What causes straight fatty molecules to form a lattice (e.g. saturated fats)? What causes multiple chains of saturated fats to pack together and form a solid? It is said that because the chains are straight, they form a lattice and line up more easily.

But if I had a bunch of straight sticks and drop them into a pile, they don't magically form a lattice where everything lines up. So is there something about the fatty chains that causes them to line up and pack closely together into a lattice?

Nature abhors a vacuum, and your sticks aren't sticky.

When you have a jumble of sticks, there's lots of voids and spaces between them - the pile of sticks takes up much more room than the bundle. For wooden sticks, this is no problem, as air can easily fill in the voids between them. With saturated fats, this isn't possible. There really isn't anything that can fill the voids. The only way to fill them is to line up the chains in an ordered lattice.

Also, when the chains are in an ordered lattice they're making a large number of van der Waals interactions between them. Because the chains are straight, the best way to maximize these favorable attractive interactions is to pack the chains together in an ordered array. The van der Waals forces are comparatively small, but over a long chain it adds up. (I haven't ever seen this done, but I'm guessing that if you coated your wooden sticks with something sticky like honey, and shook them around a bit, your wooden sticks would also want to line up with each other and stick together in an ordered array.)

Temperature, of course, can overcome both effects. The additional kinetic energy in the molecules allows them to overcome the attractive interactions between the molecules, and higher temperatures mean that the molecules are moving faster, pushing away other molecules and effectively filling in the voids to some extent. When you get to a high enough temperature, the fat melts.

In unsaturated fats, the chains are kinked. This means that it's much harder to line them up to form all those van der Waals interactions, and there are intrinsically voids in even the most ideal packing. Thus it's much easier to break those interactions and add the additional voids needed for a liquid state, so the solid to liquid transition happens at a lower temperature.

One thing to keep in mind is that things like solid/liquid transitions are not absolute things. Where the solid to liquid phase transition happens is based on the relative stability of the solid versus the liquid. It isn't that the saturated fat can form a solid and the unsaturated fat can't, it's just that the solid saturated fat is more stabilize relative to the liquid saturated fat, than the solid unsaturated fat is to the liquid unsaturated fat. As the change in stability is much greater for saturated fat, it takes a higher temperature to melt the saturated fat than it does the unsaturated one.


Does plant material oxidize from 02 gas (air or solved in water) and/or oxygen in H20? https://en.m.wikipedia.org/wiki/Curing_of_tobacco
Hi
This is my first time growing tobacco for fun, pleasent green at home, to make snuff for personal use and maybe save som money.
Ive been reading a bit on wiki, it says oxidization is what makes tobacco cure/age/mature take out the harsh.
Im wondering about if the oxidization of caratenoids and chlorophyll in tobacco curing is from O2 in air (or solved in water) and/or from O in H2O itself?
From what i remember, from school and various archeology programs on tv, wood in sea bed with no O2 solved in it doesnt rot but metals will rust. ?
At what relative humidity at room temperature will tobacco leaves sealed in a jar (in fall and winter the relative humidity in my apartment will drop too much for curing tobacco, i guess) cause them to cure? If air is required, how much?
/Johan

Okay first, it's the oxygen in the air that causes oxidation, also dissolved oxygen in water, but if you let your tobacco get wet enough for the latter to be significant you will be getting more rot than curing ...
Anyway, I don't think a sealed jar would be a good idea. Ammonia build-up, wet rot, etc would tend to be an issue.
You could build a proper curing/fermenting kiln to cure the leaf, there's DIY instructions here: https://www.seedman.com/wkiln.htm for one type.  But if you're using the finished product for dry snuff rather than smoking or chewing, you might not be too bothered about the taste/smell so careful fermenting is probably not that important to you. Then easiest is to just hang them somewhere like the garage and spray them with a plant mister when they get too dry like this guy: https://dengarden.com/gardening/Tobacco-Growing-and-Curing-at-Home or if it gets too cold where you live to use the garage, in a closet, wooden box or linen chest.


Can you carbonate a non aqueous liquid? We have a home carbonator for makeing seltzer. It blows $\ce{CO2}$ under high pressure into water. Could I use it to carbonate corn oil? Is $\ce{CO2}$ soluble in a no aqueous solution?

$\ce{CO2}$, being non-polar (due to the 180 degree bond angles),  will probably be more soluble in solvents that are non-polar as well, although the quadrupole moment might cause higher solubility in polar solvents as well (thanks to @georg for this addition). Also temperature will play a significant role, the higher the temperature the lower the solubility.

These qualitative remarks however only take you so far. If you really want to know the solubility in a specific liquid you will have to look it up or, if that is not possible, test it yourself. In this paper they test the solubility of $\ce{CO2}$ in different types of crude oil and make predictions based on the structure of the oil.

I cannot find anything specific for corn oil, although the reverse process of dissolving palm oil in supercritical $\ce{CO2}$ might set you on the right track to find what you want to know.

Just a fun addition: don't ever try to carbonate milk as these guys do - it will get messy! (it is in Dutch, but you can still watch what happens). Click here an English video with similar footage


Hund's rule & different H₂ molecules Does Hund's rule allow both of the following scenarios?

Filling each orbital with a single electron, so that a sub-shell, at first, only electrons with a negative spin
Filling each orbital with a single electron, such that a sub-shell, at first, only electrons with a positive spin?

I assume two H atoms whose electrons have different spins would form $\ce{H-H}$ and atoms where the spins are the same would form $\ce{H:H}$. Is that correct?

For H$_2$ there are only two electrons and thus you can only have singlet (electrons with opposite spin projection) and triplet states (electrons with equal spin projection).

The spin wave function of the singlet state is a linear combination (a superposition state) of the first electron with spin projection $\alpha$ and the second electron with spin projection $\beta$ and the opposite situation. Both spin states in the quantum superposition have opposite phase:
$$|\uparrow\downarrow\rangle -|\downarrow\uparrow \rangle$$
If the electrons are in the same molecular orbital then (by the Pauli principle) only the singlet state exists. If they are filling $\pi$ molecular orbitals then they could be in different degenerate orbitals (with angular momentum projection on the molecular axis $M_z = \pm \hbar$) and thus both singlet and triplet terms would exist.

There are three degenerate states belonging to the triplet term, that is, there are three wave functions that have equal energy. There spin components are:
$$|\uparrow \uparrow \rangle$$
$$|\downarrow \downarrow \rangle$$
$$|\uparrow \downarrow\rangle + | \downarrow\uparrow \rangle$$

So you can see that it doesn't matter if you fill one molecular orbital with one electron in spin $\alpha$ and a different molecular orbital with the electron with $\alpha$ spin or you do so with both $\beta$ and $\beta$ spins. However you can not put both electrons with the same spin projection in the same molecular orbital. That is against the Pauli principle, that is, against the anti-symmetry of the electronic wave function.


Will NH3 decompose into N2 and H2? In chemical kinetics , we learn that in a reversible reaction ,both the forward reaction and reverse reaction occur at the same time. At equilibrium , the rates are the same. For example:Production of ammonia NH3 .During the production of ammonia both the forward & reverse reaction is present .

$\ce{ 3H_2(g) + N_2(g) ⟶ 2NH_3(g)}$ forward reaction

$\ce{2NH_3(g) ⟶ 3H_2(g) + N_2(g)}$ reverse reaction

Now come to the Gibbs free energy.

Under standard condition(298K) , it can be calculated that the reaction is forwardly favored (forward reaction is spontaneous,and thus the decomposition of NH3 is non-spontaneous).
I understand H2 and N2 will produce more NH3.My question is at this situation will NH3 decompose into H2 and N2 to some extent(though it is a non-spontaneous process)?.But chemical kinetics tells that both reaction(not necessarily same rate) have to be present to create equilibrium in future .

Under standard conditions (100kPa, 0°C), nitrogen and hydrogen do not react with each other in any measurable way. It follows logically that also ammonia cannot decompose (as it would to a small extent in an equillibrium reaction), because otherwise it would decompose completely with time, although it is stable.

In equillibrium, the forward and backward reaction rates must be identical and finite.


Which is more basic, hydrazine or ammonia? Which of the following will be more basic, hydrazine $\ce{H2N-NH2}$ or ammonia $\ce{NH3}$?

My chemistry teacher said that $\ce{NH3}$ is more basic since after giving $\ce{H+}$ to hydrazine results in $\ce{H3N+-NH2}$, and accommodation of the lone pair of $\ce{-NH2}$ is not possible by $\ce{-N+H3}$, since it doesn't have the space. So instead, $\ce{-NH2}$ will pull electrons from it making it unstable. Ammonia has no such problem so it must be more basic.

I am not so pleased with this argument. In $\ce{H3N+-NH2}$, although the lone pair cannot be accommodated, but the positive charge present on its sides , to an extent, should neutralize the intensity of the lone pair, making it somewhat stable. And also, not to forget, hydrazine has two spots where we can get the electrons, therefore, its ambident nature should also support it's basicity.

I need a bit of clarity on this.

Obligatory link to Evans' pKa table.

Ammonia is more basic than hydrazine, by about one order of magnitude.  This is expected, because the -NH2 group is more electronegative than -H or -CH3.

The second lone pair is not involved in the acid-base reaction, it does not point towards the -NH4+ group.  You can, however, force two lone pairs into close proximity.  This destabilizes the unprotonated form.  The keyword is "proton sponge".

1,8-Bis(dimethylamino)naphthalene has a pKa of 12.3, it's one of the strongest known amine bases.  Compare that to the pKa of aniline, which is something like 4.5.


Which chlorine species can act as an oxidizing agent but not as a reducing agent? Which species can act as an oxidizing agent but not as a
  reducing agent?

  (A) $\ce{Cl2}$
  (B) $\ce{Cl-}$
  (C) $\ce{ClO2-}$
  (D) $\ce{ClO4-}$

I thought the answer was A because if $\ce{Cl}$ is reduced, it becomes very stable with a full octet and therefore cannot be the reducing agent; however, according to the 2007 Chemistry Olympiad answer key, the answer is D. Why is that?

For a species to act as a reducing agent, it should have the ability to get oxidised. Any element in its highest oxidation state, cannot oxidise anymore, and can hence not act as a reducing agent. $\ce{ClO4-}$ has $\ce{Cl}$ in $+7$ oxidation state, and cannot have a higher oxidation state, and hence cannot act as a reducing agent.


Why doesn't anyone manufacture 4.5 V cells? Every 4.5 V lantern batteries are a series of three 1.5 V cells. It would have more capacity if it consisted of single 4.5 V cell. But I don't see any 4.5 V cell.

According to Wikipedia's Standard Electrode Potential data page, there are vast possibilities of a 4.5 V cell. The one I think most plausible is:

$$
\begin{align}
\text{Anode:} &\quad &\ce{Ca + 2 H2O &→ Ca(OH)2 + 2 H+ + 2 e-} \quad|\cdot 5\\
\text{Cathode:} &\quad &\ce{2 BrO3- + 12 H+ + 10 e- &→ Br2 + 6 H2O}\\
\hline
\text{Overall:} &\quad &\ce{5 Ca + 2 HBrO3 + 4 H2O &→ 5 Ca(OH)2 + Br2}
\end{align}
$$

Have anyone tried this or anything else 4.5 V?

