text:
What is the proper Lewis structure for HCOOH?
I tried to draw the Lewis structure of $\ce{HCOOH}$ but I didn't get the correct representation.
As you can see in the picture below, at no. 1, I put $\ce{C}$ as the central atom because it has less electronegativity than $\ce{O}$ ($\ce{H}$ cannot be a central atom). I drew the structure, but it is wrong because it violates the octet rule.
Then I came up with another structure as shown at no. 2,
and it seemed correct to me, but when I searched on Google for the correct structure, I found out that I am wrong.

Why is the structure that I drew in no. 2  wrong, even though the formal charges are zero?
How do I figure out the correct order of atoms in such molecules like the one above. What I mean is how do I know that the $\ce{C}$ will have $\ce{H}$ on the left and one $\ce{O}$ up and $\ce{O, H}$ on the  right side?

With familiarity you will recognize that $\ce{COOH}$ in a formula generally refers to a carboxylic acid group. Sometimes you will also see $\ce{CO_2H}$. Either is acceptable. Vinegar, or acetic acid, is a carboxylic acid. Its formula is $\ce{CH3COOH}$ or $\ce{H3CCOOH}$.

It is important to realize that a carboxylic acid group is NOT a peroxide. Peroxides have  involve O-O bonds. You may also generalize that the oxygens in peroxides have a negative one oxidation state. A common peroxide is hydrogen peroxide, $\ce{H2O2}$ or $\ce{HOOH}$.

Also note that carbon is generally tetravalent - i.e. it is commonly found in stable molecules as having an octet of electrons. In addition, carbon generally exhibits no overall formal charge. So this suggests that most of the time you will see carbon forming four bonds - and this is the case the majority of the time. This configuration gives carbon no net formal charge and fills its octet.
Of course, you may find carbocations or carboanions, but carbocations are highly unstable and exist only ephemerally. Carboanions also exist; see the cyanide ion for an example.
The above, in addition to keeping in mind that a carboxylic acid is a Bronsted/Lowry acid - i.e. proton donor - should enable the construction of an acceptable Lewis structure.

And, of course, the correct structure (one double bonded oxygen and one O) does obey the octet rule.

As for carbon usually having zero formal charge: at least one common compound (carbon monooxide) has carbon has formal charge -1, though actual charges in the molecules are almost zero. This, however, is a very special case.

Good example. One that I thought of in the text was the cyanide anion too.

Formal charges don't tell the whole story.  What atoms are connected to what is the first thing you need to consider. Note that $\ce{HCOOH}$ gives you a hint that the $\ce{OH}$ goes together, rather than the 2nd $\ce{H}$ attaching to $\ce{C}$. (Though the linear formula is still misleading in that both $\ce{O}$ atoms are attached to $\ce{C}$, which seems not to be the case in the formula).

From a sum formula like $\ce{HCOOH}$ it is usually not possible to construct the Lewis structure. In this case, there are some obvious points, why your structures cannot be correct.

One of the oxygens only has an electron sextett. As oxygen is the most electronegative element in this compound, this is just not possible. But you saw this yourself.
In these structures the carbon has only an electron sextett. Since it is much more electronegative than hydrogen, this can also be not correct.

So in praxis sum formulas should always be accompanied with a name, or a schematic drawing.
For example, if you try to formulate a Lewis structure based on the sum formula $\ce{C2H6O}$, you will end up with dimethyl ether or ethyl alcohol/ ethanol.

as long as they don't exceed an octet for period 2 and a duet for period 1 and the charge exists in some molecules it is a valid structure. Thus the first and is valid and the second and third invalid because: 1) the one with 4 bonds to C has neither oxygen exceeding an octet and neither hydrogen exceeding a duet. An alkoxide anion does not make it invalid. Neither does an O^2+(which in fact does exist in things like OF2) and 2) hydrogen methyl peroxide does exist but not with a C^2+ and 3) methanediol exists but again C^2+ does not exist(even though C^4- does)

@caters I do not completely understand your comment. According to your first statement, all of the OP's given structures are valid in the Lewis framework (none of them is likely). In 2 there are no charges at all, just empty valencies. And I am almost certain, that the ones in 2 could be isolated in gas phase matrix close to 0 K. Number 1 is just impossible, since oxygen is the second most electronegative (neutral) element. It will therefore never have an electron sextett. Even in $\ce{OF2}$ it has an octet! And in the Lewis formalism it is not charged at all.

Formal charge in first structure: O up top: -1, O to the right: +1. Formal charge in peroxide: C: 0, O: 0, H: 0. Formal charge in diol: C: 0, O: 0, H: 0. Okay these formal charges are possible for these in other compounds. As long as they don't exceed an octet it is not a violation of the octet rule.

@caters I still do not understand your comment. Could you elaborate a little more please. How would you explain the validity of structure 1?

In the triple bond resonance structure of CO the O has 5 electrons around it giving it a +1 formal charge. The same thing is true for the O in structure 1. O has the equivalent of 3 lone pairs of electrons(that is 6 electrons) to form a bond with. If it uses 1 lone pair out of the 2,2,1,1 arrangement for O to form a single bond(which it does) than the O has used 2 electrons to form a bond leaving 4 nonbonding electrons. If you separate the O and the C by breaking the bond the O is going to have its original 6. Again it is exceeding an octet that is invalid for period 2, not less than an octet.

@caters I believe you are comparing apples with oranges, you also picked one of the most complicated molecules to substanciate your claim. Aparently you also believe, that Lewis structures represent any real bonding situation - which is of course not true, since it is only a very crude approximation. If you think my answer is wrong, then please submit one that is true and downvote mine. I do not think we can resolve this in the comments section.

CO is not one of the most complicated. However a chain of thousands of sugars with branching and everything is.

@caters If you state it like this, then you clearly did not understand the first thing about bonding in carbonmonoxide. There are research groups spending decades on that molecule and its potential. There have probably been more than a hundred publications on the structure and the electronic structure and the time dependency of it. While in a chain of sugars you can hardly find any complicated bonding situation at all, a hydrogen bond is as complicated as it gets, there are not even any pi orbitals.

The physical reason why answer 2 is wrong is because it represents a class of compounds that is very reactive (a carbene) and if the compounds in (2) actually existed, they would probably rearrange very quickly to the carboxylic acid isomers. The reason the rearrangement would occur is because of bond strengths (i.e. the energy it takes to cleave a chemical bond homolytically)--carbon-oxygen bonds are stonger than oxygen-oxygen bonds. See this link for rough numbers on this property

one of them is like methanediol but the C has a 2+ charge. The other is like hydrogen methyl peroxide but again has a 2+ charge on the carbon. Carbide anions do exist(C^4-) but C^2+ ions do not.

@caters You confuse formal charge with oxidation state. In structures 2 there are no formal charges.

Lewis structure:

The properly way to determine the Lewis structure, based on this example, is:

Total valence electrons: $4 + 6\cdot2 + 1\cdot2 = 18$
Total electrons needed for octets/doublets: $8\cdot3+2\cdot2 = 28$
Total shared/bonding electrons: $28-18 = 10$ (In other words, there are only five bonds.)
Total electrons in lone pairs: $\text{Step 1} - \text{Step 3} = 18 - 10 = 8$ (In other words, the are only 4 pairs of lone electrons (all for $\ce{O}$)

Furhter information on how to draw Lewis structures, please follow the link

Oxygen can have less than an octet. Here's why:
In the triple bond resonance structure of $\ce{CO}$ the $\ce{O}$ has 5 electrons around it giving it a +1 formal charge. The same thing is true for the $\ce{O}$ to the right in structure 1.
$\ce{O}$ has the equivalent of 3 lone pairs of electrons (that is 6 electrons) to form a bond with. If it uses 1 lone pair out of the 2,2,1,1 arrangement for $\ce{O}$ to form a single bond (which it does) than the $\ce{O}$ has used 2 electrons to form a bond leaving 4 nonbonding electrons. If you separate the $\ce{O}$ and the $\ce{C}$ by breaking the bond the $\ce{O}$ is going to have its original 6. It is exceeding an octet that is invalid for period 2, not less than an octet.
If $\ce{O}$ has less than an octet than yes it is unstable but it is still possible. This is why positive ions of $\ce{O}$ exist as well as $\ce{O^-}$. There is also a small amount of neutral atomic oxygen (which is a biradical). Carbocations exist for the same reason that oxygen ions with less than 8 electrons do. That reason is because having less than an octet is not invalid.

Formal charge is an idealization that deliberately ignores electronegativity and treats bonds as perfectly covalent, and the comparison with $\ce{CO}$ is simply not germane. Arguing that structure #1 is reasonable by analogy to $\ce{CO}$ is equivalent to my arguing that a neutral carbene is stable because the carbon has the same formal charge as it does in an alkane. The proposed structure is essentially a dioxirane with the $\ce{O-O}$ bond heterolytically cleaved. That's just not plausible. Is the structure legal per the rules of LEDS? Yes. Is it correct? Definitely not.

and that O-O single bond is a peroxide bond so dioxiranes are peroxides.


text:
What do the half-reactions need to be for balancing a redox reaction in an acidic medium?

