How do valence shell configurations affect attraction and repulsion forces between atoms? It seems clear that the electrons of one atom are attracted to the protons of another and that this is the source of attractive forces between atoms. However, it is not clear why an atom with a full valence shell such as a noble gas or covalently bonded atom is incapable of bonding or being attracted to other atoms. Are the electrons not attracted to protons in the same way? Is it because the repulsion between the electrons wins out over the electron-proton attractive forces. Furthermore, why do valence shell arrangements with few electrons bring about attractive forces?

Firstly, why should covalently bonded atoms be incapable of bonding? Bond changes are the very basis of quite a huge amount of chemical reactions. Oxygen occurs in the $\ce{O2}$ form, which is covalently bonded, but readily breaks its O–O bond in order to form other bonds.

Now, to your question. Why are elements in group 18 (generally: think of the xenon fluorides) inert? Why is sodium so willing to give up its electron, while neon doesn’t form any compounds at all, when they only differ by one proton in the nucleus?

The answer lies in the orbital energies. You can find a table listing the outer shell orbital energies page 16 of Appendix B of Miessler and Tarr's Inorganic Chemistry (5th ed.). Let us compare the outer orbital energies of Ne and Na. Neon has two 2s electrons with an energy of $-48.47\ \mathrm{eV}$ and four 2p electrons with an energy of $-21.59\ \mathrm{eV}$, while the 3s electron in the sodium atom lies at $-5.14\ \mathrm{eV}$. The difference is overwhelming. When a ionic compound is formed, an electron is removed from the less electronegative atom and added to the more electronegative atom, forming a cation and an anion. Now, the Na electron is much easier to remove than the Ne electron – you just need $-5.14\ \mathrm{eV}$ to do the task, while eliminating the Ne electron is much more difficult. The electron is then placed into an atom such as fluorine, whose outer 2p subshell lies at $-18.65\ \mathrm{eV}$: you needed energy to remove the Na electron, but energy is released when you place it in the F atom (since it is put into a low energy orbital). Placing an electron from F to Na wouldn’t release energy because it is much more stable in the fluorine atom, due to their orbital energies (if you want to simplify things my anthropomorphizing the electrons, you could say they’re “happier” in the fluorine atom). Removing an electron from the neon valence shell is very difficult because it has very low energy; adding one doesn’t bring much because since the second shell is full, you would have to add it to the third shell – its energy is similar to the Na 3s shell, so it’s not convenient.

I have to note here that while the first ionization energy corresponds more or less to the orbital energy (Na: $495.8\ \mathrm{kJ/mol}$ which corresponds to $5.139\ \mathrm{eV/atom}$), the electron affinity doesn’t (F: $-3.399\ \mathrm{eV/atom}$), because of electron-electron repulsion in the fluorine p subshell. That’s why an isolated $\ce{Na+ - F-}$ compound cannot exist: the additional energy released when forming the lattice is needed to form a stable entity.

Having said this, we can move on. You may ask: Yes, but why are the F electrons more stable than the Na ones, when Na has more protons? The reason is that the outer Na electrons are in the third shell, and thus they have higher energies. In a hydrogen-like atom (an atom with only one electron, which is clearly an oversimplification, since it doesn’t involve screening) the orbitals have the following energies:

$$E \approx -13.6 \left(\frac{Z}{n}\right)^2 \mathrm{~eV}$$

You can see that raising the $Z$ (atomic number) value from 9 (F) to 11 (Na) isn’t enough for lowering the energy, since $n$ (the principal quantum number, i.e. the shell) changes from 2 to 3.

So, this is why sodium forms positive ions, fluorine forms negative ions, and neon doesn’t form any ions at all (chemically). Nota bene: as I said, the formula above doesn’t take screening into account – when going down a group, it (usually: see d-block contraction) prevails over the growing nucleus. That’s why Ar is less inert than Ne, even though the formula above would suggest the contrary.

But what about covalent compounds? According to VB theory, bonding occurs through orbital overlap. Lower energy orbitals mean smaller overlaps, which is inconvenient if you want a stable bond: Carbon forms bonds while neon doesn’t because its orbitals are less stable (see table) and can overlap with other orbitals. Its orbitals are less stable because neon has more protons with the same principal quantum number – see the equation above to understand the concept.

In short: noble gases, especially those positioned higher in the periodic table, don’t bond because their outer orbitals have the lowest energies. Ultimately, this is the result of having the largest number of protons possible for a given outer shell in the ground state.

If you consider Hydrogen bonding to itself, the bond is stabilized both by the electric attraction between proton and electron. As well the 1s orbital of both atoms merge into one bonding orbital, this single orbial is more energetically stable because electrons like to be paired off (spin 1/2 and -1/2) in orbitals.

The electron configurations of the noble gas feature valence shells with completely filled orbital pairs. According to the Pauli Exclusion Principle (PEP), no two electrons may inhabit the same quantum state in the same system. If all available electronic states for the atom are filled then there is no stabilizing phenomena for either the loss or gain of a new electron, and the formation of bonds is not favored energetically.

An analogy commonly used for the PEP is that there is an additional repulsive "force" between elections beyond the electric repulsion, it is precisely this "force" that prevents the noble gasses from bonding. Though it's not so much a force as it's simply the inherent fermionic behavior of the electrons at play. However, exotic noble gas compounds do exist, but are not terribly stable.


Why is covalent bonding stronger than electrostatic attraction? I guess if we look at this problem from a "conjugate" perspective then the conjugate of a covalent bond is two elements with electrons lying around. On the other hand, with two electrostatically attracted things, the lack of a bond leaves two separate, charged particles.

But that still doesn't answer the question; why are electrons less stable than ions? I guess in the case of an ion though there are still electrons but also a nucleus the stabilize the electrons.

Ionic bonds can easily be as strong as covalent ones.

First off, let's be clear that almost everything has some ionic and some covalent character. Moreover, it's not an either-or situation. Some bonds have both strong ionic and strong covalent character.

For now, I'll stick to simple cases.

Let's consider $\ce{LiF}$, which seems a prototypical ionic bond. The interatomic distance is $\pu{156 pm}$ according to NIST. That gives a potential energy of $\pu{903 kJ/mol}$. Now, that assumes a full electron is transferred from $\ce{Li+}$ to $\ce{F-}$, which isn't the case (more like $0.8 - 0.9$). This simple calculation also ignores Born repulsion which is similar to the short-range Van der Waals repulsion between atoms.

A more complete analysis J. Chem. Educ. 2000, 77 (8), p 1076 gives a much better agreement. The real bond dissociation energy is $\pu{577 kJ/mol}$ (Source, PDF via the Internet Archive). A simple spherical ion model from the paper gives the bond enthalpy as $\pu{608 kJ/mol}$.

  calculations based on a simple ionic model reproduce the bond energies of $\ce{LiF(g)}$ and $\pu{NaF(g)}$ to within $5\%$.

If we look up covalent bond dissociation energies, the triple bond in dinitrogen is $\pu{945 kJ/mol}$. Ok, that's perhaps stronger, but $\ce{LiF}$ only has a "single" charge and clearly some ionic compounds have greater charge transfer.

More "prototypical" covalent bonds include $\ce{C-H}$ ($\pu{410 kJ/mol}$) and $\ce{O=O}$ ($\pu{497 kJ/mol}$), both less than $\ce{LiF}$.

Now, as I said earlier, it's not the case of a bond being only covalent or only ionic.

To quote from R. Gillespie J. Chem. Educ. 2001, 78 (12), 1688:

  The large atomic charges in $\ce{BF3}$ and $\ce{SiF4}$ are important to understanding the great strength of these bonds—which are the two strongest known single bonds, with bond dissociation enthalpies of $613$ and $\pu{565 kJ/mol}$, respectively. In comparison, the $\ce{C–C}$ bond in ethane has a bond dissociation enthalpy of only $\pu{345 kJ/mol}$. The great strength of the $\ce{BF}$ and $\ce{SiF}$ bonds arises from both the relatively large amount of shared density and the large atomic charges. So it is not strictly correct to say that these bonds are “predominately ionic”; if we are forced to use these terms we have to say that they are both strongly covalent and strongly ionic.

Finally, you can of course treat both covalent and ionic bonds using MO theory. For example, an approximate picture of $\ce{LiF}$ under MO theory might be something like this:

We readily see $\ce{LiF}$ has very high ionic character. For all intents and purposes, the "bonding" $\sigma$ orbital is localized on the $\ce{F}$ atom, there's very little energy difference between the $\pi$ and $\sigma$ here.

It really depends on which covalent bonds and ionic bonds we are comparing.

Generally, the strength of a bond, ionic or covalent, depends on two things:

The atomic radius of the elements which take part into the bond
the difference in electronegativity between those 2 elements


Halogenation of 4‐methylcyclohex‐1‐ene My problem is with the methyl group. If it wasn't there, I had no problem.
When the halonium bridge is made does the methyl group shift to give a more stable transition state or not?

In order for a methyl shift to occur, there would need to be an empty orbital resulting from a carbocation intermediate for the methyl group to occupy.

Note that in the bromonium intermediate shown above, neither carbon in the three-membered ring has an entire empty orbital available. Hence methyl shift doesn't occur and the mechanism continues as expected, resulting in halohydrin formation.

methyl groups will not shift in a non-classical carbocation mechanism. This reaction will first have a $TS$ where the $\ce{Br2}$ molecule makes a 3 membered ring with the $\pi%$ bond and the water molecule  will attach itself to $\ce{C3}$, as that is a more stable electron-deficient site due to the inductive effect of the methyl group.


Can I make a mirror on PET plastic using Tollen's reaction? I want to make a very shiny surface, like a mirror. I tried to use chrome paint but it's barely shiny. It's more like gray even though I bought one of the most expensive chrome looking paint.

I know mirrors can be produced using Tollen's reagent on glass (seen some demonstrations on YouTube).

But glass is quite expensive for what I want - I would like to make the mirror on plastic - more specifically on PET - the same material that a normal Coca Cola bottle is made from. I have big sheets of this PET material and I intend to transform them into mirrors.

Will silver deposit on the PET as it does on glass?

I need to order the ingredients online, then wait for it a couple of days - also because they are quite large quantities, it will end up being quite expensive. So, before making a mistake, I thought to ask here.

Could I use a Tollen's reaction to make a mirror on plastic?

I found a video on YouTube making a mirror on PET bottles.

Just that it consumes quite a lot of substance, and is not using Tollens reagent. My question is specifically about Tollens reagent.

What she uses is some "secret mixture" called RA-205.

It just proves that Ag can precipitate on PET bottles, which further supports the fact that will work with Tollens reagent. I'm will edit this answer once I execute my experiment, so this finding may only be somewhat helpful.


Why does chloromethane have a larger dipole moment than chloroform? Why does $\ce{CH3Cl}$, methyl chloride, have a larger dipole moment than $\ce{CHCl3}$, chloroform?

Let us consider $\ce{C-Cl}$ bond moment to be $x$ and that of $\ce{C-H}$ bond to be $y$.

