Why enthalpy is defined at a constant pressure? How absolute enthalpy and enthalpy change are defined? What is the clear difference between enthalpy and heat? Is it the constancy of pressure what makes the enthalpy a state function?

My teacher told me in very simple words that when heat goes inside a system, it is called enthalpy. Is this correct? Would constant pressure make the volume constant in a thermodynamic system?

Answers to my confusing questions would be highly appreciated.

How are absolute enthalpy and enthalpy change defined?

Enthalpy $H$ is defined as in your question. The most basic definition of enthalpy change when going from state 1 to state 2 is the enthalpy of state 2 minus the enthalpy of state 1:
$$\Delta H = H_2 - H_1$$

  What is the clear difference between enthalpy and heat?

Enthalpy is a state function and heat is not. Heat is energy transferred from a hotter system to a cooler system (not to be confused with thermal energy, that part of internal energy due to the kinetic energy of particles randomly moving, i.e. what makes a sample have a certain temperature).

A simple experiment to experience the difference: If you rub your hands together, they warm up. If you wait a while until they are back at the original temperature, they will be in the same state (no change in enthalpy). At the same time, there was a heat exchange with the environment, increasing its temperature a bit.

In essence, some mechanical energy was converted into thermal energy. There was some heat transfer $q > 0$, but the enthalpy did not change ($\Delta H = 0$) in this thermodynamic cycle. So enthalpy change and heat are different concepts, and there are cases where their values are different.

  Is it the constancy of pressure what makes the enthalpy a state function?

No, it is because of how enthalpy is defined. The internal energy is a function of state, and P and V are a description of the state. This means that by the definition of enthalpy, the same state will have the same enthalpy. (If you want to know why internal energy is a function of state: it is related to the first law of thermodynamics, but more fundamentally - it is what all measurements so far have shown).

There is a reason why looking at enthalpy instead of internal energy is often simpler, see below, related to how much of what we experience is at constant pressure (usually atmospheric pressure).

  My teacher told me in very simple words that when heat goes inside a system, it is called enthalpy. Is this correct?

This is a shortcut for the following: If you run a process in a way that the only forms of energy transferred between the system and the surroundings are heat and pV-work, the change in enthalpy will be equal to the heat transferred. (In my example above, there was mechanical work; that would break the equality of $\Delta H$ and $q$, as would electrical work done by or on the system, or electromagnetic radiation going in or out.) Also, the system has to be closed (no matter entering or leaving the system).

pV-work is a term for the work of expanding the system's volume by $\Delta V$ into the surrounding. When this happens, you work against the pressure of the surrounding. Enthalpy is defined in a clever way so that it does not change due to pV-work at constant pressure:

$$ H = U + PV$$
$$ $$
$$ \Delta H = H_2 - H_1$$
$$ = U_2 + PV_2 - (U_1 + PV_1)$$
$$ = \Delta U + P\Delta V$$

It turns out the change in internal energy due to pV-work (at constant pressure) is $-P\Delta V$ (it takes work to push against the surrounding, done by the system, so the system loses energy), which cancels out against the $P\Delta V$ term in the calculation above. Thus there is no change in enthalpy due to pV-work at constant pressure.

  Would constant pressure make the volume constant in a thermodynamic system?

No. Take an ideal gas (in a closed system) whose temperature rises for some reason. If you keep the pressure constant (by having the gas inside a floppy mylar baloon), the volume will increase. Enthalpy is a useful state function for systems at constant pressure because it does not change just because the volume changes.

  Why is this answer so long and complicated (I added this question)?

At my college, enthalpy is introduced as a "fact" in our first chemistry course. A semester later, we learn about first and second law of thermodynamics, and we discuss the tools of thermodynamics a year or two later. So it is not taught from first principles, and you have to learn some stuff without a full picture of the why and how. What I summarized here would be taught in three different courses at my college (along with other material, of course). If you have not been introduced to all of these topics yet, this answer will contain some concepts unfamiliar to you.


Why is the ether linkage breaking here? I have been asked to find the major product of the following reaction:

This is my thought process: the lone pairs of the oxygen atom connected to $\ce{-Et}$ are less delocalised than the other oxygen atom. So, that will abstract the $\ce{H^+}$ from $\ce{HI}$ leaving behind $\ce{I^-}$ and producing a positive charge on the oxygen atom. Now the oxygen atom will take up the bond pair electrons and $\ce{I^-}$ will form a bond in $\ce{S_{N}2}$ mechanism, producing $\ce{I-Ph-O-Ph}$ and $\ce{EtOH}$ as products.

But, the answer in the book states that:

The ether linkage will break as it does not have a double bond character due to the absence of resonance.

Why is the ether linkage breaking here? What does it mean by "it does not have a double bond character"?

I have used the image you used $2nd$, see carefully the markings.
The $C-O$ bond with benzene ring is stronger than the $C-O$ bond of ethyl, as there is resonance with benzene making it partially a double bond. So, that $C-O$ is difficult to break compared to the $C-O$ bond in ethyl branch.


how to derive equilibrium concentrations from initial conditions for mass action reactions? I am reading a tutorial on biochemical reactions and mass action kinetics (https://www.math.utah.edu/~keener/books/control.pdf, pp. 1-2) and would like to derive an analytic solution to confirm simulation results. This is NOT a homework question, I am simply trying to understand things.
The tutorial considers a simple reaction where A and B combine, reversibly, to produce C:
$A + B \rightarrow^{f} B$
$C \rightarrow^{r} A + B$
where $f, r$ are the forward/reverse rate constants. It is clear that the change in [C] over time is:
(Eq. 1) $\displaystyle\frac{d[C]}{dt} = f[A][B] - r[C]$
It's simple to then show that the equilibrium constant $K_{eq}$ is:
(Eq. 2) $K_{eq} = \displaystyle\frac{r}{f} = \frac{[A]_{eq}[B]_{eq}}{[C]_{eq}}$.
The question: how can we use this to derive the equilibrium concentration of one of the species (like $[C]_{eq}$) as a function of the initial concentrations of A and B, $A_0, B_0$ and rate constants $f, r$? Simulation of Eq. 1 for $A_0 = 200, B_0 = 100, f = 0.0001, r = 0.001$ shows that $[C]_{eq}$ is around $90$. How can this be confirmed analytically?
Attempt at solution: try to rewrite Eq. 2 in terms of initial concentrations. We can use the fact that $[A]_{eq}$ and $[B]_{eq}$ can each be rewritten in terms of $A_0, B_0, [C]_{eq}$:
$[A]_{eq} = A_0 - [C]_{eq}$
$[B]_{eq} = B_0 - [C]_{eq}$
this is because the equilibrium concentration of "pure" A (A that wasn't used with B to make C) has to the total amount of A we started with, minus the amount of A that went into producing C. Same argument for B.
But it's not clear if this helps get a solution? Plugging these quantities into Eq. 2 we get:
$\displaystyle\frac{(A_0 - [C]_{eq})(B_0 - [C]_{eq})}{[C]_{eq}} = K_{eq}$
$\displaystyle\frac{(A_0 - [C]_{eq})(B_0 - [C]_{eq})}{[C]_{eq}} - K_{eq} = 0$
which seems too messy to be correct. I was expecting a simpler quadratic equation for such a simple problem. Guidance on solution or references to derivations will be great.

Here is the expression from the question:
$\displaystyle\frac{(A_0 - [C]_{eq})(B_0 - [C]_{eq})}{[C]_{eq}} = K_{eq}$
Get rid of the fraction:
$(A_0 - [C]_{eq})(B_0 - [C]_{eq}) = K_{eq} {[C]_{eq}}$
Distribute the sums in the product:
$A_0 B_0 - [C]_{eq}(A_0 +  B_0)  + [C]_{eq}^2 = K_{eq} {[C]_{eq}}$
Sort the terms and you have your quadratic equation:
$A_0 B_0 - [C]_{eq}(A_0 +  B_0 - K_{eq})  + [C]_{eq}^2 =  0$


Hybridisation of Mn in potassium permanganate I'm clear with the concepts of crystal field theory. But I can't figure out the exact reason why the hybridisation of manganese in potassium permanganate ($\ce{KMnO4}$) is $\mathrm{d^3s}$. Can anyone explain please?

This is formally a manganese(VII) compound and hence there are no 3d electrons.

