text:
Polarisation of electron clouds by partially positively-charged hydrogen atoms
Premise of my question:
I was just wondering about this as I was writing a response to the question (linked here: Bicarbonate decomposition and pKa): Is it possible for partially positively-charged atoms participating in covalent bonds to polarise the electron cloud of the electron-rich part of the molecule to the extent that it becomes unstable. FYI My response explained the instability of the carbonic acid molecule by the polarisation of the "carbonate ion" by the "hydrogen ions". The reason I have placed quotation marks on the two ions is that they are no longer ions, but rather just partially positively-charged atoms in a neutral molecule. But I am calling them "ions" so as to emphasise the magnitude of the partial charges on them.

Point of contention: I believe this is contentious because we always describe the polarisation of an anion by a cation, not the polarisation of the rest of a molecule by one atom of the molecule. By the former, I am referring to cases aluminium carbonate, iron (III) carbonate etc. To clarify, polarisation here refers to the distortion of the electron cloud of molecules/ions. (This post over here discusses polarisation of anions by cations in greater detail: Polarising power.)
Rationale for my argument: But I don't see why the partially positively-charged hydrogen atom cannot polarise the electron cloud of the rest of the "very polarisable" molecule because it has been established that hydrogens bonded to electronegative atoms, such as F, O and N, tend to behave like a "bare protons", stripped of a lot of electron density. This also makes hydrogen bonding a particularly strong type of an intermolecular force of attraction.
My stand: Thus, the partial positive charge on the hydrogen atom is of quite a large magnitude, such that it is capable of strong polarisation of electron clouds.
Question to users: Firstly, I want to know if covalently-bonded hydrogen atoms can polarise (i.e. distort the electron cloud) the rest of the molecule. Based on my line of reasoning above, it should. Secondly, I would also like to know if this distortion of the electron cloud by the hydrogen atom would lead to instability of species of molecules. In particular, for the question at least, I would like to know if the polarisation of the molecule affects the stability of the carbonic acid species. However, it would be great if we could generalise this relationship between stability and extent of polarisation of the electron cloud, for all molecular species.
Further clarification: "Instability" in this case refers to the ease of decomposition of carbonic acid into carbon dioxide and water.

Well, what are you asking about exactly? If protonated species could be truly unstable because of this additional proton? You should highlight it.

@Mithoron If I am not already clear enough in my question (and title), I would like to know if a covalently-bonded hydrogen atom (especially if it is bonded to a very electronegative atom) is capable of distorting the electron cloud of the rest of the molecule, if the rest of the molecule is indeed very polarisable, such as that of a carbonate ion.

@Mithoron Please feel free to comment on my answer, if you find that it is "simply bad". Otherwise, I take that as a baseless allegation and will not bother to edit my answer.

@Those who have put my question on hold, I have already edited my questions to clarify any points of ambiguity. Please re-read my question and raise any other points of ambiguity that still persist in the existing question. Thanks.

Write what exactly you understand by polarising and if it's only about this "polarisation" or also about "instability".

@Mithoron I have clarified the definition of "polarisation" in my post. Also, I would like to discuss polarisation and stability. In particular, the polarisation of the electron cloud of the carbonic acid molecule and the stability of this molecule, upon polarisation of its electron cloud.

I still don't understand the question. It's asking about the effect of having a hydrogen on electron density, but compared to what? Nothing? Methyl group? Oxygen anion? Obviously, it will have an effect that will change the properties of the compound.

@jerepierre My question is focused on the extent of the effect compared to a metal ion, such as aluminium or iron (III). Not whether it has an effected. We know that it definitely has an effect. But would the effect be so great to cause it to be more easily decomposed.


text:
Solution to Raoult's Law for a volatile binary solution (no pun intended)
Raoult's Law
In a closed container, $$P_A=P_A^\circ x_A$$ where $P_A$ is the pressure of compound in gaseous phase, and $x_A$ is the mole fraction of the liquid component A in the liquid mixture. If the liquid is alone in the container, $x_A=1$.
Problem I'm trying to solve

$A$ moles of compound $A$ and $B$ moles of compound $B$ are stored in
  container closed with a piston. External pressure on the piston is
  maintained as $P_e$. Find moles of $A$ and $B$ that will be found in
  the gaseous phase at equilibrium. ($P_A^\circ$ and $P_B^\circ$ are
  known values)

Attempt at solving
Let $a$ moles of A and $b$ moles of B exist in gaseous phase. This means $A-a$ and $B-b$ moles of respective compounds are leftover in the liquid phase.
Now Raoult's Law gives us
$$a \frac{RT}V = P_A^\circ \left(\frac{A-a}{A+B-a-b}\right)$$
$$b \frac{RT}V = P_B^\circ \left(\frac{B-a}{A+B-a-b}\right)$$
(Ideal gas equation used implicitly)
Also total pressure of gases inside should equal $P_e$
$$(a+b) \frac{RT}V = P_e$$
These are not ordinary linear equations, so I'm not totally sure my calculations are correct, but I'm getting the solution as
$$P_A^\circ \left(\frac Aa -1\right) = P_B^\circ \left(\frac Bb -1\right) = \frac{AP_B^\circ (P_e-P_A^\circ)+BP_A^\circ(P_e-P_B^\circ)}{(A+B)(P_A^\circ+P_B^\circ-P_e)}$$
So far so good, but it turns out that on plugging in specific values for $A,B,P_A^\circ,P_B^\circ,P_e$, $a$ and $b$ attain negative values.
Mathematically it can be proved that this happens when the rightmost expression is negative. The rightmost expression can be rewritten as
$$P_e \left[\frac {\frac{\frac{A}{A+B}}{P_A^\circ}+\frac{\frac{B}{A+B}}{P_B^\circ}-\frac{1}{P_e}} {\frac{1}{P_A^\circ}+\frac{1}{P_B^\circ}-\frac{1}{P_e}} \right]$$
Let $N$ mean numerator and $D$ mean denominator of this expression. We can prove that $N<D$. This means that $\frac ND >0$ if and only if $N>0$ or $D<0$ holds.
Question 1
Assuming all the calculations are correct, this seems to be a rather arbitrary mathematical condition. Can someone explain the physical significance of this? Like a more intuitive way to understand why we must check such an arbitrary condition rather than something as simple as $P_e>P_A^\circ+P_B^\circ$?
Question 2
If $\frac ND <0$, is the solution always $a=A$, $b=B$ (which means no liquid evaporates)?

As you correctly pointed out, you can connect the moles of each compound in the gas phase, $a$ and $b$, to the composition of the liquid phase, $A-a$ and $B-b$, through their partial pressures, $p_A$ and $p_B$:
$p_A=a\cfrac{RT}{V}=p_A^o\cfrac{A-a}{\left(A-a\right)+\left(B-b\right)}$
$p_B=b\cfrac{RT}{V}=p_B^o\cfrac{B-b}{\left(A-a\right)+\left(B-b\right)}$
And, as you did, we can use the total number of moles in the gas phase to rewrite these equations in terms of $p_e$. Since
$p_e=p_A+p_B=\left(a+b\right)\cfrac{RT}{V}$
and therefore
$\cfrac{RT}{V}=\cfrac{p_e}{a+b}$
we can write
$p_e\cfrac{a}{a+b}=p_A^o\cfrac{A-a}{\left(A-a\right)+\left(B-b\right)}$
$p_e\cfrac{b}{a+b}=p_B^o\cfrac{B-b}{\left(A-a\right)+\left(B-b\right)}$
Note that this is a two-equation system with two unknowns, so, barring incompatible values of the constants $A$, $B$, $p_e$, $p_A^o$ and $p_B^o$, it's determined (as you essentially pointed out).
Let's find a relationship between $a$ and $b$. A straightforward way to do that is by dividing the equations:
$\cfrac{p_e\cfrac{a}{a+b}}{p_e\cfrac{b}{a+b}}=\cfrac{p_A^o\cfrac{A-a}{\left(A-a\right)+\left(B-b\right)}}{p_B^o\cfrac{B-b}{\left(A-a\right)+\left(B-b\right)}}$
$\cfrac{a}{b}=\cfrac{p_A^o}{p_B^o}\cfrac{A-a}{B-b}$
This expression is key; we will return to it later.
Clearing $a$:
$a=\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}\left(A-a\right)$
$a\left(1+\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}\right)=\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}A$
$a=\cfrac{\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}}{1+\cfrac{p_A^o}{p_B^o}\cfrac{b}{B-b}}A=\cfrac{b}{\cfrac{p_B^o}{p_A^o}\left(B-b\right)+b}A$
We haven't made any additional assumptions in the derivation (beyond Raoult's law itself), so this is valid for all the situations where Raoult holds. Let's interpret the numbers to see what they mean.
With this expression you can check out various situations to get a feeling of how does Raoult behave. Some limit cases:
· If $p_A^o \approx p_B^o$, $\cfrac{a}{A} \approx \cfrac{b}{B}$ and both components are distributed to the same extent between the gas and liquid phase.
· If $p_A^o \gg p_B^o$, the vapour pressure quotient $p_B^o/p_A^o \ll 1$ and therefore $a \approx A$ (the more volatile component will be almost entirely in the gas phase).
· And, contrarily, if $p_A^o \ll p_B^o$, $a \approx 0$ (the less volatile component will be almost entirely in the liquid phase).
Let's return to the key expression above:
$\cfrac{a}{b}=\cfrac{p_A^o}{p_B^o}\cfrac{A-a}{B-b}$
If you ponder carefully what this means, it's telling you that for ideal mixtures (Raoult mixtures), the relationship between the components in the gas and liquid phases is always fixed as long as there is equilibrium, and the proportionality factor of that relationship is the vapour pressure ratio of the two components. So, if you have $p_A^o=2p_B^o$, the ratio of moles of A and B in the gas will always be twice the ratio of moles of A and B in the liquid. Or, equivalently, Raoult predicts that the gas phase is enriched in the most volatile component, precisely in proportion to the relative vapour pressures. The more volatile a component is, the more "additional fraction" of the gas phase it represents.
So, what values for $p_e$ are compatible with Raoult? Let's find out:
$p_e=p_A+p_B=p_A^o\cfrac{A-a}{\left(A-a\right)+\left(B-b\right)}+p_B^o\cfrac{B-b}{\left(A-a\right)+\left(B-b\right)}$
$p_e\left(\left(A-a\right)+\left(B-b\right)\right)=p_A^o\left(A-a\right)+p_B^o\left(B-b\right)$
$\left(A-a\right)\left(p_e-p_A^o\right)=\left(B-b\right)\left(p_B^o-p_e\right)$
$\cfrac{A-a}{B-b}=\cfrac{p_B^o-p_e}{p_e-p_A^o}$
Note that this is a particular case of the lever rule and, if all the quantities are positive, it demands that $p_A^o \leq p_e \leq p_B^o$ or $p_B^o \leq p_e \leq p_A^o$. A value of $p_e$ outside that range would correspond to a value of $a$ or $b$ outside of the range $0 \leq a \leq A$ or $0 \leq b \leq B$ - which doesn't make physical sense.
But is any value of $p_e$ between $p_A^o$ and $p_B^o$ valid? No, it depends on $A$ and $B$. Every value of $p_e$ corresponds to a certain composition - so you can't set both to arbitrary values.
Take for example $p_A^o=100$, $p_B^o=200$, $p_e=125$, $A=1$ and $B=4$. Applying the lever rule expression we just derived,
$\cfrac{A-a}{B-b}=\cfrac{p_B^o-p_e}{p_e-p_Ao}=\cfrac{200-125}{125-100}=3$
So, the ratio of moles of A to moles of B in the liquid in equilibrium is 3. In other words, 3/4 of the liquid molecules are of A.
Let's introduce this value in the key expression we derived earlier:
$\cfrac{a}{b}=\cfrac{p_A^o}{p_B^o}\cfrac{A-a}{B-b}=\cfrac{100}{200}3=1.5$
So, the ratio of moles of A to moles of B in the gas in equilibrium is 1.5. In other words, 3/5 of the gas molecules are of A.
All of this is compatible with Raoult's law... but it is incompatible with a total system composition of 1 mole of A and 4 moles of B!
Hope this helps.

