Applying Huckel's rule to non-planar compounds Which of the following compounds is aromatic?

From the 4n+2 rule, it must be either option 2 or 4. The answer given in my book is option 4. Why? I think It might be due to one being planar and the other not, but to me both seem to be planar. Is there a way to check if one is planar or not?

Number 2 is not aromatic because transannular $\ce{C-H}$ interactions cause the molecule to be non-planar although it satisfies $4n+2$. This problem was solved by the synthesis of number 4, which, while a little less than planar but rigid, displays aromatic properties (NMR, etc., Ref 1).
References:

E. Vogel, H. D. Roth, "The Cyclodecapentaene System," Angew. Chem. Int'l. Ed. 1964, 3(3), 228-229 (DOI:  https://doi.org/10.1002/anie.196402282).


Why does temperature affect cell potential? The Nernst equation describes the relationship between cell potential and temperature. But why does temperature affect cell potential?

My understanding is that the collision model of kinetics is not relevant to electrochemical cells, plus an increase in temperature decreases cell potential.
So why does the potential change?

Depending on how you write the Nernst equation the temperature might increase or decrease the potential of the cell. This depends more on the reaction quotient than on anything else.

The temperature comes in to the equation as a scaling factor where RT/nF has units Volt. This essentially defines how much the voltage changes per a decade change in the reaction quotient.

The temperature dependence here is just a matter of the scaling factor.

Going back to the collision model, that is more about kinetics than thermodynamics and it is not at play for the Nernst equation.

Nernst equation describes the thermodynamics of the electrochemical system.

The temperature dependence can be explained by the temperature dependence of work-functions.

Any electrochemical reaction by definition is an electron transfer. The electron goes from the highest occupied MO of one species to the lowest unoccupied MO of the other. Both these levels are temperature dependent.


How many resonance structures does the oxalate ion have? I have drawn the Lewis structure of the oxalate ion, and I see that two oxygens are double bonded to the carbons. When finding the resonance structures, I said there are 2, but the exam solutions state 4, which was confirmed by internet images such as this:

However, I see no real difference btween structures 1 and 2, and 3 and 4. Are they considered different resonance structures?

Resonance structures are not real things. As also indicated by rbw, they have weight on the actual compound. Simpler example is ozone.

http://www.wou.edu/las/physci/poston/ch222/pdf/Ch08-s12-part2.pdf

From the first slide second image there;

you can see that the actual molecule is none of the "resonance" structures, and both bonds between leftmost and rightmost oxygen atoms with central oxygen atoms have equal lenght, and it is between 1 bond length and 2 bond length.

The general idea of drawing resonance structures is, after predicting the most appropriate ones, when you superimpose them and average the bond lengths you will be very near to see the real chemical bonds lengths.

This is straightforward with ozone, oxalate or carbonate etc. but things get complex with increasing atoms, though.

And wrapping up and also answering the question you posed, you should look atoms as if they are all unique, like oxygen 1 oxygen 2 oxygen 3 and oxygen 4. Flip the molecule and you will NOT get the other one, this is not like isomerism. It is just to predict the electron cloud density around atoms of a molecule.

Hope this also helps.


What happens to the water inside the egg when it is boiled? An egg contains 90% of water. When it is boiled, the water inside it cannot escape as water vapor because of the covering of calcium carbonate shell. After boiling, when the egg becomes solid the water cannot be seen. So what happens to the water inside the egg when it is boiled?

The water present in the egg is a colloidal solution of mainly proteins in water.  On boiling proteins get coagulate. The coagulate proteins absorb water (solvent and it disappear).


Is LiF a weak electrolyte in acetic acid? The vapor pressure of acetic acid is 0.033 atm at 25°C. If 5.00g of $\ce{LiF}$ is added to 100g of acetic acid, what would be the vapor pressure of the solution at 25°C?

The lowering in vapor pressure (VP) can be calculated with this equation:

${∆P = X_{solute} * P°_{solvent} * i}$
${∆P}$ is the change (lowering) in VP.
${X_{solute}}$ is the mole fraction of the solute.
${P°_{solvent}}$ is the VP of the pure solvent.
${i}$ is the van't Hoff factor of the solute.

What is the "correct" van't Hoff for this problem, from the perspective of a general chemistry student? I know that $\ce{LiF}$ is not very soluble in water, and I know there are papers on lithium fluoride and acetic acid, but that's all beyond the scope of a general chemistry class.
Therefore, I think it's 2, because $\ce{LiF}$ is expected to be a strong electrolyte. It's ionic, and the solvent is similarly polar (acetic acid has hydrogen bonds). The solvent even resembles water in that both have hydrogen-bonding. So I would expect $\ce{LiF}$ to dissolve fully in acetic acid.
Remember, this is a general chemistry class, and the students aren't expected to know much beyond "like dissolves like." $\ce{LiF}$is polar, and so is acetic acid. So to them, $\ce{LiF}$ should dissolve in acetic acid.
Problem is that I'm in trouble with my supervisor since I told someone that i = 2. My supervisor is telling me that the van't Hoff factor should be 1. I looked over his work to see if he accounted for the dissolving of $\ce{LiF}$when calculating the mole fraction of the solute but he didn't - he just found the mole fraction of $\ce{LiF}$, not the combined mole fractions of $\ce{Li+}$ and $\ce{F-}$.
So, what's the deal with this problem? Is there some weird exception I'm not aware of regarding $\ce{LiF}$ and acetic acid? The van't Hoff factor should be 2, correct?

According to Acid-Base Equilibria in Glacial Acetic Acid. III. Acidity Scale. Potentiometric Determination of Dissociation Constants of Acids, Bases and Salts J. Am. Chem. Soc., 1956, 78 (13), pp 2974–2979:

  It is generally agreed [references 3 and 4] that acids, bases and
  salts are only slightly dissociated in glacial acetic acid.

While the article does not specifically address lithium fluoride, it has quantitative data for lithium chloride.

Lithium Chloride dissociates with a pK of 7.08 +/- 0.02 in acetic acid.

So considering the high concentration of LiF in the problem (5 in 100 grams), if LiF is similar to LiCl and the other salts studied, 1 rather than 2 is correct.

This knowledge is not part of a standard general chemistry curriculum, so unless a specific book or lecture provided relevant information, I wouldn't expect students to know this.

Furthermore, the solubility of LiF in acetic acid is only 0.84 grams per kilogram according to Glacial Acetic Acid as a Non-aqueous Solvent for Metal Fluorides, so the problem is not realistic. (Only 0.084g of the 5.00 g would dissolve).  Also, HF is a strong acid in acetic acid according to that paper, so LiF might even be an exception to the general observation that salts do not dissociate in acetic acid.

Also, keep in mind that in the liquid phase, acetic acid exists as mostly cyclic dimers, which gives the solvent some non-polar characteristics.


Is an acid (just) a proton donor and if not, what is the correct definition of an acid? Why is acid called proton donor?

On the internet, I found:

  Acids are substances that can donate $\ce{H+}$ ions to bases. Since a hydrogen atom is a proton and one electron, technically an $\ce{H+}$ ion is just a proton. So an acid is a 'proton donor.'

However, my teacher said that it is not fully correct. If this is the case, then what's the correct definition of an acid?

Your teacher said that this is not completely correct as there are (at least) three common definitions for acids and bases (Wiki has a good article for this https://en.wikipedia.org/wiki/Acids_and_Bases):

Arrhenius: In water, acids under this definition protonate the water and form Hydronium (H3O+). Bases in this definition deprotonate the water.
Brønsted–Lowry: This is your proton donor definition. Acids in this definition donate protons to bases.
Lewis: In this definition acids are electron pair acceptors and bases are electron pair donors.

The relationship between these definitions is that as chemistry evolved we ran into more chemicals (acids and bases) with similar reactivities that the previous definitions didn't quite encompass and so chemists broadened the definitions. So, as you go down the list above you will find more and more chemicals that match your new definition of acid or base.


Pressure of carbon dioxide due to two simultaneous equilibriums $1~\mathrm{mol}$ each of barium carbonate and calcium carbonate are put in a closed vessel which is maintained at $300~\mathrm{K}$ temperature. Two equilibria are established, with the respective equilibrium constants:
  $$\begin{align}
&\ce{BaCO3(s) <=> BaO(s) +CO2} & K_p&=3\mathrm{~atm} \\
&\ce{CaCO3(s) <=> CaO(s) +CO2} & K_p&=5\mathrm{~atm}
\end{align}$$
  Find the final pressure exerted by carbon dioxide.

Let $n_1$ moles of $\ce{CO2}$ form from barium carbonate and $n_2$ moles of $\ce{CO2}$ form from calcium carbonate.

Then $$K_{p1}=(n_1+n_2)RT$$
$$K_{p_2}=(n_1+n_2)RT$$

But the above two equations are not meaningful because $K_p$ values are different. (what mistake am I making here?)

Let barium carbonate be put into the vessel first. Then there will be $5\mathrm{~atm}$ of carbon dioxide. Then If I put calcium carbonate, will it affect the equilibrium mixture? The answer he gave was $5\mathrm{~atm}$.

It is a conceptual subtlety. It is needed all the species in the hypothetical reaction to require that the equilibrium constant equation be satisfied. This is normally non understood in most chemistry textbooks. The sentence Two equilibria are established is unhappy.

Briefly you can not use this two equations system till you be sure that all the species exist in equilibrium state.

Initially, both solids will react to form the products, once a pressure of 3 atm is reached the only the second reaction will progress. But, as it will increase pressure above of 3 atm the reaction 1 will go in reverse. This situation will continue while CaCO$_3$ is present. Notice that until now there is no requirement that equilibrium constant equation be satisfied due to the states in which the system was were non equilibrium state.
This situation will continue till all BaO disappears. So, the pressure will increase to 5 atm and the first equilibrium won't take place.

Edit: "what happens if CaCO$_3$ is exhausted? then wont CaO react back to reduce the pressure of CO$_2$?"

If CaCO$_3$ is exhausted before reach 5 atm (after that it is not possible), let say at 4 atm, CaO could react back, but as a soon as a it happens (that a little piece of CaCO$_3$ enough big to have the intensive properties of a macroscopic crystal is formed), there will be more tendency to go forward again.

In such case the final pressure will be 4 atm, different from both equilibrium pressure. It will be another equilibrium state. Nothing will change for BaCO$_3$ in this analysis.

Above I exposed two possibilities: The first one, before the edition, that the vessel is small enough to allow the CO$_2$ pressure reach 5 atm. This is the supposed "right answer".
The later where final pressure of CO$_2$ greater than 3 atm and smaller than 5 atm.
If the vessel is very big, it can happen that reactives are exhausted before reach 3 atm, so other equilibrium state will take place.


What is behind of the attraction between a burned matchstick with a magnet? I do not know how to understand the reasons of why a burned matchstick is attracted by a magnet (most probably) of Neodymium.

Here are some screenshots of a video in which this natural phenomenon is presented:

(In the video, see at 1 min 25 s):

Source
Source

Some matchstick heads contain iron(III) oxide as a colorant.

The yellowish color of the burning match indicates that it has low oxygen, i.e. a reducing flame. It reduces the iron oxide to iron which is attracted by the magnet.

