Chemical formula and heat In my Physical Science class, we are learning about chemical formulas and how to balance them. I balanced this equation:

$$
\ce{Fe2O3 + Al -> Fe + Al2O3}
$$

One of the other students asked me if this chemical formula releases or absorbs heat.

Could anyone help me give a great explanation for this question?

The best way to solve this problem is to look at the $\Delta H$ charts found online. Using the simple equation below will give you the answer.

  $$\sum \Delta H_{\text{Products}} - \sum \Delta H_\text{Reactants} = \Delta H_{\text{Reaction}}$$
  If the $\Delta H$ is positive in sign then the reaction is endothermic and if it is negative in sign then it is exothermic.


Why does a reduced compound become a better electron donor? When a compound gains an electron, why are they more likely to give them up (i.e. become stronger reducing agents)?

Simple electrostatic repulsion can be used to explain this. For example, if you have $\ce{O^2-}$, there are 8 protons in the nucleus, but 10 electrons, so there is a net negative charge of 2. Like charges repel, pushing the electrons away.

BTW, compounds rarely "gain an electron", the reactive species are usually radicals.


BCS class of nifedipine So I had an exam today in which one of the questions was in what class (I-IV) of the Biopharmaceutical Classification System (BCS) the drug nifedipine falls in. Class I drugs are considered highly soluble and highly permeable, class II badly soluble and highly permeable, class III highly solube and badly permeable and class IV badly soluble and badly permeable.

  Solubility

  The aqueous solubility of a drug substance is considered as high according to the HHSFDA
  BCS criteria when:

  The ratio of the highest orally administered dose (in mg) to the solubility (mg/ml)
  is less than 250 ml
  This criterion is met over the pH range 1-7.5 at 37°C. [1]

  Permeability

  According to HHS-FDA a drug is considered a highly permeable, when more than 90%
  of the orally administered dose is absorbed in the small intestine.[1]

Obviously judging from literature it falls under BCS class II and you'd probably assume so if you look at the solubility (practically insoluble in water <1g/10000ml) and the permeability, which is good because it is absorbed well (>90%). [2] However the way the question was asked was interesting because of the information given. It explicitly said that a dose of 20 mg was the maximum dose and no exact solubility was given except that the solubility was "practically insoluble", meaning a solubility of < 1 g/ 10000 ml. Now whether the solubility is 1g /10001 ml or 1g/ 100002 ml, nobody knows... The extent of absorption given was 90% and thus high permeability was assumed. However when you calculate how much volume of water is needed to let this fully dissolve, you'd come with the answer 0,02 * 10000 = 200 ml water. This is smaller than 250 ml and thus one would be inclined to then assume the solubility is high according to the specifications stated above, which would lead you to think the drug falls under BCS class I. Of course in reality you'd have to take into account the exact solubility which probably isn't exactly 1 g/10000 ml and the exact volume of the gastrointestinal tract which probably isnt exactly 250 ml.

Now what am I supposed to believe in this situation? In this case the dose plays an important role in determining the BCS class and thus answering the question. Would you in this case have taken the rules for what they are and follow them like I did, taking into account dose and assuming the lowest solubility or would you have based it purely on the "practical insolubility in water" of the drug and conclude low solubility and thus BCS class II?
Maybe I'm thinking too hard about this, but I'm curious if anyone has something to say on this.
I personally think it would've been fair if the exact solubility (perhaps even a fictive solubility) of nifedipine and exact volume of GIT were given in this situation to make this question more clear.

Sources:

http://www.who.int/medicines/services/expertcommittees/pharmprep/QAS04_109Rev1_Waive_invivo_bioequiv.pdf
http://www.dissolutiontech.com/DTresour/200905Articles/DT200905_A01.pdf

The tl;dr answer is that, as the question is worded (and without an actual copy of the question or the data provided to you), its not possible to definitively assign a BCS class.

Nifedipine is a class II drug:

  class II is low solubility and high permeability. Formulation typically is used to enhance solubility of compounds in this class.

The data in your question is a little confusing, but you state that the compound is 'practically insoluble' which means that it must be either class II or class IV (practically insoluble definitely suggests low solubility).

With the information provided (i.e. nothing about permeability), you cannot distinguish between II and IV. Permeability, whilst often correlated to solubility, is not directly linked, such that you cannot attempt (as I think you have tried) to calculate one based on the other (how permeable something is depends upon structure, and whether it can get through the membrane).


Ratio of uncharged to charged side chains The $\mathrm{p}K_\mathrm{a}$ of the side chain imidazole group of histidine is $6.0.$ What is the ratio of uncharged to charged side chains at $\mathrm{pH}~7?$

Here's my attempt to the solution:

$$7 = 6 + \log\frac{[\ce{A-}]}{[\ce{HA}]}$$

$$1 = \log\frac{[\ce{A-}]}{[\ce{HA}]}$$

$$10[\ce{HA}] = [\ce{A-}]$$

$$\underset{\text{uncharged}}{1} : \underset{\text{charged}}{10}$$

I've used the Henderson-Hasselbalch equation to the solve this, considering imidazole group to be the acid $\ce{HA}$. However, the correct answer appears to be 10:1 uncharged to charged ratio. If I use $\ce{HA^+}$ to be the acid, the answer turns out correct.

Can anyone tell me what have I done wrong? If I was wrong in assuming $\ce{HA}$ please tell me why. Let me know if some clarification is needed.

I think you are simply misunderstanding the "uncharged to charged" part of the question. Imidazole is a base, so it gains a proton to become charged when the pH < pKa. That's why you got the right answer when you used $\ce{HA+}$ as the acid, as you put it. In this case, unlike with most general chemistry problems of this nature, the acid is charged while the conjugate base is uncharged.

Below pH 6, imidazole exists largely as a protonated imidazolium ($\ce{HA+}$) species. Above pH 6, it is primarily in the neutral ($\ce{A}$) imidazole form. When imidazole is protonated (i.e., in acid or $\ce{HA+}$ form) it is charged. When it is deprotonated (i.e., in base or $\ce{A}$) form, it is uncharged.

So the proper expression is $$7 = 6 + \log\frac{[\ce{A}]}{[\ce{HA+}]}$$ which should lead you to the correct answer:

$$1 = \log\frac{[\ce{A}]}{[\ce{HA+}]}$$

$$10 = \frac{[\ce{A}]}{[\ce{HA+}]}$$

So, the ratio of uncharged:charged species is $\frac{[\ce{A}]}{[\ce{HA+}]}$, or 10:1.


In the Nernst equation, does T refer to the temperature of the standard cell or the cell under non-standard conditions? I am doing an experiment where I am varying the temperature of the electrolyte of the anode half cell. I am trying to calculate the cell potential using the Nernst equation under standard conditions but I'm not sure what T to put in.

The Nernst equation is $\displaystyle E=E^\text{o}-\frac{RT}{nF}\ln(Q)$ where $Q$ is the ratio of activities, $a$, of products over reactants, e.g. for $aA+B\leftrightharpoons cC+D$ then $\displaystyle Q=\frac{a^c_Ca_D}{a^a_Aa_B}$. The emf of the cell is $E^\text{o}$ which is measured when the emf of the cell is zero, $E=0$, no current flows, and means that the reactants and products are at equilibrium. In this case $\displaystyle E^\text{o} =\frac{RT}{nF}\ln(K_e)$ where $K_e$ is the equilibrium constant.

The measured emf $E$ changes with temperature as indicated by the Nernst equation and this is a good way to measure thermodynamic properties; for example the reaction free energy is  $\Delta G = -nFE$ and because $\partial\Delta G/\partial T|_p=-\Delta S$ then $\displaystyle \Delta S = \frac{\partial E}{\partial T}\bigg|_p nF$ which is a convenient way to determine the entropy change in a reaction by measuring the reaction emf at two temperatures and letting the derivative be $\Delta E/\Delta T$, which will be accurate for a small change in temperature, and give $\Delta S$ at the mean of the two temperatures. Since $\Delta G = \Delta H -T\Delta S$ the change in enthalpy can also be calculated.


Identifying base in a chemical equation According to the Brønsted Lowry principle, "base" is a proton acceptor. I've tried to apply this idea to every choice as you can see these red circles in each equation below, but they are all loses protons, aren't they? How could it be a base?

However, if I choose water in a blue circle, I will get the underlined substance as a base for both choice B and C instead. Also, if you have a look at other substances involve in equation B and C, they do support the assumption that they act as acids that involved in donating protons. Hence, both B and C are correct. But the solution said the correct answer is C.

I would write down the half equations for each reaction:
A

$\ce{ HNO3 + H+ -> H2NO3+ }$ (Base)
$\ce{H2SO4 -> HSO4- + H+}$  (Acid)

B

$\ce{  HSiO3- -> SiO2 + OH- }$  (Base)
$\ce{HCN  -> CN- + H+ }$ (Acid)

C

$\ce{HNO2 -> NO2- + H+ }$ (Acid)
$\ce{HCO3- -> CO2 + OH- }$(Base)

D

$\ce{C6H5O- + H+ -> C6H5OH }$ (Base)
$\ce{CH2ClCO2H -> CH2ClCO2- + H+ }$(Acid)


How to calculate hydrogen from SMILES string? How to calculate the number of hydrogen atoms from SMILES string?

For example,

SMILES String: C1=CC(=C(C=C1C(CN)O)O)O

The known result is C8H11NO3

Other answers have done a good job of providing a description of the concepts and algorithms needed to answer the question.  Here I'll demonstrate the opposite: a black-box empirical approach.

In Python using rdkit:

from rdkit import Chem
from rdkit.Chem import rdMolDescriptors

my_smiles_string = 'C1=CC(=C(C=C1C(CN)O)O)O'
my_mol = Chem.MolFromSmiles(my_smiles_string)
print(rdMolDescriptors.CalcMolFormula(my_mol))

This shows the result C8H11NO3.  You could parse that further to get only the number of hydrogens, if you need to do so.  Alternately:

my_mol_with_explicit_h = Chem.AddHs(my_mol)
print(my_mol_with_explicit_h.GetNumAtoms() - my_mol_with_explicit_h.GetNumHeavyAtoms())

displays 11.  Slightly more cumbersome but it has the output you want.  If you take the latter approach you don't even need to import rdMolDescriptors.


Usage of ammonia as odorant in fuel gases Usually, thiols such as ethanethiol are added to fuel gases to warn the user of a leak. Why is a few percent of ammonia not used instead?

Ammonia is flammable so it will burn to nitrogen and water in the flame.
Ammonia has a strong smell, and causes a stimulation effect which may wake sleeping people in the event of a gas leak.
Ammonia produces no sulfur oxides when it burns (although the usual amount produced with ethanethiol is insignificant).

In principle, you could use ammonia, but there are practical disadvantages.

The sulfur-based odourants are some of the smelliest compounds known and are detectable in very low concentrations. In practice, they are used in concentrations of perhaps 30ppm. Several percents of ammonia would be thousands of times more than this concentration implying greater expense. At this level, the concentration of ammonia would be problematic because of the corrosive nature of certain types of pipe and would probably create other safety concerns (it might even be a direct safety hazard at the concentrations where it is readily detectable by smell).

Typical sulfur-based odourants are detectable at such low concentrations that they are not a direct safety hazard. Nor is there much worry about the combustion products at such low concentrations.