The four $\ce{O^2-}$ ions are considered to be donating two electrons each to the atomic orbitals. Tetrahedral "hybridisation" can be achieved by using the $\mathrm{4s},~\mathrm{3d}_{xy},~\mathrm{3d}_{yz}$, and $\mathrm{3d}_{xz}$ AOs. Since in Mn(VII) the $\mathrm{3d}$ AOs are lower in energy than the $\mathrm{4s}$ and $\mathrm{4p}$ AOs, the "hybridisation" is best considered as $\mathrm{d^3s}$.

I have put "hybridisation" in quotes because it is not used in advanced chemistry (it is an artifact!) but it is correct in saying that the bonding MOs contain mostly $\mathrm{4s},~\mathrm{3d}_{xy},~\mathrm{3d}_{yz}$ character but they'll also probably mix some $\mathrm{4p}$ character in there for good measure!


How to predict the basicity of organic amines? The following compounds were given, and I have to predict their order of basicity:

\begin{align}
\text{guanidine} &: \ce{(H2N)2C(NH)} \tag{A}\label{guanidine}\\
\text{acetamidine} &: \ce{(H3C)(H2N)C(NH)} \tag{B}\label{acetamidine}\\
\text{benzamidine} &: \ce{(H5C6)(H2N)C(NH)} \tag{C}\label{benzamidine}\\
\text{dimethylamine} &: \ce{(H3C)2NH} \tag{D}\label{dimethylamine}
\end{align}

After nitrogen donates a lone pair to $\ce{H+}$ in option \eqref{guanidine}, it has 2 resonance structures, whereas in option \eqref{benzamidine}, after donating a lone pair to $\ce{H+}$, it has 4 resonance structures. Therefore \eqref{benzamidine} should be the most basic or should have the highest rate of protonation. The answer given is option \eqref{guanidine}.
Whereas \eqref{acetamidine} and \eqref{dimethylamine} have 1 and 0 resonating structures, so the main confusion is between option \eqref{guanidine} and \eqref{benzamidine}.

The actual number of resonance structures is inconsequential: what matters is whether they are significant contributors to the resonance hybrid. This issue has been discussed at length, particularly in the case of carboxylic acids vs phenols: see e.g. Why is acetic acid more acidic than phenol?

Along this line of thought, the resonance structures that involve cationic carbon with six valence electrons are not really worth considering. Octet-complete resonance structures are always much more major contributors than octet-deficient ones.

If we only count octet-complete resonance structures of the conjugate acids, then guanidine (A) wins hands down with three; the two amidines (B) and (C) both have two; and dimethylamine (D) only has one.

Between (B) and (C) the choice is tougher. You might be inclined to think that this is where the minor contributors come in: surely benzamidine (C) is more basic than acetamidine (B), since its conjugate base has some of these minor resonance contributors. However, experiment shows that the order of acidity is the other way round: benzamidine (pKa 11.6 in water at 20 °C) is actually a weaker base than acetamidine (pKa 12.52),[1] making for an overall basicity order of (A) > (B) > (C) > (D).

The reason for this is not a resonance effect, but rather an inductive effect. It is true that the benzamidinium ion enjoys marginally more stabilisation by resonance. However, in general, sp2 carbons (and aryl groups) are more electron-withdrawing than sp3 carbons (and alkyl groups). So, the positive charge on nitrogen is also destabilised by the inductively electron-withdrawing phenyl group.

A rare case of inductive effects outweighing resonance effects, but there you are.

Reference

Albert, A.; Goldacre, R.; Phillips, J. The strength of heterocyclic bases. J. Chem. Soc. 1948, 2240–2249. DOI: 10.1039/JR9480002240.


Why does the stability of hydrides decrease down the group in nitrogen family? Why does the stability of hydrides decrease down the group in nitrogen family? Does this have to do with the fact that $\ce{NH3}$'s $\ce{N}$ is hybridised ($\ce{sp^3}$) while the rest of the group prefers unhybridised central atom for hydrogens leading to steric replusion in $\approx 90^{\circ}$ alignment of $\ce{H}$s?

Other possible explanation:

Perhaps,  the orbital overlap order is: $\ce{2sp^3-1s >  1s-3p> 1s-4p>1s-5p> 1s-6p}$ which leads to greater stability of $\ce{NH3}$ over the others.

The increasing size of the elements play a role here along with the overlapping factor.

Bismuth is humongous as compared to hydrogen so it's hydride is unstable. However nitrogen being small in size can form strong bonds with hydrogen. This also accounts for the fact that the reducing nature of the hydrides increase down the group.

As far as overlapping is concerned the bonds in the hydrides are - $\ce{2p-2p}$ up to $\ce{5p-2p}$.

Clearly the upper hydrides will be more stable as they have more effective overlapping between the orbitals.


How to calculate mass of nanomaterial grown on a substrate I'm growing graphene on copper substrates and need to know the mass. The graphene is grown via chemical vapor deposition; amounts of reactants are not completely known. The best approach I've thought of is recording the mass of the substrate before and after using an analytical balance, but with the nature of the material being nano, this method may be elusive. What other method could I use to find the mass?

One possible option for thin film thickness measurement is by Energy Dispersive Spectroscopy (EDS) in a scanning electron microscope (SEM).  There exist commercial packages for this such as LayerProbe by Oxford and STRATAGem (I have no financial interest in Oxford or SAMx).  If your campus/research facility has an SEM with an EDS (many do), but not a commercial package, you can still work it out yourself.  If I were tackling this in my own lab, I would probably do something along the following lines:

Option #1:

Obtain a spectrum from a known thickness of carbon on your Cu substrate.  This can be any method such as sputtering or evaporation so long as the thickness is known by other means.  Most geology labs have the ability to do this kind of deposition since they need reproducible carbon thicknesses on their samples.

For example, geo labs will typically place a polished brass object next to the sample to be coated in an evaporative coater.  The change in the color of the brass corresponds to the thickness of the carbon: e.g. blue is 250 Å but bluish green is 300 Å (Kerrick, D. M., et al., 1973).  So something like 25 % accuracy is possible this way.

Then you measure that thickness in the SEM using an exact beam current, size, distance from the objective lens.  Then you measure your graphene in the SEM using the exact same conditions.  If your sample is less than about 200 nm thick, then the number of counts/second is proportional to the number of carbon atoms present in the layer.  Using the mass for your standard layer, (assume a density of graphite), then you can compute the mass of your graphene.

If your thickness is more than about 200 nm, then you may need to do an absorption correction – i.e. the photons are being absorbed on their way out and your count rate is no longer proportional to the number of atoms.  You can compute attenuation coefficients using the CXRO calculator (and a couple other places if you google).  Then you can do a thin film correction.  At 200 nm, your error from ignoring the thin film correction is probably about 5–10 %, which is likely a bit less than the error in the determination of the standard film using the color method.  So you can think with that for your sample.

Option #2:

You can measure the mass directly using an oscillating crystal.  Commercial systems comprise a quartz crystal with electronics set to find the resonant frequency.  You place the quartz crystal in your chamber alongside your Cu substrate.  As your reaction happens, it grows both onto your Cu and onto the quartz crystal.  The resonant frequency of the crystal changes as mass is added.  These are sold commercially, though a bit pricey.  For example, at TedPella (again, no financial interest).  The resolution is typically billed as 1 Å thickness resolution (assuming you know the density of your material.)

For your application, you would have to be sure that the graphite grows on your monitor crystal the same as on your Cu substrate.  That may require checking in an SEM, and it may require prepping the crystal somehow.  Hence, I offer this as suggestion #2.

Finally, you could also figure this out using a focused ion beam and sectioning the sample from the side.  If this is too complex, there also exist labs where one can hire them to do this kind of measurement.

I hope one of these solutions works for you!

Ref:

Kerrick, D. M., Eminhizer, L. B., & Villaume, J. F. (1973). The Role of Carbon Film Thickness in Electron Microprobe Analysis. American Mineralogist, 58, 920–925.


How do I overlay spectra in an ACD/NMR Processor, version 12.01? I am having great difficulty getting spectra to overlay (the option doesn't even appear) after setting to new window mode and selecting each spectrum.

If anyone else is running into the same issue, after selecting which spectra you would like to overlay go to series -> collect spectra and then select "manual offset" if you want to adjust the height.


In a monomolecular reaction, how can a reactant experience collision to react? Monomolecularity means no molecules which experience collision with each other taking part in reaction. So how it is possible that in monomolecular reaction only one reactant experiences collision?

