Bond Lengths: Equatorial vs Axial in Trigonal Bipyramidal Why is it that the axial bond length is less than equatorial bond length in $\ce{PF2Cl3}$ and $\ce{SF2Cl2}$ even though both have trigonal bi-pyramidal geometry?

I know that in general axial bond length is greater than equatorial bond length in trigonal bi-pyramidal structures (as in the case of $\ce{PCl5}$). (See: PBP vs TBP geometry?)

For trigonal bipyramidal structure, we know that lone pairs are preferred first to be positioned in the equatorial position. So in $\ce{SF2Cl2}$ lone pairs will be positioned in the equatorial position.
After placing lone pairs in the equatorial position, double bonds are then preferred in the equatorial position but as there are no double bonds in the structures we are considering, so we jump on to the next rule i.e. atoms having higher electronegativity is positioned at axial position and atoms having low electronegativity is positioned at equatorial position. You can also interpret this in another way. You can also say that atoms having higher atomic size is positioned in the equatorial plane and atoms having lower atomic size is positioned in the axial plane.
Now coming back to your structures, $\ce{PF2Cl3}$ and $\ce{SF2CL2}$ we will see structures in this manner:

$\ce{F}$ is more electronegative than $\ce{Cl}$, hence the $\ce{P-F}$ bonds are going to be shorter than $\ce{P-Cl}$ bonds and $\ce{S-F}$ bonds are going to be shorter than $\ce{S-Cl}$ bonds. As $\ce{F}$s (fluorines) are located in the axial position, hence these $\ce{P-F}$ or $\ce{S-F}$ bonds which are actually axial bonds are shorter than that of equatorial bonds (i.e. $\ce{P-Cl}$ or $\ce{S-Cl}$ bonds).
In $\ce{PCl5}$ all the atoms around $\ce{P}$ are basically the same i.e. $\ce{Cl}$. So the electronegativity factor of $\ce{Cl}$ doesn't play any role in reasoning out why axial bonds are longer than equatorial bonds in case $\ce{PCl5}$. The reason is - in $\ce{PCl5}$ the axial bonds suffer repulsion from the equatorial bonds so they become long. The equatorial bonds providing repulsion to axial bonds also happens in case of $\ce{PF3Cl3}$ and $\ce{SF2Cl2}$, however this factor is overlooked/ not considered in cases of $\ce{PF2Cl3}$ and $\ce{SF2Cl2}$ because there this factor/ effect is not dominating. In those cases as the type of atoms attached (i.e. electronegative atoms) are different, hence the effect of bond length shortening due to the electronegative atom attached is the dominating factor.


Confused about strong/weak dibasic and tribasic acids I know that sulfuric acid is considered a strong acid, and the dissociation of its first $\ce{H+}$ ion is complete and of its second $\ce{H+}$ ion is partial.

Phosphoric acid is weak and all three of its $\ce{H+}$ ions dissociate partially.

I just wanted to clarify: are dibasic and tribasic acids considered strong if their first dissociation is complete (and subsequent ones will naturally be partial)?

If you think about each dissociate state separately, than it's easy to answer your question. $\ce{H2SO4}$ is a very strong acid $(\mathrm{p}K_\mathrm{a}\approx -3.0),$ but $\ce{HSO4-}$ is not so strong $(\mathrm{p}K_\mathrm{a}\approx 2.0).$

Phosphoric acid starts with $\mathrm{p}K_\mathrm{a}\approx 2.14,$ which as as you said is not so strong, and $\ce{H2PO4-}$ and $\ce{HPO4^2-}$ have $\mathrm{p}K_\mathrm{a}\approx 7.2$ and $\mathrm{p}K_\mathrm{a}\approx 12.3,$ so definetly weak.

So yes, we say sulfuric acid is strong because we're talking about $\ce{H2SO4},$ and what comes after should be mentioned separately.


Drinking water out of a 24 year old Ether Bottle I have an empty aluminum Ether bottle (Shipping name diethyl Ether) that I'd like to use as a large drinking bottle as this is thick aluminum. Is it dangerous? Here is some of the info on the bottle.

  Chemical Composition: (C-2 H-5)2-O f.w.74.12.

  Name:  Ether Purified (For Fat Extraction)

  Other: Mallinckrodt (Brand?)
   5.51 lb

  0884

  Diethyl Ether 60-29-7

  Ethyl Alcohol 64-17-5

Mallinckrodt is the chemical manufacturer. The ethanol is added to the ether to prevent peroxides from forming in the ether (like an antioxidant). If you wash and rinse the container thoroughly a few times, then it should be fine.


Correct stability order of resonance structures Question no 9 as far as I know none of the rules of relative order of stability apply to this( completion of octate,charge seperation etc.) So they all must be equally stable but the answer is (D)

Compound II and III have 2 aromatic rings whereas I and IV have only 1 aromatic ring, and more the number of aromatic rings, the more the compound is stable.

  Aromatic molecules are very stable, and do not break apart easily to
  react with other substances. Organic compounds that are not aromatic
  are classified as aliphatic compounds—they might be cyclic, but only
  aromatic rings have special stability (low reactivity).

Thus stability of II= stability of III and stability of I = stability of IV.

And since II and III have greater number of aromatic rings than I and IV,

Stability(II,III) > Stability(I,IV)

Thus the answer would be [Option D] and not [Option C].


Carbon dioxide in lime ground Is storing carbon dioxide in limestone a good idea? Wouldn't that lead to a reaction like

$\mathrm{C}\mathrm{O}_2 + \mathrm{H}_2\mathrm{O}\to {\mathrm{C}\mathrm{O}_3}^{-2} + 2\mathrm{H}^+$

And then the (unbalanced) reaction

$\mathrm{Ca}\mathrm{C}\mathrm{O}_3 + \mathrm{H}^+\to \mathrm{C}\mathrm{O}_2 + \mathrm{Ca}(\mathrm{O}\mathrm{H})_2$

So now the the ground turns into a water caustic water solution. Is this right?

Limestone is the natural sedimentary rock, e.g., the top of Mt. Everest that began as sea floor.  "Storing" $\ce{CO2}$ in limestone requires heating calcium carbonate white hot to dissociate it to lime, then scrubbing "other" $\ce{CO2}$ with the lime.  This is, to coin an adjective, smartless, since you also burned fuel to heat the limestone. Curing concrete imbibes and traps $\ce{CO2}$, but it was initially dissociated to release it.

Your second equation is terrible. You obtain a strong base by adding acid to a salt.  $\ce{CaCO3}$ plus carbonic acid, $\ce{CO2}$ in water, makes water-soluble calcium bicarbonate $\ce{Ca(HCO3)2}$.  That is how limestone caves are dissolved out, and why karst topography is stark (surface erosion) and unstable (sinkholes).

To vastly store carbon at a profit, plant forests, harvest, and build wood-frame buildings.  The photosynthesis bottleneck, carbon fixation, depends upon the slowest least efficient enzyme on Earth, RuBisCO.  A smart (at least, not corrupt) society then

1) Gene-gineers a better RuBisCO.  Ten times the productivity leaves it the worst enzyme on Earth.  Everybody would have a full belly.

2) Gene-gineers a 10x fast-growing forest with huge lignin content in its wood.  It won't rot, it won't feed termites, it will be immensely strong.  If wood-frame buildings become unfashionable, bury the wood in deserts to store carbon.

There is no problem so easily solved that social advocacy will not make it intractable, disastrous, and expensive.  The unabled maximize self-esteem to be credible - effectively selecting for the deluded, hence advocacy.  A society inundated with rules from religious and secular political classes ignoring productive ends will collapse.  USSR to Pol Pot, reality is what it appears to be - thermodynamics.  There is no cheating possible.


Dess–Martin oxidation work up I'm doing an oxidation of an alcohol with Dess–Martin periodinane which works really well but having trouble with the work up. Doing this reaction with about 5 g of alcohol so the reaction ends up using about 17 g of DMP. As the reaction progresses the DMP byproduct crashes out. And working up turns it into a gun that is quite difficult to remove

Running reaction in DCM and work up with bicarb. Even after filtration DMP is still present. Separates fine on a column but with how much there is it seems to be trappings some of my product.  I've read that a thiosulphate wash is also used in these reactions, but I don't think this will be useful to get rid of the gum.

Any suggestions?

It looks like you are not the first to get gummed up by the mono-acetoxy iodinane byproduct of this reaction. A seemingly simple solution is given in this article about converting an alcohol to ketone via DMP:

  The mono-acetoxy iodinane by-pdt can be converted to easier to remove by-pdts during workup by treatment with aq $\ce{NaOH}$ or $\ce{Na2S2O3}$.

This is not a procedure I have experience with specifically, so I hope this a do-able solution for your problem as I understand it.  Good luck regardless.


Which computational method/basis set should I use for small molecular ions? I apologize for a perhaps newbie question.

I would like to compute the potential energy surface for several small molecular ions, like $\ce{H2+}$ and $\ce{H3+}$, using quantum chemistry software. What are the proper method and basis set to do that and why? I am thinking about ROHF/STO-3G or UHF.

$\ce{H2+}$ or $\ce{H3+}$ have unpaired electrons because they are ions. Due to this, you should be using an unrestricted shell theoretical model which doesn't restrict electrons to paired (closed shell) orbitals. You should start with unrestricted Hartree-Fock (UHF) and can increase the level of theory to include electron correlation with something like 2nd order Møller Plesset (MP2) or 4th order (MP4).

The electron correlation models acknowledge the fact that electron locations are not totally random, repulsion between electrons is a real effect and can give more realistic results.

With regard to basis set, STO-3G would be considered a starting point. If time and resources allow, you could build up to something like a 6-31G(d,p) which adds p-orbitals to the hydrogen atoms allowing more optimization of molecular orbitals. Increasing the basis set quality approximates the radial components of the atomic orbitals with more Gaussian curves, thereby increasing the possible accuracy of the calculation.

Exploring Chemistry with Electronic Structure Methods - Foresman and Frisch


What are the reaction products of sulfur and potassium nitrate? YouTube shows videos of dropping a piece of elemental sulfur into molten potassium nitrate and applying additional heat to accelerate the process. A white precipitate falls out to the bottom. There is never any mention as to what this material is. An internet search yields not a clue. Sulfur dioxide is obviously evolved, but KNO doesn't seem to exist if one relies on an internet search.

There are at least two videos (first and second) showing the reaction between sulfur and potassium nitrate. For the chemicals:

\begin{array}{cc}
\rm Species&\rm Melting~point\\\hline
\rm Sulfur &  \rm 115~^\circ C     \\
\rm Potassium~nitrate & \rm 334~^\circ C
\end{array}

It isn't clear if nitrogen, some nitrogen oxides, or both would be produced. My guess is that it would depend on particle size, mixing and so forth, thus that a mixture of nitrogen oxides would be very likely. Also the test tube starts with a tiny amount of oxygen gas from the air. One possible balanced reaction would be:

$$\ce{4KNO3 + 5S -> 2K2O + 5SO2_{(g)} + 2N2_{(g)}}$$

Looking at the products of black powder burning $\ce{K2SO4}$ and $\ce{K2S}$ can also be formed.


What happens to gasoline when it lands on your hand? I was filling the mower with gasoline and some gas splashed on my hand. Not even 2 seconds later it was gone...

1) Does gasoline turn into vapor that fast? And if it does, why doesn't it evaporate in the gasoline can?

or

2) Does it seep into your skin and into your blood?

Also.. after it had 'dried' there was a white residue leftover, what is it?

