What is a sacrificial agent? I came across this reading a journal article. Here I have the context:

  Here we report for the first time the synthesis of $\ce{TiO2}$ NPs
  on a layered $\ce{MoS2}$/graphene (MG) hybrid for use in
  photocatalytic H2 production. It is shown that the activity of
  the $\ce{TiO2}$ NPs is significantly enhanced by the presence of this
  layered MG cocatalyst. In this case, ethanol was used as a
  sacrificial agent, as it is a sustainable and renewable source and
  showed very good performance with this photocatalyst;
  however, other sacrificial agents (e.g., glycerol) can be used
  to make this strategy feasible.

Google results came up with similar instances.

My guess is that it would be a chemical consumed in a reaction, but one that doesn't contribute to the structure of the product. Thoughts?

"Sacrificial agent" usually refers to something variable, but with little value, that can be used to produce something specific or carry out a specific function. In this case, an alcohol (ethanol or glycerol) is sacrificed to act as a source of hydrogen gas, while the titanium species is a catalyst that is activated by light to allow generation of hydrogen.


Does bicarbonation require water? In an aqueous solution of $\ce{NaOH}$ exposed to the atmosphere one can probably expect some degree of $$\ce{NaOH(aq) + CO2(g) -> NaHCO3 (aq)}$$

But, what would happen if just a solid block of $\ce{NaOH(s)}$ was placed in some completely anhydrous environment, populated only by excess $\ce{CO2(g)}$ at STP? After all,

$$\ce{NaOH(s) + CO2(g) -> NaHCO3 (s)}$$

is technically a spontaneous process with $\Delta G^\circ = -76.91\ \mathrm{kJ/mol}$, but I can only think of three possibilities:

Nothing happens—no water (i.e., as a catalyst), no reaction!
A passivation layer of $\ce{NaHCO3(s)}$ forms on the surface, preventing the rest of the $\ce{NaOH(s)}$ inside from reacting^
No passivation layer—the $\ce{NaOH(s)}$ just keeps reacting away until it's all $\ce{NaHCO3(s)}$!

Rather

$$\ce{2 NaOH(aq) + CO2(g) -> Na2CO3 (aq) + H2O}$$

and analogically on the solid $\ce{NaOH}$.

Bicarbonate in aqueous solutions cannot survive the excess of hydroxide, forming carbonate.

$$\ce{HCO3^{-}\ (aq) + OH- (aq)<=>> CO3^{2-}(aq) + H2O}$$

The same is further enforced by dehydration effect of solid hydroxide.

$$\ce{NaHCO3(s) + NaOH(s) -> Na2CO3(s) + H2O}$$

Only in excess of $\ce{CO2}$ Is formed bicarbonate:

$$\ce{CO3^{2-}(aq) + CO2(aq) + H2O <=>> 2 HCO3^- (aq) }$$


How to deterime the number of equivalents of Grignard reagent used for this reaction? I know that the amount of substance of Grignard reagent used is 5 moles. But the answer given is 2. Is this anything to do with 'equivalents'?

No you are right it should be five. there are three dissociative protons thus one from the secondary amine, and two from the hydroxy groups.  These must be reacted first then you need two to react with the ester. The question was made by a human and the book answer of 2 is probably just an error from such.


What causes the anti-siccative effect of soot (amorphous carbon) as a pigment in oil paints? Oil paint containing carbon black (soot, charcoal) pigment is known to polymerize ("dry") a lot slower than if it has other pigments. I understand this is not just due to the absence of pigments which are themselves siccatives (drying agents), but is also a direct "slowing" effect of the carbon black in the paint. The pigment supplier Kremer Pigmente confirms that it is a "slow drier", and the book "Artists' pigments, c. 1600-1835" by Rosamond D. Harley indicates:

  They [authors on the matter] invariably state that a drier should be added to it to accelerate its exceptionally slow drying time.

What is the nature of this effect? Why does carbon black slow the drying of oil paints?

According to the Wikipedia page on oil-drying agents (emphasis added):

  An oil drying agent, also known as siccative, is a coordination compound that accelerates (catalyzes) the hardening of drying oils, often as they are used in oil-based paints. This so-called "drying" (actually a chemical reaction that produces an organic plastic) occurs through free-radical chemical crosslinking of the oils. The catalysts promote this free-radical autoxidation of the oils with air.

The exposed surfaces of carbon black are highly functionalized, in many cases with oxygen-containing moieties.  The Birla Carbon website indicates that this surface chemistry "is a function of the manufacturing process and the heat history of a carbon black and generally refers to the oxygen-containing groups present on a carbon black’s surface." Per panel 3 of the figure at Asahi Carbon's page, "Carbon black properties," these functional groups can include at minimum carboxyl, hydroxyl, and carbonyl (aldehyde/ketone) functionality.  The section of the Wikipedia page on carbon black describing its surface chemistry indicates that quinonic, lactonic, and phenolic groups may also be present.

Most of the above-mentioned functional groups can participate in radical reactions, including reactions with the active species involved in paint hardening.  These side reactions of the carbon black siphon off the active radical species in the paint into pathways that do not contribute to the hardening/drying process.  The Lamp Black page on the Natural Pigments website provides support for this (emphasis added):

  However, [lamp black] is not generally recommended for use in oils, because carbon is an anti-oxidant that greatly slows the drying time of oil paint.


Is it right to compare the molarity of a substance with the probability of its molecules reacting? I was watching this video on Khan Academy (about the equilibrium constant), and the analogy that they made really got me thinking. I wanted to verify whether it was completely correct, or just an analogy to explain reaction rates to high school students. I'm well aware that this is an equilibrium constant, and the probability of molecules reacting relates to kinetics. However, this is how it was explained in the video, and I'm trying to make sense of it.

Would anyone please guide me as to whether this analogy, of the molarity and the probability of a molecule reacting, is really true and mathematically correct or whether its just a simplification for students? (I'd recommend that you watch the video first before trying to answer.)

Also, please note that I'm not saying that what Khan Academy is teaching is wrong, I'm just trying to clarify whether what they're teaching is a simplification, or a scientifically accurate explanation.

As Alchimista points out, two different principles are at work here. Concentration, or how many of the species are available, and are these species able to react when they do meet.

The transcript of the video at 3:32 states:

'some constant that takes into account things like temperature and how the molecules are actually configured. Because it's not dependent on them just being there. You have to worry about their kinetic energies...'

So, Molarity will not affect the probability that any two molecules that meet will react, but it does absolutely affect rate of reaction. If the reaction is possible (there is sufficient activation energy) then the more concentrated the substance, the more likelihood of reactable species meeting, and therefore a greater rate of reaction.


Why did the alpha-particles in Rutherford's experiment not collide with the electrons? In Rutherford's experiment to show the existence of nucleus in an atom, the alpha-particles were exposed on the surface of certain metal i.e. gold. He observed that more than 99% of these particles were able to go straight and pass through the gold atoms. Thus he was able to conclude that atoms are mainly empty space.

But why did these particles not collide with the electrons present in the gold atoms? I think that the reason behind it is that the speed of the electrons travelling in orbits did not cause any obstacle in the travelling path of the alpha particles.

Protons and neutrons have a mass approximately 1836 times greater than an electron, ignoring relativistic effects and nuclear binding energies. There are 4 of these baryons in an $\mathrm{\alpha}$-particle. There is no way a puny electron could do much to deflect an $\mathrm{\alpha}$-particle that is almost 7500 times more massive. This is like if a 150-lb human tried to block a 747 (the 747-100's maximum takeoff weight is only about 6400 times greater than the person's).


How can I prevent my spin-coated silica thin film from cracking when I heat it up? I am spin-coating a porous silica solution on a glass microscope coverslip, and then heating the resulting thin film to $\pu{350 °C}$, in an attempt to create a thin film of porous silica sol gel. I am trying to replicate only the porous sol gel portion of the work of Yantasee et al.:
  Nanoengineered electrochemical sensor based on mesoporous silica thin-film functionalized with thiol-terminated monolayer

The problem I am encountering is that my heated sol gel is cracking and partially delaminating. This is unacceptable. Can anyone offer advice on how to prevent this cracking when heating to $\pu{350 °C}$?

Relevant Information

The recipe for the silica precursor solution is:
2.578 g quantity of Pluronic F-127 surfactant was dissolved in 3.430 g of deionized water and 8.846 g of ethyl alcohol. Then 0.108 g of concentrated $\ce{HNO3}$ was added into the mixture. Finally, a 5.0 g quantity of TEOS was added to the solution.
 The solution was sonicated for 8 hours and then spin-coated at 3000 rpm onto the surface of a 12 mm glass microscope cover slip.

 Film thickness: ~10 μm.

Possible Causes

 Perhaps I am using the wrong type of glass for the substrate, and the respective thermal expansions are causing cracking at the interface. If so, what relatively cheap glass could I use?

 Maybe I need to rigorously prepare the glass substrates by cleaning them with a solution of acid (or base) and solvent. Could an ill-prepared substrate contain contaminants acting as nucleation sites for crack formation when heated? If so, what best method is used by these groups to prepare their slides? I can't seemed to find a paper on sol gel thin films where they describe their substrate preparation.

The reference you sited has used gold surface to make the thin film. If the method you wanted is the one in that reference, folloeing is how they descibed it in experimental section:

  The modification of a gold electrode array on microchip with a SH-FMS thin film consisted of two steps. First a mesoporous silica film was deposited onto the electrode surface by the spin-coating technique. Then the thiol functional groups were immobilized inside the mesopores of the film by self-assembly chemistry. To prepare the silica film solution, a 2.578 g quantity of Pluronic F-127 surfactant was dissolved in 3.430 g of de-ionized water and 8.846 g of ethyl alcohol. Then 0.108 g of concentrated HNO3 was added into the mixture. Finally, a 5.0 g quantity of TEOS was added to the solution. The molar ratio of TEOS : water : ethanol : HNO3 : Pluronic F-127 was 1 : 8 : 8 : 0.05 : 0.008. The solution was shaken for 24 h, then spin-coated at the speed of 3000 rpm onto the bare surface of a gold microelectrode array on microchip (part No. M1450110, Microsensor System, Inc.) for 30 s. Fig. 1a shows the schematic of the microelectrode array. The specific electrode array had 50 finger pairs (15 microns finger width and finger spacing) with gold as the electrode material and quartz as the substrate. The electrical connection for the array was covered with plastic tape before spin coating with the film solution. After the spin coating, the tape was removed and the film-coated electrode was calcined on a hot plate at 350 °C for 8 min in air.

Hope this is helpful.


Activity coefficient greater than 1 From Practical skills in Biology by Allan Jones et al.

  At high concentrations of certain ionic solutes, $\gamma$ (activity
  coefficient) may increase to become greater than unity.

My questions : If activity coefficient $\gamma$ $=$ $  {a}/{[C]} $ where $a$ is activity which signifies effective concentration and $[C]$ is concentration, how can $a$  be greater than $[C]$ ?

