IUPAC: Would I prioritize low numbering to highest-priority group, OR try to assign lowest numbers overall? I made up a pretty silly example of a molecule here, but I hope it illustrates my point. It is a dial, but which is the scientifically-preferred order of numbering: left to right, or vice versa?
As you can see in this example, going from left to right gives the -OH group — highest priority after the aldehydes — the lowest possible number (19 versus 21, if we started from the right). However, numbering from right to left gives us the lowest numbers cross the board, but we’re not prioritizing the lowest possible number to our -OH, and instead give it to our lowly methyl group.
I have a feeling it’s right to left, because 4- and 21- is so much more reasonable than 19- and 36-. Would you mind sharing any thoughts, please?

Very loosely speaking, if your OH group is being named as a prefix, i.e. "hydroxy-something", then it's no longer "higher priority" than the methyl group, which is itself also a prefix. The "priority" only matters when you're deciding on the suffix, and the aldehydes already occupy that spot.
Yet, that rather informal and unsubstantiated claim probably won't satisfy some readers, and it shouldn't satisfy you, as I could be any useless person on the Internet feeding you false information (which is sadly common). Let's just say that if you had to speak to your teacher about it, you would probably be embarrassed to tell them that "somebody on the Internet wrote a comment that said this".
So, I will quote a few rules in the IUPAC Blue Book (Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013) which need to be looked at to assign the name. The first step is to decide on the  "principal characteristic group". The seniority order is listed in Table 4.1, P-41:

1 Radicals
2 Radical anions
[...]
15 Aldehydes and chalcogen analogues
16 Ketones [...]
17 Hydroxy compounds and chalcogen analogues (includes alcohols and phenols) [...]

The principal characteristic groups here are obviously the aldehydes. Notably, P-33.1 states that "only one [suffix] can be placed at the end of a name to indicate the principal characteristic group", so the alcohol is now relegated to being a prefix, i.e. "hydroxy-". The next step is to decide on the "parent structure":

P-44.1.1
The senior parent structure has the maximum number of substituents corresponding to the principal characteristic group (suffix) or senior parent hydride in accord with the seniority of classes (P-41) and the seniority of suffixes (P-43).

Again, it is quite clear that the parent structure here is the chain containing both aldehydes. Lastly, we come to the numbering:

P-14.4
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
[...]
(c) principal characteristic groups and free valences (suffixes);
[...]
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
[...]

(c) doesn't help us resolve the ambiguity at hand, because the aldehydes are at both ends of the principal chain whichever way round we count. So (f) is what matters, and note that it says "all considered together": it no longer distinguishes between any kind of "priority" between different prefixes. The locant set (4, 21) is lower than the locant set (19, 36), and is consequently the preferred numbering.
If you search Chem.SE for "P-14.4" there will no doubt be similar questions and answers, which may help to clarify the situation with similar examples.
Note that the "lower" locant set is obtained by point-by-point comparison i.e. comparing the first locant of both at the same time, then the second locant of both, etc. until any ties are broken (P-14.3.5). In this case, (4, 21) is lower than (19, 36) because 4 is lower than 19; there is no need to compare the second locants. A common and baseless misconception is that the smallest "sum of locants" wins. This is not true and is not written anywhere in the IUPAC Blue Book.


Reaction of ether with HCl Most of the reactions undergo ring expansion whenever favourable.
But this reaction [1, p. 514] doesn't seems to do so:

Practice Problem 11.18 Provide a mechanism for the following reaction.

Could anybody explain why?
References

Solomons, T. W. G.; Fryhle, C. B.; Snyder, S. A. Organic Chemistry, 12e ed.; John Wiley & Sons, Inc: Hoboken, NJ, 2016. ISBN 978-1-118-87576-6.

There are 2 things to consider: What would be the energy difference between before and after rearrangement from initial $2^\circ$-carbocation (I) and each of rearranged carbocations II and III:

It is obvious that the rearrangement of carbocation I to give carbocation II by ring expansion have minimal energy change since both initial and final carbocations are $2^\circ$ and ring expansion does not relieve too much energy because it is cyclopentyl to cyclohexyl. However, the rearrangement of carbocation I to give carbocation III by 1,2-hydride transfer relieves good amount of energy since initial carbocation is $2^\circ$ and final rearranged carbocation is $3^\circ$. Although the carbocation II would undergo further stabilization by 1,2-hydride transfer to relieve extra energy, I to III happens much faster than I to II transfer to give 3 as the major product.


Is graphene a polymer? A polymer is sometimes defined as a macromolecule with MW larger than 1000 g/mol. So can graphene be considered a polymer?

Wow. Good question. Graphene is not considered a polymer in the traditional sense. While it is used a doping agent for other polymers it does not have the designation of being one itself, the allotropes of carbon, at least in my experience and that of my adviser, are typically in a class of their own. Unique structures such as buckminsterfullerene and diamond differ from graphene only by the angles between the carbon atoms, 3-d shape, (and number of carbons present). All of which are definitely not considered polymers. Also remember that a macromolecule is composed of polymers.
http://www.macrogroup.org.uk/schools/polymer_chemistry.php

https://answers.yahoo.com/question/index?qid=20100906193125AAzb9T0

So the short answer would be no, graphene is not considered a polymer.


Physical Meaning of the Difference between MO and VB Wave Function of H2? What would be the physical meaning of the difference between the MO wave function and VB wave function of $\ce{H2}$?

I'll give a somewhat simplistic answer. VB assumes that all bonding is between only two atoms. MO bonding considers bonding as a molecular phenomenon.

In $\ce{H2}$ there isn't really much difference. However consider benzene, $\ce{C6}\ce{H6}$. VB theory would have three double carbon bonds and three single carbon bonds which isn't really true. So VB bonding then hybridizes over the six carbon atom ring to predict the aromatic character of the molecule. However in MO theory the aromatic character is fundamental to the model.

From wikipedia

  "In particular, valence bond theory correctly predicts the dissociation of homonuclear diatomic molecules into separate atoms, while simple molecular orbital theory predicts dissociation into a mixture of atoms and ions. For example the molecular orbital function for dihydrogen is an equal mixture of the covalent and ionic valence bond structures and so predicts incorrectly that the molecule would dissociate into an equal mixture of hydrogen atoms and hydrogen positive and negative ions."


ORCA cannot read isotope information from XYZ files -- the OpenBabel specification does not include it and so ORCA doesn't look for it in the geometry file.

You need to copy-paste the geometry into your input file:

* xyz 0 1
  ...
  H   0.55630745024474     -2.41808278363585     -1.72738492075924 M 2
*

Alternatively, if you are only interested in the frequencies, normal modes, and/or IR spectrum, you can edit the atomic masses in the .hess file after the run is completed and use the orca_vib utility program to re-run the analysis with the substituted isotopes. See pp. 201-202 of the ORCA v3.0.3 manual for more details.


What causes KNO3 to decompose into KNO2 and Oxygen in gunpowder? I've been studying fireworks and from what I understand:

$\ce{KNO3}$  decomposes to form $\ce{KNO2}$ and $\ce{O2}$ gas, and this free oxygen reacts with carbon to react with carbon and sulfur to form $\ce{CO2}$ and $\ce{SO2}$ gases in exothermic reactions.

What I don't understand is: What causes the $\ce{KNO3}$ to decompose, what is the purpose of using$\ce{KNO3}$  in gunpowder?

KNO3 (Potassium Nitrate) is a molecule that is used with pyrotechnics and is typically mixed with Carbon and Sulfur as fuels.

When Potassium Nitrate is thermally decomposed it forms Potassium Nitrite and Oxygen.
$$\ce{2KNO3 -> 2KNO2 + O2}$$

Potassium Nitrate is known as an oxidizer for this reason (it provides Oxygen to the reaction). This free Oxygen then reacts with the fuel sources in combustion to create an exothermic reaction. Carbon is typically the main fuel source and Sulfur is added to reduce the ignition temperature required to begin combustion of the Carbon-based fuel.

If you had a mixture of Potassium Nitrate and Sucrose (table sugar), you would see something like this:
$$\ce{5C12H22O11 + 48KNO3 -> 24K2CO3 + 36CO2 + 55H2O +24N2}$$

Where you can see that you have a combustion of the two reactants (after the Potassium Nitrate thermally decomposes, it provides the necessary Oxygen to begin combustion) resulting in the products of combustion, Carbon Dioxide and water, along with the byproducts of Nitrogen gas and Potassium Carbonate.


Why are the sulfide ores generally converted to an oxide and then reduced? In the elementary extraction of metals from their concentrated ores, why is the ore generally converted to an oxide, and then reduced? Why is the reduction of these metal oxides easier than the other corresponding compounds, say sulphides or carbonates.

Here are some of the elementary metals that are converted to their oxide before being reduced:

Zinc blende (sphalerite) is roasted from $\ce{ZnS}$ to $\ce{ZnO}$
Galena $\ce{(PbS)}$ is converted to $\ce{PbO}$ (Wikipedia)

Some other ores originally in their oxides remain as they were, like bauxite $\ce{(Al2O3.2H2O)}$ and hematite $\ce{(Fe2O3)}$.

An exception happens to be copper pyrite which is roasted to form copper sulphide instead of copper oxide (Wikipedia).

So, for the metals $\ce{Al, Zn, Fe}$ and $\ce{Pb}$, why is extracting the metal from the oxide favorable? And why is copper an exception?

In my book, NCERT (12) India, it is given that:

  The concentrated ore must be converted into a form which is suitable
  for reduction. Usually the sulphide ore is converted to oxide before
  reduction. Oxides are easier to reduce. Thus
  isolation of metals from concentrated ore involves two major steps viz.,
  (a) conversion to oxide, and (b) reduction of the oxide to metal.

Why, it's simple: you heat a metal oxide with carbon, and you get the metal you were after, plus a byproduct of $\ce{CO}$ or $\ce{CO2}$ which easily flies away. Coal is still abundant on our planet.

Try to pull that with a sulfide! I guess the reaction will not go in the first place, since $\ce{CS2}$ is a great deal less exothermic(*) than $\ce{CO2}$. In the unlikely case if it will, you'll have to deal with the byproduct. $\ce{CO2}$ can be simply dumped in the air (though this is no longer free). Dump $\ce{CS2}$, and all your neighbors will pay you an extremely unfriendly visit, possibly armed with pitchforks.

(*) So much so as to be endothermic, in fact.

Copper is less active, so instead of going via oxide, we have a chance to go straight to the metal in one step. Would be stupid not to use it.

Then there is aluminum which is not reduced like that, but then again, we don't convert it from sulfide to oxide either, primarily because we don't have much sulfide around. Should it be otherwise, it isn't immediately clear would we choose to convert it or not. Arguably, $\ce{Al}$ has no form which is suitable for reduction.

So it goes.


Why is introductory chemistry so non-rigorous? In introductory chemistry books, courses, ...

Many "definitions" contain words like: typically, most, often, likely, generally.

I read since questions on this site mentioning different definitions of same concept in different books.

I just read about a concept where the explanation contained smth like "some other chemists believe it should be the other way".