Balance the following reaction:
  $$\ce{CuS + SO4^{2-} -> CuO + SO2}$$

Below, the number mentioned in the parenthesis is the oxidation number of sulfur.
I have found out that the oxidation half-reactions and reduction half-reactions are as follows:
\begin{align}
\ce{CuS^{(-2)}     &-> ^{(4)}SO2}&& \text{Oxidation half reaction}\\
\ce{^{(6)}SO4^{2-} &-> ^{(4)}SO2}&& \text{Reduction half reaction}
\end{align}
Problem:
In this, what about $\ce{CuO}$ in the product side? I have also learned that we need to balance the spectator ions before charges if the spectator ion is other than oxygen and hydrogen.
Here, there is no $\ce{Cu}$ in reduction half reaction. So, I think there is something is wrong in my reduction half-reaction.

Your "half-reactions" involve the right oxidizer/reducer (oxygen in both cases), but a half-reaction needs to preserve atoms and charges, so you need to try again.  In fact the original equation is just a shorthand -- notice that there are no charges on the left but there on the right, so something is missing. You'll need to add $\ce{H^+}$ on one side and $\ce{H_2O}$ on the other.  Which goes on which side?

Also, does the problem say you have to use the half-reaction method?  I ask because this method is not the easiest way to solve this problem (again, because O is being both oxidized and reduced).  Completing the equation with $\ce{H^+}$ and water and then comparing coefficients for each type of atom and for the charges will be easier (you might want to introduce an unknown, x, on the left side: $\ce{1 CuS + x \cdot SO4^{2-} + \cdots}$)

@SilvioLevy i know this is not the good approach but my teacher told you should learn this so i have show him by solving this reaction only by this method :-)

@SilvioLevy, I don't think oxygen changes oxidation states; it appears to be -2 in all products and reactants. Sulfur, however, is -2 in the sulfide, +6 in the sulfate, but +4 in sulfur dioxide. So, the sulfide is oxidized while the sulfate is reduced.

@Freddy, almost every sulfur redox reaction I've seen is done in acidic solution. Is that the case here? This is relevant, since the half-reaction methods of balancing equations for acidic and basic solution are different.

I have done that only take a look at both half reactions. Should i mention oxidation number in it then it will be more easier

@GregE. yes it is in acidic medium

@GregE. Of course.  And I wrote it twice no less -- clearly I shouldn't be composing answers this late at night. Unfortunately I don't seem to be able to edit those comments anymore.

@SilvioLevy, as a fellow insomniac, I sympathize. It's a trivial error.

@SilvioLevy you know at my place it's still lunch time :-)

@Freddy I see you've written down the oxidation states now.  Comparing the change in oxidation state in the first "half reaction" with the one in the second, can you figure out how many copies of one half reaction should be combined with how many of the other?

@SilvioLevy I think 3 copies of 2nd half reaction should be combined with 1st half reaction. after balancing charge, hydrogen and oxygen.

Perfect.  You're almost done... Now do the balancing and write the answer so we can vote it up!

@SilvioLevy take a look at the answer if you come through any mistake please inform me. Thanks for your help

@Freddy - see comment after the solution.  Nice job.

Below, the number mentioned in the parenthesis is the oxidation number of sulfur.

Writing down half reactions:
\begin{align}
\ce{CuS^{(-2)}     &-> ^{(4)}SO2}&& \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} &-> ^{(4)}SO2}&& \text{Reduction}
\end{align}
Adding electrons in both reactions:
\begin{align}
\ce{CuS^{(-2)}           &-> ^{(4)}SO2 + 6e-}&& \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} + 2e- &-> ^{(4)}SO2}&& \text{Reduction}
\end{align}
Balancing $\ce{Cu}$:
\begin{align}
\ce{CuS^{(-2)}&-> ^{(4)}SO2 + 6e- + Cu^{+2}}&& \text{Oxidation}
\end{align}
Balancing charges of both half reactions:
\begin{align}
\ce{CuS^{(-2)}&-> ^{(4)}SO2 + 6e- +CuO + 6H+}&& \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} + 2e- +4H+ &-> ^{(4)}SO2}&& \text{Reduction}
\end{align}
Balancing oxygen and hydrogen:
\begin{align}
\ce{CuS^{(-2)} + 3H2O&-> ^{(4)}SO2 + 6e- +CuO + 6H+}&& \text{Oxidation}\\
\ce{^{(6)}SO4^{2-} + 2e- +4H+ &-> ^{(4)}SO2 + 2H2O}&& \text{Reduction}
\end{align}
Adding 3 copies of reduction half reaction with oxidation half reaction, gives finally balanced redox reaction:
$$\ce{CuS + 3SO4^{2-}  + 6H+ -> 4SO2 +CuO + 3H2O}$$

(Thanks to Silvio Levy for helping me to solve this problem.)

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text:
Explanation for increase in the number of oxidation states for transition elements from Sc to Mn of the 4th period
From Sc, to Mn, the number of oxidation states increases from one (Sc) to seven (Mn). The explanation for this is because the unpaired 3d electrons can be lost along with the 4s electrons during bonding.
Should this be the case? Wouldn't it be easier to lose an electron when it is paired because of interelectronic repulsion in the orbital? Could someone explain this? Thanks!

After some research I have found an "answer" - apparently paired electrons are less likely to be involved in bonding than unpaired electrons. The question, now, is why?

The oxidation state of a transition metal is an accounting device.  It's not the number of "lost electrons".  In potassium permanganate, which has Mn (VII), the manganese atom shares its seven 3d electrons with the ligand (oxygen) in what are essentially covalent bonds, and the "positive charge" associated with the metal is not very different than what it is for ionic calcium compounds.
Regarding "paired electrons are less likely to be involved in bonding than unpaired electrons" -- both paired and unpaired electrons in a transition metal can participate in bonding, provided they are in the outermost shell.  The paired electrons in inner shells are too deep in energy to take part in bonding.

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text:
Removing the hydrazine byproduct from the mitsunobu reaction
Looking to remove the hydrazine byproduct left after the mitsunobu reaction. Using DIAD and polymer bound triphenyl phosphine. The reduced diad is coeluting with my product on silica. My thoughts for getting rid of it are trying an alumina column, acidic wash (I have an N-Boc so it can't be too acidic), or switching to a different azodicarboxylate.

can you provide drawing of your product?

Unfortunately I cannot include a drawing.

Well, without any idea of what your product is, it is unlikely you will get an answer.

Acidic wash is not going to help. As you said you can try a DIAD alternative for example DCAD can give good results (sry no advertising intended here). There is also the option of taking your contaminated product to the next step(s) and hope you can get rid of the impurity there. I would take option A if I have to make a bunch of similar analogues by a Mitsunobu and B otherwise for not wasting too much time

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text:
Carbonyl/Nitrosonium Complex Stability

Why is the nickel complex exhibiting the strongest pi-bonding?
From what I understand, charge density plays a role. Also we need some empty orbitals on the metal cation, since we're looking at Lewis adducts.
Also I note that some have positive overall charges; could this play a role? Metal cation <-> NO+ repulsion? So we might want a metal anion?
Also would size play a role? Size goes hand in hand with charge density. And what about "backbonding"?

Also we need some empty orbitals on the metal cation, since we're looking at Lewis adducts.

Ahem. Nope. $\pi$-bonding in $\ce{CO/NO+}$ complexes is from back-donation, from occupied metal d-orbitals onto unoccupied $\pi^*$ orbitals of the ligand.
Looks like homework question, so no further details provided.


text:
How could household vinegar create a salt on car carpet?
I am not a chemist! My chemistry classes from my molecular biology undergrad are far behind me. But I think I created a salt, and I don't know how. Here is my mystery:
I recently got some water in my car, and wanted to prevent mold growth in my carpet mats. I brought the mats inside, vacuumed off, and completely soaked them in household vinegar. Then I placed the mats near a running dehumidifier (where the warm air comes out in the back).
White crust appeared on the carpet. At first I thought my dehumidifier was spitting something out, but the white substance was not on any other nearby surface--only on the car carpet. Eventually as the further parts of the carpet dried, they also showed the white crust. It covers pretty much all of the carpet.
Is this, in fact, a salt? If so, what basic substance could have been in my car carpet mat that would cause this reaction? Thanks everyone!

Was a cleaner with baking soda ever applied to the mat?

@brinnb Not to my knowledge (at least not in the last few years). Also, there was no noticeable bubbling when I poured the vinegar over the mat. It seemed to be the application of the heat/dry air from the dehumidifier that catalyzed the reaction.

Is the powder water soluble?  If you heat it in a blowtorch flame what color is it?

@brinnb I can check solubility in water tonight. But I don't own a blow torch or anything resembling it!

The powder is water soluble.

I once had a humidifier and I ran it with tap water.  I soon noticed a build up of white dust on the carpet right in front of the humidifier.  I suspect your situation might be related to what caused my white dust.  Tap water contains salts (just like table salt), the harder the water the more salts it contains.  They are present in low concentration so you don't normally see them.  In my case, as more and more tap water evaporated it led to a visible build up of these salts.  In your case, as the water on the carpet dried, it left behind the same salts.  Maybe you really soaked your carpet with a lot of water, maybe your water is very hard, and finally, maybe salts were actually put into the carpet during their manufacturing process - and maybe washing them with vinegar (dilute acetic acid) solubilized the salts and brought them to the surface, making them visible upon drying.

I see what you're getting at, but it doesn't fit all the pieces together for me. I never applied tap water to the carpet. The only liquid applied to it was my vinegar, and before that, the rain/runoff water that flooded my car. I live down south, so the roads here aren't salted. I'm processing as I go here... I used to live way up north where winter was long and there was a lot of salt on roads and sidewalks, which my shoes would have easily transferred to car carpet. Could that have been what the vinegar solubilized?

I suspect that the runoff water that flooded your car has a lot more salt content than your tap water.  I'm guessing it was runoff salt and salts from the carpet manufacturing process.