In chemical thermodynamics, activity (symbol a) is a measure of the “effective concentration” of a species in a mixture, in the sense that the species' chemical potential depends on the activity of a real solution in the same way that it would depend on concentration for an ideal solution.
(From Wikipedia.)

So the main point is that that activity is just a measure of the deviation from the ideal solutions, because the laws you are using in physical chemistry  assume that solution is ideal. But usually it is not true. Laws can work for real life with a certain precision, but you should always remember about their "ideality" and the non-ideality of our world.


Why rotate orbitals in a CASSCF calculation? I'm trying to get the rate constant for the reaction of propene to give propenyl (radical) and H.

Using Molpro, I calculated the energy for the optimized geometry of the reactant using CASSCF.

It is required at one point to rotate two orbitals (the one with energy just below the HOMO and one in the valence shell).

I don't understand the point in doing this; what's the purpose?

The reason for rotation of orbitals (really the reordering of orbitals) is that the starting orbitals are usually not in the correct order for any sort of MCSCF calculation (CASSCF, RASSCF, GASSCF, adding correlation, ...). Doing a single CASSCF calculation with no "preparation" beforehand should force most programs to start with Hartree-Fock-like orbitals; this is the reference wavefunction $\Psi_0$. Hartree-Fock includes almost no static correlation to speak of, which can manifest in an incorrect ordering of the orbitals when compared to MOs that do have static correlation accounted for.

From here:

  Reorder your orbitals if necessary before running the MCSCF. Except for the most trivial cases, the orbitals that belong in the active space are not the orbitals that come straight out of your single-reference calculation. If they were, then you might not be running MCSCF to begin with.

Poor convergence of an MCSCF procedure is a good sign that the active space orbitals are not the correct ones. It is not a matter of the ordering within a space, since the energy is invariant to rotation/swapping within a space (inactive/frozen, active, or secondary/unoccupied). It is that an orbital Hartree-Fock predicts as a frozen core orbital should be a valence-type orbital in the active space, etc.

One of the reasons that MCSCF calculations are multi-step calculations and shouldn't be done in one shot is the generation of (hopefully) better orbitals than HF ones, inspection and any necessary rotation, then the MCSCF calculation. The best starting orbitals for MCSCF are other MCSCF orbitals, but you probably don't have those. Usual choices are MP2 or CI natural orbitals, which gives an indication of the total size of the active space and adds some dynamic correlation to "patch up" the lack of any real correlation in HF orbitals.

In my limited experience, HF orbitals are a fine starting point as long as the order is correct.


Room-conditions supercritical fluids? Are there any reasonably obtainable supercritical fluids that I could, say, run my hand through? Wikipedia makes it sound like there are plenty of room-temperature ones, but no room-pressure fluids.

As you can see on this data page on Wikipedia helium-4 has critical pressure of 2.24 atm, but helium-3 has 1.13 atm, only slightly above atmospheric pressure. Hyperbaric chambers work with 1,3-1,5 atm, so you could try there.

There is, unfortunately, question if it has sense, as you wouldn't notice any serious difference in properties between subcritical helium-3 and slightly above critical point in pressure, but so far above in temperature. It would still be lighter than air, for example.

I don't think that much higher pressure would be a huge problem - the highest pressure survived by human is supposedly 54 atm - so says Wikipedia, but I didn't find proof. If there is a compound with critical parameters sufficiently near standard, you could touch liquid-like, not gas-like supercritical fluid.

Ionic liquids have relatively low critical pressures (even only 14 bar), but too high crit. temps (source). On the other hand ethane's crit. temp. is below 30°C, but pressure over 48 atm. Maybe there's sweet spot somewhere, but it would be still high pressure and may be quite hot or cold.


Reaction between CH3CH2Li and Ethyne In the reaction CH3-CH2-Li + ethyne

Do the lithium and a hydrogen from ethyne just switch places? And how would i draw the curved arrow notation for this? I'm assuming from the C-Li bond in one to the C-H bond in the other and vice versa? Will upload a pic of my attempt when I get home.

Terminal alkynes are weak acids, but they are still stronger ones than alkenes and alkanes. This actually means they would readily react with conjugate bases of weaker acids, giving them the proton. So yes, since ethyne is a stronger acid than ethane (and ethyl anion is a stronger base than ethynyl anion - the list for bases is reverse due to obvious reasons), they will exchange H and Li.

Below you can see a mechanism I would use to describe the problem in your question. C-Li bond has significant contribution of both covalent and ionic components, so I wrote it as both in equilibrium:


How can you deduce from the structure that probenecid is an acid and ephedrine is a base? Probenecid contains a carboxyl group, indicating an acid. But it also contains a nitrogen atom, which indicates a base? How can I decide that it is an acid and not a base?

I have the same problem with the ephedrine structure.

The nitrogen atom in Probenecid is not part of a basic functional group (eg: amine), but of another kind of functional group: sulfonamide.

Sulfonamides are, just like amides, quite acidic: the presence of an electron-withdrawing group next to the nitrogen atom makes the electron pair less available, and makes the functional group quite acidic (see Lewis' definition of acids and bases).

For comparison, ammonium has a pKa of around 9, amides have a pKa above 15.

Sharing with amides the presence of an electron withdrawing group, sulfonamides can be expected to share the same acidic behaviour of amides.

You should check pKa list for the functional groups you have studied.

An example of such chart:

http://evans.rc.fas.harvard.edu/pdf/evans_pKa_table.pdf
(Be careful: values between parentheses are referred to the acidity in DMSO).

To better understand the delocalization of the electron pair in amides, see:
https://chem.libretexts.org/Bookshelves/Organic_Chemistry/Supplemental_Modules_(Organic_Chemistry)/Amides/Properties_of_Amides/Structure_of_Amides


Chemical shift value and proton NMR spectroscopy I am a bit confused about what chemical shift value signifies in proton NMR spectroscopy. If the proton of the standard substance TMS absorbs X Joules of energy to go into resonance, than a substance with a chemical shift value of 2.5ppm means it absorbs:

energy absorbed= X- (2.5X/10^6)?

Could someone please tell me if this is what chemical shift value is mathematically. In my book, it says the substance absorbs "2.5 millionth less amount of energy than TMS to go into resonance"... but I don't really understand this.

Firstly, the ppm scale is a dimensionless scale that describes the energy level of a signal relative to a reference (usually TMS). Significantly, it is magnetic field independent, and so allows comparison of signals recorded on different systems (not just a 300MHz vs a 600MHz spectrometer, but between all different 600MHz spectrometers, because no two will have been set up at identical Bo fields).

The chemical shift scale is defined as:

$$\delta =(v-v_r )*10^6/v_r$$ , where $v_r$ is the spectrometer reference frequency (of TMS, say)

Now, for a 600MHz NMR spectrum, each ppm represents 600Hz. So, if the resonance frequency of a signal at 0ppm is 600,000,000Hz, the signal at 1ppm represents an energy of 600,000,600Hz. At 2.5ppm, the frequency is 600,001,500Hz.

The chemical shift scale is often confusing in its labelling, but the thing to remember is that as you go to higher chemical shift values, you go to higher frequencies.

Higher ppm == downfield == Deshielded == higher frequency

Lower ppm == upfield == Shielded == lower frequency


Organometallic and oxygen I often read that reactive organometallics must be handled under an inert atmosphere because they react with oxygen in the atmosphere. This applies to grignard reagents, organolithium reagents, and organozinc reagents.

What is the reaction of these organometallics with oxygen? I don't see how oxygen can be electrophilic. Maybe it forms a peroxyacid?

Grignards react with molecular oxygen to form hydroperoxides. According to this paper  here the yield for alkyl hydroperoxides can be as high as 90%, yields from aryl Grignards are lower (10-20%).


Why do better π-acceptor ligands cause smaller Δ(T) d-orbital splitting? Consider a formazanate ligand. Then imagine an electron-withdrawing group added to it. I expect a small σ-donating effect, and good π-accepting properties.

My supervisor tells me that an electron-withdrawing group causes a lowering of the π*-orbital energy, leading to a smaller $\Delta_\mathrm{T}$ in the d-orbital splitting of the resulting iron-based molecular orbitals. Why would this be the case?

Since not many people know the ligand formazanate (I hadn’t heard of it before this question), let me quickly supply an image of what it looks like:

Scheme 1: General formazane structure.

Coodination happens from the lower two nitrogens, one of which is deprotonated (hence -ate) and both of which chelate the metal with $\mathrm{sp^2}$-type orbitals. The five formazane atoms form a π system, which can be extended by appropriate $\ce{R}$-groups. And one of these $\ce{R}$-groups is now modified to include an electron-withdrawing group.

I already previously gave a qualitative orbital scheme for a tetrahedral complex including σ and π interactions as given below.

Figure 1: Qualitative MO-scheme of a tetrahedral $\mathrm{d^0}$ complex with σ and π interactions. Originally created for this answer.

The σ donor orbitals transform according to $\mathrm{a_1 + t_2}$, leaving $\mathrm{e + t_1 + t_2}$ for the π ligands. Unfortunately, the picture is a tad clouded, since both σ and π interactions have a $\mathrm{t_2}$ contribution.

Adding an electron-withdrawing group to your ligand will leave the σ interaction unchanged in first approximation, since mainly the ligand’s π system would be affected. The orbital energies of the π system would be reduced. Since we are talking about a metal with a low oxidation state, its orbitals are higher than drawn in the qualitative scheme (which was designed for permanganate). And thus also, the most important π interaction of the ligand will be, as you stated, accepting electrons into its π system.

The exact nature of any effect observed will depend on the original orbital energies. However, the $\mathrm{e}$ orbitals are only affected by the π interaction while the $\mathrm{t_2}$ orbitals are also affected by the σ interaction. The magnitude of π interaction is expected to be the same for both groups of orbitals, since the start from the same value on each side. Thus, the difference between $\mathrm{e}$ and $\mathrm{t_2}$ stems mainly from σ interactions which only affect $\mathrm{t_2}$ due to symmetry.

Because the σ interaction starts with the ligand orbitals being much lower in energy than the metal orbitals, reducing their energy even more will result in a weaker interaction, therefore a lowering of the $\mathrm{t_2}$ level and thus a smaller $\Delta_\mathrm{T}$ or $10~\mathrm{Dq}$ value.


How to determine the molecular formula based on molecular weight, amount of substance, and partial masses of some, but not all contained elements? I am working on a chemistry assignment and can’t figure out part of the problem.

$\pu{0.2 mol}$ of a compound with a molecular weight of $82$ contain $\pu{9.6 g}$ of carbon, $\pu{1.2 g}$ of hydrogen and the rest is nitrogen.

The question is asking for the empirical and molecular formula which I can solve, but how do you determine the amount of nitrogen?

First I assumed a $\pu{100 g}$ sample but that didn't get me anywhere. The empirical formula answer is supposed to be $\ce{C2H3N}$ but I couldn’t achieve that. Unless the book is wrong.