Why are there many non-rigorous definitions in introductory chemistry?

I like this merriam-webster definition of rigorous:

  done carefully and with a lot of attention to detail

Chemistry is an experimental science there are a lot of patterns that work fine in most of the cases but not in all (e.g. the Octet rule) these are called rules of thumb. Chemistry is however a rigorous science, simply there are exceptions that don't follow clear patterns or easy simplifications so they are harder to predict or to comprehend for a student in an introductory text.

As in all the science (physics, mathematics etc. etc.) there are some topics that are still debated, this don't mean that introductory chemistry is non-rigorous.
I suggest you to read some books about history of chemistry (e.g. Asmiov, A Short History of Chemistry), you will find that chemistry has been for most of his history (and it is in part now days) a trial and error research, followed by the attempt to find a theory, a pattern or a sense to the observations done, this doesn't mean that there isn't a rigorous approach but simply that reality don't fit the scheme that we propose so easily.

I think maybe most of the text books don't emphasize in this aspect of chemistry, but most of the reasoning is a posteriori.


Rate and mechanism of the reaction of Mg and HCl In the reaction between hydrochloric acid and Magnesium the overall reaction is:

$$\ce{Mg + 2HCl -> MgCl2 + H2}$$

My proposed mechanism of reaction is that $\ce{HCl}$ dissociates in $\ce{H2O}$ like so:
$$\ce{H2O + HCl <=> H3O+ + Cl- }$$

And then the $\ce{H3O+}$ oxidises the magnesium like so:

$$\ce{2H3O+ + Mg -> Mg^{2+} + H2O + H2}$$

This, in an experiment to determine the rate of the reaction, measuring the $\ce{H2}$ produced would measure this mechanism, because it is the step in which hydrogen gas is evolved. Presumably these Magnesium ions are then in solution. I would therefore deduce that the step:
$$\ce{2H3O+ + Mg -> Mg^{2+} + H2O + H2}$$
Is the rate-determining step, because it is the step in which magnesium ions are formed and go into solution.
My rate law, deduced from these mechanisms, would look like this:

$$\ce{rate  = k * [H3O+] }$$
Because it is the concentration of the $\ce{H3O+}$ which determines the formation of the Mg ions. From there, I would assume, the $\ce{Mg^{2+}}$ and the $\ce{Cl-}$ would remain in solution, as the reaction is carried out in the acid, obviously consisting of water.

From this, can I say that the rate law for the reaction

$$\ce{Mg + 2HCl -> MgCl2 + H2}$$

is merely:
$$\ce{rate  = k * [H3O+] }$$
But $\ce{[H3O+]}$ is:
$$\ce{[H3O+] = (K*[H2O][HCl])/[Cl^{-}] }$$
Where K is the equilibrium constant of the acid equilibrium.

So can we write our rate law like:

$$\ce{rate = k*(K[H2O][HCl])/[Cl^{-}] }$$
Where K is the equilibrium constant and k is the rate constant and might look like this (following the Arrhenius equation):
$$\ce{k = A e^{-E_a/(R T)}}$$

To make our rate law like this:
$$\ce{ rate = A e^{-E_a/(R T)} *((K*[H2O][HCl])/[Cl^{-}]) }$$

My question is: Is my rate law correct for this reaction? If not, where have I gone wrong?

Additionally: This was one of our practicals. We ended up concluding that rate goes up quadratically or exponentially with concentration. This rate law would seem to indicate a linear increase, not exponential. I assume in my rate law there is no contribution from the solid Mg. Obviously in a powder, more surface area is exposed, and a high rate is experienced. Is there a way to factor this into my rate law?
Thanks for the help! :)

Your rate law should be
$$ r = k_\text{obs} \times [\ce{H3O+}]^2 = A_\text{obs} \exp\left(\frac{E_\text{a, obs}}{RT}\right) \times [\ce{H3O+}]^2\; ,$$
which nicely fits your experimental data of a quadratic dependence on the concentration for the rate.

The exponent is necessary because the reaction is seemingly second-order in the concentration of $\ce{H+}$.


The difference between one pot synthetic method and one step reaction I often read in the papers of organic synthesis that the reactions can be carried out in one-pot method, and the others can be realized in one-step. I want to know the difference between one pot and one step in organic synthesis.

"One step" refers to a process that produces the desired product by running a single reaction--especially in a situation where there are alternative pathways to the same product involving two or more reactions in a sequence.

"One pot" refers to a process involving a series of distinct reactions carried out sequentially and in a single reaction vessel to produce the desired product without isolating and purifying the intermediates after each step.


How do I find out what benzene containing compound this is with IR and NMR? I have the following IR and NMR of an unknown substance. Can someone tell me what is a possible structure of the compound or point me in the correct direction. So far I am thinking of a benzene ring based compound with an OH group. Something like 2-phenylethanol but not sure.

Some properties:

melting point: $\pu{116 ^\circ C}$.
Composition: white powdery solid.
Soluble in $\ce{NaOH}$ and $\ce{NaHCO3}$.

Based on the IR, the broad peak at $\pu{3424 cm-1}$ represents an alcohol while the 4 humps in the region $\pu{2100-2600 cm-1}$ is indicative of an aromatic ring (most likely a benzene ring). The medium peak around $\pu{1660 cm-1}$ is most likely a carbon-carbon double bond. NMR is where I get stuck; I know there should be 6 distinct types of protons and since most of the peaks are around $\pu{7-8 ppm}$, it is most likely the benzene ring protons, but I am not sure how to calculate J values and splitting as it was done in

This is another example of how important it is to work with good quality data, which can only come from preparing good quality samples. Sample preparation comes from good technique and lots of practice. Without it, you are in a very difficult starting position. Enough of my soapbox rant.

Let's start with IR (not my strong point)

The first and second IR spectra are so different that it is hard to be confident about which one is correct. Are they collected in different media (one a solution and one a mull perhaps)? The first looks very low concentration

~3420 is clearly an -OH stretch. Probably too high to consider a N-H group of any sort. So, it could be an alcohol or an acid.
1682 a C=O group. Probably in the acid region, likely aromatic acid, conjugated or intramolecular H-bonded. Possibly amide, but not convincing based on other peaks.
There are lots of other peaks, most of which could mean many things, but I do see 749 and 708 which support the NMR notion of a monosubstituted benzene. I usually think that IR is only useful to corroborate peaks of a known structure, but then I'm an NMR advocate.

NMR

Again, the first and second spectra are very different; the obvious concern is that they are of different batches of sample - were the first and second analyses taken from the same sample, or from a different batch. In any case, let us look at the second, which is only a partial spectrum, and omits some possibly very significant information, especially in the region above 10ppm, where I would like to see a broad acid peak.

The aromatic protons shown exhibit a classic mono-substituted benzene ring pattern, labelled above. There is a strong downfield shift of the ortho (H2) protons, consistent with an electron withdrawing group substitution (such as -(C=O)R). Ortho coupling for the H2 proton is difficult to determine accurately as there are 3 peaks picked. This may be a digitizer point average, but considering the average of each peak, and the coupling from the first spectrum, we get a ortho coupling of 7.2-7.9. Not overly useful, but less than 8, so we can exclude ethers and halides. Couplings are calculated by:

(8.1651-8.1507) x 500 = 7.2

Using additive analysis for monosubstituted benzene rings, we can determine the following shifts from benzene:

H2 (8.162) 7.34 + 0.82
H3 (7.514) 7.34 + 0.17
H4 (7.651) 7.34 + 0.31

Looking at the published tabulated data for substituent effects on monosubstituted benzene rings (I'm using Pretsch, Buhlmann and Badertscher, but lots of other sources are available), we find one possible candidate that matches closely:

Benzoic Acid H2(+0.79), H3(+0.14), H4(0.28)

The caveat that I would add to this is that the data presented is not the best quality, and the solution presented is consistent with the data only (not definitive). The key concern is the missing -OH peak in the NMR spectrum. Unless this sample was run in D2O, or washed with D2O, I would expect to see an acid peak around about 12 or so. Depending on the amount of water in the sample, this may be shifted somewhat and broadened, but should still be visible, even if a broadened hump in the baseline. Additional NMR experiments could be useful to confirm the structure.


What structural property creates stickiness? Recently, me and my mom parked her car in a spot reserved for customers of a super expensive shoe store while going to get some also fairly expensive drinks at Starbucks. When we came back, an employee at the shoe store put a sticker on my mom's window warning us he could have us towed and ticketed. Clearly they're bitter about having no customers. This sticker left a whole bunch of sticky residue on the window.

I knew from experience that water wouldn't get the job done, so I put some canola oil on a paper towel and sure enough the sticky residue dissolved into the oil and I could see it dissolved in the liquid on the paper towel.

This got me thinking. Is it generally true that a sticky substance will be nonpolar? Furthermore, is there a generalization which can be made about structures which are sticky on a macroscopic level?

My only guess is that sticky substances are usually viscous too which leads me to believe that there must be a lot of VDW/dispersion interactions involved in those materials because I understand that high viscosity is related to strong intermolecular attractions. Although, the viscosity argument doesn't work for sticky solids (unless you say that a sticky solid is actually just an extremely extremely viscous liquid).

Long story short, no, sticky substances don't have to be nonpolar at all. Pour a little wine or beer on a table, and let it dry almost completely: it will get sticky, but still be water-soluble and hence polar. Also, there are certain water-based glues, which are surely sticky and polar. On top of all that, go read about the hagfish, only make sure you're not eating at the moment, for fear of adverse reaction.

As for the structural properties that make a substance sticky (or viscous, which is basically the same thing), these are not so easy to point out. Many sticky things are polymers, i.e., very long molecules with lots of van der Waals interactions, just like you said, and probably lots of interactions of other sorts, too. But this is hardly a rule, as exceptions both ways are plenty. Polyethylene is a polymer, but is not sticky at all. On the other hand, sugar syrup can be pretty sticky, despite containing no polymers.

As for the sticky solids, I would claim that there are none. Of the possible examples, some are actually highly viscous liquids, some are sticky because they are covered with such liquids, and some would stick to things because of electrostatic effects.


What role does H2O2 have in copper (II) acetate formation? Mixing solid copper, 5% vinegar, and hydrogen peroxide ($\ce{H2O2}$) causes copper acetate to form. The process will occur very slowly without hydrogen peroxide. Adding $\ce{H2O2}$ speeds the formation of copper acetate.

How does Hydrogen Peroxide promote the formation of Copper (II) Acetate?

Presumably hydrogen peroxide causes copper metal to donate 2 electrons to form $\ce{Cu(II)}$ -- which is soluble in water -- and we'd say the hydrogen peroxide is acting as an oxidizing agent.


Calculating vinegar strength The pKa of vinegar is 4.76. If I know the  pH of the vinegar, what would be the formula for determining the percentage acid base on the relationship between pH, pKa and concentration? Can you show the formula so that I only need to plug in the ph?

I am not a chemistry student and I have not done these kinds of calculations. I am searching for a convenient way for farmers in developing countries to quickly approximate the strength of vinegar. The final test would be done in a lab but I need a way for them to test vinegar other than titration because titration can seem more like witchcraft than science to the uninitiated.