I know "salt" is very general term--but what salts would there be in runoff water if the roads are never salted due to winter weather? Do lawn chemicals contain salts?

You bet, fertilizers are a big contributor, road degradation, car emissions and degradation products from brake pad linings also contribute.


text:
Why aren't chlorides of the noble gases as prevalent as their fluorides?
It is my understanding that compound formation has only been observed for the noble gases argon, xenon, and krypton, while no compounds have ever been observed for neon and helium, and that this might have something to do with both the decreased ionization potential of larger noble gas atoms and the high value of electronegativity of the halogen (usually fluorine) that bonds with the noble gas atoms to form compounds. So, my question is why aren't compounds formed from argon, xenon, and krypton bonding with chlorine, bromine, and iodine as prevalent as those formed from those noble gases and fluorine? Could the smaller size of the fluorine atom be a reason for its apparently higher reactivity?

Molecular ions, containing Ne, He, Ar, were observed. 2) The question is more about relative stability. Many molecules, that are stable in vacuum, can't be isolated as pure compounds. 3) Noble gases have very low-energy, small outer orbitals, so they need similar partner to interact. Fluorine and to some extent oxygen have lowest-energy valence orbitals, so they give relatively stable compounds with noble gases.

The bond between fluorines in F2 is weak while bonds between fluorine and other atoms tend to be strong.

Wikipedia's list of oxidation states now documents compounds of helium and neon.

You know that the bigger the noble gas atom, the easier it is to ionize it. By the same token it is also easier for a Xe atom than for a He atom to share some electron density in a covalent bond.  That is, Xe is more reactive (a very little bit) than He (not at all).
For halogens, the tendency is in some sense the opposite: The smaller the halogen atom the more eager it is to gain electron density -- that is, the more electronegative it is.  Fluorine is so electronegative that it can persuade Xe, Kr and even Ar (with difficulty) to give up some electron density to form a covalent bond.  Cl is not quite so electronegative, and Br and I even less.  Basically you need the least "noble" of the noble gases and the greediest halogen of all to even have a chance at a bond.
One can justify this handwaving with thermodynamic calculations and molecular orbital arguments (though the latter are often handwavy too).

The smaller the halogen atom the more eager it is to gain electron density || The funny thing is that electron affinity of fluorine, chlorine and bromine is 328, 349 and 325 kJ/mol with highest for chlorine. So, assuming your logic is correct, noble gases' chlorides should be more widespread, than fluorides.

But electron affinity isn't only determinant of electronegativity, as you know.  In the present case, we must look at the energy change between (say) $\ce{X_2{+}Xe}$ and $\ce{XeX_2}$. The lower dissociation energy of $\ce{F_2}$ compared to $\ce{Cl_2}$ (alluded to in brinb's comment) is enough to insure the combination is thermodynamically possible for F but not for Cl.

Yeah, cuz orbital energies are so very well defined.  Go to http://cccbdb.nist.gov/gap2.asp and type H2O -- what is the HOMO-LUMO gap for water?

ACtually, yes, their energies may be well defined in terms of peaks of spectra of electrons, knocked out by, say, x-rays.

@permeakra Perhaps you can flesh out your comment into an answer, using actual values of the energies.  Table 3-3 in Harry Gray's "Chemical Bonds" gives the valence-orbital ionization energies as 115,000 cm${}^{-1}$ for Kr (4p), 111 for Cl (3p) and 151 for F (2p).  Are those the relevant numbers for your argument?

unfortunately, no.

So what is the notion of valence orbital involved in your argument?  Is it measurable?

Don't give up completely on noble gas chlorides. Xenon doesn't.
Xenon dichloride aside, it is true that noble gases are more apt to combine with fluorine. In that sense they are not so different from other nonmetals that form "expanded octet" compounds preferentially with fluorine.
Most readers are familiar with sulfur hexafluoride and its apparent twelve valence electrons in an "expanded octet" on the sulfur atom. In reality, of course, the four additional electrons do not actually overlap the sulfur atom; they are in ligand-based orbitals that are orthogonal to the sulfur valence orbitals. We call these orbitals "nonbonding" with their nonoverlapping the central sulfur atom, but they actually are antibonding between the ligands. Sulfur hexafluoride is stabilized by having relatively small ligands that hold their electrons tightly around a larger central atom; if we were to try to replace the fluorine atoms with larger and more polarizable chlorine atoms we'd aggravate the ligand-ligand antibonding as well as introduce unfavorable steric effects.
So it is with a noble gas as the proposed central atom, except that since the uncombined atom already has the closed-shell $ns^2np^6$ valence-electron configuration, just one pair of halogen atoms sets off the "expanded octet" condition with its potentially antibonding ligands. Xenon dichloride (which is linear) has a big enough central atom and the chlorine atoms far enough apart to get by (or, accordig to the Wikipedia article,it may be merely a van der Waals complex between a xenon atom and a dichlorine molecule), but putting those rather large chlorine atoms closer together around a noble-gas core is asking for trouble.

Also, bromides also exists albeit unstable. $\ce{XeBr2}$: https://en.wikipedia.org/wiki/Xenon_dibromide. $\ce{XeBr}$* excimer: https://patents.google.com/patent/US4143336A/en

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text:
Is it possible to remove salt from food without changing the flavor?
I love soup, even a thick soup for dinner, yet many soup products have 400 to 1000 mg of sodium in them. Is there a simple chemical procedure any consumer could do to take the sodium out of the product without 'hurting' the flavor?
If one takes some, say, Campbell's soup and heats it to boiling, then freezes it, then heats it again, then freezes it etc., would this extract some sodium?

Several brands offer low-sodium versions, but maybe you'd like even less salt that they offer?

I don't know how practicable it would be with soup, but size exclusion chromatography or dialysis can used to remove salts (although there would be no selectivity for sodium).

IF several brands offer low salt versions WHERE do you find them? Shopper's Drug Mart sells a lot of tasty foods in some of their stores but they generally have a salt content over 300 mg. for each product. THEY'RE supposed to be for health I assume. EVEN those Health Check product in  Grocery stores have over 400 mg. of salt for one person with one serving.

The salt is there FOR the flavor. Since most junk food like canned soups contain little if anything, their flavor is mostly coming from taste enhancers like salt and msg.

Thats going to be far more complicated than making your own soup. (Which is a far healthier and cheaper time killer than watching TV, so get at it! ;))

Short answer: no, there is no easy (or even moderately hard) procedure, and certainly not boiling and freezing.
The thing is that water dissolves sodium ions better than just about anything else, so almost everything will come out of solution before the sodium does. You're much better off with a product that starts with less salt to begin with.

Could some safe substance be added to the soup that absorbs the sodium and doesn't spoil the flavor?

I've added a 2nd paragraph to my answer addressing that.  Sorry to disappoint you.

@user128932 To elaborate on Levy's answer, sodium very rarely forms any type of solid compound because of its solubility. There's nothing you can do to "absorb" the sodium because elements can't be lost. Your best bet would be to react something with the sodium ions that produces an insoluble solid, and then to physically remove the solid. But that's virtually impossible with sodium.

In truth it is not inconceivable that some day we will have molecular sieves that can preferentially trap sodium ions while rejecting bigger species.  You could then add some of these magic beads to your soup, stir for 10 minutes and scoop them out, carrying the salt (assuming the soup is not too thick).  I doubt taste would be unaffected, though.  Your soup would taste sour since you'd be replacing $\ce{Na^+}$ with, presumably, $\ce{H^+}$; even if most of these got attached to bases you'd still be lowering the pH.  In any case I know of no such technology at present.

Great information! It would be nice if more chemists and food chemists could complain to Campbells soup , Knorr soup ,Healthy-choice soup, and a whole lot of other product about the amount of sodium in the product that can't be gotten rid of. Isn't it true if most people cut down their sodium levels from processed food A LOT  of money would be saved in healthcare costs?

Taste will be affected a lot even if we weren't replacing the ions. The reason why this salt is put in is for taste. The soup will taste bland without the sodium, and even if we had some magic method of making it disappear without a trace, the taste will change a lot.


text:
Calculating pH of diprotic and amphoteric solutions
I am trying to understand this problem and its corresponding solution (complete solution not shown)

First off, from the above one can see that the equilibrium constants when $\ce{H2ASO4-}$ donates and receives a proton are as follows:
\begin{aligned}
\ce{H2AsO4 &<=> HAsO4^{2-} + H+} & K_2 &= 1.1 \cdot 10^{−7}\\
\ce{H2AsO4- + H+ &<=> H3AsO4}    & K_{-1} &= 1.6 \cdot 10^{2}\\
\end{aligned}
However, if this is an amphoteric substance as the solution claims, shouldn't both $K_a$ values be comparable? It seems logical that if it is able to act as both an acid and a base, the equilibrium constants should at least be similar to each other. Instead, one is several order of magnitudes larger than the other? Why can this be true and $\ce{H2ASO4-}$ still be an amphoteric substance?

Next, how did they arrive at the statement

We approximately have the equality relation: $$\ce{[H3AsO4]} = \ce{[HAsO4^{2−}]}$$

What is the evidence that shows this to be true? Isn't the $K_a$ value of the basic reaction ($1.6 \times 10^2$) much larger than that of the acidic reaction? Why should the concentrations be equal?

