What is the seventh plane of symmetry in benzene? We get the first six planes of symmetry by joining the opposite edges and vertices. But what is the seventh plane of symmetry?

Have a look at this picture which shows some of the symmetry axes and planes. You should be able to see other similar ones. Taken from www.molecule-viewer.com.(You can also use this site to practice working out point groups.)


Rotational Energy Levels Why does the rotational constant B decrease and transition spacings decrease as the mass of a particle increase?

I understand from a purely equation perspective that since

$$B = \frac{h} {8\pi ^2 cI}$$

that as $I$ increases the denominator increases and so $B$ decreases. But what is the physical reasoning behind this? Why or in what way is the rotational constant dependent on mass?

First, take a look at classical physics. The angular momentum of a particle rotating in a plane is defined as
$$L = I \omega$$
and its kinetic energy is
$$E = \frac{1}{2} I \omega^2 = \frac{L^2}{2I}.$$

So if you formulate your energy in terms of the angular momentum of your rotating particle, you arrive at the inverse relation.

In analogy to the classical picture, the eigenvalues of the rotational Schrödinger equation,
$$ E = hcBJ(J+1),$$
likewise depend quadratically on the angular momentum quantum number, and thus have a similar inverse dependence on the moment of inertia through $B$.


Why can't I reproduce this photolytic reaction of ferrocyanide? I was so intrigued by the reported observation that sunlight precipitates the iron from $\ce{K4Fe(CN)6}$ (originally attributed to Matuschek, 1901) that I wanted to see it myself.  I put a saturated aqueous solution, uncovered, in direct sunlight for four hours at mid-day.  Absolutely nothing precipitated from the solution!

My understanding from the given reference is that in solution the $\ce{K4Fe(CN)6}$ gives $\ce{Fe(CN)6^{4−}}$ and then:

$\ce{Fe(CN)6^{4−} + 2H2O <-> Fe(CN)5 + (H2O)^{-3} + HCN + OH-}$

The photolysis causes $\ce{Fe(CN)6^{4−} ->[h\nu] Fe(CN)6^{3−} + e-}$,  so then we also have

$\ce{Fe(CN)6^{3−} + 2H2O <-> Fe(CN)5 + (H2O)^{-2} + HCN + OH-}$

Is it correct to assume that if a $\ce{Fe(CN)6^{4-}}$ anion absorbs a photon of adequate energy (in this case it appears to require $\lambda < 313nm$) then with 100% probability the anion will photolyze the Fe bond?

And are there models that predict (at least order of magnitude) the absorption probability of a photon by an aqueous anion (being irradiated with a given spectral flux)?

Furthermore, what is the reaction that would precipitate iron in this case?  When we bring the K back into the equations it looks like the HCN would react to produce KCN.  I don't know what should happen to $\ce{Fe(CN)5}$.

In fact, after several days in open sunlight the solution is subjectively getting darker and I do see a small amount of red sediment at the bottom of the dish that appears to redissolve on swirling.  Could be an iron oxide, but would also be consistent with $\ce{K3Fe(CN)6}$.  In either case I don't have any indication from the source what should happen to the $\ce{(CN)5}$ groups if iron is precipitating, or what else might take one of the K atoms from the initial compound.

(Ultimately I had hoped to quantify the rate of photolysis by weighing the precipitate.)

I would put the solution under a blacklight exposure source or a equivelant such as a mercury arc lamp at close range. The referenced writing makes no specifics about the the timeframe of prolonged. This could have been 3 months of exposure to sunlight.


Role of nodes in stability of orbitals What role does the node play in determining broadly the relative stability of various types of atomic orbitals with the same principal quantum number?

As a simple answer one might say the more nodes the higher the orbitals energy, i.e. the less stable it is. But you can have orbitals with the same number of nodes and different energies.

In general the is a node rule in quantum mechanics, which states that the ground state of a system has no nodes and higher excited states have more nodes.
Two common examples are the particle in a box and the harmonic oscillator. Both links contain the equations for the eigenfunctions and their energy as well as their plots.
A simplified example dealing with orbitals would be Hückel MO Theory.

Now for atomic orbitals the story is a bit more complicated. First of all, orbitals are not the electronic states of an atom or molecule, but something you can approximately decompose the electronic state into. Furthermore we have not 1 but 3 quantum numbers (5 if you include spin) defining an orbital: $n$, $l$ and $m$. The node rule for atomic orbitals is:

$n-l-1$ radial nodes
$l$ angular nodes:

$|m|$ nodal planes shaped like a cone around $z$-axis
$l-|m|$ planar nodal planes perpendicular to $xy$-plane

So we have $n-1$ nodes in total. The energy of an atomic orbital mainly depends on the quantum number $n$:
\begin{equation}
  E_n \propto -\frac{Z}{n^2}
\end{equation}
where $Z$ is the atomic charge. So higher $n$ means more nodes and higher (less negative) energy, which is in agreement with the general node rule for $n$. This also agrees with the quantum numbers $l$ and $m$, as they neither change the orbital energy (it only depends on $n$) nor the total number of nodes. (This is considering an atom with one electron only).

However, the orbital energies furthermore depend on the coulomb interaction with other electrons. This is what leads to $2p$ orbitals usually having slightly higher energy than $2s$ orbitals, but they still have the same number of total nodes, they are just differently distributed between radial and angular part of the orbital.

Also note, that you cannot directly compare orbitals of same quantum numbers between different elements, as the nuclear charge $Z$ goes into the orbital energy as well. For example the energy of the $1s$ orbital in the Hydrogen atom is $-0.5\ E_{\rm h}$. In a Carbon atom this would be something more like $-16\ E_{\rm h}$ for $1s$, while $2s$ with around $-1.2\ E_{\rm h}$ would be energetically much closer to the $1s$ Hydrogen orbital.


Succinic anhydride esterification won't go forward I'm trying to react succinic anhydride with a decyl alcohol to produce mono-decyl succinate. Literature and Google searches suggest that the monoesterification should proceed without catalyst. However I'm failing to get the reactions going. Reactions were in reflux, $\pu{80-140 ^\circ C}$ at $\pu{1 atm}$.
Any suggestions if anhydride esterifications even needs a catalyst? Would a solvent help?

The monoesters of succinic acid can be prepared using Steglich esterification method (Ref.1,2), which is specially suitable for alcohols with the $\ce{OH}$ functional group is in steric hindrance. In this method, dicyclohexylcarbodiimide (DCC) as the coupling reagent and 4-dimethylaminopyridine (DMAP) as a catalyst are used. The reference 1 gave a typical procedure as follows:

To a stirred solution of 10mmol carboxylic acid in $\pu{lO ml}$ anhydrous $\ce{CH2Cl2}$ (in the case of sparingly soluble acids, in DMF) is added $\pu{30-110 mg}$ DMAP and $\pu{20-40 mmol}$ alcohol or thiol ($\pu{10 mmol}$ with alcohols or thiols which are not easily removable without significant loss; $\pu{3.4 mmol}$ with glycerol). DCC is added to the reaction mixture at OT, which is then stirred for $\pu{5 min}$ at $\pu{0 ^\circ C}$ and $\pu{3 h}$ at $\pu{20 ^\circ C}$. Precipitated urea is then filtered off and the filtrate evaporated down in vacuo. The residue is taken up in $\ce{CH2Cl2}$ and, if necessary, filtered free of any further precipitated urea. The $\ce{CH2Cl2}$ solution is washed twice with $\pu{0.5 N}$ $\ce{HCl}$ and with saturated $\ce{NaHCO3}$ solution, and then dried over $\ce{MgSO4}$. The solvent is removed by evaporation and the ester isolated by distillation or recrystallization. Crystalline products can be obtained in pure form by filtration on a short silica gel column (eluent: $\ce{CH2Cl2}$ or $\ce{CHCl3}$).

For example, recently, this method has successfully used to couple Vitamin $\mathrm{D_2}$ to succinic anhydride in order to achieve succinic monoester of Vitamin $\mathrm{D_2}$ (Ref.3). However, there is an US patent publication revealing cheap solvent free method using alkali salt bases such as $\ce{NaOAc, KOAc, Na2CO3}$, and $\ce{K2CO3}$ as catalysts to efficiently prepare mono alkyl esters from cyclic anhydrides (Ref.4).
References:

Bernhard Neises, Wolfgang Steglich, "Simple Method for the Esterification of Carboxylic Acids," Angew. Chem. Int. Ed. 1978, 17(7), 522–524 (https://doi.org/10.1002/anie.197805221)(PDF).
Gerhard Höfle,  Wolfgang Steglich,  Helmut Vorbrüggen, "4‐Dialkylaminopyridines as Highly Active Acylation Catalysts. [New synthetic method (25)]," Angew. Chem. Int. Ed. 1978, 17(8), 569-583 (https://doi.org/10.1002/anie.197805691).
Victar Nyaarugwe, "Synthesis and Characterization of Hybrid Compounds for the Treatment of Bone Related Diseases," Journal of Trauma and Orthopedic Nursing 2018, 2(1:4), 16 pages (PDF).
Mark, B. Erman, Joe W. Snow, Mikhail Y. Lebedev, "Process for making monomenthyl esters," US Patent 2007, US 7,247,743 B1 (PDF).


What exactly is formed when K2Cr2O7 is reduced? I read on Wikipedia and in many books that when $\ce{K2Cr2O7}$ is reduced, it forms $\ce{Cr^3+}$ and a change is noticed from orange color to green. I want to know in what exact complex/salt/compound $\ce{Cr^3+}$ exists that it shows green color.

I know $\ce{[CrCl3].3H2O}$ shows green color but in many of these reactions, there is no $\ce{Cl-}$ ion, so I guess that is not formed. Also, $\ce{Cr^3+}$ can exist as $\ce{[Cr(H2O)6]^3+}$ but since it is violet in color, that possibility is also ruled out. So basically, how does the solution becomes green in color?

The dichromate ($\ce{Cr2O7^2-}$) ions are strong oxidizing agents at low pH. During the redox process, each chromium atom in the dichromate ions (oxidation state = +6) gains three electrons and get its oxidation state reduced to +3. In redox reactions in aqueous acid solution, the aquated $\ce{Cr^3+}$ ion is produced according to following half-reaction (Electrochemical Series):
$$\ce{Cr2O7^2- + 14H3O+ + 6e- <=> 2Cr^3+ + 21H2O} \space \space \space \pu{E^0 = 1.36 V}$$

Assume that the $\ce{Cr^3+}$ ion in aqueous solutions is in its simplest ion form: the hexaaquachromium(III) or $\ce{[Cr(H2O)6]^3+}$. This ion is violet-blue-grey in color. However, when $\ce{Cr^3+}$ ion is formed during a reaction in aquous acid solution, it is often appeared in green color. Thus, we always describe the green color due to $\ce{Cr^3+ (aq)}$, implying it as the hexaaquachromium(III) ion ($\ce{[Cr(H2O)6]^3+}$), but that's actually not the case.