Perhaps you consider the volume allotted to the batteries as equivalent to capacity. A better measure would be ampere-hours at 4.5v, or watt-hours. And weight might be a factor to consider.

An electrochemical cell with a high voltage (4.5v) that requires complicated or dilute ingredients could better fill the volume allotted, but if it runs out of power in a shorter time, it has less capacity. You could consider 6 cells in a 9v battery to be a complicated device, but it seems so simple. Three 1.5v cells in a package would give you a simple overall structure, and could give you high capacity due to the simple internal composition of each cell.

One reason that lithium ion batteries are so highly regarded is their high electrochemical capacity.
"The more energy a [rechargeable] battery can hold when charged, the less often you need to charge it. No one wants to be replacing the batteries in their wireless keyboard every week or to have their kid’s toy conk out after just 15 minutes. Whereas a common disposable AA battery’s capacity is anywhere from 800 mAh for budget brands up to 2,000 mAh for premium versions, most rechargeable batteries start at around 1,800 mAh and go up to nearly 3,000 mAh. Most trustworthy companies right now make AA batteries around 2,000 mAh. We don’t consider batteries with stated capacities lower than 1,700 mAh, which is too underpowered, or higher than 2,500 mAh, which makes us suspicious that a company is overpromising." https://thewirecutter.com/reviews/best-rechargeable-batteries/


How to measure chlorine content using ORP sensor? In my project I need to measure pH, turbidity, and chlorine level of treated drinking water. I found sensors for measuring pH, turbidity, and also chlorine.
The actual problem begins is that the cost of chlorine sensor is too high, so I googled for an alternative solution, which led me to use an oxidation-reduction potential (ORP) sensor.
Can anyone tell me about the relationship between pH, chlorine, and ORP value in millivolts and how to measure chlorine content in treated water using ORP sensor?

For a school project like yours, you can use iodometric titration. Iodometric titration is one of standard methods of water examination. The amount of free chlorine in tap water is really small on the order of < 1 parts per million therefore you need a lot of sample. One would typically take 500 mL of tap water (or more), add potassium iodide, and titrate with 0.01 M sodium thiosulfate. You can use an ORP and plot a potentiometric titration curve.
Consult your teacher for this option. He/she should explain to you the details.
There are spectrophotometric methods but the reagents may not be available in your school.


Low solubility of propanal compared to propanol, propanone, propanoic acid and methyl ethanoate I found on the internet that propanol, propanone, propanoic acid and methyl ethanoate where all completely soluble in water at room temperature. (23 degrees Celsius) But I found that propanal was, comparatively, was not completely soluble in water (Wikipedia said it was 20g/100ml). I was wondering why it was not completely soluble like all the others as small aldehydes can interact with hydrogen bonds and dipole-dipole forces with water (like all the others). Is there some other intermolecular force that influences solubility that I'm missing?

They're all on the verge of miscibility, methyl ethanoate (methyl acetate) is in fact not miscible according to wikipedia (soluble to 25%).

The acid is the most polar of all the C3 compounds you name. Butyric acid (C4) is also miscible with water, but valeric acid (c5) already not.

n-butanol is already rather badly soluble (8g in 100ml) in water. Same with butanone, etc.

Compared to the ketone, the aldehyde has a longer unpolar part, that is overall less influenced by the polar carbonyl bond. Both are clearly, i think, the least polar of the group, and so the aldehyde just drops beneath the line.

No mystery here. ;-)


Finding out pKa of acid from molar conductivity I'm reading about the electrical properties of solution where there is a problem like that:

  The molar conductivity of $0.0250\ \mathrm M$ $\ce{HCOOH(aq)}$ is $4.61\ \mathrm{mS\ m^2\ mol^{-1}}$. Determine the $\mathrm pK_\mathrm a=-\log K_\mathrm a$ of the acid. (limiting ionic conductivity of $\ce{H+}=34.96\ \mathrm{mS\ m^2\ mol^{-1}}$ and limiting ionic conductivity of $\ce{OH-}=19.91\ \mathrm{mS\ m^2\ mol^{-1}}$)

I have to solve the problem using the equation below:

$$\frac1{\Lambda_\mathrm m}=\frac1{\Lambda_\mathrm m^0}+\frac{\Lambda_\mathrm mc}{K_\mathrm a\left(\Lambda_\mathrm m^0\right)^2}$$

Here which limiting ionic conductivity should I use to solve the problem?  $\ce{H+}$ or $\ce{OH-}$?

You need to add the limiting ionic conductivities for $\ce{H+}$ and $\ce{OH-}$ together to get the limiting ionic conductivity for all the ions in solution ($\Lambda_{0}$, which will replace $\Lambda^{0}_{\mathrm m}$ in your equation).

This arises from a simplification for calculating $\Lambda_{0}$ in weak electrolyte solutions (such as yours) according to Kohlrausch's Law in which it is stated:

  Each ionic species makes a contribution to the conductivity of the solution that depends only on the nature of that particular ion, and is independent of the other ions present.

from which we can then estimate $\Lambda_{0}$ as:

$$\Lambda_{0} = \sum_{i}\lambda_{i,+}^{0} + \sum_{i}\lambda_{i,-}^{0}$$

For your problem:

$$\Lambda_{0} = \underbrace{(34.96 + 19.91)}_{54.87}\ \mathrm{mS\cdot m^{2}\cdot mol^{-1}}$$

we solve (I am omitting units here for clarity, and have confirmed via dimensional analysis that the answer is correct):

$${1\over 4.61} = {1\over 54.87} + {4.61\times 0.025\over K_{\mathrm a}\times (54.87)^{2}}$$

which simplifies to yield the answer (spoiler alert):

  $$K_{\mathrm a} = 1.926\times 10^{-4} \implies \mathrm{p}K_{\mathrm a} = 3.72$$

and can be favorably compared to the value of 3.77 given here.


How accurate is this polar mechanism for the Diels-Alder reaction? So the prof got to Diels-Alder without discussing it in the context of molecular orbital theory. Instead we got a resonance picture of what happens... this is what he told us:

1) 1,3-butadiene has both an electrophilic and a nucleophilic terminal carbon atom because we can draw resonance structures showing dipolar terminal carbon atoms ...

2) For ethene we can similarly draw a dipolar resonance structure.

From there we're told to simply use Coulomb's law - opposite charges attract - to complete the mechanism.

So, how adequate is his electrostatic explanation? Is it adequate - probably ... if one simply wants to learn it for a test. But is it conductive to further understanding of chemistry? I feel rather short-changed here since I'm paying him the big-bucks (tuition is insane) and the book goes into far greater detail ...

ETA: I asked the guy about and grilled him on the electrostatic explanation. For example I raised Martin's excellent point about the fact that the dieneophile cannot be that polarized - otherwise wouldn't it be a liquid instead of a gas due to the strong intermolecular attractions?

He instead went on a tangent on electron-withdrawing groups and how the electrostatic explanation can explain these ... what do you think?

Disclaimer: This is not a full answer. I was just too long for a comment.

I think the suggested mechanism is not entirely wrong, it might be a contributing factor of the whole process, but not a very important one. In the last fifty years, the mechanism has been discussed in almost countless publications. One of the more recent ones by Goldstein et. al. concluded that a concerted pathway is the most economical.[1]
They also included possible diradical pathways and the similar. Of course there is no uniformly applicable mechanism, it is very dependent on the substrates. Many of the reactions can be effectively described by frontier molecular orbital theory, so it is also possible to do so for the simplest Diels-Alder reaction.

There might be many more publications out there discussing the whole mechanism in pretty exhausting detail, but for a start I thought the cited one will do just fine. Just to name some more, see [2] through [4].

The electrostatic interpretation is tempting, as it is simple and it explains the simplest case very well. I believe it explains it too well for the wrong reasons:
 I would not expect the polarisation, i.e. induced dipole, in ethylene and butadiene be strong enough to make this happen at all. Both are colourless gasses at room temperature. If there were more induced dipoles, they would be liquid, which means that these molecules are basically unpolarised.

In a publication by Gian Franco Tantardini and Massimo Simonetta it was stated, that the contribution of the polarised Valence Bond Structure is very little, see [5]. If I read it correctly, the contribution is less than 1%. This alone should suffice to disregard a mechanism purely based on electrostatic considerations. In the same publication it is stated, that ethylene has quite a significant contribution from its polarised structure (about 25%). However, these consideration would in most cases not be able to describe streospecific reaction products, that are obtained from different substrates.

As another aspect, one must not forget, that the ground state of 1,3-butadiene is trans and it requires just a bit of energy to rotated it to the necessary cis formation, which should be about 10 times less present. So what are the odds having a well polarised ethylene hitting a polarised cis-butadiene in the right angle at the right speed. There has to be more to it...

E. Goldstein, Brett Beno, and K. N. Houk; J. Am. Chem. Soc., 1996, 118 (25), 6036–6043.
R.B. Woodward, and Thomas J. Katz; Tetrahedron; 1959, 5 (1), 70-89. Mirror at znu.ac.ir.
Michael J. S. Dewar, Santiago Olivella, and James J. P. Stewart; J. Am. Chem. Soc., 1986, 108 (19),  5771–5779.
J. Sauer; Angew. Chem. Int. Ed., 1967, 6 (1), 16-33.
Gian Franco Tantardini and Massimo Simonetta; Int. J. Quant. Chem., 1978, 14 (S12), 131-141.


Incense Stick vs Candle *Not sure whether to put this on Phy or chem.. *

What exactly happens when you light an incense stick?

How is this different from a candle, or embers for that matter?

From what I understand, a candle-flame is..... well.... basically a flame and embers are pieces of glowing hot coal or wood, but then where does the lit portion of an incense stick fit in?

In a candle, the wax first melts, then turns into a vapor above the wick and the oxidation region, where the vaporized fuel mixes with air, surrounds the central, cooler region. You can demonstrate this by extinguishing a candle and then relighting the smoke above the candle... the flame will flash back to the wick. This "flash-back" can be hazardous in a lab using ether, since a nearby flame can ignite the fumes which will burn back to the ether can.

Incense, smoldering cotton and similar combustion relies on thermal degradation of a solid, producing a narrow flame-front on the surface of the solid fuel. Since oxygen cannot penetrate the solid, the rate of combustion is limited. For that reason, cotton bales burn slowly but are very difficult to extinguish, since the interior remains hot and water cannot quickly penetrate. On the other hand, stretching a cosmetic cotton ball changes the burning from smoldering to an intense flame.


n-Factor of phosphorous acid It's very confusing that while $\ce{H2SO4}$ has an $n$-factor of 2, whereas $\ce{H3PO3}$ has $n$-factor 2 instead of 3. Many tried explaining it to me by molecular structures, but I don't know how to draw them. Could you please try explaining without the help of molecular structures?