If $pK_a = 4.76$, then $K_a = 1.74 \times 10^{-5}$

Then approximating that $K_a$ can be written in terms of concentration rather than the true definition which is in terms of activities:

$K_a = 1.74 \times 10^{-5}M = \frac{[A-][H+]}{[HA]} = \frac{[H+]^2}{c-[H+]} $

$1.74 \times 10^{-5}M = \frac{[H+]^2}{c-[H+]} $

where $c$ is the molar concentration of the vinegar

If you want pH explicitly in the equation, then approximating pH = -log[H+], then:

$1.74 \times 10^{-5}M = \frac{(10^{-pH}M)^2}{c-(10^{-pH}M)} $

So, for example if pH is 3, $10^{-3}$ is 0.001, so

$1.74 \times 10^{-5} = \frac{10^{-6}M}{c-(10^{-3}M)}$

$1.74 \times 10^{-5}c -1.74 \times 10^{-8}M = 10^{-6}M$

$1.74 \times 10^{-5}c = 1.02 \times 10^{-6}M$

$c = 0.059M$


Resonance stability: ester vs. thioester My problem set solution cites:

  "Although sulfur is less electronegative than oxygen, overlap is a
  more important consideration.  Sulfur is a member of the third period.
  As a consequence, a sulfur atom is larger than an oxygen  atom and the
  C–S bond is longer than the C–O bond, contributing to poorer overlap."

However, I don't quite understand why a longer bond length would contribute to poor orbital overlap, and how greater overlap contributes to more stability.

Try to understand it in a layman's way, greater the extent of overlap between the atoms, the closer their nucleus will be. The stability of the molecule will thus increase as stability of a molecule is inversely proportional to the bond length of the atoms in the molecule.

However this is applicable only upto some extent like your situation but if the nucleus of two atoms get too close a stronger force which is nuclear force starts acting on them which repels them, it is a very short length force.


From the Gaussian[1] webpage:

  6-311G: Specifies the 6-311G basis for first-row atoms and the McLean-Chandler (12s,9p) → (621111,52111) basis sets for second-row atoms [McLean80, Raghavachari80b] (note that the basis sets for P, S, and Cl are those called negative ion basis sets by McLean and Chandler; these were deemed to give better results for neutral molecules as well).

I have no idea exactly why they still call it 6-311G: I suspect it is because this configuration provides a more accurate guess for a triple-zeta Pople basis set, and to create an entirely new basis set would be impractical.

What I do know is that when you use the '6-311G' basis set, you're not necessarily getting this configuration. What you're getting is a basis set which is roughly of this quality (computational effort, accuracy).

Whilst this explains the abnormality, it still doesn't explain the exact configuration: the original paper[2] suggests a (621111,52111) set whilst the set used is (621111,42111). I hope somebody can contribute to this and complete my answer as, whilst I recognise the abnormality, further investigation has led me to this confusion.

[1] http://gaussian.com,

[2] J. Chem. Phys. 72, 5639 (1980).


Manganese(VII) oxidation state A transition element is one which forms one or more stable ions which have partially filled d-orbitals. Manganese (Mn) could have an oxidation state of 7+, which means that it will lose all of the $\ce{3d^5}$ and $\ce{4s^2}$ electrons. So how could it make a 7+ ion in that case (which means that it will lose all of the $\ce{4s^2}$ and $\ce{3d^5}$ electrons) and still be defined as a transition element?

The fact that manganese has a stable oxidation state in which it has (formally) lost all its $d$ electrons does not mean that it cannot be considered a transition metal, because it still has a lot of other oxidation states (like +II in $\ce{Mn^2+}$ or +VI in $\ce{MnO4^2-}$) with partially filled $d$ orbitals. Moreover, a naked $\ce{Mn^7+}$ ion does not exist in solution, it is only stable in the form of the oxocomplex $\ce{MnO4-}$.


What is the correct approach to obtain PES using CASSCF? I am new to CASSCF. I want to draw a PES for the ground state and the first excited sate of methyl amine for learning purpose. I want to verify if I am correct using the correct approach.

Do a RHF/STO-3G opt pop=full ... orbital calculation.
Use the checkpoint file obtained in step 1 and use CASSCF (provided I know the active space) to optimize the structure again and then use the scan keyword to map the PES by increasing the N-H bond distance in a different Gaussian job.
Take the optimized geometry of ground state obtained using CASCF and then use NROOT=2 and opt to get the the optimized geometry of 1st excited state and then scan to get the PES, similar to what was done in step 3.

Any correction would be greatly appreciated.

Yes in principle that is a correct approach, but you should be aware of a few pitfalls here and there.

The basis set you chose for your optimisation might not be large enough. I assume you are not going to use STO-3G for your CASSCF calculation. It is better if you run your initial optimisation with the same basis set that you are using for the CASSCF calculation. You will most likely be closer to the minimum and will safe a few cycles when optimising the molecular structure on the larger level of theory.
Picking the right active space might not be trivial. You may have to run a couple of single points to find the right balance of what you must include and what you may neglect.
The SCAN produces a rigid scan of the molecule, i.e. the backbone of the structure does not react with the induced change. That might not lead to the correct behaviour. More sophisticated (and more costly) is a relaxed scan via the ModRedundant keyword.

Be aware, that CASSCF calculations are quite resource demanding and you probably won't get results overnight.


Elimination of OH from spirocyclic compounds I was wondering, what would happen if I treated the following compound i.e. trispiro[2.1.25.1.29.13]dodecane-4,8,12-triol with sulfuric acid[$\ce{H_2SO_4}$] to eliminate OH group to form double bond:

Will the double bond form? if yes then where and why? If no then what other compound will form and why?

This question has been unanswered for quite long, and meanwhile, I have found an answer, so I wanted to share it.

On treating the compound with sulfuric acid, the $\ce{OH}$ groups will be protonated to $\ce{OH_2^+}$.

The $\ce{OH_2^+}$ groups are good leaving groups, so they will leave as $\ce{H_2O}$, forming carbocation.(one by one)

As cyclopropane is under significant ring strain, it will undergo ring expansion to form cyclobutane ring, which is somewhat less strained. At the same time, $\ce{H^+}$ ions will be expelled, forming double bonds. (It should also take place step by step.)

In this case I don't think that the double bond will not cause any significant barrier to the formation of the molecules, as the double bond is here the bridging bond, not at the bridgehead[formation of double bonds at bridgehead is not possible for smaller rings(<8 carbon rings), as it causes very high strain: Bredt's rule]

I have considered only elimination in this reaction scheme. Other compounds may also be formed. Feel free to post answers if you think other compounds will be formed or if you think the above mechanism is wrong in any way.


Molar enthalpy vs Enthalpy of formation? (Note I've refocused a previous unanswered question into a more general question. You can find the original much more specific question here)
There seems to be two main ways of looking at the enthalpy of substances: molar enthalpy and heat of formation. What are the differences/advantages and disadvantages of each and are either one or more or less suitable for different purposes?

Molar enthalpy is the energy released when the substance undergoes combustion. It is essentially the energy required to fragment the substance into its component sub-compounds.

Heat of formation is the theoretical difference between the energy each atom typically has at STP and the total energy that the compound actually has. Essentially, it is the energy that goes into formulating the bonds between the atoms.

Molar enthalpy is generally used a lot more in experimental chemistry because it can be measured directly, and it gives useful values on the energy that can be extracted from a substance.

Heat of formation is preferred in theoretical chemistry, as it is the theoretical maximum amount of energy able to be extracted in perfect conditions.


What are Lattice Points? What are lattice points exactly? For what purpose are they used in solid state chemistry?

There are different types of lattices. A lattice point is a point in any of these lattices. Lattices are used to describe  highly ordered systems such as crystals and some supersolids. I am unsure whether lattices can only describe periodic systems (see quasicrystals). However, I will be assuming that we are talking about periodic lattices. I will also be assuming that we are dealing with crystals.
Crystal lattice
The crystal lattice is used to describe the lattice of a real crystal. For example, in NaCl, a lattice point in a crystal lattice represents the position of a sodium ion or a chloride ion.
Bravais lattice
Bravais lattices are more mathematical and abstract than crystal lattices. They are pretty much the same as crystal lattices. Unlike the crystal lattice, however, lattice points in the Bravais lattice no longer represent a position of a particular atom. Instead a lattice point represents a position in which an atom can be placed. In other words, a lattice point in a Bravais lattice is a point, which is equal and indistinguishable from any other another point. What matters in a Bravais lattice are not the points themselves, but how they are arranged (i.e. symmetry).
Reciprocal lattice
The reciprocal lattice is the Fourier transform of either the crystal lattice or a Bravais lattice. More often than not, it is used to refer to the Fourier transform of the Bravais lattice. More of that is explained in Physics.SE. However, I personally think that this video explains it better. A lattice point here can represent an atom or a point, depending on the context.
Finally, as mentioned by other posts, these points can be vacant.


Heat given off from an electrochemical cell compared to mixing reactants My confusion regarding the difference between the heat given off in a reaction that takes place in an electrochemical cell compared to in a beaker/appropriate container when the reactants are simple mixed, arose from the following problem question:

For the reaction:

\begin{align}
\ce{HgCl2~(s) + H2~(g,1~bar) &-> 2Hg~(l) + 2H+~(aq) + 2Cl- }\\
\Delta G^\circ &=-51.64~\mathrm{kJmol^{-1}},\\
\Delta S^\circ &=-61.6~\mathrm{JK^{-1}mol^{-1}},\\
\Delta H^\circ &=-70~\mathrm{kJmol^{-1}}.\\
\end{align}

Part (a) of the question then asks: "Determine how much heat is given off at $298~\mathrm{K}$ per mole of $\ce{HgCl2}$ if the cell is operated reversibly"

The correct approach is to use $T\,\mathrm{d}S=\mathrm{d}q$ giving a value of: $18.36~\mathrm{kJmol^{-1}}$

Then part (b) asks: "Now determine the heat given off if the reaction is brought about by simply mixing the reactants"

I realised that the electrical work done in the cell would now be given off as heat so I just added the value for the Gibbs free energy to the value calculated in (a) and got: $70~\mathrm{kJmol^{-1}}$. This answer is correct but my tutor noted that this value is the value for the enthalpy change which would have been a shortcut to the answer. I didn't spot this and I don't understand why this is the case so wouldn't spot it again, should another question like this come up. Please can you clear up why this value is the enthalpy change.

I have some issues with the above working (it is all correct but I don't fully understand it).

Firstly, in part (a) I had to use the Clausius inequality (except it was an equality since the process was carried out reversibly) to find $\mathrm{d}q$. However, why must I use this equation because $\Delta H=q_p$ - the process was carried out at a constant pressure?

