Hybrid orbitals forming molecular orbitals My teacher showed me this diagram on how the hybrid orbitals of two atoms combine to form molecular orbitals. I was confused by this because I thought that VB and MO theories were two separate theories. In VB theory, atoms form hybrid orbitals that overlap, and the electrons are located in the overlap.

In MO theory, molecular orbitals are formed from the superposition of regular atomic orbitals (not hybrid orbitals), so most MO diagrams I see are drawn like this:

So are molecular orbitals formed from regular (s,p,d, etc.) atomic orbitals, or are they also formed from the hybrid orbitals? And if hybrid orbitals do combine to form molecular orbitals, are the electrons still in between the two bonded atoms or are they delocalized like in the MO theory?

It is still LCAO-MO theory, but just dumbed down a lot. The difference is that, instead of feeding the "pure" atomic orbitals into the LCAO mechanism, you carry out an additional mathematical step in order to get orbitals that have nice directional properties, and you feed those into the LCAO mechanism instead.

Let's talk about a simple example, methane. You could hybridise the $\mathrm{2s}$ and $\mathrm{2p}$ orbitals on carbon to obtain a set of four $\mathrm{sp^3}$ hybrids. If you now try to derive the symmetry of these $\mathrm{sp^3}$ hybrids, you will realise they transform as $A_1 + T_2$. And if you didn't bother doing the hybridisation step, guess what are the symmetries of the $\mathrm{2s}$ and $\mathrm{2p}$ orbitals? The $\mathrm{2s}$ orbital transforms as $A_1$ and the $\mathrm{2p}$ orbitals transform as $T_2$. In the end, you get roughly the same picture through MOT regardless of whether you use hybridised orbitals or not. (Note that the $4\mathrm{sp^3}$ in the second diagram does not indicate a principal quantum number of $4$, but rather that there are four such orbitals. I should have changed it, but I was lazy.)

There is a difference between this and conventional VB theory. Conventional VB theory tells us that each of the $\mathrm{sp^3}$ orbitals overlaps with one of the hydrogen $\mathrm{1s}$ orbitals to form four identical bonds. Once you draw a MO diagram, complete with symmetries, you are entering the realm of MO theory.

However, part of the reason why that example worked so well is that methane has a central atom. When you come to larger molecules, and have to piece together different fragments of the molecule, there is a difference between LCAO using the hybridised and unhybridised orbitals. The best example is the very first MO diagram that you drew - I cannot tell what it is, but I think it is that of an alkene. So let's consider ethene, $\ce{C2H4}$.

In fact, the diagram is "wrong". There are 8 electrons in total in the constituent AOs, but the MOs only have 4 electrons. This is not because of any elementary mistake by your teacher. It is because two of the $\mathrm{sp^2}$ orbitals on each carbon do not point towards each other; they point towards the hydrogens instead. Therefore, they are assumed to be involved only in C-H bonding and not in C-C bonding. Strictly speaking, they should therefore not be shown on the MO diagram. Nevertheless, it is this assumption that gives us some insight into the difference between this "hybridisation-LCAO" treatment and normal LCAO treatment using the pure AOs.

We'll first look at the "real" MOs of ethene (image taken from Fleming, Molecular Orbitals and Organic Chemical Reactions):

$\hspace{15 mm}$

We can try and draw some parallels between your "simplistic" MO diagram and this "proper" MO diagram here. In the simplistic diagram, the C-C $\sigma$ bonding arises from overlap of the two $\mathrm{sp^2}$ hybrid orbitals. So, this is likely to somewhat correspond to the $\pi_x$ MO above, which kind of looks like what you would expect from constructive overlap of $\mathrm{p}$ orbitals on carbon pointing towards each other. However, there's likely to be a bit of the $\sigma_s$ MO mixed into it, because you are using $\mathrm{sp^2}$ hybrid orbitals which also have the $\mathrm{s}$ orbital mixed in. The other filled MOs are not likely to contribute since they are antibonding with respect to the two carbons. In fact, they are mostly there to provide the additional C-H bonding required to make the molecule stable, which is exactly what is neglected in the simplistic diagram.

However, the $\pi$ bonding orbital in your simplistic diagram is a resounding success. It's exactly the same as the $\pi_z$ MO in the real diagram! And the $\pi^*$ antibonding orbital in the simplistic diagram is also exactly the same as the $\pi_z^*$ orbital in the real diagram! This leads us to the main point: we can do this much simpler MO treatment using hybrid orbitals and get a sufficiently accurate MO picture, without having to go through a bunch of symmetry arguments to determine which AOs overlap and which AOs don't. Furthermore, guess which are the most important MOs that control the reactivity of ethene? Exactly; it's the two $\pi$ orbitals.

Now that I've gotten that out of the way, I can proceed to answer your specific questions:

  I was confused by this because I thought that VB and MO theories were two separate theories.

What I wrote above mostly applies to this confusion. The crux of the matter is, hybridisation can be applied to VB theory and it is actually in this context that you first meet hybridisation, but it is not exclusive to VB theory. You can do a little mix-and-match in order to simplify your life, without sacrificing too much accuracy.

  Are molecular orbitals formed from regular (s,p,d, etc.) atomic orbitals, or are they also formed from the hybrid orbitals?

It is your choice. One leads to a more accurate picture, but it becomes extremely tedious as the molecule becomes more complex.

  And if hybrid orbitals do combine to form molecular orbitals, are the electrons still in between the two bonded atoms or are they delocalized like in the MO theory?

The electrons are localised. This is most obvious from the difference in the $\sigma$ orbitals shown above. Full delocalisation a la "proper" MO theory would imply that there are MOs which lead to bonding over the whole molecule, i.e. the $\sigma_s$ orbital.

  Then why do you still call it MO theory if the electrons are localised?!??!!??!

You didn't ask this question, but I anticipated you might wonder about it, since I've been going on and on about MOT only to end up saying that the electrons are localised. It's still LCAO-MO theory, just that the basis set of orbitals chosen is different, which leads to a different form of the wavefunction which happens to be localised between two nuclei. In terms of the electron density distribution, it is the same qualitative picture that is painted by VB theory. But, remember that VB theory has no idea what an antibonding orbital is, or that it even exists. So, what you have done is: you have actually arrived at the same result (localised electron density) by another means (MOT).

Disclaimer: There are people who will debate exactly what "VB" and "MO" theories are. Wavefunctions exist in VB theory as well, except that they are constructed differently, and one might very well say that the simplistic MO theory above is, in fact, VB theory. Nevertheless, I am sticking to the introductory, qualitative notions of VB and MO theory here.


What happens to energy changes in non-reversible processes? Take internal energy for example:
$$dU\leq TdS-PdV$$
Firstly, I would like to make clear that the following question/argument relies on the following statement being correct: the left hand side of the above equation represents the change in internal energy of the system and the right hand side represents the work done on the system by the surroundings. Please clarify if that is correct.

The two equations, above, are in fact inequalities where it only becomes an equality when the system is at equilibrium (when changes are reversible). However, when  the system is not in equilibrium the inequality implies that the work done on the system by the surroundings is greater than the internal energy gained by the system; how can this be? Where does the energy difference between the energy put in and the energy gained by the system go?

Secondly, at a constant $S$ and $V$, spontaneous changes occur only when $dU \leq 0$. What's going on here? Is it as simple as: the system must lose internal energy in a spontaneous change or does it mean that work must be done on/by the system in a different way (eg. electrical work)?

$\Delta U = q + w$ and $dU = dq + dw$

This is the first law of thermodynamics.  It is not an inequality.

$dU\leq TdS-PdV$ is correct when there is only PV work. It is incorrect when there is other work such as electrical or gravitational.

  the right hand side represents the work done on the system by the surroundings. Please clarify if that is correct.

The PdV term alone represents the reversible work, the TdS term represents reversible heat

  Secondly, at a constant $S$ and $V$, spontaneous changes occur only when $dU \leq 0$. What's going on here? Is it as simple as: the system must lose internal energy in a spontaneous change or does it mean that work must be done on/by the system in a different way (eg. electrical work)?

the system must lose internal energy in a spontaneous change


Suggestions for simulating transport through membranes with Molecular Dynamics (MD) and Monte Carlo (MC)? I'm intending to perform a simulation of the transport of simple ions or water through perm-selective membranes, using Molecular Dynamics (MD) and/or Monte Carlo (MC). I've read around that, in general, with Monte Carlo the polymerization of the membrane can be modeled starting from linear chains and monomers, and get the pseudo-membrane well cross-linked. Once the polymer structure is optimized, MD is performed with explicit solvent, the membrane putted in between two different solutions, so transport properties through membranes can be explored.

I never used MD nor MC. This is where I stand in the knowledge:

Monte Carlo (stochastic approach):
Useful to calculate thermodynamic macroscopic properties. The simulation follows the Markov Process, where the steps are independent one from each other (Jan Labanowski).

  The stochastic approach, called Monte Carlo, is based on exploring the energy surface by randomly probing the geometry of the molecular system. The iterations are independent of one another (i.e., the system does not contain any "memory'').
  Is accepted more by physicists than by chemists, probably because MC is not a deterministic method and does not offer time evolution of the system in a form suitable for viewing. (But) for Markov chains, there are efficient methods for deriving time related quantities such as relaxation times (…) Many chemical problems in statistical mechanics are approached more efficiently with MC, and some (e.g., simulations of polymers chains on a lattice) can only be done efficiently with MC.

Molecular dynamics (deterministic approach):
To find trajectories and get an insightful movie of the molecular movement, with parameters of interest. Normally the system composes of around 5000 atoms with explicit solvent (water molecules generally). Periodic conditions are employed. The Verlet algorithm and the SHAKE routine are commonly used; here the bond lengths are fixed and the whole molecules are allow to move between certain potential range given (in position and energy), this repeated successively through a minimum path over the potential surface. It's a leap-frog method.

Considering this, I basically don't know where to begin. I would like to know the basic howto's and softwares for MD (GROMACS, CHARMM, LAMMPS, AMBER, etc.) and ways to perform Monte Carlo for polymers. Also, besides the fundamentals, I would like to know if there is like a Handbook for the features of each MD software or modes of application (such us potentials/ functionals used, force fields), or the most important practical keys to be taken into account before simulating.

Thanks in advance,

I always find AMBER to be the most streamlined for conducting MD, however it (as well as CHARMM) are proprietary.  Note that both are Langevin dynamics based, whereas LAMMPS is brownian dynamics.

That being said, an MD simulation is only as good as the parameters for the force field; the engine is of no matter (unless we're discussing speed).  Basically, I would use a QM program (Gaussian is my goto, but nwchem or psi4 are excellent open source options) to optimize the geometry of the membrane.  Because this is rather heavy, you'd most likely be best to go with a semi-empircal method as ab-initio could take weeks (depending on your computational power).  Once optimized, use the QM software to derive force constants as well as ESP charges (partial charges) for each atom.  Take this data, and parameterize a force field with it (I'm partial to using the Amber FF's...google GAFF or FF99SB... Theres a paper out there from Cornell, Merz, et. al. that should give you insight).