$n$-Factor is basically a method to find out a relationship between the compound and what it is equivalent to in terms of acidic nature or basic nature. Note that the two hydrogen atoms in $\ce{H2SO4}$ are both attached to oxygen. But in $\ce{H3PO3}$ two of the three hydrogens are attached to oxygen and the other hydrogen is directly attached to phosphorous.

Note that only those hydrogens attached to highly electronegative elements can be donated easily in case of inorganic acids. One could explain better by using the concept of stabler structures with the $\ce{H}$ removed from $\ce{O}$ than from $\ce{P}$ by resonance. But for now know it as a thumb rule.

Therefore, since $\ce{H3PO3}$ has only two $\ce{H}$ attached to $\ce{O},$ its $n$-factor is 2 and not 3.


Why are neutralisations involving weak acids and bases less exothermic than those involving only strong acids and bases.? I read somewhere that for example the neutralisation reaction between sodium hydroxide and acetic acid is less exothermic than those of sodium hydroxide with hydrochloric and nitric acid because some energy is needed to cause the weak acid (acetic acid) to completely ionise. Is this true?

I think the answer is closer to what was stated in the original question: "some energy is needed to cause the weak acid (acetic acid) to completely ionise". Stated differently, less energy is released from making acetate ion from acetic acid than from making chloride ion from hydrochloric acid (or water from hydronium ion).

Despite the lesser tendency of acetic acid to ionize, the overall stoichiometry of the two reactions is the same--as pointed out in a comment by the OP.


Is a compound aromatic if it also has anti-aromatic rings? Is 5,6,11,12-tetradehydrodibenzo[a,e][8]annulene aromatic?

If I use Hückel's rule for the whole molecule, I get the answer as anti-aromatic. But the compound has two benzene rings which are individually aromatic.

The molecule actually optimizes to a planar, delocalized $\pi$ system

Using B3LYP/6-31G(d) optimizations, we see that the molecule optimizes to become planar. (The initial configuration was intentionally bent.)

Moreover, if I plot the HOMO, it's highly delocalized:

Similarly, the LUMO is also delocalized across the entire molecule:

Consequently, I don't see any indication that the "side" benzene rings are localized from the cycloocta-1,5-diyne ring. The bond lengths are in fairly good agreement with the crystal structure:

Here's the crystal structure from J. Am. Chem. Soc., 1975, 97 (3), 658–659. Note there is a profound triple bond character, but the bridging bonds (1.44 Å) are somewhere between a typical aromatic bond length (1.40 Å) and a full single bond (~1.54 Å).


How are electronic configurations including hybrid orbitals notated? I know the general scheme for notating electronic configurations is (with s and p orbitals) [noble gas core]nsel.$n$pel. etc., where $n$ is the principal quantum number and el. is the number of electrons occupying that subshell, e.g. carbon: [He]2s22p2.

Now, how do you include hybrid orbitals? The problem arises because some hybrid orbitals already have superscript numbers (sp2, sp3 …) – how do you differentiate between these superscript numbers and those denoting how many electrons fill that orbital? If I were to write out the electronic configuration for the hybridised carbon atom in $\ce{CO2}$, this would not be a problem: [He]2sp22p2, but what about the carbon atom in $\ce{CH4}$, which has sp3 orbitals? Would you write [He]2sp3, 4 or [He]2sp34 or something else?

The electronic configuration concept does not really extend itself well to hybrid orbitals, because these are best considered mathematical concepts only. However, the problem is easily solveable:

Always use brackets when dealing with hybrid orbitals.

Even your $\ce{CO2}$ case is potentially misleading: I misread it once as $\mathrm{2s^1\ 2p^2}$. It is best to show what belongs together with: $\mathrm{[He]\ 2(sp)^2\ 2p^2}$ or, for methane: $\mathrm{[He]\ 2(sp^3)^4}$.


What is the heat needed for an isothermal expansion of gas. Why do physics and chemistry yield different answers? One mole of a certain ideal gas is contained under a weightless piston of a vertical cylinder at a temperature T. The face of the piston opens into the atmosphere. What is the heat supplied in the process to expand the gas from volume $V_1$ to $V_2$ isothermally ? Friction of the piston against the cylindrical wall is negligible is small

Now thermodynamics is a common topic to both physics and chemistry and as per my understanding of thermodynamics I am getting different results for the heat that will need to be supplied.

Chemistry says $$∆U =∆Q +∆W $$ and defines work as the work done on the gas and this will give $$W = -P_{atm}(V_2-V_1) $$ and as for isothermal process$$ ∆U =0 , ∆Q =P_{atm}(V_2-V_1)$$

According to physics on the other hand $$∆Q =∆U+∆W$$ and here the work done is the work done by the gas. That is

$$W = \int_{V_1}^{V_2} PdV=\int_{V_1}^{V_2} \frac{RT}{V}dV = RT \ln V_2-RT \ln V_1$$ and as for isothermal process $$∆U =0 ,\, ∆Q =RT \ln \left(\frac{V_2}{V_1}\right)$$

I am also uncertain that maybe neither of these is the correct expression and the network ok is actually in that work from both these forces and we have to take the sum of these to yield $$∆Q =RT\ln \left(\frac{V_2}{V_1}\right)-P_{atm}(V_2-V_1)$$

Now the value of Q should be unique because in the real world when we perform an experiment, only one value of Q would be supplied.

So which one of these is correct and where am I lacking in my understanding of thermodynamics from a chemistry  standpoint ??

Now as this question links both physics and chemistry, I want to post it on both sites (I hope it's fine) and the link to it is here

The problem is that both the problem statement you presented and the chemistry solution you showed are fallacious.  The fallacy is that, for an ideal gas $V=nRT/P$, so that, if the pressure and temperature are both constant, the volume can't change.  If the volume does change, then either the pressure must change or the temperature must change.

In the "chemistry solution" you presented, the pressure is taken as constant (atmospheric).  So the temperature must change.  But if the temperature changes, then $\Delta U$ is not zero (as they assume).  So the solution is incorrect.  The correct solution, based on the 1st law, is $$\Delta U=nC_v\Delta T=Q-P_{atm}(V_2-V_1)$$or equivalently, $$\Delta H=\Delta U+P_{atm}(V_2-V_1)=nC_p(T_2-T_1)=Q$$Also, since $$n=\frac{P_{atm}V_1}{RT_1}$$we have $$\frac{T_2}{T_1}=1+\frac{QR}{C_pP_{atm}V_1}$$

In the "physics solution" you presented, they make the opposite assumption that, rather than the pressure remaining constant, the temperature remains constant.  For the temperature remaining constant, the solution you presented is correct.  But the final pressure is:  $$P_2=P_{atm}\frac{V_1}{V_2}$$


Why is the rate of reaction of polar EAS favored by inductive effects and the rate of less polar EAS reactions through hyperconjugation? 1. Reactions involving strong electrophiles (nitration) are
       said to be governed
       by
       Inductive effects thus the order of nitration is
                                       D>C>B>A
  2.Chlorination is a less polar reaction thus chlorination said the be governed by hyper conjugation.
                                        A>B>C>D

My question is,  why?

The factors controlling EAS in both the cases will be  the same for both the substitutes.
Hyperconjugation and inductive effects will both operate without distinction in both the cases.

So why is it that
The rate of reaction of polar EAS favored by inductive effects and the rate of less polar EAS reactions through hyperconjugation.

SOURCE   A guidebook to mechanism Peter Sykes

EDIT
I have been asking around for an answer to this question.

No one has been able to answer it conclusively , the primary school of thought   is that

  In Highly polar reactions , inductive effects dominate as more electron density is available , hyperconjugation on the other hand is just redistribution of electrons.

First of all is this correct ?   Even if it's correct but how does it explain the opposing rate orders for nitration and bromination?

I may be oversimplifying things.

A look at electron density

Let's pick up the extreme cases of the substituted benzenes . $$\ce{A -> ArCH3}$$
$$\ce{D -> Ar(CH3)_3}$$

Case A shows the maximum hyperconjugation. Case D on the other has the maximum +I effect.

For a moment let's imagine that the electrons are coins.
Case D , has more coins, more methyl groups (let's say 3).
Case A, on the other hand has 1 coin , hyperconjugation would simply mean that I am  simply moving that coin in circles. It's simply redistribution of electrons.it doesn't actually increase the density.

So the first point which I'm going to use to answer the question is that -

Inductive effects increases  the electron density, hyperconjugation simply redistributes the electron density.

A look at Coulomb's law

$$\ce{ F \propto Q_1*Q_2}$$

More polar reactants will have a greater charge on them as compared with less polar reactants.
Since inductive effects concentrate  electron density at one area the negative charge concentration will be greater.

HOMO/LUMO

Unoccupied orbitals are high in energy and occupied orbitals are low in energy. For the best overlap the energy gap between the HOMO and LUMO must be small.

Rate Order for more polar reactions

In more polar reactions the electrostatic forces of attractions play the primary role as the polar reactants is highly charged. As established earlier ,inductive effects lead to a electron density thus a greater pull. Homo/Lumo interactions are secondary as the overlapping happens after a collision occurs.

The electrostatic forces are pulling the benzene and reactant. That would by itself increases the number of collisions.
A greater electron density leads to the better stabilization of the +tive charge on the reactant.

inductive effects support more polar reactions

Rate order for less polar reactions

The electrostatic forces of attraction are lower in the case as as the reactant is less charged.
So the control of the reaction is primarily through Homo/Lumo interactions.

Unoccupied orbitals of the reactant are high in energy.( They are Unoccupied).
To bridge the HOMO/LUMO gap the Lumo of benzene carbon must be high in energy. Non bonded electrons or the electrons which are delocalized through hyperconjugation are high in energy as they are held by only one atom.
So the HOMO/Lumo gap reduces in the case of hyperconjugation thus hyperconjugation supports less polar reactions.


Did O₄ turn out to explain why liquid oxygen appeared to defy Curie's law? If not, what does? Wikipedia's Liquid Oxygen says:

The tetraoxygen molecule ($\ce{O4}$) was first predicted in 1924 by Gilbert N. Lewis, who proposed it to explain why liquid oxygen defied Curie's law. Modern computer simulations indicate that, although there are no stable $\ce{O4}$ molecules in liquid oxygen, $\ce{O2}$ molecules do tend to associate in pairs with antiparallel spins, forming transient $\ce{O4}$ units.

The paragraph is a bit ambiguous and doesn't rule out that the "transient $\ce{O4}$ units" do not explain why "liquid oxygen defied Curie's law", so I'd like to ask:
Question: Did $\ce{O4}$ turn out to explain why liquid oxygen appeared to defy Curie's law, at least in some way? If not, or not completely, what does?

The observation that oxygen did not seem to strictly follow Curie's law (where $1/\chi$ is linear with temperature)  has been known since Kammerlingh Onnes pointed it out in the 1910's. By the late 1920's, it was known that diluting the liquid with nitrogen preserved Curie's law, suggesting that it took oxygen in close proximity to other oxygen to be the issue.

