Determining the structure of N2O using hybridization I have learnt how to determine the structure of molecules where only the central atom is hybridised like $\ce{ClF3}$, $\ce{C2H2}$, $\ce{PCl5}$, but in $\ce{N2O}$ it seems as though both nitrogen and oxygen have hybrid orbitals. How do I find the hybridization of oxygen and nitrogen in $\ce{N2O}$ and finally determine its structure?

Research Effort: I watched some lectures on hybridization, but all of them included cases in which only the central atom had hybrid orbitals. I also read the topic in my book but this issue wasn't addressed there too.

Usually you would approach this problem from the other direction. First, find the most likely structure with the correct number of bonds, then deduce the hybridisation. This is also how it works in quantum chemistry: hybridisation is deduced from geometry, not the other way around.

If you try to write the structure of $\ce{N2O}$ you should come up with two equally like structures; these are mesomeric structures as given in the spoiler tag below.

   $$\ce{\overset{-}{N}=\overset{+}{N}=O <-> N#\overset{+}{N}-\overset{-}{O}}$$

From this, we can deduce the most likely hybridisation which will result in a linear molecule. The central atom in a linear molecule typically features two $\mathrm{sp}$ hybrid orbitals and two unhybridised $\mathrm p$ orbitals.

So what about the outer atoms? Well, we cannot say for sure because we do not have enough geometric information. We can tell by the type of bonds that we again need two unhybridised $\mathrm p$ orbitals on the end atoms (remember that the orientations of the double bonds are equivalent!) but we don’t know whether it is a better description for the other two orbitals to be seen as $\mathrm s$ and $\mathrm p$ or two $\mathrm{sp}$ hybrids. In these cases, I would tend to go with ‘unhybridised’ until an experimental or calculative result proves me wrong.

After we have done this, we can indeed look back and realise that we could have determined the structure a priori like this from the general rules; we would just have had to ignore the outer atoms and discuss hybridisation only for the central atom. Indeed, ignoring the hybridisation of terminal atoms gets you very far even in the simplified theory that puts hybridisation first (and is practically wrong).


Why is entropy change in an irreversible adiabatic process positive? I've read this in a book

During irreversible compression, large quantity of heat is produced due to friction, while the decrease in volume is less. Thus, increase in entropy due to rise in temperature exceeds the decrease in entropy due to decrease in volume.

I don't get this at all. Is the friction they're talking about the one between pistion and container? What if we consider an ideal condition with no dissipation?
I also want to know the physical interpretation of increase in entropy in adiabatic (irreversible) process. Shouldn't it be zero? This book tried to explain it but it was confusing and I'd love to get further insight on this.

The friction they are talking about is viscous friction of the gas itself.  This comes into play when the gas is deformed rapidly, and is roughly proportional to the square of the rate of gas deformation.
Regarding entropy, the change is entropy is equal to the integral of dq/T only for a reversible process.  In an irreversible process, in addition to this exchange of entropy with the surroundings, there is also entropy generated within the system itself.


AAS background correction with Zeeman effect I've found here the explanation to how Zeeman effect can be used for background correction. I understand everything up to this part:
"When light is polarized parallel to the field, both atomic and background absorption occurs, and when light is polarized perpendicular to the field, only background absorption occurs."

I understand that light can be polarized, and that we are trying to somehow exclude either atomic or background absorption from one of our measures. I don't see, however, why this would be valid. Why is the perpendicularly polarized light not experiencing also the atomic absorption?

The very short answer is that properly applied magnetic fields, via the Zeeman effect, can cause atomic absorption bands to split into multiple bands, and these bands have different polarization properties. This is the simulation model I use to demonstrate the spliting and polarization behavior:

This model demonstrates the transverse and longitudinal Zeeman effects.  The theory of Zeeman splitting is covered elsewhere, but the basic idea is this: in a magnetic field, atomic transitions may be “split”, because the levels associated with the atomic transitions are no longer degenerate (due to those pesky magnetic quantum numbers and the associated selection rules for allowed changes in them).  Hence, the transition splits into several transitions.  The simplest case, that of ordinary Zeeman splitting, involves a splitting into three bands: 1 $\pi$ band and 2 $\sigma$ bands.  In the transverse field case, where the magnetic field is transverse to the light propagation direction, the $\pi$ band is linearly polarized parallel to the magnetic field direction.  It does not spectrally shift as a consequence of the application of the magnetic field.  The two $\sigma$ bands are spectrally shifted, by an amount proportional to the strength of the applied magnetic field, to each side of the $\pi$ band.  The $\sigma$ bands are linearly polarized orthogonally to both the light propagation direction and to the magnetic field direction.  The light propagation is the “z” direction.  In the transverse magnetic field orientation, the magnetic field axis is along the “x” axis.  Therefore, the $\pi$ component absorbs light that is plane polarized along the x axis.  The two $\sigma$ components absorb light that is plane polarized along the “y” axis. The next figure shows the unshifted $\pi$ band in the center and the 2 $\sigma$ bands shifted out of harms way:

So the light source is linearly polarized and the $\pi$ band cannot absorb it: it could only absorb the orthogonal polarization, which is not present. The 2 $\sigma$ bands would be able to absorb the light source's polarized light, except that the magnetic field has spectrally shifted them out of harm's way. Thus, when the magnetic field is on, only the broad background absorbance absorbs light. When the magnetic field is off, the background still absorbs light and the atoms do as well.

In the anomalous splitting situation, it is possible to have rather complex splittings, even neglecting hyperfine splitting complications, isotope effects, etc.  In the simplest anomalous splitting, there are two $\sigma$ bands and two $\pi$ bands.  The $\pi$ bands are spectrally shifted, but are still polarized parallel with the magnetic field.  Their spectral shift is less than that of the $\sigma$ components.  In the longitudinal Zeeman configuration, the magnetic field is along the light propagation axis, so there is no $\pi$ component to the absorbance because light cannot be polarized along its propagation direction.  The two $\sigma$ components are circularly polarized, with opposite handednesses, and are spectrally shifted by an amount proportional to the strength of the applied magnetic field. The next figure shows the splitting for calcium, in the longitudinal Zeeman configuration:

Both of these Zeeman configurations are used for background correction in atomic absorption spectroscopy.  The basic idea behind all background correction schemes, in atomic absorption spectroscopy, is this: perform two absorbance measurements and take their difference.  Both measurements should be designed to do an excellent job of measuring the interfering background absorbance, but only one of the measurements should do an excellent job of measuring the desired atomic absorbance.  The other measurement should do a poor job of measuring the desired atomic absorbance.  Differencing then discriminates against the interfering background absorbance, because it is common mode, and it yields the desired atomic absorbance, because it is differential mode.  Just how well the scheme works depends on a variety of factors, so, as it happens, there are several viable background correction schemes in use.

The Zeeman effects are useful for background correction because, by suitable use of polarizers and magnetic fields, the desired atomic absorbance can be effectively “turned off”.  This makes it possible to perform a measurement where the absorbance measured is almost exclusively due to just the interfering background absorbance.  As noted above, this measurement is one of the two needed for the differencing involved in background correction. Hope this answers your question.


Software for viewing a structure and altering bond angles Does anyone know of any good software that would allow me to view the structure of a compound while also allowing me to change the bond angles in the structure (and give me numeric values for those angles)?

I'm looking for something freely available, but that's not a major requirement. Thank you so much!

Have a look at Avogadro!

build your own structures or import files in various formats.
optimize the geometry using different force fields
change and measure bond length, bond angles and dihedral angles

Once you've used it, you don't want to miss it again.

Geoff Hutchison, who cannot be praised enough for his child Avogadro, has given the right directions in his comment. I'm just adding some images to illustrate how easy angle adjustment is:

Draw a molecule in Avogadro
From the main menu, select Build -> Properties -> Angles or whatever the localized menu of your system shows (on my German machine, it is Ansicht -> Eigenschaften -> Winkeleigenschaften)
In the table that pops ups, select the angle you want to change. Note that the atoms of the selected angle are highlighted

Change the value for the angle. Note that the geometry of the molecule is immediately updated!

So again, I think that Avogadro is one of the best tools available and it is free and open source! Use it in education, use it in research and whenever you do, cite the Avogadro paper:
Hanwell, M. D.; Curtis, D. E.; Lonie, D. C.; Vandermeersch, T.; Zurek, E.; Hutchison, G. R.; Avogadro: an advanced semantic chemical editor, visualization, and analysis platform. J. Cheminf. 2012, 4 (1), No. 17. DOI: 10.1186/1758-2946-4-17.


Are lithium bicarbonate and magnesium bicarbonate both soluble in water? Are lithium bicarbonate and magnesium bicarbonate both soluble in water?

Source: JEE Mains 2017 (National Level Examination of India)

Please do cite a reputed source along with your answer. Currently I have found a line from JD Lee (Concise Inorganic Chemistry 5th ed):

According to this Wikipedia solubility table, lithium bicarbonate is soluble in water to $\pu{5.7 g/100mL}$ at $\pu{20^oC}$, while this Wikipedia page gives the solubility of magnesium bicarbonate as $\pu{0.077 g/100mL}$.

However, another source (found by Oscar Lanzi in his good answer), "The Solubility of Magnesium Carbonate in Aqueous Solutions of Certain electrolytes, The Journal of Physical Chemistry, Volume 7, pub. 1903, reports making two measurements of the solubility of magnesium bicarbonate and finding $\pu{3.1 g/100mL}$ and $\pu{2.0 g/100 mL}$.  Note that this source is from 1903, but also that the the Wikipedia link above does not give a source for their value at all. At least one reason for the dearth of information regarding magnesium bicarbonate is that it easily decomposes and it cannot readily be isolated outside of aqueous solutions. My feeling is that the Wikipedia value seems low, and that the values reported in the Journal of Physical Chemistry seem more reasonable.


What is the difference between fractional freezing and fractional crystallization? More simply put, what is the difference between freezing and precipitation when considering a mixture?

Take two scenarios:

First a mineral in water. If the temperature of this solution is lowered, the solubility of the mineral will decrease and eventually precipitate out of solution. This could be done in a cascade type setup so that each time, the solution is cooled further and the liquid phase becomes more and more concentrated in water.

Second, consider a mixture of ethanol and water. Here too, I could lower the temperature until crystals first form. They would be almost pure water, leaving behind a solution higher in concentration in alcohol. I could repeat this procedure until I reach the eutectic mixture.

In both cases, a drop in temperature is allowing a solid phase to form out of an aqueous solution, yet one is referred to as partial crystallization and the other is referred to as partial freezing. I am not sure I see the difference.

There is no difference. Strictly speaking, "freezing" does not necessarily include crystalline solids, and may include amorphous materials that precipitate out of solution. But, this does not really matter in the context of these eutectic (or similar) systems.

I suspect the latter is referred to as "partial freezing" because the solid is water, whereas it is more general in the former. We rarely speak of water crystallising to ice, but rather water freezing to ice. This is more of a linguistic issue than scientific.


