In a half cell reaction, Is the potential halfed? If you have an electrolyte of $\ce{Zn^2+}$ and $\ce{SO4^2-}$ (Neutral) Then you dip in it a Zinc electrode.

Zinc potential is $\pu{-0.74 V}$ (relative to SHE). So that means there is a $\pu{-0.74V}$  difference between the electrolyte and the electrode.

But When you dip in the solution the electrode makes the solution more positive and it self more negative the same amount so the overall potential of the electrode will be $\pu{-0.74 V} / 2 = \pu{-0.37 V}$, and the solution will be at $\pu{+0.37 V}$.

Is this true? (assuming the solution is not connected to any other salt bridge or porous disk)

Edit: Absolute electrode potential is really the difference between the electrolyte and the electrode potential.

And the question is just this if you have an electrode inside an electrolyte with no salt bridge will it's potential be halfed.

Those $\pu{-0.74V}$ are measured against a standard hydrogen electrode. That's the universally accepted but totally arbitrary standard.

You don't have a second electrode, hydrogen or other, so the absolute number is totally meaningless. If someone had decided to use the zinc electrode instead, as zero point for the electrochemical series, then the voltage you mean would be zero. Makes no sense.

There is an absolute cell potential (absolute electrochemical series), with a different zero point, but it's totally uncommon, because it's very tricky to measure absolute electrostatic potentials of only a few hundred millivolts. I believe the problem is the calibration, while a platinised platin electrode bubbled with reasonably pure hydrogen is probably the simplest reference standard to reproduce at all.


Reaction of carboxylic acids with primary amines $$\ce{RCOOH + RNH2 -> RCONHR + H2O}$$

$\ce{R}$ stands for any alkyl group.

I have two queries regarding this reaction:

I understand that the reaction is an $\mathrm{S_N{AE}}$ (nucleophilic addition elimination type of reaction). But during the reaction $\ce{R-(CO)-N^+RH2}$ intermediate is formed. After this step the $\ce{OH^-}$ which was eliminated acts as base takes away $\ce{H^+}$ from  $\ce{R-(CO)-N^+RH2}$, converting it into $\ce{R-(CO)-NRH}$. But why did $\ce{OH^-}$ extract $\ce{H^+}$ from $\ce{R-(CO)-N^+RH2}$  and not $\ce{R^+}$ ? Isn't $\ce{R^+}$ always a more stable carbocation compared to $\ce{H^+}$ ?
Why does this reaction require a basic medium?

After this step the $\ce{OH-}$ which was eliminated …

No. $\ce{OH-}$ is never eliminated but formally. To eliminate the oxygen required, it needs to first accept a proton from somewhere leading to $\ce{R-C(O^-)(OH2+)-NHR}$. The hydroxide ion is simply much too bad a leaving group for any $\mathrm{S_N}$ reaction; it must always be transformed into a better one.

This leads to a severe problem in the reaction. Basic conditions hinder the reaction because they decrease the degree of protonation on that oxygen. However, under acidic conditions the amine will be protonated in solution ($\ce{RNH3+}$) and a protonated amine no longer has any lone pairs that can attack nucleophilicly. You’ll want neutral to mildly basic conditions for the reaction to work out.

Another thing to consider: the ‘elimination of a proton’ is a simple acid-base reaction. Proton displacements are typically much more rapid than any carbocation eliminations because protons, being as small as they are, can tunnel. Additionally, you should never compare the stability of a naked proton against that of a naked carbocation in these eliminations. A naked proton will not be generated, it will attach to a base immediately — and there are enough bases around: $\ce{OH-, H2O, RNH2, \dots}$ Furthermore, carbon-heteroatom bonds are pretty stable if the heteroatom side does not equal a good leaving group (e.g. $\ce{NHTs}$). Thus, these are hardly ever cleaved.

In practice, you would not use a particularly acidic or basic medium in this reaction. Rather, you would use acid activation techniques such as the Steglich conditions ($\ce{DCC, DMAP}$).


How to calculate galvanic cell spontaneity? I was wondering how to calculate cell spontaneity for a galvanic cell. Do you just look up the enthalpy values of electrode reactions? Does the electrolyte solution you choose have an effect on cell spontaneity?

To determine spontaneity of a reaction, one looks at the sign of $\Delta G$ for the process. And you have the following relation between the change in standard Gibbs free energy and the (standard)EMF of the cell:

$ \Delta G^0 = -n\,F\,E^0$

where, $n:=$ no. of electrons exchanged

$F:=$ Faraday constant (Charge on 1 mole of electrons. About $96,500\,\mathrm{C}$)

So, it is sufficient to compute the $E$ and look at it's sign. A positive sign for the EMF would ensure a negative value for $\Delta G^0$ and thus, you can conclude that your cell reaction is spontaneous.

The EMF of the cell is given as follows:

$$E^0_{\mathrm{cell}} = E^0_{\mathrm{cathode}} - E^0_{\mathrm{anode}} $$

Where, $E^0_{\mathrm{cathode}}$ and $E^0_{\mathrm{anode}}$ are the standard reduction potentials for your cathode and anode half reactions, and you can easily find tabulated values for most metals, ions.

An illustrative example:

Consider the following cell reaction:

$$\ce{Zn_{(s)} + Cu^2+_{(aq)} -> Zn^2+_{(aq)} + Cu_{(s)}}$$

Your cathode half reaction is:

$$\ce{Cu^2+_{(aq)} + 2e- -> Cu_{(s)}}$$

and your anode half reactions is:

$$\ce{Zn_{(s)} -> Zn^2+_{(aq)} + 2e-}$$

the tabulated reduction potential values for the two:

$$E^0_{\mathrm{anode}}= -0.763\,\mathrm{V} \qquad E^0_{\mathrm{cathode}}= +0.342\,\mathrm{V}$$

Plugging this into

$$E^0_{\mathrm{cell}} = E^0_{\mathrm{cathode}}-E^0_{\mathrm{anode}}$$

we obtain,

$$E^0_{\mathrm{cell}} = +1.103\,\mathrm{V}$$

Thus, this cell reaction is spontaneous.


Does electron shielding increase or stay constant moving LEFT to RIGHT across a period? Does electron shielding increase or stay constant moving left to right across a period?
I have read about both, and I just want to know which one is right.

I believe that electron shielding remains constant because when you move across a period, you are essentially adding more valence electrons, not shielding electrons, in your valence shell.    Therefore, your
valence-electron-count increases from left to right in a period, but your shielding-electron-count stays the same.

So the overall shielding remains constant, just the amount of valence electrons increases

However, my explanation has a flaw:
It does not support the reason why an atom's radius increases when it becomes an anion.

When an atom becomes an anion, it adds more valence electrons to its outermost shell. Now this is controversial to what I said previously, because a gain in valence electrons, but not shielding electrons, wouldn't increase the electron shielding of the atom. So the atomic radius wouldn't increase either!

The only way for an atom's radius to increase when it becomes an anion is if the electrons that are added to the valence shell have some sort of effect on the shielding, right?

This supports the first thought in my question that if we move left to right across a period, shielding increases.

But the idea that electron shielding increases as we move left to right has a flaw as well:
It is controversial to another periodic trend: ionization energy

If electron shielding did increase across a period from left to right, then this would mean that ionization energy is lower as we move from left to right, which is not true.

The only way for ionization energy to increase across a period is if the  only the number of protons and valence electrons have effect on the amount of energy required, not the shielding electrons. So the shielding-effect must stay the same all throughout a period.

This goes with the what I had initially said about electron-shielding remaining constant.

So now we have a flaw for each possible answer to my question!
Please correct me if I am wrong in any of my context.

To answer this question, it's important to define what you mean by shielding. Generally, shielding refers to a reduction in the effective nuclear charge experienced by an electron in a given orbital due to the other electrons on the same atom.

The quantitative degree of shielding for a given electron can be approximated by Slater's rules. According to those rules, electrons within the same group of orbitals (for example 4d) have a coefficient of 0.35 (except for 1s electrons, which have 0.30). So valence electrons do shield each other, just not as much as the lower level electrons shield the valence electrons.

For example, let's consider the elements with increasing numbers of 2p electrons (B, C, N, O, F, Ne). Going from left to right, each addition of a 2p electron reduces the effective nuclear charge experienced by another 2p electron by 0.35. So the amount of shielding is increasing as we move left to right.

The apparent contradiction with the ionization energy comes about because you have not considered the increase in the actual nuclear charge. Each time we add a 2p electron, we also add a proton to the nucleus. So even though 0.35 of that charge is shielded from a valence 2p electron, that electron still experiences an increase of 0.65 effective charge and is therefore bound more tightly than a 2p electron on the element to its left.

As a specific case, let's compare the shielding experienced by a 2p electron in carbon versus neon. The carbon electron configuration is $1s^22s^22p^2$. Using Slater's rules, we can calculate that the shielding coefficient for a 2p electron is 2(0.85) + 3(0.35) = 2.75. So the effective nuclear charge experienced by the 2p electron is 6-2.75=3.25.

For neon, the electron config is $1s^22s^22p^6$, so the shielding coefficient of a 2p electron is 2(0.85)+7(0.35)=4.15. So there's more shielding. But the effective charge is 10-4.15=5.85, so the effective charge is still greater and the energy of the coulombic interacation of the electron with the nuclear charge is greater in magnitude. Hence, the ionization energy is greater.

EDIT: Original version of this answer did not address the question about anion formation (presumably mostly interested in anion formation by addition to an already partially filled shell). When an anion is formed, the additional electron increases the shielding of the other valence electrons, but there is no change in the actual nuclear charge. Thus, each valence electron experiences a lower effective nuclear charge. Since the size of an orbital is a function of the energy of interaction with the nuclear charge, the atomic size increases.


Is the electrophile in sulfonation of benzene SO3 or SO3H+? What's the electrophile in sulfonation of benzene? Is it $\ce{SO3}$ or $\ce{SO3H+}$?

According to what my professor taught me, both $\ce{SO3}$ and $\ce{SO3H+}$ may be the electrophile involved - and mechanisms with both species can be successfully written.

I'm looking for some source(s) that confirms the existence of both electrophiles in sulfonation reaction of benzene.

From Clayden, under the heading "Sulfonation of Benzene" (electrophilic aromatic substitution):

  The cationic intermediate can
  also be formed by the protonation
  of sulfur trioxide, $\ce{SO3}$, and
  another way to do sulfonations is
  to use concentrated sulfuric acid
  with $\ce{SO3}$ added. These solutions
  have the industrial name oleum.
  It is possible that the sulfonating
  agent in all these reactions is not
  protonated $\ce{SO3}$ but $\ce{SO3}$ itself.

From March's $7^{th}$ edition, under Sulfur electrophiles:

  Sulfonation is often accomplished with concentrated sulfuric acid, but it can also be
  done with fuming sulfuric acid, $\ce{SO3}, \ce{ClSO2OH}, \ce{ClSO2NMe2/In(OTf)3}$ or other reagents.

  Sulfur trioxide reacts much more rapidly
  than sulfuric acid with benzene it is nearly instantaneous

  A great deal of work has been done on the mechanism, chiefly by Cerfontain
  and co-workers. Mechanistic study is made difficult by the complicated nature of
  the solutions. Indications are that the electrophile varies with the reagent, though
  $\ce{SO3}$ is involved in all cases, either free or combined with a carrier. In aqueous
  $\ce{H2SO4}$ solutions, the electrophile is thought to be $\ce{H3SO4+}$
  (or a combination of
  $\ce{H2SO4}$ and $\ce{H3O+}$) at concentrations below $\sim 80–85%$ $\ce{H2SO4}$, and $\ce{H2S2O7}$ (or a
  combination of $\ce{H2SO4}$ and $\ce{SO3}$) at concentrations higher than this. (the changeover
  point varies with the substrate)

  In fuming sulfuric acid ($\ce{H2SO4}$ containing
  excess $\ce{SO3}$), the electrophile is thought to be $\ce{H3S2O7+}$
  (protonated $\ce{H2S2O7}$) up to $\sim 104% \ce{H2SO4}$ and $\ce{H2S4O13} (\ce{H2SO4+ 3SO3})$ beyond this concentration.

