Probability of measuring a particle in the ground state: having trouble with the integration I am working on a problem in the context of an infinite symmetric potential barrier around the origin, with barriers at $x=- \frac{a}{2}$ and $x=\frac{a}{2}$. The wave function is a symmetric triangle-function:

$$
\Psi (x) = \begin{cases} 0 & x < - \frac{a}{2} \\ \sqrt{\frac{12}{a^3}} \left( \frac{a}{2} - \lvert x \rvert \right) & -\frac{a}{2} \leq x \leq \frac{a}{2} \\ 0 & x > \frac{a}{2} \end{cases}
$$

I must calculate the probability of measuring a particle in the above eigenstate in the ground state (when $n=1$). The way to attack this problem is to calculate the coefficient $c_1$, and then square it. Here follows my (wrong) attempt.

$$
c_n = \int\limits_{-\infty}^{\infty} \psi_n^*(x)\Psi (x)\,\mathrm dx
$$

I understand from the above expression that the $\psi_n^* (x)$ is just the "regular" solution to the particle in a box. Given this symmetric potential, and that $n=1$ (odd $n$, even function), I must set

$$
\psi _1^* (x) = \sqrt{\frac{2}{a}} \cos\frac{\pi x}{a}
$$

The expression for $c_n$ becomes

$$
c_n = \int\limits_{-a/2}^{a/2}  \sqrt{\frac{2}{a}} \cos\frac{\pi x}{a} \sqrt{\frac{12}{a^3}} \left( \frac{a}{2} - \lvert x \rvert \right)\,\mathrm dx \\
$$

Simplifying gives

$$
c_n = \frac{2 \sqrt{6}}{a^2} \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) \left( \frac{a}{2} - \lvert x \rvert \right)\,\mathrm dx
$$

This can be solved by partial integration, setting $u'=\cos \left( \frac{\pi x}{a} \right)$ and $v= \left( \frac{a}{2} - \lvert x \rvert \right)$. Hence, $u=\frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) $ and $v'=-1$. The integral itself then becomes

$$
\int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) \left( \frac{a}{2} - \lvert x \rvert \right)\,\mathrm dx = \frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right)\left( \frac{a}{2} - \lvert x \rvert \right) - \int \frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) (-1)\,\mathrm dx
$$

Cleaning up

$$
\mbox{Integral} = \frac{a^2}{2 \pi} \sin \left( \frac{\pi x}{a} \right) - \lvert x \rvert \frac{a}{\pi} \sin \left( \frac{\pi x}{a} \right) - \frac{a^2}{\pi ^2} \cos \left( \frac{\pi x}{a} \right)
$$

Now we just need to insert the limits. However, we see quite easily from the above expression that upon inserting $x=\pm \frac{a}{2}$, the last term vanishes, since $\cos ( \pm\pi / 2 = 0)$. The sine terms will equal $1$ when the limit is positive and $-1$ when the limit is negative. Hence, we are left with

$$
\mbox{Integral} = \left[ \frac{a^2}{2 \pi} - \frac{a}{2} \frac{a}{\pi} \right] - \left[ \frac{a^2}{2 \pi}(-1) - \lvert \frac{-a}{2} \rvert \frac{a}{\pi}(-1)  \right] = \frac{a^2}{2 \pi} + \frac{a^2}{2 \pi} = \frac{a^2}{\pi}
$$

Plugging this into the original expression

$$
c_n = \frac{2 \sqrt{6}}{a^2} \cdot \frac{a^2}{\pi} = \frac{2 \sqrt{6}}{\pi} \approx 1.56
$$

So, you see the problem. My $c_n$ is larger than one, which will give me a rather un-physical probability. Am I wrong in this derivation in my assumption for $\psi _n^* (x)$? If not, then where?

Wolfram Alpha gives me that

$$
\mbox{Integral} = \frac{2a^2}{\pi ^2} \\
\Rightarrow c_1 = \frac{2 \sqrt{6}}{a^2} \cdot \frac{2a^2}{\pi ^2} = \frac{4 \sqrt{6}}{\pi ^2}
$$

Using this gives me a probability of $c_1^2 \approx 0.986$, which at least has physical meaning. Also, logic tells me that there is a very high probability of finding the particle in the ground state, so this further backs up WA's answer.

I think your tactics using partial integration from the start is the problem because $\lvert x \rvert$ is not differentiable at $x=0$ so working with its derivative is not quite kosher (and $v' = -1$ is certainly wrong since for $x < 0$ the gradient of $-\lvert x \rvert$ is $1$).

I suggest you go about it step by step:

\begin{align}
  c_1 &= \frac{2 \sqrt{6}}{a^2} \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) \left( \frac{a}{2} - \lvert x \rvert \right) dx \\
      &= \frac{2 \sqrt{6}}{a^2} \biggl( \frac{a}{2} \underbrace{\int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) dx}_{ = \, I_{1} } - \underbrace{\int\limits_{-a/2}^{a/2} \lvert x \rvert \cos \left( \frac{\pi x}{a} \right) dx}_{ = \, I_{2} } \biggr)
\end{align}

Now, integral $I_1$ is easy:

\begin{align}
  I_{1} &= \int\limits_{-a/2}^{a/2} \cos \left( \frac{\pi x}{a} \right) dx \\
   &= \left[ \frac{a}{\pi} \sin \left( \frac{\pi x}{a} \right)  \right]_{-a/2}^{a/2} = \frac{a}{\pi} \sin \left( \frac{\pi}{2} \right) + \frac{a}{\pi} \sin \left( \frac{\pi}{2} \right) = \frac{2a}{\pi}
\end{align}

For integral $I_2$ I suggest that you use that

\begin{equation}
 \lvert x \rvert = \begin{cases} x &  x \geq 0 \\ -x & x \leq 0 \end{cases}
\end{equation}

and split up the integral into two integrals

\begin{align}
  I_{2} &= \int\limits_{-a/2}^{a/2} \lvert x \rvert \cos \left( \frac{\pi x}{a} \right) dx \\
   &= \int\limits_{0}^{a/2} x \cos \left( \frac{\pi x}{a} \right) dx - \int\limits_{-a/2}^{0} x \cos \left( \frac{\pi x}{a} \right) dx
\end{align}

Now use integration by parts with $u'=\cos \left( \frac{\pi x}{a} \right)$ ($\Rightarrow u=\frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) $) and $v= x$ ($\Rightarrow v'= 1 $) to solve the integral

\begin{align}
  \int x \cos \left( \frac{\pi x}{a} \right) dx &= \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) - \int \frac{a}{\pi}\sin \left( \frac{\pi x}{a} \right) dx \\
  &= \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \ ,
\end{align}

so

\begin{align}
  I_{2} &= \int\limits_{0}^{a/2} x \cos \left( \frac{\pi x}{a} \right) dx - \int\limits_{-a/2}^{0} x \cos \left( \frac{\pi x}{a} \right) dx \\
        &= \left[ \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \right]_{0}^{a/2} - \left[ \frac{a}{\pi} x \sin \left( \frac{\pi x}{a} \right) + \frac{a^2}{\pi^2} \cos \left( \frac{\pi x}{a} \right) \right]_{-a/2}^{0} \\
        &= \frac{a^2}{2 \pi} - \frac{a^2}{\pi^2} - \frac{a^2}{\pi^2} + \frac{a^2}{2 \pi} = \frac{a^2}{\pi} - \frac{2 a^2}{\pi^2}
\end{align}

Plugging this into the equation for $c_1$ gives

\begin{align}
  c_1 &= \frac{2 \sqrt{6}}{a^2} \left( \frac{a}{2} I_{1} - I_{2} \right) \\
      &= \frac{2 \sqrt{6}}{a^2} \left( \frac{a^{2}}{\pi} - \frac{a^2}{\pi} + \frac{2 a^2}{\pi^2} \right) \\
      &= \frac{4 \sqrt{6}}{\pi^2}
\end{align}

and that is indeed what Wolfram Alpha gives you.


Lanthanides react with mineral oil? I have an element collection and noticed that the lanthanide (rare earth) element samples seem to be reacting with the mineral oil I've used to help prevent oxidation.

When I originally placed the samples I researched the lanthanides and found that the lower atomic numbers indeed tend to oxidize in air. Even though I've stored them in jars, I also filled the jars with mineral oil, similar to how I've stored the alkali metals. And at least for the alkali metals, the mineral oil has served as a good barrier to keep out oxygen.

So why would the lanthanides, which are a lot less reactive to oxidation, corrode so quickly? Is there another, perhaps organic reaction going on, between the lanthanides and the mineral oil?

David Hamric did a four year Rare-Earth Metal Long Term Air Exposure Test

and has a wonderful set of time-lapsed photographs on his site.

Europium is the most reactive to oxidation.

Several sources, such as Avalon Advanced Materials: Europium make statements like:

  Samples of the metal, even when coated with a protective layer of mineral oil, are rarely shiny and it readily ignites in air above 150°C.

So rather than reacting with the mineral oil, I would say that europium and possibly others are reacting with oxygen despite being covered with the mineral oil.

THE SOLUBILITY OF O2, CO, AND N2 IN MINERAL OIL AND THE TRANSFER OF CARBON DIOXIDE FROM OIL TO AIR. The Journal of Biological Chemistry (1927)
 vol. 72,  pages 545-548 gives quantitative values for the solubility of oxygen in mineral oil, but I can't understand what units are being used. It says solubility is 0.134 +/- 0.004 at 24 degrees C.  I think this means 0.134 volumes of gas phase oxygen at standard temperature and pressure per volume of oil. Whatever the units are, it does show that significant oxygen can be present in mineral oil.

Also, I'll quote your own answer to another question:

  Even when these [alkali] metals are stored under mineral oil, oxygen can dissolve in the mineral oil and react.

Original Source: Management of time-sensitive
chemicals (II): Their
identification, chemistry
and management


Does the distillation of alcohol cause the fractioning of isotopes and carbon-12 enrichment? In water, distillation causes fractioning of the $\ce{^{18}O}/\ce{^{16}O}$ oxygen isotopes (Wikipedia).

Does something similar happen to the $\ce{^{13}C}/\ce{^{12}C}$ and possibly $\ce{^{14}C}$ isotopes in alcohol when it is distilled out of a watery brew?

Does the distillation of alcohol cause the fractioning of isotopes[...]?

The short answer is it depends.

Distillation of ethanol from water can be done nearly to completion.  There are published methods for determining isotopic composition of sugars in fruit juices which involve fermentation of the juice to yield ethanol, followed by nearly quantitative distillation of the ethanol from the resulting brew.  Then, the 13C content of the ethanol is measured.  If the ethanol distillation step (or fermentation step) substantially altered the isotopic composition of the sugars, this method would not be suitable.  The main thing which determines the isotopic composition of the ethanol in this case is the type of photosynthetic metabolism used by the plants from which the juices or syrups are derived.  Quoting from the linked paper:

  The fruit juices or maple syrups are completely fermented with yeast, and the alcohol is distilled with a quantitative yield (>96%). The carbon-13 deviation (delta13C) of ethanol is then determined by IRMS. This parameter becomes less negative when exogenous sugar derived from plants exhibiting a C4 metabolism (e.g., corn or cane) is added to a juice obtained from plants exhibiting a C3 metabolism (most common fruits except pineapple) or to maple syrup.

If distillation is not done to completion, there can be a fractionation.  In fact this has been measured fairly accurately in 1990 for distillation of pure ethanol.  Here is a figure from that paper.