It evaporates, but it will first take some hydrocarbon compounds in your skin

Gasoline is a mixture of a variety of volatile hydrocarbons, some branched chain alkanes and some aromatics. The warmth of your hand is enough to cause small quantities to evaporate. But it will also dissolve some of the hydrocarbon-soluble oils from your skin. Then when it evaporates, they will be left but they will form a microcrystalline powder rather than returning to the parts of the skin they were extracted from: this is the white residue. This is a common when volatile non-polar solvents are spilled on the skin (don't do it too much or your skin will be deprived of the oils that keep your skin flexible and impervious to outside contamination). Small quantities are nothing to worry about, just don't get into the habit or it will hurt your skin.


2 modes of radioactive decay for hydrogen-5? While looking at a table of helium isotopes it said that helium-8 turns into Tritium(hydrogen-3) and helium-5 through fission and beta decay.

But the pathway to helium-5 isn't so direct. It is more like this

helium-8 -> hydrogen-3 and hydrogen-5 (fission) -> helium-5 (beta decay)

But the table of hydrogen isotopes said nothing about hydrogen-5 going through beta decay. All it had there was double neutron emission producing tritium which after 12 years turns into helium-3 via beta decay.

Does this mean that I discovered that hydrogen-5 does go through beta decay but not nearly as often as double neutron emission?

Sorry, you haven't.  I don't know where your information is from but I think there is an error in your decay chain.  For the 0.9% of $\ce{^8_2He}$ decays that proceed via spontaneous fission your equation should look more like this:
$$\ce{^8_2He ->[fission] ~^3_1T + ^5_1He + \beta ^{-}}$$
$$\ce{ ^5_1He  ->[neutron~emission]  n + ^4_2He} $$


Why doesn't this question violate the conservation of mass? Combustion of of $0.255 \space g$ of isopropyl alcohol produces $0.561
 \space g$ of $\ce{CO_2}$ and $0.306 \space g$ of $\ce{H_2O}$.  Determine the
  emperical formula of isopropyl alcohol.

I'm not asking how to solve this problem, but why this doesn't violate the conservation of mass?  Shouldn't the initial combusted mass equal the sum of the resultant masses?

Because the problem statement indirectly includes an additional reactant.

At an introductory chemistry level, the term combustion implies reaction with $\ce{O2}$. Thus, your missing mass should show up when you introduce a stoichiometric amount of oxygen to your reactants.


Long-lived, non-lethal radioisotope for fiction I'm writing something where the characters are looking for someone among a large set of people. They don't know who this person is and they don't know what they're looking for, so their best bet is to look for anything unusual about this person. I already have a method by using which they'll eventually find him, but I want them to muse over possible, reasonable alternatives. One that comes to mind is looking for any radioactive tracer that might have been injected in the subject. However, for this to make sense within the context of the plot, such tracer should be long-lived enough to be detectable at least for years, and safe enough not to appreciably increase the subject's risk of death over short timespans. (If he's more likely to get cancer at age 90, tough luck, but if the tracer will kill him within five, that's not good.) Any ideas?

Good candidates for your research would be the fallout residues of atomic bombs having exploded in the world between $1945$ and $1964$. These bombs were made by USA and the former USSR (Russia today) during the "cold war". Each time one of these superpowers had developed a "more powerful" bomb, the other one managed to create a still more powerful one. And so on for about $20$ years. This stupid course stopped in $1964$, because of the level of the fallout radioactivity measured all over the world. In $1964$, the radioactivity from the fallout had much increased all over the world. The whole planet had been contaminated by the winds to $10$% of the limit value where all food would have to be decontaminated before being eaten. So both Russians and Americans knew about this danger. They managed to meet in Geneva and decided to stop their nuclear rivalry. All that were made for "in the interest of the peace in the world". It was magnificent.
As a result the radioactivity slowly decreased all over the world. At the end of the century it was back again to the level $1945$.
One of the isotopes present in the fallout is the strontium-$90$, who is assimilated like calcium, and milk contains much calcium. So $\ce{^{90}Sr}$ was found in the teeth of the newborns overall. Its half-life is about $28$ y. It was followed in teeth all over the world from $1964$ to today. It was observed that this isotope disappears from the body with much less than $28$ y. Apparently calcium (and strontium) are naturally and continuously renewed in the body.
Another isotope that can be interesting for you is the radium-$226$ who was deposited on the watch dials before $1964$. Like strontium, radium is assimilated by the body like calcium. Workers that were painting the dials with radium used brushes, and have been contaminated by this isotope. So they are radioactive. They emit constantly gamma rays. They can be detected with a Geiger counter. The level of the radioactivity is not dangerous. But they emit rays. Is it what you want ?


Why are fats glossy? I am talking about the standard fats we see in everyday use (triglycerides). It seems that no matter what the composition, saturated or unsaturated, all fats are glossy when liquid. I don't know if this extends to other esters, as I have seen only few of them.

What gives fats this shiny look?

Sorry if the question appears unresearched, but all my attempts to combine "fat" and "glossy" in a search query resulted in nutrition advice.

Reflected light (gloss) is a maximum when the angle of observation is equal to the angle of incidence:  https://en.wikipedia.org/wiki/Gloss_(optics). The image shows reflection from a plane surface with some roughness that gives diffuse reflection.

At Brewster's angle, the reflected ray is strongly polarized. The image is intended to indicate a depth of penetration in transparent materials; the polarization is not relevant here.

In conductors, like metals, and strong absorbers, like iodine, there is essentially no transmission, and almost all the light is reflected from a plane surface, because the penetration depth is exceedingly small.

In a dielectric (i.e., a nonconductor like glass or fats), some light will be transmitted, and reflection occurs after a penetration of some depth - on the order of a wavelength of the light. Glass is atomically uniform on this scale, and  a fat molecule (e.g., glyceryl trioleate) would be a glob about 2 or 3 millimicrons in diameter, therefore essentially homogeneous on this scale. Any liquid volume larger than that would have no discernible inhomogeneity and reflection would be a uniform wavefront, specular (mirror-like) rather than diffuse.

But suspensions and runny emulsions and milk are not homogeneous on that scale. They have particles on the order of a wavelength of light (~500 millimicrons) in the liquid, so reflection occurs from sites of different composition within this penetration depth. These materials reflect similar to reflection from a roughened metal surface and give a diffuse reflection rather than a gloss.


Is it possible to prevent agglomeration and sedimentation of microparticles just by increasing the viscosity of the medium? I am considering a suspension of a fine powder in a high-viscosity liquid.  The suspension shall be stable over long time scales, i.e. the particles only move a few mm per year, and I have the following boundary conditions:

The particles sizes shall be between 0.1 and 1.0 microns, so the suspension is colloidal
The concentration of dispersed particles in the medium is high (the phases have approximately equal volume).
The viscosity of the medium must be high. I am considering >10000 cSt, which is comparable to thick honey or shower gels
The medium will also be free of water; possibly long hydrocarbons

In a first naïve treatment with Stokes' law I thought that such a suspension should be sufficiently stable. Then I read about interparticle effects, which can lead to agglomeration of the particles and sedimentation (mainly on wikipedia). I don't know yet which material will be used as the solid powder, so I have to consider the worst case of attracting forces between the particles, so that there will be agglomeration and sedimentation in my case. This has to be prevented, but for several reasons I would like to avoid steric stabilization.

However, all examples of agglomeration considered a "thin" liquid like water, while polymeric stabilization technique is based on long polymer chains in the medium. Since in my case the viscosity of the medium is very high in my case, I wondered if this might be enough to prevent agglomeration. The high viscosity is due to long molecular chains, so I'm thinking that this might count as a case of polymeric stabilization. Is it possible to prevent agglomeration and sedimentation of microparticles just by increasing the viscosity of the medium?

If you can make this, it will be stable. At equal volume of the phases, this will already be solid-like (0.5 is above percolation threshold, which is in the range of 0.25-0.3, depending on the geometry of your particles.).

See https://en.wikipedia.org/wiki/Percolation_threshold#Thresholds_for_3D_continuum_models

The absolute particle size is irrelevant, only the size distribution matters. When you reach the percolation threshold, the only thing that matters about the continous phase is that it's incompressible, which is the case for practically all condensed matter. If it's viscosity is very low (and the solid particles large, > ~100 µm), it could flow out of the compound, but surface tension would usually preclude that.

Also a very lowly viscous continuous phase could mean that the compound yields already to gravity. https://en.wikipedia.org/wiki/Yield_(engineering)

However it is not so simple to make such a suspension. If you try to mix powder and polymer, your stirrer (of whatever kind) will stop or break long before you get a homogeneous mixture, because your sample becomes solid, and there is no way of melting it or anything. A helper solvent can lower the solid fraction, below the percolation threshold, but is hard to get rid of later.

So you'll have to find out where the percolation threshold actually is, and just use a bit more. It varies a bit with temperature (your polymer phase will have a larger expansion coefficient, not sure if that is enough to make the difference). Anyway the yield stress is still tolerable just above the threshold, and processing is possible.


Does 1-phenyl-1-keto-but-2-ene show positive iodoform test? It is told often that only compounds having a methyl ketone and corresponding alcohols show positive iodoform test.

I tried to sketch a mechanism to show that 1-phenyl-1-keto-but-2-ene also gives iodoform test. However I haven't found any source that confirms that 1-phenyl-1-keto-but-2-ene shows iodoform test.
Please check if there is any mistake in the mechanism:

This mechanism is very similar to the classical iodoform reaction mechanism.

Enones react with with iodine and catalytic base under aqueous conditions to give $\alpha$-iodoenones (Ref.1). Pyridines are most often used for this, the reference quoted also uses $\ce{K2CO3/DMAP}$. They further show that reaction works for a range of enones including those with the $\ce{RCOCH=CH-CH3}$ grouping and no mention is made of $\gamma$-iodination. A "background" reaction was run with $\ce{K2CO3/I2}$ in aq. THF and no iodination was observed.

Proposed mechanism from the referenced paper:

References:

Marie E. Krafft, John W. Cran, "A Convenient Protocol for the α-Iodination of α,β-Unsaturated Carbonyl Compounds with $\ce{I2}$ in an Aqueous Medium," Synlett 2005, (8), 1263-1266 (DOI: 10.1055/s-2005-868495).


How to calculate the amount of substance of barium sulfate that is precipitated? Determine the amount (in mol) of barium sulfate that will be precipitated when $200.0~\mathrm{cm^3}$ of $0.450~\mathrm{mol\,dm^{–3}}$ barium nitrate solution is added to an excess of sodium sulfate solution, given that the equation for the reaction is:
  $$\ce{Ba(NO3)2 (aq) + Na2SO4 (aq) -> BaSO4 (s) + 2NaNO3 (aq)}$$

I have just tried doing it and I got an answer of $0.09~\mathrm{mol}$, but I am not sure if this is correct, because I have only just been introduced to $\mathrm{dm^{-3}}$ so didn't really know what I was doing. I converted $200~\mathrm{cm^3}$ to $0.2~\mathrm{dm^3}$ (is this right?), and then multiplied this value by the concentration, $0.450~\mathrm{mol\,dm^{-3}}$, and this gave me $0.09~\mathrm{mol}$. Since the stoichiometric ratio between $\ce{BaNO3}$ and $\ce{BaSO4}$ is 1:1, I just suspected that the amount of substance of $\ce{BaSO4}$ would also be $0.09~\mathrm{mol}$.

If anyone could tell me if I have done this correctly, it would be greatly appreciated.

Yes, your answer is correct.

However, since you are still learning I recommend, that you take the time to write out all the equations, with proper units and every step. This should help you to gain a better understanding of the chemistry that is actually going on.

One of the key words of the exercise is excess of sodium sulfate solution, which indicates that the reaction will be complete, i.e. entirely on the product side.

