed to integrate over all angles which is $r^2\sin(\theta)drd\theta d\phi$. The probability from $r\to r+dr$ is then for the 1s orbital $\displaystyle p=\frac{4}{a_0^3}r^2e^{-2r/a_0}dr$ whereas the 1s wavefunction itself has the form $\sim e^{-r/a_0}$. For other orbitals, e.g $2p_{\pm 1}$ the wavefunction is a function with complex numbers so cannot represent anything physical.


Which elimination is this? This is the question:
$$\ce{Cl-CH2-CH2(Cl)2->[alc KOH][Δ]?}$$
Now, I got the answer right but I think that the reaction should go by the $\ce{E1}$ mechanism because of the high temperature/heat, however, my teacher says that the mechanism will be $\ce{E2}$.  I don't understand why it will be like that. Everyone seemed to agree, so I didn't have the guts to ask in class. Please tell me why it's $\ce{E2}$? And then, what's the use of high temperature/heat? It was there to make ionization easy, right?

An E1 mechanism is highly unlikely. I don't know what level you're at, but one could invoke the Hammond Postulate in this case, and look at the high energy cation intermediate. There are three chloride leaving groups that could generate one of two intermediates. Both of these intermediates are primary cations. A chloride substituent isn't likely to be able to stabilize this much via resonance but more likely destabilize due to inductive effects.

The other potential mechanisms to consider (that do not involve a cation intermediate) are E2 and E1cb. E1cb might work for the first step depending on where the anion is formed, especially if the inductive effect of the chlorines is enough to help stabilize the carbanion. E2 just requires an anti-periplanar arrangement, which isn't hard here with free rotation.

The reason I'd favor E2 and the reason the reaction likely requires heat is the fact that hydroxide is just an OK base. Add to that the fact that it's in a polar protic solvent that might further dampen basicity. If you tried this reaction with an amide base (e.g., LDA) in toluene, I would guess no heating is necessary.

Heating is generally helpful for E1 because the dissociation of the leaving group is slow. However, the need to heat a reaction is by no means indicative that it is E1 (it might not even be an elimination!). Any number of factors could conspire to make the reaction slow, in which case, heating might help make the reaction proceed at a faster, more practical rate.


When did the atom Theory Become indisputable? We know that the Greeks were the first scientists (or better, philosophers) to introduce the idea that matter is made up of little “chunks”, and they named these chunks “atoms”. This idea that matter comes in discrete small pieces was termed wrong and right over centuries when John Dalton, in the 19th century, constructed a theoretical framework in which he could explain in an elegant way the work of scientists such as Joseph Priestley and Antoine Lavoisier on qualitative and quantitative properties of matter and its transformations.
The framework which John Dalton used to better understand the work of other scientists had in its core the “language” of atoms, that is, he used the concept of atoms to explain, for example, why reactions only occurs if we have quantities “in proportion”. Note that the work of John Dalton doesn't proof that atoms are real, it only shows that thinking of matter in terms of atoms is a suitable way to understand the results coming out of experiments.
Scientists only proved that atoms are real at the beginning of the 20th century, and that discovery gave birth to Quantum Mechanics.

Now comes my question:
When did scientists prove that molecules “are real”? In other words, what is the history of the discovery of (the shape of) molecules?

Scientists have been playing around with toy models of particles for a long time but only a short time ago IBM researchers managed to “take a picture” of an actual molecule, and its configuration matches exactly the models we've been using for a long time, for example this pentacene molecule:

When did we first learn that molecules have a “geometry”? That is, an exact arrangement in space?

Short Answer: Henry Moseley provided the first experimental evidence for the nuclear model of atoms through X-ray observations in 1913.

Long Answer:

Proving something is a rather nebulous task when there is no judge or jury and consensus is arrived at gradually over a century rather than instantaneously. Many scientists would consider the atom "proved" decades apart.

Avogadro was the first to distinguish atoms from molecules, which alleviated much confusion regarding atoms and allowed quantitative description/experimental evidence as relating to Advogadro's number (a mole). That was during the early decades of the 1800s (19th century)

Various scientists became convinced at various times over the next century.

Take Max Plank for an example. Around 1898 Plank became persuaded that the molecular approach was the only way forward for physics. He had formerly ascribed to the "pea soup" model where there were not discreet atoms. It was the molecular approach that allowed Plank to discover energy level quantization that is the foundation for quantum mechanics. E=hv

Lewis dot structures were some of the earliest electronic approaches to molecular geometry (circa 1910), but long before that there was an understanding of basic molecular geometries as demonstrated in the correct postulation of Benzene's hexagonal ring structure in 1865. Carbon was known to form 4 bonds with other carbons in the late 1800s. Linus Pauling was the first to calculate bond angles based on quantum mechanics in 1931.


Counting electrons in Ozonolysis I can't understand the electron count made on this page about Ozonolysis. (http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/ozonolysis/Oz.html)
This is the specific part of the article:

From my calculations the oxidation of an alkene to an ozonide is a 4 electrons oxidation and not 6e, in fact the two carbons in the double bond loses 2e each from bonding two oxygens.
Anyway I agree with the 4e oxidation from alkene to two carbonyl compounds. Where am I wrong?
Thanks in advance.

If you want to demonstrate that the addition of ozone to an alkene is a 6-electron oxidation of the carbon framework, here are two methods. To assign a charge to the carbon atoms of a neutral compound (sum of all charges equals zero) treat all hydrogens as +1 and all oxygens as -2, then do the math. Each carbon in ethylene (1) has a calculated (in red) value of -2. Molozonide 2 has carbons with a charge of +1 but, for convenience, they are written as zero and +2. Already one can see that 1 $\rightarrow$ 2 is a 6-electron oxidation [-4e $\rightarrow$ +2e]. Collapse of the molozonide gives formaldehyde (3) of carbon charge zero and zwitterion 4 with carbon charge +2. These species recombine to form ozonide 7 that has the same carbon charge as the molozonide 2. That the zwitterion 4 is the equivalent of formic acid 6, follow the steps 4 $\rightarrow$ 5 $\rightarrow$ 6. Thus, both the molozonide 2 and ozonide 7 are equivalent to one formaldehyde and one formic acid.
Alternatively, oxidation of ethylene (1) to ethylene glycol (8) [-4e $\rightarrow$ -2e] is a 2-electron oxidation and a second 2-electron  oxidation [-2e $\rightarrow$ 0e] affords two moles of formaldehyde. Oxidation of one of the formaldehydes to formic acid [0e $\rightarrow$ +2e) is the third 2-electron oxidation. Both pathways demonstrate that the oxidation of ethylene to formaldehyde and formic acid are 6-electron oxidations.


Frequency, the only real characteristics of electromagnetic radiation ?? My book says: frequency is considered as the only REAL characteristics of electromagnetic radiation, because it is unchangeble if the radiation comes up with some other environment, while the wavelength and the wavelength number do change. How can wavelength change and the frequency not ? The formula tells us very clear that while frequency increases, wavelength decreases and vice versa ?!

Recall that frequency $f$, wavelength $\lambda$ and the speed $v$ of a wave are all related as follows
$$
    v = f \lambda \, .
$$
So presumably, your book is talking about a situation when an electromagnetic wave crosses boundaries between different media. In such cases frequency indeed remains constant, while the speed of the wave, and consequently, its wavelength change.


Preparation of NaOH Solution How would one prepare a one molar sodium hydroxide solution?

I see a few issues with this:

$\ce{NaOH}$ absorbs carbon dioxide. So old NaOH samples might have a significant portion of $\ce{NaHCO3}$ or sodium bicarbonate. How would get get rid of the sodium bicarbonate?
I am told that $\ce{NaOH}$ is hygroscopic. So it absorbs water, apparently. What is the mechanism for this? I can see the hydroxide attacking carbon dioxide in the above bullet point, that's easy, but I can't really see $\ce{NaOH}$ attacking water. Also couldn't we circumvent the water absorption problem by cooking the $\ce{NaOH}$?

There is a difference in preparing a solution and determining its concentration. As you said, NaOH solutions absorb carbon dioxide over time. It is possible to remove the absorbed carbon dioxide by boiling, but it is just as easy to use the solution as is and re-determine its concentration when you use it.

The concentration of sodium hydroxide solutions can be determined by titrating a known amount of dried potassium hydrogen phthalate.

If you really need an exact concentration of the solution, make it slightly more concentrated, standardize it and then dilute it to get the concentration desired. It would only be accurate for a few hours.

  I am told that NaOH is hygroscopic.

The powder and pellets of sodium hydroxide are hygroscopic, but the solutions of sodium hydroxide already have water, so it is unlikely more will be absorbed.


What is meant by "fairly" when used as a quantifier for the solubility of a chemical? While doing homework and I found a statement that says a chemical is fairly soluble in water, but I do not understand what this means.

However, I know that when things are soluble in water they tend to dissolve. Is it that fairly soluble in water means that a part of it dissolves, while another part doesn't dissolve?

It seems that you understand the concept very well and that this is just a minor confusion about word fairly.

If you would characterize the solubility of a compound (solid, liquid, or gas) by words, rather than by giving the exact solubility in $\pu{mg\cdot mL^{-1}}$, you would probably come up with terms like

highly soluble
soluble
slightly soluble
insoluble

So, fairly means that the solubility isn't excellent, but it's not that bad either. It is acceptable, which of course isn't very precise.

  Is it that fairly soluble in water means that a part of it dissolves while another part doesn't dissolve?

That one part does not dissolve will happen in all aqueous solutions when you add to much of the compound that you want to dissolve. Just think in sodium chloride, which has a solubility of about $\pu{390 g\cdot L^{-1}}$ in water. When your solution is saturated, it can't take any more. If you add another spoon of the salt, it will not dissolve and precipitate.


Can't you form a ring from almost any number of sides from carbon? Recently, we have been learning Chemistry in my physics class at school. We learned that Carbon is one of the elements that has the most variety in the compounds it can formed, being able to form four covalent bonds. This got me thinking- I know you can form Carbon into hexagonal rings- graphite does this- but shouldn't you be able to form almost any kind of polygon from carbon, with each one double bonded to the two next to it?

Carbon bonds have certain preferred angles. For example, the tetrahedrally symmetrical methane, which has sp3 hybridization,, $\ce{CH4}$, has an angle of 109.5 0. In benzene, where there is sp2 hybridization, the angle is 120 0.