If you have $0.2\ \mathrm{mol}$ of a compound with $M_\mathrm{r}=82$, then you have $16.4\ \mathrm{g}$ of material. If there is only carbon, hydrogen and nitrogen present, and the carbon and hydrogen total up to $10.8\ \mathrm{g}$, then the amount of nitrogen must be $16.4\ \mathrm{g}-10.8\ \mathrm{g}=5.6\ \mathrm{g}$.


What chemicals cause the softer, less-raw taste of purified water sold in bottles? I've seen the post, "What defines an element's taste?", but it doesn't answer my question.

I took water that is supplied by the municipality corporation and put it through a gravity based water filter. The taste of the water didn't change. It still had a raw, tap-water kind of taste.

However, if I buy one of the 20-liter cans (they charge x+y for the water+can and when you return the empty can, they return y amount of money), the water tastes softer and less tap water-ish.

A 2-liter bottle of mineral water costs x. Which is the same as the 20 liter can cost. So I'm wondering if they really purify it or whether they just add a few chemicals to it to fool people into thinking the water is purified. They might just be supplying us with ordinary tap water or borewell water.

Are there any chemicals that can change the raw taste of water? Maybe the raw taste caused by chlorine.

Water issues tend to be very specific to a certain location, so it is not possible to say exactly for sure what is happening in your situation.

Pure water has no taste.  The taste comes from the impurities.  Impurities, which are almost always present, can partially be removed by the filter shown.  That filter looks like a carbon filter which is effective on chlorine, organic particles, and sediment.  This filter is ineffective on minerals and inorganic compounds.

Minerals give tap water its hard taste. This hard taste is harmless (and may actually serve as a dietary mineral supplement.) Removing the hard taste requires a process like deionization or distillation which is costly. The cost of the water can go up significantly with more mineral removal.

Chlorine, which is used to kill pathogens, can be neutralized with citric acid.  Squeezing a lemon or lime in the water can cut down on a chlorine taste.

Hard water can be chemically softened with something like baking soda; however, this is generally trading one ion for another, so the water will only taste softer.

Try adding a reverse-osmosis filter after the filter shown. The filter shown will treat the chlorine while the reverse-osmosis filter can treat the minerals. (Chlorine may damage a reverse osmosis filter, so it is best to get it out first.)


Reaction of alkenes with bromine in presence of sodium chloride Consider the following reaction
           $$\ce{CH3\bond{-}CH\bond{=}CH2 ->T[$\ce{Br2}$/$\ce{NaCl}$]?}$$

What would be the product(s) for this reaction? I am confused because there are two nucleophiles here, $\ce{Br-}$ and $\ce{Cl-}$. So, there are two alternatives for the second step. Let me explain what I am saying.

First of all, the $\pi$-electrons will attack an electrophile which in this case is a $\ce{Br+}$ which is obtained form polarisation of the $\ce{Br\bond{-}Br}$ and it is added to the less substituted carbon, which complies with Markovnikov's rule, forming the carbocation below:
 $$\ce{CH3\bond{-}CH+\bond{-}CH2\bond{-}Br}$$

Now the carbocation will be attacked by a nucleophile. There are 2 nucleophiles in this problem, as I have stated earlier.
So, my question is which one of them will attack the carbocation and subsequently what will be the product? Will it be only 1,2-dibromopropane or only 1-bromo-2-chloropropane or a mixture of both? Please give the reasons.

The first step is $\ce{Br-Br}$ approaching the pi-cloud, being polarized, loosing a $\ce{Br^{-}}$ to form a 3-cyclic bromonium cation.  Given solubility and reactively inert solvent, $\ce{Br^{-}}$ (large poor nucleophile) competes with $\ce{Cl^{-}}$ (smaller, better nucleophile) for ring-opening capture.  Now factor in relative concentrations.  Now factor in thermodynamics (reversibility, $\ce{Cl^{-}}$ wins for the stronger $\ce{C-Cl}$ bond) versus kinetics (smaller $\ce{Cl^{-}}$ vs. initial tight ion pair rapidly collapsing to product).  Temperature probably makes a product ratio difference, too.

If your linear ion intermediate is controlling, the $\ce{Cl}$ comes in at the 2-position by default.  If my cyclic ion intermediate is controlling, the $\ce{Cl}$ comes in at the 1-position by steric hindrance. And if you do it in water, you get halohydrin byproducts.

Keep the reaction way from light or you will get radical abstraction and allylic substitution.


Derivation of the form of the time correlation function in NMR spectroscopy For a given protein, I know that the NMR spectrometer magnet generates a field $B_0$ and that the interactions with the spins in the local environment generates a much smaller field $B_\mathrm{loc}$ (not necessary aligned with $B_0$).

Owing to Brownian collisions between solvent molecules and the protein, the atoms associated with these spins move, hence making $B_\mathrm{loc}$ a function of time ($B_0$ is constant).

We define a time correlation function$$G(t,\tau) = \overline{B_\mathrm{loc}(t)B_\mathrm{loc}(t+\tau)},$$ where $G(t,\tau)$ is a stationary random function (see Time evolution of correlation functions (specifically Onsager's hypothesis) in time correlation link)
Hence $G(t,\tau)$ only depends on $\tau$, the delay in measuring $B_\mathrm{loc}$.

So, for simplicity we set $t = 0$ and note that

$$G(t,\tau) = \overline{B_\mathrm{loc}(0)^2} \mathrm{e}^{-\tau/\tau_\mathrm{c}},$$

where $\tau_\mathrm{c}$ is the correlation time, the time it takes for the whole molecule to rotate by 1 radian in a process called rotational diffusion.

How do I derive the second equation from the first?

Your function $G(\tau)= \langle B_x(t)B_x(t+\tau) \rangle$ is an autocorrelation function. If it is only a function of $\tau$, the time delay, and not t, this is called the stationary assumption.

The fluctuating field has  zero average $\langle B_x(t) \rangle =0$. The bracket implies averaging over a long time for a single spin or average over many spins at a particular time, which is the ergodic hypothesis.

The magnitude of the fluctuating fields is the mean square fluctuation $\langle B_x^2(t)\rangle \ne 0$ and is positive thus the mean square is not zero.

It is also necessary to know how rapidly the fields fluctuate and this is where the autocorrelation $G(\tau)= \langle B_x(t)B_x(t+\tau) \rangle \ne 0$ comes in. We compare the field at one point t with another point $t+\tau$ some time later. If $\tau$ is small then the value of $B_x(t)$ and $B_x (t+ \tau )$ are similar and $B_x(t)B_x(t+\tau)$ is large and positive, by comparison if $\tau$ is large the product $B_x(t)B_x(t+\tau)$ is small approaching zero as $\tau$ increases.

Thus the general shape is that $G(\tau)$ is large at small $\tau$ and decays away to zero with increase in $\tau$. Often we assume that the decay is exponential thus $$G(\tau)= \langle B_x^2 \rangle \mathrm{e}^{-|\tau|/\tau_\mathrm{c}}$$ where $\tau_\mathrm{c}$ is a correlation time. This has the correct qualitative form but is difficult to justify on the basis of proper theory.

You state that $\tau_\mathrm{c}$ is caused by rotational diffusion thus it will, in general, be small in fluid solution and large in viscous ones. Of course $\tau_\mathrm{c}$ also depends on the temperature (smaller at larger $T$) and size of the whole molecule's rotation (as a prolate or oblate ellipsoid) or of a  mobile group therein, depending upon which property you are observing. Groups in a protein can undergo 'wobbling in a cone' type motion in addition to rotation of the whole protein. Note also that if the molecule is not spherical then it will have more than one rotational relaxation time so $G(\tau)$ may decay with more than one $\tau_\mathrm{c}$. Hope that this helps!


Behavior of NaCl in water with electrodes If I had some NaCl dissolved in water, it is conductive, so if I were to put two electrodes in the container (powered), what is causing the flow of electrons?
Do the NaCl ionic bonds break and the ionic atoms collect at the oppositely charged electrode (how I'd assume electroplating works) with all of the atoms themselves moving to carry current, or are electrons loose in a "sea" like a metallic bond, with some of the atoms being pulled along with the flow?

$\ce{NaCl}$ once in water dissociates, producing $\ce{Na+}$ and $\ce{Cl-}$ ions. Note: this process is possible only because immediately after dissociation ions are solvated by water, i.e. water molecules bind to ions

This is releasing energy and increasing effective size of ion, decreasing ion charge density. Once relatively free ions are produced, they can move between electrodes, transferring charge between electrodes. On electrodes ions may discharge taking/giving up electrons, and this is what usually happens with $\ce{Cl-}$ ion. However, if this is not an option, as the expected product is no stable enough in water, water molecule may react instead, producing another ion and either hydrogen or oxygen. This is what happens with $\ce{Na+}$: $\ce{Na}$ is too active to be stable in contact with water, so instead of sodium, hydrogen and $\ce{{}^{-}OH}$ are produced instead, resulting in  net reaction (assuming cold solution) $\ce{2NaCl + 2 H2O <=> 2 NaOH + Cl2 + H2}$ with chlorine and hydrogen released on different electrodes.


How can the hydrogen bonds in solid HF be best represented? The hydrogen bonds in solid HF can be best represented as:

The correct answer is supposed to be (c), but I don't understand why this is so. Does the strength of hydrogen bonding depend on the $\ce{F-H\bond{...}F}$ or the $\ce{H-F\bond{...}H}$ bond angles?

At Wikipedia you find this structure (drawn by Benjah-bmm27 on Wikimedia Commons):

So, the answer (c) is correct. Yomen Atassi correctly stated that in such a hydrogen bond the two electronegative partners and the hydrogen prefer a linear arrangement, as this maximizes the orbital overlap for the hydrogen bond. That the configuration (c) is preferred over (a) can basically be explained via the VSEPR theory: a fluoride ion in HF is surrounded by 3 electron pairs and 1 H–F bond: those 4 "ligands" should roughly be arranged tetrahedrally – approximately, not exactly, because electron pairs need more space than bonding electrons – around the F atom, and this leads to the zigzag chains from answer (c).


When would the presence of lone pair result in a polar molecule? In what cases would the presence of a lone pair result in a polar molecule? For instance, O3 is polar despite only consisting of O atoms due to the presence of lone pair of electrons. However, O2 also has lone pair of electrons but is not polar, why? Also, for HAt, the electronegativity difference between H and At is 0, but is HAt polar or non-polar?

A polar molecule must have a finite net dipole moment.

A dipole moment is a result of separation of charges.

The below explanation requires you to have a basic understanding of dipole moment, vectors and resonating structures.

Ozone is not polar because of the presence of lone pair on oxygen atoms. In ozone, all the oxygen atoms are not same. Some oxygen atom wants to hold on the electrons more than the others which leads to separation of charges. This separation of charges results in a dipole which makes a molecule polar.

Electronegativity difference in a bond causes one atom to have more affinity for the electrons. This makes that atom attract the pair of electrons more, hence developing a slight negative charge on itself(because of more electron density) and a slight positive charge on the other atom(because of less electron density). This leads to separation of charges and creates a dipole. In the figure, a dipole is shown from carbon atom to chlorine atom. A dipole is represented as a vector with its tail pointing towards the atom having more electron density.

