text:
Is a highly oxidized element more electropositive than a lesser oxidized counterpart?
My issue
I have a significant problem with the following excerpt from a general chemistry text:
Note that ON stands for oxidation number (state). The excerpt is with regard to polyatomic anions - specifically, chloride ion and its related polyatomic anions.

It now follows that as oxygen-atom content increases by one, ON for the atom of variable ON increases by two because each added O has an ON of -2 and O is more EN than the atom of variable ON - thus, bonding O to this atom makes the atom more electropositive.

If electropositivity is defined as the ability to stabilize positive charge, how does a highly partially positive central chlorine atom have this ability? How  can EP increase?

If electropositivity is more formally defined as the ability for an atom to give up electrons within a covalent framework (note that this is the converse of the definition of electronegativity), how exactly does a highly oxidized chlorine atom, such as the chlorine atom in the perchloric ion, have the ability to give up any more electrons? It already has no valence electrons to give up if we stick with an ionic model to assign oxidation states!

If we look at this intuitively, and agree that more electropositive elements are more likely to bear positive charges - i.e. sodium (Na) is highly electropositive and it is very easy for Na to be stripped of its one electron to become isoelectronic with a noble gas - then why would chlorine become more electropositive as it is further oxidized?
If we consider the successive ionization energies of chlorine, they become more and more massive! So how in the world would chlorine be more electropositive (willing to give up electrons within a covalent framework) as it is further oxidized?
(I do recognize that ionization energy of an electron is not a measure of electronegativity but still it is highly correlated with EN).

Professor's response
My professor, respectfully, didn't offer much a defense. He said something about how positive charge doesn't exist because it's the lack of valence electrons rather than the presence of something material. This, to me, is sidestepping the argument.
He also asked me which would be more likely to bear a positive charge - Cl or O? Obviously Cl; it has electrons way further from the nucleus than oxygen, so of course Cl would more easily be stripped of an electron than O. However, this again, I believe, is a mistake, because we cannot consider electronegativities of elements and expect them to hold within ions and molecules. Carbon, for example, isn't very electronegative at all. But no one doubts the electron-withdrawing powers of a carbonyl carbon, or better yet, the electron-stabilizing powers of a ... CARBOCATION.
Further, last semester, I distinctly remember correcting him with regard to this electronegativity business; he said that a more negative ion would be more electronegative. I had to correct that statement; a more negative ion would be less able to stabilize valence electron density and less able to attract valence electrons if bound within a covalent framework, so the correct term would be instead more electropositive.
Does anyone here agree with me? Sometimes I feel that it's not the argument that matters as much as your tenacity and ability to present the argument. I lacked both tenacity and presentation skills when presenting the above argument.
I came close to telling him that his entire take of electronegativity was fundamentally flawed when I told him that "I don't associate a positive charge with electropositivity and a negative charge with electronegativity" but that was met with nothing to the contrary (which would have been welcome!) - because no, positive charge has nothing to do with electropositivity; in fact, it suggests the opposite of electropositivity!

They must mean the central atom becomes more positively charged, not more electropositive. That atom becomes more electronegative!

They do mean the central atom but they insist it's more EP for some reason!!!

They are clearly not using the term correctly, or at least in the same way I understand it. Don't get up on it.

How do you define EP?

There are two definitions of electropositive:

1:  charged with positive electricity
2:  having a tendency to release electrons

The author is using the first definition, while you are thinking of definition 2, which is the one that usually appears in chemistry books.
Instead of "makes the atom more electropositive" it could say "makes the atom more electrically-positive".
For example according to the calculation data from Paul J. Schupf Computational Chemistry Lab
Starting from a -1 charge for Cl-
Adding one oxygen atom (ClO-) increases the charge on the Cl atom to -0.298.
Adding two oxygen atoms (OClO-) increases the charge on the Cl atom to +0.212.
Adding three oxygen atoms (chlorate) increases the charge on the Cl atom to +0.510
Adding four oxygen atoms (perchlorate) increases the charge on the Cl atom to +0.907.
Another example of a different author using the word "electropositivity" in the same way, but who makes the meaning explicitly clear is Interaction of indole derivatives with biologically important aromatic compounds. Part 22. Importance of simultaneous co-operation of hydrogen-bond pairing and stacking interactions for recognition of guanine base by a peptide: X-ray crystal analysis of 7-methylguanosine-5′-phosphate–tryptophanylglutamic acid complex

MNDO calculations showed that the protonation of the
  guanine N(7) atom increases the electropositivity of the 8-H
  atom [ -0.014 e for guanine and 0.079 e for N(7)-protonated
  guanine],

(bracketed text in original document)
Specifically regarding chlorine, the term "electropositive chlorine" has been used to refer to Cl+, for example in the Canadian Journal of Research
Another example to look at is from the thesis Synthesis of functionalised silanes for use in the asymmetric allylation reaction:

In silicon based C-C bond forming reactions the important factor key to their
  success is the ability of the coordination number of the silicon atom to be
  varied. As the number of bonds to silicon increases the electropositivity and
  consequently Lewis acidity on the atom increases.[reference 1]
Upon expansion (Scheme 3; 11 to 13), and ensuing reduction in the s-character
  orbital composition at the silicon centre, the electron density decreases.
Therefore the electropositivity and Lewis acidity at the silicon centre is
  increased. These tetravalent and pentavalent Lewis acidic silianes have been
  exploited in several Lewis acid-catalysed transformations. Once the shell has
  expanded to incorporate six subsituents it is unlikely for any further extension of
  the valence shell to occur.[reference 2]
Silicon becomes more positively polarised with the addition of each ligand. As
  this occurs there is a shift in electron density. While the electron density is
  increasing at the ligand, it is decreasing at the silicon centre. The magnitude of
  this polarization is also dependent on the electronegativity of these ligands.

For a very old example, showing the original meaning of "electropositive" to be the first definition, see William Henry's  1823 The Elements of Experimental Chemistry, Volume 1 where on page 229 there is a section "Electro-positive Bodies" which defines electropositive bodies as bodies attracted to a negative electrode and whose inherent electrical state is therefore positive.
Also, there is an historically-interesting article "Electropositivity" and "electronegativity" J. Chem. Educ., 1938, 15 (6), p 291 discussing the inconsistent usage of the term electropositive and advocating for saying "electron donor" instead (if meaning #2 is intended) to avoid confusion.
There is also a 1949 textbook General Chemistry which uses both meanings in the same sentence, using the older meaning to define the newer meaning:

The electropositive character (tendency to go to a more electropositive condition) of elements, in general, decreases across the periods

Overall, it is confusing to use the word "electropositive" contrary to the definition that usually appears in recent chemistry books, but not really wrong.

Right, looks like they are confusing 'electropositive' and 'Lewis acid' or 'electrophilic'.

I would say that is fundamentally incorrect to say that the addition of oxygens to an atom makes it more electropositive. Electropositivity is the measure of an atoms ability to donate electrons, and the more electrons that are withdrawn by oxygen atoms, the harder it will be for the central atom to donate more electrons to form higher oxidation states. I think what it may have been trying to explain is that a partial positive charge (denoted $\delta +$) is induced in the central atom with to the addition of oxygen atoms.
So to answer your question of “how can electropositivity increase,” I would say that this is done in the opposite manner to how your text proposes, that is by removing oxygen atoms. By decreasing the partial positive charge, the central atoms affinity for its own electrons decreases, and the easier it becomes to share those atoms in a covalent bond.
You also seem to be mixing different ideas when you talk about positive charge. There is formal charge, and there is the oxidation state. Formal charge is a number assigned to an atom in a molecule that tells you how many more pairs of electrons the atom is in control of than is usual for that molecule, and oxidation state tells you the theoretical charge an atom would have if all its electrons participating in bonding were considered completely donated (as in an ionic bond).
Interestingly enough, atoms that are good at accepting a positive formal charge are markedly bad at achieving high oxidation states. Take the molecule 3-chloropropan-1-ol, for instance. In an acidic solution, which atom is more likely to donate an electron pair to share with a $\ce{H+}$ ion? Oxygen, because it is more basic (as a result of its high electronegativity). Now consider the perchlorate ion. As you said yourself, chlorine’s valence electrons are more shielded that oxygens, and they are more available to be shared in a covalent bond with an electronegative element like oxygen, even to the extent as to give Cl(+VII). Just try to imagine $\ce{O5}$. It just doesn’t work! Oxygen is far too electronegative.
Also I am not sure in what way your teacher meant “positive charge doesn’t exist.” What then is a proton?

I think he meant that in the same way people say cold doesn't exist


text:
Chemical or simply a physical reaction? coca cola + liquid propane
In this video what is that creates the propulsion of the bottle? Is it a chemical reaction or simply a state change?
https://m.youtube.com/watch?v=VxvJcPToXb0
I think is the propane that passes from liquid to gas, nothing more.

Apparently the latter.

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sort: votes


text:
Does the ignition point of flammable substances changes when provided a pure oxygen atmosphere?
I'm wondering if autoignition point decreases for substances when provided higher oxygen concentration.

No, but because of a technical point.
From Wikipedia: The autoignition temperature or kindling point of a substance is the lowest temperature at which it will spontaneously ignite in normal atmosphere without an external source of ignition, such as a flame or spark.
So the definition of the autoignition temperature is based on air not pure oxygen.
For the ignition temperature in general the Wikipedia reference goes on: The temperature at which a chemical will ignite decreases as the pressure or oxygen concentration increases.

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text:
Can you fix rubber on a handle after it has all but turned into a liquid?
I picked up a splitting maul at a yard sale last year. I used it for the rest of the season with no problems. It has been in the wood shed and exposed to some sun, but mostly shade.
I picked it up the other day and it stuck to my hand. When I tried to let go, it slid from my hand, leaving behind black goo and rubber from the handle. Now it's so bad the whole rubber part is gooey and if I try to rub anything on it the rubber comes off. It's permanently soft and sticky and is coming off the handle.
Can I save it or is it done?

Suggestion: this should be moved to Lifehacks.

Take it off, then Plasti-Dip.

