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Can we use liquid hydrogen to power a conventional engine?
Hydrogen has the best fuel efficiency, so if we were to use it in our engines wouldn't their efficiency increase as well.
The engines would require changes and all but would it be plausible to have that?

Engine is not much of a problem. Hydrogen production and storage is the tricky part.

Does hydrogen have the best fuel efficiency?  That depends on how you measure it.  Hydrogen wins (1) on energy per mass but loses to gasoline on energy per volume (even under high pressure).  The mass advantage will be negated in most applications by the mass of the storage mechanism e.g. a high pressure cylinder v the gas tank of you car.  https://en.m.wikipedia.org/wiki/Energy_density (1) Ignoring nuclear fuels.

I'm voting to close this question as off-topic because it's about car engineering rather then chemistry.

@Mithoron I disagree not least because I was taught the essentials of this subject in a chemistry course and by a Nobel prize winning chemist. And a lot of the concepts are chemical. Yes, there is a lot of engineering too, but the basics are chemistry.

@matt_black I'm not exactly sure what's this question is about - it's obvious that you can use hydrogen as gas fuel and there are lots of actual users. If it's about efficiency then how you define it?

@Mithoron the question is plausible but a little naive. My answer explains why a simplistic view doesn’t work. And it defines the efficiency that matters in a vehicle.

The efficiency that matters is the efficiency of the vehicle not the efficiency of the engine
The problem of designing more efficient transport is bigger than the problem of designing better engines. Conventional combustion engines can be adapted to burn a wide variety of fuels (alcohols, natural gas and hydrogen are not too big a challenge). And there are alternative ways to move vehicles than combustion engines (fuel cells, for example, which run on hydrogen). So it looks like we can solve the engine problem.
But that isn't enough. To have a viable vehicle we have to solve other problems as well. How to store and deliver the fuel to the engine, for example. And how to distribute the fuel across the country so vehicles can be refilled. The second problem is probably easier to solve if the first problem–a viable vehicle–has a solution. Unfortunately, it isn't that easy.
To have a hydrogen powered car we would need a way to store enough hydrogen in the vehicle to allow it to drive several hundred miles.
This is easy to do with liquid hydrocarbon fuels in conventional vehicles (petrol/gasoline and diesel have very high energy densities and can be stored in a simple unpressurised tank). The weight of the fuel alone is larger then the weight of the tank that contains it and the volume is small as the energy density is very high.
It isn't so easy with hydrogen. While there are several existing ways to store hydrogen and some proposed solutions that are not yet operational, all of them involve big issues with volume and weight. Compressed hydrogen will involve very heavy thick-walled tanks and even then the energy density of the hydrogen gas alone will be lower than a tank full of petrol. The weight and volume of the tanks required to store as much energy as a tank of petrol will very large. Liquid hydrogen has a different set of compromises but isn't much better as it requires cryogenic system to keep the liquid hydrogen cool. There is considerable interest in clever chemical ways to store hydrogen (clathrates, metal organic frameworks and various molecular systems than can reversibly store hydrogen are all being researched).
To get an idea of the scale of the challenge it is worth looking at the mass and volumetric characteristics required to compete with conventional fuel. Diesel (slightly more efficient than petrol) carries about 12kWh/kg and about 10kWh/L. The current targets for research on hydrogen storage for the Fuel cell Technology Office are 1.5kWh/kg and 1kWh/L. This is about 10 times worse than diesel even ignoring the fact that fuel-cell power vehicles might have other advantages partially compensating the overall vehicle when compared to combustion engines. And the best technology solutions are much better than liquid hydrogen storage.
So you can use liquid hydrogen (or other ways of storing hydrogen) to power an conventional engine or vehicle (and such vehicles exist). But the cost and design compromises to make this work for a whole vehicle given that you need 10 time the volume and weight to store the equivalent energy, make such solutions not competitive for mass market vehicles.


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Is there an example where interpreting reaction coordinate as time leads to an incorrect conclusion?
I have heard that we should not interpret the reaction coordinate in an energy diagram as time, but are there any situations in which we would arrive at an incorrect conclusion if we did so?
When I say interpreting reaction coordinate as time, I really mean as $\frac{\text{time so far}}{\text{total time for reaction}}$ (I would have written this instead of "time", but I'm not sure of a word or phrase for this).

There's probably a monotonic mapping from reaction coordinate to time but there's no reason to believe this mapping is uniform or smooth. Also, it's not clear what you mean by "wrong conclusion." What conclusion can you draw from a reaction coordinate at all?

You can compute the free energy of a system as a function of a reaction coordinate. If you carry over that to time, the result is conceptually misleading.  Thermodynamic calculations don't tell you about kinetics.

@CurtF What does free energy plotted against time look like?

A reaction coordinate is never time but a description of the distance between the reacting atoms as the reaction proceeds vs the energy. Time is not linear in this respect, i.e. the reaction moves with different speed at different points on the reaction coordinate.

Transition states are very short lived species, generally lasting less than a picosecond. At the single molecule level, crossing a transition state is a very fast event: it goes from reactants to products in less than a picosecond.
Some reactions are very slow because the probability of crossing the transition state is very low, but never because they spend a long time at the transition state. Thus, interpreting reaction coordinate as time is incorrect.
In a slow reaction, a small number of molecules will react early, but the majority will spend a long time in the reactants state, waiting to cross the transition state. If one could take a static picture of an ensemble of molecules, some would be in the reactants state, some would be in the products state, and it would be almost impossible to see any molecule at the transition state.

One of the oldest reactions to be understood in modern terms qualifies.  Heat mercury in the presence of oxygen, the mercury is first oxidized only to be reduced again at higher temperature as the equilibrium constant shifts.  The reaction coordinate, defined as the amount of mercury oxidized, doubles  back as time advances in this nonisothetmal process.

It is incorrect to define the reaction coordinate "as the amount of mercury oxidized". The reaction coordinate relates to breaking or forming chemical bonds, not to the number of molecules that are in the products state.

The number of Hg-O bonds formed, maybe?

Number of molecules is not a reaction coordinate. Bond length between Hg-O is a valid reaction coordinate.

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Why the DMM (multimeter) register negative resistance when measure a drop of hydrochloric acid
I noticed that when I put the multimeter electrodes (in the ohmmeter mode) in a drop of hydrochloric acid with mild concentration (15%) the multimeter displays negative values. Can someone explain this behavior?

Because you have made an electrochemical cell.

The multimeter probe tips are dissolving and setting up a galvanic cell that augments the current flow thru the ohmmeter.
The probe tips may be of the same metal, but the ohmmeter voltage will favor the dissolution of one over the other, and the cell thus formed will assist the applied voltage.
Quickly switching to voltmeter (or milliammeter) mode will show a cell voltage (current).
Using the ohmmeter to measure resistance of a non-corrosive solution (e.g., NaHCO3) will show some (positive) resistance.
Inert electrodes (Pt, Au) will show a positive resistance.

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Why does silver plating require undercoats of copper and nickel?
I came across a question that was about silver plating (of steel), but I couldn't help but wonder why the sequence was illustrated as this:
step 1 A coating of copper is applied to the object.
step 2 A coating of nickel is applied to the object.
step 3 The coating of silver is applied to the object.
Why are these initial coats of copper and nickel required, and why this seemingly random sequence (since nickel is in between but has the highest reactivity)?

I have read that the nickel plating helps removing oxide layers when working with stainless steel which is beneficial for the silver plating process.
Sadly I do not know why the copper is used but I suspect the context could explain the copper.

I never saw an explanation, only that ( historically) that is the way that works. I expect it has something related to getting a good bond . Although the nickle makes the substrate ( copper) "silver" color so the silver layer does not need to be as thick. Both the copper and nickle also provide corrosion protection to the steel. Chromium plating is done the same way.


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Why can't amylase digest glycogen?
Amylase is an enzyme that breaks down starch in the form of amylopectin and amylose. Both amylose and amylopectin are formed by alpha glucose joined together by (1-4) and (1-6) glycosidic bonds. Glycogen is no exception, just that it has more branching.
However, why is it that a Google search shows that it is hydrolyzed by glycogen phosphorylase rather than amylase? Also, how can amylase digest both (1,6) and (1,4) glycosidic bonds?

Enzymes do magic :) I am not kidding much. Do not forget that is not " the chemistry" that changes, but if the  catalyst can fit with the substrate structure. Enzymes  and in general proteins are extremely specialised by their  primary, secondary etc structures.

Note that there is not one amylase, but three main types, dozens of different subtypes, and a number of non-amylase enzymes that also break down starch, all of them with different activity under different conditions and molecular surrounding of the bond to be cleaved.

Amylase can’t digest glycogen because of its inability to attack the branching (1→6) linkages.
Perhaps, another very important reason is controlling the rate of glycogen metabolism through glycogen phosphorylase. Just like any other biological system, regulation of metabolic substrates and/products is crucial to maintaining the balance (homeostasis) so to prevent excess glucose production from glycogen metabolism or to little according the needs of the organism.

Tl;dr
There are  actually three forms of amylases:

α-amylase (an endoglycosidase, which can hydrolyze a glycosidic linkage anywhere along the chain to produce glucose and maltose).
β-amylase (an exoglycosidase that cleaves from the nonreducing end of the polymer).
gamma amylase

Now starch consists of two main components amylose and amylopectin;

Amylose is a linear polymer of several thousand glucose residues linked by (1→4)
bonds:

Amylopectin consists mainly of (1→4)-linked glucose residues but is a branched molecule with (1→6) branch points every 24 to 30 glucose residues on average.
The primary structure of glycogen resembles that of amylopectin(not amylose), but glycogen is more highly branched, with branch points occurring every 8 to 14 glucose residues:

Digestion
The digestion of starch, the main carbohydrate source in the human diet, begins in the mouth. Saliva contains α-amylase, which randomly hydrolyzes all the (1→4) glycosidic bonds of starch except its outermost bonds and those next to branches.
On the other hand glycogen being a highly branched molecule, it is evident that amylase won’t be a good enzyme to digest it due to physical structure constraints.
 Substrate specificity of the enzymes ensure that the appropriate substrate which is highly optimized for the active site will be broken down.
Although glycogen posseses parts similar to starch's amlyopectin its structure is not optimized to be broken down by amylases. Note: even these amylases only partially break down starch to different oligosaccharides

There are alone three main types of amylase, alpha, beta, gamma. https://en.wikipedia.org/wiki/Amylase


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What is equivalents per mole?

Which has maximum number of equivalents per mole of the oxidant?
  $$
\begin{align}
\ce{MnO4- + H2O2 &-> Mn^2+ + O2}\tag{i}\\
\ce{Cr2O7^2- + Fe^2+ &-> 2 Cr^3+ + Fe^3+}\tag{ii}
\end{align}
$$

The answer is the second option. I wanna know why
If there's another term different from equivalent mass like equivalents, then I tried doing it using the formula = No.of moles of oxidant x n factor (no of electrons gained)

This is a homework question. We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.

The question does not make sense to me---what should 'equivalents' mean in this context?