Hence, you can (and you correctly did) determine from the reaction equation,
$$\ce{Ba(NO3)2 (aq) + Na2SO4 (aq) -> BaSO4 (s) + 2NaNO3 (aq)},$$
that the stoichiometric ratio is one, $n(\ce{Ba(NO3)2})/n(\ce{BaSO4}) = 1$. Therefore you can also write $$n(\ce{Ba(NO3)2}) = n(\ce{BaSO4}).$$

Now you simply have to calculate the amount of substance of barium nitrate. You have been given the volume of your solution $V(\ce{Ba(NO3)2}) = 0.200~\mathrm{cm^3}$ and the concentration $c(\ce{Ba(NO3)2}) = 0.450~\mathrm{mol\,dm^{-3}}$. Using the relation
$$c = \frac{n}{V}$$
you can calculate the amount of substance of barium nitrate.

You have also correctly converted cubic centimetres to cubic decimetres. In any case, you should get familiar with metric prefixes since they will come up in every science class you will take.


Cohesive forces in solid solutions I have searched through printed books and the Internet, but I cannot find a definitive answer on an issue that fascinates me: what forces keep the various compounds belonging to a solid solution together?

If I correctly understand, a rock is a solid solution, and I know that the atoms of a single mineral species are bound by its particular covalent and ionic bonds, but what keeps different minerals together in a rock? Do they mix by sharing "neighbouring" cations and anions? I am not sure whether intermolecular forces such as Van der Waals forces are present, but I would definitely expect stronger forces in a solid solution or between the minerals of a rock.

It seems you were on right path - for example common feldspars are solid solutions and they are mixed because they share cations, while they easily substitute one another, otherwise they exsolve - become separate phases. Also you're right that van der Waals forces aren't important here as they are ionic compounds.


Which is the major product formed on nitration of benzyl methyl ether? Which is the major and minor product formed from the reaction of benzyl methyl ether with dinitrogen pentoxide?

I know that $\ce{NO2+}$ is behaving as an electrophile here and attacking the aromatic ring. Furthermore, the substituent directs electrophilic substitution to the ortho- and para-positions. I thought that since the ortho position is more sterically hindered, the para product should be the major product and the ortho  product minor.

However, the answer I was given says the opposite, i.e. that the  ortho product is major and the para product is minor. Where am I going wrong?

1) $\ce{NO+2}$ electrophile.

Yes

  2) Benzene ring is attached to an activating group, so meta positions
  are enriched and electrophile has chances of attacking ortho/para
  positions.

The effect of the substituent in your example is probably similar to the effect of a methyl group (e.g. replacing a methyl hydrogen with a methoxy group won't have a big effect on the electron donating ability of the substituent).  Look at the relative rates of nitration in the following figure.

(image source)

We see that in toluene the ortho and para positions react 43 and 55 times (these two numbers are rather close) faster than in benzene, while reaction rate at the meta position changes little from benzene.  This is consistent with the pi-donating ability of a methyl group adding electron density to the ortho and para positions making them more attractive to attack by an electrophile.

  3) ortho position is more sterically hindered, hence para- major
  product and ortho minor product.

Look at the relative rate for the ortho position in t-butyl-benzene, here steric factors definitely play a role in slowing down attack at the ortho position. By comparison, the methyl group (and your substituent) is not very large and may slightly reduce the reaction rate at the ortho position in toluene (43 vs. 55 rather than both numbers being 55) or your example.

Another factor to consider is statistical in nature, there are two ortho positions, but only one para position. So, if everything else were equal, we would expect twice as much ortho product as para product.

In summary,

the electron donating effect of the substituent increases the rate at the ortho and para positions to a similar degree
steric factors might reduce the rate at the ortho position a small amount
statistical factors favor twice as much attack at the ortho position compared to the para position

This statistical effect is the largest of the three and is consistent with the ortho isomer being the major product.

One final cautionary note. When we discuss reaction rates we are really discussing kinetics - which product will form the fastest. However, electrophilic aromatic substitution is an equilibrium process. If you allow the reaction to run at high temperature, or for a long time, then thermodynamics can take over and we may wind up with a product composition based on relative product stability rather than kinetic factors.


Franck-Condon factors in photoelectron spectroscopy Franck-Condon factors have been used to predict / assign the vibrational structure of photo-electron spectra.

$FCF=|\langle\psi''_{vib}|\psi'_{vib}\rangle|^2 = |\langle\psi''_{vib,1}\psi''_{vib,2},\ldots,\psi''_{vib,n}|\psi'_{vib,1}\psi'_{vib,2},\ldots,\psi'_{vib,n-1}, \psi'_{vib, emitted}\rangle|^2=|\langle\psi''_{vib,1}|\psi'_{vib,1}\rangle|^2\ldots|\langle\psi''_{vib,n}|\psi'_{vib,emitted}\rangle|^2$,

Question: what is $\psi'_{vib,emitted}$? Is it just a free particle, with its wavefunction spanning all space, hence it doesn't have any effect on the integral?

$\psi^\prime_{vib, emitted}$ is an energy normalized continuum wavefunction. The Franck–Condon factors involving bound ($\psi_v$) and continuum states ($\psi_E$) are actually Franck–Condon densities that are functions of the energy.

The Franck–Condon factors and densities are related by a closure relation
$$ \sum_{v^{\prime\prime}}|\langle \psi_{v^{\prime\prime}}|\psi_{v^{\prime}}\rangle|^2 + \int |\langle\psi_{E^{\prime\prime}}|\psi_{v^{\prime}}\rangle|^2 dE^{\prime\prime} = 1$$

References:

[1] J. Tellinghuisen, The Franck-Condon Principle in Bound-Free Transitions, in Photodissociation and Photoionization, edited by K.P. Lawley (John Wiley & Sons, Ltd, 1985).

[2] R.J. Le Roy, Comput. Phys. Commun. 52, 383 (1989).


Why is the vertical part of a titration curve longer for a strong acid by a strong base than for a weak? As the title says, I would like to understand chemically why this occurs. My theory behind it is that in the former case both are completely dissociated, and so at the equivalence point (and approaching it), everything is controlled increasingly by the autoionization of water, whereas in weak acids there is also a much more sizable contribution of $H^+$ by the acid itself, so the change is not as drastic. If this is correct, I'd like further explanation as to why it is, since I don't quite feel comfortable enough with it intuitively.

This is not a duplicate. I am not asking why the strong acid curve is steep, but rather why the strong acid curve is steep while the weak acid one is not.!

Weak acid and strong acid titration curves with addition of a strong base

Compare (1) the titration of $\ce{NaOH}$ with $\ce{HCl}$ and (2) that of $\ce{NaOH}$ with $\ce{AcOH}$.

At the equivalence point of each titration, solution (1) contains $\ce{Na+}$, $\ce{Cl-}$, and $\ce{H2O}$, with negligible concentration of $\ce{H+}$ and $\ce{OH-}$. In contrast, solution (2) contains $\ce{Na+}$, $\ce{AcO-}$, and $\ce{H2O}$, and in particular $\ce{AcO-}$ will react with water to form a buffer solution of $\ce{AcO-/AcOH}$.

This will be true in general for any weak acid/base titration. The resulting buffer resists changes in pH; hence a weak-acid titration curve is not as steep at the equivalence point as is a strong-acid titration curve.


Is the effect of hydroperoxyl radicals significant in the oxidation of organic compounds in the Fenton reaction? Why or why not? In the Fenton reaction, a ferrous catalyst and hydrogen peroxide are used to generate hydroxyl radicals which in turn oxidise organic compounds. The reaction is commonly seen in the treatment of wastewater and operates using the following mechanism:

  $$
\begin{align}
\ce{Fe^2+ + H2O2 &-> Fe^3+ + ^.OH + OH-}\tag{1}\\
\ce{Fe^3+ + H2O2 &-> Fe^2+ + ^.OOH + H+}\tag{2}\\
\ce{Fe^3+ + ^.OOH &-> Fe^2+ + H+ + O2}\tag{3}\\
\ce{RH + ^.OH &-> H2O + R^{.}}\quad\text{(chain propagation)}\tag{4}\\
\ce{R^. + Fe^3+ &-> R+ + Fe^2+}\tag{5}\\
\ce{R^. + H2O2 &-> ROH + ^.OH}\tag{6}
\end{align}
$$

In the reactions, hydroperoxyl radicals $(\ce{^.OOH})$ are produced as a byproduct of the reduction of the ferrous catalyst, among other reactions. However, literature regarding the impact of the hydroperoxyl radical on the oxidation of organic compounds is limited, with most papers ignoring the effect of the hydroperoxyl radical completely. So, is the effect of the hydroperoxyl radical significant in the oxidation of organic compounds? Why or why not?

The effect of the hydroperoxyl radical in the Fenton process can be largely ignored as the oxidising ability of the $\ce{^∙OOH}$ radical is not as effective as that of the hydroxyl radical.
According to Chang et al., the reaction rate coefficients of p-nitrophenol and $\ce{OH}$ and $\ce{HO2}$ were found to be $\pu{9.05 \times 10^6 M−1 s−1}$ and $\pu{1.50 \times 10^−6 M−1 s−1}$, respectively. Considering the orders of magnitude of these two rate coefficients, the reaction pathway involved the p-nitrophenol reaction with $\ce{^∙OOH}$ radicals can thus be neglected. This can therefore be extended to cover the oxidation of organic compounds by hydroperoxyl radicals through the Fenton process.
References

Chang, M., & Chern, J. (2010). Decolorization of peach red azo
dye, HF6 by Fenton reaction: Initial rate analysis. Journal of the
Taiwan Institute of Chemical Engineers, 41(2), 221-228.
DOI: 10.1016/j.jtice.2009.08.009


Why is the mole a unit of measurement? I really can't understand why is the mole a unit of measurement and not just a constant. I mean, we use units to express things that we can measure but that are not countable, that are continuous - and not discrete. However, the number of atoms (or whatever) of something is clearly countable, discrete. It is also a huge number, which would request for a constant, a number to multiply to: Avogadro's constant is that number, like mega or kilo in SI for other units. We could still write 14 mole, where mole is the Avogadro constant (exactly the way it is), but every book I read so far states that the mole is a unit of measurement, official in the S.I., and so does Wikipedia. It just makes no sense to me that it is considered that way, but there is obviously a reason why... What is it?

I think you are confusing unit of measurement with fundamental unit.  A unit of measurement is a definite magnitude of a physical quantity.  Consider a very common grouping unit, the dozen which is conceptually the same as a mole and is a little easier for us to grasp since it is a commonly used small number.  As far as I know, the only requirements for a unit of measurement to be valid is general consensus.  If you and I agree that we will measure distance in units of 'Chemistry Textbook Thickness', then it is a valid unit of measurement.

A fundamendal unit is a part of a particular set of units upon which all other units can be derived.  In this case, the mole is related to the fundamental unit of mass and is defined as an amount of items equal to the amount of carbon atoms in 12 grams of $^{12}C$.


How pure NaCl is typical table salt? How is it purified? How pure $\ce{NaCl}$ is typical table salt? Is it pure enough for research-level experiments in chemistry? How is $\ce{NaCl}$ purified if need to be?

Table salt is often iodized (iodine is an essential micronutrient, and this is one way to introduce it into the diet in the right amount). According to Wikipedia,

  Edible salt can be iodised by spraying it with a potassium iodate or potassium iodide solution. 60 ml of potassium iodate, costing about US $1.15 (in 2006), is required to iodise a ton of salt. Dextrose is added as a stabilizer to prevent potassium iodide from oxidizing and evaporating. Anti-caking agents such as calcium silicate are commonly added to table salt to prevent clumping.

Elsewhere, Wikipedia mentions other anti-caking agents, like sodium ferrocyanide, sodium aluminosilicate, and magnesium carbonate. And they point out that salt may also be "fortified" with iron or other micronutrients.