I hope this helps give you an idea of what you're looking at for a respectable MD simulation.


About electricity and liquid products This may sound weird but could one 'zap' some liquid like milk with electricity of a certain voltage for a 'few' seconds hoping to kill any bacteria in it? I know some people suggested some type of radiation on meat ( I think) but could a small charge of electricity on liquid products kill any bacteria in them?

Yes, this is a known phenomenon. It's a process called electroporation. Here's a paper talking about using it for sterilization. Basically, when you apply an electric field to a cell membrane, you can render it porous. At low field strength, the pores are small and can be closed by the cell, but if you apply a larger potential across the cell membrane, cell lysis is the result, killing the cell. It's actually used for killing cancerous tumour cells in humans.

Unfortunately the trick with doing it with something like milk is that, to get a large enough potential across the cells in the liquid, an enormous potential (up to 1 kV/mm) needs to be applied to electrodes immersed in the liquid, even though the cell membrane is only seeing a volt or two. For this to be manageable, you would probably need a thin flow cell or something. It could be a fair bit more expensive than pasteurization. Even though only a short pulse is used, electrolysis will occur, which could affect the taste. How well it works will also be affected by the electrical properties of the solution, which you can't necessarily play with too much in foods without changing the taste, so it might not be the best technique for foods, but there are certainly other areas where it's useful.


How can the hybridisation schemes of transition metal complexes be determined? Lets say I have to find hybridisation of $\ce{[Ni(CO)_4]}$ and $\ce{[Ni(CN)_4]^{2-}}$.

The metal atom/ion in these compounds are $\rm Ni$ and $\rm Ni^{2+}$ respectively.

So the outer shell configuration for $\rm Ni$ in $\ce{[Ni(CO)_4]}$ would be $$\rm 4s^2  \ \ 3d_{xy}^2\ 3d_{yz}^1\ 3d_{zx}^1\ 3d_{z^2}^2\ 3d_{x^2 - y^2}^2 \ \ 4p^0 \ \ 4d^0$$

The electrons from s oribital will jump to d orbital and so I expect $\rm CO$ to donate electron pairs in $\rm 4p$ and $\rm 4s$ orbitals and form $\rm sp^3$ hybridisation.

The outer shell configuration for $\rm Ni^{2+}$ in $\ce{[Ni(CN)_4]}^{2-}$ would be $$\rm 4s^2  \ \ 3d_{xy}^0\ 3d_{yz}^1\ 3d_{zx}^1\ 3d_{z^2}^2\ 3d_{x^2 - y^2}^2 \ \ 4p^0 \ \ 4d^0$$

Here also I expect the electrons to jump from s to d and hybridsation to be $\rm sp^3$, but the answer given here is $\rm dsp^2$. Why ?

I am heavily confused with this hybridsation stuff, is there a systematic and logical way of knowing hybridsation of coordination compounds ?

I tried to search this topic on internet and I came across  $H = \dfrac12(V + M - I)$ where $V$ is number of valence electron, $M$ is number of single bonds and $I$ is charge on the atom. Here if $H= 2$ then it is $\rm sp$ , $H = 3$ then $\rm sp^2$ and so on.

For $\ce{[Ni(CO)_4]}$, here $V = 10$, $M = 4$ and $I = 0$, so $H = 7$ hence $\rm sp^3d^3$.

Also for $\ce{[Ni(CN)_4]}^{2-}$, here $V = 8$, $M = 4$ and $I = -2$, so $H = 7$ hence $\rm sp^3d^3$.

As it can be seen, this formula is complete failure.

Why this formula does not work for coordination compounds whereas it works for compounds like $\rm PCl_5,PCl_3$ and $\rm SF_6$ ?

In this answer the author makes the claim that $\ce{[NiCl_4]^{2-}}$ has hybridsation $\rm sp^3$ and $\ce{[Ni(H2O)6]^2+}$ has hybridsation $\rm sp^3d^2$. How did he know that ? This is my question not why hybridsation is not a valid theory to study coordination compounds.

Firstly, note that hybridisation theory as applied to transition metal complexes is an incorrect, flawed theory. It is an attempt to rationalise experimental observation (e.g. geometries), but in the process it invokes models of bonding which are highly unrealistic and demonstrably untrue. See: Why is it wrong to use the concept of hybridization for transition metal complexes?

Now, with that disclaimer out of the way, let's examine how that flawed theory works. In traditional hybridisation theory, tetrahedral and square planar 4-coordinate complexes are considered to be $\mathrm{sp^3}$ and $\mathrm{dsp^2}$ respectively. Since $\ce{[Ni(CO)4]}$ is tetrahedral and $\ce{[Ni(CN)4]^2-}$ is square planar, one can assign the hybridisations accordingly. If you have a textbook that still mentions this outdated theory, it should have a table telling you what hybridisation each geometry corresponds to. Here is an example from Housecroft's Inorganic Chemistry 4th ed. (p 667). To their credit, there is a subsequent section debunking this theory.

These are generally based on post hoc justifications. Essentially, somebody looked at the geometry and then tried to piece together which orbitals could be combined to form such a geometry. So, I do not think there is a good way to predict them a priori. This formula you have dug up does not work very well, even for main group elements. For transition metals the factors deciding geometries are much more complex, and simply cannot be captured in one line of mathematics.

The real problem with the approach above is that you need to know the geometry before you can assign a hybridisation. (Don't even try to do it the other way round - it simply does not work because hybridisation is an incomplete theory. It is like trying to predict wave-particle duality using classical mechanics.)

If you cannot predict the geometry of the complex, then you cannot predict the hybridisation. You could argue that cyanide is a strong field ligand and in conjunction with a $\mathrm{d^8}$ metal, a square planar geometry is preferred. On the other hand, in $\ce{[Ni(CO)4]}$, the central atom is $\mathrm{d^{10}}$ and hence even with a strong-field ligand like carbon monoxide, a tetrahedral geometry is favoured, simply based on steric grounds.

However, this does nothing but introduce more questions. The term "strong-field" has no meaning outside of molecular orbital theory. Hybridisation theory has no inkling of what strong- or weak-field ligands are, and consequently cannot explain why $\mathrm{d^8}$ Ni(II) only prefers to go square planar with strong-field ligands. On top of that, it is not at all obvious from hybridisation theory how the electronic configuration of the central metal should favour a particular geometry. It's simply not possible for your formula to explain any of this.

So, to answer your questions directly:

  Is there a systematic and logical way of knowing hybridisation of coordination compounds?

Yes. You use a more advanced theory (MOT) to predict the geometry, and then you use the table above to find the hybridisation. That's a step backwards though, you already used MOT and there's no reason why you should go back to hybridisation.

  Why this formula does not work for coordination compounds?

Because hybridisation theory is an incomplete theory and cannot capture the subtle factors influencing geometry of coordination compounds.

Main group compounds

I guess I will elaborate a bit more on main group compounds. The formula you quoted seems to work for main group compounds, but let's get two things out of the way.

It gives the correct hybridisations as espoused by old theory. $\ce{SF6}$ is not $\mathrm{sp^3d^2}$ hybridised because d-orbital involvement in bonding is negligible. So, this formula does not actually work.
It does not predict the hybridisation without predicting the geometry.

Point 1 is important, but let's not go into that now. Point 2 may seem confusing to you. Clearly it is predicting the hybridisation, simply from the number of valence electrons, etc. etc.?

The truth is that this formula only "works" because these quantities (number of valence electrons, number of bond pairs, ...) allow one to predict the geometry, using VSEPR theory. For example, the structure of $\ce{CH4}$ is tetrahedral, and VSEPR theory is sufficient to predict that. Once the geometry has been established as tetrahedral, the hybridisation can then be correctly assigned as $\mathrm{sp^3}$.

So, this formula is skipping a step in between. In fact, the correct sequence is to predict the geometry using VSEPR theory, and then obtain the hybridisation. Now, with transition metal complexes, VSEPR does not work. Thus, it is no surprise that that formula also does not work.


How to identify an unknown compound with spectroscopic data? I need some help identifying the following unknown compound. Any suggestions or hints regarding the possible compound would be highly appreciated.

So far I am thinking the broad peak on the IR is an alcohol that somehow mixed with possible $\ce{sp^3-CH}$ bonds. And the peak around $\pu{1600 cm-1}$ is I think a $\ce{sp-C-C}$ bond. The compound should have four distinct groups of protons. Beyond this basic analysis, I couldn't really come up with a compound.

I have updated (07/09/14) the question with a newer NMR (below) and an IR.

Some other additional info I worked out in the lab:

look: white powder like solid.
melting point: $\pu{116.3-119.0 ^\circ C}$
Solubility:

Water - insoluble
NaOH - insoluble
Sulfuric Acid - soluble

Classification Tests:

Water insoluble phenol test - positive; a dark red precipitate formed.
Chromic acid test - positive; green precipitate formed.
Bromine test for unsaturation - positive; bromine discoloration happened until about $\pu{2 ml}$ of bromine was added. White precipitate formed.

This is my first time doing these tests so please don't consider their results too strictly. My Instructor said the IR is a much safer way to determine functional groups.

Have a look at typical IR bands (other source) and identify them.

   - 3424 phenolic $\ce{OH}$ stretch
 - 1640 aromatic $\ce{C=C}$ bending or $\ce{C=C}$ alkenyl stretch
 - 1264 aromatic $\ce{C-O}$ stretch
 -  747 aromatic $\ce{C-H}$ bending

Have a look at you spectrum, identify your residual solvent signal.

    7.15 is chloroform $\ce{CDCl3}$

Have a look at the typical $\ce{{}^{1}H-NMR}$ shifts table and identify the protons.

   These are aromatic protons

What is the summary of your first conclusions?

   It's an aromatic compound with an hydroxyl group.

How many protons?

   multiple of 5

Calculate the coupling constants via $$J=\Delta\delta_{\mathrm{ppm}}\cdot\nu_0$$ and identify the splitting pattern. (I can only assume it is a 500MHz spectrometer, hence $\nu_0=500~\mathrm{MHz}$.)
Example: Doublet at 7.55 ppm:
$$J=(7.5615-7.5461)\cdot500~\mathrm{MHz}=7.7~\mathrm{Hz}$$
Solutions:

   7.55 doublet, $2\ce{H}$, $J=7.7~\mathrm{Hz}$
 7.40 doublet of doublet (fake triplet), $2\ce{H}$, $J=7.6~\mathrm{Hz}$, $J=7.5~\mathrm{Hz}$
 The last one is tricky because there is one peak not picked
 It should be at 7.30 a doublet of doublet, $1\ce{H}$,  ${}^3J=7.7~\mathrm{Hz}$, $J=7.6~\mathrm{Hz}$

Disclaimer:
Please read long's answer and follow his thoughts. They make a lot of sense and hint that the compound we are looking at is not the one I am proposing. But due to my lack of current knowledge, I cannot make another suggestion.

Combine all you see and you will get a clue which substance it is.

   7.55 ortho
 7.40 meta
 7.30 para
 Phenol.