Air recycling system(CO₂-->O₂) or (CO₂-->fuel)? I'm not a chemist but I'm working on an eco-friendly project that is supposed to recycle $\ce{CO2}$ from the air and convert it to $\ce{O2}$ or at least a usable fuel source ,the system can use the solar system as an energy source (I know that such system isn't efficient) but for the country in which the project is applicable it would be a great economical benefit (over direct sun power).

-What is the most appropriate chemical reaction to achieve such goal ?

the reaction of course shouldn't emit toxic material(the whole purpose of the project is to benefit the environment, plus the cost of any filter or disposal mechanism isn't affordable)

is there any way to construct a circular system (Regenerative carbon dioxide removal system )
Wikipedia's article doesn't elaborate further.

Please help tag this question(this is my first time on chemistry.stachexchange)

This is quite a hot topic at the moment - the current buzz-phrase is power to gas.

There are quite a few conference papers and technical papers, and at least one dedicated conference, for it.

And it's too soon for there to be a known "best" process.

Audi and others are using a modified Sabatier process.

$$\ce{CO2 + 4 H2 -> CH4 + 2 H2O }$$

Others are looking at a modified Fischer-Tropsch process.

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

(image courtesy of the RWE power to Gas Project)

Essentially, you've got to prise the Oxygen off the Carbon, which is an energy-intensive process; and then attach the Hydrogen, which is not. But the Hydrogen typically comes from electrolysis of water, which is energy-intensive.

Not only are the processes being prototyped high-energy, but they're high-exergy too, requiring very high heat and / or electricity.

So it's unlikely that this is a better option than using solar power for solar-thermal or photovoltaics, unless the application is long-distance travel, which would require an energy source with a very high energy density. In which case, there are several renewable-fuel synthesisers being prototyped at the moment, but it's too early to tell which is the best.


Rutherford's Alpha Ray Scattering Experiment & Plum-Pudding Model My teacher said, "Rutherford thought that all the Alpha particles would directly go through the gold foil without any collision if the plum pudding model was right. Because, as said in plum pudding model, positive charges are arranged in a balanced way in the atom with electrons like plum in pudding. As there is no densely accumulated charge in the center, so the Alpha particle will not deviate by anything."
But two question raised on my mind.

Is my teacher right?
If not, then why?
If yes, then why evenly accumulated positive charge will not deviate the Alpha rays? I mean, when the alpha ray will come near to the pudding like positive charges, it should deviate more sooner nah? Moreover, as there are evenly arranged positive charges, so, no empty space is in the atom. So, no alpha particles would come across the foil. Is my assumption true if thought from plum pudding model?

First of all, yes, your teacher is right for most of the part but it is not that they will go without any collision. Instead, they will penetrate through the foil like a bullet shot at paper.
It is not that the alpha rays will not be deviated but that if the charge was evenly spread throughout the atom, it would not be able to provide sufficient repulsive forces to wear out the kinetic energy already given to the alpha particles and slow them down. As a result, the alpha particles should have penetrated the atom like a bullet penetrates a sheet of paper.


On mathematics behind molecular orbital energies When explaining the mathematics behind calculating molecular orbital energy , namely HOMO and LUMO energies, using quantum chemistry, Chainer Chemistry documentation states that

  From mathematical viewpoint it requires a solution of an internal
  eigenvalue problem for a Hamiltonian matrix.

I have never heard of internal eigenvalue problem in mathematics. Is this a term specific to quantum chemistry and what it means?
Is the Hamiltonian matrix same as Kohn-Sham Hamiltonian?

An internal or interior eigenvalue problem seeks a solution somewhere in the middle of the spectrum of eigenvalues as opposed to trying to find the smallest or largest eigenvalue. Its generally trickier to solve for a particular eigenvalue than it is to solve for the smallest/largest one. The HOMO and LUMO are necessarily somewhere in the middle of the eigenspectrum.
The Kohn-Sham Hamiltonian is the orbital-DFT approximation of the molecular Hamiltonian. The Hamiltonian describes the interactions within the molecule and allows one to solve for the molecular wavefunction through the Schrödinger equation. Most of electronic structure is focused on approximations of the Hamiltonian or the form of the wave-function that make solving this problem tractable. There is a lot more that can be said about Hamiltonians than would ever fit in an answer here, so you should try to find good textbooks and papers that discuss them.


Good collection of images showing periodicity of the Table? Mendeleev's recognition of the periodicity of atomic trends was a milestone of human understanding, the detection of a pattern in what had appeared to be chaotic data. I think the significance and sheer beauty of these patterns is lost on many people (layfolk and students alike). I am wondering whether anyone knows of a coffee-table book or some other collection of images that evoke the predicted nature of atomic properties such as atomic radius, ionization energy, melting point, etc. Sure, the information is available via interactive software, but I'm looking for something more aesthetic... if it exists yet.

Particularly, my search is provoked by this striking arrangement of oxidation state information. Were it to be cleaned up a bit, it would make a beautiful poster.

I have an entire chapter regarding periodicity in my textbook. Have a look at the images in it:

Atomic Radius:

Ionization Enthalpies:

A summary of periodic trends:

These are the pictures in my textbook. As a bonus, I'll even add a few examples from our day-to-day life:

Alkali Metals:
Lithium:

Sodium:

Potassium:

Rubidium: (its enclosed in glass because it spontaneously combusts in air)

Cesium:

You can notice that properties of elements periodically repeat. All the alkali metals have silvery appearance, with cesium having a little bit of yellow-golden in it. They are soft, and have exceptionally high reactivity.

Halogens:

Fluorine:

Chlorine:

Bromine:

Iodine:

You can see that all the halogens are quite colored, and most have low melting and boiling points. Thats why fluorine and chlorine are gases, while bromine is a liquid with thick vapors. Iodine, however is a solid due to higher attractions due to increased molecular mass. All the halogens are quite reactive, with fluorine and chlorine violently so.

Transition Metals (a few examples):

Iron:

Cobalt:

Nickel:

Chromium:

You can see how the transition metals are quite similar to each other. They are lustrous in appearance, are quite hard, and also display quite a few catalytic properties. For example, iron is used in the manufacture of ammonia, nickel is used to hydrogenate vegetables oils, palladium is used in catalytic converters, et cetera.


Confusion regarding a stoichiometric formula In a book, which contains only stoichiometry problems, I came across this formula.

  For any given compound $\ce{M_xN_y}$, $x \text{ Moles of N} = y \text{ moles of M}$

According to this rule if I take $\ce{H2O}$ then, $x = 2 \text{ and } y = 1$.

So by the above formula,

$2 \text{ moles of Oxygen} = 1 \text{ mole of Hydrogen}$

But surely,

$2 * \mathrm{N_A} \not= 1 * \mathrm{N_A}$, where $\mathrm{N_A}$ is Avogadro's number.

However, in an example, the author has used this formula and successfully solved a problem? So, if it's correct, can anyone prove this and point out where my contradiction is wrong.

Example as requested:

  What weight of oxygen will react with $1$g of calcium?($\ce{Ca}= 40\mathrm{g \over mol}$)

Answer by author :
Since all the atoms of $\ce{Ca}$ have changed into $\ce{CaO}$, the amount of $\ce{Ca}$ in $\ce{CaO}$ is $1$g. Now from the formula of $\ce{CaO}$, we have,

$\text{Moles of Ca = Moles of O}$

$$\frac{\text{mass of Ca}}{\text{atomic mass of Ca}}= \frac{\text{mass of O}}{\text{atomic mass of O}}$$

therefore, $\text{mass of O} = {1\over40} * 16 = 0.4$g

You misread the formula. For compound $\ce{M_{x}N_{y}}$

$$y\cdot \ce{mol M} = x\cdot \ce{mol N}$$

This is because the formula essentially tells you that the ratio between moles of $\ce{M}$ and $\ce{N}$ is:

$$\frac{\ce{mol M}}{\ce{mol N}} = \frac{x}{y}$$

This rearranges to the equation above.


Is the first ionization energy in oxygen slightly more than nitrogen? Why is the first ionization energy in oxygen slightly more than nitrogen?

In nitrogen: $\ce{[He] 2s^2 2p^3}$
In oxygen: $\ce{[He] 2s^2 2p^4}$

This tells me that it should be easier to remove an electron from oxygen than it is for nitrogen as the electron in oxygen is slightly further away from the nucleus meaning lesser nuclear charge.

But why is it harder to remove an electron from oxygen, i.e. why is the first ionization energy of oxygen higher?

Oxygen has a lower first ionization energy as the electron that is removed is coming from a paired orbital.

Electrons within the same orbital experience maximum repulsion as the distribution of their wavefunctions is the same, so the probability density distribution is the same and the electrons can be thought of as occupying the same space. This maximizes their repulsion and increases the potential energy of the electrons in that orbital, making the electrons easier to remove. This is despite the increased effective nuclear charge experienced by the electron in oxygen and the decreased radius of the orbital.

See: "Physical Chemistry", Atkins, P.W. Section 13.4, p.p.370 (4th edition) - sorry, I have an old one!


Is there an algorithm that takes atom coordinates and produces VSEPR geometries? That is, you would give the coordinates of an atom, the coordinates of four (for example) other atoms, and it would return "tetrahedral" or "square planar" or whatever.

I'm currently writing a function that does just that, but (1) I feel like I'm making a lot of arbitrary decisions in the angle tolerance cutoffs I am using, and (2) I'm not sure it's very efficient the way I am doing it.

I have searched but haven't found a prior implementation of this - is there one?

Quantum chemistry software usually exploits molecular symmetries to speed up computations. To do so they need some way to determine the point group for a given set of atoms and their coordinates. So this is basically what you are asking about, just that get something like $D_{4h}$ instead of "square planar", or $T_d$ instead of "tetrahedral".

I am not sure whether this is actually used by the mentioned software, but it is quite simple and efficient if you only want to test for certain point groups:

Each symmetry operation can be represented as a $3\times 3$ matrix, for example mirroring in the $xy$ plane is

\begin{equation}\sigma_{xy} = \begin{pmatrix}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & -1
\end{pmatrix}\end{equation}

We can represent the atom coordinates in a $3\times N$ matrix, where $N$ is the number of atoms. The molecule needs to be properly oriented (the center of mass should be in the origin of the coordinate system and the principle axes of inertia should coincide with the $x$, $y$ and $z$ axes).
For example $\ce{H2}$ (or any other homonuclear diatomic):

\begin{pmatrix}
 0 & 0 \\
 0 & 0 \\
 1 & -1
\end{pmatrix}

To get the transformed coordinates according to a certain symmetry operation, we just apply the matrix of the symmetry operation to the coordinates matrix.

\begin{equation}\begin{pmatrix}
 1 & 0 & 0 \\
 0 & 1 & 0 \\
 0 & 0 & -1
\end{pmatrix}\times
\begin{pmatrix}
 0 & 0 \\
 0 & 0 \\
 1 & -1
\end{pmatrix}=
\begin{pmatrix}
 0 & 0 \\
 0 & 0 \\
 1 & -1
\end{pmatrix}
\end{equation}

We can then compare the resulting set of coordinates with the original one. Since for $\ce{H2}$ both atoms are the same, we can interchange the 2 columns and see, that we have identical matrices. Thus, $\sigma_{xy}$ is a valid symmetry element of our input coordinates. At this point you may implement a tolerance threshold, by allowing the norm (distance) between two "same" points being larger than 0.
The set of found symmetry operations will then directly give us the point group of that molecule (this information is available in character tables). So we can check for a certain point group by checking whether the symmetry operations of that point group can be found in our input geometry. This however means we need to define which point groups we want to test for.