The reduction reactions that occur are probably quite complex. Below are two simplified possible reaction equations. $\ce{(CH2)_n}$ in equation (2) stands for paraffin from the splint's impregnation.

$$\ce{Fe2O3 + 3 CO -> 2 Fe + 3 CO2} \tag{1}$$

$$\ce{Fe2O3 + \frac{1}{n} (CH2)_n -> 2 Fe + CO2 + H2O} \tag{2}$$


What is the chemical test for identification of benzophenone? How to identify that a unknown compound is benzophenone?

My working:-

First I added 2,4- DNP(2,4-Dinitrophenyl hydrazine) to the compound. Yellow-orange crystals were formed which indicated presence of carbonyl group. Then, I added Tollen's reagent to the compound. It did not gave silver mirror which means aldehyde group is absent. So, it must be a ketone. So, I added m-Dinitrobenzene to the compound. It did not gave violet colour. Alternatively, I added Sodium nitroprusside to the compound. It did not gave a red color. Both the test is for confirmation of ketone compound.

I asked my teacher what is the unknown compound. He said that it is benzophenone. Also, in the practical book where all the chemical test were mentioned, it was mentioned  as a side-note that benzophenone does not give the above mentioned test. So, what is the chemical test for identification of benzophenone?

After identification of ketone, to identify benzophenone you need to find the range of melting point using melting point apparatus which you should get in range of $48.3\ \mathrm{^\circ C}$ to $50.6\ \mathrm{^\circ C}$.

Then you can find IR of unknown(benzophenone).

Source: Prezi


What is the name for the relationship between two elements in the same group? I'm looking for a specific term which can be used to describe two elements in the same group of the Periodic Table. This would be useful since elements in the same group tend to have similar chemical/physical properties.
For example, is there a word that could fill in the blank: "X is a _______ of Y", where X and Y are in the same group? It has to be something that is generalisable to all groups, not just one particular group, so terms like "alkali metals" or "halogens" are too specific.

The correct term is congener. Congeners refer to elements in the same group in periodic table.
For example, congeners of the Group 1 elements are Li , Na and K etc.
Therefore , X is a congener of Y is the right statement.


Why does HNO3 not give off H2 when reacting with Cu? A dilute solution of which acid is most likely to produce a  reduction
product other than $\ce{H_{2}}$ when it reacts with a  metal?
(A) $\ce{HF}$
(B) $\ce{HCl}$
(C) $\ce{HNO3} $
(D) $\ce{H2SO4}$
$\tiny{\text{Question from 2012 local chemistry olympiad}}$

The given answer is (C), $\ce{HNO3} $ but I don't understand why.
First off, what does it mean by reduction product -- is it the product that is formed when the acid is reduced in the redox reaction between the metal and the acid?
Second, why is the answer $\ce{HNO3} $?

Whenever any metal is added to an acid, the metal is oxidised to one of its positive ions, at the cost of either the acid which is reduced or water which is reduced to hydrogen. The product formed by reduction is what is referred to as the reduction product.

The reduction potential of $\ce{NO3-}$ is $0.8$ which makes the reduction of nitrate instead of hydrogen ion $\ce{H+}$ ($E^\circ=0.0$) more favourable. Thus it is the nitrate in nitric acid which is reduced instead of $\ce{H+}$ or $\ce{H2O}$. But at very low concentration of nitrate ion and low temperature, as explained by Nernst equation, the reduction potential decreases significantly and combined with overpotential effects, it is $\ce{H2}$ which is formed and not the reduction product of nitrate which is $\ce{NO2}$.

In the case of halic acids, the halogens cannot be further reduced. In the case of the sulfate anion, although it has a reduction potential of $0.17$, owing to overpotential effects, sulfate is not reduced, instead, water is reduced yielding $\ce{H2}$. At high concentration, the reduction potential rises sufficiently to overcome the overpotential effects, allowing the reduction of sulfate to form $\ce{SO2}$ at higher concentrations of sulfuric acid. Hence under normal conditions, only nitric acid gives a product other than hydrogen gas.

Reduction potential data

$$\ce{SO4^{2-} + 4H+ + 2e^- -> SO2 + 2H2O}(E^\circ=0.17)$$
$$\ce{NO3- + 2H+ + e- -> NO2 + H2O} (E^\circ=0.8)$$
$$\text{Nernst equation } E=E^\circ- \frac{RT}{nF}\ln Q$$
$Q$= reaction quotient, $E$=redox potential, $E^\circ$=Standard redox potential

EDIT
The oxidation of metal at the cost of reduction of another substance occurs whenever it is thermodynamically favourable. Usually, (except silver, copper) metals have negative reduction potentials, which makes their oxidation at the cost of hydrogen thermodynamically favourable as the $\Delta G<0$. This leads to a spontaneous redox reaction. This is not true if there is a kinetic or physical barrier to reaction. For example, Iron is rendered passive by nitric acid due to the formation of an oxide layer at the surface which prevents any further contact and hence stops the reaction.


What are inner core electrons and how do they influence chemical and physical properties? There are six groups of p–block elements in the periodic
table numbering from 13 to 18. Boron, carbon, nitrogen,
oxygen, fluorine and helium head the groups. Their valence
shell electronic configuration is $\ce{ns^2 np^{1-6}}$(except for He).

  The inner core of the electronic configuration may,
  however, differ. The difference in inner core of elements
  greatly influences their physical properties (such as atomic
  and ionic radii, ionisation enthalpy, etc.) as well as chemical
  properties

The above lines are from my school's textbook.

How does inner core electron create such differences and and what are inner core electrons?

Inner core electrons do have a very strong affect on the chemistry of elements,for example inner d orbitals have a very poor shielding effect thus the outer electrons are bounded more strongly to the nucleus thus decreasing their reactivity and oxidation states.You will come across inert pair effect,stability of lower oxidation states down the group etc because of effect of inner core electrons.


Why does increasing the temperature for a reaction at equillibrium cause the reaction to shift to the endorthermic side? Increasing the temperature would also increase the rate of the exothermic reaction, however what drives the molecules present in the system to choose the endorthermic pathway?

Think about it this way:

There are two people. One guy likes chocolates and one guy hates them. They decide to transfer the chocolates. If you start throwing chocolates randomly to both of them, in which side will the reaction go? The guy who hates chocolates would keep giving it to the guy who likes them.

In a similar way, if you provide heat to a reaction, it would move in the direction of the where the heat is absorbed, i.e the endothermic pathway.


Can you find 5 mistakes in this organic molecule? I found two:
1) the Oxygen needs a positive charge
2) the carbon on the far bottom has 5 bonds

The five mistakes are:

Oxygen in the 5-member ring needs a positive charge [mentioned]
Carbon at the bottom should be a methylene, not methyl [mentioned]
Alkyne Group is not drawn in a linear fashion [mentioned in comments]
Carbon with $3^o$ alcohol has two substituents going out of the plane [mentioned in comments]
Chiral Carbon to the left of the oxygen in the 5-member ring has no stereochemistry indicated.

Now, 5 is not technically wrong but since the rest of the molecule is labelled for stereochemistry it is odd that this chiral center is not marked.


Why only molecules with three or more atoms undergo unimolecular reaction? Only molecules with three or more atoms undergo unimolecular reactions. A diatomic molecule cannot dissociate in this way because it has a single mode of vibrational freedom. If this mode is excited by an amount equal to dissociation energy, the molecule dissociates in $10^{-12}$ seconds.

What is the connection between the dissociation of diatomic molecules and vibrational degrees of freedom ?

I think this is a false statement because molecular hydrogen dissociates into hydrogen atoms, even its dissociation energy is given alongside other diatomic molecules. Am I missing something ?

I'm guessing this quote is given in the context of classical transition rate theory. Although the statement may seem very strong at first, it is generally true but has some important caveats.
Energy Landscapes
Lets consider a diatomic molecule.  It's geometry can entirely be described by one degree of freedom - the bond length.  As we move along that degree of freedom, stretching or compressing the bond we can develop a surface of the energy of the bond at different distances.  A common model used to approximate this surface is the Morse Potential.

These surfaces are often call "energy landscapes".  Minima in the energy landscape are stable structures - our starting materials and different products.  If you push them a bit they move back down the surface to where they started.
And maxima are our transition states between minima, push a little and they'll move down one side of the barrier or the other and into a minima.
Also, a system can easily explore the parts of the energy landscape with less energy than the system currently has - the "classically allowed region" - like a ball rolling up and down the side sides of a valley.
Over time a system tends to move along the energy landscape to positions of lower energy if allowed time to equilibrate and release energy to the outside - like a ball rolling to rest at the bottom of the valley, losing energy to friction.
So normally, below the dissociation energy, the system will roll around in the bottom of the energy landscape, vibrating the bond.
In classical transition state theory, a reaction happens when the atoms move over the energy landscape  along a particular degree of freedom or group of degrees of freedom.
In order for a diatomic molecule to dissociate, it needs to have enough energy moving along the bond vibration degree of freedom to explore parts of the energy landscape at large $r$.  Which means it has to have a lot of energy in the vibrational mode as mentioned in your quote, an amount about equal to the dissociation energy.  If it mode has that much kinetic energy though then, as mentioned above the molecule will fall apart incredibly quickly.
Changing Energy Landscapes
However, rather that adding thermal energy we can change the shape of the energy landscape.  Many different energy landscapes exist at different energies.  the Difference between different landscapes is the arrangement of the electrons in the molecule.
By shining high energy light at a molecule, you can excite the electrons in it into a different configuration of orbitals, leading to a different shape .  This is what happens in photodissociation.  You excite the molecule into a different electronic configuration, and onto different potential energy surface.  On the new surface the molecule is no longer sitting in a minimum, but near a maximum and the system releases energy by moving down the surface - this time towards extending the bond length and dissociating, rather than towards a stable minimum.


Condensation Rate is Increased with a Decrease in Temperature? I'm having trouble understanding why the rate of condensation is increased with a decrease in temperature. My reasoning is as follows:

If...

The rate of condensation (R) depends on the concentration of gas molecules (reactants), and it's defined as:

R = k[Gas]

And...

Any rate is proportionate with respect to temperature; as is k...

Then why would a decrease in temperature increase the rate of condensation (R)?

My deductive reasoning for this is as follows:

The problem in your question is the over-simplified statement that the rate of condensation is simply proportional to the concentration of the gas.  That is only part of the very complex picture here.

The primary key to condensation rate is the degree of supersaturation of the water vapor at the temperature of the object on which the water is condensing.  Under conditions of sub-saturation, i.e. the ambient water vapor concentration (vapor pressure) is less that the vapor pressure of water at the temperature of the object, no bulk (excluding molecular layers of chemisorbed water) condensation occurs.  This of course make your equation R = k[Gas] completely false under these conditions.

Once the partial pressure of water vapor exceeds the saturation vapor pressure at the temperature of the object, i.e. the relative humidity exceeds 100% (by decreasing the temperature), then bulk condensation of water can occur.  Because our atmosphere has an excess of water condensation nuclei, relative humidities of greater than 102% rarely occur.  So, at a given temperature, the condensation rate is roughly proportional to the partial pressure of water for relative humidities of 100-102% at the surface of the object.  Note that the relative humidity increases with decreased temperature. Under these specific conditions, your equation is approximately correct.  I say approximately because there are other factors like air movement (wind velocity), and kinetic limitations like the rate at which water diffuses across the nm or so layer of air at the surface of the object at which the wind velocity approaches zero.