Finally, let's say you are calculating the pH for the diprotic substance $\ce{H2S}$, which undergoes the following diassociations:
$$\ce{H2S <=> H+ + HS-}$$
$$\ce{HS- <=> H+ + S^{2-}}$$
To find the pH you would assume that the first reaction is dominant, and that the effects of the second reaction consuming $\ce{HS-}$ are minimal, and then calculate the pH from there — right?
However, why can't you treat $\ce{HS-}$ as an amphoteric substance (because it can both gain and lose a proton) and then calculate the pH as you would for an amphoteric substance (as seen in the image at the beginning)?

$\ce{HS^{-}}$ is amphiprotic, but in acidic solution it will not act as an acid. Just like $\ce{H3+O}$ is formally amphiprotic, but is will almost never act as a base. (You have changed the question a lot before your bounty, now the original answer looks completely off.)

@Martin But isn't the equilibrium constant when $\ce{H3O+}$ acts a base much higher than when it acts as an acid? Are you saying that any substance can technically be amphiprotic? If so, what governs whether it actually is in practice? Why doesn't $\ce{HS-}$ act as an acid in acidic solution while $\ce{H2AsO4-}$ can?

Of course the equilibrium constant is much higher for the reaction of $\ce{H3+O}$ accepting a proton. That does not change the fact, that $\ce{H3+O}$ still has a lone pair and could in theory accept another proton. It is a property that is in no way anyhow related to the chemical system we are talking about. For example, in an solution of $\ce{H2O}$ and $\ce{HSbF6}$ in liquid argon, you will find two equilibria $$\ce{H2O + 2HSbF6 <=> H3+O + HSbF6 + SbF6- <=> H4O^{2+} + 2SbF6^- }.$$ The equilibrium is in the middle, but you can most likely see all species.

The same applies to the other example. In acidic solution of $\ce{HS^-}$ there is an excess of protons, hence the protonation will always be favoured over the deprotonation.  The property of a molecule is not a 1:1 relation to the property of a chemical mixture containing that molecule. It does not lose its initial properties, but some of its properties will not be visible as it reacts to an external force. Acid base reactions are always equilibrium processes. Especially when working with multi proton donors you can expect to find a lot of different species.

@Martin So while $\ce{HS-}$ in an acidic solution will prefer to act as a base, in a neutral solution both protonation and deprotonation will exist, right? Will the amount of protonation be the same as the amount of deprotonation? Is a similar argument what leads to the claim that $\ce{[H3AsO4] = [HAsO4^{2-}]}$ in the image at the top?

Yes, the argument is similar. You always have to consider also the stability of the isolated species. Taken the example, the naked sulfide ion is not very stable, so even in neutral solution the amount of deprotonation will be minimal, especially when talking about aqueous solution. Do not forget autoprotolysis. The reaction $$\ce{HS- + H2O <=> H2S + {}^{-}OH}$$ will always be favourable over $$\ce{HS- + H2O <=> S^{2-} + H3+O}$$ since the hydronium ion is a much stronger acid than $\ce{HS- }$.

I'll address your questions in 3 parts:
(A) Why amphoteric
(B) How to solve the problem without making unexplained assumptions
(C) Why the solution in the book uses the approximate equality
$\ce{[H3AsO4]=[HAsO4^{2-}]}$.
PART (A)
An amphoteric species (molecule or ion) is one that can either gain a proton with relative ease, or lose a proton with relative ease. "Relative ease" means, for example, that it can happen in aqueous solution. If your species will gain a proton in the presence of enough $\ce{H3O+}$, and if it will lose a proton in the present of enough $\ce{OH-}$, it can certainly be called amphoteric.
Translated into equilibrium constants, if your species has a pKa between 0 and 14, and its conjugate acid (what you get when you protonate it) also has a (necessarily lower) $pK_a$ between 0 and 14, your species will act in aqueous solution as either as an acid (losing a proton in the presence of strong base) or as a base (gaining a proton in the presence of strong acid).
The $pK_a$ of (neutral) arsenic acid is between 2 and 3 (because the equilibrium constant for removing the first proton, $K_1$, is between $10^{-2}$ and $10^{-3}$). The $pK_a$ of the version that lost one proton, $\ce{H2AsO4-}$, is around 7 (from the value of $K_2$). So clearly $\ce{H2AsO4-}$ will gain a proton if the pH is below 1, say, and it will lose a proton if the pH is above 10. This justifies why it's amphoteric.
PART (B)
In a monoprotic acid like HF, all the F appears either as F$^-$ or as HF (two species). You can write down 1 equilibrium equation, a mass balance for F and a charge balance, and solve all three equations simultaneously; when you combine the three you get a quadratic equation that can be solved exactly.
But with a multiprotic acid, things get complicated fast. In this problem you have three equilibrium equations, plus the mass-balance and charge-balance equations.  That's 5 equations in 5 variables.
Fortunately it's possible to combine 4 of them easily, using only high-school algebra. They are the 3 equilibrium equations and the charge balance. This reduces the problem to two variables (the pH and the concentration of one of the arsenic ions) in two equations.
Let's write and combine the 4 equations. There are four arsenic species with varying charges; I'll call their concentrations $A_0,A_1,A_2,A_3$ -- the index denotes the charge (which is 3 minus the number of hydrogens). So $A_0=[\ce{H3AsO4}]$, $A_1=[\ce{H2AsO4-}]$, and so on.
By definition, $A_1[\ce{H^+}]/A_0=K_1$, or equivalently $A_0=A_1[\ce{H^+}]/K_1$. This relates two concentrations, $A_0$ and $A_1$. Similarly $A_1=A_2[\ce{H^+}]/K_2$ and $A_2=A_3[\ce{H^+}]/K_3$.
The important thing is that all these four concentrations are proportional and given one you can find all others if you know the pH. (We don't know the pH, but I'll address that in a moment.)
The charge balance (expressed in moles present in 1 L of solution) is simply
\begin{equation} 0\,A_0+ 1\,A_1+ 2\,A_2+ 3\,A_3+K_w/[\ce{H^+}]= [\ce{H^+}]+0.1.
\end{equation}
On the left we have the negative charges, on the right the positive ones. Notes:

The concentration of each arsenic species is multiplied by its charge, obviously.

On the left side, $K_w=10^{-14}$ is the dissociation product of water, and $K_w/[\ce{H^+}]$ represents the concentration of OH$^-$; it will turn out to be negligible, but we don't know that a priori, and including it doesn't make the problem any harder.

The 0.1 on the right comes from Na$^+$ ions; being an alkali metal, sodium does not exist in water solution except in ionic form, so we know [Na$^+$] exactly. This explains something that will be important later: The 0.10 M solution of $\ce{NaH2AsO4}$ is exactly what you obtain by dissolving 0.10 moles of NaOH in one beaker, 0.10 moles of $\ce{H3AsO4}$ in another, and mixing the two, assuming the combined volume is 1 L.

Now we express the left side of (1) in terms of just two
concentrations, $A_3$ and $\ce{[H^+]}$: $$
\begin{aligned}A_1+2\,A_2+3\,A_3&=(\ce{[H^+]}/K_2)\,A_2+2\,A_2+3\,A_3\\
&=(\ce{[H^+]}/K_2+2)\,A_2+3\,A_3\\
&=(\ce{[H^+]}/K_2+2)(\ce{[H^+]}/K_3)\,A_3+3\,A_3\\
&= \bigl[(\ce{[H^+]}/K_2+2)(\ce{[H^+]}/K_3)+3\bigr]\,A_3
. \end{aligned} $$
The last line is the key; it allows us to relate the concentration of one arsenic
species to the pH.  With this (and moving the $K_w/\ce{[H+]}$ to the
right)
the charge balance becomes
$$ \bigl[(\ce{[H^+]}/K_2+2)(\ce{[H^+]}/K_3)+3\bigr]\,A_3
=\ce{[H^+]}+0.1 -K_w/\ce{[H+]}$$
or again
$$ A_3
=\frac{\ce{[H^+]}+0.1 -K_w/\ce{[H+]}}{\bigl[(\ce{[H^+]}/K_2+2)(\ce{[H^+]}/K_3)+3\bigr].}$$
This look complicated, but it's easy to put into a spreadsheet. The last step is this: The right [H$^+$] is the one that, with $A_3$ calculated as above, obeys the mass balance, $$ A_3 + A_2 + A_1 + A_0 =0.1.$$
At this point we could declare the multiple-choice problem solved, by cheating a bit: take the various proposed values of  [H$^+$] and see which one works.  For answer (a) the value of the sum $ A_3 + A_2 + A_1 + A_0$ is 0.62; for (b) it is  0.128; for (c) it is 0.10165, and for (d) it is 0.100002.  So the best answer is (d).
What if you didn't have multiple choice? Then you'd want to draw a graph or a table, taking (say) every integer value of pH and connecting them.  Here is such a graph:

The horizontal axis is pH and the vertical one is total arsenic concentration.  The correct pH is where the vertical coordinate is exactly 0.1, and you can get a good idea of its value by eyeballing, or solve by successive approximations.
Note that this method allows you to solve with no more effort the following problem: what is the pH when you add 0.1 mol/L of $\ce{NaH2AsO4}$ (as in the original problem), plus 0.05 mol/L of $\ce{H3AsO4}$?  The only difference is that now you'll be looking for the point on the graph where the vertical coordinate is 0.1+0.05 = 0.15, instead of 0.1.
PART (C)
Why does the book say that $\ce{[H3AsO4]=[HAsO4^{2-}]}$, approximately? And is this a good approximation?
The equality comes from the claim that the "dominant reactivity" of $\ce{[H2AsO4-]}$ is $$\ce{2\,H2AsO4- \leftrightharpoons H3AsO4{}+HAsO4^{2-}}.$$ If $\ce{H2AsO4-}$ exchanges protons only with itself, and not with, say, water, then every proton lost by one $\ce{H2AsO4-}$ ion means a proton gained by another $\ce{H2AsO4-}$, and the result is that $\ce{[H3AsO4]}$ and $\ce{[HAsO4^{2-}]}$ are generated in equal amounts.
So given the claim about "dominant reactivity", the equality is true, and by the way, the equality implies that the pH is the average of $pK_1$ and $pK_2$. (This follows, by taking logs, from the analysis that I'm guessing your book gives right after the part of the solution you quoted.)
The problem with the claim is that it does not follow from the amphoteric nature of $\ce{H2AsO4^-}$ alone. It requires two additional conditions: You're starting from only $\ce{H2AsO4-}$, and in relatively high concentrations. I'll illustrate why these conditions are necessary:
(1) You have to start with only $\ce{H2AsO4-}$ (plus a counterion that doesn't participate further, like Na$^+$). If you started with some $\ce{H3AsO4}$ in addition, you wouldn't have the equality $\ce{[H3AsO4]=[HAsO4^{2-}]}$, even approximately. Or, to use the example you gave, if you start with H$_2$S the amphoteric character of HS$^-$ wont help you calculate the pH of the resulting mix. On the other hand if you started by dissolving NaHS alone, it would.
(2) You have to start with a strong solution. If the original question had specified a 0.1 mM solution instead of a 0.1 M solution, the resulting H$^+$ concentration would not be $2.5\times 10^{-5}$ anymore but $3.3\times 10^{-6}$, and the equality $\ce{[H3AsO4]=[HAsO4^{2-}]}$ would be off by a factor of 60!
I hope this helps answer your very interesting and perceptive questions.  Don't hesitate to ask for further clarification if necessary!

Thanks so much for taking the time to type all of this! Only one small question — in the last part of your answer to why the dominant reactivity claim holds, you mentioned that you must start with a strong solution, or otherwise the claim doe snot hold. Why doesn't it hold for weak concentrations though? Shouldn't $\ce{H2AsO4-}$ exchange protons with itself regardless of its concentration?

"$\ce{H2AsO4-}$ reacts only with itself" is at best an approximation. It only has a chance of being valid if the concentration of $\ce{H2AsO4-}$ is many orders of magnitude more than that of all other species.  At $10^{-4}$M you'd see $\ce{OH-}$ starting to play a role, even though its concentration is still only $\sim 10^{-8}$.  Or, if $pK_1\approx pK_2$ (succinic acid, $\ce{COOH(CH2)2COOH}$ - can you see why?) the equilibrium pH is close to the $pK_a$s and therefore $[\ce{HA-}]$ is no longer enormously greater than $[\ce{H2A}]$ or $[\ce{A^{2-}}]$, so again $[\ce{H2A}]=[\ce{A^{2-}}]$ fails.

In the previous comment, I realize now that instead of $\ce{OH^-}$ the major issue is $\ce{H^+}$, which at $3.3\times 10^{-6}$ would not be negligible in comparison to 10$^{-4}$ M. So the reaction $\ce{H2AsO4^-}+ \ce{H+}\leftrightharpoons\cdots$ would no longer be negligible compared to the reaction $\ce{2,H2AsO4-}\leftrightharpoons\cdots$.

First, I have read that $\ce{H2AsO4-}$ is a amphiprotic substances in
  that the dominant reactivity is $$\ce{2H2AsO4- <=> H3AsO4 + HAsO4^{2-}}$$
However, this also seems to be a polyprotic substance.

The two terms - amphiprotic and polyprotic - aren't mutually exclusive. Amphiprotic substances can simply behave as both Bronsted-Lowry acids and bases in that they can both give up and release hydrogen protons.
Polyprotic acids have multiple protons to donate. For example, phosphoric acid is triprotic.
Often times amphiprotic substances will be polyprotic. By nature of being able to lose multiple protons, these substances are also able to gain multiple protons.

Why is one $K_a$ value so much larger than the other? If this were a
  amphoteric substance shouldn't the $K_a$ values for both be
  comparable?

Good observation. This lies with Coulomb's law. Think about it. A neutral molecule kicks off a hydrogen proton. Now you have the first conjugate base. The first conjugate base is negatively charged. It also attempts to kick off a proton. Which do you think is easier? Clearly the first process is. Remember that oppositely charged particles attract. This explains why $K_{a1}$ values for polyprotic molecular acids are generally significantly larger than corresponding $K_{a2}$ values.
Just because the $K_a$ values differ by several orders of magnitude doesn't mean the ion/molecule can't be amphiprotic.

Next, I researched how to calculate the $\mathrm{pH}$ value for
  amphiprotic substances, and I found this page that details how to
  do so, and the end result is that the $\mathrm{pH} = \sqrt{K_1 K_2}$.
  Why is the $\mathrm{pH}$ independent of the concentrations?

First, amphiprotic substances don't have $\mathrm{pH}$ values by themselves. In solution, then yes.

Further, I also found another page that gets to the same result in a
  different manner. They start by saying:

We approximately have the equality relation: $$\ce{[H3AsO4]} = \ce{[HAsO4^{2−}]}$$

Regarding the validity of the quoted statement: just note the main reaction stoichiometry. And the magnitude of the main reaction's K. The main K can be found through dividing the Ka of the reactant acid by the Ka of the product acid.
\begin{aligned}
\ce{H2AsO4^{−} + H2AsO4^{−} &<=> H3AsO4 + HAsO4^2-}\\
K_\mathrm{rxn} &= \frac{K_a(\ce{H2AsO4^{−}})}{K_a(\ce{H3AsO4})}\\
\end{aligned}

Next, why can't this method of calculating pH in amphoteric acids be
  used for diprotic acids? I found another page that details how to do
  so but even though the reactions are essentially similar to those for
  amphiprotic acids, the end result is different. Why does the same line
  of reasoning as used for deriving the formula for amphoteric
  substances fail for diprotic/polyprotic acids?

For diprotic acid, just keep $K$ value magnitudes in mind. This statement applies to diprotic and polyprotic acids:
$$K_{a2}(\ce{H3AsO4}) = \ce{[HAsO4^{2−}]}$$

One type of behavior may be dominant but that doesn't rule out the possibility of another type of behavior.

I see. Note the atoichiometry of the main reaction equation.

But aren't the Ka values order of magnitudes different? Since $K_{-1}$ $>>$ $K_2$, won't $\ce{[H3AsO4]}$ be much greater than $\ce{[HAsO4^{2-}]}$?

That's true but the main equilibrium's K is much greater than any of the Ka values. This makes intuitive sense; you're reacting an acid with a base stronger than water (you're reacting an amphoteric substance with itself).

I don't think I have the time to write up a detailed answer now, being that it is past midnight in my timezone. I will give it a treatment later though. Best wishes.

Wait, the main equilibrium K value is $1.8 \times 10^{-5}$ which is quite small. How does this lead to the fact that $\ce{[H3AsO4]} = \ce{[HAsO4^{2-}]}$? Also, why can't you do something similar for diprotic acids?

Sorry to bump this again, but could you please elaborate on how the main equilibrium value of K which is still small results in approximately equal concentrations of H3AsO4 and HAsO4?

It's small but only on a relative basis.

But it's not greater than all of the $K_a$ values. The $K_{-1}$ value is much greater than the main equilibrium. So how does this mean that the concentrations of $\ce{H3AsO4}$ and $\ce{HAsO4^{2-}}$ are equal?

What is the value that you refer to?

The main $K_a$ value of $$\ce{2H2AsO4- <=> H3AsO4 + HAsO4^{2-}}$$ is $1.8 \times 10^{-5}$ right? However the value of $K_{-1}$, for the basic behavior of $\ce{H2AsO4- + H+ <=> H3AsO4}$, is $1.6 \cdot 10^{2}$ which is much higher than the net $K_a$ value. So how do you come to the conclusion that the concentrations of $\ce{H3AsO4}$ is equal to that of $\ce{HAsO4^{2-}}$?

And also, could you please answer the last question regarding to why a similar procedure as used to find the pH of amphiprotic acids cannot be used for diprotic acids?


text:
The spontaneity of a redox reaction
Suppose we have a redox reaction and we want to see in what direction the reaction is spontaneous. We do this by comparing the standard reduction potentials of the sub-reactions available in the main reaction and say if
$$E^\circ = E^\circ_\text{Oxidizer}-E^\circ_\text{Reducer}>0$$
 then the reaction is spontaneous in the direction mentioned otherwise in the reverse direction. The same result can be obtained by comparing the standard oxidation potentials. However, as every spontaneity has its root in the second law of thermodynamics I am eager to see how the spontaneity of a redox equation in a given direction can be checked based on thermodynamics.
So basically how can we derive the condition
$$E^\circ = E^\circ_\text{Oxidizer}-E^\circ_\text{Reducer}>0$$
 for spontaneity of a reaction from the fundamental thermodynamics laws?

To clarify the question better let me add that I already know that the second law for a common chemical reaction yields into the condition $$dG \le 0\,,$$ my question is about to find the condition
$$E^\circ =E^\circ_\text{Oxidizer}-E^\circ_\text{Reducer}>0$$ from that inequality. Also let me add further that I even already know that the electric work equals $-n F E^\circ_{\text{cell}}$ but I cannot justify for myself why $\Delta G$ is equal to this work. (Note. the electric work is an internal work not crossing the boundaries of the system, it is not the work of the system, and if it was then at best we had $\Delta G=W_{\text{max}}$ but the inequality $\Delta G\le 0$ no longer hold.)

