How do the three Miller indices (hkl) denote planes orthogonal to the reciprocal lattice vector? The Wikipedia page for Miller indices defines Miller indices as follows:

There are two equivalent ways to define the meaning of the Miller indices: via a point in the reciprocal lattice, or as the inverse intercepts along the lattice vectors. Both definitions are given below. In either case, one needs to choose the three lattice vectors $\mathbf{a}_1$, $\mathbf{a}_2$, and $\mathbf{a}_3$ that define the unit cell (note that the conventional unit cell may be larger than the primitive cell of the Bravais lattice, as the examples below illustrate). Given these, the three primitive reciprocal lattice vectors are also determined (denoted $\mathbf{b}_1$, $\mathbf{b}_2$, and $\mathbf{b}_3$).
Then, given the three Miller indices $h, k, \ell, (hk\ell)$ denotes planes orthogonal to the reciprocal lattice vector:
$$\mathbf{g}_{hk\ell} = h \mathbf{b}_1 + k \mathbf{b}_2 + \ell \mathbf{b}_3.$$
That is, ($hk\ell$) simply indicates a normal to the planes in the basis of the primitive reciprocal lattice vectors. Because the coordinates are integers, this normal is itself always a reciprocal lattice vector. The requirement of lowest terms means that it is the shortest reciprocal lattice vector in the given direction.

This idea of the three Miller indices ($hk\ell$) denoting planes orthogonal to the reciprocal lattice vector is something that I'm not confident that I'm understanding. My understanding is that the real (real space) lattice vector is described by the Miller indices as $h{\mathbf  {a_{1}}}+k{\mathbf  {a_{2}}}+\ell {\mathbf  {a_{3}}}$, where the $\mathbf{a}_i$ are linearly independent basis vectors. However, if I am understanding this correctly, the Miller indices are actually not (directly) defined in terms of the real lattice vector, but are actually defined directly in terms of the reciprocal lattice vector $h{\mathbf  {b_{1}}}+k{\mathbf  {b_{2}}}+\ell {\mathbf  {b_{3}}}$, where ${\mathbf  {b_{i}}}$ are the basis of the reciprocal lattice vectors; specifically, the Miller indices are defined to denote the family of planes orthogonal to $h{\mathbf  {b_{1}}}+k{\mathbf  {b_{2}}}+\ell {\mathbf  {b_{3}}}$. This is evidenced by the fact that said orthogonal plane is not always orthogonal to the linear combination of real lattice vectors $h{\mathbf  {a_{1}}}+k{\mathbf  {a_{2}}}+\ell {\mathbf  {a_{3}}}$ (see image below), since the reciprocal lattice vectors need not be mutually orthogonal (that is, although the $\mathbf{b}_i$ are basis vectors, they are not necessarily linearly independent, which I think is mathematically correct (?)).

(By Felix King, and posted to https://en.wikipedia.org/wiki/Miller_index)
I would appreciate it if people would please take the time to review everything that I've written so far. Is my understanding correct? Am I misunderstanding anything?
The last point that I'm having a bit of trouble with is how a plane can be defined in such a way that it is orthogonal to three basis vectors? I suspect that this is possible if and only if the basis vectors are not all linearly independent, as I alluded to in my explanation above, but I'm not sure. I think this is more of a mathematical question, so I will instead post it to math.stackexchange for a more mathematics-focused answer, but I would also appreciate it if people could please review my understanding of this and provide clarification.

There are two equivalent ways to define the meaning of the Miller indices: via a point in the reciprocal lattice, or as the inverse intercepts along the lattice vectors.

The reflecting plane are parallel to a plane that includes the following three points:

$$\frac{\mathbf{a}_1}{h}; \frac{\mathbf{a}_2}{k}; \frac{\mathbf{a}_3}{l}$$

To show that the plane defined by these three points is perpendicular to $h \mathbf{b}_1 + k \mathbf{b}_2 + l\mathbf{b}_3$, we can calculate the dot product between vectors in the plane and the supposedly perpendicular reciprocal lattice vector. To get vectors in the plane, we simply subtract pairs of points in the plane.

$$\frac{\mathbf{a}_1}{h} - \frac{\mathbf{a}_2}{k}$$
$$\frac{\mathbf{a}_1}{h} - \frac{\mathbf{a}_3}{l}$$

Now we multiply it with the reciprocal lattice vector:

$$(\frac{\mathbf{a}_1}{h} - \frac{\mathbf{a}_2}{k}) (h \mathbf{b}_1 + k \mathbf{b}_2 + l\mathbf{b}_3) = \mathbf{a}_1 \mathbf{b}_1 - \mathbf{a}_2 \mathbf{b}_2 = 1 - 1 = 0$$
$$(\frac{\mathbf{a}_1}{h} - \frac{\mathbf{a}_3}{l}) (h \mathbf{b}_1 + k \mathbf{b}_2 + l\mathbf{b}_3) = \mathbf{a}_1 \mathbf{b}_1 - \mathbf{a}_3 \mathbf{b}_3 = 1 - 1 = 0$$

The mixed terms vanish because they are orthogonal. Because the dot products are equal to zero, we have shown that the vector is perpendicular to the reflecting plane.

  [OP] My understanding is that the real (real space) lattice vector is described by the Miller indices as $h{\mathbf  {a_{1}}}+k{\mathbf  {a_{2}}}+\ell {\mathbf  {a_{3}}}$, where the $\mathbf{a}_i$ are linearly independent basis vectors.

No, the significance of $h, k, l$ in defining reflecting planes in terms of the unit cell vectors is as described in the textbook (inverse intercepts along the lattice vectors).

  [OP] The last point that I'm having a bit of trouble with is how a plane can be defined in such a way that it is orthogonal to three basis vectors

That is not how it is defined. It is defined by the intersections of a plane with the three axes, i.e. through three points that are in the plane.


Why are molecular reactions endo- or exo-thermic? It seems strange to me that chemical reactions should be exothermic, meaning the molecules move faster after the reaction.

Normally, in physics when two moving objects collide and stick together, the resultant velocity of the combined object is less than that of the constituent objects, due to conservation of momentum.

Yet, when some molecules combine they move faster. An extreme example of this are fast reactions like explosives in which the chemicals combine and move off at high speeds, sometimes greater than the speed of sound (a "detonation").

Why would this be? Is there a theory that positively can predict this behavior? In other words, if I were to supply you the names of two chemical species of which you had no prior knowledge, could you predict ahead of time whether their reaction would be exothermic or not, and by how much? Or, can this only be determined experimentally?

You are thinking it in a pseudo classical way. In those terms, imagine bonds like some kind of springs with some potential energy. During the reaction there can be rupture or formation of springs, changing the potential energy. In consequence, due to energy conservation kinetic energy must change.

A pictorial argument, that may help comparing your view point about collisions and the chemical reaction case, that is not always true, can be go like this: If we think of two balls bonded by an ordinary spring, in the "relaxed" position (that occurs somewhere between max and min elongation) we have the minimum potential energy. This is also true for a diatomic molecule (look some graph of potential energy curve for a diatomic molecule). Imagine that two of this balls bonded by the spring collide in the region of the (ideal, non mass, non perturbed by the collision etc.) spring and cut them out:

If you make this collision the speed of each ball will remain the same due to the spring is mass less. If molecules were like the balls no change in temperature would occurs, because kinetic energy does not change.

The key point is that molecules does not behave like that, spring never cut so: potential energy always is present. So if they where molecules, as products the separation between balls is larger and so the potential energy is greater and the kinetic energy turns to be smaller.

Now, think in the backward reaction, this reaction would be an exothermic reaction.

I am not sure if that is the kind of answer you need. If not or if something is unclear leave a comment.

Edit
I forgot answer that last questions. Yes, the temperature change can be calculated. Normally what is calculated is the enthalpy change, but they can be related trough heat capacity. Of course there are limitations in precision and size of the molecule, but in essence the answer is yes, it can be calculated following the quantum mechanics principles.


How well (or poorly) does natural leather insulate? What is the general thermal conductivity of animal skin leather?

I understand there are many different animal's skins (cow, sheep, buffalo, deer, pig), but in general, does natural leather insulate well (compared to, for scale, an esky for food insulation), or not?

Interestingley enough, there are papers entirely devoted to the subject of measuring the properties of living skin. The paper linked gives a value of $7 \times 10^{-4}~\mathrm{\frac{cal}{cm\, s\, °C}}$ for human skin, or in everyday units used nowadays: $0.29 ~\mathrm{\frac{W}{m\, K}}$. This is only for the dermis, the part of skin that then gets transformed into leather.

Now, this website has a comprehensive list of many different thermal conductivity values, and dry leather clocks in at $0.14 ~\mathrm{\frac{W}{m\, K}}$, which is about 2 times less the conductivity than that of living skin. What a difference the tanning process can make!

Given the similarity  observed in the review paper between man, beef and porcine tissue, I think it is fair to say that the value for leather will not change greatly across species.


How can I obtain crystal-clear glycerin? I'm trying to get perfect crystal clear glycerin for my experiments with decorative soap. Is there a special way to get it or should I buy a special formula of it? I have not succeeded in purifying the glycerin I have to more than a milky-translucent, semi-transparent soap.

You can buy high-purity glycerin from numerous vendors online. High-purity grades will be as crystal-clear as you can possibly get.

For instance, check out the premium grade glycerin product from Sigma-Aldrich which should be suitable for your needs but is quite expensive.

There are less expensive (but pure enough for your needs) options, so be sure to explore the other grades available from other vendors. For example, you can secure a gallon of Kosher USP food grade vegetable-derived glycerin for $14.99 from Amazon. This product (or similar) is adequate for your needs at a better price point that lab-grade reagents.


Reaction energy of oxygen and hydrogen From the bonding energy of following bonds that I found in a Wikipedia article, I calculated the reaction energy that would be released when oxygen and hydrogen are reacted to produce water. The phase of each component is unknown.

$$
\begin{array}{lc}
\hline
\text{Bond} & D_0/\pu{kJ mol-1} \\
\hline
\ce{H-H} & 436 \\
\ce{O-H} & 497 \\
\ce{O=O} & 498 \\
\hline
\end{array}
$$

Therefore,

$$\ce{2 H2 + O2 -> 2 H2O} + \pu{618 kJ}$$

However, this Wikipedia article suggests that $\pu{241.8 kJ}$ of energy is released for every mole of hydrogen, which is $\pu{483.6 kJ}$ per $\pu{2 mol}$ for the context of the equation above.

Even when I consider the vaporization of water, which is $\pu{40.65 kJ mol-1}$, the total energy will amount to $\pu{536.7 kJ}$ per $\pu{2 mol}$, which is still far away from $\pu{483.6 kJ}$ per $\pu{2 mol}$ that is suggested in the Wikipedia article.

Where could this extra energy be going?

Part of the discrepancy appears to be due to the need for two $\ce{O-H}$ dissociation energy terms to compute the $\ce{OH}$ bond energy for water: one for the $\ce{OH}$ bond in water (i.e. $\ce{H-OH}$), and another for the $\ce{OH}$ bond in the $\ce{OH}$ radical (i.e. $\ce{^.O-H}$). Values for these are reported by Ruscic et al. [1]. Using the average value of $\ce{D_0}$ for $\ce{H2O}$ (the $\ce{OH}$ bond energy for water), namely $\pu{459 kJ/mol}$, gives

$$\ce{2H2 + O2 -> 2H2O} \quad \Delta H = \pu{-466 kJ/mol}$$

This point is addressed in the Wikipedia page for "bond dissociation energy", the reference cited therein providing a slightly different bond energy for water of $\pu{461.5 kJ/mol}$.

Given various uncertainties common to the measurement of bond dissociation energies, and the potential effect of temperature on the bond energy, the final discrepancy seems reasonable.

References

Ruscic, B.; Wagner, A. F.; Harding, L. B.; Asher, R. L.; Feller, D.; Dixon, D. A.; Peterson, K. A.; Song, Y.; Qian, X.; Ng, C.-Y.; et al. On the Enthalpy of Formation of Hydroxyl Radical and Gas-Phase Bond Dissociation Energies of Water and Hydroxyl. The Journal of Physical Chemistry A 2002, 106 (11), 2727–2747. https://doi.org/10.1021/jp013909s.


Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot? Does Bent's rule have any utility? I got a vehement earful for "NO." Points raised by my professor:

Coulombic considerations can be used to rationalize bond angles, strengths, and lengths without the use of Bent's rule.
Orbitals are not real so why invoke orbitals when simpler Coulombic considerations are available? (Occam's razor)
The Coulombic effect is also real as opposed to orbitals.

Consider carbon tetrachloride and chloromethane.

Obviously, carbon tetrachloride has the stronger and shorter carbon-halide bond because in chloromethane, the methyl group is slightly electron donating, and this repels the terminal chlorine, which lengthens the carbon-halide bond. No application of Bent's rule is necessary.

Sure, you can justify this difference with Bent's rule; s-character is concentrated in orbitals directed toward more electropositive substituents; the carbon is more electropositive than chlorine; thus, the bond has more s-character around the chlorine and more p-character around the carbon in chloromethane. This greater p-character around the carbon in chloromethane weakens the carbon-halide bond relative to the bond in carbon tetrachloride. And greater p-character in a bond is also correlated with a compression in bond angles.

However, bond angles are also explained nicely through Coulombic effects. The C-H bonds in chloromethane are obviously shorter due to the electron-withdrawing nature of chlorine and the carbon-chlorine bond is obviously expanded due to methyl group/chlorine lone pair repulsion. This shortening of the C-H bond lengths explains the H-C-H bond angle expansion in chloromethane, and the lengthening of the C-X bond explains the X-C-H bond angle expansion.

Is there any case in which other elementary considerations (Coulombic, VSEPR, van der Waals repulsions) fall short in the realm of molecular geometry and Bent's rule can step in to pick up the slack? Anyone got some sort of vindication for Bent's rule?

There is a reason for everything.

  Does Bent's rule have any utility?

YES! It wouldn't be there if there was not. But I will get back to this at the end of this post.

Let me go through the points raised by your teacher first:

  Coulombic considerations can be used to rationalize bond angles, strengths, and lengths without the use of Bent's rule.

This might be true to some extent, but that only means that coulombic considerations give the same result as a complete orbital analysis. Boltzmann, Planck, Einstein, Schrödinger, Pauling, etc. worked the field for a reason - classical mechanics cannot be applied to molecular systems. Any theory that simplifies it beyond quantum mechanics, is a model to understand or comprehend - not a model to predict.

Orbitals are not real so why invoke orbitals when simpler Coulombic considerations are available? (Occam's razor)

Occam's razor would suggest picking molecular orbital or valence bond theory, as it states that from competing hypotheses the one with the fewest assumptions should be selected. Coulombic consideration is purely based on assumptions, it is a rather crude approximation to the correct picture.
The statement, that orbitals are not real is clearly wrong. There are many books that explain how one can comprehend this. Teaching that orbitals are not real is, with all due respect, just plain wrong.

The Coulombic effect is also real as opposed to orbitals.

The coulombic effect is not more real than orbitals. Orbitals should be used to explain the coulombic effect.

But let's talk about Bent's rule. The IUPAC goldbook states:

  In a molecule, smaller bond angles are formed between electronegative ligands since the central atom, to which the ligands are attached, tends to direct bonding hybrid orbitals of greater p character towards its more electronegative substituents.

It is important to know, that Bent's rule is, like Lewis rule of two (or eight, known as octet rule), a rule (not a law) based on observation. Hence its character is more or less one of a guideline. The key paper was published in Henry A. Bent, Chem. Rev., 1961, 61 (3), 275–311.

Bent's rule originates from valence bond theory (VB). Early forms of VB theory proposed that a molecule can entirely be described by a set (linear combination) of resonance Lewis structures, i.e. every bonding orbital is a directed localised orbital, which will be occupied by two electrons and the same applies to lone pairs. The advantage of this approach is that it is easily comprehensible. The disadvantage is, that it is quite inflexible. The reason for this was the use of either pure $\ce{s}$ and $\ce{p}$ functions or fixed hybrid orbitals $\ce{sp^3,~sp^2,~sp}$, which only allowed a subset of angles. A direct quote from Bent's original paper states this as follows:

  For atoms that satisfy the octet rule, and it is with such atoms (e.g., carbon, nitrogen, oxygen, fluorine) that this review is primarily concerned, the four valence-shell orbitals of an atom may be thought of as compounded from one spherically symmetrical $\ce{2s}$ atomic orbital and three mutually perpendicular dumbbell-shaped $\ce{2p}$ orbitals, designated $\ce{2p_$x$, 2p_$y$, 2p_$z$}$, ($x, y, z$ indicate mutually orthogonal axes). The $\ce{2s}$ orbital is lower in energy than the $\ce{2p}$ orbital; hence the ground state of a carbon atom, for example, is written $\ce{ls^2~2s^2~2p^2}$ (not $\ce{ls^2, 2s~2p^3}$. However, in a molecule such as methane, which is known to have the shape of a regular tetrahedron, the $\ce{2s}$ orbital of carbon may be considered to mix (or “hybridize”) with the three $\ce{2p}$ orbitals to form four equivalent orbitals that point to the corners of a regular tetrahedron and make angles with each other of $109^\circ28’$. These are called hybrid orbitals. Each contains one-quarter $\ce{s}$ character and three-quarters $\ce{p}$ character, abbreviated $\ce{s^{$1/4$}p^{$3/4$}}$ or $\ce{sp^3}$, or $\ce{te}$ (for tetrahedral) (223). In ethylene, on the other hand, the $\ce{2s}$ orbital of carbon may be considered to mix with but two of the three $\ce{2p}$ orbitals to form three equivalent hybrid orbitals that lie in a plane and make angles with each other of $120^\circ$; these three hybrids, which contain 33 per cent $\ce{s}$ character, are abbreviated $\ce{s^{$l/2$}p^{$2/3$}}$ or $\ce{sp^2}$, or $\ce{tr}$ (for trigonal); perpendicular to the plane of these three orbitals is the axis of cylindrical symmetry of the remaining pure $\ce{2p}$ orbital. The geometry of acetylene is often explained by assuming that the carbon $\ce{2s}$ orbital mixes with but one of the three $\ce{2p}$ orbitals, forming two $\ce{di}$ (for digonal) hybrids that point in diametrically opposed directions; these are abbreviated $\ce{s^{$1/2$}p^{$1/2$}}$ or $\ce{sp}$.

The original approach allowed already good results. However, Bent already draw the conclusion, that hybridisation is connected to molecular geometry:

  Molecular geometry provides then a direct clue to interorbital angles and
  (table 1) the distribution of $\ce{s}$ character. For example, from the bond angles in ammonia ($106^\circ46'$) (284) and water ($104^\circ27'$) (134, 284), it is inferred that the nitrogen atom of $\ce{NH}$, devotes slightly more $\ce{s}$ character to its bonding orbitals than does the oxygen atom of $\ce{H2O}$.

In conclusion to this (unfortenuately I ran out of time, so this may be extended soon) he suggested to use fractional linear combinations of hybrid orbitals to add more flexibility to the VB approach. Therefore Bent's rule is an extension to VB theory, that refers to the linear combination of atomic orbitals, that eventually form hybrid orbitals.

The formulated rule was later merely a result of the approach to flexibilize this theory. It is therefore a very important part of theoretical chemistry.

Of course there are now many more modifications to VB theory, and it is nowadays considered complementary to molecular orbital theory, but this would certainly exceed the scope of this question.


Why are there only 6 series (Lymen, Balmer, Paschen, Bracket, Pfund and H.Davy) in the hydrogen atomic spectrum? Why can't there be more? The Balmer series, for example, is formed when the electron jumps from $n_2 = 3, 4, 5, \ldots, ∞$ to $n_1 = 2.$

The Humphry Davy series (The last series is formed when electron jumps from $n_2 = 7, 8, 9, \ldots, ∞$ to $n_1 = 6.$

Can't there be another series formed when electron jumps from $n_2 = 8,9,10... \ldots, ∞$ to $n_1 = 7?$

As Ivan commented, there are actually an infinite number of possible series of this type. So your question is really why there are only six named series.  The reason is part of the culture of science.  Typically, a result is named if it is, to use Mithoron's term, sufficiently notable.  Thus we have the Diels-Alder reaction, the Hartree–Fock method, the Wentzel–Kramers–Brillouin (WKB) approximation, the Southern blot, and so on.

Hence the reason why the initial series of hydrogen spectra were named is because they were notable scientific results. [In addition—and perhaps someone more expert in the history of science can comment on this—it is possible that identifying the, say, fifth and sixth series weren't that scientifically notable, but a tradition had become established of naming successive spectra of hydrogen.] But beyond the sixth, finding more series was essentially deemed routine. The last published series I know of is the seventh, identified by Hansen and Strong, but not named after them [Peter Hansen and John Strong, "Seventh Series of Atomic Hydrogen," Appl. Opt. 12, 429-430 (1973)].

That's not to say it wouldn't be experimentally difficult to see higher-numbered series, since they become increasingly faint. Rather, it's that, at this point, finding them isn't interesting science.


If a sugar is nonreducing, does that mean it doesn't ionize in water? I'm thinking of sucrose, which I learned is nonreducing, what I take to mean it doesn't accept/take electrons from the surrounding solution. Does this also mean that it doesn't ionize and thus has a Van't Hoff factor ( i ) of 1?

I believe the confusion lies with the term Reducing sugar. From my perspective sucrose is not an ionic compound, it is covalently bonded. I believe that this should indicate to you that it does not "ionize" or the term that I was taught "dissociate" since it has nothing to dissociate from.

To quote wikipedia:

  "The van 't Hoff factor is the ratio between the actual concentration
  of particles produced when the substance is dissolved, and the
  concentration of a substance as calculated from its mass."

Since it does not "dissociate," I would say that the answer is no, it does not "ionize," and therefore has a Van't Hoff factor of 1 when mixed with pure water.

Like anything with electrons, they can be lost. This takes place in our bodies through more complicated reactions than I understand completely. I believe however, according to wikipedia, that sucrose has a Glycosidic bond which when broken, will separate sucrose into smaller sugars, if this takes place in your solution then I believe it will have a Van't Hoff factor > 1.


Calculating temperature change of helium due to Joule-Thomson effect We are trying to fill a rocket helium tank that is 78 liters with helium, final pressure of $\pu{10500 psi}$. We plan to pressurize high pressure tanks to $\pu{20000 psi}$ using a compressing pump system, so that on launch day we don't have to wait for the 78 liter tank to fill slowly by the compressing system. We plan to fill the rocket tanks from the high pressure tanks in a cascade manner (the number and volume of the tanks may vary, depending on the solution of the JT equation(s).

The problem is that helium has a negative JT coefficient. It will heat up during the transfer from the high pressure tanks to the lower pressure tank if I understand it correctly.

So, the concern is how hot will it get. Will the O-rings/seals melt? I have some equations, but it seems to me there may be too many unknowns.

I've seen others make certain assumptions on this site, but its not clear to me how that works. Can anyone advise how you would proceed?

To do this analysis, one needs data up to 20000 psi (1400 bars) on either enthalpy vs temperature and pressure (a graph) or compressibility factor Z vs temperature and pressure.  The only graphs I have found of the former type go up to only 100 bars, which is a factor of 14 too low.  However, I have found data on the compressibility factor Z at temperatures up to 300 K and pressures up to 1000 bars:  https://cds.cern.ch/record/1444601/files/978-1-4419-9979-5_BookBackMatter.pdf
Although this is still a factor of 1.4 too low, it might provide some idea of the temperature rise that might be expected in the valve.

So here is how the data would be used.

The effect of pressure on enthalpy (per mole) of gas is given by $$\left(\frac{\partial H}{\partial P}\right)_T=V-T\left(\frac{\partial V}{\partial T}\right)_P\tag{1}$$For a real gas, the equation of state in terms of the compressibility factor Z=Z(P,T) is given by$$PV=ZRT\tag{2}$$If we substitute Eqn. 2 into Eqn. 1, we obtain:  $$\left(\frac{\partial H}{\partial P}\right)_T=-\frac{RT^2}{P}\left(\frac{\partial Z}{\partial T}\right)_P\tag{3}$$Integrating Eqn. 3 between P=0 and arbitrary P at constant temperature yields the so-called Residual Enthalpy $H^R$:$$H^R(P,T)=-RT^2\int_0^P{\left(\frac{\partial Z}{\partial T}\right)_{P'}\frac{dP'}{P'}}=-RT^2\frac{\partial}{\partial T}\left(\int_0^P{(Z(T,P')-1)\frac{dP'}{P'}}\right)\tag{4}$$where P' is a dummy variable of integration.

If the final pressure coming out of the valve is low (so that the gas exiting the valve is in the ideal gas region), we can write:  $$\Delta H=-H^R+C_p\Delta T=0$$where, for a monoatomic gas like Helium, $C_p=\frac{5}{2}R$.  Therefore, $$\Delta T=-\frac{2}{5}T^2\frac{\partial}{\partial T}\left(\int_0^P{(Z(T,P')-1)\frac{dP'}{P'}}\right)\tag{5}$$
This expression would be evaluated using the data presented in the reference above.

If a reference can be found with data going out to 1400 bar, that would be even better.


Hydrogen displaces lead? there was a reaction in my book. It is as follows:

$\ce{PbO + H2-> Pb + H2O}$

My question is Hydrogen is below lead in the reactivity series. Then how is this reaction possible?

The activity series you're referring to is constructed for ambient conditions where the temperature is $25°\text{C}$ and water is a liquid.  But, of course, thermodynamic properties can depend on temperature and especially on phase.  At higher temperatures where water is ordinarily a gas, it has a lot more entropy than the liquid.  That makes oxidation of hydrogen to water more thermodynamically favorable than what appears in your ambient temperature series.

For high temperature reactions where water mixes with hydrogen in the gas phase, the entropy effect pushes hydrogen up the series so it roughly matches iron.


How many protons does phenolphthalein lose in basic solutions? So, I've looked it up and Wikipedia claims that two protons are released by Phenolphthalein when pH is over 8.2, but I've seen other sources that claim only one proton is lost. So, is it really just one or two? Which form is responsible for  its colour? (I would expect a fully deprotonated PhPh in more basic solutions, so I guess the PhPh that released two protons is the one responsible for the solution's colour)

ChemSpider phenolphthalein's $\mathrm{p}K_\mathrm{a}$ value: 9.7 (at room temperature). Here they've is only given one $\mathrm{p}K_\mathrm{a}$ value, but it is still reasonable to assume that there are other forms in other pH conditions, like in this image:

(source: Wikipedia)

In other sources like academic ones, it is always shown with 2 forms, and these are associated with the supposedly (especially back titration from pink to colourless is very hard to catch) sharp color change in pH 8-10 range. If one is to use an indicator for that pH region, phenolphthalein is chosen but not for the other regions so no one is interested in possible other forms.

It is also understandable if people actually don't even know the other forms, or try to use them also since pH meter(also possibly UV-Vis spectrometry) is used for a long time to track the pH, I have only seen these indicators in student laboratories for teaching purposes, not in any of research laboratory.


Hybridization in ethene I recently learned that there is such thing as a hybridization in chemistry, but I don't really get it.

For example, in $\ce{C2H2}$ (Ethene), they say the there are 3 electrons in three ${sp^2}$ and one electron in a normal $p$ orbital. How do we decide what orbitals get hybridized and what doesn't?

We do not decide either which. Hybridization is an observational theory - it is a theory crafted to fit observation, it is not made from the bottom up. That doesn't make it any less valuable (well... maybe it does...) but your question becomes rather easy to answer. We decide which orbitals become hybridized based on how we measure the molecule to be. We see the symmetry.

Hybridization is just a model. But then again so are most things we learn in chemistry. I would say you are not missing out on anything if you just skip learning it and go directly to molecular orbital theory. Your teacher might argue though, do what (s)he says.


What are disadvantages of having an impurity? I thought about this while thinking about alloys and why they are better than pure metals, even though they are technically impurities. It got me wondering, are there actually any good real-life applications where having an impure substance is completely disadvantageous?

Impurities are the UNWANTED substances that are present in the substance..
There are Impurities in most of stuff including a lot of chemicals and it's very hard to remove them too
However If impurity exceeds the limit its completely Dangerous..
Like for example :

  Surface water (river or lake)
  Surface water can be exposed to acid rain, pesticide runoff, water runoff, and industrial waste.

  Groundwater (private wells or public water supplies)
  Groundwater can be contaminated by disease producing pathogens, careless disposal of household wastes, and leaching from landfills and septic tanks.

which can lead to Nausea
Lung irritation,Vomiting, Dizziness, Skin rashes and even Death

For the Chemistry and Chemical Industry impurities might slow down the reaction or may lead to formation of unwanted bi products , Might affect the rate of Crystallization - Most common issue I think

There are lots of articles online out there here are some for you : Impurity in medicine
Impurity in Electrical cells
Impurity in water
Effect on Solder machine
you can google and find more


Chemical kinetics of a reaction, rate constant, and activation energy From the Arrhenius equation in kinetics of reactions,

$$k = A\exp{\left(-\frac{E_\mathrm{a}}{RT}\right)}$$

Which tells us about the temperature dependence on rate constant of a reaction.

Activation energy is dependent on the temperature (as it is the energy difference between the average energy of the reactants and threshold energy, and since the average energy of reactants depends on the temperature, activation energy also should depend on temperature) Is my assumption that activation energy depends on temperature correct?

Moreover, is there any case where activation energy is zero or negative? (like in the case of spontaneous reaction or $E_\mathrm{a} = 0$) I feel that whenever activation energy is zero, all the molecules undergoing collisions should form products successfully, but this case won't occur.

But when we substitute $E_\mathrm{a} = 0$, in the Arrhenius equation, we get the result as $k = A$, i.e. rate constant will be equal to pre-exponential factor.

The rate constant and the activation energy tell us how fast the reaction proceeds (moles/sec), not whether it is spontaneous. The higher the temperature, the higher the rate constant and, for specified concentrations of reactants, the faster the reaction. The rate of a reaction always increases with increasing temperature (higher rate of collisions), so the activation energy is always positive.  As the temperature increases, the rate constant approaches the pre-exponential factor (E/T approaches zero). For some reactions, the activation energy is very low, so even at lower temperatures, the rate constant approaches the pre-exponential factor. The activation energy has a weak dependence on temperature, and, in practice, this temperature dependence is usually  neglected.


How to synthesize t-Boc protected derivative of PVA I've been toying with different potential recipes for DIY photoresist. My current favored design is a chemically amplified resist based on PVA (poly(vinyl alcohol)). For those who are unaware, a CAR works like this: a polymer is functionalized with a protecting group, and is mixed with a photoacid generator (PAG), which as the name suggests, generates acid upon exposure to light. The acid then cleaves the protecting group (in this case, t-Boc), rendering the polymer soluble in the developer. Basically, my question is how to functionalize PVA with the t-Boc group.

References are scant, but I did find one, which is unfortunately extremely vague. It says to react fully hydrolyzed PVA with di-tert-butyl dicarbonate (boc anhydride) and sodium hydride in N-methylpyrrolidone. However, the paper itself says that this reaction is not very predictable, and the authors are (were?) looking for better options.

Not being satisfied, I decided to lookup the reaction used for functionalizing poly(hydroxystyrene), which is the polymer used in the original CAR, APEX. Unfortunately, I wasn't really able to find anything on this either, save for even more vague references.

Anyhow, I have looked up information on how amino acids are protected with t-Boc, but the problem here is that I'm not sure if that would even work on PVA.

Does anyone have any idea how I would I go about doing this? Even more preferable would be a method not involving dangerous/explosive and for that matter, expensive substances such as Sodium hydride, or dangerous solvents (although NMP is not really that bad).

I did find one option  that simply involves reacting amino acids directly with molten boc anhydride. Is it likely this procedure would work with PVA?

Interesting, and i like the part with the photoacid.

I can't read the full details of the cited article (paywall), but since the authors aren't very pleased with the functionalization of the PVA, how about a change in the order of events:

prepare or buy the monomer (vinyl tBOC)
polymerize it according to this article (or a better source)


Why is arsenous acid denoted H3AsO3? Inspired by this question, I'm wondering why arsenous acid is frequently denoted $\ce{H3AsO3}$, as opposed to $\ce{As(OH)3}$, which would appear to more accurately reflect its connectivity?

[edit]

I feel like such a doofus, forgetting about sulfuric acid ($\ce{H2SO4}$), phosphoric acid ($\ce{H3PO4}$), and boric acid ($\ce{B(OH)3}$), etc. Great discussion, people!

All quoted stuff copied from IUPAC Red Book, IR-4 unless otherwise specified

It's not only $\ce{H3AsO3}$ . What about $\ce{H3BO3,H3PO4,H2SO4,HClO4}$? Or indeed, any other oxyacid?

Basically, we have multiple ways of writing inline formulae of covalent compounds. One is the boring way (molecular formula), and one is the fun way (structural formula) where the formula reveals a bit of the structure. $\ce{As(OH)3}$ reveals the structure, $\ce{H3AsO3}$ doesn't.

  For compounds consisting of discrete molecules, the molecular formula, as opposed to the
  empirical formula, may be used to indicate the actual composition of the molecules. For the
  order of citation of symbols in molecular formulae, see Section IR-4.4.

  A structural formula gives partial or complete information about the way in which the atoms
  in a molecule are connected and arranged in space. In simple cases, a line formula that is just
  a sequence of atomic symbols gives structural information provided the reader knows that
  the formula represents the order of the atoms in the linear structure.

So, what are the rules for writing the molecular formula?

Well, initially these rules were basically that we just follow electronegativity--the most electropositive element is listed first and so on. Looks like they slightly changed it.

For one, there are actually two ordering standards--one is the extremely boring alphabetical one, which I've never seen used (except in emperical formulae), and one is via electronegativity.

Now, instead of following the absolute electronegativities, the IUPAC has assigned a "pseudo-electronegativity" order (my terminology). The rules are as follows (source):

  for two elements in different groups - then the element in the higher numbered group has higher "electronegativity"
  for two elements within the same group the element with the lower the atomic number has the higher "electronegativity"
  Hydrogen is fitted in to be less electronegative than polonium and more electronegative than nitrogen. Hence the formulae of water and ammonia can be written H2O and NH3 respectively.

This leads to the following "pseudo-electronegativity" order of elements:

(click to enlarge)

The motivation behind this must probably be the erratic and hard-to-quantify nature of "real" enectronegativity.

Now, we can apply these rules in the following way:

If the compound is a binary compound (only two elements, not two atoms), then we just follow the pseudo-electronegativity thingy.

  In accordance with established practice, the electronegativity criterion (Section IR-4.4.2.1)
  is most often used in binary species.

If the compound is a coordination compound, we order central atoms by pseudo-electronegativity, and ligands by alphabetical order, as written (eg $\ce{(en)}$ before $\ce{(dmg)}$, but $\ce{C2H4(NH2)2}$ after $\ce{(dmg)}$, even though $\ce{(en)}$ and $\ce{C2H4(NH2)2}$ are the same).

  The order of citation of central atoms is based on electronegativity as described in
  Section IR-4.4.2.1. Ligands are cited alphabetically (Section IR-4.4.2.2) according to the
  first symbol of the ligand formula or ligand abbreviation (see Section IR-4.4.4) as written.
  Where possible, the ligand formula should be written in such a way that a/the donor atom
  symbol is closest to the symbol of the central atom to which it is attached.

Finally (barring annoying extra rules for isotopes and hydration), we have what are called "generalized salts". This is for compounds which can be treated as being made up of ions. The ions are ordered by electronegativity, and the elements within each ion are ordered alphabetically{*}.

  If the formula of a compound containing three or more elements is not naturally assigned
  using the preceding two sections, the compound can be treated as a generalized salt. This
  term is taken to mean any compound in which it is possible to identify at least one
  constituent which is a positive ion or can be classified as electropositive or more
  electropositive than the other constituents, and at least one constituent which is a negative
  ion or can be classified as electronegative or more electronegative than the rest of the
  constitutents. The ordering principle is then:

  (i) all electropositive constituents precede all electronegative constituents;

  (ii) within each of the two groups of constituents, alphabetical order is used.

This is the rule we need to write the formula for $\ce{H3AsO3}$. First, we need to clear up what is meant by "treated as being made up of ions". Well, for this, as @Terry noted above, we look for the weaker bond. In $\ce{NaOH}$, we have relative bond strengths $\ce{Na...O-H}$, whereas $\ce{H3AsO3}$ is $\ce{As-(O...H)3}$. Thus, in $\ce{NaOH}$, the "ion" breakup is $\ce{Na+, OH-}$, while in $\ce{H3AsO3}$, the ions are $\ce{H+,AsO3-}$.

So, arsenous acid is broken up into ions, of which $\ce{H+}$ is written first. Within each ion, we order alphabetically, so we get $\ce{AsO3}$, not $\ce{O3As}$. Putting them together, we get $\ce{H3AsO3}$.

*Despite being the IUPAC rule, it seems that, within ions, most people just follow the electronegativity rules again. Selenous acid seems to be commonly named $\ce{H2SeO3}$, when it should be $\ce{H2O3Se}$. But this seems to be a minor offence:

  Deviation from alphabetical order of constituents in the same class is allowed to
  emphasize similarities between compounds.


Manifolds in quantum chemistry Sometimes the concept "manifold" is used in quantum chemistry, for instance,
in Calculation of size‐intensive transition moments from the coupled cluster singles and doubles linear response function:

  where $\tau_{2s}$ and $\tau_{2t}$ denote the singlet-singlet and triplet-triplet spin coupled double excitation manifolds where spin couplings initially are carried out on the occupied-occupied and unoccupied-unoccupied orbital indices.

However, in general, manifold refers to a geometric object that is locally Euclidian. This is not essential to the coupled cluster method.

My question is, what is the motivation to introduce manifolds in the coupled cluster method?

The use of the word "manifold" in most of the quantum chemistry literature is purely as a synonym for multiple or different. Usually it appears when the number may be very large. Nothing more complicated than that.

For a simpler example than coupled cluster, consider configuration interaction with only single excitations (CIS). One of the key quantities is the set of single excitation amplitudes, $t_{ia}$ or $t_{i}^{a}$ or $k_{ai}\hat{a}^{\dagger}\hat{i}$ depending on your preference for notation. $i$ is an occupied MO index, and $a$ is a virtual MO index. We might say that the set $\{t\}$ forms a manifold, and the number quickly becomes large as you increase the basis set size: there will be $N_{ov} = N_{\text{occ}} \times N_{\text{virt}}$ single excitation amplitudes.


What is the reason of the color difference for the cis/trans isomers of tetraamminedichloridocobalt(III) chloride? The cis and trans isomers of $\ce{[CoCl2(NH3)4]Cl}$ are quite famous as one is purple and the other one is green which is a huge difference in color. Now I can't find any information to why this is the case?

I remember they told me in the lecture on organometallic chemistry when we were discussing substituted carbonylmetallates that a trans-effect occurs where a π-donor trans to the carbonyl (which binds to the metal's d-Orbitals via its antibonding π*-orbitals) increases the electron density in the metal and therefore also increases the the bond's strength ($\ce{M-CO}$) which then changes the bond length. A π-acceptor ligand on the other hand decreases the e-density.

In case of an octahedron the trans isomer would only possess same ligands trans positioned to each other while the cis isomer has mixed ones, too.

But is this the explanation I am looking for or is there another effect I just can't think of right now?

If you search a library for a copy of Inorganic Chemistry by Miessler and Tarr, they have an excellent diagram showing which explains the phenomenon.

$\ce{Cl-}$ ligands have lone electrons that can donate into the metal, as you said yourself. The trick is to recognize what effect this has on the $t_\mathrm{2g}^*$ and $t_\mathrm{2g}$ orbitals. Remember that the five frontier orbitals in an octahedral configuration are the $t_\mathrm{2g}$ and $e_\mathrm{g}^*$ orbitals.

When a $\ce{Cl-}$ ligand pi-donates into the metal center, it can only interact with the $\mathrm{d}_{xy}$, $\mathrm{d}_{xz}$, $\mathrm{d}_{yz}$ $(t_\mathrm{2g})$ orbitals because the $\mathrm{d}_{x^2-y^2}$, $\mathrm{d}_{z^2}$ $(e_{g}^*)$ orbitals have only a sigma-type interaction with the ligand (for them, there is no node, draw this to ensure yourself). When these ligands like $\ce{Cl-}$ pi-donate, they fill the $t_\mathrm{2g}$ bonding orbitals, widening the gap between them and the open the $T_\mathrm{2g}^*$ antibonding orbitals. These antibonding orbitals become much closer in energy to the $e_{g}^*$ orbitals, so you reduce the octahedral field splitting energy $\Delta_\mathrm{O}$ which redshifts the absorbance of the compound.

The reason the trans isomer is more redshifted (green as opposed to purple) is due to the fact that in the trans isomer, the ligand-metal pi overlapping does not clash. On the other hand, if you draw the ligand-metal pi overlapping generated by two cis $\ce{Cl-}$ ligands, you will find that there is always one orbital that contains a lobe where the ligand overlap does clash. When this clash does exist, it becomes less favorable for the $\ce{Cl-}$ ligands to pi-donate into the metal via that clashing orbital -- so, the $t_\mathrm{2g}^*$ antibonding energy level is closer in energy to the $t_\mathrm{2g}$ bonding energy level, and further in energy from the $e_\mathrm{g}^*$ energy level. In the cis compound, field splitting $\Delta_\mathrm{O}$ is higher and the color should be purple instead of green.

If you are having trouble seeing the clash, draw a $\mathrm{d}_{xy}$ orbital on a metal $\ce{M}$, and then draw two $\ce{Cl-}$ ligands such that $\ce{Cl-M-Cl}$ is a right angle (cis ligands). When you try to show pi-donation from both ligands, you will find that there is one lobe of the $\mathrm{d}_{xy}$ orbital that is used by both; that is the clash that is disfavored.


What simple formal systems are there that have behavior similar to actual chemistry? I've been doing reading about formal/axiomatic chemistry and am wondering: are there attempts at making formal systems (well defined rules that can be simulated on a computer) that have similar emergent behavior to real world chemistry in some way?

Clearly this would be very difficult for reproducing all of the physics of matter but I haven't been able to find any papers about attempts to make simple systems that have meaningful 3D interactions besides cellular automata which are very different.

The Lewis model (https://www.ncbi.nlm.nih.gov/pmc/articles/PMC3285721/) meets those criteria.


Acidic character and anion stability across periods and groups of the periodic table I understand that to compare relative acidity one must consider the stability of the conjugate bases.
Across a period the electronegativity of an element increases. And that is for example why $\ce{HF}$ is more acidic than $\ce{H2O}$, as the negative charge of the fluoride anion is drawn closer to the positively charged nucleus and thus more stabilized than that of an oxide anion.
Down the group the atom radius increases. And that is for example why $\ce{HI}$ is more acidic than $\ce{HF}$, as the negative charge of the iodide anion is distributed over a greater volume and thus more stabilized than that of a fluoride anion.
To me, these two trends are contradictory.
How can the stability of an anion increase as the negative charge gets closer to the nucleus(observed trend across periods) but at the same time decreases as the negative charge gets further away from the nucleus(observed trend down groups)?
Is there a concept I have misunderstood?

There are two factors determining acidity of a compound, the electronegativity factor and the size factor. Down the period, the size factor dominates, and the reason $\ce{HI}$ is more acidic than $\ce{HF}$ is because of this factor, as in the former molecule, the orbitals of iodide participating in bonding are larger and more diffuse, hence bond is weaker.
Across the period, however, size does not vary as significantly as down the group. Hence, we compare using the electronegativity factor. If electronegativity difference is higher, there is larger ionic character, and hence it is easier to deprotonate.
Also, basicity of an ion is proportional to its charge density. So you misunderstood the first concept. You should compare fluoride with hydroxide ion, and not oxide.


reaction coordinate, kinetics, equilibrium in example In this special reaction coordinate diagram with two reaction mechanisms, I tried to analyze it in two other ways, one with kinetics, another with equilibrium.

A. kinetics, I tried to find reaction rate of forward, and reverse. By the diagram, rate determining step of forward reaction is the first one (because activation energy is the highest and let's say it is one and only factor for RDS in this case). so Vf=k1[A][B].
RDS in reverse reaction is second step of the reaction judging by the diagram activation energy (M->A+B). So I used pre-equilibrium approximation for that. That's Vr in the picture.

B. equilibrium, I wrote all the equilibrium constant and simply added the reactions.

now here is the question. Expression of the rate of forward and reverse reaction in dynamic equilibrium is quite different from each case, analyzing in a way of kinetics and equilibrium constant, as you see in the picture below. Why is it different? What's right expression? What am I missing here?

You have chosen an unusual scheme in that A appears to be an intermediate as well as a reactant. If you use the simpler scheme things are easier and so rather than try to follow yours I have examined the scheme below which is very nonetheless very similar

$$RX\underset{k_{-1}} {\stackrel{k_1}{\leftrightharpoons}}R+X$$
$$R+Y\underset{k_{-2}} {\stackrel{k_2}{\leftrightharpoons}}RY$$

where species RX looses X and is replaced by Y. The intermediate species is R. The overall reaction is $RX+Y=RY+X$ and the equilibrium constant for the two steps are

$$K_1=\frac{k_1}{k_{-1}}=\frac{\mathrm{[R]_e[X]_e}}{\mathrm{[RX]_e}}$$
$$K_2=\frac{k_2}{k_{-2}}=\frac{\mathrm{[RY]_e}}{\mathrm{[R]_e[Y]_e}}$$

and the overall equilibrium constant

$$K=K_1K_2=\frac{k_{1}k_{2}}{k_{-1}k_{-2}}=\frac{\mathrm{[RY]_e[X]_e} }{\mathrm{[RX]_e[Y]_e }}$$

and all the concentrations in square brackets with subscript $e$ are equilibrium values. [This type of equation is true if there are many equilibria one after the other, $\displaystyle K=K_1K_2K_3K_4\cdots=\frac{k_1k_2k_3k_4}{k_{-1}k_{-2}k_{-3}k_{-4}}\cdots$].

In a rate equation approach we can apply a steady state approach to the intermediate species R. The steady state assumes that the rate of change of R is zero;

$$\frac{d[R]}{dt}= k_1[RX]-k_{-1}[R][X]-k_2[R][Y]+k_{-2}[RY]=0$$

from which

$$ [R]_{ss}= \frac{k_{-2}[RY]+k_1[RX]}{k_{-1}[X]+k_2[Y]}$$

The rate $r$ can be given by

$$\begin{align}
r=-\frac{d[Y]}{dt}=k_2[R][Y]-k_{-2}[RY] &= \frac{k_2(k_{-2}[RY]+k_1[RX])[Y]-k_{-2}[RY](k_{-1}[X]+k_2[Y])}{k_{-1}[X]+k_2[Y]}\\&=\frac{k_1k_2[RX][Y]-k_{-1}k_{-2}[RY][X]}{k_{-1}[X]+k_2[Y]}\\
\end{align}$$

and the numerator is zero when equilibrium concentrations are used making the rate equal to zero also. So this connects the equilibrium to the rate equations.

Initially just after the reactants are mixed the amount of product is very small and $k_1k_2[RX][Y]>>k_{-1}k_{-2}[RY][X] $ and the the rate is

$$r=\frac{k_1k_2[RX][Y]}{k_{-1}[X]+k_2[Y]}$$

which can be confirmed by experiment. Hope this helps.


Molar Absorptivity of Copper(II) Sulfate What is the molar absorptivity of copper sulfate? I am trying to find the molarity of copper sulfate solution by absorption in a spectrophotometer with a cuvette.

However, I can't find $\varepsilon$, the molar absorptivity coefficient of $\ce{CuSO4}$. I searched it up online, and I found experimental data saying that is $\pu{2.81 L mol-1 cm-1}$, but another source says that it is $0.91$.

I have searched on various databases including NIST's, but I couldn't find anything. Is there a definite molar absorptivity for everything, or is it all experimental?

These are the absorbance spectra of copper sulfate in water (open markers) and heavy water (filled markers) (1):

The absorbance in water is maximum at 780 nm, as stated in the reference. From the spectrum, the molal absorption coefficient, at 780 nm, appears to be approximately 12.5 $\mathrm{(mol/55.51 mol \ of \ water)^{-1} cm^{-1}}$ with aquamolality equal to 0.5.

Since only ordinary water, i.e., $\ce{H2^16O}$, is of interest in the OP's question, the aquamolality is the same as the molality, so the solution is simply 0.5 molal. The measurements were reported to have been made at room temperature, with no specific temperature given. Hence, the solution molarity is approximately 0.5 M.

Reference

G. Jancso, "Effect of D and $^{18}O$ isotope substitution on the absorption spectra of aqueous copper sulfate solutions", Radiation Physics and Chemistry, 74 (2005) 168-171.


Which base accepts the proton abstracted from water in propyne hydration? I am enrolled in Organic Chemistry 1 at a junior college and I need some assistance in understanding this mechanism. I have previously taken chemistry in high school with no problems, this would be my first time not comprehending the subject. We are covering the hydration of alkynes.

In the general mechanism, the triple bond of propyne attacks $\ce{H2SO4}$. The hydrogen, the electrophile, goes to the least substituted carbon. We now have a carbocation so water comes in as a nucleophile and attacks. But, now we have a positive charge on the oxygen, so a base de-protonates the molecule. My question is, what is this base?

After the base attacks the lone pairs from oxygen travels down to the carbon and the pi bond then attacks $\ce{H3O+}$. This is the keto-enol tautomerizatiom. My second question is, where did we get this $\ce{H3O+}$? Our list of reagents is $\ce{H2O, H2SO4}$ and $\ce{HgSO4}$. We get a product called acetone. I'm just very fuzzy on the in-between steps.

Protons can basically jump on and off at any time if there is sufficient incentive for them to do so. In your reaction, you not only have the hydrogensulphate anion $\ce{HSO4-}$ that can accept a proton to go back to sulphuric acid but also water itself which can be protonated to $\ce{H3O+}$. The only important aspect is to identify the least weak base and use that in your mechanism.

Furthermore, general chemistry should have taught you that a solution containing sulphuric acid and water will contain $\ce{H3O+}$ ions by deprotonation immediately.


Confusion with Law of Equivalence Law of equivalence states that

  One equivalent of an element combines with one equivalent of the other.

Equivalent of $\ce{Na}$ is $22.99$ and that of $\ce{O}$ is $8$. By the law of equivalence, $1$ equivalent of $\ce{Na}$ combines with $1$ equivalent of $\ce{O}$. But in $\ce{Na2O2}$, $$ 23 \times 2 ~\text{parts by weight of sodium combines with } 16 \times 2~\text{parts by weight of oxygen} \\\implies 23~\text{parts by weight of sodium combines with } 16~\text{parts by weight of oxygen}$$ which means one equivalent of sodium combines with two equivalents of oxygen which is contradiction to law of equivalence. So why does this compound not follow the law of equivalence?

The idea of a chemical equivalent relies on a compound's properties with respect to the usual reactions that occur with a few select species that tend to behave in very predictable ways.  In this case, the number of grams of sodium that would react with 8 grams of elemental oxygen, 23g Na, are one equivalent of sodium for oxidation-reduction purposes.

$\ce{Na_2O_2}$ is sodium peroxide.  It contains oxygen (normally a very well behaved element, chemically speaking) behaving in an unusual way, bonding not only with sodium but also with another oxygen atom.  The oxygen can be thought of as part of a peroxide ion, $\ce{O2^2-}$.

If you know about oxidation and reduction, then you'll remember that oxygen usually has an oxidation number of -2, but when it is in a peroxide it has an oxidation number of -1.  Because this particular compound involves the -1 state, it takes twice as much oxygen to react with a given amount of sodium (which in both compounds has an oxidation state of +1.

Finally, and perhaps more important in answering the root source of your trouble here, a scientific law is a thing that describes the behavior of the universe in specific conditions.  Outside of those conditions the law does not apply.  Even inside those conditions laws are often incorrect to some degree (though still useful approximations.)  It's not unexpected that we would find some exceptions to this law.


What are good examples of questions that make college students think about chemistry concepts? I am teaching the second semester of general chemistry at the college level. The course tends to get bogged down in calculations, and sometimes we forget about the big picture and what is going on at the atomic level. What are some good questions that would make students think deeply about the following topics:

Intermolecular forces
Kinetics of chemical reactions
Equilibrium and direction of reactions
Acid/base reactions
Redox reactions (electrochemistry)

Intermolecular forces

Are there contradictions about intermolecular forces in the postulates of kinetic theory?

Why do molecules having a higher relative molecular mass have stronger inter-molecular forces?
Kinetics of chemical reactions

Does a first order reaction really get completed only at infinity?
Equilibrium and direction of reactions

What does it mean to shift equilibrium?

At equilibrium are the rates of the forward and backward reactions equal to zero?
Acid/base reactions

Is the behavior of a dilute strong acid the same as the behavior of a weak acid?
Redox reactions (electrochemistry)

Why is it important to use a salt bridge in a voltaic cell? Can a wire be used?


Why is iso-octane different from other iso-s? While learning nomenclature of organic compounds I found the common  names to be a bit difficult to learn than the IUPAC ones. One of these common name stuff is the iso-,neo-, sec-, etc. So I searched for the ways to differentiate between these on the YouTube and found this video (note that the anchor of the video uses hindi for communication so it might not be intelligible to most people here). To summarize what the anchor said:

Prefix-iso: is used when all carbon atoms except one form a continuous chain.
$$\mathcal {OR}$$
They have a structure like this:

Now in the same video the anchor says that iso-octane is an exception to this rule, as it has the following shape:

Before this came in the video I formed the view that these are just the ways of naming compounds with similar structure but now I think that that isn't the case. So:

Why iso-octane has shape different than the other iso-s?

The anchor says that the structure of iso-octane was determined experimentally. This implies that there is some way to identify these compounds without knowing their structures beforehand so:

How do we know that whether some given compound is iso-, neo-, etc?

What might be the most general guidance for using iso-, neo-, sec- and tert- prefixes?

The usage of non-standard prefixes such as "iso" has never been defined with hard, unambiguous rules such as those in the IUPAC recommendations. Instead, their usage has often more been "by convention". In Organic Chemistry (1st ed.) by Clayden et al., the authors write (p 315):

  Strictly speaking, if the standard meaning of ‘iso’ is followed, the name isooctane should be reserved for the isomer 2-methylheptane. However, 2,2,4-trimethylpentane is by far the most important isomer of octane and so, historically, it has been assigned this name.

(This phrase is also found at Wikipedia, where it is copied wholesale without any indication that it's a direct quote. Please don't do this in your work. Even if you provide a citation, you should either paraphrase the information, or clearly indicate that it's a quote, e.g. by enclosing it in quotation marks.)


Common tangent construction I'm still rather confused as to how the common tangent construction ensure the chemical potentials are equal in the two phases?
(Also, if the one-phase regions and the two-phase metastable regions both concave upwards, why do the one-phase regions not undergo phase separation upon nucleation?)

Several answers (full disclosure - mine) are available that shed at least some light. So read them first.

https://physics.stackexchange.com/questions/196686/what-causes-a-eutectic-point

Plot Binary Eutectic from Temperature of Melting and Enthalpy

OK. Now to your confusion (I think). What does a straight line across a plot of free energy vs composition actually mean? Well, mathematically that straight line represents the weighted average of the pure free energies, $x_{A}G_{A} + x_{B}G_{B}$ (perhaps more obvious using $x_{B} = 1-x_{A}$). So, $G_{A}$ is the intercept at $x_{A}=1$ and $G_{B}$ is the intercept at $x_{B} = 1$. If the free energies are a fixed value, the chemical potentials are a fixed value. So, the tie-line that is a common tangent to two free energy curves is the solution where the free energies (chemical potentials) are equal in both phases.

As for why the one-phase region does not chemically separate, the question is why should it? There is no second composition that has a common tangent, so there is not way to make the chemical potentials the same in two phases while in a one-phase region.


How to quantify caffeine content in coffee? How do I titrate caffeine from coffee to estimate the percentage of caffeine in coffee?

I have tried using iodine solution, sulfuric acid, and sodium thiosulfate to no avail.

According to a paper published named: "Chemical Test for Caffeine" for the "Hong Kong Chemistry Olympiad for Secondary Schools (2013-14)" (Hyperlink provided below). You can titrate caffeine, with the help of iodine and potassium iodide, and sulfuric acid as the acid. You will have to add an excess of iodine, which means it will be an iodometric backtitration. The iodine will react in two times the number of moles compared to your caffeine.
The only problem I have with this method, is that I cannot found any elaboration on how this reaction is possible on the internet (I haven't tested it myself). If there is someone who knows more about how this reaction if possible, please provide me with more information.

If someone knows how or why the reaction can take place, please keep me notified about it. I will provide the reaction below:
Reaction:
Reactants --- Caffeine + 2 Iodine + Potassiumiodide + Sulfuric Acid
Products ---- Potassium Bisulflate + Caffeine (in a peculiar complex with the remaining substances, notated like a hydrate. No water present.)

$$\begin{align}\ce{C8H10N4O2 + 2 I2 + KI + H2SO4 &-> C8H10N4O2 . HI . I4 + KHSO4}\\
\ce{\color{red}{I2} + 2 Na2S2O3 &-> 2 NaI + Na2S4O6}\end{align}$$

EDIT 1: This URL should lead to the paper I mentioned (if it works).
Chemical Test for Caffeine
EDIT 2: Another URL provided, this is the source used in the actual paper. The site is in Chinese (if I'm not mistaken), but can be translated via google. For anyone interested: Source used in mentioned paper, from "www.med66.com"


How does Palladium dissociate H₂ so easily? If I understand correctly, $\ce{H2}$ in the presence of $\ce{Pd}$ readily dissociates as it dissolves into the metal. With the dissociation energy for the $\ce{H—H}$ bond being so large, how is this possible?

At first I thought that the $\ce{H}$ atoms were falling to a lower energy level in the $\ce{Pd}$ that was somehow only available if they were monatomic, but if that were true wouldn't some heat be generated by the absorption?

$\ce{Pd}$ can dissociate $\ce{H2}$ because the resulting $\ce{Pd-H}$ bonds are more stable than the starting $\ce{H2}$.
But the reason why $\ce{Pd}$ is so good at dissociating $\ce{H2}$ is related to the energy barrier to bond formation. The dissociation of $\ce{H2}$ on a $\ce{Pd}$ surface (and on $\ce{Pt}$ and maybe several other metals) has no barrier. So you don't need to put the $\ce{H2}$ in an excited state to go over a barrier and create $\ce{Pd-H}$ bonds.
On $\ce{Cu}$ for example, the bonding is possible but there is a barrier, you need to excite $\ce{H2}$ to dissociate it.

Here is a good theoretical article (with a great title):

It shows the different barriers to dissociation of $\ce{H2}$ on $\ce{Pt}$, $\ce{Ni}$, $\ce{Cu}$ and $\ce{Au}$. It also gives an explanation for such differences (more physics than chemistry).


What exactly is the long chain band? I have progressed quite enough in infrared spectroscopy, and now I'm looking at the spectrum of alkanes in detail. I found out that straight chain alkanes have this long chain band in the spectrum:

The infrared spectrum of decane. Taken from: Introduction to Spectroscopy, Pavia et. al.

On the other hand, cyclic one's don't have this band:

The infrared spectrum of cyclohexane. Taken from: Introduction to Spectroscopy, Pavia et. al.

I also found this info in my book:

  $\ce{CH2}$ - The bending (rocking) motion associated with four or more $\ce{CH2}$ groups in an open chain occurs at about $\pu{720cm^-1}$ (called a *long-chain band).

I am still perplexed. Why can't the $\ce{CH2}$ groups in the cyclic chains undergo rocking vibrations? Why is this specific to the open chained ones?

The 'long chain band' described in your textbook (Pavia) is actually a vibrational mode in which the terminal methyl groups are 'rocking'.

This rocking mode is observed in open chain alkanes,  -(CH2)n-, where the number of methylene units is greater than 4.

Polyethylene is the 'classic' examples, the IR spectra and corresponding vibrational modes are given below:

Why do we see this band?

The reason we see any given IR band is as a consequence of the symmetry of the molecule, with only certain molecular vibrations IR active (we don't, for example, observe the C-C vibrations directly, they aren't active). Your question, more broadly, then becomes 'how to predict the IR active vibrational modes for a given molecule?'.

For a full explanation of the IR spectra in this particular case, using molecular symmetry and group theory, you might try reading J. Chem. Phys 1956, 25, 549.

Why don't we see the band in cyclic alkanes?

In cyclic alkanes, the methyl groups have been constrained to form one continuous ring of methylene groups. As discussed above, the band at ~725 cm-1 is caused by the terminal methyl groups of longer chain alkanes, hence, removing these methyl groups removes their corresponding vibrational modes.


Why is it wrong to calculate the amount of oxygen directly from the products of a combustion reaction? I've been stumped by a question in my textbook concerning combustion analysis.

  A $\pu{0.5438g}$ sample of a liquid consisting of only $\ce{C}$, $\ce{H}$, and $\ce{O}$ was burned in pure oxygen, and $\pu{1.039g}$ of $\ce{CO2}$ and $\ce{0.6369g}$ of $\ce{H2O}$ were obtained. What is the empirical formula of the compound?

Firstly, I calculated the amount of substance of carbon and hydrogen by dividing the masses of each product by their respective molar masses.
\begin{align}
(1.0390/44.01) = n(\ce{C}) &= 0.02360\\
2\cdot(0.6369/18.02) = n(\ce{H}) &= 0.0706\\
\end{align}

My textbook then states that I must find the sum of the masses of carbon and hydrogen, and then deduct this sum from the mass of the liquid compound to get the mass of oxygen:
\begin{align}
0.02360 \cdot 12.01 &= 0.283535\\
0.0706 \cdot 1.01 &= 0.28637\\
0.5438 - (0.283535 + 0.28637) &= m(\ce{O}) \\
m(\ce{O}) &= \pu{0.1888g}\\
\end{align}

I have no qualms with this approach, but why can't I find the amount of oxygen directly from the products?
\begin{align}
\text{I.e.}&& 2\cdot(1.0390/44.01) &= n(\ce{O}) &\text{in }\ce{CO2}\\
\text{and} && (0.6369/18.02) &= n(\ce{O}) &\text{in }\ce{H_2O}.\\
\end{align}

$$\text{Amount of oxygen in liquid compound} = n(\ce{O})\text{ in }\ce{CO2} + n(\ce{O})\text{ in }\ce{H2O}.$$

I know this doesn't work, but my question is why? It seems perfectly intuitive to me, and yet it doesn't work.

Let's first write the chemical equation for the reaction (I will leave it unbalanced because the stoichiometry is irrelevant):
$$\ce{C_xH_yO_z + O2 -> CO2 + H2O}$$
To do this question, we need to find the mole ratio of carbon, hydrogen and oxygen in $\ce{C_xH_yO_z}$, then we can find its empirical formula.

From the above equation, it is clear that all the carbon in $\ce{CO2}$ originated from the compound and that all the hydrogen $\ce{H2O}$ originated from the compound. However, all the oxygen in $\ce{CO2}$ and $\ce{H2O}$ didn't just come from the compound, it came from the compound and gaseous oxygen. Therefore, from your method, you are just finding the total amount of oxygen in the reactants, not the number of moles of oxygen in the compound.


What is the mechanism for the double bond migration in 4-methylcyclopent-2-en-1-one? The problem I was asked to answer:

Please explain using only enolisation and protonation steps:

My work so far:

Seems completely wrong... can someone please correct me?
EDIT I just proposed a new mechanism:

After Step 2, re-protonate alpha to the carbonyl (the double bond has now migrated to the beta-gamma position).  Now enolize the carbonyl - remove the proton next to the methyl group (the top one), then push the electrons all the way back, protonating the gamma position.  Job done!


What are the mechanisms for capture of As(III) and As(V) by magnetite from water? Magnetite, $\ce{Fe3O4}$ is able to capture $\ce{As(III)}$ and $\ce{As(V)}$ from drinking water through the following mechanisms.

Bidentate-binuclear complex

Monodentate-binuclear complex (both from here)

I'm assuming these are standard reactions in inorganic chemistry, but are there known mechanisms (I don't have a text available)? The first looks like an attack on the $\ce{As}$ by the $\ce{O}$, but the second seems to take place in two steps.

Unfortunately there are no (or only very hard to find) studies to the mechanism of the actual sorption of arsenic (acids) to magnetite. Most literature does describe it as a simple Ligand exchange, although I find that hard to believe. From what I have read[1-4] I will try to compose some hints on what may happen.

First of all lets have a look at arsenic acid. It is a medium strong acid and a quite well oxidising agent.
\begin{align}\ce{
H3AsO4 &\equiv AsO(OH)3\\
AsO(OH)3 + H2O &<=> H3+O + AsO(O^- )(OH)2\\
AsO(OH)3 + 2H3+O + 2e- &<=> As(OH)3 + 3H2O
}\end{align}

In presence of any electron rich material and acidic environment it will therefore not be unusual that arsenous acid will be formed.[5] We may therefore also focus on this, amphoteric molecule:
\begin{align}\ce{
H3AsO3 &\equiv As(OH)3\\
As(OH)3 + H3+O &<=> As+(OH)2 + 2H2O\\
As(OH)3 + {}^{-}OH &<=> AsO^{-}(OH)2 + H2O
}\end{align}

Depending on the pH of the solution one would expect to find many of different of these  species. Arsenic is also somewhat oxophil, meaning that the $\ce{As-O}$-bond is quite strong.

The magnetite in aqueous solution reacts basic, also forming complex like structures or hydroxides, for example this fully hydrated cluster (ref. [2], stoichiometry omitted):
\begin{aligned}\ce{
 Fe3O4 + $x$\, H2O &<=>  Fe2(OH)2(H2O)8^4+
}\end{aligned}

Of course one has to pay attention, that the used ferric nano particles for sorpting arsen(o)ic acid(s) are far larger than the used model complex. However, we can see, that there are functional groups like $\ce{-OH}$ or $\ce{-OH2+}$ (or maybe even $\ce{-O-}$) covering the surface.

Using this as prerequisites we can think about what might happen next. I generally would suggest to start with physisorption, i.e. the cluster and the acid molecules come in close range. Since we are dealing with contaminated water it is maybe safe to assume, that we can deal with soluble salts of arsenoic acid, hence focusing on $\ce{AsO3^{3-}}$.

When the complexes come into close range hydrogen bonds will form, with
$
\newcommand{\nanoFe}{\bbox[lightgrey,3pt,border:1px solid black]{\ce{Fe}}}
\nanoFe$ being the nano particle:
$$\ce{\nanoFe-OH\cdots{}O-AsO2^{3-}}$$

Since the whole surface is covered with hydroxide groups, several of these hydrogen bonds may form. As we can see, with this we may now enter the realm of chemisorption. Once these complexes have formed, it is from an entropic point of view favoured to release smaller molecules, water in this case. Also the oxygen atoms bound to arsenic are now in close range to iron atoms. Since we are dealing with nano particle, there will be non-coordinated irons in proximity. This allows for a more permanent complet to form:
$$\ce{\nanoFe-O-AsO^3- -O\cdots{}HO-\nanoFe}$$

With this monodentate complex we have an already stable complex of a bound arsenic to an iron. It is furthermore possible, and probably favoured, to form a bidentate complex with a somewhat similar mechanism.

Let us also consider a more basic scenario. In this case the surface of the magnetite will not be entirely covered with hydroxide atoms, but more freely accessible oxygen atoms will be present. Direct bond formation between $\ce{\nanoFe-O\cdots{}AsO3^{3-}}$ may occur.

After writing all this I also found a credible source that somehow proves my theory. In “Understanding Arsenate Reaction Kinetics with Ferric Hydroxides”[6] they used a model system of $$\ce{HAsO4^2- + Fe2O10H15+ -> Fe2O10H15-HAsO4-}$$ to explain the mechanism.

In this they describe that physisorption is barrierless, leading fast to the adduct that is $\Delta G^\circ_\mathrm{r}=-58~\mathrm{kJ/mol}$ more stable in energy. However, chemisorption and developing the equilibrium is a more lengthy process.

When the formed complexes become neutral, the activation barrier to release arsenate is very high, compared to adsorption. I will stop right here, please read the publication for further details.

Notes, references and further reading:

Dimitri A. Sverjensky, Keisuke Fukushi, Geochimica et Cosmochimica Acta 2006, 70 (15), 3778-3802.
Nianliu Zhang , Paul Blowers , and James Farrell, Environ. Sci. Technol. 2005, 39 (13), 4816–4822.
Arnold F Holleman; Egon Wiberg; Nils Wiberg: Lehrbuch der Anorganischen Chemie. 102. Auflage. Walter de Gruyter & Co., Berlin, New York, 2007. (german)
Egon Wiberg, Nils Wiberg: Inorganic Chemistry. Academic Press, 2001. (english)
Thompson, A. & Goyne, K. W. (2012) Introduction to the Sorption of Chemical Constituents in Soils. Nature Education Knowledge 3(6):15
Heavier halogens ($\ce{X=I, Br, Cl}$) should do the job and they will most certainly be present in usual ground water (more or less acidic solution) :
$$\ce{AsO(OH)3 + 2H3+O + 2KX <=> As(OH)3 + 3H2O + X2 ^ + 2K+}$$
James Farrell and Binod K. Chaudhary, Environ. Sci. Technol. 2013, 47 (15), 8342–8347.
Also interesting: (a) Suvasis Dixit and Janet G. Hering, Environ. Sci. Technol. 2003, 37 (18), 4182–4189 (b) Dimitri A. Sverjensky and Keisuke Fukushi, Environ. Sci. Technol. 2006, 40 (1), 263–271.


Hamiltonian: accounting for entropy I might have a bit of a conceptual misunderstanding here, but how do we account for the entropy of a system when constructing a Hamiltonian? All entropy should be accounted, because the Hamiltonian includes the total energy, including the entropic contribution. However, when considering e.g. $\ce{H2+}$, the fact that the dissociated state is more entropically favoured doesn't go from the Hamiltonian.

Any suggestion that would clean up my understanding of the relationship between entropy and energy in this and other cases would be very helpful. Thanks!

One definition of entropy is that the number of microstates that makes up a macrostate influences the probability of observing the macrostate.  So your wavefunction needs to be multi-configurational and include both $\mathrm{H}_2^+$ and $\mathrm{H}^+ + \mathrm{H}$. The usual approach is to invoke the Born-Oppenheimer approximation, which makes the wavefunction depend on a single static set of nuclear coordinates, i.e. single-configurational. This approach will just give you the energy and you must compute reaction probability using statistical mechanics.

Multi-configurational SCF does include entropy in the following sense.  Let's say you have a molecule with a HOMO-LUMO gap that is sufficiently small to allow for significant mixing. You run a CASSCF(2,2) to determine the LUMO contribution and is 5%.  Now suppose you have the same system, same HOMO-LUMO gap, but the LUMO is doubly degenerate.  Your CASSCF(2,3) would now have a larger LUMO contribution, though I am not sure it would be 10%.


What is the origin of the Quadratic factor in calculation of XRD peaks? Whenever we are to calculate the d-spacing(for cubic system) in XRD peaks with reference to Quadratic Factors we do that with the help of the formula $$\frac{1}{d_{hkl}^2}=\frac{h^2+k^2+l^2}{a^2}.$$

Please tell me how we are arriving at this form and where from are we getting ${h^2+k^2+l^2}$ as a factor. I know that things are different for hexagonal. Would be happy if both are explained with more emphasis on the cubic one.

Reciprocal space

One way to derive the formula is to use reciprocal space coordinates. The reciprocal space lattice has three unit cell vectors, a*, b* and c*. The diffraction vector d* is given by:

$$d^* = h a^* + k b^* + h c^*$$

The d-spacing is given by the reciprocal length of d*. The relationship between direct space and reciprocal space unit cell vectors is

$$a \cdot a^* = 1$$
$$b \cdot a^* = 0$$
$$c \cdot a^* = 0$$

$$a \cdot b^* = 0$$
$$b \cdot b^* = 1$$
$$c \cdot b^* = 0$$

$$a \cdot c^* = 0$$
$$b \cdot c^* = 0$$
$$c \cdot c^* = 1$$

So to get the d-spacing, we have to figure out the length of $d^*$. For a cubic system, the reciprocal space unit vectors all have the lenght of 1/a, and they are all at right angles. To get the length for the cubic system, we use Pythagoras' theorem (or the length of a vector in Cartesian space): It is the square root of the coordinates squared.

So:

$$|d^*| = |a^*| \sqrt{h^2 + k^2 + l^2}$$

To get to your formula, substitute 1/d for $d^*$ and 1/a for $a^*$. For the hexagonal system, it is a bit different because the angles are not all 90 degrees, so there will be a trigonometric factor (sin 120 or cos 120 or such) in the respective formula.

Direct space

One plane will go through the origin, the next through a/h, b/k and c/l. In the cubic case, a vector normal to the plane is (h, k, l). To get the distance of the plane from the origin, you can take any point in the plane and form the dot vector with the normalized vector (h, k, l). For example,

$$ d = \frac{(h, k, l)}{\text{length} (h, k, l)} \cdot (1/h, 0, 0)$$

For a cubic system, this will simplify to the expression for the d-spacing. The term $h^2 + k^2 + l^2$ is from the length of the (h, k, l) vector.

(h k 0), no vectors

Both derivations above use vectors. Here is a derivation for the 2-dimensional case using no vectors.

Let the origin be point A. In a cubic system, the plane hk0 intersects the x-axis at a/h (point B) and the y-axis at b/k (point C). Our goal is to calculate the altitude. We use that the area of the triangle can be calculated two ways, using the perpendicular sides or using the hypothenuse and altitude:

$$\text{Area} = \frac{1}{2} * \frac{a}{h} * \frac{b}{k} = \frac{1}{2} * \text{hypothenuse} * \text{altitude}$$

We can get the length of the hypothenuse from Pythogoras:

$$\text{hypothenuse} = \sqrt{(\frac{a}{h})^2 + (\frac{b}{k})^2}$$

Then, we can solve for the altitude:

$$\text{altitude} = \frac{\frac{a}{h} * \frac{b}{k}}{\text{hypothenuse}}$$

$$ = \frac{\frac{a}{h} * \frac{b}{k}}{\sqrt{(\frac{a}{h})^2 + (\frac{b}{k})^2}}$$

With a = b and simplifying, we get the formula the OP asked for (in two dimensions).


Why exactly is a fluoride anion a stronger base than iodine anion? First off, I've learnt that stronger acids produce weaker conjugate bases (through Brønsted–Lowry acid–base theory).

Then I looked at the $\mathrm{p}K_\mathrm{a}$ values of $\ce{HF}$, $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ and came to the conclusion that $\ce{HI}$ is the strongest among them and that explains why $\ce{I^-}$ is the weaker base.

So far so good.

But then I remembered that $\ce{F^-}$ is in fact the strongest oxdizing agent out there.
So shouldn't that mean that $\ce{F^-}$ ion is more stable than $\ce{I^-}$? And hence does not give away its electrons easily and therefore is a weaker base?

Where am I going wrong here?

I have a little bit of gut feeling that this is because of the large size of iodide ion which can hold the negative charge better than the tiny fluoride ion. But I have found nothing to back this up.

First off, I've learnt that stronger acids produce weaker conjugate bases (through Brønsted–Lowry acid–base theory).

That is correct.

  Then I looked at the $\mathrm{p}K_\mathrm{a}$ values of $\ce{HF}$, $\ce{HCl}$, $\ce{HBr}$, and $\ce{HI}$ and came to the conclusion that $\ce{HI}$ is the strongest among them and that explains why $\ce{I^-}$ is the weaker base.

I don’t agree with that explains; but the observation is correct, iodide ($\ce{I-}$ is the weakest base and $\ce{HI}$ is the strongest acid.

  But then I remembered that $\ce{F^-}$ is in fact the strongest oxdizing agent out there.

No. The fluoride anion is not an oxidising agent. If it were, it would take up electrons to become a fluoride dianion ($\ce{F^2-}$), inserting an electron into the 3s level—not happening.

What you probably meant is fluorine (the element; the diatomic molecule) being the strongest oxidising agent out there. That might be correct or it might not (I am not up to date with extremely strong oxidising agents) but it is not a statement I could dismiss at first glance. Certainly, fluorine is a stronger oxidising agent than iodine.

  So shouldn't that mean that $\ce{F^-}$ ion is more stable than $\ce{I^-}$?

Thermodynamic stability is sometimes a tricky beast, but in general a fluoride ion is more stable than an iodide ion which is prone to oxidation to give iodine or iodate. This is connected to the oxidising ability of fluorine (strong) versus iodine (weak), not to the strengths of the acids, though.

  And hence does not give away its electrons easily and therefore is a weaker base?

The conclusion is invalid. Not wanting to give away one’s electrons means not wanting to get oxidised. As mentioned, fluorine is a strong oxidising agent so to reduce fluorine to fluoride is a favourable process. But acid-base chemistry has nothing to do with gaining or losing electrons: it is an entirely electrostatic process and builds on different principles.

The high oxidising ability of fluorine can be thought of as a side effect of its high electronegativity: it has a tendency to strongly attract electrons since it has a highly positively charged nucleus that is shielded by only two core and seven valence electrons. Adding an eighth electron is comparatively easy. For iodine, there are four core shells and a valence shell between the nucleus and the incoming electron; this shielding effect can be thought of as reducing electronegativity and oxidising ability.

The high acidity of $\ce{HI}$ and low basicity of iodide is a direct consequence of its size: the negative charge is distributed across a much larger volume meaning it is more stabilised. In fluoride, the negative charge is confined to a much smaller volume meaning that positive charges such as a proton are attracted stronger. This is the underlying reason for the observed acidities/basicities.


Is there such a thing as a Boron-Carbon-Lithium alloy? I am trying to create an alloy with heat resistance for a project. Is there such a thing as a boron-lithium-carbon alloy, and are there any benefits to it?

In the study Compound Formation in the Systems Lithium-Carbon and Lithium-Boron, Journal of the American Ceramic Society, Volume 50, Issue 10, October 1967, Pages 520–523

molten lithium metal was added to $\ce{B_{4.0}C}$ in proportions ranging from 7 to 71% lithium.

  a  substitution  reaction  evidently  occurs  between
  lithium  and  boron  carbide  which  leads  to  the  formation  of
  lithium  boride  and  lithium  carbide


What is the hybridization of the carbonyl oxygen in a carboxylic acid? My professor sent out a review guide. One of the questions on this guide is

  The carboxylic acid carbonyl oxygen is:
  1. $\ce{sp^2}$
  2. $\ce{sp^3}$
  3. $\ce{H}$-bond donor
  4. rapidly equilibrating between $\ce{sp^3}$ and $\ce{sp^2}$
  5. 1 and 3
  6. 1 and 4

I chose (4) because of the resonance that can be drawn with a negative oxygen and positive carbon. However apparently the answer is (6). I don't understand how something can be equilibrating between two states and still be one of those states.

Could anyone explain?

I already pointed it out in the comments, but I believe it is time to give it some more thought and explanation. Let's deal with some conceptual issues first.

Hybridisation is a model, that can be used to describe a bonding situation. It is never cause for a certain geometric arrangement, it is always the result of a bonding arrangement. Strictly speaking it is not even necessary to be invoked, but it can be quite handy. Quoting from Aditya's answer - I could not have said it better:

  First of all, 'hybridisation' is a hypothetical concept, i.e. orbitals don't really mix to form new orbitals, but 'hybridisations' are very successful in explaining structures of molecules and their physical properties. So knowing that a compound has $\ce{sp^2}$ hybridisation will just help you arrange the constituent atoms around the central atom. Physically it has no other meaning.
  Note: Only orbitals can be hybridised, never compounds.

Having understood that, it is easy to see, that atoms in a molecule could use many different hybridisations schemes. There are some concepts, based on quantum chemical calculations, which would find the most appropriate fit between a Lewis structure and the molecular orbital approach. See for example this fine question: What is natural bond orbital theory used for? There is also a more quantitative approach, which is called Valence Bond theory, but that is beyond the scope of the answer.

Since this is just a mathematical concept, an equilibration between $\ce{sp^2}$ and $\ce{sp^3}$ would not change a thing. We can decide which way we want to look at it. Ron consequently raised the question in the comments:

  I don't understand how "6" can be correct. How can the carbonyl oxygen be $\ce{sp^2}$ (1) and also be rapidly equilibrating between $\ce{sp^3}$ and $\ce{sp^2}$ (4)?

As I pointed out, (4) should not even be an option. The premise of the question is clearly flawed. Please do not take hybridisation too seriously.

This is how I come to my next point. Are the oxygen's orbitals in carbon acids really $\ce{sp^2}$ (or even $\ce{sp^3}$) hybridised? Well.

It is often appropriate to describe oxygen's orbitals as $\ce{sp^2}$, when it is bonded to two other elements. Two $\ce{sp^2}$ lobes are needed for the bonds, the remaining $\ce{sp^2}$ orbital hosts one lone pair, the remaining $\ce{p}$ orbital hosts the other lone pair.
More often the case is that oxygen has only one bonding partner. In these cases it is more appropriate to describe the oxygen with $\ce{sp}$ orbitals due to symmetry reasons. Consider the local symmetry to have an infinite rotational axis $C_\infty$, hence there have to be two degenerate, perpendicular orbitals. In most of these cases you will also have (partial) π-bonding and/or hyperconjugation.
There are very few exceptions, like $\ce{H3O+}$, where it is appropriate to describe the oxygen with $\ce{sp^3}$ orbitals.

In all of this it should be considered that hybridisation can be fractional. See the last note here and more consequently "What is Bent's rule?" and "Utility of Bent's Rule - What can Bent's rule explain that other qualitative considerations cannot?".

An important point to remember is, that when there is a π-bond, the hybridisation can be $\ce{sp^2}$ at maximum.

I believe that answer 6 is really about a wrong interpretation of resonance stabilisation of the molecule. (Please see Why is there a need for resonance? for more information)
Carbon acids can be described in the resonance scheme below. Now it appears, that the oxygen employs $\ce{sp^2}$ orbitals in the first structure (A) and $\ce{sp^3}$ in the others (B, C). It is important to recognise, that the bonding situation in this molecule is a superposition of all (possible) resonance structures. Theses structures are not in equilibrium.

In conclusion there are several issues with this question and you might want to address this with your teacher.


What is has a higher priority in naming organic compounds, alkyl groups or halides? 2-Chloro-3,3-dimethylbutane or 3-Chloro-2,2-dimethylbutane?

According to these sources, both are correct:

NIST/TRC Web Thermo Tables (WTT)
ChemSpider

The current IUPAC recommendations (2013) read as follows:

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

Note that Rule f takes precedence over Rule g.
Furthermore,

The lowest set of locants is defined as the set that, when compared term by term with other locant sets, each cited in order of increasing value, has the lowest term at the first point of difference (…)

Therefore, the example is named as 3-chloro-2,2-dimethylbutane rather than 2-chloro-3,3-dimethylbutane since the locant set ‘2,2,3’ is lower than ‘2,3,3’.


What makes mercury vapors so harmful? I just read an experiment in which mercury has to be heated and in the same experiment there was a note warning not to inhale mercury vapors as they are very harmful.

I just can't get the point that why are mercury vapors harmful.

Mercury poisoning (also known as hydrargyria or mercurialism) is type of metal poisoning and a medical condition caused by exposure to mercury or its compounds. Mercury (chemical symbol $\ce{Hg}$) is a heavy metal occurring in several forms, all of which can produce toxic effects in high enough doses. Its zero oxidation state $\ce{Hg^0}$ exists as vapour or as liquid metal, its mercurous state $\ce{Hg2^{2+}}$ exists as inorganic salts, and its mercuric state $\ce{Hg^{2+}}$ may form either inorganic salts or organo-mercury compounds; the three groups vary in effects. Toxic effects include damage to the brain, kidneys and lungs.

Mercury poisoning can result in several diseases, including acrodynia (pink disease) and Minamata disease.

source wikipedia


Relation between thermodynamic reversible process and reversible reaction I know it seems to be a weird question. But for long I have been thinking whether there is any relation between thermodynamic reversible process and reversible reaction. Do they have any connection and if so how?

In thermodynamics, a reversible process -- or reversible cycle if the process is cyclic -- is a process that can be "reversed" by means of infinitesimal changes in some property of the system without entropy production (i.e. dissipation of energy).[1] Due to these infinitesimal changes, **the system is in thermodynamic equilibrium throughout the entire process.**

Thermodynamic equilibrium is an axiomatic concept of classical thermodynamics. It is an internal state of a single thermodynamic system, or a relation between several thermodynamic systems connected by permeable walls. In thermodynamic equilibrium there are no net macroscopic flows of matter or of energy, either within a system or between systems

dO YOU OBSERVE SUCH KIND OF equilibriums IN A REACTION. SO THEY ARE NOT RELATED AT ALL

ACCORDING TO WIKIPEDIA

A reversible reaction is a chemical reaction that results in an equilibrium mixture of reactants and products. For a reaction involving two reactants and two products this can be expressed symbolically as
 aA + bB ---> cC + dD
A and B can react to form C and D or, in the reverse reaction, C and D can react to form A and B. This is distinct from reversible process in thermodynamics.


Proportionality between number of absorbed photons and optical density I am to show that the number of absorbed photons only is proportional to the optical density at low optical densities.

I do not know how to do this, but this reminds me of the linear range of absorbance vs concentration; at high concentrations, the relationships is no longer linear. I also feel that the "number of absorbed photons" is another way of saying "absorbance", and that "optical density" is another way of saying "concentration times optical path length", but I am not sure. I cannot find references that clarifies this for me, so I would appreciate some guidance.

Yes, your statement that "the number of absorbed photons only is proportional to the optical density at low optical densities" is correct. (I'll point out that optical density is a depreciated term for what is now referred to as absorbance.)

First let's define some terms.

  T: Transmittance, i.e. the fraction of photons not absorbed. So a
     transmittance of 0.90 would mean that 90% of the photon pass through
     the sample.

  A: Absorbance of sample which is given by the Beer–Lambert law: $\ce{A = -log_{10}(T)}$ or: $\ce{T = 10^{-A}}$

Now a bit of math swizzling...

For exponentials of $e$ there is a nice series expansion. $$ e^x = 1 + x + \dfrac{x^2}{2!} + \dfrac{x^3}{3!} + \dfrac{x^4}{3!} + ... $$ and if $x < 0 $ then: $$ e^{-x} = 1 - x + \dfrac{x^2}{2!} - \dfrac{x^3}{3!} + \dfrac{x^4}{3!} + ... $$
now if also $|x| << 1$ then the higher order terms can be ignored and the function becomes linear where:
$$ e^{-x} \approx 1 - x $$

Now for a bit more math voodoo... Let's let $$10^x = e^y$$ so: $$y = x\mathrm{ln}(10) \approx 2.303x$$

So now substituting $\ce{e^{-2.303A}}$ for $\ce{10^{-A}}$ we get: $$\ce{T = e^{-2.303A}}$$ and thus only when $2.303A << 1$ is the Beer-Lambert law linear.

Now for the rest of the story let's define some more terms:

  $\epsilon$ is the molar attenuation coefficient of the attenuating species  in the sample;

  $c$ is the molar concentration of the attenuating species in the sample;

  $l$ is the path length of the beam of light through the material sample.

Now also accoding to the Beer-Lambert law: $\ce{A = \epsilon c l}$ so $\ce{A}$ is proportional to $\ce{c}$.


Is pyrene aromatic despite failing Hückel's rule? Pyrene doesn't seem to be aromatic. However, sources claim that it is aromatic.

Considerations:

Pyrene is cyclic. ✓
Pyrene is flat (planar). ✓
Pyrene has 16 π electrons.
Every atom in the ring structure of pyrene is $\ce{sp^2}$ hybridized. ✓

The problem is that pyrene fails the $4n+2$ rule. $4n + 2 \neq 16$ where $n$ is an integer.

Something leads me to suspect that Hückel's rule is an oversimplification here. Or is pyrene indeed non-aromatic?

To clarify @ron's point, the general "cheminformatics" rule for deducing aromatic fused-ring systems is whether there is one path (usually the peripheral one) that satisfies $4n+2$:

So for naphthalene:

The bold bonds give you a ring path of $\mathrm{sp^2}$ carbons with 10 π electrons, so it's aromatic. There are some arguments about whether the central bond is aromatic or not. (I say baloney, but it's fair to say aromaticity generates lots of arguments.)

So yes, the bold path gives you 14 π electrons for a $4n+2$ path. Again, I've heard people say the central two carbons aren't aromatic. (IIRC, the $\ce{^13C}$ NMR disagrees.)

I wrote the code for Open Babel and aside from some tricky nitrogen-heteroaromatic rings, this rule generally matches chemical reality. (Setting aside the question of the interior carbons.)

I think another fairly safe rule is "if a fused aromatic system is made up of individually aromatic rings, aromaticity is retained."


What is the pKa of iodous acid? Searching the web readily gives the $\mathrm{p}K_\mathrm{a}$ of hydroiodic acid, hypoiodous acid, and iodic acid, however, extensive searching did not result in any value of the $\mathrm{p}K_\mathrm{a}$ of iodous acid. So I am wondering if this value has been documented before, and if it hasn't, then what progress has been made to determine its $\mathrm{p}K_\mathrm{a}$?

According to Schmitz [1], $\mathrm{p}K_\mathrm{a}(\ce{HIO2})\approx 6$ at $\pu{25 °C}$ (numerical references for the bibliographical entries have been updated):

Stanisavljev [2] has proposed a correlation between thermodynamic values of oxo-halogen ions and has obtained $ΔG_\mathrm{f}^\circ (\ce{IO2-}) = \pu{−51.8 kJ mol−1}$. He has also estimated the dissociation constant of $\ce{IO2H}$, $\mathrm{p}K_\mathrm{a}(\ce{IO2H}) = 4$, and obtained $ΔG_\mathrm{f}^\circ (\ce{IO2-}) = \pu{ −75 kJ mol−1}$. However, this value introduced in the above equation leads to $K_\mathrm{C}^\circ = \pu{2.8e12}$ and an impossibly large value of $k_\mathrm{+C}^\circ$. His correlation supports the idea that $k_\mathrm{+C}^\circ$ is very large but overestimates $ΔG_\mathrm{f}^\circ (\ce{IO2-})$ and probably also the dissociation constant of $\ce{IO2H}$. Figure 1 shows known $\mathrm{p}K_\mathrm{a}$ values of oxy-halogen acids, including the value of $\mathrm{p}K_\mathrm{a}(\ce{BrO2H})$ [3] not considered by Stanisavljev, and suggests that $\mathrm{p}K_\mathrm{a}(\ce{IO2H})$ is closer to $6$ than to $4$.

Figure 1. $\mathrm{p}K_\mathrm{a}$ values of oxy-halogen compounds $\ce{XOH}$ (•) and $\ce{XO2H}$ (◦).

References

Schmitz, G. Inorganic Reactions of Iodine(III) in Acidic Solutions and Free Energy of Iodous Acid Formation. International Journal of Chemical Kinetics 2008, 40 (10), 647–652. https://doi.org/10.1002/kin.20344.
Stanisavljev, D. Consideration of the Thermodynamic Stability of Iodine Species in the Bray-Liebhafsky Reaction. Berichte der Bunsengesellschaft für physikalische Chemie 1997, 101 (7), 1036–1039. https://doi.org/10.1002/bbpc.19971010708.
Faria, R. B.; Epstein, I. R.; Kustin, K. Kinetics of Disproportionation and $\mathrm{p}K_\mathrm{a}$ of Bromous Acid. J. Phys. Chem. 1994, 98 (4), 1363–1367. https://doi.org/10.1021/j100055a051.