There are also flow charts (available in books about molecular symmetry or via your favorite search engine), telling you which symmetries elements to check for and then guide you to the point group.

I am not aware of a more general approach. If someone knows more, please extend.


Are these 3 molecular structures of ammonium bicarbonate the same? Are these 3 molecular structures of ammonium bicarbonate the same?

A:

B:

C:

Like Punarbasu Roy already stated, the depicted structures are certainly not the same. To be frank, only one of them (A) is correct, one of them is not completely wrong (B - but only because it is incomplete), and one is completely, utterly wrong (C).

In lieu of an actual crystal structure of ammonium hydrogencarbonate,[1] I'll refer to the structure of the parent compound ammonium carbonate monohydrate to make my point.[2] It is actually quite difficult to assign one correct Lewis structure. In its solid form ammonium (bi)carbonate is a salt, in which ammonium and (bi)carbonate ions are interacting via ionic bonds in a lattice. In solution this certainly is broken up and we will find separate ions, according to the following equilibria:
\begin{align}
\ce{(H4N)HCO3~(s) &~->C[(H2O)] NH4+ ~(aq) + HCO3- ~(aq)}\\
\ce{NH4+ ~(aq) + H2O &~<=> H3+O~(aq) + NH3~(aq)}\\
\ce{HCO3- ~(aq) + H2O &~<=> H2CO3 ~(aq) + {}^{-}OH~(aq)}\\
\ce{HCO3- ~(aq) + H2O &~<=> CO3^2- ~(aq) + H3+O~(aq)}\\
\end{align}
The same equilibria will establish for ammonium carbonate with different concentrations.

In the crystal structure of ammonium carbonate monohydrate you can see the different molecular entities (ions) held together via hydrogen bonds, reproduced from [2]:

Wikipedia[3] therefore offers the best possible Lewis structure:

Strictly speaking a Lewis structure may only contain covalent bonds, i.e. two-centre-two-electron bonds, while ionic bonds are implied by placing the (molecular) ions in close proximity to each other. However, the hydrogen bond is a well known concept in chemistry and it was already extensively discussed by Linus Pauling. In his book "The Nature of the Chemical Bond" he uses dashed lines to represent the hydrogen bond.[4] Certainly another interesting read is the approach of a more unified theory about the hydrogen bond.[5]
Another possibility is to introduce the donor-acceptor notation with arrows, to indicate a bonding interaction. Therefore one of the following structures may also be considered correct in an extended Lewis concept.

The following article reads like this is still a work in progress: R. Brooks, and T. C. Alcock, Nature, 1950, 166, 435-436. (It does also not come with any supporting information.
A. D. Fortes, I. G. Wood, D. Alfè, E. R. Hernández, M. J. Gutmann, and H. A. Sparkes, Acta Cryst. Sec. B, 2014, 70, 948-962. This quite interesting article is open access.
Ammonium bicarbonate, Ammonium carbonate
Linus Pauling: The Nature of the Chemical Bond. Cornell University Press, 1960.
G. Gilli, P. Gilli, J. Mol. Struc., 2000, 552 (1-3), 1-15.


SN1 Polar Protic Solvent Stabilization of Substrate I know that the standard textbook answer is that a polar protic solvent is able to better stabilize/solvate carbocations through hydrogen bonding.

I understand that a solvent such as water or methanol is capable of hydrogen bonding, but what about the substrate? The substrate is a hydrocarbon; how is the substrate capable of reciprocal hydrogen bonding?

Is this stabilization better described as a dipole-ion interaction rather than a hydrogen bond interaction because I don't see how a hydrocarbon can form reciprocal hydrogen bonding interactions.

I know that the standard textbook answer is that a polar protic
  solvent is able to better stabilize/solvate carbocations through
  hydrogen bonding.

In an $\ce{S_{N}1}$ reaction, a polar protic solvent can stabilize the leaving group through both hydrogen bonding and solvent dielectric effects.  While these same solvent dielectric effects can also stabilize the carbocation intermediate, I don't think hydrogen bonding plays a role in the stabilization of the carbocation itself.  First off, any protic molecule close enough to the carbocation to hydrogen bond would simply react with it instead of forming a hydrogen bond. Second, since the carbocation is positively charged it cannot form a hydrogen bond.  Again, solvent stabilization of the carbocation occurs through dielectric (dipolar) effects of the medium.


Glowing stone - Scarab carving I have an intriquing question and, after searching whole Google, found almost no information. I simply don't know where else to look for answer and so I am asking you.

When I was a child, my parents travelled with me to Egypt and there, they bought me a little scarab beetle carved of some stone, about 5 cm long, 2 cm width and height. The stone is white-ish and translucent and glows green in dark (for some time). I only remember that it might (just a speculation) have been in Luxor in Alabaster workshop. But, according to online sources, Alabaster is not supposed to glow in dark, so what is it then?
Can you please tell me?

Oh yea, right. Here are some images. The fluorescence is really weak and doesn't last longer than 5-10min so I am sorry for a bad photo, just couldn't make it better.

Daylight 01

Daylight 02

Daylight with flash

Daylight against sun 01

Daylight against sun 02

Fluorescence

Well, it doesn't look like Fluorite to me(according to Google images) but then again, I'm no mineralogist either so, maybe it's some form of it... what do you think?

Your description matches fluorite, you can see many pictures in Internet, and compare with your stone. To determine what it is precisely, you have to post a picture.

EDIT. After the pictures are posted, I take back my suggestion: the key difference
is that fluorite is not transparent, at least those pieces that I possess, and those which I can see on Internet. But in the dark, it looks exactly of the right color. There  are very many fluorescent stones, look for example, here:
 http://www.galleries.com/minerals/property/fluotabl.htm


"Carbon snakes" - what physico-chemical process causes the exothermic response? Making a "carbon snake" requires the mixing of strong sulphuric acid with sugar, as explained in this about.chemistry webpage - the reaction causes rapid dehydration of the sugar and produces steam in the process, which expands and forms the 'carbon snake', following the incomplete chemical reaction:

$\ce{C12H22O11 + H2SO4 → 12C + 11H2O}$  + mixture of excess unreacted water, sulfur dioxide and acid (and presumably a bundle of other chemicals)

My question is, what physical/chemical processes cause this rapid dehydration and exothermic reaction?

$\ce{C_mH_{2n}O_n -> m C + n H2O}$

The hydration reaction of sulfuric acid acts as the driving force because it is highly exothermic (95.33 kJ/mol H2SO4).


Typesetting of the retardation factor (Rf) What should be the typesetting for the symbol of the retardation factor as used in thin-layer chromatography?

These are the possibilities:

Rf - only f in italics
Rf - both R and f in italics
Rf - only R in italics

On Wikipedia, they have used all sorts of typesetting. There is also RF with a capital F.

According to Nomenclature for Chromatography (IUPAC Recommendations 1993) [1, pp. 843, 845] (also listed in IUPAC Gold Book), retardation factor is denoted as follows:

Column chromatography: $R$ (capital $R$ in italics):

  3.7.13 Retardation Factor ($R$)

  The fraction of the sample component in the mobile phase at equilibrium; it is related to the retention factor and other fundamental chromatography terms:

  $$R = 1/(k + 1)$$

Planar chromatography: $R_\mathrm{F}$ (capital $R$ in italics and upright capital $\mathrm{F}$ in subscript):

  3.8.04 Retardation Factor ($R_\mathrm{F}$)

  Ratio of the distance travelled by the center of the spot to the distance simultaneously travelled by the mobile phase. Using the symbols of Fig. 2:

  $$R_\mathrm{F} = b/a$$

  By definition the $R_\mathrm{F}$ values are always less than unity. They are usually given to two decimal places. In order to simplify this presentation the $hR_\mathrm{F}$ Values may be used: they correspond to the $R_\mathrm{F}$ values multiplied by 100.

  Ideally, the $R_\mathrm{F}$ values are identical to the $R$ values (see 3.7.13).

Reference

Ettre, L. S. Pure and Applied Chemistry 1993, 65 (4) DOI: 10.1351/pac199365040819.


Peculiar deprotonation of an alkenyl proton In this synthesis of Punctatin A, there is this peculiar deprotonation step. When subjected to the base $\ce {NaH}$, we would expect deprotonation at the $\alpha$-position, at the side opposite to the double bond. However, what is observed here is, in fact, the deprotonation at the double bond. Not only is the $\ce {C(sp^2)-H}$ bond stronger, there is also no delocalisation of the negative charge in the conjugate base that is produced. Note that the lone pair in the carbanion conjugate base is held in an $\ce {sp^2}$ hybrid orbital that is perpendicular to the $\ce {p}$ orbitals of the enone $\pi$ system. Does anyone have any idea what is going on here?

Decided to convert this to an answer anyway. Disclaimer: I'm not an organic chemist (anymore), and corrections/suggestions are more than welcome.

One distinct mechanistic possibility here is that you deprotonate γ to the carbonyl, then the extended enolate reacts at the position you see, and then after that there's some tautomerisation which brings the double bond back into conjugation with the carbonyl (maybe during workup).

These are specific conditions under which you might expect deprotonation at the γ-position over the other α-position. In particular, the extended enolate resulting from this is more stable than the cross-conjugated enolate resulting from α-deprotonation, so this is called the "thermodynamic enolate". Paquette et al. use this term in the paper (J. Am. Chem. Soc. 1987, 109 (10), 3017–3024):

  Toward this end, the thermodynamic enolate of 6b was exposed to [...] 1-bromo-2-methylpropane.

I won't comment on the specifics of this reaction, but will mention that it can be very dangerous if scaled up.

If you wanted to deprotonate the other α-position, you would probably have more success with something like LDA. That deprotonation relies on chelation of the carbonyl oxygen to lithium (via a cyclic transition state*), so goes for the proton nearer to the carbonyl. That would be the "kinetic enolate".

* The six-membered chair-like transition state is widely taught, but is actually of questionable accuracy in this specific case. I think the transition state is cyclic, but is eight-membered or something like that. See e.g. slides 50 onwards of this very informative presentation from David Macmillan's group: "Lies My Sophomore Organic
Professor Told Me".


Hess Cycle, determining enthalpy change of formation I am asked to find the enthalpy change of formation of the following:
$$\ce{N2 + 1/2O2 -> N2O}$$

I am given the following enthalpies of reaction:
\begin{align}
\ce{C + N2O &-> CO + N2} &\Delta H_f =-193 \tag{1} \\
\ce{C + 1/2O2 &-> CO} &\Delta H_f=-111 \tag{2}
\end{align}

How do I calculate enthalpy change of formation of the nitrous oxide?

I thought it would just be adding the two enthalpies together because I substitute the second equation into the first one, but this is wrong.