Thanks for posting the answer. Last doubt though, about the last paragraph. When this happens, do we expect only gas to be present in the container or only liquid? Or some kind of equilibrium which raoults law cannot predict?

Well, real systems never behave like Raoult, although the deviation can be minimal. Raoult is only applicable when there is an equilibrium, so when you get a result that doesn't make sense, mathematically, it's telling you that you are outside the window of applicability of the model - but the model won't depict correctly what happens there.

It might be easier to work your way from the other end - start with the moles of both components, and find out the range of $p_e$ that is compatible with them. For pressures above you expect to have only liquid, for pressures below you expect to have only gas, all within the ideal depiction of the system that is implicit in Raoult's law.

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text:
Compound class naming convention: heteropoly-acid vs isopoly-acid
There are heteropoly-acids and isopoly-acids, two classes of polynuclear compounds with and without heteroatoms within its molecular structures, respectively. These are the names widely used in the literature for a long time[1].
Why two non-homogeneous prefixes ("iso-" vs "hetero-") are used? Wouldn't it make more sense to denote these names with consistent prefixes showing parity ("hetero-" vs "homo-"; "aniso-" vs "iso")? For instance, IUPAC Red Book utilized the terms "homopolyatomic" vs "heteropolyatomic" in chapter IR-5 Compositional Nomenclature, and Overview of Names of Ions and Radicals [2, p. 68].
Wouldn't heteropoly-acid and homopoly-acids pair be morphologically more accurate, since it's a heteroatom presence that allows that distinction, just like in case of homocyclic vs heterocyclic or homopolyatomic vs heteropolyatomic species?
References

Middleton, A. R. J. Chem. Educ. 1933, 10 (12), 726 DOI: 10.1021/ed010p726.
IUPAC “Red Book” Nomenclature of Inorganic Chemistry; Connelly, N. G., Royal Society of Chemistry (Great Britain), International Union of Pure and Applied Chemistry, Eds.; IUPAC Recommendations; Royal Society of Chemistry Publishing/IUPAC: Cambridge, UK, 2005.

Wouldn't heteropoly-acid and homopoly-acids pair be morphologically more accurate (than heteropoly-acid and isopoly-acids)?

I completely agree with andselisk of this suggestion because the former makes more sense than latter in this regards. For instance, as correctly pointed out in the question, the IUPAC Red Book has utilized the terms "homopolyatomic" vs "heteropolyatomic" in the chapter, IR-5: Compositional Nomenclature, and Overview of Names of Ions and Radicals (p. 68-82, Ref.1). A few examples given for the homopolyatomic cations in the section IR-5.3.2.3 are $\ce{O2^+}$, $\ce{S4^2+}$, $\ce{Hg2^2+}$, and $\ce{Bi5^4+}$ (p.71). Likewise, few examples of heteropolyatomic cations are also given in the following section IR-5.3.2.4, which are $\ce{NH4^+}$, $\ce{H3O^+}$, $\ce{PH4^+}$, and $\ce{SbF4^+}$ (p.71). Thus, it make perfect sense of using the terms "homopolyatomic" and "heteropolyatomic" in the context, since it's the presence of heteroatom in the latter case ($\ce{N, O, P,}$ and $\ce{Sb}$, respectively) that allows the distinction.
Nonetheless, in another chapter in Ref.1 (IR-8: Inorganic Acids and Derivatives, p. 124-141) has utilized the terms "homopolyoxy acids" and "heteropolyoxy acids" to described polyoxy acids:

Examples 1, 3–6 and 14 above demonstrate that homo- and heteropolyoxoacids and their partially dehydronated forms […] (p.136, ref.1)

For instance, the example 1 and 4 mentioned in the excerpt are $\ce{H2PO7^2-}$ and $\ce{H4[SiW_{12}O_{40}]}$, respectively (Section: IR-8.4, p. 135, Ref.1). As we all know that $\ce{H4PO7}$ is an example of isopoly-acids and $\ce{H4[SiW_{12}O_{40}]}$ (Ref.2) is an example of heteropoly-acids. Thus, even if these unconventional names are frequently use in inorganic chemistry literature to date, I think IUPAC is already recognized the necessity to keep consistency in the naming.

References:

Nomenclature of Inorganic Chemistry: IUPAC recommendations 2005 (Issued by the Division of Chemical Nomenclature and Structure Representation in collaboration with the Division of Inorganic Chemistry); Prepared for publication by N. G. Connelly, et al.; RSC Publishing (for the International Union of Pure and Applied Chemistry): Cambridge, United Kingdom, 2005 (ISBN: 0-85404-438-8).
Li Min Yang, Zhi Kui Yin, and Li Qiang Wu, "$\ce{H4[SiW_{12}O_{40}]}$: An efficient catalyst for the synthesis of new spiro-[dibenzo[a,i]xanthene-14,3′-indoline]-2′,8,13-triones," Chinese Chemical Letters 2012, 23(3), 265-268 (DOI: https://doi.org/10.1016/j.cclet.2012.01.002).

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text:
Wondering how I should have graded this question (freshman lab, about buffer capacity)
I TA the intro freshman lab course. In a lab about buffers and buffer capacity, the students were instructed to make 50 mL of a 0.1 total molar pH 6.70 buffer by dissolving solid $ KH_2PO_4 $ and $ K_2HPO_4 $. The students then had to add a drop or two of NaOH and HCl to the buffer and see if/how much the pH changes. Next, they are supposed to dilute the buffer to double volume with DI water, and repeat. Then respond to the following questions:
1: How did diluting the buffer effect the change in pH when a few drops of NaOH or HCl were added?
2: Did adding water to the buffer change its pH? Explain why or why not.
For #1, obviously, diluting the buffer lowers its buffer capacity, so the pH change in the diluted buffer will be more drastic. I was a bit unsure about how to grade #2, however.  Some of the answers I got were:
"The pH raised slightly but was small enough that it was probably random  fluctuations in the pH meter"
"We didn't measure a pH change"
"The pH went up as DI water has a pH of 7 which is above the buffer, so even though the buffer resists change, it'll change a little"
If you look at the Henderson-Hasslebalch equation:
$ pH = pKa + \log(\frac{[A^-]}{[HA]}) $
Since both pH of the buffer was 6.7, and the pKa of the system is also 6.7, $[A^-]$ and $[HA]$ must be equal, and since the total concentration was 0.1M, then $[A^-]$ and $[HA]$ are 0.05 M each. So if we dilute it to double volume with water, the concentrations become 0.02 M each; the ratio remains unchanged and the pH would still be 6.7.
In reality, when you dilute the solution, the Ka and therefore pKa of the $ KH_2PO_4 \leftharpoonup\rightharpoondown K_2HPO_4 $ equilibrium will change slightly, so the pH would be expected to change. For clarification, see this question: Does dilution of a buffer affect pH?
The issue is that this is an early lab in a freshman course and they certainly aren't expected to know about activity vs concentration or anything that complex yet. There is no answer key and I am not TA for the lectures, so I am not expected to go to them, so I don't know what they were told there, or in their textbook. My supervisor seemed to give a non-answer when I asked about this. So I was a bit unsure what kind of thing I should be expecting the students to write. The lab will be the same next semester and I want to know what sort of answers to accept.

Why does dilution affect capacity? The amount of conjugate base available to neutralize acid is the same. In strong dilutions, the capacity may be affected once water autoionization becomes a major factor in the equilibrium. Also, $K_{\mathrm{a}}$ is a property of an acid and not subject to change.

Say you have a a 1 molar total (so 0.5 molar for HA and A) buffer, pKa = 6.7, pH = 6.7. Add 0.1 moles of HCl, so 0.1 moles of A- are converted into HA.
$ pH = 6.7 + \log(\frac{[.5-.1]}{[.5+.1]}) = 6.524 $
Technically in the brackets should be concentration, not moles, but since they are in the same solution the volume is the same so that cancels out.
Now say the buffer was diluted to double volume, so now the concentrations of A and HA are 0.25, and you repeat:
$ pH = 6.7 + \log(\frac{[.25-.1]}{[.25+.1]}) = 6.332 $
So the diluted buffer changes more, as it has less capacity.

That's not correct because 0.1 moles of $\ce{HCl}$ would need to be converted to a concentration if you want to add or substract from concentration in solution... And technically, you can just leave the values of $\ce{HA}$ and $\ce{A-}$ at moles because the total volume is the same and those will cancel in the fraction.

Would you say that when you take 1 mL of a 10 mL buffer, the 1 mL has a lower buffer capacity?


text:
Easiest way (software) to visualize charge density from an .xyz file with point-charges?
I have an .xyz file which contains partial charges calculated by a Quantum-Chemistry software (NWChem).
The system looks like this, here I show the partial charges calculated from the electrostatic surface potential (using CHELPG method).