  Finally, when pure $\ce{SO3}$ is the reagent in aprotic solvents, $\ce{SO3}$ itself is the actual
  electrophile. Free $\ce{SO3}$ is the most reactive of all these species, so that attack
  here is generally fast and a subsequent step is usually rate determining, at least
  in some solvents.


What kind of stereoisomerism do cisplatin and transplatin exhibit? Cisplatin and transplatin are obviously isomers, more precisely stereoisomers. Now, they aren't chiral, so they cannot be enantiomers. They cannot be diastereomers either since they don't have two stereocenters. They don't belong to cis/trans stereomers either since they are a subgroup of diastereomers and again require two stereocenters. So - what the heck?

The definition of diastereomer does not include "requires two stereocentres" anywhere. It is simply

  Diastereoisomers (or diastereomers) are stereoisomers not related as mirror images. (IUPAC Gold Book)

Therefore cis- and trans-$\ce{[PtCl2(NH3)2]}$ are diastereomers.

(E)- and (Z)-alkenes are also diastereomers of each other. Your "cis/trans stereomer" is in fact simply a subset of diastereomerism.


Why is tetraamminecopper(II) a square planar and not a tetrahedral species? In $\ce{[Cu(NH3)4]^2+}$, the $\ce{Cu^2+}$ ion has 9 electrons in the $\mathrm{3d}$ orbital with only one unpaired electron. How is it a square planar geometry? Where is that unpaired electron going?

It is very convenient to use crystal field theory to discuss this.

It is usually assumed that in octahedral coordination the energy levels of the five d-orbitals are split, with two orbitals ($d_{z^2}$ and $d_{x^2-y^2}$) well above the other three. The splitting is assumed to be large enough to overcome electron pairing energy.

The first six electrons will populate three lower orbitals ($d_{xy}$, $d_{xz}$, $d_{yz}$). The next two electrons should occupy $d_{z^2}$ and $d_{x^2-y^2}$. However, if these two electrons occupy $d_{z^2}$, two corresponding opposite ligands become well shielded and leave. This is the common case of $d^8$ complexes, such as $\ce{Ni^{+2}}$, $\ce{Pd^{+2}}$, $\ce{Pt^{+2}}$, $\ce{Rh^{+1}}$, $\ce{Cu^{+3}}$ etc.

Adding one more electron makes the remaining ligand's bonding weaker. The charge of bivalent cation $\ce{Cu^{2+}}$ is large enough to bond four charged ligands if available, such as in $\ce{[CuCl4]^{2-}}$, but the ligands are bound weakly and easily dissociate.

However, $\ce{Cu^{+2}}$ ions usually adopt a distorted octahedral geometry, with two ligands having a longer bond length than the four others. These two are very weakly bound and exchange quickly. For this reason the $\ce{NH3}$ complex is written with only four molecules; the two other are so weakly bound.

This type of distortion is a case of Jahn–Teller distortion, very common in chemistry.


Can a half-reaction be oxidation if it has the higher reduction potential? I am working on a problem that provides these two half-reactions:

$$
\begin{align}
\ce{Fe^3+ (aq) + e- &-> Fe^2+ (aq)} &\quad E^\circ_\mathrm{red} &= \pu{+0.77 V}\\
\ce{S2O6^2- (aq) + 4H^+ + 2e- &-> 2H2SO3 (aq)} &\quad E^\circ_\mathrm{red} &= \pu{+0.60 V}
\end{align}
$$

What is confusing me is the way the question is worded:

  Write the balanced chemical equation for the oxidation of $\ce{Fe^2+ (aq)}$ by $\ce{S2O6^2- (aq)}$.

Isn't this backwards though? Because the $\ce{Fe}$ half reaction has the higher reduction potential, shouldn't it be reduced while the $\ce{S2O6^2- (aq)}$ half reaction reverses to become oxidation?

Redox reactions are reversible. Despite the forward reaction is not spontaneous (as indicated by the unfavourable potential), it can still proceed.


Which ion of iron is produced in a reaction between iron and copper(II) sulfate? Using these equations:
\begin{aligned}
\text{(Fe2):}&&\ce{ Fe + CuSO4 &-> FeSO4 + Cu}\\
\text{(Fe3):}&&\ce{ 2Fe + 3CuSO4 &-> Fe2(SO4)3 + 3Cu}
\end{aligned}
Beginning with $\pu{0.78 g}$ of iron, the theoretical yields of copper are $\pu{0.89 g}$ (with $\ce{Fe^{2+}}$) and $\pu{1.33 g}$ (with $\ce{Fe^{3+}}$), and the actual yield was $\pu{0.78 g}$ of copper. I know to find of the ratio of the moles of iron used to the moles of copper formed (88% for $\ce{Fe^{2+}}$, 58% for $\ce{Fe^{3+}}$), and so the clear choice would be that this reaction formed $\ce{Fe^{2+}}$ ions.
But my teacher said this would form $\ce{Fe^{3+}}$ ions, but why is that? Shouldn't it have formed $\ce{Fe^{2+}}$? The reaction formed $\pu{0.12 mol}$ of copper, compared to the $\pu{0.14 mol}$ in the theoretical yield, and the other $\pu{0.02 mol}$ could have just been lost somewhere along the reaction process.

When "doing" science, usually you try to (in-)validate a model of the workings of the world. In the case of chemists, this world is mostly confined to the scale of molecules.

That said, let's proceed with

A Theoretical Model

Listed below are some standard reduction potentials I've gathered from the Handbook of Chemistry and Physics, 95th ed. (probably paywalled for you):
$$ \begin{align}
&&E^0_\text{red} / \mathrm{V}\\
\hline
\ce{Fe^{2+} + 2e-  & <=> Fe} & -0.447 \\
\ce{Fe^{3+} + 3e-  & <=> Fe} & -0.037 \\
\ce{Cu^{2+} + 2e-  & <=> Cu} & 0.342 \\
\end{align} $$

Now we can calculate the potential (and the resulting electromotive force EMF) for each of the reactions:
$$ \begin{align}
E^0_\text{(Fe2)} = E^{0}_\text{red}(\text{red}) - E^0_\text{red}(\text{ox}) &= 0.342~\mathrm{V} - (-0.447~\mathrm{V}) = 0.789~\mathrm{V} \\
E^0_\text{(Fe3)} &= 0.342~\mathrm{V} - (-0.037~\mathrm{V}) = 0.379~\mathrm{V}
\end{align} $$

It is clear that the reaction (Fe2) releases more energy and thus should be preferred over the other reaction. This means that we expect a yield of 100% Cu and not 150% Cu.

Now that we have gathered some theoretical understanding of the matter, let's proceed to the interesting part, which is

Confirmation With Experimental Results

You have correctly calculated the theoretical yields for copper to be 0.89 g and 1.33 g, respectively. The corresponding amounts of copper are 0.014 mol and 0.021 mol.

You have measured the amount of copper produced, and it clocks in at 0.78 g. This is already a strong indication that we should side with (Fe2) on this one. But just to really be sure, let's look at the error that we would have if it was (Fe3):
$$ \varepsilon = \frac{\Delta x}{x} = 41.4\% $$
This is a very high error, especially when compared to the error for (Fe2), which is only 12.1%. Assuming that you did not perform the reaction with pure Fe but a mixture of iron with some iron oxides as contaminants, did not use an analytical scale and performed the experiment in fairly uncontrolled conditions (are you sure the reaction was complete when you aborted?) this error is not as bad as it seems.

With the experimental results in our pocket, we can now move on to the final part:

The Conclusion

We have shown, using theoretical predictions and experimental confirmation, that the reaction observed was the oxidation of iron to iron(II) under reduction of copper(II) to copper.

By the way, most of the calculations I did with an iPython Notebook, which you can view here. I might have messed something up, you never know.


Is it possible to graph the equivalence of p[Cu²⁺] with pH? In no way I am a chemistry major, so please forgive me if my question seems stupid. I just need a way to find the "equivalent $\mathrm{pH}$" of $\mathrm p[\ce{Cu^2+}]$ if that makes any sense to you.

I'm making a paper about a transistor that can sense ions (ion-sensitive field-effect transistor); i.e. it's a potentiometric transistor. I was able to simulate voltage values with respect to $\mathrm{pH}$ activity in a simulation software named LTspice, but in order for this paper to be passable to whoever shall evaluate it, I must compare my data with a preexisting experimental data.

The only experimental data that I have is from my co-advisee. He measured the voltage values with respect to $\mathrm p[\ce{Cu^2+}]$ $(\ce{CuCl2 · 2 H2O}).$ I think it wouldn't be proper to compare the voltage vs $\mathrm{pH}$ graphs and voltage vs $\mathrm p[\ce{Cu^2+}].$

So, I figured that if I can express $\mathrm p[\ce{Cu^2+}]$ to its equivalent $\mathrm{pH}$ (I don't think this is possible and makes any sense), then I'd be able to compare the two graphs.

TTbulaski, I gather that you are developing a sensor for copper ions whose output is proportional to -log [Cu(II)]? I believe this is what you are trying to say.

What you can test is whether the output of your sensor gives any linear relationship between actual Cu concentration and the antilog of your sensor signal or not. Plot the Cu ion concentration on the x-axis and the anti-log of your sensor signal on the y-axis? Do you see a linear relationship?

pH and -log [Cu(II)] connection is pretty tricky and it depends what level of depth you want to go to. There is no point in looking for a similarity between the two measurements.
People have done calculations to show what species of Cu is present as a function of pH. It is called speciation diagram of copper.

P.S. The most important thing to test any metal ion sensor is something called as "selectivity". No analytical chemistry reviewer will accept the paper until and unless you show how selective is your sensor. For example if you have Zinc or Cd or other transition metals, do you still get a signal for copper in a correct way?

Not long time ago, NASA did a mistake in ion-selective electrodes. They designed or used a nitrate selective electrode and it turned out that is more selective towards perchlorate ion. It was an interesting news.


Why is cytosine far more vulnerable to hydrolytic deamination than both adenosine and guanosine? Why is cytosine far more vulnerable to hydrolytic deamination than both adenosine and guanosine? Here's the source for the claim (the half-life is 340 years at 298 K, which is much smaller than that of adenosine and guanosine).

Cytosine deaminates to form uracil at a finite rate in vivo

Pyrimidine bases e.g cytosine in DNA are more susceptible to spontaneous deamination than are the purine bases because there are more thermolabile.

  This was demonstrated by the previous observation of significant
  cytosine deamination but no guanine deamination upon heating of the
  synthetic copolymer poly(dG-dC):poly(dG-dC) and the homopolymer
  poly(dG):poly(dC)

Reference: Journal of Biological Chemistry -  Nitric Oxide-induced Deamination of Cytosine and Guanine in Deoxynucleosides and Oligonucleotides

Another research article: Cytosine Deamination Plays a Primary Role in the Evolution of Mammalian Isochores agrees with the theory of temperature as the major cause of sponteneous demination in cytosine as it states:

  Because a 10°C increase in temperature in vitro increases the rate of
  cytosine deamination 5.7-fold, cytosine deamination must be highly
  dependent on body temperature.