It plots the ratio $R_D$ of 13C to 12C of the distillate ($D$) normalized to the same ratio $R_Q$ for the starting material $Q$, as a function of the fractional yield or progress $\rho$ of the distillation (so $\rho = 0$ corresponds to no distillation and $\rho = 1$ corresponds to complete distillation).  It also plots the same curve for the remaining liquid $L$.

As you can see, at $\rho = 0$, when only a few molecules of ethanol have distilled, the liquid (unsurprisingly) has an unchanged isotopic composition relative to the starting material, and the tiny amount of vapor that does come over has an isotopic composition that is enriched in $\ce{^{13}C}$.  That's right, for pure ethanol at least, distillation is a process with an inverse isotope effect.
As $\rho \rightarrow 1$, the isotopic content of the distillate tends to 1.0,  consistent with the idea that for quantitative distillation, isotopic composition doesn't change.
At 50% completion, the distillate has about 0.2% more $\ce{^{13}C}$ than the starting material, and the remaining liquid has about 0.2% less $\ce{^{13}C}$ than the starting material.  So even though the composition does vary, the effect isn't very big.

  Does the distillation of alcohol cause [...] carbon-12 enrichment?

No, contrary to what I would have guessed before I read the literature, distillation of alcohol can cause carbon-13 enrichment.


Radiation time vs. half life period Is radiation time the same thing as half life period?

Also, if I'm talking about excited species in a gas, how would the gas interaction affect the radiation time, are there some formulas/approaches?

In the context of photochemistry, radiation time or half life period are not used, as far as I know.

The only relevant technical term i'm aware of -in that context- is half-life (time), which describes the time needed for the concentration of an excited species to decrease to 50% of the initial value, cf. S. E. Braslavsky, Glossary of terms used in photochemistry,
Pure Appl. Chem., 2007, 79, 293-465
[DOI].

For longer-lived excited triplet states in the gas phase, radiationless deactivation upon collision with a ground-state molecules will play a role.

Thus, the lifetime might be correlated with the mean free path of the molecules.


Alkali metals chemistry Ok my understanding of the electronic structure of atoms could be wrong but this is why I don't understand:Why alkali metals are so much electropositive.

Some high school professor in chemistry would say because if they lose an electron they have an electronic configuration of noble gas etc.

If alkali metals gained 1 electron then the wavefunction of the each electron in the s orbital would shield some of the charge of the nucleus for the other electron in the s orbital->the overall net charge on each electron would be decreased but because now we have 2 electrons instead of one potential energy would be decreased.

That is clearly not the case.Why?

Maurice notes that alkali metal diatomic molecules are at best a very minor constituent.  But alkali metals do fill their $s$ valence shells in other ways.

Most alkali metals are known to form alkalides, in which an electron is transferred from one alkali metal to another (or from barium to an alkali metal, in a more recent example).  To effect this electron transfer, a complexing agent is used to stabilize the cation, such as 2.2.2 cryptand in the compound $\ce{[Na}\text{(2.2.2-crpytand)}\ce{]^+Na^-}$.  A combination of low ionization energy, positive electron affinity and the stabilization offered by complexing the cation enable this outcome.

Lithium apparently fails to form such alkalides, but not because it refuses to adopt an $s^2$ valence electron configuration.  Quite the opposite: it tends to form polar covalent bonds with that configuration.  Organolithium compounds are a widely used example of this electron structure, and the covalent bonding component in lithium salts may contribute to their solubility in some organic solvents.


Why is logP quite a bit lower in cyclohexane than in hexane? http://www.rsc.org/suppdata/nj/b3/b303016d/b303016d.pdf

If anyone doesn't know the definition, logP is the the log of the ratio of a molecule in dioxane to water. So a high logP would mean the molecule is mostly in the organic dioxane layer rather than water. E.g a molecule with a logP of 2 would prefer the organic layer 100 times more than water, and would be considered hydrophobic/lipophilic.

Hexane logP=4.15, very hydrophobic and almost all of it will stay in the organic layer.

Cyclohexane logP=3.66, still very hydrophobic, but hexane is almost 4 times more hydrophobic than cyclohexane. Can anyone please try to explain why the cyclic molecule appears to like water slightly more than its straight chain counterpart?

Think of the hole in the water that the molecules occupies as a bubble. The bigger the bubble the greater the area over which surface tension is acting.

The highest volume to surface ratio is for a sphere. So hexane is like a rod and cyclohexane is like hexagonal pyramid. Guess which has a higher volume to surface ratio?

edit more information...

Not that it is exactly the same, but look at the densities of the liquids. The cyclohexane molecule occupies a small volume in the liquid as compared to hexane, hence it is denser, even though it has 2 hydrogen atoms less per molecule.

Now water has 18 ml/mole so we can figure out roughly how many water molecules are displaced by a molecule of each hydrocarbon.

                                                     # water
                                                     molecules
             density         molecular mass  ml/mole displaced
Cyclohexane  0.7781 g/mL     84.16 g/mol     108.2    6.0
Hexane       0.6548 g/mL     86.18 g/mol     131.6    7.3


Can different groups within a single molecule photoisomerize at different wavelengths? Azobenzene undergoes trans → cis photoisomerization when irradiated with light with a wavelength of 300–400 nm.

By adding electron withdrawing groups, the wavelength can be red-shifted.

If I were to make the molecule shown below, would it be possible to selectively photoisomerize around each of the $\ce{N=N}$ at different wavelengths?

In theory: yes

As soon as you have two distinct groups, you will have two different, discrete energy differences. If you manage to irradiate at exactly the correct wavelength, only one of the two will be excited.

In practice: possibly, but it depends

Two diazo groups are still rather similar. You will either need a beam that has a good wavelength focus or you will have to make their wavelengths sufficiently different.

By the way, note that both your average IR spectrum and your average UV/Vis spectrum show this property. At a specific wavelength, exactly one vibration (IR) or electron (UV/Vis) will be excited. With sufficient resolution, you should be able to excite the different nitro group vibrations separately. If this applies to one light wavelength band, it will work in others, too.


Describe the bonding in a tribromide anion, what is its structure? I'm thinking it will be linear with a negative charge on the middle bromine. Someone told me it is trigonal bipyramidal geometry with 3 lone pairs around the central bromine and the two other bromines above and below making it linear. Is this correct? Also is it just covalent bonding in the molecule?

In fact, tribromide ion has the generic formula $\ce{AX2E3}$ according to Lewis diagram. Where  $\ce{A}$ is the central atom, $\ce{X}$ an atom bonded to $\ce{A}$ and $\ce{E}$ a lone pair on $\ce{A}$.  Using Valence Shell Electron Pair Repulsion (VSEPR) theory,  the geometry  of this ion will be
 :

So, the molecular shape is linear, while the electron pair geometry is trigonal  bipyramid.
 And the bonds in the ion are covalent.


Monitoring H -> D exchange I was wondering how the exchange of hydrogen by deuterium, like on arenes like benzene

could be monitored for a reaction on a scale larger than say 10 g of product.  While I know that C6D6 is commercially available, used for NMR spectroscopy, I recognize

the observable signal of the protons bound to carbon is attenuated by the advancement of the reaction, and
the very deuterium installed equally serves as lock signal for the 1H-NMR spectrometer (similar to the deuterium in CDCl3).  So there might be an experimental problem, from the spectrometer, too.

But what is a typical standard compound may serve as an added inner standard in such a reaction, provided protocols accessed so far mention extended heating / autoclave or/and presence some hydrogen gas to initiate the reaction as reaction conditions?  Do commercial suppliers carry out the exchange hydrogen by deuterium, and (perhaps by virtue of even larger batches) obtain the 99%+ isotopically enriched product by sophisticated distillation / zone melting, recycling the partially deuterated product?

You've managed to squeeze a few questions into one here, but all nice questions, so let's have a go at answering them.

What standard could be used for monitoring this type of reaction?
As you indicate, any internal standard needs to be inert to any reaction process, and many industrial processes for deuteration are fairly extreme, so it is unlikely that a reliable standard could be found. Instead, what is commonly done is to add an internal standard post-reaction. That is, take an accurately measured aliquot of the reaction mixture and add a known amount of standard. This is commonly done for quantitative analysis, and is easy and accurate to do. Take 0.5mL of your reaction mixture and add 0.1ml of a stock solution of your standard. Common standards include TMS, TSP, acetonitrile, methylsulfone, benzene, 1,4-dioxane and dichloromethane and of course the one to use wil depend on the expected peak ranges of your material. Wise spectroscopists use a stock solution containing two standards.

Many modern spectrometers can use an electronic form of providing an internal integration standard. The ERETIC method (search NMR ERETIC) introduces a highly accurate and reproducible concentration measurements by applying a synthesised electronic reference, and can be postioned anywhere in the spectrum out of the way of other peaks. This method is used extensively in qNMR methodology.
A precaution that you need to take care with to ensure accurate results is in ensuring no loss of solvent volume (ie changes to concentration) over time of your reaction mixture.
Interference of 2H signal in experiment. It is entirely possible to run an experiment without an internal lock. You can use an external lock, or simply turn of field sweep for short durations (up to a few hours). Alternatively, it is possible to use 19F as a spectrometer lock on spectrometers where the hardware allows. In the event of requiring an internal lock, this too would be no problem, providing the signal was well away from your peaks of interest. Similarly, large amounts of 1H signal from other reactants could easily be eliminated by solvent suppression methods.
Commercial deuteration of common solvents for NMR typically use processes involving D2O. D2O is manufactured by a number of means, such as the Girdler sulfide process, and is pretty cheap to manufacture. Other solvents will use multi step processes to drive the exchange. DMSO, for example, is heated over CaO to induce partial exchange, and the process repeated several times before (almost) complete conversion.


Why do we use different arguments for determining the strength of hydracids and solubility of ionic compounds? HI is a stronger acid than HF. Why? Because when dissolved in water, the bigger iodide ion handles the negative charge way better than the small fluoride ion. So Iodide ion is a weak conjugate base making HI a strong acid.  Which one is more soluble in water, AgF or AgI? Answer is AgF. Why? It is an ionic compound and by Fajans' rules, an ionic bond shows more covalent character if it is more polarised. The smaller the anion, the less polarisation there will be, thus more ionic character, hence more solubility. But what if I said that AgI is more soluble in water, because what happens when AgI dissolves in water? It turns into Ag+ ion and I- ion. And I- handles the negative charge better than F-. What's wrong with that?

When pondering solubility, you can't just think about what happens to the compound, you have to think about what happens to the solvent, too.  Solubility is a balance between the loss of interactions between the particles of the compound and the gain of interactions between those particles and the solvent particles.

In this case, the small highly charged F- anion will have much stronger ion-dipole interactions with water, as reflected in its much higher heat of hydration (-524 kJ/mol for F- versus -308 kJ/mol for I-).  That suggests to me that even if the lost Ag+/F- interaction (whether defined as "ionic" or "covalent") is greater than the equivalent lost Ag+/I- interaction when AgI dissolves, the much strong interaction of the F- with water more than compensates, giving AgF the greater solubility.


Significant Figures Calculations I am confused regarding significant figures. Consider "Calculate the volume of 2.8 kg of liquid with density of 1.11 g/cm^3"

Then,
Volume = 2.8 kg * (cm^3 / 1.11 g) = 2522.522 (where 522 is recurring).

Since the original figure (2.8 kg) has only two significant figures, I thought the answer should be 2.5 kg.

The textbook says the answer is 2.523 kg, which is four significant figures.

Which is correct?

Your answer of $2500$ and the book's answer of $2523$ both have four significant figures.

If you want to keep your answer to the same degree of accuracy as the given data with the least number of significant figures, i.e. $2.8$ kg, it is best to express your answer as $2.5\times10^{3}$.


