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Why would hydrogen not be generated at the cathode when electrolysing zinc sulfate?

Describe how zinc metal can be obtained from zinc sulfate solution by electrolysis.
  A labelled diagram is acceptable. Include all the products of this electrolysis. The electrolysis is similar to that of copper (II) sulfate solution with inert electrodes.

The answer key says that zinc would be formed at the cathode. However, zinc is more reactive than hydrogen, so logically hydrogen should be given off at the cathode, not zinc.
As I understand it, the ions involved in this electrolysis are $\ce{Zn^2+}$, $\ce{H+}$, $\ce{OH-}$ and $\ce{SO4^2-}$.
Also, why can possible cathodes in this reaction be carbon, platinum OR zinc while the anode can only be carbon or platinum? What allows the cathode to be zinc?

In an electrolytic cell we are using the electrical energy from an outside power source to drive a chemical system against the direction it would spontaneously move.  In this case, we can take a solution of zinc ions and force them to become zinc metal.  As long as we put the zinc ions in an environment where they can be reduced, they will be.
Consider the half reaction that occurs when zinc metal is produced:
$\ce{Zn^{2+}(aq) + 2e- -> Zn(s)}$
This is a reduction reaction, and reductions always occur at the cathode, whether the cell is galvanic or electrolytic.
The other half reaction involves hydroxide being converted into water and oxygen.
$\ce{4 OH- -> 2 H2O + O2 + 4e-}$
This is an oxidation reaction, and oxidations always occur at the anode.
Platinum and carbon are nonreactive in this system, so they can be used for either electrode.  Zinc can be used as a cathode because that's where it is being deposited, so we're just adding additional metallic zinc onto our already existing zinc electrode.
With the idea that oxidation occurs at the anode in mind, what would happen if we used a zinc anode?

Zinc atoms would turn into zinc ions? But why would zinc ions be reduced at the cathode instead of hydrogen ions? That is the main issue I have.

It's the concentration of H+ and especially overvoltage.  Overvoltage is a fancy word for activation enthalpy.

Um, can someone please reword this for someone taking his IGCSE's to understand?

@dadadok Gotcha.  In a basic solution, the concentration of H+ is very, very small.  With a much larger concentration of zinc ions available and sufficient potential to make either reaction occur, the zinc reaction is far more likely by numbers alone.  Both reactions ought to occur, though the H+ reaction will be very slow due to its low concentration.  (Again, it's important that there is more than enough potential, the overvoltage Abel Friedman mentioned.)

You are correct that given the standard potentials in the table below, you would expect hydrogen to be reduced first under standard conditions.
\begin{array}{cc}\hline
\text{Equation} & E_0 / \mathrm{V}\\ \hline
\ce{Zn^2+ + 2 e- <=> Zn} & -0.7628\\
\ce{2 H+ + 2 e- <=> H2} & 0\\ \hline \end{array}
The key concept is standard conditions. Standard conditions assume that all relevant ions have a concentration of $1~\mathrm{mol/l}$ — and hydrogen is a relevant ion, thus standard conditions require $\mathrm{pH}\ 0$. To calculate the redox potential at non-standard conditions, the Nernst equation $(1)$ must be used. Thankfully, for the zinc equation it only depends on the concentration of zinc ions. I shall neglect zinc’s concentration for the remainder of this answer, because hydrogen is more interesting.
$$\begin{gather}E = E_0 - \frac{0.059~\mathrm{V}}{z} \lg \frac{[\ce{Red}]}{[\ce{Ox}]}\tag{1}\\[0.6em]
E = 0~\mathrm{V}- \frac{7 \times 0.059~\mathrm{V}}{2} = -0.2065~\mathrm{V}\tag{2}\end{gather}$$
Equation $(2)$ tells us, that under neutral conditions, the potential of hydrogen is already $-0.21~\mathrm{V}$. This is still not enough to explain zinc formation at the cathode, but a big step forwards. More can be achieved by further raising of the $\mathrm{pH}$ value but it won’t get you all the way.
This is where a second issue comes in: overvoltage. Unfortunately, it cannot be expressed easily numerically (at least to the best of my knowledge) but in a nutshell the phenomenon is that it is hard to form gases on certain electrode materials (think activation energy). Due to the help of overvoltage (i.e. the correct choice of cathode), hydrogen liberaion can be almost completely suppressed and zinc formed.

Well, at reversible potentials (standard conditions), Hydrogen were to be discharged at the cathode but since it is a non-reversible type cell in which hydrogen Over-voltage is dependent on the combination of hydrogen atoms to form hydrogen molecules which is a slow process kinetically (RDS of the reaction). Hence, two factors affect the overvoltage phenomena. First, the nature of the cathode surface and second is the 'current density'.
It is seen that  Platinised Platinum/palladium metal electrode act as a good catalyst for Hydrogen Evolution with lower values of over-voltage involved as compared to Lead/Mercury catalyst which not only are poor catalysts but also required high overvoltage for the reaction to occur.
Moreover, by plotting the Tafel Equation, this can be inferred that there is a parabolic relationship between over-voltage and current-density and hence as we increase the current density( upto a certain extent), deposition increases as we are supplying higher EMF as compared to what is offered by the deposited material as Back-EMF (when both become equal, deposition stops) and hence higher current density is needed for further deposition.
Also, after certain deposition of zinc metal on the cathode, hydrogen over-voltage on the zinc surface is of the order of 1 Volt. However, voltage required for Zinc deposition is 0.763V (Standard)/ non-ideally less than 1 Volt and hence, hydrogen ion will be discharged only when the zinc concentration in the solution becomes so small that the emf of zinc lies above the hydrogen overvoltage value.

Up Vote (+1). I agree with your comment " hydrogen Over-voltage is dependent on the combination of hydrogen atoms to form hydrogen molecules which is a slow process kinetically (RDS of the reaction). Hence, two factors affect the overvoltage phenomena. First, the nature of the cathode surface and second is the 'current density'".

For the record, I propose as a possible explanation, albeit, involving a more advanced depiction of the underlying chemistry, to account for why with zinc sulfate electrolysis leading to zinc metal, there is an apparent absence of significant hydrogen gas evolution.
The answer starts from the excellent presentation given by Martin on this very forum (see "Electrolysis of Water" ) as to likely mechanics involved therein.
In particular, note the cited formation of surface radicals including  formation of atomic hydrogen, or, as also referred to as the hydrogen radical, in the realm of chemistry of radicals. Normally, this step is immediately followed by hydrogen radical self-reaction liberating hydrogen gas. So for most purposes, the omission of this step is not usually problematic.
However, in the special case of zinc, and I would also reference aluminum metal, there can be a significant surface attachment of the hydrogen radical. In essence, we have now further wandered into the field of surface chemistry for which I refer the reader to a 2008 doctoral thesis by Saikat Adhikari, "Alkaline dissolution of aluminum: surface chemistry and subsurface interfacial phenomena".
So, bottom line and quite simply, I suggest there is some possible surface chemistry intervention occurring with water electrolysis in the presence of zinc metal (long noted as a powerful reducing agent with association to 'nascent hydrogen'), such that one could witness a reduction/cessation in the apparent evolution of hydrogen gas.


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What form of zinc's found naturally in vegetables & meat?
What kind of zinc naturally occurs in plants? Zinc oxide, zinc sulphide, or something else?

The most likely form is $\ce {Zn^{2+}}$ bound inside a protein, or maybe chelated to some amino acids. It is not very likely that there is zinc oxide, or zinc sulfide, or zinc sulfate in organic matter.

Different vegetables contain different types of Zinc compounds, so we can't say specifically which type of zinc compound is there in vegetables.
Zinc Oxide is the most used supplement in the U.S.
Zinc carbonate is also present, which our body cannot absorb.
Also they can contain  zinc picolinate, zinc gluconate or zinc citrate (zinc picolinate is better than zinc gluconate.), zinc glycinate, etc.

Reference: Wikipedia

I don't know why this was flagged VLQ, but since I am here: I believe with "types of zinc" you are referring to the variety of zinc compounds present/possible.

@Martin yes I have tried to list "types of zinc" and I think that is what OP want!

I did not downvote your answer. I just wanted to clarify, before upvoting, that's all. I don't think your answer deserves a downvote and not a flag either.

@Martin I know it is not perfect(not so bad that it should be flagged). This type of question can only answered by some experts. Though I have tried to give answer from stuff I have read at different places.

I polished a little the language, please check. (roll back if you think i made a mistake) As I said, I don't think it deserves a downvote and I have no idea why it was flagged

source: Stack Exchange

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Additional Questions Regarding the Auto-Ignition Temperature
I had some followup questions regarding a previous post I made here regarding the auto-ignition temperature and ASTM E659

For fuel temperature below AIT, we should still have finite reactants above activation energy, reacting, and heating the remaining mixture so additional reactants are above the activation energy. Theoretically as $t \rightarrow \infty$, could the fuel burn itself out like this, regardless of what temperature it is at (obviously the rate will be very different)? If so, is there an implicit rate requirement in defining the AIT that the above chain reaction process has to happen within a short duration?
Given that ignition is defined as when a flash and temperature rise is seen, does this mean that the actual quantity of fuel burnt is unimportant and it is assumed that a significant enough amount is used?
Is the reason for using an open flask in ASTM E659 likely just for simplicity, as opposed to a piston-cylinder arrangement that offers more control over air-fuel mixture?

I don't have time for a proper answer, but in 1, it sounds like you've got the right idea.  Think of paper browning; over centuries it will oxidize into brittle, ruined material, essentially burning to completion at room temperature air.  Here the time limit they establish for a flame to appear is explicit, 10 minutes.  They do state that some compounds have a longer delay at a given temperature before they ignite.  The reaction rate you refer to would be different for each substance, since each is releasing a different amount of energy.

To understand the answer to (1), you have to think about what combustion actually is. It isn't a single reaction, with a single set of reactants, and a single set of products. It's actually a whole bunch of tiny steps (thousands) that all occur together. Those reactions happen among a whole bunch of unstable radical species (hundreds). When you put a match to a pool of gasoline, all that energy from the match starts tearing apart bonds in the fuel, leading to the formation of radicals. That's what triggers the initial release of energy, and what creates the cascade of ignition. So theoretically, no, you could never reach combustion products, through the pathway of combustion, because you'd never have that activation energy, the big kick of energy needed to create radicals and keep them alive long enough for them to create a chain reaction.
The answer to (2) is, no, it doesn't matter how much fuel. Remember you're measuring a mass-independent property. The autoginition temperature isn't "5 degrees per gram". It's a single temperature. What matters is the ratio of oxygen to fuel.
As to (3), it's a lot easier to operate an open flask than it is to operate a piston-cylinder, and it doesn't have a substantial impact on the air-fuel ratio if you do it correctly.