I've had the same problem with other types of rubber-coated tool handles. The best solution is to completely remove the rubber coating, which takes a lot of scraping and then scrubbing (outdoors) with paint thinner on paper towels. Invariably, as you have already noticed, everything touched gets coated with hard-to-remove black goop.
Another choice would be to wipe off the loose stuff and then over-wrap the handle with a strong cloth tape, checking carefully that it will not slip when you use the maul.
You might cover the tape with heat-shrink tubing, if you have the right diameter stuff, or with tubing cut from an old bicycle inner-tube (cheap or free, and makes a good grip).


text:
Distinguishing metals from nonmetals?
Please help me. We just did an experiment wherein we used Fe, Al, Zn, Sn, Cu and HCl to see their reaction (e.g. Fe + HCl, Al + HCl...). The result of the experiment:
1.) $\ce{Fe + HCl}$ = formed bubbles (56 min)
2.) $\ce{Al + HCl}$ = formed bubbles (more than 1 hour)
3.) $\ce{Zn + HCl}$ = formed bubbles (42 min)
4.) $\ce{Sn + HCl}$ = No reaction
5.) $\ce{Cu + HCl}$ = No reaction
Now with the result, how can I distinguish metals from non-metals?

Fe, Al, Zn, Sn and Cu are all metals

Well as santiago has pointed out, they are all in fact metals.
The way you can determine if it is a metal or not is that when any metal reacts with acid, hydrogen gas is released. This is shown in this general equation (it is not balanced): $$\ce{M^{m+} + HCl -> MCl + H2}$$
Meanwhile, generally non-metals don't react with acid. This because when a substance reacts with acid, it electrons to the H+ ions created by the acid. Non-metals are generally electron-accepters, rather than electron donors so there is no way for them to donate electrons to the hydrogen ions in the acid.
In your experiment, the bubbles that formed when reacted are indication that hydrogen gas is being formed, therefore the substance is a metal. Therefore, you should be able to tell that Fe, Al and Zn are metals.
Therefore, you might assume that Sn and Cu have to be non-metals. However they are actually metals. The reason why no bubbling is observed is because these metals are unreactive and don't get oxidised by the acid.

Well, $\ce{Sn}$ does, only very slowly.

True, but I don't think you will be able to see it react given the time restrictions in a school experiment.

Yes, that's right. Could leave it overnight, though.

Does that mean that an acid with sufficiently low pH will react to release hydrogen with all metals?  Or are there truly "unreactive" metals?


text:
How can diamond have a fixed refractive index?
I was doing a lesson on solid states. There were given two kinds of solids namely amorphous solids and crystalline solids. Some differences between these two solids were given in my book. One of the difference was that crystalline solids are anisotropic in nature whereas amorphous solids are isotropic in nature.
Following is from my book

Crystalline solids are anisotropic in nature, that is, some of their physical properties like electrical resistance or refractive index show different values when measured along different directions in the same crystals.

Diamond is a crystalline solid. That means it is anisotropic in nature. This means that measuring the refractive index of diamond from different direction will give different values of refractive index. Then why we have fixed value of refractive index of diamond.
Not just about diamonds, many other crystalline solids have fixed refractive index.

Your textbook is wrong. Some crystalline solids are optically anisotropic. Many crystalline solids are indeed optically isotropic in nature. In solids like this the light beam experiences the same electron density regardless of the direction the crystal is oriented. Some examples of that are (obviously) diamond, halite and sylvite (sodium and potassium chlorides), garnets, spinels and more. Note that this is a feature derived from the electronic structure and it has nothing to do with the actual shape of the crystal.
Even in the case of anisotropic crystals, some are so weakly anisotropic they may considered isotropic in practice. Zeolites are a perfect example of those.

I would not call those really isotropic in nature. True, they're isotropic to the light; but once you try to split a crystal... Wait, it's simpler than that: just look at the crystal. Is it spherical? Usually no. Then it's not like all directions are equal.

@IvanNeretin obviously, I was referring to isotropic in the optical sense, a feature determined by the spatial electron density. The actual shape of the crystal is not important. A diamond is isotropic whether it's shaped like a sphere, cube, octahedron or carved into an elephant.

What's obvious for us can be less-than-obvious for someone reading this topic in a year from now, so thank you for putting in the word "optically". As for the features, all of them are determined by the spatial electron density, yet some would reveal certain anisotropy, while the others (including optical) would not.

Crystal can be isotropic when it is symmetrical about all the axes i.e. the cubic system. Halite (sodium chloride) and diamond form cubic crystals and therefore their optical properties are isotropic.

One property that remains anisotropic even in cubic crystals is their elastic deformation. As described here, a cubic crystal has three elastic constants, and these can be varied in ways that create different deformation characteristics along different directions.
This is because elastic deformation is a more complex physical phenomenon than other processes such as heat conduction. Thermal conductivity correlates one vector (heat flux) to another (temperature gradient) and thus is a second-order tensor. The only component of a second-order tensor that obeys the symmetry laws associated with heat conduction and also conforms with cubic symmetry is the fully isotropic component, which corresponds to equal conductivity in all directions. But the elastic deformation constant correlates two quantities that are already second-order tensors: stress and strain. So the deformation constant is a fourth-order tensor, which has more components than a second-order one.
When we work out the mathematics of this fourth-order tensor for the cubic case, we find that three independent components are allowed. Only two of these correspond to equal elastic responses in all directions; these may be expressed in terms of Young's Modulus and Poisson's ratio. The third independent component can impart different deformation characteristics, which show up as different Young's Modulus and Poisson's ratio, to (for instance) deformation in the [100] direction versus the [111] direction.
An even higher level of symmetry, then, is required to get a truly isotropic crystal with respect to elastic deformation ... and now we know it exists thanks to the rise of quasicrystals. An icosahedral quasicrystal, which must obey more symmetry relations in its properties than a mere cubic crystal, really does have equal values of constants such as Young's Modulus and Poisson's ratio in all directions.

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text:
Is the change of the Gibbs free energy always zero at constant pressure?
At constant pressure, which is the usual condition for chemical reactions, heat absorbed/released by the system is the same thing as enthalpy change. According to the Gibbs equation,
$$\mathrm dG = \mathrm dH - T\,\mathrm dS = \mathrm dH - đq$$But, $\mathrm dH=\mathrm đq$ at constant a pressure. So is the Gibbs energy change $\mathrm{d}G$ simply zero for any reaction?

You wrote:
$$\mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S = \mathrm{d}H - đq$$
There are several issues with this. I don't wish to co-opt other answers and comments, but allow me to mention these here, since my answer stands as the accepted (which I have no power to change):

You haven't assumed constant temperature, which means that there is an additional term $-S\,\mathrm{d}T$ which you did not take into account;
You haven't assumed constant composition, which likewise leads to another term $\sum_i \mu_i\,\mathrm{d}n_i$.

In fact, these considerations (particularly the second) are more important than my original point, which is retained below. See Kexanone's answer for discussion of these points. In the following discussion, one should assume constant temperature and composition.

In the second equality, where you equate $đq$ with $T\,\mathrm{d}S$, this assumes that the process is reversible. Without going into too much detail (it can be found easily elsewhere), entropy is defined by the following equation:
$$\mathrm{d}S = \frac{đq_\mathrm{rev}}{T}$$
where the subscript $\mathrm{rev}$ indicates a reversible process. It can be shown that for an irreversible process,
$$\frac{đq_\mathrm{irrev}}{T} < \frac{đq_\mathrm{rev}}{T} = \mathrm{d}S$$
This result is the second law of thermodynamics. Therefore, for an irreversible process, $T\mathrm{d}S > đq$.
You are right in saying that at constant pressure, $\mathrm{d}H = đq$. Putting it all together, we have:
$$\mathrm{d}G = \mathrm{d}H - T\mathrm{d}S = đq - T\mathrm{d}S < 0 \qquad \text{(irrev. process; closed system; }p, T\text{ const.)}$$
Therefore, the direct answer to the question is: $\mathrm{d}G$ is, in general, not equal to $0$ because $T\mathrm{d}S \neq đq$.
In fact, you may notice that the condition $\mathrm{d}G < 0$ is the condition for spontaneity under constant $p$ and $T$. This is no coincidence. The concepts of spontaneity (in the context of $G$) and irreversibility (in the context of $S$) are extremely similar, in that they both refer to a tendency for a process to proceed.

For a reversible process, $T\mathrm{d}S = đq$. Following the logic that you used in your question, this means that $\mathrm{d}G = 0$, which is correct. If $\mathrm{d}G = 0$, the forward process cannot be spontaneous (that would require $\mathrm{d}G < 0$) and the reverse process cannot be spontaneous (that would require $\mathrm{d}G > 0$).
This can only mean one thing: the system is in equilibrium. Indeed, the condition for equilibrium at constant $p$ and $T$ is $\mathrm{d}G = 0$. And at this point, it should not come as a surprise that a reversible process is, by definition, one that is at equilibrium throughout.

@orthocresol What if we want to include the number of particles? I mean $$dH=TdS + Vdp + \sum_i \mu_i dn_i $$ so now $dq\neq dH$.

For isobaric heat transfer: $$dS = \frac{dH}{T} \geq \frac{\delta q}{T}$$
$dS = \frac{dH}{T}$ is correct even for an irreversible processes.

The accepted answer is incorrect. It has nothing to do with reversibility. The underlining issue is that your $dG$ is incomplete:
$$dG = dH - TdS - SdT$$
You are missing $-SdT$, so at constant pressure you don't end up with zero.
As "ado sar"  has commented in the accepted answer:
$$dH = TdS + Vdp + \sum_i \mu_i \cdot dn_i$$
Hence, you end up with:
$$dG = Vdp - SdT + \sum_i \mu_i \cdot dn_i$$
The chemical potentials $\mu_i$ describe the change of Gibbs energy when particle $i$ get removed/added, e.g. by chemical reactions.
The nice thing about Gibbs energy is that at constant pressure and temperature, you only have to worry about the chemical potentials, i.e. the chemical reaction:
$$dG = \sum_i \mu_i \cdot dn_i$$

The answer to the original post does not require invoking systems of variable composition. The OP had not understood how dS is related to dq. As orthocresol explains they can be related when a process is carried out reversibly.

The OP was explicitly talking about chemical reactions and did not realize that they are described by the chemical potentials and he also forgot the $-SdT$ term.
I agree, the nice thing about orthocresol's answer is that he explains why OP can't just set $TdS$ equal to $\delta q$ in general, but he fails to realize that $dH=TdS$ even for an irreversible isobaric heat transfer.

Good point: $dH=dU+pdV  =TdS$ when: constant p; only pV work; constant composition

By the way it is important to note that "dH=TdS" for a microscopic  (infinitely small) isobaric heat transfer. As soon as the transfer is finite, there are no guarantees about reversibility since isobaric might refer to the end points being at the same pressure, but the path itself may not be isobaric.

Hmmmm, indeed, there are several layers to the question: I really don't remember writing this answer, but it seems that I only looked at the most superficial one, i.e. equating $\mathrm{d}q = T,\mathrm{d}S$ in general.

OK, I added a note to recommend yours at the top of my answer. I don't mean to "steal" the points you raise, but my answer is accepted which means that (for better or for worse) people will see it first.