Why methyl group is 2,4-directing? Why methyl group is 2,4-directing? I am an A level student and this isn't required at this level. I am looking for a simple explanation which a high school student can understand.

Why methyl group is 2,4-directing?

When toluene undergoes electrophilic aromatic substitution the products are primarily the ortho and para isomers; usually only a small amount of the meta isomer is produced.  In order to explain this observation there are two effects to consider, inductive and resonance effects.

Inductively the methyl group releases electron density into the benzene ring.  This is because the methyl group, being $\ce{sp^3}$ hybridized, is less electronegative than the $\ce{sp^2}$ hybridized aromatic carbon.  The carbon with more s-character is more electronegative, electron density likes to flow from low to high s-character orbitals since the more s-character in an orbital, the lower its energy.  If inductive arguments were most important we would expect the ortho isomer to predominate followed by the meta and para isomers in that order.  This is not observed, the meta isomer is only a minor isomer.

Resonance arguments are based on the concept of hyperconjugation involving the $\ce{C-H}$ bonds in the methyl group.  The following figure shows some of the possible hyperconjugative resonance structures for toluene.

Additional resonance structures can be drawn using the other methyl $\ce{C-H}$ bonds as well.  As the resonance structures show, electron density is increased at the ortho and para positions, not the meta.  This hyperconjugative resonance effect makes the ortho and para carbons more susceptible to electrophilic attack.

Hyperconjugation explains how alkyl groups can be electron releasing in a resonance sense.  Hyperconjugation explains the mild activating effect of an alkyl group in electrophilic aromatic substitution and its ortho-para directing effect.


What happens when a super-critical fluid is compressed? Suppose that I have some substance with a critical temperature of 20°C. That means above that temperature, the substance exists neither as a gas nor a liquid, but instead as a super-critical fluid. Does it mean that no matter how high a pressure I apply on the fluid, it is not going to turn into a liquid above 20°C? Why is that so?

A substance is a liquid if the particles are close enough to interact, i.e. the space between particles is similar across different liquids. It seems that if we increase the pressure (i.e. decrease the available volume), particles should get close enough to interact.

I know that the higher the temperature of a gas, the higher the speed of the particles. How does this affect the phase transition?

What is the role of intermolecular forces? They are part of the explanation why at lower temperatures, gases condense into liquids.  It seems that we are ignoring them in the case of super-critical fluids. Are we now saying I just need to apply pressure to decrease the distance between particles? If it weren't for intermolecular forces, I would have to apply much higher pressure to turn a gas into a liquid (at temperatures below the critical temperature).

So what happens when a super-critical fluid is compressed, and how is the interplay of particle speed, intermolecular forces and distance of particles different below and above the critical temperature?

When you heat up a liquid at constant volume (leaving sufficient space for the gas phase), the density of the liquid will decrease and the intermolecular interactions will weaken. Some of the liquid will transition to vapor, so the vapor above the liquid will get denser and the frequency of intermolecular interactions will increase (it behaves less and less as an ideal gas as the number of collisions increases with increasing density). At the point where the density of the liquid and the gas are the same, there is no more phase boundary (with gravity, we picture liquid on the bottom and gas on the top, but if the density is identical, that is no longer the case).

This is all pretty outlandish for most of us because we are used to constant pressure, not constant volume. Here are two videos showing the process:

Liquid chlorine: Here, it looks like the liquid level stays constant - the expansion of the liquid and the evaporation balance each other out. If you look carefully, though, you can see how the vapor phase color gets more intense, indicating more molecules in the vapor phase. At the same time, the color of the liquid gets less intense, indicating a decrease in density.
https://chem.libretexts.org/Courses/University_of_North_Texas/UNT%3A_CHEM_1410_-_General_Chemistry_for_Science_Majors_I/Text/10%3A_Solids%2C_Liquids_and_Solutions/10.13%3A_Critical_Temperature_and_Pressure

Carbon dioxide: When approaching the critical point, the system gets turbid because the difference in density between liquid and gas is no longer sufficient to keep the liquid down and the vapor up. The system is a heterogeneous mixture of liquid and vapor for a while, and then a single pure phase.

https://www.youtube.com/watch?v=GEr3NxsPTOA

Making of video: https://www.youtube.com/watch?v=-gCTKteN5Y4


Why is it considered acid rain with pH <5.6? I recently read in a book that rain is considered acid rain if the pH falls below 5.6. However a substance is acidic when the pH is below 7; so why is the boundary for acid rain 5.6?

I was thinking pH between 5.6-7 would be too diluted to have an effect on limestone and other materials, and that's why the pH has to be under 5.6. But I wasn't sure if this is true so I wanted to find out.

The $\mathrm{pH}$ of pure water (rain as well as distilled water) in equilibrium with the atmosphere ($p_{\ce{CO2}}= 10^{-3.5}\ \mathrm{atm}$) can be calculated as follows.

$$[\ce{H2CO3^*}]=K_\mathrm H\cdot p_{\ce{CO2}}$$

where $[\ce{H2CO3^*}]$ is the total analytical concentration of dissolved $\ce{CO2}$, i.e. $[\ce{H2CO3^*}]=[\ce{CO2(aq)}]+[\ce{H2CO3}]$, and
$K_\mathrm H= 3.39\times10^{-2}\ \mathrm{mol\ l^{-1}\ atm^{-1}}$ is Henry's law constant for $\ce{CO2}$.

$$\begin{align}
\log[\ce{H2CO3^*}]&=\log K_\mathrm H+\log p_{\ce{CO2}}\\
&=-1.5-3.5\\
&=-5.0
\end{align}$$

The commonly used first acid dissociation constant of carbonic acid $\mathrm pK_{\mathrm a1}=6.3$ (at $25\ \mathrm{^\circ C}$) actually is a composite constant that includes both the hydration reaction
$$\ce{H2O + CO2(aq) <=> H2CO3}$$
and the protolysis of true $\ce{H2CO3}$
$$\ce{H2CO3 <=> H+ + HCO3-}$$
For a weak acid
$$\begin{align}
\log[\ce{H+}]&\approx\frac12\left(\log K_\mathrm a+\log[\ce{H2CO3^*}]\right)\\
&=\frac12\left(-6.3-5.0\right)\\
&=-5.65\\
\mathrm{pH}&=5.65
\end{align}$$

Thus, pure rain in equilibrium with the atmosphere has about $\mathrm{pH}=5.65$. Any acid rain with lower $\mathrm{pH}$ would be caused by additional acids.


Why do acids usually completely react with bases? When a neutralisation reaction happens, for example, $ \pu{100 mol l^-1}$ of $\ce{HCl}$ with $\pu{100 mol l^-1}$ $\ce{NH_3}$, why does all of the base and acid get converted to salt? Why isn't there an equilibrium established between salt and acid-base?

Acids and bases don't necessarily completely react with each other. Consider your hypothetical reaction between hydrochloric acid and ammonia.

\begin{align}
\ce{HCl + H3N &-> H4N+ + Cl-}&
 \text{Extent:}& \approx100\%
\end{align}

The products are ammonium ion and chloride ion - the conjugate acid and conjugate base, respectively. So if the forward (above) reaction is complete, how can the reverse reaction between the conjugate acid and conjugate base be anything but insignificant? I.e.

\begin{align}
\ce{H4N+ + Cl- &<<=> HCl +H3N}&
 \text{Extent:}& \approx0\%
\end{align}

An equilibrium is established whenever the extent of reaction is not complete. Another good example would be an acetic acid solution. Acetic acid and water exist in equilibrium with acetate anion and hydronium ion. So no not all acid-base reactions go to an extent of 100%. Things to consider: stoichiometry and acid and base strengths.


Zinc Sulfate Purification In the process for electrolysis for zinc, three things must be done:

Leaching the ZnO
Purifying the ZnSO4 by means of cementation
Performing the electrolysis

However, step 2, as it is presented by Wikipedia, is quite vague. In particular, it says:

  The purification process utilizes the cementation process to further purify the zinc. It uses zinc dust and steam to remove copper, cadmium, cobalt, and nickel, which would interfere with the electrolysis process.

However, when looking it the definition of cementation, a crude summary of what happens is that one metal gets oxidized in order to reduce the other ion. I am assuming that this means that the zinc sulfate is getting reduced. However, what is reducing it? What equations are involved? I apologize if this question does not make sense, as chemistry is not particularly my strong suit, but I believe that this has gone unanswered online so far.

As with nearly all STEM-related Wikipedia pages, this one reads quite clear and concise... that is, if you already know at least 75% of the stuff involved. Otherwise it may lead you into the deepest muddle, so you did the right thing by asking this question.

Follow the logic. What is electrolysis used for? It reduces a metal (in particular, zinc) from its salts. You don't reduce what's already been reduced. This means zinc was not reduced before the step 3. In particular, it was not reduced during the cementation. Still, cementation implies that one metal is reduced by another. Which of the two is zinc if we've just concluded that zinc does not get reduced on this step?

Since the naturally found compounds are rarely pure, zinc sulfate after the step 1 contains a few percent of other metals (copper, nickel, etc.), also in the form of sulfates. Should we electrolyze it now, these metals will get reduced along with our zinc and remain in it as impurities, which we don't want. Worse yet, they are mostly less active than zinc, so they will reduce before it and thus concentrate in the produced metal. How do we get rid of them?

Now comes the idea: let's reduce them. We vaguely remember this can be done with a more active metal. But it must not be too active, otherwise it will reduce zinc as well. Also, the oxidized form of this new metal must not spoil our $\ce{ZnSO4}$. Just what metal can that be?

Guess what?

   It is zinc.
 Yes, we add metallic zinc (just a few percent, I remind) and it reduces the impurities as follows:
 $$\ce{CuSO4 + Zn -> Cu + ZnSO4}$$
 The only soluble byproduct is $\ce{ZnSO4}$, which does not upset us one bit.

So it goes.


Why is fluoride the most nucleophilic halide in aprotic solvents? I am confused on the topic of electronegative nucleophiles, and polar aprotic solvents.

From my understanding, the more electronegative a nucleophile is, the weaker of a nucleophile it will be, because atoms will be less likely to share its electron density.

However, in a polar aprotic solvent, because H-bonding cannot occur, it seems as if the trend of nucleophilicity is such that $\ce{F- > Cl- > Br- > I-}$.

Can someone please clarify how this is possible? I would have expected $\ce{F-}$ to be the LEAST nucleophilic nucleophile in a polar aprotic solvent, because it is the most electronegative atom. Therefore according to the first point I learned, I thought it should be the weakest nucleophile.

All this is what I've learned through reading Kaplan's Organic Chemistry Review.

When talking about elements of a group, the size of the atom is more effective than the electronegativity factor.
As $\ce{F-}$ size is too short so its electron density will be much higher than other hallogens.

In aprotic solvent the order of nucleophillicity:
$\ce{F- > Cl- > Br- >I-}$(No hydration occurs) In protic solvant(like $\ce{H2O}$) its reversed because of hydration of ions having much electron density.


At equilibrium, the mass of reactants and products are equal. Does that mean the reaction stops? This one is troubling me, and I even got it wrong in my exam:

  at equilibrium the mass of reactants and products are equal, does that mean the reaction stops?

Please explain this one.

At equilibrium, the reaction continues, but the rate of the forward reaction is equal to the rate of the backward reaction.

Let's consider the first order reaction $\ce{A <=>[k_f][k_b] B}$

So, the rate of the forward reaction is $k_f[A]$, and the rate of the backward reaction is $k_b[B]$. Initially, $[B]=0$, and $[A]$ is high, so the forward reaction is very fast, and the backwards reaction doesn't occur at all. As time passes, $[A]$ decreases and $[B]$ increases, so the forward reaction becomes slower and slower, and the backwards reaction speeds up. At one point, both reactions have equal rates ($k_f[A]=k_b[B]$). Here, both reactions take place, but for every time when an $A$ becomes $B$, another $B$ becomes $A$. The reaction is taking place, just that you don;t notice it since the net effect is zero.