I want something like this classic example of water (red is showing high electron density while blue shows low), e.g. to demonstrate polarity and electronic structure:

Is it possible to get this from point-charge data (as opposed to a very large set of points)? I'm familiar with VMD, Avogadro, and other visualization softwares but I don't know a way off-hand to accomplish this. Here is the .xyz file itself (charges are the 5th column):
78

 N  -0.215180  -3.611590   1.013730  -0.980426
 N  -0.907220  -3.663400  -1.269030  -1.159281
 H  -1.020120  -4.516150  -1.239180   0.451727
 H   0.592410  -3.533220   3.408420   0.466050
 H   1.138340  -4.068520  -6.662270   0.121348
Zn   1.048810   0.361560   1.461840   1.085568
 C   0.826190   0.806270   4.182580   0.885122
 C   0.417080   1.494320   5.455600  -0.158259
 C  -0.185960   2.733600   5.457730  -0.038223
 H  -0.320120   3.185220   4.658260   0.111225
 N   3.029810   0.215180   1.013730  -0.989702
 N   2.978000   0.907220  -1.269030  -1.323792
 H   2.125250   1.020120  -1.239180   0.552920
 H   3.108180  -0.592410   3.408420   0.599787
 O   0.611010   1.411960   3.076910  -0.805465
 O   1.325620  -0.316130   4.242000  -0.786727
 C  -0.826190  -0.806270  -9.141970   0.932072
 C  -0.417080  -1.494310  -7.868950  -0.116686
 C   0.185960  -2.733600  -7.866810  -0.092391
 H   0.320120  -3.185210  -8.666290   0.116237
 O  -0.611010  -1.411960 -10.247640  -0.908737
 O  -1.325620   0.316130  -9.082550  -0.838830
 C  -0.487480  -3.005900  -0.183350   1.029873
 C   0.217840  -2.574210   1.791620   1.084450
 N  -0.286910  -1.704180  -0.139370  -0.556339
 N   0.174010  -1.421260   1.164300  -0.669532
 H  -1.062620  -3.235690  -1.998680   0.479136
 N   0.588430  -2.745550   3.064650  -1.219932
 H   0.822210  -2.066800   3.539000   0.535669
Zn  -1.048810  -0.361560  -1.461840   1.115312
 C  -0.826190  -0.806270  -4.182580   0.930235
 C  -0.417080  -1.494310  -5.455600  -0.165126
 C   0.185960  -2.733600  -5.457740  -0.068129
 H   0.320120  -3.185210  -4.658260   0.125910
 C  -3.635500  -0.487480   0.183350   1.112801
 C  -4.067190   0.217840  -1.791620   1.089920
 N  -4.937220  -0.286910   0.139380  -0.669012
 N  -5.220140   0.174010  -1.164300  -0.683650
 N  -3.029810  -0.215180  -1.013730  -1.010564
 N  -2.978000  -0.907210   1.269030  -1.379275
 H  -3.405710  -1.062620   1.998680   0.465316
 H  -2.125250  -1.020120   1.239180   0.591458
 N  -3.895840   0.588430  -3.064650  -1.359821
 H  -4.574600   0.822210  -3.539000   0.462374
 H  -3.108170   0.592410  -3.408420   0.595021
 O  -0.611010  -1.411960  -3.076900  -0.846291
 O  -1.325620   0.316130  -4.242000  -0.789904
 C   0.487480   3.005900   0.183350   1.054807
 C  -0.217840   2.574210  -1.791620   1.079690
 N   0.286910   1.704180   0.139380  -0.596520
 N  -0.174000   1.421260  -1.164300  -0.651619
 N   0.215180   3.611590  -1.013730  -0.982105
 N   0.907220   3.663400   1.269030  -1.153898
 H   1.062620   3.235690   1.998680   0.472248
 H   1.020120   4.516150   1.239180   0.448303
 N  -0.588430   2.745550  -3.064650  -1.192789
 H  -0.822210   2.066800  -3.539000   0.511494
 H  -0.592410   3.533220  -3.408420   0.458606
 C   0.599050  -3.311400  -6.662270  -0.253677
 C  -0.738520  -0.893930  -6.662270  -0.060650
 H  -1.176860  -0.074380  -6.662270   0.135692
 C   0.826190   0.806270   9.141970   0.921571
 C   0.417080   1.494320   7.868950  -0.136263
 C  -0.185960   2.733600   7.866820  -0.033469
 H  -0.320120   3.185220   8.666290   0.092701
 O   0.611010   1.411960  10.247650  -0.900753
 O   1.325620  -0.316130   9.082550  -0.836472
 C  -0.599050   3.311400   6.662270  -0.299314
 H  -1.138340   4.068520   6.662270   0.134260
 C   0.738520   0.893930   6.662270  -0.048643
 H   1.176860   0.074380   6.662270   0.130420
 C   3.635500   0.487480  -0.183350   1.071205
 C   4.067190  -0.217840   1.791620   1.085316
 N   4.937220   0.286910  -0.139370  -0.660949
 N   5.220140  -0.174000   1.164300  -0.672967
 H   3.405710   1.062620  -1.998680   0.459743
 N   3.895850  -0.588430   3.064650  -1.356568
 H   4.574600  -0.822210   3.539000   0.457166

I think, the easiest way is to use Multiwfn in combination with Jmol. There you can estimate/rebuild the ESP from partial charges.
Multiwfn
Load your xyz file with the charges into Multiwfn, which should recognize the charges. If not, then remove the first two lines and rename it into file.chg.

5 Output and plot specific property within a spatial region (calc. grid data)
8 Electrostatic potential from nuclear charges
3 High quality grid  , covering whole system, about 1728000 points in total
2 Export data to Gaussian cube file in current folder

Now you should have a file called nucleiesp.cub, which is a Gaussian cube file.
Next thing that you need is a cube file, that contains the electron density. That should be possible to obtain with NWChem. If not, then somehow get a molden file and build the cube file also in Multiwfn using 5 > 1 > 3 > 2.
Jmol
In Jmol you make a right click in the black area and choose the console. There you simply write: load "path/to/files/geom.xyz"; isosurface myiso cutoff 0.002 "path/to/files/density.cub" map "path/to/files/nucleiesp.cub"; show $myiso; color $myiso "rwb"; color $myiso translucent
 and you will get an image like this:

This is the red-white-blue-colored translucent ESP on the electron density (isovalue 0.002).
If you don’t have the option to get the density, you can also use the SAS, the surface accessible surface, which would then look like this:
load "D:/calculations/mwfn-charges2esp/geom.xyz"; isosurface myiso solvent map "D:/calculations/mwfn-charges2esp/nucleiesp.cub"; show $myiso; color $myiso "rwb"; color $myiso translucent

PovRay
One way to get nice pictures out of Jmol is to render them with PoyRay, but as I am no expert with Jmol, I am a far less expert with PovRay. So the result is only kind of a "preview" on what you get without any effort.

Because I didn’t like the big atoms, I changed the atom size resp. the sphere radius macro in the Jmol.pov file like following:
#macro a(X,Y,Z,RADIUS,R,G,B,T)
 sphere{<X,Y,Z>,RADIUS/5
  pigment{rgbt<R,G,B,T>}
  translucentFinish(T)
  clip()
  check_shadow()}
#end

which ends up in an image like this:

I just saw, that in my last image the hydrogens are a little too small. So using RADIUS/2 makes the PoyRay atom size look like in Jmol.

If someone cares about Chimera, I could add sth about that.

After the successful calulation of the electrostatic surface potential, molden equally allows the display of it in a form like

(source)
A step-by-step, hopefully still functional, tutorial is this. (My last contact with Molden already dates back some years.)

Using Molden to calculate the charges is undesirable / unpreferred in my case. I've seen this tutorial before, and it was helpful, but I was looking for a way to get the surface straight from an .xyz with point-charges. I discovered ways to do that in my answers. Thanks again!

I have found one simple way to do this via Molden:

Open the .xyz file via Molden: gmolden myfile.xyz

Click the "Create Surfaces ...etc." button
Click Solv. Acc. Surface
Click Elec. Pot. Mapped
Click the "Create Surfaces ...etc." button again
Click Charges Derived

(Optionally, to save the surface, click the "Create Surfaces..." button again and click "Write Surface" and give it a filename.)
I got a visualization like this:

It's not as pretty as I would like, though, so if someone posts an answer with something that looks more like my water example I would be grateful.

I think molden can export the result as a cube file which you can probably visualize afterwards with Avogadro.

I think this qualifies as a separate answer, using VMD.
I converted the .xyz file to .pdb format (or at least some variant of the strict convention). Thus the file looks like below. The columns are mostly arbitrary, except the coordinates (columns 7,8,9), and charges (column 11).
1) I loaded the file with VMD
2) Clicked (in Main window) Graphics ->
   Representations
3) Click Coloring Method -> Beta  (the column with
   charges in the .pdb)
4) Click Drawing Method -> QuickSurf
5) Then
   set these settings: Resolution=0.5, Radius Scale=0.5, Density
   Isovalue=0.5, Grid Spacing=0.5, and Surface Quality=Max.
6) Clicked
   (in Main Window) Graphics -> Colors -> Color Scale (tab)
7) Then set
   these settings: Method=BWR, Offset=-0.03, Midpoint=0.00.
I got this image:

ATOM      1    N MOF F   1      -0.215  -3.612   1.014 14.00700 -0.98043 0.971573 34.722151 3.260689
ATOM      2    N MOF F   1      -0.907  -3.663  -1.269 14.00700 -1.15928 0.971573 34.722151 3.260689
ATOM      3    H MOF F   1      -1.020  -4.516  -1.239 1.00790 0.45173 0.413835 22.141661 2.571134
ATOM      4    H MOF F   1       0.592  -3.533   3.408 1.00790 0.46605 0.413835 22.141661 2.571134
ATOM      5    H MOF F   1       1.138  -4.069  -6.662 1.00790 0.12135 0.413835 22.141661 2.571134
ATOM      6   Zn MOF F   1       1.049   0.362   1.462 65.39000 1.08557 1.988700 62.399227 2.461553
ATOM      7    C MOF F   1       0.826   0.806   4.183 12.01100 0.88512 1.288599 52.838055 3.430851
ATOM      8    C MOF F   1       0.417   1.494   5.456 12.01100 -0.15826 1.288599 52.838055 3.430851
ATOM      9    C MOF F   1      -0.186   2.734   5.458 12.01100 -0.03822 1.288599 52.838055 3.430851
ATOM     10    H MOF F   1      -0.320   3.185   4.658 1.00790 0.11123 0.413835 22.141661 2.571134
ATOM     11    N MOF F   1       3.030   0.215   1.014 14.00700 -0.98970 0.971573 34.722151 3.260689
ATOM     12    N MOF F   1       2.978   0.907  -1.269 14.00700 -1.32379 0.971573 34.722151 3.260689
ATOM     13    H MOF F   1       2.125   1.020  -1.239 1.00790 0.55292 0.413835 22.141661 2.571134
ATOM     14    H MOF F   1       3.108  -0.592   3.408 1.00790 0.59979 0.413835 22.141661 2.571134
ATOM     15    O MOF F   1       0.611   1.412   3.077 15.99980 -0.80546 0.851973 30.193174 3.118146
ATOM     16    O MOF F   1       1.326  -0.316   4.242 15.99980 -0.78673 0.851973 30.193174 3.118146
ATOM     17    C MOF F   1      -0.826  -0.806  -9.142 12.01100 0.93207 1.288599 52.838055 3.430851
ATOM     18    C MOF F   1      -0.417  -1.494  -7.869 12.01100 -0.11669 1.288599 52.838055 3.430851
ATOM     19    C MOF F   1       0.186  -2.734  -7.867 12.01100 -0.09239 1.288599 52.838055 3.430851
ATOM     20    H MOF F   1       0.320  -3.185  -8.666 1.00790 0.11624 0.413835 22.141661 2.571134
ATOM     21    O MOF F   1      -0.611  -1.412 -10.248 15.99980 -0.90874 0.851973 30.193174 3.118146
ATOM     22    O MOF F   1      -1.326   0.316  -9.083 15.99980 -0.83883 0.851973 30.193174 3.118146
ATOM     23    C MOF F   1      -0.487  -3.006  -0.183 12.01100 1.02987 1.288599 52.838055 3.430851
ATOM     24    C MOF F   1       0.218  -2.574   1.792 12.01100 1.08445 1.288599 52.838055 3.430851
ATOM     25    N MOF F   1      -0.287  -1.704  -0.139 14.00700 -0.55634 0.971573 34.722151 3.260689
ATOM     26    N MOF F   1       0.174  -1.421   1.164 14.00700 -0.66953 0.971573 34.722151 3.260689
ATOM     27    H MOF F   1      -1.063  -3.236  -1.999 1.00790 0.47914 0.413835 22.141661 2.571134
ATOM     28    N MOF F   1       0.588  -2.746   3.065 14.00700 -1.21993 0.971573 34.722151 3.260689
ATOM     29    H MOF F   1       0.822  -2.067   3.539 1.00790 0.53567 0.413835 22.141661 2.571134
ATOM     30   Zn MOF F   1      -1.049  -0.362  -1.462 65.39000 1.11531 1.988700 62.399227 2.461553
ATOM     31    C MOF F   1      -0.826  -0.806  -4.183 12.01100 0.93024 1.288599 52.838055 3.430851
ATOM     32    C MOF F   1      -0.417  -1.494  -5.456 12.01100 -0.16513 1.288599 52.838055 3.430851
ATOM     33    C MOF F   1       0.186  -2.734  -5.458 12.01100 -0.06813 1.288599 52.838055 3.430851
ATOM     34    H MOF F   1       0.320  -3.185  -4.658 1.00790 0.12591 0.413835 22.141661 2.571134
ATOM     35    C MOF F   1      -3.635  -0.487   0.183 12.01100 1.11280 1.288599 52.838055 3.430851
ATOM     36    C MOF F   1      -4.067   0.218  -1.792 12.01100 1.08992 1.288599 52.838055 3.430851
ATOM     37    N MOF F   1      -4.937  -0.287   0.139 14.00700 -0.66901 0.971573 34.722151 3.260689
ATOM     38    N MOF F   1      -5.220   0.174  -1.164 14.00700 -0.68365 0.971573 34.722151 3.260689
ATOM     39    N MOF F   1      -3.030  -0.215  -1.014 14.00700 -1.01056 0.971573 34.722151 3.260689
ATOM     40    N MOF F   1      -2.978  -0.907   1.269 14.00700 -1.37928 0.971573 34.722151 3.260689
ATOM     41    H MOF F   1      -3.406  -1.063   1.999 1.00790 0.46532 0.413835 22.141661 2.571134
ATOM     42    H MOF F   1      -2.125  -1.020   1.239 1.00790 0.59146 0.413835 22.141661 2.571134
ATOM     43    N MOF F   1      -3.896   0.588  -3.065 14.00700 -1.35982 0.971573 34.722151 3.260689
ATOM     44    H MOF F   1      -4.575   0.822  -3.539 1.00790 0.46237 0.413835 22.141661 2.571134
ATOM     45    H MOF F   1      -3.108   0.592  -3.408 1.00790 0.59502 0.413835 22.141661 2.571134
ATOM     46    O MOF F   1      -0.611  -1.412  -3.077 15.99980 -0.84629 0.851973 30.193174 3.118146
ATOM     47    O MOF F   1      -1.326   0.316  -4.242 15.99980 -0.78990 0.851973 30.193174 3.118146
ATOM     48    C MOF F   1       0.487   3.006   0.183 12.01100 1.05481 1.288599 52.838055 3.430851
ATOM     49    C MOF F   1      -0.218   2.574  -1.792 12.01100 1.07969 1.288599 52.838055 3.430851
ATOM     50    N MOF F   1       0.287   1.704   0.139 14.00700 -0.59652 0.971573 34.722151 3.260689
ATOM     51    N MOF F   1      -0.174   1.421  -1.164 14.00700 -0.65162 0.971573 34.722151 3.260689
ATOM     52    N MOF F   1       0.215   3.612  -1.014 14.00700 -0.98211 0.971573 34.722151 3.260689
ATOM     53    N MOF F   1       0.907   3.663   1.269 14.00700 -1.15390 0.971573 34.722151 3.260689
ATOM     54    H MOF F   1       1.063   3.236   1.999 1.00790 0.47225 0.413835 22.141661 2.571134
ATOM     55    H MOF F   1       1.020   4.516   1.239 1.00790 0.44830 0.413835 22.141661 2.571134
ATOM     56    N MOF F   1      -0.588   2.746  -3.065 14.00700 -1.19279 0.971573 34.722151 3.260689
ATOM     57    H MOF F   1      -0.822   2.067  -3.539 1.00790 0.51149 0.413835 22.141661 2.571134
ATOM     58    H MOF F   1      -0.592   3.533  -3.408 1.00790 0.45861 0.413835 22.141661 2.571134
ATOM     59    C MOF F   1       0.599  -3.311  -6.662 12.01100 -0.25368 1.288599 52.838055 3.430851
ATOM     60    C MOF F   1      -0.739  -0.894  -6.662 12.01100 -0.06065 1.288599 52.838055 3.430851
ATOM     61    H MOF F   1      -1.177  -0.074  -6.662 1.00790 0.13569 0.413835 22.141661 2.571134
ATOM     62    C MOF F   1       0.826   0.806   9.142 12.01100 0.92157 1.288599 52.838055 3.430851
ATOM     63    C MOF F   1       0.417   1.494   7.869 12.01100 -0.13626 1.288599 52.838055 3.430851
ATOM     64    C MOF F   1      -0.186   2.734   7.867 12.01100 -0.03347 1.288599 52.838055 3.430851
ATOM     65    H MOF F   1      -0.320   3.185   8.666 1.00790 0.09270 0.413835 22.141661 2.571134
ATOM     66    O MOF F   1       0.611   1.412  10.248 15.99980 -0.90075 0.851973 30.193174 3.118146
ATOM     67    O MOF F   1       1.326  -0.316   9.083 15.99980 -0.83647 0.851973 30.193174 3.118146
ATOM     68    C MOF F   1      -0.599   3.311   6.662 12.01100 -0.29931 1.288599 52.838055 3.430851
ATOM     69    H MOF F   1      -1.138   4.069   6.662 1.00790 0.13426 0.413835 22.141661 2.571134
ATOM     70    C MOF F   1       0.739   0.894   6.662 12.01100 -0.04864 1.288599 52.838055 3.430851
ATOM     71    H MOF F   1       1.177   0.074   6.662 1.00790 0.13042 0.413835 22.141661 2.571134
ATOM     72    C MOF F   1       3.635   0.487  -0.183 12.01100 1.07120 1.288599 52.838055 3.430851
ATOM     73    C MOF F   1       4.067  -0.218   1.792 12.01100 1.08532 1.288599 52.838055 3.430851
ATOM     74    N MOF F   1       4.937   0.287  -0.139 14.00700 -0.66095 0.971573 34.722151 3.260689
ATOM     75    N MOF F   1       5.220  -0.174   1.164 14.00700 -0.67297 0.971573 34.722151 3.260689
ATOM     76    H MOF F   1       3.406   1.063  -1.999 1.00790 0.45974 0.413835 22.141661 2.571134
ATOM     77    N MOF F   1       3.896  -0.588   3.065 14.00700 -1.35657 0.971573 34.722151 3.260689
ATOM     78    H MOF F   1       4.575  -0.822   3.539 1.00790 0.45717 0.413835 22.141661 2.571134


text:
Need a malleable solid which dissolves in HCl, acetone, or water
I need to temporarily plug the end of a 3 mm ID copper tube with some sort of malleable solid solute that I can dissolve at later point by flushing with a solvent/acid via syringe. I need to keep the area at around room temperature, so I cannot use dry ice or excessive heat.  I'm using it to prevent liquid silicone from seeping up the end of a tiny copper tube while I seal the tube, so it needs to be relatively inert to the copper tube and the silicone.
I've tried using Styrofoam and acetone (which is almost ideal), but this leaves behind a residue of slimy gunk that is difficult to remove, and I need the tube to be clear of foreign residues.  Low melting point solid materials may be possible to use, but they can't leave behind any residue for my purposes.  I can obtain any special chemicals through my university.

Crystalbond might be a good choice.

If the temperature is low enough you could use a gallium alloy?

Gallium alloy looks promising, I'm going to run that in my next batch of tests.  Thanks!

If the cavity in which you insert silicone is not air tight I would use and air pump to blow in the tube.
The easiest chemical I could imagine using is a low melting point salt such as ammonium acétate. If the tube is removable you dip in the liquid and upon solidification you can remove the excess. It's water soluble.
Sugar toffee will have a similar behavior and is malleable at a certain proportion of water.
Plumber use bread.

These are all excellent suggestions.  I'm going to go grab all of these and run a few tests.  You might have just started a trend of grad students using sugar toffee on their thesis projects.

Bentonite clay is used to plug the ends of fireworks and is water soluble.
Waxes are available in various types, blends like hot melt glue are a possibility.
Napthalene has a melting point of around 80 degC and a strong smell.
A mechanical plug might be the easiest if possible to remove.  Using a thin plastic tube sealed at one end and pressurised from the other end to swell it may be possible to block a tube.
I would love to see what your mechanical arrangement is, there may be other solutions.

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


text:
How are Grignard reagents stable in an ethereal solution?
I have heard of the leveling effect, which mentions that highly basic ions cannot exist in a solvent if it's acidity is high enough to be deprotonated by the base.
Considering that, check out the Grignard reagent. I'll use a methylgrignard for this purpose, generated in a diethylether solvent.
This Grignard reagent furnished methanide ions ($\ce{CH3-}$). This ion can abstract protons from any compound thats more acidic than methane. Diethyl ether possesses $\beta$-hydrogens that are easily abstracted due to the inductive pull of the oxygen atom:

The inductive pull may not be much, but it certainly makes it more acidic than a regular hydrocarbon hydrogen.
However we know that Grig reagents do work in ethereal solvents. How is it so?
Edit: This isn't a duplicate of the mentioned question. My question deals with acidity of $\beta$ hydrogens.

True, the question asks something different. The answer will be largely the same. Voting to leave open (as per site policies) and waiting for someone to give practically the same answer, referencing the other question ;)

We tend to assume that acid-base reactions are instantaneous. But they need to go through an activated complex of higher energy, like every other reaction.
These reactions are usually done at - 78°C. As long as the energy provided to the system is below that which is required to reach the activated complex there should be no reaction.
I know for sure that tetrahydrofuran is fully deprononated by nBuLi at room temperature. The color turns deep yellow, if I recall correctly.

Um, is thf a typo for "the" and what is nBuli?

I think they mean "tetrahydrofuran by normal butyl lithium"

@user46680 Do you have any sources to support your assertion? For example, which sources did you get the part about the thermodynamics of acid-base reactions (i.e. "they need to go through an activated complex of higher energy")? Just want to ascertain the accuracy of your answer.