It is not the dipole but the net dipole moment that makes a molecule polar. The net dipole moment is a vector sum of all the dipole vectors. Now, why need vectors for finding the net dipole moment?
Suppose you are being surrounded by a bunch of strong guys who are pulling you through a rope from all sides. They are applying heavy forces to pull you towards them but you won't move because the net force on you is zero. Similarly, in the second figure, the four chlorine atoms are pulling the bond pair of electrons toward itself from 4 different orientations. To account for those different orientations which we know will affect the net pull of electrons by chlorine atoms, we need vectors.

Consider this case of carbon tetrachloride. The carbon can be thought of being placed at the centre of a regular tetrahedron and the chlorine atoms are placed at the corner of the tetrahedron. All the chlorine atoms are at equal distance from carbon. Even though you might say there is electronegativity difference between carbon and chlorine, the net pull on the electrons of carbon is zero. Hence, the dipole moment is zero despite the electronegativity difference.

The non-polarity of oxygen atom has got to do nothing with the presence of lone pair on oxygen atom. The oxygen molecule is not completely non-polar. There is instantaneous dipole in oxygen molecule.

No molecule is completely non-polar at every instant of time. H-At is also polar, although its dipole moment is very less. H and At are different atoms. One atom must have a little more affinity toward the electron and hence there will be separation of charges. The electronegavity difference between H and At is very less, therefore the separation of charges is very less which makes its dipole moment near to zero, but still finite.


What is purpose of sulfuric acid in iodine extraction? I am trying to extract iodine from kelp following this guide here.  It calls for sulfuric acid, but I don't see what this does do help the reaction.  I thought the reaction was simply oxidizing iodide to iodine with hydrogen peroxide, so I don't understand the purpose of the acid.  I do not have any sulfuric acid, and it is expensive to purchase, I do have a large supply of sodium bisulfate.  Could this be used as a substitute to the sulfuric acid if all that is needed is a low pH.

You need a strong acid to provide hydrogen (technically hydronium) ions, which is true for most aqueous oxidation reactions:

$\ce{2I- +H2O2 +2H+ -> I2 +2H2O}$

Unfortunately, bisulfate ion appears to be a weaker acid than water $(K_{\mathrm{a}} =  10^{-2})$, so it may not work unless you use a very concentrated amount to obtain enough hydrogen ions. Try it!

A fairly cheap alternative could be hydrochloric acid, which may be available in hardware stores as muriatic acid, and is stronger than sulfuric acid. But read the MSDS, as hydrochloric acid fumes and is nasty stuff.

Best of luck, it’s a cool experiment!


Cyanide inhibitor of Catalase Why is it that cyanide is an inhibitor of catalase but carbon monoxide isn't?
What makes cyanide bind well to $\ce{Fe^3+}$?

Jan has already covered this topic in a related question: Why is the cyanide ion toxic?

$\ce{CN-}$ and $\ce{CO}$ are isoelectronic, and one might expect them to behave exactly similarly. However, there is a key difference: the orbital energies in $\ce{CN-}$ are higher than in $\ce{CO}$, due to the lower electronegativity of $\ce{N}$ compared to $\ce{O}$.

This means that $\ce{CN-}$ is a much better σ donor, and a much poorer π acceptor than $\ce{CO}$. Consequently $\ce{CN-}$ is much happier binding to a metal in a higher oxidation state, whereas $\ce{CO}$ is much happier binding to a metal in a lower oxidation state.


Why do solutions of electron in a box (and in a ring) predict coefficients for LCAO (linear combination of atomic orbitals) in 1D systems? The solutions to the 1D particle in a box quantum mechanic system are standing waves (zero at both ends of the box) with 0,1,2... nodes for increasing energy (zero for the ground state).

If I look at the LCAO solutions for a linear system, combining AOs of the same shape and energy (e.g. conjugated double bonds, linear array of sodium atoms), the coefficients follow the pattern of a standing wave with 0,1,2... nodes. For example, when I combine 3 AO's in a row, the solutions are +++, +0-, +-+, i.e. zero, one, and two switches of signs as I go to higher energy states. For the state with one node, the node is in the center of the box, corresponding to the middle AO, which has a coefficient of zero. (More examples, i.e. hexatriene and pentadienyl cation, are on slide 126 and 128 of this document.)

The solutions to the particle in a ring system are standing circular waves (same value at 0 and 360 degrees) with 0,2,4... nodes for increasing energy. The animation shows a standing circular wave with 8 nodes:

If I look at the LCAO solutions for a circular system, again combining AOs of the same shape and energy (e.g. p-orbitals in an aromatic system), the coefficents follow the pattern of a circular wave with 0,2,4... nodes. For benzene, the solutions are ++++++, +++--- and +0--0+, +-++-+ and -0+-0+, +-+-+- (see picture):

What underlies the connection between the unbound electron in a box or corral with the formation of covalent bonds of bound electrons?

A short answer is that the functions that satisfy the Schrodinger equation for a system are largely determined by the potential energy part of the function, since that's the part of the Hamiltonian that varies with different scenarios. The particle-in-a-well (= 1D box) is defined by having infinite potential energy outside the walls and a constant finite energy within. That turns out to be a very good approximation of the potential energy function of a pi system or a line of identical metal atoms.

At the end of the line of atoms, an electron faces a steep potential energy curve. Not quite the vertical wall of the 1D box, but pretty close to it. The same is true on the sides, so the system can be treated as 1D. Within the system, the electron moves essentially freely (or exists in a broadly delocalized way) with very little change in potential across the system.

Since the potential functions are nearly identical, the solutions to the Schrodinger equation take the same mathematical form.


Winkler's method in the absence of iodide I'm currently doing an IB chemistry individual investigation on the effect of adding bleach to water on the water's dissolved oxygen content. I've used Winkler's method for this and I'm just wondering whether it is possible to do Winkler's method without the presence of iodide ions $(\ce{I-})$, i.e. without adding chemicals like potassium iodide?

I used $\ce{NaOH},$ $\ce{MnSO4},$ $\ce{H2SO4}$ in this experiment (and $\ce{Na2S2O3}$ as the titrant) and I'm not too sure whether I used potassium iodide. However, I've obtained a result and found out a relationship where the dissolved oxygen content increases as I add more bleach into the water.

So my question is: is it possible to do a Winkler's method experiment without the presence of iodide ions? If so, could I know how my experiment worked?

Bad news is that Winkler's method will fail in the presence of bleach or any other oxidizing agent besides oxygen. Neither sodium thiosulfate nor potassium iodide are compatible with bleach, whether it is a oxygen bleach or a chlorine bleach.

  I used $\ce{NaOH},$ $\ce{MnSO4},$ $\ce{H2SO4}$ in this experiment (and $\ce{Na2S2O3}$ as the titrant) and I'm not too sure whether I used potassium iodide. However, I've obtained a result and found out a relationship where the dissolved oxygen content increases as I add more bleach into the water.

As a good scientist of future, you always write whatever you are doing at the time of experiment in your notes.

Your results are skewed towards higher numbers because sodium thiosulfate is being consumed by released iodine not only from the oxygen reaction but also from the bleach.


Through bond interaction in biradical dioxo compound and its correlation diagram I'd like to investigate into the ring opening, the migratory reverse reaction of the electrocyclic ring-closure of Dimethyldioxirane:

As a first question I'd like to ask if I have drawn the through bond interaction correctly?

Following scheme is wrong, because the p orbitals are perpendicular to the O-C-O as @ron clearly pointed out in his answer in Are the p orbitals of the biradical dioxo compound in the HOMO perpendicular to the plane?

Next I tried to draw the orbital correlation diagram.
I guess for this correlation diagram I don't have to draw the through bond interaction, yes?

In the dioxo biradical product I've drawn a smaller energy gap between $\pi$ and $\pi^*$ orbitals as for $\sigma$ and $\sigma^*$. I guess this should be correct.
Because there is an additional through bond interaction for the product I've drawn a small exothermicity. I see no activation energy for the disrotatory ring opening. So the disrotatory ring opening must happens very fast (without a lot of thermal energy, may be even at room temperature).

Finally, I tried to draw the state correlation diagram:

For the photochemical reaction there is a transition from $S_0$ to $S_2$. Because of the avoided crossings the conrotatory correlation is steeper falling down, therefore I expect most of the electrons direcetly going in the pericyclic minimum and do there Interal Conversion to the $S_1$ surface and afterwards the excited product gives off a photon going back to the $S_0$ state. I expect products in excited states are not very long living and therefore undergoes emission of a photon as Internal Conversion under fluorescence here.
Another way out from the pericyclic minimum can be the directly falling to the transition state of the $S_0$ state and giving the product to a good quantum yield.
As third option some part of the $S_2$ states can go through the disrotatory pathway and directly emitting light from $S_2$ to $S_0$. As written in the first point I think this way is less probable because of the less steeper falling down the energy surface.

Is this all correct?

I would like to help here, but I am not sure where to begin. First of all, I want to point out that the reaction you have shown is not an electrocyclic reaction, because it does not involve the central saturated carbon atom. It is rather a simple homolysis reaction. The terms "conrotatory" and "disrotatory" are misapplied, since the two oxygen atoms do not have functional groups on them, and the two modes of ring opening are indistinguishable.

Dioxirane does have an electrocyclic ring opening reaction available to it - namely cleavage of a carbon-oxygen bond to form "acetone-O-oxide". Perhaps you should consider that as an alternative pathway. Does that help?


chemical equilibrium and exothermic reaction here is an interesting question about the chemical equilibrium and Le Chatelier's principle.

the following pressure show the partial that produced in the equilibrium reaction at time (t1 and t2)

$$\ce{N2 + 3H2 <=> 2NH3 \Delta H =-92 KJ}$$

at point t1 hydrogen is added to the previous equilibrium system at this point on the curve and after a period a new equilibrium state took place at point t2
so
A = ...
B = ...
C = ...

what i say is the answer is A=H2 ,B=N2 and C=NH3
becuase by adding more hydtogen thus increasing the conc. the reaction will move forward so the temp will increase increasing the pressure according to The Pressure Temperature Law so the the pressure of N2 increases but again also the pressure of NH3 should increase because the no of moles will increase as the reaction will shift forward so I am lost here

Since, H2 is added at the equilibrium state, it's initial pressure will be more and starts decreasing as it goes into the reaction with N2. This is clearly depicted by graph A. While the pressure of N2 starts decreasing from the level it was in equilibrium state, represented by graph C. Also, the reaction shifts forwards in order to gain equilibrium, pressure of NH3 will increase. So, the correct answer would be,

  A = H2, B = NH3, C = N2


Treatment of ambidentate ligands in geometrical and optical isomerism Are ambidentate ligands considered different entities when they donate through different atoms? For example will a hypothetical square planar complex like [Pt(ONO)2(NO2)2] show geometrical isomerism?