Fast forward some 80 years, and the computational tools available to us provide some key insight. As one such paper, Ab Initio Molecular Dynamics Investigation of the Structure and the Noncollinear Magnetism in Liquid Oxygen: Occurrence of O$_{4}$ Molecular Units takes a look at structure and properties of liquid oxygen. Quoting from their conclusion:

  In  conclusion, we  performed ab  initio molecular  dynamics  to  investigate  the  noncollinear magnetism  of  a system with an evolving atomic structure. Application to liquid oxygen provides us with a picture in which the large majority of colliding O$_{2}$ molecules assume structural and magnetic configurations which closely resemble those in the O$_{4}$ molecule. Formation of truly long-living molecular O4 units  also  occurs but involves a considerably smaller fraction of O$_{2}$ molecules.

So, it would seem that O$_{4}$ units, whether long lived 'molecules' or temporary short ranged order is the answer to the deviation from Curie's law in liquid oxygen.

Several comments on these O$_{4}$ units - first, as noted in the paper above, 'long lived' means under a picosecond (which doesn't really seem long-lived to me, but...). Further, the authors above point to The Infrared Spectrum of Bound State Oxygen Dimers in the Gas Phase to explain what they mean by O$_{4}$, which they call a "somewhat rigid (O$_{2}$)$_{2}$ molecule".


Nomenclature of an organic compound Which substituent should be placed first (in case of alphabetisation)?

isopropyl-
methyl-

I have come across two compounds: 1-methyl-3-methoxy-4-isopropylbenzene and 1-isopropyl-4-methylcyclohexane

So, is “iso” taken care of in case of alphabetisation? And, are there any special criteria for this operation?

(Note that the prefix ‘isopropyl’ is retained for use in general nomenclature but the preferred IUPAC name is ‘propan-2-yl’.)
Simple prefixes (simple substituent groups consisting of just one part that describes an atom, or group of atoms as a unit, for example methyl, methoxy, and isopropyl) are arranged alphabetically disregarding any multiplicative prefixes. Any multiplicative prefixes are inserted later and do not alter the alphabetical order.
For example, ‘1,2-dibromo-’ is considered to begin with ‘b’; ‘isopropyl’ is considered to begin with ‘i’.
On this matter, the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

P-14.5 ALPHANUMERICAL ORDER
Alphanumerical order has been commonly called ‘alphabetical order’. As these ordering principles do involve ordering both letters and numbers, in a strict sense, it is best called ‘alphanumerical order’ in order to convey the message that both letters and numbers are involved
Alphanumerical order is used to establish the order of citation of detachable substituent prefixes (not the detachable saturation prefixes, hydro and dehydro), and the numbering of a chain, ring, or ring system when a choice is possible.
(…)
P-14.5.1 Simple prefixes (i.e., those describing atoms and unsubstituted substituents) are arranged alphabetically; multiplicative prefixes, if necessary, are then inserted and do not alter the alphabetical order already established.

Therefore, the correct alphanumerical order in your first example corresponds to the name x-isopropyl-y-methoxy-z-methylbenzene.
In your second example, the correct alphanumerical order corresponds to the name x-isopropyl-y-methylcyclohexane.
Furthermore,

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

Note that Rule f takes precedence over Rule g.
Therefore, your first example is named as 1-isopropyl-2-methoxy-4-methylbenzene rather than 4-isopropyl-3-methoxy-1-methylbenzene since the locant set ‘1,2,4’ is lower than ‘1,3,4’.

Your second example is named as 1-isopropyl-4-methylcyclohexane rather than 4-isopropyl-1-methylcyclohexane since isopropyl is cited first as a prefix in the name.


Calculate the amount of substance of an isotope based on the percent composition Calculate the amount of substance of $\ce{^37Cl}$ in a $\pu{42.3 g}$ sample of $\ce{Cl}$ containing $24.00\%~\ce{^37Cl}$ and $76.00\%~ \ce{^35Cl}.$
  Given the atomic masses $m(\ce{Cl}) =\pu{35.45 amu}$, $m(\ce{^35Cl}) =\pu{34.97 amu}$, and $m(\ce{^37Cl}) =\pu{36.97 amu}$.

I know you would multiply $\pu{42.3 g}$ by $0.240$ to get $\pu{10.2 g}$ of $\ce{^37Cl}$, but my teacher says you then divide by the atomic mass of $\ce{Cl}$ rather than $\ce{^37Cl}$. That is the part that confuses me, she says it is because

  You don't know what those $\pu{10.2 g}$ are made of, so you need to use the atomic mass of $\ce{Cl}$.

If you have given an accurate account then your teacher is mistaken. We've all gotten confused while trying to talk and think at the same time.
\begin{align}
\pu{42.3 g} \ce{Cl}\cdot 0.24\frac{m(\ce{^37Cl})}{m(\ce{Cl})}
&= \pu{10.2 g} \ce{^37Cl} \\
\implies
\frac{\pu{10.2 g}~ \ce{^37Cl}}{\pu{36.97 g//mol}\ce{^37Cl}} &= \pu{0.276 mol}
\end{align}


Help with finding intermediates in the synthesis of Ritalin Here I have a Chemistry Olympiad question I'm doing that outlines the synthesis of Ritalin. So far I've worked out A, Anion B -, and compound C. You can see my workings, as I've attached them here too.

  Nucleophilic substitution with the nucleophile $\ce{C#N-}$ produces A:

  A proton is taken away from the $\alpha-\ce C$ to deprotonate it to form Anion B -:

  Anion $\ce{B-}$ acts as a nucleophile and substitutes $\ce{Cl}$ to produce C:

But how would I work out the structures of D, E and F?

I'm not familiar with all the mechanisms/reactions of nitriles and am unsure how to proceed.

In A and B, you should probably draw the C-C single bond in the nitrile longer, it's a bit difficult to see at the moment.
The nitrile is sequentially hydrolysed to an amide and then to an acid.
Treatment with $\ce{ROH/H+}$ is a rather obvious Fischer esterification.

You can arguably work all of this out from first principles: the nitrile carbon is electrophilic, and is therefore susceptible to nucleophilic attack by water. The reaction mechanism is given here. On top of this, the IR peaks strongly support the presence of amide and acid functional groups in D and E respectively.

With that said, I'd argue that there is always going to be an element of "I just know this because I read about it/was taught it before". The very fact that you said "I'm not familiar with the mechanisms/reactions of nitriles" indicates this - if you had happened to read a textbook chapter on nitriles, then you would probably have seen this, since it is one of the most fundamental reactions of nitriles. You would also have to read up about the IR wavenumbers - otherwise it's just gibberish.


Is carbon dioxide more acidic than carbon monoxide? Which of the given compounds is most acidic? $\ce{CO}$ or $\ce{CO2}$ ?

How do I solve this question, and how do I explain it?

I took the oxidation state of carbon in both. The oxidation state in $\ce{CO2}$ is higher, therefore it is more electronegative, so therefore $\ce{CO2}$ may be more acidic.
For $\ce{CO }$ I don't know about the charge so that I can use it to understand if it would give electron or accept electrons, so that I can apply Lewis acid/base theory.

Carbon dioxide is a much stronger Lewis acid than carbon monoxide. In carbon dioxide, the carbon has a much higher oxidation state (formally, $+4$ vs. the $+2$ of carbon monoxide), and the two oxygen atoms heavily withdraw electron density from the carbon atom, leaving a very large partial positive charge ($\delta^{+}$). This is rigorously confirmed by electrostatic potential maps and calculated orbital coefficients. As a consequence, the electron-deficient carbon of $\ce{CO2}$ is highly electrophilic, and reacts with a variety of nucleophiles (e.g., Grignard reagents, organolithium reagents, etc.).

Carbon monoxide, on the other hand, actually contains substantial electron density at the carbon atom (in spite of the bonds being generally polarized towards the oxygen atom), which possesses a fairly localized and active lone pair. Consequently, a few notable exceptions aside, carbon monoxide mostly functions as a Lewis base.


Regioselectivity in iodolactonisation of γ,δ-unsaturated carboxylic acid In the last step of the iodolactonisation of 3,3-dimethylpent-4-enoic acid,[1,2] why does the carboxylate attack the iodonium to form a 5-membered ring, when a more stable 6-membered ring is possible? Both of the sites seem to be equally favorable sterically. I can't come up with a reason favoring the formation of the 5-membered ring.

References

Takano, S.; Sato, N.; Akiyama, M.; Ogasawara, K. A Synthesis of trans- and cis-Caronaldehydes. Heterocycles 1985, 23 (11), 2859–2872. DOI: 10.3987/R-1985-11-2859.
Ding, Y.; Jiang, X. A Novel Highly Stereoselective Synthesis of the A-Ring of Taxol Via Two Aldol Reactions. J. Chem. Soc., Chem. Commun. 1995, 1693 DOI: 10.1039/C39950001693.

There are several arguments. Firstly, five-membered ring formation is generally kinetically more favourable than 6-membered ring formation. As ring size increases, the entropy of activation also increases – this is explained in greater detail in Clayden 2ed, pp 805–807. It is true that six-membered rings are less strained, but that is generally more of a thermodynamic consideration, and from a kinetic point of view the entropy factor happens to outweigh the enthalpy factor.

Secondly, for cationic ring-opening reactions (e.g. reactions of halonium ions, or epoxide opening under acidic conditions), it is usually the more substituted end that preferentially reacts. See e.g. Regioselectivity of bromination of alkenes. This is also explained in Clayden 2ed, pp 436–438.

However, probably the most relevant factor here is stereoelectronic in nature. Stereoelectronic effects refer to effects which depend on the spatial orientation ("stereo") of orbitals ("electronics"). In the case of an SN2 reaction, the key requirement is that the lone pair of the nucleophile (Nu) must be able to reach the σ* orbital of the C–X bond being broken. Since the σ* orbital points out of carbon diametrically opposite the leaving group X, this is equivalent to saying that we need a collinear Nu···C···X geometry in the transition state.

For ring-forming reactions, there are a set of guidelines – Baldwin's rules – which empirically describe whether these reactions are stereoelectronically favoured (i.e. whether the nucleophile and electrophile are capable of orienting themselves in a geometry that allows for reaction). In the case of six-membered ring formation via substitution at a tetrahedral carbon, the reactions are labelled 6-exo-tet or 6-endo-tet depending on whether the leaving group X is outside the 6-membered ring (exo), or inside the ring (endo). [Technically, 6-endo-tet reactions are not ring forming.]

It turns out that although 6-exo-tet reactions are perfectly OK, 6-endo-tet reactions simply do not happen, as the nucleophile cannot stretch itself to reach the C–X σ*.

The opening of a three-membered ring is somewhat intermediate between these two cases. Because the C–X bond is constrained in a small ring, the reaction leans towards being a disfavoured 6-endo-tet reaction, even though formally it is 6-exo (as the leaving group is outside the ring being formed):

The lack of overlap isn't obvious from the 2D diagrams here, but bear in mind that these compounds are not flat – especially in the case of the epoxide, where if the epoxide is in the plane of the paper, the alkyl substituent must be pointing up towards us (or down away from us).