Here's a diagram for how the rates of the reaction proceed for the formation of ammonia. Note that the rates never reach 0:


Confusion in the order of precedence: alkyl side-chain vs. dimethyl at different positions I've read that prefixes like 'tert-butyl, 'sec-butyl' and have been discarded while 'iso-propyl' is still retained by IUPAC. So I wrote another name according to the general rules. Is it correct? If not please correct me.

(The structure in question is framed)

EDIT: When I typed the name here it gave a slightly different structure so I tried putting 5-(1-methylpropyl) first, as in 5-(1-methylpropyl)-2,7-dimethylnonane, which gave the correct structure.

So, my question now is, since both substituents have same alphabetical order, why do we write 5-(1-methylpropyl) first?

Your starting point is correct. According to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the name 5-(sec-butyl)-2,7-dimethylnonane is no longer recommended. (The prefix ‘sec-butyl’ was still contained in the 1993 recommendations.)

The names that are proposed in the question, 2,7-dimethyl-5-(1-methylpropyl)nonane and 5-(1-methylpropyl)-2,7-dimethylnonane, are unambiguous and describe the same structure. There may be many different ways to draw the structure of such branched alkanes. If you enter these names into the Open Parser for Systematic IUPAC nomenclature (OPSIN), which is mentioned in the question, the results are as follows:

2,7-dimethyl-5-(1-methylpropyl)nonane

5-(1-methylpropyl)-2,7-dimethylnonane

The structures may look different at first glance, but they actually describe the same compound. In this respect, there is no difference between 2,7-dimethyl-5-(1-methylpropyl)nonane and 5-(1-methylpropyl)-2,7-dimethylnonane. However, the name 5-(1-methylpropyl)-2,7-dimethylnonane is not in accordance with IUPAC nomenclature since ‘methyl’ is earlier alphabetically than ‘methylpropyl’. Therefore, the name 2,7-dimethyl-5-(1-methylpropyl)nonane is a correct IUPAC name.

Anyway, the preferred IUPAC name (PIN) according to current recommendations is 5-(butan-2-yl)-2,7-dimethylnonane.


Green color substance forming during electrolysis A green-colored substance formed during electrolysis, wherein the electrolyte was magnesium sulfate and minerals in tap water.

I did not have any leads to stick into the water, so I just put a 9 volt Duracel battery in the solution. The battery didn't leak.

White flakes formed after 1 minute before green substance at 10 minutes.
So, what is the green stuff and what are the white flakes?

The green material is probably an iron(II) salt, corroded from what I assume is the steel composing the positive pole of the battery.  If it has a gelatinous appearance, and slowly turned orange-ish over time, my guess would be it's $\ce{Fe(OH)2}$, with any orange color deriving from slow oxidation to various iron(III) species such as the oxide-hydroxides.

Per the above link, the white flakes are possibly that same iron(II) hydroxide, prior to reaction with $\ce{O2}$ to form the green color.  Alternatively, it could be $\ce{Mg(OH)2}$, or a related magnesium compound, formed at the cathode (negative terminal) from the hydroxide ion produced by reductive water electrolysis:

$$
2\ce{H2O} + 2e^- \longrightarrow \ce{H2} + 2\ce{OH-}
$$

A photograph would be tremendously helpful in facilitating identification.


Why heating up things before H2C2O4 &KMnO4 titration? In my school, there is a redox titration between H2C2O4 being titrand and KMnO4 being titrant. After adding H2SO4 to the acid to give an acidic medium for redox, the whole mixture inside the conical flask is heated up to 80 degree Celsius before titration starts, what is the need of heating?

In general, the higher the temperature the faster the reaction. This is because increasing the temperature increases the number of molecules with more kinetic energy, and hence more molecules would have energy greater than the activation energy. If the reaction between oxalic acid and potassium permanganate wasn't heated then it would take extremely long to see the self-indicating titration occurring.


Which mechanism do ethers follow when cleaved by hydrogen halides? Ethers can be cleaved under acidic conditions by treatment with hydrogen halides (HX, where X = Cl/Br/I). Some of these reactions proceed by SN1 mechanisms and some by SN2 mechanisms. How can I tell which mechanism is being followed?

The first step of this reaction is always a proton transfer:

Then, the substitution follows either the SN2 or SN1 pathway depending on whether R is methyl, primary, secondary, or tertiary.

Methyl and primary ethers undergo acidic cleavage by the SN2 mechanism:

Tertiary ethers undergo cleavage by the SN1 mechanism:

Secondary ethers probably undergo cleavage by both mechanisms competing with each other.

Note that unless R' is an aryl group, the alcohol R'OH that forms will also be converted into a halide by a similar mechanism.

Read more at any of the sites in this Google search.


Why is arsenic trichloride more likely to be polar than aluminum trichloride? According to my chemistry teacher, $\ce{AsCl3}$ is more likely to be polar than $\ce{AlCl3}$. Why is this? As I understand it, the difference in electronegativity is what makes a compound more or less likely to be polar, and $\ce{AlCl3}$ has a much greater difference between the two elements' electronegativities. So why isn't it more likely to be polar of the two?

It is because $\ce{AlCl3}$ has a trigonal planar geometry (leading to no permanent dipole moment) and $\ce{AsCl3}$ has a pyramidal geometry (leading to a permanent dipole moment).


Which factors best account for the increase in boiling point? The boiling points of $\ce{CH3COCH3}$, $\ce{CH3COC2H5}$, and  $\ce{CH3COC3H7}$ are $56˚C$,
  $80 ˚C$, and $102 ˚C$, respectively.  What factor(s) best account(s) for this?

I believe london dispersion forces is one of them. Is dipole-dipole interactions another one? It seems to me that there are dipole-dipole interactions since none of the structures of symmetrical, so can they account for this increase in boiling point?

You're definitely right on London. Between the three molecules, the only thing that's changing is the carbon chain on the right hand side of the Oxygen. That plus the corresponding increase in boiling temperatures together form a good indicator that London forces are at play here.

Dipole-dipole (or Hydrogen bonding for that matter) are non-significant in this case, because you're not changing the electronegativity of anything here, only adding more of the same repeating atoms.

Source: This was on the ACS 2011 local exam, question 55.

Further information here


what are possible isomers of (meso-stilbenediamine)(isobutylenediamine) palladium (II)? It can form a square planar and a tetrahedral complex but within square planar or within tetrahedral can there be isomers possible?

The use of the term "meso" in the compound name indicates that both phenyl groups are on the same side of the molecule, "cis" to one another, not "trans".

Here are the tetrahedral and square planar representations.

Note: I used $\ce{Pt}$ instead of $\ce{Pd}$ as the central metal, but it makes no difference in terms of isomers.

The tetrahedral molecule has a plane of symmetry containing the "top" 5-membered ring, bisecting the 2 methyl groups on that ring and then bisecting the "bottom" 5-membered ring with the 2 "meso" phenyl groups. Since the tetrahedral molecule contains a plane of symmetry it is achiral and has no isomers.

In the square planar form, there is no symmetry element.  Therefore the square planar molecule is chiral and exists as an enantiomeric pair of isomers.


Why do electron donating groups decrease the acidity of aromatic carboxylic acids? If we look at the resonance structures of benzoate anion, we would see that some amount of positive charge is present on the ring and this charge can get delocalised on the ring, like this:

Now, if we had an electron donating group (EDG) at the para position, then the EDG should stabilize the ring because it can donate pi electrons to the empty p orbital of carbocation (look at third resonance structure) and form a double bond, thus resulting in more resonance and more resonance implies more stability and this should imply that EDG increases acidity as conjugate base is getting more stabilized.

But, it is a well established fact that EDG decrease the acidity of the molecule (benzoic acid). For example: $\mathrm{p}K_\mathrm{a}$ (para methoxy benzoic acid) > $\mathrm{p}K_\mathrm{a}$ (benzoic acid)

Why am I facing such a contradiction?

You seem to pay too much importance to these individual structures. However, note that there are no individual resonance structure in existence, but only the resonance hybrid - a combination of all these resonance structures in proportion to their stability.

The final three structures that you've drawn have a really negligible contribution to the final resonance hybrid of the benzoate anion. Why, you ask? Because they exhibit charge separation: creation of an additional positive and a negative charge.

Hence, your conclusion is wrong, because it's based on the invalid premise that the third (and by extension, the fifth) resonance structures have a major role in the resonance hybrid. The resonance hybrid in fact has no appreciable positive charge on the ring carbons.

The correct way to explain this is that you should look at the resonance hybrid as a whole, which is negatively charged. Adding an electron donating group to the ring would increase the negative charge further, thereby destabilizing the anion.


Ca and Mg hydrated radius in clay flocculation Ca has an atomic radius of $197\mathrm{~pm}$ and a Pauling electronegativity of $1.00$.

Mg has an atomic radius of $197\mathrm{~pm}$ and a Pauling electronegativity of $1.31$.

Since Mg is smaller and has a higher electronegativity, one would think Mg would have a smaller hydrated radius than Ca, but it doesn't.  Would someone explain to me why this isn't so?  I have read an explanation pertaining to charge density, but I'm not understanding why Ca would have a larger charge density if it is bigger with less electronegativity.

The goal of understanding this is to understand why the addition of Ca to soil will make it drain better and be looser, less sticky, etc.  And conversely, why highly respected agricultural consultants say soils high in Mg (low in Ca) are "sticky".  Any help in understanding this would be well appreciated.

Charge density is a measure of electrical charge per volume. It is independent from electronegativity, and increases with the number of charges of an ion and with decreasing radius, because a smaller ion has also a smaller volume. The number of charges for $\ce{Ca^{2+}}$ and $\ce{Mg^{2+}}$ is the same (+2), but $\ce{Mg^{2+}}$ has the smaller atomic radius. Its charge is therefore distributed over a smaller volume, and the charge density is higher.

The hydrated radius of an ion can somewhat differ from its atomic radius, because it is influenced by the charge density. If the ion charge is concentrated in a smaller volume, it exerts stronger attractive forces on the dipoles of water molecules, which results in the formation of a larger hydration shell. The smaller $\ce{Mg^{2+}}$ attracts more water dipoles and forms a larger hydration shell. Hence, its hydrated radius can be bigger than that of $\ce{Ca^{2+}}$, which has a larger atomic radius but a smaller hydrate shell.


Why does diesel have a higher energy density than gasoline? One of the primary reasons why hydrocarbons produce energy is the production of water during the combustion process.

I've been trying to find out why diesel is considered to have a higher energy density compared to gasoline and I can think of a couple of reasons:

Higher molar mass (more carbon/hydrogen to burn)
Easier to burn diesel since it is heavier (weaker bonds)

I think I'm on the right path but I don't have a clear picture of the situation.

Diesel and gasoline have roughly the same energy per unit mass (lower heating value), about $41~\pu{MJ/kg}$).

The density of diesel is about $833~\pu{kg/m^3}$ compared to $740~\pu{kg/m^3}$ for gasoline. This gives diesel about 13% higher energy density per volume.


Concept of Equivalent Weight I'm having a hard time understanding what equivalent weight means physically.  I've looked at many articles on Google but the definition always seems nebulous to me.
For example, if the equivalent of $\ce{H2SO4}$ is $\pu{49 g/mol}$, what exactly does that mean?

