What is the actual mechanism of Fenton reaction? Fenton reaction was supposedly discovered by Fenton as Tartaric Acid oxidation by hydrogen peroxide in presence of ferrous cation in 1894. Research Paper link.

I have worked on this reaction for years, not an exaggeration. Whenever I mixed $\ce{H2O2}$ with $\ce{Fe^2+}$ solution, or directly solid ferrous sulphate hepta hydrate, I instantaneously got a dark beer-like color, and a vigorously reacting mixture without adjusting $\mathrm{pH}$. $\pu{10 ml}$ of 35% $\ce{H2O2}$ and $\pu{150 mg}$ of $\ce{Fe2SO4.7H2O}$ is mostly what I used. This much peroxide generally defined the reaction as Catalyzed Hydrogen Peroxide Propagations by RJ Watts group. When mixing with 1g soil in first 12 seconds temperature escalated to $\pu{84^\circ C}$, then slowly calmed. Overall bubbling ceased in 60 seconds, what kind of soil did not make any difference. And for $\mathrm{pH}$ measurement, it is very problematic for many reasons so I simply could not get any meaningful results. This just increased my curiosity.

There are experimental and computational studies over its mechanism of reactive oxygen species generation and organic matter degradation. The second one is a bit easier since somehow catching the end products gives the idea, especially with spin trap compounds.

I have lost all of references so I can only give the categories I have encountered.

Majority went for radical species mechanisms and chain reaction, involving hydroxyl, perhydroxyl, superoxide radicals and uses as a base Haber & Weiss,
Haber, F. & Weiss, J. J. (1934) Proc. R. Soc. London Ser. A 147, 332-351.
Yet these reactions are inherently fast, they are just proposed mechanisms.
There is a considerable number of people stand for ferryl ion pathway, which mostly does not involve radicals but unstable iron (IV) oxidation state compound.
Usually computational studies propose this idea, as hydroxyl radicals lifetime is very short for making it as an efficient reactant.

The reaction of Fenton I am asking is reaction between aqueous hydrogen peroxide and $\ce{Fe^2+}$ ion without any ligand other than water. If it does not help for explanation you may very well ignore more than 30 variations like sonofenton.

The Fenton reaction

The oxidation of organic substrates by iron(II) and hydrogen peroxide ($\ce{H2O2}$ ) is called the Fenton Reaction. It was first described by Henry John Horstman Fenton in 1894 who first observed the oxidation of tartaric acid by $\ce{H2O2}$ in the presence of $\ce{Fe^{2+}}$ ions [1]. After more than 100 years, the mechanism of the reaction is still unsettled. The recent review by Dunford (2002) stated that:

  The mechanism of reaction of hexaquo iron(II) with hydrogen peroxide has been unresolved for 70 years. Most scientists, perhaps by default, have accepted the free radical chain mechanism of Barb et al. (1957)[3]. However an earlier proposal involved formation of the ferryl ion, $\ce{FeO2+}$ [4]. Recent work has favored a mechanism involving $\ce{FeO2+}$ and $\ce{FeOFe^{5+}}$ species. Similarly there are differences of opinion on the mechanism of reaction of iron(III), both hexaquo and chelated, with hydrogen peroxide. These differences have fostered a recent burst of activity, with claims on one hand that hydroxyl radicals play a key role, and on the other, that there is a non-free radical mechanism. In contrast, the mechanism of reaction of the heme-containing peroxidase and catalase enzymes with hydrogen peroxide, orders of magnitude faster than reactions of iron(II)/(III), now appears to be well established.

According to the free radical chain mechanism [3], the formation of radicals $\ce{HO^.}$ and $\ce{HO2^.}$ have been proposed. Although, this mechanism has some flaws, it has still been used to explain result in current research as recently as 2014 [5]. The proposed mechanism was depicted in Figure 1, which is followed by several other well received mechanisms for the reaction discussed in reviews sighted.

Mechanism of the redox reaction involved in the Fenton’s reagent (Figure 1)[5]:

Basic reactions and intermediates involved in the classic Fenton and the metal centered Fenton reactions (Figure 2)[6]:

Proposed non-radical mechanism for the Fenton reaction (Figure 3)[6]:

Possible reaction pathways for the Fenton reaction in absence of organic substrates (Figure 4)[6]:

Requirements of the reaction:

$\mathrm{pH}$ should be adjusted to 3-5: if the $\mathrm{pH}$ is too high the iron precipitate in $\ce{Fe(OH)3}$ and will decompose the $\ce{H2O2}$ to oxygen. Basically, the optimal $\mathrm{pH}$ occurs between $3$ and $6$. It's really important to pay attention to the double $\mathrm{pH}$ drop due to the addition of $\ce{FeSO4}$ ($\ce{Fe^{2+}}$ source) and $\ce{H2O2}$. Indeed, $\ce{FeSO4}$ catalyst which contains residual $\ce{H2SO4}$ and the $\ce{H2O2}$ addition is responsible for the fragmentation of organic material into organic acids. Thus, the iron catalyst should be added as a solution of $\ce{FeSO4}$ with slow addition of $\ce{H2O2}$, you may able to control the suddern variation of $\mathrm{pH}$ and the temperature during the reaction. It's better to complete the reaction step by step with a continuous adjustment of these factors.

References:

[1] LXXIII.—Oxidation of tartaric acid in presence of iron: H. J. H. Fenton, J. Chem. Soc., Trans., 1894, 65, 899–910 (DOI: 10.1039/CT8946500899).

[2] Oxidations of iron(II)/(III) by hydrogen peroxide: from aquo to enzyme: H.Brian Dunford, Coordination Chemistry Reviews, 2002, 233–234, 311-318 (https://doi.org/10.1016/S0010-8545(02)00024-3).

[3] Reactions of ferrous and ferric ions with hydrogen peroxide. Part I.—The ferrous ion reaction: W. G. Barb,  J. H. Baxendale,  P. George,  and  K. R. Hargrave, Trans. Faraday Soc., 1951, 47, 462-500 (http://dx.doi.org/10.1039/TF9514700462).

[4] Ferryl ion, a compound of tetravalent iron: William C. Bray, and M. H. Gorin, Journal of American Chemical Society, 1932, 54(5), 2124-2125 (DOI: 10.1021/ja01344a505).

[5] Design Polysaccharides of Marine Origin: Chemical Modifications to Reach Advanced Versatile Compounds: Nathalie Chopin, Xavier Guillory, Pierre Weiss, Jean Le Bideau, and Sylvia Colliec-Jouault, Current Organic Chemistry, 2014, 18(7), 867-895 (DOI: 10.2174/138527281807140515152334).

[6] Fenton reaction - Controversy concerning the chemistry: Krzysztof Barbusinski, Ecological Chemistry and Engineering S, 2009, 16(3), 347-358.


Symmetry Space Group Diagram I am doing an introductory course to Crystallography and I have found in the following web page: http://pd.chem.ucl.ac.uk/pdnn/symm3/sgmonoc.htm a guide for the elaboration of the diagrams of the spatial groups.

In this tutorial, they say the following: "The symmetry operator for the c-glide plane is of the form x,T-y,1/2+z, while the symmetry operator for the two-one screw axis is of the form T′-x,1/2+y,T′′-z. It is possible to show (using matrices) that the only values for T, T′, and T′′, which permit the symmetry operators to form a closed group, are 1/2, 0, and 1/2, respectively"

They could tell me how to calculate the values of T, T' and T'' using the matrices.

You are missing an important part in the quote (the first sentence):

  Fixing the point of inversion at the origin implies that one of the symmetry operators is -x,-y,-z. What are the symmetry operators for the glide plane and screw axis? (The answer is clearly listed, but suppose it was not given here!) The symmetry operator for the c-glide plane is of the form x,T-y,1/2+z, while the symmetry operator for the two-one screw axis is of the form T′-x,1/2+y,T′′-z. It is possible to show (using matrices) that the only values for T, T′, and T′′, which permit the symmetry operators to form a closed group, are 1/2, 0, and 1/2, respectively. This produces the symmetry operators listed on the right-hand side of the space-group diagram shown above.

Once you fix the origin, T, T' and T'' are indeed fixed. Otherwise, they are not. So to show that they are, you have to combine a glide plane or screw axis with the inversion operation for the chosen origin, and then make an argument.

These arguments usually go as in this example: Two symmetry operations combined give a pure translation. This translation has to be compatible with the crystal lattice, otherwise you have a contradiction.


Why rinse flask with distilled water when the flask will be used for storing aliquots of HCl for titration? In an experiment for standardisation of NaOH with titration, the lab manual instructed students to rinse conical flasks that were supposed to store 1mL aliquots of hydrochloric acid with distilled water.

Won't the excess water dilute the aliquot? Why not rinse with the HCl solution instead?

Thanks in advance.

The dilution with excess water won't affect the total amount of $\ce{HCl}$ in the solution, which is what you will be titrating. You still know how much $\ce{HCl}$ you have from your quantitative addition of the $\pu{1 mL}$ aliquiot. If you rinsed with $\ce{HCl}$, you would have no idea how much $\ce{HCl}$ you really had and therefore you couldn't calculate the concentration of your $\ce{NaOH}$.


What is the reaction between NaOH and urea? In this video it is mentioned that urea and sodium hydroxide will react to produce ammonia. However, I can't seem to find a clear explanation of the reaction and the other products generated.

One explanation I found says that the NaOH causes the urea to hydrolyze, yielding sodium carbamate, which will then further hydrolyze to sodium carbonate.

Another says that the urea tautomerizes to ammonium cyanate, which preforms a substitution with the NaOH to yield ammonium hydroxide and sodium cyanate.

Which of the two reactions occurs? Do they both occur? Neither?

Urea can be attacked by nucleophiles like any carbonyl. The immediate intermediate is a tetrahedral orthoacid which will break down to liberate ammonia once. The same process can be repeated a second time to liberate the second molecule of ammonia.

Technically, this is a hydrolysis catalysed by hydroxide. Hydroxide may be the reactive species but it is regenerated in the course of the reaction.

For urea to tautomerise to ammonium cyanate, a significant rearrangement and breaking a $\ce{C-N}$ bond would be required. Neither is likely in any way.


Reconcile the Beer Lambert Law with rate of absorption The Beer-Lambert law gives the proportion of incident light intensity that will be absorbed. Intensity isn't strictly a rate but if the light is monochromatic and the path length is constant it sort of is.
$$I=I_0\exp(-\epsilon c l)$$

But the rate of absroption of photons is given by:
$$K=B\rho$$
Where B is the Einstein coefficient and $\rho$ is the spectral energy density which is proportional to the intensity of incident radiation.

Surely the second equation should mean that the intensity decays linearly? Would the absorbed intensity double if the concentration was doubled?

To answer your second question first, just use the Beer-lambert law. The first question needs us to go back a few steps and re-define the Einstein B coefficients in terms of the intensity which means that the 'new' B or $B^I=B^{\rho}/c$ and from now on for simplicity we just use B instead of $B^I$.

The intensity I is defined (per unit frequency interval) as flux across unit area per unit time thus has units  $\pu{W/m^2.s^{-1}}$ or mass time$^{-2}$ and thus is not a rate. The energy density $\rho$ means density of energy per unit volume and is related to I as $I=c\rho$ where c is the speed of light. (The rate of absorption is unchanged but written as $B^II$ rather than $B^{\rho}\rho$.)

The energy density $\rho$ has units $\pu{J.m^{-3}/s^{-1}}$ which is mass length$^{-1}$ time$^{-1}$. In these units the Einstein B coefficient has dimensions time mass$^{-1}$ or $\pu{s kg^{-1}}$ in SI units.

