Analysis of methylphenidate (Ritalin) HCl extended release tablets For a project I have to determine the drug concentration release profile of self-made methylphenidate HCl (MHCl) extended release tablets (hydrophilic matrix). We are thus doing a dissolution test as described conform the US/Eur. Pharmacopeia. However a problem we are facing is that we are not sure how to analyse the samples that we take from the dissolution test. This is because we are not sure what detection method to use. The problem is that we cannot directly get the right concentration for our primary detection method which is UV spectrophotometry. This is because MHCl has a very low extinction coefficient and thus a relatively higher concentration is needed to get the measurable absorption range of 0.2-0.8, but since the tablet dose is only 18 mg this is difficult to accomplish. We thought of doing two things: a solid phase extraction followed by UV or picric acid ionpairing followed by colorimetry. We asked our professor if the SPE followed by UV was possible, but he said that in this case it wasnt practical cause of the such low amount of drug in the tablets. So that leaves us with the picric acid analysis method. Problem is that we havent ever performed a colorimetric analysis before, though it looks pretty similar to spectrophotometric. From what we understand we need to measure the absorbance at a specific wavelength that the picric acid-MHCl complex is absorbing at. But how can we calculate the MHCl concentration from this? How much picric acid do we need to use to know for sure that all the MHCl has formed pairs with the picric acid? It's kind of confusing still and appreciate it if anyone could point us to some literature or give us insight on how to analyse these samples. Thanks in advance.

If you are insistent on a picrate salt, I'd suggest first taking your methylphenidate and forming a known concentration of the picrate by adding a known amount of picric acid to a solution in which you know the base would be in excess. This ensures that all of the picric acids forms the picrate salt, and you would therefore know the picrate complex concentration (assuming there are no other compounds in solution that might react with the picrate). You could then measure the absorbances at a wavelength at which picric acid has only a negligible absorbance, provided one exists. Note that if no such wavelength exists, matters become rather more complex, and you will have to look for changes in absorbance on either side of an isosbestic point.

An alternative reaction to the use of picric acid is to instead use 1-fluoro-2,4-dinitrobenzene (FDNB), which reacts with primary and secondary amines at room temperature to form a highly-coloured 1-amino-2,4-dinitrobenzene (ADNB). The spectral overlap between FDNB and ADNB is typically pretty small, so using an excess of FDNB usually doesn't greatly change the absorbance at the maximal wavelength of ADNB. Again, to at least know the absorbance of the methylphenidate ADNB, you would have to first perform the reaction with an excess of methylphenidate, and assume a complete reaction to the ADNB.


Can an atom have more than 8 valence electrons? If not, why is 8 the limit? According to some chemistry textbooks, the maximum number of valence electrons for an atom is 8, but the reason for this is not explained.

So, can an atom have more than 8 valence electrons?

If this is not possible, why can't an atom have more than 8 valence electrons?

2017-10-27 Update

[NOTE: My earlier notation-focused answer, unchanged, is below this update.]

Yes. While having an octet of valence electrons creates an exceptionally deep energy minimum for most atoms, it is only a minimum, not a fundamental requirement. If there are sufficiently strong compensating energy factors, even atoms that strongly prefer octets can form stable compounds with more (or less) than the 8 valence shell electrons.

However, the same bonding mechanisms that enable formation of greater-than-8 valence shells also enable alternative structural interpretions of such shells, depending mostly on whether such bonds are interpreted as ionic or covalent. Manishearth's excellent answer explores this issue in much greater detail than I do here.

Sulfur hexafluoride, $\ce{SF6}$, provides a delightful example of this ambiguity. As I described diagrammatically in my original answer, the central sulfur atom in $\ce{SF6}$ can be interpreted as either:

(a) A sulfur atom in which all 6 of its valence electrons have been fully ionized away by six fluorine atoms, or

(b) A sulfur atom with a stable, highly symmetric 12-electron valence shell that is both created and stabilized by six octahedrally located fluorine atoms, each of which covalently shares an electron pair with the central sulfur atom.

While both of these interpretations are plausible from a purely structural perspective, the ionization interpretation has serious problems.

The first and greatest problem is that fully ionizing all 6 of sulfur's valence electrons would require energy levels that are unrealistic ("astronomical” might be a more apt word).

A second issue is that the stability and clean octahedral symmetry of $\ce{SF6}$ strongly suggests that the 12 electrons around the sulfur atom have reached a stable, well-defined energy minimum that is different from its usual octet structure.

Both points imply that the simpler and more energetically accurate interpretation of the sulfur valence shell in $\ce{SF6}$ is that it has 12 electrons in a stable, non-octet configuration.

Notice also that for sulfur this 12- electron stable energy minimum is unrelated to the larger numbers of valence-related electrons seen in transition element shells, since sulfur simply does not have enough electrons to access those more complex orbitals. The 12 electron valence shell of $\ce{SF6}$ is instead a true bending of the rules for an atom that in nearly all other circumstances prefers to have an octet of valence electrons.

That is why my overall answer to this question is simply "yes".

Question: Why are octets special?

The flip side of whether stable non-octet valence shells exist is this: Why do octet shells provide an energy minimum that is so deep and universal that the entire periodic table is structured into rows that end (except for helium) with noble gases with octet valence shells?

In a nutshell, the reason is that for any energy level  above the special case of the $n=1$ shell (helium), the "closed shell" orbital set $\{s, p_x, p_y, p_z\}$ is the only combination of orbitals whose angular momenta are (a) all mutually orthogonal, and (b) cover all such orthogonal possibilities for three-dimensional space.

It is this unique orthogonal partitioning of angular momentum options in 3D space that makes the $\{s, p_x, p_y, p_z\}$ orbital octet both especially deep and relevant even in the highest energy shells. We see the physical evidence of this in the striking stability of the noble gases.

The reason orthogonality of angular momentum states is so important at atomic scales is the Pauli exclusion principle, which requires that every electron have its own unique state. Having orthogonal angular momentum states provides a particularly clean and easy way to provide strong state separation between electron orbitals, and thus to avoid the larger energy penalties imposed by Pauli exclusion.

Pauli exclusion conversely makes incompletely orthogonal sets of orbitals substantially less attractive energetically. Because they force more orbitals to share the same spherical space as the fully orthogonal $p_x$, $p_y$, and $p_d$ orbitals of the octet, the $d$, $f$, and higher orbitals are increasingly less orthogonal, and thus subject to increasing Pauli exclusion energy penalties.

A final note

I may later add another addendum to explain angular momentum orthogonality in terms of classical, satellite-type circular orbits. If I do, I'll also add a bit of explanation as to why the $p$ orbitals have such bizarrely different dumbell shapes.

(A hint: If you have ever watched people create a two loops in a single skip rope, the equations behind such double loops have unexpected similarities to the equations behind $p$ orbitals.)

Original 2014-ish Answer (Unchanged)

This answer is intended to supplement Manishearth's earlier answer, rather than compete with it. My objective is to show how octet rules can be helpful even for molecules that contain more than the usual complement of eight electrons in their valence shell.

I call it donation notation, and it dates back to my high school days when none of the chemistry of the texts in my small-town library bothered to explain how those oxygen bonds worked in anions such as carbonate, chlorate, sulfate, nitrate, and phosphate.

The idea behind this notation is simple. You begin with the electron dot notation, then add arrows that show whether and how other atoms are "borrowing" each electron. A dot with an arrow means that that the electron "belongs" mainly to the atom at the base of the arrow, but is being used by another atom to help complete that atom's octet. A simple arrow without any dot indicates that the electron has effectively left the original atom. In that case, the electron is no longer attached to the arrow at all but is instead shown as an increase in the number of valence electrons in the atoms at the end of the arrow.

Here are examples using table salt (ionic) and oxygen (covalent):

Notice that the ionic bond of $\ce{NaCl}$ shows up simply as an arrow, indication that it has "donated" its outermost electron and fallen back to its inner octet of electrons to satisfy its own completion priorities. (Such inner octets are never shown.)

Covalent bonds happen when each atom contributes one electron to a bond. Donation notation shows both electrons, so doubly bonded oxygen winds up with four arrows between the atoms.

Donation notation is not really needed for simple covalent bonds, however. It's intended more for showing how bonding works in anions. Two closely related examples are calcium sulfate ($\ce{CaSO4}$, better known as gypsum) and calcium sulfite ($\ce{CaSO3}$, a common food preservative):

In these examples the calcium donates via a mostly ionic bond, so its contribution becomes a pair of arrows that donate two electrons to the core of the anion, completing the octet of the sulfur atom. The oxygen atoms then attach to the sulfur and "borrow" entire electrons pairs, without really contributing anything in return. This borrowing model is a major factor in why there can be more than one anion for elements such as sulfur (sulfates and sulfites) and nitrogen (nitrates and nitrites). Since the oxygen atoms are not needed for the central atom to establish a full octet, it is possible for some of the pairs in the central octet to remain unattached. This results in less oxidized anions such as sulfites and nitrites.

Finally, a more ambiguous example is sulfur hexafluoride:

The figure shows two options. Should $\ce{SF6}$ be modeled as if the sulfur is a metal that gives up all of its electrons to the hyper-aggressive fluorine atoms (option a), or as a case where the octet rule gives way to a weaker but still workable 12-electron rule (option b)? There is some controversy even today about how such cases should be handled. The donation notation shows how an octet perspective can still be applied to such cases, though it is never a good idea to rely on first-order approximation models for such extreme cases.

2014-04-04 Update

Finally, if you are tired of dots and arrows and yearn for something closer to standard valence bond notation, these two equivalences come in handy:

The upper straight-line equivalence is trivial since the resulting line is identical in appearance and meaning to the standard covalent bond of organic chemistry.

The second u-bond notation is the novel one. I invented it out of frustration in high school back in the 1970s (yes I'm that old), but never did anything with it at the time.

The main advantage of u-bond notation is that it lets you prototype and assess non-standard bonding relationships while using only standard atomic valences. As with the straight-line covalent bond, the line that forms the u-bond represents a single pair of electrons. However, in a u-bond, it is the atom at the bottom of the U that donates both electrons in the pair. That atom gets nothing out of the deal, so none of its bonding needs are changed or satisfied. This lack of bond completion is represented by the absence of any line ends on that side of the u-bond.

The beggar atom at the top of the U gets to use both of the electrons for free, which in turn means that two of its valence-bond needs are met. Notationally, this is reflected by the fact that both of the line ends of the U are next to that atom.

Taken as a whole, the atom at the bottom of a u-bond is saying "I don't like it, but if you are that desperate for a pair of electrons, and if you promise to stay very close by, I'll let you latch onto a pair of electrons from my already-completed octet."

Carbon monoxide with its baffling "why does carbon suddenly have a valence of two?" structure nicely demonstrates how u-bonds interpret such compounds in terms of more traditional bonding numbers:

Notice that two of carbon's four bonds are resolved by standard covalent bonds with oxygen, while the remaining two carbon bonds are resolved by the formation of a u-bond that lets the beggar carbon "share" one of the electron pairs from oxygen's already-full octet. Carbon ends up with four line ends, representing its four bonds, and oxygen ends up with two. Both atoms thus have their standard bonding numbers satisfied.