What's really happening in the solution is that one or more of the water molecules in each of $\ce{[Cr(H2O)6]^3+}$ ions get replaced by other negative ions in the solution, typically by $\ce{SO4^2-}$ (if the acid used is $\ce{H2SO4}$) or $\ce{Cl-}$ (if the acid is $\ce{HCl}$). For example, if one of the water molecules is replaced by a $\ce{SO4^2-}$ ion, $\ce{[Cr(H2O)6]^3+}$  becomes $\ce{[Cr(H2O)5SO4]+}$, which is green (notice also the change in the charge on the ion). You can check this by yourself: warm freshly prepared aqueous  chromium(III) sulphate hexahydrate solution, which would change its color of purple-blue or violet ($\ce{[Cr(H2O)6]^3+}$) to green ($\ce{[Cr(H2O)5SO4]+}$) due to the ligand exchange discussed above [Chem Guide].

In $\ce{CrCl3.6H2O}$, on the other hands, the green color of the solid itself or its aqueous solution is due to the $\ce{[Cr(H2O)4Cl2]+}$ ion. Hence, actually more proper way of writing the formula is $\ce{CrCl2(H2O)4Cl.2H2O}$. In contrast to previous case, the green $\ce{CrCl2(H2O)4Cl.2H2O}$ solution would slowly becomes violet in color by standing at room temperature overnight, by slow ligand exchange with solvent (water)[Chem Guide].

Interesting fact: Chromium was discovered by Louis Nicholas Vauquelin in 1797 who named his new element after the Greek word chroma — which means color.

Late Additions:

Although $\ce{Cr^3+}$ ions complexes with most ligands to give green color, it gives vivid line of colors with other ligands (note it names after chroma!). Some interesting examples are illustrated in the picture below:

Also see:

Chromium(III) nitrate: Blue-violet crystals (anhydrous), and purple crystals (nonahydrate), according to Wikipedia (see the picture below).


Heat energy of phosphoric acid evaporation Looking at a system in which phosphoric acid is concentrated in an evaporator which uses a constant recycle stream with a heat exchanger in this stream to heat the acid so that water evaporates out of the acid when it re-enters the main stream, How can I find the Q needed to remove a certain amount of water from the stream in order to concentrate the acid to a specific level?

I am given the amount of water evaporated, both concentrations, and the flow rate of both the entering acid stream and the exiting acid stream. I have also been told that you cannot simply use $Q=mC_p\Delta T$, because this is inacurate for phosphoric acid.

Any help or sources you could link me to would be greatly appreciated.

You need to know the latent heat of vaporization of the solution, which would be the same as that of water when phosphoric acid concentration is low.

See Fig. 10 of this bulletin for latent heat of vaporization as a function of phosphoric acid concentration, but keep in mind this is only strictly valid at the stated temperature.

If you are varying phosphoric acid concentration (concentrating phosphoric acid) you would need to integrate heat of vaporization as a function of phosphoric acid concentration over the appropriate concentration range.

Also, as phosphoric acid concentration gets very high, it is no longer valid to consider the evaporating solution to be only water, it will increasingly contain phosphoric acid as well.


What is the difference between a reaction of Lewis acids with Lewis bases and a redox reaction? For example I came across this equation:
$\ce{2Al_{(s)} + 6HCl_{(aq)}} \longrightarrow \ce{2Al^{3+}_{(aq)} + 6Cl^{-}_{(aq)} + 3H_2_{(g)}}$
And thought this was a redox reaction but my book says it shows basic nature of aluminium?

It's indeed redox - Lewis theory doesn't account for redox, but on grounds of Usanovich's theory ...

A Lewis acid is an electron pair acceptor; a Lewis base is an electron pair donor. The simplest Lewis acid is the hydrogen proton. H$^{+}$ has no electrons by itself. It accepts electron pairs from bases in Lewis acid-base reactions. Here's an example of a Lewis acid-base reaction; note how a pair of electrons from the Lewis base attacks the Lewis acid.

Redox - a portmanteau of reduction and oxidation - involves the transfer of electrons. Hey, there are electrons being "transferred" above, right?

At first glance, it seems that all Lewis acid-base reactions would be redox reactions. However, there is an artificial line in the sand that one would do well to heed.

Redox reactions, as taught at the general chemistry level, involves changes in formal oxidation state. What's an oxidation state? It's the hypothetical charge that an atom would have if all bonding were ionic (as opposed to covalent, or electron-sharing). This means that the more electronegative atom in a bond would have complete possession of the bonding electrons.

If we go back to the reaction pictured above, then we see that electrons are not being "transferred" in the formal sense. Hydrogen is of much lower electronegativity than oxygen; hence, in a formal sense, it has gained no electrons - remember, when assigning formal oxidation states, we take all bonding as ionic.

Obviously, in reality, no bond is 100% ionic, but this is exactly the artificial line in the sand I was talking about.

  For example I came across this equation:
  $\ce{2Al_{(s)} + 6HCl_{(aq)}} \longrightarrow \ce{2Al^{3+}_{(aq)} + 6Cl^{-}_{(aq)} + 3H_2_{(g)}}$
  And thought this was a redox reaction but my book says it shows basic nature of aluminium?

This indeed is a redox reaction, as aluminum is oxidized and hydrogen is reduced as we go from left to right.

It also shows that aluminum can react with hydrochloric acid to make hydrogen gas; this removes hydrogen protons from the solution and thereby makes the system more basic.

This is not an acid-base reaction in the Lewis sense, as electron pairs aren't being transferred. It's not an acid-base reaction in the Bronsted sense either, as hydrogen protons are not being moved around.

  On grounds of Usanovich's theory Al as reductor would indeed be basic.

Usanovich defines an acid that accepts negative charge; a base as something that accepts positive charge. Therefore, Al is a base here; it becomes positive from left to right. This theory is an even more generalized view of acids and bases than Lewis acid-base theory.


Migratory aptitude in Baeyer–Villiger reaction I found a reaction in which we have to find the product:

I tried and got to know that it is Baeyer–Villiger reaction but I got confused after seeing the answer, I thought there would be a shift of methyl group according to migratory aptitude, but it is not like that the product is:

Why?

In the following image you can see the charge distribution in the transition state of the Baeyer–Villiger reaction (image taken from Clayden Organic Chemistry):

As you can see, there is partial positive charge at the migrating C atom, therefore a group will migrate easier if it contains cation stabilizing moieties. A benzene ring significantly stabilizes positive charge in its proximity through delocalization.


How can there be decimal subscripts in a molecular formula? While learning about how batteries work I have encountered the following notation for a Li-ion cathode: $\ce{Li_{0.5}FePO4}$.[1]
According to Wikipedia, the subscript number in a reaction equation represents the number of atoms in a molecule. But what does the $0.5$ represent? Is it a fraction of atoms in a molecule?

Also, $\ce{LiFePO4}$ is a cathode. I would expect that for a fully charged battery, the anode is populated with all the $\ce{Li}$, therefore the cathode should be $\ce{FePO4}$, right? I can also understand that some $\ce{Li}$ gets stuck in the cathode, but then I read somewhere that a fully charged battery, the cathode is $\ce{Li_{0.5}FePO4}$.

I have added an exaggerated sketch for what I imagine the system would be.

I know that the structure of $\ce{Li_{0.5}FePO4}$ is more complex than this, but this is only a hypothetical sketch. Also, please note that I don't have a background in chemistry, so any additional material that can help me understand the concept better is greatly appreciated.

References:

Tan, H.; Fultz, B. Rapid Electron Dynamics at Fe Atoms in Nanocrystalline Li0.5FePO4 Studied by Mössbauer Spectrometry. J. Phys. Chem. C 2011, 115 (15), 7787–7792 DOI: 10.1021/jp200127b.

First of all, in chemistry two types of formulas exist: structural and empirical. A structural formula shows the way atoms are connected. An empirical formula only summarizes atoms and their ratios. While often an empirical formula references a molecule, there are cases when the compound in question is not molecular, or it is unknown if it is molecular. In that case, since the total number of atoms in the molecule is undefined, the empirical formula may be scaled by any number, and as long as all indexes are scaled by the same number, it will denote exactly the same information.

So, formulations like $\ce{Ti_{0.9}O}$, and $\ce{Ti9O10}$ denote the same thing: a Ti+O phase with Ti/O ratio 0.9. Usually in such cases, if ratios are fixed, corresponding indexes are kept integral. So, it is  $\ce{Al2O3}$ because the Al/O ratio is fixed here, but $\ce{Li_{0.2}MoS2}$ because the Li/Mo ratio here is not.

Returning to the $\ce{Li_xFePO4}$ issue: in this phase $\ce{FePO4}$ part forms the host network similar to quartz and thus has a fixed composition. However, Li may and does migrate in and out of the host network, so its amount in the bulk is not fixed. This is possible because Fe ions have a lot of free orbitals and some of them have rather low energy, so Li electrons are passed to iron and the small Li ions may travel through the host network. Thus the composition of the electrode changes through the charge/discharge process.

In accumulator chemistry electrodes are rarely forced towards full transfer of traveling ions, because it may lead to irreversible degradation in the electrodes. For example, the $\ce{LiCoO2}$ phase would collapse if all or even a major part of Li left, so never is all or even a major share of the Li allowed to leave it. Consequently, Li-ion battery electrodes typically contain a significant amount of Li both in the fully discharged and fully charged states.

Similarly, it's usually technically impossible to achieve complete transfer even in more traditional batteries because the voltage and consequently the travel speed of ions between electrons falls down with discharge.


How does Drain opener decompose? I used drain opener (sodium hydroxide and sodium hypochlorite) in an RV and accidentally caused it to drain out (mixed about 1:10 with water) onto concrete below.

There is a fairly strong smell.

How quickly will these chemicals decompose?

What will they decompose into?

How fast will the smell go away?

The sodium hydroxide will not decompose appreciably, and leaves an area of concrete dangerously caustic. A person or pet walking across it could suffer serious burns.

You can neutralize the alkali with a mild acid such as vinegar or citric acid. Vinegar is only 5% acetic acid, so you'd need about 20 times as much as the amount of drain cleaner. Too much acid will attack concrete, so use pH indicator strips (e.g. for fish tanks or swimming pools) to adjust pH to neutral (about 7).

During the neutralization, a bit of chlorine gas will be driven off from the hypochlorite, but that smell should no last long. Any other smell would be from waste in the drain... that might last quite a while.


What analytical method(s) to use for finding the contaminations in a fatty base material In a large batch of a fatty combustible mixture there appear to be two contaminations, since after combustion there are 2 residues clogging up the injectors and exhaust system:
- a calcium salt
- a phosphate

We can easily analyse these residues, but to find a method to remove these contaminations before they are combusted we need to find out what form they have in the base material.
From that information we could try to 'wash them out' and quantify the contamination in both the 'washing sollution' and in the combustion gasses to optimize this process.
But for starters we need to know where the contamination is comming from in the first place, so hence my question: what analytical method(s) could be used to figure out these (probably small) contaminations in this fat ?