The cell potential is related to the change in Gibbs free energy by this equation:
$$
\Delta G^\circ= -nFE^{\circ}_\text{cell}
$$
where $n$ is the number of moles of electrons transferred in the balanced reaction equation and $F$ is Faraday's constant.
The UC Davis chem wiki has a brief explanation of the connection between cell potential, the reaction quotient (via the Nernst Equation), and the equilibrium constant. You should be able to find a more thorough explanation in a general chemistry textbook. My copy of Tro's Chemistry has a derivation of all of these relations, but to summarize the first relation:
Work against any potential is given by force times distance, and in the case of moving a charge, that works out to:
$$
w = -qE^{\circ}_\text{cell}
$$
Faraday's constant $F$ gives the charge in coulombs on one mole of electrons, which means the total charge is given by:
$$
q = nF
$$
Where $n$ is the number of moles of electrons.
We can substitute this back into the work equation to get:
$$
w = -nFE^{\circ}_\text{cell}
$$
Since $\Delta G^{\circ}$ is equal to the maximum amount of work that can be done by a chemical reaction,
$$
\Delta G^\circ = w = -nFE^{\circ}_\text{cell}
$$
In other words, saying that change in Gibbs free energy is negative is the same as saying that the cell potential is positive – in both cases we are simply looking at the total usable energy change and saying that "things move downhill" – the reaction proceeds in the direction that minimizes the thermodynamic potential energy of the system.

thank you for your answer, but let me clarify my question a little more. I already knew that second law leads to $\Delta G\le0$ and also knew that the electric work equals $-n F E^\circ_{\text{cell}}$, the problem I have is why we are allowed to set equal $\Delta G$ to this work, so to have the relation between the second law and $E^\circ_{\text{cell}}$? Note that here the electric work is not a work of the system on hand, it is internal not crossing the boundaries of the system. Even if it was an external work then the 2nd law had led to $W_{max}=\Delta G$ and not $\Delta G\le 0$. Am I wrong?

From the first law, $$\mathrm dU = \mathrm dQ - P\mathrm dV + \mathrm dW_{\text{non- mechanical}}$$
Now, for reversible process with non-mechanical work, at constant $T$ and $P$.
Note for such process, $\mathrm dQ =T\mathrm dS$, but $\mathrm dQ \neq \mathrm dH$.
With this, $$\Delta G = W_{\text{non-mechanical}} = -nFE$$
The last sign is the convention depends on how we define $E$.
So second law for the spontaneous process will imply, $\mathrm dG <0 $, $W_{\text{non-mechanical}}<0$ and $E_{\text{products}} -E_{\text{reactants}} >0$.

@Jayadithya Regarding your edit suggestion. Please read, Which symbols are written in roman (upright) font and which are italicized?


text:
What stable arrangements of hydrogen-bonded water exist in the range of 4-20 water molecules?
Water clusters made up of various numbers of water molecules have been described.
Are all of these structures theoretical?
What experimental evidence supports any of these structures?
How does the free energy of a particular water cluster (per number of molecules in the cluster) vary from one thermodynamically stable arrangement to another? Is there a general trend, e.g. larger clusters are more stable?
In response to Martin's initial comment:
I am curious about water clusters in the context of receptor-ligand interactions (for example, in protein receptors, enzymes or synthetic host cavities).
In cases where the cavities of these macromolecules is small (for example, a small enzyme active site) what kinds of water clusters are likely to fill this void in the absence of ligand?
Put more broadly, does the structure of the water that is filling the cavity destabilize the receptor in such a way that ligand binding is enhanced?
Considering a hypothetical receptor that is known to bind ligands (whose sizes range between 4 and 20 non-hydrogen atoms) in water, what are likely structures for the encapsulated water cluster that will be displaced by the ligands upon binding?

What do you mean by 'stable'?  Also, many arrangements exist and experiment has been able to identify some of these low energy configurations.

Have a look at this site. There's a large section devoted to water clustering.

@LordStryker I feel thermodynamic stability would be most helpful in answering this question, but info on kinetic stability is also welcome (although I imagine kinetic barriers to cluster rearrangement would be very low).

@NicolauSakerNeto Thanks for the helpful link! Very cool site

If larger clusters of water weren't more stable, water wouldn't form droplets.

If you're interested in water inside enzymes, I know this paper did some simulations about it. I don't understand it, he just works down the hall from me.Whalen, K.L., Spies, M.A. Flooding enzymes: quantifying the contributions of interstitial water and cavity shape to ligand binding using extended linear response free energy calculations. 2013, J Chem Inf Model. 53(9):2349-59. PubMed PMID: 24111836.

The Cambridge Cluster Database can give you a start with theoretical binding energies, but only one isomer for each size is shown.
http://www-wales.ch.cam.ac.uk/~wales/CCD/TIP4P-water.html

Could you elaborate a bit more on the answer?  I realize there's probably a ton of information there, but one or two examples of what they offer would be helpful.


text:
Why do other sugars melt whereas sucrose decomposes?
Sucrose decomposes at 186 °C. It doesn’t melt but when it reaches decomposition temperature, sucrose decomposes into fructose and glucose.
After cooling and hardening it is no longer sucrose. Am I right?

There is no set melting point of sucrose. This article explains what really goes on when sucrose is heated. http://mobile.foodnavigator.com/Science/Sugar-doesn-t-melt-it-decomposes-say-researchers#.VOkOQhl6jqA

You're mostly right.  Sucrose is a disaccharide composed of glucose and fructose.

It is a crystalline material.  Heating sucrose results in a complex thermal process that involves both melting and decomposition.  Depending how fast you heat it, you will observe different things.  Basically, sucrose will start to decompose when it starts to melt.  If there is some water present during the heating process, then a reaction known as sucrose inversion will occur and the sucrose will break down into its components, one molecule of glucose and one molecule of fructose.
However if dry sucrose is heated to its melting \ decomposition point, then more complex things happen.  Significant amounts of glucose are initially detected.  Continued heating leads to "caramelization".  Caramelization is a condensation process where water is eliminated from sucrose; the initial loss of water is followed by isomerization and polymerization.  Over 4,000 products are thought to be formed during caramelization.  When this complex mixture is cooled, it is no longer sucrose.

What you are doing when you heat a sugar like sucrose is you are dehydrating it. The crystalline structure of sucrose breaks down and the molecules decompose into glucose and fructose and then lose water and then become isomers and polymerize to form caramel, a red-orange colored solid at room temperature. However there are several other products other than just fructose and glucose.Caramelization is poorly understood so I cannot give you a direct answer stating all of the products formed.
You can do this at home, just put table sugar in a pot and heat it up. The same type of process is used in candy making.
Source

Sucrose does melt at $\pu{366.8 ^\circ F}$. It decomposes once you get to hotter temperatures, which is why caramel darkens till it becomes burnt sugar. Thus unlike sodium chloride and most other substances sucrose and lots of other sugars from trioses upwards in numbers of carbons don't have a boiling point, just a melting point because heat breaks the $\ce{C-C}$, $\ce{C-H}$, and $\ce{C-O}$ bonds more in sugars than other organic molecules. The exact reason?
From $\ce{R-C-O}$ it is easy to form carbon dioxide, and there is also an $\ce{-OH}$ attached to one of the carbons which water can form easily from.

when you melt it you break the bonds and if you break the bonds then it is not sucrose So what you are saying if it is melted at the right temperature for the right amount of time with the right amount of liquid then it will melt but it has to be anything but monosaccharides or disaccharides so no sugar in the liquid except for the sucrose

Please use punctuation. This is not at all understandable.


text:
What does the hydrolysis of a salt tell us about its basic/acidic character?
I'm curious to know how basicity and acidity arises when a salt or covalent compound is dissolved in water.

Think of the salt in terms of its components: for example, sodium sulfide can be thought of as derived from sodium hydroxide (a strong base) and hydrogen sulfide (a weak acid) - resulting in a net basic character. When you dissolve sodium sulfide in water, the sulfide ion partially reacts with water to form hydrosulfide ion and hydrogen sulfide, plus hydroxide ion. Aluminum chloride, on the other hand, is derived from aluminum hydroxide (a weak base) and hydrogen chloride (a strong acid), so it is an acidic salt. Sodium chloride, from sodium hydroxide and hydrogen chloride, is a neutral salt (fully ionized). Knowing the dissociation constants of the corresponding acid and base of a salt, one can calculate how acidic or basic the salt is.

Some salts may have basic Brønsted/Lowry anions but are extremely stable in water. This belies the true nature of the salt's anion.
Consider silver sulfide as an example. This salt has a Ksp value of about ${6*10^{-51}}$. Despite the fact that the sulfide anion is a highly basic Brønsted/Lowry base, a solution of silver sulfide will actually exhibit a pH of 7.0.
Note that salts do not have to be exclusively Brønsted-Lowry bases/acids. Some salts may also exhibit Lewis acidity/basicity in water. This is another way in which basicity and acidity may arise when a salt or covalent compound is dissolved in water. Consider boric acid. This isn't a salt, but it is a covalent compound. It is an acid, but some argue it is not an acid in the sense of being a proton donor. The central boron instead utilizes its empty 2p orbital to heterolytically cleave water molecules as to form H+ in solution.