Further complicating the issue is the temperature dependence of the water vapor concentration at 100% humidity.  In other words, there is much more water vapor in the air at $\mathrm{30^oC}$ and 100% humidity vs. $\mathrm{5^oC}$ and 100% humidity.  This means that the amount of water vapor available for condensation at any given degree of supersaturation, say 101% humidity, is a function of temperature.

Because of the complexity of this issue my answer has been largely qualitative in nature.  For a thorough discussion of this issue (albeit even more complex as it includes discussions of water vapor-to-ice condensation), see this article, "Supersaturation of Water Vapor in Clouds", Journal of Atmospheric Science, December 15 2003, p. 2957.


How can iodized oil be made? I want to combine solid elemental iodine with olive oil, to form an iodized oil - what is called an iodostearic acid or iodooctadecanoic acid,  here the double bond in the oil becomes saturated by iodine. I have not done this before and I am looking for guidance on how to do this with basic kitchen tools. Would this be as simple as adding iodine and oil in the ratio of 8 parts iodine to 10 parts oil (as the iodine number for olive oil is around 80), and then waiting several hours or days for the reaction to take place? Is heat or some catalyst needed? Please advise! Thank you.

Essentialy yes, this will work for an unsaturated acid. It will work better if you dissolve your acid substrate in a volatile organic solvent (ether or dichloromethane would be good) and stir it efficiently, wash with sodium thiosulfate solution to remove any unreacted iodine, then allow the solvent to evaporate.


What did actually happen when Staudinger tried to synthesise diamond? It is a famous reaction that used to be a teaching lab experiment, but is now banned in Germany, because it is too dangerous. To quote-translate quite liberally from the German www.seilnacht.com:

  In a stone quarry near Zurich, Hermann Staudinger attempted to artificially produce a diamond by reacting the potassium-sodium melt with tetrachlorocarbon in a bomb tube. The attempt failed as the mixture tore the bomb tube apart.
  In einem Steinbruch bei Zürich versuchte Hermann Staudinger einen Diamanten künstlich herzustellen, indem er die Kalium-Natrium-Schmelze mit Tetrachlorkohlenstoff in einem Bombenrohr reagieren ließ. Der Versuch misslang, da die Mischung das Bombenrohr zerfetzte.

I was probably one of the last students who were still able to see this experiment, and while considerably down scaling the actual set-up, it was still one of the most powerful explosions I ever witnessed.

To me it was actually surprising that I found pages, while researching this reaction, that asked whether $\ce{CCl4}$ can act as a solvent in the Wurtz Reaction, or that in the Wurtz reaction of carbon tetrachloride and sodium no product would be formed, because $\ce{CCl4}$ is too stable to react.

However, due to this I found the following passage in Inorganic Chemistry:

  Tetrahaloalmethanes can be reduced by strong reducing agents, such as alkali metals. For example, the reaction of carbon tetrachloride with sodium is highly exoergic:
  \begin{align}
\ce{CCl4 (l) + 4 Na (s) &-> 4 NaCl (s) + C (s)}&
\Delta_\mathrm{r}G^\circ &=\pu{-249 kJ mol^-1}
\end{align}
  This reaction can occur with explosive violence with $\ce{CCl4}$ and other polyhalocarbons, so alkali metals such as sodium should never be used to dry them.

Unfortunately I reached a dead end with finding more literature about the reaction. I am especially interested in the mechanism, and the final products. I would assume that the above equation only represents a theoretical model, and the products will likely be carbon dioxide and other combustion related products. I'd be interested to know if there are any practical and/ or theoretical studies to the kinetics and thermodynamics of the reaction.

If the source does not cover tetra-, but tri-, or bihalogenated methane, I assume it is fine, too, but it should go beyond the Wurtz reaction ($\ce{2 R-Cl + 2 Na -> R-R + 2 NaCl}$).

This is not really an answer, but more like a historical and literature review. Anyway, I hope it can help to further understand the reaction the the ideas behind.

First of all, thaditional Wurtz reaction suggests formation of 2D bonds such as $\ce{R^1 - R^2}$, whereas the idea in Staudinger's experiment is to obtain hree-dimensional network, ideally the the $\mathrm{sp^3}$ diamond hybrid state. Reaction of alkali metals with organics with high halogen content most likely will always lead to an explosion. For this reason qualitative Lassaigne's test should always be performed carefully, and one must not use fire extinguishers with tetrachloromethane for burning sodium.

According to the great review in [1, p. 8], the ﬁrst documented record of a combustion reaction between a metal and halocarbon compound refers to the ways of obtaining aluminium carbide for technical methane production. In 1907 Camille Matignon observed a very exothermic reaction between hexachlorobenzene and aluminium [2]:

$$\ce{10 Al + C6Cl6 ->[\pu{225^\circ C}] 2 Al4C3 + 2 AlCl3}$$

In 1913 H. Staudinger published his attempts to synthesize ethylenedione $\ce{O=C=C=O}$ [3]. He suggested  abstracting the halogen atoms ($\ce{X}$) from either oxalyl chloride and -bromide $\ce{(COX)2}$ by a reaction with potassium or its sodium alloy ($\ce{NaK}$). This is one of the experiments performed by Staudinger (doing science was much more fun back then):

  In einer starken Bombe wurde aus 5 g Kalium und 5 g Natrium die flüssige Kalium-Natrium-Legierung hergestellt, dazu ein Röhrchen mit 8 g Oxalylbromid gegeben, und die Bombe nach dem vollständigen Evakuieren zugeschmolzen. Wenige Sekunden nach dem Zertrümmern des Röhrchens trat äußerst heftige Explosion unter gewaltiger Detonation ein, die nur durch glücklichen Zufall ohne erhebliche Beschädigung des einen von uns verlief.
  Liquid potassium-sodium alloy was prepared from 5 g of potassium and 5 g of sodium in a strong bomb, then a tube of 8 g of oxalyl bromide was added, and the bomb was melted after complete evacuation. A few seconds after the tube had been trimmed, and an extremely violent explosion ensued, with a great detonation, which proceeded only by chance without considerable damage to one of us.

Formation of an unknown highly reactive intermediate such as Dioxyacetylenkalium was suggested, but it was not confirmed experimentally.

A decade later, in 1922, Staudinger reported explosive reactions of alkali and alkaline earth metals with partially and perhalogenated solvents ($\ce{CH2Cl2}$, $\ce{CCl4}$). He again assumed the formation of different very instable species on contact that would be very sensitive to mechanical impact and thus trigger an explosive reaction. [4, part "5. Explosion mit Alkalimetallen"].

At the same time he also proposed usage of reaction between various alkali and alkaline earth metals ($\ce{Na, K, Ba}$) with halogenated organics with short $\ce{R}$ ("Rest des Moleküls") ($\ce{CH2Cl2}$, $\ce{CHCl3}$, $\ce{CCl4}$, $\ce{CH2Cl-CH2Cl}$, $\ce{CHCl2-CHCl2}$, $\ce{CHCl2-CCl3}$, $\ce{C2Cl6}$, $\ce{CHCl=CHCl}$, $\ce{CHCl=CCl2}$, $\ce{CCl2=CCl2}$, also corresponding $\ce{Br, I}$-derivatives) as an ammo detonating charges:

DE Patent 396,209 1922 "Verfahren zur Darstellung von Sprengmitteln" (DPMA Publikationsnummer CH 100 199 A);
DE Patent 391,346 1922 "Verfahren zur Initialzündung von Sprengstoffen" (DPMA Publikationsnummer CH 100 200 A).

He also underlined that the more halogenated the molecule is and the less other groups are present, the more vigorous the explosion will be. In addition, potassium shows better overall performance due to electropositivity ("Vergleicht man die Reaktionsfähigkeit der ver
schiedenen Alkalimetalle, so ist das elektropositivste, das Kalium, das reaktionsfähigste, Lithium das reaktionsträgste"), but various amalgams or alkali metal mixtures will outperform potassium (flux).

Based on this, later its been proposed that the explosive reaction between sodium and tetrachloromethane to make diamond. His idea was implemented in late 1980s for obtaining diamonds in TNT/RDX detonation soot [5].

Finally, an autoclave synthesis based on original Staudinger work (funny enough, they didn't cite it) implementing sodium and tetrachloromethane [4] resulted in successful nanodiamond synthesis (in a 2% yield), and, subsequently, a well-recieved report in Nature in 1998 [6]:

$$\ce{CCl4 + 4Na ->[\pu{700^\circ C, 48 h};  Ni-Co (cat)] C(diamond) + 4 NaCl}$$

  We call it the reduction-pyrolysis-catalysis (RPC) route. According to the free energy calculations $\Delta G^\circ (\text{diamond}) = \pu{-416.7 kcal mol^{-1}}$ and $\Delta G^\circ (\text{graphite}) = \pu{-417.4 kcal mol^{-1}}$,
  $$\ce{CCl4 + 4 Na -> C(diamond) + 4 NaCl}$$
  and
  $$\ce{CCl4 + 4 Na -> C(graphite) + 4 NaCl}$$
  are thermodynamically spontaneous. Formation of graphite (amorphous carbon) and diamond is possible, and the yield of graphite (amorphous carbon) and diamond may be determined by kinetics.

The role of $\ce{Ni-Co}$ catalyst is not clear yet, and athors suggested than experimentation with other transition metals can potentially improve the process.

Review [1] points out that idea of diamond synthesing is mentioned in Staudinger's monography [7], though I failed to find this book, so unfortunately I cannot provide any substantial details from there.

Bibliography

Koch, E.-C. Metal-fluorocarbon based energetic materials; Wiley-VCH: Weinheim, 2012. ISBN 978-3-527-32920-5.
Matignon, C. Comptes rendus hebdomadaires des séances de l’Académie des sciences 1907, 145, 676–679.
Staudinger, H.; Anthes, E. Ber. Dtsch. Chem. Ges. 1913, 46 (2), 1426–1437. DOI 10.1002/cber.19130460222.
Staudinger, H. Angewandte Chemie 1922, 35 (93), 657–659. DOI 10.1002/ange.19220359302.
Greiner, N. R.; Phillips, D. S.; Johnson, J. D.; Volk, F. Nature 1988, 333 (6172), 440–442. DOI 10.1038/333440a0.
Li, Y.; Qian, Y.; Liao, H.; Ding, Y.; Yang, L.; Xu, C.; Li, F.; Zhou, G. Science 1998, 281 (5374), 246–247. DOI 10.1126/science.281.5374.246.
Staudinger, H. Arbeitserinnerungen; Hüthig Verlag, 1961.


Would this recipe on stink bomb work? I found this recipe on how to make a stink bomb:

The recipe basically says to mix ammonia and match heads to create ammonium sulfide. However, I am skeptical, because

It seems that match heads have mostly phosphorus, and only a negligible amount of sulfur.
I can't find any reference for synthesizing ammonium sulfide from ammonia and sulfur; rather, I found that it is synthesized from hydrogen sulfide and ammonia.

Would this recipe still work?