In $\ce{CCl4}$, the dipole moment of any three $\ce{C-Cl}$ atoms is balanced by the dipole moment of the fourth $\ce{C-Cl}$ bond dipole moment. This means the resultant of dipole moment of three $\ce{C-Cl}$ bonds in tetrahedral structure is equal to $x$, which is equal and opposite to the dipole moment $x$ of the fourth $\ce{C-Cl}$ bond.

Now, when we replace one of the chlorines with a hydrogen, the net dipole moment equals $x+y$, (as they are in the same directions). Similarly if we carry out the same procedure for $\ce{CH3Cl}$ it can be proved that its dipole moment is also $x+y$.

Where have I gone wrong?

Your conclusion is built on the assumption that the bond lengths will be constant. A bond’s dipole moment varies with bond length and thus for the vector addition to give the identical product, all $\ce{C-Cl}$ and all $\ce{C-H}$ bonds would have to be identical. They’re probably not.

Sutton and Brockway studied the bond lengths in chloromethane, dichloromethane and chloroform back in 1935 by electron diffraction.[1] Their results indicate that the $\ce{C-Cl}$ bond lengths don’t vary between the compounds in question as indicated in the table below. (Note that all chlorine atoms in each molecule are homotopic and thus equal bond lengths are theoretically expected.)

$$\begin{array}{lccc}\hline
\text{Compound} & \ce{CH3Cl} & \ce{CH2Cl2} & \ce{CHCl3}\\ \hline
d(\ce{C-Cl})/\mathrm{pm} & 177 & 177\phantom{^\circ} & 178\phantom{^\circ} \\
\angle(\ce{Cl-C-Cl}) & \text{n. a.} & 111^\circ & 111^\circ \\ \hline\end{array}$$

These results point towards the chlorine side of things not having an impact. But there is still the hydrogen side of things. Hydrogen atoms are generally very hard to locate in diffraction type experiments. X ray diffraction is observed if photons interact with electrons in a certain manner; hydrogen atoms do not have many electrons around them and are thus hardly seen. One needs to observe diffraction at very large angles to be able to locate hydrogens. I assume, although I don’t know, that it’s similar in electron diffraction. Indeed, the paper notes:

  […] and $l_{\ce{C-H}}$ was taken as $1.06~\mathrm{\overset{\circ}{A}}$.

The two carbons in chloromethane and chloroform are, however, very much not identical. One is surrounded by three electronegative chlorine atoms which will strongly remove electron density from it, polarising the $\ce{C-Cl}$ bonds towards each of the three chlorines. The other will have only one such withdrawing partner. The different electron densities around chlorine are well represented by the corresponding $\ce{^13C}$ NMR shifts: $\delta(\ce{CH3Cl}) \approx 22~\mathrm{ppm}$; $\delta(\ce{CHCl3}) \approx 77~\mathrm{ppm}$.[2] The hydrogen shifts are also different in similar magnitudes, showing that hydrogen in chloroform is much more polarised than in chloromethane.

These different electronics must have an influence on the $\ce{C-H}$ bond length of the two molecules. I can’t imagine two electronically such different systems to have equal bond lengths. My assumption is that $\ce{CH3Cl}$ should have the shorter $\ce{C-H}$ bond lengths, since carbon is more negatively charged while hydrogen remains positively charged; this should give a slightly greater attractive force and hydrogen, being such a small element, immediately feels these. Therefore, instead of the final dipole vector $\vec d$ being equal in both cases ($\vec d = \vec x + \vec y$), we have one lesser and one stronger vector.

$$\begin{gather}\vec d_{\ce{CH3Cl}} = \vec x + \vec y + \delta\vec y\tag{1}\\
\vec d_{\ce{CHCl3}} = \vec x + \vec y - \delta\vec y\tag{2}\end{gather}$$

References:

[1]: L. E. Sutton, L. O. Brockway, J. Am. Chem. Soc. 1935, 57, 473. DOI: 10.1021/ja01306a026.

[2]: Experimental spectra found for chloromethane/chloroform on SciFinder.

It is a vector, parallel to the bond axis.This vector can be physically interpreted as the movement undergone by electrons when the two atoms are placed a distance d apart and allowed to interact, the electrons will move from their free state positions to be localised more around the more electronegative atom.

you can compare the dipole moment of both the molecules by these digrams. source:wikipedia.
CHCl3

CH3Cl

For polyatomic molecules there is more than one bond, and the total molecular dipole moment may be approximated as the vector sum of individual bond dipole moments. Often bond dipoles are obtained by the reverse process: a known total dipole of a molecule can be decomposed into bond dipoles. This is done to transfer bond dipole moments to molecules that have the same bonds, but for which the total dipole moment is not yet known. The vector sum of the transferred bond dipoles gives an estimate for the total (unknown) dipole of the molecule


Why does HCO₂⁻ exhibit delocalized bonding while CO₂ does not? $\ce{HCO_2^-}$ exhibits delocalized bonding and $\ce{CO2}$ does not?

What I am confused about is, that both of them have resonance structures. Therefore, why doesn't $\ce{CO2}$ exhibit delocalized bonding?

Here is an orbital diagram for $\ce{CO2}$.

Note how the red pi orbitals on the terminal oxygens are rotated 90 degrees with respect to one another; they are said to be "orthogonal" to one another.  This means that the p-orbitals on the terminal oxygens can only interact with the p-orbitals with which they are aligned on the central carbon; there is no way for an electron on one oxygen to interact with the orbitals on the other oxygen.

Here is the analogous drawing for the formate anion ($\ce{HCO2^-}$).  Note how the p-orbitals on the terminal oxygens are now both aligned parallel with the p-orbital on the central carbon atom so that continuous overlap (delocalization) is now possible.

For completeness, we can restate the drawings from above in terms of resonance structures.  In the top row of the following picture I've drawn the possible resonance forms for $\ce{CO2}$.  The only charge-separated structures we can draw involve charge separation on each of the two orthogonal $\ce{C=O}$ groups.

In the bottom row of the picture I've drawn 4 resonance structures for the carboxylate anion (when R=H this is the formate structure in the title of your question).  Structure II corresponds to the charge separated structure I drew above for $\ce{CO2}$.  But notice that I have drawn 2 new resonance (III and IV) where the negative charge is delocalized over the O-C-O 3 atom array.  This is because the p-orbitals on the terminal oxygens are now parallel and can overlap with the p-orbital on the central carbon.  Just like the "particle in a box" problem you may have studied in physics, it turns out the longer the path the electron has to roam over, the more stable the electron.  So resonance structures III and IV, where the electron can roam (delocalize) over 3 atoms are very important contributors to the description and stability of the carboxylate anion.  Carbon dioxide has no comparable resonance structures due to the orthogonal alignment (90 degree rotation) of the orbitals on the terminal oxygens

$\ce{CO_2}$ does not show delocalization because the p-orbitals of the carbon atom are orthogonal. They can't interact. The hydroxymethylidene ion shows delocalization because the lone pair on the carbon atom is "upgraded" to a full p-orbital, just as (one) the lone pair on each oxygen because that would stabilize the negative charge. Now you have a system of three p-orbitals linked together. There is no need for $\ce{CO_2}$ to stabilize anything, so the energy required to "upgrade" to a full p orbital (on the oxygens) is too high of a barrier.

Please correct me if I'm wrong (then I'll remove the answer). This is a very simple analysis, there could be things in MO theory that I'm not aware about.


Why are there multiple lines in the hydrogen line spectrum? Hydrogen has only one electron, yet it exhibits multiple lines in a spectral series, why is this?

As the electrons fall from higher levels to lower levels, they release photons.  Different "falls" create different colors of light.  A larger transition releases higher energy (short wavelength) light, while smaller transitions release lower energies (longer wavelength).

The visible wavelengths are caused a by single electron making the different transitions shown below.  There are even more transitions that release invisible wavelengths.

 Wavelength   Transition  Color
 (nm)
---------------------------------------------
 383.5384      9 -> 2    Ultra Violet
 388.9049      8 -> 2    Ultra Violet
 397.0072      7 -> 2    Ultra Violet
 410.174       6 -> 2    Violet
 434.047       5 -> 2    Violet
 486.133       4 -> 2    Bluegreen (cyan)
 656.272       3 -> 2    Red
 656.2852      3 -> 2    Red

*Values taken from Hyperphysics: Hydrogen Energies and Spectrum

(Why are there two different 3-> 2 transitions?  See here: Hydrogen Fine Structure )

This arises due to the electron making transitions between the various energy levels in an atom . The electrons fall from a higher energy state to a lower energy state or vice versa emitting or absorbing wavelengths of different magnitude which fall directly on the spectrum of hydrogen thereby releasing colours of various wavelengths and creating a continuous spectrum .


How to convert from w/w% to molarity? After reading a number of explanations about it after googling, I'm a bit confused on this. Some explanations are doing it using a solution density, others aren't, though most are. On one site it's showing different densities for a solution of a compound at different w/w% values. I'm now very lost as to how you convert from a w/w% to molarity. Can someone clarify?

Weight percent is essentially the same as (I arbitrarily picked grams as the mass unit):

$$
\%~\mathrm{w/w~X} = \mathrm{g~X\over g~solution}
$$

In order to get to the $\mathrm{mol\over L}$ units of molarity, you have to convert by multiplying by the solution density, $\rho$, and dividing by the molecular weight of $\mathrm X$, $M_\mathrm{X}$:

$$
\mathrm{g~X\over g~solution} \times \stackrel{\left(1\over M_\mathrm X\right)}{\mathrm{mol~X\over g~X}}
 \times \stackrel{\left(\rho\right)}{\mathrm{g~solution\over L~solution}}
 = \mathrm{mol~X\over L~solution}
$$

One key thing to remember is that $\rho$ is the density of the solution, not of the pure solvent. If you're working with dilute solutions (below $\sim\!100~\mathrm{mM}$ for aqueous) these two densities will be similar, but in concentrated solutions they will diverge.

W/W% is just the same as a any percentage but now the difference is that let say that a concentrated solution contains  67% w/w ammonia that means that there is 67g of ammonia in 100g of the solution;
now let us consider a question

  A concentrated solution of aqueous ammonia is $28.0\%\text{w/w }\ce{NH3}$ and has a density of $\pu{0.899 g/mL}$. What is the amount concentration of $\ce{NH3}$ in this solution?

Converting $28.0\%\text{w/w}$ to ratio form gives us $28/100$ then you multiply it with the density $\pu{0.899 g/ml}$ for ammonia and $\pu{17.04 g}$ is an equivalent of one mole.

$$28/100 \times \frac{\pu{0.899 g}}{\pu{1 ml}} \times \frac{\pu{1 mol}}{\pu{17.04 g}} \times \frac{\pu{1000 ml}}{\pu{1ml}}$$


What are the isomers of a complex having a structure of M(AB)2C2? Our instructor have us answer an isomerism problem of a crystal structure that has a form of $\ce{M(AB)2C2}$. I keep getting 9 isomers, which includes the optically inactive isomer, and my peers and I have different answers.

Referring to the replies of Mr. Ivan Neretin, complexes that has a form of M(AB)2C2 posses 8 stereoisomers, 2 of which are optically inactive. The other 6 isomers can be interchanged to their respective enantiomers.