The certificate of analysis for "ultra pure NaCl" from American Bio shows TRACE amounts of aluminum, arsenic, bromides, heavy metals, iron, magnesium, phosphates, and sulfates. Those impurities may well be expected in table salt, too.

From all that, you'd think that table salt is fairly impure. Let's look at the numbers to put the amounts of these impurities in perspective. The box of Morton iodized salt in my kitchen gives me the amount of sodium per serving. If all of the sodium is in the sodium chloride, I estimate the purity as

$$\frac{0.590~\mathrm{g}~\ce{Na+}}{1.5~\mathrm{g}~\text{table salt}}\frac{1~\mathrm{mol}~  \ce{Na+}}{22.989769280~\mathrm{g}~\ce{Na+}}
\frac{1~\mathrm{mol}~\ce{NaCl}}{1~\mathrm{mol}~\ce{Na+}}\frac{58.44277~\mathrm{g}~\ce{NaCl}}{1~\mathrm{mol}~\ce{NaCl}}\cdot 100\% = \fbox{99.99% NaCl}$$

where I'm assuming the label information is actually a bit more precise than reported (exactly 1.5 g of table salt contains approximately 590 mg Na). This is just a BOE (Back of the Envelope) calculation; I should round my answer to three or possibly two significant figures.

In fact there are sometimes other sodium salts present (anti-caking estimates such as Tivolex contain sodium) so this purity is a slight overestimate. One salt company (WA Salt supply) lists "typical" purities for the table salt it provides at 99.72%; this salt contains uses Tivolex. Morton Salt claims that their iodized table salt contains 0.04% dextrose and "less than 0.5% calcium silicate", so the mg Na per serving on the label is a little bit higher than it actually is; I should get something above 99.46%, considering that additional information.

Still, that's pretty pure. The assay for "ultra pure" NaCl linked above gives 99% to 100.5% NaCl. Sigma-Aldrich has reagent-grade NaCl (>98%), ACS grade NaCl (>99.5%) and TraceSelect NaCl for doing heavy metal analysis that is more than 99.999% pure.

Is 99% purity "pure enough for high level experiments in chemistry?" There is no one answer. It depends on what those experiments are. How many significant digits do you need in your results? Are the results going to be affected by the presence of tiny amounts of iodide, iodate, glucose, calcium, or silicate, or the other impurities listed above? Maybe, maybe not. If a specific impurity interferes, find a way to lock up, remove or otherwise neutralize its effects.


What are the molecular requirements for condensation polymerization? I know that in order to have addition polymerization, the monomers must have a C=C double bond. However, I don't know the molecular requirements for condensation polymerization. Must the monomers contain a functional group at each and?

Any help is appreciated. Thanks.

Monomers must contain a functional group at each side (at least bifunctional) and these functions  have to react with each other (like amine and carboxylic acid to yield amide group or alcohol and carboxylic acid to yield ester group,...) .

  We can start with two types of monomers that hold identical functional groups at each side.

One example of that is the reaction of 1,6-diaminohexane with hexandioic acid to yield Nylon-6,6.

  We can also start from one type of monomers with two different functional groups at each side.

One example of that is the reaction of 11-aminoundecanoic acid to yield Nylon-11.

The condition that the reaction should give off a small molecule is no longer valid. The counter example is the formation of polyurethane by reaction of diisocyanates and diols without giving off  small molecules. We prefer name this type of polymerization "a step growth polymerization" in opposite to "chain growth polymerization" (known before as polyaddition).


Is it possible for one specific atom in a molecule to have a non-integer oxidation state? I am wondering if fractional oxidation states of an atom are possible. I'm not referring to cases such as $\ce{Fe3O4}$ or $\ce{Mn3O4}$ where the average oxidation state is fractional, since these actually comprise a mixture of atoms which are individually in the +2 and +3 oxidation states. What I mean is, is it possible for an individual atom in some compound to have an oxidation state of (for example) 2.5?

To me it doesn't seem possible just because of the way oxidation states are defined. However I have seen some sources which state that fractional oxidation states are possible. I would be interested in knowing if there is some weird compound that has fractional oxidation states?

Note: This is not a duplicate of Are fractional oxidation states possible? I want to know if it is possible for an individual atom in some compound to have a fractional oxidation state, not its average fractional oxidation states.

It depends. Consider various radicals such as the superoxide anion $\ce{O2^{.-}}$ or $\ce{NO2^{.}}$. For both of these, we can draw simple Lewis representations:

In these structures, the oxygen atoms would have different oxidation states ($\mathrm{-I}$ and $\pm 0$ for superoxide, $\mathrm{-II}$ and $\mathrm{-I}$ for $\ce{NO2}$). That is the strict, theoretical IUPAC answer to the question.

However, we also see that the oxygens are symmetry-equivalent (homotopic) and should thus be identical. Different oxidation states violate the identity rule. For each compound, we can imagine an additional resonance structure that puts the radical on the other oxygen. (For $\ce{NO2}$, we can also draw resonance structures that locate a radical on both oxygens and another one that expands nitrogen’s octet and localises the radical there.) To better explain this physical reality theoretically, we can calculate a ‘resonance-derived average oxidation state’ which would be $-\frac{1}{2}$ for superoxide and $-\frac{3}{2}$ for $\ce{NO2}$. This is not in agreement with IUPAC’s formal definition but closer to the physical reality.


Can you approximate the change in entropy of a reaction? I found a problem in which I am given a hypothetical reaction in which from 2 liquids (reactants) I get 4 moles of gas and 1 mole of a solid substance (products):

$$\ce{A (l) + B (l)  -> 4C (g)  + D (s)}$$

I am asked about the change in entropy (an approximation of it), so I am curious if there is some kind of approximation for the change in entropy of a reaction based on the change in the amount of substance of gaseous compounds (since I know that the highest influence on entropy is that of gases).

The molar standard entropy of formation ($S_{f}^{\ominus}$) for many common substances is documented in tables. You can approximate the overall change in entropy of the reaction ($\Delta S^{\ominus}_{reaction}$) with with Hess's law:

$\Delta S^{\ominus}_{reaction} = \sum S^{\ominus}_{f,molar,products} - \sum S^{\ominus}_{f,molar,reactants}$

You notice that there is no $\Delta$ in front of $S^{\ominus}_{f,molar}$, since entropy is an absolute physical quantity. Every system has an absolute entropy.

Don't forget to multiply every $S^{\ominus}_{f,molar}$ with the number of moles that you have. This is only an approximation because the molar entropy of a substance depends on Temperature as well, and the reaction likely doesn't take place at standard temperature.

As a sidenote please be aware that the molar entropy of formation for elements is not zero, although the molar enthalpy of formation is!


How to calculate volume for NaOH and drano max gel (includes dilution and pH, pOH) I have a question on an assignment that I don't quite understand. It says:

  What volume of $\pu{0.0100 mol/L}$ $\ce{NaOH(aq)}$ would have the same number of moles of hydroxide ions as $\pu{100.00 mL}$ of drano max gel?

The pH of drano max gel is $14.7$.

My first step was to switch from pH $14.7$ into mol/L. That turned out to be $\pu{5 mol/L}$ of $\pu{OH}$ concentration. Then I multiplied the $5$ by $\pu{100 mL}$ ($\pu{0.1 L}$) and got $\pu{0.5 mol}$. Now I don't understand how to get the volume needed. Would you use the formula for dilution ($C_1V_1 = C_2V_2$)?

Yes.

If you have for drano $C_1 = \pu{5 mol/L}$ and $V_1 = \pu{0.1 L}$ that is $\pu{0.5 mol}$, and for $\ce{NaOH}$ $C_2 = \pu{0.01 mol/L}$ and $C_2 = x$. Using formula for dilution:

$$C_1V_1 = C_2V_2 \quad \implies \quad V_2 = \frac{C_1V_1}{C_2}$$

General tip: With encountering questions like these, always think about what units you have (mol/L, L etc.) and what units you want to go to. Most of the time, it gets clear when you just look at the units.


Determine products of redox-reaction (KMnO4) by Standard electrode potential I want to figure out a way how to determine the products of a redox-reaction by looking at the documented Standard electrode potential (for example here). But with my "way" I have some trouble/contradictions with common redox reaction, in my case the reduction of $\ce{MnO4-}.

Here is my procedure for the reduction of $\ce{MnO4-}$:

Collect all available reactions with their Standard Potential:*
\begin{align}
\ce{MnO4- + 2H2O + 4e- &-> MnO2 + 4OH-}&
E(\text{Red})&= \pu{0.83 V}\tag{a}\label{a}\\
\ce{MnO4- + H+ + e- &-> HMnO4^-}
& E(\text{Red}) &= \pu{0.72 V}\tag{b}\label{b}\\
\ce{MnO4- + 8H+ + 5e- &-> Mn^2+}&
E(\text{Red}) &= \pu{1.23 V}\tag{c}\label{c}\\
\ce{MnO4- + 4H+ + 3e- &-> MnO2 + 2H2O}&
E(\text{Red}) &=\pu{1.6 V}\tag{d}\label{d}
\end{align}
Then figure out the reaction with the highest reduction potential, because according to $E = E(\text{Red}) - E(\text{Ox})$, $E$ is maximal with the biggest $E(\text{Red})$ and the reaction is going to happen with the highest $E$.

In my example, $\eqref{d}$ has the highest potential, so according to my theory $\ce{MnO2}$ is formed in the reduction of $\ce{MnO4-}$ (in combination with for example the oxidation of $\ce{I-}$. But in all textbooks, internet sites, the reduction $\eqref{c}$ is named - not reaction $\eqref{d}$ which I had determined.

I am aware that these potentials are pH- and concentration dependent, I fit the $E(\text{Red})$-data with the Nernst-Equation to $\mathrm{pH} = 3$. Nevertheless, $E(\text{Red})$ of $\eqref{d}$ is always higher in $E(\text{Red})$ than $\eqref{c}$. According to my calculations (graph of $\eqref{d}$) which is dependent on the $\mathrm{pH}$ is always higher than $\eqref{c}$).

I have drawn a diagram which shows the $E(\text{Red})$ potential of the two reaction $\eqref{c}$ and $\eqref{d}$ (see legend). Reaction $\eqref{d}$ has always the higher standard potential - is the better oxidation reaction. Why does nevertheless reaction $\eqref{d}$ happen in acid solutions?

Functions calculated with Nernst-Equation:
\begin{align}
E(\text{pH}) &= 1.507 - \frac{0.059}{5} \cdot 8 \text{pH}\tag{c'}\\
E(\text{pH}) &= 1.7 - \frac{0.059}{3} \cdot 4 \text{pH}\tag{d'}
\end{align}
(other concentrations are assumed as equal)

What you have listed are standard redox potentials. You need two of them to determine a reaction. As I'm not sure what reaction you intend, I'll give an example below, which you can use to work out your own reaction. The diagram helps to sort out what can happen.

Note that a reducing agent is an electron donor, and an oxidising agent an electron acceptor and $$\ce{oxidised form + electrons <=> reduced form}.$$

For example:
The oxidiser $\ce{Sn^{2+}}$ reacts completely with reducer $\ce{Cr^{2+}}$ (at standard redox potentials).
\begin{align}
E^\circ &= \pu{-0.136 V}:&
\ce{Sn^2+ + 2e- &<=> Sn}\\
E^\circ &= \pu{-0.41 V}:&
\ce{Cr^3+ + e^- &<=> Cr^2+}\\
&&
\ce{2Cr^2+ + Sn^2+  &<=>> 2Cr^3+ + Sn v}
\end{align}
With energy
$$E= (\pu{-0.136 V}) - (\pu{-0.41 V}) = \pu{0.274 V}$$
 which corresponds to a negative free energy and hence is spontaneous:
\begin{align}
\Delta G &= -nFE= (-2\times  96485\times 0.274) \pu{J};&
(\pu{1J} &= \pu{1C*V}).
\end{align}

Although the reverse reaction, $\ce{Cr^{3+}}$ to oxidise $\ce{Sn}$ metal could occur, this happens to a negligible extent.