The Beer Lambert law derives from the differential form
$$-dI_{\nu} =I_{\nu}k_{\nu}dx$$ where for clarity  x is used to indicate length rather than l. The constant $k_{\nu} = \epsilon [c]$ at frequency $\nu$ and has dimensions of length$^{-1}$.

If the absorption band is wide we take a small element about frequency $\nu$, which is $\delta\nu$,  and calculate the rate of energy removal from the beam during absorption from state 1 to state 2 with populations $N_1$ and $N_2$ respectively, with Einstein coefficients for (stimulated) absorption $B_{12}$ and stimulated emission $B_{21}$
$$ -dI_{\nu}\delta\nu =  h\nu I_{\nu} [N_1B_{12}dx -N_2B_{21}dx  ]$$
rearranging gives
$$-\frac{1}{I_{\nu}}\frac{dI_{\nu}}{dx}\delta\nu=h\nu[N_1B_{12} -N_2B_{21}  ]=k_{\nu}\delta\nu $$

If the light intensity is low, the population of the upper level is only a minute fraction of that of the lower level and can be set to zero. In this case B can be obtained directly as $$B_{12}=\frac{k_{\nu}\delta\nu}{h\nu N_1}$$

Then integration over $\delta\nu$ gives the B value for the absorption band. The value of $N_1$ is effectively the total population.

This shows that the B values are related to the extinction coefficients as expected and that time does not come into the calculation. However, this is only part of the story.

A quantum calculation starting with the Schroedinger equation and using the dipole approximation and first order perturbation theory is necessary to explain the electric dipole interaction. The result of this calculation (see Atkins & Friedman Molecular Quantum Mechanics chapter 6 for gory details) is that the probability of absorption from state a to b is
$$P_{ab}= \frac{|V_{ba}|^2}{\hbar^2}\frac{\sin^2((\omega-\omega_{ab})t/2)}{(\omega-\omega_{ab})^2}$$
where $V_{ab}= <a|H|b>$ is the interaction energy, $\omega$ the radiation frequency and $\omega_{ab} = (E_a-E_b)\hbar$. This oscillating behaviour as a function of frequency is not ordinarily observed in thermal samples but has been observed in a cold molecular beam of HCN molecules. Time was made constant by passing the beam at constant speed through the fixed illuminated region (see Dyke et al. J. Chem. Phys. vol 57, 2277, 1972).

The rate of absorption is the time derivative of the probability which is
$$ R=\frac{|V_{ba}|^2}{2\hbar^2}\frac{\sin((\omega-\omega_{ab})t)}{\omega-\omega_{ab}}$$

and as the the absorption coefficient is directly proportional to the rate of absorption  $\gamma(\omega,t)=\beta R$ (where $\beta$ only contains constants) which indicates that near resonance absorption should increase in time. However, this is not what is usually observed for molecules at room temperature as a vapour or in solution. The reason for the generally observed behaviour is that the absorption is continually interrupted by first order processes, for example by collisions.

Suppose that the system is initially in state a and then the radiation field is turned on. The molecule is then in a superposition state involving a and b, for example $\Psi=c_a(t)\psi_a+c_b(t)\psi_b$ where the c are time dependent coefficients that move the system from a to b. A collision will destroy the superposition and return the molecule to state a or b each with a certain probability, then the radiation causes the superposition to grow again.

If we assume that the number of superposition states is n and that a first order process destroys them with lifetime $\tau$ , then $n(t)=n(0)\exp(-t/\tau)$. There is now a competition between the superposition state growing and being destroyed. The average absorption coefficient is obtained by averaging over the probability distribution of it being interrupted thus

$$\gamma(\omega)=\frac{\int_0^{\infty}\gamma(\omega,t)\exp(-t/\tau)dt}{\int_0^{\infty}\exp(-t/\tau)dt}$$
evaluating gives
$$\gamma(\omega) \propto  L(\omega_{ab}-\omega)$$
where L is the Lorenzian function
$$L(\omega_{ab}-\omega)=\frac{1}{\pi }\frac{\tau^{-1}}{\tau^{-2}+(\omega_{ab}-\omega)^2 }  $$
and examining the function the half-width  at half-height $\Delta \omega = 1/\tau$. The first order process thus explains why the absorption does not increase in time.  However, the experimentally line shapes on many molecules do not have a Lorenzian shape and the reasons for this are not that the idea of interrupting the superposition is wrong but that other effects dominate, doppler broadening for example, or in solution there is always inhomogeneous broadening. This is caused by interactions with solvent molecules which shift energy levels up and down and thus the solution effectively contains molecules with a mixture of energy levels. That this is so can be observed at very low temperatures (say 10K) where it is possible to burn a spectral hole in a sample and so remove a small part of the inhomogeneous population. The spectral hole produced is very narrow as expected from the calculation above.


Why does H bond to S in HSO3? In the structure of $\ce{HSO3-}$, we see that the H is bonded directly to the S, not to the O's, as would be the case in other oxyacids. Why is this the case?

There are two distinct forms of $\ce{HSO3-}$ (see picture below; all horizontal bars are formal $-$ signs).

Only the leftmost of these two forms can be called hydrogen sulphite (or ‘bisulphite’, if you really insist); the correct name of the rightmost form is sulphonate. The two differ in their chemical characteristics; hydrogen sulphite is, for example still acidic and has a lone pair on sulphur, while sulphonate is not.

Johansson, Lindqvist and Vannerberg from Göteborg solved the crystal structure of $\ce{CsHSO3}$ in 1980 and concluded that:

the $\ce{S-O}$ bond lengths are much closer to those in sulphate than those in sulphite ($\ce{Na2SO3}$)
the $\ce{O-S-O}$ angle is $113°$ and thus larger than the tetrahedral angle $\approx 109°$ (expected in sulphate) and much larger than the angle in $\ce{SO3^2-}$
the ion’s point group is $\mathrm{C_{3v}}$, so all three oxygens are equal
there is IR evidence of a $\ce{S-H}$ bond.

From this they deduced that $\ce{CsHSO3}$ be, in fact, cesium sulphonate, not cesium hydrogen sulphite.[1]

The first reference I found for the structure of a true hydrogen sulphite was the crystal structure of cis-$\ce{[Ru(bpy)2(HSO3)2]}$ by Allen, Jeter, Cordes and Durham from Arkansas in 1988. In their structure, $\ce{HSO3-}$ coordinates to the central $\ce{Ru(II)}$ ion via the sulphur atom: $\ce{[Ru(bpy)2(HSO3}\unicode{x2D}\unicode[Times]{x3BA}S\ce{)2]}$. This coordination is, of course, only possible if there is a lone pair on sulphur, and thus the structure must be that of a hydrogen sulphite.[2]

(In my introductory chemistry class, Prof. Klüfers stated that structures for both forms be known and published.[3] Alas, I didn’t find the crystal structure of a simple metal hydrogen sulphite anywhere quickly.)

It gets easier when dealing with the corresponding acids’ esters. The hydrogen on sulphonic acid can be replaced with organic residues creating the vast class of organosulphonic acids. Sulphonic esters are also possible; they would have an organic residue attached to both the sulphur directly (replacing the hydrogen atom) and one of the oxygens. Every mesylate or tosylate (in formulae: $\ce{OMs}$ or $\ce{OTs}$) is, in fact, a sulphonic ester. Sulphurous esters are less prevalent in organic chemistry but do exist (e.g. dimethyl sulphite, $\ce{OS(OCH3)2}$). Of course, the chemical properties of methyl mesylate ($\ce{H3C-SO2-OCH3}$) — a sulphonate — and dimethyl sulphite are substantially different.

After discussing $\ce{H2SO3}$ and its salts, it is important to mention the wrong assumption you proposed in your question:

  we see that the $\ce{H}$ is bonded directly to the $\ce{S}$, not to the $\ce{O}$’s, as would be the case in other oxyacids.

This is not generally true. Phosphorous at least displays a similar chemistry, even to the extent that ‘hydrogen phosphite salts’ cannot be isolated and phosphonate salts are isolated instead. (One can synthesise both trimethyl phosphite and dimethyl methylphosphonate, the corresponding phosphorous and phosphonic esters, though.) I am not firm in the oxyacid chemistry of heavier homologues of phosphorous and sulphur, but it would deeply astound me if this chemistry is not observed there at least to some degree.

Note that the halogens do indeed follow your proposed rule: all their oxyacids are monoprotic and contain only a single, oxygen-bound hydrogen in neutral state.

References:
[1] L.-G. Johansson, O. Lindqvist, N.-G. Vannerberg, Acta Cryst. 1980, B36, 2523. DOI: 10.1107/S0567740880009351.
[2] L. R. Allen, D. Y. Jeter, A. W. Cordes, B. Durham, Inorg. Chem. 1988, 27, 3880. DOI: 10.1021/ic00295a003.
[3] Refer to the internet scriptum of general and inorganic chemistry 1 (in German) as taught 2006 to 2008 and likely will be taught again in 2015.

(Note that the articles are most likely behind a paywall)


Do hydrogen peroxide stabilizers affect oxidizing ability I plan to use hydrogen peroxide for a redox reaction in oxidizing $\ce{I-}$ to $\ce{I}$.  I found 20 volume hydrogen peroxide as a hair developer, but it contains stabilizer.  I imagined these would have little effect on its oxidizing potential, but I wanted to make sure.  Will any of the common stabilizers used for hydrogen peroxide inhibit its oxidizing ability.

Shouldn't hinder your oxidation much, just keep in mind they will still likely be around when your reaction is finished.


Does quantum mechanics play a role in protein folding? Protein folding takes a very long time (relatively speaking) when thinking of quantum mechanical effect. However, for the initial micro-steps of folding, when an atom, or a configuration of atoms, can go (freely speaking) right or left (that is, to other possible configurations), in processes that take less than a picosecond, isn't QM relevant? that is, the atom has a probability to go to either places, in a QM sense, that it goes to both places, and due to interaction with the world, it's wave-function collapses or localizes and that is how the folding progresses?

There's several aspects to your question.

Does quantum mechanics play a role in protein folding? Yes. The origin of the van der Waals interaction is ultimately a quantum mechanical one. (At least the induced dipole portions: how the electrons move with respect to each other and an external electrical field is driven by quantum mechanics.) Also, while part of hydrogen bonding can be described by classical dipole-dipole interactions, there's a non-negligible portion which is due to non-classical electron interaction. (How much this last bit contributes, if at all, is a bit controversial, though.)

But that's all electrons. The other part of your question is asking implicitly about the quantum behavior of nuclei.

In molecular modeling we often use a viewpoint termed "the Born-Oppenheimer approximation". That is, while we allow the electrons to delocalize and exchange and be their quantum mechanical self, we treat the nuclei pretty much like classical particles. Using this approximation, we can get pretty decent correspondence with experimental results.

So the best we can tell, the quantum nature of atoms as a whole (that is, nuclei as opposed to electrons) isn't needed for things like protein folding. The nuclei are probably not substantially delocalized, they're probably not interfering with one another, and they're probably not quantum tunneling.

Probably. There's some evidence that quantum tunneling of nuclei in the process of enzyme catalyzed reactions. This is mainly hydrogens, though. In most of these experiments the quantum tunneling drops off dramatically when you go to deuterium or tritium nuclei. I'm unaware of any system where carbon/oxygen/nitrogen tunneling is seriously proposed - they're just too heavy.

So there may be some contribution to the quantum nature of nuclei to protein folding, but all indications at this point indicate that it would be minor. The quantum properties of electrons, though, are definitely important.