One reactant, e.g. O3, a pure chemical species, has many molecules that are in continual motion and can collide with each other (as well as with the walls of a container or with a catalyst), so you'll get some O2 generated. Or the converse, if you have UV photons colliding with pure O2 gas.


How much heat is required to dehydrate a hydrate? If I have a hydrate, such as copper (II) sulfate pentahydrate, $\ce{CuSO_4 \cdot 5H_2O}$, what is the amount of heat to required to dehydrate it? I assume that it is equal to or greater than the amount of heat required to evaporate the water molecules. Does the required heat vary for every hydrate?

The formula for dehydrating copper (II) sulfate pentahydrate is:

$$\ce{CuSO_4 \cdot 5H_2O ->[\Delta] CuSO_4 +5H_2O}$$

According to Wikipedia, copper(II) sulfate will dehydrate from penta- to trihydrate at 63 °C, then to monohydrate at 109 °C and then finally to the anhydrate at 200 °C. It melts at 150 °C, but won't decompose into ions until 650 °C which is beyond most benchtop rigs (your Pyrex flask would give way at about 500 °C).
The required heat of decomposition, as you can see, does vary for each hydrate (especially for the same salt), but doesn't always have to be greater than the boiling point; water has a significant vapor pressure at nearly all ambient Earth temperatures (even at freezing), and so in a dry environment, water will evaporate rapidly at temperatures well below boiling.


What is the necessary volume of concentrated HCl to prepare a solution with a pH of 1.8? I am trying to solve the following exercise:

  What volume of a concentrated $\ce{HCl}$ solution, which is $36.0\%$ $\ce{HCl}$ by mass
  and has a density of $1.179~\mathrm{g~mL^{-1}}$, should be used to make $5.00~\mathrm{L}$ of an $\ce{HCl}$
  solution with a $\mathrm{pH}$ of $1.8$.

I found the respective volumes that $\ce{H2O}$ and $\ce{HCl}$ contributed.
I then found the molarity of the $\ce{HCl}$ aqueous solution given.
I used $10^{-\mathrm{pH}}$ to get the amount concentration of $\ce{HCl}$ needed, I then proceeded to find the amount of substance of $\ce{HCl}$ by multiplying the concentration needed by $5.00~\mathrm{L}$. I then took the amount of substance of $\ce{HCl}$ needed and multiplied by the density given to get an answer of $2.4~\mathrm{L}$.

Am I correct?

Unfortunately no, you're not right. As mentioned in a now deleted comment by Bruno, your answer gives a $\mathrm{pH}$ of $-0.74$, which is not correct. You didn't give detailed steps of your calculations, and your first step doesn't seem like anything we should do.

We can solve this by making the calculations step by step. Let's start with the concentration of $\ce{H+}$, which will be the same as the concentration of $\ce{HCl}$, we will achieve:

$$\ce{[H+]} = 10^{-\mathrm{pH}} = \pu{0.01584893 mol//L}$$

Since we're going to prepare 5 liters, this means the amount of substance of $\ce{H+}$ should be

$$5 \times \pu{0.01584893 mol} = \pu{0.07924466 mol}$$

To transform this into amount of $\ce{HCl}$, we need the acid's molar mass ($\pu{36.46 g//mol}$) to convert from amount of substance to mass:

$$\require{cancel}%
g_{\ce{HCl}}
= \mathrm{\frac{36.46\ g}{\cancel{mol}}}
  \times 0.07924466\ \mathrm{\cancel{mol}}
= 2.88926 \ \mathrm{g}$$

Now that we have the amount of substance of $\ce{HCl}$, we calculate the volume, which is where the density comes in:

$$V_{\mathrm{HCl}}
= \mathrm{\frac{2.88926\ \cancel{g}}{1.179\ \cancel{g}\ mL^{-1}}}
= 2.450602 \ \mathrm{mL}$$

Now the last step: Our stock solution isn't pure $\ce{HCl}$, it's diluted in water. So actually the volume we need is:

$$V_{\mathrm{HCl}}
= 2.450602 \times \frac{100}{36}
= \pu{6.807229 mL}
\approx \pu{6.8 mL}$$

This might look that it is not much, but remember $\ce{HCl}$ is a very strong acid. For example, to prepare $\pu{1000 mL}$ of a $\pu{0.001 M}$ solution you only need $\pu{0.086 mL}$ of concentrated $\ce{HCl}$!

You can check this result by imputing the given density, weight percentage, final volume and $\ce{H+}$ concentration (first step) into Sigma-Aldrich's Molarity Calculator.


Is there any food-safe way to break down capsaicin (chilli) so food is less spicy? That is, is there a chemical or enzymatic reaction with capsaicin that break it down into other food-safe chemicals? Or alternatively, bind to capsaicin such that it doesn't trigger the same spicy receptors in our mouths.

I know about how casein can be used to wash capsaicin away, but it doesn't actually change it into another form or deactivate it.

It is an interesting problem, from my experience the capsaicin behaves in a way which is consistant with it being a liphilic substance. I know that if you fry chilli pueppers with some cheese that the cheese helps to bring out the flavour of chilli peppers.

I know that anacdotal evidence suggests that a dairy product such as milk or yougurt helps to stop your mouth burning. I think that this is reasonable if we assume that the chilli pepper toxin dissolves in the fatty droplets in the milk or yogurt. This might make it a better mouth decontamination liquid than water.

I think that the way to test this would be to put known amounts of capsaicin on filter paers and then soak them in milk and water and then measure the capsaicin concentration in the liquids after soaking. It could be done with HPLC or if we were able to get C14 labled compound we could use liquid scintillation counting.

From my student days I recall that a vegatarian or sasuage chilli was hotter than one made with minced beef. So it is possible that the capsaicin adsorbs onto the surface of the beef. Maybe you should consider using the langmuir isotherm to model such a thing.

As a simple alternative just put less chilli in the food.


why is tertiary carbocation more stable than primary allylic carbocation? According to general organic chemistry the order of stability is as follows:

aromaticity > resonance > hyperconjugation > inductive effect

Since primary allylic carbocation shows resonance as well as hyperconjugation, moreover resonance is greater than any other effect then why is primary allylic carbocation less stable than tertiary carbocation.

The decreasing order of stablity of carbocations is:

aromatic (tropylium, cyclopropenyl) > benzyl > allyl > tertiary > secondary > primary

Allylic carbocations are resonance stabilized whereas tertiary alkyl groups are stabilized by hyperconjugation (and inductive effect). The hyperconjugation (and inductive effect), in general, is quite weaker than resonance (and hyperconjugation effect). The only exceptions to the previous statement I am aware of are halogens (inductive effect is stronger than resonance).


Can protection group be used in a Wolff–Kishner reaction? How can I modify the ozonolysis of cyclohexene such that one of the aldehyde groups of the product, adipaldehyde, is efficiently protected as a base stable protecting group? I know that attempting to monofunctionalize adipaldehyde will lead to mixtures.

My aim is to then do a Wolff–Kishner reduction on the monoprotected dialdehyde.

@Halake: The protection of adipaldehyde as the monoacetal was solved by Schreiber and Claus some years ago [Also see, Schreiber, S.L.; Claus, R.E.; Reagan, J. Tetrahedron Letters, 1982, 23, 3867.] Ozonolysis of cyclohexene (1) in the presence of methanol leads to the molozonide 2, that opens to the zwitterion 3 which is trapped as the methoxy hydroperoxide 4. Addition of p-toluenesulfonic acid catalyzes the formation of the acetal 5. After neutralization with sodium bicarbonate, reduction of the hydroperoxide with dimethyl sulfide affords the monodimethyl acetal of adipaldehyde 6. Since acetals are base stable, Wolff-Kishner reduction of the aldehyde can be accomplished followed by aqueous acid hydrolysis of the acetal.


Ideal gas Law best describes the properties of which of the following gases at 0 degrees Celsius and 1 atm? a. $\ce{PH3}$

b. $\ce{HBr}$

c. $\ce{SO2}$

d. $\ce{N2}$

I believe that it is $\ce{N2}$, but I'm not too sure, since technically all of them would act the same at STP.