Why is the melting point of magnesium nitride less than that of beryllium nitride? Why is melting point of $\ce{Mg3N2}$ lower than that of $\ce{Be3N2}$ ?
I have two reasonings:

The lattice energy of $\ce{Be3N2}$ is higher than that of $\ce{Mg3N2}$ so accordingly the melting point of the former should be higher
Both of them form ionic nitrides. Since $\ce{Mg3N2}$ has more ionic character it should have higher melting point than $\ce{Be3N2}$

Since both of these are contradicting, I wish to ask which one is correct? And if both are correct in some way, which one should be prioritized over the other?
Can this be determined or is it experimental?

Melting point in a compound with a lattice structure correlates more closely with lattice energy than with ionic character, and lattice energy tends to be greater when you use smaller atoms.  Among alkaline earth nitrides it is not just beryllium versus magnesium that shows this effect; it keeps going.  Melting point data below are from the Wikipedia articles as of April 30, 2021:
$\ce{Be3N2} = 2200°C$
$\ce{Mg3N2} = 1500°C$
$\ce{Ca3N2} = 1195°C$
$\ce{Sr3N2} = 1200°C$
The alkaline earth oxides show a similar trend, for the same reason.  Here the lattice energy of beryllium oxide is slightly depressed because each atom has a coordination number of only four whereas magnesia and its heavier congeners have coordination number six; thus the peak melting point is at magnesia:
$\ce{BeO} = 2507°C$ -- four-coordinate
$\ce{MgO} = 2852°C$ -- six-coordinate, same for the following oxides
$\ce{CaO} = 2613°C$
$\ce{SrO} = 2531°C$
$\ce{BaO} = 1923°C$


Steam distillation of açaí powder I have a steam distillation setup with the flask for boiling the açaí powder with distilled water. Are you able to extract the essential oil from the boiling material?

Wiki on açaí oil

Açaí  oil is obtained by cold pressing.  Steam distillation may be able to extract compounds like vanillic acid but may not get the anthocyanins or the triglycerides.  Many of the health benefits are from the healthy triglycerides and the anthocyanins.


How does a liquid melt a hotter solid? I don't have a degree in anything, but I do know basic rules of science I believe. It seems pretty intuitive that heat will always flow from something hotter into something colder in order to reach an equilibrium.

My understanding of temperature is that it is a measure of how fast something's particles are vibrating.

I've been very interested in the mechanics of salt water and ice spontaneously dropping in temperature because it seems counter intuitive and goes against everything I seem to know.

As far as I know, pure ice begins to melt at 0°C and above. If I were to place an ice cube at -2°C in oil that is -5°C, it wouldn't melt because the oil has less energy than the ice cube. If anything, energy would leave the ice cube to warm the oil and the cube of ice would get colder.

So, if I have salt water at say -2°C, and I put an ice cube in the mix that is pure ice and is 0°C, how is that colder water going to impart energy into something that's warmer (the ice cube). This seems extremely bizarre and seems to go against basic rules of thermodynamics.

I know that adding more salt would lower the melting point even further, but isn't a temperature a temperature? Does the material matter? Do two substances with the same temperature contain the same kinetic energy on the molecular level? If not, then what is temperature even a measurement of?

Thank you.

The cold salt water is going to take heat out of the ice cube just as you say, (and so the salt water will be heated a little to an extent depending on their relative amounts  etc.) The ice surface in contact with the salt water will dissolve as the freezing point of the salt water is lower than that of the ice. Eventually all the ice  will dissolve, but this will have taken some small amount of energy to do as explained next. (I assume little ice and lots of solution but whether all the ice will in fact melt will depend on the exact composition and could be found from a temperature vs. mole fraction plot for a solid liquid equilibrium at constant pressure. There could be solid salt-ice & ice, solution & ice, solution & solid salt-ice or just solution.)

Dissolving a solid in a liquid, say salt in water, can cause a drop in temperature (provided the container is insulated) because it takes  energy to dissolve compounds and this can only come from the liquid. (The ions or molecules in the solid must be attracted to one another, by ionic or van der Waals forces, and so to be separated takes energy). The measure is called the heat of fusion (or melting), or enthalpy of fusion which is a positive quantity, for common salt it is $\approx 30 \pu{~kJ mol^{-1}}$ and of ice $\approx 6 \pu{~kJ mol^{-1}}$.


Why calcium chloride is used to melt ice over sodium chloride I'm not quite sure I understand this. My question is:

  Calcium chloride is a salt used widely to melt ice on sidewalks and roads.  Explain why one mole of $\ce{CaCl2}$ would be more effective than one mole of $\ce{NaCl}$.

So from everything I've learned, $\ce{NaCl}$ would be more effective for the melting, because calcium chloride dissociates into three ions and sodium chloride dissociates into two, so that would make the boiling point of water with calcium chloride higher.  Therefore, it shouldn't be used to melt ice over sodium chloride, as it would make the water have a higher boiling point so wouldn't melt it as well!

That one mole of $\ce{CaCl2}$ is more effective in melting ice than one mole of $\ce{NaCl}$ is explained by the van 't Hoff factor. The freezing point depression of a solution is calculated by
$$\Delta T=K_fbi$$
where $K_f$ is a cryogenic constant which is specific to the solvent, $b$ is the molal concentration of the solute, and $i$ is the van 't Hoff factor, which indicates number of solute particles. In this case, it is the number of ions produced upon dissociation. Because $K_f$ and $b$ are constant, and $i=3$ for $\ce{CaCl2 \rightarrow Ca^2+ + 2 Cl-}$ and $i = 2$ for $\ce{NaCl \rightarrow Na+ + Cl-}$ the freezing point depression is greater for $\ce{CaCl2}$.

The reason $\ce{CaCl2}$ is widely used as melting agent may have more to do with keeping the concrete intact. Concrete is composed of many calcium-containing species. When water flows over the concrete, the calcium can be leeched out, and the concrete becomes brittle. The higher the calcium content in the water, the less leeching occurs, and the concrete remains structurally sound.


Why does a complete discharge destroy a Li-Ion-Battery? I read that completely discharging a Lithium Ion Battery is a very bad idea because it will lose plenty of capacity. But why?

I know that the reaction (fot the LiFePO4 battery) is:
$$ \ce{LiFePO4 + C6 <=>[1 e-] FePO4 + LiC6 } $$

So at a complete discharge, we only have our electrodes made of $\ce{LiFePO4}$ and $\ce{C6}$. Why shouldn't this react to the products of the reactions above?

When overcharging the battery I could find a solution, why it destroys the latter. Maybe all the graphite is ionized and so, the electrode is in solution. This would be a very big problem, but the graphite will only ionize if it can react with a cation (like lithium), so a small excess of graphite should prevent this.

What about other Li-Ion-Batteries?

This is what I read on the internet:

  The discharge voltage is around 2.5 V depending on the cell type; It
  must not fall below, otherwise, the cell will be destroyed by
  irreversible chemical reactions. However, many electronic devices
  already switch off at significantly higher voltages, eg 3.0 V.

What do they mean with the irreversible chemical reactions? Which ones?

In a Lithium ion cell, the anode material can dissolve in the electrolyte, and then on recharge, precipitate in the midst of the electrolyte and insulating membrane, short-circuiting the cell. Further, the cathode material can release oxygen, which migrates away and does not get reincorporated on charging.

Another problem with most secondary (storage) cells, Pb-acid as well as Li, is physical destruction of the structure on over-charging and under-charging. During the charge-discharge cycle, there is some change in volume of the electrodes, which causes them to spall, shedding pieces of the active material. This leads to loss of capacity and sometimes a short-circuit.

Even a fully charged cell deteriorates gradually... but increased discharge increases the growth of precipitates. Li-ion cells self-discharge about 2%/mo, (half-life of ~2 years), so a charged cell is quite stable, with no perceptible growth of whiskers or precipitate for years.


Comparing Basic strengths of different species I was doing questions on organic chemistry. I came across tonnes of questions  where I was required to compare basic and acidic strengths.

Comparing acidic strengths is fine — I remove an $\ce{H+}$ and compare the stability of the anion because my teacher told me a strong acid has a stable anion and explained the reason to me from an equilibrium point of view.

When it comes to comparing basic strengths what do we do?

Can I add an $\ce{H+}$ to the species and compare the resulting stability by resonance, hyperconjugation and inductive effect and then can conclude the more basic one will have a stabler cation?

Really baffled.
Please do correct me if I am mistaken in my fundamentals.

Comparing acidic strengths is fine - I remove an H+ and compare the stability of the anion coz my teacher told me strong acid has a stable anion and explained me the reason from equilibrium point of view.
  When it comes to comparing basic strengths what do we do ?

You follow the same general idea - the stronger the base, the weaker its conjugate acid. Consider a generic base such as $\ce{B-}$ reacting with water, the acid in this case.

$\ce{B- + H2O <=> HB + HO-}$

The conjugate acid of $\ce{B-}$ is $\ce{HB}$. And the stronger that $\ce{HB}$ is as an acid, the more likely it is to protonate hyroxide ion and favor the left side of the equilibrium, thereby making the base weak.

  Can I add an H+ to the species and compare the resulting stability by resonance ,hyperconjugation and inductive effect and then can I conclude the more basic one will have a stabler cation?

Yep, that's it.


What's the name of this electrolysis device? I've seen it in a couple of videos, it's used to electrolyze water in 2 different containers at the same time to make a comparison and thus is very handy.

What is it called (besides just a hydrolizer)?

Based on this article, in which the authors describe a system that can do 25 cells simultaneously, the term 'Parallel Electrolyser' would seem appropriate.


Why do alcohols and ethers have approximately the same solubility in water but different boiling points? In Morrison & Boyd, I found this question:

Butan-1-ol (b.p. $118~\mathrm{^\circ C}$) has a much higher boiling point than its isomer diethyl ether (b.p. $35~\mathrm{^\circ C}$), yet both compounds show the same solubility ($8~\mathrm{g}$ per $100~\mathrm{g}$) in water.

How do you account for these facts?

Now, as far as I know, the boiling point depends on the inter molecular forces. Since in the alcohol, there is hydrogen bonding in the O-H group, it has a higher boiling point than the ether. But in an aqueous solution, the solubility is also related to hydrogen bonding. Since there is more hydrogen bonding in the alcohol, I would expect it to have a higher solubility in water, but that is not the case. So how do we explain it? I am completely confused by this question. What is the answer?

Ivan's comment is spot on. It take two to tango (or hydrogen bond) - a donor $\ce{Z-H}$, where $\ce{Z}$ is some electronegative atom (usually $\ce{O},\ \ce{N},$ or $\ce{F}$) and an acceptor (something with just $\ce{Z}$). 1-Butanol can hydrogen bond with itself, since it has an $\ce{O-H}$ group capable of being both donor and acceptor. Deithyl ether cannot hydrogen bond with itself since it lacks an $\ce{O-H}$ group. However, both can form hydrogen bonds with water. Thus, 1-butanol has the higher boiling point, but they both are similarly soluble in water.

Actually, Morrison and Boyd oversimplify the solubility difference (and there is one). 1-Butanol is slightly more soluble in water than diethyl ether, perhaps because it can interact with water in more than one way, although the dipole moment of the alcohol group is certainky larger than that of the ether group as well.