Cytosine residues in DNA (2) can undergo hydrolytic deamination to yield uracil residues (3)

The reaction may proceed either via attack of hydroxide on the neutral nucleobase or attack of water on the N3-protonated base:

  In single-stranded DNA, and at base mismatches in duplex DNA,
  deamination proceeds much faster (t1/2 ~ 200 years) presumably due to
  increased solvent accessibility of the base.   Deamination of guanine
  and adenine residues in DNA is much slower, occurring at only 2–3% the
  rate of cytosine deamination.

Alkylation of the N3-position of cytosine and reactions that lead to saturation of the 5,6-double bond in cytosine and 5-methylcytosine accelerate deamination.

Activation-induced cytidine deaminases catalyse the conversion of cytosine residues to uracil residues in single-stranded regions of DNA.

Reference:  An Overview of Chemical Processes That Damage Cellular DNA: Spontaneous Hydrolysis, Alkylation, and Reactions with Radicals

Another factor which apparently "contributes" to susceptibilty of cytosine to hydroysis is its presence in both DNA and RNA.

DNA is not susceptible to alkaline hydrolysis while RNA is alkali labile and is readily hydrolysed by hydroxide ions. On the other hand RNA is relatively resistant to the effects of dilute acid, but gentle treatment of DNA
with acid leads to hydrolysis of purine glycosidic bonds and the loss of
purine bases from the DNA.

These different conditions  (i.e acidic and basic conditions) to which the bases are exposed to appears to have a "double effect" on cytosine. (and additionally , as a pyrimidine it is more thermolabile)

Other reference

Biochemisty (Grisham)


How to relate a reaction barrier to the time the reaction needs to proceed? As I am writing this I am at a conference and one of the participants just asked a question where he linked reaction barriers to durations for the reaction to complete. To paraphrase:

  From our experience a reaction with an activation barrier of 15 kcal/mol should occur instantaneous at room temperature. An activation barrier of 20 kcal/mol takes about one to two minutes and an activation barrier of 25 kcal/mol needs about 10 hours.

I would like to rationalise this statement as it seems quite hand-weaving to me. How can I judge from (a possibly also computed) activation barrier how long a reaction will need to complete? For the the sake of the argument, let's only consider reactions that proceed in one step; generalisations may also be implied, but may be too complex.

In the comments (and the already existing answer) the Eyring equation is mentioned. If the connection between the energy values and the duration can be made with that, an illustrative example would be nice.

Here are some back of the envelope numbers from using the Eyring equation:

$$k = \frac{k_{B}T}{h}e^{-\frac{\Delta G^{\ddagger}}{RT}}$$

Let's just assume we're at $298\ \mathrm{K}$ for the reaction, and the reaction is relatively simple:

$$\ce{A->B}$$

I constructed the following table by plugging in values. $t_{1/2} = \frac{\ln 2}{k}$

\begin{array}{|c|c|}
\hline \Delta G^{\ddagger}\ (\mathrm{kcal\,mol}^{-1})& k\ (\mathrm{s}^{-1}) & t_{1/2} \\ \hline\hline
15 & 63.4 & 10.9\ \mathrm{ms} \\
20 & 0.0138 & 50.2\ \mathrm{s} \\
25 & 2.98\cdot 10^{-6} & 64.6\ \mathrm{h} \\\hline
\end{array}

The values for 15 and 20 $\mathrm{kcal\,mol}^{-1}$ seem pretty consistent with your rule. The top value is a bit off, but we're working with a very small numbers at this point, and there may be other sources of error that we're not accounting for in the model.


Hydroboration/oxidation vs acid-catalyzed hydration of 1-phenylpropene I am trying to predict the products of both the hydroboration/oxidation and acid-catalyzed hydration of 1-phenylpropene:

Both C1 and C2 of the double bond are secondary carbons. However, I have the feeling that there are slight differences in the electron density on each carbon, which will affect the addition of the boron electrophile and the stability of the carbocation for the two reactions respectively.

Which regioisomers are predominantly formed from both reactions?

Hydroboration

With the simplest hydroborating agent $\ce{BH3.THF}$, the hydroboration/oxidation of 1-phenylpropene proceeds to give 1-phenylpropan-1-ol as the major product.[1] The isomer of the substrate used was not reported.

There is some ambiguity in the above source, as the labels 8 and 8′ (which the paper uses to refer to the two regioisomers) are consistently swapped; hence, I have taken the names and not the labels to be accurate. However, as additional proof, there are other examples of similar regioselectivity in the hydroboration of styrenes (here, an E/Z mixture of styrenes was used):[2]

Rationalising this selectivity is not very easy, but in one of Herbert Brown's papers where the same regioselectivity is seen with 9-BBN as the hydroborating agent, he writes:[3]

  The preferred site of hydroboration in 1-phenylpropene, the site to which boron becomes attached, is the 1-position. This is not what one would expect on the basis of steric effects. One possible explanation for this is the combined effects of phenyl conjugation (-K) and methyl hyperconjugation (+K) which act to decrease the amount of electron density at the 2-position and to increase it at the 1-position.

I'm not totally convinced by this, but I also don't have any better explanation. Further evidence for this explanation is the fact that the (Z)-isomer reacts with poorer regioselectivity than the (E)-isomer. This is because there is allylic strain in the (Z)-isomer which forces the phenyl group to twist out of conjugation with the C=C double bond (depicted below). Consequently, the electron-withdrawing effect of the benzene ring (via conjugation) is reduced, and the formation of the 2-hydroxy isomer is less disfavoured.

Acid-catalysed hydration

Examples of these in the literature are harder to find and I cannot find any generic examples of acidic hydration of $\ce{R^{1}-C=C-R^{2}}$, where $\ce{R^1}$ is an aryl group and $\ce{R^2}$ an alkyl. However, I see no particular reason to go against the conventional wisdom: the more stable carbocation intermediate is formed at the benzylic position, which should lead to the 1-hydroxy compound as the major product.

References

Camacho, C.; Uribe, G.; Contreras, R. Diphenylamine Borane, A New Stable Amine .Borane with Remarkable Hydroborating and Reducing Properties. Synthesis 1982, 1982 (12), 1027–1030. DOI: 10.1055/s-1982-30051.
Allegretti, M.; Bertini, R.; Cesta, M. C.; Bizzarri, C.; Di Bitondo, R.; Di Cioccio, V.; Galliera, E.; Berdini, V.; Topai, A.; Zampella, G.; Russo, V.; Di Bello, N.; Nano, G.; Nicolini, L.; Locati, M.; Fantucci, P.; Florio, S.; Colotta, F. 2-Arylpropionic CXC Chemokine Receptor 1 (CXCR1) Ligands as Novel Noncompetitive CXCL8 Inhibitors. J. Med. Chem. 2005, 48 (13), 4312–4331. DOI: 10.1021/jm049082i.
Brown, H. C.; Nelson, D. J.; Scouten, C. G. Hydroboration. 63. Kinetics and regiospecificity of the hydroboration of isomeric cis- and trans-alkenes via 9-borabicyclo[3.3.1]nonane (9-BBN). Effects of 1,1- and 1,2-dialkyl interactions. J. Org. Chem. 1983, 48 (5), 641–643. DOI: 10.1021/jo00153a003.


Constructing a Walsh diagram of AH2 According to Wikipedia, for the $\ce{AH2}$ Walsh diagram:

and the Walsh diagram in Shriver and Atkins Inorganic Chemistry, the $a_1$ orbital sinks in energy going from 90 to 180°. Furthermore, the $b_2$ orbital at 90° is higher in energy than the $a_1$ (lone pair).

However according to the second, newer diagram for a AH2 molecule on the same page:

the $a_1$ is rises in energy and the $b_2$ is definitely lower in energy than  the $a_2$ (or $a_1$ lone pair). This matches also my own self-made diagram, using Hartree-Fock 6-31G*, calculated through Spartan. Other pages use the same Walsh diagram.

What is going on here?

Shrivers textbook explains the rise of the $a_1$ orbital through decreased H-H overlap and reduced involvement of the $\ce{X~2_{pz}}$.

I'd say the diagram from Atkins "Inorganic Chemistry" is in error.

I went to what I consider the "definitive source" for such Walsh diagrams: Orbital Interactions in Chemistry by Albright, Burdett and Whangbo.

Here's their diagram for $\ce{H2S}$ (the linear on the left, the bent on the right):

They don't readily indicate the method they used for the calculation, but on this energy scale the difference between the $1\sigma_g$ and $1a_1$ is minimal. They show that the $1a_1$ is lower in energy since the interaction with the central p orbital will be stabilizing.

As to the $1b_2$ rising above the $2a_1$ in the bent structure -- the diagram from Albright, Burdett, and Whangbo agree that there's no "crossing."


How does Melatonin put you to sleep? How is this compound radio-protective? Melatonin (IUPAC semi-formal name: N-acetyl-5-methoxytryptamine) is a hormone produced in humans, animals and some of plants. It's mainly famous for being the drug suggested to people with sleeping problems, jet lags and even insomnia (in trivial cases, still experimental). Recent studies have shown this compound's role as an anti-oxidant and a protection for mitochondrial DNA.
Melatonin interacts with the immune system; and is clinically proved to have anti-inflammatory effects. It can act as a radio-protective agent...
Hold it there. This is chemistry, not biology. First of all, how is Melatonin radio-protective? And what is in that structure that makes it radio-protective? 2nd, what reactions occur in the body, in presence of Melatonin, that cause sleepiness (resources have claimed the chemical function to be a mystery but I doubt that's true)?
 Most of the info above is available at the wikipedia: Melatonin.image credit: http://cenblog.org/the-haystack/files/2011/06/melatonin.png

Melatonin is not "radio-protective" in the sense that it blocks radiation or anything. As you note, it's an antioxidant. One of the ways in which ionizing radiation can interact with the body is by providing enough energy to break chemical bonds, producing highly reactive radical species that can cause biological damage by reacting with various things in the body. Melatonin can easily be oxidized by many of these reactive species, preventing them from reacting with anything else. This paper shows the products for the reaction with radio-generated hydroxyl radicals.

The role melatonin plays in sleep is fairly well known (though the effects of exogenous melatonin, less so). Without going into too much detail, there are melatonin receptors in the body, found in cell membranes in the brain and a few other places (G protein-coupled receptors). When these bind melatonin on the outside of the cell, their conformation changes, further activating signalling processes on the inside of the cell. The effects of almost all signalling molecules in the body work in this way. The molecules activate some kind of receptor, which then activates a sequence of other signals, eventually resulting in biological effects.


Amount concentration versus number concentration The IUPAC Gold Book defines

Amount concentration
Amount of a constituent divided by the volume of the mixture. Also called amount-of-substance concentration, substance concentration (in clinical chemistry) and in older literature molarity. For entities B it is often denoted by B. The common unit is mole per cubic decimetre or mole per litre sometimes denoted by M.

and

Number concentration
Number of entities of a constituent in a mixture divided by the volume of the mixture.

I can't see a distinction between the two terms since it seems to me that "Amount of a constituent" is the same as "Number of entities of a constituent".
Could you please explain the difference between them? Is it a conceptual difference or just a difference in the context in which they are most commonly used?

The "amount" of a substance is the proper term for the "number of moles" (Wikipedia; Gold Book). So, the two quantities are related by a factor of the Avogadro constant, $\pu{6.022\times 10^23 mol-1}$.

If you have $\pu{1 mol}$ of helium in a $\pu{1 m3}$ box, the amount concentration is $\pu{1 mol m-3}$, or $\pu{0.001 mol dm-3}$ if you prefer those units.