These are the preferred angles for bonds. In some molecules, the geometry forces them to odd angles, such as cyclopropene, with a 60 0 bend. This imposes stress, much like making Tinker toy models where you have to bend the sticks a bit. Overdo this and the stick (bond) snaps.

So, here are some cyclic (ring) compounds with the following number of carbons, and at least one alkene bond:

3: cyclopropene

4: cyclobutene

5: cyclopentene

6: cyclohexene

7: cycloheptene

cyclooctene

Notice in the drawings of the molecular structure that larger rings are increasingly asymmetrical, degenerating to a twisted mess. So yes, very large carbon "rings" are possible, but eventually cannot hold their shapes. Some compounds contain more than on C=C bond, such as 1,3-cyclohexadiene, but I know of no cyclic compounds where all carbons contain a double (alkene) bond.

You can extend this, though, to build up three-dimensional polyhedra, such as cubene or the elegant (though not double-bonded) buckminsterfullerene.


Confusing Lines About Extinct Elements In "The Greatest Show On Earth", author Richard Dawkins spends a chapter discussing radiometric dating. I find myself confused by the following lines, though I'm sure the reason is elementary. (excellent pun and emphasis both mine)

  Among all the elements that occur on Earth are 150 stable isotopes and
  158 unstable ones, making 308 in all. Of the 158 unstable ones, 121
  are either extinct or exist only because they are constantly renewed,
  like carbon-14.

What is meant by "occur on Earth" and "extinct" here? I take "extinct" to mean "decayed to the point that it cannot be detected". If I am using that word correctly, then in what sense does an extinct element "occur on Earth"?

Defintions

Occur on Earth differentiates between all isotopes in the universe and isotopes which exist or have existed on Earth. The passage identifies that there are a total of 308 isotopes occurring on Earth, but it is worth noting that there are more than 308 total isotopes in the universe. This discrepancy is because some isotopes have extremely short half-lives, making their total "lifetimes" extremely short. As an example, an isotope generated in a supernova which has a half-life of 1 second will have totally degraded in at most 32 seconds (32 half-lives degrades the sample completely no matter the original amount). That is not nearly enough time for the isotope to travel from the exploded star to Earth, so that isotope does not occur on Earth (and it never can). This hypothetical supernova isotope would exist but it would not be part of the 308 Earth occurring isotopes.

Extinct means that the isotope is no longer naturally occurring within a given environment. Isotopes have differing half-lives, and a currently extinct isotope would have degraded so completely as to no longer be present at all. The isotopes would have originally been seeded into the environment by processes like supernovas, which produce ejecta containing the elements produced by nuclear fission and nuclear fusion. When that ejecta reaches the Earth the isotopes within it begin to occur on Earth, but eventually those isotopes will disappear due to half-life degradation. There are processes which can replace degraded isotopes in a self-regenerating manner and thus those isotopes never go extinct due to self-regeneration. An example of this type of self-regenerating isotope is Carbon-14; in fact, Carbon-14 will likely never become extinct. Self-regeneration is not the only process to replenish isotopes; the aforementioned supernovas are another. Unlike self-regeneration, which is a continuous process in which degraded isotopes continuously get replaced, supernovas deposit isotopes intermittently and comparatively infrequently. In fact, if a supernova deposit is delayed for an extended amount of time, isotopes which come only from supernovas can actually disappear from Earth entirely- they would become extinct. These extinct isotopes would stay extinct on Earth until a supernova redeposited the isotope.

Clarifying the Passage

As far as your question about an extinct isotope occurring on Earth- for the isotope to be extinct it cannot be currently occurring on Earth, but it was present in the past (logically, if an isotope never appeared on Earth it couldn't be part of the 308 isotope total, so for some of that 308 total to be extinct means they were present on Earth in the past). That question is sort of unrelated to the passage however, but the passage was written poorly enough that I think you misunderstood that they were trying to say that 121 isotopes should hypothetically be extinct, but that some of them are not actually extinct. To reduce the confusion the passage is causing, I have rewritten it to be far more clear: "Among all the elements that have have ever occurred or currently occur on Earth there are 308 total isotopes. Those isotopes are split into 150 stable isotopes and 158 unstable isotopes. Of the 158 unstable isotopes, 121 should be or currently are extinct, however, there are isotopes not actually extinct because they are constantly renewed, like carbon-14."


Mixing bleach and ammonia? I know that the two should not be mixed, but i want to know that if the two become mixed in a contained unit, does it build pressure? Or can you seal the container safely untill it can be disposed of properly?

Here is the usual citation for the toxic products and symptoms created upon mixing NH3 and NaOCl (not in lab conditions, which involves employing very cold dilute aqueous solutions):

When bleach is mixed with ammonia, toxic gases called chloramines are produced. Exposure to chloramine gases can cause the following symptoms:
Coughing.
Nausea.
Shortness of breath.
Watery eyes.
Chest pain.
Irritation to the throat, nose, and eyes.
Wheezing.
Pneumonia and fluid in the lungs.

The chemistry proceeds along the lines:
$\ce{NaOCl ⇌ Na+ + OH- + HOCl}$
$\ce{NH3 + HOCl ⇌ NH2Cl (monochloramine) + H2O}$
$\ce{NH2Cl + HOCl ⇌ NHCl2 (dichloramine) + H2O}$
$\ce{NHCl2 + HOCl ⇌ NCl3 (trichloramine) + H2O}$
Wikipedia also cites possible disproportion paths in acidic conditions starting with monochloramine, to quote:

In an acidic medium at pH values of around 4, chloramine disproportionates to form dichloramine, which in turn disproportionates again at pH values below 3 to form nitrogen trichloride:
$\ce{2 NH2Cl + H+ ⇌ NHCl2 + NH4+}$
$\ce{3 NHCl2 + H+ ⇌ 2 NCl3 + NH4+}$
At low pH values, nitrogen trichloride dominates and at pH 3–5 dichloramine dominates. These equilibria are disturbed by the irreversible decomposition of both compounds:
$\ce{NHCl2 + NCl3 + 2 H2O → N2 + 3 HCl + 2 HOCl}$

Note, with respect to any formed nitrogen trichloride (appears as a yellow oily liquid), extreme caution as it is a very unstable explosive compound upon exposure to shock, heating, sunlight and organics, even while in aqueous solutions. Because of its dangerous sensitivity, $\ce{NCl3}$ is best avoided.
Also, with respect to exposure of swimmers where chloramine can be formed in non-chloramine pools from free chlorine interacting with amine groups present in organic substances, or chloramine treated pools employing NH2Cl as a disinfectant, especially in the case of indoor pools, per the same Wikipedia source:

There is also evidence that exposure to chloramine, can contribute to respiratory problems, including asthma, among swimmers.[11] Respiratory problems related to chloramine exposure are common and prevalent among competitive swimmers.[12]

On the question, can one safely disposed of aqueous NH2Cl? Be careful, as on just adding it to a water source may increase the solubility of heavy metals. This can result, for example, in Lead poisoning for animals and humans (see 'Chloramine Catch: Water Disinfectant Can Raise Lead Exposure).


Substitution at thionyl chloride sulfur What is the correct way of writing the mechanism for substitution at the sulfur atom of thionyl chloride:

Should one write the cleavage of the $\ce{S-O}$ π bond? The argument for the first option is that $\ce{S}$ is already tetrahedral and direct displacement could occur. The argument for the second option is that the π bond will be cleaved first as it is the weakest. Another possibility would be the formation of a tetravalent sulfur species (like the $\mathrm{S_N2}$ substitution at silicon). Is there any experimental or computational evidence regarding the matter?

Your second drawing would imply a $\ce{S=O}$ double bond, due to available orbitals it would be a d-p π bond. There is considerable debate as to whether these bonds even exist with most of the evidence presented here at chemistry.stackexchange.com pointing towards a ‘no’. That would rule out your second mechanism.

That said, your first mechanism suffers from comparing sulphur with carbon too much. You assume a rather short-lived transition state and direct displacement. I don’t know for sure, but I don’t think that is the case.

Rather, I would propose a two-step mechanism with a medium-lived tetravalent sulphur. In this mechanism, the attack of the nucleophile would turn pseudotetrahedral (one corner of the tetrahedron being a lone pair) into a pseudo-trigonal bipyramid or a see-saw with $\ce{Nu}$ and $\ce{Cl}$ forming a four-electron-three-centre bond. This would then collapse back into a pseudotetrahedral product liberating the better stabilised anion. Close to your first mechanism, but not identical:


Why is manganese(II) coloured although the transition should be spin-forbidden? In every basic coordination chemistry class, at some early point the crystal field theory and LFSE will be taught, explaining that there will be an energy difference between d-orbitals (typically named $\mathrm{t_{2g}}$ and $\mathrm{e_g}$) and that electrons can be excited from one to the other to give colour. The next statement is typically that transitions have to be spin-allowed (the higher-energy orbitals need to be able to accomodate for the lower electron without spin-flipping being required) and Laporte-allowed (in a nutshell: no d→d transition in octahedral complexes).

The syllabus then goes on to say that the Laporte rule is not that strict a rule (typically, all near-octahedral complexes of 3d-elements which have spin-allowed transitions have a well visible colour; e.g. iron(II), chromium(III), manganese(III), nickel(II), copper(II) etc.) since small random vibrations can destroy the symmetry making the electron transition allowed and an incoming photon can excite the electron as needed. They also remark that the colour is much less intense than for Laporte-allowed transitions — hence why the colour of the $\mathrm{d^7}$ ion $\ce{[CoCl4]^2-}$ is much more intense than that of $\ce{[Co(H2O)6]^2+}$.

A very recent question on the colour of manganese(II) sulphate then had me confused. Manganese(II) salts are typically of a very pale pink. The difference to white can be observed when they are put before a white background. But manganese(II) is a $\mathrm{d^5}$ ion, so the transition is spin-forbidden.

I think that being spin-forbidden explains why manganese(II) complexes are so weakly coloured, but what is the mechanism that relieves the spin constraint and how does it work in practice?