@Tan The concept of activated complex and energy of activation is broadly covered by most books of statistical thermodynamics. This one is a good one "Statistical Thermodynamics M. C. Gupta"

THF is tetrahydrofurane, nBuLi is butyl lithium, the n says that the carbon chain is linear

t-Butyllithium causes beta-elimination in diethyl ether. The liberated ethylene acts an electrophile for excess t-BuLi to form the primary organolithium, t-BuCH2CH2Li.


text:
What are the applications of poly-crystalline GdFe₂?
Poly-crystalline $\ce{GdFe2}$ is a ferromagnetic compound but where is it practically used for its magnetic properties?


text:
Steric inhibition of resonance vs. hydrogen bonding
I have to compare the acidic strength of these compounds:

Now, in A and B hydrogen bonding will be present which will help stabilize the negative charge of the conjugate base. However, the hydroxyl group is not bulky and won't cause steric inhibition of resonance, hence the +M effect of phenyl will still be present.
In C, methyl will cause steric inhibition of resonance and hence stop/reduce the +M effect of phenyl which will stabilize the conjugate base.
How do I compare the relative stabilization offered by these two effects (H-bonding and SIR)?

I think the main effect in play here is straight electron withdrawing potential of the oxygens. In my opinion this is a classical BSc question that tries to avoid the grey zones of pka eye balling. I would invite you to review the concept of steric inhibition of resonance since I'm not entirely sure I understand your argument there. Here's a good starting point if you do not have a text book covering the subject https://www.quora.com/What-is-steric-inhibition-of-resonance

https://chemistry.stackexchange.com/questions/7683/ortho-effect-in-substituted-aromatic-acids-and-bases

https://chemistry.stackexchange.com/questions/20190/why-does-hydrogen-bonding-in-salicylic-acid-make-it-more-acidic

I think the answer is C>B>A. In C due to SIR effect the conjugate base is more stabilised whereas in A and B no SIR takes place as OH- isn't bulky. But in B there are 2 OH- present for hydrogen bonding compared to A so conjugate base of B is more stable than A


text:
Relative acidity of p-chlorobenzoic acid and p-flurobenzoic acid
I want to compare the relative acidity of p-chlorobenzoic acid and p-flurobenzoic acid. Both of the facts below point to p-chlorobenzoic acid being more acidic:

The carboxylic acid group is too far away from the halogen for inductive effect to have any significance. The +M effect of fluorine is greater than chlorine (due to more effective orbital overlap). +M effect is not distance dependant and hence p-fluoroobenzoic acid  should be less acidic since the negative charge on its conjugate base is destabilized.
There is resonance of the negative charge of the carboxylate ion with the empty d-orbitals of chlorine which would stabilize it.

However, in reality, p-chlorobenzoic acid is less acidic than p-fluorobenzoic acid . Why?

$\mathrm{p}K_\mathrm{a}$ of para chlorobenzoic acid $\ce{-> 4.03}$.Source
$\mathrm{p}K_\mathrm{a}$ of para fluorobenzoic acid $\ce{-> 4.14}$.Source

What are the pKa values?

@orthocrestol I don't know. This was given to me as a problem, and in the answer key it's given that the latter is more acidic.

@Xashthor You're forgetting that F is more electronegative than Cl

+mockingbird the carboxylic acid group is too far away from the halogen for inductive effect to have any significance.

There are two answers there ^ so it's not OK to ask again just because you're not satisfied. BTW you can use simple rationalisations for aliphatic acids but for aromatic they just don't work properly.

@orthocresol $\mathrm{p}K_\pu{a}$ values for reference — p-chlorobenzoic acid: 4.03 p-fluorobenzoic acid:  4.14. If the values are correct, then p-chlorobenzoic acid is more acidic than p-fluorobenzoic acid.

For acidity of para substituted benzoic acids, -I effect is helpful and +R effect decreases the acidity.
In 4-fluorobenzoic acid, the +R effect is prominent in $\ce{F}$ due to stronger $\ce{2p}\pi - \ce{2p}\pi$ overlap with nearest carbon. Thus, due to stronger +R effect of $\ce{F}$ than -I effect, it creates a negative charge on your starred carbon. This charge delocalises in the carbonyl carbon  and the electrophilicity is thus decreased of that carbon. Releasing hydrogen also doesn't help much as the negative charge created over $\ce{O}$ of $\ce{-OH}$ group can only delocalise a little.
Whereas in the $\ce{Cl}$, there is much weaker $\ce{3p}\pi-\ce{2p}\pi$ overlap. So, +R effect of $\ce{Cl}$ is much weaker and due to a relatively higher -I than its +R, there is more positive character on the starred carbon due to $\ce{Cl}$.
And, thus, in reality, acidity of 4-chlorobenzoic acid is a little higher than that of 4-fluorobenzoic acid.

Why is there relatively higher -I for chlorine? Isn't F more electronegative?

I am saying that in $\ce {Cl} $, $-I $ effect is relatively stronger  than $+R $ effect. I am not comparing $\ce{F } $ with $\ce{Cl } $ in -I effect.

In the case of halogens(on ortho/para), the -I effect dominates over the +R effect and hence due to this the negative charge of carboxylate anion is better stabilized by the flourine substituted compound then the chlorine substituted compound. Therefore, the p- flouro benzoic acid is more acidic than the p- chloro benzoic acid.

Read the entire question. I've addressed this.

You rather ignored the inductive effect.

In most cases, when halogens are attached to aromatic systems, the Inductive effect dominates over the Resonance effect. Also, due to resonance, the carbon at para-position acquires a partial positive charge which the halogens destabilise due to their -I effect. So, the resonance of the C=O bond with benzene weakens and the negative charge delocalisation becomes better. Hence, the correct order should be
p-flourobenzoic acid > p-chlorobenzoic acid > benzoic acid.

the carboxylic acid group is too far away from the halogen for inductive effect to have any significance

@xasthor the carboxylic group does not have to be near to the halogen. the +ve charge comes to the para position by resonance.

Okay, that makes sense. I'll accept this answer once a few more people confirm this(via upvotes)

I think you meant "-I effect", that means electron-withdrawing by inductive effect. You wrote "+I effect" which refers to electron donation.

"due to their +I effect. " You definitely meant the -I effect instead...

The words you are looking for: negative, positive. Please don't use your lab jargon here.


text:
Do bond lengths vary across molecules of the same kind?
I only have superficial knowledge of chemistry, and today I found out about bond length for the first time. For example, on Wikipedia it says that all C-C bonds of benzene have a length of 140 picometres (both C-C and C=C, I think). But is this exactly true for all benzene molecules?
Or does bond length of C-C vary slightly for 10 molecules of benzene? Is 140pm an average distance for a million molecules, or each and every benzene molecule has bonds of length 140pm? Is this a fundamental and unvarying length of the benzene molecule, so that a different length cannot possibly occur?
I know that during studies of the bond length in the laboratory, each individual measurement has a slight error. I'm not asking about that.

"All C-C bonds have the same length" is about as precise as "All men have the same height".

''both C-C and C=C, I think''  there are no fully "single" or "double" bonds in benzene, otherwise they wouldn't have equal lengths.

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

This is very interesting. How much does a bond length fluctuate? And how fast? Does it fluctuate more-or-less the same for all molecules of a kind (like all benzene molecules)? The same, meaning they have very similar minimums and maximums and standard deviation etc. Also, temperature and gas pressure will affect the fluctuation, right?

@CamilB You might like to read about infrared spectroscopy - as that relies on those vibrations, explanations of it include information about the vibrations of interatomic bonds.

@CamilB , good idea by Andrew. Another good resource, though doesn't answer all you questions, can be found here.

@airhuff That animation of the $CH_2$ group rocking and wagging made my day.

@CamilB , likewise ;)

@airhuff How do we take into account that a molecule have different vibrational modes? I mean for $υ=0$ or $1$ we should expect different expectations values.

In a simple molecule such as oxygen, the bond length measured will alway be the same as the interaction between any two oxygen atoms is the same since one oxygen atom is indistinguishable from another. Thus by extrapolation in any molecule the same bond has the same length.
Now by length we usually assume equilibrium bond length and this is because in every molecule the bonds vibrate and, obviously, the bond length is instantaneously changing.
Note that most experiments are done on a very slow time scale such that the vibrational motion (period of a few femtoseconds, $10^{-15}$ s) is completely averaged out. The range of bond extension for the lowest vibrations is small just a few percent of the total length.  You can work this out from  spectroscopic results.
The equilibrium value, i.e that at the minimum energy is usually defined as the bond length although one could take the average which in most cases differs very little from this.

This answers all my remaining questions, thank you!


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Why does increasing the temperature of a *solid* increase its entropy?
I understand why increasing the temperature of a fluid would increase its entropy, as the particles are free to move, and therefore an increase in kinetic energy would allow the particles to move more, thus increasing the fluid's entropy.
However, this doesn't make sense to me in the case of a solid - increasing the solid's temperature would result in an increase in K.E., but this would only result in the particles vibrating more. As entropy is defined as the 'disorderliness' of a system, and the molecules in a solid always remain in the same 'order'  regardless of temperature, how is temperature proportional to entropy in a solid?

All right, someone bearing the standard of thermodynamics will give you the equations shortly... From a layman to another, here goes my attempt at a simpler explanation.
Entropy may be seen as the "disorderlyness" in some settings, but that is not a very useful way of seeing it. The metaphor is often used, but creates the wrong conclusions when looking closer at it. To me, is best to look at it as a quantity closely related to the number of possible (different) states available to a certain system. It is then obvious that if you increase the temperature, more modes of vibration are available, thus more states, ergo more entropy. The more stuff is in your room,  the less probable is the state where everything is ordered - if all states are equal in probability ;)

This pretty much summarises what happens, I would add that the entropy increases due to the fact that the number of energy levels available is increased (due to temperature increase) and so the number of ways of arranging the energy into these levels clearly increases and this is what the entropy measures so this also increases. Don't think of energy as 'disorder', which is rather a vague thing, but more correctly as the number of ways of placing the energy into all  the available energy levels. Thus it applies to everything in the same way.


text:
Reaction of iron(III) oxide with nitric acid
I'm currently working in a project in which I have to convert the iron oxides present in iron ore slime into ferric nitrate.
Iron ore slime is which has the particle size of below $\pu{0.15 mm}$ is being discarded as waste during the mining and processing stages iron ore and it will be stored at the tailing dam. It is estimated that $18~\%$–$25~\%$ of tailing will be generated during the processing of iron ore. The major compositions of iron ore slime are hematite, quartz, alumina, mica. So basically, it consists of ferric oxide and alumina, silica, magnesia, calcia, etc. impurities. I wanted to extract the iron part of the slime by converting it into water soluble ferric nitrate leaving behind the insoluble impurities. Even if some soluble impurities like clacium nitrate is formed then its not a problem.
I tried treating the slime with nitric acid both concentrated and dilute but no reaction was observed.
Please suggest me some useful ways to get ferric nitrate from iron ore slime — or any other steel industry waste product like mill scale, slag etc.

Yes, reacting iron(III) oxide with nitric acid would yield iron(III) nitrate.
$$\ce{Fe2O3 + 6HNO3 → 2Fe(NO3)3 + 3H2O}$$

Iron(III) oxide react with nitric acid to produce iron(III) nitrate
  and water. Nitric acid - diluted solution. This reaction takes place
  slowly.(source)


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Why is absorbance the negative logarithm of transmittance?
What is the physical significance of absorbance? Is it not possible to directly decipher the amount of radiant energy absorbed by the sample from transmittance? Why is that negative log of Transmittance is called absorbance?