The question can be understood as:
Which one of the following oxidants has a greater n-factor as there's only 1 mole each of permanganate ion and dichromate ion.
The n-factor of permanganate ion is (7-2)=5
while
The n-factor of dichromate ion is (6*2-3*2)=6
So, the answer is the 2nd reaction.

Oh now I understand, So in general n factor is the initial oxidation state * no.of atoms - final oxidation*no.of atoms ?

Is the equivalent mentioned here the same as equivalent weight

n-factor is basically the number of moles of electrons being exchanged in the reaction. And equivalent weight is the weight of the species required for the exchange of one mole of electrons.


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Concentration of electrons in silicon sample with a phosphorus dopant
I'm studying electrical and computer engineering and all I remember from chemistry is from high school. We have a $\ce{Si}$ sample and we want to replace one in every million atoms with a phosphorus atom.
I want to determine the concentration of conduction electrons after this process and I'm given the molar mass $\pu{0.028 kg/mol}$ and the density $\pu{2300 kg/m^3}$.
It might be very simple, but what can I get from these two? I checked molar mass since I haven't heard about it ever, but I'm getting more confused.

Unless there is something I'm missing entirely, the answer is quite straightforward. Lets assume that each phosphorus atom contributes one free electron. Then the concentration of electrons is equal to the concentration of $\ce{P}$ atoms doped to the silicon crystal:
$$c_\mathrm{e} = c(\ce{P})\label{eq:1}\tag{1}$$
Concentration of phosphorus $c(\ce{P})$ can be found from its amount $n(\ce{P})$ and volume $V$:
$$c(\ce{P}) = \frac{n(\ce{P})}{V}\label{eq:2}\tag{2}$$
On the other hand, volume can also be determined from the mass $m$, density $\rho$ and molar mass $M$ of silicon:
$$V = \frac{m(\ce{Si})}{\rho(\ce{Si})} = \frac{n(\ce{Si})\cdot M(\ce{Si})}{\rho(\ce{Si})}\label{eq:3}\tag{3}$$
Now substituting unknown parameters in \eqref{eq:1} and \eqref{eq:2} with \eqref{eq:3}, and knowing that $n(\ce{P}) : n(\ce{Si}) = 1 : 10^6$, concentration of electrons can be found using the values you provided:
$$c_\mathrm{e} = \frac{n(\ce{P})}{n(\ce{Si})} \cdot \frac{\rho(\ce{Si})}{M(\ce{Si})} = 10^{-6}\cdot\frac{\pu{2300 kg m-3}}{\pu{0.028 kg mol-1}} = \pu{8.21e-2 mol m-3}\tag{4}$$

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D6h symmetry clusters- orbital classification confusion!
I was am simulating high symmetry clusters D6h in several programs such as molcas and gaussian. I understand that the highest symmetry used in these programs is D2h and therefore there would be mixing of the different symmetry representations upon correlating from D6h to D2h. However upon simulating Mn@Si12 cluster, I'm seeing that the silicon atoms in the silicon cage have for all representations: s, px, py and pz orbitals despite there being restrictions according to D2h e.g b3u representation has only px functions and b2u only having py functions. Could anyone please shed some light on this phenomenon?

The symmetry operations are usually defined with respect to the origin of the coordinate system. For single atoms one naturally chooses $(0,0,0)$, which coincides with the symmetry center.
For example the $\sigma_{xy}$ mirror plane in matrix representation (with $(0,0,0)$ being the symmetry center) would be:
$\sigma_{xy}=\pmatrix{1&0&0\\0&1&0\\0&0&-1}$
which means: invert every $z$ component. If applied to some object centered on $(0,0,0)$ (e.g. the AO of an atom on this position), the result would still be centered on $(0,0,0)$.
In case of molecules we can have at most one atom sitting on the symmetry center. Sometimes none, e.g. in $\ce{C2}$ where the symmetry center is in the middle between both atoms. If you apply the above matrix here (assuming the molecule axis is aligned with the $z$ axis), then it would map the AO from one to the other atom. Therefore it is not symmetric (with respect to the above defined symmetry operation) anymore.
Thus you cannot relate the IRREP of an atomic orbital in the single atom, with its IRREP in the molecule. The symmetry center changed! For example take an $s$ atomic orbital of any of your $\ce{Si}$ atoms. The position of the atom does not coincide with the center of the molecule. Therefore any symmetry operation (except for $E$) would map the $s$ orbital to some other $\ce{Si}$. The AO is no longer characterized by an irreducible representation.
What the code does instead, is first generating Symmetry Adapted Linear Combinations (SALC) of the atomic orbitals. Those are then again characterized by IRREPs (of the molecule) and used for the calculation. Now each obtained MO is indeed only constructed from SALC of the same IRREP. In other words: the Hamiltonian, and therefore the MO coefficient matrix as well, become block diagonal. However, in the output the MO coefficients are transformed back from the SALC basis set to the AO basis set, which is not block diagonal and may contain contributions from any AO for each IRREP of the MOs (an exception would be the AOs of an atom located in the center of the molecule).

Thanks alot for your answer, I believe you have answered my question quite clearly, but I'm still struggling to understand a few things. Would you be kind enough to answer this query please. If it is not possible to relate Irrep of AO to Irrep of MO and each MO may contain contributions from any AO for each IRREP, how is it possible to classify basis functions according to irreps . This done for high symmetry molecules in molcas manual p438.  Many thanks for the help.

I only found a HTML version of the molcas manual, so I cannot check what p438 is about. Anyway, you do have a basis which you can classify according to irreps: the SALC, which is build based on the AO basis.

Are you referring to Table 10.5? It has a footnote: "Functions placed on the symmetry center." As there can be at most one atom in the symmetry center, this does not work in general.

Yes i'm referring to table 10.5. Thanks alot for looking at it. So for the example of NiH, Ni is on the symmetry centre? and for C2 and Ni2, one atom of each lies on the centre and by symmetry operations, the functions are applicable to the other atom?

Homonuclear diatomics, like $\ce{C2}$ and $\ce{Ni2}$ have an inversion center inbetween both atoms. That is their symmetry center, hence there is no atom sitting on the symmetry center.
For the other example: heteronuclear diatomics don't have inversion symmetry, so there is no symmetry center.

Thanks alot for the help and the detailed answers. Please let me know if there is any way to reward your answers apart from upvoting.


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Relationship between Yield and Applied Potentials
This question revolves around this question from Atkins' Chemical Principles 5th ed., Chapter 13, Question 13.116:

Consider the electroplating of a metal $+1$ cation from a solution of unknown concentration according to the half-reaction $\ce{M+ (aq) + e- -> M (s)}$, with a standard potential $E^\circ$. When the half-cell is connected to an appropriate oxidation half-cell and current is passed through it, the $\ce{M+}$ cation begins plating out at $E_1$. To what value ($E_2$) must the applied potential be adjusted, relative to $E_1$, if $\pu{99.99 \%}$ of the metal is to be removed from the solution?

How is this problem solvable? I know that at applied potential $E_1$ the cell potential is $0$ ($\Delta G = 0$), so that means at $E_2$, the equilibrium constant must demand a yield of $\pu{99.99 \%}$. However, doesn't this yield depend on what the oxidation half-cell is?
For example, the oxidation reaction could be one from cerium(III) to cerium(IV) or from chloride ions to chlorine gas, both of which have different effects on the equilibrium constant. I've gotten as far as $-nFE = RT\ln{K}$, where I do believe $E = E_2 - E_1$ and $E$ is positive to ensure negative Gibbs free energy change.
This is an even number question so there is no given answer. I would really appreciate a solution or at least explanations clearing up any misconceptions that I may have / clues to how to solve this problem. This is not a homework problem.

Well, if this was better made exercise you'd probably have no problem. BTW stop this "not a homework", it's a textbook exercise so it falls under "homework stuff" in obvious way.


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What does 'forces' mean in the output of Gaussian calculations?
In the output of a Gaussian calculation of frequencies, I can see a table listing all the 'forces' of all the atoms. It looks like this:
Center Number       Atomic Number       Forces (Hartrees/Bohr) X Y Z

  1        6          -0.001094502   -0.000768130   -0.006919164
  2        6           0.000528733   -0.000637636    0.005949765
  3        6           0.000424174   -0.000366750   -0.002904382
  4        6          -0.002619381   -0.000001885    0.005496979

...
What are these 'forces'? Are they related to those vibrations? But during vibrations, the force on each atom by other atoms is always changing, so why is only one constant 'force' value assigned to each atom?

I am not sure, but those might be related to the Hessian matrix, the partial second derivatives, which is diagonalized in a frequency calculation. Note that $F=m\frac{\mathrm{d}^2}{\mathrm{d}x^2}$ in 1D.

Did you optimize the geometry before running the frequency calculation? Frequency calculations don't have any meaning if they aren't taken at a stationary point of the potential energy surface.

@Feodoran But the forces are not calculated from that formula, right? The forces are calculated from the electronic structures of the atoms, regardless of the masses of the atoms, right?

I always took them as force constant (spring on my mind)

@Alchimista But they're not. Look at their unit - Hartree/Bohr. Obviously they are just forces.

@Ath right, I guess I confused the formulas for coulomb force and gravitational force.

@Ath is there something else you are looking for from an answer?

The forces you have listed are the derivatives of the energy with respect to a displacement of a particular atom in each Cartesian direction. These forces are useful for finding stationary points on the potential energy surface, which occur when the force on each atom is zero. Computationally, it's not possible or worthwhile to try to get all the forces to be exactly zero, so stationary points are determined by a combination of criteria involving the forces and the displacement between optimization steps.
The vibrations in the system are related to the second derivatives of the energy with respect to the atomic coordinates, which are contained in the Hessian matrix.

Thank you for your helpful answer, but I still have a question: these forces are calculated based on the atoms' electronic structures and the distances between atoms, and these calculations don't involve masses of atoms, correct?

@Ath they shouldn't depend on the masses explicitly. See this other question that references how to obtain forces from the Hartee Fock equation or the referenced source (Szabo and Ostlund's Modern Quantum Chemistry) for the full formulation.

You said '...stationary points on the potential energy surface, which occur when the force on each atom is zero', but I don't think the stationary points means forces are zero. A stationary point is a minimum energy in an area, which means it has smaller forces than other points in this area, but not necessarily zero

@Uh-Oh by definition, a stationary point of a function is one for which the derivative (in this case, the force) is zero. Now true, in implementations we never manage to completely zero out all the forces so we set some minimum RMS force criteria, etc, but that is more an issue of our implementation and approximations than with how you would find a true stationary point.

@Tyberius conceivably those forces could also be used to run AIMD, not just geometry optimizations. Also, Uh-Oh , the force constant of a harmonic vibrational mode should not be confused with forces.


