Does everything exist as a gas at p = 0? Attached is a phase diagram for carbon dioxide. At $p = 0$, it appears that the substance will exist as a gas regardless of temperature.
My questions are:

For any liquid at a fixed temperature, is it possible to vaporize it completely by simply reducing its pressure isothermally?

For any solid at a fixed temperature, is it possible to sublime it completely by simply reducing its pressure isothermally?

What if the solid is at $T \approx0\ \mathrm K$?

Regardless of whether a substance is in a vacuum, it will vaporize if some fraction of its particles have enough energy (due to their temperature) to exceed their binding energy.  Binding energy is the energy contained in the intermolecular bonds holding a substance together.  It is definitely possible for the particles of a substance to have a binding energy that is substantially higher than thermal energy will readily provide.

No.  While most liquids have fairly low binding energies (and thus, relatively high vapor pressures) there are liquids that will not evaporate in a vacuum at an appreciable rate, one example being ionic liquids.  As I wrote above, the attraction between the ions of these liquids is strong enough that they do not measurably evaporate over time.
If you know the Periodic Videos YouTube channel, they've had a couple of videos about these fluids over the past couple of years.  (Here's a link to one, if you're interested.)  One of them featured the behavior of liquids in a vacuum, but I can't seem to find the link; it may have been on a sister channel.
No, consider all of the asteroids, space ships, satellites, and such floating around in space.  For all intents and purposes they will float around until the end of time without evaporating.  Again, the reasoning is as I listed above.  We also don't see the vast majority of solid objects here on Earth slowly evaporating away, which we would expect if sublimation were a concern in a vacuum.  The reasoning is the same as above.
If the solid is very, very cold, it is less likely to experience sublimation since its particles' average energies would be very low.  Even dry ice would not vaporize if its temperature were very close to 0 K.

Here is a phase diagram of both water and carbon dioxide that might help illustrate the situation.  You can see that as the temperature of the substance decreases the vapor pressure of the substance also decreases.  At a sufficiently low temperature the vapor pressure is effectively zero (below the x-axis of this chart.)  There is always the possibility of a single atom or molecule gaining enough energy to vaporize or doing something weird like tunneling off of the surface of the substance, but the process is so slow as to be unmeasurable.  Again, asteroids in the solar system have been hanging around for several billion years without subliming.

Picture from Wikipedia.

Note: Be careful if you look up phase diagrams of common substances.  They often involve only a few points that are actually referenced and the rest is drawn with quite a bit of artistic license.  Compare the phase diagrams of water here and here to see what I mean.  They are often drawn without scales (or with minimal scales) and are sloppy at best.


How does addition of impurities in a solid decrease the melting point? In my textbook, it is written:
The addition of impurities in a solid decreases the melting point of the solid.

How does it do so? Why can't it increase the melting point?

one of the most prevalent (and simplest) model which explains this fact is colligative freezing point depression.

Freezing (or melting) point is temperature at which solid and liquid phase are in equilibrium, and their vapor pressures are equal. Adding (non-volatile) impurity in solid phase lowers the vapor pressure of solid phase (because you lower the "concentration" of your compound in solid phase), and phase equilibrium can be achieved at lower temperature.

Difference between melting points for pure and impure compounds can be quantified with following equation:
$$ \Delta T = i K_f b $$
where $\Delta T$ is melting point difference, $i$ is Van't Hoff factor which tells you on how many species does the impurity dissociate, $K_f$ is cryoscopic constant which tells you how great the effect is, and $b$ is molality of the impurity.

Basically, all impurities lower the melting point, if they do not interact strongly with the given compound. In some cases (i.e. urea/choline chloride) freezing point depression can be in order of 100 °C! Organic compounds in particular have pretty high cryoscopic constants, which can cause headaches in organic synthesis (in most cases, you end up with some kind of oil which has to be purified to obtain solid product).

Elevation of freezing point is possible only if pressure is decreased, but the effect is so minuscule that it is not practically useful.


Does a chiral centre have to have 3 or 4 different groups attached? I am seeing conflicting answers on the net

e.g.

  A carbon atom is chiral if it has four different items bonded to it at the same position. (Wiki)

compared to

  Stereocenter (chiral center): An atom with three or more different attachments (Chem.Ucla)

Why does one say 3 or more, and one say 4?

Also, there is a double bonded carbon atom here that is said to be a stereocentre, which I cannot understand.  Isn't the double bond counted as two of the "same" attachments (so not even satisfying the 3 different groups criteria). How is that still a chiral centre?

The answer is that all chiral centers are stereocenters, but not all stereocenters are chiral centers thus the words are not interchangeable.

A chiral center requires four or more  different molecular groups attached to a single center atom (asymetrical). In the case of pnictogens (nitrogen, phosphorus, etc...) or chalcogens (sulfer, selenium, etc...) an electron pair can count as one of the molecular groups.

An alkene is a stereocenter that requires 3 or 4 different molecular groups such that no two common groups are on the same side of the bond (must be able to determine an E/Z configuration). In the case of the blue stereocenter, the methyl groups are on the same side and thus switching their position would create an identical compound meaning the alkene is not a stereocenter.  For the red center switching the position of the bromine and methyl groups would create the Z-isomer from the E-isomer thus is a stereocenter, but not a chiral center.  If however, the hydrogen of the blue stereogroup were to switch positions with a methyl group then the blue alkene would be a stereocenter with three distinct molecular groups.


How do flavonoids absorb UV radiation? I know that energy is absorbed due to chromophores which enable electron transition which uses up energy. But I am a bit unsure about what happens after the absorption and where this absorbed energy went.

I am not very sure if the energy is dissipated since it would contradict with the law that energy can't be destroyed. If the energy is lost as heat, then it is a bit contradictory again according to the equation $E= hc/\lambda$.
The wavelength would be big for infra-red and thus, the energy lost must be less than the absorbed energy from UV.

So how exactly are these flavonoids shielding UV radiation? And also, if these flavonoids are in the excited state, it makes me feel that they might be quite reactive and may form free radicals.

So how exactly are these flavonoids shielding UV radiation?

Due to the structure of the flavonoids, there are many transitions to absorb the UV radiation. Also, there are many transitions to release the energy. Most of the time, it is released in 2 steps (2 photons) which equal the energy of the photon that exited the molecule.

Fluorescence emits a photon that is close to the same energy as the first photon. For quercetin, the excitation is at 440 nm and the emission is at 540 nm.


How to rationalise the difference in the melting points of acids and alcohols with inter molecular forces? Acids: arachidic acid, palmitic acid, lauric acid, etc.
Alcohols: butan-1-ol, pentan-1-ol, etc.

The melting points of acids with $\ce{-COOH}$ at the end compared to melting points of alcohols, $\ce{-OH}$ at the end, is higher. I need to relate this to intermolecular forces. I was going to say that because of many more electrons in the acids, there are greater London dispersion forces and also more polarity (dipole dipole forces) because of the $\ce{COOH}$ at one end. Is this correct?

The ultimate goal of melting-point reasoning is to explain how intermolecular forces lead to better packing of molecules in the solid phase.
Both alcohols and organic acids have hydrogen bonds, but organic acids have stronger hydrogen bonds due to the withdrawing carbonyl group, $\ce{COO-}$. This causes the acids to interact stronger with each other. More importantly, organic acids form dimers (usually) with their hydrogen bonds (acetic acid dimer, for example). For this reason, the packing of the acids in the solid phase is significantly better than that of alcohol; hence the higher melting point. Acetic acid: $16~^\circ\mathrm{C}$ vs. $-114~^\circ\mathrm{C}$ for ethanol.


What is the most rare element in periodic table What is the most rare element in periodic table? I have a lot of confusion about this.

Well, the first question to answer would be: where?  The rarest element in the universe? the sun? in meteorites? on earth? in the ocean? in humans?  We can tap in to the curated datasets provided by Mathematica to get the answer:

Elements up to atomic number 98 have been found in trace quantities; however, the dataset does not contain any information about the trans uranium elements which were once thought to be synthetic.
Here's a plot of abundance as a function of atomic number:

We can see that the elements beyond about Z = 82 (lead) have a marked decrease in abundance, and this is consistent with all of the elements with atomic number above 82 being radioactive.
Where do we come from?
We can compare the elemental abundance of these individual "spaces" to speculate on where humans come from.  Below are plots showing the correlation of the elemental abundances.  The plot labels give the best fit equation:

The best slope is observed when comparing our abundance to that of the Earth's crust; however it does appear as if the scatter is smallest for the Us vs. the Ocean plot.


Can aromaticity arise from 3c-2e bonds Electrons are delocalized between the three centers in a  3c-2e bond. If there was a molecule with 3c-2e bond in each direction, e.g. a boron cluster, would it be aromatic?

The cyclopropenyl (1) cation is believed to be aromatic as it demonstrates abnormal stability and NMR-shifts typical for aromatic compounds. Moreover, there are a good number of derived systems, where aromatic conjugation is maintained despite extra methylene insertions into the ring, such as is typical for the nornbornadienyl (2) cation. Some of these cations can be isolated in salts or observed via NMR in solution. Signs of aromaticity are found also for the cyclooctatetraene dianion and the cyclononatetraene dianion, both produced by reduction of the corresponding alkene with metallic sodium. Remember, please, that even though all systems noted above are 'aromatic' in some sense and are abnormally stable for their family of species, they are still very reactive, because they all belong to a very reactive family of species. While the neutral boron analog of the cyclopropenyl cation has not been isolated to my knowledege, a similar structure was found in organometalic compounds. Aromaticity of some ring systems may result in some weird properties of some compounds, like squaric acid (abnormally strong for ketoenol and a stable enol), high polarity and basicity of cyclononatrienol and cyclopropenone and so on.

(1) - cyclopropenyl cation

(2) nornbornadienyl cation


May I treat units (e.g. joules, grams, etc.) in equations as variables? Suppose I have two units, Tonks, $T$, and Borks, $B$, as well as the equation $\frac{50T}{20B}=x$. May I manipulate the equation as though $T$ and $B$ were the kinds of symbols that we learned to manipulate in math class? For example:

\begin{align}
\frac{50T}{20B}&=x\\
50T&=20Bx\\
\frac{50T}{20}&=Bx\\
\end{align}

Yes, you may. It is quite common to convert units into each other. The simplest conversion might be the prefixing of units, e.g.
$$\mathrm{\frac{km}{m}}=1000 \Longleftrightarrow \mathrm{1~km = 1000~m}.$$

Another example is the interconversion of units of energy. In some parts of chemistry, it is still quite common to use calories, in others Joule, as a SI unit, has taken its place.
$$\mathrm{\frac{cal}{J}}=4.184 \Longleftrightarrow \mathrm{1~cal = 4.184~J}.$$

In principle, all units are defined to be constants, special numbers in a wider sense. You can treat them as such.


Why do many transparent plastics turn opaque when deformed? Many transparent plastic materials, rigid or not, turn opaque (white) when subjected to stress, deformations or when folded (if not rigid) . The effect seems to be irreversible.

What is the cause?

I would have guessed that some internal crystalline structure is broken, but plastic (like glasses) have no crystal structure, no?

The process of forming an opaque area when you apply stress to a polymer film is known as crazing.

Polymers are long chains and, when coated, often form tangled, amorphous, randomly aligned polymer regions.  When stress is applied to the film, the stress can be relieved by the polymer chains aligning.  In other words, energy introduced into the film by stressing it can be used to align some polymer chains.

Increasing the alignment in the stressed region means that the density of polymer in this region of the film will be different from the density in non-stressed regions.  This difference in density will cause a difference in light reflection and scattering accounting for the different optical properties between these two regions.

Further, within the stressed area of the film, mechanical stresses are produced between areas that are more and areas that are less aligned.  These stresses can be relieved through the formation of micro-cracks between the regions.  These small cracks also result in different light reflecting and scattering properties of the film.

Finally, as the polymer chains align, they are more likely to crystallize.  The formation of small crystallites within the stressed (aligned) region will also alter the light reflecting and scattering properties of the film.  Often, upon heating or standing at room temperature (for a longer time) the optical changes may reverse as the thermal energy supplied will:

cause the polymer structure to randomize back to its amorphous structure,
allow the polymer to "flow" and fill the micro-cracks, and\or
the crystallites can redissolve.

References: The following links provide more information about crazing and associated phenomena at different levels of complexity:

reference 1

reference 2

reference 3


How can batteries reach equilibrium if the reactions are irreversible? My book says that the batteries we have considered so far (Daniell cells) have a low life span because they reach equilibrium too quickly.

In the next paragraph, it states they are also made by irreversible reactions and we call them "primary batteries."

How can a reaction reach equilibrium without being reversible?

Your confusion probably stems from the definition of a reversible path between two states: an infinite sequence of steps along a continuum of equilibrium states, such that at each step the equivalence condition of the second law of thermodynamics, $\mathrm dS_\text{total}=0$, is satisfied.

However, there is no requirement that a path between two states must be reversible. Many textbook problems are based on comparing the entropy changes associated with reversible and irreversible paths. For the reversible paths, $\Delta S_\text{total}=0$, for the (spontaneous) irreversible ones, $\Delta S_\text{total}\gt0$.

Your textbook explains that the discharging process of a Daniell cell is considered irreversible. The usage of the term irreversible in the book is rather similar to the common usage, which means "impossible to bring back to its original state, for practical reasons.". The answer that Karl has given you is correct. Basically it is not possible to bring the cell to its original state. The chemical and physical integrity of the recharged electrodes and electrolyte are altered compared to the original structure. There are entropic costs associated with cycling the cell.


Why does mercury form polycations? For example, mercury (I) is $\ce{Hg2^2+}$ and not $\ce{Hg+}$. What causes  the stability in covalently bonded $\ce{Hg}$ ions?

I don't have an excellent understanding of the topic, but I will try to write something up.

Polycations are found with the Hg(I) oxidation state. The mercury cation, $\ce{Hg+}$, has a $\mathrm{5d^{10}6s^1}$ configuration. The mercurous cation $\ce{Hg2^2+}$ can then be considered to simply be formed by overlap of the two $\mathrm{6s}$ orbitals on two separate $\ce{Hg+}$ ions. In this sense, the bonding is qualitatively very similar to that in $\ce{H2}$. Note that the $\mathrm{5d^{10}6s^0}$ Hg(II) cation has no chance of forming such covalent bonds.

Generally, going down a group, orbitals become larger, more diffuse, and the overlap between them decreases. As a consequence, $\sigma$-bonding tends to become weaker going down the group. A simple example is provided by the alkali metal dimers:

$$\begin{array}{c|ccc}
\text{Element} & \ce{H} & \ce{Li} & \ce{Na} \\  \hline
D_\mathrm{0}(\ce{E-E})\text{ / }\pu{kJ mol-1} & 432 & 99 & 71 \\ \hline
\text{Reference} & 1 & 2 & 3
\end{array}$$

These dissociation energies are extremely small. For the analogous $\ce{Hg2^2+}$ cation, the situation is not quite as bleak. The Group 12 metals in general have more low-lying and contracted ns orbitals than the corresponding alkali metals, due to the presence of ten poorly-shielding (n-1)d electrons. However, given that zinc does not form similar cations, it is odd that mercury with its larger 6s orbitals should do so.

The most obvious explanation is simply that Zn does not commonly exhibit the +1 oxidation state. The existence of the +1 oxidation state for Hg is directly related to the relativistic stabilisation of the 6s orbital, which increases the first two ionisation energies4 of mercury.

$$\begin{array}{c|ccc}
\text{Element} & \ce{Zn} & \ce{Cd} & \ce{Hg} \\ \hline
I_1\text{ / eV} & 9.393 & 8.992 & 10.44 \\ \hline
I_2\text{ / eV} & 17.96 & 16.90 & 18.76 \\
\end{array}$$

However, the question still remains as to why Hg(I) chooses to form polycations instead of existing simply as $\ce{Hg+}$. In fact, relativistic effects have also been implicated in this for a long time: Pyykkö and Desclaux wrote in 1979 that5

  The relativistic contraction of the half-filled valence s AO makes $\ce{Au2}$ much more strongly bound than $\ce{Ag2}$. Inasmuch as $\ce{Hg2^2+}$ and $\ce{Cd2^2+}$ are isoelectronic with these two molecules, the remarkable stability of the mercurous ion [compared to the Cd analogue] can also be understood.

Therefore, this may most simply be explained as the relativistic contraction of the 6s AOs leading to stronger-than-expected overlap to form a covalent bond. As suggested by the above paragraph, similar effects are seen in the neutral dimers of Group 11 metals:6

$$\begin{array}{c|ccc}
\text{Element} & \ce{Cu} & \ce{Ag} & \ce{Au} \\  \hline
D_\mathrm{0}(\ce{E-E})\text{ / eV} & 1.95 & 1.65 & 2.29 \\
\end{array}$$

The decrease going from Cu to Ag is expected on the basis of weakened overlap, and the dissociation energy of $\ce{Au2}$ would be even lower in the absence of relativistic effects: calculations indicate that relativity contributes roughly $\pu{1 eV}$ to the dissociation energy of $\ce{Au2}$.6 Presumably, the effect in $\ce{Hg2^2+}$ is of a similar significance.

References

Frenking, G.; Bickelhaupt, F. M. The EDA Perspective of Chemical Bonding. In The Chemical Bond: Fundamental Aspects of Chemical Bonding; Frenking, G., Shaik, S., Eds.; Wiley: Weinheim, Germany, 2014; p 128.
Velasco, R.; Ottinger, Ch.; Zare, R. N. Dissociation Energy of Li2 from Laser‐Excited Fluorescence. J. Chem. Phys. 1969, 51 (12), 5522–5532. DOI: 10.1063/1.1671979. Free version provided by the Zare group here.
Jones, K. M.; Maleki, S.; Bize, S.; Lett, P. D.; Williams, C. J.; Richling, H.; Knöckel, H.; Tiemann, E.; Wang, H.; Gould, P. L.; Stwalley, W. C. Direct measurement of the ground-state dissociation energy of Na2. Phys. Rev. A 54, R1006(R). DOI: 10.1103/PhysRevA.54.R1006.
Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed; Oxford UP: Oxford, U.K., 2014; pp 836–7.
Pyykko, P.; Desclaux, J. P. Relativity and the periodic system of elements. Acc. Chem. Res. 1979, 12 (8), 276–281. DOI: 10.1021/ar50140a002.
Pitzer, K. S. Relativistic effects on chemical properties. Acc. Chem. Res. 1979, 12 (8), 271–276. DOI: 10.1021/ar50140a001.


Equlibrium for a non-spontaneous reaction Consider a reversible reaction $\ce{A -> B}$, at standard state wher $\ce{A}$ and $\ce{B}$ are at $\pu{1 atm}$. If the free energy of products is greater than the free energy of reactants, $\Delta_{\text{R}} G^\circ > 0$, and since $Q = 1$, the overall free energy for this reaction is $\Delta_{\text{R}} G^\circ > 0$ and this tells me the reaction is non-spontaneous. Yet, being at standard state does not mean that the reaction is at equilibrium, in fact $Q = 1$ could be greater or less than $K$, the products are not at equilibrium, and hence amount of $\ce{A}$ and $\ce{B}$ will adjust to a point where $Q = K$. Wouldn't that have just mean that the reaction is not at equilibrium and hence would spontaneously adjust to reach equilibrium. But yet $\Delta_{\text{R}} G^\circ$ tells me that it would not be spontaneous. Wouldn't this two concepts be conflicting against each other?

Note: This first part isn't actually that related to the specific scenario you are giving, but I think it's a misconception that you had that led you to ask this question. If you want a direct answer, go to the last paragraph.

Just because it is non-spontaneous, it doesn't mean that no reaction will occur. In fact, the reaction will always occur to some extent, even if only 0.001% of the particles react.

Consider this formula, which I'm sure you have seen before:  $$\Delta G=-RT\ce{ln}K$$
This makes sense because if the reaction is spontaneous ($\Delta G<0$), $K>0$, so the forward reaction, which is spontaneous is favored over the reverse reaction; vice versa.

If we, for example, say $\Delta G = +100$ kJ, at 298 K, $K$ would equal $3.4\cdot10^{-17}$. Thus, the forward reaction pretty much completely dominates the reverse reaction, so we say that the reaction is essentially irreversible. However, the reverse reaction is still happening, to a very small extent.

If we happened to make $\Delta G = +10$ kJ, $K$ would equal only about $56$, so the reverse reaction becomes visible. The terms spontaneous and non-spontaneous are only generalizations; it doesn't mean that a non-spontaneous reaction doesn't occur at all. There will always be some particles in that substance that will overcome the activation energy and react.

In terms of your question, since the reaction $\ce{A->B}$ is non-spontaneous, the reaction $\ce{B->A}$ is actually spontaneous (think about their equilibrium constants). Thus, the reverse reaction will be favored until the system reaches equilibrium, or when $Q=K$.


Precise definition of atomic orbital In school I was taught that an atomic orbital is the 3-dimensional region in which the electron is located with a probability of 90%.

However, aren't there infinitely many regions of space in which the total probability of finding the electron is equal? And if so, wouldn't that imply that it is not possible to assign a shape to an atomic orbit as there are infinitely many shapes with equal likelihood to contain the electron?

In school I was taught that an atomic orbital is the 3-dimensional region in which the electron is located with a probability of 90%.

This is not correct.

An atomic orbital is a solution to Schrödinger's equation given the electrostatic potential. In the simplest case, we solve the one electron atom and perturb that with more electrons. An exact solution to the multielectron problem is intractable due to the chaotic nature of 3-body systems.

The concept you are referring to is a boundary surface. (See this question). Basically, it's a surface that has equal probability density for every point on the surface, and the surface encloses a region of space where the sum of probabilities is the desired value.

Because the surface is defined to have equal probability everywhere, and the wave function is smooth, it should be unique.

The boundary surface is usually how we represent the atomic orbitals that are familiar to us ($s, p, d, f$) but also molecular orbitals. Usually, we take something at 90% probability, for example, and plot the boundary, giving the orbital "shape." Otherwise, visualization can be tricky because the orbital only goes to zero probability at infinite distance from the nucleus, so there is a nonzero chance that an electron will be quite far from the nucleus.


Basic sites on Indene ion Answer key says that A is the most basic site but I don’t understand the concept behind that.

To assess the basic site, I’d normally look at the molecule as is and assess the current stability of the atom in question. The least stable would be my strongest base.

Looking at this question both the atoms properties are the same (C with lone pair) so resonance is the next thing I would look at. Both of these rings have resonance which moves (delocalizes) the negative charge around. So I would think since the volume of resonance is greater in the bigger ring, A is more stable thus making B the stronger base.

What am I misunderstanding or what concept am I not considering?

I believe the concept assessed here is the idea of aromaticity. You must first be able to recognise that the lone pair held by each carbon atom with the negative formal charge is in a different type of orbital. In the carbon atom of the 6-membered ring, it is in an $\ce {sp^2}$ hybrid orbital that is in a plane perpendicular to the $\pi$ electron cloud of the indene molecule. This lone pair is not delocalised into the indene $\pi$ system. On the other hand, in the carbon atom of the 5-membered ring, it is in an unhybridised p orbital that is parallel to the other p orbitals of the indene molecule. This lone pair is, in fact, delocalised into the indene $\pi$ system. As such, it is much less available for donation to  an acid.

Protonation at sites C and D would also be unfavourable due to destruction of aromatic character. Between sites A and B, I have explained above that the lone pair at A would be much more available for donation due to lack of delocalisation in to the $\pi$ system. Hence, the preferred site of protonation would be site A.


How to set up an ICE table for adding a weak acid to a strong acid I am working on setting up an ICE table for adding 0.020 mol HCl to 1.0 L of 0.10 M propanoic acid.  Since these are both acids (propanoic acid being the weaker one and HCl being strong), how would I set this up in an ICE table to calculate the pH of the resulting acid?
Edit: I was able to find an answer by taking the negative log of 0.02 to get the pH (because HCl is a strong acid).  However, I am unaware with how to even set up the ICE table.

ICE tables tabulate amounts or concentrations pertaining to an equilibrium reaction: at the initial stage (I), how they change (C), and what they are at the end or at equilibrium (E). They are an alternative to solving multiple simultaneous equations (equilibrium constant expression, mass balance), instead using a single parameter (related to the extent of reaction $\xi$).
For the problem at hand, there is one reaction that goes to completion (strong acid) and one that attains equilibrium (weak acid, both protonated and deprotonated species present at equilibrium).
The ICE table needs to have species from the second reaction only. A good estimate of the hydrogen ion concentration is the concentration of hydrochloric acid (it dissociates completely, and we are ignoring the autodissociation of water for now).
$\begin{array}{|c|c|c|c|} \hline
\ & [\ce{HP}] & [\ce{H+}] & [\ce{P-}]\\ \hline
\text{I} & 0.1 & 0.02 & 0 \\
\text{C} & -x & +x & +x \\
\text{E} & 0.1-x & 0.02 +x & x\\ \hline \end{array}\\\\$
If you do the math, you will find that very little of the propanoic acid will dissociate, as the $\mathrm{p}K_\mathrm{a}$ is far from the initial pH.


Is it possible to extract magnesium from epsom salt? Is it possible (at home) to extract magnesium from $\ce{MgSO4·7H2O}$, also known as Epsom salt (or, alternatively, from $\ce{MgSO4}$)?

Supposed you are trying to obtain magnesium metal from Epsom salt, $\ce{MgSO4.7 H2O}$:
Is is worth the effort?
Magnesium metal in the form of chips (particle size 4-30 mesh) in decent purity (99.98%) is available from suppliers such as Sigma-Aldrich for about 150 Euro per kilogram.
Can I use electrolysis to generate $\ce{Mg}$ metal?
Yes, but you wouldn't want to do that at home. An older industrial process uses electrolysis of molten magnesium chloride, $\ce{MgCl2}$ (mp. is around 700 °C). While $\ce{Mg}$ metal is formed at the cathode, chlorine gas is released at the graphite anode. Did I mention that this isn't meant to be done at home?
Can I convert $\ce{MgSO4}$ to something more usable to make $\ce{Mg}$ metal?
Yes, in principle. A group of Chinese researchers examined the conversion of  $\ce{MgSO4}$ to  magnesium oxide, $\ce{MgO}$ by heating in the presence of natural gas at temperatures around 500 °C and outlined their results in
Geochemical Journal, 2011, 45, 97-108.
The initial pressure by natural gas at room temperature was 3 MPa, but upon heating to the final temperature, the pressure in the reactor was as high as 12-20 MPa! This can't be done at home and it shouldn't be.
What if I would have $\ce{MgO}$? What could I do with it?
Magnesium oxide, $\ce{MgO}$, is indeed the most important starting material in the production of magnesium metal in the Pidgeon process. Here, magnesium oxide is heated in the presence of silicon in an evacuated reactor. Silicon is oxidized to silicon dioxide, magnesium oxide is reduced to magnesium metal. The latter boils off (bp. around 1100 °C) and is condensed. This isn't kitchen chemistry either.
In summary, you're better off buying some magnesium metal from a supplier of your choice!


Structure of Metallic Hydrogen I am simply curious what the theoretical arrangement of metallic hydrogen would be.

I found this article which gives a picture, but I don't understand what the labels mean (e.g. is H5 an isotope, some kind of charge, or just denoting the layer?).

I would like to know:

1) If the image is the correct one.

2) What the labels mean if it is correct.

As annav states, the labeling H1, H2 etc is labeling of hydrogen atoms within a graphene-like hexagonal ring structure. Another way of looking at it, is in the image below:

Image source

In an article published after the web article you link to, Aromaticity, closed-shell effects, and metallization of hydrogen. (Abstract only), the authors confirm that

  Recent theoretical and experimental studies reveal that compressed molecular hydrogen at 200-350 GPa transforms to layered structures consisting of distorted graphene sheets.

This structure, according to the article, forms a stable structure at those pressures.  As per the comments, this is form graphene-like hydrogen, the abstract goes further, discussing the process that would result in hydrogen metallization:

  Metallization occurs upon further compression by destroying the closed shell electronic structure, which is mainly determined by the 1s electrons via lowering of the bonding bands stemming from the unoccupied atomic 2s and 2p orbitals.

This metallization of hydrogen is, according to the article High temperature superconductivity in sulfur and selenium hydrides at high pressure theorised to occur at pressures of 400 GPa or greater.


How to calculate the volume of a 0.1 M NaOH(aq) solution necessary to make a 0.5 M EDTA solution with pH = 8? I was told that EDTA dissolves at a pH of $8$. Therefore I need to use a base to increase the pH to $8$ is what I understand. How exactly would I go about this?

In my head I if I wanted $1~\mathrm{L}$ of $0.5~\mathrm{M}$ EDTA I would take $186.12~\mathrm{g}$ of the EDTA, since its molar mass is $M = 372.24~\mathrm{g\,mol^{-1}}$.

Then I would add $800~\mathrm{mL}$ of distilled $\ce{H2O}$ to dissolve it, but since it wont be at $\ce{pH}=8$, it wont dissolve.

So I assumed I would just add it drop wise until it dissolved while spinning? I lack access to a burrette and only have a crappy pH meter to check pH.

One I observe the EDTA dissolved I simply add up to $1000~\mathrm{mL}$?

I just wish I had a way to figure out how much $\ce{NaOH}$ I needed to add to dissolve the EDTA so I wouldn't have to do it drop wise.

I saw this on the researchgate.net website:

I followed this procedure for preparing  0.5M EDTA ph8(100ml) . Initially I  warmed around 70ml of water and added an ml of 6MNaoh solution to shoot up the ph to 7.6  with the container on  a hot plate .Then I slowly added the measured quantity of EDTA and Simultaneously kept adding small volumes of 6M Naoh solution while measuring  the ph.Finally around ph 8 the EDTA was completely dissolved and the process was way too quick.."

Bhavani Nagarajan · University of North Texas HSC at Fort Worth


Why does hydration of ions depend on their charge to area ratio? The following is written in my book Concise Inorganic Chemistry by J. D. Lee.

Hydration energy, hydrated radius and hydration number of a particular ion depends upon charge per unit area. Hence their order is the same for the following cations in the given examples:
$\ce{Li+(aq) > Na+(aq) > K+(aq) > Rb+(aq) > Cs+(aq)}$

And other similar examples are also given. My question is, why does charge to surface area ratio matter? The force felt by the water molecules due to any of these cations should be the same due to shell theorem. Also, a cation of bigger size can accommodate more water molecules in its vicinity due to its size. Then still why does an increase in charge to surface area ratio corresponds to an increase in hydration tendency?

Many people have done experimental, theoretical, statistical, molecular dynamics, etc. studies of solvation free energies of ions. I encourage you to just search solvation free energies of ions on google scholar and you'll find a bunch of very interesting and quite readable papers.

One of these papers ref. [1] (here for pdf) provides a lot of data and a very simple model which can basically be used to answer your question. In this paper, they derive a simple model which has only one fitting parameter. From this, they can determine the effective cavity size of an ion in solution. This cavity size is given by $r+\Delta r$, where $r$ is the radius of the ion and $\Delta r$ is the thickness of the shell.

Essentially, the value of $\Delta r$ is determined by the fitting parameter, and also determines the number of water molecules within the cavity of the ion. A portion of the data table is shown below with actual experimental values on the far right and parameters determined from the model in the other columns.

Just focus on the alkali metals which you list in your question. Notice as we go further down, the value of $\Delta r$ approaches zero and the actual radius of the ion is also the size of the cavity. This, however, is very much not true for the smaller ions. The Hydrogen ion being an extreme example where a solvation number of 12.0 is obviously unphysical. This, however, does provide a clue for a reasonable interpretation of $\Delta r$. $\Delta r$ is a measure of how far away a molecule must be for it to not "notice" the effect of an ion in solution. Thus, somewhat counter-intuitively, smaller ions interact effectively with a larger number of water molecules. The reason for this is that as the ions get larger, solvating them requires a greater reorganization of the hydrogen-bonding network in liquid water. This means that one must lose hydrogen bonds in order to solvate the ion.

Thus, the reason the charge to area ratio dictates the solvation energy and the solvation number is because a smaller ion will bring the same charge into solution, but due to its smaller size, will only require a small reorganization of the hydrogen bonding between waters, which minimizes losses in already stable interactions. This lack of reorganization is what leads to large values of $\Delta r$ and hence large values of $n$ in the model described above.

Let me be clear, $n$ is not simply the number of water molecules pointing towards an ion. This would not be measurable experimentally, or well-defined theoretically. Rather, $n$ will always be some kind of average value which tells you the effective number of water monomers which energetically benefit from the presence of the ion in solution. So, now the picture is clear. The smaller cations do not cause too large of a first hydration shell, so that the second hydration shell is closer to the ion than for a larger ion. Hence, the free energy and the solvation number are larger for smaller ions of the same charge.

As a bonus, ref. [2] is an interesting paper which approaches this problem from molecular dynamics. They give plots showing the relative effects which van der Waal's forces and electrostatics have. Note that the smaller ions are the only ones which are stabilized effectively by van der Waal's interactions as these scale closer to $r^{-5}$ rather than $r^{-1}$

References:

[1] Marcus, Y. (1991). Thermodynamics of solvation of ions. Part 5.—Gibbs free energy of hydration at 298.15 K. Journal of the Chemical Society, Faraday Transactions, 87(18), 2995-2999.

[2] : Grossfield, A., Ren, P., & Ponder, J. W. (2003). Ion solvation thermodynamics from simulation with a polarizable force field. Journal of the American Chemical Society, 125(50), 15671-15682.


Safe, easily detectable gas or aerosol for testing fume hoods I've recently started undertaking a project to construct a small fume hood. So far I'm considering its use for soldering, 3d printing with ABS and polystyrene, and PCB etching using a muriatic acid/hydrogen peroxide solution, however I'd also like to leave my options open. I want to test the fume hood once completed - I've seen some videos that use dry ice for this purpose, and I figure I might also be able to test the hood by smell using bathroom deodorant, but I'm wondering what sort of concentration these would have to reach in order for me to detect them, and whether there are any other tests I might consider.

Smoke generation is preferred over smell, as the visual test is a bit more objective. See this reference for more info, although $\ce{TiCl4}$is not available for home use. There are hobbyist alternatives such as smoke matches which will help you test the fume hood performance. (The above link is just an example, I did see better prices at less-known websites.)  An anemometer might be of help as well.


How do you calculate the delta H for multiple reactions? I am having difficulty in a question:

Given that
1. $\ce{N_2(g) \rightarrow 2N(g), \Delta H = 941 kJ/mol}$
2. $\ce{N_2(g) \rightarrow N_2^+ + e^-, \Delta H = 1501 kJ/mol}$
3. $\ce{N(g) \rightarrow N^+ + e^-, \Delta H = 1402 kJ/mol}$,
Find the energy interval that produces atomic nitrogen in the flask but not any ions

I think that the energy interval should be $ 941 \leq E \leq 941 + 1402$, but the textbook says that the energy interval should be $ 941 \leq E \leq 1402$.
I think I am correct because $\ce{1402 kJ/mol}$ is needed to be supplied after the initial $\ce{941 kJ/mol}$ that is used to break the $\ce{N_2}$ bond.
Can someone please see the flaw in my reasoning?

Find the energy interval that produces atomic nitrogen in the flask but not any ions

This is the key. The question wants you to determine how much energy you need to supply just to only have atomic nitrogen, no ions.

If you supply $941~kJ/mol$, then you have supplied enough energy to move reaction 1 forward. In your flask, you would have $N(g)$ inside. Now, if you further supply energy up to $1402~kJ/mol$, then you have enough energy to make reaction 3 proceed, and that will convert your atomic nitrogen to nitrogen ions, so that's a no-no.

So you have to maintain $E$ that is in between $941~kJ/mol$ to $1402~kJ/mol$ to prevent reaction 3 from proceeding.

If you supply $1402~kJ/mol$ after breaking the bond, then you will definitely produce $N^+$ ions as you have met reaction 3's $E$ requirement.


What is the difference between "bromine" and "bromine water" in their reaction with an alkene? In question 16 of the A-Level Chemistry 2013 exam, there is this reaction scheme:

  $$\ce{CH3CH=CH2 ->[?] CH3-CHBr-CH2Br}$$

The question asks for both the name of the reagent and the product. I know that the product is 1,2-dibromopropane, but am confused about the reagent. I was told that the correct answer was "bromine", whereas "bromine water" was a wrong answer. Why is this so?

From ChemGuide:

  Using bromine water as a test for alkenes

  If you shake an alkene with bromine water (or bubble a gaseous alkene
  through bromine water), the solution becomes colourless. Alkenes
  decolourise bromine water.

  The chemistry of the test

  This is complicated by the fact that the major product isn't
  1,2-dibromoethane. The water also gets involved in the reaction, and
  most of the product is 2-bromoethanol.

In essence, the electrophilic pi bond in the presence of bromine and water attacks diatomic bromine as usual, forming the usual bromonium ion intermediate.

After this step, however, we have two competing nucleophiles which can open the ring: bromide ion and water. The water is likely present in great excess, making it the more likely attacker (statistically).

Of course, the desired reaction below to form a dibromide also occurs. But it is in competition with other reactions such as the one listed above.

So usually electrophilic halogenation of alkenes takes place in an inert medium such as methylene chloride or carbon tetrachloride. These aren't inclined to act as nucleophiles, so they don't participate in ring opening, leaving the $\ce{Br-}$ ion to do its thing unhindered.


Build the geometric structure of SnBr2 What is Sn valence in this case? Because  it depends on it if it will be linear or tetragonal.I'm talking about a molecule.

In the gas phase (molecular), SnBr2 is bent with an angle of 95 degrees.

The IUPAC definition of valence is such that there is only one valence for each element.

Valence is independent of the compound being consider.


How can an electron being a wave have such property as spin? Here's what I know about electrons.
Electrons have wave-like properties and the number of wavelengths in the $n^\text{th}$ shell is equal to $n(\lambda).$ Also, I read in my book that they have another property called spin.
But I couldn't imagine how a wave can have a property of spin. If this property is just a mathematical model, then is this property compatible for a wave i.e.  Is it appropriate to use this property of spin with waves ?

An electron is not a wave nor a particle. An electron is usually described as a quantum object with some wave-like properties and some particle-like properties.
Some of them, like particularly a spin, do not have direct counterpart in our familiar macro world. It is a kind of a mysterious, specifically quantum property of all elementary particles and  atomic kernels, that interact with angular momentum and interchange with it its values.
Non-relativistic quantum mechanics has no theoretical idea/model about the observed spin phenomena, that would come e.g. from resolving the Schrödinger wave equation. It was  added just as an ad hoc extension. OTOH, the relativistic Dirac wave equation brings spin as the natural part of the quantum mechanical model.
Note that the classical idea about a spin as a rotation related angular momentum of a fast spinning small ball is wrong. Considering the electron mass, energy and the limit of the maximum electron size, an electron simply cannot rotate fast enough to have such amount of classical angular momentum, related to its eventual spinning.
Both orbital and spin contributions to the total angular momentum have their counterparts as orbital and spin magnetic momentum, if quantum objects have nonzero electric charge or charge nonhomogenity (like neutrons ).


IUPAC naming for compounds containing carboxylic acid and anhydride together? According to IUPAC, the carboxylic acid comes first in priority order for functional groups, while anhydride has no prefix. Then how do you name a compound which contains both a carboxylic acid and an anhydride in it?

Acids are senior to anhydrides; therefore, a compound that contains both functional groups is named as acid.

The acid is substituted using the usual principles of substitutive nomenclature. The prefix ‘oxo’, denoting $\ce{=O}$, is used to indicate a carbonyl group when the group cannot be cited as suffix. The group $\ce{R-CO-O-{}}$ is indicated by an acyloxy prefix (e.g. ‘acetyloxy’, ‘propanoyloxy’, etc.).

For example: 3-(acetyloxy)-3-oxopropanoic acid

(Cyclic anhydrides, however, are named as heterocyclic pseudo ketones.)


Friedel-Crafts alkylation of five-membered heterocycles The five-membered heterocycles pyrrole, furan, and thiophene undergo Friedel-Crafts acylation, but I have not seen them undergoing a simple Friedel-Crafts alkylation reaction with a Lewis acid such as aluminium chloride. Why is that so?

Alkylation of the five-membered rings is perfectly possible - nobody says it cannot happen. But there are two problems:

Alkylation further activates the ring, making it more reactive - this is exactly the same problem that makes alkylation an unattractive reaction in benzene rings
The Lewis acid can, and will, lead to polymerisation of electron-rich rings such as the five-membered heterocycles.

With acylation, the first point is moot, and if you choose an appropriate catalyst you can avoid polymerisation. So acylation can be used in certain cases (I'm fairly sure that simply chucking in an acyl chloride with $\ce{AlCl3}$ will not be a clean reaction at all though).

From Joule & Mills, Heterocyclic Chemistry 5th ed., p 300 (chapter on pyrrole):

  Mono-C-alkylation of pyrroles cannot be achieved by direct reaction with simple alkyl halides, either alone or with a Lewis-acid catalyst, for example pyrrole does not react with methyl iodide below 100 °C; above about 150 °C, a series of reactions occurs leading to a complex mixture made up mostly of polymeric material together with some poly-methylated pyrroles. The more reactive allyl bromide reacts with pyrrole at room temperature, but mixtures of mono- to tetra-allyl-pyrroles together with oligomers and polymers are obtained.

p 327 (chapter on thiophene):

  Alkylation occurs readily, but is rarely of preparative use, an exception being the efficient 2,5-bis-t-butylation of thiophene. (J. Org. Chem., 1984, 49, 4161. DOI:10.1021/jo00196a012)

Presumably the steric bulk of the t-butyl group prevents further reactions.

p 350 (chapter on furan):

  Traditional Friedel–Crafts alkylation is not generally practicable in the furan series, partly because of catalyst-induced polymerisation and partly because of polyalkylation.


H2O2 (Hydrogen Peroxide) catalysis in clear plastic bottle? I am buying some 30% H2O2 which will be in probably black HDPE bottles. If I want to make some 3% - 5% solution, can I store that in clear bottles? How much faster will it catalyse versus having it in a dark bottle?

Will shelf life be significantly shorter? How much effect does sunlight (or other light) have on the H2O2 solution? Any way to measure this?

Regards,

An easy test for this would be to react H2O2 with yeast or baking soda (sodium bicarbonate) before and after sun bathing. Or you could simply measure gas production in the sun vs in a box for say 24 hours.

An easy way to measure the O2 released is to use a closed vessel with a ballon, but the more scientific method uses an inverted, graduated column.

You can measure light intensity with a photometer.


How does copper rich "weld thru" primer protect iron / steel? In the automotive industry, to protect steel which cannot be painted (e.g. the gap or seam between two sheet panels), a product known as weld thru primer is sometimes used.
This is a coating of zinc-rich or copper-rich material which is meant to prevent rusting of the steel, especially at the weld points.
From the reactivity table below, zinc is more reactive than iron, so I can understand how this would protect the iron via sacrificial corrosion. Similar to galvanized steel.
However, copper is less reactive than steel. When salt water get's in between the sheet metal seams, would this not corrode the steel? How would applying a layer of copper to the steel protect it?

Well there are also many different kinds of steels with different galvanic characteristics are well. For example, all of these metals have a higher reduction potential than copper:
1. Red Brass
2. Stainless steel 347 (active)
3. Molybdenum, Commercial pure
4. Copper-nickel 715
5. Admiralty brass
6. Stainless steel 202 (active)
7. Bronze, Phosphor 534 (B-1)
8. Monel 400
9. Stainless steel 201 (active)
10. Carpenter 20 (active)
11. Stainless steel 321 (active)
12. Stainless steel 316 (active)
13. Stainless steel 309 (active)
14. Stainless steel 17-7PH (passive)
15. Silicone Bronze 655
16. Stainless steel 304 (passive)
17. Stainless steel 301 (passive)
18. Stainless steel 321 (passive)
19. Stainless steel 201 (passive)
20. Stainless steel 286 (passive)
21. Stainless steel 316L (passive)
22. AM355 (active)
23. Stainless steel 202 (passive)
24. Carpenter 20 (passive)
25. AM355 (passive)
26. A286 (passive)
27. Titanium 5A1, 2.5 Sn
28. Titanium 13V, 11Cr, 3Al (annealed)
29. Titanium 6Al, 4V (solution treated and aged)
30. Titanium 6Al, 4V (anneal)
31. Titanium 8Mn
32. Titanium 13V, 11Cr 3Al (solution heat treated and aged)
33. Titanium 75A
34. AM350 (passive)
35. Silver
36. Gold

(source: Corrosion Doctors)
So, as long as you're welding any of those under copper's boiling point of 2,562°C (or 4,644°F, at which point it could just float away from your intended target), then it should be fine for your application.


Is sulfide ion a stronger base than hydroxide ion? I received these three responses. They are all, except the last response, incorrect to varying degrees. I am, however, unsure about how to grade these responses because the question itself, in my opinion, misleads the students by saying that both have the same pH and to use this information to determine the strength of the bases. Their responses also raised a few questions of my own. What are your opinions?

Question
It is an experimental fact that the pH of 1 M $\ce{Na2S}$ is essentially the same as the pH of 1 M $\ce{NaOH}$. Based on this information, is $\ce{S^{2-}}$ or $\ce{HO-}$ the stronger Bronsted-Lowry base? How can you tell?

Student 1's response:

A base's strength is measured by how much it can increase the hydroxide ion concentration in a solvent relative to its own initial concentration.
Given that both the 1 M $\ce{Na2S}$ and 1M $\ce{NaOH}$ solutions have the same pH, they both have the same $\ce{[H3O+]}$ concentration and therefore the same $\ce{[HO^{-}]}$ concentration. Therefore, the two bases are the same strength as each creates a $\ce{[HO^{-}]}$ concentration equal to their initial molarities in water.

I think this response could be improved if the student mentioned the leveling effect and recognized how all bases stronger than hydroxide ion are "leveled." It's just like the high striker game at the fair; you can hit it hard enough to make the puck go all the way to the top, and so can someone else, but that doesn't necessarily mean you guys are equal in strength; it just means that the device quantifying strength doesn't differentiate between "strong" and "super strong."
Student 2's response

$\ce{S^{2-} + H2O ->HS- + HO-}$
From this equation and the problem statement, it is clear that $\ce{S^{2-}}$ hydrolyzes water to create a hydroxide ion concentration in water equal to its own initial molarity. Hydroxide ion isn't hydrolyzing water to create hydroxide; the below equation is non-sensical. Therefore, sulfide ion is the stronger base.
$\ce{HO^{-} + H2O ->H2O + HO-}$

Is the second equation really non-sensical?
Student 3's response:

The stronger base has the weaker conjugate acid.
Sulfide ion's conjugate acid is $\ce{HS-}$.
Hydroxide ion's conjugate acid is $\ce{H2O}$.
From the $\ce{K_{a}}$ table in the beginning of the lab manual, one knows that $\ce{HS-}$ is a weaker acid than $\ce{H2O}$. Therefore, sulfide is the stronger base.

Equilibrium Expressions Related to the Solubility of the Sour Corrosion Product
Mackinawite Ind. Eng. Chem. Res. 2008, 47, 1738-1742 compiles 18 different studies concerning the second Ka of H2S and shows that they vary over 7 orders of magnitude from $10^{-19}$ to $10^{-12}$

So firstly it isn't accepted whether or not Na2S is stronger than NaOH.

Putting that aside, the only way that 1M sulfide could yield 1 M hydroxide (and hence pH be equal to pH of 1 M hydroxide) is if sulfide was infinitely strong.  If they were equally strong, it would only yield 0.5M hydroxide.

Student 1: clearly wrong.

Student 2: correct (based upon the information given)

Student 3: can't say without seeing the table, but if the table actually says that SH- is weaker than water than the answer is true, but not correct in the sense that the instructions say "Based on this information".


Normal Spinel and Inverse Spinel According to my notes, octahedral site stabilization energy (OSSE), sterical effects and electronic effects in stability create a confusion in this theory. However, we are able to solve this problem with few rules:

Tetrahedral splitting energy is nearly equal to 4/9 octahedral splitting energy.
As the OSSE increases, the $O_\mathrm{h}$ geometries will be more favored.
These are useful to decide to whether the structure is spinel or inverse spinel.

I got $\ce{ZnCr2O4}$ ($\ce{Zn^2+}$, $\ce{2Cr^3+}$, $\ce{4O^2-}$):

$\ce{Zn^2+}$; $\mathrm{d^{10}}$ ion with CFSE = $0$ and thus, there will be no effect on $O_\mathrm{h}$ holes.
$\ce{Cr^3+}$; $\mathrm{d^3}$ ion with CFSE = $\pu{-8 Dq}$ and OSSE = $\pu{8.45 Dq}$ so, this is thought to occupy $O_\mathrm{h}$ holes.

Since $\ce{Zn^2+}$ can only occupy $T_\mathrm{d}$ holes, then it suits the rules for being normal spinel.

On the other hand, if we got a group II metal with a non-zero CFSE, then what would the result be?

Let's take a (2-3) spinel ($\ce{A^{II}B^{III}_{2}O4}$) as an example. If it is normal or inverse can be predicted from the total crystal field stabilization energies. For a normal spinel, the total CFSE wold be: $$\Delta_{T}(A)+2\Delta_{O}(B)=\Delta_{T}(A)+\Delta_{O}(B)+\Delta_{O}(B)$$ and for an inverse spinel $$\Delta_{O}(A)+\Delta_{T}(B)+\Delta_{O}(B)$$ with $\Delta_{T}$ being the tetrahedral and $\Delta_{O}$ being the octahedral splitting energy of $\ce{A^2+}$ and $\ce{B^3+}$, respectively.

If $\Delta_{T}(A)+\Delta_{O}(B)>\Delta_{O}(A)+\Delta_{T}(B)$, the normal spinel structure should be favored, as the resulting CFSE is higher. If it is the other way around, an inverse spinel is to be expected.

To sum it up, if the CFSE gained by placing $\ce{A^2+}$ in the octahedral holes outweighs a potential destabilization caused by putting a part of $\ce{B^3+}$ in the tetrahedral holes, an inverse spinel can be formed. However, a normal spinel is favorable when $\ce{B^3+}$ clearly "prefers" the octahedral and/or $\ce{A^2+}$ the tetrahedral holes.


Magnetic behaviour of complex compounds Which one of the following cyanide complexes would exhibit the lowest value of paramagnetic behaviour?

$\ce{[Cr(CN)6]^{3-}}$
$\ce{[Mn(CN)6]^{3-}}$
$\ce{[Fe(CN)6]^{3-}}$
$\ce{[Co(CN)6]^{3-}}$

Here I determined central atom configuration and it is $\mathrm{3d^3}, \mathrm{3d^4}, \mathrm{3d^5}, \mathrm{3d^6}$ for each option in order. $\ce{CN-}$ is a strong-field ligand so electron will get paired hence option 2 and 4 will have least magnetic behaviour

But only option 4 is correct. Why?

Magnetic behaviour of complex compounds are determined on the basis of magnetic moment, rather the extent of paramagnetism is measured in terms of the magnetic moment μ.

μ = √n(n+2) BM , where n notates the number of unpaired electrons.

First lets give a look towards the number of unpaired electron each of the central metal atom has:

Here the question is asking to determine the complex compound which shows the lowest paramagnetic behaviour. As Co^3+ has no unpaired electron, its μ = 0, hence it is diamagnetic, and hence shows shows lowest paramagnetic behaviour among the above complexes.
And best of luck for jee'17.
Me too appearing the exam.

Hope it helps


rate determination by 1H NMR spectroscopy Currently I investigate a reaction where a starting material A reacts with an excess of thiophenol B towards products C and D.  As shown by recurrent TLC analysis of the reaction, the formation of D is slower than the one of C, while A and B are still detectable.

Perhaps the reactions are consecutive, i.e.

$\ce{A + B ->[k_1] C ->[k_2] D}$

or they are simultaneous, i.e.

$\ce{D <-[k_2] A + B ->[k_1] C}$.

A, B, C, and D are distinguishable by 1H NMR spectroscopy; NMR data of the purified compounds A, B, C, and D were  recorded successfully.

Based on a recurrent 1H NMR spectroscopy, I would like to use the integrations of the signals to determine the rate constants; perhaps there is evidence for the first case (similar to Bodenstein principle).  The advancement of the thioetherifications alter the multiplicities of the signals observed, and simultaneously indicator signals vanish.

Question:  Prior to get work on my data, where may I find a tutorial of reactions with known starting material(s)/product(s), determined rate constant(s) and NMR data-used-to-determine-these-rate-constants?

As mentioned in the comments, J. Chem. Ed is a good resource for NMR kinetics experiments. I have had success with the experiment published by J. Orvis in J. Chem. Ed 80(7), 2003, 803 link.  This publication describes the synthesis of trans-dichlorotetraamminecobalt(III) and its subsequent reaction with water to produce cis-aquachlorotetramminecobalt(III).  There are several advantages to this experiment as a model study:

The synthesis is straightforward and does not require materials unlikely to be already in the laboratory, although it is critical that the directions are followed precisely.
The aquation reaction occurs within about one hour - slow enough to easily monitor via NMR but not so slow as to take up precious instrument time
The reaction rate is temperature sensitive, which makes it a good model for extracting thermodynamic properties from the reaction

My undergraduate students are able to complete the synthesis about 50% of the time, and the yields using the published procedure are about 400 mg, so one successful synthesis is suitable for a number of NMR experiments.


Disrupting the chemistry of oobleck so that it behaves like a newtonian fluid Oobleck (a mixture of cornstarch and water) is a non-newtonian fluid, so it acts like a solid when under pressure, and like a liquid otherwise. My question is this, is there anything you could mix in with oobleck (a friend suggested vinegar, I don't know why) that would cause it to stop being non-newtonian (more specifically, to act like an ordinary liquid). The motivation for this is that I'm going to have a large amount of the stuff, and will need a practical way of disposing of it, like dumping it down a drain.

Thanks in advance for any answers!

Disposal isn't a problem - add more water as you dump it down the drain.

There are two extremes to ooblek:

Almost all cornstarch and almost no water: this is a dry solid or slightly wet starch. No big deal.
Almost all water and very little starch: this is some water with a little bit of starch mixed in. Sometimes it clumps or falls out of suspension, but it's also no big deal.

The non-Newtonian mixture occurs at intermediate ranges.

Since cornstarch will degrade in the sewer system (e.g., feed bacteria), it's safe down the drain. Alternatively, I compost it.

But if you're worried about non-Newtonian behavior, you just add more water.


Why is aniline more basic than pyrrole? In a question asking to compare the basicity of pyrrole and aniline, the explanation given for aniline being more basic than pyrrole was as follows:

Basicity depends on the availability of lone pair of $e^-$ on Nitrogen atom.
In pyrrole, lone pair of $e^–$ present on $\ce{N}$ are involved in resonance with the ring which provide aromatic character to the ring.
On protonation of pyrrole, the electron will not be above to participate in resonance, While anilines aromatic even though the involvement of line pair of nitrogen of $\ce{—NH2}$ group is not there and lone pair of aniline is delocalized in benzene ring but due to deocalization benzene loss its aromatic character so, the delocalization is not that strong as in pyrole molecule.
Lone pair is more available of protonation. Hence, aniline is more basic than pyrrole.

But, why should the aromatic character get destroyed? It is still a planar structure and should be aromatic. What is an intuitive explanation?

There would be no aromaticity in pyrrole after protonation since there is no lone pair to allow for Hückel's rule for aromaticity (the $4n+2$ rule). So, upon protonation, you are destroying aromaticity in pyrrole. Also the lone pair is completely delocalised in pyrrole and it cannot easily participate as a base.
However, in aniline, there is a partial loss of the lone pair due to resonance but the lone pair of nitrogen is still available for protonation. Aromaticity is not lost in the resonance structures of aniline.
Therefore aniline is more basic than pyrrole but is less basic compared to pyridine (where the lone pair is out of plane of resonance).


What are the rate laws for each elementary step? Overall reaction

$$\ce{2NO + H2 -> N2O + H2O}$$

Mechanism

\begin{align}
\ce{NO + NO &-> NO2}\\
\ce{NO2 + H2 &-> NO + H2O}\\
\ce{N + NO &-> N2O}\\
\end{align}

I am aware that this is not a valid mechanism for the overall reaction however I am still supposed to find the rate law for each elementary step. I am having trouble determining what this rate law would be in step 3 of the mechanism.

For elementary reactions, the rate law can be directly read from the chemical equation. In your last equation this is a bimolecular reaction, i.e.
\begin{align}
\ce{N + NO &-> N2O} & r &= k\cdot c(\ce{N}) \cdot c(\ce{NO})
\end{align}

To be more precise here, one molecule of $\ce{NO}$ reacts with on atom $\ce{N}$, and in a wider sense, that is two (bi) molecules reacting. The order of the reaction is two.

That being said, there are a couple of things you need to pay close attention to. An elementary reaction is always a reaction between the number of molecules given as reactants, with exactly one transition state leading to the products. If this number or reactants is greater than three it is highly unlikely to happen this way, since a concerted transition state is statistically very unlikely. An elementary reaction therefore also has to be balanced, so your first reaction needs to be:
$$\ce{2NO -> NO2 + N}$$
In this case we could see $[\ce{ON\bond{~}O\bond{~}N}]^\ddagger$ as a transition state.

I very highly doubt that your second reaction as an elementary reaction, from my point of view, there are too many bond that would need to be broken. Instead, the following reactions might be possible and/or necessary decompositions:
\begin{align}
\ce{NO2 + H2 &-> NO + OHH}\\
\ce{OHH &-> OH + H}\\
\ce{NO2 + H &-> NO + OH}\\
\ce{2OH &-> H2O + O}\\
\ce{NO + O &-> NO2}\\\hline
\ce{NO2 + H2 &-> NO + H2O}
\end{align}
And there are more possibilities.

In all but one the elementary reactions are bimolecular - guess which one would be unimolecular.

As for the mechanism itself, I think these are reactions that occur, just not all of them, maybe not even the majority of them.


Why is silver chloride less soluble than silver nitrate? Related: Reaction between silver nitrate and aluminum chloride

Experimentally, $\ce{AgCl}$ is insoluble in water, but $\ce{AgNO3}$ is soluble. They're pretty common in a lab (well, $\ce{AgCl}$ is a common precipitate)--so I think most of us know this.

By Fajan's rules, on the other hand, larger anion $\implies$ more polarization/covalent character $\implies$ less solubility. But, $\ce{NO3-}$ is the larger anion, yet $\ce{AgNO3}$ is more soluble.

Is there any theoretical reason for this?

In the comment to my previous answer, you asked for a theoretical reason for the solubilities, not considering energy data. Since I know from energy considerations that the issue is not the solvation of the anions, I can present a reason based on the strength of the ionic bond in the two compounds. This reference (as well as others) states the bonding in $\ce{AgCl}$ has an unusually high covalent character which makes it a tighter bond. The $\ce{Ag+}$ ion and the $\ce{Cl-}$ ion are close to the same size (with the silver ion being smaller), so they can approach each other quite closely. In silver nitrate, the $\ce{NO3-}$ ion is larger and does not allow as close an approach as the chloride ion, so the bond is weaker, easier to break up, and the salt is more soluble.


In which state does the bifluoride anion exist? Statement 1. The $\ce{HF2-}$ ions exist in solid state and also in
  liquid state but not in aqueous state.
  Statement 2. The magnitude of hydrogen bonding between $\ce{HF}$ is weaker than that between $\ce{HF}$ and $\ce{H2O}$

  Both are correct and 2 explains 1
  Both are correct but 2 does not explain 1
  1 is correct and 2 false
  2 is correct and 1 false

The answer given is 1.

I know the second statement to be true. $$\Delta H (\ce{H-F...O})=40{-}45\text{ kJ/mol},\Delta H(\ce{H-F...F})=20{-}23\text{ kJ/mol}$$

But I am completely unsure about statement 1 and the relation of the two statements. I think the existence of the afforementioned ion in aqueous solution must not be absent, as I have read somewhere that this ion acts as the conjugate base of the molecule $\ce{H2F2}$ which dissociates into hydrofluoric acid.

Wikipedia page for for bifluoride anion. The question's answer appears to be true, with explanation and references.

The fluorohydrogenate group $\ce{-HF-}$ in anions such as bifluoride can assimilate a proton by recombination:

$$\ce{HF2- + H+ -> H2F2 -> 2 HF}$$

Because of this capture of a proton ($\ce{H+}$), bifluoride has basic character. Its conjugate acid is the reactive intermediate, μ-fluoro-fluorodihydrogen ($\ce{H2F2}$), which subsequently dissociates to become hydrogen fluoride. In solution, most bifluoride ions are dissociated.

$\ce{HF2- + 3H2O}$ is in a favored equilibrium with $\ce{2H3OF + HO-}$

This makes for an interesting oxygen coordination sphere re the other lone pair of electrons.  One presumes it is hydronium fluoride in which the fluoride is bonded to a proton with a delocalized charge.

I've worked with aqueous ammonium hydrogen bifluoride as a silicate etchant. The solution has tremendous surface tension.  A drop of Fluorad surfactant was necessary to have things work.  $\ce{HF}$ will reach out and eat you.  $\ce{NH4HF2}$ is fairly benign.  No skin contact!  Don't work in Pyrex.


Do isotopic groups cause optical activity? I understood the simplified reason for optical activity of a compound based on the difference in the angular velocities of counter-rotating circularly polarized light. The difference is due to the spatial arrangement of different groups with different electronic configurations. Even though it wasn't specified in my textbook, I guess the difference in the electronic configurations cause some electrical constraints to the oscillating electric field of the plane polarized light, thereby rotating the plane of polarization.

Do isotopic groups cause optical activity?

For example, is $\ce{CHDTF}$ optically active due to the presence of different isotopes of hydrogen?

I understood that $\ce{CHTDF}$ is chiral or it exists as a pair of enantiomers. But I don't know whether different isotopes of hydrogen account for any optical activity as I think optical activity is something to do with the electronic configurations rather than the atomic masses.

From a 2007 paper::

  Here we show that instrumental advances in Raman optical activity, combined with quantum chemical computations, make it possible to determine the absolute configuration of (R)-[$\ce{^2H1}$, $\ce{^2H2}$, $\ce{^2H3}$]-neopentane. This saturated hydrocarbon represents the archetype of all molecules that are chiral as a result of a dissymmetric mass distribution. It is chemically inert and cannot be derivatized to yield molecules that would reveal the absolute configuration of the parent compound. Diastereomeric interactions with other molecules, optical rotation, and electronic circular dichroism are, in contrast to the well-known case of bromochlorofluoromethane, not expected to be measurable. Vibronic effects in the vacuum ultraviolet circular dichroism might reveal that the molecule is chiral, but the presence of nine rotamers would make it extremely difficult to interpret the spectra, because the spatial arrangement of the rotamers’ nuclei resembles that of enantiomers. The unequivocal spectroscopic determination of the absolute configuration of (R)-[$\ce{^2H1}$, $\ce{^2H2}$, $\ce{^2H3}$]-neopentane therefore presented a major challenge, one that was at the very limit of what is possible.

Reading between the lines:

The described molecule looks like this:

Optical rotation is not expected to be measurable.  Probably some fantastically minute but nonzero specific rotation could be computed with advanced quantum mechanical methods, but it would be no fun to try to measure it.
The difference in the electronic configuration of the isotopically substituted enantiomers in the ground state is negligible.
But the electronic configuration of rotationally or vibrationally excited states is not negligible.  Therefore, some technique that can measure the optical properties of rovibrational excited states might be able to discern the difference.  (I'll leave it to experts in this area to explain why Raman optical activity worked better than vUV-CD.)


Atmospheric changes relating to gases I have the following question:

Which of the following would not be used to measure changes in atmospheric gases?
i) Satellites
ii) Mass Spectrometer
iii) Digital Thermometer
iv) Fractional Distillation

I assumed it would be fractional distillation since you cannot really measure anything with it? You only separate the fractions of different materials such as crude oil.
However the answer was Digital Thermometer. I guess I can understand how this object would be useless at measuring atmospheric changes relating to gases however how would you detect atmospheric gases changes using fractional distillation?

Assuming 'change' means modification of composition, the answer is Digital Thermometer because even if temperature is linked to atmosphere composition it does not provide you any information about it.

Satelite can embed LIDAR system which will provide you crude qualitative (spectrum) and quantitative (intensities) informations on atmosphere composition, but it requires complex mathematical processing and relies on weak hypothesis;
Mass spectrography is probably the most accurate way because it is highly qualitative (retention time) and quantitative (ionization current);
Distillation will allow you to qualitatively 'separate' (boiling point) and 'quantify' (mass flow rate) distinct components.

Obviously temperature measurement is not able to provide any informations of this kind. Further more, temperature gradient in atmosphere is not only dominated by its composition, there is a lot of other phenomena involved in it.


Can PBE (and LDA) actually be a better choice sometimes? Part of my work as an inorganic chemist is to investigate the magnetic coupling between metallic centers in coordination compounds. After some time, I've noticed that the classic PBE functional is the one giving me the best results (better than B3LYP, PBE0 or even M06) and, at first, I was convinced that this was a mere coincidence.

The cases in which PBE keeps giving me better results are adding up, so, I've decided to perform a quick test using the well known $\ce{[Mn(III)2O2(NH3)8]^{2+}}$ that have a reported experimental $J_{AB} = -186.5~cm^{-1}$.

My benchmark showed this:

To me, the results are that the Local and Gradient Corrected Functionals like PBE and LDA are, actually, better for this kind of properties in which orbital superposition and spin delocalization are important.

My point here, I think, is that I struggle to believe that such simple and "old" functionals like PBE and LDA are actually better describing a complicated phenomenon like magnetic coupling.

I'd love to hear what you, guys, think.

P. S. 1: I'm using ORCA, that gives me J calculated using 3 different formulas, that's why I have 3 J values.

P. S. 2: Sometimes I'm using 2 basis sets, a bigger one for Mn and a smaller for O, N and H.

Can PBE (and LDA) actually be a better choice sometimes?

Of course, they can. This is in fact one of the major problems with DFT: there is no systematic way of improving a functional, so we never know a priori which will work better for some particular problem. The only thing we know is that some functionals perform systematically better that the others for some cases (see a bit more on that at the very end).

The choice of a functional (and even more generally the level of theory) is dictated by which property has to be calculated with which accuracy and is usually based on extensive benchmark studies. A rather quick search by "magnetic coupling dft benchmark" terms found few papers claiming that hybrids indeed work worse than pure functionals for that particular property. For instance, here, it is explicitly claimed in the abstract that

  The inclusion of exact exchange does not lead to improvements but
  worsens results drastically.

Exactly the same thing happens in your case, and this is not a big surprise for a number of reasons.

First, hybrid functionals are primarily better for energetics, and not just for any property. Many of hybrid functionals were even crafted with that particular aim in mind, i.e. to improve the description of energetics. But for other properties hybrid functionals might provide almost no improvement comparing to pure functionals and the results can indeed be even worse.
Secondly, the might be some cancelation of errors out there. In some cases you can get lucky and have a better results using a lower level of theory with a smaller basis set due to cancelation of errors.
Finally, benchmark has to be extensive. When it is said that some functionals perform better, the meaning is that they perform better systematically, i.e. for a wide variety of cases at average they spit out "better" numbers. There is absolutely no guarantee that these functionals will perform better for each and every case.

So, I recommend to:

Always start from the literature search for benchmark studies on the property of interest which were already done and reported. They very likely exist and can be of a great help.
On the basis of extensive benchmarks choose just a few combinations of methods and basis sets which are known to perform systematically well for the property of interest and benchmark them on your system, or, even better, on a number of similar systems which you might want (or even planned already) to study in the future.
Pick one or two methods and stick to them.


Ascorbic acid and aldehydes: Reaction and Influencing Metals I've researched online and I've read that ascorbic acid can both promote oxidation and reduction depending on the conditions (I've read about trace metals but I've never found which metals affect its chemistry and which ones don't).
So if I mix ascorbic acid with a bunch of different aldehydes, will it react with all if them? Some of them? Will it reduce them (into alcohols?) or oxidate them (into carboxylic acids)? Do metals have any influence on what happens? And if so, which metals do?

You would need very strongly reducing compound to be able to reduce ascorbic acid. Aldehydes are not the case.
Ascorbic acid would not reduce aldehydes like ethanal either, as can be seen by comparison of their respective redox potentials below:

Redox system
$E^{\circ} [\pu{V}]$

Ascorbic/dehydroascorbic  acid
+0.35

Ethanal/ethanol
-0.197

Transition d-orbital  metals with multiple possible oxidation numbers very frequently act as redox catalysts. Both neighbour oxidation numbers may be common ( $\ce{Fe^{+III}/Fe^{+II}}$, $\ce{Mn^{+III}/Mn^{+II}}$,
$\ce{Mn^{+IV}/Mn^{+III}}$ ), or one may be uncommon ($\ce{Cu^{+III}/Cu^{+II }, Ag^{+II}/Ag^{+I}}$)
Catalytic effect is often related to the opposite charges of reagents and catalysts. The typical example is reaction of oxalate and permanganate anions, catalyzed by manganese cations.
A good metal catalyst has standard redox potential comparable with respective redox potentials of the oxidant a/o reductant.
E.g  $\ce{Ag+}$ may act as a catalyst during oxidation by peroxodisulphate, as $\ce{Ag^2+/Ag+}$ has the similar redox potential. Similarly, $\ce{Mn^2+}$ ions act as the catalyst during  oxidation of oxalic acid  by permanganate in acidic solutions, as $\ce{Mn^2+/Mn^3+}$ has similar redox potential as $\ce{MnO4-/Mn^2+}$.


What is the chemical reaction between coins (copper and nickel) and polyvinyl chloride? I often carry coins in a cheap plastic sandwich bag. A green solid soon appears.  Is it copper oxide? How is the polyvinyl chloride (or polyethylene or polypropylene) greatly accelerating the reaction? Or are they?  Are there other chemicals being formed?
Coin collector websites all say vinyl-containing compounds, and maybe other plastics, damage coins. But they don't say how....
P.S.:  Coins are made of copper, nickel and zinc, right?

Far, far too broad a question. I don't know what is going on in your situation.
I can, perhaps, clear up some misconceptions you may have. PVC is NOT commonly used in sandwich bags. PVDC is a close relative and used to be common as "saran wrap". But most sandwich bags are made out of polyethylene (PE). Chemically both PVC and PVDC degrade due to weathering and other things. Dehydrohalogenation is the reaction of organic chlorine compounds (both PVC and PVDC) which can be expressed in the partial formula C-HC-Cl → C=C + HCl (this is a crude simplification). Note that this reaction generates HCl, also known as hydrochloric acid (or muriatic acid). HCl is a powerful acid, attacking most metals. If you live by the ocean or in a desert chloride levels in the air (dust or mist) can be quite high and can accelerate corrosion dramatically. Coins are designed to be inert. Placed clean and dry into a sealed PE bag, they should not corrode. But. Whenever two different metals (or metal alloys) come into contact, a electrical potential is created. A random bag of coins will be sorta like a bag of little batteries. That is, you shouldn't be surprised if they react (discharge), especially in the presence of moisture. Neither copper nor nickel form green oxides, but both do form green chlorides and green carbonates. You need a source of chlorine for the chlorides, but the carbonates might form when simply exposed to air and moisture when a electrical potential exists. Sweat, sweaty hands, are potentially one source of chloride/chlorine, but as I stated up front I am not familiar with your situation, I carry (mixed) coins in my pocket for days at a time without any green forming.


Striking examples where Kohn-Sham orbitals clearly have no physical meaning In Density Functional Theory courses, one is often reminded that Kohn-Sham orbitals are often said to bear no any physical meaning. They only represent a noninteracting reference system which has the same electron density as the real interacting system.

That being said, there are plenty of studies in that field’s literature that given KS orbitals a physical interpretation, often after a disclaimer similar to what I said above. To give only two examples, KS orbitals of H2O[1] and CO2 closely resemble the well-known molecular orbitals.

Thus, I wonder: What good (by virtue of being intuitive, striking or famous) examples can one give as a warning of interpreting the KS orbitals resulting from a DFT calculation?

[1] “What Do the Kohn-Sham Orbitals and Eigenvalues Mean?”, R. Stowasser and R. Hoffmann, J. Am. Chem. Soc. 1999, 121, 3414–3420.

When people say that Kohn-Sham orbitals bear no physical meaning, they mean it in the sense that nobody has proved mathematically that they mean anything. However, it has been empirically observed that many times, Kohn-Sham orbitals often do look very much like Hartree-Fock orbitals, which do have accepted physical interpretations in molecular orbital theory. In fact, the reference in the OP lends evidence to precisely this latter viewpoint.

To say that orbitals are "good" or "bad" is not really that meaningful in the first place. A basic fact that can be found in any electronic structure textbook is that in theories that use determinantal wavefunctions such as Hartree-Fock theory or Kohn-Sham DFT, the occupied orbitals form an invariant subspace in that any (unitary) rotation can be applied to the collection of occupied orbitals while leaving the overall density matrix unchanged. Since any observable you would care to construct is a functional of the density matrix in SCF theories, this means that individuals orbitals themselves aren't physical observables, and therefore interpretations of any orbitals should always be undertaken with caution.

Even the premise of this question is not quite true. The energies of Kohn-Sham orbitals are known to correspond to ionization energies and electron affinities of the true electronic system due to Janak's theorem, which is the DFT analogue of Koopmans' theorem. It would be exceedingly strange if the eigenvalues were meaningful while their corresponding eigenvectors were completely meaningless.


Why are allyl halides reactive towards both SN1 and Sn2 reactions? I know that allyl halides are reactive towards SN1 reactions due to the stabilization of carbocation due to resonance but why are they reactive towards SN2 too?

In a reaction, SN2 as well as SN1 reactions take place. It only matters that which reaction pathway gives the major product. The answer to your question depends on the type of base used, the substrate and the overall stability of the product. Allyl halides are also reactive to SN2' reactions, i.e allylic nucleophilic substitution reactions. There are a lot of possibilities.

Keep in mind that the electronegativity order of hybridized carbons is $\ce{sp > sp2 > sp3}$. So in an allylic halide, such as $\ce{CH2=CH-CH2Br}$, the alpha carbon is much more partially positive than the alpha carbon in $\ce{CH3-CH2-CH2Br}$. Therefore, the attacking nucleophile will be attracted to the alpha carbon more in the allylic halide compound than in the primary halide. Hence, allylic halides tend to follow SN2 pathways more than primary halides.


Is it true that a pH higher than the pKa of an acid make the H automatically disassociate? Will the hydrogen atom always disassociate if the pH of the solution is higher than the pKa of the acid?

I propose the following  answer:

Let's consider the following equilibrium:

$$ \ce{H}A+\ce{H_2O} \leftrightarrows A^-+\ce{H_3O^+}$$
The equilibrium constant $\ce{K}_A=\dfrac{[A^-][\ce{H_3O^+}]}{[\ce{H}A]}$. So if we take ''$-\log$'' of this equation, we can define the $\ce{pK}_A$ of an acid as: $\ce{pK}_A=\ce{pH}- \log\dfrac{[\ce{A^-}]}{[\ce{H}A]}$.

we rewrite this equation as follows:

$$
\ce{pH}=\ce{pK}_A+\log\dfrac{[A^-]}{[\ce{H}A]}
$$
From this equation, we can deduce that when $\ce{pH}>\ce{pK}_A$, $[A^-]>[\ce{H}A]$,  i.e. the concentration of the conjugated base is bigger than the concentration of the acid. But this doesn't mean that all the acid molecules are dissociated. The bigger the difference between the pH and the $\ce{pK}_A$, the bigger is the percent of dissociated acid molecules.


Can oxygen gas in air react with Mn3O4 to form MnO2? I think $\ce{MnO2}$ at high temperature can be used as a catalyst for $\ce{SO2 + O2 -> SO3}$:

$$
\begin{align}
\ce{MnO2 + SO2 &-> MnSO4}\tag{1}\\
\ce{3 MnSO4 &-> Mn3O4 + SO2 + 2 SO3}\tag{2}\\
\ce{Mn3O4 + O2 &-> 3 MnO2}\tag{3}
\end{align}
$$

I think the first two reactions are possible, but I don't know about the third one.

Manganese(IV) oxide decomposes around 500 °C (German Wikipedia lists $\pu{450 °C},$ [1, p. 396] lists $\pu{530–585 °C}):$
$$\ce{4 MnO2 ->[\sim\pu{500 °C}] 2 Mn2O3 + O2}\label{rxn:1}\tag{R1}$$
On the other hand, according to [2, p. 12], $\ce{Mn3O4}$ reacts with oxygen only above $\pu{500 °C}$ and forms manganese(III) oxide:
$$\ce{4 Mn3O4 + O2 ->[>\pu{500 °C}] 6 Mn2O3}\label{rxn:2}\tag{R2}$$
Upon further heating, the reaction in the opposite direction takes place [1, p. 396]:
$$\ce{6 Mn2O3 ->[>\pu{940–1090 °C}] 4 Mn3O4 + O2}\label{rxn:3}\tag{R3}$$
All in all, the third reaction you've proposed $(\ce{Mn3O4 + O2 -> 3 MnO2})$ is unlikely to occur and the product will be $\ce{Mn2O3},$ which unfortunately would break the catalytic cycle.
As for the first two reactions, they are described in the literature all right [1, pp. 397, 399]:
$$
\begin{align}
\ce{MnO2 + SO2 &->[\pu{450 °C}] MnSO4}\tag{1}\\
\ce{3 MnSO4 &->[\pu{850–1155 °C}] Mn3O4 + SO2 + 2 SO3}\tag{2}\\
\end{align}
$$
Reactions \eqref{rxn:1}–\eqref{rxn:3} are also described in Greenwood's Chemistry of the elements: [3, pp. 1048–1049]:

It is $\ce{MnO2},$ however, which is by far the most important oxide in this group, though it is not the most stable oxide of manganese, decomposing to $\ce{Mn2O3}$ above about $\pu{530 °C}$ and being a useful oxidizing agent. […]
$\ce{Mn3O4}$ is formed when any oxide of manganese is heated to about $\pu{1000 °C}$ in air […].
If $\ce{MnO2}$ is heated less strongly (say $,\pu{800 °C})$ than is required to produce $\ce{Mn3O4},$ then the more stable α-form of $\ce{Mn2O3}$ results […].

References

R. A. Lidin, V. A. Molochko, and L. L. Andreeva, Reactivity of Inorganic Substances, 3rd ed.; Khimia: Moscow, 2000. (in Russian)
Turova, N. Inorganic Chemistry in Tables; Springer: Heidelberg ; New York, 2011. ISBN 978-3-642-20486-9
Greenwood, N. N.; Earnshaw, A. Chemistry of the Elements, 2nd ed.; Butterworth-Heinemann: Oxford; Boston, 1997. ISBN 978-0-7506-3365-9.


Why the bond angle of PH3 is lesser that that of PF3? We can explain why the bond angle of $\ce{NF3}$ (102°29') is lesser than $\ce{NH3}$ (107°48') by the VSEPR theory, since lone pair lone pair repulsion is greater than lone pair bond pair repulsion. Then for $\ce{PH3}$ and $\ce{PF3}$, also, it is expected that the bond angle of $\ce{PF3}$ will be smaller. But I found that the thing is just reverse. Why is this so? Is it an exception?

VSEPR works by accident in the cases of $\ce{NH3}$ and $\ce{NF3}$. In reality, there are much more things to consider as shown in this answer. All four compounds should have bond angles of $90^\circ$ if there were no other effects present.

For both nitrogen compounds, the effects are the short $\ce{N-X}$ bonds which lead to steric clash of the three substituents. Thus, the bond angle is broadened from the original, hypothetical $90^\circ$ to whereever it may seem comfortable to the molecule. Since fluorines are larger atoms than hydrogens are, the $\ce{N-F}$ bond is longer than the $\ce{N-H}$ bond ($137~\mathrm{pm}$ versus $102~\mathrm{pm}$) and therefore the atoms can agree on a smaller bond angle (less crowding).

For phosphorous, we can initially assume that the bond angle of $\ce{PH3}$ is close enough to $90^\circ$ for no hybridisation to be necessary (semi-proven by this answer of Martin) and likewise for $\ce{PF3}$. The bond lengths are $142~\mathrm{pm}$ and $156~\mathrm{pm}$, respectively. The difference in bond lengths is only half of that of the nitrogen compounds ($14$ versus $35~\mathrm{pm}$). Therefore, we can assume the bond length to be mainly dictated by the central phosphorous atom, while the bond angles must be dictated more by the outer atoms. Since again fluorine is larger than hydrogen, this time the better way to stabilise the molecule is not to lengthen the bonds but to expand the angle.


Zwitterion (amino acid) equilibrium and the equilibrium constant I just had my biology test paper back from my teacher, and my teacher insisted that when functioning as a buffer in an alkaline solution, the amino acid deprotonates by the ammonium group instead of the carboxyl group, because amino acids in solution are mostly zwitterionic. I thought zwitterions only consist a small portion of the mixture, but checking Wikipedia only further confirms what my teacher said.

In an attempt to find mathematical proof, I tried to derive the equilibrium constant between zwitterionic and completely neutral forms of amino acids, with the assumption that the solution is pure and only constitutes the said acid(and its zwitterionic isomer) and water. This was my working:

Using $K_\mathrm{a}$ for carboxyl group and $K_\mathrm{b}$ for amine group.
$$\ce{H2N-CHR-COOH <=> H3N+ -CHR-COO-}$$
\begin{align}
[\ce{H2N-CHR-COOH}]   &=  [\ce{H2N\bond{-}}] = [\ce{-COOH}]\\
[\ce{H3N+ -CHR-COO-}] &=  [\ce{H3N+\bond{-}}] = [\ce{-COO^-}]
\end{align}

\begin{align}
K_c &= \frac{[\ce{H3N+ -CHR-COO-}]}{[\ce{H2N-CHR-COOH}]}\\
    &=\sqrt{\frac{[\ce{H3N+-CHR-COO^-}][\ce{H3N+-CHR-COO^-}]}{[\ce{H2N-CHR-COOH}][\ce{H2N-CHR-COOH}]}}\\
    &=\sqrt{\frac{[\ce{H3N+\bond{-}}][\ce{-COO^-}]}{[\ce{H2N\bond{-}}] [\ce{-COOH}]}}\\
    &=\sqrt{\frac{[\ce{H3N+\bond{-}}]}{[\ce{H2N-}][\ce{H+}]} \times
      \frac{[\ce{-COO^-}][\ce{H+}]}{[\ce{-COOH}]}}\\
    &=\sqrt{\frac{[\ce{H3N+\bond{-}}][\ce{OH-}]}{[\ce{H2N\bond{-}}][\ce{H+}][\ce{OH-}]} \times \frac{[\ce{-COO^-}][\ce{H+}]}{[\ce{-COOH}]}}\\
&\text{(now assuming $[\ce{H2N\bond{-}}] = [\ce{H2O.H2N\bond{-}}]$)}\\
    &=\sqrt{\frac{[\ce{H3N+\bond{-}}][\ce{OH-}]}{[\ce{H2O.H2N\bond{-}}][\ce{H+}][\ce{OH-}]} \times \frac{[\ce{-COO^-}][\ce{H+}]}{[\ce{-COOH}]}}\\
    &=\sqrt{\frac{K_\mathrm{b}}{K_\mathrm{w}} \times K_\mathrm{a}}\\
    &=\sqrt{\frac{K_\mathrm{a} K_\mathrm{b}}{K_\mathrm{w}}}
\end{align}
With $K_\mathrm{w}$ being the denominator, it is easy to see that $K_c$ would be sufficiently large that nearly all amino acids in aqueous solution exists in the zwitterionic form, that is contrary to my initial belief; bringing in the $K_\mathrm{a}$ and $K_\mathrm{b}$ values further confirms that.

However, I am not content with just that. I'd like to ask:

What are the faults (calculation errors, false/inappropriate assumptions, etc) in my working, if any?
What is the main reaction mechanism that mediates this equilibrium? Is it an intramolecular transition or is it brought about by the solvent, or something else?
Can this be considered as a form of tautomerization? If so, is it prototropy? Hence, is the $K_c$ above the so-called tautomeric ratio?

If you assume that the groups react independently, you have two equilibria with equilibrium constants $K_a(\ce{-NH3+})$ and $K_a(\ce{-COOH})$:

$$\ce{^+(H3N)-R <=> (H2N)-R + H+}$$

and

$$\ce{R-COOH <=> R-COO- + H+}$$

You are interested in the overall reaction

$$\ce{^+(H3N)-CHR-COO- <=> (H2N)-CHR-COOH}.$$

You can take a two step route to achieve this:

$$\ce{^+(H3N)-CHR-COO- <=> (H2N)-CHR-COO- <=> (H2N)-CHR-COOH}$$

The equilibrium constant for the first step is $K_a(\ce{-NH3+})$, and for the second step is $1/K_a(\ce{-COOH})$. Combined, you get a result of

$$ K_\text{combined} = \frac{K_a(\ce{-NH3+})}{K_a(\ce{-COOH})}$$

Because the $K_a$ values of ammonium and carboxylic acid are different by about four orders of magnitude, the uncharged species occurs about ten thousand times less often than the zwitter ionic form in aqueous solution. In general, comparing p$K_a$ values of two groups competing for a proton tells you which group will usually have the proton (the one with the larger p$K_a$) and how unlikely a proton transfer to the other group would be ($10^{\Delta pK_a}$).

  What are the faults (calculation errors, false/inappropriate assumptions, etc) in my working, if any?

I don't understand line 2 and 3 of your argument, and how you get from line 5 to 6. The result you get differs from mine in a square root.

  I thought zwitterions only consist a small portion of the mixture

This depends on the solvent. In aqueous solution, zwitterions are the common species, whereas in organic solvent, the uncharged species is the common one.


What exactly is causing analyte protonation in electrospray ionization MS? I commonly see explanations that the presence of acid in the mobile phase is a source of protons for the analyte, and that you should be multiple pH units below the pKa of the analyte, but in practice plenty of compounds that you end up detecting have pKa's that are well below the pH of any practical mobile phase or are not even protonatable - so how are these molecules being protonated?

[ This is a short answer, as a complete answer would require going through multiple references and books on the topic. ]
First of all, the ESI process does not require the pH to be acidic : positive ESI is routinely done in ammonium acetate or ammonium bicarbonate media, which have a pH of 7.4 or 8.1. But indeed, to produce protonated analytes it indeed requires a source of protons. So what is the source of protons?
One should go back to the initial stage of ESI, which is the production of positively (or negatively) charged droplets under the influence of a strong electric field. Let's assume that we focus on positive ESI and that the droplets carry an overall positive charge. These charges are often simply represented as + signs, but their molecular nature is important. In the case were an electrolyte is present in high concentration and in its ionic form, the major charge carrier will be the positive charge of this electrolyte. So when considering an organic acid at low pH, the positive charges will be mostly solvated protons. If the electrolyte was ammonium acetate at neutral pH, it would be ammonium ions. Even at higher pH, if the electrolyte is ammonium hydroxide, the droplets would still bear a net positive charge borne by an excess of ammonium ions.
Then come the evaporation / Coulombic fission stages whereby smaller droplets are formed from the larger one, with an overall concentration step. The concentration effect of this stage will affect the overall pH of the droplet if initially acidic, but it is not required to lead to protonated species. The important point being that within the droplets, there is still an excess of positive charges, mostly in the form of solvated proton. At some stage, as concentrations of both the analyte and the solvated proton increase, the proton will be placed in close proximity with the analyte (both not being volatile do not evaporate). At this stage, the system is no more in the bulk solution thermodynamic conditions, but not really either in the gas phase thermodynamic conditions.
But we can still at this stage approximate the situation as "in the middle" by considering the thermodynamic parameters for both ends: on the bulk side, we have pKa of the analyte, with decreasing pH, and on the gas phase side, we have the proton affinity (PA) of the analyte that should be compared to the PA of the other species competing for the proton. The two scales are related but not strictly: for instance water has a quite low proton affinity (692 kJ/mole) compared to most polar molecules (ketones are in the 800 kJ/mole range). Thus is the proton located on the analyte is more stable than on molecules from the mobile phase, it will end up protonating the molecule, without considderation of the pKa.


Is rearrangement possible in cyclic bromonium ion? I wonder what the product of the above reaction is, that is, bromination of the given alkene in non polar solvent carbon tetrachloride (proceeds through cyclic bromonium ion intermediate).

Once the cyclic bromonium ion is formed, can the ring expand as shown or is the usual product formed? Five remembered rings happily expand to six membered ones if possible - so what's going on here?

It'd be great if you could quote sources with your answer - and help me understand whether or not species involving cyclic bromonium ions can rearrange in a reaction, and why?

Thanks!

This is apparently not a likely rearrangement, even in more strained circumstances. There are a number of reactions out there in the literature like the one below.

This result makes sense if we consider the relative stability of the bromonium ion vs. your proposed rearrangement. I've redrawn the mechanism to show all the formal charges. The bromonium should be much more stable (positive charge delocalized over two carbon atoms, all atoms have full octet) compared to a secondary carbocation.

Bromonium rearrangements are not entirely impossible though, check out this fun example (J. Org. Chem., 1998, 63, 2646 DOI: 10.1021/jo9722055)


Why does reducing the pressure in an equilibrium reaction favour the side with more moles of gas? I was trying to explain it in terms of reaction rates and such; more than just 'Le Chatelier's principle states ...', but I can't figure out the underlying reason as to why. I can explain why for increasing the pressure, temperature changes and such, but not for this specific one. If there is more space for the particles to move without collided, this reduces the frequency of successful collisions, reducing the reaction rate of both forward and reverse reactions. But, in $$\ce{A + B<=>3C}$$
there is initially more moles of C than A and B combined. So, isn't it reasonable to think C will collide more frequently than A and B, meaning the reaction rate of the reverse is quicker, rather than the reaction rate of the forward being quicker?

For your reaction, $\Delta n_g = 3-(1+1)=1$ (assuming all A,B,C are gases). $K_p = \frac{[C]^3}{[A][B]} (\frac{P_{total}}{\Sigma n})^{\Delta n_g}$, where $\Sigma n$ is the total number of moles of gases in the reaction mixture(even of those species which are not taking part in the reaction). Here, it is a constant.

When pressure is decreased($P_{total}$), $ (\frac{P_{total}}{\Sigma n})^{\Delta n_g}$ decreases as $\Delta n_g >0$.

In order to keep $K_p$ constant (it is always constant at a given temperature), $[C]$ has to increase or $[A],[B]$ has to decrease. Therefore the reaction proceeds/shifts more in the forward direction, which has more number of gaseous moles.

Similarly, increase in pressure favours the side with lesser number of gaseous moles.