Secondly, if the heat given off in part (b) is $70~\mathrm{kJmol^{-1}}$ doesn't this violate the Clausius inequality? I must be wrong but it seems that this must be the case since $\mathrm{d}S \geq \frac{\mathrm{d}Q}{T}$ and using the value for the entropy change ($\Delta S^\circ=-61.6~\mathrm{JK^{-1}mol^{-1}}$) and the new calculated value for the heat given off ($70~\mathrm{kJmol^{-1}}$) then dividing this by $298~\mathrm{K}$ gives a value larger than the value for the entropy change but the Clausius inequality says that the change in entropy is greater than or equal to $\mathrm{d}q/T$. What is going on there?

I think the problem here is related to what each of these terms means, in a fundamental sense.

$\mathrm{d}S$ is the entropy of the system

$\frac{-\mathrm{d}q}{T}$ is the entropy of the surroundings - the heat change of the surroundings divided by the temperature.

$\mathrm{d}H$ is the enthalpy of the system - it is equal to $\mathrm{d}q$ only when the internal energy change occurs at constant pressure, and when there is no form of work being done besides PV work. In an electrochemical cell, moving the electrons is work, and so we can't leave it out of the total energy balance.

In other words, you can't just use $\mathrm{d}q = \Delta H$ for the 1st question because $\mathrm{d}q_p \neq\Delta H$ when you are doing any work besides PV work.

Instead you are using the Clausius inequality, which is just another way of stating the second law: the entropy of the universe must increase for any process. In this case, the universe is the system plus the surroundings, and so:

$$\Delta S - \frac{\mathrm{d}q}{T} \geq 0$$

Solving that for dq gives you the heat absorbed by the surroundings, which is equal to temperature times the entropy change of the system.

For a process that is only (possibly) doing PV work, at constant pressure, $\Delta H = \mathrm{d}q_p$, and so we can just use that. This would be an example of using the 1st law at constant pressure, when there is no other type of work besides PV.

In your case, you used:
\begin{align}
\Delta G &= \Delta H - T\,\mathrm{d}S\\
\Delta H &= \Delta G + T\,\mathrm{d}S\\
\end{align}

This works because the equation for $\Delta G$ is derived directly from the Clausius inequality at constant pressure. In other words, you are using the 2nd law to derive the same result.

Edit

To answer your last questions, and questions from the comments:

  why is $T\,\mathrm{d}S$ less than $\mathrm{d}q$ in my example? I don't really get this bit

It's not - but you are dropping the signs and that makes it hard to see. If you leave the signs in:

\begin{align}
T\,\mathrm{d}S &\geq \mathrm{d}q\\
(298~\mathrm{K})\left( -61.6~\mathrm{\frac{J}{K\cdot mol}}\right) &\geq -70~\mathrm{\frac{kJ}{mol}}\\
-18~\mathrm{kJ} &\geq -70~\mathrm{kJ}\\
\end{align}

Which is true, although it is easier to see if you divide by -1:

$$18~\mathrm{kJ} \leq 70~\mathrm{kJ}$$

  The entropy divided by the temperature gives $234.9~\mathrm{JK^{−1}mol^{−1}}$ which is not the value for entropy as it should be according to the Clausius inequality given that the change in entropy is equal to the heat given off in part (b). Why is this so?

Here I assume you meant "enthalpy divided by temperature." The second reaction is not carried out in a reversible manner, so we have no guarantee that $\mathrm{d}S = \mathrm{d}q/T$. All we know is that it will be greater than or equal to $\mathrm{d}q/T$, and if you plug in the numbers, it is.

Remember that all of this is built into the $\Delta G$ equation as well, which is why we use it when doing "regular" chemical reactions at constant T & P.


Number of CO2 in pyrolysis In the following pyrolysis reaction, we have to find how many $\ce{CO2}$ is formed.

According to me it should be 5 as there are five carboxylic acid.

But the answer is 4, how?

While it would be possible to decarboxylate almost any carboxylic acid with appropriate conditions, simple aliphatic acids are not readily decarboxylated. Carbonyl group in beta position facilitate decarboxylation due to formation of a cyclic transition state:
Now, your example has five carboxyl group, and each has a carbonyl in beta position (carboxyl also counts as a carbonyl compound). Let's look at the picture:

First we can remove the three top carboxyls marked with arrows (A). There is a carbonyl in the ring that will aid in the decarboxylation, so we will arrive to a compound at B) (I'm not saying that the reaction will proceed that way, it is just an explanation). As I said before, one carboxyl group can facilitate the removal of the other carboxyl. But once we remove one group, we get compound C). That compound does not have any groups that would help to get rid of the last carboxyl. So there is your answer.


Mechanism of acetylene formation from iodoform and silver My book mentions a synthetic route for turning iodoform into propyne.

That conversion was somewhat as follows:

  $$\ce{CHI3->[Ag] HC#CH ->[NaNH2] NaC#CH ->[CH3Br] CH3C#CH}$$

The first step is the conversion of iodoform into acetylene with silver. I am unsure how that reaction works. Could someone provide a mechanism?

What happens is that the C-I bond being weak, the iodide group is attacked by the silver atom (similar to what Na does in Wurtz reaction). As, there are 3 I-atoms and 2 molecules of iodoform usually react, 6 silver atoms are used up in the reaction and they precipitate out 6 molecules of AgI. The 2 carbon atoms form a triple bond to form acetylene.

EDIT- $$\ce{2CHI3 + 6Ag-> C2H2 + 6AgI}$$
$$\ce{C-I + 2Ag +I-C->I\bond{...}Ag + Ag\bond{...}I + C\bond{...}C}$$
As there are 3 C-I bonds, 6 Ag atoms form bonds with I-atoms, Hence, triple bond is formed between the 2 C-atoms. I hope that explains it.


How to explain two NMR spectra of the (allegedly) same compound with slightly different shifts? For my synthetic lab work I synthesised a Horner-Wadsworth-Emmons phosphonate; for the sake of the question let’s assume it to be the one presented in the image below.

When first synthesising it, I of course recorded NMR data. I performed a Horner-Wadsworth-Emmons reaction using a tiny bit of the phosphonate and kept the rest in the fridge.

One month later I retried the HWE and got very bad yields, so I decided to record another NMR spectra. The two spectra themselves look identical and the signals (those that aren’t multiplets) also show the same coupling constants. However, when referencing the spectrum to the chloroform signal ($7.26~\mathrm{ppm}$), all signals are shifted by the constant value of $\approx 0.07~\mathrm{ppm}$.

I decided to check the old literature spectrum from 1986, but it was unhelpful: It only reported multiplets such that both of my spectra comfortably fall well within the multiplets’ limits.

Considering that the two spectra look identical and that the difference between two signals corresponding to the same hydrogens is identical as are the coupling constants, I want to assume that both spectra show an identical compound. However, the $0.07~\mathrm{ppm}$ shift for everything save chloroform still irritates me.

Both spectra were recorded at $500~\mathrm{MHz}$. The parameters of the earlier one state a temperature of $25~\mathrm{^\circ C}$, the later one was apparantly measured at $25.1~\mathrm{^\circ C}$. However, I don’t think that that could explain the shift.

What could explain this shift difference?

(My first thoughts were some changes to the compound that only affect the phosphorus atom, but aside from deoxygenation being chemically unlikely shouldn’t it also cause random chemical shift differences including slightly different coupling constants? In fact, shouldn’t any chemical modification cause the shifts to change in a seemingly random fashion depending on the proximity of hydrogen in question and entity modified?)

After checking because I was asked in the comments: The later spectrum (the one that has higher shift values) has broadened peaks. That includes the chloroform peak.

That's a pretty small chemical shift difference.  Keep in mind that chemical shifts depend on temperature, concentration, pH, and other factors.  The concentration difference is the most likely, but I wouldn't rule out T effects.  If you're running it in a different batch of CDCl3 (or even the same batch, but it's aged), the pH could be different as CDCl3 becomes acidic over time.

If you used identical settings when collecting the spectra (e.g. using a default Proton) then you can measure the signal-to-noise ratio of two identical peaks and get an estimate for how different the concentrations are.  You can also simply test how sensitive your chemical shifts are to concentration by making up a more concentrated sample (or diluting your current sample).  Or try changing the temperature with the VTC and ramp it up to 25.5 or 26.0 C and see how much the chemical shifts change (referencing everything to CHCl3 for consistency).

If your compound were reacting with something, there would be a much more dramatic change in the chemical shifts than 0.07 ppm.


First Order Degradation Constant (λ)(d^-1) Can you please tell me the First Order Degradation Constant (λ)($d^{-1}$) for 1,3,5-Trimethylbenzene (mesitylene)?

If you don't have this value, can you please tell me which VOC in 35 Illinois Administrative Code 740. Appendix A. Table A. is "most similar" to 1,3,5-TMB that I may use its First Order Degradation Constant? Please provide an explanation. Thank you!

Based on a bit more than a cursory search, there doesn't seem to be a first order degradation constant readily available for 1,3,5-trimethylbenzene (mesitylene) in the open literature. This leaves us with your second option, the Title 35 Illinois Administrative Code 740$^{[1]}$, Table A (herein referred to as Table A).

By far the most similar compound to 1,3,5-trimethylbenzene that can be found in Table A, based on both chemical structure and physical properties, is the total xylenes (CAS # 1330-20-7).  This is actually a mixture of the three possible isomers of xylene.

Both the images below and all physical and chemical properties given for the total xylenes were taken from this Wikipedia page unless otherwise stated. The 1,3,5-trimethylbenzene image and all of its chemical and physical properties were taken from here.

1,3,5-trimethylbenzene (mesitylene)

In order: o-xylene, m-xylene, p-xylene

Selected properties:
Mesitylene: density = 0.86 ($\mathrm{20^oC}$), B.P. = $\mathrm{165^oC}$, M.P. = $\mathrm{-45^oC}$, and water solbility = 50 mg/L at $\mathrm{25^oC}^{[2]}$
Xylenes : density = 0.86 ($\mathrm{20^oC}$), B.P. = $\mathrm{138^oC}$, M.P. = $\mathrm{-47^oC}$, and water solbility = 110 mg/L at $\mathrm{25^oC}^{[2]}$

1) TITLE 35: ENVIRONMENTAL PROTECTION, SUBTITLE G: WASTE DISPOSAL, CHAPTER I: POLLUTION CONTROL BOARD PART 740, SITE REMEDIATION PROGRAM

2) Yalkowsky, S.H., He, Yan., Handbook of Aqueous Solubility Data: An Extensive Compilation of Aqueous Solubility Data for Organic Compounds Extracted from the AQUASOL dATAbASE. CRC Press LLC, Boca Raton, FL. 2003


Crystal field splitting pattern with strong field and weak field ligands What will be crystal field splitting pattern for octahedral complexes if some of the ligands are strong and some are weak field ligands?

Well, the split would depend on:
The identity of the metal ion as well as the identity of the ligands themselves. You'd have to consult a spectrochemical series to determine the magnitude of the split caused by the ligands.

Alternatively, you could beam white light through the complex and using the light that appears, determine the wavelength of the light absorbed (the complementary colour) to promote electron(s) from the lower split d orbital to a higher one... This would give you more information as to the magnitude of the split (sometimes, the experimental value will provide an answer that theoretically cannot be determined).