Selection rules

The intensity of the transition from a state $\mathrm{i}$ to a state $\mathrm{f}$ is governed by the transition dipole moment $\mu_{\mathrm{fi}}$ (strictly, it is proportional to $|\mu_{\mathrm{fi}}|^2$):

$$\iint \Psi_\mathrm{f}^*\hat{\mu}\Psi_\mathrm{i}\,\mathrm{d}\tau \,\mathrm{d}\omega \tag{1}$$

where $\mathrm{d}\tau$ is the usual volume element (representing integration over spatial coordinates) and $\omega$ is an additional spin coordinate (since the states $\Psi$ also specify the spin). For an electronic transition, $\hat{\mu}$ is the electric dipole moment operator, and only depends on the spatial coordinates $x$, $y$, and $z$. (I've also ignored the vibrational component i.e. the Franck-Condon factor since it's not really relevant for the purposes of this answer.)

Normally in determining selection rules, we are more interested in whether the integral in $(1)$ is zero or not. If the integral is zero, the transition is said to be forbidden. The usual way to determine whether it is zero is by using symmetry: if the integrand is antisymmetric under any symmetry operation, then the integral is zero. (It is exactly analogous to the integral of an odd function $f(x)$ over a symmetric region; for example, $\int_{-a}^a \sin x \,\mathrm{d}x = 0$.)

If we assume that both the initial and final states can be separated into spin and spatial components, then we get:

$$\iint (\psi_\mathrm{f}^\mathrm{spin}\psi_\mathrm{f}^\mathrm{space})\hat{\mu}(\psi_\mathrm{i}^\mathrm{spin}\psi_\mathrm{i}^\mathrm{space})\,\mathrm{d}\tau\,\mathrm{d}\omega$$

(I also dropped the complex conjugates - they aren't important.) Since $\hat{\mu}$ does not operate on the spin coordinate, we can separate this multiple integral:

$$\left(\int \psi_\mathrm{f}^\mathrm{spin} \psi_\mathrm{i}^\mathrm{spin} \,\mathrm{d}\omega\right) \left(\int \psi_\mathrm{f}^\mathrm{space} \hat{\mu} \psi_\mathrm{i}^\mathrm{space} \,\mathrm{d}\tau\right)$$

The first integral is nonzero if $\Delta S = 0$. The quantum number $S$ is related to the eigenvalue of a hermitian operator (the actual eigenvalue is $S(S+1)\hbar^2$), and eigenfunctions of a hermitian operator corresponding to different eigenvalues are necessarily orthogonal, making the integral identically zero if $\Delta S \neq 0$. This is the spin selection rule.

The second integral is nonzero if $\psi_\mathrm{f}^\mathrm{space}$ and $\psi_\mathrm{i}^\mathrm{space}$ have a different character under the inversion operation, i.e. one is gerade and another is ungerade. This is because $\hat{\mu}$ itself is ungerade, so for the integrand to be gerade, we need $\psi_\mathrm{f}^\mathrm{space}\psi_\mathrm{i}^\mathrm{space}$ to be ungerade ($\mathrm{u \otimes u = g}$). In turn this means that $\psi_\mathrm{f}^\mathrm{space}$ and $\psi_\mathrm{i}^\mathrm{space}$ must have different symmetry. This is the Laporte selection rule. It can be relaxed when the geometry is not perfectly octahedral, or tetrahedral for that matter, since inversion is no longer a symmetry operation. However, I digress.

Spin-orbit coupling

The problems arise because the total states are not separable into spin and spatial components. This occurs because both electron spin, as well as orbital angular momentum, are both angular momenta. They therefore interact with each other, and this is the spin-orbit coupling spoken of. Mathematically, it is treated as a perturbation to the Hamiltonian; the result is that the transition dipole moment cannot be separated into two integrals and this leads to the invalidation of the selection rules.

When considering spin-orbit coupling, the term symbols such as $^3\!F$, $^1\!D$, which imply a simultaneous eigenstate of the total spin operator $\hat{S}$ and the total orbital angular momentum operator $\hat{L}$ are no longer meaningful since both angular momenta are coupled to each other (by $\vec{j} = \vec{l} + \vec{s}$) and behave as one "combined" source of angular momentum. The quantities $S$ and $L$ are called "nearly good" quantum numbers: they are no longer strictly conserved, but the "contamination" of different quantum numbers is relatively small. Mathematically we could express this as an admixture of different spin states into the original wavefunction:

$$^3\!F_J = c(^3\!F) + c_1(^1\!F) + \cdots$$

where $^3\!F_J$ denotes a state with spin-orbit coupling (not an eigenstate of $\hat{S}$ and $\hat{L}$), and $(^3\!F)$ denotes the original $^3\!F$ state before spin-orbit coupling is considered (which is a true eigenstate of $\hat{S}$ and $\hat{L}$). If $c_i \neq 0$, then one obtains a non-zero transition dipole moment between "states" of nominally different $S$.

The fact that the selection rules are still a thing is solely because spin-orbit coupling is a small effect ($c_i$ close to 0), so the compounds behave "almost" as if they obey the selection rules. More properly, the transition dipole moment between states of different $S$ is small, and therefore the d-d bands in $\mathrm{d^5}$ spectra are extremely weak. The magnitude of spin-orbit coupling, however, does vary with $Z^4$ ($Z$ being the atomic number). Therefore, while spin-orbit coupling is rather small in the 3d series, it becomes an extremely important mechanism by which the spin selection rule in the the 4d and 5d congeners, as well as the f-block elements, is relaxed.


Factors affecting strengths of carboxylic acids For carboxylic acids, I read that electron-withdrawing groups like CF3 allow the anion resulting from dissociation to be more stabilized. However if CF3 is replaced with an electron-donating group like CH3, then the stability of the anion resulting from dissociation is less stabilized. I read that the pulling of electron density away from acidic centers will increase its stability and render it more acidic. Why does withdrawing of electron density from an acidic center result in the stabilization of the anion formed from dissociation? Or is this a misconception?

Also, what determines whether a group is electron withdrawing or electron donating?

1)"Why does withdrawing of electron density from an acidic centre result in the stabilization of the anion formed from dissociation? Or is this a misconception?"

Imagine you're an acid molecule. You can only show your "acid" characteristics if you're able to dissociate your proton (H+) when you're in water. If you lose your proton easily, you're a strong acid.If you can't lose your proton, you are not a good acid(weak acid). This is the background you need to know about acids first.

Next to the question about the importance of stability: After you lose your proton (suppose from your hand), you are now a negatively charged ion floating in water (with a negative charge at your hand). Because there is now a -1 charge in just your hand, there is a dipole between the rest of your body and your hand. This is considered not as "stable" as let's say if the -1 is dispersed throughout your body. The presence of the dipole would "encourage" the proton you lost to return to you. And if it does, you're no longer an acid.

If it requires more work for you to remain safely as a negatively charged ion in water (because of an uneven charge spread), it would be less probable that you exist in that state. Hence, it is said that you are less "stable".I've tried to simplify it.

"Also, what determines whether a group is electron withdrawing or electron donating?"

This is easier to answer. One part, Electronegativity. It is a measure of how strongly an atom pulls electron (cloud) towards itself. Highly electronegative atoms are Fluorine (F), Oxygen (O) and Nitrogen (N) in that order too. F > O > N. you can google "electronegativity chart" and look at images of the values. You can also see how the pKa changes with the atoms near the charge in your picture. (Lower the pKa the stronger the acid, btw).

Carbon and Hydrogen are much less electronegative and they both have similar electronegativity. Hence, compared to F, O, N, there can be considered to be electron donating as they are at the other end of the spectrum.


Chemical Potential of Idea gas? Let's say we have a mixture of different ideal gases. I found that the chemical potential of a single component of the mixture is
$$\mu_{i}=g_{i}^{0}+Ln(\frac{f_{i}}{P_{ref}})$$

where $f_{i}$ is the fugacity of the component. I also found somewhere else that the chemical potential of pure substances is the molar gibbs property. Would it be safe to then say..

$$\mu_{i}(T,P)=g_{i}(T,P)$$

Yes, by definition chemical potential is related to Gibbs potential in the following way:

$\mu_i = \left(\frac{\partial G}{\partial n_i}\right)_{p,T,n_{j\neq i}} = g_i$

Those two notation are synonyms. You will find very detailed and precise explanation of this in: Atkins, Physical Chemistry.


Is 1-methoxy-2-methylpropane a positional isomer of 2-methoxy-2-methylpropane? My book doesn't classify them as positional isomers but the position of the methoxy group is different in both the molecules. So, I guess they should be positional isomers.

I understand that the two compounds are metamers. But can't a metamer be a positional isomer?

2-methoxy-2-methylpropane is a synonym for MTBE (Methyl tertiary-butyl ether). The name "propane" is a little bit missleading, because it comes from the standardized IUPAC nomenclature system, which names the  ether-oxygen together with the smaller alkyl group an "alkoxy" group with the suffix "-oxy" (in our case: methoxy) and takes the larger alkyl part as its base identifier (here propane) when describing ethers. So MTBE is referred as an ether with all its chemical properties.

Now to the isomerism: Structural isomers have the same chemical formula but different constitutions (arrangement of atoms, bonds, groups). So they definitely must be constitutional isomers.

Constitutional isomerism is divided into 5 subcategories:

Chain isomerism
Positional isomerism
Functional isomerism
metamerism
tautomerism

As you already mentioned: they are metamers. Thats because different alkyl rests are attached to the same functional group (ether). This is the first reason why they are not likely to be positional isomers. And the other reasons might come from the "moving substituent" (alkoxy group), which has a carbon atom (the carbon structure has to be constant within positional isomers, therefore moving the -OCH3 could mean a difference) and is no functional group.


Why does electronegativity generally increase across a period? I've been doing some research and the only answer I seem to be getting is that the increase of protons means electrons are more attracted to the atom.

I thought though it was because as you go across a period, an atom will slowly fill up its outer shell, and hence reduce its need to gain or hog electrons?? Or is this incorrect?

Electronegativity is the tendency of atom to attract the electron cloud towards itself in a covalent bond. Across the period, since the electrons are filled in the same shell(which means it should be of the same size without any electron) and the amount of positive charge increases and as a consequence the size of the atom also decreases(Each electron is pulled more strongly by a nucleus with higher positive charge). The electrons are hence more tightly held as you go across the period.