To specify [Pt(NO2)4]2+ would mean the ligand donates electrons through the N-atom, and not the O-atom. If it were to donate through O, it would be written as (ONO) and not (NO2).

Thus, this compound would not show geometrical isomerism. Also note that, if NO2 ligands are replaced by ONO, then that is a form of structural isomerism in complexes, called linkage isomerism.


Periodically oscillating chemical reactions? I've seen before chemistry demonstrations where solutions are mixed with one another and subsequently where the resulting product goes through periodic color changes: for example blue to orange and back to blue again. The reaction seems to go on for awhile at a regular period.

My question - can this reaction be considered resonance in the sense that energy is flowing back and forth between two chemical states? In such reactions are there actually two chemical species between the color changes or just energy level/temperature differences?

A couple of points:

The demo is probably an oscillating chemical reaction, in which the concentration of a given species increases and decreases repeatedly over time. However, the energy of the overall system is dissipating as that is happening - not oscillating.

Resonance, on the other hand, is a (widely misunderstood) concept that applies to a single chemical species and not a bulk reaction mixture. Resonance does not consist of energy flowing back and forth between two chemical states, but rather is meant to represent a single energy state that is a hybrid between two (or more) classical Lewis structures. A double headed arrow between the two classical structures indicates resonance stabilization — mostly resulting from delocalization of electrons relative to the classical structures. The double headed arrow does not indicate oscillation between those two structures.


Bonding between rings and iron atom in ferrocene How are the 10 vertical bonds in Ferrocene (sandwich structure) connected, by what type of bonds?

What is the real structure of ferrocene? Concerning the 10 vertical extra long bonds?

The "real" structure of ferrocene is an iron atom sandwiched between two flat, parallel, pentagonal C5H5 rings. The diagram above shows the atomic positions from a crystal structure but the bonds are merely a convenience and don't accurately summarize the way the bonding works though the picture does accurately summarize the atomic positions in the crystal structure.
The real bonding requires some understanding of molecular orbital theory and how the d-orbitals of iron interact with the delocalized pi-orbitals in the unsaturated ring. Some pictures of the orbitals contributing to the bonding are shown here.


Stability of keto-enol vs. conjugation I am trying to figure out which tautomer of creatinine:

is more stable.

We see that the enol tautomer is more highly conjugated: the $\pi$ electrons from the alcohol to the double bond to the $\pi$ electrons of the Nitrogen to the imine to the final $\pi$ electrons of the amine.

In the keto form, we have the $\pi$ electrons from the carbonyl to the imine to the two amines.

I know the enol form is more conjugated, but in general I also know that the keto form is more stable. How do I reconcile this?

The NMR spectrum seems to show that the tautomeric state of creatinine is that of the keto-form (bottom structure).

If the ketone is the only functional group, the keto form is more stable than the enol form. One the other hand, if the carbon-carbon bond is part of a conjugated system, the enol form might be more stable (e.g. acetylacetone or phenol).

You also suggest either the imino or the amino tautomer. There is a Raman study that addresses this question and also raises the dependence on pH.


Nomenclature alkyl halides I was solving questions on the MSU website on organic nomenclature.
Out of the following questions, I got only the first and last one right.
Where am I going wrong with the others?Hints would suffice.

EDIT:I got the second one correct as well.

(2)----7,7-Dichloro-4-ethyl-3,3,8-trimethyl-4-propyldecane

Indeed, most of the names proposed by you correspond to the preferred IUPAC names in accordance with the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

(1) 3-bromo-2,5-dimethylhexane

(2) 7,7-dichloro-4-ethyl-3,3,8-trimethyl-4-propyldecane (your edit is correct)

(3) see below

(4) 2-chloropent-1-en-4-yne

(5) 1-bromo-3-ethyl-5-methylbenzene

For compound (3), you have to consider the relevant rules for the numbering of locants for substituent prefixes:

  P-14.4 NUMBERING

  When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:

  (…)

  (f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;

  (g) lowest locants for the substituent cited first as a prefix in the name;

  (…)

Note that Rule (f) takes precedence over Rule (g).

In accordance with Rule (f), compound (3) could be named as 5-bromo-1-methylcyclohexa-1,4-diene (which was your proposal) as well as 1-bromo-5-methylcyclohexa-1,4-diene since both names correspond to the locant set ‘1,5’. However, according to Rule (g), this example is named as 1-bromo-5-methylcyclohexa-1,4-diene rather than 5-bromo-1-methylcyclohexa-1,4-diene since bromo is cited first as a prefix in the name.


Electron configuration and the 2p orbital I looked through the questions already asked about this topic and they were very specific and over my head. I'm having trouble with the basics.

When drawing orbital diagrams I know electrons are placed in the 2p boxes singly first then paired because of they have parallel rotations and the Pauli exclusion principle states that an orbital can only house 2 electrons and those electrons must be of opposite spin. My question is why do electrons behave this way in the 2p orbital but not the 1s or 2s? And why does the p subshell contain three 2p orbitals?

As far as electron configurations and orbital diagrams are concerned I've gotten to Ne and the 2p configuration confused me so I though I'd ask before moving on.

first of all electrons do not rotate but they have spins
next magnetic quantum number decides how many electrons an orbital can have
s can have 2(m=1){consider 1 box}
p can have 6(m=3){consider 3 boxes}
d can have 10(m=5){consider 5 boxes}
f can have 14(m=7){consider 7 boxes}
where each box can have at max 2 electrons and each box represents different orientation of the orbital
and all the orbitals(except s) have degenerate orbitals, i.e they have the same energy so all have to filled with atleast one electron before pairing up. this rule is known as Hund's Rule of Maximum Multiplicity
there are are some more reasons fro it. they are
to maximize exchange energy
to attain symmetry


Why doesn't Calcium Carbonate Burn? I was doing my Science Fair Project and I tried to burn Calcium Carbonate. But the match went out as soon as it touched the Carbonate. I am wondering why the Carbonate isn't burning. I later tried to use a torch and a lighter but nothing worked. My science fair project is due tomorrow and I've looked everywhere on the internet but nothing really explained why the calcium carbonate didn't ignite.
Do I need HEAT or something like that to make it burn?
(I'm a 5th grader so please take it easy on me)

Calcium Carbonate cannot burn. Its already in its highest oxidation state (+2 for Calcium and +4 for Carbon), and cannot be oxidized (combusted) to a higher oxidation state. However, calcium carbonate does decompose on heating to quicklime and carbon dioxide.

$$\ce{CaCO3 -> CaO + CO2}$$

This doesnt give out any flames in the first place.

Now let's talk about how to salvage your project. Why don't you try to heat the calcium carbonate, and show that carbon dioxide is evolved by using a burning match stick, or by using lime water? $\ce{CO2}$ extinguishes the burning flames and forms a milky precipitate with $\ce{Ca(OH)2}$ (limewater). This can be used to show that perhaps egg-shells and marble have carbonate content in them.


Public API for physical properties Is there a publicly available API that serves physical property data for organic molecules?

Pubchem does not serve properties, ChemSpider closes property access to API calls, and Toxnet only serves html.  Im basically looking for a free programmatic interface to the kind of info in a CRC handbook. Thanks for any suggestions.

ChEMBL database provides API access: https://www.ebi.ac.uk/chembl/ws
properties can be retrieved from the 'molecule' endpoint, for example aspirin:

https://www.ebi.ac.uk/chembl/api/data/molecule/CHEMBL25.json

Disclaimer: I'm the author of this API.


How do metals split the oxygen double bond during corrosion? For a metal like $\ce{Mg}$ to corrode into $\ce{MgO}$, the double bond in $\ce{O2}$ must break. Since the dissociation energy of $\ce{O2}$ is 500 kJ/mol, using estimates from the Boltzmann distribution, it looks like there would not be any $\ce{O2}$ molecules in the entire atmosphere  with enough energy to break this bond at room temperature:
$$\text{Fraction of $\ce{O2}$ with at least $\pu{500 kJ}$ energy} \approx \mathrm e^\left(\dfrac{-500\,000\ \mathrm{J/mol}}{298\ \mathrm K(8.3145\ \mathrm{J/(mol\ K)})}\right) \lt 10^{-261}.$$

But my understanding is that $\ce{MgO}$ does form at room temperature. In fact, it seems that even stable $\ce{Pt}$ can break the $\ce{O2}$ bond at room temperature (for example in the context of catalyzing the ignition of hydrogen).

It seems that something like this must be happening: an $\ce{O2}$ molecule collides with $\ce{Mg}$ metal and gets stuck onto the surface, temporarily forming $\ce{MgO2}$. And once this bond is formed, the remaining bond between the two oxygen atoms suddenly becomes easy to break, and one of the oxygen atoms falls off and bonds with a different $\ce{Mg}$ atom. Is this the case?

I'd like to understand this mechanism better. How weak does the $\ce{O2}$ bond become after latching onto magnesium, and why does it become weaker? Why are metals able to split the $\ce{O2}$ bond this way, compared to other materials like hydrocarbons?

The corrosion of metals like iron is essentially an electrochemical process.

In corrosion, a metal is oxidised by loss of electrons to oxygen and formation of oxides. Corrosion of $\ce{Fe}$ (commonly known as rusting) occurs in presence of water and air. The chemistry of corrosion is quite complex, but it may be considered essentially as an electrochemical phenomenon. At a particular spot of an object made of iron, oxidation takes place and that spot behaves as anode and we can write the reaction.

\begin{align}
\tag{Anode}
  \ce{2Fe (s) &-> 2 Fe^{2+} + 4 e-} &
  E^\circ_{\ce{Fe^{2+}/Fe}} &= \pu{– 0.44 V}
\end{align}

Electrons released at the anodic spot move through the metal and go to another spot on the metal and reduce oxygen in presence of $\ce{H+}$ (which is believed to be available from $\ce{H2CO3}$ formed due to dissolution of carbon dioxide from air into water. Hydrogen ions in water may also be available due to dissolution of other acidic oxides from the atmosphere). This spot behaves as cathode with the reaction

\begin{align}
\tag{Cathode}
  \ce{O2(g) + 4 H+(aq) + 4 e- &-> 2 H2O (l)} &
  E^\circ_{\ce{H^+/O2/H2O}} &= \pu{1.23 V}
\end{align}

The overall reaction being:

\begin{align}
  \ce{2Fe(s) + O2(g) + 4H+(aq) &-> 2Fe^{+2}(aq) + 2 H2O (l)} &
  E^\circ_{\text{cell}} &= \pu{1.67 V}
\end{align}

The ferrous ions are further oxidised by atmospheric oxygen to ferric ions which come out as rust in the form of hydrated ferric oxide $\ce{(Fe2O3. x H2O)}$ and with further production of hydrogen ions.


Hydrogen peroxide / sodium or potassium permanganate: reaction type I have some people on youtube stating the steam generating reaction of $$\ce{3H2O2 + 2KMnO4 → 3O2 + 2MnO2 + 2KOH + 2H2O}$$

is catalytic, from where I'm sitting it is purely stoichiometric. Can someone please enlighten me as to why some people would consider this reaction catalytic with the presence of the permanganate the catalyst?