Why HF plus H2O is not a buffer solution? I think $\ce{HF}$ can be hydrolysis and get $\ce{F-}$ as a conjugate base, so this solution should be a buffer solution. But the correct answer is that this solution is not a buffer. Please explain it to me. I'm very confused.

What are you talking about?
$\ce{HF}$ in water solution is acting like moderately strong acid. It's because:

$$\ce{HF + H2O \rightleftharpoons H3O+ + F-}$$
$$\ce{H3O+ + F- \rightleftharpoons [H3O]F}$$
which is undissociated.Then:
$$\ce{F- + HF \rightleftharpoons HF2-}$$

As you can see the large affinity of $\ce{F^-}$ to $\ce{H^+}$ won't allow it to dissociate.
The $K_a(HF)(6.6x10^{-4})$ is strongly favoring reactants.
So the $\ce{HF}$ in $\ce{H2O}$ isn't buffer because there isn't almost any $\ce{F-}$ to be considered as conjugated acid. So if you add some acid there will not be enough $\ce{F-}$ for it's neutralisation.
So if you add some acid or base it will react with water and equilibrium constant will be changed and that will lead to drastic change of pH.

So the buffer is solution which consist of:

salt of strong acid and weak base and the weak base (for example $\ce{NH_{4}Cl}$ and $\ce{HN3}$ ).
salt of weak acid and strong base and the weak acit (for example  $\ce{CH3COONa}$ and $\ce{CH3COOH}$).

Hope i helped you. If you have any other questions please ask.


Is there a liquid crystal or similar material that gets cloudy when an electric field is applied, but is otherwise clear? Is there a way to make a liquid, such as a liquid crystal material, transition from cloudy or clear and vice-versa using only electric current?

By cloudy, I mean, similar to the effect that cholesterol or 5CB exhibits, thermally.

If so, where could this be obtained, and how quickly does it switch between states?

If not, how else might such an effect be achieved using electric current? I've read about the Kerr effect, but it didn't change opacity so much as refractive index, and it also appeared to use a great deal of electricity. I'm thinking something within the single volts.

Electrochromic materials are produced for this purpose and are typically sold as “smart glass”.


What is charge density? For example, Magnesium has a higher charge density compared to Sodium, therefore its metallic lattice is stronger and it has a higher melting point. Does that mean that Magnesium has more protons and a smaller atomic radius or is it about its charge (2+), that makes its lattice stronger?

What does charge density show?

If it was about the number of protons in the nucleus, then Mercury would hardly be liquid at room temperature, would it?

Actually, it is about the density of free electrons, i.e. the density of the "electron gas" that makes up the metallic bonding. Obviously, this density has to be the negative of the ions' charge, so you might equivalently say it's about magnesium's (2+) charge, though that doesn't make much sense physically unless you also consider the conduction electrons, as the ions themselves of course repulse each other.


How to explain the dipole moment anomaly of methyl fluoride and methyl chloride with respect to their hydracids? I was checking dipole moment orders, and came across a very peculiar result:
Methyl fluoride $(\ce{CH3F})$ has lesser dipole moment than methyl chloride ($\ce{CH3Cl}$), but hydrofluoric acid ($\ce{HF}$) has more dipole moment than hydrochloric acid ($\ce{HCl}$).

At first I thought it was about distance, but upon seeing the second one, it led me to think about polarizability. So my question shortens to this:
Which factor dominates while checking dipole moments and when?

Summary

A systematic examination of the $\ce{CH3F}$ and $\ce{CH3Cl}$ complexes with density functional techniques and high-level wave function theory indicates that the charge-separation within these complexes influences the relative corresponding dipole moments for these compounds.  Though fluorine is indeed more electronegative than chlorine, $\ce{CH3Cl}$ had a larger dipole moment due to the longer C-Cl bond with respect to the C-F separation.

Introduction

Recall that a dipole moment can be explained relatively simply by a separation of two charges.  The dipole moment, $\mu$, is dependent on two things: the magnitude of the charge ($q$) AND the separation between the two charges ($r$).  This can be expressed as:

$\mu = q \cdot r$

We know from general chemistry that fluorine is the most electronegative atom on the periodic table.  Therefore, for simple molecules such as $\ce{HF}$ or $\ce{LiF}$, we would expect a rather large dipole moment than if we considered $\ce{HCl}$ or $\ce{LiCl}$.  However, a dipole moment depends on separation as well.  This can lead to some discrepancies in our general trends that we chemists love to use.

Consider chlorine.  The atomic radius for chlorine is larger than it is for fluorine.  A C-Cl bond is going to be longer than a C-F bond.  In the case of methyl fluoride ($\ce{CH3F}$) and methyl chloride ($\ce{CH3Cl}$), we will have a difference in separation of the methyl group and the halogen (larger for the latter).  Looking back at our electric dipole equation, an increase in $r$ translates into an increase in $\mu$.  Perhaps it is this larger separation between the carbon and chlorine atoms that gives rise to these relatively similar dipole moments between $\ce{CH3F}$ and $\ce{CH3Cl}$.

Super Mini Research Project Time

Let us apply electronic structure theory to the problem.  Dave has already provided a link to NIST which is a superb resource for these types of simple computations. However, to be thorough, we will perform our own computations.  The experimental dipole moments for $\ce{CH3F}$ and $\ce{CH3Cl}$ (provided by DavePhD) are $\pu{-1.858 D} \text{ and } \pu{-1.897 D}$, respectively ([1] and [2]).  Dipole moments from electronic structure theory for these complexes can also be found on the NIST website (source3 and source4)

Methodology

Full optimizations and corresponding harmonic vibrational frequency computations are performed using the B3LYP density functional technique (sorry Martin) using the Gaussian 09 software package.  The 6-31+G(d,p) basis set will be employed along with a series of Dunning's correlation consistent basis sets: heavy-aug-cc-pVXZ (where X = D, T, Q, 5).  This basis set denotes that cc-pVXZ was employed for hydrogen atoms and the aug-cc-pVXZ basis set was employed for all heavy (i.e. non-hydrogen) atoms.  A numerical integration grid consisting of 99 radial shels and 590 angular points per shell were employed for all DFT computations.  In addition, full optimizations have been performed using the CCSD(T) level of theory using the CFOUR software package.  The haDZ, haTZ, and haQZ basis sets have been employed for these computations.  All geometries were constrained to $C_{3v}$ symmetry.  Everything is converged to 'overkill'.  Frozen-core approximation applied... 5D 7F spherical basis functions used, etc. etc.

Results

The dipole moments (in Debye) as well as select geometrical parameters ($R$ in Angstroms) for the $\ce{CH3F}$ and $\ce{CH3Cl}$ complexes at various levels of theory are given in the table below.  The deviation of the computed electric dipole moment from experiment is denoted as $\delta_{\mathrm{theory}}^{\mathrm{expt.}}$.  Using this deviation as our main measurement of error, we can see that for the $\ce{CH3F}$ complex, the B3LYP/ha5Z level of theory deviates from experiment by only $\pu{0.01 D}$.  A larger deviation from experiment is observed for the CH$_3$Cl complex with a magnitude of $\pu{0.1 D}$ at the B3LYP/ha5Z level of theory.  These magnitudes are qualitatively similar at the CCSD(T) level.  Therefore, with respect to the dipole moment, the B3LYP method with a triple-$\zeta$ quality basis set is necessary to describe this first-order property.

Clearly $\ce{CH3Cl}$ contains a larger separation between the Cl atom and the methyl group with respect to $\ce{CH3F}$ by about 0.4 Angstroms.  If you wanted to back-calculate the charge for these complexes by assuming $R$(CX) (where X is the halogen) is equivalent to $r$ in the first equation, you will see that CH$_3$F has a larger charge (by about 0.25).  However, we still see that CH$_3$Cl has a larger dipole moment.

The final column in the table above ($\mu_{\mathrm{geom}}^{\ce{CH3Cl}}$) denotes the dipole moment of $\ce{CH3F}$ when using the optimized geometry of $\ce{CH3Cl}$ at the corresponding level of theory.  The dipole moment is significantly larger, by about $\pu{1.5 D}$, which is expected due to the larger C-F separation.

Conclusions

Though we as chemists like to resort to using the general trends provided by the periodic table, we must take into account other factors to explain deviations from these trends.  Here we consider the separation of the halogen from the methyl group as an explanation for $\ce{CH3Cl}$ having a larger dipole moment than $\ce{CH3F}$, even though F is more electronegative.  Our computational results qualitatively agree with experiment.  In addition, B3LYP with a triple-$\zeta$ quality basis set is adequate in describing this first-order property according to CCSD(T) results.  Finally, these results indicate that it is the charge-separation in these small molecules that greatly impacts the qualitative ordering of these dipole moments.

References:

Marshall, M. D.; Muenter, J. The electric dipole moment of methyl fluoride. J. Mol. Spectrosc. 1980, 83 (2), 279–282. DOI: 10.1016/0022-2852(80)90051-X.
Wlodarczak, G.; Segard, B.; Legrand, J.; Demaison, J. The dipole moment of CH335Cl. J. Mol. Spectrosc. 1985, 111 (1), 204–206. DOI: 10.1016/0022-2852(85)90083-9.


Does DMT-Acetate become Polar in H2O, with added NaCl? I'm extracting DMT (N,N-dimethyltryptamine) from Acacia confusa via acid/base. I extracted the DMT with $\ce{H2O}$ and acetic acid, converting it to DMT-acetate.  Here, I used a non polar solvent, naptha, to defat the material, causing an emulsion.

I used $\ce{NaCl}$ to break the emulsion, this caused a thick layer of black tar to form on top of the aqueous solution, and a glossy black tar on the bottom of the vessel.  I need to freebase the DMT acetate; however, I'm concerned that DMT-acetate migrated into one of the two polar layers.  (Correction: Either the non polar layer on top or gunk layer at the bottom)

How can I confirm if this is so? Does $\ce{NaCl}$ cause DMT acetate  (or any acetate) to become polar in $\ce{H2O}$? Why and how? Will the DMT acetate have migrated into any of these two polar layers?

Does NaCl cause DMT acetate (or any acetate) to become polar in $\ce{H2O}$?

In aqueous acidic solution, that is in the presence of a sufficient amount of acetic acid, N,N-dimethyltryptamine is protonated. The acetate is the counter ion of the ammonium salt. The compound does not become more polar by adding sodium chloride!

  Will the DMT acetate have migrated into any of these two polar layers?

Are you having two polar layers? I doubt it. You're having a top unpolar layer, the aqueous layer and some gunk.

Take the separated aqueous layer, neutralize carefully, extract with an unpolar solvent of your choice, dry over sodium sulfate, remove the solvent in vacuum. If necessary, recrystallize and determine the yield.

Please keep in mind that DMT is a controlled substance in many countries!


Nomenclature - Common vs IUPAC Names and Alphabetical Priorities I got a problem with this:

What if we stuck with the IUPAC name and called the isopropyl group a 1-methylethyl group? That would definitely reverse the alphabetical priorities; because the substituents here on the hydrocarbon are on equivalent positions, we must give the lower number to the substituent that comes first in the alphabet.