How can I derive the preferred oxidation states of a transition metal from its electronic configuration? For example $\ce{Fe}\ (Z=26)$ Short formula: $\mathrm{1s^2\ 2s^2p^6\ 3s^2p^6d^6\ 4s^2}$. If we make the electronic formula we get 4 single electrons ($s=1/2$), which means that the valence in normal state is 4, but in the periodic table it is $2/3/6$. How can I find its valence?

Firstly you need to appreciate that the 4s orbital is at higher energy than the 3d orbitals (despite the filling order suggesting the oposite). This means that the two 4s electrons are lost first giving a valency of two (which is common to all the first row transition elements). The next point is that one of the 3d orbitals is fully ocupied so one of these two electrons will be preferentialy released due to the repulsion of its pair. This results in a valency of three. These are by far the most common valencies of iron. In fact compounds of 4, 5, and 6 oxidation states (valencies) are known but are very unstable, so of little importance.

The electron configuration of iron is 1s^2 2s^2 2p^6 3s^2 3p^6 3d^6 4s^2. Here, the 4s shell is the outermost shell. So, iron has 2 valence electrons. While reacting, these 2 electrons can go to any other electronegative element who requires these electrons to complete its outermost shell. So, this gives iron a valency of 2. Now, the one out of 2 orbitals is completely filled. So, due to repulsion this electron can also be given out. This gives iron a valency of either 2 or 3. However, iron is mostly found in valency 3 as it is mostly stable.


What is the best way to make a substantial amount of soot in a high school laboratory? I need to make about 50 grams of soot (carbon black particles) in a high school laboratory. What would be the best way to do this?

The purpose of needing 50 grams of soot is because I would like to carry out an experiment that investigates how the percentage soot in oil (e.g. diesel) affects the viscosity of the mixture. To measure the viscosity, I would have to put 500 ml of oil in a measuring cylinder and do the ball-bearing method. The highest percentage I would like to test is 10% soot and that calls for around 50 g of it.

There is an extensive literature on producing carbon black see for instance Donnet, Jean-Baptiste, ed. Carbon black: science and technology. CRC Press, 1993.

Instead of adapting the Roman recipe for producing lamp black explained on page 19, I would use olive oil rather than resin or turpentine (that is quite a nasty substance). For doing it on a small scale maximizing the yield maybe I would try (under the fume hood)  to heat the oil in a flask attached to a water-cooled Liebing condenser after reaching the smoke point the soot should start to accumulate on the walls of the condenser. It should be quite easy to recollect the carbon black from the wall of the condenser with nylon bristle and maybe from a little funnel with a slight offset fixed on the exit of the condenser. All this should be carried out using heat-resisting glassware in a lab under the supervision of an adult.

P.S.

  The highest percentage I would like to test is 10% soot and that calls
  for around 50 g of it.

This means that you will need more than 50 g for the whole experiment so if you test 10 different proportion you will need at least 80 g performing the experiment only once. It could require a lot of oil to produce 80 grams.

Try burning turpentine.
It produces a very soot rich flame, per my recollection.
Another path per the article: Preparation of Carbon Nanoparticles from Candle Soot.To quote:

Carbon nanoparticles (CNPs) were prepared by simply mixing candle soot into ethanolic Nafion solution under ultrasonic radiation for 90 min. Well-defined CNPs could be obtained after centrifugation and washing.

[EDIT] My colleague mentions other choices for a school lab, citing benzene as a non-choice. The reason, let's hear from what the CDC says about it, to quote:

Immediate signs and symptoms of exposure to benzene.
People who breathe in high levels of benzene may develop the following signs and symptoms within minutes to several hours:
Drowsiness
Dizziness
Rapid or irregular heartbeat
Headaches
Tremors
Confusion
Unconsciousness
Death (at very high levels)

Another option suggested is toluene, comments from Toluene Safety Tips from MSDS experts – MSDSonline, to quote:

Toluene is a highly flammable liquid and it can cause mild damage to the skin and the eyes. However, the most-common hazard associated with this chemical is inhalation. Products containing toluene can produce dangerous fumes which can cause nausea, headaches, unconsciousness, and even death if inhaled.

Note, my final choice, for high school labs in general, is safe ordinary candle soot with a reference citing a preparation path to nanoparticles. Now, I could make other more potentially efficient and likely problematic suggestions, but I sleep well with this conservative/responsible choice.


Will supercritical nitrogen stay supercritical if we raise temperature and pressure high above critical values? Nitrogen's critical temperature is -146.96 °C and critical pressure 33.96 bar. If we maintain nitrogen at 220 bar and 400 °C, what will be the state of $\ce{N2}$? Will it be still supercritical? I know that phase diagram says after critical temperature and pressure we have a supercritical phase, but with enough p and T, more exotic states of matter occur.

If we maintain nitrogen at 220 bar and 400 °C, what will be the state of $\ce{N2}$? Will it be still supercritical?

Yes, it will still be supercritical.  The only reasonable definition of "supercritical" is something along the lines of "a fluid phase without surface tension wherein the fluid pressure and temperature exceed the critical point".

Thus, compressed air is a supercritical fluid.  The same is true of compressed nitrogen.

You may be wondering how "supercritical" air is different from regular, non-supercritical, gaseous air.  The answer is, it isn't meaningfully different.

If you research "supercritical fluids", it will be very hard to discover this fact.  That is because the "exotic" properties of "supercritical fluids", such as having densities "in between that of a gas and a liquid", are really only properties of supercritical fluids that are near (i.e. slightly above) the critical point.

It isn't really much of an exaggeration to say that supercritical fluids are really gases.  They have no surface tension.  They expand to fill their container.  Just like gases.  Near the critical point, supercritical gases can have densities that are very unusual for a gas. At these high densities, the gas is very non-ideal, and intermolecular interactions in the gas lead to all sorts of interesting behavior.  Those high-density intermolecular interactions give rise to the interesting properties that people refer to when they say "supercritical fluid".  The strength and nature of those interesting intermolecular interactions depends strongly on density, which is why people refer to "supercritical fluids" having "tunable" properties.  Small changes in density (by changing temp. or pressure) can give rise to large changes in those interactions.

At conditions way past the critical point, those intermolecular interactions don't matter very much any more, and the "supercritical fluid" again starts to behave more and more like an ideal gas.

  I know that phase diagram says after critical temperature and pressure we have a supercritical phase, but with enough p and T, more exotic states of matter occur.

Yes, other answers and comments have done a good job of discussing this point.  If you increase the pressure enough, eventually you'll have a black hole!  Of course, long before that happens, you can get other less exotic phases of matter such as various solid phases.  In general, predicting the high-pressure phase behavior of even small molecules like nitrogen is very difficult, and as such, experimental data is required.  Unfortunately, measuring the high-pressure phase behavior of materials is also very difficult, at least past a few hundred gigapascals.  So I'm not sure where the high-pressure "exotic" phases of e.g. solid nitrogen start to form.  But it's not at 220 bar and 400 ° C, under those conditions nitrogen will just be a gas, albeit a "supercritical" gas.

Surpassing the supercritical point -- and the parameters provided by you exceed both the critical pressure and the critical temperature you provided, too -- will still yield the supercritical state; i.e. the absence of a gaseous phase in presence of a liquid phase.  It wont be a plasma, but still a supercritical fluid, as $\ce{scCO2}$ already used for decaffeination and chromatography, for example.


Elimination reactions of vicinal dibromide When a vicinal dibromide undergoes elimination reactions in vigorous basic conditions, say in ammonia with $\ce {NaNH2}$, is the product an alkyne or is it a conjugated diene? I would think that the answer depends on thermodynamic and kinetic factors. Considering bond energies of the $\ce {C-C}$ $\pi$ bonds, perhaps the conjugated diene would be more favourably formed?

After the first elimination, you have a vinylic bromide, so there are two options for the second elimination. The two acidic hydrogens that the base can abstract are one attached to sp2 carbon and one to sp3. Since sp2 hybridized carbon is more electronegative, the base abstracts the hydrogen attached to it, and then bromine leaves being a good leaving group, which gives alkyne as the major product.


What causes polarization in noble gases? According to my knowledge, polarization occurs when in a compound there is a difference of polarity its constituents.

But in the case of noble gases, there is a single component. What exactly is polarization of noble gases and how does it occur?

Although this applies to all noble gases, consider helium. There are two electrons moving around the nucleus. The movement of those electrons is symmetrical and during a defined amount of time, the atom is not polarised. However, the momentary positions of those electrons can polarise the atom. This can happen when both electrons are relatively close, on the same one side of the atom. In this case a momentary dipole is formed, since there is separation of two charges, nucleus being positive and the pair of electrons being the negative pole.


What is the net ionic equation of sodium hydroxide when it dissolves in water? What is the net ionic equation of sodium hydroxide when it dissolves in water?

For the net ionic equation I got
$$\ce{NaOH(s) + H2O(l) -> NaH+(aq) + OH- (aq) + H2O(l)}$$
but it was wrong.
Then I tried:
$$\ce{NaOH(aq) -> Na+(aq) + OH- (aq)}$$
and it was wrong as well.
Can someone please explain?

When you write down the net ionic equation, you have to consider that the educt and the products are in different phases, a solid and a liquid phase. Undissolved $\ce{NaOH}$ is a solid, denoted by the index $\ce{(s)}$, while the dissolved $\ce{Na+}$ and $\ce{OH-}$ ions are solvated in the aqueous phase (index $\ce{(aq)}$ in the equation):

$$\ce{NaOH(s) ~->~ Na+(aq) + OH^{-}(aq)}$$

You don't need the H ion with the Na ion. You were correct the second time. and if there is two of the same product or reactant of on either side of the equation it is eliminated because it is just a spectator.


How does the silver nucleate on glass in Tollen's reagent mirroring? I want to know the exact mechanism for how silver would start to grow during a mirroring with Tollen's reagent. I've made two pictures to illustrate how I see the nucleation/growth process of silver on silica:

As you can see, I didn't draw any chemical bonds, because I think that the silver adherence is due to Van Der Waals forces. Am I correct? If that is the case, then I believe a place with high permanent polarisation would be better suited for the nucleation of silver.
Now, let's follow the logic. We have a greater electronegativity difference between $\ce{O}$ and $\ce{Si}$ ($3.44 - 1.90 = 1.54$) than between $\ce{O}$ and $\ce{H}$ ($3.44 - 2.2 = 1.24$).
For the upper oxygen, assuming an angle of 109 degrees, you get a total dipole moment of: $2(1.54)\cos\left(\frac{109}{2}\right) = 1.79$
For the lower oxygen with the bonded hydrogen, assuming an angle of 120 degrees , you would get a total dipole moment of: $1.54 + 1.24 = 2.78$
(please tell me if I'm wrong to assume a straight line!)
This would mean that you have a greater accumulation of electrons at the bottom oxygen, which would favor Van Der Waals attraction (Debye interaction, I guess, since you have a permanent dipole and an induced dipole). This, in turn, would mean that it would be a better nucleation site.
If you make a surface treatment using silane groups, I believe you promote the presence of $\ce{OH}$ groups at your silica surface, though I'm unsure about this. I do know that it makes better silver mirroring though. I've made another drawing:

Following my logic, this one is wrong in many way. First of all, I drew angles on the $\ce{Si-O-H}$ bonds, which I didn't draw on my first pictures. Secondly, the silver is not even near the oxygen atoms, which I believe are the nucleation points.
I'm confused, could you please point out to me where I'm right or wrong in my logic/knowledge?
Mainly:

Is there a bond forming between the silica surface and the silver? Or is there only Van Der Waals (Debye) attraction?
Is my depiction of nucleation & growth correct (for the first two pictures)?
Is my logic regarding the electronegativity correct?
Is my logic of " higher total electronegativity difference, hence better nucleation site" correct? (assuming I'm right about the Van Der Waals Forces)
Is my depiction of silanisation correct?
Am I right to say that the higher abundance of surface $\ce{OH}$ groups by silanisation implies better nucleation?
Could I do the same when using a strong acid on my glass surface?