Regarding (1), given infinite time, every finite kinetic barrier will eventually be overcome at any finite temperature, no matter how high the barrier is, or how close to $0\ \mathrm{K}$. For example, it would take at most approximately $10^{1500}$ years for nuclei to spontaneously fuse into iron. Any chemical reaction would have a much, much lower kinetic barrier, and would happen in a much shorter timescale. Whether a puddle of fuel inside an atmosphere of oxygen at $300\ \mathrm{K}$ could have had enough time to oxidize completely during the current lifetime of the Universe, I don't know.

That's getting into the realm of metaphysics. The mass of fuel you'd need would be greater than the mass of carbon in the universe. Sounds like navel-gazing to me...

@charlesreid1 (1) A little confused. The answer implies that fuel-air require an ignition source to or else reaction will not occur. But aren't oxidation reactions spontaneous so we should expect the products to form so long as T (and thus forward rate constant) is greater than zero? My current though is that as t -> infinity, enough fuel will react such that equilibrium will reached (and depending on the heat of reaction there should be very little fuel remaining).

A fuel-air mixture does require an ignition source to combust. a spark is just a big shot of energy to the system, transferred from one set of (say, metal) molecules to another set of (say, methane) molecules. that energy starts tearing apart molecules to create radicals, which initiates combustion. it isn't a spontaneous process at all. otherwise, all fuel in existence would spontaneously combust!

@charlesreid1 What I'm confused about is that, if we have an closed system of fuel-air, T and Ea are finite, and per Arrhenius equation forward rate constant and reactant concentration are all greater than zero. From this I expect a net formation of products (given that initial product concentration is zero) until equilibrium. I get the logic in your explanation but I also don't know the flaw in my logic. Also, isn't AIT technically a point where fuel spontaneous combust?

"forward rate constants... are all greater than zero" - yes, but there are reverse rate constants too, that are much, much larger, so at equilibrium your reactants are always reactants. There's no "spontaneous" combustion, just chemical reactions.

@charlesreid1 If the reverse rate constant is greater than the forward, doesn't that imply that the enthalpy change of the forward reaction is positive (as Ea_rev < Ea_fwd), which implies an endothermic reaction (assuming T is constant) while combustion is exothermic?

no - you're confusing thermodynamics and kinetics. reaction enthalpy changes and rate constants are independent.

@charlesreid1 My apologies I worded that poorly. I was referencing to this diagram where enthalpy is shown as a difference between the activation energies of forward and reverse. Just from a rate perspective if the reverse rate >> forward rate, at the same T, from re-arranging the Arrhenius equation I arrive at Ea_rev << Ea_fwd which is the case for an endothermic reaction and the source of my confusion.

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Why does Cl- get eliminated from a trichloromethyl anion to give dichlorocarbene?
In the formation of dichlorocarbene, sodium trochloroacetate (in tetrachloroethene) is refluxed at 130 deg C resulting in decarboxylation which forms a trichloromethyl anion. (Does this step happen because of the increase in entropy? What is the driving force for this step?) Alpha elimination of Cl- then occurs to give dichlorocarbene but as carbenes are so unstable surely this final step would be energetically unfavourable? Why does it still happen? Is it just because the reaction is being heated? Does this provide enough energy for this step to happen? Is this step in equilibrium?


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Is it possible to create a 100% identical copy of a chemical?
Is it possible to create a 100% identical copy of a chemical? For example, a 100% identical copy of vitamin A would have the same shape, the same formula, and the same structure.

With such a general question, I will give you the general question: "Yes."

define 'identical'.

It is 100% possible to A) take something like naturally occurring vitamin A and purify it and B) synthesize vitamin A in a lab and purify it, and then C) mix up the two samples in such a way that nobody could figure out which was which.

Although with many biological molecules we have much more difficulty synthesizing just the one handedness that nature manages. But those with the correct chirality are just the same as any others.

By identical i mean exactly the same, with exactly the same functions and side effects.

From epidemiology studies sometimes we can find side effects using a synthetic vitamin instead of a natural one.
Or even hormones, do synthetic hormones have the same side effect risk as natural ones?

@Konrad If functional equivalence is "identical" then the answer is clearly yes. Epidemiology is not likely to be reliable in detecting a difference here compared to chemical analysis unless there is a very active impurity.

I agree with most of what the commenters and @jerepierre have written, with one or two exceptions.

Is it possible to create a 100% identical copy of a chemical?

If your question were asked in an introductory chemistry class, I believe most instructors would answer with a simple "yes", for the reasons outlined above.  On the other hand, if your question were asked in an advanced chemistry class, or perhaps in a court of law, then I believe the answer would likely be "no".  A frustrating dichotomy, but such is life.
Let me explain the "no" answer.  To begin with, realize that simple molecules carry deuterium and $\ce{^13C}$ at levels found in natural abundance. Next, note that the way an organism would synthesize a vitamin, hormone, protein, etc., is likely different than the way a chemist might synthesize it in the laboratory. The differences would lie in the raw materials used in the process as well as the actual synthetic pathway.
Each step in a synthesis that requires breaking (or making) a carbon-hydrogen bond or a carbon-carbon bond will result in isotopic fractionation.  Simply put, $\ce{C-H}$ bonds and $\ce{^12C-^12C}$ bonds are easier to break than the corresponding $\ce{C-D}$ bonds and $\ce{^13C-^12C}$ bonds.  Therefore, if we repeatedly break some number of (they can be different or the same) C-H bonds and/or C-C bonds, then our final product will be enriched in molecules with lower deuterium and $\ce{^13C}$ content, the degree of enrichment being dependent upon the difference in the number of carbon-hydrogen and carbon-carbon bond breaking and bond making steps between the synthetic process as carried out within the organism and as carried out on the lab bench.  There may be further fractionation if the starting material used by the body has required C-H and/or C-C bond making and breaking and has, therefore, also suffered isotopic fractionation.  Such fractionation can easily be monitored through comparison of the mass spectra of natural and synthetic compounds.
In addition to this isotopic differentiation between synthetic and natural products, there will also exist a difference in reaction by-products and contaminants created in these two different synthetic processes.
Both methods (isotopic fractionation and by-product/impurity differences) could be used to determine if, for example, the testosterone in an athlete's body was produced naturally within the body or synthetically outside the body.  Also, we are probably all aware of various reports were a synthetic version of a drug (often illicit, but not always) naturally produced by a plant or metabolite has produced death upon administration due to toxic by-products produced in the lab synthetic route - clearly a different side-effect profile.
So in answer to your question, no, it is not possible to synthesize a natural product in the lab that will be 100% identical to the same material produced in a living organism.

Excellent points as usual.

A very informative answer as usual. However, I feel like the correct or perhaps most inclusive answer should be that it depends on the molecule and method used to synthesize it. It seems possible to create many biomolecules by purifying the enzymes involved in the natural biosynthetic pathway. Surely then there would be no difference between the products created in vitro and in vivo.

This argument can be applied equally to the natural world also, and isotopic analysis is a well-established and widely used technique to establish the origin and authenticity of many foods and plant-based produce, such as grapes, wines, oranges, olive oils etc. SNIF-NMR has been around for over 20 years, and many other NMR/GCMS methods are being developed.....con't.

...By your argument (which makes a very good point), two identical species of grape on opposite sides of a country (or even the next valley) do not produce the same tartaric acid -See this paper as an example.
One could argue that with so much variation in the natural world no two living organisms can produce identical compounds either.

@canadianer  Good point!  If the bench synthesis, using the actual biosynthetic enzyme, also used the identical biosynthetic starting material used in the natural process, then I agree with your statement.  If the bench synthesis used a different starting material then the natural process uses, then isotope ratios in the two products will be different from each other.  Whether they will be different enough to observe the difference I don't know, but they will be different.

@long  Interesting reference! They say, "we present the results obtained for the measurement of the isotope ratios of carbon-13 and oxygen-18 of natural and synthetic samples of L-tartaric acid considering also natural samples of different geographical origin and years of production."  What was their conclusion, what is the magnitude of difference (was it measurable) between natural samples that differed only in geographical origin or year of production?

Your answer contains the right detail but the wrong conclusion. If you start with the right starting material you clearly can produce an identical starting material. Also, for small molecules, you understate the significance of any differences which will be minuscule.

@matt_black  1) "Your answer contains the right detail but the wrong conclusion", so what is the "right" conclusion?  Feel free to post your views as an answer; 2) "start with the right starting material you clearly can produce an identical starting material", did you mean to use "starting material" twice? If so I don't follow; 3) "you understate the significance of any differences which will be minuscule", understate, miniscule, sorry, I don't follow here either.

The question is about whether it can be done not whether that is what routinely happens with standard starting materials. So starting materials are relevant. And I did mean "product' not the second starting material. Also the question was about whether the molecules would be *functionally" equivalent so minor isotopic differences will be irrelevant.

Isotope ratios are reported as parts per thousand (o/oo) deviation from a defined standard. For carbon, natural samples range from -21 to -23o/oo with sd 0.37. Synthetic samples range from 20 to 28o/oo with sd 3.3. Oxygen measurements are more defined, with natural samples 25-30o/oo (sd 1.5) vs synthetic samples 12-20o/oo (sd 3.1). Plotted as a bivariate analysis (18O v 13C), the place of origin of tartaric acid is very clearly established.

A number of methods using SNIF-NMR are even more interesting. The doping of fruit juices or maple syrup with beet sugars (cane sugar) can be detected by analysing the proportion of deuteration in ethanol molecules. Due to different metabolic pathways, the deuterium content at specific positions of the sugar molecules is higher in fruit sugars than beet sugar. This results in different levels of monodeuterated methyl groups in ethanol in these groups of plants. Both ethanol molecules, but clearly distinguishable origins based on isotope ratios. Happy to provide some refs if interested.

@long  Thanks for the additional information.  I knew that processes differing in the number of "critical" bond making\breaking steps would show isotopic differences;  I'm pleasantly surprised to see that differences can be substantial.