Another more subtle insight from this figure is that since a u-bond represents a single pair of electrons, the combination of one u-bond and two traditional covalent bonds between the carbon and oxygen atoms involves a total of six electrons, and so should have similarities to the six-electron triple bond between two nitrogen atoms. This small prediction turns out to be correct: nitrogen and carbon monoxide molecules are in fact electron configuration homologues, one of the consequences of which is that they have nearly identical physical chemistry properties.

Below are a few more examples of how u-bond notation can make anions, noble gas compounds, and odd organic compounds seem a bit less mysterious:


Why do sp2 carbons tend to prefer eclipsed conformations whereas sp3 carbons prefer staggered I am a second year undergraduate and have come across this when studying stereochemistry and conformations and have been unable to find a satisfactory answer.

I know that $\ce{sp^3}$ carbons prefer anti-periplanar and gauche conformations due to the overlap of filled orbitals with empty orbitals but cannot seem to find an answer for why $\ce{sp^2}$ carbons prefer an eclipsed conformation. I was possibly thinking about the need to avoid $\pi$ electrons however I am not sure.

I think you are asking about preferred conformations for $\ce{sp^3-sp^3}$ single bonds and $\ce{sp^3-sp^2}$ single bonds.

You are correct that the staggered (or gauche) conformation is preferred in simple alkanes.  This is generally explained by noting that

adverse steric interactions are minimized in the staggered conformation, and
stabilizing hyperconjugative interactions are maximized

(From: “Isomers of Organic Compounds”, section 24.2 from the book Principles of General Chemistry (v. 1.0).)

In the case of the $\ce{sp^3-sp^2}$ single bond as found in propene, toluene, acetaldehyde, etc., the eclipsed conformation is now the more stable.

(Michael Di Maso, Jared Shaw, the UC Davis ChemWiki by University of California)

Again both steric and electronic arguments can be advanced to explain this preference.  The steric argument says let's use the bent model to describe the double bond.

Now replace one of those ethylene hydrogens with a methyl group and note that in the eclipsed conformation (build a model if you can) none of the methyl hydrogens eclipse the hydrogen on the adjacent olefinic carbon or any of the bent bonds.  In the staggered conformation all 3 of the methyl hydrogens are eclipsing something, the 2 bent bonds and the $\ce{C-H}$ bond on the adjacent olefinic carbon.  The basis of the electronic argument can be found here.  Personally, for me, the steric argument is easier to understand.


What about Miller Indices of planes passing through origin? For Miller Indices of planes if we encounter a case where the plane passes through the origin then I have heard that the only remedy can be arrived by shifting the origin to a different point.However I do not understand whether or not I am supposed to change the index of the other planes whose origin was not passing through the previous origin.Please give me a bit of justification for the case.

Since a comment does not allow the inclusion of a picture, perhaps this way is helpful:

Have a look at the following picture, derived from an illustration about the Miller Indices on wikipedia.  It depicts a simplified situation of a unit cell which may occur if your crystal is of the cubic lattice system, i.e., the three vectors defining the unit cell are orthogonal to each other, and all are of same length. In the drawing, the author choosed $\vec{a_1}$, $\vec{a_2}$, and $\vec{a_3}$ for the unit cell's dimensions.

Not only the plane closest to the coordinate system's origin drawn is a valid Miller plane, equally so the second, farther one drawn.  For each set of Miller indices $(hkl)$, the corresponding Miller planes are parallel to each other.  The distance between these planes depends both on the indices, as well as on the symmetry of the crystal.

Since a crystal consists of many unit cells adjacent to each other, it is perfectly reasonable to draw Miller planes either in "positive direction" along vectors $\vec{a_1}$, $\vec{a_2}$, and $\vec{a_3}$ -- as shown here -- or in opposite direction along $-\vec{a_1}$, $-\vec{a_2}$, and $-\vec{a_3}$ with the same set of $(hkl)$.


How does conversion of a free radical to a carbonyl compound take place? Unsure about mechanism Came across a question that talked about fragmentation of peroxyesters, about the peroxide linkage of course.
After the fragmentation of the peroxyester was done, it gave out three products (2 being free radicals, and the other being carbon dioxide)

One of the products(i.e. free radical), went under a reaction (with itself) and gave two products, one being another free radical and the other being a carbonyl compound. I am unsure about the mechanism.

Here's what I thought:
The single bond between the C1 and C2 will cleave in a homolytic manner. The (now) unpaired electron on C1 will combine with the unpaired electron of oxygen to form a $\pi$ bond. Also, the unpaired electron will remain on C2.

Why am I not sure about my reasoning? Because why would a homolytic cleavage take place between C1 and C2 after all?

Any hints or tips would be highly appreciated.

You seem unsure about the homolysis between C1 and C2, but I speculate to consider the product's stability in this suggested reaction:

For one, two previously unpaired electrons form a $\pi$ bond in acetone.  For the other, you form a benzyl radical which may be described in more than one mesomeric form, for example for a radical bromination reaction of toluene.  Plus the neuter tropyl radical, the seven membered ring structure on the bottom right, result of a rearrangement.

Of course, the benzyl radical won't be persistent, eventually either looses a hydrogen, or combines with an other radical to form a closed shell molecule.


Determining direction of equilibrium I am having a lot of trouble with this question. We have to determine if $K$ is greater or less than 1. I at first thought the equilibrium would favor the right side because this could have to do with ligand binding where the negative charge would probably cause binding to occur tighter than the left side.
I was told by someone else that it could favor the left side however due to resonance stabilization present in the molecule. Now I am not sure how to judge if $K$ will be greater or equal to 1.

I got the answer, equilibrium favors the right side because (Cd) is a soft metal and (S) is soft too, so it binds better to (Cd) than (O), in accordance with the HSAB theory.


Correlation between functional group and viscosity of an organic compound? I'm looking to explore the correlation between the type of functional group on an organic molecule and the viscosity of the compound. I know that viscosity is affected by intermolecular forces (Ion-induced dipoles, Ion-dipole, Hydrogen bonding, van der Waals), and that the type of forces present depends on the functional group. I'm aiming to explore viscosities of compounds with the following groups:

Alkenyl
Alkynyl
Hydroxyl
Ether
Aldehyde
Carbonyl
Carboxyl
Amide
Amine
Nitrile
Phenyl

How do the (absolute) viscosities of compounds with each of these groups (assuming single-functional group compounds) differ, and why (in terms of intermolecular forces present)? Are there any databases with values for viscosities that I could use?

Thank you for the help.

There are empirical methods that can give you the viscosity from the structure. For lower temperatures, Orrick-Erbar equation, and for higher temperatures, Letsou-Stiel equation is what you are looking for. Orrick-Erbar maybe more suitable for you, because it is a group contribution method, and by looking at the values assigned to functional groups, you can develop an intuition about how each group affects viscosity. Letsou-Stiel uses acentricity factor, because at high temperatures, secondary bonds become less important.

As for database, NIST may be useful. Or if not, look at some supplier webpage (I use SigmaAldrich for this).


Comparing the experimental and calculated UV/vis spectra for ethene I am trying to compute UV/Vis absorption spectrum for Ethene using Gaussian 09.
I created the ethene molecule in GaussView 5, and cleaned it. Then I optimized the molecule and checked if I got the minimum energy conformation with the following:
# freq=noraman cphf=noread b3lyp/6-31g(d) geom=connectivity formcheck
 integral=grid=ultrafine scf=maxcycle=1000

Next I performed the energy calculation as follows:
# td b3lyp/6-31g(d) geom=connectivity

As a result I got a single peak with the maximum at 134 nm (see picture below) which seems to be way off from what I googled around on the Internet, which is like 173 nm or 180 nm.

Can anyone please help me figure out if I have done it correctly and whether I am way off the experimental data?

In short, there are two obvious problems with the setup OP uses for TD-DFT calculations:

B3LYP functional is not a good choice for TD-DFT.
6-31G(d) basis is usually too small.

At M06-2X/Def2-TZVP level I get a maximum at ~160 nm, which, taking the accuracy of the TD-DFT approximations into account, is close enough to the experimental value.


Entropy change during flow of heat If heat $Q$ flows from the surrounding nature temperature at $T_\text{surr}$ to the system at temperature $T_\text{system}$, then $$\Delta S_\text{total}=\frac{+Q}{T_\text{system}}+\frac{-Q}{T_\text{surr}}$$

This definition is from my book in which heat is exchanged between system and surrounding. This formula for me would only be valid when both the system and the surrounding is undergoing reversible processes but nothing about it is said over here. Does this mean this formula assumes both the heat gained and heat lost are rebersible processes? Isn't entropy defined for reversible processes?

Entropy is a state function, which means its value does not depend on how it gets from state A to state B, rather it only depends on the final state B itself.  In this case, we can imagine a process which takes the system from state A to state B via a reversible process, and your expression for the change in entropy would be correct:
$$\Delta S = \frac{q_\text{rev}}{T_\text{sys}} - \frac{q_\text{rev}}{T_\text{surr}}$$
where I have explicitly labelled the heats as reversible heat flows (this quantity does depend on the path taken).
Even if in reality such a reversible path would not be taken, the final change in entropy would still have the same value, so it is still valid to use that expression.


Predict the standard potential at 310 K for the cell constructed from the hydrogen electrode and metal-insoluble salt electrode I've run into trouble on this question

  Predict the standard potential at 310 K for the cell constructed from the hydrogen electrode and metal-insoluble salt electrode $\ce{Ag(s)|AgCl(s)|Cl-}$. Assume $\Delta_\mathrm rS$ is independent of $T$.

I know the reaction is
$$\ce{AgCl +e- <=> Ag(s) + Cl- (aq)}$$
$$E^\circ=0.222$$

The left cell is a hydrogen electrode making the equation
$$\ce{1/2H2 + AgCl(s) -> Ag(s) + Cl- (aq) + H+(aq)}$$
forming a nernst equation of

$$E=E^\circ-\frac{RT}{vF}\ln\left(\frac{\alpha_{\ce{Ag(s)}}\ \alpha_{\ce{Cl-}}\ \alpha{\ce{H+}}}{\alpha_{\ce{AgCl(s)}}\ \alpha_{\ce{H_2}}^{1/2}}\right)$$

which reduces to
$$E=E^\circ-\frac{RT}{vF}\ln\left(\alpha_{\ce{Cl-}} \alpha_{\ce{H^{+}}}\right)$$

I know $R$, $T$, $v$ and $F$ but I don't understand where to begin to solve for $\alpha_{\ce{Cl-}} \text{and}\ \alpha_{\ce{H}}$

How would I find the activities?