  I thought though it was because as you go across a period, an atom
  will slowly fill up its outer shell, and hence reduce its need to gain
  or hog electrons?? Or is this incorrect?

A need to gain or lose electron is dictated by the nearest configuration of greatest stability(Usually Octet). And this is not closely related to Electronegativity either. Electronegativity only has to do with how strongly can an atom pull the electron cloud towards itself. Actually, as you go across the period, the tendency to gain electron increases(Indicated by increase in Electron Affinity) and not decrease as you thought.


How do nylon and polypropylene blend together in the production of a toothbrush? Why is nylon (nylon 6) and polypropylene chosen together to make a toothbrush? Specifically, how are they blended together during the production process, and what determines their price?

A toothbrush need not be made of the two materials you mention.  The two materials are choosen because of their low price of production*), ease of manufacture of the intended tooth brush (heat molding of the frame), moderate / tuneable flexibility of the material for the bristles (nylon).  In comparsion to bristles of natural origin, nylon equally is easier to cleanse by rinsing with water, and dries rapidly ‒ offering less ground for bacteria to grow on them.

*) Waste disposal is not accounted here. While the brush, consisting of polymers, may be incarcerated, why do we throw away the whole toothbrush away if apparently only the heads / bristles wear out?


What does "dep" mean? I'm looking at some data with substrate concentration on the $x$-axis and $k_\text{dep}$ on the $y$ axis. What does $k_\text{dep}$ mean? I understand that it must be a rate of some kind but what is the significance of the 'dep' part?

The data I'm looking at can be found in Fig. 9 in this paper (PDF):

Essentially this is a dose response curve for a substrate and enzyme. I suspect $k_\text{dep}$ is simply the rate but why $k_\text{dep}$ and not just $k$?

The "dep" probably stands for "depletion": read the paper you linked to for more context. Often times, subscripts on rate constants are used to refer to which reaction in a system of multiple reactions. You'll see $k_1$ and $k_2$ for rate constants of "reaction 1" and "reaction 2", etc.

I suspect in the paper, they were measuring the depletion of a substrate, and not what the substrate was converted to.  So they wouldn't know if they were measuring the rate of a single reaction $\ce{A->B}$, or the combined rates of multiple reactions such as $\ce{A->B}$ and $\ce{A->C}$.  Thus (probably) the desire to indicate $k_{depletion}$ instead of just $k$.


Why does the phosphate ion have resonance structures? My first-year chemistry teacher depicted $ \mathrm{PO_{4}^{3-}}$ as exhibiting the resonance structure inset below this question.

However, I can't figure out what diagramming rules (which is what she was teaching) she used to deduce her answer. I thought that the double bonded oxygen would be single bonded and that it would have six unbonded electrons like the other oxygens.

Why should one oxygen be different than any of the others?
What diagramming rules imply that  $ \mathrm{PO_{4}^{3-}}$ should be depicted as my teacher did?

A much better representation of phosphate is the one given below (You need to add three lone pairs to each oxygen atom which I was too lazy to do):

It does not expand phosphorous’ octet, which calculations have shown is unlikely.
It does not require phosphorous’ 3d-orbitals which are energetically too far removed to participate in hybridisation in a noteworthy manner.
It gives you four equivalent $\ce{P-O}$ bonds a priori without needing to resort to mesomery.
It gives all oxygen atoms the same charge a priori and also gives phosphorous a charge in better agreement with experiment/calculation.
The experimental bond lengths are better explained as single bonds rather than having a bond order of $\frac{5}{4}$.

This can all be summed up as ‘phosphorous obeys the octet rule much like every other main group atom (doublet rule for hydrogen/helium).’

But you probably want answers to your questions on the decidedly minor mesomeric representations your teacher drew. Well:

  Why should one oxygen be different than any of the others?

Those four mesomeric structures are entirely equal, just like in benzene. To get a ‘picture closer to the truth’ you would need to put them all on top of each other and look through them. You would see that every $\ce{P-O}$ bond has the same (minute) chance of being a double bond and thus no oxygen is different from the others.

  What diagramming rules imply that $\ce{PO4^3-}$ should be depicted as my teacher did?

Well: none, really. If you are going to stick with the double bond representation, the order of the four structures does not matter as long as they are all there. Also, you can go from every single one to every single other one in a single step by appropriate electron shifts. So, as MaxW commented, you really should add the two missing diagonal arrows to be correct.


Tensile properties of cellulose fibres In cellulose fibres, such as cotton or hemp, I've noticed different percentages of lignin and cellulose. How does this affect their tensile strength and toughness? Are there any other important factors? Thanks!

Lignin and cellulose in cotton, work together like the epoxy resin and  the glass fibers in  fiberglass composite materials (a fiberglass boat, for example). So, the cellulose is the fibrous component and is the primary load-bearing element.  While lignin is the polymeric matrix that provides stiffness and rigidity.

The more cellulose is in cotton, the higher is the tensile strength.


What does this representation of organic aromatic compounds mean? I saw this structure in some problem. What does this mean? How to express this in form of a simple structure

This is the cyclooctatetraene dianion. One way to get to it is to start with cyclooctatriene and remove one hydrogen ion each from the two $\ce{- CH2 -}$ groups. This is related to the cyclopentadienyl anion, the more common aromatic anion.

Image source: https://www.masterorganicchemistry.com/2017/05/17/frost-circles/


Conversion of benzenediazonium chloride to benzene using lithium aluminium hydride Why does benzenediazonium chloride get converted to benzene upon
reacting with $\ce{H3PO2}$ and not $\ce{LiAlH4}$ ?
Also, what is the reaction mechanism of the reaction with $\ce{H3PO2}$
?

This Wikipedia page list all kind of nucleophilic substitution reaction on aromatic compound why can't $\ce{LiAlH4}$ / $\ce{H-}$ react like one of these to give benzene and expel nitrogen.
I just wanted to add that it's up to my knowledge that $\ce{LiAlH4 / H-}$ can't reduce the double bond of Benzene, because when styrene reacts with $\ce{LiAlH4 / H-}$ it gives ethyl benzene.

Update
I was just researching (on google) for the mechanism where I came across this site suggesting the mechanism to be radical and as follow :

But still there is one problem the side products don't match this site states the side product to be $\ce{H3PO3}$ and $\ce{HCl}$ but as stated above the side products must be $\ce{H2PXO2}$.

I just wanted to add that it's up to my knowledge that $\ce{LiAlH4 /
H^{−}}$ can't reduce the double bond of Benzene, because when styrene
  reacts with $\ce{LiAlH4 / H^{−}}$ it gives ethyl benzene.

That answer is as good as any for the first part of your question.  It demonstrates that LAH will react preferentially with aliphatic pi bonds rather than the aromatic ring.  Similarly there are aliphatic pi bonds in $\ce{Ar - N^+\equiv N}$, and so the LAH will reduce the diazonium group rather than displacing it.

For the hypophosphorous acid reduction, the mechanism you found has a problem.  The principle of microscopic reversibility argues against three components coming off at once in what is shown as the first step (it would be equivalent to three independent bodies colliding simultaneously).  The first step may be correct but it likely has more than one component to it.


Algebra of conserved quantities for chemical reactions So there is the standard formation enthalpy $\Delta H^0_f$ and there is also formation entropy.

Are there more (independent) quantities? What about the heat capacity? Is there a general theoretical structure where one can show one got all the conserved quantities associated with a chemical reaction? It seems to me that one could this on a quite rigid structure and invesigate the properties of chemical processes from that point of view.

And to what extend are the procedures like using Hess's law just rules of thumb or what are some insights about he deeper structure where they come from.

What I think you are asking about is the number of thermodynamic parameters which characterize a chemical reaction. Those parameters are used to make quantitative predictions of the final state of the reactive system (at least in the context I'm familiar with, i.e. chemical engineering); for example, the amount of heat exchanged with the reactor's environment, the composition of the mixture at equilibrium, it's material phase, overall density, etc.

In classical thermodynamics, the way to address this problem is through the potentials defined on the system of interest, and studying equilibrium as the extremum of one these. The chosen potential depends on the particular configuration of the system under study; and in all cases their relation with equilibrium is a consequence of the Second Law. For example, if the system is restricted to evolve at constant volume and entropy, the minimized potential is the internal energy; if it evolves at constant energy and volume (an isolated simple compressible system) it maximizes entropy; etc.

From a practical point of view of natural systems and industrial processes -which evolve at nearly constant temperature and pressure fixed by the environment- the involved minimized potential is the Gibbs free energy. That the minimization of $G$ yields an equilibrium state for a closed system at constant pressure and temperature is straightforward (it can be found in any physical chemistry textbook). Moreover, when we say here that the system at it's minum $G$ is at equilibrium, we are saying it's at overall equilibrium: it's simultaneously at mechanical equilibrium (because of null pressure gradients), thermal equilibrium (because of null temperature gradients) and in chemical equilibrium (because of null chemical potential gradients); so there's no "force" (in a generalized sense) to drive any further process.

So, going back to your question, which are the independent variables of a reactive system as such? Well, suppose that in your system you have $P$ chemical species which can be both reactives or products of $Q$ reactions taking place in a monophase. For every species, the balance on the number of moles of each one yields:

$n_{i}=n_{0i}+\sum_{j=1}^Q{\nu_{i}^j\cdot \xi_{j}}$

Where $\xi_{j}$ is the extent of the j-th reaction.

