Waste water treatment This is a very simple question: Why are oxidized compounds generally thought to be less toxic/polluting than reduced species? This is used in waste water treatment where oxidizing agents (such as chlorine) are added to remove organic AND inorganic pollutants.

There is not such as general/ic rule but water treatment has used, waste water as inflow. It is not as we are discussing the oxidation of X to obtain Y, if you get the point.  The target is to get an outflow of water that can be reused or discharged in nature without significant impact. The lines below do not pertain to the treatment of industrial wastes nor the removal of a specific pollutant.
Normally no toxic compounds go to the sink, and even in the lab we are not supposed to flush chemicals down.

Oxidative processes in waste water treatment mainly consists of biological aerobic ones, basically the demolition of a wide range of chemicals as it would naturally occur in a natural environment by means of bacterial activity.

What is peculiar here is that the waste water is kept in a regulated physical and chemical environment able to sustains and ease the process (for instance the oxygen required by the microorganisms is guaranteed by injecting air, or the temperature is regulated, ecc.).

This aerobic biological oxidation is targeting biodegradable pollutants is not targeting any specific chemical. Again, keep in mind that we are speaking of waste water not of special industrial waste. Even a chemical toxic to the bacteria, or in other words interfering with the bacterial  metabolism, does not pose a limitation as far is getting degraded in the while.

Most of the organic pollutants are decomposed into carbon dioxide and water leaving a water with high content of nitrites, nitrates, phosphates that must be lowered in subsequent steps and is full of particulate and biomass that must be separated as well.

Let assume all the steps above have resulted in a sufficiently clear and relatively pure water, then yes a chemical oxidation is performed. This step might use chlorinated chemicals and other oxidizers, but it is just targeting the residual microorganisms and tailored just to do that, like in the case of the disinfection of a swimming pool. It is performed when virtually there is nothing to oxidize but the residual organisms that were at the start of the whole treatment.


Can the molecule POOF exist? Obviously, it really would be $\ce{O=P-O-F}$.  Since $\ce{F-O-O-F}$ exists the $\ce{-O-F}$  part is obviously possible and the rest of it doesn't strike me as unreasonable, but whether it can actually hold together is way beyond my knowledge.  That $\ce{-O-F}$ says it would be some nasty stuff if it exists at all.

There exists related anions called monofluorophosphate ion ($\ce{PO3F^2-}$) and difluorophosphate ion, ($\ce{PO2F2-}$). The hypothetical neutral compound, $\ce{PO2F}$ is called phosphenic fluoride. It is also observed that irradiating potassium difluorophosphate with gamma rays create free radicals $\ce{PO2F^{.−}, PO3F^{.−} and PO2F2^.}$.
Moreover, there also exist a polymer $\ce{(PO2F)_n}$ which is a decomposition product of a white solid of composition $\ce{P7O10F15}$. It was apparently named polymeric phosphorus oxyfluoride species. Its structure has been proposed by Wannagat and Rademachers1:

It is suggested that the parent ions of Group I result from the simple
ionization of a linear polymer with the general formula, $\ce{(PO2F)_n}$,
where each of the terminal phosphorus atoms of the chain is bonded to
two oxygen atoms, one of which is a bridging atom, and to two fluorine
atoms.

Conclusion: There is no such compound called POOF, but there certainly is a radical ion and a polymer of exact formula.
Notes and References

High molecular weight phosphorus oxyfluorides by D.W.Muenow, O.M.Uy, J.L.Margrave, Journal of Inorganic and Nuclear Chemistry, Volume 31, Issue 11, November 1969, Pages 3411-3415, DOI: 10.1016/0022-1902(69)80324-6
Other decomposition products were $\ce{PF5, POF3}$ and $\ce{P2O3F4}$: Technique of Inorganic Chemistry, Hans Boegh Jonassen, Arnold Weissberger Volume 1
Interscience Publishers, 1963


Understanding Millimeter of Mercury A Millimeter of Mercury(mmHg) is by definition the pressure exerted on a column of mercury 1 mm  in height.

I found this definition confusing because it doesn't say how the diameter of the container will affect the pressure.

Pressure is defined as force exerted per unit area. Therefore, the diameter of the mercury column is taken into account when we calculate the area of the column.  A mercury column with a large diameter will exert a greater force than a column of the same height, but with a smaller diameter - but the force divided by the area will be the same.


How to dissolve grass/leaves without affecting enclosure? I've got two questions:

What sort of compound can dissolve grass/leaves fastest without leaving "much" trace?
What sort of enclosure is required for this to be doen safely and without damaging the enclosure?

I am doing a small experiment and would like to try dissolving few leaves within a glass or something of the sort.

Plant matter is mostly cellulose which is insoluble in most solvents so it is unlikely you can "dissolve" the material.  You can react the material.  Two possibilities are nitric acid and sulfuric acid but there are a number of safety precautions you need to follow.


Is this a violation of Equipartition theorem? My book has a question asking to calculate average kinetic energy of the molecules in 8g of methane.

By the Equipartition theorem, I calculated that energy:
1. per molecule per degree of freedom = $(1/2)kT$ (by definition)
2. per molecule = $3kT$ (since methane is polyatomic => 6 degrees of freedom*)
3. per mole = $3RT$
4. for 8g methane ($=0.5$ moles) = $1.5\cdot RT$

But the answer stated by my book is $0.75\cdot RT$  - just half of what mine is.

This is not the only question where values have mismatched. I have started losing faith from Equipartition theorem.

Question:

  What did I get wrong?

SUMMARY+UPDATE: Final conclusion reached is that while total kinetic energy is indeed the one Equipartition theorem gives (and so it's correct! :D) but the textbook assumes the molecule to be at at very low temperatures (although this is NOT mentioned) implying the rotational and vibrational energies are NOT counted.

*=>my textbook says so. I understand that there's a lot of complex things in this topic, but let's please assume it is 6 as my book says.

For a non-linear polyatomic molecule like methane, there are a total of $3N$ degrees of freedom, where $N = 4$ is the number of atoms in the molecule. This is broken down as follows:

$$\begin{array}{c|c} \hline
\text{Component} & \text{# Degrees of freedom} \\
\hline
\text{Translational} & 3 \\
\text{Rotational} & 3 \\
\text{Vibrational} & 3N - 6 \\ \hline
\end{array}$$

(Note that for a linear molecule, or for an atom, this breakdown is different.)

Under "normal" temperatures (close to room temperature) the vibrational contribution to the internal energy can be neglected. (The vibrational gap $\Delta E = \hbar\omega \gg k_\mathrm{B}T$ for "normal" $T$, so higher vibrational states are not populated and the equipartition theorem does not hold.) This is probably the reason why your book states that there are six total degrees of freedom.

The key is to realise that the kinetic energy of a molecule arises only from the translational degrees of freedom. From my limited knowledge of physics, it seems to be a common thing to separate translational degrees from the "internal" degrees of freedom, i.e. rotational, vibrational, and electronic. This allows the translational motion of a body to be modelled as the movement of a point mass, located at the centre of mass of the body.

In this paradigm, rotations and vibrations preserve the centre of mass of a molecule, and therefore don't contribute to the kinetic energy of the molecule. Therefore, the correct answer is $(1.5~\mathrm{mol})\cdot RT$.

Another way of looking at it is to consider the definition of the kinetic energy

$$E_k = \frac{1}{2}m\langle v^2\rangle = \frac{1}{2}m(\langle v_x^2\rangle + \langle v_y^2\rangle + \langle v_z^2\rangle)$$

where $v$ is the velocity of the centre of mass, $v_\mathrm{cm}$. For each quadratic term the contribution to the energy is $kT/2$, for a total of $3kT/2$ per molecule. These three terms correspond to precisely the three translational degrees of freedom.


Stability Comparison between the Tropylium and Tricyclopropylcarbinyl Carbocation Why is the Tropylium carbocation less stable than the tricyclopropylcarbinyl  carbocation?  The tricyclopropylcarbinyl carbocation undergoes a sigma-tropic rearrangement whereas tropylium is highly stable due to conjugated system, that being, it is resonance stabilized and the number of canonical forms of tropylium is more. Then, why is it less stable?

It seems likely that the tricyclopropylcarbinyl carbocation would be more stable than the tropylium carbocation for the reasons I'll outline below, but if you have a reference proving this point, it would be nice to add it to your question.

Background

The cyclopropyl group is similar to an olefinic double bond in that it is very effective at stabilizing adjacent positive charge. The reason for this is that the $\ce{C-C}$ bonds in a cyclopropane ring are close to being $\mathrm{sp^4}$ hybridized (see this earlier answer for further elaboration of this point), said differently, there is a lot of p-character in the bond. In fact, the p-character is high enough that a cyclopropane compound will absorb bromine much like an olefin. If that cyclopropane high p-character bond (orbital) can align itself with the p-orbital on the adjacent carbocation center, it will stabilize the carbocation, just like a p-orbital on an adjacent double bond (the allyl system) stabilizes a carbocation.

Hart and Sandri [J. Am. Chem. Soc. 1959, 81 (2), 320–326] studied the solvolysis of the following 3 compounds (Ar = p-nitrophenyl) and found that the relative rates of solvolysis were 1 : 246 : 23,500.

The corresponding tricyclopropyl p-nitroester could not be synthesized, so to complete the series Hart and Law [J. Am. Chem. Soc. 1964, 86 (10), 1957–1959] prepared the analogous phenyl esters and found the relative rates of solvolysis to be 1 : 1,080.

Clearly, every time a cycloproyl group is added significant additional stabilization of the carbocation center occurs.

Answer

The number of substituents stabilizing an adjacent positive charge is critical in determining the stability of a carbocation. Just as the trityl carbocation is much more stable than the benzyl carbocation, we see above that the  tricyclopropylcarbinyl carbocation is much more stable than the cyclopropylcarbinyl carbocation. Three stabilizing substituents are better than two substituents which is better than one.

In the tropylium carbocation there are 2 double bonds and a hydrogen stabilizing the carbocation center.  In the tricyclopropylcarbinyl carbocation there are 3 cyclopropyl groups stabilizing the carbocation center. Three groups provide  better stabilization than two groups, hence we might expect the tricyclopropylcarbinyl carbocation to be more stable than the tropylium carbocation.

A Word about Resonance Structures

The number of significant resonance structures for the tropylium cation is 7, the positive charge can be delocalized to each of the seven carbons as shown below. However 7 resonance structures can also be drawn for the tricyclopropylcarbinyl carbocation. So actually both carbocations have the same number of resonance structures.


Alcoholic potassium hydroxide and aqueous potassium hydroxide I searched much about alcoholic KOH and aqueous KOH, but I was unable to understand it.

I mean, I want to know how ionisation of KOH in alcohol is different from that in water.

I also don't know what actually happens during its ionisation. Why is alcoholic KOH used in dehydrohalogenation of alkyl halides?

Why is alcoholic KOH used in dehydrohalogenation of alkyl
  halides?

One possible explanation is lack of solubility of the alkyl halides in water. Thus, one switches to a less polar, presumably alkyl side chains containing alcohol. (H2O is formed in minute amounts from OH and the proton from the hydroxyl group of the alcohol and can thus be neglected.) Boiling points vs. reaction conditions can also be relevant.