If you start by reversing equation 1, you get $$\ce{CO + N2 ->C + N2O~~ {\Delta}H_{f}=+193}$$

Now, add this result with equation 2 and you get $$\ce{CO~+N2~+C~+1/2O2->C~+N2O~+CO~~{\Delta}H_{f}=+193-111=82 }$$

Cancelling like terms that exist on both sides of this equation yields $$\ce{N2~+1/2O2->N2O~~{\Delta}H_{f}=82}$$


What does being an "elementary substance" mean? Which of the following isn't an elementary substance?

Ozone
Sulfur
diamond
quartz

I don't get what elementary substance means.
By seeing the options and the word elementary, I guess it means something like pure or obtained in its original form.

From LATINTOS:

  An elementary substance is a pure chemical substance that consists of atoms belonging to a single chemical element.

Now you should be able to solve your question.


Are atomic mass unit (amu), unified mass unit (u) and Dalton (Da) the same unit but with different names? I recently read up on some chemistry and found these units of mass:

atomic mass unit (amu)
unified atomic mass unit (u)
Dalton (Da)

Are they all the same unit but just with different names or are they different? I'm looking for a simple explanation as I'm new to chemistry.

It is a simple question that requires a rather complicated answer.
Origins

For technical reasons, in 1898, chemist Wilhelm Ostwald and others proposed to redefine the unit of atomic mass as 1/16 of the mass of an oxygen atom. That proposal was formally adopted by the International Committee on Atomic Weights (ICAW) in 1903. ... This suggestion was made before the discovery of the existence of elemental isotopes, which occurred in 1912. ... This definition remained unchanged until 1961.$^1$

I think the Wikipedia article is wrong about the name. I think the notion was initially conceived as "atomic weight," not "atomic mass."
atomic mass unit(amu)

The discovery of isotopes of oxygen in 1929 required a more precise definition of the unit. Unfortunately, two distinct definitions came into use. Chemists choose to define the amu as 1/16 of the average mass of an oxygen atom as found in nature; that is, the average of the masses of the known isotopes, weighted by their natural abundance. Physicists, on the other hand, defined it as 1/16 of the mass of an atom of the isotope oxygen-16 ($\ce{^{16}O}$).$^1$

unified atomic mass unit(u)

...in 1961 the International Union of Pure and Applied Chemistry (IUPAC), which had absorbed the ICAW, adopted a new definition of the atomic mass unit for use in both physics and chemistry; namely, 1/12 of the mass of a carbon-12 atom. ... The new unit was named the "unified atomic mass unit" and given a new symbol "u", to replace the old "amu" that had been used for the oxygen-based units. However, the old symbol "amu" has sometimes been used, after 1961, to refer to the new unit, particularly in lay and preparatory contexts.$^1$

I'll be less diplomatic. Since 1961 "atomic mass unit" has been conflated with the "unified atomic mass unit."
dalton

In 1993, the IUPAC proposed the shorter name "dalton" (with symbol "Da") for the unified atomic mass unit.$^1$

"And now..., for the rest of the story" as Paul Harvey used to say.
Until the 2019 redefinitions of SI units, the number of "entities" in a mole was determined experimentally. Since 1889 the kilogram was the mass of a particular platinum/iridium cylinder. With the 1961 definition of the unified atomic mass unit, exactly 12 grams of $\ce{^{12}C}$ had exactly 1 mole of carbon atoms. So the experimental measure of a mole was based on how many $\ce{^{12}C}$ atoms would be in exactly 12/1000 of a kilogram.
However in 2019 both the kilogram and the mole were redefined to give them particular values that would no longer depend on experimental measurements. This was in a continuance of an effort by scientists to define a set of primary physical constants that didn't depend on particular physical standards such as scratches on a bar to determine the meter or a blob of metal to determine the kilogram.

The mole is now defined as exactly $6.02214076\times10^{23}$ entities.
The kilogram "is defined in terms of three fundamental physical constants: The speed of light c, a specific atomic transition frequency ΔνCs, and the Planck constant h."$^2$

The point of this is that with the mole and kilogram now being given definite fixed values, exactly one mole of carbon-12 atoms no longer has a mass of exactly 12 grams by definition. Thus now $u≈\pu{1.66053906660(50)\times 10^{−27} kg}$.

To chemists any difference between the pre and post 2019 values will be infinitesimally small. This is because the average atomic masses of the elements depend on the relative abundances of the isotopes of the element which vary. Thus the isotopic variation greatly limits the precision of the average atomic mass of the elements.
To physicists the difference in measurements will make a difference since the mass of a particular isotope is known with much greater precision that the average atomic mass of the element.

tl;dr For practical purposes the three terms are synonymous as user Maurice noted. It would seem that the preferred term is now the dalton.
$^1$ Wikipedia's article on the dalton (Da).
$^2$ Wikipedia's article on the kilogram.


Do volumes and pressures calculated from Boyle's Law depend on the temperature? As the title suggests, i have a question regarding Boyle's law.

As Boyle's law states that "The volume $V$ of a given mass of a gas, is inversely proportional to the pressure $p$ applied to it when the temperature $T$ is constant."

So if we had a gas at say a pressure $p_1$, and a volume $V_1$, corresponding to this pressure at a constant temperature $T_1$.

Now if i were to increase the the temperature of this gas to a new temperature $T_2$ and make it constant. And then apply the same pressure $p_1$ on the gas would the volume still be $V_1$?

Considering the fact that volume and pressure are inversely proportional for a constant temperature, and hence they dont depend on the temperature?? Am i right?

Thanks

No,the volume would change this time  as by Gay Lussac law pressure is directly proportional to temperature  which has increased this time and can overcome the pressure you had applied earlier thus increasing the volume.
Also I think you have a misconception regarding the term constant temperature.It means that in Boyle s law pressure is inversely proportional to volume at a given temperature say 200K .But for different temperature the law will follow but with different values.For eg. at a particular temperature 200k it can be:- 200litres and 50 pascal or 100 litres and 100 pascal.For higher temperature it would be:- 200 litres and 100 pascal or 100 litres and 200 pascal.


Gold recovery under vacuum Can gold recovery with acid peroxide (ap) method be done under vacuum or left in sealed chamber, as opposed to placement in open air?

This process is typically done by extracting the gold in a mixture of something like 3 parts concentrated hydrochloric acid to one part 3% hydrogen peroxide. If this is roughly the procedure you are using, which is the common metehod posted all over the internet for recovering gold from used electronics, then the answer is an unambiguous none of the above!

Unless you have experience and training for working with these chemicals in a laboratory setting, then you should not be doing this at all; it is simply not worth it.

Given that, the best answer your question is to use a fume hood, eye protection, gloves, lab coat, etc. The biggest danger you face, other than splashing the solution into your eyes, is inhaling the fumes. As this process produces toxic gases (chlorine as well as the fumes from the reagents themselves) you don't want to use a sealed container or a vacuum system that would just pump the gases out of your pump. You need an appropriate, certified fume hood for working with these materials. Relying on diffusion or wind in the "open air" is just incredibly dangerous and you could permanently damage your lungs in one unlucky breath.

Again, the answer is: none of the above, use a fume hood!


Why do glucose, galactose and mannose differ in solubility in water? I don't understand why they would be so different, when the only structural difference is the orientation of hydroxyl groups.

The solubility of these sugars will depend on the relative energies of the crystalline form versus the dissolved form of the sugar. Without knowing the specifics, one could surmise that the crystal structures of these three sugars may differ substantially from each other - one may be more tightly or efficiently packed into a crystal lattice than another due to the specific shape of the molecule, resulting in a lower energy state than that of the other sugar's crystal form, and hence giving a lower solubility in water.


The role of halogens in electrophilic aromatic substitution I'm trying to understand what halogens do in electrophilic aromatic substitution. Consider the following.

Bromobenzene is introduced to a carbocation. Bromobenzene will likely react, but at a slower rate than unsubstituted benzene, because the bromine is inductively withdrawing. The bromine is also not resonance donating/activating, but can stabilize the intermediate product through resonance donation.

The bromine is an ortho/para director, but not because it donates electron density to these positions. Rather, when bromobenzene attacks a carbocation, the meta positions are formally positive, and through resonance donation, bromine can complete an octet in one of the resonance depictions.

Is this an accurate depiction of what a halogen does to benzene?

The bromine is also not resonance donating/activating

You can't have it both ways.  If the bromine can resonance interact with the intermediate, then it can also resonance interact in the starting bromobenzene.

Let's back up to the starting bromobenzene.

  Bromobenzene will likely react, but at a slower rate than plain
  benzene, because the bromine is inductively withdrawing

That's correct, but let's compare further.  The oxygen in phenol is electron withdrawing (oxygen's electronegativity is even greater than bromine's), yet phenol is a very reactive aromatic compound towards electrophilic aromatic substitution (it will decolorize a solution of bromine, it doesn't even need a Lewis acid catalyst).  The difference between bromobenzene and phenol is that the phenolic oxygen can interact strongly via resonance and donate electrons into the aromatic ring, while the bromine in bromobenzene also resonance interacts with the aromatic ring, but much less efficiently (the larger size of bromine's $\ce{4p}$ orbitals makes overlap much less efficient with carbon's 2p orbitals).  Although the interaction is much less efficient, there still is some resonance interaction between the bromine substituent and the ring.  Look at the analogous resonance structures for bromobenzene that you drew for the intermediate, in the contest between inductive deactivation and weak resonance activation, the ortho and para positions are the least deactivated.

When bromobenzene undergoes electrophilic aromatic substitution, the whole ring is deactivated because (unlike phenol, where the resonance effect overwhelms the inductive effect) the weak resonance effect doesn't offset the strong inductive effect.  However, since the bromine substituent does interact with the ortho and para positions by resonance, these two positions are less deactivated that the meta position which cannot interact via resonance with the bromine substituent.


Standards for drawing chemical molecules I'll preface this by saying I haven't had a chemistry class in 10 years and molecular structures are not my thing. But I'm submitting a revision to a journal paper where I need to include figures of molecules and I need to make sure I get it right so I don't get nit-picked apart for it.

Is there a standard/guide/list of rules for how to create chemical molecule figures like the one below?

I'm looking for a comprehensive list of rules but as an example of things I would need to know:

Should I include the carbon atoms in the ring drawing?
At what angle should the entire molecule be drawn (I've seen rigid-body rotated versions of the same molecule before)?

Rather than answering those specific questions, I'd like to know if anybody (IUPAC maybe?) has a standard for drawing these figures.

As it happens, there are a set of quite involved documents from the "IUPAC Chemical Nomenclature and Structure Representation Division".