Why protons combined with electrons will release heat? During water electrolysis, the electrochemical reaction on the cathode side can be expressed as protons combined with electrons to form hydrogen. I believe this electrochemical reaction is a heat release process, but I don't know the reason.

Why protons combined with electrons will release heat?

It will be better if you could provide some references.

If the process of uniting protons and electrons is assumed to occur isothermically in a vessel of constant volume, then the heat released will be equal to the change in internal energy of the system:

$$q_v=(\Delta U)_v$$

Now if the mixture were to behave as an ideal gas, the change in internal energy would be zero and therefore so would the heat. But we can expect a mixture of oppositely charged species such as protons and electrons to behave as an extremely non-ideal mixture, consistently with the large Coulombic potential between the particles. Since $\Delta U$ for the process of uniting protons and electrons (and thereby also minimizing repulsions) is negative, the process is then expected to be exothermic ($q<0$).


What is the precision of intensity data in XRD measurement? I made some relative crystallinity measurements on a zeolite (crystalline aluminosilicate). Which is measured for a specific peak area and compared with a reference sample.
The XRD of powder patterns were recorded using Malvern PANalytical X’Pert PRO Diffractometer with CuKα radiation. The patterns are collected over a range of 3-60 °2θ with a time per step of 0.017 °s-1 using a quarter° divergence slit. In this setup the sample is rotated during measurement. I use Origin for integrated area peak calculation for 21-33 °2θ and the reference sample is made to 100%. We try to reduce the sample thickness error by putting the same mass quantity in all the holders.
I obtained some discrepancy within the data, within a 4% margin. I'm trying to find in the literature what is the margin of error when measuring intensity with XRD.

In short: what does the manual of your diffractometer claim?  Were you able to replicate its standards in first place?
The peak area depends both on the peak's height and width.  But this is difficult to answer because your question does not detail how data were recorded in first place.  Beside content (elements) and quality of your sample (remaining mechanical stress, orientation / size / shape of the crystallites, sample thickness), temperature and pressure while recording the data, the peak area will depend on the source (e.g., wavelength monochromators, slits widths on source and detector), type of detector, i.e. if a point detector was used moving around the sample, or a fixed image plate / area detector and data processing (e.g., how do you account for background intensity, (how) do you deconvolute peaks in question).

(modified from Stoe's IP-PSD)
If you opted for an area detector recording all intensities along $2\theta$ simultaneously, was it the same as for your reference sample?  You are set by the geometry and the inner design of this detector (pixel size and response to the radiation used) on $2\theta$-resolution and may improve intensity resolution only by change of the time of the exposure.
With a point detector, you are less limited for angular resolution because the detector may travel as far around the capillary or rotating dish as mechanics permit in tiny steps.  Large machines

(credit)
claim increments down to steps of 0.0001° (reference) which may equate to only one experiment per day but likely better peak resolution (think peak deconvolution), potentially with a much better signal/noise ratio than the area detector, especially if the sweep is recorded more than once.


How are non-Stoichiometric compounds determined? For solids where the compound contains defects, the stoichiometric equation deviates from the "pure" one. How are these compounds determined? I am looking in my chemistry book but I do not see anything on it. Are there some resources on this?

UPDATE:

After doing some thinking, I considered the Auger Electron Spectroscopy technique used to study the composition of surfaces in materials science. These , I believe, are typically used to study the amount of defects in materials. I do not know how this would be applied to the situation I described above though (as this is only applicable to surfaces). I will continue searching.

You can measure non-stoichiometry by determining how many atoms are present.  There are a number of ways to do this.

You can dissolve your unknown material in an acid and use a reagent to precipitate away one element but not another.  By keeping careful track of the number of moles of reagent you used, you can know how many atoms were precipitated and thus determine the stoichiometry of the original sample.  Once upon a time, this approach was used to demonstrate the fact that some chemicals ARE stoichiometric!

Today, another commonly used method is to count the number of X-rays emitted by a sample in an electron microscope.  Each atom emits X-rays with different amounts of energy so you can determine not only what atoms are present, but how many relative to other atoms.

For example, you can measure iron sulfide which stoichiometrically should have one Fe atom for each S.  Fe emits X-rays with about 2.5 times the energy of S, so they are easy to tell apart.  You then place a known stoichiometric iron sulfide (for example, using chemistry as above) in the microscope and measure how many Fe and S X-rays you get.  Let's say, hypothetically, you get one X-ray from Fe for each X-ray from S (just to keep the math simple).  Then you place a non-stoichiometric iron sulfide in and suddenly you get about 0.9 Fe X-rays for each S.  Presto, you know you have about 9 Fe atoms for each 10 S.

Note: For simplicity sake, I left out all the intricacies of the experimental method which would have to be considered in a real measurement.  If you are interested you can google: Energy Dispersive Spectroscopy (EDS), wavelength dispersive spectroscopy (WDS), and wet chemical analysis.


What does this dashed and curved line mean in pyruvate molecular graph? Hi all, I came across this while researching a gene sequencing assignment and I cannot find out what the curved/dashed line means. Can anyone tell me please? Is it something to do with the negative symbol? Thanks

It actually hybrid form of the contributing resonance structures of the ion. The significance of the dashed line is that it shows delocalisation of negative charge over the two oxygen atoms.

Note that

Neither of the above two forms represent the actual compound
Actual compound is represented by the resonance hybrid :


Why P2O5 does not show oxidising properties while N2O5 is the strongest oxidising agent among pnictogen oxides? Stability of oxides of higher oxidation state decreases with increasing atomic number
Then why P2O5 is thermally stable while N2O5 not.
Infact N2O5 is the strongest oxidising agent among pnictogen oxides because N2O3 is most stable.
Please Offer a lucid explaination to understand this...

A very easy half-answer would base this on electronegativity.

Nitrogen’s electronegativity is $\approx 3$ while phosphorus’ is $2.19$. Oxygen’s is, of course, $3.5$. So the electronegativity difference between nitrogen and oxygen is much lower, meaning that oxygen is much less able to stabilise nitrogen’s high oxidation state.

Contrarily, phosphorus’ low electronegativity means oxygen can stabilise its high oxidation state rather well and the compound is much less likely to act as an oxidating agent by itself.


IUPAC nomenclature for an interesting looking organic compound How can I name this organic compound while adhering to the IUPAC standards?

It indeed can be named quite nicely using the recent "phane" nomenclature from the Nomenclature of Organic Chemistry: IUPAC Recommendations and Preferred Names 2013, here is the example*

By abstracting to the bicyclo phane composed of nine benzene nodes ("superatoms") of three kinds, we get

1,5(1,3,2),2,4,6,8,9(1,4),3,7(1,2)-nonabenzenabicyclo[3.3.1]nonaphane

Note that the "superatom" (more precise "amplificant") locant ordering matters. If the numbers in the first parentheses were sorted, it would mean a different structure, that is little bit hard to draw in 2D:

1,5(1,2,3),2,4,6,8,9(1,4),3,7(1,2)-nonabenzenabicyclo[3.3.1]nonaphane

Publicly accessible references:

Phane Nomenclature, Part I: Phane Parent Names
Phane Nomenclature, Part II: Modification of the Degree of Hydrogenation and Substitution Derivatives of Phane Parent Hydrides

*) Please note there is a mistake in the example from the book, the pyridine (4) has wrong internal ring numbering - 4 should be replaced with 5. The name is correct, though. (It is mentioned in errata, for page 310).


Are all amides also amines? I watched a Khan academy video on organic chemistry functional groups. It mentioned the amine and amide groups.

From what I see, an amide is an amine with a carbonyl group. However, the internet says that that assumption is wrong. I am guessing it has to do with the "R" groups on the amine functional group.

But in the video, he used an example of 2 ethyls (2 carbon chains, not sure what they are called) attached to the nitrogen, and the third R group is just a hydrogen.
So what exactly is an R group, and are all amides also amines? Thanks in advance for the help.

An R group is used in structural formulae as a placeholder for a range of possible subtituents, e.g. hydrogen, alkyl chains or aryl groups etc. It can also be used as an abbreviation when R is a predefined rest.

An amide is a derivative of an organic acid where an $\ce{OH}$ group is substituted by $\ce{NR2}$ (with R = H or any organic rest, see here, for example). An amine is a base, and not formally derived from an acid, but from ammonia. It has the general formula $\ce{NR3}$, with R being H or any organic rest except X=O (X = C, S, etc.) or any other rest derived from an acid. So amides and amines are different classes of compounds. Here are examples for amides of a carboxylic acid, a sulfonic acid (sulfonamide), and an amine (from left to right):

The compound from the video that you have described is diethyl amine. You can formally derive it from ammonia by substituting two hydrogens with ethyl groups:


What is the procedure of designing a salt bridge for a galvanic cell? I'm doing an experiment where I'm measuring the effect of changing the $\ce{Cu^{2+}}$ ion concentration on the voltage in a copper-zinc galvanic cell.

I'll be doing at least five trials for each of the five concentrations I'm testing for (from $0.2-1.0\:\mathrm{M}$), so in total that is 25 trials. However, do I need a new piece of filter paper as a salt bridge for each trial, or can I reuse the same piece of filter paper and dip it in a $\ce{KCl}$ solution each time? Do I really need to make 25 separate pieces of filter paper?

If you wish to do 25 trials with 5 trials per concentration, I would suggest the following:

You can use the same filter paper as long as the $\ce{Cu^{2+}}$ concentration remains constant. However, since you are changing the concentration 5 times, you will need 5 pieces of filter paper.
After every trial, thoroughly wash the filter paper with distilled water.

So I would say you need at least 5 pieces, but you would not need to go all the way to 25 pieces. Just make sure to follow step 2 after each trial.


Why does oxygen not condense onto open dewers of liquid nitrogen? I had a discussion with my supervisor today after I was a touch careless pumping down a flask with a liquid nitrogen cold trap attached. I checked, and there are reports of explosions after running air through a liquid nitrogen trap.

Given this, why does liquid oxygen not condense when you leave an open dewer of liquid nitrogen out? I mean, the air touching the surface of the nitrogen on top of the trap is the same as the air pumping through it. Is there too much nitrogen evaporating for the air to get close, or is the surface of the nitrogen too warm, or some other reasons?

The boiling point of nitrogen is 77 K, and the boiling point of oxygen is 90 K, so oxygen is liquid when liquid nitrogen is around. Wikipedia repeats the warning you mentioned:

  Liquid nitrogen has a lower boiling point at −196 °C (77 K) than oxygen's −183 °C (90 K), and vessels containing liquid nitrogen can condense oxygen from air: when most of the nitrogen has evaporated from such a vessel there is a risk that liquid oxygen remaining can react violently with organic material.

So it's not because the surface of the nitrogen is too warm.

My guess is that the pressure due to the evaporating nitrogen is pushing air away from the opening. LN2 has an expansion factor of 694, which you can compute by dividing the density of the liquid (0.808 g/mL) by the density of the gas at 20 °C (1.17 g/L). Of that expansion factor, a factor of 303 K/77 K = 3.94 comes from warming the gas from 77 K to 20 °C. The other factor of 176 comes from turning the liquid into a gas. That expansion is pushing air out of the headspace of the dewar.