Difference between Gibbs Free Energy and Exergy I'm wondering if there is any difference between Gibbs Free Energy and Exergy because their definition seems to me the same $\Delta H - T_\text{env}\cdot\Delta S$.

Exergy represents the maximum amount of work that can be extracted from a system going in equilibrium with its environment, while the Gibbs Free Energy is evaluated at $T,P$ of the system in equilibrium with the environment.

Practically what is the difference, if any?

Exergy is equal to the Gibbs free energy if you happen to be in the isothermal--isobaric ensemble $(N, P, T)$, i.e. in a closed system if the pressure $P$ and temperature $T$ are constant. This is indicated in Eq. 7 of the Wikipedia page for exergy.


In what way could benzoin give Tollen's test? In Q29 of Joint Entrance Exam (JEE) 2016 India, the official answer key mentions that benzoin gives Tollen's test. However, I saw this post which says that it doesn't: Why do α-hydroxy ketones give Tollens' test?

Benzoin:

I'm very confused now. If benzoin does give the test, how? And what is the mechanism/intermediate involved?

I thought of all possible reactions benzoin may undergo in basic medium:

I did its aldol condensation but the product was too crowded and seemed quite unlikely. It didn't have any aldehyde group.
Another possibility seemed oxidation to benzil, but I'm not sure. I don't think $\ce{Ag+}$ is strong enough.
I was thinking about cleavage of central $\ce{C-C}$ single bond — as the hydroxyl $\ce{O}$ donated its lone pair to the carbon it is attached to, but am not sure on this too.

I couldn't find any related reaction of benzoin or any rearrangement on the internet.

Of your three "thoughts", you are correct that an aldol reaction is not an option. Not only is the product "crowded", but the reaction is reversible. [BTW, an aldol condensation occurs when water is eliminated from the initial aldol product. In the case of a benzoin aldol product, elimination of water is impossible.]

Your second premise is a good one. Benzoin (1) in the presence of alkaline Tollens' reagent, can be enolized (blue arrows) and the hydroxyl group deprotonated (red arrows). The enediolate 2 is a prime candidate for oxidation by Ag+. This species is the dianion of the enediol 6 tautomer of benzoin. The enediolate 2 is formed in the acyloin condensation of ethyl benzoate (7).
A one-electron oxidation of the enediolate 2 produces resonance stabilized radical anion 3. Protonation of 3 may occur at this point but a second one-electron oxidation produces the diradical 4, which is benzil (5). Under more vigorous alkaline conditions, Liebig's benzylic acid rearrangement of benzil occurs to afford benzylic acid (8), which itself is known to oxidize to benzophenone (9) under a variety of oxidative decarboxylation conditions. Neither of these reactions, i.e., the formation of 8 and 9, is expected to occur under the mild conditions of the Tollens' oxidation. Accordingly, your third idea is unlikely under the reaction conditions.


How to balance this redox reaction How to balance this redox reaction (using half reaction method):

$$\ce{Cu + HNO3 -> Cu(NO3)2 + NO + NO2 + H^2O}$$

My attempt:

$$\ce{Cu -> Cu(NO3)2 + 2e-}$$
$$\ce{3e- + HNO3 -> NO}$$
$$\ce{e- + HNO3 -> NO2}$$

So we multiple Cu half reaction by 2 to balance electrons and we get:-

$$\ce{2Cu + 2HNO3 -> 2Cu(NO3)2 + NO + NO2}$$

I would have solved further but the correct answer happens to be different.

Correct answer: $\ce{2Cu + 6HNO3 -> Cu(NO3)2 + NO + NO2 + H2O}$

How to balance this redox reaction (using half reaction method):$$\ce{Cu + HNO3 -> Cu(NO3)2 + NO + NO2}$$

You really can't balance this equation for two reasons.

(1) There is hydrogen on the left hand side, but none on the right.

(2) There are two competing redox reactions. One produces $\ce{NO}$ and the other produces $\ce{NO_2}$. The overall reactions are:$$\ce{Cu_{(s)}  +  4HNO3_{(aq)}  ->  Cu(NO3)2_{(aq)}  +  2NO2_{(g)}  +  2H2O_{(l)}}$$  $$\ce{3Cu_{(s)}  +  8HNO3_{(aq)}  ->  3Cu(NO3)2_{(aq)}  +  2NO_{(g)}  +  2H2O_{(l)}}$$

Thus the ration of $\ce{NO2_{(g)}}$ to $\ce{NO_{(g)}}$ is a kinetics problem. In concentrated acid the top reaction predominates and you get mostly the first reaction. In dilute acid the second reaction predominates and you get mostly the second reaction.

Now as an example let's look at the redox reaction of the first reaction.

Looking at the copper first we have: $$\ce{Cu_{(s)} -> Cu^{2+}_{(aq)} + 2e^{-}}$$

Just looking at $\ce{NO3^{-} -> NO2 + e^{-} }$ the charge balances, and the $\ce{N}$ atoms balance, but the $\ce{O}$ atoms do not balance and hence this is not a valid half cell reaction. The valid reaction is: $$\ce{e^{-} + 2HNO3_{(aq)}  -> NO^{2-}_{3} + NO2_{(g)}  +  H2O_{(l)}}$$

Using "redox" I can simply balance the electrons. So I need the Cu reaction as is, but I need to multiple the nitrite acid reaction by two. $$\ce{Cu_{(s)} -> Cu^{2+}_{(aq)} + 2e^{-}}$$ $$\ce{2e^{-} + 4HNO3_{(aq)}  -> 2NO^{2-}_{3} + 2NO2_{(g)}  +  2H2O_{(l)}}$$

Now that the electrons match, I can add the two equations:
$$\ce{Cu_{(s)}  +  4HNO3_{(aq)}  ->  Cu(NO3)2_{(aq)}  +  2NO2_{(g)}  +  2H2O_{(l)}}$$


How can collission theory of chemical reaction and Law of mass action be connected? From Wikipedia,

  Collision theory is a theory ... that qualitatively explains how chemical reactions occur and why reaction rates differ for different reactions. The collision theory states that when suitable particles of the reactant hit each other, only a certain percentage of the collisions cause any noticeable or significant chemical change; these successful changes are called successful collisions. The successful collisions must have enough energy, also known as activation energy, at the moment of impact to break the preexisting bonds and form all new bonds.

From Britannnica,

  Law of mass action, law stating that the rate of any chemical reaction is proportional to the product of the masses of the reacting substances, with each mass raised to a power equal to the coefficient that occurs in the chemical equation.

How can both explain and imply each other?

My theory: Due to more availability of reacting species (due to more amount), rate of reaction of increases. But it does not explain the reason of exponentiation of stoichiometric coefficients.

For exponentiation, we can say that it is actually repeated multiplication. Comparing two cases:

$$\ce{A + B -> AB}$$
Here, the rate expression is $r=K[\ce{A}][\ce{B}]$. This is because one molecule of $\ce{A}$ and one molecule of $\ce{B}$ is needed for the product to form.
$$\ce{2A + B -> A2B}$$
First, write it in this way: $$\ce{A + A + B -> A2B}$$
Now, rate expression is $r=K\ce{[A][A][B]}$.
That's it - $r=K[\ce{A}]^2[\ce{B}]$

An intuitive starting point for me to understand why concentrations are multiplied is probably to connect it with multiplication theorem in probability (for independent events).


What are the dangers of handling 0.12N sulfuric acid? I'm handling 0.12N of sulfuric acid for a science experiment that I am carrying out for a school project. I hear about all these terrible effects if it's ingested, inhaled, or gets in contact in skin. But how dangerous is it?

My safety equipment that I'm using currently are a pair of polyvinyl gloves, goggles, long sleeve shirt, apron, jeans, and closed toe shoes. However, I do not have a respirator and wondering if I should get one.

Would it be safe to handle this without a respirator? I will not be spraying the sulfuric acid directly in the air. However, I will be adding approximately 0.5mL of the 0.12N sulfuric acid into 500mL of water, putting it in a spray bottle, and be watering some plants with it. Will it still be dangerous even spraying a low concentration in the air?

Would the 0.12N concentration still be enough to cause chemical burns on my skin? To what extent and severity? Also, just to confirm, it is safe to add drops of the acid to water, but not vice versa? I would really appreciate the advice that you can provide me.

0.12 Normal = 0.06 Molar.

For comparision, concentrated sulphuric acid is 18 Molar.

So the 0.12 N sulphuric acid is over 99% water.

  Would it be safe to handle this without a respirator?

Yes, the vapor pressure of sulphuric acid is low, and it has already been diluted with considerable water, so there is no need for a respirator.

  My safety equipment that I'm using currently are a pair of polyvinyl gloves, goggles, long sleeve shirt, apron, jeans, and closed toe shoes.

Goggles are the most important safety equipment.  An eye wash station and shower are also standard lab safety equipment.

The gloves are good because liquids will not pass through them.  Ordinary clothing isn't really helpful because if you did spill something on that part of your body, the substance would absorb into the cloth and you would just be delayed in washing that part of your body by having to take off the clothes.

  Would the 0.12N concentration still be enough to cause chemical burns on my skin? To what extent and severity?

Pronged contact may cause some damage, but nothing comparable to the concentrated acid.  If you spill some on yourself, stay calm, and wash with soap and water. If you spill on your clothes, take the wet clothes off first, then wash with soap and water.

  Also, just to confirm, it is safe to add drops of the acid to water, but not vice versa?

This is always a good rule to keep in mind.  0.12N is dilute enough that little heat will be released from 0.5mL, but just follow the rule.

  However, I will be adding approximately 0.5mL of the 0.12N sulfuric acid into 500mL of water, putting it in a spray bottle, and be watering some plants with it. Will it still be dangerous even spraying a low concentration in the air?

Now you are diluting another 1000 fold.  The solution will be around pH 4, similar to some rain water.  I would still want you to wear goggles and not spray in the direction of anyone.


Is RS- more basic than RO-? The basic strength is determined by the ability of an ion or molecule to accept a proton. How do I know whether RSH is more stable than ROH?
(R is an alkyl group)

RobChem has already pointed out in his comment that your assumption is not quite correct.

Take a look at $\mathrm{p}K_\mathrm{a}$ values for $\ce{ROH}$ and $\ce{RSH}$ in water from the CRC Handbook of Chemistry and Physics and/or other online sources, such as this or that.

\begin{array}{lrr}
\mathbf{R} & {\ce{\mathbf{OH}}} &\ce{\mathbf{SH}}\\
\hline
\ce{H} & 15.7 &7.0\\
\ce{Et} & 15.9 & 10.6\\
\ce{(H3C)3C} & 18.0 & 11.7\\
\ce{C6H5} & 9.9 & 6.6\\
\end{array}

In all the cases above, the $\mathrm{p}K_\mathrm{a}$ value for $\ce{RSH}$ is smaller than that of $\ce{ROH}$.


Why does HCl both precipitate and dissolve Silver? Adding $\ce{HCl}$ to silver nitrate solution causes silver chloride to precipitate. Adding more $\ce{HCl}$ causes the silver to re-dissolve. What is happening to the silver chloride when it re-dissolves?

The precipitation reaction is:
$$\ce {Ag+ + Cl- -> AgCl_{(s)}}$$
The next reaction is:
$$\ce {AgCl_{(s)} +Cl- -> AgCl2- }$$


Alkyl halide reaction In the following the reaction ,

According to me the product should be compound  (A) due to substitution reaction.

But the product formed is the compound (B) , what could be the mechanism for the reaction for that product.