If this reaction really is regarded as catalytic how is this so, it flies in the face of everything I've learnt about reaction types. I'm only an amateur, can someone who is proficient in their chemistry please clarify this.

Learning from the comments and from looking at this video (https://www.youtube.com/watch?v=9S6v-NMMnB4), there are two reactions:

$$\ce{3H2O2 +2KMnO4 -> 3O2 + 2MnO2 + 2K+ + 2OH- + 2H2O}\tag{1}$$

$$\ce{2H2O2 ->[MnO2] O2 + 2H2O}\tag{2}$$

The first reaction creates the catalyst $\ce{MnO2}$, the second generates oxygen gas even after the permanganate is used up. If you look closely at the video, you will see bubbles (oxygen) coming out of the solution even after the solution is colorless (i.e. permanganate has completely reacted).

For some more chemistry, I also recommend this video by Prof. Martyn: https://www.youtube.com/watch?v=OLUyeCC-2Ko

  Can someone please enlighten me as to why some people would consider this reaction catalytic with the presence of the permanganate the catalyst?

This is not accurate. What is true is that you need much less permanganate than hydrogen peroxide because only a little is required to make the catalyst, which then allows the reaction to proceed without being used up.


Stoichiometry and iodometric titrations $1.25~\mathrm g$ of a sample of bleaching powder is dissolved in $100\ \mathrm{mL}$ of water. $25\ \mathrm{mL}$ of it are treated with $\ce{KI}$ solution. The iodine so liberated required $12.5~ \mathrm{mL}$ of $\frac{\mathrm{M}}{25}$ hypo solution ($\ce{Na2S2O3}$) for complete titration. Find percentage of available chlorine from the sample of bleaching powder.

My attempt:

In $25~\mathrm{mL}$ sample:

mEq of $\ce{Cl2}$ in bleaching powder $=$ mEq of $\ce{I2}$ liberated $=$ mEq of hypo solution

$\mathrm{mEq}$ of Hypo $= NV = \left(1\times\frac{1}{25}\right)\times12.5 = 0.5$

This is where I have a problem, if we are to go by the solution provided in my book, then the n-factor for the hypo solution is 1, whereas, when I write the balanced redox reaction, that is,

$\ce{2S2O3^2- -> S4O6^2- +2e^-}$

You can see that n-factor should be $2$.

No. The $n-$factor is still $1$ because one mole of $\ce{S2O3^2-}$ furnishes 1 mole of electrons.

In your balanced reaction, 2 moles of $\ce{S2O3^2-}$ gives 2 moles of electrons. To find n-factor, you need to calculate the amount of electrones given by one mole of $\ce{S2O3^2-}$.


How can I calculate the number of branches in an alkane isomer? I am so confused. So my textbook says that

  2-Methylheptane has 1 branch
  2,2-Dimethylhexane has 2 branches

However, when I get to:

2,4-Dimethylpentane
2,2,3-Trimethylpentane
2,2,3,3-Tetramethylbutane

I have no clue how to count the number of branches from the structural formula. My textbook assumes you'd understand how to calculate the number of branches after looking at the example. Can someone please help me understand?

Each instance of a comma-delimited number in the prefix is directly indicating a branch on the "backbone" or base-chain. It's quite literally the number of numbers.

The most complicated example you have can be dissected like this:

$2,2,3,3-$ tells us to anticipate four branching groups from the base-chain on the 2nd, 2nd, 3rd, and 3rd carbons.

$Tetra$ tells us that all four branching groups are identical (since there are four total)

$methyl$ tells us the name of the branches. The $methyl$ group is a $\ce{-CH3}$ attached to the backbone.

$butane$ tells us what the backbone is; in this case, a 4-carbon ($but-$) fully-saturated carbohydrate ($-ane$).

So the molecule will look like this, with the 4 branches indicated:


What is an oxidizer? When talking about gaseous fuel / oxidizer mixture what does one mean by oxidizer if there is no oxygen involved?

Is it correct to think of oxidizer as the more electronegative specie, because it is pulling the electron? Or is this definition only valid for solutions?
The Wikipedia article is not clear whether the definition applies to gaseous reactants or only valid for solutions.

Oxidation: losing electron(s), Oxidizer / oxidizing agent: a chemical that can oxidize another reagent. Reduction: gaining electron(s) - think of it as reduction of charge! Reducing agent: A chemical that can reduce another reagent.

And oxidation can never occur without a reduction, meaning if you use an oxidizing agent to oxidize a substance, when the oxidizing agent is in turn being reduced itself. It gains the electrons of the oxidized substance.

To describe how well a substance functions as an oxidizer, we use reduction potentials, measured in volts! A substance with a large, positive reduction potential is easy to reduce, which in turn means it is a good oxidizing agent. Similarly, if a substance has a large, negative reduction potential, it means that it is hard to reduce this substance. This is because it's good at reducing other substances itself - it's a reducing agent.

To figure out if a redox reaction is spontaneous (if it can potentially proceed on its own), we use standard reduction potentials, $E_0$. This is for concentrations of 1 M and a temperature of 25 degrees celcius. You look these up in your textbook or online.

An example: Is the following reaction spontaneous?
$$\ce{Cu^{2+}(aq) + 2Ag(s) -> Cu(s) + 2Ag+(aq)}$$

To find out we first find the standard reduction potentials for each participating species:

$\ce{Cu^{2+}(aq) + 2e- -> Cu(s)}, E_0=+0.34V$

$\ce{Ag^{+}(aq) + e- -> Ag(s)}, E_0=+0.80V$

This describes how much they want to be reduced (GAIN electrons). The silver ion wants it the most! However, notice that in our reaction we're asking for it to LOSE electrons (be oxidized). THEREFORE: No, the reaction is not spontaneous under the aforementioned standard conditions. However, the REVERSE reaction is spontaneous. To calculate the potential for our reaction, we simply say:

$$E_0 = (+0.34V) - (+0.80V) = -0.46V$$

The reduction potential for the substance that we want to gain electrons (reduced), minus the potential for the substance we want to lose electrons (be oxidized). The fact that the result is negative, is what tells us the reaction will not proceed - and that it will in fact proceed in the OPPOSITE direction.

Take notice that these reduction potentials are not given the unit Volt for no reason! They are true electric potentials. You can think of electric potentials as "electron pressures". The pole with the highest electron pressure will be (-), because electrons are negatively charged, and the other will be (+) relative to it.

Remember that the reduction potential (the willingness to take electrons) for the silver ion is higher than that of the copper ion. Silver wants to take up electrons more than copper, meaning that electrons will travel from the copper to the silver. you could just as well say that the copper wants to get rid of its electrons more than the silver - copper has the higher electron pressure.

What about electronegativity?
Electronegativity does indeed correlate somewhat with reduction potentials, at least for simple reactions involving pure elements. And you can see that Ag does indeed have a higher electronegativity than Cu, which makes sense. Electronegativity doesn't take into account the oxidation states of different species etc., and quickly it becomes a better idea to just leave electronegativity out of the picture when dealing with redox reactions.


How does dissolving a hydrate affect concentration? Say I have 0.72 grams of calcium chloride dihydrate. If I dissolve it in water, the 2 water molecules attached onto it dissociates just like the $\ce{Ca^{2+}}$ and $\ce{Cl-}$, right?

And if so, wouldn't it lower the concentration of the calcium chloride, as compared to making an anhydrous calcium chloride solution?

There are two methods to dissolve the calcium chloride salt.

For the first method you'd dissolve the salt in a "small" amount of water and dilute to some particular volume. So if the solution was 0.100 molar, the amount of water would be the same regardless if you started with $\ce{CaCl2}$ or $\ce{CaCl2\cdot2H2O}$.

The other method would be to make say a 1.00 % solution by weight of the calcium chloride salt. Here it does matter. Do you want 1.00 % solution of $\ce{CaCl2}$ or a 1.00 % solution of $\ce{CaCl2\cdot2H2O}$?


If the equilibrium constant in a reaction is greater than 1 is it possible that there are more moles of reactants than products? I'm wondering if this is possible because could there not be a situation in which the reactants were pure solids/liquids and would therefore not contribute to the value of k but still contain more total moles that the product side?

The equilibrium constant does only contain the activities/concentrations of the species that are actually available to the reaction, e.g. in solution it would only contain species that are actually dissolved in the solvent or that are present at the interface between the solution and another phase. So, solids would not enter the equation with all there moles but only with the small percentage that is acutally available to the reaction: if some of the solid is dissolved and those dissolved species take part in the reaction then it is those (and only those) molecules that enter into the equation for the equilibrium constant; in the case of a heterogeneous reaction the solid can only enter the equation via its available surface sites. So, the bulk of a solid would not contribute to the equilibrium constant in a reaction taking part in a liquid phase that shares and interface with the solid.

Now, concerning the main question: If the equilibrium constant in a reaction is greater than 1 is it possible that there are more moles of reactants than products? This can certainly be the case if you already have some of the products present and keep their concentrations constant. Look at the example reaction

\begin{equation}
  \ce{A + B <=>> C + D}
\end{equation}

which shall have an equilibrium constant of $K_{c} = 10$. Let's assume you keep the products at a concentration of 1 mol/L each, so $[\ce{C}] = 1 \, \mathrm{mol}/\mathrm{L}$ and $[\ce{D}] = 1 \, \mathrm{mol}/\mathrm{L}$. The equation for the equilibrium constant is

\begin{equation}
  K_{c} = \frac{[\ce{C}] [\ce{D}]}{[\ce{A}] [\ce{B}]}
\end{equation}

Since you've already fixed the product $[\ce{C}] [\ce{D}]$ at $1 \, \mathrm{mol}^{2}/\mathrm{L}^{2}$ the product of your reactants must fulfill

\begin{equation}
  [\ce{A}] [\ce{B}] = \frac{[\ce{C}] [\ce{D}]}{K_{c}} = 0.1 \, \mathrm{mol}^{2}/\mathrm{L}^{2}
\end{equation}

So, if you use only very little $\ce{B}$, say $[\ce{B}] = 0.01 \, \mathrm{mol}/\mathrm{L}$, your concentration of $\ce{A}$ in the solution will be

\begin{equation}
  [\ce{A}] = \frac{[\ce{C}] [\ce{D}]}{K_{c} [\ce{B}]} = 10 \, \mathrm{mol}/\mathrm{L}
\end{equation}

Now, if you conduct this reaction in a vessel with a volume of 1 L you'll find a total of 2 mol of products and 10.01 mol of reactants in your reaction mixture, i.e. more reactants than products.


Why are the acidities for H-CH3 and H-OH so different? I was comparing the acidity of H-CH3 and H-OH, and a lot of things weren't making sense. Generally, the process of turning H-X into $H^{+}$ and $X^{-}$ can be broken into 2 hypothetical steps: 1.) Homolytic cleavage of the H-X bond, followed by 2.) Electron donation from H to X. The energy required for step 1 is correlated with bond length, while the energy released by step 2 is correlated with electronegativity of X.