The corresponding number concentration is $\pu{6.022\times 10^23 m-3}$.


Solubility of NOx gases Are the oxides of nitrogen (such as $\ce{NO}$ and $\ce{NO2}$) emitted from the exhaust of a diesel engine soluble? If so, in what solvents is it soluble?

I want to analyse the nitrogen oxides in the exhaust gas of a diesel engine. Is it possible to collect the gas by solvent extraction method?

$\ce{NO_x}$ can be quantitatively removed from such combustion streams by absorption/dissolution in basic aqueous solutions containing an appropriate oxidizing agent. While $\ce{NO2}$ is readily soluble in aqueous solution, $\ce{NO}$ is not. However, $\ce{NO}$ is easily converted to $\ce{NO2}$ when an oxidant is added to the solution.

The PhD thesis by Chen-Lu Yang at NJIT, "Aqueous Absorption of $\ce{NO_x}$ Induced by Sodium Chlorite Oxidation", concluded the following:

$\ce{NO2}$ is essentially quantitatively absorbed in $\ce{NaOH}$ aqueous scrubbing solutions.
$\ce{NO}$ is not absorbed by $\ce{NaOH}$ aqueous scrubbing solutions.
$\ce{NaClO2}$ containing aqueous scrubbing medium quantitatively oxidizes $\ce{NO}$ to nitrate [which is highly water soluble].

The net effect of these solubility/reactivity properties is that $\ce{NO}$ and $\ce{NO2}$ can be quantitatively removed from a gas stream using an aqueous solution of $\ce{NaOH}$ and $\ce{NaClO2}$

Additionally, according to this PubChem entry, $\ce{NO}$ is soluble in sulfuric acid, ethanol and carbon disulfide, while according to this Wikipedia article, $\ce{NO2}$ is soluble in carbon tetrachloride, nitric acid and chloroform.


Decomposition through thermolysis, electrolysis, photolysis What decides whether a compound will dissociate (break-up) under thermal decomposition or photolysis or electrolysis i.e if I am provided with a compound and I'm asked to dissociate it then how will I get to know that it would dissociate under thermal decomposition or photolysis or electrolysis.

Please try to provide an easy-to-understand as I'm in just 10th grade

You should analyse the chemical bindings in your compound, e.g. ion bindings will easily break-up in water, stronger bindings may require You to use heating or electric current to dissociate the molecule. Also if you know the structure of the compound, look for some similar structures on the internet, the more similar the structures, the more similar properties they have, so similar factors will dissociate them.


How did J. J. Thomson prove that the cathode rays were made of negative particles and not negative rays? In Thomson's experiment, he used a discharge tube to prove that the cathode rays that emanate from the cathode were made of "a stream of negatively charged particles" because they were repelled by an applied negatively charged electrical field. Why did Thomson conclude from this experiment that the negatively-charged cathode rays were made of particles, and hence the discovery of the electron? Why did this force him to conclude that the rays were really particles? What am I missing here?

Thompson was able to measure the ratio of their mass to their speed

Cathode "rays" had been known for some time before Thomson. They were first observed as experiments in gas discharge tubes started to exploit better and better vacuums (the early experiments observed the varying forms of discharge in low pressure gases; cathode rays only become significant when there is very little gas left in the tubes).

At very high vacuums, people observed that the glass at the end of the tubes would glow even though the contents of the tubes did not (as there were too few gas particles left to cause a discharge). So the idea that some sort of ray was emitted by the cathode in the tube was developed. Initially the rays were seen to travel in straight lines.

Later, scientists such as Crookes observed that the rays could be deflected by magnetic fields (though he thought they were just negatively charged gas particles). Hertz, though, claimed electric fields would not deflect them (he was wrong).

The critical experiments done by Thompson relied on having an excellent glassblower who was able to construct an apparatus where electrodes inside the discharge tube which could be charged with an electric field (while retaining a very high vacuum). This enabled Thompson to refute the claim by Hertz that electric fields didn't cause deflection (maybe Hertz's glassblower wasn't as good). Once he had this apparatus further experiments with electric and magnetic fields could reveal a great deal more about the particles.

Thompson describes the critical experiments like this:

  As the cathode rays carry a charge of negative electricity, they are deflected by an electrostatic force as if they were negatively electrified, and are acted on by a magnetic force in just the way in which this force would act on a negatively electrified body moving along the path of these rays, I can see no escape from the conclusion that they are charges of negative electricity carried by particles of matter. The question next arises, What are these particles? Are they atoms, or molecules, or matter in a still finer state of subdivision? To throw some light on this point, I have made a series of measurements of the ratio of the mass of these particles to the charge carried by it (“Cathode rays,” Philosophical magazine, 44 (1897), 293-316, pdf of original available here)

[A good review and description of modern recreations of this experiment is given here]

The point of the experiments is that the deflection of the particle in a uniform electric field depends on its mass, velocity and charge. And its speed can be measured by carefully balancing the electric and magnetic fields to give zero net deflection (why this is so involves equations, but this is Chemistry.SE not Math.SE and you can read them in the paper).

So Thompson observed that cathode rays were best explained by a charged particle (a classical wave would not have been deflected) with an observable charge to mass ratio (which implied either a very large charge or a very small mass). And he observed that the particles were always the same regardless of the cathode material or the original gas in the tube.

Later experiments clarified the mass and charge. Which correspond to what we now call electrons.

Note that this happened well before any of the ideas of quantum mechanics and wave/particle duality appeared. Thompson was concluding, in contradiction to some theorists at the time, that the properties of cathode rays were better explained by particles obeying the known laws for electricity and magnetism than by a wave. And he was right. His discovery was the first time a key component of the atom had been identified.


How is it possible to account for the fact that NH3 is a weak base in water but a strong base in HCl? I understand that the solvents differ in their acidity but why does this affect the basicity of the solute?

The difference between a weak base and a strong base is that a strong base is completely converted to ions in a given solvent, whereas a weak base is not. Although this depends on the actual basicity of the solute, it is also very dependent on the solvent involved.

In water, $\ce{NH3}$ can react to form $\ce{NH4+}$ and $\ce{OH-}$. However, most of the ammonia remains as $\ce{NH3}$. This is due to the difference in pKas:1

$\ce{NH4+}$ = 9.2

$\ce{H2O}$ = 15.7

Because the ammonium ion is more acidic than neutral water, very little $\ce{NH4+}$ will be present at equilibrium.

In contrast, the relevant pKas for the $\ce{HCl}$ solution are:

$\ce{NH4+}$ = 9.2

$\ce{HCl}$ = -8

This tells us that $\ce{HCl}$ is much more acidic than the ammonium ion, and therefore essentially all of the ammonia will be protonated when dissolved in $\ce{HCl}$. As a result, it is considered a strong base in this case.

1Evans' pKa tables


What is the smell of 'burning' metal? When a pass a construction site where someone is welding two pieces of metal together there is a very distinct smell that I associate with 'burning' metal, although I am not quite sure it is burning in the traditional sense. Also, when someone ignites a sparkler this same (or at least similar) smell emerges.

I would like to know what is causing the smell? Is it the smell of metaloxide particles that are dispersed from the weld site? Or is it something in the air that reacts and causes the smell?

According to the Canadian Centre for Occupational Health and Safety page Welding - Fumes And Gases, the welding fumes are

  a complex mixture of metallic oxides, silicates and fluorides. Fumes are formed when a metal is heated above its boiling point and its vapours condense into very fine, particles (solid particulates).

This composition varies dependent on the composition of metals being welded and fluxes.  The website lists several main types and their associated health risks.  Often associated with metallic oxides are some pungent/noticeable gases:

Gases produced from the thermal breakdown of coatings, from welding and cutting processes, stated in the article linked as being examples of "welding gases" (examples from the linked website include ozone)
And depending on the coating that may be on the metal, organic vapours (including aldehydes, diisocyanates, phosgene and phosphine)


How is aqua regia stored? How is aqua regia stored as it dissolves all metals?

How is aqua regia stored as it dissolves all metals?

No one stores aqua regia for extended periods of time, because it decomposes rapidly, yielding volatile products that escape from the solution resulting in a loss of potency.
From the Wikipedia article on aqua regia,

Due to the reaction between its components resulting in its decomposition, aqua regia quickly loses its effectiveness (yet remains a strong acid), so its components are usually only mixed immediately before use.

Upon mixing of concentrated hydrochloric acid and concentrated nitric acid, chemical reactions occur. These reactions result in the volatile products nitrosyl chloride and chlorine as evidenced by the fuming nature and characteristic yellow color of aqua regia. As the volatile products escape from solution, the aqua regia loses its potency.
$$\ce{HNO3 + 3HCl -> NOCl + Cl2 + 2H2O}$$

The nitrosyl chloride formed is, essentially, a decomposition product of $\ce{HCl}$ and $\ce{HNO3}$. This itself, is further decomposed as,
$$\ce{2NOCl -> 2NO + Cl2}$$
$$\ce{2NO + O2 -> 2NO2}$$
The mixture itself is created by combining $\ce{HCl}$ and $\ce{HNO3}$ in a 3:1 molar ratio in glass vessels immediately before use.

Aqua regia: Why does it dissolve gold but not glass?
From what I gather from this PDF, nitric acid is a notoriously strong oxidizing agent and is quite capable of oxidizing elemental gold (a "noble metal").
$$\ce{Au(s) + 3NO3^{-}(aq) + 6H^{+}(aq) <=> Au^{3+}(aq) + 3NO2(g) + 3H2O(l)}$$
However, we wouldn't see that happening even if we plopped a gold coin in a jar of nitric acid and waited a lifetime. This is because the reaction has an extremely small equilibrium constant (I'm having trouble getting the actual value of the constant myself, feel free to give me a heads up if you know where to find it), meaning only a minuscule (almost negligible) quantity of gold actually reacts/dissolves.
So for all practical purposes, nitric acid doesn't react with gold (though strictly speaking, it does).
So you want to dissolve gold, and you know (for real) that nitric acid can do it (but not very "efficiently", if you will). What do you do?
The trick is to throw in some concentrated $\ce{HCl}$ as well.
If you remember your high school chemistry lessons on equilibrium and Le Chatelier's principle, you'll find this easy to follow: The $\ce{Cl^{-}}$ generated by the $\ce{HCl}$ complexes with the $\ce{Au^{3+}}$ and essentially "plucks" out the $\ce{Au^{3+}}$ from the earlier equation, resulting in the equilibrium shifting to the right, causing more gold to react.
$$\ce{Au^{3+}(aq) + 4Cl^{-}(aq) <=> AuCl4^{-}(aq)}$$
Now according to this Quora answer, the reason aqua regia doesn't attack glass, is because silicates (the stuff that constitutes glass) are fairly inert to the oxidizing power of nitric acid (on account of their strong $\ce{Si-O}$ bonds).

Aqua regia doesn't readily react with glass. So while you could keep it in a glass retort for years, it would be nothing like the original mixture (you be hard pressed to even call it "aqua regia" after a day, much less, after a few years).
A rather famous incident regarding the "storage" of aqua regia, was when de Hevesy dissolved two gold Nobel medals in aqua regia to hide them from Nazi Occupiers, and kept the solution in a glass retort on a shelf. He returned to collect it in about 5 years (the bottle was still intact)

When Nazi Germany occupied Denmark from April 1940, during World War II, de Hevesy dissolved the gold Nobel Prizes of Max von Laue and James Franck with aqua regia; it was illegal at the time to send gold out of the country, and had it been discovered that Laue and Franck had done so to prevent them from being stolen, they could have faced prosecution in Germany. He placed the resulting solution on a shelf in his laboratory at the Niels Bohr Institute. After the war, he returned to find the solution undisturbed and precipitated the gold out of the acid.