If you are asking if a compound, such as vitamin A, can be resynthesized to give the same chemical structure, the answer is yes. That is routine and carried out naturally inside organisms and by chemists in the lab.
The effect that you suggest in the comment is more likely related to a different phenomenon. Most likely the sample is different if there is a different effect. It would be similar to eating a carrot versus taking a vitamin A pill. One could ingesting the same amount of vitamin A from either source, but the body might process them differently based on the other "stuff" in the sample. In the carrot, there's the rest of the plant. In the pill, there are excipients that are used to bind the material of the pill, make it shelf stable, and convenient to take.
When drugs are synthesized, they have to be made (and analyzed) following rigorous regulations to ensure that every sample of drug that goes out meets the specifications of the previous batch. Even though the active ingredient may be structurally identical to the previous batch, impurities must be within established specifications. The sample must have the same crystalline form so that it dissolves identically to prior batches. That's just one example of the lengths that chemists to through to replicate the composition of a sample, which in many respects is more difficult than replicating a chemical structure.


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SI Units for Isothermal Incompressibility and Specific heat capacity at constant volume
I am looking for the SI units of isothermal incompressibility and specific heat capacity at constant volume.
What I found was that isothermal compression was measured in " $\mathrm{m^2}\ \mathrm{N^{-1}}$" so I'm guessing isothermal incompressibility would be the same?
I also found that specific heat capacity was " $\mathrm{J}\ \mathrm{K^{-1}}$"so is this the same units for constant volume also?

The units of a partial derivative $\displaystyle\left(\frac{\partial Y}{\partial X}\right)_Z$ are the units of $Y$ divided by the units of $X$.  The units of $Z$ don't matter.

Isothermal compressibility is $-\displaystyle\frac{1}{V}\left(\frac{\partial V}{\partial P}\right)_T$, so the SI units are reciprocal pressure, $\rm Pa^{-1}$ or $\rm m^2\ N^{-1}$.
Heat capacity at constant volume is $ \displaystyle\left(\frac{\partial E}{\partial T}\right)_V$, so its SI units are energy/temperature or $\rm J\ K^{-1}$. When you say specific heat capacity, that means per unit mass, so the specific heat capacity at constant volume is $\rm J\ kg^{-1} K^{-1}$.

By specific heat capacity at constant volume, do you mean $c_v$? If so the units should be J/kgK.
If by isothermal incompressibility you mean the coefficient of isothermal compressibility found here, it looks like the units you have is correct (1/Pa).


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Effect of temperature on benzenediazonium during diazotization
What will benezendiazonium ion become if it is not maintained at a temperature below $5\ °\mathrm{C}$ during diazotization?

I have done to much editing as it was flagged as low quality. If you feel it's not proper you can rollback.

I don't know what is "diazodization" I think it should be "diazotization" but did not edited as i wasn't sure about it.

Diazonium salts are prone to ejection of molecular nitrogen. The process is fairly slow for aromatic diazonium salts, so the salts are usually relatively stable around 0 celsius (alkyldiazonium salsts, though, are a completely different matter). In case of benzenediazonium resulting phenyl-cation is extremely active and will react with first nucleophile present, in many cases it is water, so the product would be phenol.


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How to derive the nuclear spin of 23Na?
Is it possible to derive the nuclear spin I=3/2 for $\ce{^23Na}$ from a term scheme or from something else from spectroscopy?
I thought the nucleus spin is empirical (and cannot be calculated from J and F values) or does it?

The nuclear shell model is a useful "first approach" to determining nuclear spin.  It doesn't always work, but it is a relatively simple way to make a first attempt.
Here is a nuclear shell energy diagram.  As you can see it is somewhat analogous to an electron orbital energy diagram.

A set of shells are filled by neutrons and a separate set is filled by protons.  The protons will pair when possible, same applies to the neutrons.  Notice that shells with J=1/2 hold 2 nucleons (a nucleon can be either a proton or a neutron), shells with J=3/2 can hold 4 nucleons, shells with J=5/2 can hold 6 nucleons, and so on.  This is because the shell with (for example) J=5/2 has 3 sub-shells: J=5/2, J=3/2 and J=1/2, each capable of holding 2 nucleons.  Higher J sub-shells are "usually" filled first.
Let's apply this model to a few examples.
$\ce{^2H}$ (deuterium): 1 neutron, 1 proton; the neutron goes into the neutron $\ce{1s_{1/2}}$ shell and is unpaired, spin = 1/2; the proton goes into the proton $\ce{1s_{1/2}}$ shell and is unpaired, spin = 1/2; total nuclear spin = 1/2 + 1/2 = 1
$\ce{^{17}O}$: 9 neutrons, 8 protons; the first 8 neutrons go into the neutron 1s and two 1p shells and are all paired, the 9th neutron goes into the neutron $\ce{1d_{5/2}}$ shell and is unpaired, spin = 5/2; all 8 protons are paired in the 1s and two 1p shells, spin = 0; total nuclear spin = 5/2 + 0 = 5/2
$\ce{^{23}Na}$: 12 neutrons, 11 protons; the first 8 neutrons go into the neutron 1s and two 1p shells and are all paired, the last 4 neutrons go into the neutron $\ce{1d_{5/2}}$ shell and each set of two is paired, spin = 0; the first 8 protons fill the proton 1s and two 1p shells and are all paired, the next 2 protons go into the $\ce{1d_{5/2}}$ J=5/2 sub-shell and are paired, spin = 0, the last proton goes into the $\ce{1d_{5/2}}$ J=3/2 sub-shell and has spin 3/2; total nuclear spin = 0 + 3/2 = 3/2

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Difference between shells, subshells and orbitals
What are the definitions of these three things and how are they related? I've tried looking online but there is no concrete answer online for this question.

Related question: What are similarities and differences among shells, orbitals, subshells, and energy levels?

Here's a graphic I use to explain the difference in my general chemistry courses:

All electrons that have the same value for $n$ (the principle quantum number) are in the same shell
Within a shell (same $n$), all electrons that share the same $l$ (the angular momentum quantum number, or orbital shape) are in the same sub-shell
When electrons share the same $n$, $l$, and $m_l$, we say they are in the same orbital (they have the same energy level, shape, and orientation)

So to summarize:

same $n$ - shell
same $n$ and $l$ - sub-shell
same $n$, $l$, and $m_l$ - orbital

Now, in the other answer, there is some discussion about spin-orbitals, meaning that each electron would exist in its own orbital. For practical purposes, you don't need to worry about that - by the time those sorts of distinctions matter to you, there won't be any confusion about what people mean by "shells" and "sub-shells." For you, for now, orbital means "place where up to two electrons can exist," and they will both share the same $n$, $l$, and $m_l$ values, but have opposite spins ($m_s$).

The picture is really helpful! However, are orbits really circular?

@Abcd the s orbitals are spherical on the surface of an electron density plot. Inside the orbital the density varies with radial distance from the nucleus, with one or more "nodes" of zero electron density. You can think of it as kind of like fuzzy concentric spheres. As n increases, the number of nodes increases.

I think you didn't understand what I asked. I want to know whether the picture you have inserted is merely an abstraction or do things actually look like that.

@Abcd I think you didn't understand my answer :)
It doesn't really make sense to talk about what things "look" like at this level - just how things are. The way things are is that the electrons in s orbitals are distributed so that if you choose a given probability of finding an electron within a region (say, 90%), and plot the boundary of that region, you will get a sphere.

You said that "all orbitals with the same n are in the same shell". What's a shell/energy level?

@Abcd a "shell" is a collection of orbitals that share the same principal quantum number, and an energy level is an allowed energy state. This is too complex to give a good answer in a comment - you can create another question if you'd like, or read more about it at https://en.wikipedia.org/wiki/Atomic_orbital

May I ask what is m1?

Have a look here:

Orbitals that have the same value of the principal quantum number $n$ form a shell. Orbitals within a shell are divided into subshells that have the same value of the angular quantum number $l$. Chemists describe the shell and subshell in which an orbital belongs with a two-character code such as 2p or 4f. The first character indicates the shell (n = 2 or n = 4). The second character identifies the subshell. By convention, the following lowercase letters are used to indicate different subshells.

s:         l = 0
p:         l = 1
d:         l = 2
f:         l = 3

What is called an orbital might differ according to the context.
With orbitals in the context of shells and subshells one usually means atomic orbitals, i.e. two-electron eigenstates of an atom's Hamilton operator which are characterized by the three quantum numbers: the principal quantum number $n$, the angular quantum number $l$ and the magnetic quantum number $m$.
But often the word orbital is also used for spin-orbitals, i.e. one-electron eigenstates of the system's one-electron Hamilton operator which are characterized not only by $n$, $l$ and $m$ but also by the spin quantum number $m_{\mathrm{s}}$ which can be either $+\frac{1}{2}$ or $-\frac{1}{2}$.

Actually, I believe the term "orbital" was first used to describe the one-electron functions. So orbitals would always differ in all quantum numbers ($n$, $l$, $m_l$, $m_s$).

@tschoppi I think what you mean is more frequently referred to as spin-orbital. At least whenever the term orbital was used in most of my chemistry courses, like in $\mathrm{d}_{xy}$ orbital, my tutors always meant the two-electron state. But there might be differences between universities and lecturers.

@tschoppi That's also why I wrote 'With orbitals in the context of shells and subshells" because in that context the differentiation of states according to their spin usually isn't needed.


text:
Can acidified or neutral KMnO4 oxidise toluene to benzoic acid?
Why is alkaline $\ce{KMnO4}$ used in the oxidation of toluene to benzoic acid? Can acidified or neutral $\ce{KMnO4}$ be used in this conversion?

Related: Mechanism of arene side chain oxidation by permanganate

Here are the three equations describing the reduction of manganese (and concurrent oxidation of whatever substrate may be present) under basic, neutral and acidic conditions respectively.
$$\ce{Mn^{+7}O4- +e- ->~ Mn^{+6}O4^2-~~~~[basic]}$$
$$\ce{2H2O + Mn^{+7}O4- + 3e- ->~ Mn^{+4}O2 + 4OH-~~~~[neutral]}$$
$$\ce{8H+ + Mn^{+7}O4^{-} + 5e- ->~ Mn^{+2} + 4H2O~~~~[acidic]}$$
Acidic conditions are the most economical as 5 electrons are transferred per mole of manganese, but oxidation can be achieved under all 3 conditions. A problem with neutral conditions is the precipitation of insoluble $\ce{MnO2}$ and the need for its subsequent separation. A problem with all 3 conditions is that oxidation will only occur if the material to be oxidized is somewhat soluble in the reaction medium.
The milder basic conditions are generally preferred for the oxidation of aromatic alkyl side chains because with the milder basic conditions an organic co-solvent that is soluble in water can be employed to facilitate solubility of the organic substrate.


text:
Is the Kekule structure of benzene completely wrong?
I've found out about something called resonance but I'm getting the impression that this indicates the kekule structure is possibly correct as it uses double bonds (I thought that the delocalised model was the only correct one).
I'm a little confused on this so could someone clarify?