To reflect on the comments below:
In literature this reaction seems to actually produce product A and synthesis of product B seems to require much more exotic conditions than described in the question. As such I have to conclude that the correct answer would have be A.

Old answer:

Due to the polar aprotic solvent (DMF) one would think Sn2 is favoured as the reaction mechanism, however if the product is indeed B it would suggest an Sn1 mechanism where the aromatic ring stabilises the cation. One of the resulting resonance structures would indeed react with fluoride. Due to the stability of aromatics the product is formed by re-aromatisation.


Alkyne formation from vicinal dihalide Why must 3 equivalents of sodium amide be used here to form the alkyne? The part I specifically have issue with is in red.

From what I understand, we should just use 2 equivalents of sodium amide to get rid of the bromines and to form the alkyne. Another equivalent of sodium amide will form the acetylide anion. Removing two protons effectively forms the triple bond and eliminates the two halogens. Also, take a look at example 3 here ..

Why then does the book's solution manual want me to use 3 equivalents of sodium amide to get neutral acetylene, and another equivalent to get to the acetylide ion?

As you can see, my book's solution manual contradicts another book's discussion of obtaining alkynes from vicinal dihalides.

I think the comments basically have the answer - if you use only 2 equivalents of sodium amide, you will use 1.33 equivalents creating the alkyne and the remaining 0.67 equivalent will react with the alkyne after it is formed (an acid-base reaction) to form the acetylide salt -- probably a much faster reaction than the dehydrobromination reactions. So about 1/3 of the original dibromide will remain unreacted. You need the third equivalent of sodium amide to drive the reaction to completion.


Heating copper wire and sulfur powder In an experiment in school, we heated copper wire (Cu) and yellow sulfur powder together in a crucible. I was wondering why this reaction cannot produce Copper Sulfate. I know it produces Copper Sulfide but I want to know why don't oxygen from the air, sulfur, and copper combine to form copper sulfate.

Can anybody explain this in simple high school terminology?

By burying the copper wire with plenty of sulfur and then heating the mix, the wire becomes completely submerged in liquid sulfur. The liquid sulfur then gets to react with the copper, and any oxygen that could possibly oxidize the copper would have to diffuse into the liquid sulfur before reaching the copper. However, oxygen quickly reacts with heated sulfur forming sulfur oxides, so it never reaches the copper wire. This is all under ideal conditions, of course.

(Originally given as a comment by @Nicolau Saker Neto)


Plutonium having more oxidation states than samarium? Why does plutonium have more oxidation states than samarium?

Electron configuration of Pu: $\ce{[Rn] 5f^6 7s^2}$

Electron configuration of Sm: $\ce{[Xe] 4f^6 6s^2}$

I thought that only the valence electrons affected the oxidation states, so why does plutonium have more oxidation states (6,5,4,3) than samarium (2,3) whilst they both have the same valence electrons (including f-orbitals): ..$\ce{f^6}$..$\ce{s^2}$ ?

Elements such as scandium (Sc),Yttrium (Y) and Lanthanum (La) all have the same valence electrons (including d-orbitals): ..$\ce{d^1}$..$\ce{s^2}$ , but these elements have the same oxidation state : 3 and that doesn't change, even when going down that group and looking at actinium.

Does this have to do with the ionization energy, or the energy level or ....?

This is yet another interesting effect of the anomalous compactness of orbitals in the first appearance of each type of subshell ($1s$, $2p$, $3d$, $4f$, $5g$, etc). The solutions to the Schrödinger equation for electron wavefunctions in hydrogen-like atoms are such that these subshells are composed of orbitals with no radial nodes. This means the electrons in these subshells are closer to the nucleus than you might expect, and therefore they are more strongly bound.

The striking chemical similarity of the lanthanides comes from the fact that the $4f$ subshell is both anomalously compact and subjected to a high effective nuclear charge. The electrons end up being held so closely to the nucleus that they effective behave as core electrons, and participate very little in chemical interactions; lanthanides do not form coordination compounds using their $4f$ orbitals, and it is very difficult to go past an oxidation state of +3 in most because few conditions can compensate the large energy required to ionize a fourth electron coupled with the relatively weakly bound compounds they would form.

In the actinides, the $5f$ orbitals do not suffer from the same lack of radial node, and therefore are more chemically available. The first half of the actinides show many compounds with high oxidation numbers, and have significant coordination chemistry using $f$ orbitals. Curiously, as you go form curium to berkelium, there is a sudden increase in reluctance to display high oxidation numbers, and the actinides beyond curium tend to behave similarly (though of course exploration of their chemical properties is significantly hindered by their rarity and instability). This is likely because even though the $5f$ orbitals have a radial node, in the second half of the actinides they are subjected to such a high effective nuclear charge that become too strongly bound to display significant participation in chemistry.

This same effect explains why the metals in the first row of transition metals do not display as many stable compounds with high oxidation states compared to the second and third rows, as the $3d$ orbitals are much smaller than $4d$ or $5d$ orbitals.


Mercury assisted deprotection of dithiane 1,3 dithianes are used in umpolung reactions and can be deprotected by mercury containing compounds - but could anyone help explain what the exact mechanism is? Many texts seem to skim over this by explaining it is similar to acetal hydrolysis, but I'd really like a more detailed explanation as to the role of the compounds. A reaction I have seen involves $\ce{HgO}$, $\ce{HgCl2}$, $\ce{CH3CN}$ and $\ce{H2O}$. If it is similar to acetal hydrolysis, the $\ce{H2O}$ is probably to complete the final step of attacking the $\ce{S=H}$ bond, but what about the initial step?

I am assuming one sulfur atom forms a bond to the $\ce{Hg}$, but which Hg compound would it be in this case, the $\ce{HgO}$ or the $\ce{HgCl2}$? What is the role of the $\ce{CH3CN}$ in this case? Thanks!

$\ce{Hg^2+}$ has a high affintity to thiols and other sulfur functional groups, and therefore $\ce{HgCl2}$ forms a Lewis acid-base adduct with the dithiane. The cyclic sulfonium cation undergoes ring opening, and after nucleophilic attack by water, a second Hg-S bond under ring closure is formed. Deprotonation finally yields the deprotected carbonyl compund and the cyclic mercury dithiolate.


Can non-deuterated solvents be used for 13C-NMR (and in fact other nuclei)? Basically, my question is can you used a non-deuterated solvent for 13C NMR? I'm thinking that there might be some problems with locking?

If using a deuterium lock, then normally you would require, as minimum, about 10% deuterated solvent. Without any deuterated solvent, it is not possible to obtain a spectrometer lock (unless using a 19F lock channel, but that is unlikely, and is a hardware requirement).
It is, however, perfectly feasible to acquire spectral data for all nuclei without any deuterated solvent, and hence, without a spectrometer lock. When running without a lock, you need to turn the lock sweep off, to hold the spectrometer field constant. This does not provide the same level of field stability as a lock, but is more than adequate for short-medium 1D experiments that do not require very high resolution. Using gradient shimming, it is also possible to shim on non-deuterated solvents and achieve the same lineshape specifications.
Running long experiments (such as an overnight 13C{1H} experiment), without a lock will give you linewidths that are slightly broader than usual; the consequences of this are that (i) your resolution is reduced (rarely a problem with 13C), and (ii) your signal-to-noise will be reduced (and therefore longer experiment times may be required...leading to further broadening and longer experiment times...)
Not claiming to be the first to report about this technique, the Hoye group published two papers in Organic Letters (2004OL953 and 2004OL2567) with examples of application of no-D $\ce{^1H}$-NMR spectroscopy, covering sample preparation and data acquisition, too.  Especially the second one, outlining how to determine the concentration of LDA or Grignard reagents by addition of 1,5-cyclooctadiene as internal standard may be a reliable and faster alternative to the titrations once you are used shim manually for samples of this type.  But speak and obtain clearance in advance with your NMR manager before running your no-D sample; some interfaces to the spectrometer tend to be less puzzled to record NMR with a lock after your sample (e.g., elder Varian), than the more modern (e.g., Topspin Bruker) to prevent unnecessary delays in the queue of samples following yours.


Stereochemistry & mechanism of β-cyano enone methylation I am helping a friend with some upper-level mechanism problems. We came across this one. This is my proposal. Do you agree/disagree? Thank you for any input.

Source question:

Mechanism I developed from the paper @user55119 graciously provided. Appreciate it! Any refinements I appreciate. :)


Gibbs Free Energy and Entropy I understand Enthalpy, the exchange of energy between products and reactants. But what is Gibbs Free Energy and Entropy? I know that Gibbs free Energy is the difference between the change in Enthalpy - [(the change in entropy)*the Temperature]. However I am not able to understand the physical meaning. What does this really mean? On the molecular scale where is it stored? If you can make it less abstract than it is right now, that would help a lot.

I am assuming once I understand Entropy, I should be able to understand Gibbs Free Energy. So, my question on Entropy is, where is entropy stored? I know it is the total disorder in the ENTIRE system, not just one molecule, the units for Entropy are J/K, and by disorder I mean the preference of the system to go to a chaotic/disordered state.

So, when solving for Gibbs Free Energy, what are we finally getting by subtracting the energy from the system (dS) from the reactions exchange in energy (dH).

Like you said Gibbs free energy is the enthalpy minus TdS so basically it's the energy available (free energy) after you take the thermal energy out of the system. The TdS term is the energy due to random thermal fluctuations so sometimes it's more useful to remove that from the enthalpy and deal with the free energy especially when trying to determine spontaneity of reaction.


When is a reaction said to be under kinetic control, is it not easily reversible? If we consider this reaction: Compound 1 + Compound 2 --> Compound 3

I understand that if a reaction is under kinetic control, the less stable (e.g. a mono-substituted diene) is formed and if the reaction is under thermodynamic control the more stable stable (disubstituted diene) is formed.

However, if we wanted to go from the kinetic product to the thermodynamic product, would this reaction be considered as being under kinetic control? In other words, are the reactants under kinetic/thermodynamic control, or is it the products?

I think this is more of a terminology question.

A product formed under kinetic conditions may or not be reversible under the conditions of the reaction. Here is an example of one that is reversible under the reaction conditions. Kinetic control is determined by the difference in activation energy (ΔEa) of the two pathways. The thermodynamic equilibrium is determined by ΔΔGo. Acetoacetic ester (AAE 1) is a mixture of keto and enol forms. Bromination of AAE in an inert solvent rapidly gives α-bromoAAE 2 in addition to HBr. This is the product that is formed faster having a lower activation energy. This reaction is also  reversible, initiated by HBr (microscopic reversibility). Eventually, the more stable γ-bromoAAE 3 is formed from AAE 1 by enolization toward C4 (γ-position) of AAE under the influence of HBr.


Heat produced when magnesium is reacted with acid Calculate the heat produced when a strip of $\ce{Mg}$ metal with a mass of $\pu{0.0801 g}$ is reacted with $\pu{50.0 ml}$ of $\pu{1.0 M}$ $\ce{HCl}$ to raise the temperature of the water by $\pu{7.6 °C}$ (change in temperature is $\pu{7.6 °C})$ Calculate the heat produced when a mole of this metal is used.

I have this so far:
$$q = mc\,\Delta T$$
$m =\pu{0.0801 g}$ and the change in temperature is
$$\Delta T = T_2 - T_1 = \pu{7.6 °C} - \pu{4.184 K} = \pu{3.416 K}.$$
Is that right? I have no idea what the specific heat $c$ is. Is it the specific heat of $\ce{Mg}?$
What do I do with the $\pu{50.0 ml}$ of $\pu{1.0 M}$ $\ce{HCl}?$

It can help to imagine the experimental setup.