The way I see it, electronegativity comes as a DIRECT consequence of atomic size. Adding an electron to smaller element means the electron is added to an orbital that's relatively close to the nucleus, which is more stable. Thus, smaller elements are generally more electronegative, particularly across a row in the periodic table.

Now we consider H-CH3 and H-OH. Oxygen, being smaller than carbon, will be more electronegative; thus, $OH^{-}$ is more stable than $CH3^{-}$ and step 2 will be more favorable for H-OH. But oxygen being smaller than carbon also means the O-H bond length is smaller than the C-H bond length, and thus harder to homolytically break. Overall, I'd expect these two steps to roughly cancel, and their acidities to be at least in the same ballpark. Yet the pKa for H-OH is 15.7, while the pKa for H-CH3 is approximately 55!

How can I make sense of this? What are other factors that impact electronegativity and homolysis that I am not considering? Or are there other aspects to this problem that I am not considering entirely?

As far as the acidity of water goes, It is a very special case that observes the process of homoassociation. This is when an acid forms stabilizing hydrogen-bonds with it's conjugate base This allows water to be the more stable acid, because hydrogen bonds are observed between $\ce{OH-}$ and $\ce{H2O}$, while none are observed between $\ce{CH3-}$ and $\ce{CH4}$
This factor might not be significant enough to provide explanation for all of the difference that you are seeing, so there is still another culprit out there.


When is it true that more nodes equals higher energy? Consider all the MOs of some isolated molecule. (It could be a single
atom too; I'll use MO to refer to AOs as well.) Number them in
increasing order of the number of nodes (node = surface where the wave
function has zero density). Orbitals with the same number of nodes can
be numbered in any order. Now you have a sequence of orbitals $O_1,
O_2, ...$. Let their respective energies be $E_1, E_2, ...$.

It seems to be "common knowledge" that $E_n \le E_{n+1}$ for any such
system and any $n$. As Martin put it to me yesterday, "An orbital
with 47 nodes can never be lower in energy than one with only 46." (Follow up on Counting Nodal Planes in cyclopropane.)

For various reasons, given below, I think this cannot be true in
general, and I'd really like to know under what conditions it is
known to be true. "Known" here can mean either a rigorous statement
with a reference to a proof (a trivial example would be: It is true
for a one-electron atom; we can calculate the energies exactly) or a
precise statement with empirical justification (something like "no
counterexamples are known for class X of molecules" - again with a
reference).

Important: Please, I'm not looking for an explanation that re-states
the rule in some equivalent or even looser fashion ("more nodes means
the orbital is larger and less dense, therefore it must be higher in
energy").

Why do I think the statement can't be always true? Well, a calcium
atom has a filled 4s orbital and empty 3d orbitals. If this does not
count as a counterexample, please explain what notion of orbital
energy the statement applies to.

In general, I'm happy to believe that two MOs must satisfy the rule if
they have "comparable" sets of nodes (say, constant $n_x$ and $n_y$ in the
example I'm about to discuss), but I'd like to understand what is
meant by "comparable" in general. In a molecule with no symmetry, are
there comparable MOs at all? If so, how do we know if two given MOs
are comparable?

The case of a much simpler system, the 3D rectangular box, may
also be relevant.  The energy levels for such a box are of course
$$
\frac{\hbar^2\pi^2}{2m}\biggl(
\frac{n_x^2}{L_x^2}+
\frac{n_y^2}{L_y^2}+
\frac{n_z^2}{L_z^2}
\biggr),
$$
where $n_x$, $n_y$ and $n_z$ are one more (or one less, if you count
the walls) than the number of nodal planes in the respective
direction. If we take $L_x = L_y = 1$ and $L_z = 0.1$ (say), the
wavefunction with $n_x=5$, $n_y=1$, $n_z =1$ has energy $5^2+ 1^2 +
(1/0.1)^2 = 126$ and 4 nodes (or 10, if you count the walls), while
the wavefunction for $n_x=1$, $n_y=1$, $n_z=2$ has energy $402$ and 1
node (or 7). So clearly the rule is not true here.

Granted that molecules are not boxes, but this does show that
arguments based simply on the number of sign changes are not rigorous,
so they don't answer my question.

General case

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

  Given the self-adjoint second order (partial) differential equation

  \begin{equation} \left(\hat{L} + \lambda \rho(\mathbf{x}) \right)
 u(\mathbf{x}) = 0 \end{equation}

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

The proof is rather involved and so I won't show it here. But if you want you can look it up in [1] or here.

So, Courant's nodal line theorem tells us, that if we order the possible energy eigenvalues of the time-independent Schroedinger equation as $\lambda_1 \leq \lambda_2 \leq \lambda_3 \leq \dots$, then (depending on precisely how you set up the numbering) the  $n^{\text{th}}$ eigenfunction, $\Psi_{n}$ (the one with energy eigenvalue $\lambda_n$) has at most $n$ nodes (including the trivial one at the boundary $\mathbf{x} \to \infty$).
Unfortunately, this gives you only an upper bound for the number of nodes a wave function with a certain energy eigenvalue may possess. So, all we know is that the ground state wave function $\Psi_{1}$ cannot have any nodes within the region $G$ (in total it has one node, namely the one at $\mathbf{x} \to \infty$). Wave functions for higher $n$ may possess up to $n-1$ nodes within $G$ but may as well have less. Thus, we cannot in general say that if a wave function has more nodes than another one it will automatically correspond to a state with higher energy.

Special case: Schroedinger equation in one dimension

There is however a special case:
For the Sturm-Liouville eigenvalue problem (and thus for ordinary second order differential equations with homogeneous boundary conditions) we can strengthen Courant's nodal line theorem such that if we order the possible eigenvalues as $\lambda_1 \leq \lambda_2 \leq \lambda_3 \leq \dots$, then the  $n^{\text{th}}$ eigenfunction (the one with energy eigenvalue $\lambda_n$) has exactly $n$ nodes (including the trivial one at the boundary $\mathbf{x} \to \infty$).

This is useful since the one-dimensional time-independent Schrödinger equation is a special case of a Sturm-Liouville equation. So, in the case of the inhomogeneous radial Schrödinger equation with a local potential and node-less inhomogeneity such as the radial Schrödinger equation for the hydrogenic atom

\begin{equation}
\bigg( \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \mathrm{d}^{2} }{ \mathrm{d} r^{2} } + \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \frac{ \ell (\ell + 1) }{ r^{2} }  - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r } - E \bigg) r R(r) = 0
\end{equation}

it is generally true that a wavefunction with more (radial) nodes must always correspond to a state of higher energy than a wavefunction with less radial nodes. Also, it is clear that the wavefunctions of the one-dimensional particle-in-a-box must follow this rule. But for the three-dimensional particle-in-a-box this is not true anymore, since in that case the Schroedinger equation of the system is not an ordinary second-order differential equation but a partial-differential equation for which only the general version of Courant's nodal line theorem holds.

Some concluding remarks

For real-world system like molecules or crystals the Schroedinger equation is a partial differential equation for which the special case outlined above doesn't apply so that only Courant's nodal line theorem in its general form holds which doesn't give a strict justification for the statement that more nodes mean higher energy. Yet it is very often observed that the number of nodes indeed increases with increasing energy.
The reason for this can be motivated in the following way: The kinetic energy $E_{\mathrm{kin}}$ of a state is proportional to $\int \Psi \Delta \Psi \, d^{3} r$. Via Gauss's theorem it can be shown that $\int \Psi \Delta \Psi \, d^{3} r \propto \int |\nabla \Psi |^{2} \, d^{3} r$ and so $E_{\mathrm{kin}} \propto \int | \nabla \Psi |^{2} d^{3} r$. Now, nodes force a wavefunction to change it's sign. This often means that the value of $\Psi$ has to increase/decrease rather rapidly thus leading to areas with high absolute values of the gradient and thus to high kinetic energy. Since the potential energies shouldn't differ too much between the different states the higher kinetic energy usually also entails a higher total energy. As an example consider the bonding and antibonding wavefunctions of a homonuclear diatomic molecule whose atoms are placed at the positions $r_{\mathrm{A}}$ and $r_{\mathrm{B}}$.

The bonding wavefunction has no nodes. Its value between the atoms doesn't have to undergo a rapid change and thus the slope is rather low. The antibonding wavefunction has one node between the atoms. Its value between the atoms must change rapidly from its positive to its negative maximum thus entailing a very high slope. The slopes of the tail regions are comparable for the bonding and antibonding wavefunctions since it can smoothly fall of to zero at infinity and is not required to go from a maximum value to zero within a very confined region of space - thus even if one wavefunction has to start of at a higher maximum value the gradient will not be much higher. It follows that the antibonding wavefunction has a higher kinetic energy than the bonding wavefunction.

References

[1] R. Courant, D. Hilbert, Methods of Mathematical Physics, Vol. 1, Interscience, New York, 1953, p. 451-455.

[2] R. Courant, "Ein allgemeiner Satz zur Theorie der Eigenfunktionen Selbstadjungierter Differentialausdrücke", Nachr. v. d. Ges. d. Wiss. zu Göttingen 1923, p. 81.


What product is formed in the Wurtz reaction of 1-bromo-3-chlorocyclobutane? I want to do Wurtz reaction of 1-bromo-3-chlorocyclobutane:

Wikipedia suggests to do an intramolecular reaction. Instead, I remember reading somewhere that bromine is more reactive and hence we should do an intermolecular reaction, and attach two carbons to where $\ce{Br}$ was attached.

So, what product is formed in the Wurtz reaction of 1-bromo-3-chlorocyclobutane?

Br is a much better leaving group than Cl.First of all presuming that Cyclobutane is stabilized by substitutent groups is important since it wil gradually rearrange itself to buta diene.But since it is a theoretical question hence  we can say that it should add to the Br side since Br leaves more efficiently being larger in size than Cl and hence having less efficient orbital overlap with Carbon as well as a large electronic repulsion due to the large number of electrons it has is the reason behind this.And the major product will probably be a bicyclo [1.1.0] butane since it is the kinetic product.
so basically your answer-


What happens when butane and methane both absorb 100 J of heat? Two cylinders, one containing 1 mole of $\ce{C4H10}$ gas at $\pu{1 atm}$ and the other containing 1 mole of $\ce{CH4}$ gas at $\pu{1 atm}$, are at $\pu{288 K}$.  If each gas absorbs $\pu{100 J}$ of heat under conditions of constant volume, which of the following is true?
  (a) The temperature of the $\ce{CH4}$ increases more than the temperature of the temperature of the $\ce{C4H10}$.
  (b) The internal energy of both the $\ce{CH4}$ and the $\ce{C4H10}$ decreases.
  (c) The heat capacity of the $\ce{C4H10}$ is less than the heat capacity of the $\ce{CH4}$
  (d) The entropy of both the $\ce{CH4}$ and the $\ce{C4H10}$ decreases.
  (e) The heat transferred to the $\ce{C4H10}$ is greater than the heat transferred to the $\ce{CH4}$.