One of the defining characteristics of ideal gases is that ideal gas particles do not interact with each other, and there are no intermolecular forces between them. If a certain type of gas molecule has a molecular bipole, then each molecule will have ends with opposite partial charges. In a collection of gas particles, if the particles have a molecular dipole, these particles will attract and repel each other and thus will experience intermolecular forces. Because the two nitrogens in $\ce{N2}$ are equally electronegative, the bond and each $\ce{N2}$ gas particle will have minimal partial charges and thus minimal intermolecular forces. Thus, $\ce{N2}$ behaves very ideally.


If liquid hydrogen fluoride autoionises, how can it be an insulator? According to Wikipedia, one of the last major hurdles in the isolation of elemental fluorine was the need to mix hydrogen fluoride with another substance in order for its electrolysis to be possible (Henri Moissan eventually hit upon potassium bifluoride, which is still used for this purpose today).

From "Fluorine Isolation":

  Edmond Frémy postulated that electrolysis of pure hydrogen fluoride to generate fluorine was feasible and devised a method to produce anhydrous samples from acidified potassium bifluoride; instead, he discovered that the resulting (dry) hydrogen fluoride did not conduct electricity.

And from "Henri Moissan Preparation of elemental fluorine":

  Moissan eventually succeeded in preparing fluorine in 1886 by the electrolysis of a solution of potassium hydrogen difluoride ($\ce{KHF2}$) in liquid hydrogen fluoride ($\ce{HF}$). The mixture was needed because hydrogen fluoride is a nonconductor.

However, hydrogen fluoride is a protic solvent, and undergoes autoionisation via the following reaction:

$$\ce{3 HF <=> H2F^+ + HF2^-}$$

Pure liquid $\ce{HF}$ should, therefore, conduct electricity at least weakly, like liquid water does.

What gives?

Unlike metals which can conduct with free electrons, conduction in non-metallic liquids (ionic and covalent) is dependent on the ion concentration to carry charges. Pure water has a autoionization constant of $\rm K_w = 1\!\times\!10^{-14}$, which corresponds to a total ion concentration of $\rm 2\!\times\!10^{-7} \pu M$. At this concentration the resistivity of water is $\pu{18.2 M\Omega~ cm}$ which corresponds to a conductivity of $\pu{5.5 \mu S/cm}$ and is considered to be a poor conductor. Sea water on the other hand has about $\pu{3.4\%}$ sodium chloride and a conductivity of about $\pu{55,000 \mu S /cm}$ which is a million times improvement in conductivity which is considered conductive.

Hydrogen fluoride has an autoionization constant of $\rm 10^{-12.5} ~~(3.162\!\times\!10^-13)$ which corresponds to a total ion concentration of $\rm 1.125\!\times\!10^{-6}$ which is about five-times that of water. Five times the ions gives us five times the conductivity of $\pu{275\mu S/cm}$ or a resitivity of $\pu{2.28M\Omega ~cm}$ which is still a quite poor conductor. Thus it is necessary to add an ionic substance to increase conductivity to have appreciable conduction of the hydrogen fluoride similar to sea water.


Manipulate melting point of frozen water How to manipulate water in way that ice melting point isn't 0°C, but for example 4°C?

One way of doing so would be to significantly increase the pressure. As you can see from this phase diagram, at a pressure of 1 GPa, water ice will remain frozen at temperatures up to about 30°C.


Independant variable for Asprin pill titration with NaOH solution I am assigned to do an independently designed chemistry experiment for school. I want to do titration of Aspirin pills with NaOH solution using Phenolphthalein as indicator. I am struggling to find and independent variable for the experiment. I thought about doing heat, but is unsure about the mechanisms of that. If anyone could help me or guide me in a direction I would greatly appreciate it!

Thank You!

Independent variable means the variable which you can change easily. Direct titration aspirin is not a good idea as aspirin hydrolyzes into salicylic acid. There are a lot of non-sense high school/college level experiments doing direction titration and ignoring the side reaction of hydrolysis.

The proper procedure of aspirin analysis is by back-titration with NaOH and taking hydrolysis into account. Ask your teacher about it by showing him/her this procedure Back titration of aspirin

Coming back to independent variables...Just now you have another independent variable...comparison of two different methods of titration (direct vs. back). Think of other independent variables and share in your comments.


Energy of activation vs Bond energy In the second chapter, Methane Energy of activation and Transition state in the textbook Organic Chemistry by Morrison and Boyd, Energy of activation is given the following definition:

  The minimum amount of energy that must be provided by a collision for a reaction to occur is called the energy of activation, Eact.

The Eact for the reaction:

Cl(free radical) + CH3-H ---------> H-Cl + CH3(free radical)

is 4 kcal/mol. Therefore, on an average, it is 4/NA kcal per individual collision (where NA is the Avogadro number).

However, the bond energy of the carbon-hydrogen bond in methane is 104 kcal/mol. Therefore, on an average, it would be 104/ NA kcal for a single molecule of methane. In other words, 104/ NA kcal should be the energy provided by a collision that involves a methane molecule in order to break the carbon-hydrogen bond. Why does this not agree with the statement above which says 4/NA kcal per individual collision is what is required?

The bond dissociation energy is the energy needed for the reaction leading to the rupture the $\ce{C-H}$ bond. I.e. it is the enthalpy of the reaction:

$$\ce{CH_3-H \rightarrow CH_3^{\bullet} + H^{\bullet}}$$

In the case of the reaction in the question, formation of the $\ce{H-Cl}$ bond will counterbalance the energy cost for breaking the $\ce{C-H}$ bond.


Can soap make an liposome around virsuses? I have read that soap does not kill bacteria and viruses, but it rather removes them, that is, it strips them off the skin sort of speak by forming micelles around them, which are then rinsed off.
But isn't it more logical that soap forms liposomes for viruses instead of micelles to remove them.

Whether a mixture of phospholipids/lipids/detergents form micelles or liposome is not dependent on the hydrophobic "payload" but on the nature of the amphiphilic molecules.
Detergents form micelles (and bubbles and antibubbles, when there is air around). The micelles are fairly small (radius similar to the length of the detergent), and they don't have space for an entire virus (diameter on the order of 100 nm) or bacteria (diameter on the order of 1000 nm). Detergents that form micelles are able, however, to solubilize membrane proteins (diameter on the order of 10 nm) by forming a band of detergents around the hydrophobic parts of membrane proteins.
Phospholipids mixed with cholesterol form lipid bilayers and liposomes (the latter by extrusion or sonification). The distinct behavior has to do with differently charged head groups and the number of hydrophobic tails.
Some bacteria are killed by detergents because they break down the cell membrane, integrating the phospholipids and cholesteral into the micelle. Viruses can't be killed because they are not alive. If they are surrounded by a lipid bilayer, the detergent can also break this down, rendering the genetic material inside harmless.
No matter whether bacteria and viruses are destroyed by the soap or not, the soap allows the water to wash the germs away, and they end up in the canalization (e.g. coronavirus genetic material in Boston) instead of in people's bodies.


sn2 mechanism help please? My mechanism for this conversion is an intramolecular attack sn2 to displace Cl- anion. OH- can then attack this aziridinium intermediate, but will the nucleophile not attack preferentially at the most substituted site (where greatest positive charge resides i.e. tertiary center)? This gives the incorrect product?

Steric factors are more important here. The positive charge resides on the N and the nucleophile attacks the least hindered site hence the product is as drawn.


Wittig reaction with an alkyl dibromide Recently, I have come across a problem, in which a ketone needs to be converted into an alkene. However, in the alkene product, there is also another bromide functional group, at the end of the alkyl chain. I was thinking of using a dibromide (an alkyl chain with bromine atoms at both ends) and performing the Wittig reaction. However, I foresee that both bromide functional groups would participate in the reaction. Would one solution to this be to use a protecting group? Or would the problem be solved if I manipulate the stoichiometry of the reactants (e.g. ensure that the dibromide is in excess etc.)?

It seems that the method to circumvent this problem is indeed to use a protecting group to protect one end of the alkyl chain, which is the end to be converted to a bromide after the Wittig reaction.

As Waylander aptly mentioned, another problem which comes with using an alkyl dibromide is that there may be an intramolecular substitution reaction, in which the anionic carbon in the phosphonium ylide reacts with the other end of the alkyl dibromide.

To circumvent this problem, we could use an alkyl bromide with an alcohol functional group at the other end of the chain. This alcohol group can then be protected using various protecting groups, such as a tetrahydropyran ether or trimethylsilyl ether. After the Wittig reaction, the compound can then be deprotected using acidic hydrolysis. After which, the alcohol functional group can then be converted to a bromide using $\ce {PBr3}$. The desired compound is thus obtained.