I have found that with the ICP-MS method we can at least get a quantitative approach of the elements in these samples.
Unfortunately it is not so easy to figure out the form these contaminations are in, especially the phosphate.


Why do the halogenations of benzene and toluene lead to different products? The reaction of  benzene with $\ce{Cl2}$ in presence of sunlight breaks the double bonds and 6 chlorine atoms attach to the ring.

However, the same reaction done with toluene does not lead to removal of the double bonds. Instead, the $\ce{H}$ atom of $\ce{CH3}$ is substituted.

Can somebody please explain why does this happen. If more amount of $\ce{Cl2}$ is taken in sunlight and the reaction is carried out for long time, will the double bonds break?

Both halogenations can be, according to procedures on SciFinder, performed at ambient temperature, simple sunlight and in reasonable reaction times.

The first step in both transformations is the formation of chlorine radicals by photoinduced homolytic cleavage of the $\ce{Cl-Cl}$ bond (equation $(1)$).

$$\ce{Cl-Cl ->[$h \nu$] 2 Cl^.}\tag{1}$$

These chlorine radicals want to react with something to lose their radical configuration. In the case of toluene, the kinetically favoured attack is that generating the stabilised benzyl radical as shown in equation $(2)$.

$$\ce{Ph-CH3 + Cl^. -> Ph-CH2^. + HCl}\tag{2}$$

This benzyl radical can then react with another chlorine molecule forming benzyl chloride (equation$(3)$), and so on in two further steps to form trichloromethylbenzene (equations not shown but analogous to $(2)$ and $(3)$).

$$\ce{Ph-CH2^. + Cl2 -> Ph-CH2-Cl + Cl^.}\tag{3}$$

Once we have reached the fully chlorinated methyl group, we can again imagine a chlorine radical looking for a reaction partner. Again, we should ask us which the favoured attack is — the one leading to the most stable products. And rather than attacking the aromatic ring, the benzylic carbon can again be attacked as shown in equation $(4)$:

$$\ce{Ph-CCl3 + Cl^. -> Ph-CCl2^. + Cl2}\tag{4}$$

Thus, we are ‘destroying’ our product because this attack is still better (leads to more stable products) than attacking the aromatic system.

Why does benzene react in the way it does? Well, equation $(1)$ is identical. However, there is no benzylic group; there are only two possibilities how a radical can react with benzene:

abstract a hydrogen radical forming a phenyl ($\mathrm{sp^2}$-hybridised) radical
attack the aromatic ring generating a 6-chlorocyclohexa-1,3-diene-5-yl radical

The second reaction is clearly favoured over the first, since while it does break the aromatic system, it still generates a p-orbital centred pentadienyl-type radical which is much more stabilised than a radical in an $\mathrm{sp^2}$ orbital.

However, toluene or its reaction products will hardly ever go this route since a benzyl radical is still much more stabilised than a pentadienyl-type one. The minute quantities of toluene products that do chlorinate this way will be fully chlorinated, but their yields will be very insignificant.


Why are antibonding molecular orbitals formed? When two atomic orbitals make one molecular orbital there must be an antibonding orbital also. Why should they make also an antibonding molecular orbital, such as the $2\mathrm b_2$ orbital in the scheme below?

Why should they make also an antibonding molecular orbital?

Orbital symmetry must be conserved. Thus, when you mix orbitals, the number of orbitals must also be conserved. As a consequence, when two orbitals are operated on to produce new orbitals, two new orbitals must be created.

Think about it conceptually this way: When two orbitals that have the same symmetry and appropriately matched energy come together, a bonding orbital is formed. This bonding orbital contains all of the "good orbital stuff" from that pair of orbitals. But, because there were two orbitals originally, there has to be "orbital stuff" left over. This leftover "orbital stuff" forms the antibonding orbital.

Now, the screwy part is that orbitals are not real. They are mathematical constructs to describe the wave-like probabilistic properties of electrons. If there are no electrons in the orbital, then the orbital is not really "there". So, while an antibonding orbital is formed, it is not real unless it has electrons in it. In the case of water above, the antibonding orbitals are not populated, so they don't exist.


When does Alcohol act as Solvent and as a Nucleophile? I was recently solving problems related to Alcohol.

and its solution is

My doubt is why methanol act as a nucleophile here and when does alcohol acts as a nucleophile and as a Solvent?

The proposed solution to this roadmap problem has several errors. I am assuming that $\ce{K2Cr2O7}$, methanol and $\ce{LiAlH4}$ are employed in excess. In the oxidation of 1 (A), chromic acid, $\ce{H2Cr2O7}$, is formed from $\ce{K2Cr2O7}$. To obtain the purported hydroxy acid 6 suggested in the solution, exclusive oxidation of the primary alcohol would be required, a pathway that leads to lactone 5 via hemiacetal 4. One could argue that hydrolysis of the lactone occurs to form 6 but, if this were the case, then hydroxy acid 6 would oxidize to levulinic acid 2 in the presence of excess oxidant. Clearly, levulinic acid 2 fits the formula $\ce{C4H8O3}$ presented in the original problem. As the problem is set forth, oxidation of the secondary alcohol is faster than that of the primary alcohol of diol 1.
The Fischer esterification of levulinic acid 2 leads to methyl levulinate 3 ($\ce{C6H10O3}$). Such an esterification is invariably conducted with excess methanol. The reduction of 3 with $\ce{LiAlH4}$ is conducted in a solvent such as ether or tetrahydrofuran (step 1) and then decomposed cautiously with water (step 2) to reform diol 1 (A) by reduction of both the ketone and ester groups of 3. Clearly, this sequence of transformations is an academic exercise and not one to be conducted in a laboratory. Hopefully, the person who provided the solution was not the same person as the one who created the question.


Can a stereospecific / stereoselective reaction produce a racemic mixture of enantiomers? I think I have my terms mixed up but I'm confused about the concept of stereoselectivity / stereospecificity as they relate to reactions with alkenes.

For example, hydroboration is a stereospecific reaction. It leads to syn addition across an alkene. However, you can form a racemic mixture from it because you can add from either the top or bottom faces.

I think this is a good depiction of what I mean. It seems to me that hydroboration is called stereospecific because it does syn (rather than anti and syn). But it still forms two different stereoisomers, so how can we still call it stereospecific?

You haven't mixed up your terms: a stereoselective reaction is defined as one which preferentially forms one stereoisomer over another. Likewise, all stereospecific reactions are also stereoselective, also stereospecificity has stricter criteria then plain old stereoselectivity.

However, you have to remember that the term stereoisomer encapsulates both enantiomers and diastereomers. The hydroboration reaction is stereospecific because reacting one particular diastereomer of the alkene (either cis or trans) leads to one particular diastereomer of the product. It doesn't necessarily lead to a particular enantiomer. Basically, hydroboration leads to the formation of a racemic mixture of a particular diastereomer.

The alkene you have chosen, 2-methylbut-2-ene, is not a good example substrate that can actually show how the reaction is stereospecific. Since one of the alkene carbons is symmetrically substituted, there's no diastereomerism present in the products. If we use (2E)-3-methylpent-2-ene as an example substrate, then it becomes clearer:

Note how neither of the (2R,3R) or the (2S,3S) stereoisomers are formed. However, the two products that are formed are enantiomers of each other.


Need a malleable solid which dissolves in HCl, acetone, or water I need to temporarily plug the end of a 3 mm ID copper tube with some sort of malleable solid solute that I can dissolve at later point by flushing with a solvent/acid via syringe. I need to keep the area at around room temperature, so I cannot use dry ice or excessive heat.  I'm using it to prevent liquid silicone from seeping up the end of a tiny copper tube while I seal the tube, so it needs to be relatively inert to the copper tube and the silicone.

I've tried using Styrofoam and acetone (which is almost ideal), but this leaves behind a residue of slimy gunk that is difficult to remove, and I need the tube to be clear of foreign residues.  Low melting point solid materials may be possible to use, but they can't leave behind any residue for my purposes.  I can obtain any special chemicals through my university.

If the cavity in which you insert silicone is not air tight I would use and air pump to blow in the tube.

The easiest chemical I could imagine using is a low melting point salt such as ammonium acétate. If the tube is removable you dip in the liquid and upon solidification you can remove the excess. It's water soluble.

Sugar toffee will have a similar behavior and is malleable at a certain proportion of water.

Plumber use bread.


Why do two solutes permitted to diffuse simultaneously across a semi-permeable membrane both end up at equal concentrations on both sides? Let's say we have solute a, which is highly concentrated (let's say 99%?) on side 1 of a semipermeable membrane.

We also have solute b, which is has a much lower concentration (let's say 0.01%) on side 2 of a semipermeable membrane.

If we allow these two solutes to diffuse across a semi-permeable membrane that permits both, my reading seems to suggest that we would eventually expect 50% of solute a to occur on side 1 and 50% on side 2. We would also expect 50% of solute b to occur on side 1 and 50% on side 2.

I'm not sure that I understand this.

I completely agree that 50% of the a+b molecules will end up on side 1 and 50% of a+b will end up on side 2.

But I don't understand why different solutes travel along their own concentration gradients independent of other solutes in solution. Why would they not interact with the other solute?

Perhaps they do, and the diffusion kinetics of solute a would dictate that it reaches equilibrium much faster than solute b?

Would it be correct to predict that the larger the difference in initial global concentration, the longer the lag between when "a" reaches equilibrium vs. when "b" reaches equilibrium?

If we assume that the solvent is the same on both sides of the membrane (or that the solvent itself can penetrate the membrane) then the final equilibrium will have the same concentration across the membrane.  That only translates to "50% of solute b to occur on side 1 and 50% on side 2" if the volumes are equal.

The rate at which solute A moves from left to right is proportional to the rate at which molecules of solute A on the left side randomly collide with the membrane.  This is proportional to the concentration of A in the left side solution.  The presence of molecules for solute B does not affect this process.

The reverse process is also happening with molecules diffusing from right to left at a rate proportional to their concentration in the right side solution. As the concentration on the right side increases to be equal to the concentration on the left, so the diffusion rates become equal and there is zero nett diffusion and the system approaches equilibrium.

Note that this assumes a "perfect" system where there is no chemical reaction occurring between the solutes or between the solutes and the membrane.  In practice this means that either the interaction between solutes A and B is the same as the interaction between the solutes and the solvent or that the solute molecules are so greatly outnumbered by the solvent molecules that the solute-solute interactions are not significant.

The rate of diffusion of solute A may be different from B (i.e. the proportionality constant between rate and concentration may be different).   This means that before reaching equilibrium the relative concentrations of A and B may change but at equilibrium, the relative concentration will be the same as initially.

If we define "reaching equilibrium" as having some fraction (say 99.99%) of the final concentration then increasing the initial global concentration will increase the lag for both solutes equally and will not change their relative concentrations.


Is it true that organic chemistry as an independent research field is no longer developing? One of my professors told me that organic chemistry has become a tool for biological research, rather than an active research field of its own.

Is this assertion valid? Does it describe correctly the general attitude towards organic chemistry in research institutions today?