Predict the standard potential at 310 K for the cell constructed from the hydrogen electrode and metal-insoluble salt electrode $\ce{Ag(s)|AgCl(s)|Cl−}$

The reactions would be:
$$\ce{AgCl +e- ->Ag +Cl-}\tag{$E_1$ V}$$
$$\ce{2H+ +e- ->H2}\tag{0 V}$$
Adding:
$$\ce{AgCl +H_2->Ag +Cl- +2H+}\tag{$E_1 V$}$$

The standard potential would be:
$$E=E^\circ -\frac{RT}{\nu F}\log\ce{\frac{[Cl- ][H+]}{P_{H2}^2}}\quad (E^\circ =E_1,\nu=1)$$
H
Now you have $E_1$, also you must have $\ce{[Cl- ],[H+ ]}$ beforehand, there's no other way, other than considering the fact that (standard) Hydrogen electrode is maintained at $\text{1M}$ and $\text{1 atm}$, and to maintain charge $\ce{[Cl- ] =[H+ ]}$.In that case,
$$E=0.222-\frac{8.314\times 310}{96500}\ln(1)=0.222$$

Just for info (from Wikipedia):
$$\begin{array}{c|c}\text{Temperature}&\text{Potential} E^\circ\\
^\circ\text{K} &\text{V versus SHE at the same temperature}\\\hline
298 &0.22233\\
333 &0.1968\\
398 &0.1330\\
423 &0.1032\\
\end{array}$$
Also when you consider $\Delta S\ne\text{ constant}$:(A.J. Bard, R. Parson, J. Jordan, "Standard Potentials in Aqueous Solution", Marcel Dekker, Inc., 1985.)
$$E^\circ=0.23695 - 4.8564\times10^{−4}t - 3.4205\times10^{−6}t^2 - 5.869 \times 10^{−9}t^3$$


Reason behind conversion of acid to acid bromide in Hell-Volhard-Zelinsky reaction Why in Hell-Volhard-Zelinsky (HVZ) reaction the acid is first converted to acid bromide by using $\ce{PBr3}$ and then alpha halogenation is carried out?

The $\alpha$ halogenation occurs through the enol form.

The formation of the acyl halide is key because carboxylic acids do not form enols readily since carboxylic acid proton is removed before the $\alpha$ hydrogen. Acyl halides lack the carboxylic acid hydrogen.

The phosphorus halides form the acyl halides readily because phosphorus has a huge affinity for oxygen.
Sometimes, phosphorus tribromide is not enough to form the acyl halide but, since there is also bromine in the media, phosphorus pentabromide is formed which can form acyl halides even of solid acids.

The halogenation is performed by the $\ce{X2}$, either bromine or chlorine.
When $\ce{PBr3}$ and $\ce{Cl2}$ is used, the alpha-chloro acid is obtained and not the bromo acid.


Calculating pKa's using Gaussian I've been learning to use Gaussian recently and I found a question here that seemed like a good opportunity for me to practice. I know that I can obtain $K_a$ from $$\Delta G=-RT\ln(K_a)$$ using the general reaction $$\ce{HA(aq) + H_2O(aq)->A-(aq) +H_3O+(aq)}$$ and taking the sum of thermal and electronic free energy for each compound as their value of $G$ However, I'm not confident that the method I'm using in Gaussian is correct.

I performed an optimization/frequency calculations on all these molecules using the APFD functional, with basis set 6-311G+(2d,p) and SMD solvation using water.

After working out my calculation, I obtained $pK_a=26.10$. My main question comes down to: did I use a sufficient level of theory, at least approximately, to capture the effect of solvation? I've looked at the Gaussian white book for thermochemistry as well as a few journal articles (1  2) and couldn't get a clear sense of if I had the right level of theory or if my calculations were completely off-base.

Another finer point might be: are there any fairly similar experimental values to which I could compare my calculation to see if I'm in the right ballpark?

This is unfortunately not a real answer to your question as such, but I'd like to provide some pointers why this approach is a lot more complicated than it seems at first. It might very well be (currently) impossible.
That is probably also a reason why you don't find a lot of publications on the theory of it and so far no black-box method has been developed. Solvation is still one of the major problems in computational chemistry, hence the use of implicit solvents and counter-charges. Such methods naturally come with the implicit disclaimer that these are approximate corrections only.
The most elaborate method I know so far is COSMO-RS, which still has large error margins, at least that is the my current state of knowledge.

I think for calculations you would at least need to run molecular dynamics simulations with many explicit solvent molecules to describe the equilibrium. This is important, because the proton which dissociates will be caught in the bulk of water molecules and a highly complex network of hydrogen bonds. This is very well demonstrated by the Grotthuss mechanism, which shows how easily bond lengths can change. I would say it is impossible to describe this complex system with a single quantum chemical calculation.

Furthermore, DFT is notoriously bad at describing non-covalent interactions, and including dispersion corrections will only get you so far. I have not checked recently, but I believe there are quite a few publications discussing various sizes of water clusters on this level. I have neither used nor heard of AFPD before, so it would be wise to thoroughly search for literature (benchmarks) about its performance on these clusters. Suffice to say, that one particular functional will not give you a reliable result as you have nothing to compare it to. Especially in such complicated cases you have to calibrate your results.

Another problem here is that you are the describing heterolytic bond dissociation. More often than not this may lead to additional problems. I don't always have confidence in my results when I treat simple donor-acceptor complexes where there is no charge separation and all singlet species. A supermolecular approach would probably a better choice, but could heavily suffer from a BSSE. However, it might be worth considering $$\ce{H2O\bond{...}HA <=> H2OH\bond{...}A}.$$

I think the key for calculating acid dissociation constants is treating the whole system as one ensemble.

Recall the common definition of the equilibrium constant for the reaction
\begin{align}\ce{AH (aq) &<=> A- (aq) + H+ (aq)},&
K &= \frac{a(\ce{A-})\,a(\ce{H+})}{a(\ce{HA})}.\end{align}
We conveniently ignore water here, because it is abundant and we assume ideal solution. However, when your system is too small, you will get a large error for $K$. This is most likely due to your solvated proton, which will be described worst with solvation models. Explicitly you are writing
$$\ce{HA (aq) + H2O (aq) -> A− (aq) + H3O+ (aq)},$$
which is not really what you are calculating. Instead you are (more ore less) using already approximations to treat thermal and solvent corrections
$$\ce{HA (g, th, solv) + H2O (g, th, solv) -> A− (g, th, solv) + H3O+ (g, th, solv)}.$$
And this doesn't even consider that you are calculating two states in an equilibrium and you will get the equilibrium constant (using activities, not concentrations) for only these two states (which might not even be the ground sates),
$$K = \exp\left\{-\frac{\Delta G}{RT}\right\}.$$
You have also not accounted for the possible conformational space of $\ce{HA}$/$\ce{A-}$, which might become very important as soon as you have ethyl moieties, or even cyclohexyl moieties.

These are just some issues that immediately come to mind, there might very well be more underlying insufficiencies. I have looked into this a couple of years ago and came to the conclusion not to bother with it any more.

What can you do instead?

Assessing acid strengths does not need to be at the level of what experimental chemists understand. If they really need the $\ce{pKa}$ they should run the appropriate measurements and obtain reliable results instead of educated guessing. If they want you to classify whether an acid is stronger or weaker, use a rigorous approach to obtain the conformational space for all species and compare them with isodesmic equations. This will have the advantage that insufficiencies of the methodology will likely cancel each other.

Another purely theoretical approach would be calculating gas-phase proton affinities. From this you can judge if a compound behaves stronger or weaker. Keep in mind that this approach is better suited for bases or nucleophiles. If you have the resources, you can also evaluate Lewis acidities,
$$\ce{Base + BX3 <=> Base-BX3}.$$

This should give you at least a good idea how on a relative level your compounds behave.

You can also use a first-principle derived value for your proton, which will likely cause the largest error. A quick check turned up a few promising results. (For example: Chang-Guo Zhan and David A. Dixon. J. Phys. Chem. A 2001, 105 (51), 11534–11540.) You can then go ahead and calculate know compounds and try to find a correlation between $\ce{pKa}$ and $\Delta G$. This way of "calibration" might lead to some significant error compensation.

I would also recommend to look at COSMO-RS in more detail. There might be a lot of development that I have missed and they might have found a way to obtain more reliable results. As far as I remember, the foundation of this was solid and a good step into easier capturing solvent effects without the need to resort to full blown MD simulations.


What is the most polar covalent bond? Which polar covalent bond of the following:

Cl-F

S-O

P-N

C-Cl

It's trivia - I guess. And I got it wrong.

Electronegativity list was given for the following:
F, O, Cl, N, S, C, H, P

My wrong reasoning was:
why would two highly electronegative atoms bond with each other. Even they did, given their high electronegativity as both pull the electron cloud to each side (tug of war, weak dipole moments) with much strength by forming a weak bond. I refused to consider the difference between electronegativity of the species. (Even I did, S-O would be the highest).

I applied the same to S-O and P-N as well.
C-Cl was the pick and bad!

C-Cl has a acceptable dipole moment given C's less en 2.55 and Cl's high en. 3.16. The size of atoms wise, Cl-F is shorter compared to C-Cl. Which is questionable.

I checked few other related questions and answers, such as this peculiar one and this.

As I read ron's answer (the latter), the concept is not about a compound like CCl4 which negates the polarity between C-Cl bonds in the tetrahedral geometry. Thus C-Cl is a polar covalent bond. (isn't it?)

Where did I go wrong in reasoning, what did I miss?
PS: Not homework but some theory revision questions. Trying to reinstate/expand the understanding of concepts.

There are various electronegativity scales.

Electronegativity vales from Wikipedia

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\hline
\text{Element} & \text{Pauling} & \text{Allen}& \\
\hline
\text{H}  & 2.20  & 2.30 \\
\text{C}  & 2.55  & 2.544 \\
\text{N}  & 3.04  & 3.066 \\
\text{O}  & 3.44  & 3.610 \\
\text{F}  & 3.98  & 4.193 \\
\text{P}  & 2.19 & 2.253 \\
\text{S}  & 2.58 & 2.589\\
\text{Cl} & 3.16 & 2.896 \\
\hline
\end{array}
$$

The compounds are all binaries, so the polarity, $\mu$, is equal to the difference in charge, $\delta$, times the distance, $d$, between ions ($\mu = \delta \cdot d$). Only given a table of electronegativities, the first level of analysis would be to assume that all the bond lengths were equal and that that the polarity would simply be a function of the difference in electronegativity. The differences are thus:

$$\newcommand{\d}[2]{#1.&\hspace{-1em}#2}
\begin{array}{lrl}
\hline
\text{Element} & \text{Pauling} & \text{Allen}& \\
\hline
\Delta\text{Cl-F}  & 0.82  & 1.297 \\
\Delta\text{S-O}  & 0.86  & 1.021 \\
\Delta\text{P-N}  & 0.85  & 0.813 \\
\Delta\text{C-Cl}  & 0.61  & 0.352 \\
\hline
\end{array}
$$

So on the Pauling scale S-O should be the most polar, but on the Allen scale Cl-F should be the most polar.