Can atoms other than hydrogen be reduced to a bare nucleus? In acid-base theory, hydrogen is able to become the $\ce{H+}$ ion, or a bare proton. I'm curious if other, larger atoms can have their electrons stripped to the point of being only a nucleus. I know helium can in the form of an alpha particle, but are there any other examples? For example, could you have $\ce{Li^3+}$?

The fact that ionization energies have been measured for many single-electron ions is clear proof that bare nuclei, other than the examples mentioned in the question, have been produced. However, some might say that the process of ionization is more of a physical than a chemical one. This distinction is left to the reader.


How to obtain the Raman spectrum along every coordinate of a scan in Gaussian? I am doing a scan calculation using Gaussian09 and adding the Freq=Raman keyword to my input file. My objective is to obtain the Raman spectrum of each conformation in my compound. However, only the Raman spectrum of the last conformation in the scan is calculated. How can I solve this problem?

I understand that I will get negative frequencies, but these will be in a certain frequency range, so I can observe the variation in the signals.
I need to do these calculations, because I want to see the type of conformation that my compound adopts under heating.

No, there is no automation for this, and there should not be.

Using compound jobs is unfortunately very common among computational chemists, but they should always be viewed critically. If you are using any kind of IOp, then the probability that the second part of your job is doing what you want is low. Additionally, you do not really have any control about what happens if something happens to fail. In such cases, you also give up control over the validity of your checkpoint-file, which you might want to reuse.
Another issue is that these are basically two jobs running one after another and in most cases need different memory requirements, are more effective with another setup. It's the dream of every administrator (and co-worker) when you block resources you don't need for an extended period of time just to avoid a little more waiting.
Lastly, with the introduction of %oldchk= in in revision D.01, it is completely superfluous. Every well organised queuing system should be able to run jobs one after the other when specified (i.e. let one job wait for another to finish).

Only run compound jobs if you are 100% sure they cannot fail (and even then it's questionable).

From a theoretical standpoint, you should consider how much sense your approach makes. As TAR86 pointed out, Raman intensities are expensive. Additionally frequency calculations are only meaningful at well converged local minima structures. You are employing the harmonic oscillator approximation, which will break down if you are not using a local minimum (or at least a stationary point).

If you really need this approach, then you need to do it manually by extracting the geometries (and possibly the MO) and perform the frequency job on top of it.


Is the surface of glass a fine layer of silanol? Today we learned in class through a vivid demonstration that water sticks to glass. We were then taught that water sticks to glass because while glass is mainly composed of silicon dioxide, the surface of glass is actually silanol. Hydrogen bonds form between molecules of water and silanol, which is why water sticks to glass.

Is this correct?

Yes, it is correct that the surface of glass has silanol ($\ce{Si-O-H}$) at the surface. This is because water will react at the surface to relieve bond strain of a $\ce{Si-O-Si}$ bond to form $\ce{Si-O-H H-O-Si}$ pairs. additionally there is sodium siloxide at the surface ($\ce{Si-O-Na}$) which will release sodium hydroxide on contact with water $(\ce{Si-O-Na + H2O -> Si-O-H + NaOH (aq)})$ and is major reason why normal glass cannot be used for biomedical implants.

Additionally these surface silanols can react with chlorosilanes to produce hydrophobic surfaces $$\ce{Si-O-H + Cl-SiR3 -> Si-O-SiR3 + HCl}$$
where the R is a non-polar organic pendant group that is hydrophobic. This coating causes water to bead off of treated glass that would priory be wetting.


heat capacity at constant volume and internal energy I read that heat supplied at constant volume to a system containing ideal gas changes the internal energy of system i.e $q=n C_v \Delta T= \Delta U$

But what happens if I heat a system and change its temperature from $T_1$ to $T_2$ with system not being at constant volume,can I still say that $\Delta U = n C_v \Delta T$?
Is there any proof regarding this?

Yes, you can apply the formula $\Delta U=nC_V\Delta T$ for an ideal gas in a container whether or not the process is isochoric.

This is because the change in internal energy for an ideal gas undergoing some mechanical process is mainly affected by the change in kinetic energy of the gas. It does not depend on the change of  volume of the gas. So as long as the temperature difference is same the change in internal energy will be same. Heat capacity is the amount of heat required to raise the temperature of a substance by 1 unit so $\Delta U$ is directly proportional to heat capacity and $\Delta T$ along with the number of moles of gas.

When the process is carried with constant volume only then $\Delta W$ is $0$ so in that case $\Delta Q = \Delta U$ whereas $\Delta U =nC_V\Delta T$ holds good always for an ideal gas under some mechanical process.


Predicting product ratio by rearrangement in Dehydration Reactions I came across this question recently, which would have a clear major product. The OP has given satisfactory explanation for the product ratio, but wasn't sure about the major product. When I was looking for reference for better explanation, I show the following question in undergraduate textbook (Ref.1):

Dehydration of 2,2,4-trimethyl-3-pentanol with acid gives a complex mixture of the alkenes in the indicated percentages. Write a mechanism that accounts for each product.

I: 2,3,4-trimethyl-1-pentene, 29%
II: 2,4,4-trimethyl-1-pentene, 24%
III: 3,3,4-trimethyl-1-pentene, 2%
IV: 2,4,4-trimethyl-2-pentene, 24%
V: 2,3,4-trimethyl-2-pentene, 18%
VI: 2-isopropyl-3-methyl-1-butene, 3%

I thought it would be a good practice for our readers to predict the mechanism for the obtained product ratio and give explanation for the major products. Can anybody give a reasonable mechanism for this product ratio and explanation for why $\bf{I} \gt \bf{II} \ge \bf{IV} \gt \bf{V}$, and why $\bf{III}$ and $\bf{VI}$ are in such small amounts?
Late edit: It is evident that the textbook print had an error (Thanks Nisarg Bhavsar for the finding). The compound V is actually 2,3,4-trimethyl-2-pentene, not as 3,3,4-trimethyl-2-pentene in the print. I have corrected it now. Nonetheless, the question is still interesting.

References:

Robert J. Ouellette and J. David Rawn, “Chapter 9: Haloalkanes and alcohols Nucleophilic Substitution and Elimination Reactions,” In Organic Chemistry: Structure, Mechanism, Synthesis; Second Edition; Academic Press (an imprint of Elsevier): London, United Kingdom, 2019, pp. 255-298 (ISBN: 978-0-12-812838-1).

The reaction pathway I would propose for the obtained products is as below:

Coming to the percentage of each product obtained, I would defer from the data provided in the question. I would believe that V would be the major product because it is both thermodynamically and kinetically the best possible product.
I would be the next best as it is formed from the best intermediate.
Than II and IV would be the next major products as they are formed from the second best intermediate. Actually I would assume that IV would have a slight higher percentage due to the fact that it is thermodynamically more stable.
III and VI are produced in very less quantities because to form the carbocation required to form these products it has to undergo a very unfavourable rearrangement which would happen only in very less proportion.
So according to me the order should be V>I>IV≥II and very minor quantities of III and VI.


Which proton in cyclohexene is the most acidic? I know that $\mathrm{H^b}$ is most acidic due to conjugation (resonance).

But I am confused as to how to compare $\mathrm{H^a}$ and $\mathrm{H^c}$. I think $\mathrm{H^a}$ should be more acidic than $\mathrm{H^c}$ due to hyperconjugation.

When comparing acidities of protons, we need to find out what the corresponding conjugate bases are.

The conjugate base upon removing $\ce{H_\mathrm{b}}$ is an allyl anion. There is (and should be) no question that it is the most stable carbanion of the three we are going to consider.

The conjugate base upon removing $\ce{H_\mathrm{a}}$ is an $\mathrm{sp^3}$ configured carbanion while the conjugate base upon removing $\ce{H_\mathrm{c}}$ is an $\mathrm{sp^2}$ configured carbanion. This is nice, because it allows us to directly compare the two using their hybridisation.

In general, lone pairs (and that explicitly includes those coming from anions) want to be in the lowest-lying atomic or molecular orbital possible. Since an s orbital is lower in energy than a p orbital from the same shell, lone pairs prefer s orbitals. If there is no pure s orbital available — as would be the case in practically all carbon-centred lone pairs — an orbital with the highest possible s character is preferred instead. If we look at the configuration $\mathrm{sp^3}$ we can say that it has $25~\%$ s character while $\mathrm{sp^2}$ has $33~\%$ s character. Since $33~\% > 25~\%$, the $\mathrm{sp^2}$ configured carbanion is more stable — which is also supported by corresponding $\mathrm{p}K_\mathrm{a}$ values.

The method outlined above is a very crude approximation building on a high amount of simplifications but gives us the correct result in practically all carbon-centred anion cases.

Thus, the acidity order is:
$$\ce{H_\mathrm{b}} > \ce{H_\mathrm{c}} > \ce{H_\mathrm{a}}$$

Note: Ron noted in a comment above that the acidity of an $\mathrm{sp^2}$ hydrogen almost parallels that of an allylic hydrogen. So while most courses may teach this simple ordering, you should keep that in mind as additional information.


Is it possible to solve the equation for the diffusion of gas through a sphere analytically by applying e.g. "combination of variables"? In the case of diffusion through a thin film it is possible to combine the variables in the following way:
$\sigma = \frac{x}{\sqrt{4Dt}}$

where

$x$=length
$D$=diffusion coefficient
$t$=time
$\sigma$=a function of x and t

Thereby, x and t is substituted by the single variable sigma

Is there a similar expression for a sphere?

E. L. Cussler discribes in detail how to solve the equation for a thin film by applying "combination of variables".

Yes. Let's improve the wording here: that is not just an expression, it's the definition of a dimensionless variable that eases the understanding of the solution of the transient diffusion is a semi-infinite medium. Because there is no clear characteristic length as in a finite diffusion path problem, Buckingham π theorem will tell us the two dimensionless quantities we need are
$$\Pi_1 = \frac{z}{\sqrt{D t}}$$ (Or any function of it, like squaring it or dividing it by 4, such as your $\sigma$)

And
$$\Pi_2 = \frac{y-y_{A\infty}}{y_{As} - y_{A\infty}}$$

With $s$ being the surface index and $\infty$ indicating a position "at infinity" (often we use $y_{A\infty} = 0$).

Then we conclude $\Pi_2 =\textrm{function}(\Pi_1)$. This particular function can found by solving Fick's second law with the appropriate boundary and initial conditions. In the most common case, it yields a Gauss error function.

Had we be initially interested in a sphere instead of a planar surface, the only major difference would be the use of spherical coordinates instead of a Cartesian frame, which would result in a different functional solution. But the problem of defining the dimensionless variables would be the same, given we assume perfect angular symmetry. The $z$ position would just be substituted by a radial position $r$.


Is palladium on carbon acting as an oxidising or reducing agent? In the conversion of piperidine to pyridine, palladium on carbon (Pd/C) is used. But it is reducing agent.

Clear the confusion, please.