The closest thing to organic chemistry would now be the shape and sequencing of protiens and or DNA. That field is best left alone as it takes too much time to pay anyone to get any applicable results.
Advances in things like machine learning are promising to the unraveling of these complex structures. Humams are better at recognising the patterns hence the "fold it" game.


preparation methods of nicotinoyl glycine How can we prepare nicotinoyl glycine  from nicotinic acid ?Is that possible by reaction between nicotinic acid and glycine?

As a microscale laboratory method, it should be possible to react nicotinic acid with, e.g., thionyl chloride, yielding nicotinoyl chloride (the acid chloride of nicotinic acid). This can be reacted with glycine to yield the desired product. Unfortunately, glycine would exist predominantly in zwitterionic form at neutral pH, so I would guess that pre-treatment of the glycine with base would probably be required prior to the reaction with nicotinoyl chloride.


Gromacs: What can be reason of getting final potential energy greater than the starting one? I wanted to optimize potential energy of a molecule using Gromacs software (v. 5.0.4, single precision). I was really surprised when I discovered that after running Conjugate Gradients integrator, the final potential energy was greater than the starting one. Does anyone know what can be a reason of that?

Perhaps I don't understand what is optimized because potential energy went down and up during simulation.

Fragment of logs:

Step 0, Epot=4.840216e+06, Fnorm=2.899e+05, Fmax=4.849e+07 (atom 672)
....
Step 56, Epot=8.949906e+03, Fnorm=2.368e+03, Fmax=4.351e+05 (atom 783)
...
Step 244, Epot=8.550481e+07, Fnorm=2.406e+05, Fmax=4.658e+07 (atom 54660)

Energy minimization has stopped, but the forces have not converged to the
requested precision Fmax < 10 (which may not be possible for your system). It
stopped because the algorithm tried to make a new step whose size was too
small, or there was no change in the energy since last step. Either way, we
regard the minimization as converged to within the available machine
precision, given your starting configuration and EM parameters.

Polak-Ribiere Conjugate Gradients converged to machine precision in 244 steps,
but did not reach the requested Fmax < 10.
Potential Energy  =  8.5504808e+07
Maximum force     =  4.6581140e+07 on atom 54660
Norm of force     =  2.4062566e+05

You are right, in a normal simulation using conjugate gradient (i.e. an energy minimization) the total energy should gradually decrease towards a plateau.

Based on the numbers in the log I guess that there are some atom clashes. The potential energy of the system is too high and the forces also. I would check atom 54660 (as indicated in the log) to see if you have it too close to other atoms or in a strange configuration. If you don't see anything wrong, double check the dimension of the simulation box, maybe it is too small and it causes problem with the periodic boundary conditions.

Anyway, energy minimization using conjugate gradients (CG) should be able to solve these kind of problems. You say you run gromacs in single precision. Try to use gromacs compiled in double precision, at least at the beginning of the energy minimization. If the initial conformation of the system is very bad, single precision can be unstable and unable to minimize the energy.


Triphenylphosphine (PPh3) vs. 1,2-bis(diphenylphosphino)ethane (dppe) ligand? What about $\ce{NiCl2(dppe)}$ makes it a more efficient catalyst for Suzuki cross-coupling as opposed to $\ce{NiCl2(PPh3)2}$ or $\ce{NiCl2(PCy3)2}$? How does the dppe ligand increase the reactivity of the catalyst?

Literature claims:

The experiment I am following is from J. Chem. Educ. 2017, 94(6), 786-789 (https://doi.org/10.1021/acs.jchemed.6b00273). The supporting information contains class data which says that the turnover number using a dppe ligand is roughly 4 times that of $\ce{PCy3}$ or $\ce{PPh3}$. The paper suggests that increasing electron ability increases catalytic activity, so what about electron donating ability changes catalytic activity and what makes dppe more electron donating?

As far as I am aware, the selection of the best ligands for transition-metal catalysed reactions is still trial and error. In fact, not only is the selection of ligand trial and error but so is the choice of starting materials to result in that metal–ligand combination.

What the original discoverer of a reaction method typically do is run a number of screens keeping most parameters constant but varying one; this one varying parameter will be ligand, solvent, time, temperature, additives and metal salt. (Usually, they will have decided before running the experiments which metal they want to use for their reaction so the only reason to include another one is to directly compare the new method to an existing one and e.g. show that nickel is better than palladium in this reaction.) Whichever reference you’re looking at, the authors most certainly performed such a screening and received a slightly higher yield for one set of reactants compared to another; they then published that as their final choice.

After having found the optimal conditions, many researchers are tempted to ‘explain’ why a certain ligand is better than another ligand. Our knowledge does include how electron-rich or electron-poor the donor atom of a given ligand is, how sterically demanding a set of substituents is and so forth. If ligands of one general type form a general trend, one could well be tempted to draw from that general trend and exclaim that e.g. an electron-rich phosphane works best for this reaction. In ideal cases, that prediction is then tested experimentally with a previously unused but more electron-rich phosphane; a reaction that may or may not work. The point, however, of this paragraph is that more often than not the ‘explanation’ is little more than a post-factum rationalising ballpark and should be treated as such. Years to decades of research will likely be required until we actually fully understand the intrinsic properties at play and are able to make good predictions.


What causes positive and negative deviations when weighing analytical samples? Positive deviations are when the measured mass is more than the true mass of the sample and negative deviations are when the measured mass is less than the true mass of the sample. What steps causes this type of errors in analytical measurements?

There can be plenty. There is no definite list. I would suggest to look up the following terms in Skoog or Daniel Harris' Analytical Chemistry. Worth a trip to the library now :-) The following concepts give rise to systematic errors

(i) buoyancy error
(ii) temperature differences
(iii) electrostatic charge build up on the sample
(v) moisture
(iv) above all loss of calibration of the balance.


Opposite effects of Le Chatelier's Principle How does Le Chatelier's Principle deal with a change in reaction conditions which can simultaneously effect the equilibrium in opposite ways.
For example, here's the equation for the conversion of $\ce{N2}$ and $\ce{H2}$ into $\ce{NH3}$. It's an exothermic reaction, so I've added heat as a product on the right side.

$$\ce{N2(g) + 3H2(g) <=> 2NH3(g) +\text{heat}}$$

If I increase the temperature (add heat) that would tend to push the reaction to the left because the reaction to the right is an exothermic reaction. But if I add heat (and the volume remains constant), then I'm also increasing the pressure, which should tend to push the reaction to the right, towards fewer moles of gaseous products.
Is there a simple explanation for how Le Chatelier's Principle works in such a situation?

It is important to know the externally caused change the system is dealing with.
If the primary change is increasing the pressure, then the equilibrium shifts to the lower volume, in case of gases the lower molar amount.
If the primary change is increasing the temperature, than the equilibrium shifts away from the direction that evolves thermal energy.
If external causes are combined, like pressurizing the system AND externally initiated heat transfer, the resulting direction of equilibrium shift would depend on what effect is stronger. As the Le Chatelier's rule is qualitative, it cannot provide answer here, but it must be quantitatively calculated via the equilibrium constant at particular temperature.


Carbonic Acid Titration with Sodium Hydroxide For my school lab, I'm determining the solubility of carbon dioxide at different temperatures. I will determine the amount of dissolved carbon dioxide by titrating carbonated water with sodium hydroxide. I'm confused as to how I can determine the pH at equivalence point (for indicator)? I don't know the concentration of the carbonated water so how would I go about determining the appropriate indicator?

The two pKa values for carbonic acid are given as 6.37 and 10.25, so you may want to pick the midpoint between these (pH 8.31) as your equivalence point. At that pH, a tiny amount of base will change the pH a lot, so the amount of base needed should not depend strongly on the exact value you select for the equivalence point. But the pH range is rather narrow, so I would recommend that you use a pH meter instead of a visual indicator for the titration.


Can I boil off chlorine? Does it evaporate quickly at high temperatures? Can I boil off chlorine?
Does it evaporate quickly at high temperatures?

I am asking because I want to remove it from drinking water, and I don't want to wait 24 hours for it to evaporate naturally.

Yes, solubility of chlorine decreases drastically as the temperature rises, and it's almost insoluble in boiling water.
That's also the reason why in the areas where tap water is chlorinated, it advised to boil it before drinking.
Data from [1, p. 8]:

Figure 5. Solubility of chlorine in water, hydrochloric acid (two concentrations), and sodium chloride solutions (three concentrations) All percentages are weight percents.
In aqueous solutions, chlorine is partially hydrolyzed, and the solubility depends on the pH of the solution. Below 10 °C chlorine forms hydrates, which can be separated as greenish-yellow crystals. Chlorine hydrate is a clathrate, and there is no definite chlorine: water ratio.

References

Chlorine: Principles and Industrial Practice, 1st ed.; Schmittinger, P., Ed.; Wiley-VCH: Weinheim ; New York, 2000. ISBN 978-3-527-29851-8.


What databases or repositories are there that have generalized SMILES for functional groups? Does a resource exist that has the SMILES formula of the generalized pattern that represents the group?

For instance: I have the aldehydes group, which has the pattern: -CHO.  Is there a resource that lists each functional group with the generic pattern that defines the group and its SMILES representation?

I'm hoping for, at the very least, getting the SMILES formula for the functional groups represented in this pdf: EPA Chemical Compatibility Chart

Follow up question: Is SMILES still relevant or are there better ways to represent chemicals?

I'm currently working on a content store for a resource like this because I've realized I need this as well (and came across your question ~2 weeks ago).

Content Vairable Store

I am writing a python package called global-chem to actively support variable common storage (especially for SMILES/SMARTS) strings that can be easily distributed via pip.

https://github.com/Sulstice/global-chem

The code is fairly simple and can be a one-liner:

from global_chem import GlobalChem
functional_groups = GlobalChem().functional_groups_smiles

GlobalChem is a class and properties on that class correspond to different sets of SMILES/SMARTS strings.

As of right now, I have

93 functional group for SMILES
85 functional groups for SMARTS
19 Amino Acid SMILES strings

I suspect it will grow more over time and probably as I, and others, contribute to it.

If you want to read the docs: https://globalchem.readthedocs.io/en/latest/?badge=latest

Validation

Since these strings can often be hard to validate I've been using a blend of MolVS and RDKit to make sure the strings will match as expected. It probably needs to be also manually tested each one and verified so use at your own discretion while I get some guard rails up.


Competitive protonation between alkene and carbonyl Given this compound in an acidic medium, what would get protonated first?
Would the double bond not get protonated preferentially, since oxygen is more electronegative than carbon and less willing to donate an electron pair for protonation?

I checked the $\mathrm{p}K_\mathrm{a}$ value for the conjugate acid of an  aldehyde, and it was around -10, which is pretty low; this suggests that the aldehyde would have a very low basicity.

Also carbocation had a $\mathrm{p}K_\mathrm{a}$ value of around -3.

Going by this information does it not make sense for a double bond to get protonated first?

On the other hand, the oxonium ion is resonance stabilised, which would indicate that it is protonated in preference to the double bonds.