I don't care about points. I want people that see this question to have a more complete picture, which referring to my part in your answer exactly does. Thx for the edit.

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text:
Can benzene form catenanes?
With there being many particles reacting together forming different shapes of molecules, is there even a slight possibility of some cyclo or benzene miraculously forming into a chain?

You mean ring opening or chain of benzene rings? Biphenyl is made up of two benzene rings. There are natural macromolecules called polyphenols.

nope, I mean literally a chain. So Im not looking for a molecule formed by a bond, but two normal benzene molecules that happened to form going through each others doughnut holes.

you mean concatemers of benzene.. The ring is not huge enough to allow that. You generally need long polymeric loops to achieve that.

No way. There "hole" in benzene molecule is so small, we may safely assume there is no hole at all. On the other hand, chain-like compounds do exist and are known as catenanes.

Thanks! Catenanes is exactly the types of structure i was talking about

oh yes it probably is.


text:
Why do elements change their colour in reactions?
We are almost common with the reaction:
$\mathrm {CuSO_4+Fe = FeSO_4+Cu }$. In this reaction initially Copper sulphate is bluish in colour, which gradually changes to pale green. I want to know, why these elements have so and so specific colour, OR what determines the colour of these elements?  Further, why there is such specific change of colour as in copper sulphate observed?

Elements don't have inherent colors at all; substances do. Furthermore, while there are situations when some compounds (and even copper sulphate) change their colors, this is most definitely not one of them. Here copper sulphate is blue and never changes to pale green. It's just the reaction... wait, didn't you write the equation yourself? See, in the end there is no more copper sulphate. It is gone, and so is its color. What you see now is the color of $\ce{FeSO_4}$, or more precisely, of $\ce{Fe}^{2+} (aq)$.
As to why compounds have certain colors... well, this is a really huge body of knowledge. Long story short, color is caused by absorption of light with some specific wavelength. What wavelength is it, is determined by possible electron transitions within the substance in question. The transition may be between some molecular orbitals, if the compound is molecular. Or it may be between the orbitals of some metal ion, as is the case in your example. In many cases you may be able to predict the color of a compound you have never seen, using simple techniques.
Come to think of it, this question must have been asked before, and surely there is already a nice answer with lots of pics and stuff.

Elements have colours as well. Copper is pinkish brown, for example, and gold is golden.

These are elementary substances. Put any element in a different substance; the element is still here, but where is its color?

I didn't wanted to know why copper sulphate has changed its color, I wanted to know, why it is only of blue, not yellow or red, and why ferrous sulphate is only pale green, not blue or golden

Well, it's mostly ions, so read about the crystal field theory (second link in my post, or elsewhere).


text:
Negative oxidation states of Si
According to List of oxidation states of the elements, silicon has a possible oxidation state of $-4$.

Now, I've been looking everywhere for a compound that contains Si(-IV), but I cannot find any definitive references. Perhaps one of the synthetic carbonyls of silicon has this OS? I've seen references to $\ce{[Mo(CO)_4]^{-4}}$ and $\ce{[W(CO)_4]^{-4}}$ in my research, but silicon never shows up.
I am also uncertain as to how these carbonyls come together, and whether or not they actually complex in the same way for metalloids as for the transition metals.

Only some of d-block elements form carbonyl complexes, and well how you even found these complexes and not simple salt-like compounds of silicon?

I am not sure if Silane ($\ce{SiH_4}$) can really be considered.
But there are silicides which silicon forms with strongly electropositive metals. In these compounds, silicon has a negative oxidation state.
For magnesium silicide - $\ce{Mg_2Si}$, the oxidation state of silicon would be -4.

For silane, there seems to be some ambivalence as to whether silicon takes on a +4 or a -4 OS. It's hard to tell since silicon is a metalloid, and hydrogen changes its oxidation state between metal and non-metal hydrides.

Silane is definitely no good. See, it hydrolyzes to $\ce{SiO_2}$ and $\ce H_2$, hence it is Si that must have been with plus.

Silicides are good, you are right.

Silane can not be considered as one of these. This is mainly because the electronegativities of silicon and hydrogen are 1.8 and 2.1 respectively.
$\ce{Si^{-4}}$ compounds include compounds of the form $\ce{SiM2}$ where $\ce{M}$ is a group 2 metal as $-4 + 2\cdot 2 = 0$. Examples of these include magnesium silicide $(\ce{Mg2Si})$, dicalcium silicide $(\ce{Ca2Si})$ and distrontium silicide $(\ce{Sr2Si})$!


text:
How is the concentration of protons calculated in a sodium formate/ formic acid mixture?

$\pu{0.1 mol}$ of sodium formate was added to one litre of $\pu{0.2 M}$ solution of formic acid ($K_\mathrm{a} = 1.8\times10^{-4}$). How much will the $\ce{H+}$ concentration of the acid diminish assuming the salt to be completely dissociated?

I found the concentration of $\ce{H+}$ in formic acid.
For calculating $\ce{[H+]}$ in $\ce{HCOONa}$, I am planning to use the equation:
$$\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log\frac{[\text{salt}]}{[\text{acid}]}$$
But I don't know the $\mathrm{p}K_\mathrm{a}$ of $\ce{HCOONa}$.
How do I proceed, are there any alternative approaches?

HCOONa does not have the pKa, nor does it produce [H+].

Oh okay , but then how do i find the final [H+] concentration . There will be an increase in [HCOO-] right so the acid dissociation of HCOOH--[HCOO-] + [H+] will favour backward reaction hence [H+] should decrease

$$\ce{HCOONa + HCOOH<=> HCOONa + HCOOH}$$
The constant of this equilibrium equals $1$. The composition of the mixture will not change.
\begin{align}
\mathrm{pH} &= \mathrm{p}K_\mathrm{a} +
  \log\frac{\ce{[HCOO-]}}{\ce{[HCOOH]}}\\
  \mathrm{pH} &= 3.74 + \log\frac{0.1}{0.2} = 3.44
\end{align}
So, the initial concentration of protons is $[\ce{H^+}]_0 =  \pu{10^{-3.74} M}$.
The final concentration of protons is $[\ce{H+}]_\mathrm{f} =  \pu{10^{-3.44} M}$.
The decrease in the concentration of $\ce{H^+}$ is $\pu{10^{-3.44} M} - \pu{10^{-3.74} M} = \pu{10^{+0.30} M}$.

A rough calculation for the pH of 0.2-ᴍ formic acid gives:
$$ \mathrm{pH} = 1/2 (\mathrm{p}K_\mathrm{a} - \log(0.2)) = 1/2 (3.74 + 0.70) = 2.22 $$
You can check by calculating the equilibrium constant from the concentrations of all the species, and the estimate is pretty good.
The pH of the buffer is 3.44 using the Henderson Hasselbalch expression (see other answer). The change in pH is 1.22, corresponding to a change of a factor of about 17 in the hydronium ion concentration.


text:
Does Copper reacts with Sodium Sulfate solution?
Okay I am trying to make a capacitor in which current collector is copper foil and electrolyte is sodium sulfate. Would this work? Do these two react with each other or not?

If you can provide more information, that will help us help you get an answer to your question. You could expand on the experimental setup, for instance, and what capacitance you expect, and so on. What are the other materials in the cell?

Why, these two are perfectly indifferent to each other. Your capacitor should be fine, as long as you don't apply any voltage to it.

Ofcourse one must charge and discharge capacitor. Its like psuedocapacitor nano  material is deposited on the copper plate which acts as current collector.

To expand on Ivan's answer, this would not be a good idea. I assume you mean to use a solution of sodium sulphate in water, which has neutral pH. While this is not itself corrosive to copper (copper being far less reactive than sodium) it will however corrode badly at the positive electrode due to anodic dissolution. Copper can however be used safely as a negative electrode current collector.

If you're trying to make an electrolytic capacitor, you've picked inappropriate materials.
An electrolytic capacitor works because a material such as aluminum or tantalum develops a tightly-adherent insulating oxide layer when immersed in certain solutions and "formed". Copper does not have an adherent insulating oxide; instead, cuprous oxide is a semiconductor and $\ce{Cu2O}$ can be used to make a rectifier.
You can make your own aluminum electrolytic capacitor with common household chemicals.

Well its a kind of super capacitor using Psuedo and double layer capacitance. Its oxide and graphene based actually. So would aluminim foil work if I place my active materials onto it?

There are directi0ns for making your own ultracapacitor (AKA supercapacitor) at http://www.instructables.com/id/Make-a-Graphene-Supercapacitor/ and http://arstechnica.com/science/2012/03/a-supercapacitor-in-your-optical-drive-dvd-rw-used-to-burn-graphene-capacitors/. Not sure what you mean "using Psuedo", though.

use copper sulphate.
if you charge it,  it forms microfacets, via the copper moving from the + to collect on the -,  mix it with glue, wait for it to dry to form the dielectric separator, but dont break the chains you have to be gentle with it.


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Do enolates get protonated at the carbon or oxygen atom?
An enolate has two resonance structures. When it gets protonated, where is the $\ce{H+}$ more likely to go? On the oxygen to produce an enol, which tautomerizes to the keto, or on directly onto the carbon (no enol intermediate). I know that both lead to the same product, but when drawing the mechanism this would be good to know.

There is no difference, really.

The carbanion is stabilized by the carbonyl oxygen, but I am not quiet sure if this is really leads to the left form beeing easier protonated.

Like all ions in solution, the enolate does not exist on its own in free form. It will be solvated with maximum efficiency. Depending on the solvent itself, the counter ion might be further or closer away, but charge balance needs to be maintained in the bigger picture.
Contrary to popular belief, the enolate ion is nothing that truly exists in non-gaseous phase. Therefore the question what gets protonated first turns into the question: Does the oxgen or the carbon have the stronger interaction with the solvent? This is quite hard to judge and very dependent on the whole system. there is certainly no general rule for it and as previously stated by bon, it does not really matter, because keto-enol tautomerism is fast and the ratio is entirely determined by the relative stabilities to the involved species.
There is another issue when talking about protonation. The proton itself is not free, either. In an aqueous solution the smallest model would be the hydronium ion. But this is far from being complete either. A whole branch of computational and theoretical chemists work on describing water clusters with models. It is extensive. One of the reasons for it being hard to comprehend is the high flexibility and mobility of protons even when bound to other species. The high conductivity of an acidic or basic solution is mainly governed by the Grotthuss mechanism. This kind of mechanism can in first approximation also be applied to keto-enol tautomerism.
But before we dig deeper into any mechanism, I would like to look at the properties bon outlined for the enolate. The following results were obtained at the DF-BP86/def2-SVP level of theory for the deprotonated butan-2-one resonance hybrid.
Let's look at the obvious first: resonance. With the help of the natural bond orbital theory (NBO) I performed the analysis. The first and most obvious thing is that it finds a so called hyper-bond between the oxygen and the two carbons with an occupancy of 3.95 electrons. As we will see in the refined analysis, this is in favour of the carbon-carbon bond (59.5/40.5 considering only these three atoms).
A more refined analysis, I chose a cut-off of 5%, include secondary resonance structures, seven in total, gives us the following weights for the major contributors:

From these structures we can support the highlighted arguments in the statement by bon:

Attack on an enolate by an electrophile (in this case just a proton) is governed by the balance between electrostatic interactions, which are strongest at the oxygen atom due to its greater negative charge, and orbital interactions, which tend to be stronger at the carbon atom because the HOMO of the enolate has a greatest contribution at this atom [carbon].