There's a list here, but to avoid later link-rot, they are currently:

Jonathan Brecher Graphical representation of stereochemical configuration (IUPAC Recommendations 2006) 2006, Vol. 78, Issue 10, pp. 1897-1970
Richard M. Hartshorn, Evamarie Hey-Hawkins, René Kalio and G. Jeffery Leigh: Representation of configuration in coordination polyhedra and the extension of current methodology to coordination numbers greater than six (IUPAC Technical Report)
2007, Vol. 79, Issue 10, pp. 1779-1799
W. Mormann and K.-H. Hellwich: Structure-based nomenclature for cyclic organic macromolecules (IUPAC Recommendations 2008) 2008, Vol. 80, Issue 2, pp. 201-232
Jonathan Brecher: Graphical representation standards for chemical structure diagrams (IUPAC Recommendations 2008) 2008, Vol. 80, Issue 2, pp. 277-410
Andrey Yerin, Edward S. Wilks, Gerard P. Moss and Akira Harada: Nomenclature for rotaxanes and pseudorotaxanes (IUPAC Recommendations 2008) 2008, Vol. 80, Issue 9, pp. 2041-2068


Extracting "Crystal Radii" from a CIF file I'm currently downloading .cif files from various online databases and using openbabel to convert the fractional coordinates that are in the .cif files into .xyz files.  The typical output will look something like this:

5

Spinel
\begin{array}{c|c c c}
\ce{Mg}      &   1.01054     &   1.01054   &     1.01054\\\hline
\ce{Al}      &   1.01054     & 1.01054     &   1.01054\\\hline
\ce{Al}      &   4.04217     &   4.04217   &     4.04217\\\hline
\ce{Mg}      &   4.04217     &   4.04217    &    4.04217\\\hline
\ce{O}       &   2.11390     &   2.11390    & 2.11390
\end{array}
My problem is that I would like the ionic radii of each element, instead of their name.  Does anyone know of a way to recover this information from the original CIF file or am I just stuck with looking up the radii manually on,
https://en.wikipedia.org/wiki/Ionic_radius ?

The CIF file won't contain information on atomic radii, only the positions (in fractional space).

Consider that the atom might be ionic, covalent, etc. so you'll have to choose the atomic radius definition you feel is most appropriate. (Or run some sort of quantum calculation and use the resulting [electron density to define an atomic radius].)

For what it's worth, Open Babel does have a data table of covalent and van der Waals atomic radii, but it sounds like you want something else.


What is the difference between nuclear charge and ionic charge? There was a question asking why magnesium metal has a higher melting point than sodium.

In the answer, they said:

  Magnesium has more delocalised electrons.
  Magnesium ions have a greater positive charge.
  Magnesium has a greater electrostatic attraction between ions and delocalised electrons.

However, in the mark scheme in the additional guidance, it also said the mark couldn't be given if you said:

  Magnesium ions have a great nuclear charge rather than greater positive charge.

I was confused as to what the difference is.

Nuclear charge is the charge present inside a nucleus, due to the protons. For example, a carbon atom has $+6e$ nuclear charge.

Ionic charge is the charge due to the gain or loss of electrons in the valence shell. So, the sodium ion has an ionic charge of $+1e$ (due to loss of one valence electron). That ion's nuclear charge is still $+11e$ (there are still 11 protons only).

The reason because nuclear charge as an answer to this question is unaccetable is because nuclear charge - being inside the nucleus - has largely no affect on the intermolecular forces between the magnesium atoms. Those magnesium atoms have intermolecular forces governed only by the valence electrons only.


Why is benzenediazonium fluoroborate water insoluble and stable at room temperature? Benzenediazonium fluoroborate is water insoluble and stable at room temperature.

Why is this salt, water insoluble? Also I am told that benzenediazonium salts are stable only at low temperature(<5°C) and decompose at higher temperature, so what is the reason for such exceptional stability of Benzenediazonium fluoroborate?

After a bit of research on benzenediazonium fluoroborate, I came up with an (complete?) answer although it is based on some approximations and/or speculations.

Benzenediazonium fluoborate is a versatile organic salt used to create various other organic compounds like by helping to introduce aryl, arylazo, arylhydrazono, or amino groups; forms fluoroarenes upon heating and are building block for heterocycles. They have a much higher stability than the corresponding chlorides and shock-insensitive. Generally decomposes at decomposes at 114-116 °C (Source).

So, what causes it much higher stability? Maybe it is their ion size. The benzenediazonium cation and fluoroborate anion are of same size resulting their stability[citation needed?]. Or maybe it is due to the resonance of benzenediazonium cation.

But when you apply heat to the salt, it is bound to decompose, maybe because of two reasons. One is that exceptionally stable nitrogen is formed on its decomposition (i.e nitrogen is formed from nitrogen cation) and second is the $\ce{N-F}$ bond is long and weak. From the information regarding structure of benzenediazonium fluorobotate, it is evident (Here):

  Benzenediazonium tetrafluoroborate, $\ce{C6H5N2+•BF4−}$, crystallizes
  in space group P21/a with unit cell dimensions a = 17.347(2),
  b = 8.396(1), c = 5.685(1) Å, β = 92.14(1)°, Z = 4. The structure was
  solved by direct phasing methods using the program SHELX 76. The
  parameters were refined by full-matrix least-squares to a final
  R = 0.063 for 1346 observed reflections. $\ce{C—N}$ and $\ce{N≡N}$
  bond lengths are 1.415(3) and 1.083(3) Å, respectively, and the bonds
  of the benzene ring vary from 1.371(5) to 1.383(4) Å. There are
  three $\ce{N-F}$ close contacts of ≤ 2.84 Å and the positive charge
  appears to be shared between the nitrogen atoms.

Below is the abstract of the mechanism of benzenediazonium tetrafluoroborate decomposition to give an insight (full pdf behind paywall is here):

  The thermolysis of benzenediazonium tetrafluoroborate was studied by
  thermogravimetry in dynamic mode. The decomposition of
  $\ce{[ArNN]+BF4−}$ in the solid state with the formation of
  $\ce{C6H5F, BF3, C6H6}$, and $\ce{N2}$ starts at T>348 K. The speed of
  the thermolysis was estimated gravimetrically and by infrared
  spectroscopy, considering the change of the intensity of the
  absorption band at 1498 cm−1, which corresponds to fluorobenzene. The
  maximal rate of thermolysis observes at the 366.5 K. A kinetic scheme,
  which includes the formation of a neutral complex
  $\ce{[C6H5δ+⋯BF4δ−]}$, is proposed for the thermolysis of
  arenediazonium tetrafluoroborate. The decomposition of the complex
  with the formation of free-radical intermediates explains the chain
  character of the thermolysis.

There is large hydrophobic part in diazonium ion and in $\ce{BF4-}$ ion, size is large enough that hardly any hydration energy would be released to cause the dissolution. Hence the salt is insoluble. (Here)


Assignment of the 13C NMR spectrum of 4'-bromoacetophenone I was doing a 13C NMR prediction of 4'-bromoacetophenone on chemdraw but I can't figure out why positions 5 and 5' appear further downfield than positions 4 and 4'.

I know both Br and the carbonyl are deactivating groups, with Br being ortho/para directing and carbonyl is meta directing. However, the mesomeric effect of both groups should "shield" position 5 but this is not the case. I'd appreciate an explanation that'd help me understand that.

Maybe this isn't the answer you are looking for, but $\ce{^13C}$ NMR shifts are often not amenable towards analysis via electron-withdrawing / electron-donating effects—in my experience this is especially so for aromatic rings. Any such analysis is quite unreliable and can lead to wrong conclusions if you try to use it for assigning peaks.
The numbers that ChemDraw predicts are based on empirical tables of chemical shift corrections. These are experimentally observed chemical shift changes that occur at each position of the benzene ring upon substitution. For a ring with multiple substituents you can approximately add these up. See e.g. Hans Reich's website for such tables.
See also long's answer to "Assigning the 13C NMR spectrum of bromobenzene and 4-bromobenzophenone": although most of the discussion is on the ipso carbon (C6 in your labelling), the general point is the same: it's often not trivial to rationalise or predict chemical shift changes in aromatic rings.
[There might be more sophisticated theoretical analyses out there based on diamagnetic and paramagnetic shielding arguments, but I'm unfortunately not aware of these.]


Expression for Gibb's free energy change for a cell reaction According to Chemistry Libretexts, the Gibbs free energy change of a cell reaction is related to the cell voltage as
$$ \Delta G = -nFE,$$
where $n$ is the number of moles of electrons passed and $F$ is the charge on a mole of electrons.
But $E$ decreases in value through the course of the cell reaction, so  shouldn't
$$\Delta G = -\int_{q_\mathrm{i}}^{q_\mathrm{f}} E\,\mathrm dq,$$
where $q_\mathrm{f} - q_\mathrm{i} = nF?$

According to Nernst's equation,
$$E = E^o - \frac{RT}{nF}\ln Q$$
As you can see, $E$ is independent of $q$, therefore your proposed integral becomes,
$$\Delta G = - \int_{q_i}^{q_f}{E\cdot dq} = -E\int_{q_i}^{q_f}{dq} = -E\Delta q = -nFE$$

Alternatively, you can prove the the relation between Gibbs energy and cell potential as,

Multiply Nernst's equation by $nF$, we get,

$$nFE = nFE^o - RT\ln Q$$

Swapping the sides, we get,

$$-nFE = -nFE^o + RT\ln Q$$

As we know,

$$\Delta G = \Delta G^o + RT\ln Q$$
$$\Rightarrow \Delta G = -nFE$$


Structure of phosphorus pentachloride Why is the shape of $\ce{PCl5}$ trigonal bipyramidal? All the $\ce{Cl}$ atoms must be at equal distances forming a star like shape. However, in the actual shape, the distance between the $\ce{Cl}$ atoms in the equatorial position is less than the ones in the axial position.

The short answer is stated on the Wikipedia page for trigonal bipyramidal:

  … there is no geometrical arrangement with five terminal atoms in equivalent positions.

This website from the School of Mathematics and Statistics at UNSW explains why distributing points on a sphere is tough and that perfect distribution is possible only in certain cases $(n=1,2,4,6,8,12,\&\ 20)$. These cases have been known for a long time — they correspond to the platonic solids (plus a point and a line segment). Even trigonal planar does not represent an even distribution on the sphere.

Trigonal bipyamidal then just happens to be the minimum energy conformation. The repulsive interactions between the chlorine atoms cannot be optimized but they can be minimized.


When making a calibration curve, do we have to use the wavelength at maximum absorbance? If you are making a calibration curve for $\ce{CuSO4}$ where you plot absorbance against concentration, does the wavelength have to be exactly at the maximum of the peak? Or can it be somewhere else along the curve of the peak as long as it is consistent among all the measurements and still get an accurate result?

No it doesn’t. Measurements can in principle be made at any wavelength.

If there are two colored specimens in the same solution, it is actually better to measure at the wavelength where the difference between the spectra is as high as possible.

Measuring at a peak has some advantages: the absorbance is high, giving a good signal/noise ratio, and if the wavelength varies a little it has less effect on the absorbance when measured at a peak than at a slope of the curve.


How can I calculate the amount of HF in aquoues solution of NH4F and an aquoues solution of NH4HF2? Dissociation of $\ce{NH4F}$ and $\ce{NH4HF2}$ in water is complete, I assume:
$$\ce{NH4F -> NH4+ + F-}$$
and
$$\ce{NH4HF2 -> NH4+ + HF2-}$$

Can I calculate the amount of $\ce{HF}$ in these solutions from this equilibrium?

$$\ce{HF + F- <=> HF2-}$$

$$\mathrm{K_2}=\dfrac{\ce{[HF^{2-}]}}{\ce{[HF][F^-]}}=4,7$$

If I know the concentrations of NH4F in water and NH4HF2 in water.