There is, however, an interesting alternate interpretation of de Hevesy's heroic deed,

 It is commonly suggested, even years after his death, that he only chose to dissolve the gold for story-telling purposes as opposed to practically. Perhaps a much simpler and efficient option was for him to bury the Nobel Prizes.

 You can also read up on how aqua regia "dissolves" gold here. And for those of you who want to know a way to easily recover the said gold, have a look at this Nurdrage video

Also, your question assumes that all metals are dissolved in aqua regia, which is apparently, not the case, as discussed under this post.


Half cell method of voltage calculation in an electrochemical cell A cell consists of a copper electrode $\ce{Cu}$, in contact with $1.800\cdot10^{-2}~\mathrm{M}$ cupric ion and a platinum electrode, $\ce{Pt}$, in contact with $1.400\cdot10^{-1}~\mathrm{M}$ ferrous ion and $3.700\cdot10^{-2}~\mathrm{M}$ ferric ion in $1~\mathrm{M}$ $\ce{HCl}$.
  What is the potential of cell?

I would have an idea of what to do with this question if it had a concentration for copper in the cupric half reaction; I would just use the nernst equation, rearranged to solve for $E$:

$$E = E^\circ - \frac{\mathrm{R}T}{n\mathrm{F}}\log Q$$
Then plug in the concentration values for the cupric half-reaction into $Q$ to get the $E$ value for this half reaction. I'm assuming that the half reaction taking place here is:
\begin{align}
\ce{Cu^{2+} + 2e^- &-> Cu^0} &E^\circ &= +0.337~\mathrm{V}
\end{align}

and the other half reaction taking place would be:
\begin{align}
\ce{Fe^{3+} + e^{-} &-> Fe^{2+}} &E^\circ &= +0.771~\mathrm{V}
\end{align}

Then, I would plug the concentration values, along with their standard potentials, into the rearranged nernst equation to get the $E$ values for each half, then subtract the larger value from the smaller. I already did this, assuming that the concentrations of $\ce{Cu^0}$ and $\ce{Cu^{2+}}$ were the same, giving a logarithm of $0$ and making $E = E^\circ$. The other equation remained the same. I got an answer of $0.399~\mathrm{V}$, which is wrong.

Also, I'm not sure how the concentration for $\ce{HCl}$ factors into this question, because the half reaction for $\ce{Fe}$ doesn't have any hydrogens in it.

Can anyone give me a hand with this one?

Think about what state $\ce{Cu^0}$ is in. $\ce{Cu^2+}$ is dissolved in the solution, but is this true of the unionized copper? Try to write out the half-reactions with the physical states included. What activity (effective concentration) do solids have in an equilibrium expression?

Also, though there are some reactions that could happen involving the acid, and I don't know whether they actually contribute to the cell potential in this case, the way the question is written, it looks like the two half-cells you have are the significant ones.


Half equations for H2O2 for its reducing and oxidising nature in acidic and alkaline medium Is there a complete list of all the half equations for $\ce{H2O2}$ - both oxidation and reduction, in acidic and alkaline conditions? I've looked on the internet but can't seem to find a list with all of them.

These are my first attempts:

$\ce{H2O2 + 2e^- -> 2OH^-}$ (in alkaline conditions)
$\ce{H2O2 + 2H^+ + 2e^- -> 2H2O}$ (in acidic conditions)
$\ce{H2O2 -> 2H^+ + O2 + 2e^-}$ (in acidic conditions)

Are these correct and are there any more?

From Proceedings of the Symposium on Oxygen Electrochemistry and Water Research 2002, 36 (1), 85-94:

$$
\begin{array}{lcc}
\text{Formation}&:&~\ce{2H+ + O2 + 2e- &->& H2O2} &\quad E^0=\pu{0.440V} \text{ vs. SCE}\\
\text{Reduction}&:& ~\ce{H2O2 + 2H+ + 2e- &->& 2H2O} &\quad E^0=\pu{1.534V} \text{ vs. SCE}\\
\text{Direct water}&:& ~\ce{2H+ + 1/2O2 + 2e- &->& H2O}\\
\text{Decomposition}&:&  ~\ce{H2O2 &->& H2O + 1/2 O2}
\end{array}$$

($\text{SCE}$ stands for Saturated Calomel Electrode)

Another strange suggested reaction involves the anion $\ce{HO2-}$ (Abel E.: Über die Selbstzersetzung von Wasserstoffsuperoxyd. Monatshefte für Chemie 1952 83(2) 422–39.):

$$\text{Decomposition 2}:\ce{H2O2 + H2O- -> H2O + O2 + OH-}$$

At the cathode occur at the same time the hydrogen gas evolution:

$$\ce{2H+ + 2e- -> H2~} \qquad\qquad\phantom{00} E^0=\pu{0.242 V} \text{ vs. SCE}$$

Meanwhile, at the anode, oxygen gas is evolved:

$$~\ce{2H2O -> 4H+ + O2 + 4e-} \quad E^0=\pu{0.987V} \text{ vs. SCE}$$


Molecular orbital diagram for nitrogen monoxide, the nitrosyl cation and the nitrosyl anion As I understand it, all of them should have same order of energy of molecular orbitals as a model for $\ce{NO}$ should withstand ionization and adding an electron.

JD Lee Pg 109 shows the same diagram for $\ce{NO}$ as the one quoted in here. I couldn't find $\ce{NO+}$ except in here, where $\sigma_{2p_z}$ has a higher energy than $\pi_{2p_x}$ and $\pi_{2p_y}$.

If it is correct, then, there is $s$-$p$ mixing in $\ce{NO+}$ but not in $\ce{NO}$?

As usual with these simple molecules: It's not as simple as it seems. Here it actually doubles down quite a bit more.
The short and very dis-satisfactory answer is: It is very, very complicated. For this reason, I am reluctant to present a molecular orbital diagram. However, I will provide calculated MOs. While it may be tempting to just look at the pictures, the rationalisation behind it is more important. What will follow now are a few attempts at that rationalisation.

In the nitric oxide series, $\ce{NO+}$ is surprisingly the simplest. Why is that surprising? Because it is isoelectronic with $\ce{CO}$; and this molecule is one of the most complicated ones there are. You can read a bit more about it here: How can the dipole moment of carbon monoxide be rationalised by molecular orbital theory?
I do compare $\ce{NO+}$ more with $\ce{CO}$ rather than $\ce{N2}$ to which it is also isoelectronic, because of the asymmetry introduced by the different nuclei. The molecule is the simplest to explain, because it is a singlet ground state. All electrons are neatly paired, and spacial symmetry can easily be matched with the degeneracy of the orbitals.
The π-orbitals are not the HOMO, instead it is the σ lone pair at nitrogen.

The next more complicated is $\ce{NO-}$, which is isoelectronic to $\ce{O2}$. In this case it is hard to find a less symmetric isoelectronic version. It is well known, that the ground state is a triplet because of the degeneracy of the π-orbitals. While we would expect the orbital ordering to be the same as in $\ce{NO+}$, it is not. This is mainly due to the fact, that occupied orbitals behave differently than unoccupied orbitals.
In some way, because of the paramagnetic character, we have a situation wich cannot be described by a single molecular orbital description. You need at least two: one for α-electrons (or spin up) and one for β-electrons (or spin down). The good thing however is, that both of these descriptions may still be matched with spatial symmetry.
The bonding π orbitals become higher in energy than the σ lone pair because the anti-bonding counterparts are partially filled.

The most complicated molecule in the series is the radical $\ce{NO}$. There is absolutely no simple description. Because of the unpaired electron, you need at least two descriptions again. However, this is not enough because now one of the spin-state description cannot match the spatial symmetry. This is because one electron cannot occupy two orbitals at the same time in an MO description. Because of that (from a technical point of view) the orbital degeneracy has to be broken. Physically this is not possible.
The molecule has to be treated with a multi-reference approach, i.e. one MO description for every possible configuration. It is probably safe to assume that the orbitals of π symmetry don't split that much and are higher in energy than the σ lone pair.

Below you can find the calculated MO. The level of theory used for $\ce{NO+}$ was RMP2/def2-QZVPP, because this is equivalent to the level the other molecules were treated, which is ROMP2//UMP2/def2-QZVPP. One can clearly see how the orbital symmetry for $\ce{NO}$ breaks the degeneracy of the pi orbitals.
For $\ce{NO+}$ MOs 5,6 and 8,9 are degenerated. For $\ce{NO-}$ MOs 6,7 and 8,9 are degenerated.
Colour code: blue/orange, α, β-occupied; purple/yellow-orange, α/β-occupied; red/yellow, unoccupied; contour value, 0.1.

With a little more literature research (prior to answering the question) one would have found the following publication:
Renato P. Orenha and Sérgio E. Galembeck. Molecular Orbitals of $\ce{NO}$, $\ce{NO+}$, and $\ce{NO–}$: A Computational Quantum Chemistry Experiment. J. Chem. Educ. 2014, 91 (7), 1064–1069.


Is it possible to crystallize a solid ingot of metal from an (acidic) aquaous solution? To my knowledge, metalic ions only precipitate from a solution as salts & and a process that could possibly produce monocristalline metal would be very interesting indutrially - I've never heard of one, so I don't think it is possible.

But: is it not conceivable to dissolve a metal in an acid (or maybe base), and then (later) neutralize the acid, without resulting ions that form a salt with the metal?

What am I missing?

As an aside, I arrived at this question while thinking if there could be a biological, and thus wet-chemical, pathway from ores to pure metals (in metallic form). The last step in a such a process would be the precipitation/crystallization as a metal.

I think you've just described the general concept of Electrowinning?

There are a large number of ways to recover metals from solutions or molten salts.

Producing monocrystalline metals is a different matter - crystal grains in metals tend to be microscopic, but there are techniques that permit growing large crystals of metals such as the Czochralski process or zone melting. Monocrystalline metals sometimes have attractive mechanical properties not achievable in polycrystalline metals - for example, Callister Materials Science and Engineering 6e p. 225 has a nice photo of a single-crystal jet turbine blade that is highly resistant to creep failure with respect to conventional blades.


Chemistry Self Education I want to learn chemistry. I did not have the benefit of studying it when I was in highschool, and I was always intimidated by it (I feel it's more rote memorization rather than understanding - I'm more of a Physics and Math kinda guy).

Anyway, what are the concepts that I need to learn?

How, and what resources are you going to use to homeschool Chemistry to a teenager?

In that studying it would put the student on par with a normal highschool student.

Some Nice Stuff:

Topics for a Normal Highschool Chemistry Curriculum
and Maybe AP and Honors Program Curriculums?
Comprehensive Textbooks
Exams and Self Testing Resources

You might want to have a look at free online courses, e.g. at Khan Academy. To my opinion, this can however not replace the real experience of endless, often frustrating hours in the lab ;) - chemistry is an experimental science.


Reduction of amino acids to corresponding amino alcohols I've stumbled upon an interesting article about reduction of amino acids and from there to reduction of L-valine to L-valinol using $\ce{LiAlH4}$ in THF.

Article says that because of the cost of $\ce{LiAlH4}$, better method using $\ce{NaBH4}$ with $\ce{I2}$ in THF has been developed.

But I haven't found anything about $\ce{HI}$. Is it possible to reduce amino acids all the way down to their corresponding amines with concentrated $\ce{HI}$ solution?