The predominant contributor is is the enolate, hence in the overall description also its negative charge should be predominant. The π-system is certainly part of the HOMO in any case. Since it is mainly localised over the carbons, the highest contribution should be there.
We can further support that by looking at the natural charges of this model. More prominent is it looking at the charges obtained in terms of the quantum theory of atoms (QTAIM) in parenthesis.

The second part can be cleared up by looking at the HOMO; on this level of theory, the major contribution is C3-p of 56% and C2-p of 32%. The contribution of O1-p is below the threshold of printing, hence it is <10%. The major contribution of O1-p is found in HOMO -2 with 44% and a contribution of C2-p of 35%. It certainly looks a bit different in the picture. The contour value is set to 0.04.

All these calculations are only evaluations of the ion in the gas phase, so they can serve as an example to understand the true nature a bit more. It is in no way intended to be complete, it shall only provide a more concise educated guess of the general nature.

Now that we have a lot of information on the enolate itself, it's time to look towards a mechanism of a solvent-mediated keto-enol tautomerism. I have chosen the most simplest model possible, with one water (hydronium) explicitly included. The result is a six-membered cyclic transition state, where a proton is transferred from the oxygen of the enol to the oxygen of water, and one proton is transferred from water to the carbon of the double bond.
In theory there might well be many more water molecules involved. In principle any protic solvent is capable of doing this. The principle reaction path is the same for a base mediated mechanism, for example with ammonia.

If you look closely enough, I have borrowed heavily from the Grotthuss mechanism. A somewhat simultaneous transfer of protons. It needs to be acknowledged, that this is only an approximation of a stationary system. The animation only shows the transitional vibration mode (imaginary mode). It is not the calculated structural change or reaction path.
Unfortunately the calculations even on this low level of theory take quite some time. Currently I have no barrier height to report as an estimate, how fast this reaction would be approximately. I'll update this as soon as I can.
There is of course the possibility of a non-mediated transition state. This one, however, proceeds through a four-membered cyclic transition state. This one is highly strained, if you like the concept of ring strain, and seems unlikely compared to the above in any other medium than the gas phase.
If you are repulsed by the concept of ring strain, then you might want to look at it from a different angle. The deformation energy is high, since the carbon-carbon-oxygen is forced into a near 90° angle, hence increasing the p-content of the bonding orbitals. Higher p-orbital content means higher energy of the molecule. Therefore the activation energy will also be higher.

For a deeper insight into this, I recommend “Ab Initio Molecular Dynamics Study of the Keto–Enol Tautomerism of Acetone in Solution” (Clotilde S. Cucinotta, Alice Ruini, Alessandra Catellani, András Stirling, Chem. Phys. Chem. 2006, 7 (6), 1229-1234), which is available at researchgate.net.

I disagree on whether it matter: it may not matter in most cases, but when you're worried about kinetic/thermodynamic product formation of a pro-chiral enolate that won't readily tautomerize after C-protonation (which is a current issue for me), I think it matter which is protonated first.

Of course, my system isn't a ketone enolate, it's an ester enolate.

@SendersReagent I think I know what you are talking about. I hope I can come back to your question soon. I wanted to cover the basics first before diving deeper into the matter. In this post I'm not covering any hindrance in any way. Stereo discrimination is a completely different topic though, and it requires a close inspection of the complete system.

I had asked a question about this before. Two actually, both of which were related to this one. The molecules aren't the ones I'm dealing with, and they're not perfect, either. This analysis is good, though; I just think there are times where the protonation step needs to be considered.

Attack on an enolate by an electrophile (in this case just a proton) is governed by the balance between electrostatic interactions, which are strongest at the oxygen atom due to its greater negative charge, and orbital interactions, which tend to be stronger at the carbon atom because the HOMO of the enolate has a greatest contribution at this atom.
Since a hydrogen ion has a high charge density, I would expect the reaction to be governed primarily by electrostatic interactions and so the protonation will preferentially take place at the oxygen. However, keto-enol tautomerisation is fast, and the product is the same with either mechanism so it really makes no difference.

Tautomerisation in basic solution proceeds via the enolate, anyway.

I was also expecting hydrogen protonation first.  So in theory (if you could somehow prevent tautomerization) you would  get the enol as the major, right?

That would depend on the conditions. Since protonation is easily reversible and the keto form is more stable in most cases, it would gradually equilibrate in favour of the keto form.

First, let's look at your comment about tautomerisation.
Tautomerisation requires a proton transfer! Therefore (in basic solution) it proceeds as follows (sorry the double arrows are meant to represent equilibrium not resonance):
$$\ce{Ketone + Base- <=> Enolate + BaseH <=> Enol + Base-}$$
The Enolate is the intermediate in the tautomerisation equilibrium.
I assume you're getting the enolate from some other reaction / reagents in a synthetic pathway to the ketone?
As for which the Enolate is most likely to become, that will depend on the thermodynamics and kinetics of the two equilibria.
For example, if the solution contains 90% ketone and 10% enol, then it would be 9 times more likely for the molecule to "end up" as the ketone form.
But protonation/deprotonation at carbon is much slower than at oxygen, so you might expect most molecules to interconvert between enolate and enol several times before settling on the ketone form.
The bottom line is, all three species are in a pretty rapid equilibrium, so it doesn't matter.
If you're drawing the mechanism I would show $\ce{enolate -> ketone}$. I would ignore the enol form completely, unless you want to make some kind of point, in which case I would show it in equilibrium with the enolate, in a branch on the main mechanism pathway.

yes i was actually getting the enolate in a synthetic pathway, like eg. adol condensation or michael addition. In both I get the enolate and after acidic workup the final ketone. I wasnt sure where to place the hydrogen in the mechanism. The question is, is the enol formed first (even if it just exists for a short time), or does it go directly to the ketone. Meanwhile i think, as most of you stated, that the enol is actually formed initially.

In basic medium (or any medium for what it's worth) the enolate does probably not exist in a free form. There are so many ways how a proton transfer can happen, that do not involve protonation/deprotonation. And ignoring the enol form is most certainly wrong.

I don't think that question is very meaningful since resonance structures do not actually exist and both lead to the same product. As the enolate with the carbon-carbon double bond is more stable, it is probably the most accurate representation of the resonance hybrid so I personally always protonate enolates at oxygen when drawing mechanisms.

How does the fact that resonance structures don't exist affect the question? It is still possible to protonate at either site.

(-1) I second @bon. Additionally I would like to add, that in fewer cases the enol form is actually more stable than the keto form, and just because the enolate with the double bond appears to be more stable, it is impossible to assume that the proton will bind to the oxygen.


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Final concentration of stock solution upon dilution

If we have a stock solution that is $\pu{1 mg/mL}$, and we have $\pu{0.1 mL}$ of that stock to which we add $\pu{0.9 mL}$ of water, what is the final concentration?

For this I assumed I had to use the concentration formula:
$$\mathrm{M_{1}V_{1} = M_{2}V_{2}}$$
I thought the initial concentration was $\pu{1 mg/mL}$, since that is my stock. I thought the initial volume was $\pu{0.1 mL}$, since that is what they needed from the initial stock. I assume the final volume is $\pu{1 mL}$, since that is the total volumes of the components.
$$\mathrm{(1\;mg/mL)\cdot(0.1\;mL) = (M2)\cdot(1\;mL)}$$
$$\mathrm{M_2 = \frac{0.1\;mg}{1\;mL}}$$
$$\mathrm{M_2 = 0.1\;mg/mL}$$
I feel I am right due to the units evaluating properly out, since the original concentration is giving in $\mathrm{mg/mL}$ , the final should be also in $\mathrm{mg/mL}$, right?

You are correct.  You might find it more intuitive to think of it this way:
In the end you will have $\pu{1 mL}$ of solution. In that end solution you have $\pu{0.1 mL} \times \frac{\pu{1mg}}{\pu{mL}}$ of stock, or (cancelling the $\pu{mL}$s) $\pu{0.1mg}$ of the chemical in question.  Therefore, final concentration is $\frac{\pu{0.1mg}}{\pu{1mL}} = 0.1 \frac{\pu{mg}}{\pu{mL}}$.


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Could the net equations for burning hexene be used as a means of energy source?

$6\ce{CO_2} + 6\ce{H_2O}\xrightarrow{some\;catalyst}9\ce O_2 + \ce{C_6H_{12}}$
$\ce{C_6H_{12}} + 6\ce O_2\to 6\ce{CO_2} + 6\ce H_2$
$6\ce H_2 + 3\ce O_2\to 6\ce{H_2O}$

At one point, I had made a balanced set of equations while starting college. Those equations are listed above. I had wanted to get into chemistry to see if there was any possible way those equations could be used as a means for a new kind of energy source; one that could catalyze its fuel, hexene, that it could later use, but then also take the byproducts of the combustion of hexene and produce the same chemicals it needs to produce hexene, and repeat the process. I don't know if it's possible, and I don't know what, if any catalyst would be needed to make it possible. However, as time went on, my major changed to software development and I gave up on becoming a chemist. So this is both a question and a challenge: Can anyone figure out if the system described above is possible, and what would it require to make it available as an energy source?

To make this scheme available as an energy source, it should have an energy source. Now it has none. A catalyst, no matter how sophisticated, is not an energy source.

The energy source is in the second formula, in the same method as a combustion engine. A spark ignites the hexene fuel.

That, or I'm confused. The catalyst is definitely something relating to photosynthesis, but that's as far as I got with it.

Well, then 2 and 3 put together will bring you the exact amount of energy needed to pull off 1, minus the inevitable losses.