Would it also make malodorants other than ammonium sulfide?

If so, would these other malodorants be safe? Or would it be better to use a different process to create a stink bomb, like using actual sulfur, or something  entirely different?

Not all the matches have sulfur but only some types in The Preparatory Manual of Black Powder and Pyrotechnics
by Jared Ledgard see page 191, you can find the best matches for your needs, classic matches are okay.

You are right, no reaction is reported between sulfur and ammonia. However if you put in a closed box a compound containing sulfur and a little bit of organic matter Sulfur-reducing bacteria (SRB) reduce sulfur in sulfide anion.

$$\ce{S + 2e- -> S^{2-}}$$
$S^{2-}$ is a base and reacts with water to produce hydrogen sulfide:
$$\ce{S^{2-} +2 H2O -> H2S + 2OH-}$$

$H_{2}S$ is a conjugate acid of $S^{2-}$ so now can react with ammonia with an acid-base reaction:

$$\ce{H2S + NH3 <=> NH4SH}$$
or:
$$\ce{H2S + 2NH3 <=> (NH4)2S}$$

Of course like many ammonium salts it can easily decompose into the two odorous gasses.

Others odorous substances are  skatoles, mercaptans, indoles however I don't think that these are present in great quantities in this case. See this site for further informations.


Why the magnitude of dipole moment influences on FTIR % transmittance? In 3:50 of this video two FTIR spectras are shown. He explains that the % transmittance (double bond in his example) depends on the dipole moment, $\mu = \delta \cdot d$.

I understand that the absorbance center frequency depends on $\delta, d$, but why the percentage of transmittance depends on $\delta, d$? A private case of this would be a symmetric bond ($\delta=0$): why the transmittance is 100%?

Light has both oscillating electric and magnetic fields.  In the common UV-Visible-IR spectroscopy experiments it is the interaction of the molecule with the  electric field that is being observed. A molecule is very small compared to the size of a beam of light, so it is fair to view the molecule as being immersed in or surrounded by the light.  In other words, it is fair to view the molecule as being immersed in a uniform electric field.

The strength of the interaction ($I_s$) of the light's electric field ($E$) with the molecule is dependent upon both the strength of the light's electric field and the vector description of the molecule's electric field.  The molecule's electric field is described by the spatial arrangement of the molecule's electrons.  The spatial arrangement of the molecule's electrons is described by the dipole moment ($\mu$), a vector sum.  Hence,
$$I_s = E \cdot \mu $$

The absorption intensity of a spectral transition (the area under the absorption curve - usually it correlates with the peak height) is given by the same equation except that rather than use the molecular dipole moment $\mu$, one uses what is called the transition dipole moment.  Rather than consider the ground state distribution of electrons around the molecule, one must use the rearranged distribution of electrons around the molecule that occurs when the molecule absorbs light.

If the transition dipole moment is always similar to the ground state dipole moment, then what the video says is correct.  This may be true for larger molecules when we are looking at an absorption that excites a $\ce{C-H}$ or $\ce{C-C}$ bond, or something similar.  In such cases, the small perturbation in electron density around the specific bond probably doesn't radically change the overall electron distribution around the entire molecule.  In smaller molecules like $\ce{CO}$ it seems possible to me that light absorption could make a significant change to the overall electron distribution and that consequently the ground state dipole moment might differ substantially from the transition dipole moment.


What is going on in this reaction? I was trying to remove the electroplating from a aluminum rod using inverse electrolysis and everything went wrong. I'm definitely not a chemistry expert, so I'd like to know what happened here.

I used hydrochloric acid (30%) as the electrolyte, the aluminum rod as the cathode and a common copper wire as the anode. I ran it through a 12V 1A power supply, in a small glass jar.

When I turned the power on a lot of bubbles started to appear at the surface of the rod, as expected. After aproximately 30 seconds the bubbling was too strong, so I decided to turn off the power supply, however, the bubbling didn't stop. As I know that electrolysis of hydrochloric acid releases chlorine, I decided to put the jar to outside, with both electrodes still inside it. After aproximately five minutes, to my surprise, the jar was bubbling extremely strong, spilling over and the electrolyte had turned black, and quite frankly it was kind of scary.

I removed the electrodes and the reaction ceased, but I still have no clue of what happened.

As I recall, $\ce{Cu}$ is inert to $\ce{HCl}$, however $\ce{Al}$ will react with it to produce $\ce{H2}$ and $\ce{AlCl3}$, resulting in a turbid, dark gray solution that bubbles vigorously as the $\ce{H2}$ gas is evolved. The reaction may not have been instantaneous, especially if the $\ce{Al}$ had a passivation layer of oxides on it.

Once initiated, that reaction would've continued even after removal of the external current, and is probably the source of the black precipitate you describe.


Why are the radial wavefunction and radial distribution function different? I'm having trouble understanding the following two graphs, which are the radial wavefunction of the hydrogen 1s orbital and the corresponding radial distribution function:

Specifically, why is the radial distribution function zero at $r = 0$, but not the radial wavefunction?

Also, why does the radial distribution function have a maximum at the Bohr radius $a_0$?

(Mathematica input for the graphs: Plot[{2*Exp[-x], 4*Exp[-2 x]*x^2}, {x, 0, 6}, AxesLabel -> {r/Subscript[a, 0], Subscript[f, "1s"][r]}, PlotRange -> {{0, 6}, {0, 2}}, PlotLegends -> {"Radial wavefunction", "Radial distribution function"}, BaseStyle -> {FontSize -> 14}, ImageSize -> Medium])

The radial wave function $R(r)$ is simply the value of the wave function at some radius $r$, and its square is the probability of the finding an electron in some infinitesimal volume element around a point at distance $r$ from the nucleus.

But, the infinitesimal volume of space at radius $r$ is $4\pi r^{2} dr$ (it's a spherical shell with thickness $dr$ at radius $r$). That means that probability of finding an electron at radius $r$ is proportional to $R^{2}(r)4\pi r^{2}$. But the behavior of this function is such that the probability of finding the electron at radius 0 is also 0.


Naming of a Hydrocarbon with a Benzene Functional Group What would the name of this hydrocarbon be?

I thought it might have been $\ce{3-phenyl-1-butene}$, however, after searching through a number of chemical websites, I found that 3-phenyl-1-butene isn't the correct name (the structures coming up had the benzene ring attached at the 4th carbon from the $\ce{C-C}$ double bond).

First, by your logic, it would be 3-phenylbut-1-ene. Note that the 1 is right before the ene, indicating that is where the double bond is. It is a perfectly understandable name, one that many chemists would use for this compound.

The other way to name it is to consider it a substituted benzene. Then you get (but-3-en-2-yl)benzene. This is the preferred IUPAC name for the compound, because benzene is the base molecule by IUPAC priority rules (rule P-52.2.8 of the 2013 edition of the Blue Book: “Within the same heteroatom class and for the same number of characteristic groups cited as the principal characteristic group, a ring is always selected as the parent hydride to construct a preferred IUPAC name.”).


What can this chemical be? (Qualitative Analysis) I have an unknown white powder, which is an ionic solid, and I need to identify what it is by using qualitative analysis. I've run quite a few tests, but I am having a lot of trouble identifying the anion. The data I've collected is below.
General Properties:

It is a white powder, and I know that it is an ionic solid.
It's density is around $\pu{2g/mL}$ (this is not an exact number)
It appears to be a hydrate (as it loses mass when heated over a bunsen burner)
It is fairly soluble in water. It definitely isn't very insoluble, but I have to add a significant amount of water in order to get it to dissolve.
The powder has a very strong smell (although it's hard to describe).
A solution of the solid is colorless.

Cation Test

A flame test produced a deep orange / orange-red flame

Adding $\ce{Na2CO3}$ to a solution of the solid results in the formation of a white precipitate

Adding a base, such as $\ce{NaOH}$ to a solution of the solid results in the formation of a white precipitate

Based on this data, I am fairly confident that the cation is calcium, but please correct me if there are any other metals that fit the pattern described above or if additional tests are necessary.
Anion Test
This is what has really been messing with me, because I can't seem to get a conclusive test for any anion.

Adding $\ce{AgNO3}$, $\ce{BaCl2}$, or $\ce{Pb(NO3)2}$ to a solution of the solid did not result in the formation of any precipitate, regardless of how much I added.
Adding $\ce{H2SO4}$ results in the formation of a white precipitate (and no bubbles or anything). However, I'm fairly sure that this precipitate is not related to the anion, but rather producing $\ce{CaSO4}$ (from the cation).
Adding $\ce{HCl}$ or $\ce{HNO3}$ to a solution of the solid did not result in any bubbles (neither did adding $\ce{H2O}$ directly to the pure solid)
Adding Ferric Nitrate to the solution just retained the orange color of the ferric nitrate. A dark red color or precipitate was not observed.
Adding $\ce{HCl}$ did not result in the smell of vinegar.
Adding $\ce{AgNO3}$ followed immediately by $\ce{HNO3}$ also did not result in a precipitate.
Adding ammonium molybdate and nitric acid to the solution and heating it caused the solution to turn yellow, but did not yield a precipitate (even after centrifuging). This test was done twice with different ratios of ammonium molybdate to nitric acid, and the exact same result was observed both times.
The brown ring test was conducted, and came back with a negative result. It is possible that this test was done wrong, because the acidified ferrous sulphate was added drop by drop on top of the solution, rather than by rolling it down the side of the test tubes, which may have interfered with test results.

Are there any tests for ions that I have missed? I've done almost every test I know trying to identify the anion, but no dice. The results for the ammonium molybdate test are also confusing to me, because I expected either a yellow precipitate (to confirm phosphate) or no reaction, but the test seems to have gone halfway by just turning yellow. Any guidance on what I should do next would be very much appreciated.

What about a possible $\ce{F-}$ anion? While the nature of the cation might be guessed based on the flame test, since the $\ce{AgNO3}$ test gave no precipitate, it might have produced $\ce{AgF}$ (soluble in water).
This said, by looking at the densities of $\ce{M-F}$ systems with a cation with orange flame and density around $\pu{2 g ml-1}$, I'd say it's probably $\ce{NaF}$.
You might test for fluoride with a selective electrode, if you have one.
Otherwise, here are listed some other analytical methods for the identification of fluoride anions:PubChem: Sodium Fluoride-Analytical Methods


How to assess the strength of the acid? Which is the stronger acid in the given pairs?

  $\ce{HClO3}$ and $\ce{HBrO3}$
  $\ce{HClO2}$ and $\ce{HClO}$
  $\ce{H2Se}$ and $\ce{H2S}$

given that the number of oxygen atoms is the same, the most electronegative atom should result in a stronger acid, thus $\ce{HClO3}$ is stronger.
$\ce{HClO2}$ is stronger since it has more oxygen atoms.
Here's where my logic fails, since the structure is the same (both have $\ce{2H+}$) then it boils down to 2 things to account for: electronegativity and the size of the atom. Going with electronegativity, sulfur is more electronegative thus stronger acid  which was my original answer. But, the correct answer is $\ce{H2Se}$ is the stronger acid which implies in this case atomic size ($\ce{Se}$ is larger than $\ce{S}$) outweighs electronegativity. I do not understand why in this case $\ce{H2Se}$ is stronger than $\ce{H2S}$ given sulfur is more electronegative, and most importantly why the same logic used in 1. can't be applied here and how do I discern where to account for atomic size specifically or when to account for electronegativity?