Chiral amino alcohols such as L-valinol are generally prepared by the reduction of corresponding α-amino acids and suitable reducing reagent. Most commonly used reducing agent is $\ce{LiAlH4}$ since most common reducing reagent, $\ce{NaBH4}$ does not reduce carboxylic acid to corresponding alcohols. However, use of $\ce{LiAlH4}$ has notable disadvantages such as being expensive and highly flammable. To avoid these disadvantages on high scale syntheses, some novel methods have been developed. One such method is reduction of carboxylic acid using $\ce{NaBH4/I2}$ reducing sysrem (Ref.1). The most remarkable feature of this reagent is its selectivity over other known strong reducing agents. For example, if $\ce{NaBH4/I2}$ is used to reduce bicarboxylic acid, it would reduce both carboxylic groups to $100\%$. However, if one carboxylic acid i
 group was protected (say, ester-carboxylic acid combo), $\ce{NaBH4/I2}$ would reduce only carboxylic acid group leaving ester group alone (Ref.1).

Two years after this important discovery, McKennon, et al. (1993) found out that this same method can be used to reduce N-protected or naked amino acids to their corresponding N-protected amino alcohol or amino alcohol, respectively (Ref.2). A typical procedure is as follows:

  Preparation of (S)-tert-Leucinol: A dry 5-L three-necked round-bottom flask equipped with a mechanical stirrer and a 250-mL addition funnel was charged with sodium borohydride ($\pu{55.6 g}, \pu{1.47 mol}, \pu{2.41 equiv}$) and $\pu{1.6 L}$ of anhydrous THF. The solution was stirred while (S)-tert-leucine ($\pu{80.0 g}, \pu{610 mmol}$) was added in one portion. The flask was cooled to $\pu{0 ^\circ C}$ and fitted with a reflux condenser. The addition funnel was charged with a solution of $\ce{I2}$ ($\pu{155.0 g}, \pu{610 mmol}, \pu{1 equiv}$) in $\pu{200 mL}$ of THF, which was added dropwise to the flask over a $\pu{1.5 h}$ period with considerable gas evolution. The solution was allowed to warm to room temperature. When the brown color had dissipated to give a cloudy white solution, the reaction was brought to reflux for $\pu{19 h}$. The cloudy white suspension was cooled to room temperature with the aid of a water bath. The addition funnel was charged with $\pu{150 mL}$ of $\ce{MeOH}$, which was added drop-wise with rapid stirring. Vigorous gas evolution was observed. Small aliquots of $\ce{MeOH}$ were then added until all of the solid white material had dissolved. The solution was concentrated by rotary evaporation to give a white pasty oil that was dissolved in $\pu{1.20 L}$ of $20\% (w/w)$ aqueous $\ce{KOH}$ and mechanically stirred for $\pu{6 h}$ at room temperature. The light green solution was extracted with $\ce{CH2Cl2}$ ($\pu{3 \times 1.50 L}, \pu{1 \times 600 mL}$). To minimize the emulsion, $\pu{200 mL}$ of brine was added to the aqueous layer after the first extraction. The combined organic extracts were dried over $\ce{Na2SO4}$, filtered through glass wool, and concentrated in vacuo to give $\pu{74.1 g}$ of a white crystalline solid in a thick, slightly yellow oil. Distillation of (S)-tert-leucinol, bp $70$-$\pu{73 ^\circ C}$/$\pu{2 mmHg}$ (Lit. $117$-$\pu{120 ^\circ C}$/$\pu{57 mmHg}$) from a $\pu{30.0 g}$ reduction reaction performed in the above manner yielded $\pu{21.3 g}$ of product ($79\%$ yield) as a clear oil, which solidified upon cooling to room temperature.

References:

J. V. Bhaskar Kanth, Mariappan Periasamy, “Selective reduction of carboxylic acids into alcohols using sodium borohydride and iodine,” J. Org. Chem. 1991, 56(20), 5964-5965 (https://doi.org/10.1021/jo00020a052).
Marc J. McKennon, A. I. Meyers, Karlheinz Drauz, Michael Schwarm, “A convenient reduction of amino acids and their derivatives,” J. Org. Chem. 1993, 58(13), 3568-3571 (https://doi.org/10.1021/jo00065a020).


Can I get some pointers on what to do in calculating the hardness of water? Information provided: $\pu{1.64 g}$ of $\ce{Ca(NO3)2}$ and $\pu{0.94 g}$ of $\ce{MgCl2}$ are dissolved in $\pu{10 L}$ of water, calculate the hardness of water.

I have used these two equations: $m = n \cdot M$ and $c = \frac nV$.

What I have done is simply insert the molar mass of both $\ce{MgCl2}$ and $\ce{Ca(NO3)2}$ in the first equation, isolated $n$, then used that answer to isolate $c$ in the next equation. But I need to find the mass for $\ce{Ca^2+}$ and $\ce{Mg^2+}$ only, how do I do that? Have I found the concentration of the wrong things?

Please explain in simple language. Thanks!

The hardness of water is defined as the sum of the dissolved $\ce{Ca^2+}$ and $\ce{Mg^2+}$ species and can be reported in different ways (moles/L, g/L, and so on). The one I have seen most commonly is mg/L, so that is the calculation presented below.

You need to determine how many milligrams of $\ce{Ca^2+}$ and $\ce{Mg^2+}$ are present and then divide that number by the solution volume (here, 10 L).

For $\ce{Ca^2+}$:

$$1.64\;\mathrm{g}\;\ce{Ca(NO3)2}\cdot{1\;\mathrm{mol}\;\ce{Ca^2+}\over 164.09\;\mathrm{g}}\cdot{40078\;\mathrm{mg}\over 1\;\mathrm{mol}\;\ce{Ca^2+}}
 = 400\;\mathrm{mg}$$

For $\ce{Mg^2+}$:

$$0.94\;\mathrm{g}\;\ce{Mg(Cl)2}\cdot{1\;\mathrm{mol}\;\ce{Mg^2+}\over 95.2\;\mathrm{g}}\cdot{24305\;\mathrm{mg}\over 1\;\mathrm{mol}\;\ce{Mg^2+}}
= 240\;\mathrm{mg}$$

Thus, you have 640 mg $\ce{Ca^2+}$ and $\ce{Mg^2+}$ in 10 L of solution so the hardness is:

$${640\;\mathrm{mg}\over 10\;\mathrm{L}} = 64\;\mathrm{mg}\cdot\mathrm{L}^{-1}$$


Why do carbonyl compounds undergo nucleophilic and not electrophilic addition? Why aldehydes and ketones undergo nucleophilic addition reactions and not electrophilic addition?

Alkenes show electrophilic addition because of pi bond, which is also present in aldehydes and ketones. But there is nucleophilic addition only. Why?

They sure do. Except that the oxygen has lone pairs, which are more nucleophilic than the $\pi$ bond. Once those electrons attack an electrophile, it further deactivates the $\pi$ bond as a nucleophile.


Which part of the reaction of potassium with water releases more energy? When potassium is reacted with water, hydrogen is given off, which is normally ignited by the heat of the reaction and combusts. How much energy is given off by the burning hydrogen compared to the potassium reaction?

$$\begin{array}{c}
\ce{2K(s) +2H2O (l)->2KOH(aq) +H2(g)}\\
\ce{2H2(g) + O2(g) -> 2H2O(g)}\\
\end{array}$$

We have the two equations above, and we want to know which releases more energy. The answer is relatively simple to find: it's the standard enthalpy of formation of the products minus the standard enthalpy of formation of the reactants. This works because enthalpy is a state function and a systems enthalpy change independent of the path taken from reactants to products. So the way we do this is by imagining that we take our products, break them into their constituent elements (which have a $\Delta H_\mathrm{f}=0$ by definition), and convert those into the products. So:

$$(-482.37\ \mathrm{kJ \ mol^{-1}}) \cdot 2\ \mathrm{\ mol}-(-285.83\ \mathrm{kJ \ mol^{-1}})\cdot 2\ \mathrm{\ mol}=\frac{-393.08\ \mathrm{kJ}}{2\ \mathrm{mol}\ \ce{K}}=-196.54\ \mathrm{kJ \ mol^{-1}}$$

$$(-241.82\ \mathrm{kJ \ mol^{-1}}) \cdot 2\ \mathrm{\ mol}=\frac{−483.64\ \mathrm{kJ}}{2\ \mathrm{mol}\ \ce{H2}}=-241.82\ \mathrm{kJ \ mol^{-1}}$$

$$\begin{array}{c}
\ce{2K(s) +2H2O (l)->2KOH(aq) +H2(g)} & \Delta H=-196.54\ \mathrm{kJ \ mol^{-1}}\\
\ce{2H2(g) + O2(g) -> 2H2O(g)} & \Delta H=-241.82\ \mathrm{kJ \ mol^{-1}}^{[1]}\\
\end{array}$$

So the combustion of hydrogen gas produces more energy per mole than the reaction of potassium with water.

$^{[1]}$ Formation Enthalpies


What chemical characteristics give types of steel their properties? I am a mechanical engineer, and just like everyone else I had classes on the crystal structure of metals, phase diagrams, and the various heat treatments.

However, even after diving back into that recently and researching online, I can't make sense of how everything fits together. I know the high level, the low level, but can't link the two together.

For example:

How can the phase diagram of steel be linked to its physical properties? Say, given a certain carbon weight fraction, is it possible to tell from the phase diagram if the steel is going to be hard, resilient, ductile? E.g. cementite is hard and brittle, austenite is ductile and soft?
How do heat treatments work? The way I see it, the phase diagram is "steady state", and heat treatments are about spending a specific amount of time in a phase region to add "a little bit of that" etc. (or quickly cooling it to stop the reactions), but I can't link it back to the phase diagram (there is a 3rd dimension, time).

Apologies if this is not a very clear question, but this is exactly what I hope to find: clarification on a quite complex topic.

(Source: Roy Beardmore, http://www.roymech.co.uk/ (defunct, via the Internet Archive)

As carbon content increases, the ability to resist a sudden impact decreases, as measured by the Charpy Impact Test.

As carbon content increases, ultimate tensile strength and Brinell Hardness increases.

As far as a phase diagram:

increasing $\ce{}$ (cementite) content makes the steel more brittle and hard.

Near 0.76% carbon, the steel will be pearlite which is ductile (good for wires).  Below 0.76% carbon, there will be increased alpha-iron (ferrite). Above 0.76% there can be Ledeburite.  Austenite only exists at high temperature.  Depending upon the cooling rate of Austenite, Bainite (slow cooling) or Martensite (rapid cooling) can form.


Which metal hydroxides are soluble? Alkali metals (Li, Na, K, Rb, Cs) and the alkaline earth metals (Ca, Sr, Ba - sometimes Mg is included also) form hydroxides that are well-known to be soluble or at least slightly soluble and create strongly basic solutions. Apparently these solutions of hydroxides are also known as alkali.

Since the more well-known d metals (Cr, Fe, Mn, Cu, Zn, Ni, Co, Ti etc.) forms no soluble hydroxide, I used to assume that apart from the 9 metals stated at the top, there is no more metals that corresponds to soluble hydroxides.

However, while searching my (non-English) version of Constants of Inorganic Substances: A Handbook by Lidin, Andreyeva and Molochko, I discovered that Europium (II) hydroxide was written as if it is soluble in a chemical equation in the Europium section. Later, I did search for information on the compound with the keywords ""Europium II hydroxide" soluble" , and I found some results:

  europium (II) sulfate is scarcely soluble in water, while europium (II) hydroxide dissolves readily and gives an alkaline reaction

(Concise Encyclopedia Chemistry, page 392)

Is there any other research on the compound Europium (II) hydroxide? Is it really soluble?