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Gelatination of solution in saponification
When I performed a saponification experiment in school, I added some ethanol to the solution (glycerol ester fatty acid + sodium hydroxide) to change the equilibrium of the reaction. (Molecules of salt of fatty acid made some chunks so they don't react efficiently.) But accidentally I spilled a lot of ethanol and the solution became hard, like a hair gel. And after I added some water to it, it became liquid again.
What I want to know is why the solution became a gel. I know that ethanol is used for an efficient reaction because ethanol can be a common solvent of grease and water and it can help sodium hydroxide and glycerol ester meet, but I don't know why they became harder when ethanol was overused.
Of course, the solution was in the water bath.

Sodium stearate is not very soluble in water (above about 2%, it becomes viscous but not a firm gel). Ethanol can reduce the viscosity at first, but sodium stearate is not soluble in ethanol either, so an excess of ethanol causes a precipitation of solid that firms up the aqueous gel. More water  dissolves that precipitate and restores the fluidity.
The solubility and rheology of fatty acid soap solutions depends on temperature, concentration, solvents and the structure and purity of the fatty acid group. Stearate is very different from oleate, and both are different from laurate.

Thanks for the answer. Then the reason of gelation is because of only sodium stearate? Can sodium oleate or sodium laurate cause gelation either?

Sodium stearate is a firm solid (Ivory Soap); sodium oleate is softer (esp. when wet, like Soft Soap - which I think is the potassium salt in solution). Sodium laureate has a shorter chain, and while it is a firm solid, it needs a higher concentration in water to firm up. (The laureate is more water soluble.)


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Concentration of Solute And Solvent
Say I create a solution consisting of one liter of motor oil dissolved in three liters of acetone, and I leave the solution uncovered and allow it to evaporate.
As the solution evaporates, will the percentages of the solute and solvent remain the same?

No---look up non-ideal solutions.

You should read about azeotropes.

The mixture, originally 4L of 25% motor oil, loses acetone by evaporation until you have 1L of 100% motor oil. The % of motor oil increases from 25% to 100% as the acetone evaporates off.

It's a dire simplification. Fuel has significant volatility and solution becomes less non-ideal when concentration of acetone decreases.

I think motor oil (lubricant) is not appreciably volatile (bp ~ 570 F). If the questioner meant diesel fuel (bp ~420 F), even that is quite non-volatile compared to acetone (bp ~ 134 F). Granted, my answer is a simplification, but not so dire, considering the question.

Thank you; that's what I meant. A highly-volatile solvent such as acetone mixed with motor oil.


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Is [PdCl2(PMe3)2] diamagnetic or paramagnetic?
Please tell me where I am wrong. The complex will have $\ce{Pd^{2+}}$ ion, which has a $\mathrm{d^8}$ configuration. So, it will have $2$ unpaired electrons. Unpaired electrons will mean that it is paramagnetic.
My reference book has this line:

$\ce{[PdCl2(PMe3)2]}$ is a diamagnetic complex of Pd(II)

Shouldn't it say "paramagnetic"?

The splitting pattern for a square planar complex is simply different from that of a tetrahedral or octahedral complex. This is not about Jahn–Teller distortions.

Possible duplicate of Why is [PdCl4]2- square planar whereas [NiCl4]2- is tetrahedral?

You have not considered the d orbital splitting which has occurred due to the presence of the bonded ligands. Yes, the $\ce{Pd^2+}$ ion adopts the $\mathrm{4d^8}$ electronic configuration with two unpaired electrons, as shown below. However, when it forms the square planar complex, the d orbitals split in energy levels and the electrons now occupy the new energy levels differently, still abiding by Hund's rule and the Aufbau principle. These d orbitals no longer possess any unpaired electrons and thus, the complex is not paramagnetic, but diamagnetic.

Image source: Wikimedia Commons

Image source:Wikimedia Commons

Would there be any coordination complex of Pd that is tetrahedral for CN=4?

Tetrahedral complexes for palladium are rather rare (As for the explanation as to why they are less preferred for Pd, the post Orthocresol linked above more offer some insight). Upon searching, I did manage to find one: http://pubs.rsc.org/en/content/articlelanding/1999/cc/a904016a#!divAbstract

this is a special case of jahn teller distortion....in pd(II), weak ligands can remove the degeneracy of eg orbitals & as the energy difference between two eg orbitals are high, the ekectrons pair up in the dz2 orbitals. Thus form a diamagnetic complex

Okay.Is this correct:- So this complex is dsp2, hence square planar . For square planar there's that weird splitting.  So configuration would be dzx(2) dyz(2) , dz2(2) , dxy(2), dx2-y2 (0) . No unpaired electrons! And dx2-y2 is the d in dsp2. Am i correct?

I think it will be dz2. See the pic I post above or the diagram for square planar complex. dz2 has low energy than dx2-y2

Would [NiCl2(PMe3)2] be paramagnetic? With two unpaired electrons (as shown in the figure on left side in the image you sent as answer)?

For nickel(II),  JT distortion is only obtained by using strong field ligand. so the corresponding Ni complex will be paramagnetic(left diagram). But if any stong field ligand is present in complex, then it will be diamagnetic(right diagram)

But dz2 would be filled as it has lower energy. So dx2-y2 would be used for coordinate bonding.

Also there would not be any Pd coordinate complex with tetrahedral structure for CN=4 , would there be?

How is this a Jahn–Teller distortion from octahedral geometry when there are only four ligands to begin with?

In the pic that I post, see how a octaderal complex has been transformed into square plannar in the Ni(CN)6 case.....An square plannar complex is formed from octahedral complex due to extreme jt distortion

No this is not correct use of the term. A JT distortion leads to a change in molecular geometry, it does not lead to a complete loss of two ligands.

I agree that the orbital diagram is an extreme limiting situation of the JT distortion observed in d9 complexes, but using the term to refer to d8 square planar complexes is not something I’ve ever seen before and I don’t think it’s applicable.

It can leads to loss of one or two ligands. If it is not so,  then how will you explain that [Ni(CN)4]2-  is square planar not tetrahedral on the basis of CFT?

No, the Jahn-Teller theorem does not cover changes in molecular composition, it only covers changes in the arrangement of the atoms.

CFT is a theory with limitations. You have to argue that in CFT the lower four orbitals are sufficiently stabilised (in turn due to the strong field ligand - note that CFT cannot explain why cyanide behaves strong field) such that populating then is more favourable than the tetrahedral case. See eg https://chemistry.stackexchange.com/a/40881/16683

Now ans this, how to draw the splitting pattern of d-orbitals in square planar environment

splitting pattern of square planar environment is drawn by considering extreme jt distortion & removing two axial ligands from oct. complex. This is similar to that.

If you’re talking about CFT, you derive it based on qualitative electron repulsion arguments. And your picture is perfectly alright. There is some debate about the exact energy ordering of the lowest three orbitals, but it doesn’t affect the conclusion. Yes it is similar to the JT distortion, and as we agreed the orbital diagram can be derived logically from the JT distortion, but the state of only having four ligands is not a JT distortion.


text:
Reaction characteristics of alkaline battery
I'm currently struggling with understanding why electrochemical reactions take place the way they do.
For example, in an alkaline-manganese battery manganese-dioxide is getting reduced by reacting with water whereby the electron is actually absorbed by the hydroxide molecule (?):
$$\ce{MnO2 + H2O + e- -> MnO(OH) + OH-}$$
Essentially, my question is why does manganese give up the bond with the two oxygens and instead bonds with one oxygen which itself is bonded with an hydroxide and why does the other hydroxide absorb the electron instead of the other molecules.
Would manganese-dioxide and water also react without the transferred electron?

Chemistry can be difficult because it has lots of pieces of different importance, some of which are not depicted in equations. In this case, the important pieces are the ones that undergo the greatest change.
The hydrogens don't change much at all - they are always bonded to oxygen. The oxygens have their outer shells filled, and they sit in a crystal lattice (MnO2) or a liquid (H2O); there's a change, but it's moderate (in this case, anyway).
But the manganese changes from +4 to +3! Big change! It took a lot of energy to rip 4 electrons off the Mn, so there is a driving force to get at least one back. The MnO2 would become MnO2-; this is a nice place for a proton. The proton comes from a water molecule, leaving OH-.
MnO2 + H2O don't react together. MnO2 is stable unless it can oxidize something and water is quite inert.


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If ethyl bromide is treated with alcoholic silver nitrate, which product will be formed?
Silver nitrate promotes ionisation , so the ethyl bromide should ionise, the alcoholic species should attack (by loan pair of oxygen) so finally an ether should be formed , but the answer given in my book is nitroethane and ethyl nitrite.

Do you mean ethyl nitrile ($\ce{CH3CH2CN}$) or ethyl nitrite ($\ce{CH3CH2ONO}$)? I find it hard to see how ethyl nitrile could be formed at all.

The book probably assumes that the 'alcohol' is not a very nucleophilic one, so the nitrate ion attacks; and as the nitrate ion is bidentate... Everything adds up, except for the alcohol not participating at all; I'd also have thought it would do something, but there you go.

@user6376297 how nitrate can add to form nitroethane, this is my confusion, I can see formation of nitroethane when nitrite is used, but when nitrate is used , how nitroethane form ?

related https://chemistry.stackexchange.com/questions/88515/reaction-of-tert-butyl-bromide-with-silver-nitrate

@Mithoron, the dupe question is alkyl halide + silver "nitrite" but this question specifically asks about silver "nitrate".

@NilayGhosh It's OP's typo, pretty obvious from context.


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Draw a diagram to show the shape of Chloromethane and explain why it has this shape.

I know it has this shape, but I don't know how to explain it. Any help is appreciated, thanks.

https://en.m.wikipedia.org/wiki/Orbital_hybridisation     and https://en.m.wikipedia.org/wiki/VSEPR_theory

@AvatarShiny I think his question is clear.

Perhaps, you could tell us about your thoughts on the question. It really seems like a homework question with no effort put in by you to solve it. And such questions are not welcomed on this site.

Carbon has in his outer shell 4 electrons and needs 4 more electrons to get the full shell like the noble gas Ne.

Chlorine has in his outer shell 7 electrons and needs one electron to get the full shell like the noble gas Ar.
I give you an answer because the question about the tetrahedral structure of Chloromethane contradicts the perceptions of physics about a s- orbital with 2 electrons and 3 p-orbitals with 2 electrons in each p-orbital. To avoid the contradiction between the claimed s- and p-orbitals and the empirical founded tetrahedral structure the sp-hybridizations were postulated. The carbon in Chloromethane has four sp3-hybridizations with all atoms around.
If you have a look around you will see that the hybridizations are the norm. Let the s- and p-orbitals be a toy for physicists.
What you have to check is, always to get pairs of electrons. That electrons like to be pairwise was expressed by the Pauli exclusion principle. It is a principle and not an explanation why this happens this way in atoms. But if you need an imagination of how the electrons form pairs you could think for yourself of the electrons as tiny bar magnets (they indeed have a magnetic dipole moment) and two of them form a magnetic pair. This pairs around the carbon will be in some equilibration to each other. Since chlorine has a lot of other electron pairs it pushes the 3 hydrogen atoms. So you get this shape:

All images from Wikipedia


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Physical significance of magnitude of ΔG (change in Gibbs free energy) of a reaction
In thermodynamics, we know that if $\Delta G$ of a reaction is negative, the reaction is spontaneous.
Suppose we are given with two spontaneous reactions ($\Delta G\lt0$) with corresponding magnitudes of $\Delta G$. Can we extract any other information from these magnitudes?