There is very good video on Khan Academy you can watch it. This is kind of confusing topic, which to me was solved bu this video :)

In benzene, all of the carbon-carbon bond lengths are equal.  Therefore, the Kekule structure shown below is an incorrect representation of benzene.

It is incorrect because it suggests that there are two different types of carbon-carbon bonds in benzene, a carbon-carbon double bond and a carbon-carbon single bond. Typically, the bond length in a carbon-carbon double bond is around 135 pm, a carbon-carbon single bond is around 147 pm long, while the carbon-carbon bond length in benzene is roughly in between at 140 pm.
Benzene is often drawn as follows to indicate that all of the carbon-carbon bonds have the same length and the pi electron system is continuous and uniformly distributed.

Sometimes, when one is discussing the resonance properties of benzene, two Kekule structures are drawn with the double headed resonance arrow between them as shown below.  This is meant to indicate that these two Kekule structures are resonance structures of benzene, and that taken together (e.g. in one Kekule structure a carbon-carbon bond is a single bond, in the other structure the same bond is a double bond; so "real" benzene is a simultaneous, superposition of these two structures where each carbon-carbon bond is in between a single and double bond), they represent the true electronic and physical structure of benzene.

So is drawing the resonance structure (with the double headed arrows) equivalent to the structure with circle in the middle (with the delocalised electrons)? So the resonance structure isnt implying that theres double and single bonds present that are constantly alternating to produce bonds which appear to be the same length?

You're exactly correct, the resonance structures with the double headed arrows are equivalent to the single structure with circle in the middle.  The resonance structures are meant to imply that the bonds are 1) not alternating and 2) a simultaneous blend of single and double bonds.

In defence of Kekulé structures
The real answer is, in my view, much more nuanced than ron's answer would have one believe. I sincerely think it is too simplistic to just say that it is "wrong".
To understand what I mean, we need to look at the concept of a resonance hybrid, which is a weighted average of several different resonance contributors. Molecules, of course, do not exist as individual resonance forms (not even fleetingly); they exist solely as the hybrid. We already understand that the most accurate depiction of benzene is one which comprises two major resonance contributors, 1a and 1b:

One could argue that the only correct depiction of benzene is the resonance hybrid 1, where each pair of adjacent carbons is connected by 1.5 bonds, and indeed this position has already been argued. But benzene is not the only molecule in which resonance is crucial. Take an amide, for example: we know that the charge-separated resonance form is important to understanding (amongst others) the rigidity of the peptide bond, as well as its resistance towards hydrolysis.

Of course, here the C–O bond order is probably more than 1.5, and the C–N bond order is probably less than 1.5, but I don't have any other fancy ways of depicting non-integer bond orders, so we have to make do with this. According to the same logic as before, the only correct way of depicting the amide is the resonance hybrid 2, and using the individual resonance form 2a must be "wrong". However, it appears that the vast majority of organic chemists have no qualms about depicting amides in that form.
The same can be said of practically any other molecule, like butadiene, which is arguably even worse:

Are we not being inconsistent, then, if we claim that the Kekulé structure of benzene – i.e. one of two resonance forms of benzene – is "wrong"?

Part of the answer to this paradox is that not all resonance structures are created equal. It is obvious that in benzene, the resonance forms 1a and 1b make an equal contribution to the resonance hybrid 1. Therefore, by choosing to solely depict benzene as 1a, we are effectively throwing away 50% of the information. On the other hand, the resonance form 2a is a larger contributor than 2b to the resonance hybrid; so, by choosing to solely depict the amide as 2a, we do not discard quite as much information.
However, it still leaves us with the uncomfortable question of where we should draw the line when it comes to the supposedly "wrong" practice of using resonance forms to represent molecules. Any threshold that we come up with, naturally, can only be completely arbitrary: you could draw up an arbitrary list of what is "permissible" and what is "wrong", but you would be doing it until the cows came home.
A much healthier way of resolving this contradiction, and one that the vast majority of chemists have already adopted, is to realise that our Lewis structure depictions are always going to be approximations to the truth. With this in mind, we can comfortably choose to use one resonance structure – the Kekulé structure, for example – and trust in the reader to understand that, unless stated otherwise, we are not referring specifically to this resonance form, but rather to the actual molecule, which exists as a resonance hybrid. And naturally, we should stick to the dominant resonance form as our default depiction of the molecule: hence for benzene we could use either 1a or 1b, but for the amide we should use 2a, not 2b.
Regarding benzene in particular, I guarantee you that the "circle in hexagon" depiction does not exist in organic chemistry beyond introductory textbooks which try to teach you that the Kekulé structure is "wrong". It is virtually impossible to find the "circle in hexagon" in a high-quality chemistry journal. [As always, there are exceptions: η6-benzene coordinated to metals is nearly always drawn with the circle, for example, but that is really a different story altogether.] Furthermore, using the circle makes it infinitely harder to explain a lot of aromatic chemistry: o/m/p-directing effects of substituents, or kinetic/thermodynamic sulfonation of naphthalene, to give two very basic examples.
The alternative to this is to be a purist: we must always draw dashed lines, and everything else is "wrong". If you want to be in that camp, so be it: congratulations, you are technically correct, but at what cost? Even IUPAC, the kings of being technically correct, don't care:

Curves represent delocalization, yet lots of structures have some delocalized elements. Although all such elements could be depicted with curves, there is little gained by doing so. On the contrary, the arbitrary use of curves can draw the viewer’s attention to insignificant portions of a structural diagram, and away from areas that are chemically more important such as an active site or a reactive group. Accordingly, curves should only be used when the delocalization is specifically being highlighted as an important feature of the structure.
When curves are not used, any alternating configuration of double bonds is acceptable within the further constraints discussed in GR-3.5.

Pure Appl. Chem. 2008, 80 (2), 277–410 (emphasis mine).

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why does dimethyl dichlorosilane undergo hydrolysis to form siloxane polymers, but its carbon analog does not?
Dimethyl dichlorosilane will undergo hydrolysis to form siloxane polymers. But why would its carbon analog dimethyl dichlorocarbon not be able to undergo a similar hydrolysis reaction to form polymers?
My guess is that $\ce{Si-Cl}$ bond being much weaker than $\ce{C-Cl}$ bond would be easier to break and undergo hydrolysis, and that the geminal diol is unstable, hence it is less likely to form.
I'm unsure of my answer though. Could anyone please explain and elaborate?

Nah, it's mostly sterics and angles. Polymethyleneoxides $\ce{(CH2O)_{n}}$ is well known, methylmethyleneoxide also known as acetaldehyde forms trimer. Acetone, though, is sterically hindered enough to make it unprofitable to form polymere. As for the hydrolisys process itself, it also mostly sterics: 5-coordinated intermidiate state in hydrolisys is much, much easier to form for silicon than for carbon.

Actually a $\ce{Si-Cl}$ bond is much stronger than a $\ce{C-Cl}$ bond.  It takes about 90 kcal/mole to break a $\ce{Si-Cl}$ bond, but only around 81 kcal/mole to break a $\ce{C-Cl}$ bond.  The strengths of the corresponding $\ce{Si-O}$ (110 kcal/mole) and $\ce{C-O}$ (85 kcal/mole) bonds also need to be considered.
In the silicon reaction we need 90 kcal/mole to break the $\ce{Si-Cl}$ bond, but we get 110 kcal/mole back when we form the $\ce{Si-O}$; the reaction is exothermic by roughly 20 kcal/mole - a large amount.  In the analogous carbon case, we need 81 kcal/mole to break a $\ce{C-Cl}$ bond, and we only get 85 kcal/mole back when we form the $\ce{C-O}$ bond; the reaction is exothermic by only 4 kcal/mole - not much.  There is quite a difference in the overall driving force for the two reactions with the silicon case being energetically much more favorable than the carbon case!

Your guess is correct. The geminal diol is unstable because carbon atom can't hold more than one $\ce{-OH}$ group. Where as Si atom can hold three $\ce{-OH}$  groups. It is this property of silicon that makes the formation of organosilicon polymers possible.

well known example of stable gem-diol http://en.wikipedia.org/wiki/Chloral_hydrate

It's not that you couldn't make polyacetal.  In fact Dupont will sell it to you (under the name of Delrin) and it's a favourite with machinists because it holds shape extremely well.


text:
What chemical substance is obtained by dissolving C60 in olive oil?
What chemical substance/component would be obtained of dissolving $\ce{C60}$ in olive oil? Would buckminsterfullerene (bucky-ball) change its spherical fullerene structure (soccer ball-like) into something else ($\ce{C60OO}$) or will it stay the same?

See:

Fullerene - Safety and toxicity at Wikipedia
Fathi Moussa's experiment study on rats in 2012 which doubled their lifespan
The color of the Carbon-60 in olive oil
Buckminsterfullerene-olive oil production process

Dear God, fullerenes and anti-aging.  I recommend taking snake oil from the foothills of Appalachia instead.  It may not do any good, but on the other hand it does not form radical anions in the body either with no clear elimination pathway.

@AbelFriedman: C60OO toxicity experiment tested on rats by Baati showed that it is totally non-toxic and safe. So in your opinion how C60 would form radical anions in the body?

There is no chemical reaction between fullerene and olive oil, that is to say that the two molecules do not break or make chemical bonds to produce a new chemical species.  However, fullerene is soluble in olive oil and so there are typical solvent-solute interactions.  A homogenous solution of fullerene in olive oil produces a light purple color typical of many fullerene solutions.

Since Moussa's 2012 experiment where he fed fullerene in olive oil to rats and found that their life span roughly doubled, many companies have been marketing a dark-red liquid comprised of olive oil and fullerene.  It is likely that in many of these highly colored liquids, the fullerene is not completely dissolved and light scattering by the fine particulates are producing the observed color.

The Moussa experiment has problems, as outlined by Derek Live here.

Actually Lee states, "this image duplication does not directly bear on the most surprising and interesting results of the paper - life extension in rodents".  I can't say if Moussa's results are right or wrong. What I am saying is that they are very interesting and have generated a fullerene plus olive oil marketplace - buyer beware applies.

People already consuming C60OO for few years already and nobody died yet;) Based on the poll, most of the people reporting positive experience.