Imagine a thermally insulated reaction container, let's call it calorimeter, that is filled with 50 ml 1 M $\ce{HCl}$ ( = a solution of hydrogen chloride gas in water).

We have to measure a temperature change ($\Delta{} T$), that comes from an exothermic reaction. Where would we best measure this temperature change? Our reaction medium apparently is a good choice. Therefore, there's most likely a thermometer (insulated from the outside) in good thermal contact with the water.

Now we let the magnesium strip fall into the reaction medium (using some clever mechanic) while our calorimeter is closed), let it react and measure the temperature change until the temperature reaches some final (maximum) value.

As you wrote, the temperature change ($ \Delta{} T$) is +7 °C.

I don't see that an initial temperature was given, but that's not important. We have the $ \Delta{} T$.

Now we need a specific factor that correlates the heat liberated in the reaction with the temperature change in a specific medium.

Here, the specific heat capacity of water comes into play.

There you go.


Does the order of a written redox couple matter? My understanding is that a redox couple is an unordered pair of two conjugate species. So we could speak of the redox couple consisting of $Ag$ and $Ag^+$; depending on whether a given redox couple is undergoing oxidation or reduction, we could then write a half equation involving the species given in the redox couple.

In my textbook, it is given as a convention that all redox couples are expressed in the form Ox/Red. So for instance, it states "at anode containing the $Zn^{2+}/Zn$ redox couple, the oxidation half equation is $Zn \rightarrow Zn^{2+} + 2e^-$". This convention makes sense to me, since it doesn't seem that the couple itself refers to a half equation in a given direction. The order of the half cell line representation, however, does change the order to $Zn | Zn^{2+}$.

However, some texts reverse the order of the written redox couple, which I would interpret as either them writing the couple in the order of the current half equation (which seems wrong) or that they are ignoring the order because it is unimportant.

So I wondered whether anyone could clarify, is a redox couple just a pair of species which exist at a given half cell, or do we need to change the order of the written couple depending on the reaction occurring at that half cell?

My understanding is that a redox couple is an unordered pair of two
  conjugate species

This is conceptually perfect and there is no problem when we talk about electrode potentials of half cells because as I had mentioned in your earlier queries, the electrode potential value and its associated sign do not know nor care how you write the half cell.

What is creating confusion in your mind is that you are trying to abbreviate a half cell. Abbreviating a half cell is of no use, because it is creating a problem of ordering.

All we care about the order is when we write a full cell. For that, we must follow the IUPAC recommendation and rules in order to prevent chaos in the published work. There your half-cell is no longer an unordered pair.

I used to remember the IUPAC recommendation with a mnemonic: Write Reduction on the Right.

This has to be parsed as follows: Reduction is on the right, so you must order the copper half cell in such a way that it appears as reduction.

On the left, of course, we have an oxidation half cell and it must be written as such.

Hope that clarifies your confusion.


Why is activation energy drawn in a potential energy diagram in reactions? I learned in chemistry class that activation energy is the av. KE that the colliding particles need in order to successfully react. However, they are always drawn as a "hill" in a PE diagram.
I do not understand why they are classified as PE, when they are KE.

The activation energy is not the average kinetic energy that the colliding particles need in order to successfully react. Instead, it is the energy barrier between reactants and products. Often, the energy required to cross the barrier comes from the kinetic energy of collisions.

To describe the rate of a reaction there are two models, the Arrhenius model and the more sophisticate Eyring model:

The Arrhenius model uses the activation energy, defined as the energy required to reach the transition state; the "energy" here is of various nature, not only kinetic an includes the chemical potential.

The Eyring model uses the Gibbs energy of activation for the transition state that includes an enthalpy and an entropy term ($ΔG^‡= ΔH^‡ – T ΔS^‡$). In comparison, for a unimolecular, one-step reaction, the activation energy can be approximately expressed as $E_a = ΔH^‡ + RT$.

There are different y-axis in these diagrams as explained in a related answer:
Y-axis of the reaction co-ordinate graph. Depending on the model and reaction you want to study you can use the Gibbs energy (Eyring model), energy in general (Arrhenius model) or enthalpy in case you want to show the heat absorbed or released in a reaction at constant pressure.

Why does your graph show potential energy?

In simple terms, potential energy is stored energy while kinetic energy is the energy of motion. Potential energy is converted into kinetic energy, therefore you can account for the latter including it in the overall potential energy.


Is my electrolysis emitting compounds from additional reactions? I am separating water into hydrogen and oxygen via electrolysis.

The electrolyte solution is sodium hydroxide (NaOH) in deionized water.

Electrodes are 99.96% pure nickel.

All gases and liquids are perfectly clear.  Even after running for hours, the electrolyte solution is crystal clear and colorless.

However, the reaction seems to be emitting something that burns the nose and throat of everyone in the room.

Is that some sort of gas from a side reaction at one of the electrodes, or is it simply bits of NaOH being carried into the air by the splashing bubbles?

TL;DR: Your suggestion that the problem is "simply bits of $\ce{NaOH}$ being carried into the air by the splashing bubbles" is very likely correct.

You have a very simple system, meaning a minimal number of variables to consider in order to eliminate other possibilities. For example, if you were using $\ce{NaCl}$ as the electrolyte, then chlorine gas could be the irritant. But with $\ce{NaOH}$ as the electrolyte there really is no side reaction that could produce such an irritant.

On the other hand, $\ce{NaOH}$ itself can irritate sensitive tissues like mucous membranes, and even skin for that matter. According to one of your comments, you are using a $\ce{NaOH}$ strength of a bit over 1 molar. Droplets of this solution will eventually be neutralized by $\ce{CO2}$ in the air, but at that strength it should stay around long enough to elicit the symptoms you described.


Energy levels database Does anybody know where to find a database for atomic and molecular energy levels, with the corresponding energy values for each level? It would be needed just for small molecules. I tried the NIST website but it only has the ground state energies.

The Pitt Quantum Repository can get you what you want, at least for molecular species.

Go to the PQR at http://pqr.pitt.edu.
Search for a molecule, like benzaldehyde.
Click "Detailed" on the properties pane.
You can see HOMO and LUMO in the list.
Scroll down to the bottom of the page and click on the link to "Get JSON data".
The JSON file will open in your browser or you will download. You should be able to open it with a text editor. There will be a list of orbital energies for the set of molecular orbitals computed from the valence atomic orbitals. For example, for benzaldhyde.

"homo": -10.168,
"lumo": -0.712,
"orbitals": [
"-31.27648",
"-30.67710",
"-26.81219",
"-26.23865",
"-23.01445",
"-21.46816",
"-20.14594",
"-18.05465",
"-16.39792",
"-15.70143",
"-14.95762",
"-14.67105",
"-14.63486",
"-14.38581",
"-13.35746",
"-12.66484",
"-12.26937",
"-10.28746",
"-10.23037",
"-10.16786",
"-0.71215",
"-0.10196",
"1.01720",
"2.30473",
"3.35595",
"3.89337",
"4.34241",
"4.54954",
"4.71066",
"4.82113",
"4.99402",
"5.26380",
"5.29598",
"5.54486",
"5.70599",
"5.83103",
"5.89240",
"6.03437"
],


Why is chemical accuracy defined as 1 kcal/mol? "Chemical accuracy" in computational chemistry, is commonly understood to be $1~\mathrm{kcal\over mol}$, or about $4~\mathrm{kJ\over mol}$. Spectroscopic accuracy is $1~\mathrm{kJ\over mol}$, and that definition has intuitive sense. However, where does the $1~\mathrm{kcal\over mol}$ quantity come from?

From Wikipedia:

  A particularly important objective, called computational thermochemistry, is to calculate thermochemical quantities such as the enthalpy of formation to chemical accuracy. Chemical accuracy is the accuracy required to make realistic chemical predictions and is generally considered to be 1 kcal/mol or 4 kJ/mol.

Short answer: the goal of "thermochemical accuracy" for computational chemistry is to match or exceed experimental accuracy. Thus, ~1 kcal/mol comes from the typical error in thermochemical experiments.

The drive began with John Pople, who begin the modern effort to consider "Model Chemistries," comparing the accuracy of different methods across many molecules and often multiple properties. He realized that for thermodynamic properties, one could approach the accuracy of experiments. (See, for example his Nobel lecture).

  As the model becomes quantitative, the target should be that data is reproduced and predicted within experimental accuracy. For energies, such as heats of formation or ionization potentials, a global accuracy of 1 kcal/mole would be appropriate.

He then started work on composite methods like G1, G2, G3, etc. that could approach predicting many chemical properties to this accuracy.


What equipment is used for two-dimensional IR spectroscopy and how can the molecular structure of a compound be resolved? On the Wikipedia page about two-dimensional IR spectroscopy it is said that 2D IR spectroscopy can be used to study the

  molecular dynamics with femtosecond time resolution.

According to Wikipedia it

  has for example been used successfully to study hydrogen bond forming and breaking and to determine the transition state geometry of a structural rearrangement in an iron carbonyl compound.

Now I'm curious. What equipment is necessary to perform 2D IR experiments? Does somebody have experience with 2D IR spectroscopy and could explain how the geometry of the system is inferred just from the (2D) IR spectrum.

You can find a schematic drawing of the box-CARS 2D IR setup in the publication
Enhancing signal detection and completely eliminating scattering using quasi-phase-cycling in 2D IR experiments, Robbert Bloem, Sean Garrett-Roe, Halina Strzalka, Peter Hamm, Paul Donaldson

https://pdfs.semanticscholar.org/e134/cdbb98a8d899a17848f8242c8160d4913f0f.pdf


How could a scientist from the 1800s to the 1920s run spectroscopy of chemicals? How would such a scientist be able to get the spectroscopy of chemicals?  Without access to modern computing or electronics how would it be possible?

Any video link or reference link are much appreciated here.

Light from a flame, the sun, or a star, was passed though a narrow slit, then directed through a prism (F) and observed.  The prism is rotatable about a vertical axis to adjust the wavelength range being observed.  Bright lines in the spectrum are emission transitions and dark lines are absorption transitions.

Further progress was made using diffraction gratings instead of prisms and using photography instead of naked eye observation.  Photography permitted observation of infrared and ultraviolet lines.

By these techniques, known elements were detected in the sun and other stars, and helium was discovered in the sun before it was discovered on Earth, all before 1900.

For further information see Spectroscopy History.


Lifetime of a (reversible) hydrogen fuel cell system I am trying to determine how long a system that constantly converts water into H and O and then back will last, as in roughly how many cycles. Now this system can either be made of say a reversible fuel cell or a non-reversible fuel cell combined with a dedicated electrolysis machine (if there is a difference in lifetime). As far as I can tell, platinum is used (almost) everywhere as catalyst and as far I can determine it seems to corrode somewhat throughout use, especially in terms of splitting the oxygen in the fuel cell. I am no professional though and I can only find limited information so I would appreciate it if someone could please tell me roughly how long a system like this will last and how.

Fuel cell manufacturers quote a range of different lifetimes which depend on many factors. Backup cells tend to have lower lifetimes as the start-stop process degrades them compared to continuous operation. On the other hand, use as engines in vehicles requires a reasonably long life.

Impure fuel will damage the platinum (CO for example will poison the catalytic surface). But platinum is recyclable so either poisoning or other factors won't necessarily be a fundamental barrier to use (some companies take used cells back at the end of life and reuse the platinum). Membranes and electrodes also degrade mechanically depending on duty cycle and any mechanical stabilisation in the design.