"If you make a surface treatment using silane groups, I think that you promote the presence of OH groups at your silica surface". This is NOT correct. You siliconize glass to cap the free SiOH groups

2 silica_Si-OH + n SiCl2Me2 + HO-Si_silica -> silica_Si-[O(SiMe2)]$_n$-O-Si_silica. As a result you decrease the number of active OH groups and make it highly non-polar.

Before siliconization silica is surrounded by a hydration sphere. Ag$^+$ can bind it well, but we need to precipitate neutral Ag. In case of non-siliconized glass we have two effects:

(1)  It is easier for silver to grow on the surface that already has a silver nano-crystal
(2) Before treatment glass have inhomogeneities, so nucliation in some places is heavily favored.

Both effects are negated by siliconization.

Note that I didn't account for surface charges which change as a result of siliconization and might play a role.


What exactly happens while washing clothes I am reading about soaps , micelles and all that. I understand how soap behaves in water contaminated by oil. Soap helps in emulsifying the oil into water.

Let us suppose now we have dirt(i.e oil, grease etc) not alone in water but present on a some cloth. Why can't we put the cloth just in water to get it clean ? What precisely is the need for agitation in a washing machine or by any other method ?

I searched and found that agitation produces foam but according to this post foam does not alter the washing process anyhow.

When you wash a piece of cloth, even without soap, it's better to agitate. The vibration, friction, etc. will help speed up the process of mixing the dirt/grease with water.

Even the very definition of "agitation" uses the example of washing. From Vocabulary.com

  Agitation is the act of stirring things up, like the agitation of a washing machine that moves the water, detergent, and clothes around and gets the dirt out

And as you rightly pointed out from another question, "the froth has little or no effect on the detergent action".

Soap can mix with both water and with oil. Why? The soap molecule has
two different ends, one that is hydrophilic (polar head) that binds
with water and the other that is hydrophobic (non-polar hydrocarbon
tail) that binds with grease and oil.

When greasy dirt or oil is mixed with soapy water, the soap molecules
arrange themselves into tiny clusters called micelles.

The water-loving (hydrophilic) part of the soap molecules sticks to
the water and points outwards, forming the outer surface of the
micelle.

The oil-loving (hydrophobic) parts stick to the oil and trap oil in
the center where it can't come into contact with the water.
With the oil tucked safely in the center, the micelle is soluble in
water. As the soapy water is rinsed away the greasy dirt goes along
with it. Micelle Picture from Wikipedia
So basically, the soap is attracted to the fat/oil/grease because of
its fat-loving side but then tears up the grease by pulling it into
the water using its water-loving side. Sounds kind of like a football
play--surround the oil particles and move them away from one another.

Ever wonder why it is easier to clean dirty, greasy hands (and other
things) in hot or warm water rather than cold water? It is because the
fats and oils soften or melt in hot water, which allows them to attach
more readily to the hydrophobic end of the soap molecule. In turn,
that makes it easier to rinse away.

Credits: entire answer is copied from https://www.chagrinvalleysoapandsalve.com/blog/posts/how-does-soap-work/  Without any edits


Understanding the solubility of Ca(HCO3)2 According to wikipedia $\ce{Ca(HCO3)2}$ has the following solubility values:

$\pu{16.1 g/100 mL} \,(\pu{0^\circ C)}$
$\pu{16.6 g/100 mL} \,(\pu{20^\circ C)}$
$\pu{18.4 g/100 mL} \,(\pu{100^\circ C)}$

So I assume $\ce{Ca(HCO3)2}$ would precipitate beyond this level. However,  the wikipedia page for calcium bicarbonate states that attempts to prepare compounds such as solid calcium bicarbonate by evaporating its solution to dryness invariably yield instead the solid $\ce{CaCO3}$:

$\ce{Ca(HCO3)2(aq) → CO2(g) + H2O(l) + CaCO3(s)}$

which leaves me wondering if, when the solution is saturated beyond ~$\pu{16-18 g/100 mL}$, a precipitate of $\ce{Ca(HCO3)2}$ will exist  in solution? or if it also dissociating beyond that point while still in solution?

Calcium bicarbonate $\ce{Ca(HCO3)2}$ simply does not exist. It is impossible to fill up any container with pure $\ce{Ca(HCO3)2}$. Nobody has ever been able to produce a powder containing $\ce{Ca(HCO3)2}$. It is possible and easy to produce a solution containing the ions $\ce{Ca^{2+}}$ and the ions $\ce{HCO3^-}$, by bubbling $\ce{CO2}$ into a solution of calcium hydroxyde $\ce{Ca(OH)2}$. The following reaction will happen:

$$\ce{Ca(OH)2 + 2CO2 -> Ca^{2+} + 2HCO3^-}$$

The result is exactly what could be expected if some $\ce{Ca(HCO3)2}$ had been dissolved in water. And of course, it makes sense to admit that by evaporating the solution, the calcium bicarbonate could be obtained dry. Unfortunately it is not the case. When trying to evaporate this solution, $\ce{Ca(HCO3)2}$ is always and totally decomposed according to the following equation:

$$\ce{Ca^{2+} + 2HCO3^- -> CaCO3 + H2O + CO2}$$

And all you obtain by trying to evaporate a solution containing he ions $\ce{Ca^{2+}}$ and the ions $\ce{HCO3^-}$ is a deposit of calcium carbonate $\ce{CaCO3}$, whatever the method used for carrying out this evaporation.

Of course you can calculate the maximum amount of Calcium and of $\ce{HCO3^-}$ ions that may exist simultaneously in such a solution. You may even calculate the solubility of $\ce{Ca(HCO3)2}$. But this does not prove that $\ce{Ca(HCO3)2}$ exists as a pure compound. $\ce{Ca(HCO3)2}$ is a compound that only exist in solutions.

The CRC Handbook does not list solubility numbers for $\ce{Ca(HCO3)2}$ or $\ce{Mg(HCO3)2}$. We know that temporary hard water has Mg and/or Ca salts of carbonic acid that will precipitate in hot water heaters at ~$\pu{60 °C}$, which appears to be due to forced evolution of $\ce{CO2}$ rather than an inverse solubility curve.

The solubility of $\ce{CO2}$ ($\pu{0.058 g/100mL}$ at $\pu{60 °C}$) would appear to be insufficient to dissolve ~$\pu{10 g}$ of $\ce{CaCO3}$, so I question the solubility number of $\pu{18 g}$ at $\pu{100 °C}$.

The likely reason for being unable to isolate solid $\ce{Ca(HCO3)2}$ is that $\ce{HCO3-}$ ion is in a labile equilibrium with $\ce{CO2(aq)}$ and $\ce{OH-(aq)}$,

$$\ce{2HCO3- -> H2CO3 + CO3^2-}$$

At higher temperatures, the $\ce{CO2}$ simply escapes by evaporation rather than because of equilibrium processes. Stalactites form at cold temperatures because the $\ce{Ca(HCO3)2}$ solution is exposed to a large surface where $\ce{H2O}$ and $\ce{CO2}$ evaporate leaving $\ce{CaCO3}$ behind.


Can we break the covalent bonds of water molecules using resonance frequencies? Can we break the covalent bonds of water molecules $\ce{H2O}$ using resonance frequencies? What is the wavelegth of the water's molecule? Can we create a vibration like this and use it to separate hydrogen from oxygen?

Yes this is possible. Your question does not state the source of the frequency, but in this internet article they have used laser light.

I am not sure at what frequency you need to irradiate water to break the O-H bond, but I would speculate that a good frequency would be close to the IR stretching frequencies of the O-H bond in water.

The resonate frequency of water is 632 hz. The covalent bond can be easily broken using 12v of DC current at as little as 5amps. The Hydrogen and Oxygen can be separated or collected together in the form of Browns gas or HHO. I use it in my car.


What determines the speed of charging of a battery? I was told on physics.stackexchange that this question belongs to Chemistry, what do you think?

Recently we have seen the surge of superfast mobile phone charging. Just an example: http://www.bbc.com/news/technology-30708945

What underlying physical principle determines the speed of charging of a battery?

There are many factors limiting speed of charging. Here are just a few:

Run-away overheating -- some battery chemistries, such Li-ion, may even self-ignite.
Spalling of electrodes -- the physical stress of rapid charging can buckle electrodes and cause a composite electrode to break apart.
Side reactions -- to increase the charging current, higher voltage is required. This over-voltage causes undesired side reactions, such as lithium plating (and out-gassing in Ni-MH and lead-acid batteries).

To make cells charge more rapidly, each of these potential issues must be analyzed and both the cell and the charging system must be designed to overcome the problem. In some cases, there have been spectacular failures, even when millions have been spent on design.

As you may know, what determines speed of a chemical reaction is its kinetics.

I didn't read the BBC article, but charging a battery follows a simple reaction :

A + B(+) → A(+) + B

and the rate of this reaction is determined by:

R = K [A][B(+)]

as you can see it depends on K and concentrations.

the thing we can change to make the reaction speed really high is K.

K is determined by three things. first the nature of reaction and second the catalyst we use and we can change both and third the temperature.

so to make a super hell fast recharging battery you need to find the right material for reaction and right catalyst and right temperature.

I hope this was the answer you were looking for.


Synthesis of anhydrous sodium tungstate I am interested in preparing an anhydrous sample of sodium tungstate (Na2WO4), which is normally found as the dihydrate.  Finding an example of this in literature is proving difficult, I am interested if someone can suggest a preparation method based on prior experience or difficulties I should encounter.

Hoermann [1] prepared anhydrous sodium tungstate by melting the 1:1 mix of sodium carbonate and tungsten(VI) oxide (and, subsequently, growing single crystals); alternatively, he proposed prolonged drying of a dihydrate at 100 °C:

$\ce{Na2MoO3}$ und $\ce{Na2WO3}$ sind wasserfrei durch Zusammenschmelzen von 1 Mol $\ce{Na2CO3}$ mit 1 Mol $\ce{MoO3},$ bzw. $\ce{WO3},$ oder durch volliges Entwassern der Hydrate bei 100° zu erhalten. Die Salze zeigen ansgepragte Polymorphie.

Busey and Keller [2] obtained anhydrous sodium tungstate (reported water content 0.12%) by drying a dihydrate at 200 °C in vacuum. Purity has also been confirmed by Raman spectroscopy and powder x-ray diffraction:

The $\ce{Na2WO4}$ was prepared by heating the dihydrate at 200° in a vacuum.