Is oxalic acid dihydrate considered a primary acid standard in analytical chemistry? Using oxalic acid dihydrate as a primary standard just seems really odd to me. I'd expect a primary standard to be oven dried. It just seems weird that a hydrate would be used.

Granted, I saw numerous references to using oxalic acid dihydrate on the web to standardize $\ce{NaOH}$. I assume that for high school and freshman labs that it is "good enough." Using student grade burettes and open air pan balances would greatly limit the possible precision.

Also I'd guess that "good enough purity" oxalic acid dihydrate can be purchased much more cheaply than potassium hydrogen phthalate (KHP), which would be my choice.

The analytical method as I remember from nearly 50 years ago...

Prepare a concentrated stock solution (4 molar?) of $\ce{NaOH}$ using distilled water. That went in a jug with a spout just above the bottom. It was capped with a dedicator tube filled with $\ce{NaOH}$ to absorb $\ce{CO2}$ from the atmosphere. It sat for a couple of days to allow sodium carbonate to settle out. ($\ce{NaOH}$ will have some carbonate.)

Dried KHP in an oven at $\pu{120 ^{\circ}C}$ for four hours and then put that in a desiccator to cool.

Boiled distilled water to remove dissolved $\ce{CO2}$ and stoppered that to cool.

Using the cooled boiled distilled water made an approximately $\pu{0.1 M}$ solution of $\ce{NaOH}$ by diluting the concentrated stock solution.

Using an analytical balance, weigh out 3 samples of KHP to nearest $\pu{0.0001 g}$ into flasks and carefully dissolved the KHP in the cooled boiled distilled water with swirling to minimize introducing bubbles into the solution.

Then using $\ce{NaOH}$ as the titrant, phenolphthalein was used as the indicator. Again careful to swirl solution, not shake, to prevent bubbles.

Using class 1 50-mL burette which was marked to $\pu{0.1 mL}$s but read to $\pu{0.01 mL}$.

Jander-Blasius (14. Ed., 1995) uses nonhygroscopic sodium oxalate, dried at 230-250°C (it decomposes above 250 according to wikipedia), to standardise permanganate titer solution against. They give no other recommended standard for manganometry, so I assume this is it.

I have no idea why anybody would want to use (or recommend using) the free acid instead, except perhaps to insult his first-year students' intelligence. Can't imagine it's much cheaper, in analytical grade.

For NaOH titer solution, Jander recommends using a secondary standard, e.g. HCl solution. HCl itself is standardised against freshly precipitated and dried sodium carbonate, or HgO ($\ce{+ 4 KI  + H2O -> K2[HgI4] + 2 KOH}$)  dried over conc. sulfuric acid.


can an alloy be used to make another alloy? The answer is likely very obvious but I know pretty much nothing about this so:

Can I make an alloy, using an alloy? That is, could I make an alloy of steel (in itself an alloy of iron and carbon) and silver?

An alloy is purely a mixture of metals (and sometimes non-metals) which are non-chemically bonded.

So yes, you could mix steel and silver to create an alloy. There would be nothing stopping you from doing this, but I cannot say whether or not it would produce any desirable results. Mixing alloys with other materials is merely creating another alloy. Stainless steel is actually just regular steel combined with Chromium.

An alloy is usually created to increase certain properties of its constituents. For example, iron is mixed with carbon in order to create steel, which is far stronger and can be used to manufacture items such as girders. Or, in other cases, Chromium may  be added, if the steel is to be used in, say, cutlery.


How does WD-40 remove rust? One of the often-touted uses of WD-40 is that it "dissolves rust". The official website states that WD-40 "breaks down the bonds between metal and rust".

I can't understand how this would work, since there seem to be no rust-dissolving ingredients in WD-40 and it seems the product is more basic than acidic.

A (possibly not exhaustive) list of ingredients:

Mineral oil, I guess for lubrication and rust protection layer (but not a rust solvent?)
Decane, freeze-resisting
Nonane, water-displacing
Tetradecane, freeze-resisting and water-repelling
Undecane, freeze-resisting and water-repelling
Dimethyl Naphthalene (C12H12), likely used as a solvent
Cyclohexane, solvent
Carbon dioxide, likely used as a propellant

I can imagine that the low viscosity of WD-40 allows it to get in all nooks, making it easier to grind rust particles away because it acts as a short-term lubricant. Is that actually how it works, or am I missing something?

I'd say it works by coming out as a liquid. Then it diffuses through the porous rust layer or travels along the steel edge and expands to gas- as the mixture has a highly suppressed vaporisation point to push the rust from the inside. Steel has a high surface affinity for hydrocarbons as demonstrated by high difficult it is to clean. Partial vaporisation is plenty powerful enough to knock off rust.


Electrons distribution of oxygen atoms in ozone While dealing with the Lewis structure for ozone, most of the explanations show the electron distribution around the 3 oxygen atoms with a single and double bond.

But if you count the number of valence electrons of the middle oxygen atom, it will add to 5 valence electrons only, but the number of valence electrons of an oxygen atom is 6.

Similarly, the oxygen atom with a single bond in ozone, is being shown with 6 valence electrons plus one single bond, which accounts for 7 valence electrons, which does not tally with the expected number of 6 valence electrons.

Can someone please explain why this discrepancy has been permitted? Or what did we miss to consider?

The ozone molecule is described with two resonant structures(as in figure) in which all the oxygen atoms respect the octet rule. The oxygen in the middle has a positive charge because has only 5 valence electron  and the oxygen with the single bond has a negative charge because it has 7 valence electrons.
 In the real molecule both the bonds have the same length so there isn't one single bond and one double bond. Also both the terminal oxygen atoms have the same partial negative charge instead of being one neutral and one negative(-1).
The picture below is more realistic of the situation:

In the electrostatic potential map you can see that the two lateral oxygen are equivalent:


Notation for specific/volumetric entropy I'm working on a problem related to entropy, and for this particular use case it's advantageous to write down properties in 'specific' units. In the case of entropy (with units Joules over Kelvin, I think is commonly referred to as: $[S] = \pu{J K-1}$), I want to know how to communicate volumetric entropy, i.e. Joules per cubic meter Kelvin: $[\color{red}{?}] = \pu{J K-1 m-3 }$.

Is it something like $\bar{S}$?

Specific entropy

According to IUPAC "Green Book" Quantities, units, and symbols in physical chemistry, specific entropy is denoted as lowercase latin "s": $s$ [1, p. 56], whereas $\bar{S}$ would refer to molar entropy:

  \begin{array}{lll}
\text{Name} & \text{Symbol} & \text{Definition} & \text{SI unit} & \text{Notes} \\
\hline
[...]\\
\text{molar quantity}~X & X_\mathrm{m}, (\bar{X}) & X_\mathrm{m} = X/n & [X]/\pu{mol} & 5,6 \\
\text{specific quantity}~X & x & x = X/m & [X]/\pu{kg} & 5,6 \\
[...]\\
\end{array}
  [...]

  $(5)$ The definition applies to pure substance. However, the concept of molar and specific quantities (see Section 1.4. p. 6) may also be applied to mixtures, n is the amount of substance (see Section 2.10, notes 1 and 2, p. 47).

  $(6)$ $X$ is an extensive quantity, whose SI unit is $[X]$. In the case of molar quantities the entities should be specified.

  Example $V_\mathrm{m,\ce{B}} = V_\mathrm{m}(\ce{B}) = V/n$ denotes the molar volume of $\ce{В}$.

Just as specific heat capacity $c$, specific entropy $s$ is measured in $\pu{J K-1 kg-1}$ [1, p. 90].

General note [1, p. 6]:

  The adjective specific before the name of an extensive quantity is used to mean divided by mass. When the symbol for the extensive quantity is a capital letter, the symbol used for the specific quantity is often the corresponding lower case letter.

Volumetric entropy

Quite often there is no special notation used, and volumetric entropy is denoted with $S$. Specifically, volumetric entropy generation rate for the convective heat transfer in flowing viscous fluid is often denoted with S-prime notation, e.g. from classical paper "A Study of Entropy Generation in Fundamental Convective Heat Transfer" [2]:

  \begin{align}
& S' & [\pu{W K-1 m-1}] \\
& S'' & [\pu{W K-1 m-2}] \\
& S''' & [\pu{W K-1 m-3}]
\end{align}

References

IUPAC “Green Book”. Quantities, units, and symbols in physical chemistry, 3rd ed.; Cohen, R. E., Mills, I., Eds.; IUPAC Recommendations; RSC Pub: Cambridge, UK, 2007. ISBN 978-0-85404-433-7.
Bejan, A. J. Heat Transfer 1979, 101 (4), 718–725. DOI: 10.1115/1.3451063.