Any detailed-dumbed down explanations  on the logic are highly appreciated.

Let's start with $\ce{HXO_3}$.

In $\ce{HXO_3}$, after removal of proton, an $\ce{XO_3-}$ ion is formed which is resonance stabilised. For example, if X was $\ce{Cl}$,

Now, this shows that the $X$ needs to have a tendency to pull the electrons towards itself which decreases in the order :$\ce{Cl>Br>I}$. So more resonance is present in $\ce{ClO_3^-}$ than in $\ce{BrO_3^-}$ making the conjugate base of $\ce{HClO_3}$ more stable. Hence it is more acidic.
So acidic strength decreases as: $\ce{HClO_3>HBrO_3>HIO_3}$

Now let's compare $\ce{HXO_3}$ and $\ce{HXO_2}$. You should now know that the acidic nature depends on the stability of the resulting conjugate base. Here, the number of resonance structure for $\ce{HXO_3}$ is more. Hence, the acid strength decreases in the order:   $\ce{HClO_3>HClO_2>HClO}$. So its actually the number of resonance structures that affects the acidic strength.

Finally, let's come to hydrides of 16th group. Let me represent any 16th group element as $Y$. Now, as size of $Y$ increases, the $\ce{H-Y}$ bond length increases. This means, the $\ce{H-Y}$ bond becomes weaker and removal of $\ce{H+}$ ion becomes easier. The size of atom decreases in the order :$\ce{O>S>SE>Te}$.

Hence, the acidic strength decreases as : $\ce{H_2Te>H_2Se>H_2S>H_2O}$

This shows that electronegative is not the only reason. For example, if you considered $\ce{HF}$ and $\ce{HCl}$, then you may think that $\ce{HF}$ is more acidic because $\ce{F}$ is more electronegative and hence will be happy to get a negative charge. But, charge on a small atom is unfavourable. As atomic size increases, the charge can spread over a larger area. Hence, $\ce{Cl-}$ is more stable.

The same trend is seen in case of hydrides of 16th group. The charge in $\ce{Se-}$ is spread over a larger area. Hence, its more stable.


Does CDCl3 also Knock People Out? I know that chloroform can be used to make people unconscious, but would that also work with deuterated chloroform, $\ce{CDCl3}$?

My instinct is "yes", but I'm not sure I'm willing to try this experiment on anyone.

In almost all respects, deuterium behaves like hydrogen, chemically. You can safely drink pure $\ce{D2O}$ in reasonable quantities (though not enough to replace more than ~25% of the $\ce{H2O}$ of the body). In fact, prokaryotes can continue to grow in ~100% deuterated media, albeit more slowly. It does prevent mitosis in eukaryotes in large quantities, though.

Given that similarity, there is no reason to assume any different behavior between deuterated chloroform and that with just protium. But don't knock yourself out trying the experiment, as chloroform is hepatotoxic.


How can compounds like XeO4 form tetrahedral structures despite having double bonds? Xenon tetroxide is tetrahedral despite having double bonds. Is it something to do with $d$ orbitals? Also things like sulfur trioxide (trigonal planar with three double bonds but the sulfur is hypervalent). Is it possible for the $\pi$ bond to be formed from overlap of $d$ orbitals from the central atom with $p$ orbitals of the oxygen?

$\ce{XeO4}$ has a tetrahedral structure as you noted, but $d$-orbitals are not involved in explaining its geometry or hybridization. The central xenon atom in $\ce{XeO4}$ is best described as simply $\ce{sp^3}$ hybridized.  We have 8 xenon electrons so we need 4 orbitals to put them in.  The 4 equivalent orbitals are formed by mixing the $s$- and $p$-orbitals (just as we do with carbon) to form the equivalent $\ce{sp^3}$ orbitals pointing towards the vertices of a tetrahedron (which minimizes electron-electron repulsions between the 4 bonds).

The molecule can be represented by several resonance structures and I've drawn two of them (A and B) below.  I've only drawn the lone pairs on one of the oxygen atoms for each resonance structure, but they are the same for all oxygens in each resonance structure.  So the resonance structures tell us that the xenon-oxygen bonds in $\ce{XeO4}$ are some mix of single and double bond character.  In resonance structure A, the xenon and oxygens are neutral; while in resonance structure B, the xenon has a charge of +4 and each oxygen has a charge of -1.

The $\ce{Xe-O}$ bonds in $\ce{XeO4}$ are very weak. In resonance structure A the pi overlap is very poor due to the different atomic sizes of xenon and oxygen. In resonance structure B we just have a simple single bond connecting the oxygen and xenon and it contains 2 electrons, both from xenon and shared by oxygen, oxygen does not contribute any electrons to this bond. The single bond resonance structure B probably contributes more to the description of this molecule than resonance structure A. The bonds are so weak that $\ce{XeO4}$ is unstable and explosively decomposes around -36 C.

In summary we can say that

the xenon in $\ce{XeO4}$ is $\ce{sp^3}$ hybridized
the $\ce{Xe-O}$ bond is a partial single-double bond with the single bond probably being a more accurate description, and
the $\ce{Xe-O}$ bonds are extremely weak because of the poor pi overlap in resonance contributor A; and oxygen does not contribute any electrons to the xenon-oxygen bond in resonance contributor B.


How to assign E/Z configuration according to the Cahn-Ingold-Prelog rules when subsituents differ only by stereochemistry I stumbled-upon this compound:

I'm wondering how to determine the configuration of the double bond. Is it E or Z? How do I decide on the priorities when the substituents can only differ in the configurations on the chiral centers?

In your example, the stereodescriptors R and S are already given. The remaining question is about the double bond which gives rise to cis-trans isomerism.
In the hierarchical Sequence Rules, we finally find

P-92.1.3.5 Sequence Rule 5
An atom or group with descriptor ‘R’, ‘M’, and ‘seqCis’ has priority over its enantiomorph ‘S’, ‘P’ or ‘seq Trans’.

Source: Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book)
Therefore, the configuration would be ‘E’ since ‘R’ precedes ‘S’.
However, the ‘E’ and ‘Z’ stereodescriptors do not distinguish between double bonds whose descriptors are reflection invariant from the geometrically enantiomorphic double bonds whose stereodescriptors are reflection variant.
Following the recommendations for enantiomorphic double bonds, the configuration is ‘seqTrans’.


Melting and boiling points of benzene and fluorobenzene This species is a derivative of benzene, with a single fluorine atom
  attached. Its melting point is -44 °C, which is lower than that of
  benzene, indicative of the remarkable effect of fluorination on the
  intermolecular interactions as seen throughout organofluorine
  chemistry. In contrast, the boiling points of PhF and benzene differ
  by only 4 °C.

Why is this true? Wikipedia doesn't elaborate on why.

From what I understand the melting point is the point at which the solid and liquid forms are in equilibrium.

Boiling point is the point at which the liquid and gas phases are in equilibrium.

I know that intermolecular forces play an integral role in melting and boiling point. However, why does it seem that the strength of the intermolecular forces in the solid phase is stronger than that in the liquid phase for fluorobenzene?

The answer really doesn't have much to do with fluorine.  Beacall's 1928 The melting-points of benzene derivatives explains on the first page that it has long been known "that the more symmetrical the constitution of a benzene derivative, the greater the resistance it offers to the passage from the solid to the liquid state".

In the more modern Effect of molecular symmetry on melting temperature and solubility:

  As a general rule, symmetrical molecules in crystalline form have higher melting temperatures and exhibit lower solubilities compared with molecules of similar structure but with lower symmetry. Symmetry in a molecule imparts a positive amount of residual entropy in the solid phase (i.e., more possible arrangements leading to the same structure). This means that the entropy of a crystal of symmetric molecules is greater than the entropy of crystal of a similar, but non-symmetric molecule.

Benzene and hexafluorbenzene have almost the same melting point, while monofluoro, monochloro, monobromo benzene all have much lower melting points.

Both benzene and hexafluoro-benezne have a very stable "herringbone" crystal structure, where the edge of one ring approaches the center of another ring in a perpendicular manner.

So realizing that the cyrstal structure of benzene is unusually stablized, it is understandable that the melting point of benzene is higher than monofluorobenzene, but the boiling point is not.


In order to be a gas does the pressure exerted by the gas has to be more than the atmospheric pressure? In order to be a gas does the pressure exerted by the gas has to be more than the atmospheric pressure?
Really simple, I think yes to this question but teacher says no. So what is your view into this?

Your teacher is right. For example, at room temperature, the pressure of water vapor over a pond or lake might be around 25 torr; that's less than atmospheric pressure, but the water vapor is undeniably a gas.

Maybe your confusion stems from a belief that gases must be produced by boiling liquids (since a liquid must have a vapor pressure equal to the external pressure before it begins to boil). But a liquid can also become a gas through evaporation, no matter what the external pressure is.

In evaporation, molecules escape from the liquid's surface and become a gas.
Molecules in the liquid are in constant motion, and they frequently collide with each other. Every now and then one will get knocked off the surface of the liquid by the others. This can happen at any temperature or pressure the liquid exists at.

In boiling, bubbles of gas form throughout the liquid. This can only happen when the temperature is high enough to make the vapor pressure equal to the external pressure.


Old samples of magnesium oxide become contaminated with magnesium carbonate. Suggest how this contamination takes place Old samples of magnesium oxide become contaminated with magnesium carbonate. Suggest how this contamination takes place.

I don't understand what I'm meant to put for this.

Carbonate contamination in oxides compounds often happens after a certain time and air exposure because of the $\ce {CO2}$ in the air.

For your magnesium oxide it follows the reaction

$\ce {MgO + CO2 -> MgCO3}$

with $\ce {MgCO3}$ being the magnesium carbonate.

A solution to prevent this kind of polution is to keep the oxide in a dessicator or to treat them in a furnace at a temperature around to $500 °C$ wich is the decomposition temperature of most of the carbonate compounds.


Why is tin(II) fluoride more effective in turning apatite into fluorapatite? In more powerful toothpastes, like Crest Pro-Health, stannous fluoride is used in place of sodium fluoride.

According to Wikipedia:

  Stannous fluoride converts the calcium mineral apatite into fluorapatite, which makes tooth enamel more resistant to bacteria generated acid attacks. In toothpastes containing calcium minerals, sodium fluoride becomes ineffective over time, while stannous fluoride remains effective in strengthening tooth enamel

Is the stannous fluoride more effective in this reaction simply due to having two fluoride ions per mole, or is there something more complicated in the mechanism of the reaction in which the tin spectator ion is beneficial?

Found an answer to this question in a book called 'Pediatric Dentistry'.

Apparently, when Stannous Fluoride reacts with dental enamel (hydroxyapatite), both the fluoride and the tin react to form Stannous Trifluorophosphate $(\ce{Sn3F3PO4})$ -- This is highly resistant to acid attack.

At high concentrations of stannous fluoride (I don't know what counts as "high", the book doesn't say), apparently the following compound is formed as well: Trifluorostannate -- This is also highly acid resistant.