Wow, equivalent weights keep popping up again and again. It is not a nebulous term, the point is you are checking recent webpages. You have to look up chemistry books from 1930 to 60. If corona were not an issue I would have suggested a library trip to read Vogel's Quantitative Chemical Analysis (oldest edition).
Equivalent weight has nothing to do with mol. It predates moles story because the unit originated centuries ago. Mixing mole with equivalents is perhaps the main source of confusion.
Equivalent is the weight of X in grams, that will furnish or combine with 1 gram H or 16 gram O atoms.
Let us take sulfuric acid, ask yourself, what weight (=mass) of sulfuric acid will provide 1 gram H. Note I am not involving any mole concept here. The answer is that if we take 49 g of sulfuric acid we will be able to "generate" 1 g hydrogen atoms.


Strength of nucleophiles among SN1 and SN2 reactions As per the theory of chemistry, we say that unstable carbocations undergo SN2 reactions as it involves strong nucleophile. As a consequence, it does not lead to the formation of reaction intermediate, i.e. a carbocation.
Similarly, in stable carbocations undergo SN1 reaction by substitution of a weak nucleophile.
Why necessarily weak nucleophiles are involved? In case of SN2, since the carbocation is unstable, a strong nucleophile which prevents the formation of carbocation. But in SN1, since stable carbocation is involved, why shouldn't both strong and weak nucleophiles be involved?

This seems to be true only because SN1 reactions prefer polar protic solvents which stabilize both the leaving group and the carbocation. But such solvents also surround strong nucleophile and hinder it from the carbocation.
Weak nucleophiles on the other hand don't interact with protic solvents as much and thus stand a better chance of getting close to carbocation.


Why was atomic mass scale changed from Oxygen - 16 to Carbon - 12? Why was unified atomic mass scale introduced and why was Oxygen - 16 replaced by carbon - 12 for standardizing atomic scale?

The mass scale has changed over time, largely due to different isotopes of the "baseline." Not surprisingly, there's a good Wikipedia article on the matter.

  In the 20th century, until the 1960s chemists and physicists used two different atomic-mass scales. The chemists used a "atomic mass unit" (amu) scale such that the natural mixture of oxygen isotopes had an atomic mass 16, while the physicists assigned the same number 16 to only the atomic mass of the most common oxygen isotope (O-16, containing eight protons and eight neutrons). However, because oxygen-17 and oxygen-18 are also present in natural oxygen this led to two different tables of atomic mass. The unified scale based on carbon-12, $\ce{^12C}$, met the physicists' need to base the scale on a pure isotope, while being numerically close to the chemists' scale.

In short, Dalton suggested $\ce{^1H}$ as the basis of the mass scale, but Ostwald pushed later for $\ce{^16O}$. Unfortunately, no one knew about isotopes yet.

The problem was that physics stuck to isotopically pure O-16 for their mass scale, and chemistry used "oxygen" (i.e., with some O-17 and O-18 present). There's 0.028% difference between those scales. That may not sound like much, but with molecules of ~200-500 amu, it starts to add up.

Using C-12, it's isotopically pure (making physics happy) very close to the previous chemistry "O-16 + O-17 + O-18" scale.

From IUPAC:

  In April 1957 at the bar in the Hotel Krasnapolski in Amsterdam, Nier suggested to Mattauch that the $\ce{^12C}$ = 12 mass scale be adopted because of carbon's use as a secondary standard in mass spectrometry. Also, $\ce{^12C}$ = 12 implied acceptable relative changes in the atomic weight scale, i.e., 42 parts-per-million (ppm) compared to 275 ppm for the $\ce{^16O}$ = 16 scale (which would not acceptable to chemists). Enthusiastically, Mattauch made a worldwide effort in the late 1950s to publicize the $\ce{^12C}$ = 12 scale and obtain the physicist's approval, while Wichers obtained the chemist's approval.


What are the possible changes that might occur if I add an inert gas to a chamber containing another few different gases? As an example, let's assume that we have a gas $\ce{P(g)}$ inside a chamber and when the temperature inside it increases to a certain value there is an equilibrium that prevails inside the system.

So for the sake of argument, let us assume that this equilibrium is:

$$\ce{P(g) <=> Q(g)  + R (g)},$$

And let's assume the figure below shows that chamber when equilibrium is established:

What I want to know is, what are the possible changes that might occur if I add an inert gas $\ce{I(g)}$ to this chamber?

Quoting from the Wikipedia page on Le Chatelier's principle, we have to distinguish two cases:

The total volume is constant:

Adding an inert gas into a gas-phase equilibrium  does not result in a shift of the equilibrium in this case. This is due to the fact that adding an inert gas does not change the partial pressures of the other gases in the container. While it is true that the total pressure of the system increases, the total pressure does not have any effect on the equilibrium constant; rather, it is a change in partial pressures that will cause a shift in the equilibrium.

The total volume is not constant:

If, however, the volume is increased, the partial pressures of all gases will decrease which results in a shift of the equilibrium towards the side with the greater number of moles of gas.


How to determine the total pressure of a gaseous mixture of xenon tetrafluoride and hydrogen? Gaseous xenon tetrafluoride at partial pressure of $\pu{2 kPa}$ and hydrogen at partial pressure of $\pu{10 kPa}$, are exploded in an enclosed container producing xenon and hydrogen fluoride:
  $$\ce{XeF4 (g) + 2 H2 (g) -> Xe (g) + 4 HF (g)}$$
  Find out the total pressure of mixture of gases in the enclosed container.

Assuming $T$ and $V$ are constant, we can assume that the pressure is directly related to the amount of substance, hence $\pu{1 kPa}$ of $\ce{XeF4}$ reacts with $\pu{2 kPa}$ of $\ce{H2}$.

This means that only $\pu{4 kPa}$ of $\ce{H2}$ is needed to react with $\pu{2 kPa}$ of $\ce{XeF4}$.

Now, to find total pressure of mixture of gases I did

$$p = \pu{2 kPa}~\text{(from Xe)} + \pu{8 kPa}~\text{(from HF)} = \pu{10 kPa}$$

Why is this wrong? Why must I add $\pu{10 kPa}$ by $\pu{6 kPa}$ which is the unreacted hydrogen?

You are correct up to the last part. The total pressure would be the sum of all the partial pressures in the reaction vessel.

We know that $$P=\left(\frac{RT}{V}\right)n$$ and therefore $n$ varies directly with $P$ as $T$, $R$, and $V$ are constant. For the sake of the problem (it won't matter) we can just say that $x=\left(\frac{RT}{V}\right)$. Therefore, $P=xn$ and the starting amounts of substance are equal to $\frac{2}{x}\pu{ mol}$ $\ce{XeF4}$ and $\frac{10}{x}\pu{ mol}$ $\ce{H2}$.

From the balanced equation and the starting materials and stoichiometry, we know that $\frac{2}{x}\pu{ mol}$ $\ce{XeF4}$ would react with $\frac{4}{x}\pu{ mol}$ $\ce{H2}$ to form $\frac{2}{x}\pu{ mol}$ $\ce{Xe}$ and $\frac{8}{x}\pu{ mol}$  $\ce{HF}$ to make $\frac{10}{x}\pu{ mol}$ product.

However, you only reacted $\frac{4}{x}\pu{ mol}$ $\ce{H2}$ of the $\frac{10}{x}\pu{ mol}$ total, so there is $\frac{6}{x}\pu{ mol}$ unreacted $\ce{H2}$ left in the reaction vessel.

$$\begin{multline}
  \frac{10}{x}\text{ amount of substance of product }
+ \frac{6}{x}\text{ amount of substance of unreacted }\\
= \frac{16}{x}\text{ total amount of substance}
\end{multline}$$

From $P=xn$, the fact that this occurred instantly, and $T$, $R$, and $V$ are constant, we then know that the final pressure is $\ce{16 kPa}$.


Why is picric acid more explosive than TNT? I've recently read that trinitrophenol (picric acid - relative effectiveness factor $1.20$) is more explosive than trinitrotoluene (TNT - relative effectiveness factor $1.00$).

Personally, I think that the mesomeric effect of the oxygen's lone pairs in picric acid should keep the nitro groups 'happy', because of the increase in electron density on them. This should enhance the overall stability of the molecule, and reduce the relative effectiveness factor. However, this is not the case. Why is this? My guess is that the electron withdrawing nature of oxygen somehow causes this.

You're trying to define the stronger explosive by comparing the reagents, but in this case comparing the products is far more important.

Trinitrotoluene is a rather oxygen-poor explosive, to the point that when detonated as a pure substance, it produces carbon soot and hydrogen gas:

$$\ce{2C7H5N3O6 -> 3N2 + 5H2 + 12CO + 2C}$$

The production of $\ce{CO}$, $\ce{CO2}$ and $\ce{H2O}$ is highly exothermic, so a lot of extra energy could be obtained if there were more oxygen to burn the soot and hydrogen. However, in detonations there's not nearly enough time for atmospheric oxygen to partake extensively in the reaction. This means that a fair amount of the explosive potential of TNT is wasted, decreasing its detonation yield per gram of substance. One way to compensate this yield loss in oxygen-poor explosives is to mix them with an oxidiser, usually an oxygen-rich substance such as $\ce{KClO3}$.

Now, on to picric acid. The structure and molar mass are very similar to TNT, but replacing the methyl group with a hydroxyl means the oxygen content of picric acid is higher. While still overall oxygen-poor, the detonation proceeds with much less soot and hydrogen generation, meaning a more complete burn, a higher energy release, and consequently a greater explosive yield.

$$\ce{2C6H3N3O7 -> 3N2 + H2 + 2H2O + 12 CO}$$

In reality, the decomposition reactions are more complex. The balanced equations above follow a rule of thumb presented here. However, a more thorough analysis should agree with this simplified result.


Electrolysis of Beer Electrolysis of water, means that the water is separated into hydrogen and oxygen. As beer also includes water, I initially thought electrolysis of beer would cause the water to separate, and the hydrogen and oxygen given off gas, leaving behind a higher concentrate alcohol.

However then realise that ethanol is also in beer (and is what makes you drunk it seems), and like water has an OH bond, so would the electrolysis also cause that to break?

(also does the rest of the molecule affect the strength of the OH bond, either stronger or weaker?)

It's hard to say what exactly you would end up with, but it probably wouldn't much resemble beer. The trick with electrochemistry is that a lot of things are electroactive and what reactions actually occur in a given reaction are not always easy to predict and depend on electrodes, solution composition, pH, applied potential, etc. Depending on your electrodes, ethanol may be significantly more easily oxidized than water. A quick literature search shows that platinum and gold are both decent catalysts for ethanol oxidation (platinum is commonly used in fuel cells for this reason), but even with other materials, I can't imagine any conditions in which one could easily enrich an ethanol/water solution in this way. Many organic compounds, probably including some that impart flavour, are likely to be oxidized as well, even if the ethanol wasn't, so it would certainly affect the taste.

Additionally, the electrolysis of water requires a large amount of energy, far more than distillation, so even if it would work, it would be an expensive way of doing it.


Confusion about the oxidation state of copper in copper cyanide re-dissolved with excess cyanide $\ce{CuSO4}$ when reacts with $\ce{KCN}$ forms $\ce{CuCN}$ while it is insoluble in $\ce{H2O}$. It is soluble in excess of $\ce{KCN}$ due to the formation of the following complex:
a) $\ce{K2[Cu(CN)4]}$
b) $\ce{K3[Cu(CN)4]}$
c) $\ce{CuCN2}$
d) none

Answer given is b) but we know that $\ce{Cu^2+}$ is more stable hence anwser should be a), shouldn’t it?

but we know that $\ce{Cu^2+}$ is more stable hence anwser should be a).