As a result of this, ring opening of the iodonium ion to give a five-membered ring is much more favourable.

A similar example is seen in acid-catalysed epoxide opening.[1,2] The formation of the 6-membered ring (the tetrahydropyran) is not observed at all:

References

Janda, K.; Shevlin, C.; Lerner, R. Antibody catalysis of a disfavored chemical transformation. Science 1993, 259 (5094), 490–493 DOI: 10.1126/science.8424171.
Na, J.; Houk, K. N.; Shevlin, C. G.; Janda, K. D.; Lerner, R. A. The energetic advantage of 5-exo versus 6-endo epoxide openings: a preference overwhelmed by antibody catalysis. J. Am. Chem. Soc. 1993, 115 (18), 8453–8454. DOI: 10.1021/ja00071a067.


What would carbon-oxygen polymer be called? So I have a chain molecule made of carbon and oxygen, and I'm not sure what it would be called. Would any of you lovely fellows be able to help me?

The best name I've been able to come up with so far is "polyquadcarboxate" but I'm pretty sure this isn't right.

I'm looking for an IUPAC name if possible, but I suppose other names will do.

Thank you very much!

The generic name is oxocarbon. In particular, "Carbon suboxide spontaneously polymerizes at room temperature into a carbon-oxygen polymer, with 3:2 carbon:oxygen atomic ratio." This is not quite the formula you provide, but the polymer has been synthesized.


Cooking with vinegar and lemon juice (Carpaccio) I know the Maillard reaction to cook at 140–160 °C. I know the low temperature cooking reaction to cook at 70 °C. But which is the reaction that allow us cooking meat or fish with vinegar or lemon juice?

Apparently this question has been covered on the Cooking SE site - in this post, it is reported that:

The Oxford English Dictionary defines to cook as:
To prepare or make ready (food); to make fit for eating by due application of heat, as by boiling, baking, roasting, broiling, etc.
However, from a chemical point of view, what happens when you are cooking meat is that you are using heat, amongst other things, to denature the proteins of the meat. Denaturation can also be achieved by other means, such as marinating the meat in an acid, such as citric acid (contain in lemon juice) or acetic acid (contained in vinegar).
You can also use other agents to induce denaturation, but they are not always fit for eating.
However, it is important to understand that from a purely cooking perspective the two things are not the same: "cooking" with lemon (I guess it is fine to use the verb to cook as long as you put it in quotes), gives a completely different result than cooking with heat. This is because other chemical reactions happen during application of heat and because, obviously, you do not end up with an hot product!

which indicates that the application of acidic media to meat accomplishes the act of denaturing proteins therein in a way mechanistically similar to what cooking with applied heat does.
From Kitchen Daily:

Acids like lemon juice, lime juice and vinegar break down raw meat, allowing the marinade's oil and spices to deeply penetrate and infuse the meat with flavor and moisture. They dig out little pathways in the meat and allow the marinade to seep in. The scientific term for this breakdown is "denaturation," which is the process in which protein loses its structure by application of a strong acid, base, salt, alcohol or heat. Basically, the scientific process that cooks meat with heat in the oven is the same one that cooks it with acid. The results may taste a little different -- fish cooked in the oven is flaky, whereas fish cooked by lemon juice (a.k.a. ceviche) has a more raw-tasting consistency -- but the process is the same.


Why are polysaccharides not sweet in taste? Polysaccharides are defined as polyhydroxy aldehyde or ketone which on hydrolysis yield many units of monosaccharides.
I got one answer(to my question above) as:

  On our tongue, we have things called taste receptors. These receptors are loosely categorised into sweet, sour, bitter and salty. Our sweet-receptors bind to specific types of molecules, namely monosaccharides and disaccharides. Polysaccharides are not as sweet because they do not readily bind to the sweet-receptors on our tongue, as the other smaller molecules do!

My question is that in our body (due to the presence of water),polysaccharides and oligosaccharides are hydrolysed to simpler units of monosaccarides...

So why does the latter respond to the receptors(indicating sweetness) while the former does not(while both yield monosaccarides on hydrolysis)?

Well, while indeed polysacharides are metabolised to simpler units, even in mouth (saliva in the mouth can account for 30% of initial starch digestion), it usually doesn't happen fast enough, so full digestion follows later in you digestive system.

You can`t sense the sweetness of polysaccharides as they do not fit the receptors.


What is the platinum doing in a Zippo hand warmer? My Zippo hand warmer requires lighter fluid, platinum catalytic burner, and a heat source to start the flameless catalytic process. What is the platinum doing in this reaction?

Platinum is a catalyst and decreases the activation energy, so the process can start at lower temperatures.


How large an energy difference in Hartree can be used to distinguish two different structures? For example, I use two different geometries of the same molecule to do optimisation in Gaussian, and the two resultant geometries look roughly the same. The energy difference between the two optimised geometries is only 0.0000001 a.u.. So can I view them as the same geometry? Or is the energy difference large enough to distinguish them? How large an energy difference is large enough for you to see two geometries as different geometries?

I wouldn't use energy as a general criteria to discriminate conformations; a root mean squared deviation (RMSD) of superimposed structures would be a better metric. This is because two very different conformations may have very similar energy. For example, if you perform a relaxed energy scan of one of the dihedrals of cyclopentane, you get nearly the same energy for most of the dihedral angles, bu very different ring conformations.

If you found an energy difference of 0.0000001 a.u. and the superimposition of the conformations is nearly perfect, I'd say they are in the same minimum of the energy surface.


What is the reason for the acidic character in phenols? I came across this concerning phenols:

  The hydroxyl group, in phenol is directly attached to
  the $\ce{sp^2}$
   hybridised carbon of benzene ring which acts as an
  electron withdrawing group. Due to the higher electronegativity of $\ce{sp^2}$
   hybridised carbon
  of phenol to which $\ce{–OH}$ is attached, electron density decreases
  on oxygen. This increases the polarity of $\ce{O–H}$ bond and results
  in an increase in ionisation of phenols than that of alcohols.

My concepts seem to be flawed, and hence I have few questions on the same.

When the $\ce{sp^2}$ carbon atom withdraws electrons from the oxygen atom, the oxygen's electron density decreases. How does this increase the polarity in the $\ce{O-H}$ bond? Shouldn't it be the opposite as oxygen becomes more electropositive due to the carbon atom? I'm missing something here.

Also, does the carbon atom in this case act as electron withdrawing because it is less electronegative than the oxygen atom?

Note that this question isn't about the comparison of acidic character between alcohols and phenols, and the relative stability of alkoxides and phenoxides.

You are confusing the two terms electronegativity/electropositivity and higher/lower electron density.

The source you read states that the $\ce{sp^2}$ hybridised carbon in phenol is more electronegative than the $\ce{sp^3}$ hybridised one in (say) ethanol. This means, that this carbon has a greater desire to draw electrons towards it.

Therefore, even though any carbon is still more electropositive than oxygen, the phenolic oxygen will be able to draw the electrons towards it less effectively and therefore has a lower electron density than in ethanol.

But the oxygen atom in itself hasn’t changed; specifically, it is still hybridised the same way (assuming that one can speak of hybridisation for oxygens). Thus, oxygen’s electronegativity is still the same.

The combination of still having the same electronegativity but having less electron density means that any other electrons will be drawn towards it more strongly — specifically, this concerns the electrons in the $\ce{O-H}$ bond.

In a nutshell: The oxygen’s electronegativity didn’t change (it sounds weird to talk of electropositivity for oxygen if there is no fluorine around) it only has less electron density on one side due to more pulling. It combats that by pulling more from the other side.

This can be called an inductive effect.


Help in drawing resonance structures in case of polybasic acids While drawing resonance structure of dibasic or tribasic acid, e.g. should I take out two $\ce{H+}$ (in dibasic) or three $\ce{H+}$ (in tribasic) together or one by while drawing resonance structure?

Do dibasic acids give both $\ce{H+}$ at once or one by one?

Actually, I was drawing resonance structure of $\ce{H3PO4},$ but got confused. Should I remove all three protons at once or only one?

It depends on what specifically the question is asking you to do.

In reality, a polyprotic acid such as $\ce{H3PO4}$ becomes deprotonated in a sequential way; the first proton is the easiest to remove (most acidic), then the second, then finally the third is hardest to remove (least acidic).

So there are four different species that the question might be asking you about: $\ce{H3PO4}$, $\ce{H2PO4-}$, $\ce{HPO4^2-}$, and $\ce{PO4^3-}$

Each of the three anionic species above has multiple resonance structures.


Proof of kinetic energy equation Is there a way to prove that $\frac{1}{2} mv^2 $$= \frac{3}{2}RT$
The first one is kinetic energy and the second is the average kinetic energy.
Can this be proved by component of velocity.

https://physics.stackexchange.com/questions/45838/how-to-deduce-e-3-2kt

http://hyperphysics.phy-astr.gsu.edu/HBASE/kinetic/molke.html#c1

http://en.wikipedia.org/wiki/Kinetic_theory#Properties

These contain proof at various level . Choose according to your mathematical level


Why cannot nitrosonium ion attack aniline in electrophylic aromatic substitution? This is the nitrosonium ion, which is produced by the reaction of $\ce{NaNO2}$ and $\ce{HCl}$:

[Image source: Wikipedia]

And this is the reaction between aniline and nitrosonium:

[Image source: Wikipedia]

Cannot it take place as an EAS, and with substitution at the ortho or para positions? The amino group is a very good activating group for EAS after all.

Update: I forgot to mention that EAS is observed in case $\ce{-NR2}$ is attached to the ring in place of $\ce{-NH2}$.

What actually happens is that the nitrosonium ion always adds faster to the amino nitrogen.  But that reaction is rapidly reversible unless a proton can dissociate from the amino nitrogen to give a stable N-nitroso product.

A primary or secondary amine has the required N-H proton, and in the case of a primary amine the presence of a second N-H proton enables further reaction making a diazonium ion (as shown for aniline by the OP) or even molecuar nitrogen.  But a tertiary amine has no N-H proton for the elimination step.  A tertiary amine then undergoes no net nitrosation reaction at the nitrogen atom, but if it's aromatic then EAS emerges as an alternative.


How does molecular orbital overlap affect energy? If there is more overlap of the same sign wave function, this apparently leads to lower energy. I don't understand why. My professor said that there would be fewer nodes, and fewer nodes means lower energy. However, why would overlap in general cause the energy to decrease?

Although the nature of orbital and wave functions can be puzzling, by the end of the day chemical bonds are hold together by the familiar electrostatic forces.  Although the electrons don't have a well-defined position, the forces felt by the nuclei is dictated by the spacial probability distribution of the electron.

To form a covalent bond, electron density has to accumulate between the two atoms.   The energy is lowered because the electron is close to both nuclei.   In a chemical bond such electron-nuclear attraction cancels out the repulsion between nuclei and holds the atoms together.