In addition, Calcium Fluoride is also produced, which then further reacts with hydroxyapatite forming Fluorhydroxyapatite (which is also rather acid resistant, although it seems that the stannous trifluorophosphate is more resistant.)

Sodium fluoride only produces Fluorhydroxyapatite as its reaction.

It appears that Stannous Fluoride is a superior agent because it produces 1 or 2 additional compounds (that become part of the surface enamel) that are more acid resistant than just the one that Sodium Fluoride produces. These other compounds (which become a part of the surface of the tooth) also have an anti-microbial effect, thus inhibiting the growth of the very bacteria which are directly responsible for producing the majority of the acid found in dental decay.


One-step oxidation of hex-1-ene to hex-2-en-1-ol Convert the following and show a Mechanism

It has to be just in a single step and on the arrow head you can't use multiple reagents altogether. Can anyone do it? I am unable to do it in a single step.

This is my attempt ..! But it is multistep, one step is required ..!

You're supposed to turn an alkene into an an allylic alcohol in one step. The "classical" reagent for this conversion is selenium dioxide, $\ce{SeO2}$. Have a look at some older references, such as J. Org. Chem., 1979, 44, 4683-4689.
DOI

As kindly pointed out by orthocresol, this elegant method will unfortunately furnish the the wrong allylic alcohol. Whether in situ isomerization is possible (allylic hydroxylation in acidic medium) is feasible can't be decided until I have another coffee ;)

(Please note that a procedure using singlet oxygen will need two steps: the Schenck ene reaction furnishes a hydroperoxide which you would have to reduce.)


Is there an 'easy' way to determine if a ligand is a pi acceptor, donor or neutral? First of all I am confused how something can be pi neutral? something either does or doesn't have filled p orbitals for example.
given something I have never seen acting as a ligand before like trimethylphosphate how can I work out whether this is pi accepting, donating or neutral? I am thinking the donor atom would be the oxygen bonded to the Me group and P but do I only consider the orbital of the donor atom?
For a tetragonal Cu(II) complex how can a ligand even act as pi donating if since the t2g orbitals on the metal are filled?
Sorry about this messy question, I am very confused about this in general, I would be so grateful if someone could help me understand this

If you ask for an "easy" way to determine if a ligand is a pi acceptor/donor or none, look at the spectrochemical series. A quick-and-dirty rule (which means that it is correct most of the time but not always) is that weak ligands (iodide, bromide, hydroxide etc) are pi-donor ligands. The medium ligands (water, ammonia etc) are pi-neutral, and the strong ligands (cyanide, carbonyl, bipyridine, etc) are pi-acceptor.
But this method is some kind of reversing here. Because we normally use a ligand's pi character to guess if it is a strong or weak ligand.
If you can, ligand field theory is your best friend in this problem.
Generally, a pi-donor ligand has an extra electron pair that lies in an orbital that has approximately the same symmetry as the metal center's orbital (usually t2g). This allows the two orbitals to interact with each other and creating two new sets of orbitals: t2g and t2g* (as in the molecular orbital theory). The ligand's pi electron will reside in the  lower t2g orbital, whereas the metal's original d-electrons will reside in the t2g* one. That explains your question of the copper complexes.
In the other hand, a pi-acceptor ligand has an empty molecular orbital of low energy and approximately the same symmetry as the orbital of the metal center. In this case, it will interact with the metal's orbitals and the d-electrons of the metal will be delocalized to the ligand. These ligands usually have a highly conjugated system that greatly decreases the energy of the LUMO (bipyridine, triphenylphosphine etc). In the cases of carbon monoxide (called carbonyl as a ligand) and cyanide, they have low-energy LUMO for other reasons.
A pi-neutral ligand (also called sigma-donor ligand) does not process any of the above characteristics. For example, ammonia has only one electron pair on the nitrogen, and that pair is already used in the formation of the sigma bond, so no more pi electrons to donate. The LUMO of the ammonia is also high in energy, so no pi accepting here.
I think trimethyl phosphate is mostly a pi-neutral with the donor atom is the oxygen double-bonded to phosphorus. The molecule is not conjugated at all, and the oxygen can only donate one pair of electron since the other's orbital does not have the same symmetry.


While filling liquid oxygen into a tank, why do they do open one end? While filling liquid oxygen into a tank, why do they do open one end?
Then what is use of filling?

I don't think people who are filling this truck are beginners.

The end is opened to allow oxygen gas to escape. The flow of liquid oxygen into the tanker will not be pumped, just undergoing the natural tendency to flow from an area of high pressure to low pressure. As such, there would come a point when the residual gas in the tanker would start to provide back pressure on the flow input, slowing it down & potentially stopping it once pressures equalize between the tanker and the source.

By allowing oxygen gas to escape the tanker, the pressure will always be lower in the tanker, and the flow will be faster more adequately fill the tanker. The same technique is used when topping up liquid nitrogen in an NMR spectrometer, for example.


Mechanism of Diels-Alder reaction I have a question about the mechanism of the Diels-Alder reaction. Is this a two-electron or one-electron process as described by Fig (a) or (b) respectively or does not matter? How do we know that?

First of all, full arrow represents a movement of 2 electrons. Half arrows represent movement of 1 electron.

Diels-Alder is a 6π electron pericyclic reaction. The diene provides 4π electrons and the dienophile 2π electrons to the formation of the bonds. The Diels-Alder reaction belongs to [4+2] cycloadditions according to the Woodward-Hoffmann nomenclature.

A purely formal way of determining the participating electrons consists of simply counting the electron flow arrows that are used to illustrate the mechanism and multiplying by two. Each arrow moves two electrons.
You have 3 arrows representing a reaction, so 3*2=6 electrons

The number of electrons participating in the cycloaddition determines wether the reaction follows either a supra-supra (s+s) or a supra-antara (s+a) mechanism.

This is a nice source for Diels-Alder mechanisms


Why is 3-chloro-5-(1-methylethenyl)cyclopenta-1,3-dien-1-ol named so? I'm afraid I can't find an image for this organic compound, but when applying lowest locant rule in the numbering (1-methylethenyl) gets 2 and Cl gets 4. (2,4) is less than (3,5). Why then is lowest locant rule not applied here? Also here (1-methylethenyl) is an alkene, and isn't priority order of alkenes before halogens which have no priority?

The most important simplified criteria for the numbering are:

lower locants for the group that is expressed as suffix
lower locants for multiple bonds
lower locants for prefixes
lower locants for substituents cited first as a prefix in the name

The corresponding actual wording in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book) reads as follows:

  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:

  (…)

  (c) principal characteristic groups and free valences (suffixes);

  (…)

  (e) saturation/unsaturation:

    (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;

    (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);

  (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;

  (…)

The principal characteristic group of the compound that is given in the question is the $\ce{-OH}$ group. It is expressed as the suffix “-ol”. According to Rule (c), a low locant is assigned first to this group, i.e. “-1-ol”.

In the next step, low locants are assigned to the double bonds according to Rule (e), i.e. “cyclopenta-1,3-dien-1-ol”.

That leaves only one possible locant set for the remaining substituents, i.e. “3-chloro” and “5-(prop-1-en-2-yl)”. The rules (f) and (g) are not needed to make a decision.

Therefore, the complete name for the compound that is given in the question is 3-chloro-5-(prop-1-en-2-yl)cyclopenta-1,3-dien-1-ol.

Note that the preferred prefix for the “isopropenyl” or “1-methylethenyl” (completely: “1-methyleth-1-en-1-yl”) is now “prop-1-en-2-yl”.


Instability due to neutrons Why does an excess of neutrons leads to instability?
For example, both $\ce{^{3}H}$ and $\ce{^{4}H}$ are unstable, with respective half-lives of $3.89\cdot 10^8$ and  $1.39\cdot 10^{-22}$ seconds.
By my reasoning,  neutron-rich nuclei should be strongly bound by the strong nuclear force, and should therefore be more stable. Additionally, neutrons, unlike protons, are neutral and therefore do not contribute to repulsive electromagnetic forces.

The key here is that a system will tend towards the lowest energy state; in other words, for a process to be spontaneous, the final state must have lower energy than the initial state.

The mass of a neutron is slightly higher than the combined mass of the proton, electron and neutrino that result from beta decay, meaning that the decay will be energetically favoured. This might be easier to understand by considering the decay of a free neutron:

$$\text{n}^0 \rightarrow \text{p}^+ + \text{e}^-+\bar{\nu}_\text{e}$$

Whether this process will occur in a nuclear neutron is determined by the $N/Z$ ratio and the atomic number. In small neutron-rich nuclei, the process above will predominate, but as the nucleus gets larger, more neutrons are needed to counteract the electromagnetic repulsion between protons (which has a longer range than the strong nuclear force). Again, the reason is energy: the unbound nuclei collectively have higher energy than a nucleus, due to nuclear binding energy. Hence, for larger nuclei, neutron decay is not energetically favourable.


Can I synthesize iron acetate like this? The oxidation state of the iron doesn't matter too much; I am just trying to make an attractive red-brown ink. Here are some instructions I prescribed to a friend.

  Here's an approximately decent iron acetate recipe. I don't know if it'll work, though. Maybe you'll want to heat the acetic acid. The optional first step is to oxidize the iron by hydrogen peroxide or by heating in air. Pour enough heated acid to corrode the iron plenty. Wait. Filter what is now the aqueous solution of iron acetate and acetic acid through some filter paper to remove the undissolved iron. Let the (I hope so) red-brown solution dry: this is because you want to evaporate the acetic acid. When it is dry, you may dissolve what I hope to be a dark red-brown powder in as little distilled water as possible and some alcohol if you so desire. Adjust dilution to suit your hue/shading preference.

Would this synthesize iron (I or preferably II) acetate?

Red iron oxide, Fe2O3, from a paint or art supply store, or as polishing rouge (no additives).  dissolve in aqueous acetic acid, filter off any solids, and go on from there.

Dissolving iron in acetic acid will give you Fe(II)acetate.  Air or peroxide oxidation may (probably will) give you Fenton chemistry and muck up your organics.


How is Persian blue rock salt made? How is Persian blue rock salt made and what does it contain that makes it look blue?

Persian blue salt is formed by evaporating sea water, as are most if not all salt deposits. In the geosciences, we call salt deposits like this evaporites. Persian blue salt is sodium chloride (the mineral name is halite).
According to this description:

Blue salt is one of the rarest salts known to mankind, and is extracted from a salt mine in the northern province of Semnan in Persia.
The intriguing blue color occurs during the forming of the salt's crystalline structure, as intense pressure is exerted on the salt deposits. The individual crystals fracture the light in an unusual way and the resulting blue (which is caused by an optical illusion), becomes visible.


Mass of strontium iodide contained in a 350 mL solution What mass (in mg) of strontium iodide is contained in $350\ \mathrm{mL}$ of a strontium iodide solution that has a concentration of $1.93\ \mathrm{µM}$?

What I have done:

$0.350\ \mathrm{L}\ \ce{SrI_2}\cdot \frac{1.93\cdot 10^{-6}}{1\ \mathrm{L\ Solution}}\cdot \frac{341.42\ \mathrm{g}\ \ce{SrI2}}{1\ \mathrm{mol}\ \ce{SrI2}}=2.31\cdot 10^{-4}~\mathrm{g}~\ce{SrI2}=0.231~\mathrm{mg}~\ce{SrI2}$

What is not correct here?