References

Hoermann, F. Beitrag zur Kenntnis der Molybdate und Wolframate. Die binären Systeme: $\ce{Li3MoO4-MoO3},$ $\ce{Na2MoO4-MoO3},$ $\ce{K3MoO4-MoO3},$ $\ce{Li2WO4-WO3},$ $\ce{Na2WO4-WO3},$ $\ce{K2WO4-WO3},$ $\ce{Li2MoO4-Na2MoO4},$ $\ce{Li2WO4-Na2WO4},$ $\ce{Li2MoO4-K2MoO4}.$ Z. Anorg. Allg. Chem. 1929, 177 (1), 145–186. DOI: 10.1002/zaac.19291770117.
Busey, R. H.; Keller, O. L. Structure of the Aqueous Pertechnetate Ion by Raman and Infrared Spectroscopy. Raman and Infrared Spectra of Crystalline $\ce{KTcO4},$ $\ce{KReO4},$ $\ce{Na2MoO4},$ $\ce{Na2WO4},$ $\ce{Na2MoO4 · 2 H2O},$ and $\ce{Na2WO4 · 2 H2O}.$ The Journal of Chemical Physics 1964, 41 (1), 215–225. DOI: 10.1063/1.1725625.

I have actually, to my surprise and that of others, successfully dissolved the filament from an incandescent oven bulb (see before and after pictures here, click to expand on the final picture), which is composed of, per a reference, to quote:

Incandescent light bulbs consist of an air-tight glass enclosure (the envelope, or bulb) with a filament of tungsten wire inside the bulb, through which an electric current is passed.

in a galvanic cell setting. The piece of tungsten metal serves as the anode and carbon (as graphite in my experiment) can act as the cathode. In the current case, I would recommend an electrolyte of NaOH plus 6% NaOCl (which relates to the so-called Bleach battery cell).
Heat in a microwave replenishing the electrolyte as needed. When dissolution is evident and Sodium hydroxide exhausted, remove electrodes and evaporate down the solution to effect salt separation.
As to mechanics, the NaOCl is expected to result in the anodic oxidation of the tungsten metal (as generally reported in the noted Bleach Battery reference) as follows:
$\ce{W + 3 NaOCl -> WO3 + 3 NaCl}$
Then, per Wikipedia, to quote:

$\ce{Na2WO4 + 2 HCl → WO3 + 2 NaCl + H2O}$

$\ce{Na2WO4 + 2 HCl → WO3.H2O + 2 NaCl}$

This reaction can be reversed using aqueous sodium hydroxide.

Hence, the presence of added Sodium hydroxide acting on $\ce{WO3}$ is expected to result in Sodium tungstate.
Now, why a microwave approach? To quote a 2013 paper: Microwave-Specific Enhancement of the Carbon−Carbon Dioxide (Boudouard) Reaction:

This changes the position of the equilibrium so that the temperature at which CO becomes the major product drops from 643 °C in the conventional thermal reaction to 213 °C in the microwave. The observed reduction in the apparent enthalpy of the microwave driven reaction, compared to what is determined for the thermal reaction from standard heats of formation, can be thought of as arising from additional energy being put into the carbon by the microwaves, effectively increasing its apparent standard enthalpy. Mechanistically, it is hypothesized that the enhanced reactivity arises from the interaction of CO2 with the steady-state concentration of electron−hole pairs that are present at the surface of the carbon due to the space-charge mechanism, by which microwaves are known to heat carbon. Such a mechanism is unique to microwave-induced heating...

Also, interesting cited chemistry on the microwave action on carbon reputedly creating some hydroxyl radicals (${.OH}$). Reference: see per this 2007 paper, Generation of hydroxyl radical in aqueous solution by microwave energy using activated carbon as catalyst and its potential in removal of persistent organic substances. Note: this is not unexpected, as per the prior reference especially in alkaline conditions, the action of an electron-hole on ${OH-}$ should produce ${.OH}$.
Note: The presence of radicals is expected to accelerate the oxidation of the tungsten metal.
Bottom line, this microwave-assisted electrosynthesis is relatively safe, effective, and avoids high temperatures. It requires only generally available inexpensive reagents.
Note, one can use alcohol to source an anhydrous Sodium tungstate as per Wikipedia, to quote: "insoluble in alcohol", but any Sodium chloride presence is sparingly soluble in methanol (14.9 g/L).


Alkoxide as a leaving group In the reduction of esters to alcohols, the RO group is pushed off. However, the negative charge on the O atom is destabilized by the electron donating R group. Why is the RO group able to function as a leaving group given that the resultant alkoxide is unstable?

Link to reaction mechanism.

The intermediate before forming the alkoxide is a tetrahedral intermediate which is also unstable. You're going from a high energy species to another high energy species.


How do things glow in the dark? What makes something glow in the dark? The only explanation that I can come up with for what makes things glow in the dark is that there is probably a chemical reaction slowly releasing the light that was previously absorbed. Am I correct?

How does this phenomenon work?

The spontaneous emission of light by a substance not resulting from heat is luminescence (a type of cold-body radiation).  $\ce{^1}$

A few types of luminescence are:

Bioluminescence: Made by living creatures such as fireflies, glow-worms, and many marine creatures.
Chemoluminescence: made by a chemical reaction. Glow sticks work this way.
Electroluminescence: made by passing electricity through something like in a light emitting diode (LED).
Photoluminescence: made by shining light at "luminous" (phosphorescent) paints.
Röntgenoluminescence: made by shining X-rays at things. (The curious name comes from Wilhelm Röntgen (1845–1923), the discoverer of X-rays.)
Sonoluminescence: made by passing energetic sound waves through liquids.
Thermoluminescence: made when photons are emitted from hot materials.
Riboluminescence: made by rubbing, scratching, or physically deforming crystals.
Phosphorescence: photoluminescence as a result of triplet-singlet electronic relaxation (typical lifetime: milliseconds to hours)

A chemical absorbs and then re-emits a photon of light, where the energy of the photons matches the available energy states and allowed transitions of the chemical takes place. In phosphorescence, the electron that absorbs the photon undergoes intersystem crossing $\ce{^2}$  into a triplet state.

The excited in triplet state returns to the lower energy singlet state. Such transitions, are kinetically unfavored and progress at a slow rate. Most phosphorescent compounds are fast emitters with triplet lifetimes in milliseconds. If triplet lifetimes are minutes or even hours, very slow degradation occurs.$\ce{^3}$ $\ce{^4}$ In phosphorescent materials, there's a delay between them absorbing energy and giving out light.These substances are so-called "glow-in-the-dark" materials.

To make glow-in-the-dark toys manufacturers mix zinc sulfide or strontium aluminate with plastic giving in glow-in-the-dark toys.$\ce{^5}$ $\ce{^6}$

Fluorescence

Fluorescent materials produce light instantly, when the excited atoms return to normal, in as little as a hundred thousandth of a second.$\ce{^4}$ $\ce{^7}$

Chemoluminescence

In the case of chemiluminescence, two chemicals interact which giving energy as a byproduct in the form of light( heatless glow).

When glow sticks are bent, an inner glass tube is broken releasing hydrogen peroxide. Hydrogen peroxide reacts with diphenyl oxalate giving 1,2dioxetanedione, which is unstable. It decomposes to carbon dioxide releasing energy. The energy is absorbed by electrons in dye molecule, that falls down to ground state releasing energy in the form of light.$\ce{^8}$ $\ce{^9}$

Chemiluminescence can happen naturally as well. In such instances — like fireflies — it is bioluminescence.

Some creatures, such as fireflies and jellyfish, contain chemicals within them that cause them to glow. Some of these creatures glow for protection, camouflage or to attract mates.

Ex 1.Bobtail squid $\ce{^12}$ $\ce{^13}$

Ex.2 Mycena lampadis ( mushrooms )$\ce{^14}$

The principal chemical reaction in bioluminescence involves some light-emitting molecule and an enzyme, generally called the luciferin and the luciferase.$\ce{^10}$ $\ce{^11}$ $\ce{^15}$

Electrochemiluminescence $\ce{^16}$is a special form of chemiluminescence in which the light-emitting chemiluminescent reaction is preceded by an electrochemical reaction.$\ce{^17}$

References

1 https://chem.libretexts.org/Bookshelves/Analytical_Chemistry/Map%3A_Principles_of_Instrumental_Analysis_(Skoog_et_al.)/15%3A_Molecular_Luminescence_Spectrometry

2 https://www.shsu.edu/chm_tgc/chemilumdir/JABLONSKI.html

3 https://en.wikipedia.org/wiki/Intersystem_crossing

4 https://teaching.shu.ac.uk/hwb/chemistry/tutorials/molspec/lumin1.htm

5 http://www.phys.ubbcluj.ro/~dana.maniu/OS/BS_7.pdf

6 https://science.howstuffworks.com/innovation/science-questions/question388.htm

6 https://www.wonderopolis.org/wonder/how-do-things-glow-in-the-dark

[7] https://en.wikipedia.org/wiki/Fluorescence#Lifetime

[8] https://www.compoundchem.com/2014/10/14/glowsticks/

[9] https://en.wikipedia.org/wiki/Glow_stick#Fluorophores_used

[10] https://en.wikipedia.org/wiki/Bioluminescence#Chemical_mechanism

[11] https://www.nationalgeographic.org/encyclopedia/bioluminescence/

[12] https://www.nationalgeographic.org/encyclopedia/bioluminescence/

[13] https://en.wikipedia.org/wiki/File:Sepiola_atlantica.jpg

[14]https://www.timeforkids.com/g34/bioluminescence-2/

[15] http://photobiology.info/Branchini2.html

[16] https://www.sciencedirect.com/topics/chemistry/electrochemiluminescence

[17] https://www.sciencedirect.com/science/article/pii/S0008622317304943

Reactions emits energy in different forms. Could be heat, light or electrical. If the form emited is light energy form it surely will glow irrespective of the external environmental characteristics in this case dark!


Can ethanol be both a Lewis base and a Lewis acid? I'm asked to determine if ethanol is a Lewis acid, Lewis base, or both.

Since $\ce{O}$ has two lone pairs of electrons I would think that it could act as a base, since they could easily be donated. If it was a Lewis acid, it should have an empty orbital, and I can't find such one. All of the carbons are bonded and oxygen's orbitals are full. Still my book tells me it can act as both.

Is the textbook wrong or could one say that the 1s orbital to $\ce{H}$ (bonded to $\ce{O}$) acts as "empty", since $\ce{O}$ is more electro-negative than $\ce{H}$?

Lewis acidity and basicity require some amount of context because they can't function in those roles without at least a putative base or acid, respectively. Certainly the oxygen here has a lone pair so it could be basic, and the $\sigma^{*}$ orbital of the O-H bond (on the hydrogen side) can accept an electron pair. What matters more is the context of what acids and bases are present to interact with these moieties. There is no reason why these pieces can't separately fulfill their roles independent of each other.

Ethanol  donate protons to the strong bases forming alkoxide ion. Ethoxide ion is a strong conjugate base which shows that it is a weak acid.