WRONG SOLUTION: OP clarified that there are two separate solutions, not a mixture of the salts.

Assuming:

(1) that concentrations can be used instead of activities (bad assumption...)
(2) The concentration of the ions $\ce{H2F-}$ and $\ce{F-}$ is so large that the final concentrations will be the same as the initial concentrations.
(3) Since HF is a relatively weak acid, little $\ce{HF}$ will dissociate into $\ce{H+}$ and $\ce{F-}$.

Using equilibrium:
$$\mathrm{K_2}=\dfrac{\ce{[HF^{2-}]}}{\ce{[HF][F^-]}}=4.7$$

From 5 molar $\ce{NH4F}$, $\ce{[F-] = \pu{5 molar}}$

From 0.18 molar $\ce{NH4HF2}$, $\ce{HF_2^- = \pu{0.18 molar}}$

So plugging the number in:

$$\mathrm{K_2}=\dfrac{\ce{[HF^{2-}]}}{\ce{[HF][F^-]}} = \dfrac{0.18}{\ce{[HF]}\times 5}=4.7$$

$$\ce{[HF] = 0.008 molar}$$

For the final concentrations $\ce{[HF] \ll [H2F-]}$, so the assumption that the dissociation of $\ce{H2F-}$ can be ignored is valid. The $\ce{HF}$ will dissociate and acidify the solution, but still $\ce{[HF] + [H+] \ll [H2F-]}$


Reaction between an acid and acid My initial thought was that the stronger acid will act as an acid and the weaker acid as base like the reaction between $\ce{H2SO4}$ and $\ce{HNO3}$ but in this reaction there is a lone pair available for protonation, what if the lone pair was not present , will the reaction still occur? Is the answer dependent on what the acids exactly are?

In the case of $\ce{HNO3}$ and $\ce{H2SO4}$, the proton from sulphuric acid protonates the $\ce{-OH}$ group of $\ce{HNO3}$ , which gives $\ce{NO2+}$ , $\ce{HSO4-}$ and water.
This mixture of nitric acid and sulphuric acid is widely used as a nitrating mixture for nitration of aromatic compounds.

In case of other acid mixtures, no reaction happens, but there is an equilibrium. The stronger acids dissociates almost completely , decreasing the dissociation of the weaker one by common ion effect.


Is cyclopropyl methyl carbanion also exceptionally stable? I know that cyclopropyl methyl carbocation is exceptionally stable and in fact more stable than benzyl carbocation but I was wondering if it was the same case with cyclopropyl methyl carbanion. Is it also more stable than benzyl carbanion?

Sigma resonance only works on cyclopropyl methyl carbocation (and its cyclopropyl derivatives). So I don't think that cyclopropyl methyl carbanion will be exceptionally stable.


Supersaturation and Crystallization I had a fairly simple question regarding supersaturation and crystallization. When a supersaturated solution is crystallized, is all of the solute crystallized or just the excess solute (such that the solution remains saturated)? After crystallization, would you expect to find some quantity of the solute still dissolved?

Think of it this way: A saturated solution is the equilibrium. Say you had a supersaturated solution, formed a crystal out of it and somehow managed to get your solution below saturation.

But once that happened, your solution is no longer saturated and there is undissolved material on the ground (your crystal). So, always striving towards equilibrium, some of your crystal would get redissolved to re-reach the saturation.

In fact, crystallisation and equilibrium are not static but rather dynamic processes. So even if there is a crystal sitting at the bottom and it looks like nothing is happening, there is a constant flux of compounds redissolving from the crystal surface or recrystallising onto the surface out of the solution. Which is also why everything seems to strive towards the equilibrium: It is the state, where the speed of one direction matches that of the other direction.

So short answer: The solution above a crystal will be saturated, but no longer supersaturated.


Is cyclohexene an alicyclic or aromatic compound? Classification of organic compounds have open and closed chain compounds. Closed chain further have homocyclic and heterocyclic componds. In homocyclic compounds there are two types: alicyclic and aromatic compounds.

In alicyclic compounds, only singles bond are allowed, whereas in the aromatic category the ring of carbon atoms has alternate single and double bonds (i.e. conjugated double bonds). So what happens to the compounds which have a lone double or triple bond?

For example, cyclohexane is classified under alicyclic, but where would cyclohexene or cyclohexyne be classified?

Setting aside the severe steric  strain that cyclohexyne would require, an aromatic all-carbon ring must have a full cycle of unsaturated atoms.  Cyclohexene and (if you can make it) cyclohexyne have saturated carbons in the ring and thus must be alicyclic instead of aromatic.


How is electron-impact mass spectrometry data formally reported? NMR data has the following format when reported in a paper:

How would I go about reporting my EIMS data formally? I tried looking for documents online but couldn't find anything.

Third edition (latest offline edition) of the ACS Style Guide (currently known as The ACS Guide to Scholarly Communication) has a section devoted to reporting analytical data, including mass-spectrometry and EIMS in particular [1, p. 275]:

Mass Spectrometry
MS $m/z$ (relative intensity): $238.2058~(44.8\%),$ $195.1487~(100\%),$ $153.1034 ~(21.2\%).$
GC–MS $m/z$ (% relative intensity, ion): $202~(9,~\ce{M} + 4),$ $200~(32,~\ce{M} + 2),$ $198~(23,~\ce{M+}),$ $142~(35,~\ce{M} - 2\ce{CO}),$ $321~(95,~\ce{M} - \ce{Me}),$ $415~(\ce{M+} - \ce{Cl}).$
HRMS–FAB $(m/z):$ $[\ce{M} + \ce{H}]^+$ calcd for $\ce{C21H38N4O6S},$ $475.259;$ found, $475.256.$
EIMS $(\pu{70 eV})$ $m/z:$ $\ce{M+}~420~(15),$ $241~(15),$ $201~(59),$ $135~(14),$ $69~(23).$
Abbreviations: $m/z$ is the mass-to-charge ratio, $\ce{M}$ is the molecular weight of the molecule itself, $\ce{M+}$ is the molecular ion, HRMS is high-resolution mass spec-
trometry, FAB is fast atom bombardment, and EIMS is electron-impact mass spectrometry.

The recommendations for ACS journals also align with the general recommendations given in RSC Experimental data policy — Guide to the presentation of experimental data (accessed 2021-04-22):

Mass spectrometry data
Given in the form: $m/z~183~(\ce{M+},~41\%),$ $168~(38),$ $154~(9),$ $138~(31)$ etc. The molecular ion may be specified as shown if desired. Relative intensities in parentheses (% only included once). Other assignments may be included in the form $m/z~152~(33,~\ce{M} − \ce{CH3CONH2}).$ Metastable peaks may be listed as: $\ce{M^*}~160~(189\to 174),$ $147~(176\to 161),$ etc. The type of spectrum (field desorption, electron impact, etc) should be indicated. Exact masses quoted for identification purposes should be accurate to within $\pu{5 ppm}$ (EI and CI) or $\pu{10 ppm}$ (FAB or LSIMS).

If you are a $\mathrm\LaTeX$ user, check out chemmacros package (TeX.SE) and its spectroscopy module (PDF) which assists with the input of various experimental data.
Reference

The ACS Style Guide: Effective Communication of Scientific Information, 3rd ed.; Coghill, A. M., Garson, L. R., Eds.; American Chemical Society; Oxford University Press: Washington, DC; Oxford; New York, 2006. ISBN 978-0-8412-3999-9.


Anodisation of aluminium in sulfuric acid - Electrolysis reactions As an experiment for electrolysis and to show how a metal can be protected of corrosion, I am trying to anodize an aluminium plate, using sulfuric acid as electrolyte.

I know the aluminium plate is the cathode and, thus, connected to the positive terminal of the generator, while for the cathode I am using aluminium foil.

Supposing I have $H_{2}SO_{4}:80ml, 3M$. I have two questions:

$1.$ Knowing:
$$\epsilon^{o}_{Al^{3+}/Al} = -1.67V\\
  \epsilon^{o}_{O_{2}/H_{2}O} = 1.23V, pH = 0\\
  \epsilon^{o}_{2H^{+}/H_{2}} = 0V, pH = 0\\
  \epsilon^{o}_{2H_{2}O/H_{2}} = -0.83V, pH = 14$$
I want an explicit explanation using those potentials as to why the reaction at the cathode is $$2H^{+} + 2e^{-}\rightarrow H_{2}$$
and why the reactions at the anode are ( If i am not mistaken )
$$Al \rightarrow Al^{3+} + 3e^{-}\\
  2Al^{3+} + 3O^{2-}\rightarrow Al_{2}O_{3}\\
  2Al^{3+} + 3OH^{-} \rightarrow Al_{2}O_{3} + 3H^{+}$$
  Especially the second and third reactions confuse me

$2.$ Knowing:
$$\epsilon^{o}_{Cu^{2+}/Cu} = 0.34V\\
\epsilon^{o}_{Pb^{2+}/Pb} = -0.13V\\$$
Could I use a plomb or copper blade instead of the aluminium foil?

As the experiment is described, in the solution we have :
$$H^{+}, H_{2}O, Al_{(s)}, SO_{4}^{2-}, OH^{-}$$
Taking that into account and seeing how the environment is acidic ( $pH<7$ ), we can see that the reaction of reduction reaction that can take place is $H^{+}/H_{2}$
Following the same kind of reasoning, reaction $(1)$ in the anode is caused by the odydation of the aluminium plate when placed in an acidic environment, $O^{-2}$ is present because of the oxygen in the atmosphere and $OH^{-}$ exists in the solution anyway(water autoionization).

As far as the second question is concerned, notice that the aluminium foil in the anode does not participate in any reaction. It is the $H^{+}$ that is reduced, while the aluminium foil serves as a conductor for the current. Knowing that, we could replace the aluminium foil with either $Cu_{(s)}$ or $Pb_{(s)}$, that may also be used as conductors ($Cu^{2+}, Pb^{2+}$ will not be present in the solution and we will have the same reduction reaction in the anode), no matter their standard electrode potentials.

Finally, it is interesting to point out that in this experiment the anode and cathode are in the same solution which is used at the electrolytic reactions (acidic environment ) and at the same time plays the role of the electrolytic bridge constantly providing $H^{+}$ and keeping the solution electrically neutral.


Can we interpret the "extent of reaction" as the number of reactions that happened? Consider this reaction:
$$\ce{\alpha\ A + \beta\ B -> \omega\ C + \delta\ D}$$

Where $\ce A$ and $\ce B$ are the reactants, $\ce C$ and $\ce D$ are the products, and $ \alpha, \beta, \omega, \delta$ their respective stoichiometric constants

We can then define the extent of reaction at instant $t$ as: $\xi (t)$.