Which reasoning is correct? And more importantly, in these situations how are we supposed to predict which factor dominates?

In the scheme below I have shown the three most probable protonations of the molecule in question. I have considered that a tertiary carbocation is typically more stable than a secondary one and that a resonance-stabilised carbocation is typically more stable than a non-stabilised one.

From left to right, let’s discuss the structures.

The one on the left is not good. The carbocation is only stabilised by being tertiary. It would take a very strong acid to sufficiently protonate in this way, or a good nucleophile that can capture the positive charge on the tertiary carbocation.
The one in the middle doesn’t look as bad initially, because it is resonance-stabilised. However, if you draw the other principal mesomeric structure, you arrive at a positively charged oxygen with a sextet — a very unstable thing to have. It is very improbable that protonation would occur like this.
There is another (undrawn) structure that would protonate the same double bond but on the other side. However, that carbocation would destroy the unsaturated ketone’s resonance and is therefore unfavored.
Finally, the protonated carbonyl. This is clearly the best cation, since it is an all-octet structure and the positive charge can be stabilised by resonance. Futhermore, if you draw the resonance structure, you realise that it localises the cation on a tertiary carbon — the best possible case.

Thus there should be no discussion; protonation of the carbonyl oxygen leads to the most stable cation and is therefore expected to prevail.

(Your lower $\mathrm{p}K_\mathrm{a}$ of carbocations is likely due to the fact that they can often deprotonate to form a double bond giving additional stabilisation.)


For group 7 elements, are anions a lower energy state than being neutral atoms? Given that electron affinity is positive, why don't these atoms attract free electrons and "prefer" to exist as anions?  Never made sense to me — I could be missing something obvious.

Yes, the electron affinity for group 17 elements (group VII by older IUPAC nomenclature) is generally exothermic, so the uptake of an electron should be favourable.

But as Ivan mentioned in a comment, there is no such thing as a free or lonesome electron. Each electron must first be displaced from another atom — ionised; the corresponding thermodynamic property is the ionisation enthalpy. For every element, the ionisation enthalpy is positive, thus it is always unfavourable to displace an electron from a neutral atom. No exceptions.


Bombarding atoms with electron gun? Just to explain this question in a better way just think about the Rutherford's experiment(the alpha particle bombarded on Gold foil) be conducted using a electron gun in place of the Alpha one.. So what will happen when electron is targeted on a gold foil?

Either the atom rather the electrons of Ag atom become excited and therefore emit some sort of waves(if yes,then what kind of waves?), or in return electrons are ejected from the atom along with some waves,  or it may be possible that the electron pass by the gold foil without any effect...

Well, I think, in principle, depending on the energy of electrons, different events can happen, but in the classical experiment electrons will diffract from the spacing within crystal lattice of gold in the foil.

P.S. The experiment is actually very well known in the history of science due to the following reason (quoted from here).

  It is interesting to recall that G.P. Thompson, who shared the 1937
  Nobel Prize with Davisson for these experiments which proved that
  electrons are waves, is the son of J.J. Thompson who received the
  Nobel Prize in 1906 for proving that cathode rays were actually
  particles - electrons! And the amazing thing is that they were both
  right.


What causes the lowering of vapour pressure in volatile/nonvolatile solvent mixtures? "Based on Figure 13.18, you might think that the reason volatile solvent molecules in a solution are less likely to escape to the gas phase, compared to the pure solvent, is that the solute molecules are physically blocking the solvent molecules from leaving at the surface. This is a common misconception. Design an experiment to test the hypothesis that solute blocking of solvent vaporization is not the reason that solutions have lower vapor pressures than pure solvents."

The explanation (in the main chapter, not as an answer to this question) given in Chemistry: The Central Science (13th Edition) by Brown, LeMay, and Bursten (abbreviated as CTCS) is: "A solution consisting of a volatile liquid solvent and a nonvolatile solute forms spontaneously because of the increase in entropy that accompanies their mixing. In effect, the solvent molecules are stabilized in their liquid state by this process and thus have a lower tendency to escape into the vapor state. Therefore, when a nonvolatile solute is present, the vapor pressure of the solvent is lower than the vapor pressure of the pure solvent, as illustrated in ▲ Figure 13.18."

However, this is, I believe, the only textbook which gives this explanation for lowering in vapour pressure on adding a non-volatile solute. The other (reputed) textbooks I've read, like Zumdahl and NCERT Grade XII Chemistry, all give the explanation that CTCS calls a "common misconception". So, which explanation is correct? If possible, please give an experiment also.

I think that the second explanation is the correct one. A simple experiment can prove it: if you place a porous cover over the pure solvent, its vapour pressure will not change. In alternative, you can place a few corks and observe the same result (no variation in the vapour pressure). In both cases surface sites are blocked, but the vapor pressure will obviously remain unaltered. Therefore the reason cannot be that the solute molecules are hindering the solvent molecule escape by occupying the solution surface.

The force that is driving the solvent molecules to the vapour phase is the difference between the entropy of the liquid and the vapour phases: when the molecule escape the liquid there is an entropy gain. Adding a non-volatile solute will increase the entropy of the liquid phase without having an effect on the entropy of the vapour phrase (given that the solute is non-volatile). This will lower the difference in entropy between the two phases, which is the driving force that make vaporisation happen, therefore there will be a lower number of solvent particles in the vapour phase and a consequent lower vapour pressure.

Textbooks like Chemistry: The Molecular Nature of Matter and Change (Silberberg, Amateis) or Physical Chemistry for the Biosciences (Chang) provides the same explanation. In addition, there is a short paper/article on this topic[1] where the author highlights how only very few freshman chemistry texts give an adequate explanation of this phenomenon by making use of a entropy-based argument.

     1 Vapor Pressure Lowering by Nonvolatile Solutes, G. D. Peckham, J. Chem. Educ., 1998, 75 (6), 787


Where can I find a complete and accurate table of CPK colours? I have been using several chemical drawing softwares including MarvinSketch, Jmol, Accelrys DS Visualizer, Avogadro, etc., all of which I have set to CPK colouring. Despite this supposedly common setting I have noticed that fluorine atoms are coloured rather varyingly from software to software. For example, MarvinSketch uses orange, Accelrys uses cyan as does Avogadro and Jmol uses algae green for fluorine atoms. I have tried Googling for the answer but I have only found incomplete tables, aside from on the Wikipedia website but I am dubious as to using Wikipedia for this information due to its reputation for inaccuracy.

Indeed, this is annoying. Because of this problem, several years ago, one of the Jmol developers and I sat down and worked out a color scheme. It was similar to Accelrys, although we tried to make certain "known colors" match (e.g., rust-like for Fe, golden for Au, etc.)

Certainly, there's been an effort to keep color consistency between Jmol and Avogadro

These colors are open source and available, e.g., through Open Babel

If there are differences between Jmol and Avogadro, I'm sure both the Jmol developers and the Avogadro developers (including myself) would be happy to reconcile them.


Octahedral Crystal Field Splitting Orbital Energy Levels (image source)

In the above diagram, where do the 3/5 and 2/5 come from? Also, how are the $\ce{e_g}$ orbitals degenerate with each other?

The total energy of the d orbitals before and after the application of the octahedral (cubic) field has to be the same. As three levels fall and two rise, the ratio is as shown. The $\Delta_0$ arises from the solution of the equations. (In older literature the total gap is $10Dq$ with the levels $6Dq$ and $-4Dq$.)

If $V_0$ is the interaction energy $3\Delta_0/5=\int\psi(e_g)^*V_0\psi(e_g)d\tau$ and $-2\Delta_0/5=\int\psi(t_{2g})^*V_0\psi(t_{2g})d\tau$. In practice $\Delta_0$ is found empirically.

The order of levels is reversed in a tetrahedral field and the size of $\Delta_0$ is also changed to $-9/4\Delta_0$.


Position of nodal planes in molecular orbitals From the attached picture (MO diagram of cyclopentadienyl anion) I can figure out where the nodal planes are, but why are they exactly at these positions?

Why is this nodal plane invalid?

First a descriptive explanation:

The Molecule is symmetric. Thus the resulting mocular orbitals are symmetric too. If you are constructing different MOs where the nodal planes are different than in the picture above they wouldnt be symmetric.

Mathematical Explanation:

These nodal planes result from a LCAO-MO consideration with hückel approximation. You have to form the secular determinant for such a problem at which the coulomb integrals are Alpha and the resonance is only dissimilar zero when the two considered atoms are neighbors. After solving this hückel-matrix you can determinate the coefficents of the several molecular orbitals. Doing that you will see that the nodal planes exist as you drew them above.
(More Information about the Hückel-Method: here)


How to treat pressure in computational chemistry? I have a reaction where a gas at high pressure and a solution with all kinds of species is involved. How do I take pressure correctly into account to get reasonable values for $G$?
Do I calculate thermochemical corrections at the given temperature and pressure (frequency calculations) and then add additional corrections for concentration ($\Delta G = RT \ln(p)$, where $p$ is (partial) pressure or mole fraction)? Or should I calculate the thermochemical corrections at the given temperature, but standard pressure? For the additional term, should I use the overall pressure of the gas, and the mole fraction of the species in solution, or should I check for the solubility of the gas at given temperature and pressure in the solvent?
Usually, I would approach it this way:
Geometry optimization at lower level of theory, then computation of $G$ as the sum of:

single point energy at higher level
$G$ thermodynamic corrections (at $\pu{298.15 K}$ and $\pu{1 bar}$)
$G_\mathrm{solv}$ corrections from COSMOtherm (at infinite dilution)
$\Delta G_p = RT \ln(p)$, where $p$ is either the molar fraction of the compound in solution or the partial pressure in gas phase

This works quite well, however, I'm not sure how to adjust the different terms when dealing with non-standard conditions. I think that as only dissolved gas participates in the reaction, I should be interested in this as a reference point for $\Delta G_\mathrm{r}$, etc., but with what temperature/pressure do I then calculate 2? And do I take the pressure of gas above the solution in 4 or the molar fraction of gas in solution?

Old question, but still seems to attract some interest.
In general, the total pressure of the system is not taken into considerations for corrections. The pressure at which you do the corrections is the partial pressure / concentration of the molecule.
I recommend calculating the thermal correction for $G$ at the temperature of interest and the default pressure ($e.g.$ 1 atm for Gaussian). You can then add the correction for the isothermal expansion/compression to the actual concentration $c_f$ / partial pressure $p_f$:
$$\int_{p_i}^{p_f}\bigg(\frac{\partial G}{\partial p}\bigg)_T dp = RT\ln\bigg(\frac{p_f}{p_i}\bigg) = RT\ln\bigg(\frac{c_f RT}{p_i}\bigg)$$
where $p_i$ is the pressure for the thermal correction ($e.g.$ 1 atm). Note that these formulas are based on the ideal gas approximation, but that's what software like Gaussian does under the hood anyway. Hence, you could in principle directly set the correct pressure and should get the same result, but I find it more convenient to have the same route section for all calculations.
Regarding your multiphasic systems: As you noted, the reaction will most likely take place in the solution, so you have to use concentration of the gas in the solution.