Orbital overlap changes electron distributions.   Orbital with different signs cancel each other in and lower the relative probability electron appear between the bonding atoms.  In the most extreme case of nodal plane, electron density is completely annihilated.  Because the total number of electron is unchanged during interaction, the lowering of electron density between nuclei is accompanied by an increase of density to the far side.

In other words, when you have orbital with canceling phases interacting, electron density will move away from the bonding area, resulting in higher energy.

On the other hand, a strengthening interaction increase the electron density between nuclei, thus lowering the energy and creating a pulling force between nuclei.


How does Ostwald-Walker method work? From searching online I found this:

  The apparatus consists of two sets of bulbs. The first set of three bulbs is filled with solution to half of their capacity and second set of another three bulbs is filled with the pure solvent. Each set is separately weighed accurately. Both sets are connected to each other and then with the accurately weighed set of guard tubes filled with anhydrous calcium chloride or some other dehydrating agents like $\displaystyle P_2O_5$, conc. $\displaystyle H_2SO_4$ etc. The bulbs of solution and pure solvent are kept in a thermostat maintained at a constant temperature.

  A current of pure dry air is bubbled through. The air gets saturated with the vapours in each set of bulbs. The air takes up an amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of solution, i.e. $\displaystyle p_0 – p_s$. The two sets of bulbs are weighed again. The guard tubes are also weighed.

  Loss in mass in the solution bulbs $\displaystyle \propto p_s $

  Loss in mass in the solvent bulbs $\displaystyle \propto (p_0 – p_s) $

  Total loss in both sets of bulbs $\displaystyle \propto [p_s +(p_0 – p_s)] \propto p_0 $

  Total loss in mass of both sets of bulbs is equal to gain in mass of guard tubes.

  Thus, $\displaystyle \frac{p_0-p_s}{p_0} = \frac{\text{Loss in mass in solvent bulbs}}{\text{Total loss in mass in both sets of bulbs}}
= \frac{\text{Loss in mass in solvents bulbs}}{\text{Gain in mass of guard tubes}}$

  Further we know from Raoult’s law

  $\displaystyle \frac{p_0-p_s}{p_0} = \frac{\frac{w_A}{m_A}}{\frac{w_A}{m_A} + \frac{w_B}{m_B}}$

  The above relationship is used for calculation of molecular masses of non-volatile solutes.

  For very dilute solutions, the following relationship can be applied.

  $\displaystyle \frac{p_0-p_s}{p_0} = \frac{\text{Loss in mass of solvent bulbs}}{\text{Gain in mass of guard tubes}}= \frac{w_Am_B}{w_Bm_A}$

My questions:

Why "The air takes up an amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of solution"?
How can we add $\propto p_s$ and $\propto p_0-p_s$ when we don't know what the proportionality constants are or even if they are equal?
It would be better if you could explain in detail the process because I'm not getting it properly from here.For your information the text after "From Raoult's Law" and about "the setup" is purely clear to me.

How are the vapors transported?

The additional question from the comments:

  Why [is] the amt. of vapours taken up proportional to vapour press. Why directly proportional? Why not nothing else?

First things first, let's get a definition (Wikipedia, italics added):

  Vapor pressure [...] is defined as the pressure exerted by a vapor in thermodynamic equilibrium with its condensed phases (solid or liquid) at a given temperature in a closed system.

So basically, before you start the experiment (see below) the vapors have already formed, meaning that thermodynamic equilibrium has been achieved. Then you start bubbling through dry gas, which then mixes with the vapor gas already in the bubbles. If you set the gas flow appropriately, you get a near-equilibrium situation (see the graphic below) and the gas that exits the bulb assemply has exactly the same amount of vapor in it as the vapor "layer" (which can not be thought of as a layer anymore because of the bubbling) above the liquid surface.

We do not have an equilibrium situation, which is why so much care has to be taken that we get as close as possible to the equilibrium situation. (Also, we do not have a closed system, so everything starts to get slightly problematic at this point. But to a first degree of approximation, this will hold up fine.)

This is also why it is directly proportional to the vapor pressure: There is no other mechanism that moves solvent molecules from the condensed phase to the vapor phase. Of course you could add a supersonic vibrating membrane and force more solvent molecules into a vapor phase, but that is not what you want to do, because you want to quantify the relative lowering of the vapor pressure.

Is relative vapor pressure lowering a thing?

  Why "The air takes up an amount of vapours proportional to the vapour pressure of the solution first and then it takes up more amount of vapours from the solvent which is proportional to the difference in the vapour pressure of the solvent and the vapour pressure of solution"?

I guess the question you really want to ask is: Why is the vapor pressure of a solution lower than the vapor pressure of a pure substance?

The answer is not simple, but the explanation of the vapor pressure lowering effect can be given by assuming a simple model: Imagine that in the solution, there are two types of molecules; the solvent and the solute. In the pure solvent, there is only one kind. Statistically speaking, that means that there is also a smaller amount of solvent molecules at the boundary between the bulk phase and the vapor. Since solute molecules don't go over into the vapor, there is a smaller area of solvent molecules that can go over into the vapor, which means that the vapor pressure of the solution is smaller when compared to the pure solvent.

What about the proportionality constant?

  How can we add $\propto p_s$ and $\propto p_0−p_s$ when we don't know what the proportionality constants are or even if they are equal?

While we don't know the magnitude of the proportionality constants, we know that they are equal. The proportionality constant is most likely dependent on ambient pressure, temperature and flow rate of the gas. These three variables are being controlled during the experiment and kept constant for all glass bulbs. (Well, I don't know about keeping ambient pressure constant. But if you perform the experiment fast enough, that shouldn't pose much of a problem.)

The Ostwald-Walker Process

  It would be better if you could explain in detail the process because I'm not getting it properly from here.

Why do we do this?

First, we have to remember: Why are we doing this process; what do we want to find out? The answer is: We want to quantify the relative lowering of the vapor pressure.

How do we do this?

Here's what happens during the process:

Dry gas arrives at the first set of bulbs, filled with a solution.
The dry gas gets saturated with solvent vapor.
The solvent-vapor-saturated gas arrives at the second set of bulbs which contain the pure solvent.
Because of the vapor pressure lowering effect (here now applied in reverse) the vapor pressure of the pure solvent is higher. As such, the gas that enters is not saturated. When it exits the second set of bulbs, it is saturated again.
The pure-solvent-saturated gas enters the desiccator and loses all solvent vapor to it.
Dry gas exits the apparatus.

This might still be hard to understand, so here a graphic that shows you what happens to the solvent vapor partial pressure in the gas:

At time 1 the gas enters the first bulb. It exits the first cluster of bulbs at about time 4. Then it travels to the next cluster of bulbs (containing the pure solvent) and loads up there, up until time 8. At time 9 the gas enters the desiccator where it loses all the solvent vapor and exits dry at time 12.

Since we weigh all components, we know exactly how much mass we lost from solution and solvent, and how much mass the desiccant gained (both those numbers [the total of the mass loss of course] should be equal):
$$ \Delta m_\text{solution} + \Delta m_\text{solvent} = \Delta m_\text{desicc.} $$

The following proportionalities hold:

$ \Delta m_\text{solution} \propto p_s $
$ \Delta m_\text{solvent} \propto (p - p_s) $
$ \Delta m_\text{desicc.} \propto p $

The relative lowering of the vapor pressure can now be calculated, it is simply
$$ \frac{p-p_s}{p} = 1 - \frac{p_s}{p}$$

If you know the vapor pressure of the pure solvent, you can simply plug it in the equation, get the proportionality constant and from that calculate the vapor pressure of the solution.

Conclusion

In conclusion, I can only say that I hope to have clarified things for you. I myself spent quite some time researching this method, as I have never heard of this question before. Thank you for the opportunity!


Why aren't there any triangular molecules? Look at the molecular structure of benzene:

It's a perfect hexagon. Why aren't there any molecules arranged in a triangular fashion with bonds forming the edges and the molecules the vertices? Or, for that matter, square molecules? And what about other n-gonal molecules? octagons?

There exists triangular molecule, eg. cyclopropane, but they are highly reactive due to ring strain.

  Ring strain results from a combination of angle strain, conformational strain or Pitzer strain, and transannular strain or van der Waals strain.

C-C-C bond angle in rings should be approximately 109.5° degrees which in case of small rings like cyclopropane, bond angles are 60° whereas tetrahedral 109.5° bond angles are expected. The intense angle strain leads to nonlinear orbital overlap of its sp3 orbitals. Because of the bond's instability, cyclopropane is more reactive than other alkanes. Since any three points make a plane and cyclopropane has only three carbons, cyclopropane is planar.

Due to this strain they exists in higher energy state and therefore Because of their high strain, the heat of combustion for these small rings is elevated.

Kindly see Ring Strain, it will help you to understand about these kinds of molecules.


Alkali metal enolates Why do lithium enolates add to an electrophile in direct addition manner, while the sodium and potassium enolates tend to add in conjugated additions? The sodium and potassium enolates are well dissociated, making oxygen a hard base (which is why these enolates participate easily in O-acylations), but still, the alpha carbon atom is made softer, to react well in conjugated additions. This seems to me like a contradiction.

I'll assume we're talking about aldol reactions, ie. the electrophile is a aldehyde, or less commonly, a ketone. Being a more strongly coordinating cation, lithium can actually coordinate to both the enolate oxygen atom AND the electrophile oxygen. This serves to "pre-organise" both reactants for a direct addition via a six-membered, cyclic transition state. The favoured transition states are often chair-like. This is known as the Zimmerman-Traxler model.

The model explains diastereselectivity when that is possible, ie. E enolates give predominantly anti products, while Z enolates give mainly syn products.

Lithium also stabilises the initial aldolate product, by chelation via both oxygens.

The more highly dissociated Na and K cations presumably don't greatly facilitate reaction via a cyclic transition state, nor form a stable chelate of the product. The reaction is also more likely to be reversible, and the products of conjugate addition more stable.

Sent from my Samsung Note 8


How to name Hg[Co(SCN)4]? I have basically written the entire name except for the oxidation state of $\ce{Co}$.

  Mercury tetrathiocyanatocobaltate(  )

to know what is the oxidation state of cobalt, I must know the oxidation state of the mercury cation. But mercury can have $+1$ and $+2$ oxidation states. How do I know which one is it in this compound?

TL;DR: There is no inner- or outer coordination spheres in this complex. Both cobalt(II) and mercury(II) have nearly ideal tetrahedral coordination environment with $4$ $\ce{N}$ and $4$ $\ce{S}$ atoms, respectively, hence the proper name would be cobalt(II) mercury(II) tetrathiocyanate. Mercury(I) readily undergoes disproportionation, so assuming mercury(II) is the safest option.

Actually, denoting an inner coordination sphere is not correct for this compound. First crystal structure [1] has been assigned to cobalt(II) tetrathiocyanatomercurate(II) $\ce{Co[Hg(SCN)4]}$:

  It appears reasonably certain that the $\ce{S}$ atom is attached to the $\ce{Hg}$ in tetrahedral co-ordination, with $\ce{S-Hg-S}$ angles of 120° and 104°.