The mistake is that you didn't write "mol" after $1.93 \times 10^{-6}$.

Some teachers/professors might let that go, others wouldn't, depending upon what they are really trying to teach you.

Units and dimensional analysis are important.


What exactly is net ionic equation? I once saw a problem, iron ions(II) titrated with potassium dichromate to produce Fe(II) and Cr(III). If I am to write the net ionic equation of this reaction, I got to make potassium dichromate on the right into dichromate ion since potassium is a spectator.
I got very confused over here. Why should you see potassium as a spectator if K(II) will be balanced as 2K on the right.

The molecular equation for this titration is $$\ce{K2Cr2O7 + 6Fe(NH4)2(SO4)2 + 7H2SO4 ->
3Fe2(SO4)3 + Cr2(SO4)3 + K2SO4 + 6(NH4)2SO4 + 7H2O}$$

The complete ionic equation is $$\ce{2K+ + Cr2O7^2- + 6Fe^2+ + 12NH4+ + 19SO4^2- + 14H+ ->
6Fe3+ + 14SO4^2- + 2Cr^3+  + 2K+  + 12NH4^+  + 7H2O}$$

When you cross out everybody who shows up on both sides in the same form, you get the net ionic equation for this titration: $$\ce{Cr2O7^2- + 6Fe^2+ + 14H+ -> 6Fe^3+ + 2Cr^3+ + 7H2O}$$

Just as the hydrogens in $\ce{H2SO4}$ become $\ce{2H+}$ so the potassiums in  $\ce{K2Cr2O7}$ become $\ce{2K+}$ and they don't change on the other side of the equation where they form $\ce{K2SO4}$.


Which salt precipitates from a solution of ammonium oxalate and silver nitrate? I am supposed to write the precipitation reaction for

$\ce{(NH4)2C2O4 + AgNO3 ->[\ce{H2O}]}$

Is my equation correct? I left out the Spector ions.
$$\ce{4(NH4) + 2 C2 +O4\rightarrow 4(NH4)2CO3}$$

Your current solution is wrong and you don't seem to be familiar with the concepts involved.

Let's walk through this together and change that!

You take solid ammonium oxalate $\ce{(NH4)2C2O4}$ and silver nitrate $\ce{AgNO3}$ and dissolve both salts in water.
The salts dissociate into their ions. Let's do that separately for both salts first.

$$\ce{ (NH4)2C2O4 ->[\ce{H2O}] 2 NH4+_{(aq)} + C2O4^{2-}_{(aq)}}$$

$$\ce{AgNO3 ->[\ce{H2O}] Ag+_{(aq)} +  NO3^{-}_{(aq)}}$$

Since you are dissolving both salts in the the same beaker with water, your solution now contains all the ions of both equations.

Forget about the spectator ion fanciness.

It is much more important to identify the chemical units (= the ions from which the salts are composed). Leave these ions intact and don't rip them apart into weird combinations of elements.
Precipitation will occur when the combination of these ions (chose a cation and an anion) yields a salt with a very low solubility in water. There are only two cations and anions here, so the number of possible combinations isn't too big. Look up the solubilities of the possible products to figure out which one will precipitate.
Do not forget to balance the reaction!


Should we expect Jahn–Teller distortion in bis(ethylenediamine)nickel(II)? (ii) The stepwise stability constants of the following complexes in aqueous solution at $\pu{25 °C}$ are given below:
$$
\begin{array}{cll}
\hline
\ce{M} & \ce{[M(en)2(H2O)2]^2+} & \ce{[M(en)3]^2+} \\
\hline
\ce{Cu} & \pu{1E9} & 0.1~\text{(estimated)} \\
\ce{Ni} & \pu{1E6} & \pu{1E4} \\
\hline
\end{array}
$$

I got this data from an assignment problem. In the case of copper it makes complete sense for the stability constant of formation of $\ce{[Cu(en)_3]^{2+}}$ to be that low given that we have $\mathrm{d^9}$ configuration and strong Jahn–Teller (JT) distortion leading to strain in the rings formed in the complex.
But I can't figure out if there'll be any distortion in the $\ce{[Ni(en)_3]}$ ($\mathrm{d^8}$ configuration). Initially, I assumed there'll be no distortion and the complex would be satisfied with the chelation alone (compromising on the stability provided by JT distortion). But then some of my friends pointed out that $K$ reduces from $10^6$ to $10^4$ so there must be some kind of distortion. After some googling, the only article with some data I could find is by Sohail et al. [1].
In the linked article, there are no comments on JT distortion. Instead, packing effects have been held responsible for the natural distortion of the complex (the last line under Comment section). But I am not entirely convinced; so my questions are:

Would the $\mathrm{e_g}$ orbitals be symmetrically or asymmetrically filled?

If asymmetrically, then why is the JT distortion so profound in the copper(II) complex but weaker in the nickel(II) complex? And if it's symmetrically filled, then what is responsible for the decrease in the stepwise stability constant?

References

Sohail, M.; Molloy, K. C.; Mazhar, M.; Kociok-Köhn, G.; Khosa, M. K. Tris(ethylenediamine)nickel(II) tetraiodocadmate(II). Acta Cryst E Struct Rep Online 2006, 62 (2), m394–m396. https://doi.org/10/b82dng. (Open Access)

There is a key difference between the $\mathrm d^8$ state and the $\mathrm d^9$ state which lies in exactly that extra electron. The result of the extra electron is that instead of nickel(II)’s $\mathrm{(t_{2g})^6\ (e_g^*)^2}$ state copper(II) has a $\mathrm{(t_{2g})^6\ (e_g^*)^3}$ state. There are two $\mathrm{e_g^*}$ orbitals, thus occupying each with a single electron with parallel spins leads to no degeneracy problems. An additional electron will mean that one of the two $\mathrm{e_g^*}$ orbitals is doubly occupied while the other is singly occupied—but this is a degenerate state as the energy is the same whether it is $\mathrm d_{z^2}$ or $\mathrm d_{x^2-y^2}$ that is doubly occupied.

This degeneracy of possible states in copper(II)—or any other $\mathrm d^9$ system—is the underlying reason for Jahn-Teller distortions. By lowering symmetry from $O_\mathrm h$ to $D_\mathrm{4h}$, these two orbitals are no longer degenerate and they can (and must) diverge in energy. Thus, you have a clear ‘winner’—the lower-energy orbital which receives an electron pair—and a ‘loser’—the higher-energy orbital which receives the single electron. Looking only at $\mathrm d_{z^2}$ and $\mathrm d_{x^2-y^2}$ it doesn’t matter whether the symmetry reduction is performed by moving the $z$-ligands further away or closer but it does make a difference for the other three d orbitals ($\mathrm d_{xy}, \mathrm d_{xz}, \mathrm d_{yz}$) for which lengthening the $\ce{M-L}$ bonds along the $z$ axis proves favourable overall.

A $\mathrm d^8$ system at first glance has nothing to gain by this distortion. Rather than having a stabilisation that benefits two electrons at the expense of one, it can only stabilise one electron at the expense of one. This is energetically not favourable as there is nothing to be gained (but probably something to be lost) so a ‘simple’ Jahn-Teller distortion does not occur; nickel(II) prefers to remain octahedral in $\ce{[ML6]}$ systems.

Note however that there is one possibility when such a stabilisation is beneficial: if the ‘distortion’ is of such a large magnitude that the energy difference between $\mathrm d_{x^2-y^2}$ and the next-lowest orbital is larger than the spin-pairing energy. In this case, we could say that nickel(II) drops from a high-spin state to a low-spin state, although these terms are not usually used in this context. Geometrically, this ‘distortion’ is so significant that we term the resulting complex square planar rather than distorted octahedron; in general there is no longer any evidence of coordination bonding along the $z$ axis. A number of nickel(II) complexes with strong mid-field or strong-field ligands are indeed square planar for this reason such as $\ce{[Ni(dmg)2]}$, the dimethylglyoxime (or diacetyldioxime) complex.


Why do impacting electrons have a low probability of causing ionization? In a section on ionization efficiency and ionization cross section, my mass spectrometry textbook, Mass Spectrometry by Jürgen Gross [1, pp. 38–39], says the following:

  The ionization energy represents the absolute minimum energy required for ionization of the neutral concerned. This means in turn that in order to effect ionization, the impacting electrons need to carry at least this amount of energy. If this energy were then to be quantitatively transferred during the collision, ionization would take place. Obviously, such an event is of rather low probability and therefore, the ionization efficiency is close to zero with electrons carrying just the IE of the pertinent neutral. However, a slight increase in electron energy brings about a steady increase in ionization efficiency.

Why is such an event of "rather low probability"? It seems to me that if the impacting electrons are carrying at least the necessary ionization energy, as the author suggests, then shouldn't ionization occur (be a certainty or, at least, a high probability event)?

I would greatly appreciate it if people could please take the time to clarify this.

References

Gross, J. H. Mass Spectrometry: A Textbook, 3rd ed.; Springer International Publishing: Cham, Switzerland, 2017. ISBN 978-3-319-54397-0.

Ionization has a low probability of occurring if the energy of the incoming electron is exactly equal to the ionization energy. Using a classical analogy, if  ionization occurred the incoming electron would be left with no energy, ie, stationary. The incoming, now stationary, electron would not scatter away but rather remain trapped by the nuclear attraction, such that no net ionization would have occurred. It would also imply (classically) a colinear collision with the departing electron, a seemingly unlikely event. On the other hand, if the incoming electron has a significant energy surplus, such that it has sufficient energy to escape the nuclear attraction (scatter), then ionization is more likely to occur.

There are obviously more accurate non-classical ways to describe the process, but the analogy illustrates the problem. There are also intermediate scenarios, such as those exploited in Auger spectroscopy.


Jahn Teller distortion and magnetic moments. It is relatively well known that the discrepancy between the observed and experimental magnetic moments in the first row transition metals is attributed to spin-orbit coupling when it comes to metal ions with $A_2$ and $E$ ground states. For example, Cu(II) has an $E_{2g}$ ground state term has the observed moment is slightly higher (1.96 BM for a phenanthroline complex) than the theoretically predicted "spin only" value (~1.73 BM, attributed to 1 unpaired electron). I also know that this is calculated by a formula which makes use of the spin-orbit coupling coefficient:

$μ_{eff}$ = $μ_{s.o.}$(1-$α λ$ /$Δ_o$)

where, $α$ takes up the values 2 (for E ground state) and 4 (for $A_2$ ground state), $\lambda$ is the spin orbit coupling coefficient, $Δ_o$ is the crystal field splitting energy.

My question relates to the possible effect of a Jahn Teller distortion on the magnetic moment of, say, Cu(II). Intuitively, a distortion should reduce the orbital contribution to angular momentum by the loss of orbital degeneracy. On the other hand, for Cu(II), for example, the $e_g$ set is magnetically non active. So does splitting basically have no effect on the orbital part of the angular momentum? The $t_{2g}$ set should be magnetically inactive as well because of being fully filled for Cu(II).