Should double or triple bonds have preference for low locants in IUPAC nomenclature? Both double and triple bonds are present in the hydrocarbon shown above, which has the name hexa-1,3-dien-5-yne. Why is the numbering chosen such that the alkene has lower locants than the alkyne? Is there a rule which stipulates that the alkene is a more 'senior' functional group?

The appropriate passage in the 2013 Blue Book is as follows:

P-31.1.1.1 The presence of one or more double or triple bonds in an otherwise saturated parent hydride [...] is denoted by changing the ending ‘ane’ of the name of a saturated parent hydride to ‘ene’ or ‘yne’. Locants as low as possible are given to multiple bonds as a set, even though this may at times give ‘yne’ endings lower locants than ‘ene’ endings. If a choice remains, preference for low locants is given to the double bonds. In names, the ending ‘ene’ always precedes ‘yne’, with elision of the final letter ‘e’ in ‘ene’. Only the lower locant for a multiple bond is cited, except when the numerical difference between the two locants is greater than one, in which case the higher locant is enclosed in parentheses.

Therefore, the preferred IUPAC name for the above compound is hexa-1,3-dien-5-yne, as the double bonds receive the lower locants (1,3) than in the alternative name hexa-3,5-dien-1-yne.
Note that this preference for double bonds over triple bonds is only relevant when there is no difference between the locant set for the multiple bonds. In the following example, the locant set (1,4) is lower than (2,5) and hence hex-4-en-1-yne is the preferred name, even though it has a higher locant for the double bond:


Reaction of perchloric acid with cyclopropenol Question:

Identify A and B.

My attempt
The first step will be the attack of $\ce{PhMgX}$ on carbonyl carbon, that is, 1,2 position followed by hydrolysis, yielding an alcohol.
But how does the reaction proceed with $\ce{HClO4}$?

The first step is indeed 1,2-addition of the Grignard reagent to the keto group, and product A is the alcohol:

Treatment of the alcohol with the strong acid $\ce{HClO4}$ cannot yield the usual dehydration product (alkene) in this case, as the corresponding alkene would have two double bonds in an already highly strained three-membered ring, which is highly unfavorable. Instead, after protonation of the $\ce{OH}$ group and subsequent loss of $\ce{H2O}$, the carbocation:

is obtained as the final product B (with $\ce{ClO4-}$ being the counteranion). The planar cyclopropenyl cation is a Hückel aromatic compound with $(4 \cdot 0+2) = 2$ $\pi$ electrons, and delocalization of the positive charge over the two phenyl substituents can provide further stabilization.


Does concentration of salt increase or decrease rate of rusting? I want to know whether increasing the concentration of salt, specifically $\ce{NaCl}$, increases or decreases the rate of rusting. There are conflicting theories that explain opposite outcomes:

Increasing concentration of salt decreases the amount of dissolved oxygen and thus    decreases the rate of rusting
Increasing concentration increases electron mobility and thus increases rate of rusting

I haven't been able to how rusting works in full detail nor have I been able to understand why there are two contradicting theories present. It seems that people write it more out of opinion than with factual basis.

Can someone please clarify?

In general, salt (particularly NaCl) will increase the rate of corrosion (rusting).

To understand why, consider metallic iron $\ce{Fe}$ which rusts (oxidises) to iron(II) oxide $\ce{Fe2O3}$ in the presence of oxygen $\ce{O2}$ and water $\ce{H2O}$.

$$\ce{4Fe +3O2 +6H2O->4Fe(OH)3}$$

Corrosion (rust) is a 'redox' reaction, which means it involves reduction of oxygen into hydroxide ions and oxidation of metallic iron to iron cations.

For any redox reaction to take place, electrons are transferred. In this case, from the metallic iron which loses electrons to oxygen. The presence of salt (or any electrolyte) in the water accelerates the reaction because it increases the conductivity of water, effectively increasing the concentration of ions in the water and so increasing the rate of oxidation (corrosion) of the metal.

The situation is complicated by the fact that salt dissolved in water actually reduced the amount of dissolved oxygen present in the water. This is because the water molecules are attracted to the dissolved ions from the salt (solvation), which has the tendency to decrease the weak affinity of non-polar oxygen molecules to water, thereby deriving dissolved oxygen out. So if the metal is totally submerged in water, higher salt concentrations can actually reduce the rate of corrosion, if the water is not aerated.

So it depends upon the amount of oxygen getting into the water and onto the metal. For example, underwater plants can release oxygen in salty water which will accelerate the rate of corrosion of the submerged metal. Of course, if the metal is not completely submerged in salt water but only partially submerged, or is affected by salt water 'spray', it will be still be exposed to oxygen and will rust faster.


How to distinguish between benzyl chloride and chloro benzene? If I add $aq. \ce{KOH}$ to both of the compounds substitution takes place in the case of benzyl chloride and then if I oxidize with $\ce{KMnO4}$ I will get benzene carboxylic acid which will give effervescence on reaction with $\ce{NaHCO3}$. But, does $\ce{KMnO4}$ have any effect on chloro benzene?

When benzyl chloride is treated with $\ce{NaOH}$ and $\ce{AgNO3}$ , a white precipitate is formed.
Chlorobenzene does not form white precipitate reacting with $\ce{NaOH}$ and $\ce{AgNO3}$.


Feasibility of using Magnesium, Beryllium, Lithium as light sources? First off, I'm asking this for worldbuilding purposes, so don't worry, nobody will get hurt ;)

Magnesium, Beryllium and Lithium burn relatively bright. If they are somehow thinned out in some other powder so they burn weaker and last longer, would it not be possible to make torches or lamps out of them?

Berillium is toxic and quite hard to obtain, so it's off.

Lithium is more common but still quite rare. It still is used in some pyrotechnic compositions, but rarely in form of metal.

Magnesium IS used in some composition that are intended is light sources. At the dawn of photography photoflashes used magnesium-based compositions. But it is rarely used nowdays, both thanks to electrical flashes and because it generates too much UV light, so it may harm eyes.

BTW more recent photoflash compositions use aluminium to avoid avoid the bane of magnesium-based composition: extremely fine smoke.

BTW, to obtain any metal above a lot of electricity is required, and in a setting with such metals Yablochkov candle should be available and it has quite vintage/steampunk look.


Which one has higher priority according to Cahn–Ingold–Prelog rules? Which one has higher priority according to Cahn–Ingold–Prelog rules?
(1) $\ce{-CH2OH}$ and $\ce{-CH=CH2}$
(2) $\ce{-CH2OH}$ and $\ce{-C(CH3)3}$

The answer says $\ce{-CH=CH2}$ has higher priority, but I don't understand why.
Shouldn't the first one have higher priority since oxygen has a greater atomic number than carbon?
Same for the second question as well.

You are correct. In both cases $\ce{CH2OH}$ has higher priority. the Cahn-Ingold-Prelog system uses atomic number as its primary determinant of priority. However, a popular misconception is that the CIP system uses something related to "size", like mass, number of carbon atoms, length of chain, etc.

Here is a way that I use when I teach the subject. In the following compound, the chirality center has two bonds to carbon atoms. The carbon atom on the left is bonded two an oxygen atom and two hydrogen atoms. The carbon atom on the right has two bonds to another carbon atom and one bond to a hydrogen atom. Write these connections vertically below the carbon atoms with highest atomic number at the top. Compare each row, and the first difference is where you judge. In this case, the first difference is the first row: oxygen has a higher atomic number than carbon. $\ce{CH2OH}$ has higher priority than $\ce{CH=CH2}$.

You can do the same analysis for $\ce{CH2OH}$ vs. $\ce{C(CH3)3}$.


Standard curve of absorption and concentration I just wanted some quick help with a lab I'm doing, based on absorption spectroscopy.  I'm given an aqueous stock solution of Yellow Dye #5 with a concentration of 54.5 micro moles. I'm asked to create a standard curve.  It gives me the diluted volume as 50 mL, and has a data table with the volume of stock solutions and absorbance values for each.  It asks me to create a standard curve (absorbance versus concentration) in excel for the Yellow Dye #5.  First off, would I just use the constant concentration of 54.5 micro moles for this standard curve graph and the absorbance values in the data table?  I'm not familiar with what a standard curve is, or if the volume of stock solution affects the concentration value in the graph. If anyone replies I'll probably comment on the right answer with a couple more, thanks for the help!  It means a lot!

Supposed that Yellow Dye #5 refers to the FD&C numbering, this is a food colorant more known as tartrazine outside the U.S.

A standard curve correlates absorbance at a particular wavelength with the concentration of the dye.

Supposed that no distortion (e.g. due to aggregation effects) is present, the relation is given by the Lambert-Beer law: $$E_\lambda = \epsilon_\lambda \cdot c \cdot d$$

where $E_\lambda$ is the absorbance, $\epsilon_\lambda$ is the molar absorption coefficient (specific for a particular dye at a particular wavelength $\lambda$ in a certain solvent), $c$ is the concentration of the dye and $d$ is the path length of the cuvette. Please pay attention to the units in your calculations!

In order to create a standard curve,

Record the uv spectrum of the stock solution. If you have two-channel spectrometer, put a second cuvette with the neat solvent in that channel.
Get a couple of volumetric flasks and dilute the stock solution.
Record the uv spectra for each of these solutions.
Plot absorbance (probably at around 420 nm) vs concentration.


What is the difference between ΔfH and H-H° in NIST–JANAF tables? Can someone explain the difference between $Δ_\mathrm{f}H$ and $H-H^\circ$ in NIST–JANAF tables?

As a side note, "JANAF" stands for Joint Army–Navy–Air Force; these tables were designed to provide quick access to the thermodynamic data for rocket scientists.
Here, $Δ_\mathrm{f}H^\circ$ is the standard enthalpy of formation. The difference $H-H^\circ(T_\mathrm{r})$ is required to calculate the standard enthalphy of reaction $Δ_\mathrm{r}H^\circ$ if in the given temperature range a reaction takes place.
There is a thorough explanation in Silbey's Physical Chemistry [1, p. 62]:

The JANAF tables give standard enthalpies of formation at a series of temperatures, and so these values may be used directly to calculate enthalpies of reaction. [...] Some thermodynamic tables give values of $\bar{H}^\circ_T - \bar{H}^\circ_{298}$ to assist in the calculation of $Δ_\mathrm{r}H^\circ$ for a chemical reaction or phase transition:
$$\bar{H}^\circ_T - \bar{H}^\circ_{298} = \int_\pu{298 K}^T \bar{C}^\circ_P \mathrm{d}T$$
Depending on the table, $Δ_\mathrm{r}H^\circ$ for phase transitions in the intervening temperature range may be added to the right-hand side of the equation.