KOH forms homogenous mixtures with H2 at any ratio, this may not be true for some alcohols (For the amounts needed for the reaction this should not be relevant). In terms of ionisation, one can suspect that K+ is not as well solvated in alcohols as in water because the alcohol molecules will have a harder time arranging optimally around the ion. Also, the solvent shell structure may be a lot less pronounced. For OH-, one basically expects the same. In alcohol, one might also see some ion pair formation (or rather retention?) for KOH.


Rate of aromatic nucleophilic substitution reaction on ortho and para fluoronitrobenzene Why rate of aromatic nucleophilic substitution of Ortho-fluoronitrobenzene is greater than para-fluoronitrobenzene?
Kindly note question is about why rate of ortho is greater than para position. Not meta position.
This is previous year KVPY question but they had not asked reason and just asked comparison of rate.

Reaction mechanism of aromatic nucleophilic substitution is as given -

Source : https://www.masterorganicchemistry.com/2018/08/20/nucleophilic-aromatic-substitution-nas/
The rate determining step is the formation of the C-Nu bond, which causes the development of a negative charge on the carbon atom at ortho position to leaving group (fluorine). Presence of electron-withdrawing groups helps to stabilize the negative charge, and hence increase the rate of reaction.
The -NO2 substituent exerts not only a -M effect (resonance) but also a strong -I effect (inductive effect, which pulls the negative charge away from the ring). Inductive effect decreases rapidly as the distance increases, so its effect is much more pronounced for ortho than for para (Note that the initial negative charge is developed at the ortho position). As a result orthofluoronitrobenzene has a higher rate of reaction than parafluoronitrobenzene.


Can you have addition polymerisation with monomers other than alkenes? Every example of addition polymerisation I have seen so far has used alkenes as the monomers. I was wondering if it is possible to have some other molecule. I don't think the following molecule can exist due to stability reasons, but as an example could you not use something like

$$\ce{H-N=N-H}\,?$$

Surely all you need is the presence of a double bond?

$\ce{N2H2}$ is a very unstable molecule, prone to disproportionation into hydrazine and nitrogen. $\ce{N-N}$ bonds except one in $\ce{N2}$ molecule are unstable and generally unfavorable. So, no there are no polymers with nitrogen as a sole component of the main chain.

Several classes of compounds of 3-row elements demonstrate tendency for oligomerization, namely $\ce{(HP)_{n}}$ derivatives usually exist as a ring. It is possible to obtain metastable at room temperature plastic sulfur, containing linear sulfur chains terminated by free radicals.

However, generally when in search of easily polymerizing molecule that produce main chain containing elements other than carbon, one usually have to look into compounds producing chains with two elements. The most common and known one is formaldehyde $\ce{CH2=O}$, producing $\ce{HO-(CH2-O)_{n}-H}$ chains. A less known, but interesting example is hexachlorotriphosphazene, the source material for production of Polyphosphazenes.

In  general, metastable double bonds are rare for elements other than carbon (sometimes with nitrogen or oxygen partners), so usually reactions intended to produce such bonds, say, $\ce{Si=Si}$ usually end with olygomer instead, typically of ring structure. Moreover, single homoelemental bonds are usually much weaker for elements other than CHNOB, so typically when a chains containing other elements is occurred, it is usually hetero-atomic chain, produced by means other than olygomerization of double-bond containing monomer. For example, an extremely wide-used family of polymers (polysiloxanes, or silicones) is produce by hydrolisys of dialkyldicholorsilanes $\ce{R2SiCl2}$.


Loss of aromaticity vs low entropy of activation Even if reaction has low entropy of formation does that mean it is favorable to loss aromaticity and form highly strained cyclopropane ring? Why this reaction happens? Maybe someone can explain that in more details? Is this equilibrium reaction? I assume cyclopropane ring can be as easily opened with nucleophile as it closed? It does not make any sense for me here. Thank you for comments.
This is quoted from "The Art of Writing Reasonable Organic Reaction Mechanisms", Second Edition by Robert B. Grossman, DOI: 10.1007/b97257.

The reaction proceeds by an $\mathrm{S_N2}$ mechanism. The reaction has a very low entropy of activation, so it proceeds despite the loss of aromaticity. The product is a model of the antitumor agent duocarmycin. DNA reacts with duocarmycin by attacking the $\ce{CH2}$ group of the cyclopropane ring in an $\mathrm{S_N2}$ reaction.

There are several factors we can point to before resorting to entropy of activation.

1)  At the end of the "chain" is displacement of an iodide ion.  Iodine, with its diffuse valence orbitals, forms rather weak bonds with carbon making this displacement thermodynamically and kinetically favorable.  Presumably the nucleophile that re-opens the cyclopropane ring forms a stronger bond with the carbon.

2)  We may no longer have an aromatic ring, but a carbonyl group flanked by two conjugating carbon-carbon bonds still has plenty of resonance stabilization.  We certainly do not lose all that much stabilization.  Moreover...

3)  Upon further review, there is still an aromatic contribution after all.  The cyclopropane ring is well known to stabilize carbocations by hyperconjugation, and with the carbonyl oxygen tending to withdraw electrons we may argue that the quinone-like ring has some carbocation character.  Adding a pair of pi electrons to this ring through hyperconjugation will then give aromatic contributing structures in which the cyclopropane ring is positively charged and the oxygen regains its negative charge.

All told, we don't have to reach for "entropy of activation".  The reaction is better off than it looks just in terms of energy.


Does minimizing ANY type of energy ALWAYS predict a state of equilibrium? My intuition says no, but I am having trouble coming up with concrete examples.

I know that minimizing Gibbs free energy predicts a state of equilibrium, while minimizing kinetic or internal energy does not correspond to equilibrium. However, I am not so sure about potential energy--- when PE is at a minimum, is the system also at equilibrium?

When you find a minimum in the thermodynamic potential energy of a system, you will have found an equilibrium point.

The key is the thermodynamic potential part. Depending on the constraints of the system (i.e. which thermodynamic state variables are held constant) the thermodynamic potential energy function will be different.

For example, in a system of a fixed number of particles at constant temperature and pressure (we would call it NPT), the Gibbs free energy is the thermodynamic potential, and it will be minimized at equilibrium.

For NVT (constant volume and temperature) the Helmholtz free energy is the thermodynamic potential, and that is what will be minimized at equilibrium.

In a system consisting of a rock dropped from a height above the ground, gravitational potential energy is the thermodynamic potential (I am abusing the language a little bit here) and it will be minimized at equilibrium (when the rock hits the ground).


Why don't alloys self-galvanically corrode? Total chemistry newbie here (so this may be a fairly obvious question). With that said, I've had a niggling question that I can't quite resolve myself;

Given that galvanic corrosion is driven by the different electronegativities of different elements, why don't alloys made of significantly different elements self-galvanically corrode very quickly?

Essentially, if I were to place two electrically connected lumps in a conducting bath, one of copper and one of tin (with an anodic difference of 0.3V), I would expect them to corrode relatively quickly. Yet if I make these lumps small enough (i.e. make an alloy of them) the resulting bronze mass is fairly resilient. At what size do the individual lumps of copper and tin change from being rapidly corroding to something stable?

They certainly can. Most cases where the alloying compound is more noble this is a hazard. Nickel in Iron is a popular example, you can get corrosion acceleration on the iron by reduction happening on the Ni-surfaces.

Where the alloying element is less noble you usually have superficial corrosion to begin with but in the end the resulting surface is protected by the main element - which is common for bronze(s). If you just have lumps, then you do not have a bronze and slowly but certainly, nature will reclaim the tin.

One has to care about the homogeneity of the alloy.


How does an acid prevents or discourges oxidation of iron? In the titration of $\ce{KMnO4}$ with Mohr's salt, I have learned that sulphuric acid prevents oxidation of $\ce{Fe^2+}$ to $\ce{Fe^3+}$ in the solution before the actual titration. How does the acid hampers oxidation?

This is largely due to the extreme insolubility of iron (III) hydroxide, i.e. $\ce{Fe(OH)3}$.

One can write the oxidation reaction, somewhat simplified, as:

$$\ce{Fe^2+(aq) + 6 H2O(l) <=> Fe(OH)3(s) + 3 H3O+(aq)} + e^-$$

The $K_{sp}$ (solubility product) value for $\ce{Fe(OH)3}$ is very small: $2.79 \times 10^{-39}$ (Wikipedia).

From the equilibrium reaction it's easy to understand that acidic conditions (high $[\ce{H3O+}]$) push the equilibrium to the left, thus preventing the oxidation.

It also explains why Mohr's Salt resists oxidation much better than simple ferrous sulphate, as the ammonium ion provides some acidity.

In alkaline conditions the oxidation of ferrous ions to ferric ions is very fast.


not understanding buffering region vs equivalence point "During the titration of a weak polyprotic acid with a strong base, the first stoichiometric equivalence point occurs when the moles of added base are equal to the moles of the initial acid species.  Prior to the equivalence point, a buffering region forms in which both the weak acid and its conjugate base are present and the pH is determined by the pKa and the ratio of the acid and base species concentrations.  When the concentrations of the acid and base species are equal, pH = pKa."

The bolded statements seem like they're saying the same thing. What is the difference between the buffering region and the equivalence point?

Isn't the buffering region when [HA] = [A-], so that in the HH equation, pH = pka? The first bolded statement makes it sound like the first equivalence point is also when [HA] = [A-].

This is a matter of definitions which you must learn. Let's say that the titrant is NaOH and that we have a monoprotic acid.

Equivalence point - The equivalence point occurs when the moles of added base are equal to the moles of the initial acid species.

Notice that this doesn't infer anything about the pH of the solution.

Now for a polyprotic acid there are multiple equivalence points. The first equivalence point occurs when the moles of added base are equal to the moles of the initial acid species. The second occurs when twice as many moles of added base are added as compared to the initial acid species. The third three times as much base as acid and so on.

Again notice that this doesn't infer anything about the pH of the solution at the multiple equivalence points.

When the concentrations of the acid and base species are equal, pH = pKa."

This is a poor statement. Let's let the acid be $\ce{HA}$ so the conjugate base is $\ce{A^-}$. The statement would have been much better expressed as:

When the concentrations of the acid and its conjugate base are equal, pH = pKa."

Now to wrap this up, imagine doing this experiment. We start off with $\ce{HA}$ in solution. As we add more base we get to the pH where $\ce{[HA] = [A^-]}$ and where (1) there is maximum buffering capacity, and (2) the pH = pKa of the acid. We now add more base and we finally get to the equivalence point.

PS - For titrations you should also know the endpoint which is where the chemist gets some sort of signal to stop the titration.


Thermodynamics adding salt to water changes the temperature $5\:\mathrm{g}$ of an unknown salt are dissolved in $325\:\mathrm{g}$ of water. Both the water and the salt are initially the same temperature. The water's temperature falls by $11.4\:\mathrm{^\circ{}C}$. Explain how it is possible for the salt and water to change temperature even though both substances are initially at the same temperature.