Nedstack quotes lifetimes ranging from 2,000 hours for backup cells (which presumably means years of actual operation as they only get used when other power is cut which isn't a routine happening) and 8,000-16,000 hours for continuous operation (which is two years). Tests of fuel cells in vehicles have reported lifetimes of 7,500 hours (which presumably relates to several years of use in practice), though they seem to need more maintenance than diesel engines.


Why cyclohexane-1,4-dicarbaldehyde does not give aldol condensation? I was recently came to know as a fact that cyclohexane-1,4-dicarbaldehyde does not give the aldol condensation reaction in spite of possessing α-hydrogens. I was told that this is due to steric reasons, but I am trying to come up with a more specific reason, which does not require prior knowledge of this fact. Can anyone guide me towards a better explanation?

Edit 1: I am able to understand that there is absolutely no scope for an intramolecular aldol condensation. But what I am asking is whether even the aldol REACTION would take place, as basically the exact wording of the book is " cyclohexane-1,4-dicarbaldehyde gives zero aldol reaction products". So I am confused that what part is aldol reaction and/or condensation

Your claim is worded very poorly, and taken in the broadest sense, is not true. However, it's more complex than an outright yes or no.

If we're interested in fundamental reactivity, it suffices to look at cyclohexanecarbaldehyde, as there is no reason that the dialdehyde should be any different (apart from the likelihood that it gives very messy reactions in practice). The main issue is that "aldol reaction" is a very unspecific term, which is why you (and everybody else) is having trouble explaining it.

Firstly, it can refer either to condensations where you eliminate a molecule of water after the initial addition of an enolate, or to additions where the product is a β-hydroxyketone.

Formally, any condensation involving cyclohexanecarbaldehyde as a nucleophilic component is not possible, because the α-carbon is tertiary. This includes the self-condensation, where it acts as both the nucleophile and electrophile. After forming a new bond, the α-carbon will become a quaternary centre, which cannot take part in elimination of water.

However, if CyCHO is the electrophilic component (i.e. a different enolate reacts with CyCHO), condensations are still possible. The crossed aldol condensation with acetone is well-documented in the literature:

If we go back to simple aldol additions, which still count as aldol reactions, there are many examples of CyCHO acting as the electrophilic component in an aldol addition. Here is a nice example (Tetrahedron: Asymmetry 2000, 11 (15), 3211–3220):

On the other hand, aldol additions with cyclohexanecarbaldehyde acting as the nucleophilic component are fairly rare. This reflects a general trend that aldol reactions generating quaternary centres (i.e. aldol reactions involving α,α-disubstituted enolates, such as the enolate of CyCHO) are traditionally very difficult to accomplish. This is mainly because of steric hindrance – the reacting centre of the enolate (the α-carbon) is bulky.
The above is a good rule of thumb, but there still are examples of CyCHO acting as the nucleophilic component in aldol reactions. Many of them involve secondary amine catalysis (example below taken from Org. Lett. 2003, 5 (23), 4369–4372), although there is one example of the lithium enolate of CyCHO attacking a carbonyl group.

There are also plenty of examples where the silyl enol ether of CyCHO reacts with π-electrophiles. Silyl enol ethers are weaker nucleophiles than lithium enolates, so generally Lewis acid catalysis is required to form more powerful electrophiles. Still, these are considered to be aldol-type reactions (specifically Mukaiyama aldols).


Positive work in the Carnot cycle In the Carnot cycle the overall work done on the piston is positive. I am having trouble with the real meaning of that. Does positive work mean the piston is moved up and has, over the course of the cycle, gained potential energy?

The Carnot cycle uses an idealized heat engine to illustrate basic thermodynamic principles. Net work is done either by or on the system, depending on the direction in which it is run. As part of this idealization, the piston is massless and frictionless. The piston merely serves as a moveable impermeable boundary between the system and the surroundings—i.e., as an interface that allows the system to do work on the surroundings, and visa versa. The piston exerts no force of its own, so no work is done by or on the piston; the piston merely serves as a conduit for force exerted by the system or surroundings, and is not part of either.
To more specifically address your question, since the piston is massless, its gravitational potential energy is zero, independent of position.
Also, as this is a cyclic process, the system is returned to its original state at the end of each cycle, and thus the piston is returned to its original height. [Hence, as Chet Miller points out in his comment, even if the piston had mass, its PE would be unchanged at the end of each cycle.]


Why does molten plastic (PLA) stick to metals (AL)? I know this is broad so I will narrow:
Why does molten PLA stick to Aluminum?

I ask because I'm looking for a way to prevent/reduce the friction of the plastic inside the aluminum barrel of a 3D Printer. I believe that I can treat the surface of the aluminum with a High Temp enamel however I need to be sure of the physics of why the plastic sticks in the first place. Is it the surface finish, or is the molecular attractiveness? If it is the finish, then I can make it smooth. If it is static attraction then I can charge the barrel etc.

Aluminum reacts with oxygen in the air to produce aluminum oxide, which forms a fine coating on any exposed aluminum surface protecting against further oxidation.

$$\ce{ 4Al + 3O2 -> 2Al2O3}$$

Aluminum oxide is polar, amphoteric, and hygroscopic: it absorbs moisture from the atmosphere. The surface of aluminum oxide is likely coated with $\ce{Al-OH}$ groups. The polar aluminum oxide and aluminum hydroxide surface is attracted to the polar ester functional groups in PLA.

Aluminum surfaces can be rendered nonpolar by silanization. This article may be behind a paywall for you. I apologize. I could not find a good open access reference. The Wikipedia article on silanization is pretty terrible. The surface is rendered nonpolar, which may decrease the interaction with PLA.


Find the boiling point of a compound given a pressure I need to find the boiling point (in degrees centigrade) of ethanol on a day when the atmospheric pressure is $780\text{ torr}$.

I know that the molar mass of ethanol is $46.07\text{ g/mol}$ but I have no idea how to continue. Can you tell me what equations should be used or how is it solved, please?

Thanks!

You could use the Clausius Clapeyron equation:

$$ \frac{\mathrm{d}p}{\mathrm{d}T} = \frac{\Delta H_\mathrm{v}}{\Delta V \cdot T}$$

Since the change in pressure is rather small, it is fair to assume the molar enthalpy of evaporation to be constant and use the value given for standard conditions, which should be easy to obtain.

The other simplification you can make is to assume that the volume of the liquid is very small compared to the volume of the gas:

$$ V_\mathrm{l} << V_\mathrm{g}$$
so that
$$\Delta V = V_\mathrm{g}-V_\mathrm{l} \sim V_\mathrm{g}$$

Now by using the ideal gas law and expressing volume as molar quantity,

$$v=\frac{p}{RT}$$

you get the ODE

$$ \frac{\mathrm{d}p}{\mathrm{d}T} = \frac{\Delta H_\mathrm{v}\cdot p}{T^2}$$

from whose solution you get a nice equation to solve your problem.


Software for compound database creation\enumeration The only program I have found is Plexus Suite from chemaxon. (I asked for trial, but no answer to me yet).

The task is: create variation of compounds, where is scaffold and different substituents:

And I need to create dataset/database of all possible variations.

How to deal with this problem? Or I need to use some scripting to make the database?

UPD1: best option - creation of SMILES (Simplified molecular-input line-entry system), 3d structure is automated and easy step if there is correct SMILES/connectivity of molecule (which is the same, as smiles)
Using XYZ - it means, you should make substituent database in correct position for adding.  I am asking for software with functionality

https://www.youtube.com/watch?v=s4CgtUgGuzk

Vega ZZ has the module for database enumeration, but it is working as Python script.

This is typically called library (or scaffold) enumeration. Doing it in SMILES is usually pretty easy by script, but there are a few other options:

KNIME
SMILIB

But it's very easy to write a script like this, e.g. (in python)

r1 = [ "", "C", "N" ]
r2 = [ "H", "C", "c3ccccc3", "C(=O)O" ]
r3 = [ "", "C", "C#N", "O" ]
scaffold = "Xc(cc1c2)ccc1c(Y)cc2-C(C=C1)C=C1Z"

for x in r1:
    for y in r2:
        for z in r3:
            print scaffold.replace('X', x).replace('Y', y).replace('Z', z)

You can then process the resulting SMILES strings with your program of choice (e.g., Open Babel).


Evaporation of concentrated formic acid and semi-concentrated sulfuric acid at room temperature I recently looked at a reaction where CO is synthesized using formic acid and sulfuric acid, which seems to be one of the easier to control reactions for this, and obviously ventilation etc is necessary.

However I was wondering, is ventilation only necessary because of the dangerous nature of CO, or also because of the vapors or formic and sulfuric acid? I was quite into chemistry in school and after, but I do not remember that much, so I feel like I should know this but I cannot recall anymore.
So at room temperature, if you would take a small amount of each of these acids and keep them either separately or mixed together, how fast/how much would evaporate and create dangerous acid vapor?

I might sound like a kill joy but I strongly suggest that you do not attempt to make carbon monoxide. The reaction you describe is a not one for the faint hearted. I would like to point out that aerosols of sulfuric acid are very harmful to the lungs, eyes and the rest of your body.

I think that any work done in the lab with concentrated sulfuric acid should be done in a fume hood to protect you against aerosols of the acid. The fumes from the formic acid are going to be corrosive.

In my youth we used to work on the open bench with glacial acetic acid, this is not a good thing to do as the whole lab soon becomes like a viniger factory. I fear that with formic acid being handled outside the hood the lab will soon become rather unpleasnent.


How to calculate the number of moles present in a litre of saturated solution? The solubility product of a salt having same molecular formula as $\ce{Ca_3(PO_4)_2}$ is $1.08\cdot10^{-28}$. Then find the number of moles of ions present in the saturated solution.

My approach: I found out the solubility of the salt i.e.: $x=10^{-6}M$

This means $x$ moles of salt are present in a litre. As 1 mole of the salt yields 5 moles of ions I thought the answer would be $5x$.

Help required.

For your salt, calcium phosphate, or $\ce{Ca_3(PO_4)_2}$, this is the correct dissolution equilibrium:

$\ce{Ca_3(PO_4)_2 \leftrightharpoons 3Ca^{2+} + 2PO_{4}^{3-}}$

Upon dissolution of one unit of calcium phosphate, we get five units of ions total, true, but the solubility product is defined as follows (from the IUPAC Gold Book):

The product of the ion activities raised to appropriate powers of an ionic solute in its saturated solution expressed with due reference to the dissociation equilibria involved and the ions present.

For our purposes, we will use concentrations instead of activities. Concentrations are a rough gauge of activities in very dilute solutions (such as this one involving the semisoluble salt, calcium phosphate).

Also since you mentioned this was a multiple-choice test, I highly doubt that the following hydrolysis reaction will be relevant, even though at such low concentrations of phosphate anion, the phosphate anion should be nearly completely hydrolyzed.

$\ce{PO_{4}^{3-} + H_2O ->HOPO_3^{2-} + HO^-}$

Back to the problem. Upon dissolution of one unit of calcium phosphate, we should get 3 units of calcium ion and 2 units of phosphate ion.

Therefore, based on the IUPAC definition:

$\ce{K_{sp}~=~[Ca^{2+}]^3[PO_{4}^{3-}]^2}$

Taking $\ce{x}$ to be one unit of calcium ion:

$\ce{K_{sp}~=~[Ca^{2+}]^3[PO_{4}^{3-}]^2}=[3x]^3[2x]^2$

Solving the above yields $x = 10^{-6}$.