Which metal hydroxides, apart from those of those group 1 and 2 metals, are soluble? Is there any specific reason why europium (II) is such an anomaly? If there are other such cations, is there a specific reason for that particular cation to form such a soluble hydroxide?

Europium (II) is an anomaly because it is, well, "II".  Most metals in period 4 or higher, apart from the alkali and alkaline earth metals, favor higher oxidation states which lead to highly insoluble hydroxides or oxides.  Europium is different because the $+2$ ion is stabilized by having a spherically symmetric electron configuration $[\ce{Xe}]\mathrm{4f^7}$.

So we have a $+2$ ion similar in size and spherically symmetric electron configuration to the heavier alkaline earth metals, and $\ce{Eu(II)}$ compound tend to resemble their alkaline earth counterparts -- including the solubility of the hydroxide.


How many grams of sodium phosphate must be added to precipitate as much of one ion as possible? Problem text:

Solid sodium phosphate is slowly added to $\pu{200 mL}$ of a solution containing $\pu{0.002 mol}$ of aluminum chloride and $\pu{0.001 mol}$ of calcium chloride (assuming no volume change occurs). How many grams of sodium phosphate must be added to precipitate as much of one ion as possible, while the other ion just does not precipitate? What percentage of the ion that will precipitate first is still left in the solution at that time?

My approach:

Two salts will be produced as the reaction goes: $\ce{AlPO4}$ and $\ce{Ca3(PO4)2}$

$K_\mathrm{sp} (\ce{AlPO4})=9.83\times10^{-21}$ and $K_\mathrm{sp} (\ce{Ca3(PO4)2})=2.07\times 10^{-33}$

Therefore, I can conclude that $\ce{Ca3(PO4)2}$ will precipitate first.

I need to find concentration of $[\ce{PO4^3-}]$ via
$$[\ce{Al^3+}][\ce{PO4^3-}]=K_\mathrm{sp}$$
$$[\ce{Al^3+}]=\frac{\pu{0.002 mol}}{\pu{200 mL}}=0.01M$$
$$[\ce{PO4^3-}]=\frac{9.83\times 10^{-21}}{\pu{0.01M}}=9.83\times 10^{-19}$$

Then I can find $m_\ce{Na3PO4}$ via finding number of moles first and multiplying by molar mass:

$$n_\ce{Na3PO4} = [\ce{PO4^3-}]\times V = 9.83\times 10^{-19}\times \pu{200 mL} = 1.966\times 10^{-19}$$
$$m_\ce{Na3PO4} = n_\ce{Na3PO4} \times \mathrm{Molar Mass} = \pu{3.22\times 10^{-17} g}$$

At this point, my result does not make much sense to me because the mass I obtained is extremely small. Unfortunately, I do not have the correct answers for that task, therefore, I would highly appreciate any help that can be useful in spotting mistakes I made in the solution.

In an answer that is now hidden user Maurice pointed out the complexities of the chemistry. So let's assume:

All Al present in solution as $\ce{Al^{3+}}$ and any hydoxy complexes or precipitates will be ignored.
All Ca present in solution as $\ce{Ca^{2+}}$ and any hydoxy complexes will be ignored.
All $\ce{PO4^{3-}}$ added to solution stays as that ion and doesn't hydrolyze to $\ce{HPO4^{2-}}$
Two salts that will be produced are $\ce{AlPO4}$ and $\ce{Ca3(PO4)2}$
$K_\mathrm{sp} (\ce{AlPO4})=9.83\times10^{-21}$ and $K_\mathrm{sp} (\ce{Ca3(PO4)2})=2.07\times 10^{-33}$

Concentration of $[\ce{PO4^3-}]$ that will cause $\ce{Al^{3+}}$ to precipitate:
$$[\ce{Al^3+}]=\dfrac{\pu{0.002 mol}}{\pu{200 mL}}=0.01M$$
$$[\ce{PO4^3-}]_\ce{Al}=\dfrac{K_\mathrm{sp}}{\ce{[Al^{3+}]}} = \dfrac{9.83\times 10^{-21}}{\pu{0.01M}}=9.83\times 10^{-19}$$

Concentration of $[\ce{PO4^3-}]$ that will cause $\ce{Ca^{2+}}$ to precipitate:
$$[\ce{Ca^2+}]=\dfrac{\pu{0.001 mol}}{\pu{200 mL}}=0.005M$$
$$[\ce{PO4^3-}]_\ce{Ca}=\sqrt{\dfrac{K_\mathrm{sp}}{\ce{[Ca^{2+}]^3}}} = \sqrt{\dfrac{2.07\times 10^{-33}}{(\pu{0.005 M})^3}}= 1.2890\times 10^{-13}$$

$\therefore \ce{AlPO4}$ will precipitate first.

The molecular mass of $\ce{Na3PO4}$ is 163.941 g/mol.

Note here that you can't just assume a stoichiometric amount of $\ce{Na3PO4}$ to say three significant figures. The solubility of the $\ce{Ca3(PO4)2}$ is so low that the $\ce{Na3PO4}$ would have to be weighed to an absolutely unrealistic precision.

So we'll go father down fantasy lane and suppose that:

There are $0.02\bar0$ moles of  $\ce{Al^3+}$
That we are using God's balance which can weigh an infinitely precise amount of $\ce{Na3PO4}$.
That the molecular mass of $\ce{Na3PO4}$ is exactly 163.941 g/mol (this is so precise that isotopic ratios limit the value...)
$K_\mathrm{sp} (\ce{AlPO4})=9.83\bar0\times10^{-21}$ and $K_\mathrm{sp} (\ce{Ca3(PO4)2})=2.07\bar0\times 10^{-33}$
That the volume changes in the solution when adding $\ce{Na3PO4}$ or precipitating $\ce{AlPO4}$ are insignificant.

So if the $[\ce{PO4^3-}] = 1.\bar0\times 10^{-13}$ then no $\ce{Ca3(PO4)2}$ will precipitate. At that concentration the amount of $\ce{Al^3+}$ left in solution will be:

$$\ce{[Al^{3+}]}=\dfrac{K_\mathrm{sp}}{\ce{[PO4^{3-}]}} = \dfrac{9.83\times 10^{-21}}{1.\bar0 \times \pu{10^{-13} M}} =9.83\times 10^{-8}$$

So $\pu{0.001999998034000 mol}$ of $\ce{AlPO4}$ will precipitate.

For 200 ml of a $1.\bar0 \times \pu{10^{-13} M}$ solution there are $2.\bar0 \times \pu{10^{-14} moles}$

So the total moles of $\ce{Na3PO4}$ needed is $\pu{0.001999998034020  mol}$ or $\pu{0.327881677695273 g}$.

% Al left in solution $= \dfrac{0.002\bar0 - \pu{0.001999998034000 mol}}{0.002\bar0} =0.0000983\% $

Again the problem requires such a precise amount of $\ce{PO4^{3-}}$ to precipitate "all" the $\ce{Al^{3+}}$ yet leave "all" the $\ce{Ca^{2+}}$ in solution that it is totally unrealistic.


How to compute 2-electron integral for Hartree-Fock code? I'm working on writing my own code for $\ce{H2}$ in an STO-3G basis set using Hartree-Fock (HF), and I am currently stuck on how to construct the two-electron integral matrix.  I know how to evaluate the primitive two-electron integrals, but I can't figure out how to transform that into the $4\times4$ matrix of possible values for this system.

For my previous integrals, I determined the sum of the primitives by constructing a $6\times6$ matrix of the primitive integrals, and did a transformation using a $6\times2$ matrix of the coefficient values from the fit to the Slater function, for example:
\begin{align}
\mathbf{D} &=
  \begin{bmatrix}
     d_{11} & 0 \\
     d_{12} & 0 \\
     d_{13} & 0 \\
     0 & d_{21} \\
     0 & d_{22} \\
     0 & d_{23}
  \end{bmatrix}\\
\mathbf{S} &=
  \mathbf{D}' * \mathbf{S}_{prim} * \mathbf{D}
\end{align}

I tried doing a similar technique for the two-electron integrals but with a $12\times12$ matrix, but the values didn't come out right.  It seems that I need to make a $4$-dimensional array of all the primitive integrals, but I don't know how to evaluate that in order to determine the sum and hence the final
$4\times4$ matrix.

I've been using Szabo and Ostlund's Modern Quantum Chemistry book as my main reference.

You can also write this down in a similar way as for the one-electron integral. You already had:

\begin{equation}
\mathbf{S} = \mathbf{D_2}' * \mathbf{S}_{\rm prim} * \mathbf{D_2}
\end{equation}

where $\mathbf{S}$ is $(2\times2)$, $\mathbf{D_2}$ is $(6\times2)$ and $\mathbf{S}_{\rm prim}$ is $(6\times6)$.

For the 2-electron integral we now need

\begin{equation}
\mathbf{V} = \mathbf{D_4}' * \mathbf{V}_{\rm prim} * \mathbf{D_4}
\end{equation}

where $\mathbf{V}$ is $(2\times2\times2\times2)$, $\mathbf{D_4}$ is $(6\times2\times6\times2)$ and $\mathbf{V}_{\rm prim}$ is $(6\times6\times6\times6)$.

You can get $\mathbf{D_4}$ from $\mathbf{D_2}$:
\begin{equation}
\mathbf{D_4} = \mathbf{D_2} * \mathbf{D_2}
\end{equation}
where you do not sum over any index in order to get the 4-index tensor.

The equation for $\mathbf{V}$ is a bit more tricky, since you need to pay attention over which indices you sum (which is actually not specified by the notation). I add an example on how you can do this with Python/Numpy with Einstein summations.

# 1-electron integral
S = numpy.einsum('ba,bc,cd', D2, Sprim, D2)

# 2-electron integral Option 1: form (6x2x6x2) tensor
D4 = numpy.tensordot(D2, D2, 0)
V = numpy.einsum('badc,bedf,egfh->agch', D4, Vprim, D4)

# 2-electron integral Option 2: transform indices one-by-one
V = Vprim.copy()
V = numpy.einsum('ba,bcde->acde', D2, V)    # i
V = numpy.einsum('abcd,be->aecd', V, D2)    # j
V = numpy.einsum('da,bcde->bcae', D2, V)    # k
V = numpy.einsum('abcd,de->abce', V, D2)    # l

You will need to give D2, Sprim and Vprim as input.

(If anyone knows how to write this down in a formal notation, please edit.)