Ammonia is also much more reactive than the typical sulfur-based odourants and would be much more likely to be stripped from the gas during transmission which defeats the purpose.


Conformer generation for radicals Is there an automatic way to generate conformers for radical like systems (spin multiplicity is doublet)? My system for the study is a substituted fullerene with a missing H atom from the functional group.

I have tried using different conformer generators for this purpose (Baloon, RDKit, Obabel, Fullmonte, CCDC's Conformer_generator) but none of them successfully yielded out any conformers. I do not know how to run MD of such a system for which there is no general force-filed available. I have also tried to go through the AMBER tutorial but was unsuccessful: http://ambermd.org/tutorials/basic/tutorial4b/

I would really appreciate if anyone can suggest something that actually works for the system in consideration.

Thanks in advance.

tldr: Manually or script dihedral angle changes, then use a GFN or another DFTB method to find any/all conformers

Without seeing the compound, it's really hard to know the problem. Most programs, including those you mentioned, have a set of rules to identify "rotatable bonds" and suggest likely dihedral angles.

As you might expect, such rules are derived for neutral, ground-state organic molecules.

There are a few options:

As suggested by Jan Jensen's comments above, you can use RDKit, Open Babel or another tool to drive your own dihedral angle search.
As suggested by Martin's comments above, there are some tools in Grimme's xtb  program to sample conformations. These will use the GFN1 or GFN2 approximate density functional (pretty much DFTB + dispersion correction) to evaluate energies.

I'd probably suggest a combination of both. From what I understand of Grimme's conformer search scripts, it calculates the vibrational modes of the molecule and samples geometries along vibrational modes. This is a good idea, although in some molecules it may not sample all minima (e.g., if you have many conformations in a large molecule, you might never "reach" the whole potential energy surface from any given initial geometry).

So you can change the dihedral angles using your program of choice, generate a few initial structures, then use xtb to optimize the geometries and confscript to sample a few more geometries.

I feel pretty confident that in small to medium geometry molecules, this will do a good job.

A few notes:

If you're sampling a radical, you definitely want to use a good DFT or quantum method to evaluate the energies. I doubt any current force field will do what you want.
We haven't finished our evaluation, but right now, GFN2 is performing better than any current method on conformer energetics for neutral organic molecules. We haven't checked radicals, but I would trust it more than any other current "fast" method.
If you don't have the xtb package, the method and conformer sampling scripts are now available as part of ORCA 4.1 too. There's a description of this kind of conformer sampling in the manual when discussing prediction of NMR spectra.


How can I emulsify Beeswax with Aloe Vera Gel? I was looking thru ingredients for non-silicone based makeup primers and two ingredients always seem to be used: aloe vera gel and some type of wax, like jojoba wax, plus some type of oil to thin out the consistency like sweet almond oil or apricot kernel oil.

I thought it'd be interesting to see if it'd be possible to see if I could make my own, not least the companies that make them charge an arm and a leg for them.

My question is: what procedure and chemicals would you use to emulsify aloe gel with beeswax to make a cosmetic?

I vaguely remember that borax is used to make cold cream, maybe that'd come in useful?

Aloe vera gel is water based, whereas beeswax is, well, wax-based. As they say, "oil and water don't mix" -- at least, not on their own.

So, the trick is to make them willing to mix. To do this, you need what's called an emulsifier. These are chemicals (natural or synthetic) that have both oily and watery parts to them, and thus are able to make oil and water play nice(r) with one another.  Soap is one well-known, everyday example of an emulsifier; in particular, it's an example of a surfactant.

Anyways, you need an emulsifier. If you're looking to stick with natural products, probably the most readily available option is lecithin, commonly sourced from soy but also available derived from other plants such as sunflower.  While I don't have detailed recommendations, I would melt the beeswax together with lecithin and the aloe vera gel over very low heat, mixing thoroughly, then allowing the blend to cool.  You'll have to experiment until you find a ratio among these that gives the consistency you want in the final product.


Chemical formula of ethanol and other carbon compounds Should chemical formula of ethanol be written as $\ce{CH3CH2OH}$ or $\ce{C2H5OH}$? And whats the logic behind it? I have seen the former being used more often, so why is ethane represented by $\ce{C2H6}$ than $\ce{CH3CH3}$?

$\ce{CH3CH2OH}$, and $\ce{C2H5OH}$ or $\ce{EtOH}$ (Ethyl OH) tells you specifically what are the bonds in the molecule. $\ce{CH3H6O}$, which would be the equivalent of writing ethane like $\ce{C2H6}$, on the other hand, tells you nothing, since it can be both ethanol or dimethyl ether. Bigger formulas can represent thousends of different compounds if written like that.

The reason why $\ce{C2H6}$ is used instead of $\ce{CH3CH3}$ is because it can only represent ethane - it's a very simple compound with no isomers, so you can write it both ways.

You can write ethanol both ways you suggested, although I've never seen $\ce{C2H5}$ instead of $\ce{Et}$ anywhere but in examples. What you shouldn't do is write it like $\ce{C2H6O}$.

If you were asking what is the logic of $\ce{CH3CH2OH}$:

What it's telling you is that the compound it is representing [ethanol] consists of a carbon atom bonded to 3 hydrogen atoms and to another carbon atom, which is bonded to 2 hydrogen atoms and 1 oxygen atom, which is finally bonded to 1 hydrogen atom, as follows:


Disproportionation reaction of dimeric sulphur monochloride I read that group-16 monohalides exist in dimeric forms ex. $\ce{S2Cl2,Se2Cl2,Se2Br2}$ etc.
It also states that these dimeric halides undergo disproportionation reactions and gives an example as follows: $\ce{2Se2Cl2 -> SeCl4 + 3Se}$
Now I thought what would be the disproportion reaction if we consider the case of $\ce{S2Cl2}$ Since $\ce{S(VI)}$ is more stable than $\ce{S(IV)}$ would the disproportion reaction in case of $\ce{S2Cl2}$ yield $\ce{SCl4}$ or $\ce{SCl6}$?
EDIT:  As pointed out in the comments. $\ce{SCl6}$ doesnt exist hence the same question is now considering the case of decomposition of $\ce{S2F2}$ with the essence of question still remaining the same

Compiling my comments into an answer, sulfur monohalide disproportionate to sulfur dihalide and/or sulfur tetrahalide and elemental sulfur.
Wikipedia has the reaction:
$$\ce{2S2F2 ->[180 ^\circ C] SF4 + 3S}$$
Also, there is a reaction scheme from a 1974 entry1 which indicates formation of octasulfur and sulfur difluoride as sideproduct.
$$\ce{12S2F2 -> 3S8 + 8SF4}$$
$$\ce{8S2F2 -> S8 + 8SF2}$$
It also further noted that this disproportionation reaction is useful in the synthesis sulfur tetrafluoride and the decomposition of two moles of $\ce{S2F2}$ into one mole of $\ce{SF4}$ and sulfur yields $\pu{\Delta H^\circ = -24.6 kcal/mole}$.
Furthermore, sulfur difluoride may disproportionate to sulfur tetrafluoride and sulfur or some other oligomer/polymer of sulfur or back to sulfur monofluoride or isomerize to various sulfur compounds like $\ce{FSSF}$,  $\ce{SSF2}$ or $\ce{SF3SF}$. These compounds are extremely unstable and prone to hydrolysis:
$$\ce{16SF2 -> 8SF4 + 8S8}$$
$$\ce{3SF2 -> SF4 + S2F2 isomers}$$
The stability of $\ce{SF2}$ can be considered thermodynamically. The decomposition of two moles of $\ce{SF2}$  to yield $\ce{SF4}$ and $\ce{S8}$(gas) is $\pu{-75.7 kcal/mole}$. Another reason for this decomposition is polymerization which forms new sulfur-fluorine bonds. Also, it is noted that $\ce{SF4}$ is thermodynamically unstable and disproportionate to sulfur hexafluoride and sulfur:
$$\ce{24SF4 -> 16SF6 + S8~~~~\Delta H^\circ = -62.9 kcal/mole}$$
So, sulfur compounds in these intermediate oxidation state are found to be thermodynamically unstable and it is also influenced by its kinetic stability.
Sulfur monochloride disproportionation occurs in a similar way as noted in this patent:

Elevated temperature as used herein refers to temperatures between
about 75 and about 150C; and preferably between about 85 and about
138C. Higher temperatures can be used but the sulfur monochloride will
tend to disproportionate to form sulfur and sulfur dichloride. lf
higher amounts of sulfur in the sulfur monochloride product are not
objectionable, higher temperatures can be utilized

Notes and References

Advances in Inorganic Chemistry and Radiochemistry, Academic Press, 1974
There is a different pathway for disproportionation of selenium monochloride/bromide where it is seen to yield $\ce{SeX2}$ and $\ce{Se_{n}X2 (n =3,4)}$ unlike the general pathway you mentioned which yields $\ce{SeX4}$ and elemental selenium. See this paper for more details.


Making Ferric Chloride using HCl and Ferrum I am not chemist: I live in the part of the world that you can't find many things, but I have HCl and bunch of nails. I there way to make ferric chloride from these things?
If this doesn't work what else can I do to make ferric chloride?

You can find a glass beaker somewhere and simply drop in iron and diluted HCl. That's enough to make $\ce{FeCl2}$.

$\ce{Fe + 2HCl -> FeCl2 + H2}$

That forms hydrogen gas, which is very flammable, so it would be wise to do this outside. You might have to heat or reflux (heat the hell out of it while condensing whatever tries to evaporate out) depending on what concentration you use. Anyway, the resulting solution is gold-colored.

You'd then need to oxidize the Fe from a +2 state to a +3 state. I found this guide that recommends hydrogen peroxide (probably as a safer alternative to potentially stronger oxidizing agents.) It says it can take several days, so that's something to watch for.

Good luck! Again, I'd definitely do this outside or in a lab.

http://www.instructables.com/id/Make-Ferric-Chloride-for-etching-printed-circuit-/

Also, I've heard that not all nails are made of iron, so a density test could help ensure that.


How to determine the stereochemistry after elimination of anti-1-bromo-2-methylcyclohexane? The question is asking for the product and the answer is C.
My expected answer was E. Here's how I tried to solve it:
Since Br (the leaving group) is at the equatorial position, the structure would 'ring flip'.

Since $\ce{CH3}$ is pointing downwards, I assumed it would have methyl group point outwards (dashed), however, the correct answer is a wedge.

After elimination, how would we know whether the substituent would be pointing inwards, outwards, or even right on the plane? (no wedge or dash)

You diaxial intermediate is good and you can clearly see that there is no way to achieve a Zaytsev product in an $\mathrm{E2}$ elimination since that would require eliminating the methyl group. Carry on with the elimination and then you have a double bond in the top half of the molecule and the methyl group pointing downards — molecule E if inverted by a vertical plane of symmetry traversing the $\ce{C-Me}$ bond.

But the product molecules are drawn with the double bond on the bottom. Thus, you need to rotate the molecule by $180^\circ$. If you do this with a rotation axis in the paper plane (again one parallel to the $\ce{C-Me}$ bond), the double bond is in the correct position but the methyl group is pointing upwards.

You can verify this by determining the absolute configuration of the ipso-methyl carbon: it should be (R) according to your chair drawing and it is (R) in answer C.


Rationalising the order of reactivity of carbonyl compounds towards nucleophiles This is immediately following ron's answer from Why is a ketone more nucleophilic than an ester?