Does hyperconjugation only arise from C–H bonds? I have to an extent understood hyperconjugation but I don't understand why the donation can only occur from a $\mathrm{sp^3}$-hybridised orbital bonded with hydrogen? What about other bonding partners?

No.
Hyperconjugation is the donation of electrons from a one orbital to another, where at least one of those orbitals is a bonding orbital, and there is a net overlap between the orbitals.
A typical first example taught of hyperconjugation is that which occurs in the conformations of simple alkanes (e.g. ethane). When ethane is in the staggered conformation, electrons from the $\sigma_\ce{CH}$ bonding orbital are donated to the properly aligned $\sigma_\ce{CH}^*$ antibonding orbital of the adjacent carbon atom. The interaction of these two orbitals, the $\sigma_\ce{CH} \to \sigma_\ce{CH}^*$ lowers the energy of those electrons, and thus lowers the energy of the molecule.
Next up is to look at a longer alkane chain (e.g. butane), which will have $\sigma_\ce{CH} \to \sigma_\ce{CH}^*$ interactions like in ethane, but will also have additional interactions involving the $\sigma_\ce{CC}$ and $\sigma_\ce{CC}^*$ bonding & antibonding orbitals. Depending on which conformation of butane you are considering, it is possible to have $\sigma_\ce{CC} \to \sigma_\ce{CH}^*$, $\sigma_\ce{CC} \to \sigma_\ce{CC}^*$, and $\sigma_\ce{CH} \to \sigma_\ce{CC}^*$ interactions.
Continuing in a more general sense, these interactions are not limited to $\sigma$ bonding orbitals. In carbocations, which in principle have an empty p orbital ($\ce{C}_\mathrm{2p}$) on the electron-deficient carbon, there can be hyperconjugation in the form of $\sigma \to \ce{C}_\mathrm{2p}$. (Note that I didn't specify the atoms involved in the $\sigma$ bond; as long as the atoms form a stable $\sigma$ bond, this hyperconjugation interaction can occur, though how stabilizing it is will depend on the size of the orbitals involved.)
There are a variety of hyperconjugation interactions that can occur:

$\sigma \to \sigma^*$
$\sigma \to \pi^*$
$\sigma \to$ empty hybridized orbital
hybridized orbital $\to \sigma^*$
$\pi \to \sigma^*$

For example, in protonated formamide ($\ce{H-C(=O)-NH3+}$), there is hyperconjugation in the form of $\pi_\ce{CO} \to \sigma_\ce{NH}^*$ and $\sigma_\ce{NH} \to \pi_\ce{CO}^*$, as well as $\ce{O}_\mathrm{2p} \to \sigma_\ce{CN}^*$.


XRD-Shorter distance between planes than actual radius of atoms/or cell parameter I'm solving problems related to X-ray diffraction, and I get these results which apparently are correct. My question may be dumb, but I just can't see how the interplanar distance might be shorter than the atoms radius.

What's up with this?

I think it might make sense that its shorter than the cells parameter, but it's not immediately obvious to me, so if someone could prove it to me, i'd be quite glad.

Edit:

For example:

The atomic radius of chromium is 124pm. THe distance between (310) planes is .0912nm.
See problem 6

Does this help?

The distance between planes will be less than the distance between atoms in many cases because the atoms don't have to stack directly on top of one another in each plane.


How does oobleck work? Oobleck is basically a mixture of cornstarch and water. It has the interesting property of going hard when pressure is applied. For example, it feels like a liquid when you swirl your fingers through it, but it becomes hard for a second if you punch it.

The common explanation for this is that the starch-water mixture is a suspension, and the  starch particles "lock" when pressure is applied.

What exactly is the mechanism of this "locking"?

Wikipedia already offers a decent summary of two theories of shear thickening behavior - namely the generation of transient 'defects' in a high-density suspension that either resist further growth or are incompressible in the direction of shear.

A potential macroscale analogue of shear thickening is the dramatic increase in viscosity of large particle-air mixtures upon pneumatic removal of the interparticle fluid, as is exploited in jamming skin enabled locomotion. (paper here) This should be roughly comparable to the transient decrease in particle spacing in a shear thickening fluid that is being compressed beyond its relaxation rate.


How to calculate the concentration of the diluted solution given a source concentration in ppm? A student adds $37.4~\mathrm{ml}$ of a $50.0~\mathrm{ppm}$ solution to a $100~\mathrm{ml}$ volumetric flask and dilutes to the mark. What is the concentration of the diluted solution?

  $13.7~\mathrm{ppm}$
  $7.5~\mathrm{ppm}$
  $37.4~\mathrm{ppm}$
  $23.6~\mathrm{ppm}$

I used $$M_1V_1 = M_2V_2$$ which worked for all the others and I assumed would work for this considering I had all of the variables except for the 2nd concentration, but those calculations gave me $18.7~\mathrm{ppm}$, which is not an answer choice.

Let us assume for a moment, that the parts-per-notation is actually unambiguous. We then know that we can convert the concentration given in ppm $c^\dagger$ into the concentration $c$ in SI units $\mathrm{mol\,L^{-1}}$, or $c^\dagger\propto c$. Then we know that the concentration is defined as the amount of substance per volume of solution.
$$c = \frac n V$$

When diluting a solution the amount of substance stays the same, hence we have two equations, that we can combine.
$$\begin{align}
  c_0 &= \frac{n}{V_0}  &&\text{source solution}\tag1\\
  c_\mathrm{dil} &=\frac{n}{V_\mathrm{dil}}  &&\text{diluted solution}\tag2\\
  c_\mathrm{dil} &=\frac{c_0V_0}{V_\mathrm{dil}}  &&\therefore\tag{1=2}
\end{align}$$

Which is exactly the formula you used (when I assume that $M$ stands for molarity).

Ergo your calculation result, as orthocresol already pointed out, of $c^\dagger_\mathrm{dil} = 18.7~\mathrm{ppm}$ is correct.

The error therefore has to be in the question, or in copying the same.


Finding the valency of transition metals in coordination complexes I understand how to find the valency of atoms in small organic compounds. For instance, for CH4, the valency of carbon is 4, while the valency of each hydrogen is 1. But for a coordination complex, for instance:

How do I find the valency of the center metal?
And if the center metal was bound to another metal:

How would the valency calculation change, if at all?
I need the valency of these transition metals because I wish to ultimately find the d^n configuration of the transition metal in the complex via the formula:
n = group number - valency
EDIT: If you have a different approach (apart from using valency) for finding the d^n configuration of the transition metal, feel free to suggest that instead.

For the first example, you are given that there are $6\, \ce{CN}$ ligands which are -1. That means that there is a total of 6- from the ligands. We also know the overall charge of the complex is -3. That means the charge of the central metal must be +3. In the second example, we can look at the charge again. Because the $\ce{CH3-}$ has a  charge of -1, the molybdenum metals must have a +3 oxidation state each to counteract the -6 total charge from the $\ce{CH3-}$ groups. To do these problems, you should look at the charges of the ligands/metal(s) and the total charge of the complex and know that they have to match. After you figure out the oxidation state, you can find the $\mathrm d^n$ configuration by removing electrons from the $4\mathrm s$ subshell first and then 3$\mathrm d$. For example, For $\ce{Fe^{3+}}$, a neutral $\ce{Fe}$ atom has $6$ $3\mathrm d$ electrons. You first remove 2 electrons from the $4\mathrm s$ subshell and then a final one from the $3\mathrm d$ subshell. Finally, you will find that $\ce{Fe^{3+}}$ is $\mathrm{d}^5$. Again, this is based on the oxidation state of the metal.


Is tin and tinware toxic? I tried to find info on tin toxicity but there is not much.
Wikipedia here says that it is not very poisonous; but there could be cases.

My question is: is it dangerous to say have a coffee boiled in a tin container? I know that coffee has some acidity, so it can probably react with tin.

The Wikipedia article you quoted clearly says:

  The low toxicity is relevant to the widespread use of tin in dinnerware and canned food.