Peltier Bismuth-Telluride toxicity From what I understand, the working agent inside a peltier cooler is Bismuth-Telluride. I've read a toxicity data-sheet regarding Bismuth-Telluride here.

If I were to use a peltier cooler inside a sealed cooling system, for example a thermoelectric cooler, will any groceries stored in it be "infused" with Bismuth-Telluride?

Yes there are little cooling boxes, for camping, for example, where one of the walls (partially) is made of a sheet of a food-safe-metal, back-to-back to one side of a Peltier element as a cooling device.  This works fine, for example if attached to the car's battery.  Less efficient (by rate of heat transfer) may be an separation of the Peltier element viz. the inner of the cooling box by glass or food-safe plastic.
Of course, a complete immersion of the Peltier device into a cooler or cooling system won't work, as necessarily one of the flat sides of the element is going to become cool, but the other going to be warm.


Should the anode potential be increased or decreased to generate current? I am taking a course on fuel cells and we were talking about the activation energy needed to start a chemical reaction.

I would think that the answer to this question is that the anode potential should be increased to generate current so less energy is needed to begin the reaction.

Is this right? Or am I way off track.

Use Nernst equation. If you increase anode potential, the cell potential becomes more negative, thus decreasing Gibb's energy. This makes the reaction less feasible.


Looking for linear non polar compound which is used heavily due to its boiling or melting temperature or its density So my son was given an exercise to find some linear non polar compound that is used heavily due to convenient boiling or melting temperatures or its density.

Having no training in chemistry I have no idea how to help him. I tried to look for a kind of search engine for chemicals just to see what kinds of compounds fit the bill, but I cannot find anything.

Any help is appreciated.

A possible answer is n-octane,

(image credit)

for what it

consists of linear organic molecules (vide infra)
has a boiling point of about $125\,\pu{^\circ{}C}$ / $257\,\pu{^\circ{}F}$
is a hydrocarbon you either find in, or generate from natural oil

A bit on the n in n-octane:  This is a shorthand for normal, because the carbon atoms are consecutively bound to each other in an open chain.  Which is not the same as if they would form branched molecules, then denoted iso-octane.  A prominent iso-octane (iso as in isomer, i.e. same number of atoms per molecule, but different spatial arrangment) is 2,2,4-trimethylpentane:

(image credit)

Blend in mixtures of the two, they serve as in octane rating of fuels.


Why does hydrogen fluoride have a boiling point so much lower than that of water? $\ce{F}$ has more unshared electron pairs and is very electronegative, so $\ce{H}$ of another $\ce{HF}$ molecule can $\ce{H}$-bond with it.

$\ce{HF}$ has normal boiling point of $\pu{19.5^oC}$ while $\ce{H2O}$, as you know, has normal boiling of $\pu{100^oC}$.

I think there are a variety of qualitative ways of looking at this:

Perhaps the most obvious is that $\ce{H2O}$ can form a greater number of hydrogen bonds due to having an equal numbers of hydrogen bond acceptors and donors. Each of the hydrogen atoms can be hydrogen bond acceptors; each of the lone pairs on the oxygen can be donors. In $\ce{HF}$ however there is only one hydrogen bond acceptor and theoretically three hydrogen bond donors. This imbalanced ratio inevitably leaves some acceptors without a donor. Imagine this as two dance parties; one party has two men and two women but the other has one man and three women. Obviously fewer bonds can be successfully formed in the latter, assuming that everything is monogamous.
As mentioned by the other poster, both O and F are very electronegative elements. F however is more EN than O, which means that F stabilizes electrons better than O. As a result, electrons localized on F are weaker hydrogen bond donors; they are already fairly well stabilized. Remember, electrons "want" to be stabilized by nuclei. If they already are fairly well stabilized then they will "feel" less of a need to be associated (and therefore stabilized) with (by) other nuclei. This suggests that the hydrogen bonding present among $\ce{HF}$ molecules might be weaker than the hydrogen bonds present among $\ce{H2O}$ molecules.

Experimental results are the gold standard, of course, as opposed to qualitative reasoning, and qualitative reasoning here can also lead us in the opposite direction; one might argue that by having three lone pairs, F has a lot on its plate, so to speak; it might be able to stabilize one lone pair of electrons very well but three lone pairs is a bigger ordeal, and perhaps big enough that the hydrogen bonds among $\ce{HF}$ molecules are stronger than those among water molecules. Again, this is all qualitative, but this is the sort of reasoning introductory chemistry teachers desire.

If we limit our thinking to just electrostatics then we might surmise that $\ce{HF}$ should have the stronger hydrogen bonds because F is more electron withdrawing and therefore hydrogen should be more positively polarized in $\ce{HF}$ as opposed to hydrogen in water. However, hydrogen bonding is more than just electrostatics. Hydrogen bonding actually has a covalent component; this, however is usually ignored by introductory treatments of chemistry. The bond angle of elements involved in a hydrogen bond is critical. The closer the elements involved in a hydrogen bond are to 180 degrees, the stronger the bond (this specific angle is the case with hydrogen bonding in water; not necessarily other molecules). If hydrogen bonding were purely electrostatic then this would not be the case; angles would not matter - only distance would.

Other issues must be explored as well, such as the number of viable hydrogen bonds and the electron-donating/releasing tendencies of the involved elements.

After scouring the web some common "explanations" that would not explain why water has a higher boiling point than $\ce{HF}$ would be:

  water can form 4 per molecule while HF can only form 2.

Incorrect because if we only look at hydrogen bonding as involving the polarity of atoms, then how can water form four hydrogen bonds per molecule? It has two positively charged hydrogens, and a negatively charged oxygen. Seems like it should only form three hydrogen bonds. One must understand that lone pairs can each be hydrogen bond donors.


Why aren't the amine groups in the nucleic acid bases protonated? Nucleobases G, C and A carry an amine group ($\ce{-NH2}$). In amino acids, the amine groups are considered protonated at physiological pH. Why isn't it the case for the amine groups of nucleobases?

I found some references discussing the protonation of nucleobases, but I am not enough expert in either DFT (http://www.ncbi.nlm.nih.gov/pubmed/19049279) or Mass Spectrometry (http://www.ncbi.nlm.nih.gov/pubmed/17920289/) to  get a rational from this. Plus, neither they nor the references they cite and that I have briefly examined actually consider protonation of the amine groups.

There are 2 answers:

As f' answered, the amino (NH2) group in a nucleobase is conjugated
with the heterocycle. Therefore, its pKa is closer to the one of
aniline. In other words, it is much less basic than an alkylamine.
Any drawing of a molecule in chemistry or biology is only a mere representation (agreed on by scientists) of the reality. For example, amino-acids are most usually drawn as H2N-R-COOH. We all know that it does not reflect reality but we all use that representation because it is much more readable than +H3N-R-COO- (which itself is also wrong), and because the real form of the amino acid will greatly depend on the experiment conditions. So you must always consider in what medium the molecule is to determine, more or less, its major form.


Wavenumber range reversed in IR spectroscopy I recently stumbled upon a result table from an IR spectroscopy, that is an image full of peaks. The goal was to decide which peak belongs to which group by looking up the intervals in a related table.

While the task itself is pretty straightforward, I can't explain to myself why the axis, as well as the intervals in the table, are "reversed" - meaning the higher values are on the left and the lower are on the right. In my opinion, this makes it much harder to quickly estimate if the given peak is in an interval or not:

e.g. for a number 2053, compare those two sets of intervals:

2100 - 2000
2070 - 2010
2050 - 2030

vs

2000 - 2100
2010 - 2070
2030 - 2050

The estimation seems to be easier with the second, at least for me.

Can someone provide a background on why the ordering is high-to-low and not low-to-high?

NOTE: I don't have any real background in the topic and so my view is rather naive.

Sorry, I can't comment yet. I thought I'd add a historical perspective, which I don't know if it is true, but it occurred to me as a good reason for why this might be the case.

UV should historically preceed IR. In UV, the energies axis has been historically reported in wavelength (nm), which is inversely proportional to energy, meaning you have high energies on the left and low energies on the right.
It might be that for consistence with UV, the wavenumber axis has been reversed so to have higher energies on the left.

This is highly speculative, though.


Theoretical principles of molecular simulations I am pharmacy student and would like to work in the field of drug design. We learn to work with the programs but thats not enough for me. I want to learn the theoretical principles of computational chemistry and develop software themselves, or at least parts of it.

However, I have a few questions:

Where do I start with the chemistry part?

solving the Schrödinger equation
Hartree–Fock method
Density functional theory

Which mathematical principles do I need?

Are there comprehensive literature for beginners and what would you recommend? A good book would certainly be the best start.

I need a good starting point, especially in terms of mathematics and hope you can help me.

Computational chemistry is a synthesis of many different subjects, each of which is very complex on its own. So, if you really want to learn the theoretical basis, you should be prepared to put in a lot of time and effort. I don't want to discourage you, I just want you to have realistic expectations.

The fundamentals of computational chemistry are physical chemistry and computer science. The fundamentals of those are math, physics, and logic. So, depending on what your background is, you might have to start with math and physics.

Math:

Calculus
Vector Calculus
Differential Equations
Statistics

Physics:

Classical Dynamics
Classical Electrodynamics

Once you have taken those courses (if you haven't already) you can start with physical chemistry. You could conceivably teach yourself those subjects, but it would be very difficult - a free online course would be a good option if you can't take a university course. I believe Khan academy covers all of these subjects.

Physical Chemistry

Statistical Mechanics
Quantum Mechanics

I always liked Atkin's physical chemistry book (although I don't know if the newer editions are still good). McQuarrie's quantum chemistry and statistical mechanics are very good as well and are classics in the field.

You asked:

  Where do I start with the chemistry part?

  solving the Schrödinger equation
  Hartree–Fock method
  Density functional theory

Quantum mechanics in physical chemistry will cover solutions to Schrodinger's equation for all of the cases where a closed-form solution exists. This will give you an understanding of the basic principles, so you should start with that. To solve "real" systems, you need more advanced approximation techniques, and Hartree-Fock and DFT will get you started. That is a good order to study them in, as well.

For computer science, a lot depends on which software you will be using or what languages you want to use. Most of the core, highly optimized algorithms are written in fortran or C. Some go as far as assembly for the most important parts. Most of the software I have seen (non-core) is written in fortran or C++. My experience is a little out of date (I last worked in the field about 7 years ago) so that may no longer be true. At any rate, you will need at least a basic programming course in the language you want to use (C would be good to start with, since you will really learn the fundamentals). I would also take some object-oriented courses, so courses involving more advanced C++ and java would be good as well.

Key concepts (beyond programming and architecture) that you will need are:

Stochastic methods (Monte Carlo)
Linear Algebra
Numerical integration

With all of the above, you will have a good set of base skills and knowledge on which you can build.

You asked:

  Are there comprehensive literature for beginners and what would you recommend? A good book would certainly be the best start.