The NIH PubChem entries for diethyl ether and 1-butanol list several solubilities in water, but I want to focus on two entries that share the same source for both compounds:

  Solubilities in water at $25\ ^\circ \mathrm{C}$ from the NIH TOXNET Hazardous Substances Database

$$\begin{array}{c|c|c|c}
\mathrm{substance} & \mathrm{Solubility\ in\ water\ (g/100\ mL)} & \mathrm{density\ (g/mL)} &\mathrm{volume\ percent}\\ \hline
\text{1-butanol} & 6.8 & 0.8098 & 8.4\\
\text{diethyl ether} & 6.0 & 0.7134 & 8.4
\end{array}$$

  Solubilities in water at $20\ ^\circ \mathrm{C}$ from the International Labor Organization's International Chemical Safety Cards

$$\begin{array}{c|c|c|c}
\mathrm{substance} & \mathrm{Solubility\ in\ water\ (g/100\ mL)} & \mathrm{density\ (g/mL)} &\mathrm{volume\ percent}\\ \hline
\text{1-butanol} & 7.7 & 0.8098 & 9.5\\
\text{diethyl ether} & 6.9 & 0.7134 & 9.7
\end{array}$$

Note that I have used the solubility data to calculate percent compositions by volume in both cases. It was and still is common to report low solubilities as percents composition. It also is and was common to see percents composition listed without specifying whether they were mass/mass (which should be the default), mass/volume (convenient for dilute solutions), or volume/volume (convenient for solutions, but unhelpful due to colligative properties).

At both temperatures, the data supports that 1-butanol and diethyl ether have nearly the same percent solubility by volume. Morrison and Boyd is an older generation of textbook, first published before the world wide web. If they found a source of solubility data for these compounds that listed percents (but did not specify volume/volume), then the authors made a reasonable assumption that those percents were mass/mass or mass/volume and reported the same solubility.

The slight solubility edge for the n-alkanols persists over a larger range of compounds. The table below lists the solubilities in grams per 100 mL of water of the n-alkanol and the symmetric ether, with data drawn from the NIH Hazardous Substance Database.

$$\begin{array}{c|c|c|c|c}
\text{sol. in water in g/100 mL} & \ce{C2} & \ce{C4} & \ce{C6}  & \ce{C8}\\ \hline
\text{n-alkanol} & \text{miscible}  & 6.8 & 5.9  & 0.54 \\
\text{symmetric ether} & 4.6  & 6.0 & 4.9 & 0.30
\end{array}$$


Why was my soap measured as acidic? I've always been told soap is basic in pH. A few years back in a high school chemistry class, we measured the pH of various substances with universal pH indicator. We measured some liquid hand soap and it came out as acidic, which puzzled even the teacher. Unfortunately I have no idea what brand it was and I still don't know why it came out as acidic.

I learned yesterday that the fundamental reaction that produces soap happens when a base reacts with an oil chain which breaks down the glycerol backbone. Depending on which exact reaction path is being used to make the soap, some base may be added to ensure no oils remain unreacted. After the reaction is complete, some acid can be added to neutralize the base. Could it be likely that extra acid was added to the soap to neutralize the remaining base and the solution was left acidic? What other possible ideas could explain why the soap was acidic?

It has been claimed that skin has a pH of about 5.5. Of course, this must be the sweat coming out of your skin, since skin isn't a liquid. So many body washes add citric acid to their formulas to be milder to tender skin.

For washing after poison ivy or car grease, pH 11 is much more effective.


Oxide Anions in Redox Consider the following unbalanced redox equation:

$\ce{NO_2^- +e^- -> NO3^-}$

This occurs in acidic media.

My professor would note that an negative two oxidation state oxygen "disappears" from left to right.

Oxide anions are strong bases, and since this process occurs in acidic media, and since an oxide anion has "disappeared," it must have been consumed by the strongest acid present - hydronium ion. Specifically, two equivalents of hydronium ion. These two equivalents of hydronium ion react with oxide anion to produce three waters.

Thus we balance the equation as such.

$\ce{NO_2^- +2e^- +2H_3O^+ -> NO3^- + 3H_2O}$

Now, what is the validity of this method, given that redox reactions likely don't involve an oxide anion intermediate? I can find no mention of any $\ce{O^2-}$ intermediates in redox literature.

If that is the case, how can this method still yield a redox equation that is valid in terms of mass and charge? Is there some likely intermediate that is close enough in character to the oxide anion?

Your equation is wrong. In acidic medium, water is the one which gives an oxygen to the nitrite. And that should be an oxidation half-reaction (as nitrite oxidizes to nitrate), but you wrote a reduction where two electrons are spent, not released.


How can you use ICE tables to solve multiple coupled equilibria? If I have a problem involving multiple coupled equilibrium reactions, such as

  Calcium fluoride, $\ce{CaF2}$, has a molar solubility of $\pu{2.1e−4 mol L−1}$ at pH = 7.00. By what factor does its molar solubility increase in a solution with pH = 3.00? The p$K_{\mathrm{a}}$ of $\ce{HF}$ is 3.17.

The relevant reactions are:

$$\ce{CaF2(s) <=> Ca^2+(aq) + 2 F-(aq)}$$ and
$$\ce{HF(aq) <=> H+(aq) + F-(aq)}$$

They are coupled because fluoride occurs in both of them.

Is there a way to use ICE tables to organize the information (stoichiometry, initial concentrations, mass balance) as a way to solve the problem?

For example, I could try to set up one ICE table for each reaction (the column for $\ce{H+}$ is strange because in the problem, the pH is set to a value through an unspecified mechanism):

$$
\begin{array}{|c|c|c|}
\hline
 &[\ce{Ca^2+}]  & [\ce{F-}]  \\
\hline
I &  \pu{2.1e−4} & \pu{4.2e−4} \\
\hline
C &  +x  & +2x  \\
\hline
E &  \pu{2.1e−4}+x & \pu{4.2e−4}+2x \\
\hline
\end{array}
$$

and

$$
\begin{array}{|c|c|c|}
\hline
 &[\ce{HF}]  & [\ce{H+}]  & [\ce{F-}]  \\
\hline
I &  0 &  \text{N/A} & \pu{4.2e−4} \\
\hline
C &  +x &\text{N/A} & -x \\
\hline
E &  +x & 10^{-3.00} &  \pu{4.2e−4} - x\\
\hline
\end{array}
$$

However, how do the fluoride concentrations "talk to each other" in the two tables? Is the $x$ in one table the same as the $x$ in the other table?

The way to use ICE tables in this case would be to combine the ICE tables, and use "$x$" for the changes due to one reaction and "$y$" for the changes due to the other. For each reaction, you have one unknown (how far it reacted, i.e. $x$ and $y$), so you need two pieces of information to solve it, in this case the two equilibrium constants.

Here is the combined ICE table:

$$
\begin{array}{|c|c|c|c|c|}
\hline
 &[\ce{Ca^2+}]  & [\ce{F-}]  & [\ce{H+}]&[\ce{HF}]   \\
\hline
I &  \pu{2.1e−4} & \pu{4.2e−4} & \text{N/A} & 0 \\
\hline
C &  +x  & +2x-y & \text{N/A} & +y  \\
\hline
E &  \pu{2.1e−4}+x & \pu{4.2e−4}+2x-y & 10^{-3.00} & +y \\
\hline
\end{array}
$$

Now, you can use the equilibrium constants for the two reactions to solve for $x$ and $y$, giving you the equilibrium concentrations.

What the combined table shows is that the fluoride concentration depends on both reactions, so you can't first deal with one equilibrium and then with the other but have to solve a system of two equations with two unknowns.

Once you combine the tables, you could also have initial conditions where nothing is dissolved yet, to get easier expressions (I'm using $p$ and $q$ because they will be different from $x$ and $y$):

$$
\begin{array}{|c|c|c|c|c|}
\hline
 &[\ce{Ca^2+}]  & [\ce{F-}]  & [\ce{H+}]&[\ce{HF}]   \\
\hline
I &  0 & 0 & \text{N/A} & 0 \\
\hline
C &  +p  & +2p-q & \text{N/A} & +q  \\
\hline
E &  +p & +2p-q & 10^{-3.00} & +q \\
\hline
\end{array}
$$


Solvents for tartrates / tannins that won't attack glass At the bottom of my wine decanter (which due to its shape is fantastically hard to clean mechanically) a layer of reddish brown precipitate forms. I believe this is probably potassium bitartrate crystals, which are coloured reddish brown by the presence of tannins, which in essence is a polyphenol.

I would like to know whether there is a chemical that I could use to clean the decanter with minimum mechanical action. I have tried water (with and without soap), alcohol (in the form of vodka), isopropanol without success. I have yet to try acetone. Preferably the solvent should not be greatly toxic (at least I'd like to be able to wash it out with water and reuse the decanter), and not react with high lead-content glass (which is what decanters are made out of) - hydrofluoric acid is thus to be avoided (as usual). It is unsurprising alcohol doesn't work as the reason tartrates precipitate is apparently because they are not soluble in alcohol. It would also be useful if I could purchase the solvent outside a lab environment.

My understanding is that potassium bitartrate is more soluble in acid solutions. I've tried citric acid (made up from powder) and vinegar, but the effect if any is modest.

I don't much care if the method used is to react with (rather than dissolve) the precipitate.

(Note this question is in part a thought experiment. I realise that in fact the best way to clean decanters is generally mechanical despite the difficulties, and I'm familiar with the copper beads method and the egg-shell method, as well as no end of sponges on bendy sticks. However, I'd like to know whether there is a chemical way. Also note that I added an 'equipment' tag as it's possible that some know-how from cleaning chemistry glassware might be useful here)

The CRC Handbook indicates that potassium d-tartrate and dl-tartrate are soluble to the extent of 100-150 grams per 100 mL H2O vs about 0.5 grams for the bitartrate. Rather than using acid to clean the decanter, alkali would neutralize the bitartrate to the much more soluble tartrate . Washing soda (Na2CO3) should not be too alkaline, but even sodium bicarbonate (NaHCO3) might work. Neither is expected to be harmful at the level that might exist after rinsing.

The tannins or other organic might slow things down. Bleach (NaClO) is alkaline, and although bleach is toxic if ingested, it should present no problem if the decanter is rinsed. Water could be flushed around inside the decanter, then tested with litmus paper. Or the decanter could be rinsed with vinegar and a little honey to remove the bleach: the vinegar makes the liquid acidic (HClO) and the honey gets oxidized to gluconic acid, consuming all the bleach - leaving a sweet, tart taste.


How to manage complex equilibria when studying a mecanism? Consider a complex multistep reaction mechanism with several intermediates. There are multiple possible pathways as well as equilibria in solution. One wants to study the reaction mechanism using computational chemistry.

It is easy to calculate the $\Delta G$ and $\Delta G^\ddagger$ of individual reactions, but how do you take in consideration all the possible equilibria?

For example, consider an aldol reaction in presence of water (taken from Wikipedia):

Acetone can be either as a ketone, a protonated ketone, an enol or an enolate. This means that there are 16 possible pairs (although only 4 which make sense). These will yield different intermediates, which can dehydrate in different ways (ex. by protonation of the hydroxyde first or not, through a ketone or an enol or an enolate...)

In this case, many possibilities could be eliminated as too high in energy just by general knowledge of chemistry, but if you don't know enough about the system, how do you tackle this problem? How do you handle all the equilibria possible?

If you know what the final equilibrium products are or may be, and you know how much mass you start with, either as elements or molecules, you can calculate the Gibbs energy of the products as a function of (unknown) concentrations (assuming constant temperature and pressure) and then minimize the Gibbs energy with respect to those concentrations. You have to constrain the solution to conserve all of the elements that you began with, and constrain it to have all concentrations be non-negative.


Aromaticity-Bromonium Ion Is the bridged ion intermediate aromatic or not?
What is the way to quickly identify aromaticity?