Ah, but do we? I hope you don’t ‘know’ that, because that ‘knowledge’ would defy years of chemical research.

The stability of an oxidation state, especially that of a transition metal, cannot be deduced easily or by a set of memorised rules. Many a time, a species will only be stable in one environment and unstable in another one. Take copper: In aquaeous solutions, copper(I) is usually unstable concerning oxidation to copper(II). However, copper(II) on the other hand is unstable in the presence of iodide and the two will react:

$$\ce{2Cu^2+ + 4I- -> I2 + 2CuI v}$$

So copper(I) is more stable in the presence of iodide while copper(II) is more stable in the presence of water. Notice a pattern? I don’t.

In your question specifically, the first step is a redox reaction between copper(II) and cyanide according to the following equation:

$$\ce{2 Cu^2+ + 4 CN- -> NCCN ^ + 2 CuCN v}$$

At least I notice a pattern here: Cyanide behaves much like a halide or the iodide introduced above.

By this reaction, we have established that copper(I) is more stable in the presence of cyanide than copper(II). So by adding more cyanide there is no way that we can reverse the reaction — not only because there is nothing that copper could reduce with its extra electron. So the redissolution and complex formation cannot be a redox reaction, it must be a simple complexation reaction which only allows for the correct solution to be:

$$\ce{CuCN (s) + 3 KCN (aq) <=> K3[Cu(CN)4] (aq)}$$


What is Chevron with Techron fuel? I've heard the commercials for this stuff for years, but what exactly does it do?  Does it do what the commercials say it does (clean out deposits in your tank)?  How well?

In other words, is this just an advertising campaign making a huge deal out of nothing, or is there actually an advantage to Chevron fuel?

Techron is a detergent based on PEA, polyeter amine. It is a complex synthetic molecule that will basically 'loosen' and 'solve' deposits in your fuel system. They then become part of the fuel-air mixture and burn in the combustion chamber with it. Be it in the carburettor, the injectors, the valves (for indirect injection vehicles) or the combustion chamber.

The Techron fuel you buy at the Chevron or Texaco stations has only a small amount of Techron added, the rest is just gas/petrol. You can also buy the Techron in seperate high concentrate bottles (typically 12oz.) These bottles contain up to 35% of PEA.

But the working ingrediënt PEA is just as well used in Gum-Out, Amsoil and a bunch of other fuel system cleaners.

So yes, it is a difference to fuel which doesn't have these detergent but every major gas producer (Shell, BP, Chevron, Texaco, ...) will add detergents. They ARE different detergents so they do make a difference.

In fact they are just auxiliary tools to keep your engine working well. It is like a band-aid. It does help tune up your engine, but it helps a lot more to do proper maintenance, change coolant and oil regularly, make sure your radiator works well so your engine operating temp isn't too high, clean your air filter and fuel filter regularly,...

Using these fuels does have an effect eg for emissions testing, but If you ask me. I just buy the cheapest fuel and I will add a bottle of High concentrate Techron or Gumout or another cleaner before I go to an emissions test (make sure you drive about 60 miles before turning up since the deposits have to be burnt before the test). And if you really want to use it, buy a bunch of high concentrate bottles and add them to supermarket gas, that's basically the same


Why this redox equation occurs on basic medium? I have the following redox equation that I need to balance:

$$\ce{MnO4-(aq) + SO3^2-(aq) → MnO2(s) + SO4^2-(aq)}$$

The problem says that it happens on basic medium, but I don't see the $\ce{OH-}$ which I usually use to identify the basic medium.

How do I identify that this reaction happens on basic medium? Is it something that the problem should clarify?

You are given a net ionic equation, which is the reduced form of a complete ionic equation: that's why you see neither $\ce{H2O}$ nor $\ce{OH-}$.
You can deduce that the medium should be slightly alkaline a priori only if you:

either know that manganate $\ce{MnO4^2-}$ formed in alkaline medium hydrolyses when there is a lack of reducing agent (here, sulfite $\ce{SO3^2-}$) to manganese(IV) oxide $\ce{MnO2}$ and that $\ce{MnO2}$ is also formed near neutral $\mathrm{pH}$ values;
or you compose the redox reaction as you were asked to and notice that two $\ce{OH-}$
are produced:

$$
\begin{align}
\ce{\overset{+7}{Mn}O4- + 2 H2O + 3 e- &→ \overset{+4}{Mn}O2 + 4 OH-} & |\cdot 2 \tag{red}\\
\ce{\overset{+4}{S}O3- + 2 OH- &→ \overset{+6}{S}O4^2- + H2O + 2 e-} & |\cdot 3 \tag{ox}\\
\hline
\ce{2\overset{+7}{Mn}O4- + 3\overset{+4}{S}O3- + H2O &→ 2\overset{+4}{Mn}O2 + 3\overset{+6}{S}O4^2- + 2 OH-} \tag{redox}
\end{align}
$$

As for why your problem states the redox reaction takes place in basic medium, it's just an extra hint assisting you in balancing the oxidation and reduction half-reactions (number of oxygens in oxides and oxoanions, for instance), but you could've balance the equation all right even if there were no information about the alkalinity of the medium.


How small is the smallest known carbon ring containing only double bonds? How small is the smallest known carbon ring containing only double bonds?

The stress from the double bonds makes small rings impossible to make something like cyclohexahexaene so I want to know how large the ring would need to be so the stress is small enough such that the ring can be made.

TL;DR: Isolated derivatives of cyclo-$\ce{C18}$ most likely have the smallest isolated cumulenic 18-membered ring, though upon direct complexation of transition metal in/outside the ring the strain as well as its size can be reduced further.

In the past two decades a research group of Prof. Dr. François Diederich performed numerous studies on cyclo[n]carbons and monocyclic $\mathrm{sp}$-hybridized C-rings. He also published several reviews, e.g. see chapters "Cyclic and Linear Acetylenic Molecular Scaffolding" in [1, p. 43], and more recent "Oligoacetylenes" in [2, p. 443] devoted to versatile chemistry of cumulenes.
As @R.M. referred in the comment, early theoretical work [3] predicted special Hückel-aromatic cyclo-$\ce{C18}$ compound stabilized via two orthogonal $(4n + 2)$ $\pi$-electron systems, but later computational studies predict somewhat controversial electronic structures (from [2]):

Self-consistent field (SCF) calculations with a 3-21G or larger basis set predicted that the cyclic acetylenic $D_{\mathrm{9h}}$ structure 1a with alternating bond lengths represents the ground-state geometry[4].
But optimizations at the Møller-Plesset second-order perturbation theory (MP2) level including valence electron correlations as well as density functional theory calculations favor the cumulenic $D_{\mathrm{18h}}$ structure 1b as the most stable planar monocyclic geometry [5].

More from [1, p. 46]:

Accordingly, cyclo-$\ce{C10}$ is predicted to prefer a cumulenic $D_{\mathrm{5h}}$ structure, cyclo-$\ce{C14}$ is borderline, and the energies of its cumulenic $D_{\mathrm{7h}}$ and polyynic $C_{\mathrm{7h}}$ structures are nearly degenerate. It is clear that this fascinating theoretical controversy over the past years can only be definitively solved with the synthesis and characterization of cyclo-$\ce{C18}$.

cyclo-$\ce{C18}$ appeared to be a very elusive compound to isolate (on macroscoping scale typical products are anthracene and polymers), therefore only its derivatives were synthesized and characterized. Synthetic methods include the following paths: Retro-Diels-Alder; 3-Cyclobutene-l,2-dione route;  transition metal complexation [2]:

Generally, cyclic cumulenes have two types of strains: distortion from linearity imposed by $\mathrm{sp}$-hybridization and distortion from orthogonality of adjacent double bonds (cyclo[2n]carbons). [6]
Attempts to liberate the cyclocarbons strain in cyclic cumulenes is similarly relieved by complexing one of the double bonds directly to a transition metal. For example, like in this zirconocene-hexapentaene complex, where $\ce{C=C=C}$ angle is reduced to approx. $130^\circ$ by complexation with $\ce{Zr}$ [7]:

Unfortunally, up to date there are only mononuclear metal-cumulene complexes isolated in a large scale, and none of them are cyclo[n]carbons, but theoretically polynuclear transition metal complexation can significantly reduce the ring size.
References

Carbon Rich Compounds II, Macrocyclic Oligoacetylenes and Other Linearly Conjugated Systems; de Meijere, A., Ed.; Topics in Current Chemistry; Springer Berlin Heidelberg: Berlin, Heidelberg, 1999; Vol. 201; ISBN 3540653015.
Stang, P. J.; Diederich, F. Modern Acetylene Chemistry; Wiley-VCH: Weinheim, 2008; ISBN 9783527615261.
Hoffmann, R. Tetrahedron 1966, 22 (2), 521–538. DOI 10.1016/0040-4020(66)80020-0.
Diederich, F.; Rubin, Y.; Knobler, C. B.; Whetten, R. L.; Schriver, K. E.; Houk, K. N.; Li, Y. Science 1989, 245 (4922), 1088–1090. DOI 10.1126/science.245.4922.1088.
Hutter, J.; Luethi, H. P.; Diederich, F. J. Am. Chem. Soc. 1994, 116 (2), 750–756. DOI 10.1021/ja00081a041.
Stone, F. G. A.; West, R. Advances in organometallic chemistry; 1998; Vol. 42. ISBN 0120311429.
Suzuki, N.; Hashizume, D.; Yoshida, H.; Tezuka, M.; Ida, K.; Nagashima, S.; Chihara, T. J. Am. Chem. Soc. 2009, 131 (6), 2050–2051. DOI 10.1021/ja8077472.


Regioselectivity in electrophilic aromatic substitution of benzofuran and indole Why does benzofuran react preferentially at position 2 when nitrated or formylated, while indole reacts preferentially at position 3 (source: Joule and Mills, Heterocyclic Chemistry 5th ed., p 371)?

In electrophilic substitution of aromatic systems sigma complexes are typically formed as reaction intermediates.  Let's look at the sigma complexes formed in the reaction of benzofuran and indole with an electrophile ($\ce{NO2+}$) at the 2 and 3 positions and compare their stability.

With both benzofuran and indole attack at the 2 position produces a sigma complex where the positive charge is nicely stabilized by the benzene ring (1a and 1b) . The stabilization is analogous to that found in a benzyl carbocation.

Attack at the 3 position produces a sigma complex where the positive charge can be stabilized by the lone pair of electrons on the adjacent heteroatom (resonance structures 3a and 3b), but because nitrogen is less electronegative than oxygen we would expect resonance structure 3b to be a bigger contributor than resonance structure 3a. Therefore we would expect the sigma complex produced upon attack at the 3 position to be more stable in the case of indole than benzofuran.

The overall result would then be to produce more of the 3 isomer in the case of indole as compared to benzofuran.


Mechanism for the synthesis of 2-pyridone from pyridine N-oxide I'm specifically interested in the pathway from pyridine N-oxide, where there is some kind of reaction and rearrangement with acetic anhydride.

I'm assuming there is first a substitution (by the negative oxygen) at the carbonyl. The -OAc group can then attack at the alpha-position on the ring.

We could continue and say that the loss of H and the -OAc group on nitrogen happens next to restore aromaticity, but then we're stuck with that OAc group instead of the carbonyl.

From what I could find, it doesn't seem that there's a radical mechanism here, but if there is some intramolecular rearrangement, I cannot seem to find it.

Pyridine N-oxide itself is generally fairly resistant to nucleophilic attack as the ring is already 'electron rich' and doesn't want to accept additional electron pairs.