One of the most simplest questions you can ask, how can you rationalise the order of reactivity towards nucleophiles, which is given as

acyl halide > acid anhydride > aldehyde > ketone > ester ~ carboxylic acid > amide > carboxylate ion

There might be many concepts out there, that explain this order, the above linked question could only halfway satisfactory answer it in the kind of generality I am looking for.

I am looking here for any model that can be used to explain this reactivity.

I will also later on post an answer to this question using quantum chemical calculations, trying to generalise from example. The key points that will be included are stated in ron's answer linked above.

  I like Martin's posts, they make me think - and that is always a good
  thing. Here is where my thinking has taken me:

  I believe Martin's approach (basically frontier MO) is fundamentally
  correct. I would be more convinced that his actual numbers are
  meaningful if he were to calculate 1) the carbonyl carbon LUMO
  coefficient and 2) the HOMO-LUMO separation for the following series
  of compounds in order to see if his calculated values correlated with
  this order of reactivity of carbonyl compounds towards nucleophilic
  attack (we could use water as the nucleophile) - but Martin, I think
  such calculations might be a poor use of your time and resources.

While I agree with most of this, I do strongly disagree with the last statement. Which I will hopefully be able to prove soon.

In the meantime I welcome everyone to share their thoughts on the matter.

Here is my "old school" explanation.

Below is a drawing of the reaction coordinate for nucleophilic attack at a carbonyl carbon. The energy well for the starting carbonyl compound is shown on the left.  As the positively-polarized carbonyl reacts with (forms a bond with) the nucleophile we pass over a transition state and fall into a second potential well representing the tetrahedral intermediate.  $\ce{E_{act}}$ is the activation energy required to pass over the transition state.  To whatever extent a substituent might stabilize the starting carbonyl compound, the energy of the carbonyl compound will be lowered, and consequently $\ce{E_{act}}$ will increase.  The central carbon (it was the carbonyl carbon) in the tetrahedral intermediate is fully saturated (just $\ce{sp^3}$ bonds) and without charge, so the substituent X cannot interact through resonance to stabilize the tetrahedral intermediate.  In other words, while substituent X can stabilize or destabilize the starting carbonyl compound, it should have little effect on the relative energetic position of the tetrahedral intermediate.

Below are the resonance structures we can draw for our carbonyl compound.

If structure III contributes significantly to the overall description of the compound, then we would expect that carbonyl compound to be more stable, lower in energy, than a carbonyl compound where structure III contributes little.

Let's start by comparing the ester to the amide.  In both cases structure III involves a 2p-2p pi bond, but the resultant positive charge is placed on oxygen when we start with the ester and on nitrogen when we start with the amide.  Clearly, for electronegativity reasons, the amide case is better with the positive charge on the less electronegative amide nitrogen.  Our prediction: the amide is more stabilized than the ester and therefor should be the slower reacting of the two.

$$\ce{ester > amide}$$

In the case of the carboxylate anion, resonance structure III is equivalent to resonance structure I (note: there is no positive charge in structure III for the carboxylate case).  Since they are equivalent, they carry high weight stabilizing and in describing the carboxylate anion.  The carboxylate anion is so stabilized by resonance structures I and III that it is basically unreactive to nucleophiles

$$\ce{ester > amide >> carboxylate}$$

There's really no difference between the ester and carboxylic acid in terms of resonance structure III contribution so we would expect them to have similar reactivity towards a nucleophile.

$$\ce{ester \sim acid > amide >> carboxylate}$$

Let's compare the ester to the anhydride.  Resonance structure III is equally effective for both compounds, BUT, there are two carbonyls in the anhydride for the oxygen to interact with.  Therefor, the oxygen will only be half as effective at stabilizing any one anhydride carbonyl as it was for the ester carbonyl.

$$\ce{anhydride > ester \sim acid > amide >> carboxylate}$$

In the case of the acid chloride, resonance structure III involves a 2p-3p pi bond, not very effective overlap due to the size difference (also chlorine is electronegative so placing a positive charge on it is not desirable, but the poor pi overlap is what limits the contribution from structure III in the acid chloride case).

$$\ce{acid~ chloride > anhydride > ester \sim acid > amide >> carboxylate}$$

For the ketone, our resonance structure III has a slightly different look, it involves hyperconjugation as shown in the following figure.

Since there are 3 hydrogens, there are three such hyperconjugated structures,  They are not super significant resonance structures, but there are 3 of them, they do place the positive charge on a proton, so they do count for some stabilization.  To be honest, I would have guessed that it creates more stabilization than resonance structure III in the acid chloride case, but I wouldn't have known where it fits between the acid chloride and the ester.  Wherever I put it, I'd put the aldehyde to its left.  There is no resonance structure III for the aldehyde.

I wind up with
$$\ce{acid~ chloride > anhydride > ester \sim acid > amide >> carboxylate}$$
and $\ce{aldehyde > ketone}$ fitting in somewhere between the acid chloride and the ester


Is combustion considered a redox reaction? When carbon combusts with oxygen, is this considered a redox reaction since the oxygen atoms gain electrons and the carbon atoms lose them?

In general, yes.  If the reaction involves oxygen going from the oxidation state of zero in $\ce{O_2}$ to an oxidation state of -2, then there is oxidation and the reaction is a redox reaction.


Can we oxidize aliphatic alkane to produce carboxylic acid? Toluene can be oxidized to form benzoic acid. do aliphatic alkanes give similar kind of reaction?

It's tricky, because the bond enthalpies of sp3 C-H bonds are high (so they are hard to break) and because alkanes typically have a large number of CH bonds so selectively oxidizing any one of them can be challenging. Generally speaking the more electron-rich CH bonds (i.e. tertiary positions) are oxidized more easily that less-substituted, less electron rich ones.

TFDO will do it, see here: https://en.wikipedia.org/wiki/Oxidation_with_dioxiranes and there are a bunch of other methods (just try googling "aliphatic CH oxidation").

It's quite a hot area of research, since if you can control it then in theory you can turn feedstocks from crude oil into more useful functionalized molecules.


Does a difference in IR signal size from 'normal' provide information on amount of material? I have an IR spectrum that has signals for my expected product but also signals from what I believe to be some reactant (an alcohol). Typically, an O-H signal is pretty strong, but this O-H signal has an absorbance of only ~35%. Can I make a conclusion about the relative amount of reactant?

No. Frankly, in organic chemistry, IR is usually just used as support for what functional groups you think should be on your molecule. I only obtained IR spectra because you need to include them with supporting information when you want your paper reviewed.


How can we produce gasoline from electricity without fossil fuels? The other day, I was wondering, if cheap and easy nuclear fusion were developed tomorrow and fossil fuels disappeared, would there be a way to fuel all the gas-powered cars out there? After all, we have the Haber-Bosch process for producing ammonia. But then gasoline is not a chemical but a mixture of different organic compounds. So, let me structure a question to get at the heart of my curiosity:

How might we produce gasoline from electricity on an industrial scale without the use of fossil fuels as input?

Assume the electrical power came from an unlimited source, such as nuclear fusion, solar arrays in space, or very specific magic.

You may also assume that the gasoline doesn't need to be replicated perfectly -- that which can fuel anything that uses gasoline, or which satisfies an industry definition of gasoline, is sufficient. Ethanol does not count, since most cars cannot use it solely.

Feel free to comment on the environmental friendliness of the solution, excepting the electricity generation itself.

This basic question was originally asked here, and where it was suggested to migrate to chemistry SE.

Historically, this sort of process has been done with the Fischer–Tropsch process.

The core of this process is to use "syngas" (a mixture of $\ce{CO}$ and $\ce{H2}$) in the presence of a catalyst to make higher molecular weight hydrocarbons:

$$\ce{(2n + 1) H2 + n CO → C_{n}H_{2+2n} + n H2O}$$

Under normal conditions, this will actually make mostly straight-chained hydrocarbons, which are great for diesel fuel, but not so great for gasoline/petrol engines. However, there are standard catalytic cracking techniques in the petrochemical industry which can adjust the branchedness of hydrocarbons, making them more "gasoline like".

Now, the only question is where to get the $\ce{CO}$ and $\ce{H2}$ from? The $\ce{H2}$ is easy, as it is readily obtained from the hydrolysis of water.

The $\ce{CO}$ can be obtained from $\ce{H2}$ and $\ce{CO2}$ by the water-gas shift reaction (here technically run in reverse):

$$\ce{CO2 + H2 \rightleftharpoons CO + H2O  }$$

As I mentioned, the Fischer–Tropsch process was actually used during WWII by the Germans to produce motor fuel, as conventional sources were blocked off by the Allies. Their syngas didn't come from $\ce{CO2}$ and water, though, but from reformed coal, which they had a ready supply of.

This process is not done much because it's hugely inefficient, compared to just digging oil out of the ground. But if you have a "too cheap to meter" supply of electricity, the efficiency concerns go away. The main issue is getting a ready supply of $\ce{CO2}$, and there are a number of techniques being proposed at the moment for scrubbing $\ce{CO2}$ from the atmosphere - either by capturing it from point-source generators like power plants and cement manufacturers, or from "artificial trees" which capture $\ce{CO2}$ directly from standard atmospheric concentrations.

These "artificial trees" tend not to use normal photosynthesis for carbon capture, but instead often use something like a (strong) base in aqeous solution or $\ce{CaO}$. In these systems the base or the $\ce{CaO}$ reacts with dissolved $\ce{CO2}$ to form a carbonate salt, which either stays in solution or precipitates out. This carbonate salt formation is favorable, and happens spontaneously. The energy intensive portion of the process is the regeneration of the basic solution/$\ce{CaO}$. This normally happens by heating the carbonate salt, which causes $\ce{CO2}$ release. If done correctly, this can produce a stream of relatively pure carbon dioxide. But again, the regeneration step is quite energy intensive, and so would not be feasible without a cheap source of power.

Other environmental considerations would be minimal, aside from the land needed for the $\ce{CO2}$ devices (you would need a fair number to get a suitable steady volume of $\ce{CO2}$), and any disposal concerns about the process catalysts (which shouldn't be any more onerous than dealing with the catalysts now used in the petrochemical industry).


What is the reasoning behind the increasing pKa for organic compounds? Arrange the following according to increasing $\mathrm{p}K_a$.

Cyclohexanol
Phenol
Cyclohexane carboxylic acid

Please tell me the logic along with the solution.
I am confused between phenol and cyclohexanol because i guess phenol is resonance stabilized-the phenoxide ion

Cyclohexanol is an alcohol. An alcohol contains an $\ce{OH}$ group connected to a tetrahedral carbon.
Phenol is slightly different; it contains an $\ce{OH}$ group connected to a trigonal planar carbon that is part of an aromatic ring.
cyclohexanecarboxylic acid contains a very different functional group, carboxylic acid.

Cyclohexanol has a $\mathrm{p}K_a$ of about 18. It is less acidic than water. It can give up a proton, but the proton is much more likely to be bound to the oxygen than disscociated. Phenol has a $\mathrm{p}K_a$ of about 9 and is considered mildly acidic. cyclohexanecarboxylic acid has a $\mathrm{p}K_a$ of about 4.9; its acidity is similar to hydrofluoric acid.