I am only familiar with one book that might be suitable for beginners in computational chemistry, so perhaps others can give better suggestions. The one I used was Molecular Modelling by Leach. It gives a good overview of common techniques, but doesn't go too in-depth. However, it was last updated in 2001 - so it may be out of date. The state of the art moves pretty quickly in terms of detailed application, although the fundamentals tend to stay pretty consistent over time. Another classic text which covers the basics in some depth is Allen & Tildesley's Computer Simulation of Liquids.

Both of these books focus more on the modeling aspects of molecules as opposed to chemical reactions, so if you are more interested in simulating reactions as opposed to statistical mechanics or protein-ligand interactions, you would need different books.


Find the color of the solution The acid-base indicator methyl red has a $K_\mathrm{a}$ = $1 \times 10^{-5}$. Its acidic form is red while its alkaline form is yellow. If methyl red is added in a colorless solution with a $\mathrm{pH} = 7$, the color will be :
a) Pink
b) Red
c) Orange
d) Yellow

Seeing the $K_\mathrm{a}$ = $1 \times 10^{-5}$, I concluded that it is a weak acid. Thus I think that when it is added to a neutral solution ($\mathrm{pH}=7$), the resultant solution must be weakly acidic and thus its color must be red.
However, contrary to my logic, the answer given in my book is option

d) Yellow

Is there a flaw in my argument?

A pH indicator is generally a weak acid and often work in a fashion similar to buffers in that most of the time both the acid and its conjugate base exist in significant concentration. From the Ka of the indicator you should therefore be able to find the "halfway" point of the indicator or the middle of the range of pH values over which the indicator operates.

Think about how indicators work. Indicators work simply by mixing colors. If I have an indicator which is black at pH = 0.0 and white at pH = 2.0, then I know that the protonated form of the indicator must be black-colored, and that the deprotonated form of the indicator must be white colored. This should make sense. As pH increase, hydroxide ion molarity increases, and hydroxide ion deprotonates other molecules.

Thus, at pH = 1.0, we should have a 50/50 mix of black and white molecules. In other words, the indicator shows a gray color at pH = 1.0. At pH = 1.0 we also have a buffer solution - we have the acid and conjugate base - and specifically in a 1:1 ratio.

By application of the Henderson-Hasselbach equation we will realize:

$\ce{pH = pK_{a} + log\frac{base}{conjugate~acid}}$ simplifies to ...

$\ce{pH = pK_{a}}$

I'll let you take the problem from here. Don't forget to apply common-sense considerations.


Are there plastic materials that turn into ceramics? Are there other materials that like clay, are plastic (as in they can be shaped by hand) but set to form a hard ceramic, especially ones harder than the fired clay?

The set of compounds you are looking for are known as Thermosetting Polymers. A few examples of these are:

Bakelite: This is used for making switches and electrical enclosures
Melamine: This is commonly used for making food grade and microwaveable utensils and is an alternative to ceramic (at least in the food packaging industry)

You might also want to take a look at Thermoplastic Polymers, as they behave a bit like wet clay (they become soft on heating and harden on cooling and this can be done as many number of times as required)


Why is the benzyne triple bond distorted? In many places I've seen the "extra" bond in benzyne being labelled as $\ce{sp^2-sp^2}$ overlap or distorted (not parallel) $\ce{p\pi-p\pi}$ overlap. But I've failed to see why we can't have a normal, parallel $\ce{p\pi-p\pi}$ overlap.

Wikipedia says:

  In benzyne, however, the p orbitals are distorted to accommodate the triple bond within the ring system, reducing their effective overlap.

That makes no sense to me — why can't a normal, parallel $\ce{p\pi-p\pi}$ overlap be "accommodated" within the system.

So, what is the reason for the distorted $\pi$ bond in benzyne?

I'm fine with answers involving a bit of QM/MO theory.

First, Wikipedia doesn't really say what is the extent of this change. Here are superimposed a benzene and a benzyne molecule (own HF calculation, don't ask for the details):

The carbon atoms involved in the triple bond are displaced by less than $\pu{0.2 Å}$. In benzene, we have with this level of calculation:

$\ce{C–C}$ distance = $\pu{1.39 Å}$
$\ce{C–H}$ distance = $\pu{1.08 Å}$
$\ce{C–C–C}$ angle = $120^\circ$

while in benzyne, we have:

$\ce{C–C}$ triple bond = $\pu{1.22 Å}$
other $\ce{C–C}$ bonds = $\pu{1.38 Å}$ (almost all equal)

So, merely because the triple bond is stronger, the distance will be shorted and the benzyne structure is distorted.

Regarding orbitals, well, the p orbitals of the carbon atoms involved in the triple bond are “naturally” directed towards where the hydrogens would be. Slightly distortion of the system is possible, and indeed evidence above on the structure, but they will not be parallel, as that would disrupt the structure too much and diminish the $\sigma$ bond. So it's a competition between: making the new p orbitals “more parallel” increases overlap, but it weakens the $\sigma$ bonding pattern, which is much stronger. As such, while some distortion happens, you stay mostly close to the benzene-type orbitals.


What kind of math should I expect in first year chemistry? The course calendar for my university describes Chem 101 and 102 as "math intensive". However, it doesn't say what kind of math is involved.

What kind of math should I expect in first year chemistry?
Would it be helpful to have first year calculus under my belt first?

Let's answer three ways:

Specific expected math  Important mathematical skills will include algebra (including understanding variables and constants conceptually), interpreting graphs, and logorithms. There is very little chance you will need to know calculus in first year chemistry. If you have specific concerns, like those described in the comments, your advisor or department assistant should be able to directly address those.
First year chem philosophy  A good goal for first year chemistry is to understand fundamental chemistry concepts so that you can apply them in future courses and work. To do this, you need to know enough math that it doesn't prohibit you from understanding the concepts.
Continuing Education This one is just my opinion and I'm sure many will disagree. If you want to work in chemistry, take more math and physics than the department requires. It will give you a more thorough and complete understanding of chemistry. It will give you an edge in the future in solving interesting problems. This can be done in parallel however- the department won't expect that you're doing this.

Bonus: As many people before me have suggested, go to 6-8 classes on syllabus day even if you're only signed up and intending to take 4. Its always better to know your options and theres no risk to going and seeing whats available. In this case, you'll learn if you've got the background the professor expects for this course and can compare it with other options.


When converting between a hydrated electrolyte amount in mass and in milliequivalents (meq), why are the water molecules taken into account? A textbook I'm reading called "Ansel's Pharmaceutical Calculations, 13th edition" defines the milliequivalent (meq) thus (p. 187):

This unit of measure is related to the total number of ionic charges in solution,
and it takes note of the valence of the ions. In other words, it is a unit of measurement of the
amount of chemical activity of an electrolyte.

It then goes on to demonstrate how can one convert the amount of electrolyte by mass to the amount in meqs and vice versa, and it gives the following example (pasting as a picture to keep all the original formatting):

Now see the note below the exercise, highlighted in yellow. The text does not explain any further, and I'm a bit confused.
If the amount in milliequivalent represents the number of ionic charges in solution or "chemical activity", why do we need to use the atomic weight of the hydrated ionic compound? Do the water molecules contribute anything to the chemical activity of the solution? If one meq means essentially one charge, the amount of meqs obtained when using a molecular weight that includes the hydration molecules is not representative of the amount of ions in the solution (as given by molar amounts).
So what is the reason for including the weight of water molecules in the calculations?

When you want to obtain a solution of $\ce{CaCl2}$ in water, you can do it by dissolving anhydrous compound $\ce{CaCl2}$, or  the hydrated compound $\ce{CaCl2·2H2O}$. It does not matter which one to choose, provided you add enough water to compensate the difference and obtain at the end the wanted total volume.
If you want to obtain $1$ mL of a final solution that has to be $4$ meQ/ml, you must dissolve $4$ mEq of $\ce{CaCl2}$ or $4$ mEq of $\ce{CaCl2·2H2O}$ in $1$ mL . It does not matter which one you use. The only requirement is that the number of Equivalent has to be $4$ mEq. But the Equivalent weight  $\ce{CaCl2}$  is $111/2 = 55.5 $ g, and the equivalent weight of $\ce{CaCl2·6H2O}$ is $73.5$ g
To get the final solution with $\ce{CaCl2}$ you must choose to dissolve $4$ mEq $\ce{CaCl2}$, which weighs $\ce{(4·55.5}$ mg) = $222$ mg $\ce{CaCl2}$.
To get the same final solution with $\ce{CaCl2·6H2O}$ you must choose to dissolve $4$ mEq $\ce{CaCl2·6H2O}$, which weighs $\ce{(4·73.5}$ mg) = $294$ mg $\ce{CaCl2}$,
Do you understand ?


Calculate water hardness from grams of CaCO3 We have to figure out what the water hardness in mg/L or ppm is for a $\pu{20ml}$ solution of $\pu{0.400M}$ $\ce{CaCl2}$.

We learned that the formula for hardness is mg/L of calcium carbonate per liter. I started by calculating that there would be $\pu{0.801 g}$ of $\ce{CaCO3}$ precipitate if reacted with $\ce{Na2CO3}$.  I then converted $\pu{0.801 g}$ to mg and got $\pu{801 mg}$. I finally divides this by the original $\pu{20ml}$ or $\pu{0.02 L}$ but this gives $\pu{40050 mg/L}$ but the correct answer is $\pu{400 mg/L}$.

I feel like I might be making a mistake with units as my answer is 100 times greater than the correct answer but I don't see the problem.

Unless I'm missing something, I don't see the problem with your working either. Perhaps check whether you got the original concentration of $\ce{CaCl2}$ given in the problem right (the volume of the solution does not affect your final answer). The given answer may well be wrong, it's irritating but it does happen.

There is one thing I want to comment on though: never round off your answer too early. Since you obtained the intermediate answer of $801 \text{ mg}$, I am assuming you used a precise molar mass for $\ce{CaCO3}$ - I will take this to be $100.09 \text{ g mol}^{-1}$. (If you used the value $100$ you would have obtained $800 \text{ mg}$.) Your calculations should be:

Number of moles of $\ce{CaCO3}$ formed upon addition of excess $\ce{CO3^2-}$

$$\eta_{\ce{CaCO3}} = (0.400 \text{ mol dm}^{-3})(0.02 \text{ dm}^{3}) = 0.008 \text{ mol}$$

Mass of said $\ce{CaCO3}$

$$m_{\ce{CaCO3}} = (0.008 \text{ mol})(100.09 \text{ g mol}^{-1}) = 0.80072 \text{ g} = 800.72 \text{ mg}$$

Hardness of water in $\text{mg L}^{-1}$

$$[\ce{CaCO3}] = \frac{800.72 \text{ mg}}{0.02 \text{ L}} = 40036 \text{ mg L}^{-1}$$

This is your final answer and if you want to round to the appropriate number of significant figures, you should do so only at this point. Your problem is that you rounded too early (after step 2) and this means you obtained an imprecise final answer, $40050 \text{ mg L}^{-1}$.