Therefore, with respect to your original question, the number of moles of calcium phosphate present initially can be number greater than 0 moles. For the amount of calcium phosphate present at equilibrium, then yes, $5x$ appears to be the right answer.


Why does two phenylhydrazine molecule react with glucose in osazone formation? From Wikipedia:

  It involves two reactions. Firstly glucose with phenylhydrazine gives glucosephenylhydrazone by elimination of a water molecule from the functional group. The next step involves reaction of one mole of glucosephenylhydrazone with two moles of phenylhydrazine (excess). First phenylhydrazine is involved in oxidizing the alpha carbon to a carbonyl group, and the second phenylhydrazine involves in removal of one water molecule with the formyl group of that oxidized carbon and forming the similar carbon nitrogen bond.

Why doesn't it first react with the formyl group and stop with just one phenylhydrazine? That is, why doesn't the nitrogen of phenylhydrazine attack the $\mathrm{sp^2}$ carbonyl carbon (a nucleophilic addition reaction)?

Why does it oxidize glucose first and then proceed?

As according to modern mechanism it has first reacted with formyl then goes for tautomers and to form an intermediate..which further form iminoketone ,as phenylhydrazine is in excess then it again react with the c=o formed on C2 and then with carbon first and give osazone .  And as further reaction stops by H- bonding between H and Nitrogen of each phenyl hydrazine which forms chelation and makes it stable.


Hybridisation of the negatively charged carbon atom in a carbanion Given the carbanion, $\ce{R3C-}$, the carbon is $\mathrm{sp^3}$ hybridized, unless it is participating in resonance. This is clear from its steric number.

However, I am having trouble in drawing its orbital diagram. I know that if this was a neutral carbon atom, it would look like this:

$\mathrm{3s}$: _
$\mathrm{2p}$: ↑ ↑ _
$\mathrm{2s}$: ↑↓
$\mathrm{1s}$: ↑↓

Which would hybridize to this:

$\mathrm{3s}$: _
$\mathrm{2sp^3}$: ↑ ↑ ↑ ↑
$\mathrm{1s}$: ↑↓

Now if I was to examine the carbanionic state, would I simply be adding one electron to the top most diagram, then re-hybridize that? Or would I add two electrons because the carbon is not sharing the electrons with any other atom? That's another question I have with orbital diagrams. Do we break each bond homolytically, and base our diagram off of that? This would give:

$\mathrm{3s}$: _
$\mathrm{2p}$: ↑ ↑ ↑
$\mathrm{2s}$: ↑↓
$\mathrm{1s}$: ↑↓

I am not sure how I would rehybridize that. Have I drawn the radical form?

Hybridization is a concept, which is often misconceived. One has to understand, that it is a mathematical concept, which explains a certain bonding situation (in an intuitive fashion). In a molecule the equilibrium geometry will be formed respecting steric and electronic interactions and possibly interactions with a medium or external field. The geometric arrangement will not be formed because a molecule is hybridized in a certain way.

Hybridisation is therefore only a result of a certain geometric arrangement of atoms in a molecule. A linear combination of all available (atomic) orbitals will form molecular orbitals (MO). These can then be rearranged (recombined) in a certain way to form localised molecular orbitals (LMO). These can then be used to interpret a bonding situation in a simpler theory.

Each LMO is expressed as a linear combination of the atomic orbitals, hence it is possible to calculate contributions of the atomic orbitals and describe these also as hybrid orbitals.

In a very simple case methane, the $\ce{C-H}$ bonds are formed from contributions of carbon's $\ce{2s, 2p}$ and hydrogen's $\ce{1s}$. For the carbon this may be interpreted as a $\ce{2sp^3}$ orbital.

In the simple case of $\ce{{}^{-}CH3}$, as Uncle Al already pointed out is isoelectronic to ammonia, the bonding situation may be also described in terms of $\ce{2sp^3}$ orbitals. As well as ammonia (or amines in general) it may also undergo (nitrogen) inversion.
(source)

In this inversion scenario the hybridisation of the carbon (nitrogen) changes from $\ce{2sp^3}$ to $\ce{2sp^2}$ to $\ce{2sp^3}$.

With larger carbanions this becomes a little more complicated. As long as $\ce{R=Alkyl}$, hyper conjugation is not strong enough to prevent pyramidalisation. However, the inversion berrier is furthermore reduced.

If any of the $\ce{R}$ groups are different from alkyl substituents, then I would expect that the environment around the carbon becomes (more) planar, i.e. the contribution of carbon's $\ce{2s}$ orbital increases. That may have various reasons, but some kind of conjugation or hyperconjugation will be involved. When these $\ce{R}$ groups become pretty bulky, I would expect the same, due to steric interactions.

Summary

Please understand hybridisation as a tool to understand bonding situations. Therefore it is a result of a geometric alignment of atoms in a molecule. It is not the driving force to form a certain geometric constellation.


The electron configuration and electron density distribution in singlet oxygen The molecular orbital schemes for singlet ($\mathrm{^1\Delta_g}$) and triplet oxygen ($\mathrm{^3\Sigma_g^-}$) are typically given as shown in the image below.

Figure 1: Molecular orbital schemes of two types of singlet oxygen and triplet oxygen with the highest energy electrons highlighted in red.

It is pretty clear that the dioxygen molecule, if just observing the position of the nuclei, must be $D_\mathrm{\infty h}$ symmetric. This implies that both $\unicode{x3c0}^*$ bonds are symmetry equivalent. Therefore, it seems logical that in the $\mathrm{^3\Sigma_g^-}$ triplet state the two orthogonal $\unicode{x3c0}^*$ bonds are each populated by a single electron, leading to an overall rotation-symmetric linear molecule.

In the $\mathrm{^1\Delta_g}$ singlet state, this degeneracy seems to be destroyed, just by looking at the MO scheme. We now seem to have one populated $\unicode{x3c0}^*$ orbital and one unpopulated one, going by my understanding of this scheme.

What does this imply for the electron density distribution of $\mathrm{^1\Delta_g}$ singlet oxygen? Is it valid to assume that two (arbitrary) opposite sides have a higher electron density than the two other orthogonal sides? How can this be interpreted?

Probably not a full answer, but hoping to point readers in a useful direction.

If we denote the two $\pi^*$ orbitals as $\pi_x^*$ and $\pi_y^*$ respectively, then the spatial wavefunctions for each state can be written as follows (all lower orbitals are ignored):

$$\begin{align}
{}^1\Sigma_\mathrm g^+ &: 2^{-1/2}[\pi_x^*(1) \pi_y^*(2) + \pi_x^*(2) \pi_y^*(1)] \\
{}^3\Sigma_\mathrm g^- &: 2^{-1/2}[\pi_x^*(1) \pi_y^*(2) - \pi_x^*(2) \pi_y^*(1)] \\
{}^1\!\Delta_\mathrm g &: \begin{cases}\pi_x^*(1) \pi_x^*(2) \\ \pi_y^*(1) \pi_y^*(2) \end{cases}
\end{align}$$

For the ${}^1\!\Delta_\mathrm g$ case, one would expect a spatial degeneracy of 2, because the letter $\Delta$ in the term symbol indicates that the quantum number $|\Lambda| = 2$. Therefore, there is one state with $\Lambda = +2$, and one state with $\Lambda = -2$. On the other hand, the $\Sigma$ term has $|\Lambda| = 0$ and hence is spatially non-degenerate (the $^3\Sigma$ term is triply degenerate due to spin).

(Digression: $\Lambda$ represents the projection of the angular momentum along the internuclear axis. As such, there is no further projection quantum number ("$M_\Lambda$") that can take values $-2, -1, 0, +1, +2$. In comparison, in atoms, $\ell$ indicates the total angular momentum and $m_\ell$ the projection of this angular momentum onto the $z$-axis. For more information about term symbols of diatomic molecules, I suggest reading this link. There are many other sources on the Internet, but many are sloppy with notation, which only leads to confusion down the road.)

So, the MO diagram above - which ties the ${}^1\!\Delta_\mathrm g$ state to only one wavefunction - is incomplete. The ${}^1\!\Delta_\mathrm g$ state corresponds to two possible wavefunctions, which are symmetry-equivalent; there is no "preference" for the $x$-axis over the $y$-axis.

If you want to go further than what I've written, then I'd point you to the case of the boron atom which I asked about earlier, which is exactly analogous to this. The point about boron is that it has one electron in the 2p subshell. Does this go into the $\mathrm p_x$, $\mathrm p_y$, or $\mathrm p_z$ orbital? In the case of dioxygen, do the paired electrons go into the $\pi_x^*$ or $\pi_y^*$ orbital?

As you said, it does not make sense for us to assign the electron(s) to any of the orbitals in particular, as that leads to an asymmetric electron density distribution. However, I must admit that I don't fully understand the CASSCF calculation there, and I never quite got round to following up on that question, so you'll have to ask somebody else about that.


Solubility of lead phosphate I've read that lead phosphate is very slightly soluble and so any addition of protons shifts the equilibrium back to the solvated ions. What I can't understand is why the lead phosphate is more soluble in hydroxide or basic solutions also.

Source- Vogel's qualitative analysis

The precipitation of lead phosphate would be due to the following reaction.

$$\ce{3Pb^{2+} + 2PO4^{3-} <=> Pb3(PO4)2}$$

In acidic solution the phosphate ions would be protonated, thus reducing the concentration of $\ce{PO4^{3-}}$ and increasing the solubility of $\ce{Pb^{2+}}$.

But the precipitation of the phosphate is not lead's only reaction in alkaline solution.  Lead can also form a hydroxide precipitate. In strongly alkaline solution lead can also form various soluble hydroxide complexes which would reduce the concentration of $\ce{Pb^{2+}}$ thus increasing the solubility of $\ce{Pb3(PO4)2}$.


Producing methane, reactants and products How do I balance this reaction when producing methane using the reactants:

$$\ce{CH3COONa(s) + NaOH(s) + Ca(OH)2 -> ?}$$

The products are partly

$\ce{CH4(g) + Na2CO3(s)}$

What else is missing in the products? Furthermore, how do I balance this?

Thanks in advance.

The answer is:

$$\ce{2CH3COONa(s) + 2NaOH(s) + Ca(OH)2(s) <=> 2CH4(g) + Na2CO3(s) + CaCO3(s) + H2O}$$

You have to get same number of equal atoms in each side of the reaction (this technique is called mass balance). If there were charges, they must be balanced also.

Example

There are $2\times 3 + 2\times 1 + 2\,$ hydrogen atoms in left side, and there are $2\times 4 + 2$ $\ce{H}$ in the right side. So both sides give $10$ hydrogen atoms.

I hope it helps.


How could a rechargeable battery recharge itself? I'm an engineering student preparing for my graduation, in one step of my graduation project I used a rechargeable battery(1.2 volt), i was shocked when this battery is starting to recharge it self until it reach 1.04 volt (not fully) for many many cycles and immediately when the voltmeter reads 0 volt from it, when i asked my doctors they told me that maybe its chemical issue.

so is there any explanations please?

As the comment indicated, it is probably equilibrating. When you are discharging the battery, you are causing the materials in the electrodes and the electrolyte to rearrange. The surface of the electrodes are where the reactions are happening, so the products of the discharge reactions tend to accumulate at or near the surfaces of electrodes. This causes the measured potential to go lower.

Once the discharge stops, the contents start to homogenize within the cell bringing the discharge products away from the electrodes and bringing the material from the charged state closer to the electrodes. This in turn increases the measured potential.