Are the reactions simultaneous or are they independent .

@ChakravarthyKalyan I was thinking about 2 independent reactions.

It is only a rule of thumb that a negative value for $\Delta G^0$ of a reaction means that a reaction is spontaneous.  If you start out with pure reactants (and no products present), all reactions are spontaneous, irrespective of the $\Delta G^0$.  A negative value for $\Delta G^0$ only signifies that the equilibrium constant is greater than 1.

I never liked "spontaneous" when discussing equilibrium, only when discussing Q, for exactly the reason you point out. My understanding is that dG std simply states whether the reaction is product or reactant favored under the specified standard state conditions.

The change in the Gibbs free energy for a process at a constant temp. and pressure is equal to how much non-expansion work it can do.

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


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Charge on an atom

I am really confused at this part.The resonance as shown above  causes the generation of a positive charge on nitrogen atom but the fact is that it's octet is still complete even after that resonance....what is the significance of this positive charge on nitrogen?What will it do to the nitrogen atom?

https://en.wikipedia.org/wiki/Formal_charge

It doesn't result in the generation of a positive charge on the nitrogen. It results in the generation of a formal charge on the nitrogen. The formal charge is defined as
FC = V - N - B/2
with V being the number of valence electrons in the neutral, ground state atom, N being the number of valence electrons not involved in covalent bonds and B/2 being half the number of shared electrons in covalent bonds. Formal charge is the charge on the "ion" that is formed when it is assumed that all electrons "lost" are shared equally with the "ion" that captures them and it, in turn, lends an equal share of its bonding electrons.
At the left side nitrogen has 5 valence electrons. The pair is unshared and the other 3 participate in bonding. Were the bonds ionic and the electrons completely lost nitrogen would have a charge of +3. But we assume that the three that are lost are equally shared with the carbons that also contribute shares of their valence electrons so the formal charge on the nitrogen "ion" is 5 - 2 - 6/2 = 0.
Turning to the carbon which is marked in the drawing with a formal charge I get
4 - 1 - 3 = 0 so unless I'm not seeing something that's an error.
Now when the unshared pair moves over to form the second bond nitrogen loses two electrons but gets back a half share in each (and still has the half share in the pair from the original single bond) so FC = 5 - 0 - 4 = 1
For the carbon in this case: FC = 4 - 0 - 4 = 0

"the ion that capures them"...are you implying for a general case or just for this resonance

Any case where formal charge is being computed.

how will valence shell of an atom for eg nitrogen here vary as compared to a neutal nitrogen atom?

Nitrogen is unusual in that of its 5 valence electrons usually only 3 participate in bonding. In this resonance the unshared pair is donated to form a coordinate bond with the adjacent carbon atom. An remember, that formal charge is just that - a formality based on a formalized assumption which may or may not represent reality.


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Bond strength of pi acceptors and metals with filled d(pi) orbitals
I was reading in The Organometallic Chemistry of the Transition Metals (Robert Crabtree) that lewis acids like BF3 often accept electrons from d(pi) orbitals. Is this generally true for all pi acceptor ligands? That is, do metals with filled d(pi) orbitals form thermodynamically stable bonds with pi acceptor ligands? I would imagine that d(pi) electrons would only be available for donation if said metal was in a low oxidation state. Correct me if I'm wrong.
If this post merits two different question threads, I'd be happy to separate them.

TL;DR Yes, you need low oxidation states unless there is a significant $\ce{M\bond{->}L}$ σ contribution to the bonding.

There are a few kinds of π acceptor ligands to distinguish between. These classifications are quite arbitrary and the line is not always clear, but their behaviour is sufficiently different and it is hopefully an instructive endeavour.
Firstly you have ligands which are mostly σ donors, but are also capable of acting as π acceptors, albeit weak ones. Two common examples are the cyanide anion and alkyl phosphines. Because the $\ce{M-L}$ bond is mostly donation from ligand to metal, you don't necessarily need particularly low oxidation states for these complexes to form: so $\ce{[Fe(CN)6]^3-}$ is OK despite having a relatively high oxidation state of iron, $\ce{Fe(III)}$.
Next up you have ligands which are poor σ donors and pretty good π acceptors. Carbon monoxide falls into this category: it is not really all that good a σ donor and backbonding is often required for a stable $\ce{M-L}$ bond to be formed (for some exceptions, google "non-classical carbonyl complexes"). Effective backbonding, in turn, requires high-energy d electrons on the metal. So, this usually precludes higher oxidation states. Many metal carbonyls have the metal in a negative oxidation state or zero, and $\ce{[Fe(CO)6]^3+}$ most certainly doesn't exist.
Lastly you have weird ligands such as $\ce{BF3}$ which are both σ and π acceptors. In the case of $\ce{BF3}$, it is a σ acceptor via the empty p orbital on boron, and a π acceptor via the $\ce{B-F}$ σ* orbitals. These are pretty obscure - I have never heard of them before - and are called Z-ligands (Wikipedia). Obviously any bond that forms must involve significant contribution of electron density from the metal. It turns out that as of 2011 there were no examples of unsupported $\ce{M-BF3}$ bonding,1 but there are quite a few examples of similar molecules $\ce{EX3}$ where $\ce{E}$ is a heavier Group 13 element.

In all these cases the metal fragment is very electron-rich and is again in a low oxidation state (zero or negative). I am sure the electron density on the metal required for these to form are even stricter than for carbonyls; otherwise we would undoubtedly have heard about $\ce{[M_pZ_q]}$ complexes, since there are already lots of $\ce{[M_p(CO)_q]}$ carbonyls known.

Reference

Amgoune, A.; Bourissou, D. σ-acceptor, Z-type ligands for transition metals. Chem. Commun. 2011, 47 (3), 859–871. DOI: 10.1039/C0CC04109B.


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Negative gradient graphs
So basically I have a graph of Absorbance on y-axis against Time on x-axis and it is a negative curve. So this will give me a negative slope gradient. My question is how to reason the negative gradient. Do I write down the rate of concentration as a negative value and just reason it by saying that by being negative it means that the concentration is decreasing?

Welll, if you have reaction and check out for the right wavelength... But you didn't tell it in your post.

The question may be rather naive but doesn't deserve to be down voted. What does the wavelength have to do with it?

Yes. Beer's law tells us that the absorption is the product of path length, extinction and concentration. Thus concentration is absorption divided by the product of path and extinction. The rate of change of concentration (first derivative with respect to time of concentration is the rate of change of absorption divided by the product of path and extinction. If the rate of change of absorption is negative the rate of change of concentration is negative. Concentration is decreasing.


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VSEPR theory, chemical bond and quantum mechanics
VSEPR theory correctly predicts the shapes of many symmetry-broken molecules such as $\ce{H2O}$ and $\ce{NH3}$. Take $\ce{NH3}$ for example. In VSEPR theory, the nitrogen atom is (approximately) at the center of a tetrahedron, the three $\ce{N-H}$ bonds point to three of the four vertices of the tetrahedron, and the lone pair of nitrogen points to the $4$th vertex. But quantum mechanically speaking, the electrons should all be delocalized in the entire $\ce{NH3}$ molecule. How do I unify the two pictures to understand the concept of chemical bonds and VSEPR theory in quantum mechanics? Does VSEPR correspond to some kind of trial wave function (e.g. antisymmetrized geminal power (AGP))?
Note: when I say how to understand chemical bonds in quantum mechanics, I mean in the chemical bond description of molecules, electron pairs are localized at the bonds while quantum mechanics again says everything can be delocalized. So it's the same discussion as VSEPR v.s. QM. If there are only two atoms and one bond, the quantum mechanical meaning of the chemical bond is clear.

If you  already know a bit about QM then ignoring VSEPR altogether seems like a good idea to me.

You can obtain localised MOs by a unitary transformation of the canonical ("delocalised") MOs, https://en.wikipedia.org/wiki/Localized_molecular_orbitals The LMOs can in turn be related to the simple VSEPR ideas of bond pairs / lone pairs.

VSEPR isn't exactly good even for water https://chemistry.stackexchange.com/a/41206/9961

@orthocresol, I think before that we need some kind of independent pair approximation to allow us think about one pair of electrons at a time. AGP seems to be a good choice, but I'm not yet sufficiently familiar with it.

If your problem with VSEPR theory is that it assumes some independence between electron pairs, then yeah, of course; it cannot replicate the full correlation in the true wavefunction. Then again, nothing does it perfectly, so.

I accept VSEPR as an approximate theory. I just wish to see what many-electron wave function in QM would agree with its description of the electron configuration. Hartree-Fock doesn't have electron pairs so I think about AGP.

VSEPR is a simple and generalised approach based (mainly) on empirical observation. It is a great theory in predicting, in a first order approximation, the geometrical shape of molecules. That is a lot harder with other methods. It is of course based on a physical foundation, not only empiricism. You might want to explore electron-domains and the Pauli principle in this context.
There have been a lot of post-rationalisations that cause more harm than good, because they clearly go beyond the limitations of the theory. One of the most miss-taught ones is the involvement of d-orbitals in hybridisation schemes, but that comes from the lack of understanding of some instructors, and its confusion with valence bond theory.
One of the most prominent examples where it (basically) fails is outlined in my answer here: Are the lone pairs in water equivalent?  At this point it should be noted, that the general topology of the electron density is quite well reproduced.
With that in mind, a reconciliation of this empirical approach with quantum theory to understand bonding is dangerous, if not futile. It should be used for providing a reasonable guess for a molecular structure and its principle explanation. Anything beyond that might lead to wrong conclusions. It cannot, in any way, be used to generate a guess for a wave function, because it is not based on it.
You will have to use another method for that. Often confused with VSEPR is the valence bond theory (VBT). It cannot be stressed enough, that the two are completely independent (even at its crudest level). There has been a lot of development in that field, and what is often taught as VBT can only be classified as a very crude approximation. It is in principle an exact theory, but carried out at the theoretical limit is not as easily understood as that what is taught commonly. (Just have a look at resonance, and its misconceptions as are outlined here: What is resonance, and are resonance structures real?)
Another way to describe bonding is molecular orbital theory (MOT), which already comes with the necessary properties of electron delocalisation. Unfortunately, this theory is more difficult to get started in, and does not provide an easy picture to follow. The good thing about this is, that it doesn't get much more complicated at its theoretical limit.
The two approaches are at their respective theoretical limit (VBT decribes electron correlation via resonance structures, MOT need multiple determinants for this) equivalent.
Understanding bonding is not easy and it is by far not without controversy. Even at approximate VBT or MOT there can be many misconceptions, and incorrect deductions, conclusions, rationalisations. For everything in the "grey" areas, there are opinions, interpretations, and opinions about interpretations.
In any and every case one should always be aware of the limitations of the model used. One should also be critical about the found results. One should always expect everything to be a lot more complicated than expected (point in case: CO2).
If you keep all of that in mind: VSEPR is an awesome model system.