My question is: can we interpret the extent of reaction as the number of reactions that happened at an instant $t$ considering that say, for reactant $\ce A$ at instant $t$: $$n_{t}(\ce A) = n_{i}(\ce A) - \alpha\times \xi(t)$$

If I understand this correctly you want to measure $\ce A$ to infer the extent of reaction. I think you should have started with mathmatically defining the extend of reaction:

$$\xi(t) = \left[\frac{n_i(\ce A) -n_t(\ce A)}{n_i(\ce A)}\right]\tag 1$$

Where $n_t$ is the concentration at a given time $t$. Here $n_i(\ce A) -n_t(\ce A)$ gives the consumed concentration of $\ce A$ after time $t$ which is normalized by the initial concentration to yield  an extent of conversion.

  Assume $\mathrm{2\ a}$: If the stoichometric constants are normalized $({\alpha} + {\beta}= 1)$ to be taken as the initial concentrations then we can substitute: $\alpha = n_i(\ce A)$ for example $$\ce{0.33A + 0.67B -> 0.5C + 0.5D}$$
  $$\xi(t) = \left[\frac{\alpha -n_t(\ce A)}{\alpha}\right]\tag {2 a}$$
  otherwise:
  $$\left(\frac{\alpha}{\alpha + \beta} + \frac{\beta}{\alpha + \beta} = 1\right) \implies \frac{\alpha}{\alpha + \beta} = n_i(\ce A)\tag {2 b}$$

Which we can rearrange to :

$$\xi(t) = \left[\frac{\alpha -n_t(\ce A)}{\alpha}\right] \implies \alpha-n_t(\ce A) = \alpha\ \xi(t) \tag{3} $$

Subtracting $\alpha$ and then negating gives:

$$n_t(\ce A) = \alpha - \alpha\ \xi(t) = n_i(\ce A) - \alpha\ \xi(t) \tag{4 a}$$
Which would make you correct, or in the case of assumption $\mathrm{2\ b}$:
$$n_t(\ce A) = n_i(\ce A) - \frac{\alpha}{\alpha + \beta}\ \xi(t) \tag{4 b}$$
Which would make you almost correct.


Bonding in Zeise's salt I am not sure I completely understand the bonding in coordination compounds formed by donation of pi electron cloud by the ligands to the central atom.

For example, consider Ziese's salt:

I came across a question which asked the total number of sigma bonds in the salt. I expected the bond between ethene and platinum to be a sigma bond too because it appears as if it is formed by overlap of the orbitals along the axis and hence my answer was 9. But the answer given was 8 and the answer key didn't consider that bond as a sigma bond.

So is the bond between platinum and ethene a $\pi$ bond? If yes, then how?

What orbitals are actually involved in that bond?

The bond between $\ce {Pt} $ and ethene is not purely a $\sigma $ bond and also not a pure $\pi $ bond. The bonding  actually consists of a combination of both bonding and back-bonding effect.

If you look into the Molecular Orbital Diagram of Ethene, the picture will be more clear. For the formation of bond between the metal and the ligand, electrons from filled $\pi $-bonding MO of ethene ($\pi_b $) is donated to the vacant orbitals of Platinum. And electrons from the filled orbitals of $\ce {Pt} $ is back-donated to the vacant $\pi^*$ orbitals of ethene. Thus the bonding would have been a $\sigma $ bond if the electrons are donated from a $\sigma_b $  or $\sigma_{nb} $ orbitals. But here the bonding mainly is $\pi $ donation and $\pi$ acceptance ( known as Synergic effect).

Thus, it would be better not to consider the bond as $\sigma $ bond, but may be considered more of a $\pi $ type but also not a pure one due to the back-bonding.


Is there any database, or reference, for retardation factors (Rf)? I'll like to compare the Rf of myrcene, obtained in Thin Layer Chromathography, for different species.  Toluene:Ethyl Acetate (97:3) would be fine, but is not a necessity.

No there is not database for retardation factors, because Rf values are useless outside your lab. It is not a fundamental property like melting point or boiling point. One batch of silica is slightly different from another. It also depends on exact solvent composition and history of the treatment of the stationary phase.


opening valve separating two volatile solutions Let there be two containers A and B containing two volatile solutions (ideal) of  pure liquids P and Q  respectively. The containers are connected by a tube such that only vapours can travel through it but it is initially closed by a valve. Now the valve is opened at $t=0$.
What will be the final equilibrium pressure over the containers?

My  Approach:
I think that final pressure will be somewhere between the pressure of pure liquid P and pure liquid Q, might their average but I can't say for sure.
vapour pressure is a function of temperature so might be there is no change ,if the liquids would have been mixed inn a single container , forming ideal solution with specified mole fraction of each component then I know how to use Raoult' s law but here I don't see the situation being an equivalent one.

There is not enough information to decide, and the question is not clearly worded.
As to the wording, "two volatile solutions (ideal) of pure liquids P and Q respectively," implies there is no vapor present, yet "vapours can travel through". What is the volume of each volatile liquid, and the total volume of the containers, including the connecting tubes? If the whole apparatus has no place for vapor, then the pressure is determined by uncompressible liquids.
As to insufficient information, do the two chemicals react? For example, if the liquids are $\ce{HCl(aq)}$ and $\ce{NH3(aq)}$, the resulting $\ce{NH4Cl(aq)}$ would have a far lower vapor pressure than either of the initial reactants.


The amount of citric acid for golden syrup I want to make a good golden syrup at home. I've tried it few times, but all of them were bad. I think the problem was the amount of citric acid. I want to know how to calculate the amount of citric acid which is needed to make golden syrup from 1 kg of table sugar and 500 ml of water. And also: how much time does take all of the sucrose molecules to break down into glucose and fructose?

Golden syrup is a type of inverted sugar syrup, like maple syrup, treacle and molasses.

It is made by splitting the disaccharide sucrose (sugar) into a mixture of glucose and fructose components, typically by carefully controlled heating, sometimes with a small amount of citric or potassium bitartrate (cream of tartar).

For example, if you mix 1 kg of fine granular sugar with 2 cups (500 mL) water and 1 g of citric acid, bring to the boil and simmer for about 30 mins, you make inverted sugar syrup. If you simmer this for another 2 - 2.5 hrs or so until it caramelises, you end up with 'golden syrup'.


Can I use an acid other than sulfuric for hydrogen peroxide determination? I would like to perform a permanganometric titration of hydrogen peroxide at home following a procedure similar to http://www.titrations.info/permanganate-titration-hydrogen-peroxide, i.e. the reaction:

2MnO4- + 5H2O2 + 6H+ → 2Mn2+ + 5O2 + 8H2O

I do not have sulfuric acid or a licence. Can I use any other acids? I think HCl would release chlorine, and other than that I only have weak household acids such as acetic, ascorbic, citric and tartaric to hand. I can probably get hold of others, but sulfuric is regulated as an explosives precursor here.

Don't use the organics!

This will severely mess up your titration because both permanganates as well as H2O2 will oxidize the carboxylic acids into explosive or corrosive things like peroxyacids (-C(O)OOH). Carboxylic acids are also usually weak, and the peroxy versions can be up to 1000 times weaker.

On the bright side, HCl is a strong acid, and you shouldn't have to worry about the production of chlorine gas; the only chlorine product should be MnCl2. However, since so much O2 Is being produced, your MnCl2 might contain a small amount chlorate impurities, so be wary of that if you plan to use the manganese chloride in the future.


Conductance of the mixture of two solutions with given resistance values If $R_1$ and $R_2$ are the resistances of two solutions of equal volume what is the conductance of the mixture in the same conductivity cell?

In general, I grasped neither the approach, nor the meaning very well.

Do we add the resistances as now both the solution are there to resist, or do we add the conductance or conductivity of them as now both conduct?

As now the volume is doubled, do I divide by 2 or multiply? The area is doubled, but the ions per unit volumes is reduced to half (or maybe not, I'm not sure).

Also, how can we directly add and ignore the inter-particular interaction?

The answer given is

  $$\frac{1}{2}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$

There is an excellent treatment of electrolytic conductivity in Wikipedia. However, the OP’s main question involved a specific equation, pertaining to the conductance of a mixture of equal volumes of two electrolyte solutions, and this was not specifically discussed in the Wikipedia article:

$$\text{conductance} = \frac{1}{2}\left(\frac{1}{R_{1}} + \frac{1}{R_{2}}\right)$$

For simplicity, assume the electrolyte concentrations in the two solutions are in the “Goldilocks” zone: not so high that activities and fancy theory are needed, but not so low that the conductivity contribution due to water’s autoionization has to be taken into account. Neither solution is concentrated, so the ions just bounce around like they would in the unmixed solutions. This would certainly be incorrect in relatively concentrated solutions, but that is not the case at hand.

In this idealized scenario, a solution’s conductivity, i.e., specific conductance, is directly proportional to the mobile charge carrier (i.e., ion) concentrations in the solution. The conductance of a solution, denoted by $G,$ is directly proportional to the solution’s conductivity. The resistance of the solution, denoted by $R,$ is simply the reciprocal of the conductance and, likewise, the resistivity, i.e., specific resistance, is the reciprocal of the conductivity.

Now consider two electrolyte solutions that can be mixed without causing any chemical reaction. In particular, no precipitate forms, no gas is evolved, no complexes form, and so on. Measurements on the first solution provide values of $G_1$ and $R_1$, while measurements on the second solution provide values of $G_2$ and $R_2$. If equal volumes of the two solutions were well mixed, and a conductivity measurement was performed on the mixed solution, the conductivity of the mixed solution, $G_\mathrm{mix}$, would be

$$G_\mathrm{mix} = \frac{G_1}{2} + \frac{G_2}{2} = \frac{1}{2}(G_1 + G_2)$$

since each solution, when mixed with an equal volume of the other solution, had its constituent ion initial concentrations halved. But the total conductance is just the sum of the two conductances, $G_1 = 1/R_1$, and $G_2 = 1/R_2$, so

$$G_\mathrm{mix} = G_\mathrm{tot} = \frac{G_1}{2} + \frac{G_2}{2} = \frac{1}{2}\left(\frac{1}{R_1} + \frac{1}{R_2}\right)$$

Like all approximations, this one is only as good as the foundational assumptions upon which it is predicated.


Resonance of carbonyl Carbonyl has two resonance structures, where the $\pi$-bond between C and O breaks and the electrons "move" to the oxygen atom. This gives an empty p-orbital in carbon and a p-orbital in oxygen with two electrons, leading to a positive charge on carbon, and a negative charge on oxygen. To my understanding, oxygen gets the lone electron pair because it has a higher electronegativity than oxygen.

My question: is it possible that the two electrons "move" to carbon instead of oxygen, leading to a third resonance structure (I know that these resonance structures don't really exist, and that the real structure is somewhere in between), so that carbon has a lone electron pair? It would probably contribute little to the resonance energy compared to the other structures. Or is it not possible at all?

Yes, resonance should never stop at the most likely resonance structure. If you’re really willing to play the game, you can take it even further and consider singlet diradical structures where the bond has been broken into a spin up and a spin down electron, each one of which resides in a different orbital. And more, and more.

(Side note: this is why it is useless to ‘count’ resonance structures. The only correct answer to ‘How many resonance structures does $\ce{X}$ have?’ is ‘Always one more.’.)