Let's compare stability of the conjugate bases of cyclohexanol, phenol and cyclohexanecarboxylic acid:

In cyclohexanol, the anion formed by loss of a proton is localized on the oxygen. There is no resonance stabilization.
In phenol, the anion formed by loss of a proton is delocalized. Resonance structures show the negative charge can be shared between the oxygen atom and three of the carbons in the benzene ring.
In cyclohexanecarboxylic  acid, the anion is also delocalized. This time the negative charge is shared between two different oxygen atoms.

Resonance delocalization plays a clear role in stabilizing the conjugate base formed after loss of a proton. Despite the similar bond polarity in cyclohexanol and phenol, the proton is much more tightly bound in cyclohexanol. The greater stability of the phenolate anion compared to the cyclohexanoxide anion makes it easier to remove a proton from phenol than cyclohexanol.

The atoms onto which the charge is delocalized still plays a major role in determining stability. Delocalization onto an additional oxygen atom may be more stabilizing than delocalization onto several carbons.

For more details, please visit: http://employees.csbsju.edu/cschaller/principles%20chem/acidity/acid%20distal.htm


Spectroscopy pyrazine molecule Considering pyridazine and pyrazine molecules, I am asked to say if I can distinguish one from the other in a sample.

I supposed that is possible because pyrazine belongs to D2h and so it has the inversion center. Thus, in IR spectroscopy I can appreciate the antisymmetric stretching while in Raman I can identify the total-symmetric ones.

Then I am asked to say if this spectroscopic technique allows me to predict quantitive results. and how?

This question floored me because I have never done spectroscopic experience!

You need to understand some details of using point group tables. The pyrazine  belongs to  $\ce{D_{2h}}$ point group and pyridazine belongs to $\ce{C_{2v}}$
In the third column of the point group are found x, y, z which are used to identify dipole transitions, e.g. vibrational symmetries that can exhibit  IR transitions and in the fourth $x^2, z^2, xz$ etc. which are used to identify Raman transitions. If there is a squared term in column 4 and in the top line of the table (totally symmetric representation) labelled $\ce{A_g, A_1}$ then a Raman transition is totally polarised. Raman transitions from other symmetry species vibrations are not fully polarised. (Ignore the $\ce{R_x, R_y,R_z}$ they do not refer to IR or Raman transitions)
The point groups are shown below

In $\ce{C_{2v}}$ all vibrational species $\ce{A_1}$ etc. can have Raman transitions but in the IR the $A_2$ does not have an IR transition, no x, x, z in column 3. The other three types of vibrations could have both IR and Raman transitions. In the $\ce{D_{2h}}$ you can see that, because of the centre of inversion, Raman and IR vibrational transitions are mutually exclusive (compare symmetry species of $x^2, xy$ etc. with x, y, z ) because there is no vibration with a symmetry species that can generate both IR and Raman transitions.
Thus, with your compounds, the one with $\ce{C_{2v}}$ symmetry will have IR and Raman transitions with the same frequency, but for the $\ce{D_{2h}}$ compound there should be no common IR and Raman frequencies.
(See the answer to this question Understanding group theory easily and quickly more details on using point groups)


Allotropy in Nitrogen and Bismuth of Group 15 Which elements of Group 15 of the periodic table have allotropes?

P, As and Sb have well-known allotropes:

P: White, Red and Black;
As: Yellow, Black and Gray (Grey);
Sb: White, Yellow and Black.

What about Nitrogen and Bismuth?

The same source suggests that Nitrogen has an allotrope named Dinitrogen while Bismuth has none.
Chemwiki says:

  $\ce{N2}$ does not have any allotropes.

and doesn't comment on Bismuth.
Royal Society of Chemistry says yes for Nitrogen and no for Bismuth.
Inorganic Reactions and Methods: The Formation of Bonds to N, P, As, Sb, Bi (Part 1), Volume 7 has an article which suggests that Bismuth allotrope exists.
This not so reliable source says no for Bismuth.
Wikipedia doesn't mention anything for Bismuth but for Nitrogen:

  At atmospheric pressure, molecular nitrogen condenses (liquefies) at $\pu{77 K}$ ($\pu{−195.79^\circ C}$) and freezes at $\pu{63 K}$ ($\pu{−210.01^\circ C}$) into the beta hexagonal close-packed crystal allotropic form.

  Unstable allotropes of nitrogen consisting of more than two nitrogen atoms have been produced in the laboratory, like $\ce{N3}$ and $\ce{N4}$. Other (as yet unsynthesized) allotropes include hexazine and octaazacubane.

According to me, nitrogen has allotropes and bismuth doesn't, but the above sources contradict. Any clarifications and reasons for them showing or not showing allotropy would be appreciated.

Bismuth certainly has other known crystal structures at elevated pressures and temperatures, at least 4 others in addition to the rhombohedral structure stable at room temperature and pressure.  One place to start would be an article from NIF on shock physics of bismuth. For a scholarly article, you might start with 'Phase Diagrams of Arsenic, Antimony, and Bismuth at Pressures up to 70 kbars', W. Klement et al., Phys Rev 131(2) 632-637 (1963).

As for nitrogen, it too would appear to exhibit a variety of solid diatomic phases under pressure.  A starting point to explore might be 'High-pressure amorphous nitrogen', E. Gergoryanz et al., Phys Rev. B64 052103 (2001), which mentions a variety of phases in addition to the titular amorphous phase they explore.

Both Physical Review articles should be available to the general public...


Dissolve Silver Oxide I was looking at how to dissolve silver oxide (in a solvent). I know that monovalent silver oxide will not dissolve in water, but are there any solutions in which monovalent silver oxide will dissolve? I was looking whether they would do so in alkaline solutions, but I could not find anything.

Well, $\ce{Ag2O}$ is just a basic oxide. As such, it would dissolve in suitable acids ($\ce{HNO3}$ would do), but I guess that's not quite what you want. Well, some metal oxides ($\ce{ZnO}$, for instance) are amphoteric and thus would dissolve in $\ce{KOH}$ as well, via formation of hydroxo complexes. Sadly, this is not the case with $\ce{Ag2O}$. Then we must resort to some specific complex which would be stable in alkaline media.

Indeed, there is such complex: $\ce{Ag(NH3)2+}$, the active component of Tollens' reagent, which is prepared by interaction of $\ce{Ag2O}$ with aqueous ammonia, like I said. If you are actually going to do it, read the page carefully, including the "Safety" section, or at least don't make that solution by gallons and don't leave it unattended for years. Also, note that it is normally made using freshly prepared $\ce{Ag2O}$; dry reagent off the shelf would likely dissolve as well, only much slower.


Why do chloroform and benzene exhibit negative deviation from Raoult's law? Chloroform is a polar molecule, and benzene is nonpolar. Shouldn't the chloroform-chloroform and benzene-benzene intermolecular forces be stronger than chloroform-benzene interactions (like dissolves like), which would result in a positive deviation from Raoult's law?

This is because when these liquids are mixed, H-bonding type interactions are formed between hydrogen atom of chloroform (partial positive charge due to 3 Cl atoms) and 'pi' electron cloud of benzene ring.
Thus, chloroform-benzene interactions are stronger than chloroform-chloroform and benzene-benzene interactions.


Why does this region fit NMR spectroscopy technique? I was reading a paper and came across a sentence (highlighted one) about NMR spectroscopy.

  Several Leishmania Rabs, like Rab1, Rab5a, Rab5b, and Rab7, have been functionally characterized. LdRab5a is a GTPase that has been shown to
  bind to GTP and hydrolyze it to GDP. It specifically regulates
  fluid-phase endocytosis. Although the G1–G5 loop sequences
  are well conserved in LdRab5, it has unique
  interswitch region with substitutions and a 20-residue insertion
  with respect to the human Rab5. The unique insertion
  feature of LdRab5a prompted us to choose NMR spectroscopy as the most suitable technique for determining
  its structure.

  It is pertinent to mention that NMR solution
  structure has not been determined for any Rab protein
  from any organism. This would also provide the opportunity
  for exploring the residue-specific dynamics for LdRab5a in
  particular and reveal the general implications for the Rab
  family of proteins. Hence, we have undertaken the determination
  of solution structure and characterization of backbone
  relaxation dynamics for LdRab5a by using NMR
  spectroscopy.
  However, stabilization of LdRab5a through systematic
  and specific mutations and deletions was necessary to arrive
  at a construct that was stable enough for recording long
  multidimensional NMR experiments.

Why do we need to have a unique interswitch region so as to use the NMR spectroscopy technique?

The insert in the Ld ortholog occurs in a beta sheet and suggests the presence of additional strands in that sheet compared to the human variant. The structure of the human protein has been determined by crystallography. It would therefore appear that it should be possible to determine the structure of the Ld protein the same way (via crystallography), provided a crystallizable construct is identified.

Based on the available reference provided, the insertion feature from LdRab5a appears to make the protein unstable in vitro and difficult to crystallize, leaving NMR as the only good alternative for a structure determination. However, it seems the insertion feature also rendered the protein too unstable to be able to complete long NMR experiments, so the authors chose to remove the insert in order to stabilize the protein and perform NMR.


What is the velocity of an ejected electron in lithium when subjected to UV radiation? If UV radiation strikes an atom of lithium with one electron in its lowest allowed energy state with enough energy to eject the electron from orbit, what is the electron's velocity upon ejection? The wavelength is 7.1 * 10^-9 meters.

If this was hydrogen, I would use the fact that the energy of the radiation is the sum of the binding energy and kinetic energy of the electron, but the conditions are for lithium. How would I solve this problem given that it's lithium and not hydrogen?

The energy is still the sum of the kinetic and the binding energy. The only difference is how you calculate the binding energy, but you could just look it up


Baking Soda and Vinegar: Are there any secondary reactions? I want to run a Lego Pneumatic Engine on the CO2 that is produced, but before I do, I want to be sure that that's the ONLY gaseous product from a real-world reaction.  Everyone seems to assume that it is, just by omission, but no one actually says it.

I'm sure it would run, but I'd rather not damage the ABS plastic or probably-steel parts (shiny metal piston shaft), or a rubber seal or the factory lubricant or something else that I can't see.  CO2 seems inert enough by itself, but is that really ALL I'll get?
(assuming of course, that I keep the foam out)

It seems to me, according to the comments on the question, that there are NO additional reactions or products to worry about.

However, because of the foam, some of the original reactants can be tossed up, and find their way into the later parts of the apparatus as a haze/fog/cloud that is not completely inert.

Is that more-or-less accurate?  I'm trying to get an answer here, and not just a string of comments.


Ammonium Hydroxide Name Why is an aqueous solution of $\ce{NH3}$ (ammonia) often referred to as $\ce{NH4OH}$?
I know that dissolved ammonia gas reacts with water in the following equilibrium:

$$\ce{NH3 + H2O <=> NH4+ + OH-}$$

However, $K_\mathrm b = 1.8\times10^{-5}$. Since only a small amount of $\ce{NH3}$ dissociates, why is the whole solution called "ammonium hydroxide"?

Is ammonium hydroxide soluble, or is it an actual aqueous species?

The fictitious substance “ammonium hydroxide” $(\ce{NH4OH})$ was suggested in accordance with the Arrhenius theory. In 1882–1883, Svante Arrhenius had researched into the conductivity of electrolytes. His theory of electrolytic dissociation was discussed in his dissertation in 1884 and published in refined form in 1887. In 1903, the Nobel Prize in Chemistry was awarded to Svante Arrhenius “in recognition of the extraordinary services he has rendered to the advancement of chemistry by his electrolytic theory of dissociation”.