Planar complexes may be affected by this effect- not orthogonal complexes .  ref-(2017)
https://aip.scitation.org/doi/pdf/10.1063/1.4974805

    "The most striking result is that the planar copper(II) (3d9
    ) and nickel(II) (3d8
    )
    chelation afforded very strong ferromagnetic coupling often with 2J ≥ 300 K, owing to the orthogonal orbital
    arrangement between metal ion 3dx2-y2 and oxygen 2pz orbitals. Such a structure-magnetism relationship has
    recently been extended to coordination to rare earth metal ions."


Why should a system try to reverse any change that has been done to it? I understood Le Chatelier's principle perfectly but my question is: Why should a  system try to reverse any change that has been done to it?

I understand equilibrium constant, mole concept and partial pressures but I do not understand quantum mechanics yet. If you use any other technical terms in your answer, please also explain them. Answers with intuitive examples will be highly appreciated.

Well, lets first write out the general equation of a chemical reaction:

$$\ce{aA + bB <=> cC + dD}$$

As you obviously know, since this is at equilibrium, not all the reactants react to form the products. Hence in a solution which is at equilibrium, all species of $\ce{A, B, C, D}$ will be present. However the special thing is that the ratio of the concentration of these species will usually be constant, no matter how much of each species you start with. This ratio is known as the equilibrium constant. The only thing that effects this ratio is temperature. The equilibrium constant is given by:

$$K_\mathrm{c} = \ce{\frac{[A]^a[B]^b}{[C]^c[D]^d}}$$

One more equation you need to know is the reaction quotient which gives the ratio of products to reactants at a certain time during a reaction and is given by:

$$Q_\mathrm{c} = \ce{\frac{[A]^a[B]^b}{[C]^c[D]^d}}$$ Note how this is similar to the equilibrium constant, but here, $\ce{[A]}$, $\ce{[B]}$, $\ce{[C]}$ and $\ce{[D]}$ are the concentration of the substance at a specific time during the reaction and doesn't necessary have to be when the system has reached equilibrium. This means that if $Q$ is smaller than $K$, the reactants will react to form more products so that $Q$ will increase so that it will eventually equal $K$ and when $Q$ is larger than $K$, vice versa. When $Q$ equals $K$, the system is at equilibrium.

So now lets apply this concept to help us explain Le Chatelier's Principle by considering different scenarios. Lets consider a mixture of hydrogen and nitrogen gas in a container. They react to form ammonia gas. The equation of this reaction is:

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

Also the equilbrium constant is:

$$K_\mathrm{c} = \ce{\frac{[NH3]^2}{[N2][H2]^3}}$$

Scenario 1: Taking or adding more reactants or products

Lets say that this mixture of gases have finally reached equilibrium, so $K = Q$. Now I suddenly take some hydrogen has out.This would decrease the concentration of hydrogen gas and hence increase $Q$. So now $Q>K$. Therefore for the system to adjust itself, it needs to decrease the value of $Q$. It does this by creating more products and decreasing the amount of reactants. So in short, the equilibrium shifts to the left. This is what Le Chatelier's Principle predicts. Now by using the exact same logic, we can predict what happens if I were to change the concentration of ammonia, nitrogen, etc.

Scenario 2: Changing pressure or volume

Pressure and volume are inversely proportional to each other. Meaning that an increase is pressure will result in a decrease in volume and vice versa. This is due to the ideal gas law:

$$pV = nRT$$

Therefore, increasing the pressure of a system has the effect on the equilibrium  as decreasing the volume.

Now, back to example of hydrogen, nitrogen and ammonia gas in a container. If I were to suddenly decrease the volume of the container by two (equivalent to doubling the pressure) this would mean that the concentration of each species would double. Now, how would this effect our $Q$ value, would it be larger or smaller than $K$? Well the numerator would increase by fourfold while the denominator would increase by sixteenfold. Hence $Q<K$. Therefore as mentioned above, the equilibrium will shift to the right.

This can be applied to if the pressure was decreased, volume was decreased, etc.

Scenario 3: Temperature is changed

Now this one isn't really explained by using equilibrium constant. Instead will use a more intuitive approach.

For endothermic reactions, heat is required for the reaction to go forward. So by increasing the temperature, there is an increase in heat energy available to the system. Therefore the forward reaction will occur more readily. Hence it is expected that the equilibrium will shift to the right. Meanwhile, a decrease in temperature means less heat energy, hence equilibrium will shift to the left. The same logic can be applied to exothermic reactions.


Why do the physical properties of an egg shell change when the egg shell is exposed to vinegar for a week? When an egg is kept in vinegar for one week, its hard calcium carbonate shell changes into a soft rubbery membrane. As vinegar is weak acetic acid, how does vinegar change calcium carbonate into a soft rubbery membrane? What is the reason for this?

It doesn't make calcium carbonate rubbery, it removes the calcium carbonate.  Egg shells are not purely calcium carbonate they are more like a composite with a continuous matrix of calcium carbonate and a smaller continuous matrix of protien.   When you put the egg in vinegar, you etch away this calcium carbonate matrix leaving the formerly less noticeable protein matrix as a gel filled with vinegar and creates a gel.


Theoretical comparison of critical constants of real gases We know that for a real gas, its critical constants are: $T_\mathrm c=\frac{8a}{27Rb}$, attempt to $P_\mathrm c=\frac{a}{27b^2}$ and $V_\mathrm c=3b$. Now, let's compare the value of these critical constants for three real gases, say, $\ce{CO2}$, $\ce{NH3}$ and $\ce{H2O}$.

Knowing that $V_\mathrm c=\frac38\cdot\frac{RT_\mathrm c}{P_\mathrm c}$ (for one mole), experimental data from Wikipedia is:

$$
\begin{array}{|c|c|c|c|}\hline
&T_\mathrm c&P_\mathrm c&V_\mathrm c\\\hline
\text{Carbon dioxide}&\pu{304K}&\pu{7380 kPa}&\pu{0.1284L}\\\hline
\text{Ammonia}&\pu{406K}&\pu{11280 kPa}&\pu{0.1122L}\\\hline
\text{Water}&\pu{647K}&\pu{22060kPa}&\pu{0.0914L}\\\hline
\end{array}
$$

We will now attempt to justify this regular trend in the values qualitatively.

For $T_\mathrm c$, we know that it is the maximum temperature above which a real gas cannot be liquefied. If a gas has stronger intermolecular forces (IMFs), its molecules possess greater attractive forces to be able to overcome separatist forces that might separate them, and thus they'll still be able to liquefy. Thus, if a gas has higher IMFs, its $T_\mathrm c$ would be higher.
We know that H-bonds formed by water are much effective than those by ammonia (read), and carbon dioxide is a non-polar molecule with only London dispersion forces. Hence, we would expect $T_\mathrm c(\text{water})>T_\mathrm c(\text{ammonia})>T_\mathrm c(\text{carbon dioxide})$ by our qualitative reasoning, which agrees with the experimental order.

By definition, the critical pressure is the pressure needed to liquefy a real gas at its critical temperature. Now, since water molecules have the highest attractive forces, one would expect that they require the least external "support" (aka force per unit area) to be able to liquefy themselves (by bringing the molecules closer). Hence, we would expect that $P_\mathrm c(\text{water})<P_\mathrm c(\text{ammonia})<P_\mathrm c(\text{carbon dioxide})$. However, this does not match with the experimental trend! (which happens to actually be in the reverse order)

Moreover, I am unable to come up with a way to predict the trend in the critical volume at all.

Hence, my questions are:

Since $V_\mathrm c$ depends on the ratio of $T_\mathrm c$ to $P_\mathrm c$, am I correct in saying that its trend cannot be determined without actually being given the values for the other two constants? Or is there some clever way that I am overlooking?
Is there any other way to justify the trend in $T_\mathrm c$?
What is the fault in my reasoning for the trend in $P_\mathrm c$? And what is the correct reasoning?

In essence regarding this problem
$$T_\mathrm c \propto \frac{a}{b}$$
$$P_\mathrm c \propto \frac{a}{b^2}$$
$$V_\mathrm c \propto {b}$$

Values of $a$:

$\ce{NH3} \to 4.225\ \mathrm{l^2\ bar/mol^2}$
$\ce{CO2} \to 3.640\ \mathrm{l^2\ bar/mol^2}$
$\ce{H2O} \to 5.536\ \mathrm{l^2\ bar/mol^2}$.

This trend as justified by the OP through hydrogen bonding is absolutely correct.

Values of $b$:

$\ce{NH3} \to 0.0371\ \mathrm{l/mol}$
$\ce{CO2} \to 0.04267\ \mathrm{l/mol}$
$\ce{H2O} \to 0.03049\ \mathrm{l/mol}$.

$b$ or effective molecular size is essentially calculated by multiplying the size of a molecule by four. The trend over here is obviously determined by molecular size only.

Now:

The critical temperature and critical pressure are in direct /inverse proportion with $a$ as well as $b$ .An interplay of two factors. Predicting the orders of both of these cases become extremely difficult  without data. The opposing orders of $P_\mathrm c$ and $T_\mathrm c$ are seen as the order of proportion changes i.e $P_\mathrm c$ is inversely proportional to the square of $b$ and $T_\mathrm c$ is inversely proportional to $b$.
Trends in $V_\mathrm c$ on the other hand can be safely predicted by estimating the size of molecules as it only dependent upon one variable that is $b$.

All data is from this Wikipedia page


Ways to determine the concentration of alkyl halides in organic solvents I have a volatile alkenyl bromide dissolved in hexanes. I have to use the halide in a Grignard reaction but separating it from hexanes is hard (the bromide is light sensitive and very volatile). I would like to use the solution of the halide in hexanes but I don't know how to determine its concentration. I tried taking an NMR of the solution but all the peaks interfere and I can't choose two characteristic peaks to take a ratio. Is there a titration technique or some other technique to determine its concentration?

You could form the Grignard and titrate it. The Grignard should be fairly straightforward to titrate (e.g. as described by Knochel). This also has the advantage of giving you precisely the concentration of your reactive specie. If you knew only the concentration of your alkenyl bromide, you would not know precisely the quantity of Grignard formed (as the reaction is not always perfect).


Molecular orbital theory and the conservation of molecular orbitals I am not clear with the molecular orbital theory. I have read:

  If the atomic orbitals are combined with the same phase they interfere constructively and a bonding orbital is formed.
  If the atomic orbitals are combined with different phases, they interfere destructively and an anti-bonding molecular orbital is formed.

So as per my understanding when two s-orbitals are combined either a bonding orbital can be formed or an anti-bonding orbital can be formed depending upon the type of interference.
This contradicts the following statement:

  When two atomic orbitals combine they form two new molecular orbitals.

This means that when two s-orbitals combine they produce both bonding and anti-bonding orbitals. Where have I gone wrong?

The two statements are not contradictory. There are indeed two ways to combine the two atomic orbitals to form new molecular orbitals. It is not either one way or the other--in fact both occur - one way creates a bonding molecular orbital and the other creates an antibonding molecular orbital.

So for example, if you have two ground state hydrogen atoms, each with an electron in the 1s orbital, you end up with a hydrogen molecule, with 2 electrons in the lower energy (sigma) bonding molecular orbital and 0 electrons in the (sigma*) antibonding molecular orbital. The same thing is true for formation of pi bonds from p orbitals.