A chemist would say that the reaction proceeds through an "aromatic transition state". The bridged, chloronium ion transition state drawn in your question involves 2 electrons.  No electrons from the electrophile were used, just 2 electrons from the pi-bond that it added to (draw the resonance structures for the bridged intermediate, each has 2 electrons in the $\ce{C-Cl}$ bond and no electrons in the carbocation p-orbital).  If a transition state contains 4n+2 electrons, then it is said to be aromatic. This case has n=0.

Here is another example. The top line in the figure below shows a suprafacial 1,5 hydrogen shift.  6 electrons are involved here, so the transition state is also referred to as aromatic (4n+2, n=1).

image source (look far down on the page)

Like the bridged halonium ion, this reaction is also very common because it too has an energetically favored (aromatic) transition state.

The second line in the figure depicts a suprafacial 1,3 hydrogen shift.  4 electrons are involved here, so the transition state does not fit the 4n+2 rule.  The transition state is said to be anti-aromatic.  Because it is disfavored energetically (anti-aromatic), the suprafacial 1,3 hydrogen shift does not occur (at least not via a low-energy concerted pathway).

Extra Credit: I don't want to jump to far ahead, but let me briefly say, that what I've written above applies to Hückel transition states, opposite rules apply to Möbius transition states.  See this earlier answer, and references therein, for an explanation of the Hückel and Möbius concepts.


Why is decomposition of ammonia gas on quartz a 1st order reaction? In general, decomposition of a gas (like ammonia) on a metal surface (like platinum) is a zero order reaction.
This is because the gas is adsorbed on the metal surface due to pressure and the initial concentration is immaterial.
But, why and HOW does it behave differently with quartz?

Firstly a correction and general clarification: your first general assertion is not true. The metal surface acts as a heterogenous catalysts (supported by paper linked later), and the kinetics can be sufficiently modeled by the Langmuir-Hinshelwood model for a monomolecular reaction in which the rate determining step is the surface reaction

$$ \ce{A_{ad} -> B_{ad}} $$

so

$$ r = \frac{\mathrm{d}[B]}{\mathrm{dt}} =k_r [A_{ad}] = k_r \theta_A $$

$\theta_A$ is the surface coverage of A. Applying the steady state approximation to $\theta_A$ over the adsoprtion, desorption and reaction steps and using the Langmuir adsorption model gives

$$ \theta_A = \frac{k_{ad}\:p_A}{k_{ad}\:p_A+k_{de}+k_{r}}$$

so the rate is given by

$$r = k_r \; \frac{\bar{k} \: p_A}{1+\bar{k} \: p_A}$$

where $\bar{k}$ is a complex rate constant. For low partial pressures of A, the reaction thus approximately follows first order kinetics and zeroth order for high $p_A$. Here is a nice paper verifying this for several metals.

Now lets get to the quartz. The authors of the paper also mention a possible bimolecular reaction mechanism, of which the key step is the formation of hydrazin from two adsorbed ammonia molecules. They disregard this mechanism due to experimental evidence, but lets apply some metal organic chemistry here.

$\ce{NH3}$ is a great $\sigma$ donor, so a $\sigma$ acidic metal center will happily bond. A simplistic view might be that the nonbonding lone pair is being donated to the metal - taking a look at a MO diagram of $\ce{NH3}$ you will however find that the HOMO (from which we donate) is slightly bonding.

Source

Therefore, this particular $\sigma$ donation will decrease N-H bond strength, which supports the monomolecular reaction mechanism for such a surface (this does put us somewhere in the grey zone between heterogenous and homogenous catalysis).

$\ce{SiO2}$ is not a substance that comes to mind when we are looking for a $\sigma$ acid. So maybe, under these conditions the bimolecular mechanism becomes the preferred pathway. Bimolecular reaction kinetics of this sort can be modeled by Langmuir-Hinshelwood (both reacting species adsorbed) or Eley Rideal (one adsorbed, one in gas phase). Langmuir-Hinshelwood would give us the boundary cases of second order rate at low $p_A$ and zeroth order at high $p_A$, and I could find no experimental results supporting that anywhere. An Eley-Rideal model would give us a rate law of the form

$$ r = k_r \; \frac{\bar{k} \: p_A^2}{(1+\bar{k} \: p_A)}$$

which would give us first order behaviour for high partial pressures of ammonia! Here is a paper finding experimental data supporting this kinetic model. Possibly the reaction product in the rate-determining step is not hydrazin, but an intermediate hydrazinium which then decomposes.

It thus seems plausible that ammmonia decomposition on a quartz surface follows a bimolecular Eley-Rideal mechanism, compared to the monomolecular mechanism on a metal surface - as a result of the different interactions between surface and adsorbent.


Molecular orbital diagram and irreducible representations for dinitrogen I'm trying to understand how to draw molecular orbital diagrams, since I'd like to use the molecular orbitals to determine the total state of the molecule (i.e. $A_1$, $B_1$ etc.), so I prefer the notation with irreducible representations of orbitals like it's shown in this video for $\ce{H2O}$.

The problem is that I'm not sure how to make molecular orbitals for $\ce{N2}$ and $\ce{N2^+}$.

I know that $\ce{N}$ has atomic orbitals $\ce{1s^{2} 2s^{2} 2p^{3}}$. $\ce{N2}$ is a linear molecule, so it belongs to the $D_{\mathrm{\infty h}}$ symmetry point group, but we can descend in symmetry, so we'll use $D_{\mathrm{2h}}$.

But now I have no idea how to proceed. Should I work with both nitrogen atoms together, i.e. with group orbitals, similarly to the case with two hydrogen atoms in $\ce{H2O}$? Or am I supposed to work with them separately?

So could you please write down one of the diagrams using irreducible representations and explain it step-by-step? I think it would be really useful here, as I couldn't find any source with some general explanation.

Because this is a diatomic molecule, there are no group orbitals. Put another way, the group orbitals are the molecular orbitals. Knowing the nitrogen atomic orbitals (AOs) and their irreducible representation (irrep) labels is enough.

Since we'll work in the $D_{\mathrm{2h}}$ point group, we need its character table:

$$\begin{array}{c|cccccccc|cc} \hline
D_\mathrm{2h} & E & C_2(z) & C_2(y) & C_2(x) & i & \sigma(xy) & \sigma(xz) & \sigma(yz) & & \\ \hline
\mathrm{A_g} & 1 & 1 & 1 & 1 & 1 & 1 & 1 & 1 & & x^2,y^2,z^2 \\
\mathrm{B_{1g}} & 1 & 1 & -1 & -1 & 1 & 1 & -1 & -1 & R_z & xy \\
\mathrm{B_{2g}} & 1 & -1 & 1 & -1 & 1 & -1 & 1 & -1 & R_y & xz \\
\mathrm{B_{3g}} & 1 & -1 & -1 & 1 & 1 & -1 & -1 & 1 & R_x & yz \\
\mathrm{A_u} & 1 & 1 & 1 & 1 & -1 & -1 & -1 & -1 & & \\
\mathrm{B_{1u}} & 1 & 1 & -1 & -1 & -1 & -1 & 1 & 1 & z & \\
\mathrm{B_{2u}} & 1 & -1 & 1 & -1 & -1 & 1 & -1 & 1 & y & \\
\mathrm{B_{3u}} & 1 & -1 & -1 & 1 & -1 & 1 & 1 & -1 & x & \\ \hline
\end{array}$$

Using the nitrogen atom's electron configuration $\ce{1s^{2} 2s^{2} 2p^{3}}$ as our minimal basis, no d-orbitals will be present, so we can assign irreps to each AO right away, in part because we assume the principal rotation axis (the one of highest order) is aligned along the z-axis:

$$\begin{array}{cccc}
\hline
\mathrm{s}     & \mathrm{p}_x    & \mathrm{p}_y    & \mathrm{p}_z    \\
\hline
\mathrm{A_{g}} & \mathrm{B_{3u}} & \mathrm{B_{2u}} & \mathrm{B_{1u}} \\
\hline
\end{array}$$

Then, proceed by starting to form the standard MO diagram for a diatomic, but add the irrep labels to each AO:

Note that I haven't spaced the energy levels properly; the $\ce{1s}$ should be much lower than it is relative to the $\ce{2s}$. More on this later. Regardless, because this is a homodiatomic, all AOs will mix at each energy level, even the core $\ce{1s}$ AOs. Then, attempt to form the MOs:

This is probably wrong, but it's a starting point. Again, energy levels will be discussed later. Finally, add symmetry labels to each MO, remembering that

MO symmetry is derived from AO symmetry,
numbering is consecutive within each irrep, not with respect to the set of all MOs, and
lowercase is used to signify that these are MOs; uppercase is for the irreps themselves.

This is our final MO diagram for $\ce{N2}$. To form the diagram for $\ce{N2^+}$, remove an electron from $\mathrm{4a_g}$. This ignores orbital relaxation effects, but for the purposes of working this out on paper, it should be fine.

Now for the matter of relative and absolute energy levels. It is probably possible to get the correct relative ordering of the MO energy levels. Here, I assume that since AO mixes with an identical partner on the other atom, the splitting for $\ce{1s}$ would be the same as $\ce{2s},~\ce{2p_z}$, etc. Since I drew the $\ce{1s}$ too high, the $\mathrm{3a_g}$ is almost certainly too high, and perhaps should even go below what is labeled as $\mathrm{2a_g}$. The way to confirm this, and the only way to get absolute energy levels, is to perform a quantum chemical calculation. Since we've used a minimal basis for the drawing, we'll stick a minimal basis in the calculation. Here is a Psi4 input file:

molecule {
    N 0.0 0.0 0.0
    N 0.0 0.0 1.0975
}

set {
    basis sto-3g
    scf_type direct
    df_scf_guess false
    cubeprop_orbitals [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]
}

e, wfn = energy('hf', return_wfn=True)
cubeprop(wfn)

and from its output:

Orbital Energies (a.u.)
-----------------------

Doubly Occupied:

   1Ag   -15.518067     1B1u  -15.516124     2Ag    -1.442840
   2B1u   -0.722491     1B2u   -0.573123     1B3u   -0.573123
   3Ag    -0.539495

Virtual:

   1B2g    0.281319     1B3g    0.281319     3B1u    1.123476

It looks like I might have messed up the diagram, because the ordering isn't what's expected (why isn't $\mathrm{B_{1u}}$ degenerate with the other two p-orbitals?), and the virtual degenerate p-orbitals have gerade symmetry. Time to plot!

In my haste, I forgot an important point. When you make the MOs from AOs, you're making two linear combinations:

\begin{align}
\psi_{s} &= \frac{1}{\sqrt{2}} (\chi_{l} + \chi_{r}) \\
\psi_{a} &= \frac{1}{\sqrt{2}} (\chi_{l} - \chi_{r}),
\end{align}

which means that an antisymmetric combination of two s orbitals will look like a $\mathrm{p}_z$ orbital. This doesn't fully explain everything, but is a good starting point. The lesson up to now is that while drawing the diagrams by hand is a useful exercise, it is not enough to do only that when the goal is to perform a correlated ab initio electronic structure calculation.


What forces hold ink on paper? On regular paper, ink is absorbed, while on photopaper ink stays at the surface and dries for a longer time. What forces act in these two different cases?

(see also related questions about pencil marks and colorants)

There's actually two questions being asked here:

  What is holding ink on paper?

and

  Why do some inks take longer to dry?

Ink sticks to paper because it has a high affinity for the medium it's being applied to. To group things into simple categories we could say, there are hydrophobic inks (non-polar inks that don't like water) and hydrophilic inks (polar inks that do like water). As with the inks, surfaces can also fall into this category. When ink is thrown onto a surface that it doesn't like, it aggregates, clumping together to avoid all interaction with the surface.

A simple illustration showing why this is unfavorable can be seen with water interacting with lipids. In the picture below, water is forming ordered cages around the hydrophobic part. These cages cause a decrease in entropy, and by the 2nd law of thermodynamics, this is unfavorable. You can think of the ink as the water, and the hydrocarbon chain of the lipid as the surface.