What kind of energies are present in an electron? I learnt that an electron loses energy as it goes closer to the nucleus and gains energy as it goes further from it. But what type of energy is involved in this process? If it's kinetic energy, then the electron must also lose its speed, but does this happen? In addition, how do I calculate the energy gained or lost due to the electron's movement?

From your question I assume you are talking about an electron bound to a nucleus, probably hydrogen. Since the mass of the nucleus is ~1836 times larger than the mass of the electron, we may assume the nucleus to be stationary. In a classical model of the problem, we thus have an electron with a certain velocity $v_n$ and kinetic energy $\frac{1}{2}m_ev_n^2$ at a position $r_n$ from the nucleus where it is bound by a Coulombic potential $-Ze^2/(4\pi\epsilon_0r_n)$ (where for hydrogen $Z=1$). The total energy is given by the sum of the two

$$
E_n=\frac{1}{2}m_ev_n^2-\frac{Ze^2}{4\pi\epsilon_0r_n}
$$

In order for the electron to make a (classical) stable trajectory around the nucleus, the electrostatic force should be equal but opposite to the centripedal force, hence

$$
\frac{m_ev_n^2}{r_n}=\frac{Ze^2}{4\pi\epsilon_0r_n^2}
$$
from which it follows that $\frac{1}{2}m_ev_n^2=Ze^2/(8\pi\epsilon_0r_n)$ and the total energy is given by
$$
E_n=\frac{Ze^2}{8\pi\epsilon_0r_n}-\frac{Ze^2}{4\pi\epsilon_0r_n}=-\frac{Ze^2}{8\pi\epsilon_0r_n}
$$
This equation shows that the total (average) kinetic energy in the hydrogen atom is half the size and opposite to the (avarage) potential energy, i.e., $\langle E_\text{kin}\rangle=-\frac{1}{2}\langle E_\text{pot}\rangle$, a result that is also predicted by the virial theorem.

Using Bohr's postulate concerning the quantization of angular momentum, the last expression can be recast in the familiar expression for the hydrogen energies.


How many gauche butane interactions are present in cis-1,2-dimethylcyclohexane? We were asked to find out the number of Gauche-Butane interactions for this compound:

I proceeded by drawing the Newman Projection for this chair conformation as follows and marked the Gauche-Butane interactions (as shown by red curves):

They told me that I was almost correct and asked me to try once again but I can't find out where I probably went wrong.
Can someone guide me out in this regard?

While cyclohexane does have gauche interactions, they are irrelevant because chair cyclohexane is taken as the reference point. The axial methyl group in cis isomer 1 has two gauche interactions with the ring as illustrated in red in 1a and 1b. A third gauche interaction is between the two methyl groups in 2. By comparison the trans isomer 3 only has gauche interactions between the methyl groups.

The difference in the number of gauche interactions - namely, two -  between these epimers is approximated by the difference in their heats of combustion[1] which is $\pu{1.54 kcal/mol}$. [1,2 cis: $\Delta H^\circ_\mathrm{comb} = \pu{-1248.31 kcal/mol}$; 1,2 trans: $\Delta H^\circ_\mathrm{comb} = \pu{-1246.77 kcal/mol}$.] A typical value for a gauche interaction is $\pu{0.9 kcal/mol/interaction}$.
References:

Walter H. Johnson, Edward Prosen, and Frederick D. Rossini, Heats of Combustion and Isomerization of the Eight $\ce{C8H16}$ Alkylcyclohexanes. Research Paper RP1S12, Volume 39, July 1947. PDF via NIST


I have an idea through utilizing soda lime or other basic substances to help solve climate change I have an idea through utilizing soda lime or other basic substances to help solve climate change. I want to see if there are some possible errors or other negative implication that I am disregarding.

I came across soda lime during one of my biology textbooks and noticed that it has the ability to absorb $\ce{CO2}$. The first thing that popped into my head was utilising soda lime ($\ce{CaHNaO2}$) to help solve global warming.

My idea is having large or wide buildings that are made with soda lime on the edge that will "suck up" the carbon dioxide. This carbon dioxide can then be used for various different ways in which the community and the society can benefit from.

convert it to electricity or fuel (it has been said that there is a way of doing this as in Cornell University Research - YouTube video). Essentially, the electricity made would be redistributed to the city  to use.
export the gas to bread industries (the yeast produces carbon dioxide that may be used for brewing and for bakery)
if possible, change the carbon dioxide to oxygen and release oxygen.

I know I might not be aware of some factors and it may seem idealistic, but I want to hear from the community. Is what I am proposing realistic enough? Or are there some practical problems with this approach?

My worry is that the energy requirements for making soda lime, caustic soda or some other carbon dioxide adsorber will be so high that overall it will result in the emission of more green house gases.

Years ago I was touring a waste disposal site where they react waste sulfuric acid with limestone. The boss of the site told me that some years ago someone had asked if he could scrub out the carbon dioxide from the reaction tank with sodium hydroxide or a similar reagent. He gave the same answer as I am giving to us when he recalled the tale.


cross exchanging the coefficients and valences of cations and anions with each other in chemical formulas Why are the coefficient and valences of iron and bromine cross exchanged with each other in the formula for ferric bromide, FeBr3?

The formulas for ionic compounds can be predicted from the charges of their ions. Cations and anions combine to form ionic compounds with a ratio of cations to anions that will produce a neutral (uncharged) unit. Generally, compounds are uncharged species.

Most binary ionic compounds are comprised of a metal and a nonmetal. In the most generic case, a metal cation with a charge of $+m$ and a nonmetal anion with a charge of $-n$ will combine in a $n:m$ ratio to balance charge. The example below, I am using $\ce{M}$ to represent a generic metal and $X$ to represent a generic nonmetal.

$$\ce{nM^{m+} + mX^{n-} -> M_n N_m}$$

This ratio guarantees charge balance. In the generic case the total charge of all cations in the formula unit is $+m\cdot n$, while the total charge of all the anions in the unit is $-n\cdot m$. These two values are equal and we get a compound. It is possible that $n:m$ is not the simplest whole number ratio and we can simplify. See some examples:

Calcium $\ce{Ca^2+}$ chloride $\ce{Cl-}$ gives $\ce{CaCl2}$.
Iron(III) $\ce{Fe^2+}$ sulfide $\ce{S^2-}$ gives $\ce{Fe2S3}$.
Magnesium $\ce{Mg^2+}$ oxide $\ce{O^2-}$ gives $\ce{Mg2O2}$, which can be simplified to $\ce{MgO}$.

We need to be careful to only think this way for ionic compounds. Covalent compounds can be more complex, and their are often multiple binary compounds based on varying oxidation state. Additionally, for covalent compounds, we do not simiplify the coefficients to the lowest whole number ratio, as the formula unit describes discrete molecules with fixed composition.

For example, $\ce{C4H8}$ (butene) is a different compound from $\ce{C2H4}$ (ethylene).


The nomenclature of mineral acids My basic knowledge about nomenclature is: A mineral acid is called hypo when it has one less oxygen than the original one,
a pyro acid when 2 molecules of acid add up with release of $\ce{H2O}$
a meta acid when one molecule releases $\ce{H2O}$ and
a per acid when it has one extra $\ce{O}$ in the structure connected with a peroxide bond.

Question 1

I was taught in my class that this structure:

is correct because when we are drawing structures of a hypo-,per-,meta- or pyro- oxyacid, one must try to maximise the covalency. Why is that so?

Also, usually in pyro- acid we find a $\ce{M-O-M}$ bond where $\ce{M}$ is the central atom(s). Like $\ce{P-O-P}$ in pyrophosphoric acid as show below:

Then shouldn't the structure of pyrosulfurous acid be like this:

 Why is the structure not similar to pyrophosphoric acid?

Question 2

We know that in hypo-acids, there is one less $\ce{O}$ in the empirical formula of the acid i.e. Hypophosphoric acid should have one less $\ce{O}$ than in $\ce{H3PO4}$ i.e. it should be something like $\ce{(H3PO3)_$n$}$. If we put $n=2$ we have $\ce{H6P2O6}$.

But why is its formula $\ce{H4P2O6}$?

I might lack some information about it or maybe I've understood the concept wrong. Please help.

The English names are not very systematic, they just give hints.  In principle you have to either memorize the name-structure correspondence, or know "all" acids of sulfur, so based on the name you can guess what the structure it corresponds to.

In addition, the hypo having one oxygen less is just a shortcut. What the names refer to is the oxidation state of central atom from lowest to highest.
Broadest range of acids are for chlorine: hypchlorous - chlorous - chloric - perchloric.

Answer 1: Please refer to the Wikipedia page on sulphur oxo-acids
.
The $\ce{H2S2O5}$ acid drawn is hypothetical one, but it's fine to have a name for it. Note, that sulfur is in low oxidation state.
In the pyrophosphoric acid, the phosphorus is in high oxidation state and has extra oxygen in between them.

For the summary formula $\ce{H2S2O3}$ you try to mix these properties, putting "extra" oxygen between sulfurs in low oxidation state. This won't work. The acid you should get when you attempt to obtain this one will be the thiosulfuric acid, where one sulfur mimics oxygen as $\ce{S^2-}$.

Answer 2: Chemistry of phosphorus is distinctly different from sulfur.
Phosphoric acid and Phosphorus acid for the acids with one phosphorus atom.
Your formula $\ce{(H3PO3)_{$n$}}$ with $n=2$ is not really justified, the name corresponds just to the formula on hypophosphoric acid.


What gas has the highest refractive index? Coming from this answer on the Worldbuilding stack, I'm trying to find a gas with a high refractive index in the visible light wavelengths. The highest index on this site, is chlorine at 1.000773. Is this the highest a gas can be? Or is there a fundamental limit on the refractive index of a gas?

As a note, I am not looking for the densest gas or the densest inert gas. Density does not seem to be correlated with refractive index in this case, as chlorine is not especially dense and the very dense perfluorobutane is not especially refractive.

The question has to be more specific to get a specific answer.

Even in the table linked in the OP, bromine gas (1.001132) has a higher index of refraction than 1.000773.

At zero degrees C, 5462 Angstroms, and atmospheric pressure, The Refractive Index Dispersion and Polarization of Gases (1936) reports:

  $\ce{Si2H6}$ 1.0016574

$\ce{Si2H6}$ , disilane, is a gas at room temperature and atmospheric pressure.

The index of refraction is extremely dependent upon pressure.

As an introduction, see INDEX OF REFRACTION AND DISPERSION OF SEVERAL GASES IN CERENKOV COUNTER USE where indexes of refraction of gases as high as 1.02963 (for 182.79 psig ethane) are measured.


Why might my limelight experiment have failed and how can I improve it? I was teaching the lime cycle to some children yesterday. I had a limestone chip (about 10g) held at the tip of a blue Bunsen flame for several minutes. It decomposed and disintegrated but only glowed very slightly, and was mostly orange with a bit of white at the tip.

It certainly gave off white light, but nowhere near the quantities required to light a theatre stage, which is what this stuff is famous for, not by a few orders of magnitude.

I tried sitting it on gauze and bringing in a second Bunsen flame from the side, but this did not improve the effect. I recall seeing quite a bright light when this was shown to me as a child.

Any theories on what went wrong? Any gotchas for doing this kind of experiment?

Perhaps your flame wasn't hot enough. Real limelights used oxyhydrogen torches, not wimpy natural gas flames.

Or perhaps your limestone chip wasn't just $\ce{CaCO3}$. If it was dolomitic limestone, there was a lot of $\ce{MgCO3}$, too, and $\ce{MgO}$ doesn't do the trick like $\ce{CaO}$ does.