The correct answer is (a). I am trying to eliminate answers, but I seem to get stuck.

We know that (b) is clearly false as we've added heat and no work is done, therefore the internal energy for both gases must increase. Likewise, (d) is false as the entropy should increase as the molecules have more heat and become more disordered. Finally, (e) is incorrect as we transferred $\pu{100 J}$ of heat to both systems.

But here is where I get stuck. I can't seem to eliminate (c) from just the information given.

First, consider this from a nice write-up on heat capacity:

  The amount of energy needed to raise the temperature of a substance by one degree in temperature will depend upon how many different ways the atom or molecule can store the energy in the form of these thermal motions and how much energy each kind of thermal motion involves.In general, for molecules with similar bonding the bigger the molecule the more heat has to be added to raise its temperature by one degree.

Now, examine the degrees of freedom for each molecule: there are more ways to store energy in the larger butane molecule (with 36 vibrational modes in addition to the translational and rotational modes) than there are in methane (9 vibrational modes plus translational and rotational ones).

Therefore, we'd expect a greater increase in temperature for the smaller molecule than the larger one for an equivalent amount of heat added to the system.


Are the rules proposed by Slater for the calculation of effective Z right? Today I studied Slater's rules for calculating the effective nuclear charge. One particular line, "The shielding power of the d and f orbital is lesser in comparison to the p and s orbitals", seems to appear in every single book and website I have studied from.
Slater's rules does not seem to take into account the orbitals for calculation of $Z_\mathrm{effective}$. Nevertheless, I don't disagree with the logic behind the statement.
Thus my question is: Do Slater's rules actually account for this logic? If they don't, is it faulty? And, how do Slater's rules account for the shielding effect of the individual orbitals?

Slater's rules are pretty good, but they're not perfect. Consider that he published his work in 1930: Phys Rev 1930 v. 36, 57
There are a number of people who have modified Slater's scheme to better reflect shielding effects and correct for experimental ionization energies and electron affinities (and similar properties) in later rows of the periodic table.
If I remember correctly, one such modification is that d and f orbitals shouldn't have as much shielding as suggested by Slater, but the correction is small. There's another scheme that uses an exponential correction instead.
For example, Wikipedia gives effective nuclear charges from Clementi and Raimondi, who did full self-consistent quantum calculations and you can compare these to Slater's.
The main take-home from Slater's rules is that they work pretty well for a wide range of elements, including 1st, 2nd, and 3rd row. I think the corrections generally tweak things when we get into the 4d and 5d transition metals and some of the lanthanides and actinides. Not bad for an 84-year old paper!


Listing errors in experimental section I am fairly new to report writing, and am currently writing an experimental section.

Obviously we are required to list measurements, but I was wondering whether or not we are supposed to list the corresponding errors with these?

E.g. 5 mL $\pm$ 0.01 mL sodium hydroxide was added to...

Or if not, why this isn't included?

It depends a lot on the context.

For example if we are talking about synthetic chemistry the exact amount often isn't relevant, for example in case of solvents for the reaction. It doesn't really matter if you do it in 200 mL or 198 or 201 mL. Therefore no one cares to measure it exactly in the first place and no one really cares about those errors, especially since you would need to measure temperature as well.

In other cases, like many analytical methods, you need to know the exact amount which was used and in those cases you should definitely include errors.

In the end you need to include it if it's relevant information, it's common to omit it if it's not relevant.


Hydrolysis of vinyl bromide vs alkyl chloride Which of the following compounds, 1-bromocyclopent-1-ene 1 or chlorocyclopentane 2, hydrolyses faster than the second under comparable conditions? It may be acid catalysed or base catalyzed hydrolysis but the conditions must remain the same for each pair.

In my view because of the double bond in first compound hydrolysis should be easier for second, but I am not sure about it.
See part 1 here

Hydrolyzing the saturated alkyl chloride could occur by SN2 but I'd be inclined to expect SN1 in this situation. Chlorine would leave behind a (secondary) carbocation, which would be attacked by the nucleophile: hydroxide (basic conditions) or water (acidic conditions). The rate limiting step of SN1 is loss of the nucleophile, and the activation energy might be lowered in an acidic environment because the hydronium could coordinate to the chloride to stabilize it. The real issue is the resulting carbocation though. In basic conditions, hydroxide could backside-attack, kicking out the chlorine before it has the chance to leave. This concerted SN2 mechanism would avoid the carbocation entirely, but it is slowed by the steric hindrance of the secondary alpha carbon. So here acidic favors SN1, limited by formation of the secondary carbocation, and basic favors SN2, limited by secondary sterics.

Hydrolysis of the vinyl bromide I think must occur by SN1 in basic conditions or protonation-addition-elimination in acid. SN1 requires formation of a vinyl cation when the bromine leaves, which is quite unstable and will severely limit the rate of the reaction. The rate limiting step of the acidic P-A-E sequence is probably the protonation (slower than addition, and I would expect also slower than elimination because protonation is bimolecular). I would expect protonation of an alkene (PAE pathway) to occur much faster than loss of a nucleofuge (leaving group) to form a vinyl cation (SN1 pathway); forming a secondary carbocation would surely have a much lower activation energy barrier than forming a vinyl one. So I expect the acidic hydrolysis of the vinyl bromide via hydration-elimination (P-A-E) to occur much faster than the SN1 hydrolysis in basic conditions.

So the pH conditions may play a significant role here, with base severely slowing the hydrolysis of the vinyl bromide. In basic conditions, I would almost certainly bet the alkyl chloride will hydrolyze faster. Acidic conditions will speed up hydrolysis of the vinyl bromide, with the RLS being protonation to form secondary carbocation or possibly elimination of bromide to restore the double bond. Acidic conditions for the alkyl chloride promote SN1, whose RLS is loss of chloride to form the secondary carbocation. It's hard for me to predict with only theory whether vinyl bromide acidic RLS or alkyl chloride acidic RLS will occur faster.

An interesting note, hydrolyzing your vinyl bromide yields an enol, which will spontaneously tautomerize to a ketone. So your hydrolysis products are a ketone (from vinyl bromide) and an alcohol (from alkyl chloride). The difference of oxidation state in these products makes me wonder why you are asking this question.

Update: Here are my predicted mechanistic pathways. The details vary slightly from my analysis, especially how tautomerization factors in. I also included a couple of alternative pathways I was trying out (Acidic #2 and Basic #2 for vinyl bromide, and Basic #2 for alkyl chloride; these are probably negligible).


Acidic and alkaline salt solutions: why do some salt form acid/bases while others don't I have a question regarding salts and their solutions. How come some salts, like sodium carbonate and ammonium chloride, produce acids/bases when dissolved in water? Moreover, how can you predict whether these solutions will act as an acid or as a base just by looking at the salt?

Without trying to go into too much detail this has a two part answer.

First a salt by definition has an anion and a cation. So ordinary table salt, $\ce{NaCl}$ is a chemical salt, but table sugar is not a salt. So lets think of the cation as $\ce{C^+}$ and the anion as $\ce{A^-}$.

So we start out with solid salt which in chemistry is written as $\ce{CA_{(s)}}$. When the salt dissolves the two ions are hydrated which means that water molecules surround each ion and separate the pair. the chemical reaction is written like this: $$\ce{CA_{(s)} -> C^+_{(aq)} + A^-_{(aq)}}$$

Typically the (aq) subscripts are left off, but implied.

The second part of the answer involves the nature of the cation or anion. To understand if the pH of the solution will change depends on whether the cation is able to "donate" a hydrogen ion, or the anion is able to accept one.

$\ce{NH_4^+}$ is a cation and is able to donate a hydrogen ion to the solution. So the pH will drop and the solution becomes more acidic. $$\ce{NH_4^+ + H_2O -> NH_3 + H_3O^+}$$

$\ce{CO_3^{2-}}$ is a anion and is able to accept a hydrogen ion from the solution. So the pH will increase and the solution becomes more basic. $$\ce{CO_3^{2-} +H_2O-> HCO_3^- + OH^-}$$

When water, $\ce{H_2O}$, makes or breaks bonds the process is called hydrolysis.


Does a positive electron affinity actually correspond to the release of energy? I read about the periodic properties — ionisation energy and electron affinity — in my book. It was quite intuitive that energy is required to remove an electron from the atom but I wonder why energy is released when an electron is added to an atom? Is it actually released or is it because of our convention that energy of electron at infinite distance from nucleus is 0, that we consider it to be released?

It may help to take a look at the two processes you are comparing in terms of reactions. Ionisation follows the following reaction:

$$\ce{A ->[][$\Delta E_\mathrm{i}$] A+ + e-}\tag{1}$$

This process involves tearing charges apart, separating a positive and a negative one. Even if the original atom was somewhat unstable, it will be unfavourable to remove that electron simply due to electrostatic interaction. You write in the question that you were able to deduce this already.

The electron affinity on the other hand describes the process in this following reaction:

$$\ce{A + e- ->[][EA] A-}\tag{2}$$

Here, a charged particle is approaching a neutral particle and we have the same number of charged particles (and the same charges) on both sides of the equation. This means that we don’t have a strong, obvious physical effect that will make all electron affinities sign-equal for very simple reasons. Instead, we need to consider the details more.

The process often involves a release of energy because most atomic configurations have space for extra electrons at little or no extra cost. Except for the noble gases and the second group, atoms don’t have completely populated sub-shells so there should always be a position available in a subshell at an acceptable exothermic energy level. While adding an electron will increase electron-electron repulsion and increase all energy levels by a certain amount, this increase is typically less than the first ionisation energy which corresponds to the energy level of the highest occupied atomic orbital. Thus in electron affinity, we in principle gain the entire ionisation enthalpy minus the destabilising contribution of the new electron. (Caution! This is a very simplified picture and not suitable for generalisation or even strictly for explanation. It gets the basic point across though.)

Therefore, most electron affinities are exothermic with the notable exceptions of group 2, group 18 and nitrogen.


What exactly is the requirement (in terms of saturation) for a reaction to be at equilibrium? A saturated solution is a solution that contains the max amount of solute able to be dissolved. However, can't a solution still be at equilibrium if you add more solute to a saturated solution as long as it doesn't reach supersaturation? What exactly is the requirement (in terms of saturation) for a reaction to be at equilibrium?

If you add solid sodium chloride to a saturated sodium chloride solution, it is still at equilibrium. If you increase the sodium ion concentration or the chloride ion concentration (by e.g. adding a high concentration of NaOH or LiCl), then the solution is supersaturated and will not be at equilibrium.

  What exactly is the requirement (in terms of saturation) for a reaction to be at equilibrium?

A reaction is at equilibrium when the reaction quotient is equal to the equilibrium constant. For a precipitation reaction, the solid species is not part of the equilibrium constant expression, so adding more solid does not change the situation. Changing the concentration of the ions in solution, though, will change the reaction quotient, and the reaction will no longer be at equilibrium.