Then, we can define the chemical potential as the partial molar Gibbs free energy per species:

$\mu_{i}=\displaystyle\frac{\partial G}{\partial n_{i}}$

And also, given that $G$ is an extensive property (or better phrased, a homogeneous function of first degree in mass) by Euler's Theorem it follows that:

$G=\sum_{i=1}^P\mu_{i}n_{i}=\sum_{i=1}^P\mu_{i}\cdot (n_{0i}+\sum_{j=1}^Q{\nu_{i}^j\cdot \xi_{j}})$

Every chemical potential can be expressed as the sum of it's specific Gibbs free energy of it's pure form, at the temperature and pressure of the mixture, plus a term which ponderates the influence of concentration and chemical affinity of each particular species towards the rest of the present compounds which constitute the mixture (for example, the solvent matrix, if one species is so concentrated that it can be considered so). The latter term contains an adimensional parameter known as the activity of the species:

$\mu_{i}=\mu_{0i}(P,T)+RTln(a_{i})$

The zero index in the first term just means that it's referred to a pure system. This specific free energies can be related to the formation enthalpies and entropies which are tabulated through:

$\mu_{0i}=h_{0i}(P,T)-T\cdot s_{0i}(P,T)\\\mu_{0i}=\left\{{h_{fi}^0(P_{0},T_{0})+\displaystyle\int_{(P_{0},T_{0})}^{(P,T)}dh_{0i}}\right\}-T\cdot \left\{{s_{fi}^0(P_{0},T_{0})+\displaystyle\int_{(P_{0},T_{0})}^{(P,T)}ds_{0i}}\right\}$

Where $h_{f}^0(P_{0},T_{0})$ and $s_{f}^0(P_{0},T_{0})$ are the standard formation enthalpy and entropy, respectively. The integrals evaluate the variation in enthalpy and entropy of the pure compounds due to a difference in temperature and pressure in reference to standard values (25 ºC and 1 bar). The evaluation of the integrals in terms of measurable properties must be done using Maxwell relations and definitions of potentials:

$dh_{0i}=\displaystyle\frac{\partial h_{0i}}{\partial {T}}dT+\displaystyle\frac{\partial h_{0i}}{\partial {P}}dP=C_{p}dT+\frac{dP}{\rho}\\ds_{0i}=\displaystyle\frac{\partial s_{0i}}{\partial {T}}dT+\displaystyle\frac{\partial s_{0i}}{\partial {P}}dP=Cp\frac{dT}{T}-\frac{\alpha}{\rho}dP$

What about the activities? These are non-linear functions of the composition, temperature and pressure of the mixture, and usually that non-linearity is expressed through an activity coefficient (it may be the case that further coefficients are introduced theoretically, but all they do is to "partition" the contributions to non-ideality into a product of these, and so here I talk about a unique coefficient without loss of generality). So:

$a_{i}=x_i\cdot\gamma_{i}(P,T,x_1,x_2,...,x_P)$

And:

$x_i=\displaystyle\frac{n_i}{\sum_{k=1}^P n_k}=\displaystyle\frac{n_{0i}+\sum_{j=1}^Q{\nu_{i}^j\cdot \xi_{j}}}{\sum_{k=1}^P\left\{{n_{0k}+\sum_{j=1}^Q{\nu_{k}^j\cdot \xi_{j}}}\right\} }$

Finally, the (enormously complicated) expression for G turns out to be dependent of the variables:

$G=G(P,T,h_{fi}^0,s_{fi}^0,Cp_{i},\rho_{i},\alpha_{i},n_{0i},\xi_j) \ \ \ \ \ \ i: 1 < i < P \ \ \textrm{and}\ \ \ j: 1 < j < Q$

Those are, a priori, the "independent quantities" involved in the overall reaction's equilibrium. Once you've fixed the temperature, pressure, the original amount of substances and get the intensive properties stated for all the chemical species present in the system along it's evolution, G becomes only a function of the extents of the reactions:

$G=G(\xi_j) \ \ \ \ \ \  j: 1 < j < Q$

And the equilibrium is reached when $G$ is at minimum; that's a problem solvable through optimization. The values of the extent of every reaction at equilibrium is enough to calculate every other useful parameter of the system: it's composition, the heat involved, the change in overall volume, etc.

Summary: the conserved quantities in a chemical reaction are physical constraints (conservation of total mass, balance of moles according to stecheometry, conservation of energy exchanged with the environment, strict increase in entropy, etc) but that's not enough information to uniquely evaluate the equilibrium parameters of a general reaction (or reactions). Most of the expressions I left in an undefined closed form in this answer (such as the activity coefficients, all of them functions of compositions) must be experimentally correlated, or modelled with some theoretical formalisms, generally highly complex. The standard thermodynamic potentials might be tabulated though, as it's the case with the intensive properties of pure substances (specific heats, thermal expansion coefficients, etc).


How does the radial distribution function of Vanadium differ from that of Calcium and how does this affect the ionic electron configurations? When Vanadium is ionised it loses the 4s electron first, meaning that it's 3+ ion has a different electron configuration to Calcium despite it being isoelectronic. Can it be explained in terms of radial distribution functions? Maybe the maxima of the 3d and 4s become further apart as the effective nuclear charge increases?

I think @NicolauSakerNeto mentioned a number of excellent points in his linked answer. Frankly, I think he deserves a bounty.

There are several effects occurring, which give different electron configurations to the transition metals. A very nice review article is available: "The Full Story of the Electron Configurations of the Transition Elements" J. Chem. Educ., 2010, 87 (4), pp 444–448

"d orbital collapse": Because of the difference in angular-momentum "centrifugal force" in quantum mechanics, incomplete shielding of core electrons, and nuclear attraction, the $Z_{eff}$ - effective nuclear charge - for 3d and 4s electrons are different. As a rule of thumb, electrons are typically lost from the 4s first,
electrostatic repulsion in 4s vs. 3d. It's interesting you mention V, since $\ce{V^0}$ is $[3d^3 4s^2]$ and $\ce{V+}$ is $[3d^4]$. The usual explanation is that while 3d is closer to the nucleus and energetically favorable, once you start filling the 3d electrons, the electrostatic repulsion is high, and for the neutral atoms, it's favorable to have one or two 4s electrons, which are more diffuse.

There are other things going on, particularly later in the periodic table, where relativistic and spin-orbit coupling effects occur. Also, many times the electron configuration can change in a free atom or ion compared to a complex.


Understanding the derivation of Gibbs Free Energy In the equation: $$ΔG=ΔH−TΔS$$

I am having trouble understanding my professor's derivation of it from
$$ΔS_{\mathrm{univ}}=ΔS_{\mathrm{sys}}+ΔS_{\mathrm{surroundings}}$$

where he says $$ΔS_{\mathrm{surroundings}}= \frac{-ΔH_{\mathrm{sys}}}{T}$$
which I have learned is only true for a reversible process.
How is it possible that this heat can be reversed? And why can't we say that the change of entropy for the system is also equal to the equation. I understand that it would mean the reaction is at equilibrium, but would it not be since heat is reversible?
$$ΔS_{\mathrm{sys}}= \frac{ΔH_{\mathrm{sys}}}{T}$$

The surroundings are treated as a constant volume reservoir, so the first law of thermodynamics,
$$
\mathrm{d}U_\mathrm{surroundings}
  = \mathrm{d}q_\mathrm{surroundings}
  + \mathrm{d}w_\mathrm{surroundings},
$$
yields
$$
\mathrm{d}U_\mathrm{surroundings}
  = \mathrm{d}q_\mathrm{surroundings}
  \text{, or }
\Delta U_\mathrm{surroundings}
  = q_\mathrm{surroundings}.
$$
Notice that $q_\mathrm{surroundings}$ is equal to a state function, so it behaves as one, that is, it is independent whether the process is reversible or not. If the process is reversible
$$
q_\mathrm{rev,surroundings}
  = \Delta H_\mathrm{surroundings},
  \tag1\label{eq:rev}$$
if it is not reversible
$$
q_\mathrm{surroundings}
  = \Delta H_\mathrm{surroundings},
  \tag2\label{eq:irrev}$$
from \eqref{eq:rev} and \eqref{eq:irrev} follows
$$q_\mathrm{surroundings} = q_\mathrm{rev,surroundings}.$$
Therefore
$$
\Delta S_\mathrm{surroundings}
  = \frac{q_\mathrm{rev,surroundings}}{T},
$$
in particular
$$q_\mathrm{rev,surroundings} = -q_\mathrm{sys},$$
then
$$\Delta S_\mathrm{surroundings} = \frac{-q_\mathrm{sys}}{T}.$$
If the process occurs at constant pressure
$$q_\mathrm{sys} = \Delta H_\mathrm{sys},$$
therefore
$$\Delta S_\mathrm{surroundings}  = \frac{-\Delta H_\mathrm{sys}}{T}.$$


How long does it take to saturate H2O with O3 Let’s assume I have one liter of distilled $\ce{H2O}$ in an stainless steel container – from the bottom side of the container I do bubble ozone trough the $\ce{H2O}$.

Of cause it is not pure O3, so let’s assume that the gas consists of:

7% $\ce{O3}$

83% $\ce{ O2}$

10%. $\ce{ N2 / NO2 / NO}$

The gas flow is about 5 liters per minute and the corona discharge consumes about 500W.

How long will it approximately take, till full saturation of the
  $\ce{H2O}$   with   $\ce{O3}$  will take place?

The solubility of ozone in water as per one website is:

First thing is to establish which curve you need. In 1 m$^3$ gas you have 70 L ozone:

$\frac{70\ L\ \ce{O3}}{1\ m^3}\frac{1\ mol}{0.082\cdot 298\ L}\frac{16\cdot 3\  g\ \ce{O3}}{1\ mol}= 138\ g\ \ce{O3}/m^3$

Let us take the curve of 150 g/m$^3$ which has an ozone solubility of 30 mg/L water.
Then:

$\frac{30\cdot 10^{-3}\ g\ \ce{O3}}{1\ L\ \ce{H2O}}\frac{1\ mol}{16\cdot 3\ g\ \ce{O3}}\frac{0.082\cdot 298\ L}{1\ mol}\frac{1\ min}{5\cdot 0.07\ L\ \ce{O3}}= 0.044\ min$

Which is about 3 seconds.


In terms of remediation, what is sequestration? I am reading about use of zero valent iron in remediation of heavy metals.
Is sequestration just a broader term for sorption?

Sequestration is any way of preventing the heavy metals from being able to harmfully react with biologically important molecules. In the case of zero-valent iron nanoparticles, the heavy metals are sequestered by sorption as well as reduction. However, there are other methods of sequestration, such as chelation.


Dichromate in acidic medium We were taught that $\ce{Cr_2O_7^{2-}}$ (dichromate) in $\ce{H_2SO_4}$ gives $\ce{CrO_3}$

$$\ce{Cr_2O_7^{2-} +H_2SO_4 -> CrO_3 + ...}$$

Link: wikipedia

But we know that dichromate in acidic medium converts to $\rm Cr(III)$.

$$\ce{ Cr_2O^{2−}_7 + 14 H3O+ + 6 e− → 2 Cr^{3+} + 7 H2O}$$

Link: wikipedia

How is this possible?