According to the Arrhenius definition, acids are compounds that dissociate and release hydrogen ions $(\ce{H+})$ into the solution:

$$\ce{HCl -> H+ + Cl-}$$

Bases are defined as compounds that dissociate and release hydroxide ions $(\ce{OH-})$ into the solution:

$$\ce{NaOH -> Na+ + OH-}$$

The need for a hydroxide group in bases according to the Arrhenius definition led to the proposal of $\ce{NH4OH}$ (i.e. “ammonium hydroxide”) as the formula for hydrated ammonia in water. (Note that this problem does not exist in the definition of bases according to the Brønsted–Lowry theory, which was proposed in 1923.)

However, ammonium hydroxide cannot be isolated; the fictitious solid compound does not exist. Nevertheless, the misleading traditional name “ammonium hydroxide” is still widely used for solutions of ammonia in water.

Actually, a solid compound (melting point: −79 °C) with the correct atomic composition (molecular formula) can be crystallized from aqueous solutions of ammonia. But even this compound is not ammonium hydroxide $(\ce{NH4OH})$. It actually is a real hydrate $(\ce{NH3.H2O})$ in which the molecules are linked by hydrogen bonding.


Are there alternative ways to find n or number of e- in problems like the one pictured? I need help understanding how we are to come up with 12 e- because according to the directions, we are supposed to focus on any element, but it doesn't tell us what to look for. I see how carbon has a -2 value in the reactants side and a +4 on the products side. the difference is 6 e- and we are suppose to multiply that by the number of carbons.

But is there another way to calcualte to find 12 e- ?

Couldn't we add the charges from the reactants and products after multiplying them respectively with the subscripts? For instance: -2 X 2 = -4 or 4 e- and +4 X 2 = 8 or 8 e- and so 4 e- + 8 e- = 12 e- ?

Isn't this a valid substitute than the explanation below? Because the directions below ask us to look for the difference in any of the elements. But this doesn't explain how to calulcate for hydrogen. But in the explanation I prepared above does, doesn't it?

But then again, this doesn't explain how we can calculate for oxygen, which has a 0 charge in it's own gas phase and a -2 in water and -2 in carbon dioxide.

So what would be another way to calculate for -12 e- ?

In this case, we've to calculate the oxidation states of carbon in the reactant and product side separately, the difference between those two and finally multiply it with the number of carbon atoms. While this is may be a seemlingly simple calculation, things can get uber-tricky because of the mammoth number of exceptional cases. Let's start with the basics; here's a gift for you:

$${\begin{array}{|l|c|} \hline
\pu{Element} & \pu{Oxidation Number} & \pu{Remarks} \\\hline
\pu{Alkali Metals} & \pu{+1} & \pu{Always} \\
\pu{Alkali Earth Metals} & \pu{+2} & \pu{Always} \\
\pu{Flourine} & \pu{-1} & \pu{Always} \\
\pu{Halogen Family} & \pu{-1} & \pu{Generally} \\
\pu{Oxygen Family} & \pu{-2} & \pu{Generally} \\
\pu{Hydrogen} & \pu{+1} & \pu{Generally} \\\hline
\end{array}}$$

Whoa! I see three exceptional cases already (remarked "generally" in the table) and things haven't started yet. What's up with them?

Oxygen: Oxygen doesn't always exist as an oxide ion $(\ce{O^2-})$ in its compounds, it can often-times exist as

Peroxide $(\ce{O_2^2-})$: Oxidation state in this case is $-1$. Popular examples include $\ce{H2O2}$, $\ce{Na2O2}$, $\ce{BaO2}$, etc.
Superoxide $(\ce{O_2^-})$: Oxidation state is $-0.5$. Common examples $\ce{KO2}$, $\ce{RbO2}$, $\ce{CrO2}$
Ozonide ($\ce{O3-}$): $\frac{-1}{3}$ oxidation state; for instance in $\ce{KO3}$.

Halogen atoms (except flourine): When $\ce{Cl, Br or I}$ combine with elements of higher electronegativity ($\ce{O and F}$), then these halogen atoms can show positive oxidation numbers; commonly $\pu{+1, +3, +5, +7}$. In $\ce{BrF5}$, that's an inter-halogen compound, $\ce{Br}$ is in $+5$ oxidation state.

Hydrogen: In case of metal hydrides,  it can show $-1$ oxidation state. As in $\ce{NaH}$.

Phew, er any more exceptions? You dare ask? Metals. Wait! Aren't metals electro-positive!? Yes, they are, but in some cases they can show zero or even negative oxidation numbers. For example in $\ce{CsAu}$, gold shows a $-1$ oxidation state. Let's bring in coordination-compounds: nickel in $\ce{[Ni(CO)4]}$ finds itself in zero oxidation number.

There are plenty of them, I'm saving them for another day.

Let's try calculating things you've asked.

For ethanol ($\ce{C2H6O}$) we don't know the oxidation state of the carbon atom. Why don't we assume it to be $x$. Now, the oxidation state of a compound is either zero (if it's a neutral species) or it's equal to the charge it bears. Is ethanol neutral? Yes it is. So the individual oxidation numbers of elements (times the number of elements bearing that oxidation state) should sum up to be zero.

$$2x+6(+1)+1(-2)=0$$

Solving, we get $x=-2$. Doing the same thing for carbon-dioxide, we can get $x=+4$. So a carbon atom changed its oxidation state from $-2$ to $+4$, that's a jump of $+6$. And how many carbon atoms were doing this? Two carbons. That gives us $n=12$.


Calculate the molarity of Na+ ions in 20 mL of a .130 M solution of Na2Cr2O7? The answer is $\pu{0.260 M}$, but I keep getting $\pu{0.0052 M}$ because I thought the right way to do it was by multiplying $\pu{0.130 M}$ by $\pu{0.02 L}$, then multiplying by $2$ moles of $\ce{Na+}$. What would be the right way?

$0.0052$ is the number of moles of $\ce{Na+}$ ions in solution, but to find the molarity, you need to divide by $\pu{0.02 L}$. Therefore, the final answer is $\ce{2\cdot 0.130 \cdot 0.02 \cdot \frac{1}{0.02} = 0.260 M}$


Is methanol more acidic than water? Methanol is slightly more acidic than water. Their $\mathrm{p}K_\mathrm{a}$ values, in water, are $15.5$ and $15.7$, respectively. All other aliphatic alcohols, however, are less acidic than water.

Is the following reasoning correct? This is my best rationalization; is there anything better or anything that can be added?

In both hydroxide and methoxide, we have an $\ce{O}$ bearing negative charge. In the former, we have a hydrogen attached to the $\ce{O}$. In the latter, we have a carbon attached to the $\ce{O}$. Carbon is more electronegative than hydrogen, therefore, carbon should be able to better withdraw electron density via induction from the $\ce{O}$. This is a stabilizing interaction.

We often think of methyl groups as inductively donating, but that's molecular profiling. They can inductively donate to a carbocation, partly because a carbocation is highly electronegative, so the carbocation pulls the electrons toward itself pretty well (so there's both pushing and pulling of electrons going on). When you have a methyl group attached to an $\ce{O}$, however, the methyl group is often inductively withdrawing. NMR data supports this.

Now the question is why aren't other aliphatic alcohols more acidic than water? In longer-chain aliphatic alkoxides, you don't just have a methyl group attached to the $\ce{O}$ bearing the negative charge - you have a bunch of methyl groups strung together. These lessen the amount of inductive withdrawal that the alpha carbon can do. Each$\ce{-CH2 -}$ unit attached to the alpha $\ce{-CH2 -}$ is somewhat inductive donating to the alpha $\ce{-CH2 -}$.

The relative acidities of these molecules in the aqueous solution$^{[1][2]}$ is:

$$\text{methanol > water > ethanol > isopropanol > t-butanol}$$

The relative acidities of these molecules in DMSO$^{[3]}$ is:

$$\text{methanol > ethanol > isopropanol > t-butanol > water}$$

The problem with only considering the acidities of these molecules in aqueous solution is that it assumes only one explanation for acidity. The inversion of the order of acidity in DMSO indicates that there are multiple factors at play.

The acidity trend in DMSO is explained by polarizability and the anionic hyperonjugation effect$^{[4]}$ of the conjugate bases of the molecules. The larger size of the alkyl substituents in the conjugate bases allows for increased distribution of negative charge over a larger volume. This reduces the charge density, and thereby the Coulombic repulsion between the conjugate base and the solvent. In DMSO, water is much less acidic than even methanol, which is also consistent with the relative polarizability of a methyl group and a hydrogen atom, but fails to explain the relative acidity of t-butanol and methanol.

The trend in alcohol acidity is explained by anionic hyperconjugation, which is weakened in larger alkyl substituents.

  For methanol and the methoxide anion, the primary interaction occurs between a filled lone-pair ($n$) orbital on oxygen and an unoccupied $\pi^*_\ce{Me}$ orbital of the methyl group, resulting in a two-electron stabilizing effect.

  $\hspace{3.8cm}$

  Deprotonation of the alcohol leads to an increase in the energy of the n orbitals on oxygen, and a decreased energy separation between the interacting orbitals $n$ and $\pi^*_\ce{Me}$, and hence a greater stabilizing interaction$^{[5]}$.

Although DMSO does solvate the solute to an extent, because is an aprotic solvent, it is incapable of hydrogen bonding with the solute. This allows DMSO acidities better reflect the intrinsic acidities of the molecules.

In aqueous solution, however, the conjugate bases are stabilized by hydrogen bonding. Because smaller ions are better solvated by water, water's acidity is increased more than methanol's, but because methanol is intrinsically more acidic due to its polarizability, it is still slightly more acidic than water, even in aqueous solution. As the size of the alkyl substituent increases, the alcohol becomes weaker due to the decreasing ability of water to solvate the conjugate base. This has a greater affect on the alcohol's acidity than the dispersion of the negative charge, and overrides the the intrinsic acidity, inverting the order of acidity.

$^{[1]}$ Evans $\mathrm{p}K_\mathrm{a}$ Table, Alcohols
$^{[2]}$ Wikipedia, Ethanol
$^{[3]}$ Bordwell $\mathrm{p}K_\mathrm{a}$ Table, Alcohols
$^{[4]}$ ACS, Acidities of Water and Methanol in Aqueous Solution and DMSO
$^{[5]}$ ACS, Does a Methyl Substituent Stabilize or Destabilize Anions?


Are there amide versions of lactides and lactones? In my book, I read that Lactides and lactones are cyclic esters which contain two and one ester group in them, respectively. They are formed when $\alpha$-hydroxy; and $\gamma$- or $\delta$- hydroxy carboxylic acids; respectively, are esterified.

Now, there exists a certain similarity between amides and esters (both being carboxylic acid-derivatives), such as the similarity between ester bonds in polyesters and amide bonds in polyamides.

In this context, I have a question: Do amides have compounds similar to lactides and lactones? What are they commonly known as?

Please let me know a good source that I could refer to in addition to any answers posted, or please cite any sources used.