Palladium on carbon ($\ce{Pd/C}$) is neither an oxidant nor a reductant. The reaction is the dehydrogenation of piperidine: $\ce{C5H11N -> C5H5N + 3H2}$ and here, $\ce{Pd/C}$ is merely a catalyst.

Even in standard hydrogenation reactions using $\ce{H2(g) + Pd/C}$, the $\ce{Pd/C}$ is not a reducing agent. The $\ce{H2}$ is the reducing agent and Pd/C is again only a catalyst.


Why are DOM lithiations sometimes conducted at −78 °C and othertimes conducted at elevated temperatures? I'm looking specifically at n-butyllithium. I've noticed a number of lithiation reactions are conducted at −78 °C, whereas I've encountered papers conducting these reactions at elevated temperatures, up to 60 °C or 70 °C in some cases, under an inert atmosphere of course.

Is there a particular reason for lowering the temperature in some cases and not in others? Does it relate to minimizing side reactions of other groups in the molecule, or does it primarily have to do with stability of the ending lithium reagent at various temperatures? If the latter, does having multiple directing groups assist in stability?

The reason a given set of conditions used in any given reaction is likely fairly arbitrary unless you're following a paper where the conditions have been extensively optimised (for instance in an OPR&D paper where a process group has rigorously screened) - as has been discussed here before, −78 °C  is a convenient temperature to run reactions, so if it works at −78 °C, its unlikely anyone will ever bother to warm it up.

That said, in Clayden's Organolithiums: Selectivity for Synthesis (which is probably the most comprehensive reference of R–Li chemistry I can think of), a table of directing groups and representative conditions is presented. I've included the table below, but the actual text has a discussion of the various groups and a discussion of the representative conditions used. As a general guide, it seems that temperature is largely dictated by:

the activating/deactivating nature of the directing group
the sterics of the directing group
the stability of the directing group, with low temperatures required in the case of electrophilic groups like the carbamate to prevent nucleophilic addition that would occur at higher temperature

As a general point, as temperatures get higher, the stabilisation of the anion by the directing group becomes less important to the overall energy and as such at higher temperatures, one would (generally) expect selectivity to drop.

Summary of directing groups used in DoM reactions with typical reaction temperatures. Taken from Clayden, J. Organolithiums: Selectivity for Synthesis; Pergamon: Oxford, U.K., 2002; p 34.


The road to solvation To understand the behavior of solubility, it is helpful to look at the chemicals that misbehave, and apply the rules, and look for underlying explanations and common threads. Generally speaking, the solubility of some chemicals is lower at higher temperature.

The problem is, the explanation(s) for this is that entropy is negative for some chemicals, and solvation is difficult, and sometimes- the hydrated phases change (which is more like a chemical reaction, where the resultant chemical precipitates). So, I would like to identify the chemicals which misbehave and identify the explanation(s) or the excuse(s).

Here are a couple of links to other questions on this SE about this phenomenon:

Why do some solid chemicals dissolve better in colder water? and Why does sodium sulfate have an unusual solubility-temperature curve?.

What are some chemicals that dissolve better in cold water than hot water?

Calcium acetate, calcium hydroxide, calcium phosphate, cerium(III) selenate, cerium(III) sulfate, and lithium sulfate are more soluble in cold water than hot water.

Also, calcium selenate seems to have it's highest solubility at around 10 °C; and sodium sulfate has it's highest solublity at around 32 °C. These two chemicals have phase changes.

Gases (like oxygen) are also generally more soluble in colder water. The primary explanation is that colder temperatures and higher pressure results in gas becoming a liquid.


Does IUPAC nomenclature have the ability to name all organic compounds? We all know about IUPAC nomenclature. There are rules to name straight chain compounds, cyclic ones, polycyclic ones, ones with functional groups, and so on. But I have come across several examples of compounds which I simply cannot name, most of them being pharmaceutical drugs. Take the heme group complex in hemoglobin:

I'm not sure, but systematic nomenclature appears not suitable to name this structure. A popular antacid, Zantac, also known as ranitidine has this structure:

Can compounds like these even be named systematically using IUPAC rules?

Definitely not.

You got yourself in trouble specifying all organic compounds, because there is a truly immense, mind-boggling number of possible compounds. No one even knows how to accurately determine such a quantity. A very rough estimate, making some incredible simplifications such as the use of only carbon, hydrogen, oxygen, nitrogen and sulfur atoms, is that there are some $10^{63}$ "reasonable" unique structures for compounds with a molecular weight below $\rm{500\ g\  mol^{-1}}$. Thanks to combinatorics, chemical space is enormous. We (or all sapient species in our observable volume, for that matter) will never come close to scratching its surface.

Furthermore, IUPAC nomenclature is largely created a posteriori. That is, though there are many rules trying to cover as many bases as possible (in the process becoming quite unwieldy at times), eventually some unexpected compound with unusual connectivity is discovered and becomes of wide interest. Thus standardising its nomenclature and that for closely related structures becomes fundamental to allow communication between scientists. A recent example of this occurred with fullerenes, which quickly jumped to prominence after 1985. IUPAC just had to create an entire new section of nomenclature for this class of compounds, which is not at all uncommon.

The closest thing to an absolute method of describing a compound's structure is to have a table of positional data (XYZ coordinates) giving the relative positions between the atoms determined from X-ray/neutron diffraction. Any attempt at simplifying this data will be lossy, whether the structures are drawn (not too lossy) or named (very lossy).

The structures you show have comparatively simple IUPAC names, in fact. Heme is a type of porphyrin, which is a widely occurring framework in biomolecules. The central framework can have its positions numbered and the substituent in each one read off separately. Regarding Zantac, the Wikipedia page for the compound states its IUPAC name in the "Identifier" section in the box at the right, namely N-(2-[(5-[(dimethylamino)methyl]furan-2-yl)methylthio]ethyl)-N'-methyl-2-nitroethene-1,1-diamine.

As some interesting examples of the relations between available nomenclature and chemical space, consider the following:

1,1,1,2,2,2-Hexaphenylethane : A molecule with a simple structure and a simple IUPAC name which likely cannot exist in reasonable conditions.
Maitotoxin : an awe-inspiring biomolecule with a rather large structure but containing fairly simple connectivity between atoms, whose IUPAC exists but is quite complex - disodium (2S,3R,4R,4aS,5aR,6aS,7aS,8R,9R,10R,11aR,12R,12aR,13aS,14aR)-10-[(2R,3R,4R,4aS,6S,7R,8aS)-6-[(1R,3R)-4-[(2S,3R,4R,4aS,6R,7R,8aS)-6-[(1R,3S,5R,7S,9R,10R,12R,13S,14S,16R,19S,21R,23S,25S,28R,30S)-25-[(1S,3R,5S,7R,9S,11S,14R,16S,18R,20S,21Z,24R,26S,28R,30S,32R,34R,35R,37S,39R,42S,44R)-11-[(1S,2R,4R,5S)-1,2-dihydroxy-4,5-dimethyloct-7-en-1-yl]-35-hydroxy-14,16,18,32,34,39,42,44-octamethyl-2,6,10,15,19,25,29,33,38,43-decaoxadecacyclo[22.21.0.0³,²⁰.0⁵,¹⁸.0⁷,¹⁶.0⁹,¹⁴.0²⁶,⁴⁴.0²⁸,⁴².0³⁰,³⁹.0³²,³⁷]pentatetracont-21-en-34-yl]-9,13-dihydroxy-3,7,14,19,30-pentamethyl-2,6,11,15,20,24,29-heptaoxaheptacyclo[17.12.0.0³,¹⁶.0⁵,¹⁴.0⁷,¹².0²¹,³⁰.0²³,²⁸]hentriacontan-10-yl]-3,4,7-trihydroxy-octahydropyrano[3,2-b]pyran-2-yl]-1,3-dihydroxybutyl]-3,4,7-trihydroxy-octahydropyrano[3,2-b]pyran-2-yl]-2-[(2S,3R)-2,3-dihydroxy-3-[(1S,3R,5S,6S,7R,8S,10R,11R,13S,15R,17S,19R,21R,22S,24S,25S,26R)-6,7,11,21,25-pentahydroxy-13,17-dimethyl-8-[(2R,3R,4R,7S,8R,9R,11R,13E)-3,8,11,15-tetrahydroxy-4,9,13-trimethyl-12-methylidene-7-(sulfonatooxy)pentadec-13-en-2-yl]-4,9,14,18,23,27-hexaoxahexacyclo[13.12.0.0³,¹³.0⁵,¹⁰.0¹⁷,²⁶.0¹⁹,²⁴]heptacosan-22-yl]propyl]-4,8,9,12-tetrahydroxy-hexadecahydro-2H-1,5,7,11,13-pentaoxapentacen-3-yl sulfate
This hydrocarbon : A molecule which likely exists, with a seemingly very simple structure, but with slightly quirky connectivity which makes naming it a challenge. Add a few more bridges and I'm sure you can break any existent nomenclature rules.


What are the resonance structures of 3-aminocyclohex-2-en-1-ylium? How would you draw the resonance structure for the compound below:

This is one of the resonance forms I think... but I'm sure there is another one too.

The resonance structures and how to construct them for 3-aminocyclohex-2-en-1-ylium are shown below. The second structure in your question is none of them, as it is a different molecule: cyclohexa-1,3-dien-1-amine. For this molecule it is also possible to construct a resonance structure, as also shown below.

However, the two structures can be interconverted via an acid-base reaction.

  And for the stability of 3-aminocyclohex-2-en-1-ylium, the last structure would be most stable (with the positive charge on the EN nitrogen)?

There is no such thing as a most stable resonance structure. Resonance is only a concept that breaks down a highly complicated electronic structure into imaginable contributions of Lewis structures. All of these contributions are present all of the time. However, there are usually a few, that would describe the overall character best. Sometimes there is even only one predominant one.
The concept of resonance is necessary to describe delocalisation effects, hence the Lewis-like representation below is probably much closer to the actual structure than any of the resonance forms depicted above.


Best place to take off vacuum for distillation I have seen vacuum take-off adapters that are essentially plugs, which could be placed at the top of a three-way above the fractionating column, and also straight and bent vacuum take-offs with an inner stem that seem meant to be placed right before the collection vessel.

Would there be any difference in which place is better to vacuum from? Would it be best to maybe use a T-connector on the vacuum line and vacuum the top of the fractionating column as well as the collecting flask?

Assuming that you are using a typical distillation set up (boiling container, fractionating column, distillation head, condensation tube, collecting adapter, and collecting vessel), and the the "three-way" you are talking about is the distillation head. It is much better to put the vacuum take off adapter right before your collecting vessel. The reason is this set up creates a vapor flow that goes through your condensation tube, and most of your fractions will be condensed and collected.

If you attach the vacuum take off adapter on top of your distillation head, most of your vapor will take shortcut to your pump through your adapter instead of going to the condensation tube and the collecting vessel (this direction is a dead end).

Actually the top of the distillation head is the place to put your thermometer to measure the temperature of vapor, not for anything else.