My approach:

We know that $$q = mc \Delta T$$

Since $q, m,$ and $\Delta T$ aren't changed by this action, this process must result in the raising of water's specific heat. Is this correct?

The keyword is that the salt dissolves. Dissolution entails at least two steps:

1) Overcoming solvent-solvent interactions and bonds. An extreme example: your dinner plate doesn't dissolve in your kitchen table. One reason is that there is an extremely high energy barrier to overcoming the hypothetical solvent's (in this case, the kitchen table's) intermolecular forces. You'd have to take an ax to the table to overcome the strong intermolecular forces that hold the molecules of your dinner table together.

On the other hand breaking these solvent-solvent "bonds" or intermolecular forces in water is easy; you can jump into a swimming pool just fine. I wouldn't suggest jumping into a table.

Also, as this step's suggests, this is an energy intensive process or an endothermic process. Energy must be consumed to break these solvent-solvent bonds or interactions.

2) Breaking solute-solute interactions and bonds. This however isn't required. For example, some unionized $\ce{NaCl}$ may be solvated by water. You'll find this more true with some of the less soluble salts. But most of the sodium chloride you toss in water will have dissolved (we'll get to why later), so what is solvated generally isn't $\ce{NaCl_{(s)}}$ or $\ce{NaCl_{(aq)}}$ (this is somewhat misleading terminology) but rather $\ce{Na^+}$ and $\ce{Cl^-}$.

This again is an endothermic process as bonds are being broken.

3) Solute-solvent stabilization (solvation). This steps entails the formation of solute-solvent intermolecular forces. For example, the sodium ion, with its positive charge, may form a hydration shell of water molecules.

This is an exothermic process, as "bonds" or more accurately, stabilizing intermolecular forces are formed.

Couple all three of these processes together and you have the actual dissolution process of $\ce{NaCl}$.

What's interesting about $\ce{NaCl}$ is that even with the third, exothermic step, the dissolution (which comprises all three steps) is still a tiny bit endothermic. So that's what the question is premised on - the addition of sodium chloride to water kicks off a spontaneous process which is endothermic. So one might ask why might this endothermic process still occur. The answer lies with entropy; the dissolution of sodium chloride is not enthalpically favorable (dissolution consumes energy - specifically - from the water) - yet the dissolution is entropically favorable.

Remember that all processes tend toward greater disorder over time, and the dissolution of sodium chloride is a perfect way to increase disorder; we go from unionized sodium chloride to two ions! The two ions are definitely going to have a lot more degrees of freedom than a single, unified molecule.

$$\ce{NaCl_{(s)} + H_2O ->Na^+_{(aq)} + Cl^{-}_{(aq)}}$$


Single crystal x-ray diffraction experiment backwards: from CIF to genuine set of raw data without performing actual experiment Single crystal x-ray diffraction stands among the most complex yet trustworthy analytical techniques when it comes to structure determination of solids. But how hard it would be to simulate(fake) the entire raw dataset, including all reflections, possible artifacts, experiment strategy etc. in order to fit a certain given structure? Is it even theoretically possible? Wouldn't it happen at some day that with all the computational power and trained neural networks XRD suddenly looses its good reputation?

Here
is a brief report from the editorial section of Acta Cryst. E which overviews fraudulent content, which is still relatively easily detectable with the help of tools like CheckCIF even without looking at raw data. But can one commit a "perfect crime" by faking raw data entirely?

Structure determination by X-ray diffraction, regardless if powder diffraction, or single crystal diffraction analysis, is a spatial mapping of electron density.  After solving a crystallographic structure (phase problem), the subsequent structure refinement includes steps to complete the crystallographic model, and to minimise the differences in the Fourier maps mapping observed electron density ($F_O$), with electron density predicted by your current model ($F_C$).  This minimisation is done until this said difference is below a threshold value, and if the experimenter is satisfied with the result. For the later, this includes the complete and reasonable attribution of atoms (does this electron density belong to an atom of Carbon, or Nitrogen?), checking if bond distances and angles included in the model are consistent with the body of data determined in other experiments.  Some of the critera applied are interdependent to others.

If the crystallographic model is completed (in layman's terms, if there is a *.cif file), than it is easy to predict a theoretical powder diffraction pattern.  Programs like CCDC Mercury just ask about the wavelength to consider and offer graphical and numerical output like

Such a theoretrical diffractogram is then compared with an experimentally recorded one; for example to check the newly prepared sample's phase identity (polymorphism) and purity.

In PXRD, spatial arrangement and conformations of subunits in the crystallographic model are optimised (refined) until yielding a model that chemically and mathematically is reasonable. This is the reason you see in publications a difference trace below the the superposition of experimental and predicted diffraction pattern:

(source)

For single crystal diffraction, the generation of a 3D diffraction model computationally might be more costly than for PXRD, on the other hand, as said, programs like ShelX routinely compare experimentally recorded $F_O$ with predicted / calculated diffraction intensities $F_C$ derived from the current set of atoms (their coordinates, occupancy, etc.) of the model. It is done internally. ShelX' *.lst listings for example include a section like:

Hence I answer your question, if diffraction pattern may be simulated / faked with a "yes it is possible, provided there is a good data set about how the atoms / molecules are spatially arranged".  Initiatives like the recurrent CSP blind tests to predict crystal structures and databases like PCOD may provide a foundation for this.

Then how to spot an artifical and wrong diffraction model?  Checkcif / Platon are tools to check for consistency and integrity of such models against the raw data (hkl-files), structure factors. The recent installation of checksums into *.cif files (currently under watch as Platon errors PLAT012, 013, and 014), may be seen as one element of protection if few entries were changed after completing the work described above.  These bells probably remain silent if data were compromised very early on.


How does NaCl maintain its crystalline structure? My understanding is that $\mathrm{NaCl}$ is an ionic compound, in which $\mathrm{Cl}$ becomes (effectively) $\mathrm{Cl^-}$ and $\mathrm{Na}$ becomes $\mathrm{Na^+}$. So I understand why I would get a "sea" of particles that would stick together.

But why does the above mean that it will have a face centered cubic structure with the ions held in place so rigidly?

Crystals have inspired a great many chemists because they are fascinating for a good reason. Not only are they aesthetically pleasing, but they serve as an excellent subject to tour a variety of theoretical subjects important for understanding high-level chemistry.

Crystalline materials are made up of periodic structures. We’re only going to primarily focus on binary compounds where there is not a high degree of covalency. There are several ways to think about this problem, but let’s start with the melting of a crystal.

We say that at some definite temperature a highly ordered crystal will melt into a liquid. Those of us familiar with the language of equilibrium thermodynamics might recognize that the change in free energy for this phase change can be written, at constant temperature, as,

$$ G_\text{liquid} - G_\text{crystal} = H_\text{liquid} - H_\text{crystal} - T ( S_\text{liquid} - S_\text{crystal} ) $$

$$ \Delta G = \Delta H - T \Delta S $$

If we suppose that this process is spontaneous then we would say that the change in Gibbs’ free energy is negative, i.e. $\Delta G < 0$. This is true if and only if,

$$\Delta H < T \Delta S$$

Traditionally we interpret this as saying that there is a thermally-driven increase in entropy when we melt a highly ordered crystal into a liquid which more than offsets the energy cost associated with the enthalpies of the interactions holding that crystal together. A chemist tends to learn early on that the reverse is not necessarily true: at some definite temperature a perfect crystal rarely forms from the liquid. This inability to just heat up any substance and always produce a perfect crystal by cooling illustrates how crystal formation is a case of kinetic- rather than thermodynamic- control. So the process by which you form your crystal could possibly result in a different crystal structure. Sometimes crystal structures change just by altering the temperature of the chamber you’re measuring the crystal structure in! Now neither of these cases apply to sodium chloride to the best of my knowledge.

The formation of an ionic crystal such as sodium chloride is a delicate balance between electrostatic attraction and Pauli repulsion. Electrostatic attraction says that between two different charges, $q_+$ and $q_-$, there is a Coulomb force given by, $$F= \frac{k q_+ q_-}{r^2}$$

where $r$ is the distance between the two charges. If one plays with the numbers then it’s easy to see that at short distances the force is strongest, but there is a limit to how close they may come together. Eventually a repulsive force due to a quantum mechanical principle called the Pauli Exclusion Principle overpowers the attraction. An equilibrium results in which the atoms sit a certain distance from one another so that, if you will humor me, the “forces” between them balance out. This is why we traditionally represent crystal packing using marbles with a unique radii. The radii of the hard marble represents where the Pauli repulsion overpowers the attraction.

You might say, “Sure, we have these kinetic, electrostatic and quantum mechanical factors to consider, but how do these help with the final crystal structure?” Hold your horses, we’re getting there.

A famous mathematician and scientist thought about the most efficient ways to pack spheres of the same size together. By most efficient I mean this in terms of what FedEx considers efficient, fitting things together into the smallest possible volume. This is also what electrostatics want. Kepler suggested that the best way to pack spheres with this in mind can be maximized in two ways: fcc and hcp.

This conjecture, however, rests on all the spheres being uniform. We can’t assume this is the same for the atoms in table salt because the cation, sodium, is said to be smaller than the chloride anion. A tool that is useful as a guide for helping predict the structure is based off of the relative size of the cation, $r_\text C$, to the anion, $r_\text A$. This radius ratio, $r_\text C / r_\text A$, is mostly only useful for simple binary species. These $r_\text C$ and $r_\text A$ are tabulated as ionic radii in many chemistry books; they are contrived by various rules proposed by researchers, such as Linus Pauling, applied to the experimentally determined interionic distances. In this system we find it useful to discuss how many neighbors an anion or a cation has of the opposite charge.

The coordination number ($\mathrm{CN}$) tells us how many neighbors a type of atom has around it. A certain $r_\text C/r_\text A$ gives us a feel for the probable CN. From one source I was able to calculate the ratio for $\ce{NaCl}$ as $r_{\ce{Na+}}/r_{\ce{Cl-}} = 0.564$. Most textbooks will say that if your ratio is $0.414{–}0.732$, then you have $\mathrm{CN}=6$. I’ll show you how to calculate the minimum value for $\mathrm{CN}=6$. The easiest way to obtain the maximum is to obtain the minimum for $\mathrm{CN}=8$.

Briefly, if we put a small cation on a $XY$ plane and surround it in the manner shown below then we could also place two circles in the $Z$-axis above and below the central cation for $\mathrm{CN}=6$. (I won’t show those two in the $Z$-axis for clarity.) We would say that a given $\mathrm{CN}$ is stable only if the spheres are all touching each other. We would say that for the ratios drawn that a $\mathrm{CN}=8$ is not stable because our cation’s sphere would be too small and wouldn’t touch all eight of its neighbors. A higher density is always preferred, so $\mathrm{CN}=4$ is not preferred when $\mathrm{CN}=6$ is stable.