Amide analogue of lactons are lactams, their tautomeric forms are lactims. Citing from the IUPAC Nomenclature of Organic Chemistry (Preferred names 2013):

  P-66.1.5.1 Lactams and lactims

  Intramolecular amides of amino carboxylic acids, $\ce{-CO-NH\bond{-}}$, are
  called ‘lactams’ and their tautomers, $\ce{-C(OH)=N\bond{-}}$, are ‘lactims’. Lactams named in two ways:

    (1) as heterocyclic pseudoketones;

    (2) by substituting ‘lactam’ for the ‘ic acid’ ending of the systematic ‘oic acid’ name for the parent acid without the amino substituent, and inserting a locant designating the position of the amino group between the ‘o’ and the ‘lactam’. Method ‘lactams’. Lactims are named in the same way, using ‘lactim’ in place of ‘lactam’.

  (1) generates preferred IUPAC names.

  Examples:

  pyrrolidin-2-one (PIN)
  butano-4-lactam

  (…)

  3,4,5,6-tetrahydropyridin-2-ol (PIN)
  pentano-5-lactim

(Note that the numberings depicted are for the preferred names (PINs) based on nitrogen heterocyclics)

There's no (at least IUPAC) term for lactides analogue (‘lactide’ is not used in IUPAC names themselves anyway), but they exist.

In traditional or general names, the Greek letter numbering is used, e.g. ε-caprolactam, or β-lactam four-membered ring part in bicyclic penicillin skeleton.


What is the Italian ‘Acido Psammico’ (Psammic acid?) I am reading an Italian detective novel — Lo stato delle anime by Giorgio Todde — in which someone is poisoned with ‘acido psammico’.

  Marini trova nello stomaco della vittima un ostia con all’interno dell’acido psammico (usato per conciare le pelli), l’acido ha ucciso la donna avvelenandola.

Which I think translates into English as:

  In the stomach of the victim Marini finds a communion wafer, inside of
  which is ‘psammic acid’ (used to tan hides): the acid had killed the
  woman by poisoning.

I have spent quite a time with dictionary and Google searches using the Italian, French (acide psammique) and English (psammitic acid comes up but is not defined) terms without success. The term ‘psammico’ seems to come up in a geological context, but infrequently and without clarifying things.

I should add that the novel is set in rural Sardinia in the 1890s and the implication is that shepherds would have access to it.

Does anyone know what the modern name and chemical composition of this is?

Addendum: Further Information

I am now reading further in the book (slowly, very slowly) and on p. 101 the protagonist (Marini) is in the capital city (Cagliari) and enters a chemist’s  (pharmacist’s) shop to ask for ‘acido psammico’. He receives the following reply:

  La sabbiolina mortale?

In which ‘sabbiolina’ (not in my dictionaries) appears to be a diminutive related to ‘sabbia’ — sand. (‘mortale’ = ‘deadly, fatal’, of course.)

Consistent with this, further down the page we have:

  La sabbia velenosa…

“The poisonous sand…”

Actually, the meaning is:

used to tan leather.

The acid used traditionally in tanneries is tannic acid, from there its name.

Tannic acid can be found in several plants one of which is the Sicilian Sumac (Rhus coriaria). This plant is also found in Sardegna (as per Flora Italiana), where the novel is placed.
The Italian name for this shrub is Sommacco siciliano.
The description of this plant is:

piccolo albero tipico di luoghi sassosi e rupestri, con fiori giallastri in
pannocchie, le cui foglie e corteccia, ricche di tannino, si adoperano nella
concia delle pelli (fam. Anacardiacee)

[the last part says: the leaves and bark, rich in tannin, are used in the tanning of leather.]

From sommacco to psammicco there is not too much distance.
But I agree that it is speculative.

Tannic acid is toxic, but the oral LD50 in rats is 2.26±0.083 g/kg body weight, so too much to hide in an ostia.
However, that can be a writers license.

Tannic acid is a complex polyphenol with structure:


Why does Co(I) have a 3d8 configuration? Why is the electron configuration of $\ce{Co^+}$ $[\ce{Ar}](\mathrm{3d})^8$?
Since neutral $\ce{Co}$ itself has a $[\ce{Ar}](\mathrm{4s})^2(\mathrm{3d})^7$ configuration, wouldn't the ionised electron be lost from the $\mathrm{4s}$ orbital, leading to an electron configuration of $[\ce{Ar}](\mathrm{4s})^1(\mathrm{3d})^7$?

For lighter elements, the shells fill in order. Starting at the transition metals, an outer s orbital may fill before an inner d orbital, so the electron configuration of unioninzed cobalt is written $\ce{[Ar]}4\mathrm s^1\,3\mathrm d^7$, rather than $\ce{[Ar]}3\mathrm d^7\,4\mathrm s^1$.
There is a video diagramming the electron configuration of $\ce{Co}$, $\ce{Co^{2+}}$ and $\ce{Co^{3+}}$, thought it does not explain the reasoning, nor does it cover the less common $\ce{Co^{+}}$ ion, produced by photoionization or as found in some esoteric metal-organic compounds, or in the theoretical $\ce{CoCl}$.
That said, the situation becomes murky for transition elements, and downright turbid for lanthanides and actinides, where f orbitals are added. For example the outermost shells of $\ce{La, Ce and Pr}$ are $\ce5\mathrm d^1\,$, then $\ce4\mathrm f^1\,5\mathrm d^1$, and then $\ce4\mathrm f3$. What happened to the Pr d electron? My understanding is that energy levels are quite close for the larger outer shells, and it is not intuitive which orbital fills first.


Why does a certain number of moles of gas occupy a set volume? In class, I was taught that 1 mole of an ideal gas occupies a set volume eg 22.4 L at STP. Why does the gas only occupy a certain volume and not just keep expanding forever, having limitless volume?

Ideal gases neglect inter-molecular forces so shouldn't they just keep expanding?

The reason the gas doesn't expand to infinity is because it is in a vessel.

Or, to be more precise, the ideal gas equation defines the relationship between temperature, pressure and volume for a fixed amount of gas.

What is normally done to understand the relationship is to put the gas into a vessel of fixed volume and measure the pressure it exerts at a given temperature. The pressure changes with temperature according to the formula (real gases may deviate slightly).

Alternatively, we can use a more complex apparatus that allows us to measure the volume occupied by the gas as we vary the temperature while maintaining a constant pressure (say atmospheric pressure). Same equation, different constraint.

Your misunderstanding is that you have not thought through the conditions under which you are applying the rules. The thought experiment you are doing is releasing the gas in the vacuum of space where there is no external pressure. In those circumstances the gas will expand to infinity but this doesn't violate the ideal gas equation as there is no constraining volume (you could say volume is infinite).

When applying the rules you have three variables (you usually keep the amount of gas, the number of moles, fixed) and you have to constrain two of them. If you don't, are an infinite range of possible combinations of the variables (e.g. constrain temperature and there is still nothing that tells you what the volume and pressure are as they can vary together over any values that satisfy the equation).


Decreasing Powder Size I have a small amount of a powder that is roughly 100 microns but I need to get it down to less than 10 microns. It is a small amount so I can't send it to any factory specializing in that as far as I know. In your opinion, what's the best way (cheapest) way to decrease the particle size (so far I've only been able to find coffeee grinders but it is unlikely they'll be able to get the particle size down to 10 microns) to less than 10 microns?

A ball mill can get you to sub-micron particle size with a minimal capital investment - less than 500 USD on eBay.


Deamination of Spermidine? I was making a solution of spermidine, and noticed that Sigma Aldrich warns that spermidine solutions deaminate over time. I tried to think of how that would happen, and made a diagram:

Assuming some of the spermidine has the secondary amine in an uncharged state, the lone pair electrons can attack either charged terminal amine group. I don't know which is more likely, I have drawn the 4 membered ring, but the other amine would make a 5 membered ring, which would probably be more stable. Either way, this would kick off an ammonia, which would be stable enough in water. Then a water molecule would attack a carbon on the ring and open it, making an OH2+ charged group, which would give up a proton. Of course the protons would be balanced because the ammonia would pick up a proton.

Does this make sense? It reminds me of nitrogen mustard chemistry, but must not be as reactive because spermidine is used to condense DNA, and nitrogen mustards are used to damage DNA.

Additionally, spermidine seems hard to work with. I tried to make 10mL of a 10mM solution, and calculated what volume to use to get the right mass, because spermidine melts at 20-25C. However, when I used that volume, I had about 2.5 times as much mass as I wanted. I assume this is because the spermidine was not totally thawed out, and the density was higher than I assumed (925mg/mL at 25C). Has anyone worked with this compound and has suggestions for how to handle it better?

Here are some links to the mechanism of the degradation of spermidine and some analogues in living systems.

reference 1
reference 2
reference 3
reference 4

Based on these reactions I would guess that an oxidative pathway is followed here as well, once the sample has been exposed to air.  Instead of a nice enzyme catalyzed oxidation, here we would have a free radical oxidation. Here is a general outline of such a process.  The first step would involve generation

of the more stable secondary nitrogen radical (as opposed to the primary terminal nitrogens) and formation of the nitrogen hydroperoxide.  Elimination of hydrogen peroxide could produce the two possible imines which could be further oxidized by oxygen to produce both possible diamines and both possible amino-aldehydes.  You could test your material for the presence of hydrogen peroxide, but the smell of putrescine ought to be a "dead" giveaway.


Show which group the element with Z=31 belongs and I understand it belongs to the 13 th group, but my teacher wants to express it as IIIA, so how should I convert it in my mind?

This is the old periodic table in which are still used these groups. You should avoid its use, but if your teacher needs you to learn them anyway, here they are:


How can HCl be liquid if it has a boiling point at -85°C? Shouldn't it at least be boiling? When water reaches its boiling point it makes bubbles, but that doesn't happen with HCl, why?

Did I misunderstand some basic chemistry?

All halogen compounds with hydrogen ($\ce{HX}$) are gases at STP except for $\ce{HF}$, which is a liquid (b.p.: $\mathrm{19~^\circ C}$). When dissolved in water, $\ce{HX}$ dissociates to ions ($\ce{H+}$ & $\ce{X-}$), which are readily hydrated by water molecules as $\ce{H+ (aq)}$ and $\ce{X- (aq)}$, hence not exists as a gas any more.

To answer your side question, a lot of gases have this feature. For instance, sulfur trioxide ($\ce{SO3}$) is a gas but gives sulfuric acid ($\ce{H2SO4}$) when it come to contact with water. Similarly, nitric acid ($\ce{HNO3}$) can be considered as an aqueous solution of nitrogen dioxide ($\ce{NO2}$). In reality, during an industrial process, $\ce{NO2}$ react with $\ce{H2O}$ in air to give $\ce{HNO3}$.
$$\ce{4NO2(g) + O2(g) + 2H2O(l) -> 4HNO3 (aq)}$$
Exception to this trend is carbonic acid ($\ce{H2CO3}$), which decomposes under ambient conditions:
$$\ce{H2CO3(aq) -> CO2(g) + H2O(l)}$$


Different shapes of fluorescence spectra in Raman Note: I haven't owned any of these spectrometers. I'm asking so I can decide whether to buy 532nm or 785nm wavelength raman to probe molecular structure (especially solutions) and I'm concerned about fluorescence versus signal intensity. The first picture is from 16:00 minute time of this video https://www.youtube.com/watch?v=VHjmlQdzpW4&t=1099s

Why is the shape of fluorescence spectrum different between systems using 532nm and 785nm wavelength laser in Raman Spectroscopy? In the following using the 785nm RMX, the fluorescence is maximum at the left (starting at 100 cm-1) and minimum at the right (ending at 2750 cm-1).