No, I meant hexene, since hexene is a combustible fuel. The formulas can't work with glucose.

related http://chemistry.stackexchange.com/questions/5445/photosynthesis-science-fair

possible duplicate of What is the mechanism of photosynthesis? or maybe of http://chemistry.stackexchange.com/questions/696/energy-efficiency-aside-what-are-the-chemical-constraints-on-co%e2%82%82-capture-and-me ?

@Mithoron not a duplicate: this isn't about the mechanism of photosynthesis or the capture of $\ce{CO2}$. However, those posts are definitely helpful!

There is no such catalyst that makes 1. work without investing extra energy in the process, and this energy is as much (but generally much more ) than it is gained in 2,3., as already said by Ivan Neretin. So it is a no go.


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Which base is stronger? Ammonia of phosphite?
$\ce{NH4+}$: $K_\mathrm{a} = 5.6 \times 10^{-10}$;
$\ce{HPO3^2-}$: $K_\mathrm{a} = 4.5 \times 10^{-13}$
My understanding is that $K_\mathrm{b}(\ce{A-}) = K_\mathrm{w} / K_\mathrm{a}(\ce{HA})$. Thus:
$\ce{NH3}$: $K_\mathrm{b} = 1.8 \times 10^{-5}$; $\ce{PO3^2-}$: $K_\mathrm{b} = 2.2 \times 10^{-2}$
Since the $K_\mathrm{b}$ for phosphate is greater than the $K_\mathrm{b}$ for ammonia, phosphite is a stronger base. However, my professor's solution lists ammonia as the stronger base. Is his solution incorrect?

That's phosphite, not phosphate.

Also, in my book the pKa of ammonium is about 9. Not that it would change much, though.

A bit late but welcome to chemistry.stackexchange.com. Feel free to take a [tour] of the site. I improved the formatting of your post by using MathJax; for more information on how to do so yourself, check out the [help], this meta post and this one. I also corrected the $\mathrm{p}K_\mathrm{a}$ of ammonia to what it says in my and Ivan’s books and Wikipedia. Ivan was right, it doesn’t change much.

Your professor could be wrong, or rather could mean it in some subtly different context that would change everything. The way you put it here, you are right. Normally the constants would tell you the truth. I see no reason to believe this case is an exception.
Besides constants, there is actually a simple way of comparing two acids: let them fight each other to the death. Let us consider $\ce{NH_4^+}$ vs. $\ce{PO_4^{3-}}$ (I know $\ce{PO_4^{3-}}$ is not what you asked; this is just an illustration). Who would end up having that proton? It turns out that ammonium phosphate is unstable, which implies that the proton would go from $\ce{NH_4^+}$ to $\ce{PO_4^{3-}}$, meaning that $\ce{NH_4^+}$ is stronger acid than $\ce{HPO_4^{2-}}$ (in accordance with what the constants would predict), and hence ammonia is a weaker base than phosphate. Now, that was phosphate, the salt of phosphoric acid, while you ask about phosphite, the salt of phosphorous acid, which by all accounts should be weaker. Hence its anion must be an even stronger base.
Simple as it may seem, there is one more thing to mention. Phosphorous acid is actually diprotic, that is, capable of ionizing only two of its hydrogens. The third hydrogen is connected directly to the phosphorus atom. Organic esters $\ce{(RO)_3P}$ are known, but that's another story. The very existence of ion $\ce{PO}_3^{3-}$ seems somewhat dubious. By any means, even if you manage to get one in water solution, it will be instantly hydrolized. This puts it in the ranks of bases too strong to exist in water, which is of course way stronger than our lowly ammonia.


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Why does cooking oil lower the vapor pressure of water?
I added different amounts of oil to different containers with 50mL of water and weighted the containers. Then I heated the containers for 15 minutes in a water bath and weighted again. I noticed that the mass of each container was less as I added more oil to the mixture.
So how does oil increases the boiling point/decreases the vapor pressure of water?

I have two questions here. 1) If you added more oil, but the container weighed less after the heating, it seems like more water evaporated from the containers with more oil. Thus it seems like the oil has increased the vapor pressure. 2) Was there any particular reason you were doing this?


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Find ΔHr for HCl + Mg Reaction
I am having trouble calculating the value of $\Delta H_\mathrm r$ for the reaction: $$\ce{Mg + HCl}$$ The total heat capacity ($C_\text{cal} + C_\mathrm r$) is $6.88\ \mathrm{kJ/{^\circ C}}$, and the $\Delta T$ is $10.5\ \mathrm{^\circ C}$. When I multiply the two I get $72.24$ which is $Q_\mathrm r$ (Total heat produced). Finally when I divide $72.24$ and $0.005$ moles of $\ce{Mg}$ I get $14\,448\ \mathrm{kJ}$. This seems to me to be inaccurate. Can anyone help me solve $\Delta H_\mathrm r$


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Thermodynamics, pressure volume work
In thermodynamics work done by the system is negative but in case of expansion of gas it is positive because force and displacement are in same direction, why?

Because force and displacement are in same direction, just like you said. Really, what a strange kind of science the thermodynamics would be if the work were always negative?

But it should be negative as per convention adopted

You are correct. The system does positive work. Textbooks introduce the negative to show that the system loses energy as it does expansion work. As the gas expands, it does positive work but it loses this energy in the process. Work, like heat, is basically a transfer of energy.

I think it is due to pressure which is negative in the pressure volume work equation actually here two types of pressure act at a time one is external other is internal so net effect of external is less then internal which is taken negative.

Your very long answer including just one sentence is not easy to read. Could you split it up into shorter sentences?


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Work done by the System
In thermodynamics work is defined as $~-p_{ex}\Delta V~$ for an ideal gas. If the total work is a negative value, that is in case of expansion of gas, we say that the work is done by the system and is negative.
I don't understand that why we use the pressure exerted on the gas in expression of work that is the external pressure and say that the work is done by the system ? If work is done by the system we should use internal pressure and if we use that in case of expansion the work done by the system will be positive which is contradicting. Please provide an explanation

There is also an issue here with sign convention. Some developments use the symbol w to represent the work done by the system on the surroundings, and others use the symbol w to represent the work done by the surroundings on the system. In the first case, $\Delta U= q - w$, while in the second case, $\Delta U= q + w$. In working with the first law, one must carefully specify which sign convention is being used.

Well, think of a gas that is expanding against a piston shown below:
We want to find the total work which is given by: $$w = F\Delta d$$ So we need to know what is the force that is applied by the gas onto the piston. To do this, we must realise that, the force that is stopping the gas from expanding freely into the atmosphere is the external atmospheric pressure. It is this pressure which is pushing downwards onto the gas, preventing it from expanding. Therefore, the force that is required for the gas to expand must be equal to the total pressure pushing down it and has nothing to do with its internal pressure.
Consider this example. I want to push a box across my room which is covered in carpet. To move the box, obviously I need to apply a force to it. The force that I need to apply to must be equal or greater than the resisting force which is the friction between the box and carpet. This force that is required has nothing to do with the speed of the box.
So now that we know the force, we get that $$w = -p_{ex}A\Delta d$$ but the area times the displacement is actually equal to the volume, therefore we get: $$w =-p_{ex}\Delta V$$
So, as you can see, we have derived the equation for work done by a gas. The main thing that you need to recognise is that the internal pressure is irrelevant when considering the force required for the gas to expand. All you need to consider is the force that is preventing it from expanding which is the external pressure.

If force by the gas is equal to the external force how can the gas expand

Why negative sign is added before external pressure? The direction of force by gas is in same direction as the the displacement of the piston

Yes that is true, the direction of the force and displacement are in the same direction. But the actual magnitude of the force is negative. This is because the force applied by the gas is negative of the value of the external pressure (hence a negative number)

Another way of looking at it is, work can be thought as the change in kinetic energy. So, when a gas (the system expands, it obviously needs to exert a force to do so. Therefore, it is using up its own energy so that it can expand against the external pressure and hence it loses kinetic energy. Therefore, its work will be negative.

Nanoputian says that "But the actual magnitude of the force is negative" but the magnitude of a vector can't be negative

So how come every thermodynamics textbook I have says that the work done by a gas on its surroundings is $+p_{ext}\Delta V$ and the work done by the surroundings on the gas is $-p_{ext}\Delta V$?  In your figure, the change in volume is negative.  That's why you end up with the incorrect sign in your equation.

How would you explain the same formula for Compression of gas?

This is an argument I have with the sciences. An expanding system does work onto the surrounding atmosphere. Consider the expanding system as a subsystem of our atmosphere, hence as it expands it does work onto the surrounding atmosphere i.e. it is as if the atmosphere has expanded, hence the work onto the atmosphere is [(Psys)(dVsys)]. Since the system's final pressure is 1 atm and the volume change is such that (dVsys)=(dVatm), we can write that the work done onto the atmosphere is [(Patm)(dV)atm].
The confusion originates from not writing the enthalpy equation correctly because simply writing  H=E+PV does not tell us what is happening where. So clarity is given by writing
dH=(dE)sys + (PdV)atm   and since the work is done exterior to the system then we can rewrite it as
    dH=(dE)sys-(PdV)sys
 But the above can be confusing. Imagine we are heating water bring it to a boil, and dQ is the energy in (due to heating. Now we would write
dQ= (dE)sys + (PdV)atm  = (dE)sys + (W)atm
The reason is that the heating must do two things
1) change the system's internal energy
2) do work onto the surrounding atmosphere
The above equation is really latent heat of vaporization with the systems energy change being it changes in the bonding energy i.e. (dE)sys = (dU)sys
Now I argue (this is published in peer review [1,2,3] but remains controversial) that the work done onto the surrounding atmosphere: (W)atm = (PdV)atm is lost work because it is irreversible i.e. in order for the atmosphere to work on a system, then the atmosphere must be at a higher pressure than the system. This is controversial because it can become a challenge to the second law, namely lost work into the atmosphere means that the second law is no longer necessary to explain why we cannot have perpetual motion, or why energy is lost by expanding systems like a steam engine or why the Carnot cycle is ideal rather than realistic.
[1] Mayhew, K.W. "Second Law and Lost Work" Physics Essays 28,1 (2015) pg 152-56
[2] Mayhew, K.W. "Resolving Problematic Thermodynamics" Hadronic Journal vol 41 no 3 (2018) pg259

Hi Kent, I've removed the link to your website because it seemed like the purpose of the post was advertising for the blog rather than answering the question.


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While filling liquid oxygen into a tank, why do they do open one end?
While filling liquid oxygen into a tank, why do they do open one end?
Then what is use of filling?
I don't think people who are filling this truck are beginners.

Don't you agree that in order for something to get in the tank, something should get out (to avoid increasing pressure to up levels where it wouldn't be possible to fill)?

I really hope that's nitrogen and not oxygen.

Moreover, you need to allow the vapours to get out.