Clearly, $\angle \mathrm{DAC} = 45^{\circ}$. Moreover, we can say that $\overline{\mathrm{DC}} = r_\text A$ and $\overline{\mathrm{AC}} = r_\text A + r_\text C$. Trigonometry will tells us that $$\sin 45^{\circ} = \frac{\overline{\mathrm{DC}}}{\overline{\mathrm{AC}}}$$ Substituting in those values just determined,
$$\frac{\sqrt{2}}{2} = \frac{r_\text A}{\overline{r_\text A + r_\text C}}$$

Solving for $r_\text C/r_\text A$ gives,

$$r_\text C/r_\text A=\frac{2}{\sqrt{2}} -1 = 0.414$$

which is the lower bound desired.

So we can say that the $\mathrm{CN}=6$ for the cation. Similarly you can figure out that $\mathrm{CN}=6$ for the anion. This is why we say it has a 6:6 coordination pattern, which is indicative of a sodium chloride type structure. Cesium chloride, on the other hand, is 8:8 and indicative of its structure type. We tend to say that these patterns are consistent with a certain crystal structure because all the pieces fit together consistently. This geometric concept is based purely on sphere packing and is not without its limitations.

A more quantitative way to figure out which geometry is preferred is to calculate the electrostatic energies for the different geometries. A Russian by the name of Kapustinskii derived a formula to do this,

$$E_\text{lattice} (\mathrm{kJ/mol}) = \frac{\alpha}{r_\text C + r_\text A} $$

From here you need to put it in terms of the radius ratio with a constant $r_\text A$,

(I should note that this is evil algebra: $ \frac{a/b}{d/b + 1}= \frac{a}{b+d} $ where $a=\alpha$, $b=r_\text A$, $d=r_\text C$.)

$$E_\text{lattice} = \frac{\alpha/r_\text A}{\frac{r_\text C}{r_\text A} + 1} $$

Basically $\alpha$ is different for a change in the crystal structure type and from there you compare which has the preferred energy. You can reproduce the geometric trends using this more quantitative model. There are problems with this approach too, such as the neglected dipole and quadrupole and covalency. Yes, even in $\ce{NaCl}$ there is some covalent character. I think we say that $\ce{NaCl}$ $67\ \%$ ionic character. These become important when two different crystal structures are close in energy or when more exact calculations are required.


Assigning Carbon NMR I have a molecule and I need to assign the carbon NMR to each of the carbons.

My molecule is

The peaks I have for the carbon NMR are shown below. I have also used the DEPT to work out if it's a $\ce{CH3}$, $\ce{CH2}$, $\ce{CH}$ or quaternary C.

17.7ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH3}$)
19.5ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH3}$)
25.5ppm $\ce{CH2}$
25.7ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH3}$)
32.1ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)
36.7ppm $\ce{CH2}$
39.7ppm $\ce{CH2}$
66.0ppm $\ce{CH2}$
122.1ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)
124.4ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)
128.2ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)
128.3ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)
128.6ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)
131.5ppm quaternary carbon
136.2ppm quaternary carbon
149.0ppm $\ce{CH3}$ or $\ce{CH}$ (I think this is $\ce{CH}$)
166.4ppm quaternary carbon

I think I can assign the quaternary carbon with the $\ce{C=O}$ in the ester as the 166.4ppm.

Out of the three methyl groups, would the lowest ppm of 17.7ppm be the one at the top of the molecule by itself and then would the other two at the bottom of the molecule on the alkene be 19.5ppm and 25.7ppm? I don't know which one is which, I know it is to do with cis/trans but I'm not sure which way around it is.

I also know that on the benzene ring, only three $\ce{CH}$ peaks show up as the two ortho and the two meta carbons are equivalent.

If this is a real sample, and not a provided coursework spectrum, then the best recommendation is to run a HSQC and/or HMBC. This will provide complete data to allow this assignment fully, and in much the same time as it takes to run a DEPT135. I don't understand the urge to run DEPTS any more - I haven't run one for years. A phase edited HSQC will give you all of the information that a DEPT can provide, as well as connectivities between proton and carbons. Anyway, off my soapbox.

Below is a simulation of this molecule, which has reasonable alignment with your peaks.

A methyl centre trans to a alkyl chain will be downfield shifted to an equivalent methyl group trans to a proton. So 19 will be downfield shifted from 20, and be fairly similar in chemical shift to 16. Simulations have their limitations, and carbon shifts are solvent dependent to +/- a few ppm at least. To avoid any uncertainty, run HSQC and HMBC. These should be part of your normal procedure for assigning structures.


Is the order of orientation of electron box diagrams meaningful or arbitrary? Here is my interpretation when asked to:

  By drawing arrows in the appropriate boxes, complete the outer electron structures for Cu and Cu2+

I had no problem in drawing out the electron structure, but am a little confused by the orientation of my arrows.  I understand that each normal box has to have arrows in opposing directions to represent the opposite spins on the electrons, but the mark scheme defines which way round they want them to go:

Source: OCR GCE Chemistry B F334 June 2011 Markscheme pg. 11

Ordinarily I would have said that whether you go up or down first (or even consistently) is irrelevant and arbitrary, but the comment in the guidance

  ALLOW single arrow in either direction

suggests to me that with the paired arrows, the order of directions matter.

Is there any chemical reason for putting the downwards spin arrow first?  I would happily accept by convention as a reason but I just want to be sure that I'm not missing anything chemically relevant.

Raise hell with the grader. There's no reason whatsoever for this answer to be rejected. A Google book search can substantiate your argument, and the first example from a textbook is actually in your order:

I think the guidance “allow single arrow in either direction” is not exclusive with “allow double arrows in either order”, and maybe only single arrows were explicitly mentionned because it was obvious to the writer that order in double arrows doesn't matter.


Is graphite more stable than diamond? The change in enthalpy of carbon in diamond form to graphite is negative. This suggests that graphite should be more stable than diamond. Is it true?

What I thought was that graphite has van der Waals forces while diamond has covalent forces so diamond should be more stable. But the melting point of graphite is more than diamond which implies that it is thermodynamically more stable. I am thoroughly confused. So which is more stable?

At normal room temperature and pressure graphite is (slightly) more stable than diamond. But the melting point is not a good indicator of this.

Melting points are determined by the bonding structure of the solid and any potential liquid (indeed there may not be a liquid phase at some pressures). They don't necessarily tell you whether a different bonding structure is more thermodynamically stable.

The thermodynamic stability of different carbon structures depends on subtle tradeoffs among several factors in the bonding of the various solids (allotropes) than can be formed with different bonding arrangements. There is no easy way to tell which is the most thermodynamically stable other than by measurement or some very high-powered calculations about the quantum mechanics of the bonding.

The different allotropes have radically different structures (tetrahedral in diamond, a ball of hexagons and pentagons in buckminsterfullerene and flat plates of hexagons in graphite). Changing one into another requires a lot of bond breaking which isn't going to happen much at room temperature. Interconversion is possible at high pressures and temperatures when carbon is dissolved in liquid rock or metal and the carbon allotrope that forms is more likely to be whatever is stable at that whatever the pressure is (diamond is more stable at higher pressure which is why the earth or industry needs very high pressures to create diamond).

So, for most practical purposes, it doesn't matter than graphite is the more stable allotrope: your diamonds are not going to spontaneously turn to ash during your, or the universe's, lifetime.


What research methods can I use in identifying Aluminum in a solid sample? I do have a cube of solid aluminum. But I need to verify that it is really Al to validate my research. What simple scientific techniques do you recommend?

If you want a simple one, then you might want to just perform qualitative analysis (QA) using aqueous sodium hydroxide and aqueous ammonia. Firstly, dissolve a portion of the cube using a small volume hydrochloric acid. Then, add a small volume of sodium hydroxide to a small volume the metal chloride solution. To another small volume of the metal chloride solution, add a small volume of aqueous ammonia.

For the first test, aluminium ions would first give a white precipitate but will dissolve in excess. For the second test, aluminium ions would first give a white precipitate but will not dissolve in excess.

Lastly to confirm, add aqueous potassium iodide to the sample. If no precipitation occurs, the sample is indeed aluminium chloride solution and the cube does indeed contain aluminium.

Alternatively, this paper here also provides some insights on verifying if something is indeed aluminium:

http://www.iaea.org/inis/collection/NCLCollectionStore/_Public/08/346/8346019.pdf

Unfortunately, the method I have mentioned and the methods in the paper are destructive as they require you to chip of a piece of metal from the cube for analysis. So... if you can't afford to do that then they cannot be used.


How is the position of a peak on an IR spectrum determined? In an IR spectrum, all sources describe the position of a peak as a range. The ranges seem to be defined arbitrarily.  Take, for example, the IR spectrum of ethanol (which I can't upload for some reason); why is the $\ce{O-H}$ stretch defined as having a range of $3200$ to $3750~\mathrm{cm}^{-1}$. If it was up to me, I would say that the trough goes from $3100$ to $3650~\mathrm{cm}^{-1}$. And if the $\ce{C-H}$ stretch wasn't present, I would say that the $\ce{O-H}$ trough extends to below $3100~\mathrm{cm}^{-1}$.

Moreover, sometimes there are very specific values given. These also seem to be arbitrarily defined. How do I determine the exact specific value that corresponds to a peak?

And finally, what is the correct way to identify the position of a trough on an IR spectrum? Is it a range? Or do I have to give a specific value?

P.S. I know I asked three questions, but they all serve the overall purpose to help me define how a trough is defined within an IR spectrum.

Generally, the width of IR absorption peaks depends on the environment of the target molecule. Intermolecular forces alter vibrational modes.

If intermolecular attractions are fairly strong (like the hydrogen bonding present in liquid ethanol), the number and strength of interactions can vary greatly from molecule to molecule, causing the vibrational modes observed in an IR absorption spectrum to widen.

As for the reported width of the peak, the convention is to report the width at 1/2 the height of the peak.


Electrophilic substitution on Borazole As Borazole ($\ce{B3N3H6}$) is aromatic, it shows electrophilic aromatic substitution. Now, my question is when borazole undergoes electrophilic aromatic substitution reaction, on which atom does the electrophile attack? Is it boron or nitrogen? Also, how to compare it's rate with the same type of reaction when carried on benzene? Should I consider stability of Wheland intermediate formed when comparing the rates?

Chiavarino, et al. [1] report  that where electrophilic substitution occurs with carbocations (borazole more often undergoes addition), it does so on nitrogen.  Nucleophiles such as methanol prefer boron.
We can explain that result in terms of both molecular orbitals and the Wheland intermediate.  In the molecular orbital explanation, recall the familiar delocalized pi orbitals in the benzene ring.  With the polar bonds of borazole the occupied bonding orbitals whose electrons will add to an electrophile are shifted towards the nitrogen atoms, so the electrophile will tend to bond there rather than on the boron (the latter has instead mostly the empty antibonding pi orbitals and is thus favored in nucleophilic substitution).
In terms of the intermediate, adding the electrophile to the ring leaves the remaining pi electrons in a deficient condition -- only four electrons cover the remaining five conjugated atoms, with most of the electron deficiency falling on atoms ortho and para to the attack site.  Attacking on nitrogen puts the less electronegative boron atoms at these deficient sites leading to a more stable EAS intermediate.
Reference

Barbara Chiavarino, Maria Elisa Crestoni, Simonetta Fornarini, "Electrophilic Substitution of Gaseous Borazine," J. Am. Chem. Soc. 1999, 121(11), 2619-2620 (https://doi.org/10.1021/ja983799b).