What is the optical nature of the tris(ethylenediamine)cobalt(III) cation? The molecule I've drawn has neither a plane of symmetry nor a center of symmetry, even though some books refer to this as optically inactive.

Can you please elaborate why is this so, or those books are wrong.

Ethylenediammine or $en$ is an symmetrical didentate ligand. You can remember a rule of thumb for this if you like:

  In the case of $3$ symmetrical didentate ligands, i.e. $[\ce{M(AA)3}]$, the compound exists in only one form, which is optically active

So the complex you've mentioned will be optically active, no doubt.

Also, as a reference, I'm quoting the wikipedia article at https://en.wikipedia.org/wiki/Tris(ethylenediamine)cobalt(III)_chloride#Stereochemistry

  The complex can be resolved into enantiomers that are described as $\Delta$ and $\Lambda$.

The only symmetry elements that can be found in $\ce{[Co(en)3]^3+}$ are the three $C_2$ axes of rotation, one of which is shown in this image (colored in red) on wikipedia: https://upload.wikimedia.org/wikipedia/commons/e/e5/Coen3lel3.png

As a side note, here's another rule of thumb:

  In the case of $3$ unsymmetrical didentate ligands, i.e. $[\ce{M(AB)3}]$, the compound exists in two forms, namely fac (stands for facial) and mer (stands for meridional), and both of these forms are optically active.

Important note: There's a minute error in this answer; and that's the distinction between optically activity and chirality: I used the word "optically active" while it should've been "chiral". It's bad I missed it, but I'd like you to read Ben Norris' comment on this below. I'm leaving the answer untouched, editing it rip the comment of its value. I encourage you to learn from my mistake.


Why is 2-pyridone so much more acidic than a regular amide? I am comparing the structures of 2-pyridone and a regular (cyclic) amide-what is the main factor contributing to the increased acidity of 2-pyridone compared to a normal amide. For pyridone, I thought the proton was orthogonal to the pi-system, so no delocalisation of the negative charge could take place, so I am confused how it is more acidic?

The main factor is aromaticity.

Let's look at the main mesomeric forms of DMAc and 2-pyridone.

For DMAc the true structure is somewhere in the middle. We know the C-N bond cannot rotate freely but the barrier is quite low. If we look at structure 2 we can imagine that this structure contributes a lot to the acidity of the amide, since we got a protonated imine like structure, while on the left it's more like an amine.

For 2-pyridone the right structure (4) is aromatic, so it is quite stable. In this case this mesomeric structure will contribute to the real structure much more, which means it is closer to the right structure than in case of DMAc. But in structure 4 we basically got a protonated pyridine. Which is very acidic. It also tautomerizes to 2-Hydroxypyridine, in that case we have a phenol like structure, which is quite acidic too.


Insight into covalent bonds from electrostatic perspective I try to understand, from the electromagnetic side only, how the molecule could be more stable than each atom, and concluded that it could not. Where is my mistake?
Here is an illustration :
By using a ruler in order to have an "example" of numerical values, we conclude, by taking the numerical values of distance in the electric field formula (I removed the common factor with $\epsilon_0$, $\pi$, etc.), that the repulsion force is bigger than the attraction force.
So where is the mistake in my reasoning ?
(I try to stay at a very basic approach of fundamental law of electromagnetism, so not using LCAO with quantum mechanics)

I assume you mean from the classical electrodynamics side only, not from quantum electrodynamics side. By the former, even H atom cannot exist, as the electron would fall along a spiral curve on the nucleus, continually emitting radiation being radially accelerated.
For a hydrogen molecule, both electrons move around both protons, they are not dedicated to the one they came with.
Consider a similar case of charged macro objects, where applying the pre-relativistic classical electrostatics is justified. Imagine 4 respectively charged balls in gravityless vacuum.
The scenario in the bottom picture is one of the less preferred positions, having rather higher energy due short distance of negative charges. But there is plenty of different positions where the whole system would have lower energy than if they were 2 and 2 separately.
Note that even classical system have no general analytic solution for the motion equations for more than 2 objects.


Does this 'Belousov-Zhabotinsky Reaction' kit contain the correct reagents? I want to run the Belousov–Zhabotinsky reaction as a demonstration. As far as I'm aware, you need sodium bromate, sodium bromide, water to $\pu{100 ml}$, malonic acid, phenanthroline and ferrous complex.  This is the recipe that I know, but I can't get hold of the right ingredients.

I found this kit entitled 'Belousov-Zhabotinsky Reaction' for £15.00. I am not sure it has the right ingredients:

  Kit contents: Iodine sodium, malonic acid, Sulfamic Acid, sodium thiosulfate
  Manganese sulfate, starch solution, cups 5 pieces, chemicals, racks etc.

Does this kit have the right ingredients for the Belousov–Zhabotinsky reaction?

No, it doesn't, at least not strictly.
According to Wikipedia (emphasis added):

One of the most common variations on this reaction uses malonic acid $(\ce{CH2(CO2H)2})$ as the acid and potassium bromate $(\ce{KBrO3})$ as the source of bromine. The overall equation is:
$$ \ce{3CH2(CO2H)2 + 4BrO3− → 4Br− + 9CO2 + 6H2O} $$
[ $\ldots$ ]
Many variants of the reaction exist. The only key chemical is the bromate oxidizer.

The kit doesn't contain any bromide or bromate, so by definition it will not allow demonstration of a Belousov–Zhabotinsky reaction.

However: It does appear that the kit may allow demonstration of an iodine clock reaction:

Hydrogen peroxide variation
This reaction starts from a solution of hydrogen peroxide with sulfuric acid. To this is added a solution containing potassium iodide, sodium thiosulfate, and starch.

The ingredients are not a perfect match, and actually appear to be something of a mishmash between the iodine clock reaction and the Belousov–Zhabotinsky reaction, but they're in the right ballpark. Metal ions and their oxides are well known as catalysts for hydrogen peroxide decomposition, and "Sulfamic acid" could be a typo that was supposed to be "sulfuric acid".


Mechanism of oxidation by dichlorotriazenetrione How does dichlorotriazenetrione supply chlorine to a swimming pool and how does it oxidize organic material?

Dichlorotriazenetrione, also known as dichloroisocyanuric acid, is slowly hydrolyzed in water to isocyanuric acid and hypochlorous acid, which serves as the oxidation agent. The N-Cl bond is easily hydrolyzed because of its relative weakness, which is also reflected in the stability of nitrogen halides. The slow hydrolysis releases a low but constant amount of oxidizers. $\ce{HOCl}$ itself oxidizes organic compounds, being reduced to $\ce{HCl}$ in the process.

In aqueous solution, $\ce{HOCl}$ is in equilibrium with its anhydride $\ce{Cl2O}$ and elemental chlorine, which also serve as oxidizers. Furthermore, it slowly decomposes to $\ce{HCl}$ and oxygen, so that a mixture of different oxidation agents is present in its solution.

$$\ce{HCl + HOCl \rightleftharpoons Cl2 + H2O}$$

$$\ce{2HOCl \rightleftharpoons Cl2O + H2O}$$

$$\ce{2HOCl \rightarrow 2HCl + O2}$$


Spin Operator algebra I am trying to teach myself some QM.

In Christopher J. Cramers textbook Essentials of Computational Chemistry: theory and models, in Appendix C, he goes over Spin algebra.

I am unable to calculate myself that $S^2\alpha = \frac{1}{2}\left( \frac{1}{2}+1 \right) \hbar^2 \alpha$ as well as that $S^2\beta = \frac{1}{2}\left( \frac{1}{2}+1 \right) \hbar^2 \beta$

Where $S^2$ is the spin squared operator and $S = S_x + S_y + S_z$.

This seems like it should be trivial. It certainly is to him, he says that  repeated application of the following equations lead to the above

$S_x\alpha = \frac{1}{2}\hbar\beta$

$S_x\beta= \frac{1}{2}\hbar\alpha $

$S_y\alpha = \frac{1}{2}i\hbar\beta$

$S_y\beta = -\frac{1}{2}i\hbar\alpha$

$S_z\alpha = \frac{1}{2}\hbar\alpha$

$S_z\beta = -\frac{1}{2}\hbar\beta$

my understanding is that $S^2=S_x^2 + S_y^2 + S_z^2$. However, when using this operator on $\alpha$ or $\beta$ I am not sure what the eigenvalues would be..., relatedly, I am not sure even how to operate just $S_x^2$ on $\alpha$.

One possible way:

As you say
$${\bf{S}^2} = S_x^2+S_y^2+S_z^2$$
Thus
\begin{align}
{\bf{S}^2} &= \left( \frac{\hbar}{2}\right)^2
\left[\begin{pmatrix}
0  & 1 \\
1  & 0 \\
\end{pmatrix}\begin{pmatrix}
0  & 1 \\
1  & 0 \\
\end{pmatrix}
+
\begin{pmatrix}
0  & -i \\
i  & 0 \\
\end{pmatrix}
\begin{pmatrix}
0  & -i \\
i  & 0 \\
\end{pmatrix}
+
\begin{pmatrix}
1  & 0 \\
0  & -1 \\
\end{pmatrix}
\begin{pmatrix}
1  & 0 \\
0  & -1 \\
\end{pmatrix}
\right] \\
&=
\left( \frac{\hbar}{2}\right)^2\left[\begin{pmatrix}
1  & 0 \\
0  & 1 \\
\end{pmatrix}+\begin{pmatrix}
1  & 0 \\
0  & 1 \\
\end{pmatrix}+\begin{pmatrix}
1  & 0 \\
0  & 1 \\
\end{pmatrix} \right]\\
&= \left(\frac{3}{4}\right)\hbar^2
\end{align}


Basicity of heterocyclic compounds Pyrrol, furan or thiophene don't have any bond pair electrons free to release which is why they shouldn't be basic, but the lecturer of organic chemistry is saying that they are basic as they react with hydrochloric acid to form salts. Which one of the two explanations are correct?

Furan and thiophene are going to be even less basic than their saturated counterparts, which can act as Lewis bases but aren't particularly well known for acting as Brønsted bases (very strong acids can significantly protonate ethers and sulfides, but not weak ones). They do, however, have a free $\mathrm{sp^2}$ lone pair with which they can bind a proton, but only one, unlike the saturated analogues, and an $\mathrm{sp^2}$ lone pair is going to be even less basic than a an $\mathrm{sp^3}$ lone pair because the electrons are closer to the nucleus (i.e. $\mathrm{sp^2}$ is "smaller" and closer to the nucleus than $\mathrm{sp^3}$).

Pyrrole, however, might be expected to be basic, but it also is not, as you rightly expected. This is due to delocalization of the nitrogen lone pair to create aromaticity. The $\mathrm pK_\mathrm a$ of the conjugate acid of pyrrole is $-3.8$, which is lower than that of even sulfuric acid, suggesting pyrrole is a very weak base.

All of those heterocycles are less basic than $\ce{Cl^-}$ or $\ce{H2O}$, so either your lecturer is incorrect or has been misunderstood.


What exactly is Indium's ability to stabilize non-ferrous metals? I was reading the History section about Indium on Wikipedia when I came across the following sentence:

  In 1924, indium was found to have a valued ability to stabilize non-ferrous metals, which was the first significant use for the element.