Releasing oxygen gas right over the exhaust pipe seems like a wonderful idea.

The end is opened to allow oxygen gas to escape. The flow of liquid oxygen into the tanker will not be pumped, just undergoing the natural tendency to flow from an area of high pressure to low pressure. As such, there would come a point when the residual gas in the tanker would start to provide back pressure on the flow input, slowing it down & potentially stopping it once pressures equalize between the tanker and the source.
By allowing oxygen gas to escape the tanker, the pressure will always be lower in the tanker, and the flow will be faster more adequately fill the tanker. The same technique is used when topping up liquid nitrogen in an NMR spectrometer, for example.

The liquefied gas is transported without pressure, like you would carry it around in an open Dewar, in contrast to a pressurized gas bottle that has about 200bar, which would be difficult to build on such a scale.


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Justify the value of the product of the uncertainty in position and momentum
The Heisenberg uncertainty principle states that:
$$\Delta x\Delta p_x\ge\frac12\hbar$$
But more generally for operators $A$ and $B$:
$$\Delta A\Delta B=\frac12|\langle[A,B]\rangle|$$
However, $[x,p_x]=i\hbar$. With this information, I don't see how to get from the second equation to the first.

What you think you'll get upon this trivial substitution if not $1/2 \hbar$? What's your result?

The commutator $[\hat{A}, \hat{B}]$ of two operators $\hat{A}$ and $\hat{B}$ is also an operator, so when using the Robertson uncertainty relation we have to take the expectation value of the commutator first and then take the absolute value of it.
But for $\hat{x}$ and $\hat{p}$ the commutator $[\hat{x}, \hat{p}]$ is simply a constant $\mathrm{i} \hbar$, so that taking the expectation value is trivial: $\langle [\hat{x}, \hat{p}] \rangle = \mathrm{i} \hbar$. Then we take the absolute value of this result which is $\hbar$ and divide it by two to get $\frac{1}{2} \hbar$ as expected.

I think that perhaps I am being stupid but why is the absolute value of $i\hbar$ equal to $\hbar$?

The absolute value of a complex number goes as the square root of  (the complex number multiplied by the complex conjugate of the complex number), or $\hbar = \sqrt{i\hbar\cdot -i\hbar}$ in this case.

@RobChem, the absolute value of $\mathrm{i} \hbar$ is $\hbar$ by the very definition of the absolute value of a complex number.


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Sp5 hybridization in cyclopropane?
I have never before heard/read about something as a $sp^5$ hybridization. Today, Henry Rzepa's blog post made me aware of the existance of such a bonding system. That made me search a little bit and I found an entry in a german chemistry forum, where this question was also asked ... they answered it with a mathematical construction$^\ast$:

Cyclopropane has the following angles:
$\angle \ce{HCH}=118^\circ~\text{resp.}~\gg 120^\circ$
  $\angle \ce{CCC}~\text{with bent bonds:}~60 + 2 \cdot 21 = 102^\circ$
The orbitals towards the protons are $sp^2$ because of the $120^\circ$ angles.
   The orbitals towards the carbons originate in the following relation:
$$1 + a \cos~\alpha = 0$$ ... where $\alpha$ is the bond angle and $a$ is the p-amount in sp$^a$ for the orbitals, which make up the angle.
This means for the orbitals, which span the 102 degree angle: $$1 + a \cos 102^\circ = 0$$ $$1 + a \cdot (-0.20) = 0$$ $$a = \frac{-1}{-0.20} = 5$$
  $$\Rightarrow \text{sp}^5 \text{-orbitals}$$
Test:

In a single sp$^a$ orbital, the s-amount is: $\frac{1}{1+a}$, because $1+a$ equals the sum of all amounts of s and p
In a single sp$^a$ orbital, the p-amount is: $\frac{a}{1+a}$
For s:

In the orbitals that are oriented towards the protons, the s-amount is $\frac{1}{1+2} = \frac{1}{3}$
In the orbitals that are oriented towards the carbons, the s-amount is $\frac{1}{1+5} = \frac{1}{6}$
Addition of all s-amounts at a single carbon with all four bond orbitals yield: $\frac{1}{3}+\frac{1}{3}+\frac{1}{6}+\frac{1}{6}=1$, which is correct, because there is only one single s-orbital at every carbon atom.

For p:

In the orbitals that are oriented towards the protons, the p-amount is $\frac{2}{1+2} = \frac{2}{3}$
In the orbitals that are oriented towards the carbons, the p-amount is $\frac{5}{1+5}=\frac{5}{6}$
Addition of all p-amounts at a single carbon with four bond orbitals yield: $\frac{2}{3}+\frac{2}{3}+\frac{5}{6}+\frac{5}{6}$, which is correct, because there are 3 p-orbitals at every carbon atom.

This means, that the bent bonds with $21^\circ$ from the $\ce{C-C}$-bond are spanned by sp$^5$ orbitals.

So math-magically this seems to make sense, but is there another explanation that might base more on chemical intuition or "real" chemical concepts?
A fast calculation ($\omega$B97X-D/def2-TZVPP) and a subsequent analysis of the isosurface of the Laplacian of the electron density, showed at least the expected "nonlinear", slightly curved bond between the carbon atoms.

Whoever might want to see the electron localization function (ELF), which also shows the bent bonds quite good:

$^\ast$ While I tried to translate it to my best, some errors might have been introduced by this . . . please correct me, where I'm wrong.

https://en.wikipedia.org/wiki/Isovalent_hybridization

Hybridisation isn't real - it's always only mathematical concept - you're talking about it's more complicated version, which is rather incompatible with normal version.

Coulson's theorem is the key here. It can be used to interconvert bond angles and hybridization indices.  See this earlier answer where it is applied to cyclopropane and ethylene.  As pointed out in this answer, the bent-bond analysis (Pauling) is equivalent to the more common pi bond treatment (Hückel).

The notation is just a short form. It just states, that there is much more p character in the bonds than in the traditional ones, which we usually use. $$\mathrm{sp}^n\equiv\mathrm{s}^{\frac{1}{n+1}}\mathrm{p}^{\frac{n}{n+1}}$$ So this proof is just restating the original definition. Coulson's theorem (as stated in this proof) only works for equivalent hybrid orbitals, the more general approach is covered by Bent's rule (=observation).

Using hybridization for more complicated examples seems like embedding circular orbits on circular orbits to explain retrograde planetary motion in a geocentric model. If you just admit that the model is flawed and use a better heliocentric model, the complication goes away.

Until this post, $\mathrm{sp^1,~sp^2~\text{and}~sp^3}$ meant for me, that the hybrid orbitals would consist from one s orbital and one, two or three p orbitals. I thought of it being so, as this seems to be what the most textbooks suggest, as can be seen in the following image:

One 2s orbital and three 2p orbitals hybridize to four $\mathrm{sp^3}$ orbitals.
This scheme seems to work but it also seems to be a working simplification as can be seen by my question about $\mathrm{sp^5}$ hybrid orbitals, which simply cannot consist from one s orbital and five p orbitals, as there are only three p orbitals in reach.
Then what does this nomenclature actually mean?
As Martin commented, it is more like the following:
$$\mathrm{sp^n \equiv s^\frac{1}{n+1}p^\frac{n}{n+1}}$$ or to say it slightly different, a $\mathrm{sp^n}$-bond has
$\frac{100}{n+1}~\%$ s-amount and $\frac{100\,n}{n+1}~\%$ p-amount.
\begin{array}{lll}
\hline
\text{bond} & \text{s-amount} & \text{p-amount}\\
\hline
\mathrm{sp} & 50~\% & 50~\%\\
\mathrm{sp^2} & 33~\% & 67~\%\\
\mathrm{sp^3} & 25~\% & 75~\%\\
\mathrm{sp^4} & 20~\% & 80~\%\\
\mathrm{sp^5} & 17~\% & 83~\%\\
\ldots & \ldots & \ldots \\
\mathrm{sp^n} & \frac{100}{n+1}~\% & \frac{100\;n}{n+1}~\%\\
\hline
\end{array}

But this cannot work. If we make 3 sp5 orbitals we are only using about 50% of the s-orbital and 2.5 p-orbitals. So there would be left over s and p orbitals. We get a total of 7 (partial) orbitals, how do we distribute our 6 electrons here? It also doesn't work out if we make 6 sp5 orbitals (and how do we distribute 6 electrons and make 3 bonds with 6 orbitals?)

@DSVA \begin{align}
1\times\mathrm{s}, 3\times\mathrm{p}
&\leadsto 4\times\mathrm{sp}^3 \
&\leadsto 3\times\mathrm{sp}^2, 1\times\mathrm{p} \
&\leadsto 2\times\mathrm{sp}, 2\times\mathrm{p} \
&\leadsto 2\times\mathrm{sp}^3, 1\times\mathrm{sp}, 1\times\mathrm{p} \
&\leadsto \text{etc. pp.}\
&\leadsto 2\times\mathrm{sp}^4, 1\times\mathrm{p}, 1\times\mathrm{sp}^{(2/3)}\
&\leadsto \color{\red}{3\times\mathrm{sp}^5, 1\times\mathrm{sp}}\
&\leadsto 2\times\mathrm{sp}^5, 2\times\mathrm{sp}^2
\end{align}

@DSVA A slightly different approach to the sp5 bonds can be found in this answer of mine. Hybridisation is a simple concept, obviously when you form linear combinations of the orbitals while maintaining the total number of orbitals, and the orthonormalisation. Also note that in the given description, every carbon utilises $\ce{2\times sp^5}$ orbitals for two C-C bonds, and $\ce{2\times sp^2}$ for two C-H bonds.

I think the simplest way to go at this is to start with the thought that we have no clue about the hybridization of the orbitals used in the C-C bonds in cyclopropane. On the other hand, the H-C-H on any of the carbons is 'opened' up, to almost $\pu{120^{\circ}}$ (it's not quite 120, but we're going with the big picture here). So, gross assumption here - those C-H bonds are each using $\mathbf{sp^2}$ hybrid orbitals (or 33% s, 67% p, twice). We still have overall one s and 3 p orbitals to work with at each carbon, so to get carried away a bit,
$$
\pu{3.00 p - 0.67 p - 0.67 p = 1.66 p is left.}
$$
For the amount of s remaining,
$$
\pu{1.00 s - 0.33 s - 0.33 s = 0.33 s is left.}
$$
Since we have two bonds (the (C-C)'s) left over to describe, divide what's left by two; that's
$$
\pu{0.83 p (or 83\%) and 0.17 s (or 17\%).}
$$
That's 5 times as much p as s (and like Answer 1 reads), $\mathbf{sp^5}$. And yes, $\mathrm{sp^5}$ corresponds to about $\pu{101.6^{\circ}}$ angles between the orbitals, so cyclopropane in this model still has bent bonds.


text:
Is the Gibbs standard free energy always constant?
I am a biochemistry student and we are learning about thermodynamics. Is the Gibbs standard free energy for a reaction always constant? The equation below suggests that it changes with temperature:
$$\Delta G^{\circ\prime} = -RT\ln K'_\mathrm{eq}$$
It is important to note that we are using the biochemistry version of the Gibbs standard free energy equation. To answer my own equation, I believe that it is always constant. Changing the temperature would probably just change the equilibrium constant.