The typical reactions of C60 happen at the double bond: [4+2] cycloaddition, [3+2] cycloaddition, oxidation.  Just dissolving it in olive oil (will it even go up, I know it does go up in aromatic solvent) won't do anything.


text:
Free energy and kinetic energy for reacting gases
I am confused about the energies we are talking about when speaking in regards to the Maxwell Boltzmann Distribution and activation energies.

Is there an relationship between kinetic energy and free energy for ideal gases? I am wondering if it’s possible to plot the x-axis of a Maxwell Boltzmann Distribution in terms of free energy as opposed to kinetic energy or velocity.
Is the activation energy the difference in the free energy of the activated complex and the average free energy of the gas (or any reactant)?
From the answer here, the activation energy mark on Maxwell Boltzmann Distributions like this one refers to a minimum kinetic energy level. Since the activated complex/transition state is a state, I do not understand why having a minimum kinetic energy (and right collision orientation) means reactants will reach the transition state. How come there is no free energy requirement?

@Crafter0800 I fail to see how removing "confusion" from the title made it better. It is still as unspecific as it can get. Unearthing a three year old question for this is probably not worth it.

My initial thoughts on this are that you are viewing the Maxwell-boltzmann $velocity$ distribution the wrong way. Let me explain,
In nature there are three main distributions that "things" follow:
1) Boltzmann
2) Bose-Einstein
3) Fermi-Dirac
The first one is a classical result while the latter two are the result of quantum mechanical effects (symmetry of the wavefunction with respect to permutation). These three distribution functions describe the "way" energy is distributed among the energy levels of a molecule/atom/generic particle.
Perhaps a better way to phrase it would be that the macrosystem can be separated into lots of tiny microsystems. Each of these tiny microsystems can have their own particular "state". The individual microstates will combine in a collection or $ensemble$ to give the overall macrostate observed on our scale.
At this point we can vary our definition of how to define the ensemble and indeed there are quite a few buzz terms to learn. A $canonical$ ensemble is one in which the number of particles in each microstate ($n_i$), the volumes of the microstates, and their temperatures are all the same. If you choose the $micro-canonical$ ensemble then you would have $n_i$, $V_i$ and the energy of the $i$th state as being kept constant but not the temperature. A $grand-canonical$ ensemble has constant $n_i$, $V_i$ and chemical potential $\mu _i$.
Now depending on what you choose you will get slightly different equations, but they are all variations on a general unifying theme:
"The frequency distribution of a particular state or outcome out of all those states or outcomes possible, is proportional to
\begin{equation}
e^{-\frac{E}{KT}}
\end{equation}
Which is know as the Boltzmann factor. $E$ is the energy of the state, which varies from state to state."
You have already come across this factor in kinetic theory. !!AS A PARTICULAR EXAMPLE OF A NATURAL PHENOMENA THAT OBEYS THE BOLTZMANN DISTRIBUTION!!.
You can clearly see that the velocity distribution function, the frequency of having a particular velocity, is proportional to the Boltzmann factor, in this case the particular energy we are considering is the kinetic energy, so let $E=\frac{1}{2}mv^2$.
\begin{equation}
f(v)=\sqrt{\frac{m}{2\pi KT}}e^{-\frac{mv^2}{2KT}}
\end{equation}
The Maxwell velocity distribution is just a special case of a Boltzmann distribution.
With this in mind I'm sure you can see that many different phenomena will be described by this distribution. If you wanted to put the Gibbs energy in there (indeed as we do in transition state theory (a personal favourite of mine)) then we can.
This brings me to your next question which is a little easier to answer. The Activation energy can readily be described by a Gibbs potential as indeed it usually is. I think this inherently answers your third question.
If you have any questions I will try to answer them when I can! :)

The Boltzmann factor basically says that any energy term (u, h, g, KE etc.) for ideal gases follows the same frequency distribution? My confusion lies in fact that KE is not a state function (I am likely mixing things up here), but E above is described as energy of the state.

Just to be sure I understood 2 and 3 correctly, activation energy is the difference in free energy between the transition state and reactant. Only particles with free energy above the transition state can react, actual velocity/KE of collision doesn't matter?

Not really ... You use the Boltzmann factor to obtain the thermodynamic potentials/state functions /.... its all about relating the micro properties to the macro ... the molecular details that give rise to the classical thermodynamic quantities ... perhaps I mislead you with the transition state comment ... that is quite an advanced topic which mixes kinetics with thermodynamics which is indeed its own question :)

That is why you can mix a bit of Gibbs in there ... but in general you want to have the energy of the state in the Boltzmann factor ...

I was referencing this diagram that shows activation energy as a difference in free energy between transition state and reactant average energy, is this correct? Is gas KE here basically a part of the free energy and internal energy (it is microscopic, not the system KE as in first law)? In simple terms, is it correct to think that a reaction cannot happen unless the reactants has free energy above the transition state (but even if this is satisfied other factors like KE, orientation also plays a role).

I think the main problem is that your x-axis is the reaction coordinate ... the reaction coordinate is an extremal path through a potential energy surface for the reaction... every other point is a higher order transition. I wouldn't personally try to link the two ... The latest figure you linked is the type of stuff I would be more comfortable talking about ... in truth I don't know if it would be a correct thinking path to try and link the two as you are trying ....

The reaction free energy you will remember goes to zero at equilibrium ... you want to talk about the standard change between reactants and products ... talk instead about the "stability gained"

I feel like I am getting more confused. I have some followup questions here and I would really appreciate some additional help on this. Thanks!


text:
Greatest Energy Output Reaction to Atmosphere Air
First of all i'm no chemist, this is not homework and if you need further clarification please ask. I'll be happy to provide!
Imagine that you:

have 1 kg of any given chemical X
A car battery
A semi-controlled combustion chamber (chamber temp is initially 20ºC, pressure is 1atm, the chamber is either vacuum or filled with atmosphere "air"...)
As much atmosphere "air" as you want, initially at 20ºC...
A not so potent air-pressurizer... you might be able to get 3atm out of it...

The goal here is to get the most energy out of reaction as possible. So the question is: What chemical is X?
Further details:

X doesn't have to be consumed all at once as in a explosion reaction... Damage/Destruction is not the goal
If you can combine X with something in the air, and then use it to fuel the reaction, increasing the energy output, great!

AFAIK in practice good old gasoline has most energy-per-gram as long as you include storage tank into weight. Granted, theoretically hydrogen should be the most effective in the case, but it is too hard to store at average temperature.

this is a way too generalized question. you certainly need experimental data or extensive screening to make sure you are on the top of the volcano plot. have you heard the term adiabatic flame temperature? The ratio of inert to gasoline/fuel is a determining factor to get high combustion energy. For example the adiabatic flame temperature for methane is higher for a mixture of methane and air with 5-15 mol% methane. 5-15 mol% methane can ignite without any external source of fire.
Please see the attached link to compare fuels according to their energy density:  http://en.wikipedia.org/wiki/Energy_density
And for your question you can't have two variables to optimized with one equation, either you have to specify the chemical X or you have to specify what is the output energy of this process. If you know the chemical X with it's thermodynamic property than using basic heat balance equation you can calculate the output energy.

isn't there a way (possibly a equation) that given a molecular structure and a reagent it outputs the "delta" energy of the reaction? If so, as a developer i could "brute-force" it... of course i would use a few heuristics and possibly some machine-learning along with backtracking...

That's a good point. Electronic structure theory can be applied for preliminary screening. Actually i am working on computational catalysis and one of my main objective is to find better catalysts for reaction. Here i optimize each reactants, intermediates and products with appropriate level of theory and then locate the transition state. This will give you both the reaction energy and activation energy. Moreover if you put them in a suitable microkinetic model you will get the reaction rate with rate limiting step. You can use DFT for faster calculation or you can use Hartree-Fock approach.

hmmmmmmmm thats very interesting!!! i smell nobel prizes...


text:
Why doesn't the Brady's test (2,4-DNPH) work on carboxylic acids?
Why does the Brady's test, the reaction of 2,4-dinitrophenylhydrazine with aldehydes and ketones, fail with carboxylic acids?

2,4-Dinitrophenylhydrazine (2,4-DNP or 2,4-DNPH) reacts readily with aldehydes and ketones via a condensation reaction (the lone pair of electrons on the terminal amino group in 2,4-DNPH makes it a strong nucleophile and the condensation starts by the nucleophilic 2,4-DNPH attacking the electrophilic carbonyl carbon) to produce the corresponding hydrazone.  The hydrazone is usually a brightly colored yellow, orange or red compound, so the reaction is often used to test for the presences of an aldehyde or ketone

2,4-DNPH does not react with amides, esters or carboxylic acids.  As shown below for the case of an ester, an extra resonance structure can be drawn for these 3 types of compounds as compared to a ketone.  This extra resonance structure delocalizes some of the positive charge away from the carbonyl carbon onto the adjacent hetero-atom (oxygen in the case of the ester or carboxylic acid, nitrogen in the case of an amide).  This makes the carbonyl carbon less electrophilic in these compounds, and consequently attack by the nucleophilic 2,4-DNPH is less favored.

It's for the same reason that you can't make amides from a carboxylic acid and an amine: the amine deprotonates the acid.  The carboxylate anion is resonance-stabilized and the carbonyl carbon is insufficiently electrophilic.  You couldn't form a tetrahedral intermediate.
Same here: the hydrazone is formed from a ketone through nucleophilic attack on the carbon, the intermediate being a hemiacetal.  You won't get a hydrazide derivative here, although from an ester with sufficiently forcing conditions you might.


text:
How to calculate the detection limit?
I am having trouble figuring out how to calculate the detection limit. Here is the question:
You and your friend working in a different lab have been given an unknown water sample that contains potassium.  You are both asked to measure the potassium concentration in the solution (10 replicate measurements).  The methods that you and your friend are using are different.  The results that were obtained are listed below.  Determine:
(a) the concentration, standard deviation and relative standard deviation for the unknown as measured using the two methods (check for outliers!),
(b) calculate the detection limit (3sigma) for each method,
(c) compare the standard deviations and evaluate whether the two averages are significantly different (or not) at the 95% confidence level.
RESULTS:
Standard    Method 1    Method 2
Concentration   Intensity   Intensity
(mg/L)  (nA)    (mV)
0.000   0.624        1.955
0.000   0.488        2.490
0.000   0.522        2.166
0.000   0.355        1.500
5.000   9.245       15.644
10.000  17.069      31.220
15.000  26.200      44.266
20.000  33.881      62.394
25.000  43.826      75.611

Replicate
1       26.544      46.977
2       25.449      49.120
3       21.053      50.998
4       24.353      46.615
5       23.899      49.326
6       24.010      46.666
7       25.554      45.291
8       23.549      42.995
9       26.008      43.678
10      24.404      49.012

So my problem is with part (b). I know the detection limit is calculated with the formula DL = 3sigma/slope and that I can find the slope by calculating Sxy/Sxx but I'm not sure what points I'm supposed to use? What points do I use for the standard deviation (sigma) and for Sxy/Sxx?
I've tried looking at examples in the textbook but nothing was really similar. I also tried searching online, but I couldn't find anything that helped me.