On the other hand, ethanol donate electrons to hydrogen in a reaction with Hydrogen halide and hence it can be considered as strong lewis bases. This is evident from the fact that there are lone pairs on the oxygen atom.


Do acids really donate a proton? Will acids really donate a proton? In an atom, the proton is inside the nucleus and I don't think a proton will go all the way out of atom dodging all electrons and enter into the nucleus of the base.

It seem impossible but the Brønsted–Lowry theory says acids donate protons.

I am confused—could someone explain what is actually meant by acids donating protons?

Yes, according to the Arrhenius theory, acids dissociate in aqueous solution and release a proton ($\ce{H+}$). The Brønsted–Lowry defines acids  ($\ce{HA}$) and bases ($\ce{B}$) in such a way that their interaction is characterized by the exchange of a proton according to $\ce{HA + B <=> A- + HB}$.

However, this is about the reaction of molecules in which real hydrogen atoms are bound to other atoms.

To illustrate the idea, let's think about the reaction between acetic acid, $\ce{CH3COOH}$, and benzylamine, $\ce{C6H5CH2NH2}$. If we combine both liquids in our imagination, the following will happen:

$\ce{CH3COOH + C6H5CH2NH2 -> CH3COO- + C6H5CH2NH3+}$

The acid donates a proton to the amine, which serves as the base. In the course of this reaction, the acid is converted to its corresponding base, while the base is converted to its corrsponding acid.

You are confusing dissociation and proton exchange with a type of radiocative decay known as proton emission.

It donates a Positive Hydrogen Ion. It's basically a Proton but i find this formulation to be less confusing when trying to understand it for the first time.


Partial Atomic Charge in GAMESS In GAMESS, a calculation of optimisation gives data about partial atomic charges (like Mulliken or Lowdin charges). Can you please let me know how to calculate the other types of partial atomic charges (like Hirshfeld, Maslen or Politzer charges)?

FYI: I have used B3LYP and MP2 for optimization.

Unfortunately GAMESS does not support any other than the named population analyses.

You will need to use a different program. For that matter I suggest MultiWFN, which comes with a variety of different population analyses and is still in active development. From the website:

  Population analysis. Hirshfeld, VDD, Mulliken, Löwdin, Modified MPA (including three methods: SCPA, Stout & Politzer, Bickelhaupt), Becke, ADCH (Atomic dipole moment corrected Hirshfeld), CHELPG, Merz-Kollmann and AIM methods are supported.

The currently developed version MultiWFN 3.4 should be able to even read GAMESS output directly as stated in the update log:

  GAMESS-US output file now can be used as input file (not comprehensively tested, currently only single point task at HF/DFT level is formally supported). The suffix of output file should be changed to .gms so that Multiwfn can properly recognize it

As a workaround for older versions, and a reliable way to import this data, you can export a .wfn file with GAMESS. I will give a brief summary for this, since I am still using MultiWFN 3.3.8. From the corresponding manual:

  For example, the wavefunction represented by GTFs is enough for Hirshfeld population, so you can use .fch/.molden/.31~.40/.wfn/.wfx file as input, but .pdb, .xyz, .chg, .cub and .grd files do not carry any wavefunction information hence cannot be used.

As I have previously stated, .wfn files are technically outdated, especially when it comes to larger basis sets. However, for the time being, they are our only option to make it work.

I am using GAMESS version % DEC 2014 (R1) and the following input to demonstrate the procedure. I am using a Linux environment, so you might have to adjust this guide accordingly.

 $CONTRL SCFTYP=RHF            ! Restricted calculation.
         RUNTYP=OPTIMIZE       ! Geometry optimisation.
         COORD=ZMT             ! Z-matix specifies coordinates.
         AIMPAC=.TRUE.         ! Requests wfn file to be written.
         $END
 $BASIS  GBASIS=STO NGAUSS=3   ! STO-3G minimal basis
         $END
 $GUESS  GUESS=HUCKEL          ! Start with Hückel MO.
         $END
! Following group contains molecule specification.
! Blank lines are important.
 $DATA
Water
Cnv  2

O
H  1 rOH
H  1 rOH  2 aHOH

rOH=1.09
aHOH=110.0
 $END

Note that the indentations before the keywords are important. ! indicates a comment. Any additional output will be written to the punch file. This is dependent on how you set up GAMESS, but it normally is located in ~/scr/<rootfilename>.dat.

To extract the .wfn file you can use the following little script, or extract it manually from the .dat file.

#Script to extract wfn file from GAMESS dat (punch) file
#! /bin/bash

[ ! -z $1 ] && inputfile="$1" || exit 1
outputfile="${inputfile%.*}.wfn"

startpattern="----- TOP OF INPUT FILE FOR BADER'S AIMPAC PROGRAM -----"
endpattern="----- END OF INPUT FILE FOR BADER'S AIMPAC PROGRAM -----"

sed "/$startpattern/,/$endpattern/!d;//d" "$inputfile" > "$outputfile"

This should give you the following .wfn file for analysis:

Water
GAUSSIAN              5 MOL ORBITALS     21 PRIMITIVES        3 NUCLEI
  O    1    (CENTRE  1)   0.00000000  0.00000000 -0.14560943  CHARGE =  8.0
  H    2    (CENTRE  2)  -1.43257773  0.00000000  1.05592484  CHARGE =  1.0
  H    3    (CENTRE  3)   1.43257773  0.00000000  1.05592484  CHARGE =  1.0
CENTRE ASSIGNMENTS    1  1  1  1  1  1  1  1  1  1  1  1  1  1  1  2  2  2  3  3
CENTRE ASSIGNMENTS    3
TYPE ASSIGNMENTS      1  1  1  1  1  1  2  2  2  3  3  3  4  4  4  1  1  1  1  1
TYPE ASSIGNMENTS      1
EXPONENTS  1.3070932E+02 2.3808866E+01 6.4436083E+00 5.0331513E+00 1.1695961E+00
EXPONENTS  3.8038896E-01 5.0331513E+00 1.1695961E+00 3.8038896E-01 5.0331513E+00
EXPONENTS  1.1695961E+00 3.8038896E-01 5.0331513E+00 1.1695961E+00 3.8038896E-01
EXPONENTS  3.4252509E+00 6.2391373E-01 1.6885540E-01 3.4252509E+00 6.2391373E-01
EXPONENTS  1.6885540E-01
MO  1                     OCC NO =   2.00000000 ORB. ENERGY = -20.25158070
  4.22735195E+00  4.08851077E+00  1.27421022E+00 -6.18804850E-03  8.27701479E-03
  6.24678653E-03  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  6.97684832E-03  4.38722945E-03  6.95010854E-04
 -1.54630881E-03 -1.49552256E-03 -4.66089059E-04 -1.54630881E-03 -1.49552256E-03
 -4.66089059E-04
MO  2                     OCC NO =   2.00000000 ORB. ENERGY =  -1.25754111
 -9.93961353E-01 -9.61316148E-01 -2.99600254E-01 -2.02173727E-01  2.70423693E-01
  2.04092795E-01  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  2.05800516E-01  1.29412887E-01  2.05011755E-02
  4.30889918E-02  4.16737970E-02  1.29879022E-02  4.30889918E-02  4.16737970E-02
  1.29879022E-02
MO  3                     OCC NO =   2.00000000 ORB. ENERGY =  -0.59384261
  0.00000000E+00  0.00000000E+00  0.00000000E+00 -0.00000000E+00  0.00000000E+00
  0.00000000E+00  1.02654522E+00  6.45519180E-01  1.02261083E-01  0.00000000E+00
  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
 -1.24405583E-01 -1.20319664E-01 -3.74983838E-02  1.24405583E-01  1.20319664E-01
  3.74983838E-02
MO  4                     OCC NO =   2.00000000 ORB. ENERGY =  -0.45973131
 -4.42352599E-01 -4.27824176E-01 -1.33334109E-01 -1.28843153E-01  1.72338126E-01
  1.30066155E-01  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00 -1.26642108E+00 -7.96359559E-01 -1.26156734E-01
 -8.17262213E-02 -7.90420432E-02 -2.46339524E-02 -8.17262213E-02 -7.90420432E-02
 -2.46339524E-02
MO  5                     OCC NO =   2.00000000 ORB. ENERGY =  -0.39261686
  0.00000000E+00  0.00000000E+00  0.00000000E+00 -0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00  1.67545020E+00
  1.05356804E+00  1.66902879E-01  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00  0.00000000E+00
  0.00000000E+00
END DATA
RHF      ENERGY =      -74.9659012162   VIRIAL(-V/T)  =   2.00600297

You can now use this file and load it with MultiWFN. After the start-up process you should find the key values printed to the screen like this (I abridged it a little bit):

 Multiwfn -- A Multifunctional Wavefunction Analyzer (for Linux)
 Version 3.3.8, release date: 2015-Dec-1
[...]
 System energy:     -74.965901216200 Hartree,   Virial ratio:  2.00600297
 Total/Alpha/Beta electrons:     10.0000      5.0000      5.0000
 Net charge:     0.00000    Expected multiplicity:    1
 The number of orbitals:     5,  Atoms:      3,  GTFs:     21
 This is restricted close-shell single-determinant wavefunction
 Title line of this file: Water
 Formula: H2 O1
 Molecule weight:        18.01528

 Loaded water-sto3g.wfn successfully!

                    ------------ Main function menu ------------
[...]
 7 Population analysis
[...]

You may now choose 7 and enter the following menu:

                 ============== Population analysis ==============
 0 Return
 1 Hirshfeld population
 2 Voronoi deformation density (VDD) population
 10 Becke atomic charge with atomic dipole moment correction
 11 Atomic dipole corrected Hirshfeld population (ADCH)
 12 CHELPG ESP fitting charge
 13 Merz-Kollmann (MK) ESP fitting charge
 14 AIM charge

Follow the guidance of the manual and the on-screen instructions to produce the required charges. For comparison here are the computed Hirshfeld charges, written to water-sto3g.chg with the following command sequence 7; 1; 1; y.

  O     0.000000    0.000000   -0.077053   -0.296017
  H    -0.758088    0.000000    0.558771    0.147959
  H     0.758088    0.000000    0.558771    0.147959

The element identifier is followed by the Cartesian coordinates, which are followed by the charges.

In principle all other implemented charges are equally as easy to compute. Keep in mind, that basin based charges (e.g. AIM) need more computing power and depend heavily on molecule and basis set size.

Apart from this procedure, you can obtain natural charges with the NBO6 program, which conveniently interfaces with GAMESS. Unfortunately you have to buy it.

Partial charges are tricky because they are not observables, therefore it is difficult to define them rigorously, however in same cases it can be done.
I usually never use analysis of the population-based charges since they can be very arbitrary in assigning electrons to a given atom.
I know with Gaussian you can obtain Hirshfeld charges (therefore fitting the ESP) but honestly in GAMESS I do not know.
A more rigorous way of dividing the electron density is to use the Quantum Theory of Atoms in Molecules. I cannot go into details since it would take really a lot, however it is possible to obtain all kind of atomic properties partitioning the molecular electron density in basins of charge containing the nuclei, following rigorous topological arguments. From the atomic densities it is possible to obtain the expectation values of observables (atomic dipole/quadrupole moments, energy, magnetic properties etc...) and integrating the atomic density it is possible to know "how many" electrons are contained in the atomic basis, therefore the atomic partial charge.
The free software AIMALL can definitely help you.
Using GAMESS it is possible to generate wavefunction files (file.wfn) that contain the information needed to AIMALL to perform the decomposition of the molecular density. It is very easy and free, I strongly suggest you this approach.