But two decades later the structure has been refined again, by the group including the author of the original publication. This time the geometry has been established more precisely (ICSD#36062), and it turned out that there is an infinite network of $\ce{Co^2+}$ and $\ce{Hg^2+}$ cations cross-linked via thiocyanate ligands, and the new suggested name was cobalt mercury thiocyanate $\ce{Co(SCN)4Hg}$ [2]:

  The combination of two tetrahedrally coordinated atoms, $\ce{Hg}$ and
  $\ce{Co}$, has produced a most unusual arrangement in which the $\ce{Hg}$ and $\ce{Co}$ atoms are held apart by four spirals, each containing $4$ $\ce{SCN}$ bridges, which are interlinked so that any one $\ce{SCN}$ bridge takes part
  in eight spirals.

  [...] Each such spiral is a spring holding the $\ce{Hg}$ and $\ce{Co}$ atoms apart and straining the bonds in the process. It is almost certainly this strain which flattens the tetrahedral coordination round $\ce{Hg}$ and $\ce{Co}$ in the $c$ direction. If the arrangement could be reproduced mechanically it would probably provide the ideal spring mattress!

References

Jeffery, J. W. Nature 1947, 159 (4044), 610. DOI: 10.1038/159610a0.
Jeffery, J. W.; Rose, K. M. Acta Cryst B 1968, 24 (5), 653–662. DOI: 10.1107/S0567740868002980.


Is there any special rules for checking the aromaticity of polycyclic compounds? This question arose in my mind because it's given in a book that pyrene is aromatic. How can it be with 16 π electrons? So, it should be actually anti-aromatic or is there any special rule for these types of compounds?

First, Hückel's rule is only valid for monocyclic systems:

Hückel's rule is not valid for many compounds containing more than three fused aromatic nuclei in a cyclic fashion. For example, pyrene contains 16 conjugated electrons (8 bonds), and coronene contains 24 conjugated electrons (12 bonds). Both of these polycyclic molecules are aromatic even though they fail the 4n + 2 rule. Indeed, Hückel's rule can only be theoretically justified for monocyclic systems.

Pyrene is a commonly used counter-example of Hückel's rule. A refinement specific to Polycyclic Aromatic Hydrocarbons (PAH) is Clar's rule.
The case of pyrene is explained in detail in the literature by a 2007 paper in Petroleum Science and Technology:

This representation agrees with Clar’s sextet rule in which the molecule is represented as having two aromatic sextets, one for each of the rings a and d, and two fixed double bonds on rings b and c (Hernández-Trujillo et al., 2005).


Acidities of (dimethylamino)acetophenone and (diphenylamino)acetophenone How would you compare the acidities of these two compounds?

The deprotonation is obviously alpha to the carbonyl/amine. I think that the diphenyl substituent will be more electron withdrawing than the di methyl, just because the lone pair of nitrogen will have conjugation with the benzene rings, vs methyl groups electron donating onto N.

Yeah, you're thinking in the right way! In the first structure, the hydrogen atom will not have a tendency to leave since the negative charge on the  corresponding conjugate base is destabilized due to lone pair of nitrogen atom(the N atom intensifies the negative charge on the  carbanion formed by donating its electrons in the p orbital of carbanion). Whereas in the second structure the lone pair of N atom is already dispersed in the two phenyl rings so the lone pairs have very less tendency of going into the p orbital of carbanion. Therefore, the second structure is more acidic since its conjugate base is more stable( or less destabilized) than the first structure


Caustic and Toxicological Determination I need to determine whether or not a product that I am importing is "toxic" and or "corrosive" as defined below. But I am a chemistry n00b — that is to say I have virtually zero experience with and or knowledge of chemistry — which is why I am here. I am hoping that one of you guys can help me out.
Corrosive is defined as being

capable of inducing necrosis or ulceration of epithelial
tissue;
capable of causing an erythema or edema of the skin,
corneal or iris damage or conjunctival swelling or redness; or
let us not worry about this right now.

Toxic is defined as being

capable of causing a lethal effect on a human;
capable of causing a serious and irreversible but non-lethal effect on a human, such as a depressed level of consciousness, muscular weakness or paralysis, acute renal or hepatic failure, arrhythmia, hypotension, dyspnea, respiratory depression, pulmonary edema or optic neuritis; or
let us not worry about this right now.

Product Ingredients

Water
Water-soluble vinyl acetate polymer
Water-soluble natural wax
Water-soluble silicone
Natural glue

Please let me know if any information is missing or any other information is required. Also, please feel free to tag or re-tag the question however you see fit.
Thanks in advance for any help whatsoever.

The way to get the information you seek is to obtain a safety data sheet (SDS) for either your product or each of its ingredients.
Safety data sheets have slightly different content by jurisdiction, but most follow a common format, especially those that follow the Globally Harmonized system (GHS).
In a SDS that follows the GHS, Section II identifies the hazards of the substance using tightly defined common phrases. If the product or one of its compounds is toxic in some way or corrosive, Section II will have that information.
Section II contains "Hazard Statements" and "Precautionary Statements". Here is a SDS for acetic acid. There are three hazard statements:

H226 - Flammable liquid and vapor
H314 - Causes severe skin burns and eye damage
H402 - Harmful to aquatic life

There are numerous precautionary statements. The Hazard statements provide the information you want. Here is a list of GHS hazard statements.


Why does the test tube face downwards in gas production? In the preparation of $\ce{NH3}$ from $\ce{NH4Cl}$ and $\ce{Ba(OH)2}$, my teacher said that when heating the test tube, the test tube faces downwards as opposed to upwards. Why is this so? Wouldn't the solid just fall out?

If you mean the test tube containing the solids (reaction tube), then that tube faces downwards to promote the flow of gas into the receiving tube.  The idea is to let the gas sink away from the reaction zone and towards the receiving tube as it cools and becomes more dense.

The receiving tube, of course, faces down so that the ammonia gas, lighter than air, can float to the top after it meets the air.  So you have the ammonia flowing down with cooling through the reaction tube and then flowing up with buoyancy into the receiving tube.

Now for my bonus question:  Had you been producing carbon dioxide by heating magnesium carbonate, would you change the reaction tube?  How about the receiving tube?


Finding the isotherm with given minima Get the temperature of the isotherm for water for which the local minimum is at $\pu{100 Pa}.$ Use the values of $a$ and $b$ of water.

My approach

In the van der Waals equation, set

$$\frac{\mathrm dP}{\mathrm dV} = 0$$

and get the value of $T.$ Plug this value in the initial van der Waals equation and and solve for $V$ from the biquadratic.
The calculations get very messy and the values come out to be unrealistic. (For example, temperature of the order of $10^6$)

My doubt is whether the $\pu{100 Pa}$ as provided by the problem setter a realistic value?

Using Mathematica, I obtained $T=546 K$, as follows.  One of the nice things about using Mathematica for physical calculations is that it has the ability to understand/keep track of/cancel out/convert units.

The volume is non-physically small -- about 750,000 x smaller than what we'd expect for a mole of, say, ideal gas at $P=100 Pa, T = 546 K$).  One does expect a non-physically small volume, since this is at a non-physically low point on the isotherm, but I would not have predicted it would be this much smaller.

If you were doing it by hand, you'd probably want to use a different approach:


What is the alternative to chromyl chloride test? Chromyl chloride test is by far the best test to determine the presence of chloride ions in a salt because no analogous compounds are formed with fluorides, bromides, iodides and cyanides, so this test is therefore specific for chlorides. This test is actually considered selective for $\ce{Cl-}$.

But this test possess certain drawbacks. This test does not give positive results to covalent chloride like $\ce{HgCl2, SbCl3, SnCl2 }$ etc. owing to their slight ionisation and only  partial conversion occurs when added to chromyl chloride solution. Thus, this test is only positive to ionic chlorides. There are also other tests like conc. sulfuric acid test, $\ce{MnO2}$ test or lead acetate test which will almost determine the presence of chloride ions though it possess some drawbacks (see comment section).

For a particular cation or anion, a selective test is chosen in such a way that it will give positive result to any salt, no matter how complex it is and has the capability to differentiate from other analogous salts. So, why Chromyl chloride is still considered as the selective test for $\ce{Cl-}$ despite its drawbacks?

Why chromyl chloride test is used as confirmatory test for $\ce{Cl-}$:

According to this Wikipedia page:

  The chromyl chloride test entails heating a sample suspected of
  containing chloride with potassium dichromate and concentrated
  sulfuric acid. If chloride is present, chromyl chloride is formed and
  red fumes of $\ce{CrO2Cl2}$ are evident. If there is no chloride
  present, no red fumes are produced. No analogous compounds are formed
  with fluorides, bromides, iodides and cyanides, so this test is
  therefore specific for chlorides. [1]

Most chemistry lab manuals mention that only red vapour is not the confirmation for presence of chloride ion. Rather, if the red vapour dissolves in a solution of $\ce{NaOH}$ and turns the solution yellow (due to $\ce{Na2CrO4}$), and that solution gives yellow precipitate with lead acetate solution, then only, presence of chloride is confirmed.

However, this paper mentions that chromyl fluoride is produced on heating calcium fluoride with chromate and sulfuric acid. The compound chromyl fluoride is a red-vapour at normal temperatures, and is also produced by reaction of dichromates and $\ce{HF}$.[2].

Vogel's Textbook of Macro and Semimicro Qualitative Inorganic Analysis[3] mentions that for chromyl chloride test:

  Fluorides give rise to volatile chromyl fluoride, $\ce{CrO2F2}$, which
  is decomposed by water, and hence should be absent or removed.

These suggest that there is a possibility of production of chromyl fluoride by reaction of potassium dichromate, concentrated sulphuric acid and a fluoride sample. As chromyl fluoride is red vapour, this would result in a false positive. Moreover, chromyl fluoride is also hydrolysed in water, [4]  $$\ce{CrO2F2 + H2O->CrO3 + 2HF}$$ and thus it should also turn $\ce{NaOH}$ solution yellow, and produce yellow precipitate of $\ce{PbCrO4}$ with lead acetate. So, there is no way they can be distinguished without performing some other test for chloride or fluoride.

It is not possible to know whether chromyl fluoride is actually produced in large quantities like chromyl chloride without performing the actual experiment. But, I think there is a chance of error with fluoride sample.

Also, bromide salts produce red vapours of bromine when heated with $\ce{K2Cr2O7}$ and conc. $\ce{H2SO4}$(this is from my own experience). I don't know why the Wikipedia page specifically mentions that no red fumes are produced in absence of chlorine, and no analogous compound is formed with fluoride. It appears to me that the information is questionable.