Here, the appropriate number of sf is technically 3, since the original concentration of $\ce{CaCl2}$ is given to you in 3 sf. Since you obtained the imprecise answer $40050 \text{ mg L}^{-1}$, you would have rounded this to $40100$. But if you stuck to the precise value at step 2 and obtained the answer $40036 \text{ mg L}^{-1}$, you would round this to $40000$.


Why does higher temperature affect the rate of electrolysis? I understand that more heat energy= higher rate of electrolysis, but can someone explain using higher-level terms why this occurs and if there are any theories or rules that explain this?

Rate of electrolysis is only dependent on the current (Faraday's law) passing through the solution. Remember that in an electrochemical experiment you cannot independently vary voltage or current. One factor has to be fixed. These two modes are called constant current electrolysis or constant potential electrolysis.
Once you have fixed voltage, you can ask you yourself, what factors affect current rather than electrolysis. The key factor is solution's resistance or better conductance (reciprocal of resistance). Conductance increases with temperature.


Most reactive site for electrophilic aromatic substitution I was given a problem in which I had to identify the most reactive site for electrophilic substitution out of the positions 1,2,3,4 (shown in the figure).

My Approach:  Upon drawing the resonating structures of the molecule, negative charge appears on position 3 and 2.

In the resonating structure in which the negative charge is under 3, there is a positive charge under -NO2 group making that structure less stable.
This does not happen with the structure in which the negative charge is under 2.

So I concluded that 2 would be the most reactive site.
But the answer given is '4'.

What is wrong in my reasoning?

As you know, in deciding which position undergoes aromatic electrophilic substitution, three elements must be taken in account: steric effect, mesomeric effect and inductive effect.

Let's call the ring containing atoms 1 and 2 "A", and the one containing atoms 3 and 4 "B".

Before deciding the importance of any of those effects, some considerations about the geometry of the molecule should be taken in account. It could be guessed, by thinking about the spatial arrangement of such molecule, that the high steric effect between rings A and B prevents their rotation, thus keeping them in a constantly "skewed" position in respect to the nitrated ring.

A representation of the molecule, with the Van Der Waals radius of the elements, follows:

Steric Effects: Because of the presence of  a nearby phenyl (the nitrated one) to any of the ortho positions of rings A and B, and because of the presence of the other phenyl in the middle of the possible electrophile's path, all the ortho positions might be considered as non favourable.

Mesomeric effect: One might guess that the resonance structures of the nitrated ring could extend to ring B, while leaving ring A untouched.
Actually, such a resonance structure requires a "double bond behaviour" for the aromatic carbons between ring B and the nitrated ring. But the molecule is extremely unlikely to behave that way, since the copresence of two aromatic rings prevents their own rotation, and locks them in a skewed position. This conformation, in other words, renders the pi system "less conjugated", and the mesomeric effect much less relevant.

Inductive effect: This is the final and, in this specific molecule, most important consideration. Simply, everything boils down to the proximity of the nitro EWG: this makes ring A more electron-poor, and ring B more electron-rich.

The answer is, hence: the meta position of ring B is attacked, or the "4" position in your image.


How could household vinegar create a salt on car carpet? I am not a chemist! My chemistry classes from my molecular biology undergrad are far behind me. But I think I created a salt, and I don't know how. Here is my mystery:

I recently got some water in my car, and wanted to prevent mold growth in my carpet mats. I brought the mats inside, vacuumed off, and completely soaked them in household vinegar. Then I placed the mats near a running dehumidifier (where the warm air comes out in the back).

White crust appeared on the carpet. At first I thought my dehumidifier was spitting something out, but the white substance was not on any other nearby surface--only on the car carpet. Eventually as the further parts of the carpet dried, they also showed the white crust. It covers pretty much all of the carpet.

Is this, in fact, a salt? If so, what basic substance could have been in my car carpet mat that would cause this reaction? Thanks everyone!

I once had a humidifier and I ran it with tap water.  I soon noticed a build up of white dust on the carpet right in front of the humidifier.  I suspect your situation might be related to what caused my white dust.  Tap water contains salts (just like table salt), the harder the water the more salts it contains.  They are present in low concentration so you don't normally see them.  In my case, as more and more tap water evaporated it led to a visible build up of these salts.  In your case, as the water on the carpet dried, it left behind the same salts.  Maybe you really soaked your carpet with a lot of water, maybe your water is very hard, and finally, maybe salts were actually put into the carpet during their manufacturing process - and maybe washing them with vinegar (dilute acetic acid) solubilized the salts and brought them to the surface, making them visible upon drying.


Dissolving 304 Stainless Steel We installed 20 2" drop in anchors to anchor several signs in place near the entrance of our pool. We have to remove the signs about twice a year and after the end of the past season, I noticed there is a small amount of rust forming. I purchased the 304 because it was not going to be in direct contact with the water and because it was solid, not coated, as opposed to the 314, which was only coated. It seems as though the 304 was also just coated so it is beginning to rust. I need to provide some sort of plan as to how to remove the 304 anchors and replace with 316.

Drop in anchors are not made to come out of concrete as the driving pin expands the bottom into place. I had thought about drilling them out but I am hoping to use that as a last resort because I don't want to damage our new concrete and need to keep the holes in tact. I read that Nitric acid can dissolve stainless steel. Will this work in my situation? Or will that also damage the concrete?

You are not going to dissolve 304 and leave the concrete. Use sandpaper to remove the surface rust as needed : Or mechanically ( cut/grind) the 304 to remove it.


Need clarification on pH titrimetry . I found somewhere in this site that question The reason behind the steep rise in pH in the acid base titration curve

when I tried to follow accepted answer, at some point I got stuck.

Answer says, that steep rise in PH must happen when d/dV([H+])( V is volume of added base in this case) is maximum. I disagree. It also depends on current [H+] concentration; maybe at some point d[H+]=1, and at another d[H+]=2, but if [H+]=0.1 in first case and [H+]=3 in second, then d/dV(pH) will be maximum in first case.
He differentiates this :

and gets this :

I tried to solve, but got different results.So, if I'm correct, can you give me mathematical proof why d/dV(pH) is maximum at equivalence point ?

Since $pH = -log(a_{[H^+]})$, $[H^+]_{initial}$ is irreverent when discussing the equivalence point. You can always show a full titration curve over all pH for a given acid-base titration regardless of initial proton concentrations. Therefore, equivalence point is independent of initial concentration. That being said, the fact that the equivalence point occurs when $\frac{pH}{dV}$ is at a maximum  has to do with the shape of the pH vs Volume graph. Since titration curves look similar to $y = tan^{-1}(x)$, the maximum slope will occur at the equivalence point (that would be (0,0) for inverse tangent). This can be further confirmed by taking the second derivative and comparing where the graph crosses 0 and comparing to the maximum of $\frac{pH}{dV}$.

As for the derivative, using a derivative calculator to ensure the correct answer, I got:

$2[H^+]-(\frac{C_{A}V_{A}}{V_{A}+V_{B}})+(\frac{C_{B}V_{B}}{V_{A}+V_{B}})$ which simplifies to $2[H^+]+\frac{C_{B}V_{B}-C_{A}V_{A}}{V_{A}+V_{B}}$. A simple way to determine if these are the same is to use a CAS (computer algebra system) and divide the answer provided by one of these answers. If the result is 1, the difference is only in simplification.

I hope this answers the question adequately.


Radiationless energy transfer: how to arrive at the equation that give R0 between two chromophores? I have a homework question where I should calculate  $R_0$ of an energy transfer between two chromophores. However, I don't understand how my professor arrived at the equation that we should apply (which is important to understand for the exam).

  In a radiationless energy transfer experiment (RET) it was observed that the efficiency of transfer was $0.0185$ at $\pu{5 nm}$ distance between the two chromophores. Estimate $R_0$ for the two chromophores, assuming the energy transfer efficiency is $0.7$ at short distances.

In our lecture we are given the formula:
$$[E_\mathrm{eff}] = \frac{R_0^6}{r^6 + R_0^6}$$

I understand that $E_\mathrm{eff}$ is the efficiency of the transfer, and $r$ is the non-variable distance between the two chromophores. However, I am not able to solve for $R_0$. I tried to ask my professor but he thinks faster than I do and skips steps. So I was not able to follow completely, but he somehow mathematically solved the equation to (where $Eff_\mathrm{max}$ is maximum efficiency of energy transfer):
$$r = R_0 \sqrt[6]{\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}}}$$

When I apply his formula I "plug in" the given numbers and solve for $R_0$ as following:
\begin{align}
R_0 &= \frac{r}{\sqrt[6]{\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}}}}\\
R_0 &= \frac{\pu{5 nm}}{\sqrt[6]{\frac{0.0185}{1}}}\\
R_0 &= \pu{9.72 nm}
\end{align}

I would like to understand how he arrived at the final formula that I apply. Is it correct to solve for $R_0$ like I have shown above?

I think I found the answer to my own question through some help from some classmates. What the professor did, was to solve the equation mathematically by converting $R_0^6$ to $1$.

So first, the efficiency of the transfer $E_\mathrm{eff}$ is actually a ratio of the actual transfer efficiency divided by the maximum efficient (which we should assume is "one", unless we are told otherwise.

The next step is a bit hard to explain, as I am not that good in math, but I will try my best. What I understand is that you can "gather" the two $R_0^6$ by multiplying both the numerator and denominator by what we should think of as the number one, but we actually divide by $R_0^6$ and in this way we don’t have to include $R_0^6$ that we divide by on both sides of the equal sign - as it is technically the number one (mathematically allowed), as following:
$$\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}} = \frac{\frac{R_0^6}{R_0^6}}{\frac{R_0^6}{R_0^6}+ \frac{r^6}{R_0^6}}$$

We then get as following:
$$\frac{E_\mathrm{eff}}{E_\mathrm{eff,max}} = \frac{1}{1 + \frac{r^6}{R_0^6}}$$

So by applying some math and assumptions we can solve the equation like this, and what I have explained in my question should therefore be correct.


Reduction of alcohol to alkane I have been looking everywhere for hours now, and can't find anything on a simple lab reduction of an alcohol to an alkane that doesn't involve any "fancy" reagents such as tosylates, metal catalysts etc.
I would have expected that under acidic circumstances, the oxygen donates an electron to the proton from the acid, and a hydride from a general reducing agent such as LAH or NaBH4 then attaches itself to the carbon, pushing out a molecule of water, and alkane.
However, I have found no proof or mention of the existence of such a reaction. If this is is impossible, then why, and what would be the best general purpose way to reduce an OH to a H

May I introduce you to the Barton-McCombie reaction for the reduction of alcohols to alkanes.

image and further information here


Balancing a redox equation with only one product Is it possible to balance this equation using the half reaction method?

$$\ce{Au + Cl2 + HCl⟶HAuCl4}$$

I know that the equation balanced would be:
$$\ce{2Au + 3Cl2 + 2HCl⟶2HAuCl4}$$

However, how would the half reactions end up?