In case you were dealing with planar molecules, there is another rigorous way to obtain information about the partial charge of atoms via the Atomic Polar Tensors, defined as
$$P_{\sigma I}=\frac{\partial \mu_{\sigma}}{\partial r_{I}}$$
where sigma can be the x,y,z molecular dipole component and I is one of the $3N$ cartesian coordinates of the atoms defining the molecule. This tensor is analytically calculated by almost every Quantum Chemistry software, GAMESS included.

I know it is difficult to be absolutely exhaustive within an answer, there are many more things and details to say, however I give you some useful link where you can pick up more theory if needed

AIMALL web page: http://aim.tkgristmill.com/

Paper of Dinur of the charge in planar molecules: http://pubs.acs.org/doi/abs/10.1021/j100169a030

I really hope it can be helpful!


Is this a possible explanation as to why sigma bonds are stronger than pi bonds? Sigma bonds form due to the axial overlap of two orbitals whereas pi bonds form due to their lateral overlapping. Since the former is more effective, we can say that sigma is stronger than pi. We mean that the orbitals overlap to a further extent in sigma bonds and not as much in pi bonds(...or at least that is what I infer). To me this sounds like a circular proof/explanation, so I thought up another way to understand it.

In a sigma bond, the electron cloud of one atom is attracted to the nucleus of the other atom's nucleus.This happens because of the electrostatic forces of attraction between the positively charged protons in the nucleus of one atom and the negatively charged electron cloud of the other atom. The attractive force causes the electron clouds to the atoms to overlap, despite the fact that the electron clouds should repel each other.  However, the electrons start to repel each other after a certain extent of overlapping has occurred. (Because the repulsive forces between the two nuclei and the electron clouds negate the attractive forces)

In a pi bond, there is no positively charged nucleus that would attract the electrons, so there must be another factor favoring the bond formation.
Since I haven't studied the quantum mechanics of bonding yet, my best guess is that the pi bond (or lateral overlapping) occurs because of the pairing of electrons.
So this means that sigma bonds are stronger because of electrostatic forces, which consequently causes more effective overlapping than pi bonds.

Is my reasoning correct to some (if any) extent?

This is a very short answer and it might as well have just been a comment. It is probably also tempting in it's simplicity. For starters, let's assume, that nuclei are point charges.

When two atoms form bonds, all the electrons experience the attractive force of the nuclei, of both of them. This is independent of the type of orbital the electrons occupy.

In a σ orbital, the path of maximum electron density is between the nuclei. It is also where the electronic field of the nuclei is strongest. For a π orbital the situation is different.

Let's take a step back and consider an atom. The s orbital is spherical symmetric. The maximum electron density is (formally) in the nucleus. For the p orbital, there is a node at the nucleus. The maximum electron density is further away from the opposite charge, causing attraction to be lower. The energy of the orbital will be higher.

Similarly we can see that for π orbitals. The node of these kind of orbitals is exactly where the field of the nuclei is strongest. Hence the attraction is lowered, the energy raised, because the maximum electronic density is further away from the nuclei.

In very short terms: In σ orbitals the maximum electron density is on the bonding axis, while in π (and δ) orbitals it is off the bonding axis.

The answer to this question is simple:
Since sigma bonds are the first bond that is formed b/w 2 atoms, its formed by head-on overlapping of orbital, this kind of overlapping gives us a strong bond as compared to sideways overlapping(which happens in case of pi bonds).
Also, If you observe the two ways orbitals arrange themselves in the two bonds.....in sigma bonds the orbitals are closest to the 2 nuclei as compared to the orbitals of pi bond.
Being closer to the nuclei makes them strong as each nucleus pulls the e- of the opposite atoms towards itself and the orbitals overlap.
In a more relaxed language:
In sigma they come straight at each other whereas in pi they bend themselves towards each other.
Also, another imp. point: In sigma the 2 orbitals simply overlap whereas in pi bond the 2 have to bend themselves towards each other and that takes energy. Since pi bonds require certain extra energy for formations this can also be another reason why they are weaker.


Safely mixing methanol and chloroform Sorry for what is likely a naive question. I haven't worked with chemicals at all and have only had some brief lab safety instruction.

I need to create a mixture of chloroform and methanol 2:1 for a lipid extraction from bone.

I have read the safety data sheets for both chemicals but can't find any other information about how to safely create this mixture other than doing it in a fume hood. I need about 200 ml total.

If anyone has any information or can direct me to where to look that would be MUCH appreciated.

Thank you!!!

I am sorry to give my opinion here but I feel compelled to do so!

Security officers or the people writing the safety data sheet are motivated by only one thing: they don't want to be held liable! For them, everything is deadly and they don't even try to mitigate that (true) fact according to their audience.

As a result, all the risks are presented at the same level and nobody can really adjust his/her practices according to that.

I vividly remember one of my students at Oxford University. The rule there is(was?) that the student should copy the whole safety sheet on a form submitted to the Safety Officer before he/she can even get the chemical. My student had to do that for a ridiculous amount of cyanide and he was so stressed out that he ended up shaking and putting cyanide everywhere! I mean on the balance, around the balance, on the floor etc... I had to be besides him "It's OK, it's no big deal" while making signs to the others to not come around because there was cyanide everywhere.

So @Heather, I really advise you to ask your co-workers about the true risks in your field. You won't die because you had a whiff of chloroform or methanol, even though you should always work under a hood so you don't have a whiff of nasty materials. Also the safety sheets must be adapted to your own risks.


How are these two organic resonance structures different? I've been working through some questions online (self study, not homework) predicting the various forms for molecules. One of the questions involved the following molecule (see image below). It had four structures listed in total, but the two below have me baffled. They appear to be identical if rotated by 180 degrees along the horizontal axis.

Please explain the difference between these two structures.

Resonance structures do not describe the conformation (=energy state) of a molecule, but are didactic. They show how you can just flip schemactic $\pi$ electrons around the structure, without breaking any connection. The bond lenghts and angles are constant in the actual molecule, at values somewhere between those indicated by the resonance structures.

Between resonance structures, you don't look at a possible symmetry of the location of atoms, but regard them as unique and fixed (see also https://en.wikipedia.org/wiki/Born-Oppenheimer_approximation). Resonance structures can have different energy or be symmetric (=conformationally identical) to each other. If one resonance structure has higher energy than the others, that is just a hint that the actual structure might by more closely described by the other resonance structures.

While drawing resonance structures, we keep the orientation of the molecule same and assume that the bonds have changed their positions. This is done to pictorially show the movement of electrons. The two structures given are 'equivalent' since they can be superimposed but not the same since the double bond is now between two different atoms.


Why vinegar + water to clean stains? The advice for cleaning salt stains (for example from shoes) with a vinegar + water solution is fairly ubiquitous on the internet. Is there a chemical explanation for why the vinegar is important? Why not just water?

Edit: I don't actually know that vinegar helps clean salt stains. There might be actually be no chemical reason for the advice to be valid.

The acid in the vinegar helps to a) dissolve substances or b) makes the substance easier to remove. The water is neutral (more or less) so won't really dissolve any stain or such


Latent Heat, Vapor pressure and Equilibrium So after reading about this topic i wanted to share what i learned and i wanted you guys to tell me if what i am writting is correct:

A pressure vessel has propane inside and it is pressurized. The liquid is at atmospheric temperature meaning that it is in equilibrium with the outside temperature. The pressure inside the pressure vessel will change according with ambient temperature.

If I open the valve of the pressure vessel, the gas will come out and the pressure inside the vessel will decrease and the liquid propane inside the vessel will immediately star to boil so it can re-establish the equilibrium. When boiling the liquid goes into the gaseous state and to do this it requires energy in the form of heat (latent heat of vaporization).

Since the liquid has the same temperature that the outside it cannot take heat from the outside and it will need to take this energy from itself, decreasing the temperature of the liquid.

Now there is a ΔT between the liquid and the outside temperature, so the liquid will steal heat from the outside (sensible heat) so it can incease his temeprature in order to have enough energy to go to the gaseous state.

If the flow of gas that is coming out of the vessel is constant the pressure and the temperature of the liquid and of the gas will initially decrease a little but it will stabilize, because the heat that is being taken from the outside is enough to keep the liquid boiling without decreasing more of his temperature and pressure.

If the flow of gas that is coming out of the vessel increases the outside heat may not be enough to maintain the flow of gas required, so the temperature of the liquid will start to decrease (auto refrigeration) (heat absorbed by the liquid < heat used by liquid to go to the gaseous state due to the high gas demand)

If the temperature of the liquid keeps decreasing so will the pressure and if it decreases below the atmospheric boiling point, the liquid will stop boiling and the gas will stop to come out (negative pressure inside the pressure vessel).

All is about  3 basic principles.

Liquid evaporation consumes thermal energy, taken from liquid itself or surrounding. Vapour condensation releases this energy.
Liquid saturated vapour pressure (about) exponentially raises with temperature.
If saturated vapour pressure exceeds external pressure, liquid boils. If vapour pressure exceeds the saturated vapour pressure, vapour condensates.


Can you determine when the orbtials of an atom will hybridize? We know that when it comes to C its orbitals hybridize (they become sp3 , sp2 or sp) and the energy levels of an hybridized orbital of this kind is lower than that of the p orbital that formed it but higher than the s orbital so for example energetically speaking is more convenient for an atom to have its electrons (even if they are non-bonding and they don't do too much) in an hybridized orbital if possible.And I know that O and N also can hybridize so the energy barrier between their orbitals (the s and the p orbitals) is not that high to be unable to do this so why don't they just hybridize first in O2 or N2 and then form the molecular orbitals like is the case with C-C bonds , why in the case of O-O bonds we have no hybridization? I mean ok I get the perspective of being able to form a double bond but O2 can have a double bond even if it was sp2 hybridized and then all the non-bonding electrons would be in lower energy orbitals (the other sp2).

The molecule $O_2$ does not have a double bond O=O. It is a biradical which has a single bond O-O, and two independant electrons.

Anyway in diatomic molecules, it is not necessary to consider hybridization. It does not help. Hybridization is an approach that has been invented by Pauling to explain the geometry of polyatomic molecules. Hybridization is not really necessary. It allows to easily describe the shape of the molecules, but the same result may be obtained without hybridization, although it is much more complicated to get to the same result.


Reduction of glucose to hexane with hydroiodic acid My book (NCERT India) states that n-hexane is formed when glucose is heated in the presence of hydrogen iodide. That is, the following reaction takes place:
$$\ce{C6H12O6 + HI ->[{\Delta}] C6H14}$$
This seems a bit strange to me because I only know of three possibilities that can occur when a compound containing hydroxyl groups is treated with a strong acid:

A substitution reaction, where all the $\ce{-OH}$ groups are replaced by $\ce{-I}$
A dehydration reaction, where $\ce{H2O}$ molecules would be eliminated and it would end up with few double bonds.
Formation of an ether.