Entropy change upon mixing cold and hot water A closed, well-insulated container is filled with $\pu{454 g}$ of water at $\pu{94.4 ^\circ C}$. To the hot water, $\pu{200 g}$ of water ice at exactly $\pu{0 ^\circ C}$ is added. The mixture reaches an equilibrium temperature of $\pu{41.1 ^\circ C}$. Assume the molar heat capacity is constant and all the processes are at constant pressure. The standard enthalpy of fusion for water at $\pu{0^\circ C}$ is $\pu{6.008 kJmol–1}$. The constant-pressure heat capacity for water is $\pu{75.291 JK–1mol–1}$. Water has a molecular weight of $\pu{18.015 gmol^-1}$.
Calculate the entropy change (in $\pu{JK–1}$ ) for the system that happened because of this mixing.

I know the entropy change equals to $q/T$ because $q$ equals to the enthalpy exchange in the system as it is constant pressure, so what I did was:
$$q=(454/18.015)\times 75.291\times (41.1-94.4)+(200/18.015)\times 75.291\times(41.1)+(200/18.015)\times6008 = \pu{172.2 J}$$
Change in entropy = $172.2/(41.4+273) = \pu{0.55 JK-1}$
That is apparently incorrect, what have I done wrong?

The change in entropy doesn't "equals to" q/T.  For a specified closed system, the change in entropy between an initial state and a final state is given by $$\Delta S=\int{\frac{dq_{rev}}{T}}$$ where the subscript "rev" refers to a reversible path between the initial and final states.  The process you described is not reversible.  You need to devise and employ a reversible path to get the change in entropy.

Here is a hint:  For the system comprised of the 454 gm of water initially at 94.4 C, the change in entropy is $$75.291\frac{454}{18.015}\ln{[(273.2+41.1)/(273.2+94.4)]}$$
If you would like a cookbook recipe for how to determine the change in entropy for a system experiencing an irreversible process, see the following link: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/


Why does the graph of compressibility factor(z) vs pressure increase after reaching the minimum? I understand why it decreases for most of the gases but i cannot understand why it increases after reaching a certain minimum.

After a certain amount of pressure, the molecules start repelling each other and this leads to a rise in compressibility factor.

At high pressures, the van der Wall constant $b$ which represents the finite volume occupied by the gas molecules plays a greater role than the van der Wall constant $a$ which represents the attractive forces between the gas molecules.


Reduction and Oxidation - Alternate definitions Why is reduction defined as the gain of hydrogen?

Is it because hydrogen has a very low electronegativity and easily donates its electrons to other elements?

From what I understand reduction is formally defined as a decrease in oxidation state; oxidation state assignment involves supposing that all bonding is ionic and that electrons are therefore parceled to the most electronegative atom in a bond. And if something gains H - an atom of exceedingly low EN - then that element is reduced because that element will likely be more EN than H and therefore "take" the hydrogen atom's electron.

The "official" definition I've always used is that reduction is the gain of electrons. IUPAC seems to use the same definition. It looks like organic chemistry books often use the addition of oxygen to identify oxidation, and the addition of hydrogen or the removal of oxygen to identify reduction. This is in line with the historical definitions (at least according to wikipedia) that defined reduction in terms of the change in mass on heating (if oxygen was lost, the weight would be reduced).

When using these shorthand rules, I think it's important to distinguish between the purposes of the definitions. The more formal "official" (according to IUPAC) definition is intended to get as close to the underlying physics as possible. The less formal "shortcut" definition used here is intended to quickly allow chemists to identify electron transfer just by looking at the structural formulas. In fact, in my copy of Wade's Organic Chemistry (4th ed) it says

  Most organic chemists habitually use the following simple rules, based on the change in formula of the structure:

It then goes on to define redox reactions in terms of gain or loss of oxygen and hydrogen. This implies that the author doesn't intend it to be read as a formal definition - just a shortcut you can use in a pinch.

This is a case of "it works well enough" being used to make life easy - but it is important to remember that reduction isn't really being defined as gain of hydrogen in the formal scientific sense. It is still defined as a gain of electrons - it just so happens that in organic reactions, it is very difficult to add electrons to something without also adding hydrogen, and so the shortcut works (due to electronegativity differences between O, C, N, P, and H, as you pointed out).

Here's an example of a redox reaction where this rule won't work, but looking at electrons will:

$$
\ce{Li(s) + 1/2H2(g) -> LiH(s)}
$$

We are adding hydrogen to lithium, and the oxidation state of Li goes from 0 to +1 - it is oxidized, not reduced.

I have been looking for an example of an organic reduction that does not involve the addition of hydrogen, but haven't been able to find one - which is probably why this shortcut rule works.


Soil conductivity: How do I get a near equivalent between meq/100 g to mS/cm? I have two soil reports, one that reports CEC as 45 meq/100 g and another that reports 6 mS/cm using 1:1 water soil.

I can see that the two are, if not apples and oranges, they are at best tangerines and grapefruit — the meq/100 g is the source for the ions in a water extract, but an equilibrium solution may not contain all the ions.

Can a meaningful conversion between meq/100 g and mS/cm be made?

E.g: few plants will grow in soil that has conductivity greater than 2.5 mS/cm and generally you start getting concerned at 1.  What is the equivalent measure in meq/100 g?

FWIW main ions of concern are $\ce{K+},$ $\ce{PO4^3-},$ $\ce{Mg^2+},$ $\ce{Ca^2+},$ $\ce{CO3^2-},$ $\ce{HCO2-}.$

The mineral ions (nutrients) in soil can be expressed as two ways: (i) Cation Exchange Capacity (CEC), which has units of $\pu{meq}$/$\pu{100 g}$; and (11)  Electrical Conductivity (EC), which has units of $\pu{mS/cm}$.

When the conductivity is measured in 1:1 mixtre of soil:water, it is called $\mathrm{EC_{1:1}}$ (This is a way of measuring electrical conductivity of soil water or $\mathrm{EC_{sw}}$).

According to USDA Publication, $\mathrm{EC_{1:1}}$ readings less than $\pu{1 dS/m}$ ($\pu{1 mS/cm}$), soil are considered non-saline and do not impact most crops. This publication also gives you a easy procedure for how to measure $\mathrm{EC_{1:1}}$ of your soil.

My best bet is you don't have to worry about your second soil sample, which has CEC of $\pu{45 meq}$/$\pu{100 g}$, because none of the cations (exchangeble cations in the soil) in concern is $\ce{Na+}$ (the given CEC is the sum of all exchangeable cations in equivalents). According to above publication, soils with a high concentration of sodium salts (sodic conditions) have given problems.

To my understanding, there is no direct relationship between $\pu{meq}$/$\pu{100 g}$ and $\pu{mS/cm}$. For more info on Cation Exchange Capacity (CEC) and units of $\pu{meq}$/$\pu{100 g}$, read:

J. P. Reganold, J. B. Harsh, “Expressing cation exchange capacity in milliequivalents per 1 100 grams and in SI units,” J. Agronomic Educ. 1985, 14(2), 84–90 (https://www.agronomy.org/publications/nse/archive/vol-14).


Do crystals have stronger intermolecular forces than amorphous solids? Do crystals have stronger intermolecular forces than amorphous solids.

Are the properties of amorphous solids vs. crystals explained by their structures?

Diamond crystal has much stronger bonds than amorphous red phosphorus. Amorphous sodium/calcium silicate, known as glass, has much stronger bonds than crystalline metallic sodium. (Also, in all four cases the word intermolecular doesn't really apply, because these compounds are not made of molecules.)

You see that the answer depends primarily on the particular compound, rather than on its being crystalline or amorphous. In cases when something can be either crystalline or amorphous with the same chemical composition, the bonds in both are pretty much the same, as explained in the comment by Jon Custer.

Oh, and yes, chemical and physical properties of pretty much anything are explained by its structure.


Why is it necessary to consider infinitesimal changes in p,V,T for H,U and G given that they're state functions? State functions such as $G$ only depend on the state of the system and are not dependent on the "path" that took the system to that state (which would be the case for work, for example, which is not a state function.

We know that:

$$\mathrm{d}G=V\mathrm{d}p-S\mathrm{d}T$$

So…

$$\mathrm{d}G=\left(\frac{\partial G}{\partial p}\right)_T\mathrm{d}p+\left(\frac{\partial G}{\partial T}\right)_p\mathrm{d}T$$
Consequently, by comparing coefficients:

$$V=\left(\frac{\partial G}{\partial p}\right)_T$$ and $$-S=\left(\frac{\partial G}{\partial T}\right)$$

Just taking the equation involving $V$ now to save time and space:
$$\int_{p_1}^{p_2}V\mathrm{d}p=\int_{p_1}^{p_2}\left(\frac{\partial G}{\partial p}\right)_T\mathrm{d}p$$

Using the perfect gas equation and integrating leaves the result:

$$G(p_2)-G(p_1)=n\mathcal{R}T\ln\left(\frac{p_2}{p_1}\right)$$

But if $G$ is independent of the path taken to get to the final state, why shouldn't I use the equation

$$\Delta G=V(p_2-p_1)?$$

It is important to make clear any requirement that a variable be constant for each equation you write.

To go from

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

to

$$\mathrm{d}G = V\,\mathrm{d}p,$$

you would need to assume temperature is constant.

And to go from

$$\mathrm{d}G = V\,\mathrm{d}p$$

to

$$\Delta G = V(p_2 - p_1)$$

you would need to assume volume is constant.

If temperature and volume are both constant, the pressure is constant also.

So your final equation is only saying: when temperature, pressure and volume are constant, there is no change in Gibbs free energy.


How are orbitals displayed in computational Chemistry software? In typical software like Gaussian or GAMESS, the molecular orbitals (or even the basis functions) can be displayed as surfaces.

I understand that the surface is probably constructed from a series of points that give a fixed value when substituted into the wavefunction. However, how are these points calculated? Since typical software usually use GTOs, I’d like a formula if possible using those.

Thanks!

Only the program that you used to calculate the electronic structure knows all the idiosyncrasies of the computation to correctly evaluate the orbitals. Therefore, what most people do is to export the evaluation of the orbitals in a grid. Then, a plotting program performs an extraction of a polygonal mesh with a marching cubes (or similar) algorithm.


Low or high enthalpy of combustion in relation to stability I know that this is a very basic question, but I am new to chemistry and I cannot figure it out on my own:

A combustion is always exothermic, so ΔH is negative. It says in my textbook that more stable molecules have a lower heat of combustion (I am assuming here that this is basically identical to stating that it has a lower enthalpy of combustion).

What I do not understand is the following: more stable molecules have a lower enthalpy of formation than less stable molecules. When undergoing combustion, stable molecules should be expelling less heat than less stable ones. So shouldn't the enthalpy of combustion of more stable molecules be higher (i.e. less negative)?

N.B. The question is related to the following:
I was asked to determine which of the following two compounds has a lower enthalpy of combustion. I understand that B is more stable than A, but I am not able to translate that to "a lower enthalpy of combustion" because of my lack of understanding of the above…

Many thanks!