You actually have different possibilities of setting up half-reactions. One way is to note that formally, gold goes from a $\pm 0$ to a $\mathrm{+III}$ oxidation state, thus it can be written as:

$$\ce{Au -> [Au^3+] + 3 e-}\tag{Ox}$$

$$\ce{Cl2 + 2 e- -> 2 Cl-}\tag{Red}$$

Which gives you:

$$\ce{2 Au + 3 Cl2 -> 2 [Au^3+] + 6 Cl- -> 2 AuCl3}\tag{Redox}$$

And, if additional $\ce{HCl}$ is present:

$$\ce{AuCl3 + HCl -> H[AuCl4]}\tag{Aurate}$$

Since $\ce{HCl}$ does not participate in the redox part of the reaction, it does not occur until after the redox part.

Naturally, $\ce{[Au^3+]}$ is not isolateable under these conditions, hence the square brackets.


Why did the red colour in a flame test for strontium disappear and appear again on longer heating? We did a basic flame test, and I understand that different salts produce different spectra because of atoms present in the salts.

When I was doing the test, I noticed that when I dipped the nichrome wire into the solution of $\ce{SrCl2}$ and placed it on the tip of the flame, I got the expected red color. However, what would happen is I would get the red color, and then it would disappear, going back to the typical bluish flame, the nichrome would glow orange hot, and the red colored flame would return.

Essentially, the red color appears, disappears, and then reappears. I thought that it might be "wicking" the strontium chloride from another part of the dipstick, but I was careful to only submerge a small part in the solution.

What's happening here?

Good, careful observation! Here's my assumption of what happened:

The $\ce{SrCl2}$ solution on the wire boiled violently, releasing a large amount of the salt into the flame.
Some remaining anhydrous $\ce{SrCl2}$ stuck to the nichrome after all water boiled off, and melted when the wire reached ~$\pu{874^\circ C}$, coating the loop with a thin layer of the salt.
Additional heating, after melting the salt, went into raising its temperature until it approached the boiling point, ~$\pu{1250 ^\circ C}$. Nichrome can withstand ~$\pu{1400 ^\circ C}$ before melting, but you don't even need to reach the b.p of $\ce{SrCl2}$ to have a bit evaporate and color the flame. Ergo, the flame turned red again.

By the way, this implies you could melt $\ce{SrCl2}$ in a ceramic crucible, insert two nichrome wires connected to a battery, and electrolyze the molten salt to see a bit of strontium (which will catch fire as soon as it rises to the surface, with its characteristic flame color). Do this only under a teacher's supervision.


With a partition coefficient how do you decide which solute concentration is the numerator? Kpc = [X(in solvent A)]/[X(in solvent B)]

I realise that Kpc is actuallly a ratio but when you need the numerical value it has to be treated as a fraction.

When it has to be treated as a fraction how do you decide which solvent is A and which is B?

The proper way to express the partition coefficient of compound X distributed between an aqueous and organic phase, for example, is:

$$\mathrm{(K_P)_X = \frac{[X](org)}{[X](aq)}}$$

This unambiguously expresses the mathematical fraction to be used in calculating the partition constant for a solute, X between two solvents, aq and org.


Extracting silver from Tollens test When we add tollens reagent (diamminesilver(I) cation, $\ce{[Ag(NH3)2]+}$) to glucose it forms a silvery mirror on the walls of the container. Since the sides are coated with actual Silver,

is there any possible reagent/method to take the silver out from the sides of the test tube.?
will it be in its pure form or will there be organic/inorganic  impurities?

Silver's pretty soft—it's possible to just scrape it off with a metal spatula or something. Also, not all of the silver deposits on the glass. You might be able to adjust the conditions of the reaction to make more silver precipitate into the liquid where it's easier to get at. You can, of course, redissolve the silver in nitric acid or the like.
Since the byproducts of the reaction are soluble, it should be pretty pure, depending on how pure the starting materials are. If other easily reducible metal cations are present, you might get some impurities from that.

As a side-note, this isn't a particularly efficient way of getting elemental silver. The typical films produced in a test tube probably contain a couple hundred milligrams at most. Even a whole undergrad lab's worth of tubes is probably not worth recovering.


Prevent crystal growth by adding a chemical to a solution? If I have a solution of dissolved $\ce{NaCl}$ or sugar (or some other ionic compound), is there a chemical I could add to it that would prevent crystal growth if I boil it down? I'm looking for it to remain as a very fine powder or to keep the crystals as small as possible.

Spray it as a fine mist into a drying chamber - the solvent will evaporate and you are left with a powder.
No chemical involved though,  but still process chemistry.


Reducing temperature on dissolved Pb(NO3)2 I have pretty much zero experience in chemistry whatsoever but I wonder about how much crystal will I get from fully dissolved element when I reduce water temperature.

Variables

at $\mathrm{40^\circ C,\ 171}\pu{g}$ of $\ce{Pb(NO3)2}$ is fully dissolved in water
at $\mathrm{18^\circ C,\ }\ce{Pb(NO3)2}$ - How much is dissolved now? Everything that was not dissolved turned to crystals?

I know that it probably has something to do with this graph right here but I can't turn my mind to right direction. Can someone explain this problem and how can I find solution?

Assuming that the solution is saturated, meaning that if you add more salt to the solution it will not dissolve, and neglecting eventual evaporation of water, we have a mass of water of about 228g for 171g of salt (assuming solubility of 75g in 100g at 40°C).

Now, when cooling down to 18°C the solubility drops to ~55g / 100g water. This means that in our solution we still have
$$228~\text{g}~\ce{H2O} \times \frac{55~\text{g}~\ce{Pb(NO3)2}}{100~\text{g}~\ce{H2O}}=125.4~\text{g}~\ce{Pb(NO3)2}\; . $$

The remaining 45.6g of the compound will have precipitated. Whether they form nice crystals or not depends on various other things, see this.

Now, if the solution is not saturated (meaning your salt can still be fully dissolved, but there may be room for more in the solution) the calculation is impossible if you don't know the mass of the water at the beginning.


Eigenvalues of Hessian matrix when the molecule is not at its equilibrium configuration Given the potential energy surface (PES) of a system, we can calculate the Hessian matrix $H$ of each point $x$ on the PES.
$$H_{ij}(x) = \frac{\partial^2 E}{\partial R_i \partial R_j}|_{R = x}$$

When the structure of a molecule is optimized, the eigenvectors of $H$ are displacement vectors, the corresponding eigenvalues are the 'force constants' and they are all positive. On the other hand, when the structure corresponds to a transition state, some of the eigenvalues will be negative.

When a structure is neither the optimized one nor a transition state, the first derivate of the PES is not zero. However, we can still calculate the Hessian matrix for this structure. Moreover, as long as the PES has continuous second partial derivatives, the Hessian matrix is symmetric and thus diagonalizable (the proof). Can we interpret the eigenvalues here?

I am quite sure a much more thorough answer can be provided, but I will give a brief answer.

The answer is both yes and no. The answer is no in the sense that a harmonic analysis should always be performed at the true minimum on the potential energy surface (PES) because the eigenvalues have a physical interpretation as the force constant, and this must be a positive value. This is true because if one wishes to calculate a quadratic approximation to the PES, one will Taylor expand the potential about the minimum. In this case, the leading term is quadratic in the displacement and the coefficient is the second derivative with respect to the potential. This form of the potential is identical to the usual one for a harmonic oscillator, so the coefficient is a force constant and it is positive because the PES is strictly concave up at the minimum.

On the other hand, the answer is yes both approximately and exactly. First, I have seen people sometimes perform a harmonic analysis away from the minimum and simply trust the frequencies which they get out. This can be justified by noting that if the displacement is mostly along an eigenvector of the Hessian, then one can expect to basically still be at the minimum with respect to all other coordinates because these eigenvectors are mutually orthogonal. You will see this if you do a harmonic analysis slightly away from the minimum. That is, the modes whose eigenvectors do not describe a motion similar to the displacement you performed will have basically the same frequency as at the true minimum. However, if you make many displacements (i.e. randomly jiggle the atom positions) then you have no guarantees the answer you get out means anything.

On the other hand, these frequencies can be made rigorous by projecting out the eigenvectors which correspond to negative eigenvalues and re-diagonalizing the new Hessian. This guarantees all positive frequencies and can be interpreted physically as the instantaneous frequency when the system reaches a particular point on the PES. I think people have tried to use this type of procedure to determine linewidths of vibrational transitions. I am not sure how effective of a method it is. This kind of thing can at least give you some insight into how sensitive a particular frequency is to motion in other vibrational coordinates.


How can atoms be influenced by light? I was watching a SciShow video "Most dangerous Chemicals". It said that a compound called Azidozide azide explodes on being exposed to light. My question is: how does the the atom feel light. Heat can be felt (let us not go into the philosophical question why do we feel warm while standing below the sun) and can excite molecules. How can light excite molecules?

Light can be ‘felt’ because light is transmitted by photons. Depending on how you choose to observe light, it will either manifest itself as waves — for example in interference experiments — or as particles — for example when describing the photoelectric effect.

When described as a wave, the energy of a certain source of light will be determined by its wavelength by equation $(1)$ in which $h$ is the Planck constant and $c$ is the speed of light.
$$E = \frac{hc}\lambda\tag{1}$$
It may not immediately be obvious how this translates to the energy of a photon, but the key is the de Broglie wavelength as given in equation $(2)$.

$$\lambda = \frac hp\tag{2}$$

This equation connects all particle and wave properties of types of matter. (Indeed, due to the wave-particle dualism all particles can theoretically behave as waves and vice-versa; the wave nature of particles as large as entire molecules has been shown in interference experiments.)

When a photon with a certain energy travels through an atom or molecule, it may interact with the electrons present; often this interaction causes the photon to be absorbed and the electron to be excited to a higher energy level. As chemists, we describe this as an electronic transition from one orbital to another. The entire process can also be reversed, i.e. if an electron is excited to a higher energy level by any means, it can relax following certain quantum-mechanical selection rules back down to a lower energy level and emit a photon of a definite wavelength. The latter process can be observed in the flame colour given by sodium atoms: their 3s electron is excited to a 4p level thermally due to the heat of the flame. Upon relaxation, the electron emits a photon of $\pu{589nm}$ wavelength which we perceive as yellow. We could write that in an equation:
$$\ce{4p ->[][$\lambda = \pu{589nm}$] 3s}\tag{3}$$

If the target of a photon happens to be a molecule rather than an atom, we need to consider what type of orbitals we are exciting from and exciting into. Typically, the occupied orbitals will be bonding orbitals while unoccupied orbitals will be antibonding. Thus with the correct photon wavelength, an electron can be excited from a bonding into an antibonding orbital which we can, in layman’s terms, call ‘breaking a bond’. Whether or not the bond actually breaks depends on whether or not the atoms can dissociate apart.