You were asking how to get $\mathbf{V}$ as a 2-index $(4\times4)$ matrix. For this we can simply reshape the above 4-index tensors. So $\mathbf{V}_{\rm prim}$ becomes $(36\times36)$. Now this means $\mathbf{D_4} $ needs to be $(36\times4)$, and not $(12\times12)$. To these reshaped 2-index matrices we can apply again the same transformation equation from above. Here is the code for my example:

# 2-electron integral Option 3: reshape to 2-index matrices
Vprim = Vprim.reshape((Nprim**2, Nprim**2))
D4 = numpy.swapaxes(D4, 1, 2)    # making sure to combine the correct axes
D4 = D4.reshape((Nprim**2, Ncntr**2))
V = numpy.einsum('ba,bc,cd', D4, Vprim, D4)

Now $\mathbf V$ will be:
\begin{bmatrix} (\phi_1\phi_1|\phi_1\phi_1)  & (\phi_1\phi_1|\phi_1\phi_2) &  (\phi_1\phi_1|\phi_2\phi_1) & (\phi_1\phi_1|\phi_2\phi_2)\\      (\phi_1\phi_2|\phi_1\phi_1)  & (\phi_1\phi_2|\phi_1\phi_2) &  (\phi_1\phi_2|\phi_2\phi_1) & (\phi_1\phi_2|\phi_2\phi_2) \\      (\phi_2\phi_1|\phi_1\phi_1)  & (\phi_2\phi_1|\phi_1\phi_2) &  (\phi_2\phi_1|\phi_2\phi_1) & (\phi_2\phi_1|\phi_2\phi_2) \\      (\phi_2\phi_2|\phi_1\phi_1)  & (\phi_2\phi_2|\phi_1\phi_2) &  (\phi_2\phi_2|\phi_2\phi_1) & (\phi_2\phi_2|\phi_2\phi_2)   \end{bmatrix}

And here is the structure of $\mathbf{D_4}$:
\begin{equation}
\mathbf{D_4} = \mathbf{D_2} * \mathbf{D_2} = \begin{bmatrix}
d_{11}*\mathbf{D_2} & 0*\mathbf{D_2} \\
d_{12}*\mathbf{D_2} & 0*\mathbf{D_2} \\
d_{13}*\mathbf{D_2} & 0*\mathbf{D_2} \\
0*\mathbf{D_2} & d_{21}*\mathbf{D_2} \\
0*\mathbf{D_2} & d_{22}*\mathbf{D_2} \\
0*\mathbf{D_2} & d_{23}*\mathbf{D_2} \\
\end{bmatrix}
\end{equation}


Stability of α, β, γ allotropes at different temperatures While reading about elements or compounds I often come across allotropes (or forms) being referred to as α, β, γ forms.
Examples:

Sulfur has two allotropes: α rhombic and β monoclinic. Out of these two α is considered more stable at ordinary temperatures.
$\ce{SO3}$ has three forms out of which again α is considered more stable.
α-Form of iron is considered the most stable at ordinary temperatures.

Based on this small data, it makes me think whether α forms are always in general the more stable ones? Is it true? How does this nomenclature signify stability at different temperatures?

I think most of the time, they sort according to the order which temperature the allotrope/polymorph is more stable. Let's say element Z has two allotropes A and B. A is stable below 100 °C and B is stable above 100 °C. Thus allotrope A would be denoted alpha-Z and allotrope B would be denoted beta-Z.
Gray tin (alpha-tin) and white tin (beta-tin) is an example. Gray tin is stable only below 13.2 °C and white tin is stable above this point. So that's why gray tin is called alpha-tin because it is stable at a lower temperature, and ditto with white tin. As you can see in this case the beta allotrope would be the one stable at room temperature one.
Quartz (a crystalline form of silicon dioxide) is another example. See this from Wikipedia:

Quartz exists in two forms, the normal α-quartz and the high-temperature β->quartz, both of which are chiral. The transformation from α-quartz to β-quartz >takes place abruptly at 573 °C (846 K; 1,063 °F).


Naming of ester alkoxide I'm trying to name the following compound:

I believe the carboxyl group has priority and should make up the parent name. I'm not sure how to name the alcohol derivative portion. My first thought is:

  potassium ethyl 2-oxyacetate

or

  potassium ethyl oxyacetate

I've seen that there is a compound without the alcohol derivative and ester side chain called glycolic acid, but I have seen no examples of this common name being used as part of an alkoxide salt.

What is the correct way of naming a compound like this?

As you have already mentioned, the key to the answer of this question is the seniority order for classes that are expressed by suffixes.

Esters are senior to hydroxy compounds (including alcohols). Therefore, the preferred IUPAC name for the corresponding glycolic acid ethyl ester with an intact hydroxy group is ethyl hydroxyacetate. (Note that according to the current version of Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book), the locant ‘2’ in ‘ethyl 2-hydroxyacetate’ is not cited.)

However, anions are senior to esters. Therefore, the anion formed by subtracting a $\ce{H+}$ from the hydroxy group of ethyl hydroxyacetate is named as ethanolate. The ethanolate is substituted using the usual principles of substitutive nomenclature. The prefix ‘oxo’, denoting $\ce{=O}$, is used to indicate a carbonyl group when the group cannot be cited as suffix. The group $\ce{-O-CH2-CH3}$ is indicated by the prefix ‘ethoxy’. This leads to the systematic name 2-ethoxy-2-oxoethan-1-olate for the anion and potassium 2-ethoxy-2-oxoethan-1-olate for the salt.


Are 3-center-2 electron bonds and and banana bonds same? While reading about diborane, I came to know that the two bridging hydrogen atoms in diborane form 3 center 2 electron bonds. On account of repulsion between the two hydrogen nuclei, the delocalised orbitals of bridges are drifted away from each other giving the shape of a banana. Such bonds are called banana bonds. In diborane 3 center 2 electron bonds and banana bonds are same.

What if a compound has only one bridging atom which forms 3 center 2 electron bond. As there is only one bridging atom, it would not face repulsion so it won't form a banana bond.

The phrase "banana bond" is usually reserved for situations where steric conditions prevent the orbitals of two atoms from aligning properly so that the bond appears "bent" and is weaker than an optimally aligned bond would be.  Cyclopropane is the poster child for such bonds, diborane is not because this description of banana involves only two atomic centers.

It's a different concept from three-center two-electron bonds, where typically the bonds are well aligned but the three atoms bonded together are not always collinear.  This is what diborane does.  The atoms in the diborane bonds are not collinear because we need two such bonds, not because either bond suffers from imperfect overlap.


Why is cell potential defined as E0[Reduction]-E0[Oxidation] and not the reverse? Why is the cell potential defined as:

$$ E^\circ_{cell} = E^\circ_{cathode} - E^\circ_{anode} $$

or
$$ E^\circ_{cell} = E^\circ_{reduction} - E^\circ_{oxidation} $$

as opposed to
$$ E^\circ_{cell} = E^\circ_{anode}  - E^\circ_{cathode} $$

or
$$ E^\circ_{cell} = E^\circ_{oxidation} - E^\circ_{reduction} $$

A few definitions which relate to the problem:

  Cell potential is the measure of the potential difference between two half cells in an electrochemical cell

  The potential difference is caused by the ability of electrons to flow from one half cell to the other

  The difference between the potential for the reducing agent to become oxidized and the oxidizing agent to become reduced will determine the cell potential.   (Chem.Libretexts)

But why is this the case?

Also, why does it then say here that
$ E^\circ_{cell} = E^\circ_{reduction} +  E^\circ_{oxidation} $ rather than $ E^\circ_{cell} = E^\circ_{reduction} -  E^\circ_{oxidation} $

It is universally accepted to express potentials as reduction potentials. However, one of the reactions is always going to be oxidation so you need the reverse. Normally cell potential is the SUM of reduction and oxidation potentials but the value you get for the oxidation reaction from tables is actually the reduction potential so you would have to change the sign. So you actually get:

Cell potential = (Reduction potential for the reduction at the cathode) - (Reduction potential for the oxidation at the anode)

or

Cell potential = (Reduction potential for the reduction at the cathode) + (Oxidation potential for the oxidation at the anode)

because Reduction potential + Oxidation potential = 0 when the same species are involved


Cleaning butyl rubber In almost every butyl rubber product (stopper, gloves, masks) a thin white, powdery spots appear over time. I have tried scraping it, cleaning it with detergent, etc. But it comes back a day or maybe three after cleaning it. What is causing this? Is debris sticking to the rubber? Is the butyl rubber reacting in some way?

According to this MSDS, Astlett Rubber, Inc.'s BUTYL RUBBER BK-1675N product contains multiple stabilizers and antiaglomerates, each of which is described as a white powder. I suspect it is the leaching or other means of migration of these stabilizer's to the rubber surface that causes the white spots. The following is an excerpt from the above MSDS:

  Butyl rubber contains moisture, stabilizer (Agidol-2 or Irganox-1010)
  and antiaglomerate (calcium stearate or zinc stearate). Stabilizer and
  antiaglomerate are not discharged into the atmosphere during
  processing butyl rubber into finished goods.

  Stabilizer Agidol-2 is a white crystalline combustible powder. Flash
  point in the open crucible is 172°C, ignition point is 208°C, the
  standard temperature of spontaneous ignition is 352°C. Agidol-2 is
  permitted for use as a stabilizer of rubber that is used in food
  industry, in water  pipelines, and children toys production.

  Stabilizer Irganox-1010 is white crystalline combustible powder.
  Ignition point is 288°C. Meets all requirements for use as a
  stabilizer in rubber th at is used in medicine.

  Zinc Stearate is white crystalline powder or paste. The standard
  temperature of spontaneous ignition is 900°C. Maximum permissible
  concentration in work area is 4 mg/m3.

  Calcium Stearate is a white powder. The standard temperature of
  spontaneous ignition is 560°C. Maximum permissible concentration in
  work area  is 10mg/m3.


XeF4 - why are there only 2 dihedral mirror planes? I started studying molecular symmetry and I became somehow confused about dihedral mirror planes. I found a definition, which says,  that it is a vertical mirror plane, which bisects two $C_2$ axes.

But when I look at $\ce{XeF4}$ molecule, I see, that it has 4 $C_2$ axes and 4 vertical mirror planes. All of those planes are bisecting 2 $C_2$ axes and both all of them also go through some atom.

Still, there are only 2 dihedral mirror planes $\sigma_\mathrm{d}$ and the remaining two planes are just common vertical ones $\sigma_\mathrm{v}$.

So, could you please tell me what's the real difference between vertical and dihedral mirror planes?

OK, in a link given in comments by Tyberius is said, that dihedral planes are such planes, which bisects as many bonds as possible, while "normal" vertical planes bisects as many atoms as possible. Then I don't get, why ethene (see the picture below) doesn't have one dihedral plane, as one of them clearly bisects its central bond and we can't symmetrically bisect more bonds in this molecule.

$\ce{XeF4}$

Ethene

This is a situation of coordinate systems. In the $D_\mathrm{4h}$ point group, as you've noticed, there are 4 $C_{2}$ rotation axes in the plane. The easiest way to set the coordinate system is to have one pair be coincident with the $x$ and $y$ axes and the other pair coincident with the functions $x=y$ and $x=-y$.

Herein lies the issue: which pair is which? I think most people would agree that a reasonable choice is that the primary set should be coincident with the Cartesian axes. Indeed, an example character table for this group makes this choice as well (these are the $C_{2}'$ elements). This then defines which pair of mirror planes are $\sigma_\mathrm{d}$ and which pair are $\sigma_\mathrm{v}$.

This covers $\ce{XeF4}$ only, not ethene, because ethene belongs to the $D_\mathrm{2h}$ point group.

Here's the character table for that group. There's no dihedral mirror plane because the mirror planes perpendicular to the plane of molecule are coincident with the rotation axes.


Choosing hyperconjugation when resonance doesn't give a reason for stability Which of the following structures would be least stable?

I initially approached it like this:
As both are allylic carbocations, and that both are in fact resonating structures of the same molecule, so resonance can't be used to judge them. So I went for hyperconjugation. The first molecule has four alpha hydrogens and the second has only one alpha hydrogen.
Since there is more hyperconjugation in the first, it should be more stable. But I also have an intuition that there might be something more that we should have a look upon but I'm not able to figure it out. Should we always choose hyperconjugation as our second option when we can't make judgements using resonance?

My teacher says that while comparing the stability of the species concerned, we should remember that stability provided by Resonance>hyper conjugation>inductive effect. Of course there are exceptions(like in the case of chloro benzene where according to resonance the ring should be activated but inductive effect dominates it and the ring gets deactivated) to these but I think we can follow it in majority of these cases and there is no reason to not be confident about it.So likewise in your case resonance is not enough so you look upon the next one i.e hyper conjugation