Why is 2-(3-aminomethyl-2-hydroxypropyl) cyclohexan-1-one named so and not 2-(2-hydroxy-4-aminobutyl) cyclohexan-1-one? What is wrong with naming it as 2-(2-hydroxy-4-aminobutyl) cyclohexan-1-one and taking the side chain to be having 4 carbon atoms with NH as substituent on the last? Would someone please explain this? This is the exact picture given in my book labelled 2-(3-aminomethyl-2-hydroxypropyl) cyclohexan-1-one.

As drawn the compound does not contain a butyl side chain with four carbon atoms; the side chain is only a propyl chain with three carbon atoms.

The propyl side chain has two substituents: a hydroxyl ($\ce{-OH}$) group and an methylamino ($\ce{-NH-CH3}$) group. Since these groups are not the principal characteristic group in this case, they have to be expressed as prefixes. Note that the correct prefix for the secondary amine is “methylamino” and not “aminomethyl” as proposed in the question.

Therefore, the correct name is 2-[2-hydroxy-3-(methylamino)propyl]cyclohexan-1-one.


How does the inductive effect, explain the reactivity of tertiary alcohols with halogen acids? The reactions of primary and secondary alcohols with halogen acids ($\ce{HX}$, where $\ce{X}$=halogen) require the presence of a catalyst, $\ce{ZnCl_2}$. With tertiary alcohols, the reaction is conducted by simply shaking with concentrated $\ce{HCl}$ at room temperature. Which mean that, tertiary alcohols are more reactive than primary and secondary alcohols. The reason is said to be, due to the inductive effect.

I don't know how the inductive effect would explain that, tertiary alcohol is more reactive than primary and secondary.

Secondary and tertiary alcohols react with strong acid (and bases) such $\ce{HX}$ following SN1 mechanism.

$$\ce{ROH <=> HX + ROH_2^+ + X^-} \tag{1}$$
$$\ce{ROH_2^+ <=> R^+ +H_2O} \tag{2} $$
$$\ce{R+ +X- \rightarrow  RX} \tag{3} $$

In these reactions, the rate determining step is the formation of the carbocation (2). So the successive attack of the nucleophilic chlorides is dependent on that.

Here is how the inductive effect comes into play. The inductive effect of $\ce{R}$ groups of the alcohols stabilize the carbocation (figure show the stability of carbocations from the more stable to the less) and then increase the chances of a nucleophilic attack!


Is N-methylimidazole an unusual solvent, or are there more common substitutes for it? I want to test a process that was documented in the literature using N-methylimidazole.

I am not doing academic research, but rather testing a documented reaction for application to an industrial process, so I don't believe I need lab-grade chemicals.

When I look for this solvent online I can only find lab suppliers (e.g., Matrix Scientific, Acros Organics, Alfa Aeser) selling tiny quantities at very high prices.

Is there something unique to this chemical as a "donor solvent" that is not reproduced by any more common or inexpensive solvent?  I have seen it mentioned frequently in organic chemistry papers, but since I can't find industrial suppliers I assume it's peculiar to research.

The application I'm interested in is chemical stripping of lead using Rauchfuss's Elixir, which consists of sulfur as an oxidant and a donor solvent.  I haven't been able to learn why they used methylimidazole as the solvent in the study.  Obviously I would prefer one more widely and/or cheaply available.

http://www.chemicalbook.com/ChemicalProductProperty_EN_CB1316726.htm
http://www.lookchem.com/cas-616/616-47-7.html
http://www.guidechem.com/cas-616/616-47-7.html  2000 RMB = $(USD)320
http://ahjinao.en.alibaba.com/product/595605676-209523668/Time_honored_Supplier_of_N_Methylimidazole_CAS_No_616_47_7_.html
http://www.alibaba.com/countrysearch/CN/n--methylimidazole-supplier.html
http://www.buyersguidechem.com/chemical_supplier/N-Methylimidazole.php

Info and suppliers,  Note that a Chinese contract is often not the legal or philosophical equivalent of a Western contract.


Configuration of chiral center in cocaine I couldn't get the absolute configuration for the chiral center I marked. I got R, but according to my answer key it is apparently S. Would really appreciate an explanation.

You are actually right, that stereocenter does have R stereochemistry. The four substituents to the chiral center in question are:

1 - $\ce{CO(O)}$ = $\ce{C(O)3}$
2 - $\ce{C(O)(C)(H)}$
3 - $\ce{C(N)(C)(H)}$
4 - $\ce{H}$

These are ranked and numbered according to the Cahn-Ingold-Prelog priority rules. Adding the numbering to the structure:

Orienting the lowest priority group away from us and following the direction of the numbering, we confirm that it does indeed have R stereochemistry:


Can our chemical elements be differents on other universes? Let's pretend that in other universe we have an element with atomic number 6, like our carbon, but in this other universe the element is completely different from our carbon, like a new element with other properties.

Is this a possible scenario?

I admit this question is far out of my depth, but it's interesting enough that I feel I should give it a shot.

TL;DR - The question is so hard as to be almost unanswerable in detail. Things can be very different, though not arbitrarily different, and the exact line between the two is very difficult to define. There have been some mild investigations on the chemistry of toy universes, however, one of which I describe below.

The problem with thinking about other universes is that there are potentially a lot of different universes which can arise. Famously, string theory currently posits that there are something like $10^{500}$ "valid" universes, each with different details regarding physical laws and physical constants. Furthermore, universes are very complex objects in the sense that there is a ton of interrelation between everything, so even minor changes to details can create wildly different results. Not just that, but there apparently are many parameters to fiddle with, which would create wildly different periodic tables, often apparently breaking any possibility of life. This all leads to fine-tuning problems.

For example, if the attraction between two protons were about 2% larger, then every proton in the Universe would be bound in the form of a diproton nucleus, i.e. the nucleus of helium-2 ($\ce{^2He}$, which is just barely unstable in our Universe), meaning the entire periodic table would collapse to a single element. Apparently there's no very interesting chemistry in that universe! Now, what exactly would have to be tweaked in order to get a 2% stronger attraction between protons while not breaking everything else, I don't know.

So it seems thinking about chemistry in different universes is far too complicated to approach in almost any situation, but still there are some investigations. One particular case I have come across is what would happen in a universe without weak interactions.

Most of Chemistry ultimately has to do with studying different aspects of electromagnetic interactions. However, the stability of atomic nuclei depends immensely on strong interactions (aka strong nuclear force), and to a lesser extent on weak interactions (aka weak nuclear force). The authors in the article linked above analyzed the effects of taking a universe similar to ours, but removing the weak interaction. There are several differences with respect to the evolution of the universe (big bang nucleosynthesis, stellar nucleosynthesis, etc) and the abundance of the elements, but thinking just about their stability, it seems that in such a universe, a large portion of the periodic table would still exist.

Additionally, given that the weak interaction mediates the decay of neutrons in our Universe, its absence means neutrons are completely stable subatomic particles (in our Universe free neutrons decay with a half-life of around 8 minutes). This also means that neutron-rich nuclei are more stable to decay in the hypothetical weakless universe. In our Universe, deuterium ($\ce{^2_1H}$)is the most neutron-rich stable hydrogen isotope, while tritium ($\ce{^3_1H}$) is stable enough, with a half-life of about 11 years, to be chemically important (heavier hydrogen isotopes can essentially be neglected). However, in the weakless universe, it seems that even hydrogen nuclei as rich as $\ce{^6_1H}$ are indefinitely stable, and there are possibly even more neutron-rich isotopes. See the following figure:

(Source)

For comparison purposes, the most neutron-rich stable isotopes for each element in our Universe are $\ce{^2_1H}$, $\ce{^4_2He}$, $\ce{^7_3Li}$, $\ce{^9_4Be}$, $\ce{^11_5B}$, $\ce{^13_6C}$, $\ce{^15_7N}$ and $\ce{^18_8O}$.

Due to this added nuclear stability, the periodic table in the hypothetical weakless universe could actually have more stable elements in it. In our Universe, elements beyond about the 110th are quite unstable, being produced naturally in small amounts, if any, and they will not likely have great significance in Chemistry. However, highly neutron-rich nuclei could make ultraheavy elements more accessible, even in natural conditions such as supernovae.


Why is the CASSCF method multi-configurational, while the CI method is not? The CASSCF method is perhaps the most commonly used theoretical method for studying difficult chemical systems exhibiting multi-reference character or non-dynamical/static/strong correlation. CASSCF is quite similar to CI, in the sense that a full CI wavefunction is generated within the active space used in the CASSCF method.

However, the CI method is a so-called ground state method, while the CASSCF method is multi-configurational. My question is simply "Why?". I feel it may be related to that during the CASSCF optimization, both the orbitals and CSF expansion coefficients are optimized, while in CI only the expansion coefficients are optimized. But I fail to understand how optimizing the orbitals gives the CASSCF method its multi-configurational nature.

When you say multiconfigurational I think you mean that the ground state wavefunction does not have one single dominant MO configuration.

CI can be multiconfigurational depending on the type of excitation included.  For example, Full CI is of course multiconfigurational since it is the exact solution for the basis set used.  However, CIS (single excitations) or CISD (single + double excitations) calculations are the only two CI methods that are computationally efficient enough to be used in practice.  So that’s probably what you mean when you say CI. You also probably mean CIS or CISD based on an RHF wavefunction, to distinguish it from multi-reference CI.

The lowest energy solution of CIS based on RHF is the RHF energy, so you can think of the SCF procedure as a form of CIS.

So a CASSCF(2,2) calculation (e.g. where the active space is HOMO, LUMO) is a bit like a CISDT calculation.  For example, there will be contributions to the wavefunction where the LUMO is doubly occupied and another MO outside the active space is “singly excited”.  Thus, the LUMO is allowed to change when electrons are put into it, which becomes more and more important as the HOMO-LUMO gap decreases.  CISD cannot account for this, so in that sense it is not multiconfigurational.


Difference between Sodium Bicarbonate and Sodium Carbonate reacting with HCl In lab, we performed the Solvay process where we had to produce sodium carbonate as the final product. There is a certain step where we put something in the oven and after this, we expect that the product is sodium bicarbonate ($\ce{NaHCO3}$). This product does not get named in the instructions; it is simply referred to as "dry product". After this step, we were asked to add 2 drops of $\ce{HCl}$ to a small amount of the dry product in a watch glass. Finally, we should repeat the watch glass step using $\ce{NaHCO3}$ and $\ce{Na2CO3}$.

After doing these 3 experiments, I observed the same thing (bubbles). I think that the reactions were:

$$\ce{NaHCO3 + HCl -> NaCl +  CO2 + H2O}$$

and for the third one (since I added $\ce{HCl}$ drop-wise to $\ce{Na2CO3}$, $\ce{Na2CO3}$ was in excess):

$$\ce{Na2CO3 + HCl -> NaHCO3 + NaCl}$$

and then:

$$\ce{Na2CO3 + HCl -> NaCl + CO2 + H2O}$$

I believe the purpose of the 3 experiments was to identify that the dry product was $\ce{NaHCO3}$, but after adding $\ce{HCl}$, I observed exactly the same thing (bubbles) for both known reagents. So I'm confused as to what should be my answer.

It seems to me that the purpose of the 3 experiments with the HCl was not to identify the dry product. You can just write the reactions you observed, that was probably the point.