In short, no, the standard Gibbs free energy change is not constant; it is a function of temperature. The same is true for practically all other standard-state quantities.
This gets a little confusing because of how standard-state properties are often explained in lower-level (high-school and college-freshman level) textbooks. The standard Gibbs free energy change, $\Delta G^\circ$, is the value of $\Delta G$ at a certain constant pressure (usually 100 kPa or 1 atm) and constant concentration of solutions (usually 1 molal or 1 molar) and a specified temperature (usually $0~^\circ \text{C}$ or $25~^\circ \text{C}$). What is intended but not immediately obvious is that the pressure and concentration are constants but temperature is a variable; since the temperature is a variable, $\Delta G^\circ$ is a function of temperature. When the value of $\Delta G^\circ$ is given, the temperature must also be specified.
In your equation, then, $\Delta G^\circ$ is the standard Gibbs free energy change of the reaction at the specified temperature ($T$ in the equation). Change $T$, and the value of $\Delta G^\circ$ also changes; but it is still the standard Gibbs free energy change at the (new) specified temperature. (This equation cannot be used to calculate $\Delta G^\circ$ at a given temperature because  $K_\text{eq}$ is also a function of temperature.)
The IUPAC "Green Book" (ed. 3) has a good explanation of what is meant by the term "standard state"; the discussion starts on p. 61 (as printed on the page; it's p. 76 in the pdf file). It explicitly expresses the standard chemical potential ($\mu^\circ$), which is very closely related to $\Delta G^\circ$, as a function of temperature.

Note that "standard state" is not the same as STP; this is probably where a lot of the confusion comes from.

Thanks! I think I am getting somewhere with this. I blame most of the confusion on my textbook. My textbook does not address some of these more nuanced ideas. You mentioned "When the value of ΔG∘ is given, the temperature must also be specified." My textbook never actually does this which is ridiculous.

See also Notation for states and processes, significance of the word standard in chemical thermodynamics, and remarks on commonly tabulated forms of thermodynamic functions
(IUPAC Recommendations 1981):
“The above definitions of standard states make no reference to fixed temperature. Hence, it is possible to have an infinite number of standard states of a substance as the temperature varies.”

Yes, the standard Gibbs energy of a reaction is constant. You have to keep in mind however that "standard" refers to a standard state or reference state, in which by definition, things such as temperature, pressure, phase composition, concentration of all species, etc. are fixed and unchangeable.

Changing the temperature would probably just change the equilibrium constant.

It would also mean that the system would no longer be at standard conditions.
Allowing just the temperature to vary, while holding species concentrations, phase compositions, pressure, etc. constant, the variation in $\Delta G^{\,\circ \prime}$ is given by the Gibbs-Helmholtz equation.

So you are saying that if I am given a (biochem) Gibbs standard free energy value, I am only able to calculate Keq for 298 K? What if I wanted Keq for a different temp (say 325 K). Could I plug 325 K into the equation I posted above and solve for Keq? Thanks for your help!!!

Yes that is what I am saying..  (Subject to the assumptions of the  Gibbs Helmholtz equation)


text:
Why can't helium nuclei (alpha particles) react with fluorine?
Alpha particles are basically just helium nuclei, so it will accept an electron pair to become stable.
Will fluorine, being highly electronegative, not just donate an electron pair but form a bond with helium?

So are you asking about the existence of $\ce{HeF2}$?

Don't confuse the extreme nuclear stability of the $\cf{^4He}$ nucleus, i.e., an alpha particle, with the electronic stability of naked $\cf{He2+}$ cations.

Nobody said a word about $\ce{He}_2^+$; as for $\ce{He}^{2+}$, it is of course electronically stable, as it has no electrons at all.

@marcus - Please edit your question for clarity and content. As it stands, it's very unclear what you are asking and might get closed.

@ToddMinehardt I am just trying to ask why cant alpha particles react with fluorine atoms?

@IvanNeretin I couldnt undertand how could He^2+ be electronically stable?

Sorry Ivan, I made a typo and of course meant $\cf{He^{2+}}$.  It is not electronically stable at all, at least not for definitions of electronically stable I am familiar with.

How could it not be? Imagine a lone alpha particle deep in outer space; how is it unstable? What would happen to it, electronically or otherwise? (Sorry, just kidding.)

@marcus - I have edited your post to reflect your comment.

@marcus Can you clarify exactly what molecules/ions you are thinking about. $\ce{HeF2}$, $\ce{HeF^2+}$ or something else?

@bon Actually i am not thinking about any ions or molecules as i searched and found them to just not exist. I am thinking WHY couldnt they exist? Fluorine is very electronegative and if alpha particles are projected into a fluorine environment, surely they will attract the electrons of fluorine. But it would be really tough for fluorine to give up its electrons and so a covalent or a dative bond might be formed. Since they do not, i really want to know why ?

Why, many atoms would readily create a bond with helium, when it comes in a form of alpha particle (just cooled down enough to chemically interact with). There is comparatively stable $\ce{HeH}^+$ and other similar particles, too. But a positive particle is not a compound yet. And when you try to form a compound out of it, that is, to balance it with some negative ions - well, that's where the problem arises. It would violently react with absolutely anything else, form some compound, and happily release the neutral helium atom.

Thats why i asked what if alpha particles were to form bonds with fluorine? In a case like that the alpha particles couldn't just take away the electrons of fluorine and leave a positively charged fluorine?

Of course it can. The ionization energy of helium is by far the largest among all elements, so it will take electrons from anything. A covalent bond may be formed, too; I expect a particle like $\ce{HeF}^+$ to appear, which would be stable for just as long as it doesn't encounter anything negatively charged.

You mean as long as long as it does not encounter anything, except another Helium atom, right?

I said "negatively charged". Come to think of it, most neutral particles (except another helium atom, you are right on that) would react with it just as well. Positive ions, though, probably would not.

Will fluorine, being highly electronegative, not just donate an electron pair but form a bond with helium?

Yes a fluorine atom and helium atom can form a "molecule" $\ce{HeF}$. Such a molecule would be very unstable chemically. Two such molecules bumping into to one another would yield helium atoms and fluorine molecules.
An alpha particle is an "energetic" bare helium nucleus. Think of the speed that a fluorine molecule would be moving in fluorine gas. The charged alpha particle is traveling much much faster so it is ripping through the gas bumping into fluorine molecules thus creating ions and free electrons. So in fluorine gas as the alpha particle slows down it forms a helium atom. Thus overall you'd get uncharged helium atoms and $\ce{F2+}$ ions.


text:
Colour intensity of transition metal complexes
Even though they are both $\mathrm{d^7}$ complexes, the colour of $\ce{[Co(H2O)6]^2+}$ is pale whereas that of $\ce{[CoCl4]^2-}$ is more intense. Does the octahedral or tetrahedral geometry matter in determining the molar absorption coefficient of the two species?

You bet it does. Look up the crystal field theory.

@IvanNeretin unfortunately, that Wikipedia page does not explain why one colour is intense and the other is weak.

At first, it is necessary to ask why some transition metal complexes are colourful at all — remember that alkaline metals, alkaline earth metals and earth metals usually form colourless solutions, e.g. sodium, magnesium, aluminium. The transition metals’ colour stems from the d-electrons they have and the shape of the complexes they create. A typical orbital scheme of a 3d-metal in an octahedral complex will look something like this (image originally taken from Prof. Klüfers web scritptum to his coordination chemistry course at the LMU Munich):

On the left are the metal orbitals, from top to bottom it is 4p, 4s and 3d. On the right are the ligand group orbitals, assuming six ligands each only donating electrons in a $\unicode[Times]{x3C3}$ fashion only. Any remaining metal electrons will be filled into the $\mathrm{t_{2g}}$ orbitals which correspond to $\ce{d}_{xz}$, $\ce{d}_{xy}$ and $\ce{d}_{yz}$ and into the $\mathrm{e_g}$ orbitals corresponding to $\ce{d}_{z^2}$ and $\ce{d}_{x^2-y^2}$
The energy difference marked as $10\,\mathrm{Dq}$ in the image would then be the energy of a photon required to excite an electron from $\mathrm{t_{2g}}$ to $\mathrm{e_g}$; this photon would thereby be absorbed, which is where we get our absorption coefficient from. It is usually in the visible range of light.
There are a few quantum mechanical principles that allow or disallow excitation from one energy level to another. One of these is the spin-allowance rule: Excitations are only allowed if the spin of the entire system remains unchanged. (So excitation cannot involve a spin exchange.) This is why $\ce{Mn^{II}}$ or high-spin $\ce{Fe^{III}}$ complexes are generally colourless: all d-electrons are of identical spin so there is no freedom to be excited.
A second rule is the Laporte-rule whose formal definition I never properly understood, but which states that d to d excitation is forbidden in centrosymmetric species such as octahedra. This rule would mean that all d-element complexes should be colourless. However, complexes are not static but the ligands vibrate back and forth, often in a non-centrosymmetric way. Whenever a photon of the correct energy hits a complex that is not ideally centrosymmetric, the excitation is allowed and the photon absorbed. Because this is a rather rare phenomenon, the colour of such a complex is weaker than it could be.
The $\ce{[Co(H2O)6]^2+}$ case is more complicated again, because in a perfectly centrosymmetric world it would be degenerate: Since it is $\ce{d^7}$, two spin-down electrons share three $\mathrm{t_{2g}}$ orbitals. It would be favourable to somehow make these orbitals dissimilar. This can happen by distorting the octahedral shape in a Jahn-Teller manner: The two water ligands on the $z$ axis are moved slightly further away, creating a slight stabilisation of all orbitals which contain a $z$ component, most notably $\ce{d}_{xz}$ and $\ce{d}_{yz}$. But since the final product is still centrosymmetric, this paragraph only explains the wavelength of absorption slightly better and can be disregarded.
Next question is why the chlorido complex of cobalt(II) is not octahedric but tetrahedric. This is due to a hypothetical $\ce{[CoCl6]^4-}$ being unbalanced in terms of large chloride ions around a small cobalt ion that only has a low charge to combat the $-6$ the chlorides have to offer. The tetrahedral complex $\ce{[CoCl4]-}$ has everything that the octahedral $\ce{[Co(H2O)6]^2+}$ has except the centre of symmetry. Because centrosymmetry is missing, the Laporte rule is no longer valid and every photon of the correct energy can excite electrons. More excitation equals a stronger extinction.
Also, the colour of both complexes is notably different because the orbital scheme is, too. $\mathrm{t_2}$ and $\ce{e}$ are now much closer together and their order is flipped. Combining that with chloride ions that usually generate low values of $10\,\mathrm{Dq}$, the energy difference is even smaller therefore the excitation wavelength longer therefore the colour we see blueshifted (complementary colour).