  Alternative to chromyl chloride test:

A feasible (and in my opinion, easier) alternative is this procedure—

Prepare the soda extract or sodium carbonate extract of the salt in question. This would produce $\ce{NaCl}$, even from covalent chlorides. For insoluble chlorides, it has to be done in melt.
Make a water solution of the extract, and filter it.
Add $\ce{AgNO3}$ soln. to this solution. A white precipitate of $\ce{AgCl}$ would form.
To confirm that it is indeed $\ce{AgCl}$ and not $\ce{AgBr}$ or $\ce{AgI}$, add sodium arsenite($\ce{Na3AsO3}$) solution, and shake. Yellow precipitate of $\ce{Ag3AsO3}$ is formed[3]. This does not occur in case of $\ce{AgBr}$ or $\ce{AgI}$. [I am not sure about $\ce{AgCN}$ though.]

Another alternative is—

Add $\ce{AgNO3}$ solution to the sodium carbonate extract.
$\ce{AgCl}$ is white and soluble in $\ce{NH4OH}$. $\ce{AgBr}$ is pale yellow and partially soluble in $\ce{NH4OH}$. $\ce{AgI}$ is yellow and insoluble in $\ce{NH4OH}$. To improve colour detection, use a grooved tile made from black granite, and perform spot test[5].
For detecting cyanide, add iron(II) sulphate to the soda extract and then add a mineral acid. If a precipitate of prussian blue does not appear, then there is no cyanide.

References:

[1] https://en.wikipedia.org/wiki/Chromyl_chloride

[2] Pure Chromyl Fluoride. Alfred Engelbrecht, Aristid V. Grosse. J. Am. Chem. Soc., 1952, 74 (21), pp 5262–5264 |http://pubs.acs.org/doi/abs/10.1021/ja01141a007?journalCode=jacsat

[3] Vogel's Textbook of Macro and Semimicro Qualitative Inorganic Analysis, 5th edition, Longman Group Limited, 1979.

[4] https://en.wikipedia.org/wiki/Chromyl_fluoride

[5] Greener Qualitative Analysis for Anions in a Mixture.
Indu Tucker Sidhwani, Sushmita Chowdhury, Veena Tucker, Shweta Bansal, Elipsha Sharma and Parul Johar. DU Journal of Undergraduate Research and Innovation. Volume 2, Issue 2 pp 70-79, 2016| http://journals.du.ac.in/ugresearch/DU%20e-Journal%20-%20UG%20Research_files/DUJ4%20Vol%202%20Issue1/J%201.8.pdf


The Krieger-Li-Iafrate Approximation I am having trouble understanding what KLI approximation is. Could someone explain it to me in an intuitive/simple way?

The optimized effective potential (OEP) is a natural connection between local density-functional theory and many-body perturbation theory, which is derived from the Sham-Schlüter equation. Converging the full set of OEP equations is quite a challenge. Therefore an approximation was made in order to reduce the integral equations to an analytically solvable one via a dominant orbital approximation, called The Krieger-Li-Iafrate (KLI) approximation.

$$\sum_{i,j\neq i}\frac{\phi_j^*(\mathbf{r})\phi_i^*(\mathbf{r})}{\epsilon_i-\epsilon_r}\left(f_i \langle \phi_i|v_x|\phi_j\rangle - \sum_{k, \alpha}\frac{d_{ik}^{\alpha}d_{kj}^{\alpha}}{2}(\epsilon_i-\epsilon_k)\left[\frac{f_i(1-f_k)}{\epsilon_i-\epsilon_k+\epsilon_{\alpha}}+\frac{f_k(1-f_i)}{\epsilon_i-\epsilon_k+\epsilon_{\alpha}}\right]\right) + c.c.$$

$$=\frac{1}{2}\sum_{\alpha,i,k}\omega_a\frac{\left|d_{ik}^{\alpha}\right|^2 f_i(1-f_k)}{(\epsilon_i-\epsilon_k-\epsilon_{\alpha})^2}\left(\left|\phi_k(\mathbf{r})\right|^2-\left|\phi_i(\mathbf{r})\right|^2\right)$$

Source: Maybe they can see comments above.


Bonds and pairing of spins I was reading Peter Atkins' Elements of Physical Chemistry where I came across this line:

  Bonds do not form because electrons tend to pair; bonds are allowed to form by the electrons pairing their spins.

What does this line mean? What does Atkins want to say through this line & why did he italize the words I bolded above?

I dunno if the author meant exactly this, but it is a well-known fact that pairing of electrons actually consumes energy.

For example, free nitrogen atom in ground state has three unpaired electrons. To form triple bond in dinitrogen molecule, three pairs of electrons must be, well, paired. The process of pairing does consume energy.  However, the energy freed in the bond formation is so big, that it overwhelms pairing energy by huge margin. On the other hand, in oxygen molecule two highest-energy electrons has two orbitals to occupy, and they stay unpaired, so dioxygen is paramagnetic.

So, by no means it is the fact of being unpaired that drives electrons to form bonds. In fact, it prevents formation of weak bonds. However, when two atoms come closer to each other and have two half-occupied orbitals that can form a bond, the fact that electrons may be paired allows this two orbitals to combine into occupied bonding orbital, forming a bond.


Reaction of ethylene glycol with PI₃ (P + I₂) We know that $\ce{2P + 3I2 -> 2PI3}$

Ethylene glycol reacts with $\ce{PCl3}$ to give ethylene dichloride, but why does it gives ethylene when it reacts with $\ce{P + I2}$?

I don't really know why it happens so as it is quite contradictory because in case of primary alcohols, corresponding iodoalkane is formed.

User @tato_mazo already pointed out the reason that due to two iodine atoms being adjacent to each other, ethylene diiodide is sterically crowded and decomposes to produce ethylene and iodine. My answer basically points out the actual reaction involved and some properties of ethylene diiodide.

$$\ce{OHCH2CH2OH + PI3 -> [CH2I-ICH2] -> CH2=CH2 + I2}$$

Ethylene diiodide is an unstable compound which is made by mixing ethene and iodide (damp) under light or slight heating crystallizing into solid, colorless needle like structure (m.p. 75 °C). It again decomposes into corresponding reactants on stronger heating and occasionally at room temperature, though slow. It is only be sublimed or preserved colorless under ethene gas.

References

Ahluwalia, V. K. Textbook Of Organic Chemistry; Ane Books Pvt. Ltd, 2010; Vol. 2. ISBN 978-93-80156-02-6.
Strecker, A. Short Text Book of Organic Chemistry, 1881.


Sweet non-toxic salts I have recently read one post over here in chemistry exchange about tastes of ionic salts, some described lead and beryllium ones to be sweet but toxic, some my point is there any salt that is relatively non-toxic (NaCl for example) but tastes sweet (unlike NaCl)

after some searching and a little help from google I have found that Potassium (K) Bromide (Br). In a low concentration of an aqueous solution it will taste sweet. In higher concentrations it will bitter and even salty. wiki
The toxicity of Bromide isn't very high and it can be tolerated by a healthy human body under some doses. Yet it's long half-life in the human body make it hard to avoid intoxication (as written in the wiki page).

A question arises, why does the change in concentration affect the taste? I would guess that in higher conc. the KBr distorts the water structure and thus affect the interaction of the ions and cations with the taste buds.
I wouldn't use it as a sweetner, and wouldn't suggest tasting it to see the effects of concentration change on the taste.

If we don't restrict ourselves to heteroatomic salts then Sucrose Na is a sweet salt as can be seen here: http://dx.doi.org/10.1016/0031-9384(80)90077-3


Why are we still using cis/trans system while the E/Z system is better than it? I am currently learning about Alkenes. It is given that cis/trans isomerism is either ambiguous or doesn't even apply in cases where the alkene is trisubstituted or tetrasubstituted. To solve this problem, we make use of the E/Z system of classifying the diastereomers of alkenes. In some cases both the methods of classifying diastereomers works.

My doubt is, why are we still using cis/trans system while the E/Z system is better than it? Is the cis/trans system better than E/Z system in some way?

That is because the E/Z-system is an extension of the cis/trans-system. So both systems can be applied to a certain amount of structures and the E/Z-system is then able to treat more advanced systems.

From my perception, it seems that Chemistry is a very stubborn … or maybe just very conservative … field where especially new naming conventions get hardly accepted. At least not very fast. So we are still using Angstrom instead of nanometer or picometer and so on. This also means that we stick with the cis/trans-system as long as it fits and we only change in case we can’t apply it.


Why Gibbs free energy is equating to 0? I have studied thermodynamics, in which the infamous Gibbs equation came.

$$ \mathrm{d}G = \mathrm{d}H - T\,\mathrm{d}S $$

Then my book said the criteria of spontaneity based on gibbs energy is at constant pressure and temperature.

Since $T\,\mathrm{d}S = q = \mathrm{d}H$ (since at constant pressure $\mathrm{d}H = q$), shouldn't $\mathrm{d}G$ all always be equal to $0$?

That is not the Gibbs Free Energy at constant pressure and temperature.

Alphabetic soup definitions:

$G$ - Gibbs free energy
$H$ - enthalpy
$T$ - temperature
$S$ - entropy
$P$ - pressure
$V$ - volume
$w$ - work
$q$ - heat
$U$ - internal energy

Definition of $G$ is $$G = H - TS$$

Definition of $H$ is $$H = U + PV$$

The total differential of $G$ is

$$dG = dU + PdV + VdP - TdS - SdT$$

The definition for the change in $U$ is $$dU = dq + dw$$

There are two kinds of work: pressure-volume work (pv) and non-pressure-volume work (non-pv)

$$dw = dw_{\mathrm{pv}} + dw_\mathrm{non-pv}$$

Thus, substituting we get

$$dG = dq - PdV + dw_\mathrm{non-pv} + PdV + VdP - TdS - SdT$$

If the process is (approximately) reversible:
$$T =\dfrac{dq}{dS}$$

Substituting $q = TdS$ and simplifying, we get:

$$dG = VdP - SdT + dw_\mathrm{non-pv}$$

At constant pressure, $dP=0$, and at constant temperature $dT = 0$.

Therefore:
$$dG = dw_\mathrm{non-pv}$$

Hence, Gibbs free energy is sort of a measure of useful work we can get out of a system.

$$w_\mathrm{non-pv\ by\ system} = - w_\mathrm{non-pv\ on\ system}$$


Isn't the mechanism of aqueous sodium hydroxide and 1-bromobutane involves both heterolytic bond fission and attacking by a nucleophile? What is involved in the mechanism of the reaction between aqueous sodium hydroxide and 1-bromobutane?

A. attack by a nucleophile on a carbon atom with a partial positive charge

B. heterolytic bond fission and attack by a nucleophile on a carbocation

C. homolytic bond fission and attack by an electrophile on a carbanion

D. homolytic bond fission and attack by a nucleophile on a carbocation

I think that A and B would be correct answers, but the correct answer is only A. Bromine accepts both two electrons from the carbon atom and the nucleophile attacks on that carbocation of the carbon atom. Why choice (B) is incorrect? I also searched the internet and found that this reaction is SN1, but why it occurs with the primary carbon, which is usually a reactant in SN2 reaction.

Well your thinking is absolutely right but only for tertiary halogenoalkane as this is a prinary halogenoalkane it will follow SN2 mechamism. In which necleophile attacks on a partial positive carbon