I think you should post one question per one post.

The are all closely related, stemming from the main question.

Absorbance is useful because it is additive. That is, it's absorbance which is used in Beer's law:
$$A = \epsilon \cdot c \cdot l$$
While you can certainly make a version of this law which uses transmittance instead of absorbance, the math becomes much less straightforward.
The explanation of why absorbance is additive falls out of the derivation of Beer's law. The official derivation relies on calculus, but you can get a sense of things by thinking of the solution not as a single thing, but as a succession of "slabs":
Take a single thin slab of a colored material. This absorbs a certain amount of light at a given frequency. But it doesn't absorb a fixed amount of light. Instead, each photon that passes through has a certain probability of encountering an absorbant molecule and being absorbed. The more light, the more absorbed. For a given slab under typical conditions, we can define the percent transmittance (%T), which is the fraction of light that goes through that material.
But what happens if you start varying the width of the material. Or equivalently, if you stack several of such slabs together. The amount of light going through one slab would be %T. The amount going through two slabs would be %T times %T. (For example, if the slab had a %T of 50%, then the first slab would absorb half the light. The second slab would only "see" half the original light, and would then absorb half of that light. So the amount being transmitted out the back of the second slab would be only 0.5*0.5 = 25% of the total amount.) Three slabs gives you %T * %T * %T. Generally, for N slabs, you'd get only $(\%T)^N$ of the light coming through. (Note the necessity of converting from percentages to fractions to do the calculation here.)
Okay, but how about varying concentration? Well, since the chance that a photon is absorbed is proportional to the chance it interacts with an absorbing molecule, then if you double the concentration, you double the number of interactions that could happen. If you work out the math, it turns out that (in dilute solutions) doubling the concentration is equivalent to just doubling the path length - you effectively have two "slabs", just compressed into the space of a single slab. The effect is the same as increasing the path length. So for a solution at twice the concentration, you's have $(\%T)^2$ the transmittance.
Note the use of exponents. This is all because when you combine two things which absorb, you have to multiply their percent transmittance. Exponents are hard to work with. So instead, we do a trick which can convert mulexponentstiplication into addition: take a logarithm. If we work with logarithms of the transmittance, all of those multiplications are now additions, and instead of using exponents for path length and concentration, we multiply by each. Taking the negative log is just another trick to make things work out reasonably: it's more convenient to have the measurement move in the same direction as concentration and path length (such that more absorbance means a larger concentration), so we stick a negative sign on the result of the logarithm to make things easier.

That's basically it. You certainly could do all your calculations with transmittances, but the equations and reasoning gets more complicated. If you use absorbances, you get the nice Beer-Lambert law, which states the concentration and path length are proportional to absorbance.

Beer's law relates  transmitted intensity to concentration $[c]$ of the solution at each wavelength $\lambda$ as
$$ I_{trans_\lambda}=I_0e^{-\epsilon_\lambda [c]L}$$
where L is the sample path-length and $\epsilon$ the extinction coefficient of the molecule at wavelength $\lambda$ for an incident intensity $I_0$. (The extinction coefficient measures how strongly the molecule absorbs at each wavelength). Thus taking logs produces $$ \ln(I_{trans_\lambda}/I_0)=-\epsilon_\lambda [C]L$$  The product $\epsilon [C]L$ is also called the optical density which can be called the absorbance A. Thus Beers law can be written as
$$ I_{trans_\lambda}=I_0e^{-A_\lambda}$$
It is clearly matter of convenience to use absorbance rather than transmittance as this is directly proportional to concentration. (In ir spectroscopy however, transmittance is still commonly used.)
[ Note that it is usual in practice to define the Beer Lamber law with a power of 10 and not e (as above) thus $\log_{10}$ is used rather than $\ln \equiv \log_e$]

Transmittance (T) is the fraction of incident light which is transmitted. In other words, it’s the amount of light that “successfully” passes through the substance and comes out the other side.
It is formulated as T = I/Io, where I = transmitted light (“output”) and Io = incident light (“input”).
Source: https://www.sigmaaldrich.com/
From the above logic, absorbance (A) should have been defined as the fraction of incident light which gets absorbed provided other phenomena such as reflection, refraction etc. are neglected. In other words, "A = (Io-I)/Io" or "A = 1-T". However, this is not the case in reality for reasons unknown to me.
Optical density (OD), a related but different concept than absorbance, measures the ability of a material to slow the transmission of light due to overall absorption, reflection, refraction, diffraction and scattering of light when it interacts with the medium.
Source: https://www.azooptics.com/Article.aspx?ArticleID=798
https://science.nasa.gov/ems/03_behaviors
"In 1888, Hurter defined the (optical) density as the natural logarithm of the opacity, which is defined as 1−(I/Io) and equal to the absorbed fraction of light, if reflectance is neglected."
Source:https://www.ncbi.nlm.nih.gov/pmc/articles/PMC7540309/
If I understood Hurter's definition correctly, then opacity i.e. 1−(I/Io) = 1 - T is equivalent to "absorbed fraction of light" if reflectance is neglected. In this context, I would have defined absorbance (A) = 1 - T.
From Hurter's definition it further follows that,
OD = ln (opacity) = ln[1−(I/Io)] = -ln (I/Io) = -ln (T), if reflectance and other light interacting phenomena are neglected.
Absorbance (A) on the other hand is solely concerned with the ability of material to absorb light. In spectrophotometry, one generally excludes events like scattering, refraction and reflection. Lambert formulated absorption of light as:
-ln (T) = αL, where α is its Napierian absorption coefficient and L is path length traversed by light through medium.
Later Beer Lambert's law was further developed and took the form as follows and absorbance (A) was defined (if reflectance and other phenomena were neglected):
-ln (T) = ϵCL = A (mathematically equal to OD),
where ϵ molar extinction coefficient of material for a given wavelength, C is concentration and L is path length.
Conclusion:
If I went by intuivity, I would have defined absorbance as fraction of absorbed light. It would be represented by the formula A = 1 - T if reflectance and other events are neglected.
However, absorbance is defined as A = -ln (T) and is equal to OD if reflectance and other events are neglected.
This is my level of understanding and I may be wrong. It would be great to have a real expert to explain these concepts.


text:
Why do lower oxygen concentrations lead to an increase the rate of corrosion?
Is this only the case in pitting corrosion?
Is it because metal which has access to the oxygen becomes cathodic leading to the inaccessible parts of the metal becoming anodic?


text:
Effect of Base on E2 product

One base is bulkier than the other, that's my observation. How does it affect the product? Could someone please give proper reasoning and the product? Thanks a lot.
P.S.
I guess both follow E2

With methoxide there is a reasonable chance of getting nucleophilic substitution as well as elimination generating a mixture of products.  With the much bulkier t-butoxide this is pretty well non-existent so you only get elimination.

It's also possible that the stronger base will abstract the most accessible proton, producing a less stable (kinetic) product.
The Zaitsev (thermodynamic) product comes from the least accessible proton in your case.

You might be under E1 conditions and not E2.

Under $E_2$ reaction conditions, the MeO$^-$ product will be

while the Me$_3$CO$^-$ product will be
.
This is due to the bulkiness of Me$_3$CO$^-$ being unable to reach the hydrogen at the substituted carbon so instead it goes to the primary position instead.

Do you have references that support this black/white distinction?


text:
What should you couple with the reference material in testing galvanic corrosion?
I am currently studying the effect of metal coating to steel. I want to test the galvanic current between steel and aluminum by getting the current that passes through a wire that connects them in an electrolyte solution, to know if the specific aluminum will fit as a good coating for the steel. What should I use as the couple material for the steel? Should it be the coated steel or the metal to be coated? I'm using ASTM G71 as the reference method.

Wikipedia states:"Galvanic corrosion (also called bimetallic corrosion) is an electrochemical process in which one metal corrodes preferentially when it is in electrical contact with another, in the presence of an electrolyte." What are the two metals? How would we know??!! The general answer is:"They are the two metals which form the galvanic cell (in contact with the electrolyte)." Depending on your exact circumstances, this might be the metallic coating and the base metal, or it could be the metallic coating and the metallic coating,or even some other metal component in electrical contact. Of course, if the metallic coating is perfectly homogeneous and completely encloses the base metal, then there will be no galvanic corrosion (at least, not until other corrosion mechanisms have changed that situation).

I edited my question to one of your inquiry. But I think that you haven't answered my question.


text:
Shape of sp³ hybrid orbital
I have a doubt regarding the shape of the $\mathrm{sp^3}$ hybridised orbital. I looked up for it on stackexchange and found a post where the guy said that where ever the orbitals interfere with opposite signs of the wavefunction they cancel out and hence we get a small lobe .
He further said that when the right side of the p orbital overlaps with s orbital( i have linked the post) they both add up.
But the $\mathrm{s}$ orbital wavefunction can also have negative sign (in the post the guy has assumed that s orbital wavefunction is always positive) . So in all four interference combinations are possible : ++ , +- , -+ , - - . Hence the shape should be different  but its not.
I am really confused about this.
Any help is appreciated.
Thanks in Advance.
Link to original post : https://chemistry.stackexchange.com/a/15909

Don't put too much trust in the absolute signs of wavefunction, for they all are arbitrary anyway. Look at it this way: an s orbital has one sign (*). One lobe of p orbital has the same sign, so when they add up, it grows huge. Another lobe inevitably has the opposite sign, so when they interfere, it is reduced to a tiny pig tail.

(*) This is not quite true if we consider radial nodes. That's when things get really hairy.

I think this answer says what I meant to say. I'll delete mine, you were 33 seconds earlier, after all =D

Great minds think alike, they say.

but what will happen if s has two signs ( positive and negative ) . Thanks

That can't happen. Every s orbital looks from the outside exactly the same: as a nice furry ball of one color (one sign, that is).

@user8167818 https://en.wikipedia.org/wiki/Atomic_orbital#Qualitative_understanding_of_shapes is a good way of trying to understand orbitals, IMO.


text:
Must a good leaving group be a weak base, or a weak nucleophile?
In LG Wade's textbook, he has written that a good leaving group must be a weak base. Since substitution reactions (not too sure about elimination reactions) tend to be kinetically controlled, i.e irreversible, why has the concept of a weak base (which is defined for thermodynamically controlled reactions) been used here? Why does the leaving group have to be a weak base rather than a weak nucleophile?

It is a common misconception that good leaving groups should be weak nucleophiles. However, there is a simple counterexample: iodide is known to be both a good nucleophile and a good leaving group in $\mathrm{S_N2}$ reactions.
However, it does make sense for the leaving group to be defined in thermodynamic properties. Consider the reaction equation for a nucleophilic substitution following an $\mathrm{S_N2}$ mechanism:
$$\ce{Nu- + R3C-Lg -> [Nu\bond{...}CR3\bond{...}Lg]^\ddagger -> Nu-CR3 + Lg-}\tag{1}$$
Nucleophilicity, i.e. the ability of $\ce{Nu-}$ to attack the reaction centre is, as you noticed, a kinetic property: the nucleophile has to approach the electrophilic carbon atom from a certain direction in a certain angle with appropriately constructed orbitals. The leaving group itself does play a slight role here in determining the energy of the $\unicode{x3c3}^*$ orbital and thus whether it is energetically favourable for the nucleophile to attack.
However, the leaving group’s principal characteristic is to diffuse away after the reaction. If it is diffusing away, its kinetic properties no longer influence how the reaction proceeds as it is not attacking anything anymore. Rather, a favourable overall reaction enthalpy is achieved if $\ce{Lg-}$ is able to stabilise the negative charge well (or lone pair, if e.g. $\ce{Lg} = \ce{NR3+}$). This is indeed a thermodynamic property, not a kinetic one. And stabilising a negative charge directly correlates with being a weak base.