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How is the avagadros constant derived?
İ heard that :
Charge of a mole of electrons/charge of single electron =avagadros constant
How is this derived ?

What is a mole, come to think of it?

Charge of a mole of electrons is the Faraday electrolysis constant. So the Avogadro constant can be measured in an electrolysis experiment.

The key is to determine the charge of a mole of electrons vs the charge of one electron.
We get the charge of one electron from the Milliken oil drop experiment.
For a mole of electrons you can start uf you firzt get the molar mass of water, call this $18.016$ grams per mole.  Now electrolyze water with an inert electrolyte such as sodium sulfate.  Work out the charge transfer required to decompose one mole = $18.016$ grams.  This is two moles of electrons based on the stoichiometry of the electrochemical reaction.  Half of this divided by the single electron charge is Avogadro's number.

I believe it is the number of carbon atoms in 12 grams.

How do you count the atoms?

OP meant his specific calculation with electrons.


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How can the number of moles in the product side be more than reactant?

A rigid stainless steel chamber contains $\pu{0.32 bar}$ of methane, $\ce{CH4}$, and excess oxygen $\ce{O2}$, at $\pu{200.0 °C}$. A spark is ignited inside the chamber, completely combusting the methane. What is the change in total pressure within the chamber following the reaction? Assume a constant temperature throughout the process.

I wrote out the following chemical equation:
$$\ce{CH4 + 2O2 -> CO2 + 2H2O}$$
I know I have $\pu{32 kPa}$ of methane, so using up all the moles of methane would mean that pressure decreases all the way to $0$.
The solution writes the following: it multiplies the 1 mol of $\ce{CH4}$ by $1$ to get the number of moles of $\ce{CO2}$ and by $2$ to get the moles of $\ce{H2O}$ produced.
Why can we do that? I would assume that by multiplying in such a way, you're implying that all of the $\ce{CH4}$ is being used up to create $1$ mol of $\ce{CO2}$. This means that you have no more moles to spare for creating $\ce{H2O}$.
I know my conceptual understanding is wrong, but this is what I thought while doing the question and the solution doesn't intuitively make sense to me.
I was wondering if someone could clarify my misunderstanding for me.

A mole is a way of counting things, in the case of chemistry those things are molecules.
But the number of things is a chemical reaction are not necessarily preserved: some reactions create more molecules (electrolysis of water (H2O), for example, starts with two molecules of water but delivers one molecule of oxygen gas (O2) and two molecules of hydrogen gas (H2)).
So moles are not conserved if the structure of the molecules changes because of the reaction. The point of writing equations is to match conservation of matter (atoms don't disappear) to the structures of the molecules involved in the reaction. Sometimes reactions just rearrange the atoms and give the same number of moles of product, sometimes they create more molecules, sometimes they create fewer molecules.
In an ideal gas there is a neat relationship between the number of moles and the volume or pressure (at constant pressure the volume is proportional to the number of moles of gas; at constant volume the pressure is proportional to the number of moles of gas). So your question is asking you to compare the number of moles resulting from the reaction to the number going into the reaction. since they are the same, the pressure won't change.

Are you supposed to consider that these are not ideal gasses? I ask this because if you are allowed to treat them as ideal each mole of methane reacts with 2 moles of oxygen to create 1 mole of CO2 and 2 of water. At 200°C and 0.32 bar the water isn't going to condense so 3 molecules react to form 3 molecules and as PV = nRT for an ideal gas there would be no pressure change.

No, I'm supposed to consider that these indeed are ideal gases. I don't really understand the stoichiometric logic though. You multiply 1 mole of methane by (2/1) to see how much water you produce. In that case, haven't you produced no carbon dioxide? Or is the logic that, if you use 1 mole of methane, you produce both 1 mole of carbon dioxide, and 2 moles of water?

Yes. Where would the carbon go if it didn't get 'burnt' along with the hydrogen?


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How is the water escaping?
I have a number of toy balls filled with a liquid-presumably water or something else non-toxic. Over about a year, the balls have gone from firm to indented as shown. The water level in one is visibly low but four others have become indented as if some water were removed without allowing air to take its place. There are no indications of any leaks, balls have all been maintained at room temperature.
I posted this to chemistry because my hypothesis is that there is some sort of diffusion through the ball surface or in to the floating toy contained within. In the case of the ball on the right, the toy inside appears to have a hole so maybe it was originally full of air but the liquid has migrated inside pushing the air out and appearing to lower the water level. Explanation doesn't work for the other balls.

Thoughts?
-Engineer Dad

Are they perfectly 100% sealed against water evaporation over time?

Not sure, but if water could evaporate through a microscopic gap in the surface, wouldn't air also leak in to relieve the negative pressure?

Capillarity could block the rntrance.

Can be diffusion of the liquid in (just in, not through) the ball walls

As of 12/19/2020, the water in the octopus is almost completely gone

Counter question, is there any way to fill them back up a little? Ours has a permanent dent and I really want to make it spherical again.

There is a general belief that plastics are impermeable, probably because many liquids are stored in plastic containers. This is, however, not correct. Plastics are permeable to water and also to gases. The permeability is low but is not zero and therefore, when very long periods of time are involved, a perceptible change occurs. If absolute impermeability is required, then metals should be used.
The process involved is diffusion and depends basically on the type of plastic. The ones containing fluorine (Teflon and the like) are the less permeable and the polyesters the most permeable, probably because they are more polar.

What has happened to the plastic balls is that some water has permeated out and in some cases some air has diffused in.

Are the balls exactly the same size as when purchased? Perhaps the skin of the ball has expanded by absorption of some of the contents. If this has happened, an indentation could occur even without escape of contents.
On the other hand, I have stored aqueous solutions in PETE bottles (from Diet Coke), and over a couple of years, water has evaporated out and the bottles have collapsed a bit.

I never took an initial measurement but they "seem" about the same size

This is diffusion thru the plastic. Water MW 18 diffuses faster than air. To plump them up try placing them in a helium or hydrogen atmosphere at some high pressure for several months.


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Zero filling leading to increases in resolution (NMR)
Zero-filling involves adding data points with zero intensity to the end of an FID. According to this website https://www2.chemistry.msu.edu/facilities/nmr/handouts/DH%20NMR%20Basics.pdf zero filling can take a spectrum with low apparent resolution to one that is much more helpful like so
In this instance, going from bottom to top corresponds with more zero filling. I do not understand how this happens. How does adding data points with zero intensity lead to increases in apparent resolution? I am certain I am missing some crucial point, but I can not pinpoint what that may be. For what its worth, I understand that we are not increasing the resolution of the instrument's measurements, rather we are manipulating the data to view information that would otherwise not be displayed.

This does not increase resolution but in effect interpolates between points already there. The effect is superficial, no extra information is extracted from the data, but nice smooth signals are produced.

@porphyrin nailed it. This topic pops up at signal processing.SE, where zero padding is the name for zero filling. See the following if you want to see the math: https://dsp.stackexchange.com/a/24426/41790

This is pretty interesting. This is actually not a property of NMR or even something physical. Rather, this is a property of doing a discrete Fourier Transform. This is just a Fourier transform when the function is composed of points (as all data will be) rather than a continuous function.
I will give as intuitive an explanation as I can and point you to these lecture notes which I found helpful for the more mathematical treatment.
Basically, if we take data which measures some amplitude versus time (the raw NMR signal) and Fourier transform to frequency space, then we get peaks which are the frequencies of the motions involved in the physical process. Now, when we do this Fourier transform, we usually have some resolution in mind that we would like to have. This resolution is essentially how far apart we are willing for each of our points to be. If the points are quite far apart, this means we did a very low resolution Fourier transform, and we will get plots looking like the bottom spectrum you have in the question.
Intuitively, the faster we take samples, the better resolution we are going to have after we Fourier transform our data. There is a theorem that says we actually only need to sample twice as fast as the highest frequency signal present in our data, in order to map the discrete data to a continuous signal of finite bandwidth (what we want our spectrum to simulate).
Now, what we really want is to choose a resolution, $F_0$, and determine the number of data points needed from this resolution. This resolution, in practice, is determined by the quality of the instrument you are using. So, because we are thinking about our own NMR, what this means is that in order to get all of the resolution our machine can offer, we need to take data for a time $T_0$, given by,
$$
T_0=\frac{1}{F_0}
$$
Now, here is where the physical processes enter. We have no control over how quickly nuclear spins flip, so the actual process we are measuring may be quite fast. We already said, however, that we need to have $T_0$ amount of signal in the time domain in order to get the full resolution from our NMR after the Fourier transform.
Thus, the solution is simply to append a bunch of zeros onto the end our signal, and then Fourier transform. We haven't added any data, or gone beyond the resolution of our instrument. All that we've done is performed a discrete Fourier transform properly.
Also, note that you don't actually need to append zeros until you reach a power of two number of data points, but the algorithms used (called Fast Fourier Transforms) work in such a way that they work best on numbers of data points which are a power of two.
I hope this is somewhat clear.

"Intuitively, the faster we take samples, the better resolution we are going to have after we Fourier transform our data." - Hmm, I might be misunderstanding what you're trying to say, but the sampling rate affects the spectral window (Nyquist theorem) and not the resolution. The rest of your post explains that very well, so I'm wondering if you meant something else there?

The key idea behind zero padding or zero filling is that the "resolution" or the step size of the frequency axis in the Discrete Fourier Transform is dependent on the number of points you have in the time domain. Zero filling is done before doing the Fourier transform. Thus after the FFT the resolution appears to be improved because you have more points in the spectrum and the spacing between each point appears to be decreased.
This is a result of a beautiful theorem (although this is casually mentioned in most textbooks), but never shown as a theorem. I found in Cooley's original paper from 1967 [1, p. 80].
Cooley & Tukey are the key persons who made Discrete Fourier Transform (DFT) possible by computers for lowly mortals like us, otherwise it was an elitist subject among the mathematicians. Here is a snippet of it. It took me several years to find an original paper which showed this theorem. However, math researchers told me this was known in 1754. Full discussion here: History of Integral Transform Theorem
References

Cooley, J.; Lewis, P.; Welch, P. Application of the Fast Fourier Transform to Computation of Fourier Integrals, Fourier Series, and Convolution Integrals. IEEE Transactions on Audio and Electroacoustics 1967, 15 (2), 79–84. https://doi.org/10/bmw8z8.

Hmmm, what about Tukey, or Gauss? Do they not deserve mention :-) https://en.wikipedia.org/wiki/Cooley%E2%80%93Tukey_FFT_algorithm

Good catch. Both Cooley & Tukey later called their method a rediscovery. I really appreciate how honest and humble those scientists were during the golden era of modern science (60-80s) in the US. One of a leading scientist once lamented that "we gave up curiosity driven research long time ago". The impact factor, h-index, and citations counts have made science a business and a race of numbers.