Both statements on the behaviour of $\ce{Cr2O7^{2-}}$ are true, but you are confusing two different reactions.

The formation of chromium trioxide from (sodium) dichromate in sulfuric acd is not a redox reaction. There is no partner that could be oxidized by $\ce{Cr2O7^{2-}}$. Concentrated sulfuric acid is an oxidant itself and metals like copper are dissolved (oxidized to $\ce{Cu^{2+}}$) while not hydrogen is generated.

Remember also that sulfuric acid is a strong, oxidizing and dehydrating acid. This effect is used in the production of $\ce{CrO3}$, which can be considered the anhydride of chromic acid ($\ce{H2CrO4}$). Note that chromium in $\ce{Cr2O7^{2-}}$, $\ce{H2CrO4}$, and $\ce{CrO3}$ always has the same oxidation state.

The second reaction that you mentioned does happen too, and this is indeed part of a redox reaction in which $\ce{Cr(VI)}$ is reduced to $\ce{Cr(III)}$. However, this can only happen when a suitable reaction partner is present, which in turn is oxidized by $\ce{Cr2O7^{2-}}$.


How were the first molecular formulas determined? How were molecular formulas determined before spectroscopy and spectronomy? I know that John Dalton thought that the formula of water was "HO" but we now know it to be "H2O." How did Dalton even know that water had hydrogen and oxygen in the first place?

Lavoisier burned hydrogen and oxygen together and found that water was produced.  Dalton incorporated this information into his thinking that the most stable arrangement of any two atoms involved a 1:1 combination, therefore water would be "OH".

In slightly more recent times, but prior to modern-day spectroscopy, chemists would chemically chop molecules into identifiable fragments.  They would also combust the molecule and get a C,H,N,O ratio.  Putting all of this together they would (often) come up with remarkably accurate molecular formulas.


Phosphates Functional Group In my text book it states that a phosphate group is considered to be $\ce{-OPO3^{2-}}$. However, online it states that phosphate is $\ce{PO4^{3-}}$. Why do these contradict? Am i confusing phosphate the chemical with the phosphate group? Please see image for reference.

Phosphate $\ce{PO4^3-}$ is an anion of the phosphoric acid $\ce{H3PO4}$. Term phosphate comes from a salt, for instance a sodium phosphate $\ce{Na3PO4}$. $\ce{Na+}$ are bound with a $\ce{PO4^3-}$ through ionic bonds. No electron sharing.
To annotate specific moieties of a molecule skeleton, a chemical group term is used. The phosphate has donated one pair of electrons on the oxygen to have a covalent bond with other molecule, for instance sugar in the image, and becomes a phosphate group in the molecule. Since the electron pair does not sit on the oxygen any more, but is shared with carbon, the phosphate charge is lowered to −2.


What's the definition of multi phase materials? Does it mean the different phases represent different materials? I'm from a CS background and just trying to understand the real meaning behind multi phase materials. Following is the experiment which drove me towards this question.

With the help of DFT and machine learning, we identified that the material LaBiO3 could display interesting dielectric properties. Hence, a team here mixed the oxides in stoichiometric ratio in a ball mill and observed the XRD patterns of the formed composite powder.

The XRD data suggest that LaBiO3 is unstable and form Bi1.4La0.6O3 R-3m phase and Bi2La4O9 R3m phases.

So my question is, is this a composite that has nothing to do with LaBiO3 and they are just two different materials? Or could this be categorized as a metastable phase of LaBiO3?

Could anyone explain this with the definitions?

The definition of a substance that is "multi-phase" is that there is more than one distinct compound in it and the compounds form distinct regions in the substance with different properties.

Examples make this clearer. Salt dissolved in water forms a uniform mixture: a salt solution. There is more than one compound there but they are perfectly mixed.

Fatty organic compounds are not soluble in water but can form small dispersed fatty particles. We call the result milk (I'm simplifying a little as there are other things in there as well). The result has more than one phase: a microscope would reveal very small, dispersed particles of fat floating in water. Milk looks like a single uniform substance, but isn't when you look more closely.

Fibreglass is, essentially, a bunch of small fibres made of glass embedded in a matrix of some polymer. There are two distinct phases with different compositions in the substance.

The characteristic of multi-phase compounds is that different regions of the substance contain different compounds (these regions can be very small as in some metal alloys). In your case the reaction seems to produce two distinct compounds which may be finely mixed together. The XRD pattern shows two different crystal structures. This may result because this specific reaction does not create the compound you wanted under these conditions. Which could be because the compound you want is not stable or could be because the reaction conditions allow more than one product to form. Either way the reaction is giving a mixture not a single compound.


Isopropyl and Butyl Groups - Relative Priorities This does not compute for me:

Why would the four-carbon butyl group have a lower relative priority than the 3 carbon isopropyl group?

If we compare the carbons one by one between the isopropyl and the butyl group, we'd run out of carbons in the isopropyl group first! How then can isopropyl possibly have a higher priority than the butyl group?

Comparing IsoPropyl and n-Butyl

On first position both have C.
On second position IsoPropyl has two C's whereas n-Butyl has only one C.
Thus IsoPropyl has higher priority and we do not see the further chain.


Can someone confirm and/or correct my logic regarding this mechanism? In the picture, Ethyl 2-oxocyclopentanecarboxylate (I) reacts with ethyl 2-(4-chloromethylphenyl)propionate (II) in the presence of KOH in hot DMF to afford ethyl 2-[4-(1-ethoxycarbonyl-2-oxocyclopentan-1-ylmethyl)phenyl]propionate (III), which is then hydrolyzed and decarboxylated by treatment with 47% HBr in refluxing dioxane. I'd just like to walk my way through this and see if my theories regarding the mechanism are correct.

(I) has an $\alpha$ carbon that is deprotonated (the one beside the ester group). This creates a carbanion that is stabilized both by resonance as an enolate and inductively by electron withdrawing from the ester group. This stability leads to marginal nucleophilicity; chlorine is a marginal leaving group here (although the carbocation is resonance stabilized by the aromatic ring), and no inversion is observed, which leads me to believe an $S_N1$ rxn occurs between (I) and (II) to yield (III). DMF is used as a solvent so as not to inhibit nucleophilicity by using a polar protic solvent.
It's not specified, but I suspect (III) can be  captured via distillation to due boiling point/density differences between it and (I) and (II).
(III) is reacted with HBr in dioxane (I'm not sure why 47% is used; I'm assuming at STP, it's the maximum concentration of HBr(g) in water) is used, as dioxane is relatively stable at extremely high and extremely low pH values. The hydronium acts as an impetus for hydrolysis on the ester groups on both rings. I think the Bromide anion acts as just a counterion here (but I'm not sure why).
The step where (III) is reacted with sodium hydroxide is where I get lost. I'm not sure how the sodium hydroxide spurs decarboxylation and selective decarboxylation, at that (why not the ester on the aromatic ring as well?).

The decarboxylation occurs specifically at that position, because that is a $\beta$-keto acid. In presence of slight amount of heat only the decarboxylation occurs due to the presence of keto group adjacent to it. Actually, in acidic medium only,it decaboxylates in presence of heat as follows,

But no keto group is present at the carboxylic acid group attached at the benzylic position so it cannot decaboxylate.


Do bond lengths vary across molecules of the same kind? I only have superficial knowledge of chemistry, and today I found out about bond length for the first time. For example, on Wikipedia it says that all C-C bonds of benzene have a length of 140 picometres (both C-C and C=C, I think). But is this exactly true for all benzene molecules?

Or does bond length of C-C vary slightly for 10 molecules of benzene? Is 140pm an average distance for a million molecules, or each and every benzene molecule has bonds of length 140pm? Is this a fundamental and unvarying length of the benzene molecule, so that a different length cannot possibly occur?

I know that during studies of the bond length in the laboratory, each individual measurement has a slight error. I'm not asking about that.

The atoms within a molecule are not stationary, but are constantly vibrating, twisting, etc. Thus, the value given for a bond length between two atoms must be an average. Specifically, the average distance from the nucleus of one atom to the nucleus of the other is what constitutes the definition of bond length.


Standard solution and extracted solution : same concentration? Summary :
I am currently testing how a certain type of foam adsorbs HAP, PCB and OCP, and to this end I am depositing some on the foam, then extracting it. To know how much was extracted, I need to compare with a standard solution of HAP, PCB and OCP. The problem is that when I look at the chromatograms, the extracted solution's peaks' areas are about 10 times the standard solution's peaks' areas. I don't know where in the experiment I am doing something wrong.

Details :
I have a solution of HAP, PCB and OCP at 10 mg/L. I put 100 µL of this solution on the foam, so that the deposited mass is 1 µg.
The foam is ASE-extracted with acetonitrile, then the acetonitrile is rotary-evaporated. What is left in the round-bottom flask is then diluted in 1 mL acetonitrile. So at maximum I have 1 µg pollutants in 1 mL acetonitrile.
To this 1 mL I then add distilled water until I have a 20 mL solution. The concentration of this solution should at maximum be 50 µg/L, right ? (If all pollutants are extracted from the foam and none lost to evaporation.)

The standard solution is made by also taking 100 µL from the 10 mg/L solution (so 1 µg pollutants), and adding distilled water until I have a 20 mL solution. So this solution's concentration should also be 50 µg/L, right ?

But when I put the two solutions through GC-MSMS, the peaks' areas of the extracted solution are about ten times greater than those of the standard solution. (Not always ten, but it is usually the case.)

Have I made a mistake somewhere ? Are my calculations/manipulations wrong somewhere ?
The internal standard used is naphtalene d8, and in each 20 mL solution I added 10 µL of a 10 mg/L naphtalene d8 solution. (So its concentration is 5 µg/L : is it a problem that the standard is not at the same concentration as the pollutants ?)

The experiment was repeated several times, with different foam extracts and different standard solutions, and the results were the same. It is not a problem of contamination, as everything has been extensively washed with distilled water, detergent and acetone.

In case it is relevant : when a solution is done, I sample 4 mL that I extract with a SPME fiber, and then I inject into the GC-MSMS.

Thank you in advance, I am still testing and calculating, so any help is appreciated.