When inks do favor their substrate (surface) they will attach via h-bonding, london dispersion forces, and/or other inter-molecular forces.

As for the second question, the answer relies more on what the ink is made out of. According to the wiki on inks:

  Ink can be a complex medium, composed of solvents, pigments, dyes, resins, lubricants, solubilizers, surfactants, particulate matter, fluoresces, and other materials.

Generally, the ink, which has become mobile in its solvent, will remain mobile until the solvent evaporates. So you'll find that the dry time is dependent upon the rate of evaporation. That's why the dry times can be longer for some oil-based pens.


Why do phase changes occur? So I was doing a Gizmo for my Science class and I came across this question: "Why do phase changes occur?" I could not find anything about it in my resources. Does anybody here know the answer?

They occur because that's just how physics works. There isn't anything profound going on. Let's take water as an example. Water boils because enough energy has been introduced into the system such that molecules are able to break away from the interactions keeping them in the liquid state.


Why does heating transform the aquacomples of slightly acidified cobalt(II) chloride solution to the chlorido complex? There is a famous demonstration where, if you mix a $\ce{CoCl2}$-solution with the right amount of $\ce{HCl}$ you will end up with the pink $\ce{[Co(H2O)6]^2+}$ complex but when heating it to boiling temperatures it will change to blue as the $\ce{[CoCl4]^2-}$ complex is formed. This reaction is reversible as the mixture is cooled again. Now literature always describes that this works because that shift towards $\ce{[CoCl4]^2-}$ is endothermic, but why is this the case?

What causes one to be more stable here than the other one but being substituted when heated?

I have some ideas but I can't really compare them:

First of all there should actually be an entropic advantage towards the chlorido-side, as 6 water ligands leave, meaning that the exothermic step here is against the natural direction of the entropy(?).
Second, perhaps the chlorido ligand is the better leaving group. I know usually $\ce{H2O}$ is a good leaving group in organic molecules but then its usually a $\ce{H2O+}$. So perhaps chloride is the better leaving group here.
Maybe there is a kinetic effect, that it is harder to displace 6 ligands than just four ligands
Or, as much more water is present it just shifts towards the water side because of the huge excess of water?
Chloride is a weaker ligand than water. But on the other hand the iron-fluorido complex is quite stable in water and towards other ligands.
So could it be due to the different ligand fields? The splitting in the tetrahedron is much smaller than in the octahedron. Although I don't know how this actually affects the stability.

So those were the ideas I had but I can't really find a common thing among them besides the excess of water which often causes an equlibrium to shift.

By the way $\ce{[Cu(H2O)6]^2+}$ and $\ce{[CuCl4]^2-}$ do the exact same thing.

Does anyone have an idea why this reaction is endothermic?

You may know the formula for Gibbs Free Energy $\Delta G$. This is taught as:
$$\Delta G = \Delta H - T\Delta S\tag{1}$$

In this case, we would be observing $\Delta G$ for equilibrium $(2)$.

$$\ce{[Co(H2O)6]^2+ + 4 Cl- <=> [CoCl4]^2- + 6 H2O}\tag{2}$$

Within the temperature range between $\pu{25^\circ C}$ and $\pu{100^\circ C}$, we should not expect the enthalpic term $\Delta H$ to change significantly. In fact, it is usually correct at first or even second approximation to consider it constant. If we do, the only thing in $\Delta G$ that changes with temperature in equation $(1)$ is the $-T\Delta S$ term. It basically implies that a higher temperature will increase the entropic contribution.

You already noted that the product side is favoured entropically since five free-moving particles are turned into seven in the course of the reaction. Higher temperatures should thus favour the product side.

One more thing: since we have established that $\Delta S$ is positive, we can immediately see that $\Delta H$ must be positive, too; i.e. the reaction must be endothermic. This is because $T$ is positive by definition and therefore $-T\Delta S$ is always negative. However, the only way to tip an equilibrium to the opposite side is to change the sign of $\Delta G$ — which can only work if a positive $\Delta H$ can offset a negative $-T\Delta S$.


Standard state of a substance - definition In chemistry, the standard state of a material (pure substance, mixture or solution) is a reference point used to calculate its properties under different conditions. IUPAC recommends using a standard pressure $p^\circ=10^5\ \mathrm{Pa}$. Strictly speaking, temperature is not part of the definition of a standard state. For example, the standard state of a gas is conventionally chosen to be unit pressure (usually in bar) ideal gas, regardless of the temperature.

Source: Wikipedia

Now, as we can see the definition for the standard state of gas is independent of its temperature as well as volume. But the properties of gas depend on those and standard state is a reference point for calculating properties. So shouldn’t the definition of standard state for an ideal gas include temperature or volume?

The IUPAC definition of standard state is:

  State of a system chosen as standard for reference by convention. Three standard states are recognized: For a gas phase it is the (hypothetical) state of the pure substance in the gaseous phase at the standard pressure p = p°, assuming ideal behaviour. For a pure phase, or a mixture, or a solvent in the liquid or solid state it is the state of the pure substance in the liquid or solid phase at the standard pressure p = p°. For a solute in solution it is the (hypothetical) state of solute at the standard molality m°, standard pressure p° or standard concentration c° and exhibiting infinitely dilute solution behaviour. For a pure substance the concept of standard state applies to the substance in a well defined state of aggregation at a well defined but arbitrarily chosen standard pressure.

Specifying that the standard state is a specific temperature and pressure, would mean that the standard state is not even remotely realizable for many gases.  For example, how could you have pure water vapor at standard temperature and pressure?


How to designate the position of an amine substituent in an alkyl-chain? Why is 1-phenylethanamine named like it is using IUPAC rules? Why isn't it 1-phenylethan-1-amine?

I thought you had to designate where the amine group would be? I.e. the first position on the ethane-chain.

In general usage, the practice of omitting locants when there is no ambiguity is widespread. In particular, the locant ‘1’ is often omitted when there is no ambiguity.

Clearly, the locant ‘1’ for the amine group in unsubstituted ethanamine can be omitted since the isomer ‘ethan-2-amine’ does not exist.

Furthermore, the locant ‘1’ for the amine group can be omitted in substituted ethanamines provided that the amine group is the principal characteristic group (i.e. the characteristic group chosen for citation at the end of the name by means of a suffix or a class name), for example 2-chloroethanamine.

Hence, the omission of the locant ‘1’ for the amine group in 1-phenylethanamine is permissible in general usage. Since locants are placed immediately before that part of the name to which they relate, it is clear that that the remaining locant ‘1’ in 1-phenylethanamine refers to the phenyl group.

However, for absolute clarity in preferred IUPAC names, the current IUPAC recommendations are prescriptive about when omission of locants is permissible. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), locants are omitted in preferred IUPAC names in the cases that are mentioned in Subsection P-14.3.4, in particular:

  P-14.3.4.2 The locant ‘1’ is omitted:

  (a) in substituted mononuclear parent hydrides;

  (b) in monosubstituted homogeneous chains consisting of only two identical atoms;

  (…)

Thus, for example, methanamine (a) and ethanamine (b) are preferred IUPAC names.

Generally, however, no locants can be omitted if any locants are required to identify the specific isomer.

  P-14.3.3 Citation of locants

  In preferred IUPAC names, if any locants are essential for defining the structure of the parent structure or of a unit of structure as defined by its appropriate enclosing marks, then all locants must be cited for the parent structure or that structural unit. For example, the omission of the locant ‘1’ in 2-chloroethanol, while permissible in general usage, is not allowed in preferred IUPAC names, thus the name 2-chloroethan-1-ol is the PIN. (…)

Therefore, the omission of the locant ‘1’ for the amine group in 1-phenylethanamine, while permissible in general usage, is not allowed in preferred IUPAC names. The preferred IUPAC name is 1-phenylethan-1-amine.


Hybridisation of complex compound having octahedral geometry $$\ce{K[Co(NH3)2Cl4]}$$

I have problem finding the hybridisation of this compound. The strong field ligand $\ce{NH3}$ forms the minority among the ligands, so do I consider the hybridisation to be $\ce{sp^3d^2}$ ? Next, in this compound, $\ce{[Co(NH3)Cl3]}$, the number of strong field ligands is equal to the number of weak field ones. Can someone please explain the basis on which I can find out the hybridisation using crystal field splitting.

If you calculate the oxidation state of the central metal atom in both of these cases, you'll find it to be $+3$. With $\ce{Co^3+}$, all ligands behave as strong field ligands except in the cases of $\ce{[CoF6]^3-}$ and $\ce{[Co(H2O)3F3]}$. Thus, in both of the cases you've mentioned, the ligands will cause pairing of electrons, the complex will be of a diamagnetic nature and the hybridization will be $\ce{d^2sp^3}$.


Why does Iodine form Cations sometimes? It is reported that iodine forms compounds like $\ce{(CH3COO)3I}$, $\ce{I(ClO4)3}$ and $\ce{IPO4}$ which contains $\ce{I+}$ and $\ce{I^3+}$ and are not bonded covalently as in interhalogen compounds.

Generally, Halogens form anions of the type $\ce{X-}$ to achieve stability by attaining noble gas configuration. However iodine loses electrons when it forms a cation and does not attain noble gas configuration. So why does iodine form cations while no other halogen does?

Iodine is the biggest halogen and by extension the most polarizable.

Diatomic iodine can be oxidized with nitric acid to facilitate electrophilic aromatic substitution.

Iodide ion, as found in sodium iodide, can also be oxidized by concentrated household bleach to facilitate EAS as well.

In the examples you give, there are strong electron withdrawing groups (such as oxygen, which is found in each of your examples). It is plausible that the oxygen has stripped the iodine of its electron(s) in each of your examples.


Does trans 1,3-dichlorocyclobutane have zero dipole moment? Does this compound have zero dipole moment?

I think it has a non-zero dipole moment. If I assume the ring to be planar, then the dipole moments of each $\ce{C-Cl}$ bond cancel out. (One is upward and the other is downward.)

But if I take the actual puckered structure of cyclobutane, one $\ce{C-Cl}$ bond is axial and the other is equatorial and hence they won't cancel out and the compound has a non-zero dipole moment.

But the answer given in my book says the compound has zero dipole moment.

Cyclobutane and its substituted derivatives readily undergo a ring flip (or ring inversion) as pictured below.

The barrier to ring flipping is very low, around 1.5 kcal/mole, so at room temperature the flipping process is very rapid.  The lowest energy conformation of cyclobutane exists in a puckered geometry as depicted in $\ce{A}$ and $\ce{B}$. In the ring flipping process, the molecule passes through a transition state where the cyclobutane ring is planar.

Only molecules that belong to symmetry classes (point groups)

$\ce{C_{n}}$ (the molecule only contains a $\ce{C_{n}}$ axis)
$\ce{C_{nv}}$ (the molecule contains a $\ce{C_{n}}$ axis and a $\ce{\sigma}_{v}$ plane)
$\ce{C_{s}}$ (the molecule only has a plane of symmetry)

can have a dipole moment.

Conformers $\ce{A}$ and $\ce{B}$ both have $\ce{C_{s}}$ symmetry (the only symmetry element is a plane that bisects the ring and contains the two cyclobutane carbons bearing the substituents) and therefore do have a dipole moment. However, the dipole moments of conformers $\ce{A}$ and $\ce{B}$ are equal and opposite, so when flipping is rapid the dipole moment averages out to zero.

Therefore at room temperature, where flipping is rapid, the molecule has no measurable dipole moment. If you cooled the system down to a very low temperature where conformers $\ce{A}$ and $\ce{B}$ were not rapidly interconverting, then you could measure a non-zero dipole moment.