What exactly do you mean by "diffuse away?" Also, if the leaving group is a strong nucleophile but weak base like iodide, what stops is from attacking the nucleophilic centre?

@xasthor Nothing — which is why nucleophilic attacks that insert an iodide functionality are often re-attacked by iodide. However, in most substitutions that involve iodide, whatever the nucleophile is will be a much worse leaving group and thus the iodide cannot re-attack to induce the back reaction. ‘Diffuse away’ just means ‘stays in solution’, i.e. does not participate in the immediate reaction any more.

Is this correct: The entire reason substitution reactions are irreversible is because having a much weaker base than the attacking nucleophile as a leaving group, drives the reaction entirely to the right & hence good leaving groups are weak bases. This LG is so much worse a base than the nucleophile that the reaction is always driven to completion and only the kinetics of the reagent(i.e its nucleophilicity) matter. Hence we check the nucleophilicity of the reagent, but basicity of the LG (But this statement also implies that bad leaving groups make substitution reactions reversible)

Side question: are elimination reactions irreversible too?

@xasthor Let me start by pointing out that $\text{kinetically controlled} \ne \text{irreversible}$. Substitution reactions are typically irreversible either because the leaving group is a bad nucleophile or because the nucleophile is a bad leaving group. But forward and backward reactions need to be examined separately. It may even be that upon change of conditions (e.g. lowering of pH) a substitution can be selectively reversed.

Also, the same thing applies for elimination reactions, too. Depending on the conditions you can eliminate a leaving group or add one onto the double bond.

"Substitution reactions are typically irreversible either because the leaving group is a bad nucleophile." Why did you use the word nucleophile here rather than base?

@xasthor Because once we look at the reverse reaction, the roles are flipped.

Is this correct:
Let's take two compounds A, B. Both are weak bases relative to the reagent i.e the reaction is irreversible. A is a stronger base but less nucleophilic(perhaps due to sterics) and vice versa for B. Now when A is the leaving group, the TS is less stable compared to B hence its speed of reaction is decreased. Their relative nucleophilicity, in this case, doesn't matter because both reactions are irreversible.

@xasthor I don’t get that. By the way, you can come to [chat] to discuss this further.

Let us continue this discussion in chat.

@Jan Then , Does the solvent nature effect the leaving group ability ? as solvent will be able to stabilise the leaving group that just diffuses off.

LG Wade has put those concepts in another direction, but his explanation is valid and have put it across in the following discussion:
What makes a good leaving group?
When the C-X bond breaks in a nucleophilic substitution, the pair of electrons in the bond goes with the leaving group. In this way, the leaving group is analogous to the conjugate base in a Brønsted-Lowry acid-base reaction. (In order to act as a proton acceptor, a base must have a reactive pair of electrons)

Image illustrating the congujate base-conjugate acid formation
When evaluating the stability of the conjugate base that resulted from the proton transfer, basic concepts used include:

resonance - delocalization of electron density has a stabilizing effect, and the greater area over which the delocalization is possible, the greater the stabilization.
inductive effects e.g. electron withdrawing groups such as chlorine help to further spread out the electron density of the conjugate base, has a stabilizing effect.

These are the same principles when dealing with leaving groups.
In other words, the trends in basicity are parallel to the trends in leaving group potential - the weaker the base, the better the leaving group. Just as with conjugate bases, the most important question regarding leaving groups is this: when a leaving group leaves and takes a pair of electrons with it, how well is the extra electron density stabilized?

In laboratory synthesis reactions, halides often act as leaving
  groups. Iodide, which is the least basic of the four main halides, is
  also the best leaving group – it is the most stable asa negative ion.
  Fluoride is the least effective leaving group among the halides,
  because fluoride anion is the most basic.

Anomaly trend of fluorine explained
The more electronegative an atom, the better it is able to bear a negative charge. Because fluorine is the most electronegative halogen element, we might expect fluoride to also be the least basic halogen ion.
But in fact, it is the least stable, and the most basic! It turns out that when moving vertically in the periodic table, the size of the atom outdoes its electronegativity with regard to basicity. The atomic radius of iodine is approximately twice that of fluorine, so in an iodine ion, the negative charge is spread out over a significantly larger volume)
Some examples:
Chlorides, bromides, and tosylate / mesylate groups are excellent leaving groups in nucleophilic substitution reactions, due to resonance delocalization of the developing negative charge on the leaving oxygen.

Acknowledgements
Organic Chemistry with a Biological Emphasis (T Soderberg)


text:
What term should I use when I am quantifying the amount of matter, mass or weight?
I am writing an academic article but I got confused as I am writing about the weight a specific metal got by using balance scale. Should I use the term mass, which has a unit of kg, or weight, which is a force created by gravity and expressed in N?

As long as your substance never leaves Earth, who cares?

Well, maybe it is about E Musk's travel towards space, Moon, Mars ...

If you're being technical, note that amount of matter (or substance) is already a term in its own right. We generally measure this value in moles.

Weight is not expressed in N, but gram or kg. Weight is measured via the gravitational force, but it is not one. A common anglosaxon unit for a force is gf, gram force. Weight and mass use the same units, although they are only identical on the surface of an idealised earth.

@Karl any physics textbook would say you're talking rubbish. Weight is a force. Force = mass x acceleration. A mass of 1kg weighs 9.8N on earth and 1.6N on the moon, due to the respective gravitational accelerations of 9.8m/s^2 and 1.6m/s^2. If you're going to be a Pedant, get it right. All that said, in nonscientific English (and even in chemistry, where we don't need to make the distinction) it would probably be acceptable to say something has a weight of 1kg instead of a mass of 1kg. In particular, saying "this weighs 1kg" is much less awkward than the more correct "this has a mass of 1kg".

Because of the reasoning provided by you, i.e. mass is universal, and weight is the product of mass and magnitude of the local gravitational acceleration, I suggest to stay at mass.  (A purist's view, taking @Ivan's comment into account.)
In addition, referring to the International System of Units, their brochure not only mentions the basic units, but the relevant section 2.1.1.2 starts with:
Unit of mass (kilogram)

It's up to you, ultimately. The Wikipedia entry for weighing scales states that:

The balance (also balance scale, beam balance and laboratory balance) was the first mass measuring instrument invented.

Further:

A balance or pair of scales using a balance beam compares masses by balancing the weight due to the mass of an object against the weight of a known mass or masses

and

the balance or pair of scales using a traditional balance beam to compare masses will read correctly for mass even if moved to a place with a different (non-zero) gravitational field strength (but would then not read correctly if calibrated in units of force).

In comparison to a spring balance:

Either type can be calibrated to read in units of force such as newtons, or in units of mass such as kilograms.

Finally:

Technically, a balance compares weight rather than mass, but, in a given gravitational field (such as Earth's gravity), the weight of an object is proportional to its mass, so the standard "weights" used with balances are usually labeled in units of mass (g, kg, etc.).

I think you have a good reasoning, but with the fact that you had used Wikipedia as a reference source disappointed me. But still, thanks.

Well, the reference is as good as any, and does in fact address the specific issue of using a balance scale, which is not addressed by others. Your query is actually something that will elicit answers that are opinion-based. The first sentence in my answer is the relevant one: it is up to you as the author of the publication. You might get feedback from referees concerning what choice you make.

Mass. Weight is mass in a gravity field or acceleration frame.


text:
When is it necessary to check wavefunction stability in density functional theory?
In Gaussian, there is a stable keyword that checks the stability of the wavefunction. Using stable=opt reoptimizes the wavefunction until a stable one is found if there is an instability.
My question is - when do you use this stability check? Presumably one must have some hunch in advance that the wavefunction may be unstable in order to use this keyword, but what insight would lead one to this conclusion?
To add some context, this question came to mind after following instructions on how to model a system with antiferromagnetic coupling (e.g. as shown here). When I tried this procedure on cupric acetate hydrate, which has a dinuclear structure of copper(II) atoms and is known to be antiferromagnetic, I had to use the stable=opt keyword to stabilize the wavefunction resulting from the "fragment guess job" otherwise it erroneously appeared that the antiferromagnetic singlet was higher in energy than the triplet. This made me wonder when I should be using stable=opt in general (perhaps outside the context of antiferromagnetic coupling calculations) and also what it is specifically doing.

"Gaussian" says: Note that analytic frequency calculations are only valid if the wavefunction has no internal instabilities.

Yes, this definitely points to its importance. That being said, rarely have I seen people check the stability of the wavefunction, which led me to ask this question since people must have some idea the wavefunction may be unstable in order to check its stability.

No idea how standard it is, but in my old group, they checked for stability every single time for their open-shell systems, a routine maintenance like checking for imaginary frequencies. These were mononuclear iron or nickel complexes with a ~30 atom, pyridine/phenylene-diamine-based hexadentate ligand.

The very short answer is: whenever you are in doubt.
The stability check is probably unnecessary when you are dealing with closed shell molecules, where you can be pretty sure that a restricted methodology is reasonable. For example, when you calculate water at the RHF level of theory, you won't really need to look for a more stable wave function, because it would not be physically sensible. An UHF approach in principle will still lead to a lower energy since it has more flexibility.
Whenever you are calculating molecules in which you would assume some open shell character you should be considering checking wave function stability. It usually doesn't hurt to perform one check too many. Discovering instabilities later can be a lot more frustrating. That does not necessary mean that you are gunning for the correct ground state, as you are still restricted on spin. I'll include a small, but well-known example below: $\ce{O2}$. I am using Gaussian 09 Rev. D.01 for the calculation, but it should work with anything more recent. (It won't work the same way with anything older.) Also note, that I am not conducting geometry optimisations, but use $\mathbf{d}(\ce{O-O})=\pu{120pm}$ to keep it simple.
First let's run an ordinary, restricted calculation. If you are not using the SCF(xqc) keyword, you should immediately notice, that something is wrong.

%chk=bp86svp.rhf.sing.chk
#p RBP86/Def2SVP/W06
DenFit             ! Use density fitting to spead up calculation
gfinput            ! Prints basis set in a form suitable for use as general basis set input
gfoldprint         ! Prints the basis set information in the Gaussian format
iop(6/7=3)         ! Switches on printing of all MO
symmetry(loose)    ! Use correct symmetry
scf(xqc,conver=10) ! Quadratic convergence after conventional (for tough cases)
INT(SuperFineGrid) ! Largest grid, most accurate

title card unused

0 1
O  0.6 0.0 0.0
O -0.6 0.0 0.0

You'll find a summary of the energies at the end, if you want to check and run your calculations.
Next up, we'll check and optimise the wave function.