Have a look at this: http://www.chem.utoronto.ca/coursenotes/analsci/stats/LimDetect.html
It is possible to calculate the limit of detection from the standard error of the regression:
$$C_{\mathrm{LOD}} = \frac{3s_{y/x}}{b}$$ These values are calculated from the regression of all the data points, not the standard deviation of any subset of data.
However, in this case we have replicate blank measurements (four measurements at C=0), so we can also calculate the limit of detection as simply 3 times the standard deviation of the blank measurements:
$$y_{\mathrm{LOD}} = y_{\mathrm{blank}}+3s_{\mathrm{blank}}$$
(in instrument response units—convert to concentration from a known standard or from the regression equation)

So I find the standard deviation of the intensity values with concentrations of 0, right? But I'm not quite sure what is yblank?

Oh, it's the mean, correct?

Yeah, the mean blank response.

So what's the point of the other measurements (the non-zero concentrations)? Those are what really threw me off :S

The other measurements establish the response of the instrument at different concentrations. It's not possible to answer question a) or get a regression equation without them.

Uh oh.. I think I messed up question (a) then... I was referring the the standard measurements (5, 10, 15, 20, 25), just to be clear. I used the replicate concentrations for question (a), is that wrong?

You use both: The replicate concentrations are repeat measurements of the unknown. You make a linear regression from the standard measurements and use your replicates with the regression equation to calculate the unknown concentration.


text:
Are carbocations that bad?
My professor has an anti-fetish for carbocations. In any mechanism that we draw with a carbocation, the maximum earnable credit is automatically halved. In any case, he draws all mechanisms involving carbocations as concerted reactions; he never wants us to draw an carbocation by itself. If a carbocation could be drawn (as in the case of the formation of an acetal; specifically the step in which water leaves) then we must have electrons flowing from somewhere else to give the carbon an octet.
Now, is this valid? Are all mechanisms involving carbocations concerted? Excluding the obvious case of SN1; I'm pretty sure he'd accept that carbocation (and he hasn't touched on SN1 ever).

@Dissenter Rereading your question, I am more convinced that this is just a policy in grading, but there isn't necessarily anything incorrect about what you are being taught. There is no difference in whether the reaction is concerted or not. It's all about what resonance structure the instructor wants to see.

Are all mechanisms involving carbocations concerted?

Many carbocations are stable, observable entities.  Many large, delocalized carbocations like trityl perchlorate form very stable salts (reference see p. 196), other carbocations are easily observed by nmr.

If free carbocations didn't exist as discrete intermediates how could you explain the stereochemical results from an SN1 reaction where a chiral starting material generates a planar, long-lived carbocation that can be attacked from both sides to produce an achiral product?  Any type of concerted process would be expected to proceed with inversion of configuration (SN2).

(image source)

In addition to this most excellent answer: In 1994 the Nobel prize went to George A. Olah "for his contribution to carbocation chemistry." (read more)

@ron can you imagine a reaction in orgo i/ii that a) involves a carbocation intermediate and b) isn't concerted; i.e. the C+ does exist by itself if only ephemerally? Excluding SN1.

@Dissenter, taking orgo i for now. I can't think of any excluding Sn1.

@Dissenter  E1, some HX additions to alkenes, some electrophilic aromatic substitutions.

@Dissenter afaik, Friedel-Crafts alkylation of hexamethylbenzene gives salts that can be isolated, but I do not have an article on hand.

@Jan  I'll change it to a carbon-hydrogen based example.  If you google "stabilized carbocation", you'll find that the term is used to describe many examples with heteroatoms.  The crystal violet example came from a UC Davis webpage when they were discussing "very stable carbocations".

@ron There are some concerted substitution mechanisms which are able to retain configuration. The Sni mechanism for retaining glycosyltransferases being one such example. - That said, such mechanisms are typically special cases.

The mention of acetal chemistry leads me to have some sympathy for your instructor, although this person is taking it to the extreme with the grading policy. Below are two mechanisms for a trans-acetalization reaction.

I am guessing that mechanism A is one that your instructor would give half credit (since it goes through a carbocation) and mechanism B is one that your instructor would give full credit. I prefer mechanism B, and I think it's fine for your instructor to agree. However, these are identical mechanisms. How is that? Because the third structure in each mechanism are the same thing, just different resonance structures.
So if they are identical mechanisms, why do I prefer mechanism B? Because that resonance structure explains something about the reaction, particularly in relation to a reaction that you have learned before, the SN1 reaction. In the typical SN1 reaction, formation of a carbocation is slow. However in the trans-acetalization reaction here (really just a variant of the SN1), that carbocation formation is much faster. Why? Because the carbocation formed is stabilized by the ability of the adjacent oxygen to donate electrons, giving an intermediate that has a resonance structure where all atoms obey the octet rule. I think that's an exceptionally important consideration, and one that helped me understand and recognize reactions of carbonyls.
I do think that it would be unfair to give different levels of credit for mechanisms A and B. However, I do appreciate what your instructor is trying to get across, although s/he may not be doing a good job at it.

Your professor does have a case in that many carbocations can be thought of as resonance stabilised, as was depicted by jerepierre in his answer. I once had a professor who would prefer any resonance structure that has complete octets on all p-group elements over any resonance structure that doesn't have that (although he didn't go as far as downgrading for sextet cations).
However, there are also certainly reactions that cannot be thought of as concerted, and also cannot be resonance-stabilised to give an all octet structure: See for example the electrophilic aromatic substitution as depicted by this mechanism (sheepishly stealing the image from Wikipedia):

The mechanism as shown here is for unsubstituted benzene. I did it in an organic practical course with nitrobenzene. There is no way in which nitrobenzene will stabilise a carbocation (indeed, the nitro group is meta directing due to that). And the mechanism can — to the best of my knowledge — cannot be thought of as concerted.


text:
Do non-redox reactions exist?
Redox reaction is a type of chemical reaction, and is the result of electron transfer between chemical species. But, all chemical reactions somehow involve electron transfer!
So, are there chemical reactions without electron transfer?

All chemical reactions involve attractions between nuclei and protons, but that doesn't mean that all chemical reactions involve electrons moving between species (electron transfer.)

There are many, here are some examples

Ion interchange reactions of various types, such as precipitation
$\ce{NaCl + AgNO3 -> AgCl \downarrow + NaNO3 }$
Polar molecule insert/ejection. The most common case is hydrolysis and reverse reactions.
$\ce{PCl5 + 4 H2O ->  H3PO4 + 5 HCl}$
Reorganization of bonds between atoms of same type, such as catalytic benzene synthesis from acetylene
$\ce{3 C2H2 -> C6H6}$
Ligand exchange
$\ce{ [Fe(H2O)6]^{3+} + 6CN- -> [Fe(CN)6]^{3-} + 6 H2O}$
polar oligomerization
$\ce{ H2O + n CH2O -> HO-(-CH2-O-)_{n}-H}$

Can you elaborate on why these each are not redox reactions?

@spacetyper Just write down the molecules involved and produced and write down all oxidation states.

Yes, there are precipitation reactions (which don't have electron transfer), e.g. silver cation and chloride anion combining to make silver chloride, a slightly soluble compound. Also barium sulfate will precipitate.
Hydrolysis of polysaccharides and proteins: with the assistance of enzymes, water is use to break up polysaccharides into simple sugars.
There are others.

Well it sort of depends on how you define electron transfer.

@Dissenter Silver chloride is still a salt, a cation and an anion; there is no transfer of electrons.

Of course there are non-redox reactions, for example, acid-base reactions. They do not involve any electron transfer only protons move around (so no oxidation numbers change). Here are some basic examples:
$$
\begin{align}
\ce{HCl + H2O &<=> Cl- + H3O+}\\
\ce{HNO3 + H2O &<=> NO3- + H3O+}\\
\ce{CH3-COOH + H2O &<=> CH3-COO- + H3O+}\\
\ce{HCOOH + H2O &<=> HCOO- + H3O+}\\
\ce{HCl + NaOH &<=> NaCl + H2O}
\end{align}
$$

correct : $\ce{CH3−COO+}$ to $\ce{CH3−COO-}$

For us to understand the redox nature of the above examples, we have to use the definition of redox as gain or loss of electron density, not just electrons. For instance, in the reaction between sodium hydroxide and hydrochloric acid to give sodium chloride and water, chlorine is reduced by gaining electron density in its bond with sodium compared to its bond with hydrogen. Why? Because sodium releases electron with more ease than hydrogen. Consequently, chlorine binds the electron more tightly thereby creating higher electron density. This principle applies to all chemical reactions and is the result of the concept of electronegativity.  It is very important in biochemical reactions. I have also seen its importance in the molecular basis of all diseases, which are as a matter of fact the result of oxidation reactions in the body.

Since redox is defined as gain or loss of electron or electron density it is impossible to have a non-redox reaction. If a reacting species must bond with a different species post-reaction, the electron density must shift due to differences in electronegativity.

" it is impossible to have a non-redox reaction. " Three answers above yours demonstrate several examples of non-redox reactions, while you say they're impossible. It'd help if you describe how each of their examples is wrong.


text:
How to change color of urine safely?
I am trying to find a SAFE way to change the colour of my urine, from what I have read online, Methylene Blue is relatively safe, and will change urine to blue or green. Is it safe? Can I put this in a food or drink? If so what are safe amounts? If not does anyone know a way to change the colour of urine safely?

This would definitely be a question For S.E. Biology, But I have no idea what would be the motivation for this.

is that for lightsaber battles? just have a day with less food and drink, and then drink red coloring derived from beetroots to an equivalent of about 4-5 beetroots and food and drink lots and lots... of water. )

Lots of common foods can colour urine; beets, berries, carrots, asparagus (as well as making it smell) and rhubarb to name a few.
A google search will give you plenty of references, but to get you started, try the urine colors website.
Methylene Blue (CAS 61-73-4) in concentrated form is harmful if swallowed and is a known irritant. Methylene Blue solution (~1%) is also harmful if swallowed, and has reported warnings which include:

May cause gastrointestinal irritation, discoloration of oral mucosa, irritation of lips,
mouth and throat, paleness of complexion, lack of coordination or drowsiness
Effects may be delayed. Laboratory experiments with methylene blue have resulted in mutagenic effects. Repeated contact may cause corneal damage. May cause adverse eproductive effects.