When treated with acetic anhydride, however, the carbons at the 2 position become electrophilic (the acetate is electron withdrawing making the C=N bond more like an iminium), hence it is possible to have acetate attack, with subsequent elimination to re-aromatise the ring, leaving us with a 2-substituted pyridine.

If the acetate is hydrolysed (water, or mild aqueous base), the 2-hydroxy pyridine is formed, which may tautomerise to the more stable 2-pyridnone.

It's worth noting that this mechanism differs significantly from the Boekelheide reaction (the initial pyridine has an additional methyl group at the 2-position), whereby the acetate bound to nitrogen is migrated by means of a [3,3] sigmatropic rearrangement. This rearrangement isn't possible/favourable in this case, as the [3,3] requires a 6 membered transition state, whereas without the methyl group only a 5 membered TS exists.


Calculating [HI] given Kc, [H2], and [I2] for the reaction 2HI  H2+I2 In the reaction in equilibrium

  $$\ce{2HI <=> H2 + I2}$$

  The value $K_c$ is $0.0198$ at $\pu{721 K}$. Calculate $\ce{[HI]}$ at $\pu{721 K}$ if $\ce{[H2]} = \pu{0.0120 M}$ and $\ce{[I2]} = \pu{0.0150 M}$.

We use an ICE box. Since $K < 1$, the equilibrium is to the left. We are spending products to create reagents, so...

$$\begin{bmatrix}&\ce{[HI]} & \ce{[H2]} & \ce{[I2]}\\
\text{Initial} & 0 & 0.0120 & 0.0150\\
\text{Change} & +2x & -x & -x\\
\text{Equilibrium} & 2x & 0.0120 - x & 0.0150 - x
\end{bmatrix}$$

The equilibrium is given by

$$0.0198 = \frac{(0.0120-x)(0.0150-x)}{(2x)^2}$$

Solving yields:

$$x = 0.01907$$

Therefore

$$\ce{[HI]} = \pu{0.0381 M}$$

But the options are

  $\pu{0.00211M}$

  $\pu{0.133M}$

  $\pu{0.0334M}$

  $\pu{0.006567M}$

  $\pu{0.0953M}$

While it is pretty close to $\pu{0.0334M}$, I can't see how could I have made my calculation any more accurate (or if they were the ones who were inaccurate).

Was my procedure correct?


Polar/ non-polar nature of HgI2 Is $\ce{HgI_2}$ polar or non-polar?

The electronegativity difference between Hg and I is about 0.66, so the compound should be almost covalent. However, I cannot find the structure of $\ce{HgI_2}$ as a single molecule, so, I cannot find if it is polar or not, as a whole.

The electronegativity difference using the Pauling scale is not accurate in heavy transition metals. This is because the Pauling electronegativity involves the measure of bond energies, while the bonding of several metals (including "noble" metals like gold, silver, platinum, mercury, etc.) with large halides are diffuse and do not have sigma character . This is because the frontier orbitals of these elements are in the $d$ block, so molecular orbital interactions, if any, are diffuse especially in larger 4d and 5d metals. In reality, compounds like $\ce{HgI_2}$, as well as ones like $\ce{PbI_2}$ and $\ce{BiBr_3}$ (which I work with) are ionic salts.


White creamy substance (not calcium) observed when boiling water is filtered through coffee paper filter We have hard water from our well. It is filtered twice before boiling it: by the water filter and with a manual on-the-counter filter. When boiled in a clean kettle a white substance floats in the water, and it isn't calcium which is hard and sinks at the bottom. Then we filter that hot water through a coffee paper filter and a white creamy substance remains. I called water related companies but no one knows. Anyone have a answer to this one?

I assure you it is calcium carbonate.

The particles are so small they do not just sink to the bottom like a stone. Instead, they aggregate at the nearest phase boundary. All small impurities do that, unless they have a very low interfacial tension with water. It's energetically much more favourable than sinking.

Over time, this creamy substance aggregates further, up to a point where the aggregates are large and dense enough to dissociate from the surface, and sink to the bottom.


Formation of pi-allyl Pd cation complex (from pg. 1334 of Clayden Organic Chemistry 1st ed.)

In the above reaction catalyzed by Pd, I'm not exactly sure what's going on at the second step. It says in the book that Pd is using its electrons to push out X, and the electrons are returned to Pd when Nu comes and pushes out Pd.

My understanding is that in the first intermediate, the double bond's 2 pi electrons are complexed to Pd. When X leaves, the 2 pi electrons (that are now spread out between 3 C's) are still complexed to Pd. By the drawing of the mechanism, it seems to me that Pd has also used 2 of its electrons to bond with the 3 C's. So in the second intermediate, are the Pd and the allyl cation sharing a total of 4 electrons? If so, shouldn't there be a double bond, not a single bond, from Pd to the allyl cation?

This is a classic oxidative addition. You're kind of outside the classic world of single and double bonds. You have 4 electrons spread of 3 carbon atoms that link stabilize a bonding interaction with the palladium. Two of those electrons came from the palladium, so we think of it as having been oxidized, hence the name of the process.

See reference for general $\mathrm{S}_{N}2$ type oxidative addition. The only difference from your example is that the electrophile has a double bond that serves as an additional ligand for the palladium.


Are crystallographic omit maps only used with molecular replacement? (This is in the context of macromolecular X-ray crystallography.)
From Lamb et al. 2015:

Model bias is the result of how maps are calculated: because the
phase estimates [of the structure factor] for the calculation come from the model, maps will
tend to show electron density for an atom in the model whether it is
truly there or not. A simple omit map is normally a difference (Fo −
Fc) map calculated after omitting specific atoms from a model, like a li-
gand or a functionally important loop.

From this the purpose of an omit map is rather clear.
But what I don't understand is when omit maps are used w.r.t. the source of the phases. From what I can tell, if the phases have been determined experimentally (e.g., using MAD), there cannot be any extraneous electron densities in the atomic structure model. This drives me to the conclusion that omit maps are used when the source of the phases is molecular replacement (I presume this is what the "model" alludes to).
Is this correct? Are omit maps only used with molecular replacement?
Edit: I am also confused with regards to how an omit map actually works. If the phases are what determine the final electron density irrespective of the amplitudes used, how does deleting a part of the atomic model, getting the phases from that and recalculating a new map (with these phases and the experimental amplitudes) actually help?
Edit 2 (from comments): I got the impression that if we don't have the phases (say for a part of a model), it just wouldn't get built even if we have the amplitudes for that part, right? Or is it that in this case (of omitting a part), there would be an electron density (because the amplitudes are still there), just that it would be featureless?!
I would be very glad to receive some clarifications and/or pointers.

Experimental phasing and molecular replacement provide initial phases used to start model building. Later on (refinement, validation) phases are calculated from the model.

If the phases are what determine the final electron density irrespective of the amplitudes used

Both amplitudes and phases are used to determine the electron density.
If you calculate the phases omitting part of the model, the electron density in this place won't be biased by that part of the model. (It may be biased by the absence of the model in this place).


What materials are resistant to hot chlorine? I'm looking to do an experiment involving hot chlorine gas (900 Celsius), however at the moment the only material I'm aware of that could cope with this is Hastelloy's C-22 alloy. What other alternatives exist?

Short answer — none. Hastelloy C276 or C22 are the best alloys, also considering availability, presumably cost is no object. At that high temperature there is no readily available data. At 500 °C for a limited time experiment, other nickel alloys like 625 or 201 and even 316 SS have some limited life. But for 900 °C, I expect the Hastelloy C family was recommended because they have the best life at about 500 °C. Your conditions are where it would be best to get a manufacturer of high alloys or ceramics to work with you.


Why thermal decomposition of iron(II) nitrate and sodium nitrate yields different products? \begin{align}
\ce{4Fe(NO3)2 &-> 2Fe2O3 + 8NO2 + O2}\tag{1} \\
\ce{2NaNO3 &-> 2NaNO2 + O2}\tag{2}
\end{align}

Why do nitrates of different metals decompose in different ways?

Why do we get a nitrite with sodium, but an oxide with iron(II)?

Iron(II) cation is moderately reducing- an iron(II) cation has a valence electron configuration of $\mathrm{3d^5 4s^1}$, and by giving up an electron in its $\mathrm{4s}$ orbital the cation will transform into an iron(III) cation with $\mathrm{3d^5}$ valence electron configuration, which, with every $\mathrm{d}$ orbital half filled, has a higher degree of symmetry over $\mathrm{3d^5 4s^1}$ electron configuration and hence is more stable (thermodynamically favored).
Sodium(I) cation, on the contrary, has a highly stable valence electron configuration of $\mathrm{2s^2 2p^6}$, with all its $\mathrm{2s}$ and $\mathrm{2p}$ orbitals filled, identical to the electron configuration of the chemically inert noble gas neon, and either removing or adding one electron to the sodium(I) ion will lead to an electron configuration much less stable and hence thermodynamically unfavored.
You may notice that in the decomposition reaction, the iron(II) cation is oxidized into iron (III), which is favored, and the sodium cation remain intact.
Put it in another way, we may imagine a first step thermal decomposition reaction of iron(II) nitrate resembling decomposition of sodium nitrate and yields iron(II) nitrite and oxygen gas as products. Then we easily notice that iron(II) nitrite is reducing, and oxygen gas is a strongly oxidant, hence the two products of the first step decomposition can further react with each other to produce iron(III) oxide and nitrogen dioxide.


The highest melting point Inspired by this question, what is the substance or compound with the highest melting point? At standard atmospheric pressure and ever-increasing temperature which substance a) eventually becomes liquid b) does so at the highest temperature?

I don't know much about glasses or related semisolids, so perhaps a reference viscosity? Glass transition temperatures are kind of like melting points, but some materials have a glass transition temperature and a melting point.

Some initial casting about on the internet finds that rhenium and tungsten have melting points at about 3400 °C, and hafnium carbide claims to be 'one of the highest' at 4100 kelvin. Does it actually melt or just decompose and then sublimate? 4100 kelvin is its 'refractory temperature', but I'm not certain how that's defined.

Well, according to this Wikipedia article, it might be an alloy of tantalum carbide and hafnium carbide, specifically $\ce{Ta4HfC5}$.  A melting point of 4215 °C (4488 K) is listed.  But who really knows the answer to your question with so many new, proprietary alloys with complex compositions being fabricated these days.


What properties a compound should have to be called an "ice"? I wonder what compounds are usually called "ice" and what properties do determine it?

I can bring some examples of ice and not.

Not ice:

$\ce{H2S}$
cellulose
sugar
graphite
lithium
solid mercury
polyvinylchloride
solid fats
...

Known as ices:

solid water
solid $\ce{CO2}$
solid nitrogen
solid oxygen
solid methane (?)
solid ethanol (?)

It seems they all have some common properties:

Melting point below room temperature
Transparency
White or blueish color
Chystalline structure (?)
Easy sublimation and high saturation pressure
Low strength and other mechanic properties
Low surface friction (?)

Have I missed anything? What is the accepted definition of ices, say in astronomy or chemistry?

Ice is a term usually given to $\ce{H2O}$ in the solid state. Solid carbon dioxide, what is sometimes called "dry ice" is actually a misnomer — although the gas is frozen, it is not truly ice, in the chemical sense. It is called dry ice because instead of melting into liquid carbon dioxide (like actual ice, which is frozen water to liquid water), dry ice sublimates into it's gaseous state, skipping the liquid phase (almost — still goes through liquid phase, but this happens so fast it's basically negligible) entirely.

As for properties, you basically have all of them listed.


What is the true depiction of the periodic table? Normally with the periodic table the lanthanide series is separated out because it's long and would make the table wide. I looked for an expanded version and found this:

I found it kind of strange that scandium and yttrium are above the lanthanide and actinide series on the far left.

Does it make sense for those two elements to be there? I always assumed each column had certain characteristics like same number of valence electrons, that there were certain characteristics that held true for the whole table, but in this format I don't know how to understand the patterns or why the table ends up taking this shape.

La and Ac have $d^1$ electrons in their valence shells, rather than $f^1$ electrons.

The long table you found looks like that for a several reasons.

The trends going down Sc-Y-La are like those seen in groups 1 and 2. The trend going down Sc-Y-Lu is like that of groups 4 to about 10. Since lanthanide chemistry is basically that of trivalent alkali or alkaline earth metals, this tips the balance in favour of group 3 as Sc-Y-La.

Also helping is that group 4 is the first in which the really characteristic properties of transition metals (variable oxidation states; colour; paramagnetism) are seen. Ditto, same thing happens with group 12, which is why they're shown as post-transition metals.

And there's the lanthanide contraction, which starts at Ce and finishes at Lu. If you show Lu as being a transition metal it mushes the end of the Ln contraction into the d-block.

And there's the periodic law, which says that the chemical elements, if arranged according to their atomic numbers, show an approximate repetition of properties after certain regular but varying intervals. Here, La is the first element after Y that shows the approximate repetition in properties.

This research supports the periodic law outcome:

Glawe H, Sanna A, Gross EKU & Marques MAL 2016, “The optimal one dimensional periodic table: a modified Pettifor chemical scale from data mining”, New Journal of Physics, vol. 18, 093011, https://iopscience.iop.org/article/10.1088/1367-2630/18/9/093011/pdf
Restrepo G 2017, "Building classes of similar chemical elements from binary compounds and their stoichiometries", in MA Benvenuto (ed.), Elements old and new: Discoveries, developments, challenges, and environmental implications, American Chemical Society, Washington DC, pp. 95-110

Both sources point to La being more distinct from the Ln than is the case for Lu.

Finally, if you count the differentiating electron discrepancies in each block of the periodic table, noting the above table has a split-d block, you'll find that an Sc-Y-La table has 12 such discrepancies whereas an Sc-Y-Lu table has 13. The current agreed-within-IUPAC table has 14 differentiating electron discrepancies!

As many people have a hard time accepting the idea of a split-d block, I personally feel that a better solution would be a Sc-Y-Lu table with group 3 being shown as Sc-Y-Lu-Lr and La-Ac. That would have the extra benefit of resolving long-standing discussions about the composition of group 3, which have been going on since about 1982, when Jensen published his paper in JChemEd, arguing for Sc-Y-Lu:

http://www.che.uc.edu/jensen/w.%20b.%20jensen/reprints/018.%20la%20vs%20lu.pdf

Nothing much has to change. Just add a 3 above La-Ac in the f-block, in addition to the one over Sc-Y-Lu-Lr in the d-block.

A chemistry book chapter on this group 3 would make fascinating reading (a good thing given, to date, that group 3 is supposed to be the least studied group).

Hope that helps.


Catalytic poison in Pt catalysed manufacture of H2SO4 I came across a question that goes as:

  In manufacturing of sulfuric acid in presence of platinum catalyst, which metal impurity acts as a catalytic poison?

  Fe
  Cr
  Cu
  V

I searched everywhere I could but did not find anything that talks about this.
I read about $\ce{As2O3}$ being a catalytic poison in this reaction and recommendations for using the Contact process using $\ce{V2O5}$ instead of this reaction.

I can understand that this is more of a fact based question but reasons or explanation (if any) should please be included.

$\ce{Fe}$ and $\ce{V}$ are used in the contact process.
$\ce{Cu}$ is used in the preparation of chlorine gas.
Thus $\ce{Cr}$ can be used as a catalyst poison.


Why is predicting products of chemical reactions difficult? I was going to ask whether there are software that could be used to predict the products of any given chemical reaction. However, I then noticed these two earlier questions

Chemical software for solving reactions
Software for predicting chemical reactions

where it is said such predictions are too difficult to make. What makes the prediction difficult?

Because you need would need to solve the Dirac equation (relativistic Schrodinger equation) for all the particles in the system to get a completely accurate result.

Even solving the time independent non-relativistic Schrodinger equation for $\ce{H2+}$ in isolation, where there is only one electron and two protons involves approximating the protons as fixed relative to each other.

Now imagine trying solving a system of $10^{23}$ molecules each having multiple nuclei, and dozens or even hundreds of electrons, and including time dependence!


Gas chromatography analysis - number of peaks? This is an example in a Princetown Review prep book. I am not understanding the statement "his separation was only partial (because two different peaks are recorded)." This is implying that Compounds A and B did not separate, but um... there were two peaks? Doesn't that imply that it was separated?

Unless there is a solvent peak? But I did a lot of Googling, and didn't find any indication that there should be a solvent peak under normal circumstances.

Why did the example say that the 2 compounds weren't separated despite the fact that there are 2 peaks?

The GC is the instrument used (in this case, although for very small amounts, there is no reason GC couldn't be used to separate materials - assuming collection after detection (via, say, refractive index - for obvious reasons flame ionization, etc. wouldn't be usable)) to determine if his primary separation technique was effective. It wasn't.


How does molecular orbital theory explain CH4? I have understood the formation of $\ce{CH4}$ by valance bond theory, but I'm having trouble understanding it through molecular orbital theory.

The energy level diagram of molecular orbitals of $\ce{CH4}$ is not clear to me.

Methane doesn't illustrate the difference very well, but the difference is nicely illustrated with benzene. Valence bond theory would have the electrons in the "double bonds" isolated between atoms as so:

But the electrons in the "double bonds" are not isolated in three double bonds but distributed equally over all 6 carbon atoms. Valence bond theory tries to wiggle out of this dilemma by posing the Kekule resonance structure:

which still belies the situation. The double bonds don't "resonate." The extra 6 electrons are distributed in one aromatic molecular bond (with three degenerate orbitals) evenly over the 6 carbon atoms.


Why iron reacts differently with concentrated and dilute sulfuric acid? When $\ce{H2SO4}$ is dilute,

  $ \ce{Fe +H2SO4 -> FeSO4 + H2} $

and when $\ce{H2SO4}$ is concentrated,

  $\ce{3Fe +8H2SO4 -> FeSO4 + Fe2(SO4)3 + 4SO2 +8H2O}$

What made the reaction with dilute acid different from the reaction with concentrated acid?

How did the second equation form, and why there are 2 salts - one with Fe[II] and the other with Fe[III]?

In the first equation, $\ce{Fe}$ is oxidised to $\ce{Fe^2+}$ by $\ce{H+}$; the $\ce{SO4^2-}$ ions play no role in the reaction. You could replace $\ce{H2SO4}$ with $\ce{HCl}$ and get the same reaction.

In the second equation, the same oxidation of $\ce{Fe}$ to $\ce{Fe^2+}$ by $\ce{H+}$ also occurs. However, on top of that, some of the $\ce{SO4^2-}$ ions are reduced to $\ce{SO2}$, and some of the $\ce{Fe^2+}$ is further oxidised to $\ce{Fe^3+}$. Clearly, in the dilute acid, $\ce{SO4^2-}$ was not strong enough to act as an oxidising agent - but in the concentrated acid it is.

The reason for this can be somewhat explained by Le Chatelier's principle. Consider the redox reaction:

$$\ce{2Fe^2+ + SO4^2- + 4H+ -> 2Fe^3+ + SO2 + 2H2O}$$

In the dilute acid, this equilibrium lies far to the left because the concentrations of $\ce{SO4^2-}$ and $\ce{H+}$ are rather low. However, making the acid more concentrated shifts the equilibrium to the right so that you get some of the products.

As to why you get a mixture of iron(II) and iron(III), the simple answer is that the equilibrium above doesn't lie entirely to the right. $\ce{H2SO4}$ is not that strong an oxidising agent to begin with.

More quantitatively, the redox reaction above can be investigated using standard reduction potentials and the Nernst equation. The following data is taken from Wikipedia (you will get pretty similar values from more "reliable" sources):

$E^\circ(\ce{Fe^3+}/\ce{Fe^2+}) = +0.77 \mathrm{~V}$
$E^\circ(\ce{SO4^2-}/\ce{SO2}) = +0.17 \mathrm{~V}$

Under standard conditions, $E^\circ$ for the full reaction is therefore $0.17-0.77 = -0.60 \mathrm{~V}$; the negative value indicates that the forward reaction is non-spontaneous.

The Nernst equation for this reaction can be written:

$$E = E^\circ - \frac{RT}{nF}\ln Q = E^\circ - \frac{RT}{nF}\ln\left[\frac{(a_{\ce{Fe^3+}})^2(a_{\ce{SO2}})(a_{\ce{H2O}})^2}{(a_{\ce{Fe^2+}})^2(a_{\ce{SO4^2-}})(a_{\ce{H+}})^4}\right]$$

I will make several assumptions here:

Activity coefficients are taken to be unity, meaning that the activities can be replaced by concentrations.
The activities of all the products, as well as $\ce{Fe^2+}$, are unity.
The only source of $\ce{H+}$ is from the dissociation of the $\ce{H2SO4}$, and the dissociation is complete. That means that if $c$ is the formal concentration of $\ce{H2SO4}$, we have $[\ce{H+}] = 2c$ and $[\ce{SO4^2-}] = c$.

Substituting in the values of $E^\circ = -0.60 \mathrm{~V}$, $R = 8.314 \mathrm{~J~K^{-1}~mol^{-1}}$, $T = 298 \mathrm{~K}$, $n = 2$, and $F = 96485 \mathrm{~C~mol^{-1}}$, this allows us to simplify the Nernst equation:

$$E / \mathrm{V} = -0.60 + 0.0128\ln (16c^5)$$

Strictly speaking, I should use $c/c^\circ$ instead of $c$ in order to make the logarithm dimensionless, but I'm fed up of technicalities. Now for the actual calculation. Let's assume that you're using $10\mathrm{~M}$ sulfuric acid, such that $c = 10$ (or $c/c^\circ = 10$, if you want to insist). This gives a value of $E = -0.42 \mathrm{~V}$ - as expected, it is less negative than before, which corresponds to a larger value of $K = \exp(nFE/RT)$. The fact that it is still rather negative just means that there is a difference between the real world and my theoretical treatment. Presumably, those assumptions above are not entirely valid.


How dangerous is an industrial scale styrene spill? A few years ago a cisterna truck full of styrene crushed on the road near a small town nearby, spilling 4 tons of the chemical. It apparently has caused a small ecological catastrophe killing fish in the local river with concentrations of 3.7 ml/l in the water, and half of the population - some 2000 people were evacuated in the next 2 days, but it was not mandatory, and what seems more strange to me, I've seen all the firefighters not wearing any gas masks while pouring some white foam on the truck. Also, no one mentioned that styrene is a cancerogen.

Initially, there was no info on the chemical spilled, and I didn't know or care much then, but today I incidentally found out it was styrene, and I got curious. So, were those measures somehow inadequate, or on the contrary - far-fetched, and just how bad this is?

According to this safety data sheet (SDS), styrene monomer has a Category 3 respiratory classification, where 1 would be most serious. Category 3 is defined "substances produce transient (short duration or temporary) target organ effects."

Add to that these hazard, precautions and carcinogenicity statements:

H335: May cause respiratory irritation.
H372: Causes damage to organs through prolonged or repeated exposure if inhaled.
P260 Do not breathe dust/fume/gas/mist/vapor/spray.
NTP Reasonably anticipated to be a human carcinogen

Read the cited SDS or others for statement on environmental precautions and cleanup. Frankly, they're a bit vague to me, "If the product contaminates rivers and lakes or drains inform respective authorities."