I'm wondering if the ability is to improve crystallinity, structural stability of alloys, resistance to corrosion, or something else.

All I did was click on the reference from the History wikipedia link above for the sentence you were talking about. The sentence comes from a paragraph of a manuscript which is found here: http://pubs.acs.org/doi/abs/10.1021/ed011p270

Note the bolded part, that was what is meant.

" Its[Indium's] identification as well as its name came from the beautiful indigo-blue line which it so prominently displays in its spectrum. Its commercial development came as the result of the knowledge that it should be a valuable stabilizer of non-ferrous metals much as chromium is a stabilizer of the ferrous metals. "

Above is quote from:

French, Sidney J. (1934). "A story of indium". Journal of Chemical Education 11 (5): 270. Bibcode:1934JChEd..11..270F. doi:10.1021/ed011p270.


Mechanism and type of polymerization of chloroethene Ethene reacts with hydrogen chloride to form a monomer used in the production of a polymer. This monomer is chloroethene.

My questions are:

(1) How is the polymer formed from the monomer?

and

(2) Is this an addition or a condensation polymerization?

You got it wrong - monomer is chloroethene (vinyl chloride) and nothing is eliminated in addition polymerization in contrary to polycondensation. This monomer is made via addition of chlorine molecule to ethene and elimination of hydrogen chloride. Alternative methods are addition of hydrogen chloride to acetylene or oxychlorination - in this reaction $\ce{HCl}$ and $\ce{O2}$ react with ethylene to form the monomer and water.


difference between hexane and hexane fraction i'm doing a chemistry assignment and cant find the difference between hexane and hexane fraction. i think it means there are impurities but not sure so it would be nice if someone could help. for the the fuels i am doing hexane fraction and 1-hexanol. if it means there are impurities that will impact the experiment.

when i search google "hexane fraction" i get stuff on hexane not hexane fraction.

The difference is simple. Hexane, if that is what the label says it is, should mostly be n-hexane. But hexane fraction is any hydrocarbon that distills at close to the same temperature as n-hexane. This mostly means a mixture of other six-carbon hydrocarbons (e.g. 2-methylpentane) most of which boil between 55 and 70 °C. You might find that the mixture contains some C-7 hydrocarbons as well.

For many reactions what you need is some volatile hydrocarbon and it isn't that important that it is a mixture. Occasionally, though, this will affect a crystallisation (especially if the hydrocarbon is bound in the crystal lattice). Pure hexane will work better in that case.

The reason why they are sold as different things is that pure hexane is a lot more expensive to produce as it will require careful distillation to separate it from the other, similar boiling point, hydrocarbons. So you don't want to use it if the cheaper mixture works. The cheaper mixture will just be a fraction obtained from crudely distilling petroleum at around 60–70 °C.


How does water flows through pipes in cold countries? This question occurred to me when I looked into the fridge, where the water had frozen due to very low temperatures. If the water freezes due to low temperatures (0 degrees Celsius), then how can it flow through pipes in the cold countries (especially near the north pole) where the temperature is almost always below 0 degrees Celsius, especially in the cold months?

From the Wikipedia page on Frost Line

  The frost line—also known as frost depth or freezing depth—is most commonly the depth to which the groundwater in soil is expected to freeze. The frost depth depends on the climatic conditions of an area, the heat transfer properties of the soil and adjacent materials, and on nearby heat sources. For example, snow cover and asphalt insulate the ground and homes can heat the ground (see also heat island). The line varies by latitude, it is deeper closer to the poles.

So they just dig the pipes in deep.

  Canada: Can use table of estimates based on freezing index degree days.
  Ontario: Map of frost penetration depths for Southern

  Ottawa: 1.8 metres (5.9 ft) [7][8]
  Windsor: 1.0 metre (3.3 ft) [7][8]

Here's a  soil temperature vs depth graph.

Source


Why do Hydrogen and Helium have molar volumes higher than an ideal gas? In my chemistry textbook, there is a table of the molar volumes for different gases. Most of them are below 22.42 l/mol, the ideal gas molar volume, but I noticed that hydrogen and helium do not. Hydrogen is 22.433 l/mol and helium is 22.434 l/mol. Why is this? What makes these particles, with mass, have higher molar volumes than the basically mass-less particles of an ideal gas? And why does the molar volume increase when going from Hydrogen to Helium?

As hydrogen and helium's molecular weight are too less, the intermolecular attractions are also too less. So the a/V^2 is negligible.  So the v.d.o equation becomes P (V - nb) =RT . By which, it can be said as the pressure of these elements are too less and from Boyle's law , the volume is higher.


Which free radical is more stable, allyl free radical or benzyl free radical? Can someone pls tell me which one is it? Benzyl free radical has 4 resonance structures compared to allyl free radical which has only 2. More the resonance structures should mean more stable the radical.

But in the book I am referring it is written that free radical stability depends on bond dissociation energy.

According to this Allyl free radical should be more stable but according to resonance structures Benzyl free radical should be stable. Which one is correct?

Yes, generally more resonance structures mean more stability but this case is an exception.

In three of the four resonance structures of the benzyl radical, the aromaticity is being broken which makes then less stable whereas in the allyl radical both the canonical forms are stable.

Therefore the allyl radical is more stable than the benzyl radical.


Question about the Law of Definite Proportions If we know that:

$1g$ of carbon + $1.33g$ of oxygen = carbon monoxide

$1g$ of carbon + $2.66g$ of oxygen = carbon dioxide

My question is: why is that in the second case where we doubled the amount of oxygen we don't get double the amount of the first case (2 times carbon monoxide)

Or why don't we get the same amount of carbon monoxide and the remaining mass of the products is simply oxygen that didn't form any bonds.

Basically, what forces the reaction to occur in a certain way just because the proportions are different but the elements are the same.

In reality, there is not a sudden changeover of $\ce{CO}$ to $\ce{CO2}$ production when the oxygen supply is increased. This is the reason that all dwellings should have carbon monoxide detectors.

As an example, along with $\ce{CO2}$, gasoline (petrol) engines commonly produce 3-7% $\ce{CO}$ before additional air is injected.

The amounts you cite are how much oxygen is in pure samples of each gas, but producing either gas often requires filtering out traces of other gases that were also made in side reactions.

Producer gas is a fuel made by the incomplete oxidation of carbon (from coal or other source, sometimes with the addition of water) and contains both flammable $\ce{CO}$ and nonflammable $\ce{CO2}$ (and $\ce{N2}$ from the air supply). During fuel shortages, gasifiers were cobbled onto vehicles to enable them to run on this type of fuel.


Difference between S and I in the context of magnetic resonance experiments In the lecture our professor told us that nuclei with $S>1/2$ have a quadrupole and are therefore detected with NQR instead of NMR.
The majority of all elements have nuclei with $S>1/2$.

I wondered why he used $S$ (electron spin) instead of $I$ (nuclear spin) in the slides?

The nomenclature you mean has nothing to do with the electron spin S, it is something completely different. In heteronuclear experiments with two spins it is common to call them $I$ and $S$, with $I$ being the more sensitive nucleus. They're still both spin 1/2 nuclei.

$I$ is one spin, and $S$ is a different spin in this nomenclature, with a scalar coupling between them.

This nomenclature is usually introduced with the INEPT transfer and experiments like HSQC or HMQC.


Which conductivity law should I use to calculate the conductivity of a solution with high/low concentration? I have a set of NaCl solutions with concentrations ranging from 0.05 M to 0.1 M.
I have been trying to calculate the conductivity of these solutions and compare the obtained values with the experimental ones obtained with a conductometer.

For this, I used the Kohlrausch's Law. But I observe some deviation on the values. I realized that the law may not apply for high concentrated solutions such as 0.1M. I read that since NaCl is a strong electrolyte, this law should still apply.

Is there a specific law or method to use to calculate the conductivity of concentrated solutions?

You can use the Debye-Hückel-Onsager limiting law. For a strong electrolyte without the formation of ionic pairs, with $z_+ = |z_-|$, just like $\ce{NaCl}$:
$$ \Lambda_m = \Lambda_m^\infty - (az_+^3 + bz_+^3\Lambda_m^\infty)(c/c^\circ)^{1/2}$$
For aqueous solutions at $25^\circ$C and $1$ atm: $a = 60.6 \, \Omega^{-1}\mathrm{cm}^2\,\mathrm{mol}^{-1}$, $b = 0.230$. For $\ce{NaCl}$, $\Lambda_m^\infty = 126.4 \, \Omega^{-1}\mathrm{cm}^2\,\mathrm{mol}^{-1}$.

References

Ira N. Levine, Physical Chemistry, 6th Ed.


Can any other common reagent other than than H3PO4 cause this conversion? Can any other common reagent other than than $\ce{H3PO4}$ cause this conversion? What about (conc) $\ce{H2SO4}$ or (conc) $\ce{HNO3}$ or (conc) $\ce{HCl}$?

Eaton's reagent - $\ce{P2O5}$ in Methanesulfonic Acid - is a good alternative.
JOC 62(1997) 3552

Conc. Nitric acid risks nitrating the aromatic ring. I have never seen conc. Nitric or conc. $\ce{HCl}$ used for this


Energy of an electron (why negative?) Why is the energy of an electron in an orbit is always negative?
What does it mean?
Why isn't it positive?
Does it mean that energy is released by revolving electron?

Let me explain you in the simplest manner.

For a two object system,

If they repel each other when you move them towards each other, that means it takes work to do this, and the potential energy is positive.

If they attract each other when you move them towards each other, then potential energy of the two object system goes on decreasing.

Potential energy of a two charged object system at a particular orientation is defined as work done by external agent to bring the second body from infinity to desired point in presence of the field created by the first body. Now think the first body to be nucleus and the second body to be electron.  Now to find the potential energy, according to potential energy defination I just said above, we need to being the electron from infinity to any desired location, lets say to a distance of "x" from the nucleus. As we have to bring electron from infinite that means initially electron is infinitely far away from nucleus and hence at zero potential.

Potential Energy of electron:

When we are bring the electron towards the nucleus, offcourse they are going to attract each other and as said earlier if two objects attract each other, then potential energy of the system goes on decreasing. So as we move electron towards the nucleus, energy goes on decreasing. Note that initially as electrons and nucleus were at infinite distance from one another, hence initially at zero potential and any further decrease in potential energy implies negative potential energy.
So potential energy of the electron is negative inside any atom.

Kinetic Energy of electron:

The electron also has kinetic energy. Kinetic energy can never be negative (because in K.E expression we have squared of velocity which is always positive) and it is smaller in magnitude than the potential energy.

Total Energy of electron:

So the total energy i.e kinetic energy  + potential energy is negative.


Addition of Br2 to Double Bond and Carboxylic Acid In the above reaction, I remember reading that Br2 in CCl4 develops a partial polarity and adds on to the double bond, resulting in addition of Br on 2 adjacent carbons.

But, as per the text, the end product is a ester. (As given below)

Why is this the final product, and how to judge the products in cases like this?