Eat lots of carrots or liquorice instead.

Nevermind the CNS effects of methylene blue.  They are perceptible, and it's what gave people the idea to develop phenothiazine drugs.


text:
Finding the irreducible representation which an orbital transforms under
I have some major confusion on irreducible representations. An example of a problem I have is

Based on the character table for the point group of each of the follwing molecules, which irreducible representation would be used to classify the Cu $\mathrm{d}_{xz}$ orbital in $\ce{[CuCl5]^3-}$?

Since this complex probably has a trigonal bipyramidal geometry, I'm starting with the $D_\mathrm{3h}$ character table:
$$\begin{array}{c|cccccc|cc} \hline
D_\mathrm{3h} & E & 2C_3 & 3C_2 & \sigma_\mathrm{h} & 2S_3 & 3\sigma_\mathrm{v} & & \\ \hline
\mathrm{A_1'} & 1 & 1 & 1 & 1 & 1 & 1 & & x^2+y^2,z^2 \\
\mathrm{A_2'} & 1 & 1 & -1 & 1 & 1 & -1 & R_z & \\
\mathrm{E'} & 2 & -1 & 0 & 2 & -1 & 0 & (x,y) & (x^2-y^2,xy) \\
\mathrm{A_1''} & 1 & 1 & 1 & -1 & -1 & -1 & & \\
\mathrm{A_2''} & 1 & 1 & -1 & -1 & -1 & 1 & z & \\
\mathrm{E''} & 2 & -1 & 0 & -2 & 1 & 0 & (R_x,R_y) & (xz,yz) \\ \hline
\end{array}$$
But i don't know what exactly I need to do.
Is the irreducible representation just the row label, i.e. the Mulliken term? Or is it one of the operations?
Looking at the column with quadratic functions, $xz$ is paired with $\mathrm{E''}$, so that's my current best guess.


text:
What is the meaning of electrode potential?
When we say that the electrode potential for $\ce{Co}^2$$^+$/$\ce{Co}$ is -0.28 V and $\ce{Cu}^2$$^+$/$\ce{Cu}$ +0.337 V, what is the physical meaning?
Can you conceptually explain when two electrons react with $\ce{Cu}^2$ or $\ce{Co}^2$to form the metal, what happens and what is the source of these different potentials?

It compares with the standard hydrogen electrode. http://en.wikipedia.org/wiki/Standard_hydrogen_electrode

@t.c I know that, but what is the physical meaning of the potential. I know that it is presented in reference to something. How the potential of this electron transfer is different between Co and Cu, and what makes this difference.

The potential difference between two electrodes represents the voltage that is generated between the two electrodes that will allow the transfer of current/electrons. Physically it's basically a measure of how strongly one electrode attract electrons over the other which leads to the flow of current.


text:
Is the volume of a protein substrate equal before and after an enzymatic reaction? (pepsin degrading egg white suspension)
I am trying to determine the reaction rate of pepsin at different pH values by performing an experiment.
I am using egg-white suspension as the substrate and am adding differently concentrated $\ce{HCl}$ to the substrate to vary the pH of the reaction. Eventually, I add the enzyme. However, for the reaction rate I require the time and the substrate disappeared and product formed.
Is the volume of the egg-white suspension different before and after the enzymatic reaction so that I could use it to calculate the reaction rate or does the protein's volume remain the same after the reaction?

The volume of the protein will probably change (although not by much) because it will be denatured as it is degraded and as the pH decreases. As it denatures, the hydrophilic surface area and hydrophobic interior volume will change, leading to a change in the local water liquid structure and density. Denatured, short polypeptides also pack into smaller volumes than native proteins do, and so that would also cause the volume to decrease in most cases. The amount of water in the system will affect the magnitude of the change. I would predict that the volume would decrease slightly, probably on the order of what happens with a few degrees of temperature increase of water (around 0.1%).
So, the short answer is "Yes, the volume will change, but probably not consistently or by enough to be used to reliably measure the extent of the reaction."
This is not my area of expertise, so I don't know what the current standard protocols are for determining reaction rates for pepsin reactions. However, I was able to find a couple of research papers that talk about using ninhydrin to measure reaction extent:
The pH-Dependence of Pepsin-Catalyzed Reactions
Kinetics of Hydrolysis of Egg White Protein by Pepsin
In the second study, they quenched the reaction at a specific time by adding NaOH to bring the pH up, then centrifuged the products before measuring the reaction extent.
If this is not a professional scientific study (meaning it's a school project, or something like that) then you might look at the method used in this report:
The Effect of pH on the Reaction between Pepsin and Egg – White
Basically, they just added pepsin and watched to see how long it took for the solution to become "clear". This won't give you very good accuracy or precision, but it might work for a science project or school report.


text:
What holds atoms together?

I know atoms are held together by Ionic and Covalent bonds and i understand the ionically bonded atoms are held together by electrostatic forces. What about covalent bonds?
How are molecules of bonded atoms held together?Are ionic compounds just aggregates of oppositely charged ions and hence don't have any definitive "molecules". What about molecules of covalently bonded atoms?

Detail answer is given here

In non-nuclear chemistry, everything is electrostatic interactions. This is why you can learn and predict so much just by "following the electrons"

Covalent bonds are also formed because of electrostatic interactions - they are just more complicated conceptually than ionic (actually, ionic bonds are more accurately described by wavefunctions, we just try to keep things simple in the beginning). Since electrons exist as waves, when you confine them they start to do weird stuff (or what seems weird to us). For example, the shapes of atomic orbitals are all the way they are because the electrons act like standing, 3-dimensional waves trapped between a nucleus with a positive charge and a "zero-point" at infinite distance. When you put two atoms close together, the electrons on different atoms interact with each other, and the wavefunction becomes much more complicated. The result is that in some pairs of atoms, the wavefunctions combine to form bonding orbitals.

So the short answer to your first question is: "Molecular orbitals hold atoms together in covalent bonds, and those are a result of electrostatic interactions and the quantum nature of electrons."

Yes, ionic compounds are large collections of ions, and you can't really define "molecules" for them - instead we talk about "formula units" which are the lowest possible whole-number ratio of elements that represent the compound. Groups of covalently bonded atoms are also held together by electrostatic interactions, but since the covalent bonds are so much stronger, a molecular compound can exist "on its own" as a single molecule. Collectively, the forces that hold collections of molecules together are called van der Waals forces if they don't involve ions. In any atom or molecule, there is never a completely uniform charge density on the surface. For some molecules, this is extreme (water is a good example) and we say it is very polar, or that it has a large dipole moment. This is just another way of saying that one part has a negative charge and the other has a positive charge. In water it looks like this (from wikipedia):

In this picture, red means "more electrons" and blue means "less electrons." Water can form hydrogen bonds, which are very strong electrostatic interactions. Some atoms and molecules have an almost uniform charge density on the surface. We call these "non-polar" molecules - noble gases are good examples. However, even noble gases have what is called an induced dipole due to statistically correlated fluctuations in electron density when the atoms are near each other. As a result, even noble gases can be cooled to the point where they become liquid - the very, very weak electrostatic interactions will hold them together at low temperature, when they are not moving very fast. These forces are called London Dispersion Forces - after the guy who first described them. London dispersion forces are important, because they are found in all molecules - polar or not. In fact, this is what makes most plastics solid. Polyethylene, for example, is made of very long chains of essentially non-polar molecules (from wikipedia):

Each chain is attracted to the others through weak London dispersion forces, however, since each chain has tens of thousands of atoms, those tiny forces quickly add up to make large forces holding the polymer together. This is why polyethylene is solid at room temperature, and can be used to make things like shopping bags!

I don't like "electrostatic" in the sentenec "In non-nuclear chemistry, everything is electrostatic interactions." I think "coulombic" would be a much better fit.

@MaxW they are synonyms - Coulombic interactions are those described by Coulomb's Law, which is a model of electrostatic forces.
It might be a little misleading because it sounds like the charges aren't moving, when electrons certainly are. However, the original poster used the term, and it is a common way to describe the forces, so I used it as well.


text:
Orbital angular momentum
For hydrogen atom, L^2 and Lz can be obtained as eigenvalues for a particular wave function. But that does not completely specify the angular momentum vector. How to get about this problem?
Also, in the orbital approximation (ie the solutions obtained after ignoring electronic repulsions in the Hamiltonian) of the many electron atom( say helium) how does one go about finding total orbital angular momentum?

Welcome to Chemistry.SE! To acquaint yourself with this page, take the tour and visit the help center. Furthermore this tutorial shows you how math and chemical formulae can be nicely formatted on this site. Finally, we have an important policy: your questions (especially homework questions), should show your own work or thinking that you have already done in an initial attempt to answer the question.

Has my answer been helpful?

For hydrogen atom, L^2 and Lz can be obtained as eigenvalues for a particular wave function. But that does not completely specify the angular momentum vector. How to get about this problem?

Well, you can't. The different components of the angular momentum operator $\hat{\mathbf{L}}$, i.e. $\hat{L}_{x}$, $\hat{L}_{y}$, and $\hat{L}_{z}$, do not commute, so you cannot simultaneously determine all of them. But the squared angular momentum operator, which gives you the squared length of the angular momentum vector commutes with the components of $\hat{\mathbf{L}}$, so you can simultaneously determine the length of the angular momentum and one of its components. Which component you choose is immaterial, usually $\hat{L}_{z}$ is chosen. What you get out of those two operators can be pictured like this: you know the length of the angular momentum vector and its z-component; the other two components are undetermined but because you know the length they can only take on values that lie on a circle centered around the ${L}_{z}$ axis, whose offset is determined by the value of ${L}_{z}$ and whose radius is determined by $| \sqrt{\mathbf{L}^{2}}|$. For more details have a look at this Wikipedia article or any general physical chemistry or quantum mechanics textbook.

Also, in the orbital approximation (ie the solutions obtained after
  ignoring electronic repulsions in the Hamiltonian) of the many
  electron atom( say helium) how does one go about finding total orbital
  angular momentum?