But that's the rub! (No isopropyl alcohol pun!)

Isopropyl - "I" comes before "M" as in methyl.

But if we name the isopropyl group systematically,

Methylethyl comes after just methyl in the alphabet.

So could the name of this also be:

2,2-dimethyl-7-(1-methylethyl)bicyclo[3.2.0]heptane?

No, 2,2-dimethyl-7-(1-methylethyl)bicyclo[3.2.0]heptane is not an acceptable name.  Like the rule you quoted says, start numbering at a bridgehead carbon and first pass through the largest ring (in this case the 5-membered ring) and number that ring so that a substituent will have the lowest number.  So this molecule would be named 2-isopropyl-7,7-dimethyl-bicyclo[3.2.0]heptane.  Usage of "1-methylethyl" is not acceptable.  If you weren't going to use the term "isopropyl", you could call the fragment "2-propyl". So you could write 2-(2-propyl)-7,7-dimethyl-bicyclo[3.2.0]heptane, but the term "isopropyl" is much more common usage.

Edit: After reading Greg E's comment, I looked around the internet and found that "1-methylethyl" is both correct and the official IUPAC name.  Here is a paraphrasing of the relevant IUPAC rule...

These groups (those attached off the main chain) are always numbered so that the point of attachment to the rest of the molecule is C1, with the longest continuous chain beginning from the point of attachment taken as the parent.

Then all sites I visited would add something like this...

"For historical reasons, some of the simpler branched-chain alkyl groups also have nonsystematic (or common, or trivial) names."

Most of these sites seemed to (strongly) favor the trivial name.  As one site put it, "Common names are used because they are easier to say, shorter, and save paper
and ink in scientific publications when they have to be used repeatedly in manuscripts."


What is the formula of manganese oxohydroxide: MnOOH and MnO(OH)2? I found various formulae of manganese oxohydroxide. Some site says it is $\ce{MnOOH}$, other say it is, $\ce{MnO(OH)2}$. So, which one is correct?

The mineral manganite is considered manganese oxide-hydroxide $\ce{MnOOH}$, but yahoo answer claims that it is $\ce{MnO(OH)2}$.
In Winkler titration, $\ce{MnO(OH)2}$ is formed:

$$\ce{2 MnSO4(s) + O2(aq) → 2 MnO(OH)2(s)}$$

There is some uncertainty about whether the oxidised manganese is tetravalent or trivalent. Some sources claim that $\ce{Mn(OH)3}$ is the brown precipitate, but hydrated $\ce{MnO2}$ may also give the brown colour. So, is $\ce{MnO(OH)2}$ hydrated $\ce{MnO2}$?

So which one is manganese oxohydroxide- $\ce{MnOOH}$ or $\ce{MnO(OH)2}$?

First, the difference the substances is the oxidation number of manganese. In $\ce{MnOOH}$, the oxidation number of manganese is 3+. While in $\ce{MnO(OH)_2}$, the oxidation number for manganese is 4+.

However this doesn't mean $\ce{MnOOH}$ is manganese (III) oxyhydroxide and $\ce{MnO(OH)_2}$ is manganese (IV) oxyhydroxide, not for $\ce{MnO(OH)_2}$ at least. If you move the element around, you will get $\ce{MnO(OH)_2}$ is actually $\ce{H_2MnO_3}$. $\ce{H_2MnO_3}$ is dihydroxy(oxo)manganese which is an acidic substance.

For the question regarding the structure for both substances, I'm sorry that I don't haven't the image but I can describe it (Hope that would do). For $\ce{MnOOH}$, manganese is single bonded with $\ce{OH}$ and manganese is double bonded with oxygen. For $\ce{MnO(OH)_2}$, manganese have two single bonds with both of the $\ce{OH}$ and manganese is double bonded with oxygen.

In conclusion, $\ce{MnOOH}$ is manganese (III) oxyhydroxide and $\ce{MnO(OH)_2}$ is dihydroxy(oxo)manganese. In structure wise, manganese is single bonded with $\ce{OH}$ and manganese is double bonded with oxygen for $\ce{MnOOH}$. For $\ce{MnO(OH)_2}$, it likes $\ce{MnOOH}$ but have one more single bond with $\ce{OH}$.


What would the name of this organic compound be? I am a little lost in naming this compound.

I think it would be something like 2 bromo 4 methylene pentene??

Is that close?

It is a five carbon chain, so pentane, with a Br on carbon two, a methylene substituent on carbon 4, and a alkene on carbon 1.


What is resonance, and are resonance structures real? My teacher told me about resonance and explained it as different structures which are flipping back and forth and that we only observe a sort of average structure. How does this work? Why do the different structures not exist on their own?

This answer is intended to clear up some misconceptions about resonance which have come up many times on this site.

Resonance is a part of valence bond theory which is used to describe delocalised electron systems in terms of contributing structures, each only involving 2-centre-2-electron bonds. It is a concept that is very often taught badly and misinterpreted by students. The usual explanation is that it is as if the molecule is flipping back and forth between different structures very rapidly and that what is observed is an average of these structures. This is wrong!  (There are molecules that do this (e.g bullvalene), but the rapidly interconverting structures are not called resonance forms or resonance structures.)

Individual resonance structures do not exist on their own. They are not in some sort of rapid equilibrium. There is only a single structure for a molecule such as benzene, which can be described by resonance. The difference between an equilibrium situation and a resonance situation can be seen on a potential energy diagram.

This diagram shows two possible structures of the 2-norbornyl cation. Structure (a) shows the single delocalised structure, described by resonance whereas structures (b) show the equilibrium option, with the delocalised structure (a) as a transition state. The key point is that resonance hybrids are a single potential energy minimum, whereas equilibrating structures are two energy minima separated by a barrier. In 2013 an X-ray diffraction structure was finally obtained and the correct structure was shown to be (a).

Resonance describes delocalised bonding in terms of contributing structures that give some of their character to the single overall structure. These structures do not have to be equally weighted in their contribution. For example, amides can be described by the following resonance structures:

The left structure is the major contributor but the right structure also contributes and so the structure of an amide has some double bond character in the C-N bond (ie. the bond order is >1) and less double bond character in the C-O bond (bond order <2).

The alternative to valence bond theory and the resonance description of molecules is molecular orbital theory. This explains delocalised bonding as electrons occupying molecular orbitals which extend over more than two atoms.


Is activated carbon classified as organic or inorganic? Organic compounds are typically defined as “molecules containing carbon”. Wikipedia states that there for some historical (read: non-logical) reasons,

  a few types of carbon-containing compounds such as carbides, carbonates, simple oxides of carbon, and cyanides, as well as the allotropes of carbon such as diamond and graphite, are considered inorganic.

I thus wonder: is activated charcoal (also known as activated carbon) typically classified as an organic or inorganic material?

If I follow the list of exceptions given by Wikipedia, it should be organic (it's not an allotrope of carbon, in particular), but I get the feeling that most people in the field of porous materials would classify it as inorganic. So, I'm looking for an authoritative reference on this question.

While nomenclature is of particular interest to organic chemists to specify an exact compound, the classification of X into broad category Y or Z isn't a precise science, and not really of practical use.  The article cites a textbook by Seager to this effect, stating

  The distinction between "organic" and "inorganic" carbon compounds, while "useful in organizing the vast subject of chemistry... is somewhat arbitrary"

Even if you find a source that says "charcoal is (in)organic", you may just as well find one stating the opposite.  Just like the coal from which it may have been produced, it was once biomass and decidedly organic, but so was graphite and diamond, or CO2 and CO32−. I think it's overly pedantic and unproductive to try to come up with definitive judgements for these decidedly edge case scenarios.

After all, it's just a chemical on the shelf, what one does with it is far more relevant.  I don't use it on a daily basis, but it seems more like a tool than a reagent.  The fact it contains carbon seems beside the point; it's value isn't in the chemical composition but rather its extraordinary adsorptive properties.


How to determine the equation for the reaction of europium with sulfuric acid based on experimental measurements? I was given a set of results by a fellow researcher where he reacted $0.608~\mathrm{g}$ of europium ($M = 152$) with an excess of $\ce{H2SO4}$ and collected $144~\mathrm{cm^3}$ of $\ce{H2}$ gas at room temperature and pressure.

I was then asked to derive the equation for the reaction using these figures.

This is what I did.

I know that $1~\mathrm{mol}$ of gas is $24000~\mathrm{cm3}$.
I found out that $144~\mathrm{cm^3}$ is equivalent to $0.006~\mathrm{mol}$ of $\ce{H2}$.

Then, I did $0.608 / 152$ to get $0.004~\mathrm{mol}$.

This meant that europium and hydrogen gas were in the ratio $2:3$.

The equation I wrote was $$\ce{2Eu + H2SO4 -> Eu2SO4 + 3H2}$$

But I am wrong and I don't know where.

Your equation isn't balanced; the oxidation state for europium is +3 and not +1; and europium sulfate does not exist in the solution you describe.
From this link:

Reaction of europium with acids
Europium metal dissolves readily in dilute sulphuric acid to form solutions containing the very pale pink aquated Eu(III) ion together with hydrogen gas, $\ce{H2}$. It is quite likely that $\ce{Eu^3+}$(aq) exists as largely the complex ion $\ce{[Eu(OH2)9]^3+}$

$$\ce{2Eu(s) + 3H2SO4(aq) -> 2Eu^{3+}(aq) + 3SO4^{2-}(aq) + 3H2(g)}$$


A misunderstanding about the energy profile of reactions with a catalyst involved All of us are aware of the importance of the catalysts in bio-chemistry. For a high school learner like me, catalysts ,and therefore, enzymes play a bridge-like role that connect high school bio to high school chemistry.

Yet, I got baffled about the two common energy profiles drawn for an anonymous reaction with a presence of a catalyst. The question may seem rudimentary, I agree, but the possible answers to my question were too technical for me to understand.
One of the energy profiles had several curves in the pathway for $E$, and the other one has only one curve in the pathway where the catalyst should have affected.

My guess is that the first energy profile is about a net change in $E_\mathrm{a}$ but the second one demonstrates the reality when an exothermic reaction occurs.
 Now, here are the questions:

Why are the two energy profiles different? Please be as if you're teaching thermo-chemistry to a little kid!
Why do those curves occur in the latter energy profile and is there a way to know how many of those curves will occur in the presence of a catalyst?

Image credits: chemwiki.ucdavis.edu (page here) and wiki (page here).

The upper graph is just the most simple way to visualize the effect of a catalyst on a reaction $\ce{S -> P}$:

The activation energy is lowered.

The activation energy for the reaction in that direction is the difference of the energies of the starting material $S$ and a transition state $TS^\#$.

Since it is the same starting marterial in the presence or absence of the catalyst, the energy of the transition state is different.

Can the same transition state have two different energies - just through the remote distance magical action of a catalyst located somewhere? Probably not!

It is much more plausible that - in the absence and presence of a catalyst - two completely different (different in structure and different in energy) transition states are involved.

Exactly this situation is described in the second graph!