@M.Farooq Nice work (+1)!  But I have a small question: in the theorem statement, what is upper case F (in the second summation)?

I believe that it represents the maximum value frequency.  Will send you the paper. Correct me if that is wrong.

@M.Farooq Many thanks for the paper! From it, F is the constant sampling frequency, i.e., twice the Nyquist frequency.

As has already been rightly pointed out, this is a property of the discrete Fourier transform. To supplement jheindel's answer, I will try to give a taster of the mathematics behind this. The conventional NMR books typically don't cover this in sufficient detail, so for a proper treatment it is a good idea to look in engineering books, specifically on signal processing. MIT OCW also has an excellent course.
We start our journey at the Fourier series, instead of the transform. Consider a continuous-time Fourier series. (Obviously, NMR signals are discrete-time signals, and we will revisit that at the end of this post.) The point here is to express a periodic time-domain function, $f(t)$, as a sum of complex exponentials (or equivalently, sinusoids, although we will stick with the exponentials here). If this function has a period of $T_0$, then its Fourier series representation is
$$f(t) = \sum_{k = -\infty}^{\infty} c_k \exp\left(\frac{\mathrm{i} 2\pi kt}{T_0}\right) = \sum_{k = -\infty}^{\infty} c_k \exp(\mathrm{i}k\omega_0 t)$$
where we define the fundamental frequency $\omega_0 = 2\pi/ T_0$. The Fourier coefficients are given by
$$c_k = \frac{1}{T_0}\int_{T_0} f(t) \exp\left(-\frac{\mathrm{i} 2\pi kt}{T_0}\right) \,\mathrm{d}t.$$
We can plot these coefficients in the frequency domain. Each complex exponential $\exp(\mathrm{i}k\omega_0 t)$ is associated with a frequency $k\omega_0$. Since $k$ is restricted to be an integer, we only have discrete values of the frequency, and so the Fourier series can only be graphically represented as a series of points with associated heights $c_k$. In other words, we have a discrete function $F[\omega]$ which is only defined for $\omega = k\omega_0$.

(n.b. this is just for illustration purposes; the coefficients are not meant to be drawn accurately.) Now imagine the effect of padding your periodic time-domain function with extra regions between each period where $f(t) = 0$. That would increase the period $T_0$ and decrease the fundamental frequency $\omega_0$. So our discrete function $F[\omega]$ becomes more and more closely spaced.

In the limit of infinite padding, we have $T_0 \to \infty$ and $\omega_0 \to 0$, such that the function $F[\omega]$ actually becomes a continuous version $F(\omega)$. The sum becomes an integral and what we have is now a Fourier transform.

What's important to notice here is that if you draw the Fourier transformed function $F(\omega)$ over the Fourier series coefficients $F[\omega]$, you will find that it fits perfectly over them. In other words, the Fourier series is a frequency-domain sample of the Fourier transform. This only holds true if your time-domain function doesn't get stretched or otherwise modified, so this framework is not directly capable of dealing with infinitely long time-domain functions, but it suffices since our time-domain FIDs have finite durations.

So, when we pad the periodic function with zero-valued regions, we are not fundamentally changing the frequency-domain information that is present. All we are doing is taking more closely spaced samples of the Fourier transform $F(\omega)$, and this comes about only because the samples have a spacing of $\omega_0 = 2\pi/T_0$.
This is exactly analogous to zero-filling in NMR. Imagine that our NMR spectrum is (one period of) the periodic function at the beginning. When you zero-fill an NMR spectrum, you are not actually generating any new information about the peaks that are present. If the (continuous) Fourier transform $F(\omega)$ does not display a small coupling, then no amount of zero-filling your spectrum is going to reveal that small coupling.

Broadly speaking, the same considerations apply to discrete-time signals. The biggest thing to clear up is that the discrete Fourier transform (DFT) is not analogous to the continuous-time Fourier transform. Instead, the DFT is analogous to the continuous-time Fourier series, the sole difference being that it operates on a discrete-time signal. The discrete-time analogue of a continuous-time Fourier transform is instead called the discrete-time Fourier transform (DTFT).
This terminology is partly to blame for the confusion surrounding zero-filling. When you zero-fill a FID and "Fourier transform" it, you are not actually changing its Fourier transform (in the continuous-time sense); you are just calculating a more closely spaced Fourier series.
To sum up and to make the analogy clear:

In continuous time, as you pad a periodic function with zero-valued regions, the Fourier series becomes more closely spaced, tending towards the Fourier transform.
In discrete time, as you pad a periodic function with zeroes, the DFT becomes more closely spaced, tending towards the DTFT.

I wonder if the answers so far represent the complete picture. I remember reading that zero-padding up to a factor of 2 recovers information from the imaginary domain into the real-valued power spectrum and therefore can increase the resolution of Fourier-transform spectroscopic data by a factor of up to $\sqrt{2}$. Unfortunately, I can't find the relevant reference anymore.
Pelczer and Szalma write in Chem. Rev. 1991, 91 (7), 1507–1524: "Zero filling once prevents loss of half of the information acquired." They cite, among others, Bartholdi, E.; Ernst, R. R. J. Magn. Reson. 1973, 11 (1), 9–19. Unfortunately, based on my limited physics background, I can't claim any deeper insights into topic.

Hi, this is pretty interesting. I notice you mention power spectrum, which iirc means $S_\text{pow} = S_\text{re}^2 + S_\text{im}^2$ where the FT spectrum is $S_\text{re} + \mathrm{i} S_\text{im}$ - do you know if similar considerations hold if you don't process in this manner? I don't think it's very common at all to perform power mode processing, especially for 1D spectra - often the real part of the FT is shown and the rest discarded. (Ordinarily I'd try and look in the references you provided, too, but alas I'm a bit short on time right now.)


text:
Why do we use J(J+1) instead of J² in the rigid body rotor energy equation?
I looked for the reason, but it seems to be taken for granted that it must be $J(J+1)$ instead of $J$.
I'm making use of the analogy of the particle in a box energy where
$$E=\frac{n^2h^2}{8mL^2}$$

You can't compare the eigenvalues (or eigenvectors, for that matter) for the particle-in-a-box Schroendinger equation with that for a rigid rotor, for starters.

Because it is $J(J+1)$.  Those are the eigenvalues of the squared angular momentum operator in quantum units, given three dimensions of space, $J^2$ just isn't except for $J=0$.  And that's the truth.

This arises from the quantum mechanical properties of the angular momentum operator (vide infra). Since the operators for angular momentum and translational energy are different, there is no reason to expect a similarity between the form of the eigenvalues.

Spectra of the angular momentum operators
First we note the classical relationship between the angular momentum $L$ and the rotational kinetic energy
$$E_\mathrm{rot} = \frac{1}{2}I\omega^2 = \frac{L^2}{2I} \tag{1}$$
since $L = I\omega$, with $I$ being the moment of inertia and $\omega$ being the angular velocity. By analogy, the quantum mechanical Hamiltonian for rotational motion is
$$\hat{H} = \frac{\hat{L}^2}{2I} \tag{2}$$
where $\hat{L}^2$ is the operator for (the square modulus of) orbital angular momentum. Because $L$ is a vector, we have
$$\hat{L}^2 = \hat{L}_x^2 + \hat{L}_y^2 + \hat{L}_z^2 \tag{3}$$
where $L_x$ is the operator for the projection of orbital angular momentum onto the $x$-axis, and so on. The classical angular momentum is given by $\vec{L} = \vec{r} \times \vec{p}$; by expanding the cross product we can obtain the equations $L_x = yp_z - zp_y$, $L_y = zp_x - xp_z$, and $L_z = xp_y - yp_x$. Using these relations as well as the classical commutation relationship $[x,p_x] = [y,p_y] = [z,p_z] = \mathrm{i}\hbar$, we can derive the commutators of all four operators in equation $(3)$:
$$\begin{align}
[L_x,L_y] &= \mathrm{i}\hbar L_z; & [L_y,L_z] &= \mathrm{i}\hbar L_x; & [L_z,L_x] &= \mathrm{i}\hbar L_y; \tag{4} \\
[L^2,L_x] &= 0; & [L^2,L_y] &= 0; & [L^2,L_z] &= 0 \tag{5}
\end{align}$$
Conventionally, angular momentum eigenstates are chosen to be simultaneous eigenstates of $L^2$ and $L_z$: since $[L^2,L_z] = 0$, it is guaranteed that there is a complete set of states which are eigenstates of both operators. We denote these states by $|\lambda,\mu\rangle$ where $\lambda$ and $\mu$ are related to the eigenvalues of these states:
$$\begin{align}
L^2|\lambda,\mu\rangle &= \lambda\hbar^2 \tag{6} \\
L_z|\lambda,\mu\rangle &= \mu\hbar \tag{7}
\end{align}$$
In this paradigm, we then define the raising and lowering operators $L_+$ and $L_-$ (the meaning will become clear in due time)
$$L_\pm \equiv L_x \pm \mathrm{i}L_y \tag{8} $$
Now consider the effect of $L_\pm$ on any arbitrary state $|l,m\rangle$ which is simultaneously an eigenstate of $L^2$ and $L_z$. Since $L^2$ commutes with both $L_x$ and $L_y$, we have
$$\begin{align}
L^2L_\pm|\lambda,\mu\rangle &= L_\pm L^2|\lambda,\mu\rangle \\
&= L_\pm \lambda\hbar^2|\lambda,\mu\rangle \\
&= \lambda\hbar^2 L_\pm|\lambda,\mu\rangle \tag{9}
\end{align}$$
or in other words, $L_\pm|\lambda,\mu\rangle$ is also an eigenstate of $L^2$ with the same eigenvalue $\lambda\hbar^2$ as $|\lambda,\mu\rangle$ itself. Also, using the commutators in equation $(4)$,
$$\begin{align}
L_z L_\pm |\lambda,\mu\rangle &= (L_zL_x \pm \mathrm{i}L_zL_y)|\lambda,\mu\rangle \\
&= [(\mathrm{i}\hbar L_y + L_xL_z) \pm \mathrm{i}(L_yL_z - \mathrm{i}\hbar L_x)]|\lambda,\mu\rangle \\
&= [\hbar(\mathrm{i}L_y \pm L_x) + (L_x \pm \mathrm{i}L_y)L_z]|\lambda,\mu\rangle \\
&= [\pm\hbar(L_x \pm \mathrm{i}L_y) + (L_x \pm \mathrm{i}L_y)L_z]|\lambda,\mu\rangle \\
&= L_\pm(L_z \pm \hbar)|\lambda,\mu\rangle \\
&= L_\pm(\mu \pm 1)\hbar|\lambda,\mu\rangle \\
&= (\mu \pm 1)\hbar L_\pm |\lambda,\mu\rangle \tag{11}
\end{align}$$
which tells us that $L_\pm|\lambda,\mu\rangle$ is also an eigenstate of $L_z$, but with a different eigenvalue: the new eigenvalue is $(\mu \pm 1)\hbar$. The raising operator (for example) therefore has the effect of conserving the total angular momentum $L^2$, but increasing the $z$-projection of this angular momentum $L_z$.
However, for this to make physical sense, there has to be a limit to how far this can go: or else we will have no maximum value of $L_z$, even for a given value of $L^2$. Note that since $L_x^2$ and $L_y^2$ are both real and non-negative (which follows from the Hermiticity of $L_x$ and $L_y$), equation $(3)$ sets a maximum limit on $L_z$:
$$L_z^2 \leq L^2 \tag{12}$$
Mathematically, the only way this can be fulfilled is if there is a "highest" state $|\lambda,\mu_\mathrm{max}\rangle$ which the raising operator turns into zero:
$$L_+|\lambda,\mu_\mathrm{max}\rangle = 0 \tag{13}$$
Now consider
$$\begin{align}
L_-L_+|\lambda,\mu_\mathrm{max}\rangle &= (L_x - \mathrm{i}L_y)(L_x + \mathrm{i}L_y)|\lambda,\mu_\mathrm{max}\rangle \\
&= [L_x^2 + L_y^2 + \mathrm{i}(L_xL_y - L_yL_x)]|\lambda,\mu_\mathrm{max}\rangle \\
&= [L^2 - L_z^2 + \mathrm{i}(\mathrm{i}\hbar L_z)]|\lambda,\mu_\mathrm{max}\rangle \\
&= (L^2 - L_z^2 - \hbar L_z)|\lambda,\mu_\mathrm{max}\rangle \\
&= (\lambda\hbar^2 - \mu_\mathrm{max}^2\hbar^2 - \mu_\mathrm{max}\hbar^2)|\lambda,\mu_\mathrm{max}\rangle \tag{14}
\end{align}$$
However, we already know that $L_+|\lambda,\mu_\mathrm{max}\rangle = 0$. When $L_-$ acts on zero, it simply returns zero, and so the "eigenvalue" in equation $(14)$ must be equal to zero:
$$\begin{align}
\lambda\hbar^2 - \mu_\mathrm{max}^2\hbar^2 - \mu_\mathrm{max}\hbar^2 &= 0 \\
\lambda &= \mu_\mathrm{max}^2 + \mu_\mathrm{max} \\
&= \mu_\mathrm{max}(\mu_\mathrm{max}+1) \tag{15}
\end{align}$$
Fundamentally, that is where the term comes from. There are several steps now to close some loopholes. I will not go through them in detail because it is far too long, but I will outline what is needed.