Moreover, tin cookware, which used to be very common in earlier centuries, is still sold for outdoor cooking purposes. It is thus without a doubt deemed safe to use under current health regulations.

As a side note, the pH of coffee is around 5, which is only weakly acidic (same as unpolluted rain, and less acidic than tomato juice or beer):


What is the IUPAC name of [Co(NH3)5H2O]Cl3? I am not sure if it is Pentaammineaquo cobalt(III) chloride or Aquopentaammine cobalt (III) chloride.
Should the prefix be ignored while naming the ligands in alphabetical order? Besides how should the name be written with proper spacing?

Multiplicative prefixes are not considered in determining the alphabetical order of ligands.
According to the current version of Nomenclature of Inorganic Chemistry – IUPAC Recommendations 2005 (Red Book), the systematic names of coordination entities are generated by using the following principles:

IR-9.2.2.1 Sequences of ligands and central atoms within names
The following general rules are used when naming coordination compounds:
(i) ligand names are listed before the name(s) of the central atom(s),
(ii) no spaces are left between parts of the name that refer to the same coordination entity,
(iii) ligand names are listed in alphabetical order (multiplicative prefixes indicating the number of ligands are not considered in determining that order),
(…)

Therefore, the systematic name of the coordination entity in $\ce{[Co(NH3)5H2O]Cl3}$ is “pentaammineaquacobalt” since the ammine ligands are named before the aqua ligand according to alphabetical order. Note that, since 1970, the recommended name for water $(\ce{H2O})$ as neutral ligand in coordination complexes is “aqua” (formerly “aquo”).
The ionic composition of the complex salt $\ce{[Co(NH3)5H2O]Cl3}$ may be indicated in several ways:

IR-9.2.2.4 Charge numbers, oxidation numbers and ionic proportions
The following methods can be used to assist in describing the composition of a compound:
(i) The oxidation number of the central atom in a coordination entity may be indicated by a Roman numeral appended in parentheses to the central atom name (including the ending ‘ate’, if applicable), but only if the oxidation state can be defined without ambiguity. When necessary a negative sign is placed before the number. Arabic zero indicates the oxidation number zero.
(ii) Alternatively, the charge on a coordination entity may be indicated. The net charge is written in arabic numbers, with the number preceding the charge sign, and enclosed in parentheses. It follows the name of the central atom (including the ending ‘ate’, if applicable) without the intervention of a space.
(iii) The proportions of ionic entities in a coordination compound may be given by using
multiplicative prefixes. (See Section IR-5.4.2.1.)

Therefore, correct systematic names for the compound $\ce{[Co(NH3)5H2O]Cl3}$ are:

pentaammineaquacobalt(III) chloride
pentaammineaquacobalt(3+) chloride
pentaammineaquacobalt(III) trichloride
pentaammineaquacobalt(3+) trichloride

Note that multiplicative prefixes are usually not used in binary names if there is no ambiguity about the stoichiometry of the compound.


Huckel's rule and aromaticity I am learning to determine whether a monocyclic conjugated alkene is aromatic or anti-aromatic or not.
question 1:
cyclobutadiene: according to Huckel's rule, 4 * 1 = 4, it should be antiaromatic, like what is said here. However, Wikipedia says the 2 pi bonds have a 'more double bond' character, making it a rectangular shape (electrons not delocalized). Therefore, it is more like a conjugated diene, hence non-aromatic. What is the truth?
question 2:
planar / non-planar: for [18]annulene, it is aromatic (4 * 4 + 2 = 18). From Wiki,  ''it is large enough to comfortably accommodate six hydrogen atoms in its interior, allowing it to adopt a planar shape, thus satisfying Hückel's rule.'' So in this molecule, there is an extremely small steric effect. So I am wondering, is size of the inner space also a factor in determining whether a cyclic molecule is aromatic or not? If the inner lumen of the molecule is larger, it can follow Huckel's rule? Like Cyclodecapentaene, as the 2 hydrogens cannot be in planar (not comfortably accommodate) (figure 3 in Wikipedia). It is non-aromatic rather than aromatic (do not follow Huckel's rule).
question 3:
cyclooctatetraene is non-aromatic rather than antiaromatic (4 * 2 = 8) since it avoids by changing into a 'tub' shape (avoiding being the unstable planar conformation). I am wondering if other supposedly antiaromatic molecules can change their conformation to be non-aromatic (become more stable) as cyclooctatetraene does? Are there any criteria for changing the shape, or cyclooctatetraene is the only example? If yes, why? If no, what monocyclic conjugated alkene molecule is actually antiaromatic then?
Thank you very much.

Steric effects play an important role in determining the planarity, aromaticity of a molecule. Cyclobutadiene is sterically unstable(90 degrees angle between bonds), hence having high energy, and thus  anti-aromatic.
A lot of compounds change their conformations for reducing their overall energy. Check for the conformations of glucose, for instance. A compound will always prefer its low energy conformation, which occurs when Cyclooctatetraene is non-planar depicted in the right hand perspective drawing

(reference)
This compound looks like the big brother of benzene, which may incorrectly induce the notion of it being aromatic, or, by Huckel rule, anti-aromatic. But the inner angle of a regular octane shape is 135 degress, while a much stable angle nearby it is 120 degrees(an sp2 config.), which only occurs in the tub config. of Cyclooctatetraen. 120 degrees is not achieved, but ~126 degrees is achieved.
Frozen (at 129 K) cyclooctatetraene's structure was characterized by single crystal diffraction analysis by Claus and Rüger. You find the corresponding interactive model here.


Calculating solubility in water with other species present I'm very confused about calculating solubility.

Everything is in temperature of $298\ \mathrm K$

From Atkins's book, exercise 16.83:

Based on solubility constant, calculate solubility of $\ce{Al(OH)3}$ in acid solution of $\mathrm{pH}=4.5$, $K_{\rm s}=1\cdot10^{-33}$.

Using stoichiometry and equilibrium constants I proceed as follows:

$\mathrm{pH}=4.5$ corresponds to $\mathrm{pOH}=9.5$ and

$$\ce{Al(OH)3 (s) \rightarrow Al^3+ (aq) + 3OH- (aq)}$$

$$K_{\rm s}=\ce{[Al^3+][OH- ]^3}$$

Let's call $\ce{[Al^3+]} = x$.

From stoichiometry and the assumption that $\ce{[OH- ]} >> 10^{-7}$ we know that $\ce{[OH- ]} \approx 3x+10^{-9.5}$.

We get a equation like

$$x(3x+10^{-9.5})^3=K_{\rm s}$$

which I solve as $x=2.39\cdot10^{-9}$, but Atkins's answer is $3.0\cdot10^{-5}$.

Atkins's solution looks like:

$$K_{\rm s}=\ce{[Al^3+][OH- ]^3}$$

$${K_{\rm s}\over[\ce{OH-}]^3}=[\ce{Al^3+}] = 3.0\cdot10^{-5}$$

It looks like Atkins assumes that $[\ce{OH-}]=10^{-9.5}$.

I have two questions:

Why does Atkins assume that $\ce{Al(OH)3}$ doesn't affect $\mathrm{pH}$?
Which one is correct?

Since the solution is acidic you have to consider $\ce{[OH- ]}$ concentration to be $10^{-7}$

Then solubility product: $$k_\text{sp} = \ce{[Al^3+ ] [OH- ]^3}$$

And if $x$ amount dissociated: $$k_\text{sp} = x \cdot (3x)^3$$

Then we have,

$$10^{-33} = 27 \cdot x^4$$

So $$x = 2.466 \cdot 10^{-9}~\mathrm{mol/L}$$


What is the appropriate procedure for determining the quantum numbers of an electron? How do I find the set of quantum numbers for a specific electron in an element? For example,

  Calculate the set of quantum numbers for the 19th electron in chromium.

The electronic configuration of the 19th electron is 1s2 2s2 2p6 3s2 3p6 4s1. From here on, how do I decide what would the principle quantum number be?

1s2 2s2 2p6 3s2 3p6 4s1 is the electronic configuration of the 19th electron of chromium.

since the last shell is 4, therefore n=4,

l=0(as there is only one orbital in the s sub shell) ,

m=0 (since we're still in the s sub shell only)

and s=+1/2 or s=-1/2(arbitrarily assigned values)

note:+1/2 or -1/2 are used to distinguish between electrons as no two electrons can have the same electronic configuration, but here the question isn't that specific, so we need not bother.