I have redrawn your two structures A and B in their most stable conformations. Note that A has all equatorial methyl groups while B has two equatorial groups and one that is axial. The estimated difference in the heats of formation and combustion is 1.8 kcal/mol, the same as the difference in energy between the equatorial and axial conformations of methyl cyclohexane. Because A and B produce the same amount of CO2 and H2O upon combustion, the more stable A's heat of combustion is 1.8 kcal/mol less than that for B. Compare your A and B with their enantiomers S and T at this ChemSE site at which time I addressed a similar question.


Pauli-forbidden term symbols for atomic carbon Carbon has a $\mathrm{p^2}$ configuration, and within the Russell-Saunders coupling scheme, we have

$$\begin{align}
s_1 = s_2 &= \frac{1}{2} & S &= 1,0 \\
l_1 = l_2 &= 1 & L &= 2,1,0 \\
\end{align}$$

which gives rise to $^3D$, $^1D$, $^3P$, $^1P$, $^3S$, and $^1S$ terms. Out of these six, only $^1D$, $^3P$, and $^1S$ are allowed by the Pauli principle.

It's not hard to show that the $^3D$ term is forbidden, since this term has to contain one state with $m_{l1} = m_{l2} = 1$, which would correspond to a spatial wavefunction that is symmetric with respect to interchange of the two electrons. Pairing this with a triplet spin wavefunction, which is also symmetric, would lead to an overall symmetric wavefunction, which is not allowed.

But that's all the textbooks say - they don't describe why the $^1P$ term is forbidden, for example. I know how to construct a table of microstates, and I know I can use the direct product table of the full rotation group to find that $P \times P = S + [P] + D$, and I know that both of these methods show that the allowed term symbols are only $^1D$, $^3P$, and $^1S$.

Essentially, I am looking for a more intuitive explanation of why the $^1P$ and $^3S$ states are forbidden that preferably links back to the requirement that the wavefunction be antisymmetric - if that is possible.

Alternatively, an explanation of why $P \times P = S + [P] + D$ would also work. I know that from the Clebsch-Gordan series we get $S$, $P$, and $D$ terms, but not why the $P$ state should be antisymmetric. I do know the general idea of how to obtain the vector coupling coefficients (by repeated application of raising/lowering operators), and these do show that the $P$ state is antisymmetric, but I'd like a more interpretative answer than just "that's how the maths works out".

Okay, actually I managed to find somewhat of an explanation myself.

We could construct a state with $M_L = 1$ and $M_S = 0$, for example:
$$\begin{align}
\psi = \frac{1}{2}\left[\mathrm{2p_1}(1)\mathrm{2p_0}(2) + \mathrm{2p_0}(1)\mathrm{2p_1}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right]
\end{align}$$

where $\mathrm{2p}_m$ represents the 2p orbital with $m_l = m$.

However, such a combination of $(M_L,M_S)$ can only occur once, since the two electrons are indistinguishable. In other words, the following two other wavefunctions are not permissible because permuting the two electrons changes the wavefunction:
$$\begin{align}
\psi &= \frac{1}{\sqrt{2}}\left[\mathrm{2p_1}(1)\mathrm{2p_0}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right] \\
\psi &= \frac{1}{\sqrt{2}}\left[\mathrm{2p_0}(1)\mathrm{2p_1}(2)\right]\left[\alpha(1)\beta(2) - \beta(1)\alpha(2)\right]
\end{align}$$

and the $^1D$ state, which has $S = 0 \Rightarrow M_S = 0$ and $L = 2 \Rightarrow M_L = 2,1,0,-1,-2$, already has to encompass one state with such a combination of $(M_L,M_S)$. Therefore, a state with $(M_L,M_S) = (1,0)$ cannot contribute to a $^1P$ term.

It is logically not the strongest argument (sufficiency vs necessity), but it is kind of what I was looking for. If anybody can provide another viewpoint though I will be glad to accept the answer.


Where to find compounds IC50 on an enzyme? I am looking for half maximal inhibitory concentration (IC50) values on a enzyme.

I already searched through ChemBL, PubChem and BindingDB. The three seem to give consistent results.

However, looking in the literature, there are lots of compounds which are not in these databases. These later may provide higher level of curation. I am still interested in the data available in literature. Any other available repository or text mining approach will be great.

To my knowledge, there are no data bases accumulating  half maximal inhibitory concentration values on a enzyme ($\mathrm{IC_{50}}$). According to Wikipedia:

  Half maximal inhibitory concentration ($\mathrm{IC_{50}}$) is a measure of the potency of a substance in inhibiting a specific biological or biochemical function (i.e. an enzyme, cell, cell receptor, microorganism, etc.).

$\mathrm{IC_{50}}$ is commonly used as a measure of antagonist drug potency in pharmacological research. According to the FDA, $\mathrm{IC_{50}}$ represents the concentration of a drug that is required for 50% inhibition in vitro. It is comparable to an $\mathrm{EC_{50}}$ for agonist drugs.

The major reason for not to accumulate $\mathrm{IC_{50}}$ values is their abundance. For example, for given specific enzyme, there may be vast number of inhibitors present. All of these inhibitors have their specific $\mathrm{IC_{50}}$ value. Thus, if you need to find a $\mathrm{IC_{50}}$ value of specific inhibitor on a particular protein (or enzyme) or a receptor or other biological system, you should search using both names of the inhibitor and biological system as key words. Thus, you would find any literature, which may have $\mathrm{IC_{50}}$ values for sought combination.

For example, Galectin-3 is  a protein that is a member of the lectin family (carbohydrate-binding proteins). Galectin-3 is approximately $\pu{30 kDa}$, and contains a carbohydrate-recognition-binding domain with about 130 amino acids that enable the specific binding of $\beta$-galactosides, such as N-acetyllactosamine (14; the parent substrate; $\mathrm{IC_{50}} = \pu{158 \mu M}$ under the conditions; see insert red box). Several derivatives of N-acetyllactosamine (15 - 18) have displayed inhibitory activities with different $\mathrm{IC_{50}}$ values ranging from $13$ to $\pu{31 \mu M}$ (See the plot below):

On the other hand one inhibitor can be active against more than one biological systems with different $\mathrm{IC_{50}}$ values.

Relevant References:

P. Sörme, B. Kahl-Knutsson, U. Wellmar, B.-G. Magnusson, H. Leffler, U. J. Nilsson, “[12] Design and Synthesis of Galectin Inhibitors,” In Methods in Enzymology, Volume 363 – Recognition of Carbohydrates in Biological Systems, Part B: Special Applications;   Y. C. Lee and R. C. Lee, Eds.; Academic Press: Amsterdam, The Netherlands, 2003, pp. 157–169.
P. Sörme, Y. Qian, Per‐G. Nyholm,  H. Leffler, U. J. Nilsson, “Low micromolar inhibitors of galectin-3 based on 3'-derivatization of N-acetyllactosamine,” ChemBio Chem 2002, 3(2-3), 183–189 (https://doi.org/10.1002/1439-7633(20020301)3:2/3<183::AID-CBIC183>3.0.CO;2-%23).
J. Hynes, Jr., S.  K. Nair, “Chapter Nine - Advances in the Discovery of Small-Molecule IRAK4 Inhibitors,” Annual Reports in Medicinal Chemistry 2014, 49(1), 117–133 (https://doi.org/10.1016/B978-0-12-800167-7.00009-2).


In free radical substitution of propane with halogens, how do you know which carbon ends up with an unpaired electron after fission of C-H bond? In free radical substitution of propane with a halogen (eg- chlorine), how do you know which carbon out of the three ends up with an unpaired electron after fission of C-H bond?

This image shows methane, so there is only one possibility since there is only one carbon, but what about alkanes like propane with multiple carbons? How do we know which carbon atom ends up with the unpaired electron and why?

This is to do with the relative stabilities of the free radicals produced. Here is a schematic diagram showing the two possibilities for propene (similar mechanism, also with two free radical possibilities like propane):

As you can see, a more stable and less stable free radical is produced. The stability depends on how many carbons are attached to the carbon with the unpaired electron. In the diagram, the free radical where two carbons are attached is more stable than the free radical with one carbon attached.

As a general rule, the stability will be greatest in tertiary free radicals, then in secondary, and least stability in primary.

This is due to the inductive effect of carbons bonded to the carbon with the unpaired electron:

Carbons bonded to the free radical carbon are able to 'move' some of their electron density over to the free radical carbon, stabilising it. This is due to something called sigma conjugation, which we needn't go into.

But in summary, both free radicals are formed, as seen in the first diagram. Both free radicals go onto generate products, but much more come from the more stable species. Therefore, in fact both carbons possible in propane end up with the unpaired electron, but the route involving the secondary free radical species is more successful and gives the major product. The route involving the primary intermediate gives a minor product.


How is the tetrachloridocobaltate(II) anion stable if its crystal field stabilization energy comes to be positive? I tried to calculate the ligand field stabilisation enthalpy of $\ce{[CoCl4]^2-}$ (as shown in the image) but found it to be positive. What could be the reason behind this? How is it stable with positive crystal field stabilization energy?

You didnt put the right values. Higher orbitals have energy value of $+0.4 \Delta_t$ and the lower orbitals have the energy value of $-0.6\Delta_t$.

This complex has a coordination number of 4, so it can have tetrahedral or square planar geometry. Tetrahedral complexes are favored where attainment of regular complexes is important. For the tetrahedral complexes $\mathrm{d^0, d^2, d^5, d^7}$, and $\mathrm{d^{10}}$ the configurations are regular.

In tetrahedral complexes, mostly $\Delta t < P\ (\Delta_t \approx 4/9 \Delta_o)$ and hence they generally form high spin complexes.

The splitting will be into $3\ \mathrm{t_2}$ (not $\mathrm{t_{2g}}$) higher energy orbitals and $2\ \mathrm e$ (not $\mathrm{e_g}$) lower energy orbitals. The effect of the ligand wont matter here because the value of $\Delta_t$ is small. So the filling will be same for strong field and weak field.

$$\begin{align}\text{CFSE} &= +0.4(3) -0.6(4)\\
&= -1.2\end{align}$$


How does dehydration of an alcohol occur in alkaline medium? During aldol condensation reaction the aldol formed reacts further to give alkene (which is probably called enone)

I am really struggling with this part of the reaction as my textbook says nothing about the mechanism of this particular step
the medium during aldol condensation reaction is alkaline but the end product is a dehydrated product and the conversion of alcohol into alkene is dehydration. Do we need to make the medium acidic at later stage of reaction, or does some mechanism causes dehydration in alkaline medium?

Also please suggest if there is some other reaction, or examples depicting similar phenomenon that I could go through for better understanding

(I am really new to chemistry as well as to asking questions. thank you)

The commonly accepted mechanism for the base-catalysed aldol example here is that the alpha hydrogen (shown in blue on your scheme) is removed by OH- present in the reaction medium, and the negative charge on the enolate then forms the alkene by loss of OH-. This mechanism falls under the general heading of E1cB, further information on this mechanism may be found here and here 3.