You need to show that for orbital angular momentum the value of $\mu_\mathrm{max}$ is restricted to non-negative integer values; we define this quantity to be $l$ and we can replace the quantum number $\lambda$ with $l$ because $\lambda$ is related to $\mu_\mathrm{max}$, which is in turn equivalent to $l$. The eigenvalues of the total angular momentum operator $L^2$ are $\lambda \hbar^2 = \mu_\mathrm{max}(\mu_\mathrm{max}+1)\hbar^2 \equiv l(l+1)\hbar^2$.
Note that for orbital angular momentum only integer values of $l$ are allowed; half-integer values are not allowed. This condition can only be derived by solving the differential equations corresponding to the eigenvalue equations of the angular momentum operators (see further reading). For example, you need to solve this:
$$\begin{align}
L_z f(x,y,z) &= (xp_y + yp_x)f(x,y,z) \\
&= \left(-x\mathrm{i}\hbar\frac{\partial}{\partial y} - y\mathrm{i}\hbar\frac{\partial}{\partial x}\right)f(x,y,z) = \mu\hbar f(x,y,z) \tag{16}
\end{align}$$
Half-integer values for spin are allowed because the exact form of the spin angular momentum operators are not known. However, they obey the same commutation relations as the orbital angular momentum operators, so the rules governing them are the same except for the constraint on integer values.
You need to show that the allowed values of $\mu$ are integers between $-l$ and $l$ respectively. We have already shown that $\mu$ increases in integer steps, and that $\mu \leq l$, so what remains is to show that $\mu \geq -l$: the proof is exactly analogous to that in equations $(13)$ through $(15)$, except that you need to consider $L_+L_-|\lambda, \mu_\mathrm{min}\rangle$.
Conventionally the quantum number $\mu$ is renamed to $m_l$ and the eigenstates are labelled with $|l,m_l\rangle$. So now we have shown that $l$ is the maximal value of $m_l$, and that $m_l$ ranges from $-l$ to $l$ in integer steps, as we expected
The total angular momentum $J$ is the sum of the orbital and spin angular momenta, i.e. $J = L + S$. For the rigid rotor system where you ignore spin, $J = L$ and hence we can simply replace $L$, $l$, and $m_l$ with $J$, $j$, and $m_j$: we therefore have $J^2|j,m_j\rangle = j(j+1)\hbar^2|j,m_j\rangle$.
Finally, the energies are then given by
$$\begin{align}
E &= \langle j, m_j|H|j,m_j \rangle \\
&= \left< j,m_j \middle| \frac{J^2}{2I} \middle| j,m_j \right> \\
&= j(j+1)\frac{\hbar^2}{2I} \tag{17}
\end{align}$$
where $j$ is a non-negative integer (i.e. $j = 0,1,2, \ldots$).

Further reading
Chapter 4 of Molecular Quantum Mechanics, 5th ed., Atkins.
Chapter 7 of The Physics of Quantum Mechanics, Binney & Skinner for an alternative derivation.

Why define $L_x = yp_z - zp_y$? Is it just one of those "we do it because it works" sort of things? Or is there a deeper rationale? (Should I ask this as a new question?)

@hBy2Py because $\vec{l} = \vec{r} \times \vec{p}$, evaluating the cross product gets us to the three equations. Good point, though: in hindsight "define" was a really poor word choice so I've edited it out.

This is exactly the sort of thing that drives me bonkers about a lot of quantum chem derivations: things get asserted that do actually have a rational basis, but that basis isn't even mentioned in passing. Thanks!

why can't I upvote this twice?


text:
Equilibrium, Kp constant, Partial Pressures
A system in equilibrium contains
$$I_{2_(g)}\ \text{and} \ I_{(g)}$$
under pressures $$P_1=0,21 \,\text{atm. and}\  P_2=0,23 \,\text{atm.}$$
While its temperature remains constant, we reduce the system's volume to half its original value.
I have to determine $P_1'$ when the system's equilibrium is reestablished.
I know I can calculate $K_p$ and I know I somehow have to use $PV =NRT$, but from there on I am lost, I cannot figure out how to come up with enough equations to find all of my unknowns. Can anyone show me an explicit solution to the problem?
What I tried, but got really nowhere:
$$2I_{(g)} \leftrightarrow I_{2_{(g)}}, K_p=\frac{P_1}{P_2^2}=3,97.$$
$$P_t=P_1 + P_2 = 0,44 \,\text{atm.}$$
When the new equilibrium is established :
$$Kp=\frac{P_1'}{P_2'^2}=3,97$$
$$\frac{2P_1 }{P_1'}=\frac{n_1}{n_1'}$$
and same with $P_2, P_2', n_2, n_2'$, but I don't know where to go from there.

Is that the entire question?

Yeah, that's the part where i am stuck. Nothing else is given that is of any interest, if that's what you're asking and i have to find a numerical value for P1'.

What are your equations and what are your unknowns?

Edited the question, hope you understand where my problem is now.

For your own understanding, please tell explicitly how many equations do you have, and how many unknowns do you have, and what are these unknowns.

The new partial pressures P1', P2', the relationship between the number of moles in eq1 (n1 and n2) and eq2(n1' and n2') and that's pretty much all i can think of.

The method referred to in the comment is not correct and is a common mistake. One may think,
$$p_1'+p_2'=p_t\frac{V_t}{V_t'}=2p_t=0.88 \rm atm,$$
but this is invalid since $n_1+n_2\neq n_1'+n_2'$, thus $p_tV_t\neq p_t'V_t'$
The correct way is to use the fact that the number of iodine atoms is constant, from which we get,
$$2n_1+n_2=2n_1'+n_2',$$
$$(2p_1+p_2)V_t=(2p_1'+p_2')V_t',$$
$$1.3 {\rm atm}=4p_1+2p_2=2p_1'+p_2',$$
Then you can go on by combining this with the equation you have already obtained from $K_p$ (listed in your question) and solve for $p_1'$ and $p_2'$.


text:
Formal charge and valence shell
Is there a relationship between valence shell and formal charge of the atom......
Let's take an example.

A carbocation for eg $methenium$ has a formal charge of +1,can we tell how many valence electrons it has using this.A carbon atom always forms 4 bonds but here it is only forming 3 bonds then surely there will be change in it's valence shell and the formal charge will be related to it?How about other atoms such as  ammonium ion.What will this positive charge signify....decrease in lone pairs or increase.Stuffs like that

I don't know much about formal charge so that's why I am asking such confusing questions.My teacher said to me that carbon always forms 4 bonds and since it is forming 3 bonds in cases like this , a positive charge will come to it.Same for nitrogen but he didn't tell me about the valence shell so that's why i am asking

Have you tried wikipedia? https://en.wikipedia.org/wiki/Formal_charge

@Raditz_35 yea i have referred it but it doesn't answer my doubt.Wiki just tells the formulae for calculating it but I am more concerned about the valence as well as octet change caused due to the formal charge.How the atom's valence shell will be affected by the formal charge

And what is the concept here? You can have different formal charges written in every mesomeric structure. Formal charges aren't physical - that's why they're called "formal".

@Mithoron  so will every mesomeric structure with different formal charges have same valence shell for that atom for different formal charges or will it remain same........can you post an answer explaining that pls

https://chemistry.stackexchange.com/a/1008/9961 https://chemistry.stackexchange.com/a/13856/9961

Possible duplicate of https://chemistry.stackexchange.com/questions/4828/formal-charges-how-to-work-them-out

@Mithoron so it is all determined by the lewis structure?

Carbon doesn't always form 4 bonds. Consider neutral methylene, $\ce{H2C:}$ in which it forms two leaving two unshared electrons (:). Now suppose we protonate that (with a super acid?). We'd have $\ce{H2CH+}$ or $\ce{H3C+}$, a cation just as when we protonate $\ce{H3N:}$ we get the ammonium cation $\ce{NH4+}$. In both cases the bond to the proton is made from the unpaired electrons on C or N and not by sharing an electron from C or N with one from hydrogen.
In computing formal charge we assume that all bond electrons are equally shared. The formula for formal charge is V - N - B/2 in which V is the number of valence electrons in the neutral atom (4 for carbon), N is the number of valence electrons not participating in a bond (0 in the case of methenium) and B the number of electrons in bonds (6 in the case of methenium) so that the formal charge is 4 - 0 - 3 = +1.
It's a different way of indicating charge distribution. If we write methenium as $\ce{H2CH+}$ it implies that the carbon retained the electrons for the most part so that the bond to the proton is largely ionic. In the formal charge scheme we assume that this bond is totally covalent, that the + charge of the proton has transferred to the carbon because the proton got a half share in two electrons thus neutralizing its charge and write it $\ce{C^\oplus H3}$. It's a formality.


text:
Why does potassium have the electronic configuration of 2.8.8.1?
We all know that potassium, the 19th element on the periodic table has the electronic configuration of 2.8.8.1.  However, why not 2.8.9?  In the element scandium, it has the electronic configuration of 2.8.9.2.  I would like to know why potassium is not 2.8.9.

i think it’s because that the subshell 3d is of a higher energy then 4s, so by the aufbau principle it will be filled up first. nice to see another asecondary school student in s’pore

Related: Why isn't the scandium(2+) cation isoelectronic with potassium? and Why do 3d orbitals have lesser energy than 4s orbitals in transition metals? and Why does the 3rd electron shell start filling up with scandium?