Mostly correct, but there is a problem; the Laporte rule still forbids d-d transitions, even in tetrahedral complexes. The g-g or u/u requirement doesn't apply for tetrahedral complexes, but $\Delta L = \pm 1$ still does. If the transition were fully allowed then you would expect a colour as intense as that of KMnO4 (LMCT is fully allowed), but the extinction coefficient of a typical tetrahedral complex is ~2 orders of magnitude smaller than that of a complex where LMCT occurs.

If I remember this correctly (and this has been some time) in order to excite one electron from one energy level to another using a photon the transition dipole moment has to change. You can show this for the vibrational and also for the rotational excitation if you want to look this up. I think for scattering processes like Raman the polarizability has to change and for absorption and emission the transition dipole moment has to change. So if you want to describe this using quantum mechanics you will require an operator, and your interaction operator is in this case one for the transition dipole moment, which is an odd function (x). Now in quantum mechanics, when you want to describe the probability of something happening, you need the two energy levels, the operator and they are all multiplied and then integrated over the total space. So imagine now you need to integrate something that consists of three entries a * b * c, one is already an odd function because of the transition dipole moment operator being odd (x) then the transition has to be from a symmetrically odd orbital into an even one or vice versa. Because otherwise the product would be an odd function. Imagine a simple 2-dimensional function like x³, it has the same area in the negative region as in the positive one therefore the negative contributions and the positive one cancel each other out and the probability of the transition happening is zero.
The octahedron is such a case, because the lower $t_2g$ and the higher $e_g$ both have this 'g' which stands for even. In a tetrahedron this is not the case. The only reason why the octahedron is colored at all is because the ligands are still able to move or vibrate at bit and distort the octahedron taking some symmetry.
That was now a very short summary but it should contain most things.

Yes, you remember correctly: $x,y,z$ transforms as $\mathrm t_1$ and $\mathrm e \times \mathrm t_1 \times \mathrm t_2$ contains the totally symmetric irrep, so the transition dipole moment is nonzero. There is probably a second factor, related to mixing between $n\mathrm d$ and $(n+1)\mathrm p$ orbitals, which helps to relax the $\Delta L = \pm 1$ requirement as well. In octahedral symmetry the d and p orbitals transform as different irreps and cannot mix. The other (downvoted) answer is trying to get at this, but doesn't seem to explain it very well.

Really? I always thought there is at least a slight mixing of d and p in order to overcome that forbidden d-d-transition?

The way I understand it, at least, is that the forbidden aspect of the d-d transition only strictly applies to isolated atoms. In a metal complex you have MOs which can be considered "mostly d", but they aren't d orbitals because there is contribution from ligand orbitals as well as other AOs on the metal. How much the d-d rule applies is therefore a function of how much d character the MOs have; for Oh where d/p cannot mix this is more, but for Td where d/p can mix this is less.

I checked some of the newer books by Mingos and you are right, they do not mix because of symmetry issues. I didn't find an explanation anywhere but it seems like this only applies to σ-ligands, π-ligands seem to have the $t_2g$-symmetry as well and then I'd expect some mixing again, therefore, allowing the d-d-transition.

On the other hand, how many pure σ-complexes exist in an octahedral geometry? For alkyl with such high oxidation states, you'd expect hydride-elimination, if you protect it by using norbornene derivates you run out of space and for pure hydrides I guess you'd favor the redox-chemistry. So perhaps these very rare complexes just don't have a d-d-transition?


text:
Ice under very high pressure phase transitions
If you put a piece of ice into a very high pressure chamber and increase the pressure, I can understand how the ice turns to a water state.  But what happens after that?  Does the pressure push the water together into a crystal structure, once again, to make ice, or does the water just stay in its liquid phase?

The former. See the phase diagram of water. Also, chances are that you will go from one crystal structure to another directly, not touching  liquid water.

possible duplicate of Can water be liquefied or solidified just by adjusting the temperature, regardless of the pressure? or maybe http://chemistry.stackexchange.com/questions/9803/what-happens-to-water-under-high-pressures-without-possibility-of-escape

also http://chemistry.stackexchange.com/questions/20009/hypothetical-what-happens-to-water-as-pressure-increases-to-infinity

Mithoron, is that spot at 0C the point where you can press water and it will melt?  And at those triple and double points can the H2O switch between phases just with the tiniest changes?

I don't understand, you can press ice to melt it and in triple points you can have equilibria of 3 phases


text:
Why is lutetium part of the lanthanide series?
I’ve been looking at the periodic table and began to wonder - why is lutetium part of the lanthanide series? Or, for that matter, why is lawrencium part of the actinide series?
This isn’t readily apparent when looking at the traditional periodic table because the lanthanides and actinides are separate and below the main table. However, on the extended table, the two elements don’t appear that they should be in the groups that they are in: lutetium contains the $5d$ electron shell whereas the lanthanides do not, and lawrencium contains the $6d$ electron shell whereas the actinides do not.

Lawrencium actually has an electron configuration of $\mathrm{[Rn] (7s^2) (5f^{14}) (7p^1)}$. There are multiple anomalies in the electron configurations of heavy elements and it is not simple to classify them into groups by their electron configuration. In fact it is not always simple to put them into groups at all!

@bon So the boundaries for the groups are actually fuzzy and not strict as taught?  Or I could ignore them entirely?

For heavy elements, and particularly transition/rare earth metals, there is much less similarity between the reactions of elements in the same column of the periodic table. It's not like the halogens, or alkali metals, which basically all have the same reactions.

This video might interest you.

It's a bit of a controversial topic but if I remember correctly, chemists currently lean more towards Lu and Lr belonging in group 3 based on their reactivity, and La and Ac in the lanthanoid/actinoid series. However I'm too lazy to find a suitable reference. I would guess Eric Scerri has probably written something on it somewhere.

See also Which Elements Belong in Group 3 of the Periodic Table? in CI 2016, Vol. 38, No. 2, pp. 22–23.

It's not just heavy elements having "anomalous" electron configurations.  This article suggests that we are getting to a point where the whole electron shell theory breaks down.  The Periodic Table, like the atomic structure on which it's based, dissolves into a fog.

There are some arguments that Lutetium and Lawrencium should be directly below Scandium and Yttrium: https://link.springer.com/article/10.1007/s10698-015-9216-1

The assignment of elements to groups or series in the periodic table is not based exclusively on the electron configurations concerned. Helium for example is an s-block element yet is placed above Ne in the p-block due its noble gas properties.
Lutetium is part of the lanthanide series, notwithstanding its electron configuration, because its properties and those of its compounds are similar to those of the other lanthanide elements.
Lawrencium is assigned to the actinides for basically the same reason. Experiments have so far failed to find any chemistry-based evidence for Lr having a p electron. While it may have a p electron in the gas phase it probably has a d electron in the condensed phase.
Whether either Lu and Lr, or La and Ac are assigned to group 3 will not make any difference to the composition of the lanthanides or actinides, since both of these are horizontal series rather than vertical groups.

Why is Mars a "planet" while our Moon and Pluto are not?  Mars has a certain combination of properties, internally and within its orbit, that our Moon and Pluto do not.  Astronomers now label that combination of properties with the word "planet".  Same thing here, basically, so +1.


text:
Guidance on Jacobian computation for stiff chemical mechanisms
I'm writing an implicit solver that will involve evaluating the chemical source terms in the system for a given reaction mechanism. It's safe to assume that these mechanisms may be stiff. I know, in general, how to approach the problem but I am looking for some very specific guidance.
The scheme requires the computation of the source-term Jacobian. The mechanism takes in the state variables (pressure, temperature, density) and the species mass fractions then returns the change in partial density of each species. The Jacobian will be computed numerically and I know how to do that in general.
Specifically though, are there any guidelines or requirements on how large (well, small really) the perturbations should be when evaluating the Jacobian numerically? Let's say the perturbation is defined relative to the size of the variable, so let's say the perturbation is X% of the variable. How should X be chosen? Should each variable have its own X or will a global one work? It should work in a mechanism-agnostic way, but I know most mechanisms will be very sensitive to temperature because it is in the exponential and less sensitive to changes in density or mass fraction since those are raised to exponents which are not excessively large.
I understand the idea behind the Jacobian and the Taylor Series expansions, so I'm looking more for practical advice in terms of accuracy, performance, and stability.

Are you using a library or writing something that doesn't have an adaptive step size for solving the system iteratively?

@ToddMinehardt Well, this is for the source terms in a flow solver, so the time step adapts to the flow (and I suppose I could make it adapt to the chemistry as well), but the intention at this phase is that the chemistry does not affect the time step directly. Indirectly, it will alter the state and velocities and change the time step.

@tgp2114 - It sounds a little outside my area of knowledge (which is, at best, dated anyway in re: state of the art in numerical solvers). I suppose there's always some trial and error for determining the magnitude of perturbation you can get away with, although I hope some other user(s) on chem.SE can help you out here

@ToddMinehardt I think trial and error may be what I end up with anyway, I was just hoping for some guidance/bounds from best practices. I found an example source code online that set it to 1e-8 but had a comment that said "Note, this could be determined much better." I'm hoping somebody knows the much better specific to chemistry :)

This question would probably better be posted on Computational Science.SE. There's much greater expertise there with regard to this sort of algorithmic question.

@Brian I thought that also and am familiar with that community. I was hoping to get something more targeted for chemistry though since I have a feeling a chemistry solver might want to be treated differently than a general ODE solver. If it doesn't get any interest here, I'll ask for a migration.

@tpg2114 There is nothing typical about chemistry here. In the end everything is numbers. If you are just interested in finding solutions to your equations (and not design a new algorithm) then matlab has a good collection of stiff solvers. There are some freely available FORTRAN codes as well (ODEPACK). Finally, you can use some approximations in your mathematical model to convert it into a non-stiff system.

I'm voting to close this question as off-topic because it is better asked at Computational Science.SE.