This question is directly related to the inter relatedness of Lewis and Brønsted acidity.
Typical metal salts (let's say MCl3) undergo hydrolysis (presumably in an aqueous solution) as Lewis basic water (which donates electrons to the acid in this context) coordinates (binds) to the Lewis acidic metal (which accepts electrons from the Lewis base). After, before or during coordination (it's hard to tell which without extensive experiment), water will give up one or two protons (i.e. hydrogen atom minus an electron), thereby generating a metal oxo/hydroxide (in this case, e.g. MO2H) or metal hydroxide (e.g. M(OH)3), either of which will be accompanied by acid in solution in the form of 3 eq of HCl (which would be completely dissociated into H+ and Cl-).


text:
How do I measure the density of a cellulose acetate membrane?
I've been playing around with filtration methods, and got my hands on a casting machine for this purpose.
My first mission was to find out how porosity can affect filtration. To see this for myself, I just ran some titratable solvent through the membrane I produced, and analyzed the difference in concentration.
Looking through some textbooks, I found a neat formula for calculating porosity, but there's one thing that's been bugging me. The formula requires the density of the membrane.
This isn't something that I can just use the Archimedes Principle to get. I need to look at the wet weight and dry weight of the membrane or whatever.
My dilemma right now is I can't find a single resource that lets me measure the density of a membrane without requiring either some pricey machine or extremely esoteric methods that's beyond my abilities. I'm really at a loss right now for this particular method.
There's this equipment here that's usable, but that looks like one high-end son of a gun. I really don't want to throw any more money than I already have.
How do I measure the density of a membrane with tools available in the home?

Why can't you use Archimedes' Principle (by that I assume you meant measuring volume by displacement)? Can you elaborate on why you need the wet and dry weight?

The film pores are in sizes of microns. Putting the film in water just like that will give a horrendously inaccurate result. Also, uh, the formula needs it. The wet weight is basically the weight of the membrane after you use some water to remove it from the plate. The dry weight is when it's...dry. The difference of the weight gives you how much water is occupied by the pores.

Can you post the formula and its source? There seems to be some information missing here.


text:
What makes a ligand stronger than another?
A ligand is a neutral molecule or negative ion with at least one lone pair of electrons which forms a dative covalent bond with a complex ion that has a vacant, energetically accessible orbital.
It is said that in terms of ligand strength, or 'binding power', the cyanide ion is stronger than ammonia, which is in turn stronger than water, which is in turn stronger than the chloride ion.
What determines the strength of a ligand?

Another factor is metal ligand backbonding. The metal's electrons can enter the ligand's antibonding orbitals.

Thanks for the answer, but I haven't studied antibonding orbitals yet.

One definite factor is the ability of the ligand to stabilize its lone pair. You observe that the cyanide ion is a better ligand than the nitrogen ligand. Let's think about this. The lone pair that reacts on the cyanide ion is the lone pair on the carbon atom. Not the lone pair on the nitrogen atom.
Carbon is less electronegative than nitrogen. Thus carbon is less able to stabilize its valence electron density, which in turns enhances the reactivity of its valence electron density through increased negative charge density.
Same with sulfide and oxide anions. Sulfur is less electronegative than oxygen, and thus less able to stabilize valence electron density. Hence the much greater stability of silver sulfide as opposed to silver oxide.
Other competing factors include charge density; the Lewis acid the ligand is coordinating to, medium (solvent), etc. For example the oxide anion is more basic than the sulfide anion in water solvent; i.e. the oxide anion forms a stronger bond with the hydrogen proton in water. But we just saw that silver sulfide is much more stable than silver oxide.


text:
Are salt substitutes good alternatives to sodium chloride?
I bought a low salt Ketchup from Heinz and it says it uses ‘ALSOSALT’ substitute (potassium chloride). It says it does have 5 mg of sodium. Is potassium chloride (with possible additives) a good alternative to sodium chloride and does it not cause problems for the heart or general health?

Small quantities of potassium are fine. Windsor Salt of Canada makes a 50% blend of sodium chloride and potassium chloride called Half Salt. You could use it like regular salt, although it would be good to decrease the amount you use.
Nerve impairment and muscle function are affected from too much potassium.

That's the problem right there.  Someone who needs to cut down on salt does it usually because of reduced kidney function.  That demographics is at high risk from excess potassium intake.


text:
When is it true that more nodes equals higher energy?
Consider all the MOs of some isolated molecule. (It could be a single
atom too; I'll use MO to refer to AOs as well.) Number them in
increasing order of the number of nodes (node = surface where the wave
function has zero density). Orbitals with the same number of nodes can
be numbered in any order. Now you have a sequence of orbitals $O_1,
O_2, ...$. Let their respective energies be $E_1, E_2, ...$.
It seems to be "common knowledge" that $E_n \le E_{n+1}$ for any such
system and any $n$. As Martin put it to me yesterday, "An orbital
with 47 nodes can never be lower in energy than one with only 46." (Follow up on Counting Nodal Planes in cyclopropane.)
For various reasons, given below, I think this cannot be true in
general, and I'd really like to know under what conditions it is
known to be true. "Known" here can mean either a rigorous statement
with a reference to a proof (a trivial example would be: It is true
for a one-electron atom; we can calculate the energies exactly) or a
precise statement with empirical justification (something like "no
counterexamples are known for class X of molecules" - again with a
reference).
Important: Please, I'm not looking for an explanation that re-states
the rule in some equivalent or even looser fashion ("more nodes means
the orbital is larger and less dense, therefore it must be higher in
energy").

Why do I think the statement can't be always true? Well, a calcium
atom has a filled 4s orbital and empty 3d orbitals. If this does not
count as a counterexample, please explain what notion of orbital
energy the statement applies to.
In general, I'm happy to believe that two MOs must satisfy the rule if
they have "comparable" sets of nodes (say, constant $n_x$ and $n_y$ in the
example I'm about to discuss), but I'd like to understand what is
meant by "comparable" in general. In a molecule with no symmetry, are
there comparable MOs at all? If so, how do we know if two given MOs
are comparable?
The case of a much simpler system, the 3D rectangular box, may
also be relevant.  The energy levels for such a box are of course
$$
\frac{\hbar^2\pi^2}{2m}\biggl(
\frac{n_x^2}{L_x^2}+
\frac{n_y^2}{L_y^2}+
\frac{n_z^2}{L_z^2}
\biggr),
$$
where $n_x$, $n_y$ and $n_z$ are one more (or one less, if you count
the walls) than the number of nodal planes in the respective
direction. If we take $L_x = L_y = 1$ and $L_z = 0.1$ (say), the
wavefunction with $n_x=5$, $n_y=1$, $n_z =1$ has energy $5^2+ 1^2 +
(1/0.1)^2 = 126$ and 4 nodes (or 10, if you count the walls), while
the wavefunction for $n_x=1$, $n_y=1$, $n_z=2$ has energy $402$ and 1
node (or 7). So clearly the rule is not true here.
Granted that molecules are not boxes, but this does show that
arguments based simply on the number of sign changes are not rigorous,
so they don't answer my question.

Martin - very generous of you to offer a bounty.  I too am puzzled that nobody has replied or made comments so far!

I really want a canonical answer myself. However, I believe you are correct and my previous assumption was too narrow minded. But I really would like to have a little more prove. And I know that there are people in this community that could give more insight.

Particle in a 3D box follows this rule where energy scales with nodes.  So does particle in a 1D box.  But that is because the waves you are comparing are comparable.  With the 3D rectangle with different values of L (for x, y, and z), you can only compare the x-, y-, and z- components with each other.  You can't sum over all the nodes and make this comparison.  Just like you can compare 1s, 2s, 3,s, etc. with each other, p-orbitals with each other, etc. but not p- and d- orbitals.  The correlation breaks down when you do that.

@LordStryker - thanks. Reply: "Particle in a 3D box follows this rule" -- By "THIS rule" you don't mean the rule I wrote but your rule: "you can only compare the x-, y-, and z- components with each other." Fair enough. But if drive a nail into the box and destroy the symmetry of this toy example you no longer have x-, y-, and z- modes.  So in this situation WHAT rule is followed?  Regarding "You can't sum over all the nodes and make this comparison." -- Again: Can you state what makes two nodes comparable, i.e, how one can tell that two nodes are comparable?

@SilvioLevy You bring up a good point.  I've never worked PIAB examples in anything other than the simple 1D/2D examples, cubes, and rectangles.  Particle in a 'funhouse'?  I'd love to see someone work out an example like that.

While we still have not received a nice answer here, we at least enhanced the awareness of the whole problem, which I think is a good thing. This might be a question, that will bother me from time and again. Just like the $\alpha$ effect one. http://chemistry.stackexchange.com/q/7460/4945

@SilvioLevy A counterexample to the assumption that more nodes lead to higher energy might be found in homonuclear diatomic molecules of elements that are heavier than $\ce{N}$. In those you have very little s,p-mixing and/or relatively little p,p-$\pi$-overlap. In such cases you get qualitatively the MO scheme shown here. There you can see that the $\sigma$ orbital formed by 2 p-orbitals is lower in energy than the $\pi$ orbitals formed by 2 p-orbitals. Yet the $\sigma$ orbital has 2 nodes whereas the $\pi$ orbital has only 1.

Well, thinking about my counterexample again, I'm not so sure anymore that it actually is a counterexample. Because the nodal planes of the 2 p-orbitals won't remain parallel in the $\sigma$-MO. It is likely that they bend towards each other. I'm not sure whether they will actually melt into one another above and below the bond mid and form some kind of oval-shaped nodal plane or whether they will bend towards each other in an asymptotical fashion meeting only at infinity (this second possibility seems to be more likely to me). Nevertheless, the example is not as clear cut as I thought.