You could actually have the 2.8.9 configuration, putting the last electron in a 3d rather than 4s orbital.  But the 2.8.8.1 one is lower in energy and thus more stable.  How electrons fill subshells in neutral atoms to make the lowest energy configuration can be seen here.


text:
Separating sugar from soda
I want to do a lab with my students in which they will try to prove how much sugar is in a can of coke. I know that distillation is the go-to method for separating out the syrupy sugar mix, but does anyone know of a method that goes further to separate out the pure granulated sugar and not just a syrup?

The simplest approach would be to degas a sample and measure its specific gravity with a vintner's hydrometer. It will have scales of specific gravity and Brix (sugar content) on it.The Brix (Bx) reading is the percent of sucrose, w/w, in a solution with the same density/specific gravity as the one you are testing. There are, of course, other things in coke besides sucrose (fructose, coca extract, phosphoric acid) but they should be in small enough concentration that the sugars will swamp everything else and all sugars behave amazingly the same with respect to density of their solutions).
Based on the reported sugar content of Coke it contains 108 grams of sugar per liter. Thus its specific gravity should be about 1.041 and the corresponding Bx reading about 10.4. The density of water at 20 °C is 0.998203 so the density corresponding to 1.041 SG is 1.039 and 10.4% of 1039 grams (weight of a liter of 10.4 Bx sucrose solution) is 108 grams.
If you don't have a hydrometer with the Bx (or Plato) scale or if you determine density by weighing a known volume (convert to specific gravity) you will need to be able to determine the sucrose content of a solution of given specific gravity. There is lots of opportunity here to go into the history of wine and beer making, which you male students will love (though I understand that in modern schools discussion of things related to alcohol may not be politically correct) but in any case determining the w/w sucrose content of solutions of given SG is a well studied problem. There are various tables but the easiest way to get w/w sucrose from specific gravity is a polynomial promulgated by the ASBC (American Society of Brewing Chemists). It is
°P = ((135.997*S -630.272)*S + 1111.14)*S -616.868
I believe that separating out the pure sugars at anything close to 100% recovery would be difficult (barring HPLC or something of that sort). Sugars seem to be assayed by indirect methods such as the gravimetric ones I proposed or measurement of refractive index.

Degassing will be fast(er) with an ultrasound bath, or a foamer as «popularized» e.g., by Guinness as surger (video).

The Journal of Chemical Education has a lab experiment that fits here.  The calibration goes uses a linear regression with water, and samples of 5, 10, and 15 weight% of sugar and from this infers 11.9 weight% of sugar in (then available) Minute Maid apple juice.  Henderson, S. K.; Fenn, S. A.; Domijan, D. D. Determination of Sugar Content in Commercial Beverages by Density: A Novel Experiment for General Chemistry Courses. J. Chem. Educ. 1998, 75, 1122-1123; doi 10.1021/ed075p1122.

I suppose another tractable, if tedious, method would be to (i) assume nearly all of the organic dissolved solids in the Coke is sugar, (ii) to dry the degassed Coke, and (iii) to perform some sort of calorimetry on the combustion of the solid residue.

If the amount of sugar is to be determined precisely in a given solution, the simplest thing is to mix it with a $0.2$ M $\ce{HCl}$ solution, to heat it up to boiling for ten minutes, and cooling it down to room temperature. This will hydrolyze the saccharose to a mixture glucose + fructose. Then, the glucose is determined by Fehling's solution, producing a $\ce{Cu2O}$ precipitate, which can be weighed or titrated. The procedure is described in all books of analytical chemistry.

In North America at least, most soft drinks are sweetened with high fructose corn syrup (which doesn't suppress hunger and metabolizes into lipids, not glucose: hello obesity), so there might be no sucrose and not much glucose there.  How will that affect this procedure?

@Ray Butterworth: one of the deceptive names in food is "high fructose corn syrup". Sucrose is 50%/50% fructose/glucose. HFCS is 42% or 55% fructose on a dry basis (https://en.wikipedia.org/wiki/High-fructose_corn_syrup). What makes the fructose high is the low water content, ~24% in the syrup. Of course, that also makes the glucose high. Coke claims to use sucrose in its Coke, but there are varieties of Coke, so maybe some use HFCS.

How about precipitating the sucrose? Acetone and CaCl2 can precipitate up to 98% of the sucrose. There are other precipitation schemes on the web that use complex organic compounds of transition metals, but they seem more useful for analyses than for actual separation.
I can't do any better than to copy and paste an abstract (the whole article is available at the reference url):
"The use of methanolic CaCl2 for the non-aqueous extraction of sucrose from dried sugar beet (Beta vul garis L.) cossettes was investigated, with emphasis on the effects of CaCl2 concentration, time of extraction, and operating temperature. Solubility of sucrose from dried cossettes was optimal in range of 10-15% CaCl2 in methanol and increased by a factor of four as the temperature was raised from lC to 60C. Nearly complete extraction (98%) was achieved in batch experiments with four successive extractions. The methanolic CaCl2 extract was found to have both higher purity and greater sucrose stability compared with aqueous diffusion juice. The sucrose could be precipitated completely from the solution by the addition of 5 volumes of acetone at 20C, leaving 9~/o of the CaCl2 in solution. Steady state conditions could not be reached in simulated continuous column experiments, probably because of calcium exchange with other cations in the beet pectin. Pure sucrose was precipitated from these effluent solutions on cooling. The amount of sucrose recovered was pro portional to the decrease in calcium content on pas sage through the column. From this, the molar ratio of calcium to sucrose in the methanolic CaCI2 sucrose complex was calculated to be 2:1." Ref 1.
Ref 1. https://bsdf-assbt.org/wp-content/uploads/2017/04/JSBRVol30No1and2P57to69SucroseExtractionFromBeetByMethanolicCalciumChloride.pdf


text:
In vibrational spectroscopy, what affect do overtones have on the spectra?
I've been in a lecture about vibrational overtones in anharmonic oscillators. How would these affect an IR spectrum?

https://chemistry.stackexchange.com/questions/71334/aromatic-overtones

After reading the Q and A linked by @Mithoron consider the intrinsic colour of water. It is a sort of blue turquoise as red light is absorbed in a long enough path. This absorption is not electronic in nature but it is due to overtones and combinations . It is a nice example with an evident effect. Without overtones liquid water would be intrinsically colorless.

Homework: What's the difference between a harmonic and anharmonic oscillator (in QM) ? (just add to your post above)

Since a spectrum, including its overtones, is essentially a property of a molecule, one cannot really say how the overtones affect a spectrum. So, I will answer this question by first describing what an overtone is, giving an example of an effect that overtones sometimes participate in, and then an example of how overtones can be useful.
Overtones are charatceristic vibrational frequencies of a molecule which take place at approximately integer multiples of the $v=0$ vibrational modes. The existence of these overtones demonstrate that molecular vibrations are anharmonic in nature. This follows from the fact that for a purely harmonic vibration, the only non-zero transitions are those for which $\Delta v=\pm 1$. That is, harmonic oscillators can only move on state at a time. Thus, the fact that we observe these overtones tell us that the vibrations are not harmonic or anharmonic.
Now, in IR and Raman spectroscopy, there is an effect called Fermi Resonance. This effect basically allows for the mixing of multiple vibrational states. If I'm not mistaken, this also only happens for anharmonic vibrations because energy needs to be allowed to be shared between the two states which is not possible if the vibrations are pure harmonic, normal modes. I bring this up because it is quite common that one observes a Fermi resonance between an overtone and a higher frequency ground state mode. To quote the Wikipedia page,

High resolution IR spectra of most ketones reveal that the "carbonyl band" is split into a doublet. The peak separation is usually only a few cm$^{−1}$. This splitting arises from the mixing of νCO and the overtone of HCH bending modes.

Thus, if you wish, one effect that overtones have on IR spectra is their ability to participate in Fermi resonance, which leaves the appearance of a peak being split.
Sometimes it can be difficult to know if a peak being observed is in fact an overtone. Generally, overtones have low intensity, but if they participate in a Fermi resonance, they gain intensity and will look like a normal ground state vibration (depending on what mode they come from).
One place I have seen something likely similar to this is in the following paper[1]. They compute the IR spectrum of a water cluster, including the bending overtone, and show that it is likely the experimental spectrum was mis-assigned. In this example, being able to compute a spectrum independent of the bending overtone is quite useful.
Experimentally, overtones can be quite useful if there is a large number of peaks around the same frequency as a particular mode you are interested in studying. For instance, it is thought that the water dimer may play a significant role in absorption of radiation in the atmosphere and thus contribute to global warming. It is very hard, however, to determine the abundance of water dimers in the atmosphere because of the large amount of water vapor which covers up its entire infrared spectrum. Thus, ref. [2] used the fourth overtone of the $\ce{O-H}$ stretch to experimentally measure water dimers in the atmosphere.
Another example of the same use of overtones is in ref. [3]. There, the authors are interested in studying a possible intramolecular hydrogen bond, but one cannot just look at the lowest energy vibrations because they are swamped by a regular $\ce{O-H}$ stretch. Thus, the authors record infrared spectra in the regime where one would expect to find the first, second, third, and fourth overtones. It is confirmed that these are the peaks observed by simply computing the spectra from first principles. Thus, one also learns which geometric configurations cause which overtone peak.
In summary, overtones are not so much something that "affect" a spectrum, but are a necessary part of the spectrum because we live in a world where molecules vibrate anharmonically. I have also given examples of when overtones are quite a useful tool.
Just to get you thinking a bit, if and when you learn about two-photon or two-dimensional spectroscopies, imagine how overtones can then be used to gain even deeper insight into the molecules of interest.

[1]: Wang, Y., & Bowman, J. M. (2013). IR spectra of the water hexamer: Theory, with inclusion of the monomer bend overtone, and experiment are in agreement. The journal of physical chemistry letters, 4(7), 1104-1108.
[2]: Pfeilsticker, K., Lotter, A., Peters, C., & Bösch, H. (2003). Atmospheric detection of water dimers via near-infrared absorption. Science, 300(5628), 2078-2080.
[3]: Howard, D. L., Jørgensen, P., & Kjaergaard, H. G. (2005). Weak Intramolecular Interactions in Ethylene Glycol Identified by Vapor Phase OH− Stretching Overtone Spectroscopy. Journal of the American Chemical Society, 127(48), 17096-17103.