NIST-JANAF tables are given for a reference temperature $T_\mathrm{r}$ and a standard state pressure $p^\circ$. Unless stated otherwise, $T_\mathrm{r} = \pu{298.15 K}$ and $p^\circ = \pu{0.1 MPa}$ [2].
References

Silbey, R. J.; Alberty, R. A.; Bawendi, M. G. Physical Chemistry, 4th ed.; Wiley: Hoboken, NJ, 2005. ISBN 978-0-471-21504-2.
NIST-JANAF Thermochemical Tables, 4th ed.; Chase, M. W. J., Ed.; American Institute of Physics: Washington, DC : New York, 1998. ISBN 978-1-56396-831-0.


How can transition metals form so many bonds with ligands? For instance, copper can form $\ce{[Cu(H2O)6]^2+}$ so it accepts 6 electron pairs from $\ce{H2O}$ molecules to form dative covalent bonds. However it has one electron in the 4s subshell and 10 in the 3d subshell. To which shell does it accept these electrons? Is it to the 4d subshell? And how are these accepted electrons arranged?

Copper is  in +2 oxidation state so it has configuration $\ce{[Ar] 3d^9}$. Since 4s, 4p and 4d orbitals are vacant, Cu will show $\ce{sp^3d^2}$ hybridisation and will accept these elctron pairs in its 6 $\ce{sp^3d^2}$ hybridised orbitals. Read more about VBT here http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch12/valence.php


Why is lanthanum a lanthanide and actinium an actinide? I do know that the lanthanides start with the element lanthanum, but why? Lanthanum doesn't even have an $f$ orbital, so why isn't it considered a transition metal?

It's the same way with actinium, which doesn't even have any electrons in its $5f$ orbital, why isn't it also a transition metal?

According to www.webelements.com, "There is no one single or best structure for the periodic table..."

I believe that the table was split at lanthanum since it kept the elements of the d orbitals together. You could argue that the table should be split with Sc, Y, La and Ac together, with the elements Ce through Lu and Th through Lr as the f orbital elements. The next column would have Ti, Zr, Hf and Rf.

Actually, all elements have f orbitals, below lanthanum they are not usually populated. Lanthanum is the first element that could have an electron in the f orbital in the ground state. Since the transition energy of the electron between in the d orbital and the f orbital is quite low, there should be a significant number of electrons occupying the higher energy orbital, even at room temperature.


Why does the Most Stable State of an Atom Tend to be One with Full s and p Subshells? I'm new to posting on stack exchange, although I've read a lot of it before. This question seems like it might end up being marked as a duplicate, but I've looked through a lot of the similar questions and I feel like I'm still not getting it (most answers end up just explaining that the octet rule is wrong).
I'm currently taking AP Chemistry and we've been talking about electron configuration. As I understand it, atoms generally are at a lowest energy state when they fill their outermost s and p subshells (resulting in the octet rule, as the 2 + 6 electrons in these shells add up to 8). As far as I can tell, this is mostly due to Coulomb's Law being stronger with a smaller radius and the shielding effect impacting how much protons pull on electrons. But what I don't understand, then, is why an atom would tend to gain electrons in order to fill its valence shell. Wouldn't the lowest energy state be to lose a whole shell and decrease the radius and the shielding effect? It would make the atom very positively charged, but that doesn't seem to be a problem for something like Si4+. For that matter, why have any electrons at all? Wouldn't the most stable state be for an atom to repeatedly lose electrons and just be a nucleus? Presumably this has to do with the ionization energy being too high, but why is that the case? If it's just as simple as because the electrons are attracted to the nucleus, why can any electrons be removed at all? What am I missing or where is the flaw in this reasoning?
P.S. I've found electron configuration really interesting so far, but we haven't been doing much of anything with the math that explains it. I'm taking multivariate calculus right now, and I would really like to learn about the math behind it all. Is it too early for me to start? If not, does anyone have any good introductory sources to learn about that?

The major factor is electrons with the same quantum numbers $\mathrm{n}$ and $\mathrm{l}$ do not mutually shield the nucleus charge well.
Elements approaching the noble gas group in the Periodic table ( $\ce{N, O, F}$ ) get nearly  fully filled respective $\mathrm{p}$ orbitals. The effectively perceived nucleus charge grows for valence electrons. It gets progressively more difficult to ionisate these electrons, and at the same time the energy released by capturing an extra electron grows.
Elements at the opposite table side ( alkaline metals and alkaline earth metals ) have the opposite situation. They start to fill orbitals at the new, higher quantum number $\mathrm{n}$ level. The lower, now fully filled $\mathrm{p}$ orbitals shield the nucleus well. Additionally, the new $\mathrm{s}$ orbital  is farther from nucleus with lower attractive force. Both effects lead to low ionization energy of such atoms and very low affinity to extra electrons.
This leads to the octet rule, which is consequence of the fact, if chemical bonds lead to completing octets, the total electron energy is lower.
It has its limits. Ionization of electrons leads to progressively increasing ionization energy for every next electron. Similarly, accepting too many electrons leads to negative electron affinity, so the electron is released at the nearest convenience. So ions with high positive charge occur only in heavily ionisating environment, respectively the one with highly negative charge need source of electrons. Even in solid matrices there is partially covalent bond.
It may be challenging, but this explains a lot about the screening of the nucleus charge: Slater's rules


Photosynthesis-Equation Everyone knows the simplified photosynthesis-equation from school:

$\ce{6CO2 + 6H2O -> C6H12O6 + 6O2}$

Most people assume that the carbondioxide is being split into carbon and oxygen during photosynthesis, and the equation seems to confirm this. But we know that is not true, and that the oxygen derived from photosynthesis actually comes from the watermolecule being split into hydrogen and oxygen (see e.g. this course note and wikipedia).

Clearly there is much more going on as the formula most of us learned in school suggests, so my question is: what is the correct and complete equation for photosynthesis?

If you labeled the oxygen atoms in carbon dioxide in green and the oxgyen atoms in dioxygen in red (because they derived from photosynthesis, while $\ce{CO2}$ is used up in the Calvin cycle), you would have to add some water molecules to be able to write an equation like this:

$$\ce{6C\color{limegreen}{\ce{O2}} + 12H2\color{red}{\ce{O}} -> C6H12\color{limegreen}{\ce{O6}} + 6\color{red}{\ce{O2}} + 6 H2\color{limegreen}{\ce{O}}}$$

This is problematic because oxygens in carbonyls can interchange with oxygens in water fairly rapidly (e.g. hydration of carbonyl to diol, and reverse reaction losing the "original" oxygen on the carbonyl). It would be fine to write the net reaction of the Calvin cycle separately from the net reaction of photosynthesis. This would also nicely show that hydrolysis of ATP is required for some of the steps, and that NADPH is involved.

On the other hand, if you are talking about the relationship between photosynthetic organisms vs other organisms, it is nice to have a simple net reaction for the entire organism (which also ignores that plant cells do break down glucose as well).

  Clearly there is much more going on as the formula most of us learned in school suggests, so my question is: what is the correct and complete equation for photosynthesis?

There is no official complete equation. You can add as much or as little detail you want. It really depends on where you want to go with the chemical equation. The links in the comments are great starting points.

  [OP, from comments:] Found the answer as to why i was confused by the simplified equation: "Hexose (six-carbon) sugars are not a product of the Calvin cycle. Although many texts list a product of photosynthesis as C6H12O6, this is mainly a convenience to counter the equation of respiration, where six-carbon sugars are oxidized in mitochondria. The carbohydrate products of the Calvin cycle are three-carbon sugar phosphate molecules, or "triose phosphates", namely, glyceraldehyde-3-phosphate (G3P)."

Great, it feels good to figure things out! In biochemical pathways, it can be difficult pinpointing products. The Calvin cycle and pentose phosphate pathway can be tweaked to make a variety of products, and those tweak affect what the "real" net reaction is.


Compare the boiling points of isobutane and 1-butyne Question: Compare between the boiling points of isobutane and 1-butyne

The answer to this question is given as 1-butyne has higher boiling point than isobutane.

My approach: I have been taught a few concepts (which I learnt for granted), in my organic chemistry class:

The carbon with a longer chain has a higher boiling point than its isomer with more branching or a shorter hydrocarbon chain. (Kindly correct me if I am wrong here and provide an explanation)

From here I decided that butyne will have a greater boiling point (though I made an incorrect assumption), and luckily my answer came out to be correct, but I am not convinced about the reason for it. Please help with the correct explanation for this question.
I have a proposition that since the electronegativity of $\mathrm{sp}$ hybridisation carbon is very high (only a little bit less than nitrogen), will it show hydrogen bonding? This justifies the given answer.
If it does not show hydrogen bonding, why so? All conditions are fulfilled, a highly electron deficient hydrogen is present and electrons are present in the $\pi$ bond between the carbons. Please help.

The boiling points of non-polar hydrocarbons are determined by the extent of van der Waals forces in between them. More specifically, molecules with a larger surface area have larger van der Waals forces of attraction between the molecules.

From the structures, it is obvious that but-1-yne has a larger surface area than isobutane, and hence has the higher boiling point.
Your idea of H-bonding in but-1-yne is plausible, according to this source1:

Quantum chemical calculations indicate hydrogen bond energies in the range $\pu{1–2.2 kcal mol^–1}$. Interconnected contacts such as $\ce{C#C–H ··· C#C–H ··· C#C–H}$ exhibit a pronounced cooperativity effect, with energy increases of $\approx \pu{0.5 kcal mol^–1}$ compared with isolated contacts.

However, while this bonding may be sufficient to explain a higher boiling point compared to isobutane, it is still a very weak bond compared to the H-bonds in hydrogen fluoride or water.
References:

Thomas Steiner, Evgeni B. Starikov, Ana M. Amado, José J. C. Teixeira-Dias, “Weak Hydrogen Bonding. Part 2. The Hydrogen Bonding Nature of Short $\ce{C–H ⋯\pi}$ Contacts: Crystallographic, Spectroscopic and Quantum Mechanical Studies of Some Terminal Alkynes.” J. Chem. Soc., Perkin Trans. 2 1995, (7), 1321–1326 (doi:10.1039/P29950001321).


Do I need a license or certificate to practice chemistry or what are the advantages if did have one? What can I do as a non-professional chemist that is legal?

If a license is not required, then what are the advantages to having one, such as ordering compounds, getting a job, etc.?

Note that being a chemist doesn't require any kind of state license or the like in Germany, neither.

But buying dangerous* substances and running a chemical lab requires that you** have certain licenses and implement safety standards.
Some of these exams a chemist has to pass during studies (e.g. on toxicology and chemical substances legislation) so having a Master/Diplom in chemistry automatically means that you have those licenses.

Having this license (Chemikaliensachkunde) and being reliable (legal meaning) and of age are legal requirements you must meet in order to be allowed to buy dangerous substances. In practice however, the suppliers put another very practical additional need: they usually do mail orders only B2B and require you to have VAT/USt number.
So if I as a small-scale freelancer who fulfills all the legal requirements want to buy from say, Sigma, I don't get anything because I don't have the tax number (for my size of side-business it would be too much hassle to go through this, so I use the small business exemtion law and for the supplier it would be too much hassle to make sure the buyer fulfills the legal requirements if it isn't at least a certain business size).