The catalyst "reacts" with the starting material, either by forming a covalent bond, by hydrogen bonding, etc. and thus opens a different reaction channel with different intermediates and a transition state not found in in the non-catalyzed reaction.

In the end, the same product $P$ is formed and the catalyst is regenerated, but this doesn't mean that the catalyst wasn't heavily involved in the reaction going on.


Defining and testing custom made DFT functionals? What is a good and free software for this endeavor? Any recommended articles on the subject?
I want to design dft functionals, trying out different parametrizations and optimizing mixing constants for exchange and correlation contributions for specific data sets. Even add PT2 perturbation energy into the mix if possible.

I am aware of Gaussian's user-defined functional options, but would like to have a bit more control in the parameters and contributions, instead of just 6 parameters.

Depending on what you're after, you might want to look for programs using libxc, so you can completely define your own functionals.

From the libxc section of the Octopus wiki, these include:

  Abinit – plane-wave code
  APE – an atomic code
  Atomistix ToolKit – numerical orbitals code
  AtomPAW – projector augmented wave functions generator
  BigDFT – wavelet code
  CP2K -  A program to perform atomistic and molecular simulations of solid state, liquid, molecular, and biological systems.
  DP – Dielectric Properties, a linear response TDDFT code
  Elk – FP-LAPW code
  ERKALE – a DFT/HF molecular electronic structure code based on Gaussian orbitals
  exciting – FP-LAPW code
  GPAW – grid-based projector-augmented wave method
  JDFTx – plane-wave code designed for Joint Density Functional Theory
  MOLGW - a small, but accurate MBPT code for molecules
  octopus – real-space (TD)DFT code
  Yambo – solid state and molecular physics many-body calculations code

On a more personal note, and this is just my opinion, if you're just arbitrarily mixing functionals, you should really make sure you have a scientific justification for doing this, otherwise you're just generating numbers without basis.


Measuring turbidity using a spectrophotometer I am conducting a lab investigating on the effect of aluminium sulfate concentration on turbidity. In my high-school, we only have a spectrophotometer. Is there a way to measure turbidity with a spectrophotometer? Do I need to dye my solution or alter it in some other way? How do I prepare my blank?

You can use a spectrophotometer to measure turbidity since the suspended particles will scatter light and so a reduced amount of light is detected dependent upon how much scatter occurs. The relationship is $I_{trans}=I_0\exp(-k\cdot C)$ where k is a proportionality constant depending on the particle size and shape, particle refractive index, sample path-length and C is the mass concentration in what ever units you choose. Usually in a spectrophotometer  the log is produced by the machine as optical density ( OD ). Thus $OD = -\log_{10}(I_{trans}/I_0)= kC$.

It is essential when comparing samples that the particle size is the same otherwise k varies between measurements and as C is also varied this confuses any measurements. Gum arabic or gelatin are often used to stabilise a suspension. The more finely divided is the suspension the greater is the ability to scatter light. Often experiments are made by comparison to a standard suspension say of Ca oxalate, $\ce{BaSO4}$, or Zn ferrocyanide.


Acid-Base determination of organic compounds I'll start this question with the problem that led me to ask this question...

USNCO 2009 Problem 8 part c:

Part c. states that all four of the compounds above are more basic than ammonia, and the solution explains that it is because the "carbon-containing groups" that are attached to the nitrogen are "better electron donors" than just hydrogen (in ammonia). This "inductive effect" causes the lone pair of the nitrogen to attract the proton ($\ce{H^+}$) better, or as the solution said, "donate the electron pair" to the proton better.

This got me thinking.

My thought process was that the polarity of the hydrocarbon group and the nitrogen overall makes the atom extremely polar towards nitrogen. What I mean here is that the nitrogen has a large negative dipole. This also, in corollary, means that nitrogen has more electrons. In turn, it would attract the proton more strongly.

However, I do not know if my thought process is correct.

This is the official ACS solution to the problem in question:

What does it mean they are "better electron donors?" What is this inductive effect? If you can provide it, are there any other rules that are important pertaining to this and is there a source where I could learn these rules?

Thank you in advance.

Note: this is NOT homework, and my thought process for part c HAS been specified.

Source: https://www.acs.org/content/dam/acsorg/education/students/highschool/olympiad/pastexams/2009-usnco-exam-part-ii.pdf

There is a slight error in your sentence. The nitrogen does not have more electrons, but it has a higher electron density because of it's neighbours. But I think this is how you meant it. Often it is described in terms of donating electrons, whilst these electrons are not donated in a cation/anion sense, so I think electron density is a better to grasp description.

Anyway, with being "better electron donors", it is meant that they are more readily able to donate part of their electron density, just like you describe with your description of polarization.

The inductive effect is the ability of an atom or functional group to donate part of it's electron density through $\sigma$-bonds. So in this we are talking about the electronegativity of a group. It is something you have already encountered in chemistry, but this time it has received the proper name 'inductive effect'.

What you would want to be reading, is about Hammett plots. Hammett plots are a tool to picture the ability of substituents on benzoic acid to influence the acidity. Originally Hammett used hydrogen as the reference substituent. The cool thing is that you can compare the effect of substituents on a different molecule, with the effect they have on benzoic acid. I can advise you to get to the library and find 'Modern physical organic chemistry' by Anslyn and Dougherty. Especially chapter 8.2 about Substituent Effects and chapter 8.3 about Hammett Plot's as a tool will be useful.


Is it possible to functionalize silica gel as Activated Carbon can be? How? Activated carbon and Silica gel both have a high surface area.
If I understand correctly it is possible to trap stuff inside activated charcoal pores to give it specific retention properties (by exposing it to functionalizing conpounds while heated to open up the pores?).
Is it possible to likewise trap stuff in silica gel pores? What can be trapped in there and how is this done?

Yes, silica gel can be modified to make its adsorption more specific. E.G., epichlorohydrin dimethyl amine copolymer (Epi-DMA) can be used to make silica gel effective for adsorbing anionic dyes.


Aromatic and aliphatic amines Is benzylamine $\ce{C6H5CH2NH2}$ an aromatic or aliphatic amine? I've tried searching for it however different sources have conflicting information.

The nitrogen is bound to a benzyl group, so it would most aptly be classified as a benzylic amine. The benzyl group is a common protecting group for amines, and can be removed via hydrogenation with a palladium catalyst. As for the basicity of the nitrogen, it would be most similar to an aliphatic amine.


Relationship between Ka and Kb Considering the $K_\text{a}$ of the weak acid and the $K_\text{b}$ of the weak base that form the salt, we conclude that the solution will be acidic if $K_\text{b}$ > $K_\text{a}$ and basic if $K_\text{a}$ > $K_\text{b}$

I was wondering if anyone could clarify what the book is saying.
My question is If we mix a solution with a $K_\text{b}$ of 0.1 with a solution with a $K_\text{a}$ of 0.001 would the final solution be acidic or basic?

If there is a solution of mix of a weak acid/base and it's alkaline/acidic salt in molar ratio 1:1, then

The solution is acidic, if $K_\mathrm{a} \gt K_\mathrm{b}$, like acetic acid + sodium acetate(alkaline base)
The solution is alkalic(ine)/basic, if $K_\mathrm{a} \lt K_\mathrm{b}$, like ammonium chloride ( acidic salt) + ammonia(base)

Note that $$\begin{align}
K_\mathrm{w}&=\ce{[H+][OH-]} \\
K_\mathrm{a}&=\frac{\ce{[H+][A-]}}
{\ce{[HA]}}  \\
K_\mathrm{b}&=\frac{\ce{[HA][OH-]}}
{\ce{[A-]}}  \\
&=\frac{\ce{[HA]  K_\mathrm{w}}}{\ce{[A-][H+]}}  \\
&=\frac{\ce{  K_\mathrm{w}}}{K_\mathrm{a}}  \\
K_\mathrm{w}&=K_\mathrm{a} \cdot K_\mathrm{b} \\
\end{align}$$

Similarly, if there is such a 1:1 (molar amount) solution of an acid and a base, that are not in the mutual relation acid/salt or base/salt, the above is valid as well.


Why is pyridine a monodentate ligand, but pyrrole a tridentate ligand? Is it correct to say pyridine’s non-bonding electron pair does not participate in electron resonance while pyrrole’s non-bond electron participates in electron resonance? Would therefore pyridine’s resonance not be disturbed so act the molecule can act as a monodentate ligand?

From what I gathered in the comments you confused a few terms. Both compounds are monodentate, i.e. they can only coordinate one metal at a time. However, pyridine usually donates one electron pair, while pyrrole can donate three (to one metal) — I haven’t seen a structure where pyrrole does that, though.

The reason is pretty simple: Pyrrole is much like the cyclopentadienyl anion except that one carbon atom is replaced by nitrogen. If it wants to coordinate in any way, it will have to do that like a slice of bread from its side, forming a $\eta ^5$ complex. Much like ferrocene in the picture below.

Pyridine, on the other hand, has no hydrogen on the nitrogen atom; instead it has a lone pair pointing outwards, but still in the plane of the ring. This allows pyridine to donate two electrons in a $\sigma$-like fashion to a metal centre, $\eta ^1$-style. Of course, pyridine could also perform a sandwich-like coordination much like benzene sometimes does. However, the direct two-electron $\sigma$ donation will be much stronger and therefore largely outweigh a sandwich-type donation in most compounds.


Diffusion of multiple gases at once Graham’s law of diffusion states the rate of diffusion between two gases is inversely proportional to their molar masses. Now, this is helpful when we are focused on two gases. But that is rarely the case in a beaker or even the atmosphere.
So how do you measure the rates of diffusion of many gases moving at once? Do you measure each rate of effusion (inversely proportional to molar mass) and just order them, then use stoichiometry to determine which one will be dispersed quickest based on the amounts present?

If the gases in a mixture are ideal then their diffusion is independent*.
The relationship described by Graham's law is transitive, therefore the relative effusion rate of multiple substances with respect to a reference substance also provides a ranking of their relative diffusion coefficients (an ordered list).
*Outside of condensed or supercritical phases it seems unlikely that two fluid substances would display sufficiently strong interactions to alter the relative order of their diffusion coefficients.


What is the color of elemental silver? From my lab book it says it is black? I don't understand how we know the color of it is black. Can anyone explain?

Here is the reflectance spectrum for silver

It reflects the entire visible spectrum, absorbing light only in the ultra-violet region.  An ingot of elemental silver will have a silvery-white color.  In fact silver is said to be the "whitest" of all metals.  If you grind the silver up into smaller and smaller particles, the silver will appear progressively grayer or darker.  This is because smaller particles become less effective at reflecting light and produce more scattering.  Eventually when the particles are very small, even less light is reflected to the eye and the material appears black.  So depending upon the particle size, the color of silver can change from white to black.

Also, silver can react with chemicals present in air and tarnish.  It is the reaction of silver with sulfides in the air to produce silver sulfide that turns silver black, but this is through a chemical reaction.