Ignoring relativistic effects the Hamilton operator of a system consisting of $M$ nuclei and $N$ electrons, which are considered to be point charges takes on the form
\begin{align}
  \hat{H} &= \underbrace{ \sum_{j=1}^{M} - \frac{ \hbar^{2} }{ 2 M_{j} } \mathbf{\nabla}_{ \mathbf{R}_{j} }^{2} }_{ = \, \hat{T}_{\mathrm{K}} } + \underbrace{ \sum_{i=1}^{N} - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} }_{ = \, \hat{T}_{\mathrm{e}} } + \underbrace{ \frac{1}{2} \sum_{i \neq j}^{M} \frac{ e^{2} Z_{i} Z_{j} }{ 4 \pi \epsilon_{0} | \mathbf{R}_{i} - \mathbf{R}_{j} | } }_{ = \, \hat{V}_{\text{K--K}} }  \\
  & \quad \underbrace{- \sum_{i=1}^{N} \sum_{j=1}^{M} \frac{ e^{2} Z_{j} }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R}_{j} | } }_{ = \, \hat{V}_{\text{e--K}} } + \underbrace{ \frac{1}{2} \sum_{i \neq j}^{N} \frac{ e^{2} }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{r}_{j} | } }_{ = \, \hat{V}_{\text{e--e}} }
\end{align}
where $ \mathbf{R}_{j} $, $Z_{j}$, and $M_{j}$ are the coordinate, the charge number, and the mass of the $j^{\mathrm{th}}$ nucleus, respectively and $ \mathbf{r}_{i} $ is the coordinate of the $i^{\mathrm{th}}$ electron and $m_{\mathrm{e}}$ is the electron mass.
Furthermore $e$ is the elementary charge und $\epsilon_{0}$ is the vacuum permittivity.
The different terms of this Hamilton operator can be explained as follows: $\hat{T}_{\mathrm{e}}$ and $\hat{T}_{\mathrm{K}}$ are the terms representing the kinetic energy of the electrons and the nuclei, respectively, $\hat{V}_{\text{K--K}}$ describes the coulomb interaction between the nuclei, $\hat{V}_{\text{e--e}}$ is the interelectronic repulsion, and $\hat{V}_{\text{e--K}}$ represents the coulomb interaction between the electrons and the nuclei.
Using the Born-Oppenheimer approximation you can seperate the electronic degrees of freedom from the ones of the nuclei and treat the electronic problem independently and for the many-electron atom you also only have one nucleus to consider so the terms $\hat{T}_{\mathrm{K}}$ and $\hat{V}_{\text{K--K}}$ drop out of the Hamiltonian. If you now also ignore the electron-electron interaction you are basically left with a Hamilton operator describing two independent electrons moving in the electrostatic field of the nucleus:
\begin{align}
  \hat{H} &= \underbrace{ \sum_{i=1}^{N} - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} }_{ = \, \hat{T}_{\mathrm{e}} } + \underbrace{ \sum_{i=1}^{N}  \frac{- e^{2} Z }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R} | } }_{ = \, \hat{V}_{\text{e--K}} }
\end{align}
For this you can easily seperate the $N$-electron problem into $N$ one-electron problems
\begin{align}
  \hat{H} &= \sum_{i=1}^{N} \hat{h}_{i} \qquad \Rightarrow \qquad \Psi(\mathbf{r}_{1}, \ldots, \mathbf{r}_{N}) = \psi(\mathbf{r}_{1}) \cdots \psi(\mathbf{r}_{N})
\end{align}
where each one-electron Hamiltonian $\hat{h}_{i} = - \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2} - \frac{e^{2} Z }{ 4 \pi \epsilon_{0} | \mathbf{r}_{i} - \mathbf{R} |}$ and one-electron eigenfunction $\psi_{n}(\mathbf{r}_{i})$ basically satisfy a Schroedinger equation for the hydrogen-like atom
\begin{align}
  \hat{h}_{i} \psi_{n}(\mathbf{r}_{i}) = \varepsilon_{n} \psi_{n}(\mathbf{r}_{i}) \qquad \text{with} \qquad \varepsilon_{n} = -\frac{m_{\mathrm{e}} e^{4} Z^{2} }{ 32 \pi^{2} \epsilon_{0}^{2} \hbar^{2}} \frac{1}{n^{2}} \qquad n = 1, 2, ...
\end{align}
Using spherical coordinates, $x, y, z \to r, \theta, \phi$, you can rewrite the kinetic energy operator $- \frac{ \hbar^{2} }{ 2 m_{\mathrm{e}} } \mathbf{\nabla}_{ \mathbf{r}_{i} }^{2}$ in terms of $\hat{\mathbf{L}}^{2}$ so that the Hamilton operator becomes
\begin{align}
  \hat{h}_{i} = \frac{ - \hbar^{2} }{ 2 m_{\mathrm{e}} r^{2}} \frac{ \partial }{ \partial r } \biggl( r^{2} \frac{ \partial }{ \partial r } \biggr) + \frac{ 1 }{ 2 m_{\mathrm{e}} r^{2} } \hat{\mathbf{L}}^{2} - \frac{ Z e^{2} }{ 2 m_{\mathrm{e}} r }
\end{align}
where $r$ is the distance between nucleus and electron. The operator $\hat{\mathbf{L}}^{2}$ commutes with the Hamilton operator $\hat{h}_{i}$ because $\hat{\mathbf{L}}^{2}$ commutes with itself and does not involve the variable
$r$. Likewise, the operator $\hat{L}_{z}$ commutes with $\hat{h}_{i}$ because it
commutes with $\hat{\mathbf{L}}^{2}$ and also does not involve the variable $r$.
Thus the operators $\hat{\mathbf{L}}^{2}$, $\hat{L}_{z}$, and $\hat{h}_{i}$ have simultaneous eigenfunctions:
\begin{align}
  \hat{h}_{i} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &= \varepsilon_{n} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \\
  \hat{\mathbf{L}}^{2} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &= \hbar^{2} \ell_{i} (\ell_{i} + 1) \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \qquad \qquad \ell = 0, 1, 2,..., n-1 \\
  \hat{L}_{z} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) &= \hbar m_{i} \psi_{n}(r_{i}, \theta_{i}, \phi_{i}) \qquad \qquad m_{i} = -\ell_{i}, -\ell_{i} + 1, \ldots , \ell_{i} - 1, \ell_{i}
\end{align}
Thus, you have a well defined angular momentum for each electron in the system.


text:
If caffeine is an alkaloid, why is coffee acidic?
I have just put together on my mind these two facts: caffeine is an alkaloid and brewed coffee is slightly acidic (pH = 5).
My Biology teacher and my Chemistry teacher could not elaborate satisfying answers, but my Physics teacher said it could be because of substances to the industrial coffee powder (hillbillies in Brazil, for instance, do not display gastritis attained from too much coffee).

Saying something's an "alkaloid" isn't that informative as to it's pH.  Sure, if it's a free base, it likely is, but more often than not people use HCl salts of the drugs for their improved solubility, and those salts are acidic; morphine HCl as an example. The final step of methamphetamine production is often acidifying the organic, free-base form to have it precipitate out of the organic phase.

I would also add that gastritis is not caused by the acidity of coffee, or in most cases by the acidity in any particular food.  Gastritis is generally caused by bacterial infection or by overuse of NSAIDs.  If you're referring to the more commonly linked to coffee "heartburn" or GIRD, that is caused by the caffeine, not because of acidity or alkalinity but because it relaxes the muscle that controls the connection to the esophagus.

The lethal dose of caffeine is around 10-15 grams. If the main constituent of ground coffee was caffeine then anyone who had more than a couple of cups of coffee would be dead

Are you confusing Alkaloid and alkali? An alkaloid is a type of organic compound, but is not necessarily a base (or alkali).

Alkaloid means "alkali-like". Steroid means "solid-like". But factoid does NOT mean "fact-like".

Coffee contains hundreds, if not thousands, of other compounds in addition to caffeine.  Included among these other compounds are many acids.  Many small, organic acids such as  citric, malic, lactic, pyruvic and acetic acid are present, but both quinic acid and chlorogenic acid (and their derivatives) are usually present in even higher concentration.  Phosphoric acid, an inorganic acid, is also present. The exact concentration of these various acids depends upon processing variables such as roast conditions and grind size.  Here is a link to an interesting, one-page discussion of the subject.
quinic acid

chlorogenic acid

Edit: In Season 3, episode 6 of "Breaking Bad", we meet Walt's new lab assistant Gale, who is brewing coffee using a lab glassware set-up.  Walt says that it is the best coffee he has ever tasted and Gale responds by mentioning "quinic acid" as something to be considered in brewing great coffee.

Nice. But can you find a mistake in your claimed formula of chlorogenic acid?

@mykhal Thanks for pointing that out.

Alkaloids are low molecular weight nitrogen-containing compounds, mostly from plants. The nitrogen is usually in the form of amines and this often results in a basic compound, but there are many alkaloids where this is not the case. In fact, caffeine is a pretty weak base. I calculated the pKa of the imine nitrogen with MarvinSketch as -.92, so caffeine is not going to affect the pH of coffee much.
Also, caffeine isn't even a major component of brewed coffee, only accounting for ~1% of the dry mass. If you look here you'll see that brewed and green coffee naturally contains a fair bit of soluble acids, including things like nicotinic acid (vitamin B3), which probably accounts for the pH of brewed coffee, rather than the caffeine or any processing.

PubChem has the pKa as 10.4 : https://pubchem.ncbi.nlm.nih.gov/summary/summary.cgi?cid=2519

Weird, the source they cite says "Caffeine is both a weak acid and a weak base with pKa values of 14.0 and 0.7. Although partial ionization to cation and anion forms may occur, electrochemical studies have found that the neutral form of caffeine was predominant in the pH range of 5.5 to 9." http://toxnet.nlm.nih.gov/cgi-bin/sis/search2/r?dbs+hsdb:@term+@rn+@rel+58-08-2 Anyways, even if it were a strong base, its concentration is still very low (~2 mM assuming 100 mg/250 mL)

As far as pinning down a specific, standalone pH, caffeine is, if trivially, an acidic contributor to coffee:

pH = 6.9 (1% solution)
O'Neil, M.J. (ed.). The Merck Index - An Encyclopedia of Chemicals, Drugs, and Biologicals. Cambridge, UK: Royal Society of Chemistry, 2013., p. 289
from HSDB
HSDB
Record Name: CAFFEINE
URL: https://toxnet.nlm.nih.gov/cgi-bin/sis/search/r?dbs+hsdb:@term+@rn+@rel+58-08-2