As a private person, you can buy chemicals:

at the pharmacy
at specialized stores, e.g.

photography chemicals at a good photography stores
carbide at a speleology outfitter's
gases (possibly including dry ice) at a gas merchant's, etc.

you basically cannot buy dangerous substances by mail/online order (too much burocratic hassle: the seller would need to make sure of your identity and Chemikaliensachkunde etc.)
I've never tried to buy privately anything that requires the Chemikaliensachkunde e.g. at a pharmacy so I cannot tell how difficult that is in practice.

* As a rule of thumb, there are restrictions for substances (or mixtures) that are extremely flammable, oxidizing, toxic or very toxic, (suspected of being) carcinogenic, mutagenic or teratogenic (plus a bunch of listed things)

** or some employee of yours

However, you can of course work as a chemist without (much or even any) need to handle dangerous substances. For example, I'm analytical chemist / spectroscopist / chemometrician. About 90 % of my work is in front of a computer analysing measurement data, developing improved ways of doing so and writing reports/papers. At least half of the rest is in front of a computer programming instruments. I've handled exactly one dangerous substance according to chemicals legislation at two days (actually a sample of few μg I got for analysis) during the last year. The everyday stuff I handle (methanol, ethanol, isopropanol, disinfectant solution and cyclo-hexane and paracetamol=acetaminophen as calibration standards) are not legally restricted dangerous chemicals.
I do handle biohazard material far more often - but that is not what the Chemikaliensachkunde is good for. So for my work, the more relevant safety issues are biological/human material safety and laser safety... It would be quite easy not to have any chemical lab exposure (is seldom enough as it is... - but hey, what did I become chemist for?)

Also I have some colleagues doing the same type of work who are not chemists but e.g. physicists and thus do not have the Chemikaliensachkunde.

For getting a job it is obviously of advantage if you can show some kind of education/training that makes you fit for the job. For jobs in chemical industry/academia that is obviously a chemistry degree or training as chemical lab technician, but also related professions such as physics/biology/pharmacy/toxicology depending on what the job in question is.


How do Chiral centers determine the number of isomers? I think I read somewhere that if an organic compound has n number of chiral centers, there must exist 2^n number of isomers. Is that true? And why?

That 2^n number of isomers are exactly stereoisomers ( Not "total no. of isomers"). There is no direct formula for calculating the number of structural isomers.Total number of isomers = (number of structural isomers + number of stereoisomers).
Though there are direct formulas for obtaining number of stereoisomers, but there is no direct formula for finding out the total number of structural isomers.
As I said above, we do have formulas for finding out the number of stereoisomers. Here is the exact formulas which you can use to determine number of stereoisomers:
a) When the molecule is unsymmetrical and contains 'n ' chiral carbon atoms:
Total no. of stereoisomers = 2^n
b) When the molecule is unsymmetrical and has even number of stereogenic centres or chiral carbon atoms:
Total no. of stereoisomers = Number of optical isomers + Number of meso forms = 2(n-1) + 2(n/2-1)
c) When the molecule is symmetrical and has odd no. of stereogenic centres
Total no. of stereoisomers = [2(n-1)-2(n/2-1/2)] + 2(n/2-1/2)]


Why does increase in pressure favor formation of water in water-ice equilibrium (at 273K)? The question:

Ice and water are placed in a closed container at a pressure of 1 atm and a temperature of 273 K. If the pressure of the system is increased to 2 atm while keeping temperature constant, why does the system's volume decrease?

To answer this question, it is best to look at a phase diagram of $\ce{H2O}$.
Note the axes are kilopascal and celsius, but we know that $0^{\circ}\text{C}=273\text{K}$ and $101\text{kPa}=1\text{atm}$. We can see that if the pressure goes up at a constant temperature, the system will move from the phase boundary into the liquid phase.

The generally cited reason for why this occurs is that hydrogen bond network of ice makes it less dense than liquid water, and since higher pressure will favor the more dense state, the ice will melt to become liquid. This leads to a lower overall system volume.


Why is the conjugate of a strong oxidizing agent a weak reducing agent? I am a grade 12 student. While I was reviewing some electrochemistry, I had the following question. Take this reduction half-reaction:

$$\ce{F2 (g) + 2e- <=> 2F- (aq) + $\pu{2.87V}$}$$

I know that $\ce{F2 (g)}$ is a strong oxidizing agent, but why does this make $\ce{F- {(aq)}}$ a weak reducing agent? I thought that since the electrical potential of the reduction half-reaction:

$$\ce{2 F- (aq) <=> F2 (g) + 2 e-}$$

was $\pu{-2.87 V}$. Therefore, should $\ce{F- (aq)}$ not be a strong reducing agent?

It is not the magnitude of the standard cell potential that indicates whether or not the cell reaction will be spontaneous, but the sign. In the same way a reaction is only spontaneous if $\Delta G < 0$, a substance only oxidizes/reduces if the corresponding cell potential is positive.


What are ω and ν in this expression? In the picture below a proportionality equation is shown:

$$\omega \propto \nu \exp\left(\frac{-\Delta G_m}{RT}\right)\propto \dots$$

While $\Delta G_m$ is the Gibbs free energy, $R$ is the gas constant, and $T$ is the temperature, what are the parameters $\omega$ and $\nu$ exactly?

You can see the context on the picture: Yttria-stabilized-zircona.

$\Delta G_m$ is the free energy barrier for the motion of atom from one site to another. The equation overall looks like a transition state approach to this activated process, and thus $\nu$ is the frequency at which this motion is attempted, i.e. a $\nu$ is a characteristic lattice frequency. In such models, it is often assumed to equal the Debye frequency.

Finally, $\omega$ is the rate of transition from one site to another, or something proportional to it (because the proportionality symbol is used rather than an equality). $\omega$ can be thus be the ionic conductivity, or a diffusion constant.

You can find the same expression and graph in this chapter of undergrad lecture notes, for confirmation.


Why does hydrogen so easily take the place of oxygen in organic molecules during reduction reactions? It goes without saying that every reduction agent has different reducing properties and strengths in terms of what sort of bond it can reduce/hydrogenate. But, why does it generally follow that hydrogen takes the place of an oxygen molecule during a hydrogenation/reduction reaction (in the presence of the necessary hydrogen or $\ce{H}$-donor of course)?

What is it about the properties of oxygen that makes it so ready to disassociate from the molecule when it gains an electron, allowing hydrogen to take its place? I'm sure there is something simple that I'm missing, but any explanation is appreciated.

But, why does it generally follow that hydrogen takes the place of an oxygen molecule during a hydrogenation/reduction reaction [...]

Ok, if at all, it's an oxygen atom ;)

I don't see that general trend towards complete deoxygenation, neither in catalytic hydrogenations, nor in hydride transfer reactions.

It usually takes

rather drastical conditions
quite some special reagents
a particular type of derivatisation

or a combination of any of these to convert oxygen-containing compounds to alkanes.

Let's have a look at some examples:

$\ce{R1-CO-R2 -> R1-CH2-R2}$

Ketones and aldehydes may be directly reduced to alkanes in the Clemmensen reduction. The "mildest" variants use anhydrous $\ce{HCl}$ and amalgamated $\ce{Zn}$ in organic solvents.
The clasical Wolff-Kishner reduction, using hydrazine and sodium in diglyme, involves the in situ functionalization of the carbonyl and isn't a prototype for a mild reaction either. Admittedly, milder variants exist, using N-tert-butyldimethylsilylhydrazone and catalytical amounts of a Lewis acid, e.g. $\ce{Sc(OTf)3}$
Tosylhydrazones may be converted to alkanes using $\ce{NaCNBH3}$, $\ce{Na(CH3COO)3BH}$, or boranes.
Catalytic hydrogenation over Raney nickel is possible after converting the carbonyl to a dithiane; so it's actually a desulfurization.

$\ce{R1-CH2-OH -> R1-CH3}$

I'm not aware of any transformation that does not involve functionalization.

Alkyl tosylates can be deoxygenated using $\ce{LiEt3BH}$.
Acylation of alkohols with thiocarbonyl chlorides furnish thiocarbonates, which may be deoxygenated under using either tin hydrides ($\ce{R3SnH}$, Barton deoxygenation) or $\ce{B(CH3)3}$ in water/benzene.
Alkanes may also be obtained from alkohols under Mitsunobu conditions ($\ce{Ph3P}$, $\ce{DEAD}$, o-nitrobenzenesulfonylhydrazine).

$\ce{R1-CH2-COOH -> R1-CH3}$

Direct transformations of carboxylic acids to alkanes typically involve decarboxylation, i.e. the loss of $\ce{CO2}$. This may be achieved either by the Barton decarboxylation, or via photoinduced electron transfer reactions with electronically excited (electron) acceptors. In the course of the latter reactions, a carboxylate is oxidized to an acyloxy radical, which then undergoes decarboxylation.

To sum it up: It's not that easy ;)


Reaction direction change based on phase of reactants and products I am curious specifically how the entropy of a reaction will change if the phases of the reactants and products at different conditions change.

For instance, a reaction

$$\ce{O2(g) + C(s) -> CO2(g)}$$

has a positive entropy because there is no net change in the moles of gas in the system and the product is more complex in its degrees of freedom (secondary). If one were to change the conditions (temperature and pressure) such that oxygen is a gas but $\ce{CO2}$ is a liquid or solid, would the change in entropy for the system become negative?

$$\ce{O2(g) + C(s) -> CO2(l)}$$
Or
$$\ce{O2(g) + C(s) -> CO2(s)}$$

If the $\Delta H$ is still negative but the $\Delta S$ is now positive, isn't the Gibbs energy now able to be positive or negative, making this a valid method of manipulating the directionality of a given reaction?

Quite correct.  And manipulating which direction a reaction tends to go by changing the phase of reactants or products is a common trick in chemical engineering.


Energy in a Chemical Reaction In a chemical reaction, is energy always either lost or gained by the reactants? As much as I am concerned, changes in energy can occur during the absorption of heat or the emission of light or heat (these are the most often occurring). Is every chemical reaction accompanied by these conditions? Aren't there any chemical reactions without these conditions? Usually when we observe common chemical reactions (sugar that is melting in the water), we don't see any emissions or absorptions. Are these changes so trifling that we can't feel them (while they actually exist), or there are reactions with no emissions or absorptions?

your question is good yet lacks the specifics.
there are different so called "energy fields" in which changes occur (entropy, potential ,gibes and so forth).
the thermodynamic field is endless and can be spanned with any variables (using maxwells equations and Legendre transformation)
there is
a sort of an axiom that states that all matter will try to find a way to a lower state of energy.
but that said its not to scientific.

now for the question:
every reaction creates some amounts of electromagnetic waves (as a price for electron "movement") some can be seen with the naked eye as they emit waves in the Uv-vis spectrum some are just infrared radiations of inner states of electron decay to a lower energetic state via vibrational relaxation.

ideally no reaction will occur in the ideal zero as there is no energy to "pay" for any kind of a reaction.

for further read:
quantitive chemical analysis by harris
chapter 17 fundamentals of spectrophotometry.