Yes, the 3$\sigma_g$ < 1$\pi_u$ gives a counterexample, thank you. The alternative hypothesis would mean that the blue lobe is spatially bounded, i.e. there is a sphere outside of which the wavefunction has the same sign everywhere.  But this is only possible if is spherically symmetric.  I'll try to come up with a reference or a proof.

I appreciate, that this might not have been the answer we were looking for. I assigned the bounty anyways, since it gives at least some more information on the topic. We'll might have to revisit the whole thing in some time :D

Thank you very much!  I'm a newbie and wasn't sure who should assign the bounty, but I agree that Philipp's answer certainly deserves it.

General case
There is indeed a mathematical theorem that deals with the number of nodes an eigenfunction corresponding to a certain eigenvalue can possess.
It was laid down by Courant$^{[1, 2]}$ and it states the following:

Given the self-adjoint second order (partial) differential equation
\begin{equation} \left(\hat{L} + \lambda \rho(\mathbf{x}) \right)
 u(\mathbf{x}) = 0 \end{equation}
(where $\hat{L} = L(\mathbf{\Delta}, \mathbf{x})$ is a linear,
  hermitian differential operator, $\rho(\mathbf{x})$ is positive and
  bounded, and $\lambda$ is the eigenvalue) for a domain $G$ with
  homogeneous boundary conditions, that is $u(\mathbf{x}) = 0$ on the
  boundary of the region $G$; if its eigenfunctions are  ordered
  according to increasing eigenvalues, then the nodes of the $n^{\text{th}}$
  eigenfunction divide the domain into no more than $n$ subdomains. The
  nodal set of $u(\mathbf{x})$ is defined as the set of points
  $\mathbf{x}$ such that $u(\mathbf{x}) = 0$. No assumptions are made
  about the number of independent variables.

The proof is rather involved and so I won't show it here. But if you want you can look it up in [1] or here.
So, Courant's nodal line theorem tells us, that if we order the possible energy eigenvalues of the time-independent Schroedinger equation as $\lambda_1 \leq \lambda_2 \leq \lambda_3 \leq \dots$, then (depending on precisely how you set up the numbering) the  $n^{\text{th}}$ eigenfunction, $\Psi_{n}$ (the one with energy eigenvalue $\lambda_n$) has at most $n$ nodes (including the trivial one at the boundary $\mathbf{x} \to \infty$).
Unfortunately, this gives you only an upper bound for the number of nodes a wave function with a certain energy eigenvalue may possess. So, all we know is that the ground state wave function $\Psi_{1}$ cannot have any nodes within the region $G$ (in total it has one node, namely the one at $\mathbf{x} \to \infty$). Wave functions for higher $n$ may possess up to $n-1$ nodes within $G$ but may as well have less. Thus, we cannot in general say that if a wave function has more nodes than another one it will automatically correspond to a state with higher energy.
Special case: Schroedinger equation in one dimension
There is however a special case:
For the Sturm-Liouville eigenvalue problem (and thus for ordinary second order differential equations with homogeneous boundary conditions) we can strengthen Courant's nodal line theorem such that if we order the possible eigenvalues as $\lambda_1 \leq \lambda_2 \leq \lambda_3 \leq \dots$, then the  $n^{\text{th}}$ eigenfunction (the one with energy eigenvalue $\lambda_n$) has exactly $n$ nodes (including the trivial one at the boundary $\mathbf{x} \to \infty$).
This is useful since the one-dimensional time-independent Schrödinger equation is a special case of a Sturm-Liouville equation. So, in the case of the inhomogeneous radial Schrödinger equation with a local potential and node-less inhomogeneity such as the radial Schrödinger equation for the hydrogenic atom
\begin{equation}
\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} }  - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R(r) = 0
\end{equation}
it is generally true that a wavefunction with more (radial) nodes must always correspond to a state of higher energy than a wavefunction with less radial nodes. Also, it is clear that the wavefunctions of the one-dimensional particle-in-a-box must follow this rule. But for the three-dimensional particle-in-a-box this is not true anymore, since in that case the Schroedinger equation of the system is not an ordinary second-order differential equation but a partial-differential equation for which only the general version of Courant's nodal line theorem holds.
Some concluding remarks
For real-world system like molecules or crystals the Schroedinger equation is a partial differential equation for which the special case outlined above doesn't apply so that only Courant's nodal line theorem in its general form holds which doesn't give a strict justification for the statement that more nodes mean higher energy. Yet it is very often observed that the number of nodes indeed increases with increasing energy.
The reason for this can be motivated in the following way: The kinetic energy $E_{\mathrm{kin}}$ of a state is proportional to $\int \Psi \Delta \Psi \, d^{3} r$. Via Gauss's theorem it can be shown that $\int \Psi \Delta \Psi \, d^{3} r \propto \int |\nabla \Psi |^{2} \, d^{3} r$ and so $E_{\mathrm{kin}} \propto \int | \nabla \Psi |^{2} d^{3} r$. Now, nodes force a wavefunction to change it's sign. This often means that the value of $\Psi$ has to increase/decrease rather rapidly thus leading to areas with high absolute values of the gradient and thus to high kinetic energy. Since the potential energies shouldn't differ too much between the different states the higher kinetic energy usually also entails a higher total energy. As an example consider the bonding and antibonding wavefunctions of a homonuclear diatomic molecule whose atoms are placed at the positions $r_{\mathrm{A}}$ and $r_{\mathrm{B}}$.

The bonding wavefunction has no nodes. Its value between the atoms doesn't have to undergo a rapid change and thus the slope is rather low. The antibonding wavefunction has one node between the atoms. Its value between the atoms must change rapidly from its positive to its negative maximum thus entailing a very high slope. The slopes of the tail regions are comparable for the bonding and antibonding wavefunctions since it can smoothly fall of to zero at infinity and is not required to go from a maximum value to zero within a very confined region of space - thus even if one wavefunction has to start of at a higher maximum value the gradient will not be much higher. It follows that the antibonding wavefunction has a higher kinetic energy than the bonding wavefunction.
References
[1] R. Courant, D. Hilbert, Methods of Mathematical Physics, Vol. 1, Interscience, New York, 1953, p. 451-455.
[2] R. Courant, "Ein allgemeiner Satz zur Theorie der Eigenfunktionen Selbstadjungierter Differentialausdrücke", Nachr. v. d. Ges. d. Wiss. zu Göttingen 1923, p. 81.

Thanks for the response, Philipp.  Courant's theorem is beautiful but it's far weaker than the general claim, since usually in  dimension d > 1 the number of eigenwaves with n nodes seems to grow roughly with $n^d$.   I was hoping there might be some rigorous results that have practical use.  Your concluding remarks are quite interesting but again the intuitive justification seems to be rigorous only in 1D. Also your first comment to the question provides a counterexample when the symmetries differ, even for homonuclear 2-atom systems.

Another question for @Philipp: IIRC the hypotheses of Courant's theorem are satisfied by the time-independent, nonrelativistic Schrödinger equation, but not by relativistic versions such as the Klein-Gordon equation.  Is this right, or am I confused?

@SilvioLevy I fear you can't do better than Courant's theorem. It strongly suggests that it is simply not generally true that more nodes mean higher energy. But the connection between kinetic energy and gradient of the wavefunction gives some justification why it is true most of the time. How much more practical use would you want?

@SilvioLevy Also, why do you think that my intuitive justification is only rigorous in 1D? It involves the gradient, not the first derivative. I only chose a 1D example because it was easier to visualize.

"How much more practical use would you want?" Something that allows you to make a prediction in a case of interest, or even some kind of statistical measure -- you say "most of the time" but I don't see how your argument  that "more nodes means higher kinetic energy" justifies that (if it were really true it would be true all the time; if it's true under some conditions, I'm looking for the conditions; apart from that I don't understand what it means for it to be true "most of the time").

"why do you think that my intuitive justification is only rigorous in 1D" - because the consequence of the justification fails in d>1, that being the whole point of my question.

@SilvioLevy Ah, I see. I'm not a mathematician by profession (not at all), so I'm certainly not up to date with developments on that topic. Maybe there are some stronger theorems for "practical" Schroedinger equations, but I haven't heard of them. As for "the consequence of the justification fails in d>1": the proportionality between kinetic energy and $|\nabla \Psi |^{2}$ should be sound. But of course potential energy might play a bigger role in some cases, so the consequence for the total energy is a bit handwaving - that's true.

To be clear, the issue is the $E_{\mathrm{kin}} \propto |\nabla\Psi |^{2}$, which should have an integral on the right side. You could have high $|\nabla \Psi|$ in a small region only, and still get a lower overall energy than on another wavefunction with slightly lower $|\nabla\Psi|$ over a larger region.

@SilvioLevy Oh, yes. Sorry, I was sloppy there. Of course there should be integrals. I will change this in my answer.

Thats something I think about as well on occasion. Maybe one can drive the intuition a bit further. Consider $d=3$ in central symmetric fields. Here symmetry helps to conjecture that in such cases one might partition the wave functions into families (of course now corresponding to irreps). Then one think about if the relation holds within each family. Another way to see it is that the we have three degrees of freedom to distribute nodal planes, that will somehow find an appropriate symmetry adaption.

However its plausible that between different families there will confusion of the energy ordering v.s. number of nodes. Another educating toy model in that sense, might be a particle in a 3D box with different edge-lengths $a,b,c$.  In that context it would be much fun to see if there are deeper functional relations within a family, like a ladder-up/down generator, be it only some approximate ones ....