While in the following where the sample is the coke itself using 532nm raman spectrometer, the fluorescence has the peak at center with each side at minimum (also started a 100 cm-)?

This is with the coke removed and spectra saved. The peak is at 2400 cm-1.

My question is that both started a 100 cm-1. But in the case of the 785nm raman, the fluorescence spectrum started at maximum already compared to the 532nm raman which has the peak at higher cm-1. Why?

There is no reason why fluorescence emission background should look similar at all with green and red excitation.

Different substances

You say the green excitation spectrum is of Coke.
According to the video, the red excitation spectrum is of the peel of a tangerine.

Those samples consist of totally different substances so they will have different fluorescence spectra.

Different emission wavelength ranges

Assume this black line is the fluorescence emission spectrum of one sample:

the 532 nm excitation (dark green line) Raman spectrum is in the wavelength range 532 to 781 nm (green shaded area)
the 785 nm excitation (red line) Raman spectrum is in the wavelength range 801 to 1014 nm (red shaded area)

There isn't even any overlap between the spectral ranges you look at.

Different Excitation wavelengths

In first approximation (slight) changes in the excitation wavelength won't change the shape of the fluorescence emission, only the intensity.
However, if the excitiation wavelength changes sufficiently to excite a different fluorophore, also the shape of the fluorescence emission spectrum can (will) change.

About buying a Raman system

Before going deeper into the buying decision I suspect you'll need to learn more about the basics of Raman spectroscopy (and maybe about fluorescence). I recommend starting maybe with
some textbook chapters about spectroscopy for chemical analysis in general e.g. Skoog et al.: Principles of Instrumental Analysis if you directly like to have a focus on instrumentation (possibly also get some introductory text about spectroscopy for structure determination), and then to get deeper into Raman McCreery: Raman Spectroscopy for Chemical Analysis, ISBN: 978-0-471-23187-5.

Any serious advise about which system is better for you will need a whole lot of further background information: e.g.
what samples will you measure (phase, analytes, matrices), what applications (qualitative, quantitative, spatially resolved, restrictions on measurement time, lab use vs. field use, need for polarized measurements/depolarization ratio, ...),
will these specification change in future (other research projects).

As you comment that there is no Raman instrument in your country,

maybe you could visit a reputable lab that does Raman on similar systems as the one you want to study?
Raman system manufacturers are usually quite willing to demonstrate the ability of their instrument to measure what you need with a few of your samples, either by you sending them a couple of samples or even by you visiting them with your samples so you can try out the handling of the instrument yourself.
Do this at least with the few instuments are in your shortlist so you do have a basis for comparing actual performance for your samples.

A few comments about the Raman experiments in the linked youtube-video

Vodka-in-bottle spectrum: measuring through or on glass usually doesn't disturbn Raman spectra for green (e.g. 532 or 514 nm excitation) and also 633 nm red excitation is OK. With 785 nm excitation you'll get very bad background.
The reason why the handheld measurement works is its long working distance: the focus is deep inside the bottle, thus hardly any scattered light from the bottle gets to the detector.
Sugar Raman measurement background: not all background in Raman spectroscopy is fluorescence. White powders (that are actually clear substances with lots of refraction and reflexion due to the small particle size) will give a lot of elastically scattered light (i.e. still excitation wavelength). Depending on how good the spectrograph is, such scattered light may end up in the detector at the wrong pixel and thus be interpreted to have another wavelength.
Such scattering background will usually decrease the further you get from the Rayleigh line. A conclusive experiment to show that it is scattering and not fluorescence is: dissolve (or melt) the powder and deposit the substance as a clear film (after drying). Scattering will then (mostly) be gone.
Sugar vs. Sugar in fruit and the all-same-food spectra: solid sugar vs. sugar dissolved in water won't have the same vibrational spectrum (the spectra of some substances do not change much when dissolving in certain solvents, but sugar has tons of interaction with the surrounding water molecules due to its -OH groups - and that does affect the spectrum.)
As for the "almost all food has the same NIR spectrum": right in that wavelength range of 940 - 99 nm are overtones of -OH (water, carbohydrates, alcohols, H-bridges to oxygen). As all those fruits and veggies (including the peel) are in first approximation a mix of water and carbohydrate, there shouldn't be any surprise in those spectra being very similar.
With a good spectrometer and good data analysis (which includes good design of the calibration experiments) you can approach questions like "is this apple ripe for harvesting yet", but this requires the spectra to be available with a noise level (reproducibility) that allows measuring small shoulders of the absorption band.
Besides what the video discusses, there are a some other important things Tellspec claimed but cannot measure for physico-chemical reasons: mono-atomic ions such as Na⁺. They don't have molecular vibrations (again with super-good spectroscopic equipement you may be able to measure the effect of the ion on the water, but that's not going to work in practice). Which means, if you scan a soup and then add salt, it won't see the difference. Yet they claim to tell you sodium intake...


Chemistry of heat and egg whites I tried to make Italian buttercream icing the other day and followed this recipe.

Unfortunately, the icing was completely runny / liquid. So I know I made a mistake somewhere.
But since I don't understand the chemistry of egg whites I am not sure how to figure it out.

I think the key to the change from liquid to creamy/spreadable is when you pour hot syrup down the side of the bowl while keeping on mixing at high speed.

So what exactly happens there? I think this is where something went wrong but I am not sure what. I had my perfect egg whites beaten to nice firm froth and my no longer boiling syrup that I poured down the side of the bowl while furiously mixing and yet the result was close to water in consistency, even after adding the butter!

I appreciate any help in understanding how something liquid can be turned into something of high viscosity with this process! Thank you so much in advance for giving me insight into egg whites in the kitchen!

The recipe linked to the question is not very detailed. It does not even say to beat the egg whites (which you did and which needs to be done). Here is a snippet of another recipe:

  You just need a sugar thermometer so you can cook the sugar until it reaches a temperature of 113°C/235°F. Once the sugar syrup reaches that temperature, immediately take it off the heat – or you’ll risk bringing the sugar up to ‘hard-ball’ stage – and carefully drizzle it into the foamy egg whites, mixing continuously. You do remember the foamy egg whites, right? I bet you do. Drizzle the hot (!) syrup slowly into the egg whites, making sure not to pour it directly onto the whisk attachment of your mixer, or the syrup could splatter against the sides of the bowl or into your face.

Source: https://thetoughcookie.com/2014/09/11/different-kinds-meringue-make-italian-meringue/

I think the key is "drizzle into, mixing continuously". And here is a picture of what it should look like at that stage (from the same source):

  [OP} But since I don't understand the chemistry of egg whites I am not sure how to figure it out.

There is some chemistry and some physics here. You want to add a lot of sugar and little water. You need the sugar in solution so you can add it quickly, and if the syrup cools down, it is no longer a liquid. On the other hand, you don't want the egg white to heat up too much, so that's why you drizzle in the syrup while mixing vigorously. That way, the syrup cools down while being diluted with the water in the beaten egg whites, effectively removing water from the bubbles surrounded by the protein from the egg white, making the foam more stable.

If everything else was correct (water content of syrup, using bowls that are not greasy, correct ratio of egg whites to syrup), I think the key is to drizzle the syrup into the egg white where it gets mixed right away. You describe adding it to the side of the bowl, but the recipe does not say that. If I were to experiment with the recipe, this step would be my focus.


While calculating the effective nuclear charge, why don't we consider effect of outer electrons on inner ones? When calculating the effective nuclear charge for an electron, why do we only consider the repulsive effect of the inner electrons on the outer electrons? The outer electrons also repel the inner ones (towards the nucleus) and so, when determining the effective nuclear charge for inner electrons, don't we need to account for this effect as well?

As chemists, we are typically more directly concerned with the effective nuclear charge on outer electrons than inner electrons, because it's only the outer electrons that interact with other atoms and thus determine an atom's chemical properties. [Though the effective nuclear charge on inner electrons will affect their radial probability distributions and thus create a second-order effect on the effective nuclear charge of outer electrons.]
Nevertheless, this is an interesting question.
When we are calculating the effective nuclear charge seen by electrons in an atom (say, for outer electrons) (which directly impacts, for instance, their ionization energies), we start by doing so for isolated atoms, i.e., atoms in the absence of an external field.  Adding an external field (from the presence of other atoms, especially bonding atoms) is an added complication we won't consider here.
And in isolated atoms, the probability distributions of all electrons are spherically symmetrical.
So let's consider an inner electron in an isolated atom.  According to Gauss's Law, any charge that is spherically distributed about the nucleus, and is farther from the nucleus than that electron, "cancels itself out" (it integrates to zero), i.e. it has no net effect on the effective charge felt by the electron.
So an inner electron is only affected by outer electrons to the extent that the outer electrons spend some time closer to the nucleus than the inner electron.  [Again, when they are further from the nucleus than the inner electron, they have no effect on it, because of the spherical symmetry of their probability distributions.]
Based on the relative probability distributions of inner and outer electrons, the percent of the time this happens (outer electrons being closer to the nucleus than inner ones) is low.  Having said that, when it does happen, it will cause the inner electrons to feel a reduced nuclear charge.  I.e., to the extent that outer electrons affect inner electrons (in isolated atoms), they don't push them towards the nucleus, they push them away!
Hence there is no "reverse shielding" effect by outer electrons on inner electrons (again, in isolated atoms).  There is only a (small) shielding effect.


Difference between biosynthesis in organism versus lab Are there different terms for biosynthesis of a compound naturally in an organism versus the biosynthesis using chemicals in the laboratory or is it just "biosynthesis"?

More often than not, the synthesis in organisms involves the action of enzymes. In the lab, you often avoid these for various reasons. The enzymes may not be available, or you're interested in a more general procedure, that allows to synthesize a number of derivatives of the target structure or just a part of it.

When the synthesis aims to obtain a particular molecule, just as found in nature by any means, you'll often find a term like natural product synthesis.

When a procedure aims to yield a product by a reaction similar to processes in living  organism, the term biomimetic is frequently used.


What is the correct way to verify a structure's geometry, for example for benzene? My goal is to compare a calculated bond length with experimental data. Benzene is obviously common enough to expect sufficient experimental data to be available for it on the internet.

I have found two sources for the C-C bond length in benzene, and the numbers differ very slightly. This source says 140 pm for benzene, while this source says 139 pm for benzene. Both claim their number comes from x-ray diffraction.

Is the discrepancy possibly attributable to a difference in solvent? Is a difference of 1 pm even significant? What is a good way to know for sure which quantity is correct?

A deviation within the low picometres is nothing to worry about, there are many reasons for this.

Primary literature, like peer-reviewed journals, will always publish an analysis of the obtained values along with any error bars, i.e. how accurate the measurement is. When you have disagreeing values it is worthwhile to have a look at these, too.
Secondary literature, like encyclopaedias, will usually only quote values and might not include all details due to their general reference nature.

Quite often if you have different values, the conditions of the measurement simply were not the same. Also keep in mind, that within the recent years a lot of progress has been made for spectroscopic measurements. Hence you can find values from older publications that are significantly different from more modern values.

For example Tamagawa et al. reported $r_g(\ce{C-C}) = 139.9 \pm 0.1~\mathrm{pm}$.[1] They used a method, which combines diffraction and spectroscopic data. This article was published in 1976 and it was introduced with the following paragraph. I think that pretty much summarises the predicament of modern structural analysis.