Thanks to @orthocresol for the note about this


How to separate and analyse a sample of cations of alkaline and earth alkaline cations and ammonium? I am teaching a section of qualitative analysis and I am trying to get a better grasp on the assignments the students are doing. These are the cations they may be presented with and must identify:

$\ce{Na+}$
$\ce{K+ }$
$\ce{NH4+}$
$\ce{Ca^2+ }$
$\ce{Al^3+ }$
$\ce{Mg^2+}$

Now, this lab hasn't changed for the last 20 years, and there is only one way the students perform the separations, which happens to be the same method found in an old lab report that is easily found with Google. The method is:

Add excess 3 M ammonia to the sample.
If a gel-like precipitate forms, then $\ce{Mg^2+ }$ and/or $\ce{Al^3+ }$ ion is present. They have been converted into their respective hydrated hydroxides. Separate the precipitate from the supernatant liquid.
Since hydrated $\ce{Mg^2+  }$ hydroxide is not readily amphoteric, while hydrated $\ce{Al^3+ }$ hydroxide is, addition of excess  $\ce{NaOH}$ should cause the an $\ce{Al^3+ }$ precipitate to dissolve. If the precipitate from the previous step does dissolve with excess $\ce{NaOH}$, then $\ce{Al^3+ }$ ion is present in the original sample. If the precipitate does not dissolve, then $\ce{Mg^2+  }$ ion is present.
Time for the supernatant analysis (if there is a supernatant). If there is no ppt, then we can immediately rule out the presence of $\ce{Mg^2+}$ and $\ce{Al^3+ }$. Either way, the solution, non-precipitate portion can only contain $\ce{Na+}$, $\ce{K+ }$, $\ce{NH4+}$, and/or $\ce{Ca^2+ }$ since none of these form insoluble hydroxides at the concentration of ammonia used. Yes, calcium hydroxide is insoluble, but it doesn't form with 3 M $\ce{H3N}$; I've done the $Q_\mathrm{sp}$ vs. $K_\mathrm{sp}$ calculation.
Add sodium carbonate to the liquid portion. If a solid forms, the solid must be calcium carbonate, and therefore $\ce{Ca^2+ }$ was present in the original sample.
If no solid forms, or even if a solid forms, separate the solid from the liquid and perform further tests on the liquid to determine the presence of $\ce{Na+}$, $\ce{K+ }$, and $\ce{H4N+}$. A yellow flame tests tells us $\ce{Na+}$ is present and $\ce{K+}$ may or may not be present. A purple flame test tells us $\ce{Na+}$ is absent.
Heat the remaining solid sample (not any precipitate, but the original sample itself) to ascertain the presence of $\ce{NH4+}$. Heating a sample that contains $\ce{NH4+}$ will result in the release of white smoke. After heating the solid sample until smoking ceases or until one is certain the sample will not start generating white smoke, perform a sodium coboltinitrite test if necessary to determine whether $\ce{K+ }$ is present.

What other valid separations are there? Is this the easiest method, or are there easier ones?

What about initially adding $\ce{NaOH}$?

Add $\ce{NaOH}$. If a white precipitate forms, the sample contains $\ce{Ca^2+}$ (the white ppt is calcium hydroxide). If a gel-like ppt forms, then the sample contains either $\ce{Al^3+}$ or $\ce{Mg^2+}$. Perform the necessary tests described above to determine the nature of the gel-like ppt.
The liquid part of the solution must now only contain $\ce{Na+ }$ or $\ce{K+ }$. Ammonium ion has been converted to ammonia thanks to the hydroxide ion. $\ce{Na+}$ or $\ce{K+ }$ can easily be tested for using a flame test. The sodium coboltinitrite test can also be employed to detect $\ce{K+ }$ if the color of $\ce{K+ }$ has been masked by the color of Na+ in the flame test.

This method seems a bit faster. But are there any pitfalls?

In a qualitative analysis, the fast way is always the wrong way. There are already so many issues with the "long" version that I am surprised that this is remotely successful.

The cations of groups 1-3 are always the most difficult to separate and during my time as a teaching assistant were by far the biggest source of error.

The proper course of analysis is

dissolve sample (completely)
precipitate ion
dissolve precipitate
specific reaction
repeat

The reason for separating the cations in the first place is, that they often act in similar ways and the reactions are not very specific. Certain cations should be tested multiple times, too.

All the reactions you are performing are equilibria and very dependent on the overall concentration. None of them form a precipitate quantitatively. Sometimes you might have problems recognising precipitates in the first place.

You need to perform double blind tests for all used chemicals, since already small contaminations may lead to false positives.

The main issue with your proposed strategy is that you will not precipitate aluminium with this strategy, since this is extremely pH dependent.
$$\ce{Al^3+ + 4OH- <=> Al(OH)3 + OH- <=> [Al(OH)4]- }$$
Without a buffered solution at around pH 9 and ageing the solution, there should not form a precipitate. I learned this by adding ammonium sulfide first and raising the pH slowly with ammonia. (This will co-precipitate cobalt, nickel, manganese, zinc, and iron [as sulfides], and chromium [as hydroxide]. This might be useful if you have to analyse for more ions later on.)
If a precipitate forms you can dissolve it with sodium hydroxide and use cobalt(II) nitrate to form Cobalt blue.
You can also test the presence of aluminium with morin as it forms a fluorescent complex with it. Note that $\ce{Mg(OH)2}$ is slightly fluorescent, too.

I cannot remember precipitating magnesium with that strategy in any sufficient amount. I would prefer precipitating it as an ammonium phosphate, which you can try to (re-)crystallise as struvite and look at them with a microscope.
$$\ce{Mg^2+ + NH4+ + PO4^3- <=> MgNH4PO4 v ->C[H2O] MgNH4PO4.6H2O}$$

If you want to precipitate out aluminium and magnesium (and almost every other heavier element) you can use oxine, $\ce{C9H7NO}$ (abbr. $\ce{Hox}$), solution.
$$\ce{ M^{$x+$} + $x$\,Hox <=> [M(ox)_{$x$}] + $x$\,H+ }$$

Then precipitating calcium is straight forward with carbonate solution. Other earth alkaline metals will precipitate, too. You can dissolve this precipitate with acids or you can use the solid to perform flame tests. Since there can only be calcium, strontium, or barium, this should be conclusive. (Of course you could precipitate barium sulfate.)

The main pitfall with your proposed strategy is, that you will not quantitatively precipitate calcium hydroxide and it is as mentioned above quite concentration dependent. The solubility constant is not too small, i.e. $K_\mathrm{sp}=5.5\times10^{-6}$.

If you have precipitated everything so far, then the last thing left over are ammonium, potassium, sodium(, and lithium). The biggest problem is, you cannot mask them properly and you cannot precipitate them. You can use flame tests to determine their presence, but as long as you have sodium in it you might not see anything else. You should use a prism, or a spectrometer to determine the characteristic spectral lines. Apparently phones can do it now, too. Also see publiclab.org... makes you think why separate at all.

You will need to analyse ammonium from the sample. You can add sodium hydroxide, which releases ammonia. You can detect the presence of ammonia in the sample by holding a wetted universal indicator paper over the sample. Additionally, you should be able to smell it.

In other words ...
... be patient with your analysis as there are more pitfalls, false negatives, false positives, contaminations. Don't use short cuts, except atomic emission spectroscopy.


Question regarding $Z$ (Compressibility factor) Oh! This problem has been bugging me for a long time. According to Wikipedia, the Compressibility factor $Z$ is defined as the ratio of the volume occupied by a real gas to the volume occupied by an ideal gas at the same temperature and pressure.
$$Z=\frac{V_{real}}{V_{ideal}} \text{              ........(1)} $$
Now the Ideal Gas Equation is
$$P_{ideal}V_{ideal}=nRT   \text{        ....(2)}$$
Now from here we can substitute the value of $V_{ideal}$ into the above expression $(1 )$ to get
$Z=\frac{PV_m}{RT}$
Where $V_m$ is the molar volume of the gas and $P$ is the value of $P_{ideal}$ that was in the ideal gas equation $(2)$.
OK, So far so good!
Deriving the compressibility factor value from the van der Waals' gas equation:
We know for real gases the pressure exerted by them is always less than the pressure exerted by an ideal gas so we correct the pressure by the term $a\frac{n^{2}}{V^{2}}$, we get
$$P_{ideal}=P_{real}+a\frac{n^{2}}{V^{2}}$$
And the volume occupied by the gas molecules can be written as
$$V_{ideal}=V_{real}−nb$$
On substituting the above values of $P_{ideal}$ and $V_{ideal}$ in the ideal gas equation, we get our van der Waals' Gas Equation
$$(P+a\frac{n^{2}}{V^{2}})(V−nb)=nRT$$
Please note that in the above equation $P$ is $P_{real}$ and not $P_{ideal}$. Now I'm going a step further to calculate the value of Z from this equation by shifting the terms and I got this
$$Z=\frac{PV_m}{RT}=\frac{V_m}{V_m−b}−\frac{a}{RTV_m}$$
Now here's the problem that I'm facing: In the above expression the value of $P$ is not the $P_{ideal}$ but it is the value of $P_{real}$ that we had in the van der Waals' gas equation.
Doesn't that contradict the whole definition of Compressibility factor that we had learned in the first place? How this expression for $Z$ can be correct?
Thanks for reading.

Forget about the ideal gas.  The definition of the compressibility factor is just $$Z=\frac{PV_m}{RT}$$


How can the density of water be maximised? What is the maximum possible density of water at room temperature?
The density of the water in the Dead Sea at 37°C is 1.24. Would it be possible to increase the density using salts we add to our food, or possibly by using different salts than are present in the sea?

Clerici solution is a solution of equal parts of thallium formate and thallium malonate in water, density is 4.25 g/ml,     Thoulets solution is solution of potassium-mercury iodide in water with density 3.2 g/ml. Don't even think of using them.

Sodium polytungstate $\ce{Na6[H2W12O40]}$ is a relatively safe compound and allows densities up to 3.1 g/ml. $\ce{ZnCl2}$ allows densities up to ~ 1.9 g/ml, but has acidic solutions.

Unfortunately, common nontoxic salts usually either are not dense enough or not soluble enough to make really heavy liquids, but densities around 1.1-1.2 g/ml are quite easy to achieve even with common salt. Note, however, that concentrated salt solutions 'siphons' water from human body through damaged skin and mucous membranes, and to some extent even through undamaged skin, which may lead to desiccation, trauma of mucous membranes and very unpleasant sensations.


Naming of elements 113, 115, 117, and 118 Does anyone know of a blog, website, news server, or listserv to which one may subscribe to be notified when these elements get their official names?

IUPAC will certainly have a press release on their news page, given that they are officially responsible for the names.

Currently, that's at this address: http://www.iupac.org/news/latest-news.html

Sadly, they don't appear to have an RSS feed. (Not even an Atom feed [rimshot noise].)


Why does the carbocation in this scheme react under elimination conditions rather than being attacked by water? In the final step of the mechanism, elimination occurs over substitution, why is this?

The simple answer is that water is a very weak nucleophile and substitution does happen reversibly, but usually dissociates immediately. After which, the final reaction shown happens.