Do azeotropic mixtures evaporate proportionally? Do azeotropes evaporate, without boiling, in proportionate amounts, or does this property describe their behaviour only at boiling point?
Concretely, if 91% isopropyl alcohol/water (by volume, i.e. an azeotropic mixture) is left open at say 25 °C, will it remain in that proportion as it evaporates, or become more dilute (until just water) as the alcohol evaporates faster?
Further, above this point, would the alcohol evaporate faster 'degrading' it until 91% at which point it behaves as the answer to above? (Which would incline me to think the answer is no, if not boiling the water and alcohol would evaporate 'independently' and the alcohol concentration would fall no matter the starting point, it's not magic.)

Azeotropic composition is a function of the boiling pressure and therefore of the boiling temperature. This can be seen in looking up the composition of the constant boiling $\ce{HCl}$ azeotrope with changing atmospheric pressure. However, information on this is hard to find and the effect seems to be small in most cases. Wikipedia does discuss this a bit. Just thinking, this might be a possible research if one had access to an FTIR with a ATR trough cell so evaporation could be observed in situ.


What is the Ksp of Al(OH)3? Top hits on google are:

This site
 Aluminum hydroxide $\ce{Al(OH)_3}$ >> $ 1.3×10^{–33}$

and

This site
 Aluminum hydroxide $\ce{Al(OH)_3}$ >> $ 3×10^{–34}$

which is one order of magnitude different.

What can be the reason for this variation? Experimental conditions?
Where can I get accurate or reliable results/data of $K_{\mathrm{sp}}$ in general?

Let me add to the confusion:

https://books.google.de/books?id=imBzjH6SHi8C&pg=PA303&lpg=PA303&dq=solubility+product+constant+aluminum+hydroxide&source=bl&ots=jXoPiVihnF&sig=eSK8saszXshTXVEKGOlThUiKVr0&hl=de&sa=X&ei=Xl5TVcepDuj_ywOXmYDQBA&ved=0CG8Q6AEwCTgK#v=onepage&q=solubility%20product%20constant%20aluminum%20hydroxide&f=false
$2\times10^{-32}$
http://faculty.ncc.edu/LinkClick.aspx?fileticket=smTKdcQlPY4%3D&tabid=1903
$1.8\times10^{-33}$
http://www.austincc.edu/dlyon/gen_chem_1_summer04/Chapter%2020%20solubiliity%20product%20f08.pdf
$1.9\times 10^{-33}$
http://www.csudh.edu/oliver/chemdata/data-ksp.htm
$4.6\times10^{-33}$
http://www.periodni.com/solubility_product_constants.html
$3\times 10^{-34}$
https://users.stlcc.edu/gkrishnan/ksptable.html
$3\times 10^{-34}$

Usually I'd first have a look at the CRC Handbook of Chemistry and Physics. Unfortunately, I couldn't find the data for aluminium hydroxide.


Do alkanes evaporate faster than alcohols? If so, is it because alchohols have hydrogen bonds?

Which type of compound evaporates faster, and why?

Yes, alkanes do evaporate faster than their corresponding alcohols (i.e. an alkane boils at a lower temperature than an alcohol with the same number of carbon atoms). This effect is due to hydrogen bonding between the alcohol hydroxyl groups.

Wikipedia says:

  Because of hydrogen bonding, alcohols tend to have higher boiling points than comparable hydrocarbons and ethers. The boiling point of the alcohol ethanol is 78.29 °C, compared to 69 °C for the hydrocarbon hexane (a common constituent of gasoline), and 34.6 °C for diethyl ether.


Why is bromine produced at the anode when aqueous sodium bromide undergoes electrolysis? The electrode potential for Bromine is $\pu{+1.07V}$ whereas for Oxygen to be produced it is $\pu{+0.40V}$.

$$\ce{O2 +2H2O + 4e- -> 4OH-}$$

As the electrode potential for bromine is higher, it means bromine should be reduced rather than $\ce{OH-}$ ions.

I'm aware that the concentration affects the electrode potentials, but the question simply states 'an aqueous solution of NaBr' with no reference to concentration whatsoever, so I can only assume standard solutions.

Also, why do we not look at the electrode potentials of other possible reactions at the anode, like;

$$\ce{O2 + 4H+ + 4e– -> 2H2O}~~~~~~~(E=\pu{+1.23V})$$ or

$$\ce{HO2- + H2O + 2e– ->3OH–}~~~~~~(E=+\pu{0.88V})$$ or

$$\ce{H2O2 + 2H+ + 2e– -> 2H2O}~~~~~~~ (E=\pu{+1.77V})$$.

Why are these not viable?

This is  how I see it:

In the electrolysis of NaBr, water is reduced at the cathode. This occurs because water is more easily reduced than are sodium ions. This is reflected in their standard reduction potential.

At cathode reduction of water occurs:

$$\ce{2H2O(l) + 2e- -> H2(g) + 2OH-(aq)}$$

And hydrogen gas is produced

At the anode, where oxidation occurs, the standard oxidation potential of water is –1.23 volts, while that for bromide ions is -1.07 volts.

The production of bromine itself has a negative electrode potential.
One of the half-reactions must be reversed to yield an oxidation.

$$\ce{Br^2- -> Br2 + 2e-}~~~~~~~ Eº -1.07$$

Remember that when one reverses a reaction, the sign of Eº (+ or –) for that reaction is also reversed.

This means that bromide ions are more easily oxidized than water.

It is important to note: When current begins to flow, the distribution of ions around the electrodes changes, and the equilibrium electrode potentials no longer accurately apply.

I think other factors also apply;

Concentration
Temperature
Nature of ions


Clarifying question about chloride ions, water softening & corrosion I have a question regarding the use of NaCl as a water softener.

Recently i was asked to comment on a series of test results from a water softening device were a lot of corrosion had occurred post softening. Just to be sure we are on the same page, NaCl was added to the water to dissolve* the CaCO3, Ca(CO3)2 & CaSO4.

*Ion exchange is not my field of expertise, so I'm not quite sure if this is the correct wording

Now, according to the folks who ran the test, there were no signs of ‘Chlorides’ in the water sample, what they mean by this is still unclear. I was also informed that no precipitate was collected. Now to the actual question: they wanted to know why the pipes used, post softening, had crevice corrosion inside the pipe. My first intuition was that there were still $Cl^-$ ions in the water which is causing the steel to corrode. However, the folks running the test assured me that the chloride had bonded with the calcium ions.

I’ve been trying to figure out what in the world could cause the pipes to corrode, assuming that they are correct, and the chloride have formed CaCl2.

Have you guys got any ideas? Have I misted something, or am i right to assume that the folks who ran the test don't know what they are talking about.

If you need any more information, feel free to ask. As stated earlier, this is not my field of expertise so if I'm wrong about something please correct me.

A traditional softener has a zeolite resin that "holds" positive ions. When it is charged the Na ions fill the resin. Then in service , the positive ions - Ca , Mg , etc , exchange places with the Na ions, putting Na in the water . In theory the back flush of the softener after charging, removes the Cl ions. So that in service the only Cl present would be what is in your raw water as the Cl is not affected by the softening. Water corrosion results from conductivity ( any ions) and oxygen; Although Cl  ions are especially active. Generally "waters" leave a CaCO3 type scale that forma a protective layer ( scaling ) that  reduces corrosion. Serious evaluation of water corrosion looks at the water "scaling index" to get a initial idea of how corrosive it may be. Softening water reduces the scaling index and promotes corrosion . I removed a softener from one house I owned for that reason; However it was Lake Michigan water which has a very good scale index - enough to be protective but not enough to restrict water flow. I also had to replace a couple sections of copper pipe that had corroded through in the 20+ years the softener had been in service. On the other hand , softeners are useful in very hard natural waters.


Would gold have the negative charge in gold phosphides? It should make sense that phosphorus has the positive charge,and gold itself
should have the negative charge in gold phosphide(and for any other gold phosphides) because phosphorus has a lower electronegativity than gold (phosphorus is 2.19, gold is 2.54),but then, they can both just share electrons. Is this true or not?

Because their electronegativity difference is less than 0.4, I think the bond would have a lot of covalent character. I personally don't think it is a good idea to treat this substance as an ionic compound and assign a positively charged ion and a negatively charged.
According to a chemical supply company, $\ce{Au}$ is +3 and $\ce{P}$ is -3.  We typically write the cation first, so from the formula $\ce{AuP}$, you should get another clue that the gold ion is positively charged (sort of if you have to assign a positively charged ion).
I don't think you can assign charge signs solely on electronegativities. Atomic radius and position on the periodic table also play roles in determining which atom is more positively charged.
For electronegativity, more reference here.


Why is 2H2O2 -> 2H2O + O2 a first order reaction? My book says that the reaction $\ce{2H_2O_2 -> 2H_2O + O_2}$ is a first order reaction and that it's rate equation is $\ce{k[H_2O_2][I^{-}]}$.

However i don't see why that would be first order? Isn't the order of the reaction supposed to be the sum of powers of the concentration of the reactants?

Summarizing from the excellent comments:

  Isn't the order of the reaction supposed to be the sum of powers of the concentration of the reactants?

Yes, that is how you get the overall order of reaction. If you compare the rate with a given set of concentrations to the rate with all concentrations doubled, the rate goes up by two to the power of the overall order.

  I don't see why that would be first order?

It is not, but it is pseudo-first order. If you start the reaction using certain initial concentrations, you will find an exponential decay of the reaction rate over time (i.e. there is a constant half-life).

As JM97 comments, the iodide is a catalyst, so its concentration does not change over time. So while the concentration of hydrogen peroxide drops over time and causes the reaction to slow down, the iodide concentration in the rate law does not contribute to a slowing down (no matter what order the reaction is in iodide).

  [OP responding to Ivan...] Actually the textbook statement is "_This reaction is first order with respect to both H_2 O_2 and I^- "

As Ivan comments, that means the reaction rate is proportional to the first power of the hydrogen peroxide concentration (if the iodide concentration is unchanged). It is also proportional to the first power of the iodide concentration (if the hydrogen peroxide concentration is unchanged). If both are changed at the same time, you use the rate law to see what the effect is.


Organic molecules with fast non-radiative decay rates What structural features in an organic molecule would promote fast non-radiative decay to a ground state, for instance in an aromatic π-conjugated system ?

Anything that has potential energy surfaces that cross or get close in energy. Offhand things like s-indacene and the indenofluorene family of compound, there are many others as well (azulene mentioned below). In general, if the $S_0$ and $S_1$ state for a $\pi$-conjugated organic come close in energy there is the possibility for all of that energy to be used on vibrations instead of luminescence. As far as predicive rules, that is a difficult thing to generalize. Perhaps you could argue that a low $S_0 \ce{->} S_1$ transition energy opens the possibility for non-radiative decay since these potential energy surfaces are not seperated by much energy, biradicaloid $\pi$ systems fall in this category.This is the same reason there is typically not emission from higher energy state, $S_n \ce{->} S_1$ transitions are usually quite close in energy and have crossings to get to $S_1$ efficiently through molecular vibrations. For further examples and explanations see, for example:
http://www.sciencedirect.com/science/article/pii/S0009261414008847?via%3Dihub
Or Josef Michl's excellent book "Excited States and Photochemistry of Organic Molecules" for more details on the topic.