  It is often found that the structural data of the most basic molecules are not very up-to-date because such molecules were studied in the early stages of the development of experimental methods for structure determination. Benzene is one such molecule.

And since then many more studies on benzene have probably been published. Chemspider.com lists (currently) seven data sets[2] and I am confident this is just the tip of the iceberg.

In a quite recent publication by Budzianowski and Katrusiak the crystal structure of benzene has been studied at various pressures. They pressure froze the molecule(s) in situ in phase I.[3]
We learn from this, that there is more than one phase in which benzene crystallises. And a little more digging leads us to a publication by Piermarini et al. dealing with the crystal structure of benzene II at 25 kbar.[4] But I will not go into that and rather focus on the former publication.
This paper illustrates nicely that there cannot be only one true value for the compound. I will reproduce a part of table 2 from the publication here.
$$
\text{Bond lengths (}\mathring{\mathrm{A}}\text{) and angles (}^\circ\text{) for benzene in phase I.}\\
\begin{array}{llll}
             & 0.3~\mathrm{GPa} & 0.7~\mathrm{GPa} & 1.1~\mathrm{GPa} \\\hline
\mathrm{C1—C2}     & 1.40 (4)   & 1.397 (8)        & 1.420 (16)       \\
\mathrm{C2—C3}     & 1.38 (8)   & 1.384 (10)       & 1.41 (3)         \\
\mathrm{C3—C1^i}   & 1.38 (5)   & 1.354 (10)       & 1.32 (2)         \\\hline
\end{array}\\
\text{Symmetry code: (i)} 1-x; -y; -z.
$$

As you can see even in a single crystal structures, the bond lengths are not equal. This is obviously due to packing effects and intermolecular interactions. When we talk about a single value for the carbon carbon bond length in benzene, then we assume an idealised, even rigid, molecule.

And there are more ways to obtain the geometry of a molecule: IR, microwave, Raman to name a few.[5] Nowadays we can employ computational chemistry and try to predict bond lengths, but there is a catch to that, too. These calculations take place at absolute zero and no pressure and we have a couple of other approximations in the mix. You can have a look at the Computational Chemistry Comparison and Benchmark DataBase, which lists values from 138.3 to 144.3 pm for the carbon carbon bond.[6] For example, my own calculation on the MP2/def2-SVP level of theory gives me a carbon carbon bond length of 140.0 pm. But that is a value that corresponds to a stationary point on the potential energy surface. We all know that this molecule has thirty vibrational modes, which are not yet accounted for. Below is just one, one that I find particularly funny.

TL;DR Don't worry about deviations of about 1 pm. When there are larger differences, look at the original publications about the conditions and trustworthiness of their measurements. Always consider the error margins on your own experiments when you compare values. It also depends on how accurate you need your value to be before you go and dig deep into publications from 50 years ago. (For what it is worth, you could round 139 pm to 140 pm.)

Notes and References

Tamagawa, K.; Iijima, T.; Kimura, M. Molecular structure of benzene. J. Mol. Struct. 1976, 30 (2), 243–253. DOI: 10.1016/0022-2860(76)87003-2.
CSID:236, http://www.chemspider.com/Chemical-Structure.236.html (accessed November 18, 2018)
Budzianowski, A.; Katrusiak, A. Pressure-frozen benzene I revisited. Acta Crystallogr Sect B 2006, 62 (1), 94–101. DOI: 10.1107/S010876810503747X.
Piermarini, G. J.; Mighell, A. D.; Weir, C. E.; Block, S. Crystal Structure of Benzene II at 25 Kilobars. Science 1969, 165 (3899), 1250–1255. DOI: 10.1126/science.165.3899.1250.
Wikipedia has a longer list: Molecular Geometry.
Unfortunately I am unable to generate a permalink from the database: http://cccbdb.nist.gov/ I welcome you to play around a little by yourself.
However, I'll include the summary from the geometry comparison between the C1–C2 bond length:

  distance is atom 1 (C) to atom 2 (C)
  Experimental bond length is 1.397 Å     Calculated bond lengths range from 1.383 to 1.443 Å
  The bar charts show differences (theory - experiment) and the range is −0.046 to 0.046 Å


Questions about combustion and isotopic abundance the combustion of 0.214g of sulfur provided exactly 0.428g of sulfur dioxide. After a second try, 0.782g of SO2 was obtained. What mass of sulfur was used?
The combustion of lithium forms lithium oxide. Combustion involving a 1.451g sample of oxygen results in obtaining 2.710g of Li2O. What mass of lithium was involved in this reaction?
Natural argon has three isotopes: Ar-36 (35.967u), Ar-38 (37.962u), and Ar-40 (39.962u). if the isotopic abundance of Ar-36 is 0.006%, find the isotopic abundance of the other two

After pulling my hair out:

I still have no idea how to do this question.
I balanced the equation $4Li + O_{2} -> 2Li_{2}O$ then found the moles of both oxygen and lithium oxide, which was 0.090688 mols and 0.090333 mols. Used the smaller number of moles, set up the ratios $\frac{4Li}{2Li_{2}O} = \frac{x}{0.090333mols}$
x = 0.1807 mols, which equals to 1.2647g of Li
Is this correct?
35.967(0.00006) + 37.962(x) + 39.962(1-x) = 39.948
0.00215802 + 37.962x + 39.962 - 39.962x = 39.948
-2x = -0.01615802
x = 0.00807901
which mean Ar-38 is 0.8%, but the answer clearly says it is 0.588%


The Rayleigh-Ritz variation method I am currently reading Atkins and Friedman's "Molecular Quantum Mechanics" (4th ed), looking at the Rayleigh-Ritz variation method. Starting from the Schrödinger equation $\hat{H}\psi = E \psi$, we get the "Rayleigh ratio"

$$
E = \frac{\int \psi^*\hat{H}\psi d\tau}{\int \psi^*\psi d\tau}
$$

Setting a trial function to be the following linear combination (assuming real coefficients)

$$
\psi_{trial} = \sum_i c_i\psi_i
$$

we find that

$$
E = \frac{\sum_{i,j} c_ic_j H_{ij}}{\sum_{i,j} c_ic_j S_{ij}}
$$

where $H_{ij} = \int \psi_{trial}^*\hat{H}\psi_{trial}d\tau$ and $S_{ij} = \int \psi_{trial}^*\psi_{trial}d\tau$.

Now the goal is to minimize the expression for $E$. We should therefore "differentiate with respect to each coefficient in turn and set $\partial E / \partial c_k = 0$ in each case". Using the quotient rule, I get the following

$$
\frac{\partial E}{\partial c_k} = \frac{ \sum_{i,j} c_ic_jS_{ij} \cdot \frac{\partial}{\partial c_k} \left( \sum_{i,j}c_ic_jH_{ij} \right)  }{\left( \sum_{i,j}c_ic_jS_{ij} \right)^2} - \frac{   \sum_{i,j} c_ic_jH_{ij} \cdot \frac{\partial}{\partial c_k} \left( \sum_{i,j}c_ic_jS_{ij} \right)   }{\left( \sum_{i,j}c_ic_jS_{ij} \right)^2} = 0
$$

However, I am not sure how to simplify this to obtain the correct expression, which is

$$
\frac{\partial E}{\partial c_k}_{Correct} = \frac{\sum_j c_j (H_{kj} - ES_{kj})}{\sum_{i,j} c_ic_jS_{ij}} + \frac{\sum_i c_i (H_{ik} - ES_{ik})}{\sum_{i,j} c_ic_jS_{ij}} = 0
$$

Clearly I can cancel a factor in the first term of my expression, but I don't know how to handle the derivatives with respect to $c_k$.

Instead of doing the whole derivation, I will provide the following two hints usefull in the derivation:

Use the quotient rule (as is was already done by OP).
$$
  \left({\frac {f}{g}}\right)'={\frac {f'g-fg'}{g^{2}}} \, ,
$$
where in this particular case $f = \sum_{i,j} c_i c_j H_{ij}$ and $g = \sum_{i,j} c_i c_j S_{ij}$.
Then, when taking the derivatives of the above mentioned $f$ and $g$, take into account that for each and every $c_i c_j X_{ij}$ term (where $X$ is either $H$ or $S$) there are only three possibilities:

$i = k$ in which case $\partial (c_i c_j X_{ij})/\partial c_k = c_j X_{kj}$;
$j = k$ in which case $\partial (c_i c_j X_{ij})/\partial c_k = c_i X_{ik}$;
$i \neq k$ and $j \neq k$ in which case $\partial (c_i c_j X_{ij})/\partial c_k = 0$.


How to determine the source of PVA: crude or plant-based? We're a small startup company looking to make products using ingredients solely derived from plants. One of the main ingredients in one of our products is polyvinyl alcohol (PVA). We have found a company that claims that their PVA is made exclusively from plants, and that no natural gas or crude oil is used in its production.

To provide some extra clarification based on the discussion below this question: PVA is a synthetic polymer prepared by hydrolysis of polyvinyl acetate. Polyvinyl acetate is made from acetic acid and ethylene. Ethylene is often made from natural gas or petroleum. Is it possible to determine whether the original ethylene used to make our PVA is derived from petroleum or plants? If so, can this be done given just the PVA itself, and no other information?

The reason I'm asking is because I want to independently verify the manufacturer's claim that their product is indeed completely derived from plant sources.

It certainly would be possible to get an indication of some of the sources of carbon through mass spectrometry and through radiocarbon dating. For example, fossil carbon sources would have little 14C, sources predating the 20th century, e.g. older trees, would have more. Post-Atomic Age sources might have more yet, beyond that due to natural decay of 14C.

That said, it's a nice question, but are you prepared to pay for the analysis to have it answered? Perhaps your government might perform such testing to verify a claim of being "organically produced".


Sulfur trioxide - vacant d-orbitals Sulfur trioxide violates the octet rule. Upon drawing the Lewis dot structure for sulfur trioxide, we see that the central sulfur atom is bonded to three other oxygen atoms by double covalent bonds. Therefore, sulfur is in fact surrouded by 12 electrons and not 8 as per the octet rule. This is explained by the fact that the excited state of sulfur has the following electronic configuration: $1s^22s^22p^63s^13p^34s^03d^2$. As there are 3 oxygen atoms, $sp^2$ hybridised orbitals take part in single $\sigma$ covalent bonds with the unpaired $p_y$ orbitals of the oxygen atoms. In sulfur, there still exist one unpaired $p$ orbital and 2 unpaired $d$ orbitals. These remaining unpaired orbitals take part in $\pi$ covalent bonds with the unpaired $p_z$ orbitals of the oxygen atoms. My question is, why do the electrons in sulfur from the $3s$ and $3p$ orbitals in the ground state promote themselves to the $d$ orbitals rather than one electron going into the intermediate $4s$ orbital and the next electron going into the first $3d$ orbital? After all, according to the Aufbau Principle, $4s$ is less energetic when compared to $3d$.

I feel you are confused on atomic orbital and molecular orbital.

In the excited state of a single sulfur atom, electron do fill 4s first. You are right here. However, it is a different story in molecules.

In sulfur trioxide the bonding electrons are filling the molecular orbital, not the atomic orbital such as 4s or 3d. It is forbidden that both 3s and 4s participate any hybridization or formation of molecular orbital. So no matter how low is the energy of 4s, because 3s is already participating, 4s can not be part of the bonding.