To tschoppi (since I don't have enough reputation to comment) : I haven't tried with another kind of foam, but it's a good idea (to see at which step the problem lies). I'll see what we have in the lab. I am also pretty sure the foam I used isn't contaminated, because it was given to us straight from the package, and I washed it three times with different solvents and analysed the wash-solvent (which was clean)(but I may check again to be sure).

I found the answer to my question, and I am going to share it just in case.
Apparently, when extracting the pollutants from the foam, I am also extracting some foam-specific materials that influence the pollutants' behaviour during the SPME part of the extraction.
I tested that by extracting the blank foam, then adding a standard amount of pollutants to the extraction solvent. This gave the same result as extracting a spiked foam, meaning higher peak areas than a standard solution.

In short : matrix effect from the foam. I don't yet know why or what, but it's a matrix effect.


How is the potential energy between two atoms measured? In class, we learned about the interatomic potential graph.

How does one actually measure the potential energy between two atoms experimentally to ensure the graph is correct? What device is used?

The morse potential (equation see e.g. wikipedia) basically contains the bond dissociation energy, a "force constant" and the bond length at ground state. It does not give any measurable reality, but is just a mathematical model describing (approximating) the same.

The lower part of the curve (usually the ground state and perhaps the first two vibrationally excited states) are approximated by the harmonic oscillator, the energies of which you can get from  IR or Raman spectroscopy. The harmonic oscillator (= an ideal spring) has only one parameter, the force constant.

The bond lengths you can get either directly from microwave rotational spectroscopy, or from the rotational fine structure of you IR spectrum. They give you the reduced mass of the structure, which gives you the bond length if you know the absolute mass of the two atoms.

For the math and actual application, I refer you to any standard physical chemistry textbook.


Why aqueous KI is used when collecting Iodine? In room temperature solid di-iodine pentoxide($I_2O_5$) reacts with carbon monoside to form carbon dioxide & Iodine. This can be used to measure the amount of carbon monoxide in an air sample. $5dm^3$ air sample is passed through a tube containing solid $I_2O_5$. $\bbox[yellow]{Discharged \space I_2\space is \space sent\space into\space a\space container\space which\space has\space excess\space aqueous\space KI\space solution.} $ This solution is titrated with $0.005mol/dm^3$$Na_2S_2O_3$ solution( Starch indicator was used). If$10cm^3 of Na_2S_2O_3 $ was used, find the concentration of carbon monoxide in ppm. (density of air sample =$1.4\times10^{-3}gcm^{-3}$

Question: Why have they used KI solution when collecting $I_2$?

My View: Is it because $I_2$ is readily dissolving in KI ?

P.S. I already did the calculation. Just interested in the following phenomenon.
Thanks

Your guess is right. Pure water does not dissolve iodine very well, but a solution of $\ce{KI}$ does, thanks to the formation of $\ce{I3-}$ and other polyiodide anions. And of course we would prefer to have our iodine in a dissolved form when we are going to titrate it.

When it comes to the starch reaction or redox reaction with thiosulfate, $\ce{I3-}$ is as good as $\ce{I2}$.

So it goes.


Aluminium sulphate solubility as function of the pH earlier I asked about how to know the aluminium sulfate solubility as function of the ph (see the previous question here). I found this diagram on the page 57 of

  Rubin, A. J., & Hayden, P. L. (1973). Studies on the Hydrolysis and
  Precipitation of Aluminum (III).

But I'm a little confused about the interpretation. Is the
$\mathrm{Al_{2}(SO_{4})_{3}}$ insoluble between pH 5 and 7?. What will be the correct interpretation of this diagram?

The lack of continuity for the curve indicates that the concentration of $\ce{Al2(SO4)3}$ was below the detection limit of the analytical technique being used to measure the concentration of $\ce{Al2(SO4)3}$ in the solution.

Theoretically there should be a continuous curve.

+1 for asking this as a new question.


Order of lewis basic strength My teacher told me that order of lewis basic strength is inversely proportional to electro negativity. Comparing between iodide ion and chloride ion, chloride ion is more electronegative than iodide ion therefore iodide ion should be more basic than chloride ion but the order is reverse that is chloride ion is more basic than iodide ion. Why this is so? Is it related to localised and delocalize lone pair?

My teacher told me that order of lewis basic strength is inversely proportional to electro negativity.

I agree with your teacher's guideline. Now, note that this cannot be your sole guideline. Lewis basicity (and acidicty) depend on a number of factors including:

Strength of the corresponding Lewis acid or Lewis base and the strength of the resulting bonds in the Lewis adduct.
Solvent effects
Geometry of the interacting species
Electronegativity

  Comparing between iodide ion and chloride ion, chloride ion is more electronegative than iodide ion therefore iodide ion should be more basic than chloride ion but the order is reverse that is chloride ion is more basic than iodide ion.

I think you are confusing Bronsted basicity with Lewis basicity. It is true that in water (a polar, protic solvent) chloride ion is a stronger Bronsted base than iodide ion.

In terms of Lewis basicity, however, things become a lot more complicated. Iodide ion has the more readily available lone pair electrons for donation since iodide ion is less electronegative than chloride ion.

On the other hand, the strength of bonds to iodine are generally not as strong as bonds with chloride ion. This is a result of poor orbital overlap between iodide ion - which is a big atom - much bigger than many Lewis acids.

  I asked my teacher today the same question and he told that since both iodide and chloride ion have delocalize lone pair both iodide and chloride ion have delocalize lone pair in their own vacant orbital therefore the electron donating tendency decreases. here we can't compare using electro negativity. And comparing them since iodide ion has more big size than chloride ion therefore the delocalization tendency is better in iodine therefore chloride ion is better lewis base. Any comments?

I mostly agree; both have electrons in their orbitals but I'm not a fan of the term "delocalization" - that implies that the electrons are spread across multiple nuclei - but in chloride ion and iodide ion there's only one nucleus.

Again, there are multiple factors to consider - not just electronegativity.


The equilibrium constant of a reaction only involving pure solid or liquid How to calculate the equilibrium constant of a reaction only involving pure solid or liquid.
For example reaction
$$\ce{S(s) + Hg(l) -> HgS(s)}$$
or
$$\ce{S(s) + Mg(s) -> MgS(s)}$$

Concentration of  pure solids and liquids is conventionally taken to be 1. Since reaction involves only solids and liquids the equilibrium constant will be 1


Significance of Law of reciprocal proportions The laws of conservation of mass, definite proportions multiple proportions makes sense and were definitely useful for chemists in the early days of modern chemistry.
The law of reciprocal proportions states that

If element A combines with element B and also with C, then, if B and C combine together, the proportion by weight in which they do so will be simply related to the weights of B and C which separately combine with a constant weight of A.

I don't understand how the law of reciprocal proportions is useful. It seems too unnecessarily complex for someone to intuitively come up with.
Is there any intuition/specific reason that Jeremias Richter came up with it?

The comments certainly clarify the law of reciprocal proportions and are an excellent tutorial on how to write a good question but sheds little light on your question “Is there any intuition/specific reason that Jeremias Richter came up with it?”
From reading a number of short bibliographic sketches about Richter it appears his primary interest was mathematics.  The title of his doctoral thesis was, roughly translated, The use of Mathematics in Chemistry  [https://www.encyclopedia.com/science/dictionaries-thesauruses-pictures-and-press-releases/richter-jeremias-benjamin].  His study of quantitative relationships in the neutralization of acids by bases lead to the law of equivalents. [https://encyclopedia2.thefreedictionary.com/Jeremias+Benjamin+Richter ]
Based on these and similar biographical notes one might argue that he arrived at the law of reciprocal proportions not through “intuition” at all but rather through an empirically based and numbers driven approach.  When a mathematician sees a pattern or consistency in “the numbers” they then might be seek to understand the natural forces giving rise to their observations.
A modern and well written example of “math driving science” can be found here. https://www.quantamagazine.org/geometry-reveals-how-the-world-is-assembled-from-cubes-20201119/ “Scientists Uncover the Universal Geometry of Geology”


How do I prevent oxygen interfering when I prepare a FeCl2 solution from anhydrous FeCl2 powder? I have a bottle of anhydrous $\ce{FeCl2}$ powder (from here; it is 98% pure and was kept in a glovebox after opening it for the first time to minimize reactions with oxygen) and want to make this into a solution. I work in a glovebox, but what I have done prior is that I prepared the reagents outside of the glovebox and then put them inside. I let nitrogen gas bubble through water and then add this water to a certain amount of the anhydrous $\ce{FeCl2}$ powder. After this, I put the solution into the glovebox.
However, I get nowhere near to the green color of the $\ce{FeCl2}$ solutions that I find online. My solution has a slight green hue, and after some time I find a little bit of orange precipitate at the bottom of my reagent. So I fear that there is some reaction between the $\ce{FeCl2}$ and $\ce{O2}$. What can I change about my method to prevent this?
The purpose of my research is to produce the mineral tochilinite ($\ce{6Fe_{0.9}S\cdot5[Mg,Fe](OH)_2}$). I mix the $\ce{FeCl2}$ reagent with $\ce{NaOH}$, $\ce{MgCl2}$ and $\ce{Na2S}$. I do not want any $\ce{Fe^{3+}}$ in my $\ce{FeCl2}$ reagent, because it is also not present in the mineral tochilinite. If it is present in the reagent, other minerals might start to form (such as magnetite).

See if your group or an other (e.g., organic / organic catalysis) has a Schlenk line.  With some training, this allows you to reduce the interference of oxygen greatly, because you may evacuate your reaction flasks and to replace the air by an inert gas such as nitrogen or argon.  Nitrogen is relatively cheap because it equally is used for welding (as gas) or for cooling (think e.g., NMR spectrometers).  Argon gas, on the other hand has the advantage of a density greater than the one of air, thus you may cover and protect your substance under a blanket of argon.
Then, search for Schlenk flasks.  You identify these quickly for their connector to said Schlenk line. Besside round bottom flasks and tubes (a gallery, a second one), there equally are Schlenk flasks to perform a pressure filtration while working under argon enclosing a filter plate (example).
Then, degas your solvent.  In HPLC, bottled solvents for example are placed in the basket of the ultrasound bath while purging them with a steady, slow flow of argon to remove dissolved oxygen.  Do this for your deionized water.
Weigh in $\ce{FeCl2}$ in your dry Schlenk flask, and add the deionized water prepared under the inert gas.