There is likely a (very) minor product that follows along the substitution pathway with an alcohol final product.


Which has the highest enol content? Which of the following compound has highest enol content?

The answer is an option (c) and I think the reason behind it is that there are two phenyl groups attached to it, hence there will be a good conjugation. Also, since phenyl is an electron-withdrawing group, it will provide stability to the structure (as the $\ce{C-C}$ double bond formed after tautomerism will have π-electrons). But I am not quite sure about my logic and I don't know the reasons to eliminate other options.

The extent of enol content can be explained on the basis of acidic hydrogen, intramolecular hydrogen bonding.

The stability of the carbanion formed is dependent on its stability after it is formed. In compound $\ce{(c)}$, the carbanion is stabilized by extended conjugation due to presence of phenyl group. Such stabilization does not occur in $\ce{(d)}$. In $\ce{(d)}$, methyl substituent at carbanion increases negative charge via inductive polarization. The resulting carbanion is unstable.

After enolization , enol is stabilized via intramolecular hydrogen bonding, which is a common feature in all examples provided.
The enol in compound $\ce{(c)}$ is stabilized via extended conjugation.

Taking the above factors into account, $\ce{(c)}$ has highest enol content.


Why water won't dissociate more? In a given sample of water some molecules of water dissociate to give hydronium and hydroxide ion. They say one in 10 million will be dissociated (approx.). What are the reasons behind this? Why water won't dissociate more?

The energy required to break the first $\ce{H-O}$ bond in water is $\pu{493 kJ/mol}$. For comparison, the energy required to break the first $\ce{H-C}$ bond in methane is only $\pu{435 kJ/mol}$. As a point of reference for just what these numbers mean in terms of bond strength, we can think about how long it takes for methane to be destroyed in the atmosphere. The first step of the destruction of atmospheric methane is to break that first $\ce{H-C}$ bond; because this is so difficult to do (again, requiring $\pu{435 kJ/mol}$) it takes an average of about 9 years for a molecule of methane to be destroyed! And the energy required to break the $\ce{H-O}$ bond in water to form hydroxide and hydronium ions is even greater.

I realize that my analogies involve very different processes, but the underlying issue in both cases (atmospheric destruction of methane and self ionization of water) is bond strength. I hope that this gives some insight as to why the self-ionization of water is as low as it is. Don't hesitate to ask for any clarifications in the comments.


How to calculate mass of water of crystallisation? How do you work out the mass of water of crystallisation for 13.5 g of

$$\ce{Al2(SO4)3*6H2O}$$

The correct answer is 3.24 g of water but I'm unable to derive the answer.

What I have tried:

I worked out the mols of aluminium sulfate which is approx 0.038 mol by dividing 13.5 g/353.5.

Since the ratio of aluminium sulfate: water of crystallisation is 1:6, I multiplied the number of mols by 6 to result in 0.228 mol of water.

To work out the mass, I multiplied mols of water by molar mass to result in 4.101 g which is NOT the answer of 3.24 g .

What am I doing wrong and what is the correct method?

The mass of aluminium sulfate has to be taken along with the water of crystallization.

Now molar masses
$$
\begin{array}{c| c}
\text{compound} & \text{ mass (g)} \\
\hline
\ce{Al2(SO4)} & 342 \\
\ce{6H2O} & 108 \\
\ce{Al2(SO4)3\cdot 6H2O} & 450\\
\end{array}
$$

After that it's simply ratio and proportion find out the moles of hydrated salt then find out the moles of water finally get the mass of water.

$$\frac{13.5}{450} \times 6 \times 18 = 3.42$$


Difference between flame test and ability to show photoelectric effect My textbook says magnesium doesn't give flame test. But it also says that it is a good photoelectric material. Are both concepts using "visible" light to remove or excite electrons?  What is the basic difference between these two and why is magnesium showing only one of them?

I think you have a misconception. Observing an emission spectrum and photoelectric effects are different things. Photoelectric effect requires not only the electron is ejected with some kinetic energy, but it is ejected by light of a suitable wavelength. No heat or flame is involved.
In (Bunsen burner) flame emission spectroscopy, your temperature is not high enough to remove electrons from Mg atoms in the gas phase. Eventually, the flame temperatures can be made high enough that you do start to observe emission from Mg atoms, but it is in the UV range. Your eyes cannot see UV light.


Will equilibrium ever establish if one of the products starts to disassociate? When $\ce{N_2O_5}$ is heated at a certain temperature, it disassociates as $$\ce{N_2O_5 (g)<=>N_2O_3 (g) + O_2 (g); K_c = 2.5}$$ At the same time $\ce{N_2O_3}$ also decomposes as $$\ce{N_2O_3 (g)<=> N_2O (g) + O_2 (g)}$$If initially 4 moles of $\ce{N_2O_5}$ were taken in a $\pu{1 L}$ flask and allowed to disassociate, and at equilibrium $\ce{[O_2]=2.5 M}$ then find equilibrium concentration of $\ce{N_2O_5}$

Lets say the first reaction reached equilibrium and then $\ce{N_2O_3}$ started decomposing. This means more amount of $\ce{O_2}$ will be produced and the equilibrium for the first reaction will get disturbed.
The first reaction will then proceed backwards as predicted by Le Chatelier's principle (since the concentration of $\ce{O_2}$ is more than the equilibrim conc.).
Due to this the oxygen level will drop and equilibrium for the second reaction will get disturbed and $\ce{N_2O_3}$ will decompose further to form oxygen.
But this process will go on forever. There is definitely wrong with this. Will this really happen or is there some other mechanism at work?

Let's make this an abstract problem with species A, B, C, and F:
$$\ce{A <=> B + F}\tag{1}$$
$$\ce{B <=> C + F}\tag{2}$$
We can also write down the sum of the two reactions, which will also be at equilibrium:
$$\ce{A <=> C + 2 F}\tag{3 = 1+2}$$
And let's use a for the concentration of A and so on.
Finding the solution using reactions 1 and 3
The concentration of F is equal to the equilibrium constant for reaction (1), so the concentrations of A and B have to be equal:
$$a = b$$
For every mole of B made, we also get a mole of F. For every mole of C made, we get two moles of F. We know the concentration of F is 2.5 M, so we get:
$$b + 2c = 2.5 M$$
Finally, we started with 4 M of A, which makes equimolar B and makes equimolar C, so the sum of all three has to be 4 M:
$$a + b + c = 4 M$$
Now we have three equations and three unknowns. We can combine the first and third equation to eliminate $a$:
$$ 2b + c = 4M$$
We can multiply the second equation in preparation for solving for $c$:
$$ 2b + 4c = 5 M$$
Subtracting the latter equation gives the solution for $c$:
$$ c = \frac{1}{3} M$$
$a$ and $b$ come out as 1.833 M.
Le Chatelier

Lets say the first reaction reached equilibrium and then $\ce{N_2O_3}$ started decomposing. This means more amount of $\ce{O_2}$ will be produced and the equilibrium for the first reaction will get disturbed

Without the second reaction, the first will attain equilibrium when $a$ is about 1.85 M. The second reaction makes F and uses up B, so the first reaction almost stays at equilibrium, and the concentration of A does not change much.
Will it ever stop?

Due to this the oxygen level will drop and equilibrium for the second reaction will get disturbed and $\ce{N_2O_3}$ will decompose further to form oxygen.
But this process will go on forever. There is definitely wrong with this. Will this really happen or is there some other mechanism at work?

If you look at this step-wise (figure out how much one reaction would go, disregarding the other), you are never done. This is similar to Zeno's paradox. However, the corrections get smaller and smaller, and your approximation gets better and better.
In general, reactions approach equilibrium, they don’t reach it. Nothing special about this system.


Melting Point of Pure Gold The melting point of gold was changed from 1063 deg C to 1064 deg C in an ASTM procedure that uses the melting point of gold as the end point for a test.  Did the melting point change somehow or was there an error in the ASTM method?

As with any experimental measurement of physical properties, the techniques involved improve with time, and sometimes new, better ones appear. That also means some values for some physical properties sometimes are slightly adjusted to reflect these better measurements and less error.

In some very rare occasions, the basic definition of some measurement is changed, like the recent change in the definition of the kilogram (see https://en.wikipedia.org/wiki/Kilogram), and it sometimes reflect in some minute adjustments made to another measurements based on the old definitions.

In other, more mundane measurements, the standard (for the measurement) is based on a set of conditions (pressure, atmosphere composition, purity, and a lot of other parameters) and a specific procedure for the measurement that sometimes change in newer versions of the procedure, sometimes resulting in a slight change of the expected measured values.

That doesn't mean the material changed, just that the observed value for the physical property using the described set of parameters and procedures were adjusted.


How does the bite angle of a ligand affect the efficiency of its catalyst? I know that optimising the bite angle of a diphosphine ligand to its metal increases catalytic efficiency, and for cross-coupling reactions larger bite angles are favoured, but how does this have an impact on the effectiveness of a catalyst?

The bite angle of a bidentate ligand often has a pronounced effect on reactivity, but it can be difficult to predict a priori whether a given reaction will be enhanced by increasing or decreasing bite angle - you can see this if you ever read a paper where new ligands / catalysts have been developed, which are often accompanied with large tables of screening in order to find an optimised set of reactions.

In general however, bite angle does correlate with the rates of oxidative addition and reductive elimination.

Reductive elimination

The rate of reductive elimination is often correlated to the bite angle of a bidentate ligand. Upon reductive elimination, the two coordinate Pd0 complex would preferentially adopt a linear geometry however with the bidentate ligand coordinating (and the chelate effect preventing it from 'falling off') this is not possible.

Larger bite angles therefore favour reductive elimination as the L-Pd-L (P-Pd-P) bond angle is closer to the desired 180o. The classic example of this was an experiment carried out by Brown in which the rate at which PhMe was eliminated from a variety of complexes was measured.

Pd(dppf)PhMe with a bite angle of 99.1o underwent rapid reductive elimination, even at low temperatures, while Pd(dppe)PhMe with a bite angle of 90.6o was stable at low temperatures, with elimination needing elevated temperatures.

Oxidative addition

The effect of bite angle on oxidative addition is less straightforward, as it depends largely on the mechanism by which the oxidative addition takes place. In general however, complexes with larger bite angles undergo faster rates of oxidation despite the fact that a smaller bite angle increases the rate of the C-X bond cleavage.


Oil :water partition coefficients Is there a database or a website with a list of oil : water partition coefficients of substances ?

I actually want the oil:water partition coefficients of :

cholesterol, starch , methyl citrate , citric acid , iso- citrate , methyl urea , carotene

Two places where you can find predicted values of LogP (the oil:water partitioning coefficient) are chemicalize.org and Wolfram Alpha.  Both provide a predicted value for most of the compounds you are interested in (chemicalize provides a complete set, assuming you mean betacarotene).
Examples
This link provides a direct link to the output of the cholesterol page from Chemicalize.org and shows the logP value on the left hand side (7.11)
To get the logp from Wolfram alpha you must search for logp  and then chose the predicted link that appears when no results are returned.  This link should get you to the value obtained from Wolfram Alpha (7.02).