%chk=bp86svp.rhf.sing.stab.opt.chk
%oldchk=bp86svp.rhf.sing.chk
#p RBP86/Def2SVP/W06
DenFit
geom=check guess=read
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
scf(xqc,conver=10) INT(SuperFineGrid)
stable=opt

title card unused

0 1

You should find the following in your output:

 The wavefunction has an RHF -> UHF instability.

Followed by another SCF run. Pay close attention to the following:

 SCF Done:  E(UB-P86) =  -150.201660211     a.u. after    6 cycles
            Convg  =    0.3033D-11                    52 Fock formations.
              S**2 =  1.0016                  -V/T =  2.0049
 = 0.0000 = 0.0000 = 0.0000 = 1.0016 S= 0.6187
 = 0.000000000000E+00
 Annihilation of the first spin contaminant:
 S**2 before annihilation     1.0016,   after     0.0126

You are still finding a singlet state, but now as a UHF wave function. You should later find:

 The wavefunction is stable under the perturbations considered.

What you have found is a broken symmetry singlet. We do know that the ground state should be a triplet. Therefore we'll perform a restricted open shell calculation. We already know this one will have and RHF -> UHF instability, and since we cannot use the quadratic convergence, we cannot use stable either. But it'll provide us with a good starting point for the UHF calculation, or at least an energy to compare others to.

%chk=bp86svp.rohf.trip.chk
%oldchk=bp86svp.rhf.sing.chk
#p ROBP86/Def2SVP/W06
DenFit
geom=check guess=alpha
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
scf(conver=10) ! Quadratic convergence not available for RO
INT(SuperFineGrid)

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The important part here is, that it generates a pure spin state wave sunction:

 = 0.0000 = 0.0000 = 1.0000 = 2.0000 S= 1.0000

Another calculation is using the the mixing of HOMO and LUMO to generate a broken symmetry ansatz wave function directly. You'll find the same result as with the approach above.

%chk=bp86svp.uhf.sing.chk
%oldchk=bp86svp.rhf.sing.chk
#p UBP86/Def2SVP/W06
DenFit
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
geom=check guess=(read,mix)
scf(xqc,conver=10)
INT(SuperFineGrid)
stable=opt

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You'll therefore find again:

 SCF Done:  E(UB-P86) =  -150.201660211     A.U. after   11 cycles
            NFock= 11  Conv=0.88D-12     -V/T= 2.0049
 = 0.0000 = 0.0000 = 0.0000 = 1.0016 S= 0.6187
 = 0.000000000000E+00
 KE= 1.494731832724D+02 PE=-4.114759744301D+02 EE= 8.357834648903D+01
 Annihilation of the first spin contaminant:
 S**2 before annihilation     1.0016,   after     0.0126

The wavefunction is stable under the perturbations considered.

Now for the last part, run a real UHF calculation:

%chk=bp86svp.uhf.trip.stab.opt.chk
%nproc=8
%mem=8000MB
%oldchk=bp86svp.rhf.sing.chk
#p UBP86/Def2SVP/W06
DenFit
geom=check guess=read
gfinput gfoldprint iop(6/7=3)
symmetry(loose)
scf(xqc,conver=10)
INT(SuperFineGrid)
stable=opt

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You'll find as expected

 Annihilation of the first spin contaminant:
 S**2 before annihilation     2.0031,   after     2.0000

 The wavefunction is stable under the perturbations considered.
 [...]
 The wavefunction is already stable.

For giggles, why not consider a quintet?

Command file              Functional            Energy / Hartree ( cycles )
bp86svp.rhf.sing          E(RB-P86)       =       -150.155030787 (      4 )
bp86svp.rhf.sing.stab.opt E(UB-P86)       =       -150.201660211 (      6 )
bp86svp.rohf.trip         E(ROB-P86)      =       -150.213771982 (     10 )
bp86svp.uhf.sing          E(UB-P86)       =       -150.201660211 (     11 )
bp86svp.uhf.trip.stab.opt E(UB-P86)       =       -150.215679340 (      9 )
bp86svp.uhf.quin.stable   E(UB-P86)       =       -149.651916836 (      9 )

In conclusion, whenever you suspect some open shell character, check for stability. If you have well-behaved molecules (singlet), like most organics are, you can probably skip it.
Be aware that a stability optimisation will not necessarily give you the ground state.

Thank you so much! This is extremely helpful and just what I was looking for. I never thought about checking for stability, but all of my systems are open shell, so I suppose I should get in the habit of just checking it seems.

Speaking anecdotally, I once ran a bunch of calculations on an open-shell nickel system,  only to find out after my first 20-30 calculations that half of them weren't stable. The difference post-correction was also quite noticeable. The good thing was that the stable calculations went pretty fast relative to the main calculations, and after correcting an instability the system reliably remained "stable" through further optimization, if that's any consolation.

That's exactly what I'm looking to avoid since I'm just starting this project on an open-shell copper system, so the anecdote is quite helpful, thank you for sharing. It will certainly be worth me just running the stability check and playing it safe!


text:
Measuring Cu2O Thickness in Oxidized Copper Sheet
I have Cu2O fabricated using thermal oxidation of Cu sheets in air at  high temperatures. The thickness of Cu2O is expected to be at least 100 microns. There is still plenty of unoxidized Cu metal inside the Cu2O, so cleaving it to perform cross-sectional SEM is not trival. It is also too thick for spectroscopic ellipsometry and XRR.
What are some other methods of measuring the thickness of this Cu2O thermal oxide layer?

I like your idea to follow up the oxidation of metallic copper by analysing a by means of X rays. If you find a way to obtain a nice cross-section of your plates, I would suggest to analyse them however by means of EDX, where electron beams are used to excite the material locally, which relaxes by the emission of X rays of characteristic energy.  This will offer you both a qualitative, as well as a quantitative characterisation of your sample.  Monitoring corrosion is a recurrent task (cf. here) for this technique, recording concentration profiles, too.
In the easiest case imaginable, your scan trace would simply recognise "high level" (for pure copper), "low level" (for the adlayer of $\ce{Cu2O}$, lowered because of the presence of oxygen) and "background level" (around your plate, practically equal to zero), as schematically depicted in the following drawing.  Assuming you follow the concentration of copper as indicator, the higher the concentration of copper, the higher the read-out signal (the intensity of the $K_\alpha{}$ X-ray generated):

You mentioned the layer of $\ce{Cu2O}$ is not homogeneous, likely contains grains of unreacted copper.  Tentatively I tried to depict this situation in the following drawing, where the darker polka dots are meant to be representative for remaining starting material:

You may, depending on the scan trace chosen, encounter earlier or later such a grain.  Depending on the relative size of this grain, the beam diameter, the distance towards the pure copper matrix, the read-out signal will be influenced more, or less; hence the attenuation of the signal likely will not be as smooth as idealised here. Yet while scanning along the $x$-direction, it is possible to record $i$) $x_1$ when the high-level read-out decreases, and $ii$) $x_2$ when you reach the surface of the mixture of $\ce{Cu2O}$ and $\ce{Cu}$ towards the vacuum chamber, and hence to measure the thickness of the oxide layer.


text:
Mendeleev's predictions
While studying about Mendeleev's periodic table in my textbook( class 11 NCERT chemistry), I have studied that Mendeleev has predicted some elements like Eka-boron(scandium), Eka-Aluminium(Gallium),Eka-Silicon(Germanium)etc. He also predicted their physical properties which were most accurate. One such element about germanium is given in the below link.
How could he predict such accurate values of such physical properties?

Why, that's simple, and you can do it yourself: he just averaged the properties of the neighboring elements which were already known.

Mendeleev's predictions were not accurate, but rather *not too far off".  As expected. When you find something surprising in natural sciences, that usually means your previous thinking was stupid, and you should have known. ;-)

This is an interesting question. I hope to see someone explain how Mendeleeev got the boiling points.

Mendeleev did little more than a bit of educated guessing. He was the only one able to do that, not because he was so very educated (he probably was), but because his periodic table gave him the basis to guess from. ;)

Well, guess what, even I study the exact same textbook (class 11 NCERT, from India). I had asked my teacher the exact same doubt, and the answer was:

Due to its proximity with neighbouring substances, like Arsenic (which was already discovered by then - see here), Mendeleev was able to accurately calculate the average values of density and melting point of Germanium, or eka-Silicon.
The formula of its Oxide and Chloride was predicted with respect to the formula of the Oxides and Chlorides formed by other elements of the group IVA (as per old IUPAC numbering), such as $\ce{CO2}$, $\ce{SiO2}$, $\ce{CCl4}$ and $\ce{SiCl4}$. Their densities were calculated using the knowledge of Chlorine's and Oxygen's atomic weight, and the predicted density and/or atomic mass of the new element.

Although I don't know how the boiling point of a substance can be predicted, my best guess for it is that it was approximated using the boiling point of other similar compounds near it in the Periodic table. For more information on how Boiling point can be possibly approximated, here is a link.

Germanium is under silicon $\ce{Si}$ $(Z = 14, A = 28.1)$ and above tin $\ce{Sn}$ $(Z = 50, A = 119.7)$. The average of these values may define Ekasilicium. They are : $Z = \frac{14 + 50}{2} = 32, A = \frac{28.1 + 119.7}{2} = 73.9$.
This is not "too far" from Germanium's value : $Z = 32, A = 72.63$.
Same calculations may be done for density, boiling point of chloride (which is rather low), melting point of oxide (which is rather high), etc. The order of magnitude is always correct.

There could be done in some cases even two-dimensional interpolation from values of the element grid 3x3, like Al-Si-P/Ga-Ge-As/In-Sn-Sb


text:
What is the mechanism of halogen displacement?
We teach (ad infinitum....) that a more reactive halogen will displace a lesser halogen from a compound.  And yes, the halogen is a halide ion in the compound.  Now I think about it, it's usually in solutions, so the halide is nominally solvated and surrounded by water (as hydronium ions?).
But how?
Does the halide lose the electron to give a radical, or does the halogen dissociate to radicals, or to ions, or is there an intermediate Cl2- ion, for example?  I realise this is several questions, but the major question is unaffected by my speculation.  All I have found is the bald statement about displacement and reactivity and the table we always draw.

Can you give an example of the type of compound undergoing this displacement? It seems that you're talking about inorganics, but I can't tell for sure.

I'm guessing your talking about the halogen displacement reactions:
$$\ce{2NaBr + Cl2 -> Br2 + 2NaCl}$$
which is a redox reaction:

$\ce{2 Br- -> Br2 + 2 e-}$, bromine loses electrons (is oxidised)
$\ce{Cl2 + 2e- -> 2 Cl-}$, chlorine gains electrons (is reduced).

For more detail look in to the redox potentials, this explains what will reduce what.

Welcome to ChemSE.  For future reference, it may be interesting for you to familiarise yourself with the formatting tools made accessible, which are pretty helpful for chemical equations (and other aspects, too):  https://chemistry.meta.stackexchange.com/questions/86/how-can-i-format-math-chemistry-expressions-here/88 and https://mhchem.github.io/MathJax-mhchem/.

This doesn't answer the question, which is about mechanism of this redox reaction.