Highest osmotic pressure I’d like to ask a question and clear my misunderstanding.

  Which of the following 0.01 M solutions have the highest osmotic pressure?
  -$\ce{NaCl}$
  -$\ce{C12H22O11}$
  -$\ce{BaCl2}$
  -$\ce{CO(NH2)2}$
  -$\ce{[Cr(NH3)4Cl2]Cl}$

I know that the equation for osmotic pressure is
$$\Pi = cRTi$$
I have eliminated sucrose, sodium chloride, and urea from my answer, because sucrose and urea are non-electrolyte. And sodium chloride has smaller van ’t Hoff factor than barium chloride. I ended up in barium chloride and the complex compound. I’m not really sure with the complex compound since from what I learnt, a complex compound is easier to dissolved in water (my teacher in solubility chapter taught me that $\ce{AgI + NH3 -> [Ag(NH3)2]I}$ have a high solubility because it is a complex compound). However I’m not sure if it has a high van ’t Hoff factor as well.

You are correct in thinking that urea and sucrose would have van't hoff factor (i) = 1, since they are non-electrolytes and don't undergo disassociation(or ion pairing).

What you are left with are the following: $\ce{NaCl}$ , $\ce{BaCl_2}$ and $\ce{[Cr(NH_3)_4Cl_2]Cl}$.

Let us begin with the the two chloride salts: $\ce{NaCl}$ , $\ce{BaCl_2}$. we assume 100% disassociation for the two (strong electrolytes) and we have i = 2 and i = 3 respectively for both.

Now, the complex would indeed disassociate but not completely (the extent would be determined $K_f$ for the complex). Thus 1 < i < 2.

Based on this, I believe you can order them on the basis of increasing osmotic pressure.


Reactivity of magnetized materials Does the magnetisation process affects the reactivity of a magnetic material?
Does the electronegativity change if a metal is magnetised?

Magnetization as a process condition (during synthesis procedure) can affect the crystal structure and nanoparticle size and morphology which in turn can affect accessibility of reactive sites. See the following papers:

Hu, Dawei, Yanmin Wang, and Qi Song. "Weakly magnetic field-assisted synthesis of magnetite nano-particles in oxidative co-precipitation." Particuology 7.5 (2009): 363-367.

Xiao, W., et al. "Magnetic-field-assisted synthesis of magnetite nanoparticles via thermal decomposition and their hyperthermia properties." CrystEngComm 17.19 (2015): 3652-3658.

Li, Yang, et al. "Fe 3 O 4@ PSC nanoparticle clusters with enhanced magnetic properties prepared by alternating-current magnetic field assisted co-precipitation." Colloids and Surfaces A: Physicochemical and Engineering Aspects 520 (2017): 348-354.


Is this a homogenous reaction? $$\ce{HCl(aq) + NaOH(aq) ->  NaCl(aq) + H2O(l)}$$
Would this reaction be considered homogenous even though there is a liquid in the products?

Yes, it would be considered homogeneous, since all of the components are in water. We don't write water as being aqueous because it would be somewhat redundant to say the water formed in water.


How do you find the enthalpy change for a given reaction? $0.500\,\mathrm g$ of magnesium chips are placed in a coffee-cup calorimeter and $100.0\,\mathrm{mL}$ of $1.00\,\mathrm M$ $\ce{HCl}$ is added to it. The reaction is:
$$\ce{Mg(s) + 2HCl(aq) -> H2(g) + MgCl2(aq)}$$
The temperature of the solution increases from $22.2\,^{\circ}\mathrm{C}$ to $44.8\,^{\circ}\mathrm{C}$. What’s the enthalpy change for the reaction, per mole of $\ce{Mg}$? Assume specific heat capacity of solution is $4.20\,\mathrm {J\,g^{-1}K^{-1}}$ and density of the $\ce{HCl}$ solution is $1.00\,\mathrm {g/mL}$.

I don't understand how to solve this; does this have to do with heat capacity equation? (Or is that just a completely different thing).
$$q = mc\Delta T$$
(Also note this is not a homework problem. The answer should be $\Delta H = -4.64 \cdot 10^{5}\,\mathrm{J\,mol^{-1}}$ $\ce{Mg}$, but I don't know how to get there and would like to understand.)

Step 1 Calculate $q_\text{solution}$. To do so, you have to have in mind that the mass of the solution is aproximatly equals to $100\ \mathrm{ml}$ of $\ce{HCl}$ plus the mass of $\ce{Mg}$.

\begin{align*}
   q_\text{solution}&=100.5\ \mathrm g\cdot4.20\ \frac{\mathrm J}{\mathrm{mol\cdot K}}\cdot\left(318.0\ \mathrm K-295.4\ \mathrm K\right)\\
   &=9.54\cdot 10^3\ \mathrm J
  \end{align*}
Step 2 Calculate $q_\mathrm{rxn}$

\begin{align*}
 q_\mathrm{rxn} + q_\text{solution} &=0\\
 q_\mathrm{rxn} + 9.54\cdot 10^3\ \mathrm J &= 0
\end{align*}
Step 3 Calculate the value of $\Delta H$ per mole.

\begin{align*}
 \Delta H&= \left(\frac{-9.54\cdot 10^3\ \mathrm J}{0.500\ \mathrm g\ \text{of}\ \ce{Mg}}\right)\left(\frac{24.31\ \mathrm g\ \text{of}\ \ce{Mg}}{1\ \mathrm{mol}\ \text{of}\ \ce{Mg}}\right)\\
 &= -4.64 \cdot 10^5\ \frac{\mathrm J}{\mathrm{mol}}\ \text{of}\ \ce{Mg}
\end{align*}


Why the valence shell configuration of Lead is 6s2 6p2? Acoording to energy rule the electronic configuration is in order of 6s 4f 5d 6p but when writing valence shell configuration we write 6s after 4f and 5d.Why is it ?

Take Lead as an example.

This is convention. To be perfectly honest, this order is not always kept up. But it makes sense to think of the orbitals by shell (first 1, then 2, …) and not by energy level.

For lead and all elements of the p block, whether the d and f orbitals are fully populated or not does not matter, they do not take part in chemistry (only the s and p orbitals of the outermost shell do). Considering that, we can write the d and f orbitals right at the beginning where it is ‘clear’ they are not needed.

For elements of the d block, we actually are doing chemistry with the d electrons. So it is maybe more customary to actually put 3d behind 4s, just to state that they are important.

There is no physical meaning behind the ordering in either example.


Mass of a gas sample What method did John Dalton use to measure the mass of one mole of gases, for example, chlorine and hydrogen, to get their relative masses?

The concept of a mole(c. 1805), and Avogadro's number(1811) occurred after Dalton formulated his law (1801). So he would not have set out to measure a "mole of gas" as the modern formulation sets out. This history of Dalton may be useful to you: Google Book Search Most of his work centered around pressure readings, as his early interests were meterological.


Acetone - Can it already be anhydrous when purchased? If I have 99.9% Reagent ACS, Lab grade Acetone, is it safe to say that it's free of moisture/Anhydrous?

As correctly pointed out by @MaxW, the technical acetone still contains water.  In addition, it is better to quantify "dryness", so by passing a mark you may draw the line of "dry enough" for the application you aim for; especially as Karl Fischer analysis may provide you a reproducible figure of merit about the water content.

Now, if you need dry acetone, stirr the acetone with potassium carbonate or calcium chloride and subsequently distill it off into a dry container.  In the instance of acetone, however, I refrain from molecular sieves because they catalyze aldol reactions.


Ring contraction in a carbocation due to ring strain and back bonding Today our teacher told us that the following carbocation rearrangement occurs due to back bonding. I could not really follow what he meant. Can someone please explain what is actually happening during the following carbocation rearrangement, and why it happens?

The rearrangement has little to do with ring strain and a lot to do with resonance stabilization.

The ring strain in a cyclobutane ring is ~26.3 kcal/mol, while the ring strain in a cyclopropane ring is ~27.5 kcal.mol (reference).  This isn't much of a difference, certainly not enough to drive a reaction one way or the other.

The reason that cyclobutyl carbocations generally rearrange to cyclopropylcarbinyl carbocations is due to resonance stabilization.  Here are some drawings that may help, if you build a model and look at the model that will help even more. The $\ce{C-C}$ bonds in a cyclopropane ring are approximately $\ce{sp^{3.74}}$ hybridized (see here), said differently, they have a lot of p-character in the bond.  In fact, the p-character is so high that a cyclopropane compound will absorb bromine much like an olefin. If that cyclopropane high p-character bond (orbital) can align itself with the p-orbital on the carbocation center, it will stabilize the carbocation, just like a p-orbital on an adjacent double bond (the allyl system) stabilizes a carbocation.

here's another representation

Typically we discuss two conformations of the cyclopropane ring with the carbocation p-orbital, a bisected and a perpendicular conformation.  In the bisected conformation the cyclopropane $\ce{C-C}$ bond lies in the same plane as the carbocation p-orbital.  The carbocation p-orbital and the $\ce{C-C}$ orbital which resembles a p-orbital (because it is high in p-character) overlap!  In the perpendicular conformation, the cyclopropane $\ce{C-H}$ bond is in the same plane as the carbocation p-orbital, the $\ce{C-C}$ bond is not - there is no carbocation stabilization by the cyclopropane ring in this conformation.  Molecules where the bisected conformation can be attained show stabilization due to the stabilizing overlap between the carbocation p-orbital and the p-like $\ce{C-C}$ cyclopropane ring bond.  Sometimes we use resonance structures such as the following to denote this stabilization.


How are reaction rate laws derived? Let's take the reaction $2A + B \rightarrow C$, with the assumption that this is an elementary reaction. If I'm only given this information, how would I derive the following forward rate law from scratch?

$$
\frac{dC}{dt} = k[A]^2 [B]
$$

I tried to find an answer online but most sources jump straight to the above format for a rate law. I also understand that there are many experimental considerations which can affect what is actually observed (e.g. pseudo $0^{th}$rate law behavior) but I just want to know if there's a way to derive these rate laws given just the reaction. Thank you.

I am not a professional or anything close, but I think the answer to your question is simply that $C$ is jointly proportional to $A$ and $B$. The following link explains joint-proportion (https://www.mathwords.com/jk/joint_variation.htm). Just in case, the link gets broken in the future: