Equilibrium Bond Dissociation The equilibrium constant for the dissociation of dinitrogen tetraoxide is $K_c=5.9 \times 10^{-3}$.

If $\mathrm{1.00\ mol}$ of $\ce{N2O4}$ and $\mathrm{0.500\ mol}$ of $\ce{NO2}$ are initially placed in a container with a volume of $\mathrm{4.00\ L}$, calculate the concentrations of $\ce{N2O4(g)}$ and $\ce{NO2(g)}$ present when equilibrium is achieved at $\mathrm{25\ ^\circ C}$.

I created an ICE Table for $\ce{N2O4 <=> 2NO2}$:
$$\begin{array}{|c|c|c|} \hline
\ & \ce{N2O4} & \ce{NO2} \\ \hline
I & \dfrac{1.00}{4.00} & \dfrac{0.500}{4}\\ \hline
C & \dfrac{-x}{4.00} & \dfrac{+2x}{4.00} \\ \hline
E & \dfrac{1.00-x}{4.00} & \dfrac{0.500+2x}{4}\\ \hline
\end{array}$$

So then I let
$$5.9\times 10^{-3}=\dfrac{\left(\dfrac{0.5.00+2x}{4.00}\right)^2}{\dfrac{1.00-x}{4.00}}$$
So products over reactants, and I divided by 4 since to get the concentration, the number of moles needs to be divided by the volume which is 4dm^3.
After doing that I got $x=-0.167$ and $-0.339$, which is clearly wrong.

Correct answers: $\ce{[N2O4]}=0.292$ and $\ce{[NO2]}=0.0418$.

Your values for $x$ are perfectly correct and lead to the correct concentrations, though I can understand why you are perturbed by negative numbers. Let's see what's going on.

First, remember that $x$ is not the final concentration. It has to do with the change in concentration. If $x$ is negative, then that means the concentration is changing in the opposite direction than what you thought it was. We'll come back to that.

Let's take one of those values of $x$ and plug it into your equilibrium concentration expressions. Let's start with $x=-0.167$.

For $\ce{N2O4}$:

$$[\ce{N2O4}]=\dfrac{1.00-x}{4.00}=\dfrac{1.00--0.167}{4.00}=\dfrac{1.167}{4.00}=0.292$$

For $\ce{NO2}$

$$[\ce{NO2}]=\dfrac{0.500+2x}{4.00}=\dfrac{0.500+2(-0.167)}{4.00}=\dfrac{0.164}{4.00}=0.041$$

Notice anything? The first concentartion is spot on with one of the answers, and the second is probably just a rounding error somewhere (yours or mine).

  Now about that negative value for $x$.

Notice that you have the concentration of $\ce{N2O4}$ decreasing in your ICE table, but the final concentration is higher than the initial (0.292 vs 0.250). Your assumption about the direction of the shift in the reaction toward was not correct. Not all reactions shift toward the right. How could we have known that the reaction would have shifted toward the left? We could have calculated the reaction quotient $Q_c$ at the initial conditions. If $Q_c > K_c $, then the products are in excess and the reaction shifts to the left. If $Q_c < K_c$, then the reactants are in excess and the reaction shifts to the right. We calculate $Q_c$ by putting the initial concentrations into the law of mass action.

$$Q_c =\dfrac{[\ce{NO2}]^2}{[\ce{N2O4}]}=\dfrac{\left(\dfrac{0.500}{4.00}\right)}{\dfrac{1.00}{4.00}}=\dfrac{0.125^2}{0.250}=6.25\times 10^{-2}>5.9\times 10^{-3}$$

Thus, your ICE table should have the signs of $x$ reversed: the concentration of $\ce{N2O4}$ is increasing and the concentration of $\ce{NO2}$ is decreasing.

$$\begin{array}{|c|c|c|} \hline
\ & \ce{N2O4} & \ce{NO2} \\ \hline
I & \dfrac{1.00}{4.00} & \dfrac{0.500}{4}\\ \hline
C & \dfrac{+x}{4.00} & \dfrac{-2x}{4.00} \\ \hline
E & \dfrac{1.00+x}{4.00} & \dfrac{0.500-2x}{4}\\ \hline
\end{array}$$

$$5.9\times 10^{-3}=\dfrac{\left(\dfrac{0.5.00-2x}{4.00}\right)^2}{\dfrac{1.00+x}{4.00}}$$

Now $x=+0.167$ and $x=+0.338$. $x=+0.167$ leads to the same answer since the signs are switched everywhere.

The other value $x=\pm 0.338$, whether negative or positive, is not valid since $0.500-2(.338)<0$.


pKa of Tris corrected for ionic strength I have a question regarding the molar fraction calculation of Tris buffer at different temperatures.

Tris dissociates according to the following equation:

$$\ce{Tris + H2O -> Tris+ + OH-}$$

So, in order to calculate the amount of $c(\text{Tris}^+)$ and thus the amount of strong acid needed for titration, one needs to use derivation of Henderson-Hasselbalch eq.:

$$\mathrm{pH} = \mathrm{p}K_\mathrm{a}' - \log_{10}\left(\frac{[\text{Tris}^{+}]}{[\text{Tris}]}\right)$$

However, $\mathrm{p}K_\mathrm{a}'$ is $\mathrm{p}K_\mathrm{a}$ corrected for ionic strength ($I_c$) at a certain temperature.

The correction of $\mathrm{p}K_\mathrm{a}$ is done via the Debye-Hückel eqation:

$$\mathrm{p}K_\mathrm{a}' = \mathrm{p}K_\mathrm{a} -0.509\times\left| z-z^{+}\right| \sqrt{I_c}$$

Now, for Ic itself one needs to calculate the molar fraction of $[\ce{Tris+}]$:

$$I_c = 0.5\times \left([\ce{Tris+}] + [\ce{OH-}]\right)$$

which requires the use Henderson-Hasselbalch equation making this problem somewhat circular. or am I missing something?

First decide the Ic you want to work with. Second, you should correct the pka with the Debye Huckel equation and finally you use the pka corrected with the Henderson Hasselbach equation.


Why the concentration with respect to time equation of a first-order reaction does not depend on the stoichiometric coefficient of reactant? According to my Chemistry textbook (Chemistry by Raymond Chang 10th edition, pg 570), for a first-order reaction of the type $$\ce{A->product}$$
the rate is $$rate=-\frac{[\Delta A]}{\Delta t}$$
From the rate law we also know that $$rate=k[A]..... (13.2)$$
Combining the two equations for the rate we write $$-\frac{[\Delta A]}{\Delta t}=k[A]$$
And finally, using calculus, $$\ln\frac{[A]_t}{[A]_0}=-kt...... (13.3)$$
All of this feels okay to me. However, later on, the book uses the same equation (13.3) to solve a first-order reaction of the type $$\ce{2A->product}$$ and I don't understand why. Because the rate of reaction would now be
$$rate=-\frac12\frac{[\Delta A]}{\Delta t}$$ and equating this relation with (13.2) gives $$-\frac12\frac{[\Delta A]}{\Delta t}=k[A]$$ and finally $$\ln\frac{[A]_t}{[A]_0}=-2kt$$ I know I'm probably misunderstanding something. Please can someone explain this to me. I've been trying to figure it out for days now.

[Edit] The problem that I'm referring to is a Practice Exercise, the book only gives the problem and the answer, not the solution. However, based on the answer is obvious that equation 13.3 was used to solve it.

Problem: The reaction $\ce{2A->B}$ is first order in A with a rate constant of $\pu{2.8 \times 10^{-2}s^{-1}}$ at $\pu{80 ^{\circ}C}$. How long (in seconds) will it take for A to decrease from $\pu{0.88M}$ to $\pu{0.14M}$?

Answer: $\pu{66s}$

If equation $$\ln\frac{[A]_t}{[A]_0}=-2kt$$ were to be used, the answer would ended up being $\pu{33s}$.

The key point which you will study in advanced classes, is the concept of elementary reactions. When a given reaction is an elementary reaction, the stoichiometric coefficients become the power of the rate law or in better words, the reaction order corresponds to the stoichiometric coefficients.

Definition of elementary reaction: An elementary reaction is a chemical reaction in which one or more chemical species react directly to form products in a single reaction step and with a single transition state (more details on elementary reactions).

Rate of reactions are determined by experiments. We never write the rate law based on the balanced equation. Compare this with equilibrium constant where coefficients are important. Even $\ce{2 A -> B}$ may be proceeding in several elementary steps. The observed rate may be the effect of several combined elementary reactions.

In those kinetic problems like yours, the writer will tell you the order of the reaction. Don't proceed to write the rate law on the basis of balanced equation.

Reaction rate and writing the rate law are two different things. For a reaction

$$\ce{a A + b B → p P + q Q}$$

Reaction is written as

$$r = -\frac{1}{a}\frac{\mathrm d[\ce{A}]}{\mathrm dt} = -\frac{1}{b}\frac{\mathrm d[\ce{B}]}{\mathrm dt} = \frac{1}{p}\frac{\mathrm d[\ce{P}]}{\mathrm dt} = \frac{1}{q}\frac{\mathrm d[\ce{Q}]}{\mathrm dt}$$

but we cannot write the rate law i.e. you do not know what would be the order of the reaction in general.

The relation above only shows how the rate of disappearance of reactants and rate of appearance of products are related. It should not be used anywhere else. Your book answer is correct.


Providing more than required amount of energy to an electron in an energy level I know that an atom has energy levels i.e the energy of each electron in the atom falls in a discrete energy level.

I am also aware that, if the exact difference of energy between two levels is provided to an electron, then it jumps to the higher energy level, and that is a lower amount of energy is provided, the electron stays where it is.

My questions are:

What happens to the energy in the second case? Does it get radiated back as heat? If not, where does that energy go?
What happens if more than the requisite amount of energy is provided? Does the electron stay where it is, or does it go to the higher energy level?

If the energy exceeds the minimum needed, there are a few possibilities:

More than one electron may be affected.
The electron(s) can be moved to yet higher levels within the atom.
The electron can be expelled (raised to a very high level) with the electron(s)' kinetic energy making up the difference.
At sufficiently high energy, pair production can occur.
Also, rather than expending all its energy on the electron, the photon can be scattered as a lower frequency (less energetic) particle (since it interacted at a point, I'd consider it a particle for that duration).

So -- the excess energy, in effect, can be shared between the photon and electron in varying proportions.

If less than the minimum amount to raise the electron's level is supplied, then the electron stays put... except for quantum phenomena such as tunneling. The Esaki diode exhibits negative resistance, where electrons are more likely to pass through the p/n junction with lower bias.


ionising atom(s) with gamma rays if a gamma ray hits an electron and transfers energy, does it hit that electron (ionising the atom), transfer all its energy and stop or does it pass through multiple electrons, transferring a portion of its energy each time?

i.e.) if a gamma ray has enough energy to ionise multiple atoms, will it, or will it just give all its energy to the one electron?

also if is has the energy to ionise multiple atoms, about how many electrons can it remove (for a range of ray energies)? (i know it is a bit general, but lets say for the first ionisation of potassium)

also, why is the dose equivalent so low? is it because it can only ionise 1 atom by +1, or because its ionizations (and hence energy) are dispersed throughout many atoms (making the meaning less energy transferred per ionisation)?

(i know this has a lot to do woth physics and so i posted it on the physics SE to, but i was wondering weather a fellow chemist could help me out, thanks)

Various mechanisms for interaction of gamma radiation with matter are possible. Depending on the energy and the composition of the absorbing material, the most important types for gamma radiation are photoelectric effect, Compton scattering, and pair production.

In absorption of gamma radiation of lower energy by the photoelectric effect, the photon is absorbed completely by the atom and an energetic electron is ejected. The energy of the emitted electron is the difference between the energy of the gamma-ray and the binding energy for that electron in the atom.

Gamma radiation of higher energy directly interacts with one electron by the Compton effect. Usually, this is the most probable interaction of gamma radiation. The incoming photon transfers a portion of its energy to the electron and is deflected through an angle with respects to its original direction. Since all angles of scattering are possible, the transferred energy varies from zero to a large fraction of the energy of the photon. The scattered photon may have sufficient energy to interact further by Compton effect or photoelectric effect.

If the energy of the gamma radiation is very large (i.e. it exceeds $1.02\ \mathrm{MeV}$, which is twice the rest-mass energy of an electron), the process of pair production is energetically possible. In the interaction, the photon disappears and is replaced by an electron and a positron. The process may be considered as the inverse of positron annihilation.

Your additional question about the equivalent dose is slightly misleading. The resulting equivalent dose from gamma radiation is not particularly low. By definition, the radiation weighting factor of gamma radiation is $w_\text{R}=1$. Thus, an absorbed dose of $D=1\ \mathrm{Gy}$ leads to an equivalent dose of $H=1\ \mathrm{Sv}$. However, you may ask why some other radiation types have a higher relative biological effectiveness.


Non-integer hybridization I've known that hybridization in distorted geometries is not exactly $sp^3$ or $sp^2$ or whatever. For example, $\ce{PH3}$ has nearly pure $p$ orbitals in the $\ce{P-H}$ bond, and the lone pair is in a nearly pure $s$ orbital.

Basically, since hybridisation is an addition of wavefunctions, instead of a perfect symmetrical addition of the kets, we get something else.

While answering this question, I realised that it's not that easy predicting the numbers.

Take the same image of $\ce{B2H6}$:

At first, seeing the 97°, I thought "well, the inner $\ce{B-H}$ bonds will be almost pure $p$". But, with that, I couldn't figure out where the 120° came from, because that is for perfect $sp^2$.

I then realised that I was being stupid and it wasn't that simple--just because you have a 120° doesn't mean pure $sp^2$. But, I was at a loss trying to find out approximate hybridizations for $\ce{B2H6}$.

How does one generally go about predicting such "nonuniform hybridizations" if one knows the bond angles?

Approximations are OK--I believe the exact mixture ratios will require some knowledge of the exact wavefunctions.

Classical hybridization theory does not allow for noninteger hybridizations. However, ab initio calculations can be interpreted using a bond order analysis method such as NBO, where the MO coefficients are used to provide the closest analogue possible to a classical hybridization picture.

For example, one of the pure $sp$ orbitals in Pauling's valence bond theory (where hybridization was first introduced) has wavefunction

$$ \phi_{sp} = \frac{1}{\sqrt 2} \phi_{s} + \frac{1}{\sqrt{2}} \phi_{p_x} $$

Taking the square of the coefficients, this orbital is $\frac 1 2$ s character and $\frac 1 2$ p character, i.e. it is an $sp$ orbital.

The basic idea of a bond order analysis method is to reexpress a molecular orbital into a form similar to

$$ \phi = c_1 \phi_{As} + c_2 \phi_{Ap_x} + c_3 \phi_{Ap_y} + c_4 \phi_{Ap_z} + \dots$$

If the other coefficients are very small, the ratio

$$ n = \frac {c_2^2 + c_3^2 + c_4^2}{c_1^2} $$

would yield a number that could be used interpret $\phi$ as a $sp^n$ orbital on atom $A$.

Edit: As for inferring hybridization states from a direct inspection of nuclear geometries, there is in principle no such direct relationship in electronic structure theories that are more sophisticated than VSEPR. The former is an electronic property whereas the spatial arrangement of atomic nuclei are not, and the relationship between the two becomes much more complicated.


Kröger-Vink notation for defect reactions In the Kröger-Vink notation the effective charge $c$ is marked as superscript while the site location $s$ is marked as subscript on the atom itself $\ce{A}$:

$$\ce{A^{c}_{s}}$$

As far as I have learned the effective charge is the charge difference from the perfect lattice situation. The Wikipedia definition agrees:

  $c$ corresponds to the electronic charge of the species relative to the site that it occupies

While neutral charge (no difference) is usually an $x$, I have always worked with a slash $/$ or similar for one negative charge and a dot $\bullet$ for one positive charge.

I have run across a notation in a teacher's lecture material that says:

$$\ce{Zn^{/}_{Zn}} \quad \text{ and } \quad \ce{Zn^{\bullet}_{Zn}}$$

As I understand it, adding a $\ce{Zn}$ to it's own natural position makes no charge change, so it should be $\ce{Zn^{x}_{Zn}}$. How is the above notation then possible? Does it imply that the $\ce{Zn}$ has lost or gained one electron, respectivily, and thus has a net charge?

Yes, the Kröger-Vink notation you are asking about, implies that the metal has lost or gained one electron, relative to the case of the perfect crystal. This situation can be explained by non-stoichiometry.

A non-stoichiometric compound is one of which the proportions of the atoms can not be represented by integers.
Consider the case of $\ce{Fe}_{0.95}\ce{O}$. Indicated by the subscript 0.95, there are some metal atoms missing, which is expressed as the defect $\ce{$V$^{''}_{Fe}}$. This has a double negative charge, which can be compensated by 'replacing' $3~\ce{Fe^2+}$ by $2~\ce{Fe^3+}$ ions. That is, 2 iron ions need to be oxidized and 1 needs to disappear.

This can be accomplished by dissolving $\ce{Fe_2O_3(s)}$ into $\ce{FeO(s)}$, as in:

$$\ce{Fe_2O_3(s) ->[FeO] 2Fe^{.}_{Fe} + 3O_{O} + $V$^{''}_{Fe}}$$

Alternatively:

$$\ce{2Fe_{Fe} + 2O_{O} + 1/2 O_2(g) ->[FeO] 2Fe^{.}_{Fe} +3O_{O} + $V$^{''}_{Fe}}$$

More examples are described in my source, chapter 11 of the book 'Ceramic Materials' by Carter & Norton.


Why does the pH of soft drinks get lower as they go flat? I am doing an investigation about the chemistry of soft drinks and as part of this I observed the effect of time on the pH of soft drinks fully expecting the pH to increase as the carbonic acid, a weak acid, decomposed and CO2 escaped. Instead, time after time, the pH decreased, not by much though in fact it hardly made a difference to the acidity of the drinks but it the pH still decreased every time.

So I did a control experiment with soda water and found that the pH did increase when the CO2 escaped the drink. I originally did it with Coke and Pepsi, which aside from carbonic acid also have phosphoric acid (Pepsi has citric acid as well) so the increase in acidity probably has to do with an interaction between the the weak carbonic acid and the strong phosphoric/citric acid. Can anyone help me to explain this?

I think you said it... phosphoric acid. Carbonic acid is "buffering" phosphoric acid... test seltzer water (composed of water and carbon dioxide with no other additives).

Another way of thinking about this is to consider carbonic acid (nominally a weak acid) to actually be a base. Looking at the pKa (the pH at which half of the acid is dissociated) is the best way to describe this.

The pKa of Carbonic acid is 6.35

The pKa Phosphoric acid is 2.16

pH is basically a measure of free (dissociated) H+ in water... so as CO2 is removed, it is not available to act like a buffer and bind with H+... so the free H+ increases.


how to measure the efficiency of carbon filter [or of activated carbon/charcoal]? Fake HEPA filters are easy to spot using DIY PM sensor and DIY Air Purifier (we just need to turn the fan ON, place the sensor's tube in front of it, and read the data).

But what about carbon filters?

I bought a filter like this to clean the air, but I wonder if it really works (there are many fakes out there).

I'm looking for some tricks to know if it works, and (much better) to evaluate its efficiency (using Arduino sensors or any basic ways).

1/ The iodine method:

I found the answer on this video (from Cody's lab youtube channel). Cody's show how to make activated carbon/charcoal and how to test its efficiency.

It's quite an easy task. The test below measures the absorption's efficiency of your charcoal filter (activated or not).

● Take a gram of your active charcoal (from your filter)
● Put it in a test tube
● Add 2ml hydrochloric acid (to make sure the charcoal is acidified, because otherwise, the alkaline ash in the charcoal will react with the iodine)
● just warm it a little to help the reaction.
● Add 25ml of iodine (iodine with alcohol).
● Wait for 24h, look at the color of your solution: the lighter the color the more effective is the charcoal (the lighter, the more it has absorbed iodine which is dark). You could compare it with some active carbon you bought, or normal charcoal

To measure more precisely the efficiency of your charcoal:
● Take 10ml of the solution you made above.
● Add some thiosulfate to it (drop after drop) until your solution becomes clear.
● Measure the quantity of thiosulfate that was needed.

See Cody's video for more details.

2/ The Formaldehyde (or VOIC) method:

If you want to know if your activated carbon filters works, you can also test it using Thomas Talhelm method (he is the most famous air pollution DIY expert). But this technique require a solution of formaldehyde, and a device to mesure the quantity of formaldehyde in the air.

Buy a formaldehyde solution.
warm it (eg. using a rice cooker) in a closed room.
measure the level of formaldehyde
turn ON your activated carbon filters, and keep measuring.

If your filter is working you should see that:

  With a fan only (red line), formaldehyde levels stayed high. But with
  a carbon filter on the fan (blue line), formaldehyde levels went down.


Database for molecular dipole transition moments? I'm looking for the dipole transition matrix elements for electronic transitions between the ground and excited states of the carbon dioxide cation, $\text{CO}_2^+$, that is, objects like
$$\langle\text{CO}_2^+ \,\tilde{\text{A}}\,\Pi_\text{u}|\mathbf{r}|\text{CO}_2^+ \,\tilde{\text{X}}\,\Pi_\text{g}\rangle.$$
I feel this is a standard enough quantity that it ought to be easy to find in some appropriate database of chemical data. However, I've searched for quite some time and I can't even find something similar.

Where should I be looking?

You can find some experimental information on dipole transition moments in the NIST Chemistry Webbook in the section on constants of diatomic molecules.

http://webbook.nist.gov/cgi/cbook.cgi?ID=B3000043&Units=SI&Mask=1000


Differences between KOH and phosphorus pentoxide as drying agents? The most common drying agent I encountered when using a vacuum dessicator was phosphorus pentoxide in the form of Sicapent. But I've recently encountered people using KOH as a cheaper alternative (in a practical course for students, where such costs are amplified due to the high number of students).

What are the differences between KOH and phosphorus pentoxide as a general purpose drying agents to remove water? Is there a difference in effectiveness? Are there classes of substances where you can't use one of those drying agents?

I found this page, which recreates information from Purification of Laboratory Chemicals, by Perrin and Armarego. It provides a qualitative ranking of drying agents, listing phosphorous pentoxide higher than KOH. The two drying agents work by different means:

KOH

Anhydrous KOH is hygroscopic. It absorbs water from the atmosphere. It much better at doing so than more common hygroscopic drying agents like calcium chloride or calcium sulfate (drierite). A pellet of KOH left out in a beaker or dish open to the air will have absorbed so much water vapor within an hour that it will now be a solution of KOH (that will continue to be hygroscopic, at least until the solution is 50% water). Calcium chloride and calcium sulfate only continue to absorb water until they reach their maximum hydration.

P4O10

Phosphorous pentoxide (really P4O10) is an anhydride of phosphoric acid. Phosphorous pentoxide reacts with water to form phosphoric acid:
$$\ce{P4O10 + 6H2O -> 4H3PO4}$$
Phosphoric acid is also hygroscopic and absorbs water to form a solution.

Since P4O10 is water reactive it is more difficult and dangerous to handle, especially for large groups of students. I might not use P4O10 if the substance being dried is very acid sensitive. Eventually, the vacuum will start volatilizing the acid. I might not use KOH for compounds that are incredibly base sensitive, although it is unlikely that KOH will sublime. In an undergraduate lab with multiple students, KOH should be fine if you want to save money and hassle.


Non-bonding molecular orbital in triatomic hydrogen I am slightly confused about how non-bonding MOs form, for example in H3+  ion as shown below.

Why does the central atom not contribute an orbital?

The source mentions symmetry, but I can only see that the central orbital has a positive or negative phase, so it will inevitably cancel somewhat on one side and add with the wavefunction on the other side. This orbital will have a distinct appearance.

Could someone explain why there is no contribution, despite us adding s orbitals whose symmetry always allows them to be added (and even if we had p orbitals, surely the p orbital which can contribute will contribute?)

In order to draw the MO diagram for the H3+ system, it helps to first thing about the MO diagram for a simple diatomic case, and then mix in the additional s-orbital.

The diagram below hopefully illustrates this technique (and can be found in many introductory textbooks).

First consider two s-orbitals (the right hand side of the diagram). They can either mix in a symmetric (in phase) or anti-symmetric (out of phase) combination.

The two atomic orbitals we initially combine can give rise to two molecular orbitals, as shown on the right hand side.

We can give them symmetry labels to help us identify what kinds of things these molecular orbitals can then combine with to form the new molecular orbitals for the triatomic system.

The labels g and u are related to whether the molecular orbital is symmetric or anti-symmetric with respect to an inversion centre.

Once our two molecular orbitals for the diatomic system are formed, we can mix these orbitals with another hydrogen s-orbital in order to generate the final MO diagram as per your question.

Only the g molecular orbital from the diatomic system is of the correct symmetry to interact. This mixing creates two molecular orbitals (the top and bottom ones) for the triatomic system. The middle, non-bonding molecular orbital is simply the orbital of u symmetry from the initial diatomic system, which is unable to interact since there is nothing of the correct symmetry for it to do so.


Which is more activating among Phenol, Sodium Phenoxide & Anisole? Also arrange them in increasing order of their activating effect This is in reference to Kolbe's Reaction, actually I think that Sodium Phenoxide is the most activating because, in Kolbe's Reaction we don't make react $\ce{CO2}$ directly with phenol for the electrophilic substitution(Maybe because of it's comparatively less activating effect), but we do make Sodium Phenoxide react with $\ce{CO2}$ (Maybe because of it's comparatively more activating effect) so that it will get attach to the Ortho position of Benzene Ring.
Along with this, please also tell me why in Kolbe's Reaction, the main product is Ortho directing & in most of other reactions, the main product is para directing?

Sodium Phenoxide is much more activating than phenol itself. Because here directly one negative charge comes and bond with the C below and the pi bond shifts to the ortho position as negative charge. And it also answers your doubt that why ortho product is major.
You can refer to this mechanism. After reacting with OH- then the tautomerization takes place but before it no effect where as in case of phenoxide it can perform the same reaction even without OH-
So activation order
Anisole


Why is phenolphthalein an appropriate indicator for titration of a strong acid with a strong base? How does the pH at the equivalence point, as well as the pH range over which the colour of phenolphthalein changes, make it a suitable indicator for titrations between strong acids and strong bases?

If you look at the titration curve, which plots the volume of base added vs pH (source):

you can see that the equivalence point occurs at pH = 7.

Phenolphthalein is fuchsia in pH's roughly between 8.2 and 12, and is colorless below pH 8.2. When the number of moles of added base is equal to the number of moles of added acid (or vice versa; example valid for strong monoprotic acids/bases assuming 100% dissolution), the pH is equal to 7.

You might say, if the pH needed is 7, and phenonphthalein changes at pH's around 8.2, how can you use this indicator?

Well, again looking at the curve, from pH = 11 to about pH = 4, pH changes very rapidly with from an infinitesimally small change in the volume of base added. Since one drop of added titrant will cause this large change, even though the change in color of phenolphthalein does not occur right on the equivalence point, it is within approximately one drop. This kind of uncertainty is "acceptable uncertainty" in using titration to volumetrically determine concentrations.

To clarify what I mean by "acceptable uncertainty", you should realise that each of your measurements has some kind of uncertainty to them:

When you weighed out the primary standard to titrate against, was the balance perfect?
Was your solution made up precisely to the graduation in the volumetric flask?
Did you pipette the exact volume of the aliquot or were you off by a drop or two?
Are you able to state the volume added from the burette to arbitrary precision or is there some uncertainty beyond the two decimal places given by the graduated lines?

In the scheme of things, ~1-2 drops will not be a significant factor in getting an accurate result, but you should most definitely acknowledge that there is uncertainty in your answer.


Assigning R/S configuration Question:
Assign the R/S configurations to the carbons which are attached to different rings of this molecule (carbons 1 and 3 according to IUPAC numbering):

My Solution
I first started out by numbering the carbons. The one at the top would be 1 and the bottom chiral carbon would be 3. Next, According to CIP rules, I have included the phantom atoms by replacing the double bonds while deciding the priority order. Moving on, at carbon 1, My priorities would be:

$\ce{-OH}$
Cyclopentene ring (on the left)
Cyclopropene ring (on the right)
Carbon 2

This gives me a counter-clockwise turn and since the least priority group is on the vertical line (away from us) the configuration I assign would be S.
Following a similar procedure for carbon 3, The priorities would be :

Bottom 5-member ring (contaning $\ce{O-O}$ bond)
3 member ring on the right
5 member ring on the left
Carbon 2

Again, this gives me a counter-clockwise turn and since the least priority group is away, I would assign the configuration: S.
The answer key, however, says that the answer is 1R, 3S.

Where am I going wrong?

I have separated C1 and C3 into unique structures. For C1, it is clear that the oxygen of the hydroxyl group has the top priority while the methylene group has the lowest priority. A digraph is constructed such that we delineate two paths around each of the unsaturated rings. Thus, non-duplicate atoms C3 and C5 terminate with their respective duplicate atoms C(3) and C(5). The cyclopentene ring has locants in sphere 3 of {(7),7,H} which is a tie with {(4),4,H} in sphere 3 of the cyclopropene ring. But the locants {8,H,H} of the cyclopentene lose to {(3),3,H} of the cyclopropene ring. The priorities are O>cyclopropene>cyclopentene>methylene. C1 has the (R)-configuration. continued...

A different digraph for C3 is constructed by dissecting the three rings into a total of six pathways, two for each ring. The rings are designated with red, blue and black dots. The red and blue rings take precedent over the black ring because each one has locants {O,O,H} at the branchs while the black ring has {O,C,H}. To distinguish between blue and red, we move to the next atoms on the chain. Blue has two oxygens while red has two carbons. The priorities are blue>red>black>methylene. C3 has the (S)-configuration.

Related issues:

What are the CIP rules for cyclic substituents?
How to check for geometrical isomerism in cyclic compounds?
More on Rings and Digraphs


Use of combinations to determine number of isomers Our teacher asked us how many geometrical isomers are possible with the formula $\ce{c2ClBrIf}$.
I answered with the logic $^4C_2=6$ and it turned out to be correct.

Is the use of combinations this way generally helpful in determining the number of possible isomers? For example, if I have a rigid molecule with three carbon atoms and six substituents (perhaps a cyclopropane), is the correct number of possible isomers 20?

$$^6 C_3 =\dfrac{6!}{(6-3)!\cdot 3!}= 20$$

For the following I've numbered the positions on ethylene where the four halogen atoms could go. (The Carbon-Carbon double bond doesn't "spin.") It doesn't matter which halogen goes to 1, so let's make it Fluorine. If Fluorine were in any other position then you could flip or rotate the molecule to get Fluorine to position 1. Now there are three choices left for position 2, which leaves 2 choices for position 3. So the number of possible isomers is:

$3 \times 2 = 6$

Note that if you just willy nilly drew molecules then you could draw 4! or 24 permutations. However when you rotate and flip the molecules, then only 6 unique molecules exist.


What is the number of unpaired electrons in Fe 2+? Specifically, how can I find out the number of unpaired electrons in any ion?

First draw the electronic configuration for Iron.

Now since we are given with Iron(II) we take out two electrons from the outermost orbit so as to give the iron a +2 charge.

As evident from the configuration, in total there are 4 unpaired electrons.

Using the same logic you can count the number of unpaired electrons for any ion.


What is the major product on chlorination of 2-methylbutane? I came across a question asking for the major product on chlorination of 2-methylbutane. The answer in the back of the book says it will be 2-chloro-3-methylbutane. However, I thought the major product would be 2-chloro-2-methylbutane. The later is formed through a tertiary free radical with eight hyper-conjugative structures, whereas the former forms through a secondary free radical with only four hyper-conjugative structures. How is this possible?

Because of the great reactivity of the chlorine radical, abstractions of primary, secondary, and tertiary hydrogen atoms are all exothermic. Therefore, the stability of the product radical has less influence on the activation energy of the reaction. Thus, according to the Hammond postulate, the transition state is more reactant-like.

According to R. Brückner Advanced Organic Chemistry, the chlorination of isopentane has four monochlorination products:

22 % 2-chloro-2-methylbutane
33 % 2-chloro-3-methylbutane
30 % 1-chloro-2-methylbutane
15 % 1-chloro-3-methylbutane

The statistical factor is responsible for this.
If each hydrogen atom of isopentane could be substituted at the same rate, the ratio would be

1 hydrogen atom (8 %) → 2-chloro-2-methylbutane
2 hydrogen atoms (17 %) → 2-chloro-3-methylbutane
6 hydrogen atoms (50 %) → 1-chloro-2-methylbutane
3 hydrogen atoms (25 %) → 1-chloro-3-methylbutane

In terms per hydrogen atom, the experimental yield is

22 % / 1 = 22 % → 2-chloro-2-methylbutane
33 % / 2 = 16.5 % → 2-chloro-3-methylbutane
30 % / 6 = 5 % → 1-chloro-2-methylbutane
15 % / 3 = 5 % → 1-chloro-3-methylbutane

Thus, 2-chloro-2-methylbutane is indeed slightly preferred – as you have expected.


Equilibrium: Equations with complex roots I'm studying civil engineer in my first year. We have a course called "General Chemistry". One of the chapters was about chemical balance of a reaction (called equilibrium in English, I think). We got exercises on this were we had to calculate the amount of chemicals got converted to others, using acidity constants Ka.

For example, the equations we had to solve were easy and looked like this:

$$K_a = \frac{(x)^2}{(0.1 - x)(0.2 - x)}$$

These exercises were easy. But then came the point were we combined two reactions at the same time and had to calculate the balance for both at the same time. The equations we get were:

The tutor used to make simplifications by getting rid of the "$-x$" and "$-y$" in the denominator, assuming x and y would be so small it would almost make no difference. I don't like making such simplifications: I want the correct answer. So, what I did was putting the equations in WolframAlpha and let it compute it, since I don't know how to solve such a system of equations. The result shocked me: the solutions for x and y were complex (so, with a real and imaginary part).

WolframAlpha says this:

Whereas the tutor found (due to simplification):

The tutor's solution is in the complex plane not really far away from the correct solution to the equation (WolframAlpha's solution #3).

So, my question is: How do you explain these roots are in fact complex? I don't think a complex amount of chemicals can react right?

Complex roots to mass/concentration balance equations should be discarded as they have no physical interpretation. If there are only complex roots, then odds are there's some error in the manipulation of variables that lead to the equation.

Your case seems different though. Notice the very small magnitude of the imaginary part of the complex roots; the number is "almost real". This happens at times simply because of Mathematica's numerical root search algorithms, which aren't always guaranteed to reach the exact value. I'm actually a bit surprised that it got caught up in such a simple calculation.

As proof that the imaginary part of the root is nothing but an issue with the calculation, I simplified the equations until I found a better representation for Wolfram Alpha to parse, hoping the complex root would disappear.

First, let $A=1.77\times 10^{-4}$ and $B=3.3\times 10^{-4}$ for concision. Isolating $x$ in the second equation gives:

$$x=\frac{B(0.1-y)-y^2}{y}$$

Replacing $x$ for the above expression in the first equation then yields, after some tinkering:

$$(A+B)y^3+(B^2-0.1A-0.1B-AB)y^2+(0.1AB-0.2B^2)y+0.01B^2=0$$

Aha, a simple polynomial. Alpha should be able to deal with that just fine. Replacing the values of $A$ and $B$ produces the following equation, which Alpha solves with no imaginary part. The physically meaningful solution out of the three roots is $y\simeq 0.00458505$. Replacing that value in the equation for $x$ completes the problem, with $x\simeq 0.00228225$, very close to the approximate solutions.


Why is the 5d orbital instead of the 4f orbital filled in lanthanum? For orbitals with the same value of $n + l$ (e.g. the $\mathrm{3d}$ and $\mathrm{4p}$ orbitals), we usually expect the orbital with the lower value of $n$ to be filled earlier. Hence, for example, the $\mathrm{3d}$ orbitals are filled before the $\mathrm{4p}$ orbitals in the transition metals.

However, the electronic configuration of lanthanum is $[\ce{Xe}](\mathrm{5d})^1(\mathrm{6s})^2$, and not $[\ce{Xe}](\mathrm{4f})^1(\mathrm{6s})^2$.

I know that the $\mathrm{5d}$ and $\mathrm{4f}$ orbitals are of similar energies, but is there any better explanation for this?

According to Ground-state configurations of ionic species I through XVI for   Z = 57-74 and the interpretation of 4d-4f emission resonances in laser-produced plasmas Phys. Rev. A 25, 275 (where "I" means the neutral atom) :

  We begin therefore by showing in the left-hand
  columns of Table I the configurations given by
  Martin et al. for lanthanum through lutetium.
  Here it is seen that at the fourth spectrum the 4f
  level has in all cases fallen below 5d and 6s. This
  is due to the well-known phenomenon of 4f wavefunction
  collapse [reference 5]. The effective potential in
  which the 4f electron moves is made up of an attractive
  Coulomb term and a repulsive centrifugal
  term, which, in general, combine to give a double
  well potential, the height of the intervening barrier
  depending on the nuclear charge Z. For neutral
  atoms the field in the region of the outer potential
  is essentially hydrogenic. The occurrence of the
  neutral lanthanides coincides with the appearance
  of bound 4f levels in the inner well, which, in turn,
  is due to the deepening of the well and the lowering
  of the potential barrier as a result of the increasing nuclear charge.

Here, "reference 5" is Spectral Distribution of Atomic Oscillator Strengths Rev. Mod. Phys. 40, 441 which goes into quantitative equations for the attractive Coulomb term and a repulsive centrifugal term.


Comparing between SN2 and E2 Question
I am confused between the questions B. And C. The earlier is SN2 and later is E2. But what makes the difference?

The most important factor in these reactions are nucleophiles. Here, the use of strong base (Sodium methoxide) in both of the reactions favors the bi-molecular mechanisms ($\ce{S_N2}$ and $\ce{E2}$).

Now, the second factor would be degree of halogen-linked carbon. Lower degree carbons ($\ce{\text{deg.} < 2}$) favor $\ce{S_N2}$, while higher degree favors $\ce{E2, E1, S_N1}$. So, $\ce{S_N2}$ is favored in first and $\ce{E2}$ in second.

Temperature conditions and nature of solvent are least important factors, and should be looked up when above two factors are similar.


Why do dianions (such as malonate) bind cations more strongly than anions? Why does a dianion (such as malonate) bind cations more strongly than its equivalent anion (acetate)?

Is it simply because of the proximal availability of another $\ce{O-}$ group that can bind to cations? Does the second $\ce{COO-}$ group on malonate distribute charge in a better way that can allow for stronger binding?

This is known as the chelate effect.

The main reason why you observe this is that cations in solution have an ordered solvent shell around them, especially in polar solvents where there will be defined solvent geometry around the shell; often octahedral for metal ions in water. The formation of a complex with a bidentate ligand such as malonate is more favourable than that with a monodentate ligand such as acetate because the same number of solvent ligands are lost with half the number of ligands complexed. This results in a greater increase in entropy - you can see this in terms of the degrees of freedom gained by the water molecules more than compensating for those lost by the new ligands.

  Is it simply because of the proximal availability of another $\ce{O-}$ group that can bind to cations ?

This is likely also a factor, although in most cases less important. In the case of the monodentate ligands there is a small energetic cost associated with bringing the anionic oxygen centres close to complex the metal centres. In the case of the bidentate ligand, some of this cost is "already paid" in the enthalpy of formation of the ligand - it is preorganized. This contribution becomes dominant for macrocyclic ligands.


Carbocation rearrangement I am dealing with some basic reactions in organic chemistry. One of these is the acid reflux of alcohols. Something I came across was hydride shifts. I know how and why they occur, but I don't understand why they are called 1,2 hydride or methyl shifts. What do these numbers indicate? I initially thought it would be the position of the carbon atoms involved in the shift, in the principal chain, but that was probably wrong because a shift occurring at the 3rd carbon atom was also called a 1,2 shift. Help?

I believe that in the examples where a 1,2 hydride shift was mentioned, the hydrogen atom would have 'shifted' between two adjacent carbon atoms. I think the 1,2 refers to the fact that the carbon atoms involved in the shift are adjacent.
A 1,3 shift would mean a the hydrogen atom shifts to the second carbon from the original carbon atom it was bonded to, if you get what I mean.
The same applies to methyl shifts.


Do almonds smell like they do because of cyanide? I am guessing so otherwise it is quite a coincidence that cyanide smells of bitter almonds. And if so, are large quantities of almonds over a long period safe?

Almonds smell like they do mostly due to the presence of benzaldehyde:

  This colorless liquid has a characteristic almond-like odor. Benzaldehyde is the primary component of bitter almond oil and can be extracted from a number of other natural sources.

Hydrogen cyanide also has an almond-like odor, but it is not as pronounced as that of benzaldehyde. From CDC|Facts About Cyanide:

  Cyanide sometimes is described as having a “bitter almond” smell, but it does not always give off an odor, and not everyone can detect this odor.

Benzaldehyde and hydrogen cyanide are both byproducts of the enzymatic catalysis of amygdalin, which is present in almonds (and other foods). Have a look at this abstract (Toomey VM, Nickum EA, Flurer CL. Cyanide and amygdalin as indicators of the presence of bitter almonds in imported raw almonds. J Forensic Sci. 2012 Sep 57 (5): 1313-7) for more about the cyanide-amygdalin-almond connection.

In terms of safety:

  For a 70-kg human, the lethal dose is estimated at 50 mL. Benzaldehyde has been classified as a hazardous substance by the United States Environmental Protection Agency.

and

  Benzaldehyde does not accumulate in any specific tissues. It is metabolized and then excreted in urine.

Safety issues from benzaldehyde consumption $\it{via}$ the almond consumption route appear to be minimal.

From our sister SE site Skeptics in re: "How many almonds does it take to get cyanide poisoning?":

  With bitter almonds, 8 - 32 almonds will give you the lethal dosage of cyanide. Bitter almonds yield about 6.2 mg of cyanide per almond and the LD50 for cyanide is 50 mg - 200 mg.

  This applies only to bitter wild almonds: considering that you're not dead, you most likely ate domesticated sweet almonds, which apparently do not have this problem. The sale of wild almonds may be illegal, in fact, though I can only find blogs and internet comments alluding to this.

  From wikipedia

  Wild almonds are bitter, the kernel produces deadly cyanide upon mechanical handling, and eating even a few dozen at one sitting can be fatal.
  ...

  While wild almond species are toxic, domesticated almonds are not

Given the evidence and the expectation that you won't ingest several kilograms of almonds in a day or so, I'm comfortable in endorsing the almond as safe for consumption.


Molecular Oxygen in Suzuki Homocoupling I'm curious as to the role of molecular oxygen in Suzuki homocoupling reactions, as discussed here.

In a standard Suzuki cross-coupling reaction, molecular oxygen is not required. However, this paper claims that the homocoupling reaction will not proceed in an Argon environment. Is there a (relatively) straight forward justification for this that I am missing?

This is the reaction mechanism from the paper:

I don't remember a direct palladation of arylboronic acids or their derivatives in any of the couplings. Typically, $\ce{Pd(0)}$ inserts into the $\ce{C-Hal}$ bond of arylhalides and is thus oxidized to $\ce{Pd(II)}$. Ouite often, $\ce{Hal}$ is then displaced with $\ce{OR}$ by the action of a base. Think in $\ce{KOt-Bu}$ often used in the Kumada coupling.

In the homo coupling that you described here, there is - for obvious reasons - no $\ce{Ar-Hal}$ for the oxiative addition step. Without $\ce{O2}$ for the oxidative activation of $\ce{Pd(Ph3P)2}$, the mixture will thus remain inactive.


What's an "amagat"? How is it different from the Loschmidt Number? At class, we were (briefly) taught about the Loschmidt Number/Constant ($n_0$), which isn't really different from the Wikipedia definition,

The Loschmidt constant or Loschmidt's number (symbol: $n_0$) is the number of particles (atoms or molecules) of an ideal gas in a given volume (the number density).

Furthermore, it says,

It is usually quoted at standard temperature and pressure, the 2010 CODATA recommended value is  per cubic metre at $0 °C$ and 1 atm.

 (Yes, I know STP is currently defined at 1 bar pressure, and not 1 atm. Since this isn't my main question, I'm bringing it up here: Is the Wikipedia definition of the Loschmidt number inaccurate? Should it be defined at 1 bar, instead?)
Reading on, I saw,

Being a measure of number density, the Loschmidt constant is used to define the amagat, a practical unit of number density for gases and other substances:
1 amagat = $n_0$ = $2.686 7805(24)$ x $\mathrm{10^{25}}$ per cubic meter,
such that the Loschmidt constant is exactly 1 amagat.

To begin with, I've never heard of an "amagat" [<--- I mean the unit, not the law named after the same person]. Yes, this may be the result of my shortcomings as a high-schooler with limited (almost pitiful, really) exposure to physical chemistry...but what bugs me more, is that I still can't wrap my head around it even now.
Including the Wikipedia definition for the amagat before I continue,

An amagat is a practical unit of number density. Although it can be applied to any substance at any conditions, it is defined as the number of ideal gas molecules per unit volume at 1 atm and 0 °C.

We already have the good ol' Loschmidt Number...so why introduce another term/unit that is (seemingly) defined the same way, has the same value and carries the same (SI) units?
Besides, what's with the "...practical unit of number density." bit?

I haven't seen any of these particular units before so I could be mistaken, but I imagine it is similar to how we don't say "I have 3 Avogadro's Numbers of water" we would say "I have 3 moles of water". We very well could describe amounts by explicitly naming multiples of the constant, but we instead defined a new unit so that we could describe things more easily.


Why is dinitrogen tetraoxide less stable at high temperature? I have calculated the enthalpy of dimerisation for the reaction at low temperature:

$$\ce{2NO2 <=> N2O4}$$
$$\Delta_\mathrm{dimerisation}\ce{H}= \pu{-13.87 kcal mol^{-1}}= \pu{-58.03  kJ mol^{-1}}$$

Also for this reaction, $\ce{\Delta S}$ is negative due to presence of more gaseous moles on LHS.

My book mentions that $\ce{N2O4}$ is stable only at low temperature. Why is it so? I couldn't find any reliable sources on the internet.

I figured out the answer using Van't Hoff Equation:
$$\ln\left(\frac{\ce{K_2}}{\ce{K_1}}\right)=-\frac{\ce{\Delta H}}{\ce{R}}\left({\frac{1}{\ce{T_2}}-\frac{1}{\ce{T_1}}}\right)$$

Let's assume that $\ce{\Delta H}$ is constant for a limited range of temperature.

Since $\ce{\Delta H}$ is negative for the reaction $(=\pu{-13.87 kcal mol^{-1}})$ if we increase the temperature i.e. $\ce{T_2>T_1} \to\ce{K_1> K_2}$. Thus, equilibrium shifts towards the left as the temperature increases implying the decreasing stability of $\ce{N2O4}$ with increasing temperature.


Effect of pressure on transition temperature I was wondering if someone can explain why the answer to the following question is volume instead of the others.

If pressure has no effect on the transition temperature between two
crystalline forms of matter the two forms have the same molar
(a) volume
(b) energy
(c) enthalpy
(d) entropy
(e) heat capacity

I know the answer is volume but I'm not sure which equation that comes from. I was thinking maybe the law of reduced properties but at this point I'm just guessing based on the answer and I know that that particular equation is for gases. Does anyone know if there's an analog for crystalline forms of matter or if there's a more intuitive explanation? Thanks!

The mathematical answer is based on Clapeyron's equation (written in what follows in form inverted relative to the more usual):
$$\frac{dT}{dp}= \frac{\Delta V}{\Delta S}$$
which must hold on any point on the phase coexistence line.
If $\frac{dT}{dp}=0$ then it follows that $\Delta V=0$.


Pressure exerted by the mixture of nonreactive gases $\pu{2.0 L}$ of $\ce{O2 (g)}$ and $\pu{8.0 L}$ of $\ce{N2}$, each at $\pu{0.00^\circ C}$ and $\pu{1.00 atm}$, are mixed together. The nonreactive gaseous mixture is compressed to occupy $\pu{2.0 L}$ at $\pu{298 K}$. What is the pressure exerted by this mixture?

I tried to solve this problem by first using

$$P_\mathrm{tot}=\frac{n_\mathrm{tot}RT}{V}\quad(T=\text{constant}; V=\text{constant})$$

then using

$$V_\mathrm{tot}=\frac{n_\mathrm{tot}RT}{P}\quad(T=\text{constant}; P=\text{constant})$$

Plugging in the numbers for the second equation

$$10=\frac{n_\mathrm{tot}\cdot 8.314\cdot 273}{101325}$$
$$n_\mathrm{tot}=\pu{0.45 mol}$$

Then plugging in the numbers for the first equation

$$P_\mathrm{tot}=\frac{0.45\cdot 8.314\cdot 298}{2}=\pu{6.76 atm}$$

Whereas the answer is $\pu{5.5 atm}$. What am I doing wrong here?

You did too many unnecessary steps and also didn't use unified units (e.g. liters instead of cubic meters, since you decided to use $R=\pu{8.314 J mol-1 K-1}$). I denoted both total pressures as $P_1$ and $P_2$ for simplicity since we don't need partial pressures here. From the ideal gas law:

$$P_2=\frac{n_2RT_2}{V_2}\tag{1}$$

Since both gases are not reacting at these conditions,

$$n_1=n_2=\frac{P_1V_1}{RT_1}\tag{2}$$

and (1) can be rewritten as following:

$$P_2=P_1\frac{V_1T_2}{V_2T_1}=\pu{1.00 atm}\cdot\frac{\pu{10.0 L}\cdot\pu{298.13 K}}{\pu{2.0 L}\cdot\pu{273.13 K}}\approx\pu{5.46 atm}\tag{3}$$


Gas phase chiral molecules Are there any molecules that are chiral that are also gases at room temperature?

Failing that, what chiral molecules have a high vapor pressure? Propylene oxide has a boiling point of only 34 C, so that seems like a good one. Are there others?

EDIT: I appreciate all of the great suggestions! I should have mentioned before, but I am looking for something that can be purchased in both R and S enantiomers.

The molecule has to be small. trans-1,2-difluorocyclopropane is certainly chiral, and its predicted boiling point is ~ 5 deg C.  trans-1,1,2,3-tetrafluorocyclopropane is another likely candidate.  This one will be easier to prepare (from perfluoropropylene oxide and trans-1,2-difluoroethylene).

Separation of enantiomers is the hard part.


Can Michaelis-Menten constant be zero in the Michaelis-Menten Equation, or is this a non-zero constant? Can Michaelis-Menten constant be zero in the Michaelis-Menten Equation, or is this a non-zero constant?

I am applying the equation in epidemiology so I wanted to assume that it is zero but I am not sure if it is possible for this constant to be zero?

It does not make sense for the constant to be zero. The constant is usually expressed as follows:

$$ K_M = \frac{k_{\mathrm{reverse}} + k_{\mathrm{catalytic}}}{k_{\mathrm{forward}}} $$

Since the $k$ values are strictly non-negative, the constant would only be zero if $k_{\mathrm{reverse}} = k_{\mathrm{catalytic}} = 0$. In theory, $k_{\mathrm{reverse}}$ could just be $0$ (but that would not be called a Michaelis-Menten model if I want to nitpick), but if $k_{\mathrm{catalytic}}=0$ then you have no overall reaction at all.

Another way to see the same thing is to recall that the Michaelis-Menten constant has the same numeric value as the substrate concentration at the half of the maximum reaction rate. Can this be zero? It would not make any sense to me.

In any case, if you would extend your question to contain information about the epidemic model you would like to create, I am sure a lot of people (including me) would love to help finding a good reaction kinetics analogue for it.


Transmission, absorption, and reflection of light My understanding is: when light hits an object, it can do one of three things: transmit through the object, be absorbed into the object, or reflect from the object.

However, I have also learned that when light hits atoms, it can excite electrons, leading to another emission of light as the electrons go back down to their ground states.

The first paragraph, while intuitive, does not seem to be logically explained by the second. Can someone please help me understand the first in terms of the concepts of the second?

Specifically, is absorption of light exciting an electron with more energy than necessary, and some energy remains in the atom, or is the entire energy of the photon absorbed?

Is reflection just hitting the atom with a photon that is just the right amount of energy to get it excited, so no net energy is gained?

While I can't explain the phenomenon of light transmission, I can say a few things. The energy levels correlating to physical states and the gaps between the states are quantized. This an all or nothing reaction. The frequency of light must exactly match the energy (E=hv) needed for the excitation event (electron is pushed into a higher orbital, a vibrational mode to expressed, etc) for absorption to occur. When the light frequency is mismatched and does not absorb, it will reflect. The frequency that induces excitation is theoretically the frequency of emission, but phenomena such as intersystem crossing may create multiple paths by which the excitation energy will disperse. Hope this helped.


What happens to a greenhouse gas after it absorbs infrared radiation? We say that a greenhouse gas increases the temperature of the surface of the earth but how does it do that ? Only by re emitting the absorbed IR towards the earth ?

Actually a greenhouse gas absorbs light in the infrared wavelengths of light. One of two things will happen after the gas absorbs the IR radiation: 1) the molecules to move around and vibrate faster, which is really the definition of temperature, so the air gets warmer, and 2) the gas molecules can re-emit the IR radiation. In the second case, the radiation is emitted isotropically, so that some of it just goes out into space as it would have had the gas not been there in the first place, but some of it will be directed back towards the surface of the earth, warming the surface by absorbtion of the IR (heat).

Note that most of the heat being absorbed by greenhouse gases is coming from the ground after the ground absorbs sunlight and warms up, not from the sun shining through the air and being absorbed by the greenhouse gases directly.


What would a diagram of the metal lattice structure of CuSO4 look like? Just as $\ce{NaCl}$ would look like this:

What would a diagram of $\ce{CuSO4}$ look like?

From the WebElements.com page on copper sulphate, the metal lattice structure looks like:

The yellow pyramids are cooper and the other structures are sulphates.

An alternate view comes from the Wikipedia page of copper (II) sulphate:

The pink spheres represent copper, yellow are sulphur and red are oxygen.

Below is a 2D representation of copper sulphate, from the Royal Society of Chemistry

A similar question with some more insight is found in How do copper sulphate and the like change crystal structure on adding water of hydration?


Confusion about Ethinyl estradiol IUPAC name According to Wikipedia, the IUPAC name of ethinyl estradiol is:

  17α-Ethynylestra-1,3,5(10)-triene-3,17β-diol

However, in PubChem, its IUPAC name is:

  (8R,9S,13S,14S,17R)-17-ethynyl-13-
  methyl-7,8,9,11,12,14,15,16-octahydro-6H-cyclopenta[a]phenanthrene-3,17-diol

My questions:

Which one is correct by latest IUPAC blue book?
How are the locants assigned for this molecule? Specifically, which carbon was assigned the locant "10", since it does not appear in the locants of "octahydro"?

For natural products and derivatives, three levels of nomenclature are recognized in the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book): trivial names, systematic names, and semisystematic names. A new compound that can be isolated from a natural source is commonly given a trivial name. These trivial names are usually related to the biological origin of the material.

When the full structure of a natural product or derivative is known, a systematic name may be generated in accordance with the IUPAC recommendations for nomenclature of organic chemistry. However, the systematic name “(8R,9S,13S,14S,17R)-17-ethynyl-13-methyl-7,8,9,11,12,14,15,16-octahydro-6H-cyclopenta[a]phenanthrene-3,17-diol”, which is proposed in the question is not correct. This name implies that the carbon atoms 13 and 17 of the cyclopenta[a]phenanthrene parent structure remain sp2 hybridized; i.e. that a double bond remains between atom 13 and atom 17, which is not possible since both atoms have four bonds (to other skeletal atoms or to substituents). The correct systematic name is (8R,9S,13S,14S,17R)-17-ethynyl-13-methyl-7,8,9,11,12,13,14,15,16,17-decahydro-6H-cyclopenta[a]phenanthrene-3,17-diol.

Such systematic names may be too cumbersome. To overcome this difficulty and show the close similarity to related compounds, a semisystematic name may be formed. The semisytsematic name 17α-ethynylestra-1,3,5(10)-triene-3,17β-diol is based on the retained name estrane for the parent structure (8R,9R,10S,13S,14S)-13-methylhexadecahydro-1H-cyclopenta[a]phenanthrene. (Names of fundamental stereoparent structures are given in Table 10.1 of the 2013 IUPAC recommendations. The corresponding structures are shown in Appendix 3.) Addition of the three required double bonds to the estrane parent structure leads to the name estra-1,3,5(10)-triene, and addition of the missing substituents yields finally yields the name 17-ethynylestra-1,3,5(10)-triene-3,17-diol. To indicate the configuration at chiral centers, the ‘α/β’ system is used as described in IUPAC-IUBMB recommendations for the nomenclature of steroids.


How should one check an instant-equilibration algorithm for conservation? We have a system of equations that describes the equilibration between several phases $A, B, ...$.  If we define the total concentration $T = A + B + ...$, then $\partial_t T=0$ must hold.

We are further assuming equilibrium at all times, so we have devised an algorithm for this that transforms $A\rightarrow A'$, $B\rightarrow B'$ etc.  Finally, the total concentration must be conserved, so we must check if $T'=T$.  But should one also check if $\partial_t T'=0$, i.e. the conservation equation with the equilibrated variables?

Does it make sense to test both $T'-T=0$ and $\partial_t T'=0$?

I would think one of those conditions would always be sufficient. If you have shown that $T'-T=0$ for all $t$ and that $\partial_tT=0$, then  $\partial_tT'=0$ must be true by default. That is $\partial_tT=0$ shows that $T$ is a constant and if $T'-\text{constant}=0$ for all $t$, then $T'$ is also a constant.


Software for creating the QSAR model? There is the .csv file with a large amount of descriptors and known experimental activity.

Which software can find and describe model for prediction. Most important - numerical representation of model - which descriptor is taken with coefficients.

I am looking for any software, not necessary free.

There are multiple packages capable of designing QSAR models.

At a basic level, QSAR is about building a statistical model fitting some property/activity based on a set of multiple descriptors. You can use any multivariate statistical package to do that, including R (as indicated in the other answer) or Python.

If you're new, I would highly suggest KNIME an open "platform" for connecting multiple software packages for analyzing data. It's heavily used for QSAR.

There are several examples for KNIME QSAR, e.g. using CheS-Mapper - a quick search turns up several.

A critical concern, though is validation, either through separate training and test sets, cross-validation / bootstrapping methods, or other techniques. It's easy to design models that fit noise.


How can I carry out a titration to identify mercury in wastewater? First, I would like to do a some form of extraction to only get the mercury, then followed by a titration to identify the concentration or the presence of mercury.

The potentiometric titration of mercury with dithiooxamide has been successfully applied after preliminary separation of mercury by reduction (Comprehensive Water Analysis, Volume 2, pp.186-187, Elsevier).

Selective identification of mercury in waste water can be accomplished by fluorescence enhancement chemosensors (SENSOR LETTERS 2010, 8, 1–6). In this recent paper, the ﬂuorescent emission intensity of the probe is enhanced upon binding to mercury ions. The ﬂuorescence enhancement is attributed to a 1:1 complex formation between the probe and Hg(II) ion. The chemosensor can be applied for the quantiﬁcation of mercury; it shows a high selectivity toward Hg(II) ions in comparison with common metal ions.


obabel doesn't split molecules I have a .xyz input file

8
Energy =
C     0.0354972    0.7111560   -1.7500000
C    -0.0354972   -0.7111560   -1.7500000
C     1.1823040   -1.4277968   -1.7500000
C     2.3594716   -0.7301614   -1.7500000
C     0.0354972    0.7111560    1.7500000
C    -0.0354972   -0.7111560    1.7500000
C     1.1823040   -1.4277968    1.7500000
C     2.3594716   -0.7301614    1.7500000

(this is shortened, problem stays the same) which I would like to split with obabel. However, with

obabel coord.xyz -o split.xyz --seperate -m

I only get one outputfile split1.xyz containing the whole supermolecule. This is also the case if I enlargen the z-distance between the two molecules.

Does anybody either:

Know why this doesn't work
Has a different program which I can use

I don't have the time to manually split 500 molecule files.


Does all fire only emit light on its outermost shell? This question is about light emission, which may overlap with physics, but I am most interested in combustion and types of flame (incandescence, petrochemical fuel flames, nuclear ractions).
Consider this article about candle flames. In it they describe how a candle flame only emits light where it has contact with oxygen (emphasis mine):

However the reaction can only occur where the air meets the wax vapour.  This cannot happen all the way though the flame, just around the outside.  So all flames are hollow.

How accurate is this? What parts of a flame emit light?
If flames have complex scattering and absorption properties distinct from the burning substance, does this only apply to the outside regions, with the interior being goverened by whatever fuel is burning?
I suppose "fire" is many different phenomena, so are the following flames also hollow?
A: Campfire
B: Rocket exhaust
C: Oil platform explosion
D: The Sun

The answer is NO.
The article you quote makes a completely unwarranted generalisation that "all flames are hollow". This is true of some flames but only because the fuel that is burning is only able to burn when mixed with oxygen from the air. In those flames, the flame is "hollow" because only in regions where air can mix with the fuel can the fuel burn and the rate of mixing is dependent on the flow of fuel and the rate at which air can mix or diffuse into the fuel rich region. Bunsen burner flames are like this when the air is not premixed with the fuel gas (because the air intake on the bae is closed) and candle flames are like this because the wax has to evaporate to mix with air.
A bunsen flame when the air intake is open is not like that as the fuel gas and air are pre-mixed and can burn throughout most of the region of the flame.
Another factor worth considering is what causes the emission of light. In candles and many other poorly controlled flames, there is a lot of light from incandescence not from the underlying burning reaction. Incandescence involves black-body emissions from hot solid particles. Those particles are usually small particles of solid carbon produced from partial, incomplete burning of hydrocarbons. They emit light because they are hot. Candles are tuned to create a lot of these particles as this gives far more light than complete combustion. In contrast, a well-controlled bunsen flame emits very little light (the light is not black-body emission but comes from specific excited states in the reactants).
As to how this applies to your specific examples here are some analyses.
1. Campfires
Campfires are a mixture of various phenomena. They have flames  a little like candle flames. But they also have very hot solids which emit a lot of black-body radiation because they are hot. Some of the emitted gasses have little emission of light as they are like a bunsen flame, but the overall effect is dominated by the hot solids. And, of course, some camp fires have far better mixing of air and fuel than others so it is hard to generalise.
2. Rocket exhaust
Most rockets need to be as efficient as possible. So they tend to have very carefully designed mixing of fuel and oxidiser to achieve maximal energy output. They are not hollow and tend not to emit huge amounts of light (compared to the equivalent candle flame anyway).
3. Oil platform explosion
Impossible to generalise. But unintended explosions don't tend to be well controlled so will have some areas where mixing is good and others where it is not so some areas are like a candle and others like a bunsen. Incidentally, movie explosions emit a lot more light than real explosions because, well, hollywood likes visual spectacle not realism (see this youtube explanation and illustration of how movies deliberately add fuel to create poorly mixed candle like flames for more visually interesting explosions).
4. The sun
The sun does not emit light because of burning (at least not in the normal chemical sense). The sun emits light because the surface is very hot. And this happens because of nuclear fusion deep in its core (not what a chemist would call burning, though a nuclear physicist might, but what do they know?). The light we see is mostly black body emissions because the surface is hot. It is certainly not hollow either.


Addition of cyclohexene with itself? So I was practicing for exams, and this is the specific question.

I get that structure 2 is possible through polymerization and 3 doesn't fit to be the polymer of cyclohexene. I'm kinda consfused about 1 tho, I see that the double bonds have broken and joined with another cyclohexene, but I'm not sure about how that works. Like, is it similar to the addition of electrophiles..if so why isn't this seen with non-cyclic alkenes?

I'm fairly sure this question is merely testing one on their ability to count atoms. If the structure has a chemical formula of $\ce{(C6H10)_m}$ where $m \in \mathbb{Z}^+$, then it is – at least in principle – a possible polymerisation product of cyclohexene $\ce{C6H10}$. Whether it is formed in real life is a different question, but my belief is that that is not part of the question, so you're probably overthinking this if you're wondering about how exactly 1 is formed.

1 ($\ce{C18H30} \equiv \ce{(C6H10)3}$) and 2 ($\ce{(C24H40)_n} \equiv \ce{(C6H10)_{4n}}$) fulfill the above criterion, but 3 ($\ce{(C21H36)_n}$) does not.


What is the Gibbs free energy difference between conformational states In statistical mechanics, for a system of $N$ particles $x_1, \ldots, x_N$ in three dimensions, the Gibbs free energy is defined in terms of the Hamiltonian $H$ as
$$ G = -k_\mathrm B T \log \int_{\mathbb{R}^{3N}} \mathrm e^{-H(x_1, \dots, x_N)}\,\mathrm dx_1 \dots \mathrm dx_N.$$
People talk about the "free energy difference between conformations". For instance, for some region $A \subset \mathbb{R}^N$, the particles could be in a certain conformation, and for some other region $B \subset \mathbb{R}^N$, the particles could be in another conformation.

Question

  In this paper, equation (3), what is the (implied) definition of $F_A$ or $F_B$ in terms of equation (2)?

Equivalent Question

  Is the free energy difference between the two conformations defined as
  $$G_B - G_A = -k_\mathrm B T\log \frac{\int_B \mathrm e^{-H(x)}\,\mathrm dx}{\int_A \mathrm e^{-H(x)}\,\mathrm dx}?$$

Auxiliary details

The above definition made sense to me, until I was told that "it does not make sense to talk about the free energy of a conformation". An alternative I see is to think of two Hamiltonians $H_A$ and $H_B$, and instead define the free energy difference as

$$G_B - G_A = -k_\mathrm B T \log \frac{\int_{\mathbb{R}^{3N}} \mathrm e^{-H_B(x)}\,\mathrm  dx}{\int_{\mathbb{R}^{3N}}\mathrm e^{-H_A(x)}\,\mathrm dx}. $$

Are either/both/neither of these definitions correct?

I apologize if the question is exceedingly obvious. I am unable to find a straightforward answer by Googling. Any references are welcome.

Your question implies that you define the Gibbs free energy as $G=-RT\ln(Z)$ where $Z$ is the partition function, in which case there will be a difference in $G$ only if there is a difference in the partition function for your two configurations, i.e differences in energy levels and/or in the population in these levels. That will depend entirely on your system (and also the temperature).


Plasma creation and condensation Every chemical has a decomposition temperature.  My understanding is that above that temperature molecular bonds are broken.  And if we raise the heat high enough then all molecular bonds will break and we will be left with a "plasma" of ionized atoms, right?

Can the decomposition temperature of bonds be predicted by models, and if so on what data do those models depend?

Now suppose we heat a chemical in an inert (noble) atmosphere, or in a vacuum, to the point of total decomposition.  Now we cool the plasma back to STP.  Are there models that can predict the products we would expect to find – i.e., the right hand side of the reaction equation?

I'm guessing that thermodynamics stipulates we should find the lowest-"energy" compounds that can be created from the constituent elements on the right hand side.

In practice I imagine that achieving that result could require that cooling be done at some "sufficiently low rate."  So, to increase the complexity of the question: Are there any models that can predict what products will be found as the cooling rate is increased?  I believe that in the limit (a virtually instantaneous quench) one ends up with a "glass" (or combination of glass, condensate, and gas) but even then one can't be left with free ions so can we predict the composition of the quenched plasma?

To begin with, dissociating molecules to a plasma at the atomic level will not only create ions but also neutral atoms. These will indeed reassociate on cooling, and the question is how to predict what the products will be. In the extreme case of a simple system such as pure hydrogen gas (dissociating, then recombining to H2) or, say chlorine plus bromine dissociating/recombining to give BrCl, it should be easy to predict the products formed and perhaps more about the behavior of the system as a function of temperature.

If you have a more complex system, including (for the sake of argument) atoms such as carbon that can form complex ring and chain structures, predicting the outcome will be impossible - since (1) the different ways to combine atoms may lead to a huge number of possible intermediates, many of whose energies/reactivity are not known; (2) even if you could predict these intermediate structures, the number of further possible reactions between them becomes astronomical, and (3) kinetic barriers (activation energies) for  reactions are generally harder to predict that ground state energies. As is often the case when theory comes up short, you may have to go into the laboratory and do some experiments. Possible approach: aim the plasma (along with a stream of inert gas such as argon) at a supercooled probe in a vacuum and spectrally characterize the species in the resulting matrix. I imagine that you could end up with some pretty amazing chemical species that way.

I would urge those among my colleagues who are familiar with the current state of theoretical chemistry to add to this discussion. I love a question that makes you go back to basics and rethink everything from the ground up.


Is a negative pH level physically possible? A friend of mine was looking over the definition of pH, and wondering if it is possible to have a negative pH level. From the equation below, it certainly seems possible—just have a $1.1$ (or something $\gt 1$) molar solution of $\ce{H+}$ ions:
$$\text{pH} = -\log([\ce{H+}])$$
(Where $[\ce{X}]$ denotes the concentration of $\ce{X}$ in $\frac{\text{mol}}{\text{L}}$.)

If $[\ce{H+}] = 1.1\ \frac{\text{mol}}{\text{L}}$, then $\mathrm{pH} = -\log(1.1) \approx -0.095 $

So, it is theoretically possible to create a substance with a negative pH. But, is it physically possible (e.g. can we create a 1.1 molar acid in the lab that actually still behaves consistently with that equation)?

One publication for you: “Negative pH Does Exist”, K. F. Lim, J. Chem. Educ. 2006, 83, 1465. Quoting the abstract in full:

  The misconception that pH lies between 0 and 14 has been perpetuated in popular-science books, textbooks, revision guides, and reference books.

The article text provides some counterexamples:

  For example, commercially available concentrated HCl solution (37% by mass) has $\mathrm{pH} \approx -1.1$, while saturated NaOH solution has $\mathrm{pH} \approx 15.0$.


Increase in bond length of BF₄⁻ compared to BF₃ Why does $\ce{BF4^-}$ have a larger bond length than $\ce{BF3}?$ Usually in molecules the bond angle increases to reduce repulsion but, here the bond length increases. Why is it so? Does the bond length increase with the bond angle in other cases of such kinds of molecules?

There seems to be a bit of confusion between bond length and bond angle here.  It is true that VSEPR theory predicts that bonding pairs of electrons (and lone pairs, for that matter) will repel one another as much as possible, thereby increasing bond angles, but this has little to do with bond length.

One way to explain the different bond lengths in $\ce{BF_3}$ and $\ce{BF_4^-}$ is by looking at the molecular orbital diagram of $\ce{BF_3}$:

The linked diagram isn't great since it does not show the filled orbitals: the $1a_2'$ $1e''$ and $4e'$ orbitals are all filled, making the lowest unoccupied orbital (LUMO) the $2a_2''$.  If you haven't wrapped your head around molecular orbital diagrams of this complexity, don't sweat it.  The bottom line is, the $2a_2''$ orbital has antibodning interactions with the $2p_z$ orbitals of boron and the $2p_z$ orbitals of the surrounding fluorines.  Thus, adding electrons to this orbital makes the $\ce{B-F}$ bonds weaker, and therefore longer.


Nucleophilicity of bonding vs lone pairs Does the bonding pair in $\ce{FeCl4^-}$ attack a proton or does a lone pair on the chlorine attack the proton?

My prof has an argument for the first one; the first one involves fewer steps, so it is probably more favorable. My proposed mechanism has three steps of electron pushing versus two, so okay.

However, I can see an argument for the second mechanism - the first one involves a bonding pair acting as a nucleophile rather than a lone pair on chlorine. The bonding pair is stabilized by the chlorine and the iron atom, while the lone pair is stabilized by only the chlorine.

So, which one wins? I'm guessing both mechanisms are operative here.

I don't know the answer to your question, but we can speculate.  A lone pair is more sterically exposed than a bonding pair of electrons and therefor more accessible.  Also a lone pair is generally more nucleophilic because the electrons are concentrated in a smaller region since they are not being shared.  These two arguments would favor the second mechanism.  Finally, when we consider the SN2 reaction where, for example, water is the attacking nucleophile; I've never seen the mechanism drawn where the attacking electrons are from the O-H bond, but rather they always seem to show involvement of the lone pair electrons.  All of these arguments favor the second mechanism.


Why are there drastic changes sometimes in radius of isoelectronic ions? $\ce{S^{2-}}$, $\ce{Cl-}$, $\ce{K+}$, and $\ce{Ca^{2+}}$ each have the same number of electrons. Their effective ionic radii are 182, 181, 138, and 100 picometers, respectively, and the number of protons for each are 16, 17, 19, and 20. My question is, why does the effective ionic radius drastically go from 181 to 138 picometers between chlorine and potassium? Is it really necessarily because they are in different periods or is there something that I am missing here?

The whole concept of ionic radius is not sharply defined one. Note, that the same is true for the notion of atomic radius, as well as for any other kind of radius or any other notion of size. In general, the notion of size of an object looses its usual (i.e. classical) meaning in the microscopic realm since objects there do not have well-defined physical boundaries. So from the very beginning one should not expect a perfect trend in a not so well defined property.

Anyway, values of the ionic radii are based on crystallographic data, but what is actually determined in X-ray crystallography is the distance between two ions, not their radii. Then it is basically up to you how to divide this distance into the radii of ions in question. But whatever way the division is done, the ionic radius of an ions is not a fixed, perfectly defined property, since for the same ion, the radius can differ in different crystal lattices due certain variables such as coordination number and electron spin.

For instance, you mentioned the ionic radius of 138 pm for $\ce{K+}$. But this value is for 6-coordinate compounds only while for 8- and 12-coordinate compounds of $\ce{K+}$ the ionic radius is 151 pm and 160 pm respectively. Same story for $\ce{Ca^2+}$: 100 pm, 112 pm, and 135 pm for 6-, 8-, and 12-coordinate compounds respectively.1

Interestingly, for anions there is a little variation in the ionic radius depending on coordination number, so that for many of them only one value independent of coordination number is specified. To be honest I could not come up with a quick explanation why, so it feels like I have to dig into the matter deeper. But I'm quite reluctant to do so, since, as I said in the beginning, the whole concept of ionic radius looks suspicious, so I expect to find just some obscure rule arriving out of the blue which will explain this little variation in the anionic radius relatively to cationic ones.

Slater's rules to the rescue!

Now, I guess, I know which rule can be used to explain this drastic change in ionic radii. Let us examine the application of the good-ol' Slater's rules to the matter. These rules allows us to calculate the effective nuclear charge  $Z_{\mathrm{eff}}$, which is the net positive charge experienced by an electron in a multi-electron atom.
$$
    Z_{\mathrm{eff}} = Z - s \, ,
$$
where $s$ is the screening constant. Slater's rules provide a way to calculate the values for the screening constants, and consequently, to obtain the values of effective nuclear charges.

All the ions $\ce{S^2-}$, $\ce{Cl-}$, $\ce{K+}$, $\ce{Ca^2+}$ are isoelectronic: the electron configuration of them is that of $\ce{Ar}$, i.e. $\mathrm{[Ne] 3s^2 3p^6}$. Thus, we have to compare effective nuclear charges for $\mathrm{3p}$ electrons in these ions. And as a first approximation we could compare effective nuclear charges for $\mathrm{3p}$ electrons in the corresponding atoms. Now without going into the details (listing the actual Slater's rules & applying them) I will just leave here the final numbers, effective nuclear charges for $\mathrm{3p}$ electrons for the corresponding atoms:

$$
\begin{array}{c|cccc}
     & \ce{S} & \ce{Cl} & \ce{K} & \ce{Ca} \\
\hline
Z & 16 & 17 & 19 & 20 \\
Z_{\mathrm{eff}}(\mathrm{3p}) & 5.482 & 6.116 & 7.726 & 8.658 \\
r_{\mathrm{ion}} & 182 & 181 & 138 & 100 \\
\end{array}
$$

Now you can see that despite a relatively small change in the nuclear charge going from $\ce{Cl}$ to $\ce{K}$
$$
   (19-17)/17 * 100 \% \approx 11.8 \% \, ,
$$
there is a rather drastic change in effective nuclear charge
$$
   (7.726-6.116)/6.116 * 100 \% \approx 26.3 \% \, .
$$
And the drastic change in ionic radius is exactly in line with this drastic change in effective nuclear charge
$$
   (138-181)/181 * 100 \% \approx -23.8 \% \, .
$$
To summarize: the effective nuclear charge for $\mathrm{3p}$ electrons increases by about 26.3% when going from $\ce{Cl}$ to $\ce{K}$ and the ionic radius accordingly decreases by 23.8%.

Note that we used effective nuclear charges of the atoms just as the first approximation for that of the ions. For anions an additional electron will occupy an outer orbital and there will be an increased electron-electron repulsion. This will push the electrons further apart and increase the shielding, so that $Z_{\mathrm{eff}}$ for anions will be smaller than for the corresponding atoms. For cations the effect of the removal of an electron is the opposite: when an electron is lost, electron-electron repulsion and therefore, shielding decreases, so that $Z_{\mathrm{eff}}$ for cations will be bigger than for the corresponding atoms. It would not change the above mentioned trend though; in fact, it will make it even more prominent.

1 Ionic radii values are taken from here.


Convergence of spectra calculated from molecular dynamics trajectories How can we check whether or not the power spectra (vibrational density of states) calculated by taking the Fourier transform of the velocity auto-correlation function (VACF) from a molecular dynamics trajectory is converged? How to estimate the statistical error in it?

What is the meaning of the word "convergence" in this context?

There are multiple ways in which the VACF, along with other properties derived from molecular dynamics trajectories like the dipole ACF to give IR spectra, can be considered "converged".

The molecular dynamics trajectory itself must be converged by some time-dependent statistical metric, such as measuring fluctuations in the energy or temperature over a number of steps and ensuring the average is below some threshold. This will be related to the ensemble chosen, the thermostat and barostat and associated parameters (if relevant), the integrator used for time evolution, and the size of the simulation, among other things.
Molecular simulations are layers of approximations piled on top of each other. In the case of purely classical dynamics, this usually refers to the force field parameters for the solute and solvent. Force fields are designed for different systems and may or may not be transferable, but the goal is usually to accurately predict dynamics. In the case of the VACF, if the dynamics of a system are described correctly, then the resulting power spectrum should be as well. However, in the case of the dipole ACF, the force field parameters enter directly as the nuclear charges (rather than just indirectly through nuclear velocities), which places stricter requirements on the quality of the force field. One hopes that newer force fields are better, but improvement over time isn't necessarily systematic and there is no quality hierarchy.
As an extension of 2, if the simulation is quantum mechanical in any way (such as in QM/MM, Born-Oppenheimer MD, or Car–Parrinello MD), then there are approximations in how the electronic structure is treated. For spectroscopic properties, it is more clear how to systematically improve results with regard to electronic structure than force fields due to well-established quantum chemical method hierarchies. Benchmarks of this hierarchy have been performed on many properties, including geometric parameters, dipole moments, and vibrational frequencies. In theory, the error in both dynamics and spectra should decrease up going from Hartree-Fock to MP2 to CCSD to CCSD(T), or by using more advanced density functionals (say, BP86 to PBE0 to M06-2X to B2PLYP). In practice, this would require benchmarking to draw any conclusions.
As an extension of 3, there are approximations in QM/MM of how the classically-treated and QM-treated regions interact with each other. Rather than go into more detail, here is a review of different QM/MM methods covering different definitions of the boundary and how one region may interact with the other.

Regarding estimation of the statistical error in resulting spectra, I am only aware of direct comparison (absolute error) between experimental and computed spectra, because it is assumed that the MD trajectory is converged (point 1). Performing multiple independent MD simulations and calculating the VACF from each should give an error measure closer to true statistical error, though I am unsure of how this would be computed.

Here is good almost-review (with good examples) of using Born-Oppenheimer molecular dynamics to calculate power spectra, IR spectra, and Raman spectra. I am not aware of any comparisons between classical MD and BOMD for computing spectra.

On the surface, it may appear as though the introduction of quantum mechanics makes things more difficult; certainly things become more time-consuming, however it is much clearer how results will change by treating electronic structure explicitly and with less severe approximations, rather than with a force field.


How can f-f transitions happen? The question might seem quite easy to answer but still I am unable to find any useful information. I am aware of laporte- or symmetry-forbidden transitions and for f-f also spin-forbidden transitions in lanthanides but still I can't seem to find any good solution how they can happen.

For a d-d transition there are many exceptions depending on the arguments you come up with. From inverse symmetrical polyhedra that use ligand vibrations to become distorted to s- and p-mixing with the d-orbitals.

But f-orbitals, at least for the lanthanides, are said to not contribute to chemical bonds. Then there is often the argument about spin-orbit coupling but that would probably solve the spin-forbidden transitions. I'm not sure how it would affect the actual f-f transition being forbidden. In an old paper I also read about df- and fg-mixing. That would explain it but I am not sure if this is really the case.

So the question remains can the f-f transition in lanthanide complexes be somehow influenced to become at least a tiny bit more probable and therefore cause the transition to happen?

So I am currently reading the book: Physical Inorganic Chemistry A Coordination Chemistry Approach, S. F. A. Kettle

One of the chapters deals with Lanthanides and Actinides and their spectra. After a long introduction on different energetic levels and spin-orbit coupling there are a few remarks on $\ce{f->f}$ transitions although, as the author chooses to explain different ideas and phenomena during his explanation I am not sure if I fully understood his final conclusion. I guess it's probably a sum of many contributions. To summarize it quickly the aspects he is discussing are:

We can see that the intensities for $\ce{f->f}$ transitions in coordination polyhedra with high symmetry tend to be weaker than in low symmetries. That means that there are probably mechanisms such as vibrations and mixing with other orbitals, presumably d-orbitals (like in $\ce{d->d}$ transitions).
For the vibronic coupling (in order to compensate for a parity forbidden transition) he gives $\ce{NaCs2[YCl6]:M^3+}$ as an example, where at low temperatures both electronic and vibrational transitions seem to be observable for the trivalent lanthanide dopant.
Then there are also 'bands' under high-resolution conditions that cannot be explained by vibronic coupling. This is where he introduces magnetic dipole allowed transitions. In his explanation he compares how an electric dipole will invert at an inversion center while, much like in the detectors for mass spectrometers, electrons will be forced to move in a spiral along the magnetic vector and a spiral or circle does not invert here. According to his explanation the magnetic dipole is an even function and as f-orbitals are odd an $\ce{f->f}$ transition is $$u * u = g$$
and therefore magnetic dipole allowed.
Then he also refers to the fact that even though light may not be absorbed in matter it will still travel slower through a medium than through vacuum since it causes oscillations that have to propargate through the medium first.
I'll just quote the book at this part:

  It has been suggested that the light wave used in absorption studies
  polarizes the ligands and that the lanthanide ion feels the dipoles created.
  That is, it is suggested that the ligands provide a non-centric field at the
  lanthanide ion not by virtue of any vibrational movement of their nuclei but
  because of the instantaneous polarization of their electron density caused by
  the light wave itself.

Then there was a remark on the environment as well. Since the f-orbitals don't really take place in any bonding for the lanthanides, do they actually feel their environment? And for this they considered weak nephelauxetic effects.
And then there are so called hypersentisitve-transitions (transitions with environment sensitive intensities). And those seem to be related to quadrupole-allowed transitions. The two dipoles in a quadrupole may arise from a polarization by the electric vector as mentioned above. So one dipole is the light vector itself (with the energy corresponding to the hypersentisitve-transitions and the other one being the induced dipole).
The difference in environments would arise from different polarizabilities of the ligands in different environments.

The other mechanism seems a bit more complicated. It considers low-symmetry crystal fields with an inherent dipole. So high-symmetry crystal fields distort via a vibration to a low symmetry. The ligand environment is distorted to a point where it become dipolar.
And then the light itself is the other dipole.
The final remark was on the bonding in f-orbitals. Here the complex $\ce{[UF8]^3-}$ was considered. By repulsion a square antiprism would be favored if there was no interaction at all with the ligands. What we find however is a cubic arrangemenent.
And the $\ce{f_{xyz}}$ orbital can contribute to a bond only in the cubic case.


Why is it that all strong bases contain at least one hydroxide bond? Why is it that all strong bases contain at least one hydroxide bond? What is going on here at the chemical level? For example, we know that $\ce{NaOH}$ is a strong base. I know it is strong because it completely dissociates in water. And under the Bronsted-Lowry definition of bases, I know that is is considered a base because it accepts a proton. So what is it about having a hydroxide bond specifically that lets it dissociate?

All strong bases you know contain at least one hydroxyl group because we live in a world dominated by water, almost like in that old movie with Kevin Costner. Water chemistry is of paramount importance to us, and acid/base reactions in water pretty often boil down to $\ce{H+ + OH- ->H2O}$.

There is nothing special about the hydroxyl group that lets it dissociate. Come to think of it, the group itself does not even dissociate in bases - I mean, $\ce{O}$ and $\ce{H}$ remain together, the bond between them does not break. (It does in acids, but that's another story.) If you mean the dissociation of metal hydroxides, there is nothing special here either. $\ce{NaOH}$ dissociates like any water-soluble ionic compound. On the other hand, $\ce{Fe(OH)3}$ refuses to dissociate, like any water-insoluble ionic compound, of which there are plenty.

In a wider chemical world outside, there are bases which are many orders of magnitude stronger than that, and they don't contain any hydroxyl groups whatsoever. $\ce{NaNH2}$, $\ce{NaH}$, $\ce{BuLi}$ come to mind. You don't see them much in everyday life, though, because they don't survive in our world. Water act as an acid on them, and instantly transforms them to mere $\ce{NaOH}$, etc.

So it goes.


How can you identify the anode and cathode of a car battery of indistinct shape by using a potassium iodide solution? Now I know that the anode in a car battery is lead, the cathode is lead dioxide and the electrolyte is sulfuric acid, but still: how could a potassium iodide solution be used to differentiate between the anode and the cathode?

I mean, lead can't replace potassium in its salt because it lies below it in E.C.S., so how are we supposed to identify them by using potassium iodide?

By using the battery as a source of electricity in an electrolytic cell that contains potassium iodide as an electrolyte.
Electrolysis takes place where iodide ion is oxidized at anode and brown colour of iodine solution appears at anode.
At cathode hydrogen gas evolves which is indicated by formation of bubbles at cathode.


Carbocation formation in electrophilic addition of hydrogen halides to alkenes Question:

My attempt:

I made the carbocation on the carbon labelled $1$ because I think that the positive charge would be slightly more stable on $1$ compared to $2$ due to the inductive effect of $\ce{Cl}$.

But my teacher made the carbocation on $2$ giving the explanation that after rearrangement, the positive charge will be stabilised by resonance with $\ce{Cl}$.
His approach:

Now, my question is, comparing the stabilities of the carbocation intermediates without rearrangement, the positive charge will be stable on $1$ compared to $2$. After rearrangement too, I think that the tertiary carbocation will be more stable than compared to the one stabilized by resonance with $\ce{Cl}$.

So, what is wrong in my reasoning?

Or as suggested by Mithoron, will there be intramolecular attack by $\ce{Cl}$?

The rate determining step is carbocation formation. After which, depending on conditions, the trapping of the cation may or may not be slow enough for [1,2]-shifts.

The [1,2]-shifts are probably under thermodynamic control if there is time for them to take place. This means that if there's time for the shifts, we're hunting for the most stable cation. Otherwise, you trap the most readily formed cation in a kinetic sense (which is not necessarily the most stable).

Assuming shifts are fast, I think your solution is better because the tertiary cation is more stable. Dichloromethane doesn't undergo substitution as easily as t-butyl chloride, so I would have to argue that the tertiary carbocation is more stable.