A beautiful reaction to exemplify this is the reaction of chlorine and hydrogen in equimolar mixtures. Under ambient conditions, these two are metastable. If the activation energy is supplied, they will react to form hydrogen chloride.

$$\ce{H2 + Cl2 -> 2 HCl}\tag{4}$$

Activation energy can be supplied not only by a flame but also by light as long as its wavelength is short enough. A flash of blue light will work, a flash of red light will not. What happens? First, a photon excites an electron in a chlorine molecule. It will reach an antibonding, unoccupied orbital, the $4\sigma$ orbital. This causes the $\ce{Cl-Cl}$ bond to break liberating two chlorine radicals.

$$\ce{Cl2 + $h\nu$ -> [Cl2^*]^\ddagger -> 2 Cl^.}\tag{5}$$

These chlorine radicals are then energetic enough to react with hydrogen molecules to give hydrogen radicals and so on — a radical chain reaction.

$$\begin{align}\ce{Cl^. + H2 &-> HCl + H^.}\tag{6}\\
\ce{H^. + Cl2 &-> HCl + Cl^.}\tag{7}\end{align}$$

And this goes on until all the reactive species have been consumed and the radicals recombine. The reaction releases $\pu{92.3kJ/mol}$ and is therefore strongly exothermic, even explosive.

This mechanism of course also works for infrared light which we perceive as heat. However, since the energy of infrared light is very low, few reactions can be initiated by it.


Reduction by Activated Carbon Does reduction by carbon itself occur when filtration is performed using activated carbon? Is it part of the central mechanisms involved in activated carbon filtration?
One of the specificities of all activated carbon, per my understanding, is it's large surface area, which allows it to be treated in a way or another and maximize the surface area on which the respective reaction occurs. Now, just as all activated carbon has a high surface area (with some variations), can it be said that all activated carbon acts as a highly potent reducer on some ions?

The only case I know of where activated carbon causes reduction at room temperature would be filtration of ozone, $\ce{O3}$. Even fluorine will not oxidize plain carbon:

Carbon is stable in a fluorine atmosphere up to about $\pu{400 ^\circ C}$ .

Oh, well perhaps dioxygen difluoride, $\ce{FOOF}$, could also oxidize carbon (or could be reduced by carbon, should you prefer a passive voice). Read Derek Lowe's Things I Won’t Work With: Dioxygen Difluoride, if you'd care to test that. Please let me know so that I could arrange to  be elsewhere.


Does dissolving salt in water change the pH? I am particularly interested in the effect or lack of effect of dissolved NaCl in regards to pH in water. It would be interesting also if anyone has any insight for other salts and their effects. thanks.

When you dissolve a salt in water, you have to distinguish 4 cases:

The salt results from the reaction of a strong acid with a strong base, the pH will be 7. This is the case, for example, of $\ce{NaCl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ and $\ce{Cl^-}$ are spectator ions and don't react with water. So, the pH doesn't change.
The salt results from the reaction of a strong acid with a weak base, the pH will be less than 7. This is the case , for example, of $\ce{NH4Cl}$ that results from the reaction of $\ce{HCl}$ and $\ce{NH3}$. The ions $\ce{Cl^-}$ are spectator ions and don't react with water. While the ions $\ce{NH_4^+}$  do hydrolyze and give ions  $\ce{H^+}$ to water. So, the pH will decrease.
The salt results from the reaction of a weak acid with a strong base, the pH will be more than 7. This is the case , for example, of $\ce{CH3COONa}$ that results from the reaction of $\ce{CH3COOH}$ and $\ce{NaOH}$. The ions $\ce{Na^+}$ are spectator ions and don't react with water. While the ions $\ce{CH3COO^-}$  do hydrolyze and accept ions $\ce{H^+}$ from water. In this case, water act as an acid leaving a hydroxide ion. So, the pH will increase.
The salt results from the reaction of a weak acid with a weak base, the calculus of pH is a little bit complex. We need to take into consideration the $K_a$ and the $K_b$ of the weak acid and the weak base and whichever is the stronger acid or weak will be the dominate factor in determining whether the solution is acidic or basic. We can investigate three cases:

$K_a= K_b$, the aqueous solution is neutral.
This is the case, for example, of $\ce{NH4CH3COO}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{CH3COOH}$. The ions  $\ce{NH4^+}$ and $\ce{CH3COO^-}$ do hydrolyze. As the hydrolysis constant of ion acetate equals the hydrolysis constant of ion ammonium, the same concentarion of ion hydronium and ion hydroxide is produced, and the pH of the solution is neutral.
$K_a> K_b$
This is the case, for example, of $\ce{NH4F}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{HF}$. The ions  $\ce{NH4^+}$ and $\ce{F^-}$ do hydrolyze  and as  ammonium ions hydrolyze to a greater extent than  fluoride ions, i.e more hydronium ion is produced, and the pH of the solution is acidic.
$K_a< K_b$
This is the case, for example, of $\ce{NH4CN}$ that results from the reaction of $\ce{NH4OH}$ and $\ce{HCN}$. The ions  $\ce{NH4^+}$ and $\ce{CN^-}$ do hydrolyze  and as  cyanide ions hydrolyze to a greater extent than  fluoride ions, i.e less hydronium ions are produced, and the pH of the solution is basic.


Way to confirm dichromate ion presence where other ions are messing with the color I'm currently working on an electrolytic setting where a stainless steel (430 series, mainly made of iron and chromium, but no nickel) anode is dissolved in acidic environment (sulfuric and phosphoric acids at moderate concentrations).

My concern is the hypothetical creation of dichromate ions, since they are carcinogenic. But I can't confirm only color-wise because of the presence of other ions that mixes all the colors.

The experiment runs for long (3 hours). At first an orange color appears, which suggests dichromate ions (but could be Fe III). But after only a little while, the solution turns to a very dark and deep green, almost blue depending on the light. This suggests chromium III ions.

Several things could be happening, such as reduction of Cr VI to Cr III at the cathode, or just creation at the anode of Cr III whose color overwhelms that of Cr VI, or even Fe II or Fe III messing with the color.

Does anyone have an idea about a means to confirm or infirm the presence of dichromate ions despite not being able to judge by the color ? And if so, how could I determine the concentration of it ?

Thanks a lot in advance.

The best reagent for detecting Cr(VI) in aqueous solution is the diphenylcarbazide, which produces a red color in presence of Cr(VI). The sensibility limit is as low as 0.2 ppm Cr(VI). This method has been developed one hundred years ago. See
1) M. A. Moulin, Dosage du chrome, Bull. Soc. chip. 31, 295 (1904).
2) H. Crossley, Diphenylcarbazide. An indicator for the titration of iron with dichromate. Analyst 61, 164 (1936).

The reaction with alcohols to produce aldehyde and acids does work. But it is not very sensitive. It needs at least some millimoles of chromate to be detected. The reaction with diphenylcarbazice is about 1'000'000 times more sensitive.


Synthesis of Pseudopelletierine, Mannich reaction base catalysed The synthesis of Pseudopelletierine starts from glutardialdehyde, which reacts with methylamine and acetonedicarboxylic acid in a two-fold Mannich reaction.

This is the reaction:

I need to write the mechanism for this reaction. It needs to be under basic conditions also.

.

MY ANSWER SO FAR:

There will be a Mannich reaction followed by a second intramolecular Mannich reaction in order to form a ring by joining al three reagents together. Finally there will be a decarboxylation to remove the carboxylic acid groups.

This is the mechanism I have done:

Is this right? Any help would be appreciated.

You are very close. The only issue I see is that the Mannich reaction takes place between an enol(ate) and an imine (iminium). Your mechanism goes to the hemiaminal, and then the enolate must displace a hydroxide leaving group. Essentially that's an SN2 with a hydroxide as a leaving group, which does not occur. Instead, from the hemiaminal, the nitrogen kicks out the hydroxide to form an iminium, which is a much more reactive electrophile. The enolate attacks readily. Repeat that process, decarboxylate & tautomerize twice, and you're there.


Perovskite reaction equation (KMnF3) For the synthesis of $\ce{KMnF3}$ I said the reaction equation was

$\ce{3KF + MnCl2 -> KMnF3 + 2KCl}$

Is it this simple or are am I missing something with these particular types of reactions?

Quoting from Weller, M.; Overton, T.; Rourke, J.; Armstrong, F. Inorganic Chemistry, 6th ed:-

  $\ce{KMnF3}$ is prepared by the addition of $\ce{KF}$ to $\ce{Mn^2+}$ solution. [...] Fluorination of $\ce{MnF2}$ in presence of an alkali metal fluoride leads to the complex Mn(IV) fluorides of composition $\ce{M2MnF6 (M = K, Rb, Cs) }$.

So, the reaction is all right.


How can I find the bond angles for carbamic acid? I need to find the bond angles for carbamic acid (as a self-given practice). As I understand it, the structure looks like this:

        O
        ||
H - N - C - O - H
    |
    H

I'm trying to find the O-C-O bond angle, but I'm not sure how the - H on the end will factor in to the answer. Everything I read about it is dense and fairly hard to understand.

I've tried finding the number of electron domains around the C atom (I think it's 4), but I'm not sure where to go from there.

What is the general procedure for doing this? How would it look if I were trying to find a different angle (like H-N-H)?

A quick google of 'carbamic acid bond angles' reveals many papers in the literature measuring/computing the bond angles in this and similar molecules. See here, for example.

That said, chemists are often not concerned with the exact bond angle, instead, we often look at the structure and make predictions based upon bonding and the substituents around the atoms.

In carbamic acid, the simplest carbamate, we can consider the central carbonyl to be sp2 hybridised, giving it a planar structure with bond angles of 120. Due to conjugation with the nitrogen lone pair, the N can also be considered to be sp2 hybridised, and also have bond angles of around 120.

Although this is quite a crude analysis, it holds up quite well for sp2 centres, with very little variation (except in cases of exceptional steric bulk forcing groups to move closer/further apart, or in strained ring systems). The paper referenced confirms that in this system (carbamic acid), the deviation is as little as 4 degrees.


Are coefficients in a reaction taken as the exponents in the rate law expression? In my book it is written that if we have an reaction equation such that
$$\ce{a A (g) + b B(g) + c C(g) -> product}$$
The rate expression of the reaction will be
$$R = k[\ce{A}]^a [\ce{B}]^b [\ce{C}]^c.$$

But, when I researched on expression of rate law, I understood that $a$, $b$ and $c$ are not coefficients of the reaction. But, in most part of my book, the coefficients are taken as the powers $a$, $b$ and $c$.

Is my book wrong or is there something that I don't know about rate expression?

But when i researched on rate of expression i undertood that a, b and c are not coefficients of the reaction

You are correct. $a,b$ and $c$ summed together represent the order of the reaction. Note that this is very different from the molecularity of a reaction.

  But in most part of my book the coefficients are taken as the powers a b and c

Without knowing the reactions for which this is done, it is impossible to comment further. However, assuming that the book is correct, we can say, that in those specific cases, your book was dealing with elementary reactions.

Recall that molecularity of an elementary reaction is the same as its order. However, most natural reactions occur through complex mechanisms, and in those cases, molecularity is rarely equal to its order.

For more details on the difference between molecularity and order of a reaction, have a look at this question.