Note - If I look at the answer and work backwards I would say that first Br+ attacks, and a carbocation is formed, on which hydroxy group attacks to form the final product. Is this right? And of it is then, why is this happening?

The first intermediate is indeed Br+ atom attached to the double bond forming a cycle (see step one):

 On the next step, however, oxygen of carboxylic acid attacks the intermediate and forms the product you show on second figure. This happens because carboxylic acid group is in the right position for intramolecular reaction.

First step follows the diagram. In second step the attacking nucleopline is not Br$^-$, but R-COOH or RCOO$^-$ depending on the reaction condition. To obtain high yield of the cyclic product on your second figure you need to slowly add dilute solution of Br$_2$ to a solution of organic reagent. If you add Organic reagent to neat (pure) Br2, then you get a significant amount of the product you predicted on figure 1. Don't do it in real life because a reaction of neat bromine with alkenes is violent and dangerous.


Choosing the right condenser What are the advantages and disadvantages of the various types of condensers commonly found in the laboratory? Obviously more intricate pieces of glassware are more costly but assuming they are all available why would one use a Dimroth condenser vs. a Friedrichs when refluxing something? I've heard that Graham condensers are to be avoided when refluxing due to the possibility of clogging, and yet it is still very common - when is it appropriate to use one?

Illustration made with ChemDraw. Feel free to reference other designs but please attach a schematic for clarity.

Edit: This question was motivated by an organic preparation which involved the bromination of an alkene in boiling water. Bromine has a boiling point of 58.8 °C and on top of that the reaction was exothermic. It was difficult to avoid the loss of Br2 gas with your typical Allihn condenser, but the increased cooling capacity provided by a Friedrichs returned Br2 back to the flask as it was produced.

Also, while the Friedrichs forces vapors up a spiral path, the path itself is wide, and in my condenser at least, there was a bit of leeway for liquid to drop down the sides, helping to prevent blockages. I didn't try a Graham condenser but I imagine a much slower rate of addition would be supported by this condenser.

@Mart 's comment impelled me to return to this question and correct my answer. I've deleted incorrect material and expanded the discussion to, hopefully, provide correct information. There is a good discussion (better than the reference previously cited) of the issue here.

  Reflux is the process of boiling reactants while continually cooling the vapor returning it back to the flask as a liquid. It is used to heat a mixture for extended periods and at certain temperatures...A condenser is attached to the boiling flask, and cooling water is circulated to condense escaping vapors.

If you are refluxing a mixture, as you might in organic synthesis to increase the speed of the reaction by doing it at a higher temperature (i.e., the boiling point of the solvent), then any of the condensers that worked well enough to avoid the loss of solvent and avoid "flooding" would work equally well. When you're refluxing, you want the "reflux ring", the place where the vapor is visibly condensing into a liquid, to be no more than 1/3 of the way up the reflux column.

You have two different basic types of condensers shown, Graham-type condensers (the first 3) and coil condensers (the last two). In the coil condensers (the left condenser in the picture below), the water flows through the coil and the vapor moves up in the larger, outside  area of the condenser, condenses onto the cooled coils, then drips back into the pot. In a Graham-type condenser (the right condenser in the picture below), the water flows around a tube (whether straight or coiled) that contains the vapor/condensed liquid.(picture source) The Graham-type condensers clog (or flood) more easily since they have a more restricted path for the liquid to return to the pot.

Graham-type condensers: The Liebig condenser is simple, but has low cooling capacity and can be fairly easily clogged as the condensed liquid flows back into the flask and blocks the vapor that is trying to escape. The Allihn improves on this design by having a wider bore at the bottom and condensing the liquid on the "bubbles" where it can run down the sides and avoid blocking the vapor. (I've used this to good effect in refluxing many reactions.) The Graham condenser is the same basic design as the other two, but the condensation tube is coiled which provides more surface area for cooling...but also tends to send the condensed liquid right into the path of the vapor trying to move up. It is particularly prone to flooding.

Coil condensers, such as the Dimroth and Freidrichs, have high capacity for cooling with fewer problems from flooding since the vapor condenses on the coils and drips back from the little prominence at the bottom of the coils into the center of the pot. The vapor has an easy time getting past the drops falling into the pot. If you can afford it, this seems like a good choice for most applications. The Freidrichs condensers, which incorporate a cold-finger with the spiral, are higher capacity, quite bulky and heavy. I have seen them used with rotovaps where you are taking a lot of solvent off quickly, but not with an ordinary reflux apparatus. This would be over-kill for a simple reflux reaction situation.

Sorry for the incorrect information (for those of you who looked at this before) and hope this is helpful.


Is ion activity in a salt solution pH-dependent? :) Without being a chemist myself, I need some basic understanding of the ion activities in my buffer solutions.

Now I am wondering whether the ion activity of the K$^+$ and Cl$^-$ ions is different for the buffers with different pH-values.

When I google for that, all I get is ion activity for the hydrogen ion, which -I think- does not interest me at all.

If the ion activity actually IS different, could you point me to some table to look up the correction values?

Thank you!

.

edit: tl&dr:

The buffer solutions contain:
Buffer A: 250mM KCl, 10mM MOPS
Buffer B: 020mM KCl, 10mM MOPS

pH is set to 5,6,7 resp., using unknown amounts of Tris.

Is the ion activity of K$^+$ and Cl$^-$ affected by the pH?

Considering the effect of $\text{pH}$ on ion activity, one should take into account that $\ce{K+}$ and $\ce{Cl-}$ are respectively conjugated acid of a strong base and conjugated base of a strong acid. They are therefore both weak, with no influence on the $\text{pH}$.

As a consequence, I would basically assume that also the other way round is true and that $\text{pH}$ has no effect on their activity.


Order of internal pressure resistance to expect from laboratory glassware? I have a need to use more pressure differential than can be afforded by vacuum vs atmospheric pressure alone, and so I've come to consider using pressurised laboratory glassware for part of my setup.

I've made a little bit of research here on stack-exchange and on google and the general information I've gathered is that the physical shape of laboratory glassware and reaction vessel lends better to resisting external pressure (when we use them with internal vacuum), than to resisting internal pressure, though I'm left wondering what to expect to be doable in terms of pressurising glassware.

I have two pieces of glassware I'm considering using on the pressurised end of my setup: One is a 5,000mL flat bottom flask and the other is a 20,000mL flat bottom flask. Both are borosilicate glass, neither are perfect (they both have one or two tiny air bubbles in the glass at places) but both have shown very good resistance while being used with internal vacuum.

Is it realistic to expect them to be fine while adding an internal pressure up to 3 Bars? (I would put the glassware inside of a big metal cylinder for safety in case of "explosion")

What will be the best Resistance vs Glassware Shape / Glassware Volume / Temperature / etc. ?

Does anyone have any tips on the topic?

Put the flask in a metal container, with a large (at least three times the volume of the flask, for 3 atm pressure) bladder or sylphon (metal bellows) in the supply to the inner flask. Pressurize the outer metal container with air or inert gas.

The absolute pressure in the reaction vessel should be the same as the outer vessel (unless the reactants outgas rapidly), but there would be no differential pressure difference, ergo, no additional stress on the glass.

Or just conduct the whole experiment in a hyperbaric chamber... not a wimpy 0.1 atm plastic bubble, but something like this 10 atm unit, with margin for error. ;-)


Draw the four isomers of C2H2Cl2Br2 I have to draw the 4 four isomers for the molecule  $\ce {C2H2Cl2}$  when $\ce {Br2}$ is added to it. At last the four isomers for  $\ce {C2H2Cl2Br2}$.
I have found three of them:

I can't find the last one.

You have only three isomers with the configurations (1R,2R), (1S,2S) and (1R,2S). The one with the configuration (1R,2S) is achiral, as it has a symmetry plan, and it's superimposable on its image by a mirror (i.e. it is identical to its enentiomer (1S,2R)).

This isomer is known as "meso compound".

Meso compounds are a special case of stereoisomers which occur when a molecule has two chiral centers and each chiral center has identical substituants.


How much CO2 is emitted by burning coal to produce 1 kWh of energy I'm a little embarrassed to ask this question at this stage of the climate game but, in searching, I seem to get contradictory answers.

For example, here: https://www.eia.gov/tools/faqs/faq.php?id=73&t=11
we see 215.4 lbs CO2 giving 1 million btu heat. Since 1 million btu = 293 kWh, we end up with 0.735 lbs/kWh.

But at this site: http://shrinkthatfootprint.com/electricity-emissions-around-the-world
we see 1 kg/kWh = 2.2 lbs/kWh

New York times https://www.nytimes.com/2007/11/07/business/businessspecial3/07carbon.html
gives 1.9 lbs/kWh

These are very different numbers (too different for the problem to be type of coal, at least for eia vs NYT) and I have found other numbers at other sites.

Can anyone point me to a site that I can trust for this number and others like it (related to energy production)?
Or else point out the elementary error I am making?

The value from EIA is pounds of $\ce{CO2}$ produced per kWh of thermal energy produced, while those from the latter two sources are pounds of $\ce{CO2}$ produced per kWh of electrical energy produced.  Since the thermal efficiency of coal-fired plants is ~37% (~37% of the coal's thermal energy is converted to electrical energy), converting the EIA figure to pounds of $\ce{CO2}$ produced per kWh of electrical energy produced gives us:

$$ 0.735 \frac{\text{lbs }\ce{CO2}}{\text{kWh thermal energy}} x \frac{\text{1 kWh thermal energy}}{\text{0.37 kWh electrical energy }} = 1.99 \frac{\text{lbs }\ce{CO2}}{\text{kWh electrical energy}}, $$

i.e., about the same as the NY Times and shrinkthatfootprint figures.


Which method can be applied to separate ferrofluid liquid from oil? I want to know if there is any substance, liquid or any other stuff to separate ferrofluid from oil after mixing them.

Typical ferrofluids are suspension of nano-sized magnetic particles in oil. If you add more of the same oil and mix thoroughly, separation is impossible.

If you add a chemically different oil, you might be able to remove this compound from the mixture again by distillation in vacuum.


Why are solutions electrically neutral? I keep reading that solutions are electrically neutral, but I've never read an explanation. I've read after the fact justifications, such as one problem which asked me to calculate the force of repulsion between two beakers with a tiny, tiny bit of excess charge (it ended up being over 10,000 pounds).

So why are solutions electrically neutral? Not interested in after the fact explanations like the above example.

It is more acurate to say that they aren't very far from being electrically neutral. If you take a charged sphere and drop it into a beaker of some electrolyte sitting on an insulating counter, the result is a charged solution, albeit slightly.

Also, in spite of your amusing ceiling analogy (it made me chuckle), the argument about electrostatic attraction and repulsion does justify why solutions (and solids) can't be too far from being electrically neutral.  The flaw in the analogy is that humans don't need ceilings to exist; but you can' t have two charged beakers without a force between them. If each has a charge of 1 C -- which is a mere $10^{-5}$ times the charge of a mole of electrons -- and they're sitting 1m apart the force between them is about a million tons -- clearly this situation won't last very long!