There are also other sources which mention this reaction (thanks to Karan Singh, Tyberius, and user55119 for pointing these out). A Michigan State University webpage states that:

  Hot hydriodic acid (HI) was often used to reductively remove oxygen functional groups from a molecule, and in the case of glucose this treatment gave hexane (in low yield).

Also, Dietary Sugars (edited by Preedy, V. R.) states (p 80) that

  Glucose is reduced with concentrated hydriodic acid in the presence of red phosphorus to form a trace amount of n-hexane. This indicates an unbranched chain of six carbon atoms [...]

It goes on to cite the 5th edition of Principles of Biochemistry by Nelson et al. (published 2008), but I could not find this statement in any recent edition of this book.

Lastly, a similar reaction was carried out by Kiliani (Ber. Dtsch. Chem. Ges. 1885, 18 (2), 3066–3072) where the cyanohydrin of fructose was reduced to to 2-methylhexanoic acid using hot HI and phosphorus.

I would like to know:

Is phosphorus required for this reaction, or not?
What is the mechanism by which this reaction proceeds, with or without phosphorus?

Hydroiodic acid is a reducing agent.  As Wikipedia says:

  Although harsh by modern standards, HI was commonly employed as a reducing agent early on in the history of organic chemistry. Chemists in the 19th century attempted to prepare cyclohexane by HI reduction of benzene at high temperatures, but instead isolated the rearranged product, methylcyclopentane (see the article on cyclohexane). As first reported by Kiliani,[10] hydroiodic acid reduction of sugars and other polyols results in the reductive cleavage of several or even all hydroxy groups, although often with poor yield and/or reproducibility.[11] In the case of benzyl alcohols and alcohols with α-carbonyl groups, reduction by HI can provide synthetically useful yields of the corresponding hydrocarbon product (ROH + 2HI → RH + H
  2O + I2).[8] This process can be made catalytic in HI using red phosphorus to reduce the formed I2.[12]

When $\ce{HI}$ acts as a reducing agent, the oxidized product is $\ce{I2}$.  To regenerate $\ce{HI}$ from the $\ce{I2}$, phosphorus can be used.  Thus, phosphorus is not strictly required for the reaction, as long as you are OK with using $\ce{HI}$ in superstoichiometric amounts and/or getting very low yields.
The overall scheme for $\ce{HI}$ reduction of an alcohol to an alkane is first displacement of the (protonated) alcohol by iodide, leading to formation of alkyl iodide.  Then, $\ce{HI}$ catalyzes the reduction of the alkyl iodide to the alkane, leading to $\ce{I2}$ as a byproduct.  The reduction step apparently follows a radical mechanism, at least under some conditions.
I'm not 100% sure how $n$-hexane would form from glucose.  It is unclear to me how $\ce{HI}$ would catalyze the reduction of the glucose aldehydic (or anomeric) carbon to an alcohol.  But, that is only one carbon out of six.  The other carbons have bona fide hydroxyl groups, and as such can be deoxygenated and reduced by $\ce{HI}$ to form alkyl carbons.  For example, Lv et al. report that hydrocarbons can be obtained from the five-carbon sugar alcohol xylitol in yields above 60%.  Unlike glucose, xylitol has no aldehydic/anomeric carbon, and only has alcohol groups.  This process did use phosphorous acid as an in situ iodine recycling agent.  Also, Yang et al. were able to obtain 5-methylfurfural from fructose by using $\ce{HI}$, without using an in situ agent to recycle $\ce{}$ to $\ce{HI}$.

Yes, phosphorus is required for this reaction. As stated in your literature and in the comment, red phosphorous is the stoichiometric reducing agent. I assume that $\ce{PI3}$ and $\ce{P2I4}$ are formed. Both are known to be deoxygenating agents (#1, #2).

I propose the mechanism to be like this, where the oxophilicity of phosphorous is the driving force of the reaction. The mechanism starts with an electrophilic attack of phosphorous to an oxygen, which is then substituted by iodide. $\ce{C-O}$ bond cleavage leaves an alkyl iodide. These must then be reduced somehow.


Titanium in HF vs HCl So we tried the reaction between Titanium metal with hydrochloric acid and hydrofluoric acid at a similar concentration. While boiling temperatures and constant heat over a long period of time are required to etch $\ce{Ti}$ using $\ce{HCl}$ even cold $\ce{HF}$ is enough to immediately break the protective oxide layer and the metal vigorousely reacts in the solution afterwards.

My question is now why $\ce{HF}$ is so much better at etching and dissolving $\ce{Ti}$ compared to aqua regia or $\ce{HCl}$?

Of course there is a protective oxide layer on the metal but boiling it in $\ce{HCl}$ will also form a purple solution after a while and the color intensifies if heat is applied over a long period of time. So even after the layer is destroyed the metal itself continues to react with the $\ce{HCl}$. For $\ce{HF}$ it reacts way faster so I assume it's not an issue with the protective layer itself but rather the reaction with the pure metal.

I know for glass and $\ce{HF}$ it's a combination where $\ce{H2F+}$ reacts with $\ce{Si-OH}$ endgroups and increases the electron density on the bridging oxygen which makes them more basic and the $\ce{Si-O-Si}$ bond breaks to form $\ce{Si-OH}$ bonds. But this is all for the oxides and I assume the titanium dioxide will be dissolved in a similar manner.

For metals I couldn't find much information on the topic. I know there are some similar elements, Niobium, Silicon, etc. they all require $\ce{HF}$ to be etched or dissolved.

Maybe it's the case because a $\ce{Ti-F}$ bond, much like a $\ce{Ti-O}$ "bond" is way stronger than a $\ce{Ti-Cl}$ bond? Which would explain why $\ce{HF}$ starts the reaction faster as this unique $\ce{HF/H2F+}$ mechanism seems to work pretty well for tough oxides.

Maybe it's also just the fact that more energy is produced when a $\ce{Ti-F}$ bond forms but the fast reaction starts immediately so it shouldn't just be the heat that is produced enhancing the reaction.

The process itself is a redox-reaction but the essential part here is the $\ce{H+}$ not the counter-ion. So from a redox perspective it should be pretty much the same, unless the redox potential and the overall redox reaction is influenced by the $\ce{Ti-F}$ bond formation, like the hydration enthalphy adds to the oxidative potential of oxygen and suddenly $\ce{Cu(I)}$ turns into $\ce{Cu(II)}$ once the insoluble $\ce{CuCl}$ is added to water.

(Also I'm sorry, but I cannot create a suitable tag with my score at the moment)

TL;DR. Are you asking a question about energy (thermodynamics) or a question about rate? What possible relation is there between boiling HCl(aq) and HF at room temperature as far as rate? I see no value in comparing those two reactions, but I'm probably missing something. So, if I understand you correctly, you are claiming that once you see some "purple" color, that therefore the Ti° is "bare", right? That, of course, doesn't (necessarily) follow. I'm not familiar with the reaction of Ti° + H+ → Ti(IV) + H2. (I'm assuming the +4 oxidation state) but almost certainly (I'd guess) it is not a single step reaction. Where the intermediate steps occur and how the presence of F- or Cl- ions impact one or more of those steps is the question. They could be absorbed on the metal surface, form complexes with intermediates, compete with oxide or hydroxide or oxyhalide species. It seems to me that the slow reaction rate of HCl is telling you that there continues to be something (some barrier) on the surface of the Ti° (by "continues" I don't mean to imply that it was there prior to the oxide films dissolution). About the only thing the energy can tell you about a reaction is that if its exothermic then you'd expect hotter and if hotter then faster (but there are (fairly rare) exceptions to this where hotter means slower). Since the value of comparing reaction A at 25°C to reaction B at 100°C isn't particularly informative (except if B is slower at 100° than A at RT, then we'd predict A will be even faster that B at 100° and B be even slower at RT) - especially if the reaction conditions are so different that we can't assume the mixing is the same (viscosity, acid dissociation, diffusion (turbulence?, H2 behavior?) all will be substantially different). Again, I'm not familiar with this reaction, but since it is a redox reaction, you can't necessarily even assume that the reduction is occurring near (on the molecular scale) to the oxidation, electrochemical cell formation is another possible complication... Could be complicated, IDK.


Is that correct to say that osmosis happens with water only? Is that correct to say that osmosis happens with water only (based on the definition below) or it can happen with other liquids as well?

I've read the following definition in a book called "BRS - physiology" (6th edition, p.5):

  "Osmosis is the flow of water across a semipermeable membrane from a
  solution with low solute concentration to a solution with high solute
  concentration."

No, it is not correct because it happens with other liquids (solvents) also.

Taking Wikipedia's definition:

  Osmosis is the spontaneous net movement of solvent molecules through a
  semi-permeable membrane into a region of higher solute concentration,
  in the direction that tends to equalize the solute concentrations on
  the two sides.

Only as extra information: osmosis occurs until chemical potentials of every substance on each side of the membrane are equal. As the substance cannot move through the membrane, the liquid (solvent) do.

According to the definition, the book isn't correct, because it says with water only, but take into account that in biology the solvent is always water. So, as an internal definition of biology (or physiology) maybe is not wrong.

I'm pretty sure by definition, osmosis refers to the diffusion of water across a semipermeable membrane along a concentration gradient (ie from where water is in high concentration to where it is in relatively lower concentration). Diffusion refers to the movement of any particular substance along a concentration gradient, not necessarily through a semipermeable membrane (but can be). So yes, it is correct to say that osmosis refers to a process that is specific to water.


Why does ice cream make soda fizz? I've noticed that adding a chunk of ice cream to soda makes the soda fizz slightly near the soda-ice cream interface. I thought it was a physical effect due to the temperature, but adding ice has no effect.

It still can be a physical effect due to solubility, or it may be a chemical effect.

I'd like to know which it is, and the details of the mechanism.

Soda contains $\ce{CO2}$. Over time the gaseous $\ce{CO2}$ comes to the surface. The ice cream increases the rate of this  $\ce{CO2}$ from the soda causing increased fizzing. This increased rate is a result of the solvation of ice cream particles in the liquid. The particles serve as nucleation sites for the growth of gas bubbles. The bubbles get large quiet fast thus you observe fizzing.

After a bit of experimentation I found the following:

Effect of adding milk — There is a bit of extra bubbling and a curdy precipitate is formed. (Couldn't dare to drink it.)
Effect of adding salt (yes ice cream does contain salt) — Lots of bubbling around the salt. After the salt settles to the bottom, bubbles collect around it become big and then come to the surface.

Only thing left to try is sugar and vanilla essence but I ran out of soda.

So we come to know that the fizzing effect may be due to milk or salt or a combo of both.

I would say that it is a chemical reaction due to the fact that it is irreversible. Aside from that, the carbon dioxide in the drink reacts to the fat and the bubble (which make ice cream light and fluffy) and wha-la. There is still not solid research as to how it works, but this is my guess due to the fact that most soft drinks have no fat in them, and ice cream has a fair amount. There is research however that states that the carbon dioxide is able to grow in the small air bubbles in the ice cream and this causes the froth. Personally, I think it's both