Textbook Claim: "... in all cases it is the electrostatic force acting between charged particles that is responsible for all the forms of bonding." I am reading an introductory semiconductor physics textbook. The textbook states the following:

  Several types of atomic bonding have been identified, including ionic, covalent, van der Waals, hydrogen, and metallic. Whatever the name given to the type of bonding, in all cases it is the electrostatic force acting between charged particles that is responsible for all the forms of bonding.

My concern is with the validity of the latter sentence; that is, that all cases of bonding are electrostatic forces acting between charged particles.

Wikipedia says the following:

  A chemical bond is a lasting attraction between atoms, ions or molecules that enables the formation of chemical compounds. The bond may result from the electrostatic force of attraction between oppositely charged ions as in ionic bonds or through the sharing of electrons as in covalent bonds.

If electrostatic forces act between charged particles (ions), and atoms are, by definition, neutral particles, then, since atoms do bond, how can it make sense to say that all bonding is electrostatic forces acting between charged particles?

And my interpretation of the Wikipedia article is that the electrostatic force of attraction is responsible for bonding between oppositely charged ions, as in ionic bonding, whereas it is not responsible for bonding through the sharing of electrons, as in covalent bonding.

So is the textbook's claim that all types of bonding are electrostatic forces acting between charged particles false?

I would appreciate it if people could please take the time to clarify this.

You ask how bonding between two neutral atoms can be electrostatic. In the case of covalent and van der waals the electrostatic basis is not so obvious as it is for ionic compounds. A covalent bond forms when two neutral atoms establish a shared molecular orbital which is much larger than the separate atomic orbitals that the bonding electrons contributed by each atom formerly occupied. Electrostatic repulsion favors electron pairs being in larger orbitals, because this increases average distance between the electrons. In the case of Van der Waals attraction, the neutral atoms are electrostatically attracted because when their electrons are not symmetrically distributed, there is a partial positive charge at one end of a molecule and a partial negative at the other, even though overall charge is zero. Electrons in adjoining molecules will tend to synchronize their movements so that the positive end of one molecule will be near the negative end of the adjoining molecule; this is called induced charge.


In an Electrotlytic cell using MgSO4 electrolyte, Cu(OH)2 and Mg(OH)2 are visible products, Where Does SO4 go? I have an electrolytic cell with $\ce{MgSO4}$in an aqueous solution at full saturation, running at a potential of $3.6v$ with a copper anode because I need some $\ce{Cu(OH)2}$.
Production of $\ce{Cu(OH)2}$, $\ce{CuOH}$ and $\ce{Mg(OH)2}$ are all visible.
There's no gas production at the anode but I expect that the oxygen is remaining in solution due to $\ce{OH}$ production.
There's no sign of $\ce{SO2}$ or $\ce{SO3}$, so
what Happens to $\ce{SO4-}$?
EDIT
$\ce{SO4-}$ eventually rejoins $\ce{Mg}$ because the more reactive base will always oxidize a less reactive neutral or acidic salt, not to mention that $\ce{H2SO4}$ is a species that will form under these conditions as $\ce{Mg}$ and $\ce{Cu}$ both become insoluble (which should actually yield $\ce{H2SO5}$ because $$\ce{Cu + MgSO4 + H2O -> Cu(OH)2 + Mg(OH)2}$$ @cathode, $$\ce{H2O + SO4- -> H2SO5}$$ @anode, which doesn't hang around too long $$\ce{H2SO5 -> 2H2SO4 + O2(g)}$$ and the obvious reformation of $\ce{MgSO4}$ via $\ce{H2SO4}$ + $\ce{Mg(OH)2}$ and $\ce{CuSO4}$ + $\ce{Mg(OH)2}$ so I was eventually left with $\ce{Cu(OH)2}$ as the only insoluble product.
I have since been able to produce $\ce{Mg(OH)2}$, omitting $\ce{Cu}$ entirely of course, but coming up with a process wasn't easy with my 3 months of experience. I however have been able to completely separate $\ce{Mg+}$ from $\ce{SO4-}$, yielding $\ce{H2SO4}$and $\ce{Mg(OH)2}$using perlite.

It forms $CuSO_4$, which is soluble in water. $Cu(OH)_2, CuOH$ and $Mg(OH)_2$ prefer to precipitate, while $CuSO_4$ just floating there, completely invisible. Actually, even when ppl try to obtain $CuSO_4$ in such electrolytic cells, they simply collect $Cu(OH)_2$, add $H_2SO_4$ and crystallize sulfate. So it is there, but there is no way you can obtain it directly.

[remark] You can also try to crystallize it from solution, since it's there. But maybe you won't succeed.


Hyperconjugation, stability of molecules, and heat of formation I recently asked a question about why the heat of formation of organic radicals and positive ions decreases with their size and degree of branching at the radical or ionic site. The user "Buttonwood" made the comment that hyperconjugation is one of the forms stabilising cations and radicals.

This begs two questions:

What is the reasoning behind why hyperconjugation increases the stability of cations and radicals, and therefore decreases the heat of formation? Point 3. of this section of the Wikipedia article for hyperconjugation  says that "the heat of formation of molecules with hyperconjugation are greater than sum of their bond energies and the heats of hydrogenation per double bond are less than the heat of hydrogenation of ethylene."; does this not imply that hyperconjugation increases the heat of formation of molecules?
Why does an increase in stability imply a decrease in heat of formation? To a novice who is ignorant on this subject such as myself, it would intuitively seem that greater stability implies an increase in heat of formation, since the molecule is more "stable" in its current form and therefore it takes more energy to affect changes in its structure?

I would greatly appreciate it if people would please take the time to clarify these points.

Hyperconjugation is an electronic effect which is similar to resonance (one of the most fundamental concepts of organic chemistry). Here's how it works:
$$\ce{H3C-CH2+ <-> H+H2C=CH2}$$

You can write the above structure three times for each hydrogen replaced. Note that only α-hydrogens (on the adjacent sp3 carbon atom) can be replaced.

Now, answering your first question, hyperconjugation leads to stability because of the delocalisation of charge. The positive charge gets diffused away from the carbon atom, and starts to occupy a larger volume, decreasing the overall energy of the system caused by electronic repulsions. As a matter of fact, the greater the number of structures you can draw due to hyperconjugation, the greater the stabilisation. That explains why tertiary carbocations are much more stable than primary ones.
$$\ce{(H3C)3C^+ <-> H+H2C=C(CH3)2}$$

If you look closely there are nine replaceable hydrogens, so you can draw nine structures (explaining its stability).

The Wikipedia page says:

  the heat of formation of molecules with hyperconjugation are greater than the sum of their bond energies

Here, heat of formation refers to the heat released on the formation of these molecules. So it is negative.

In the absence of hyperconjugation, you would expect that the heat released on formation is equal to the sum of all bond energies. But because of stabilisation due to hyperconjugation, the product molecule has lower energy than expected, so overall, more heat is released (higher magnitude enthalpy of formation).
It's important to understand that enthalpy of formation (including its sign) decreases on increasing the stability of the molecule no matter what. But in the case that the enthalpy of formation is positive (energy absorbed), a decrease in the magnitude of heat of formation is observed, while if the enthalpy of formation is negative (energy released), an increase in magnitude is observed. In your other post, you include a list of carbocations, and their heats of formation are seen to be positive, while those of the non-ionic species are negative. So all the ionic species need to absorb heat in order to form (indicating a loss in stability), but that loss in stability (and hence the heat of formation) would be less if they were stabilised due to hyperconjugation.


Why can we use standard entropy when determining temperature at which a reaction becomes spontaneous? In this chemistry problem, $\Delta H$ and $\Delta S$ are given at $298\text{ K}$, and it asks for the temperature at which the reaction becomes spontaneous:

  For the reaction $2\text{SO}_3 \rightarrow 2\text{SO}_2 + \text{O}_2$, $\Delta H_r^\circ = +198 \text{ kJ/mol}$ and $\Delta S_r^\circ = 190 \text{ J / mol}\cdot\text{K}$. At what temperature will the forward reaction will become spontaneous?

I understand to solve this problem; the Gibbs free energy equation,
$$\Delta G = \Delta H - T \Delta S,$$
can be used to determine the temperature by assuming $\Delta H$ and $\Delta S$ are constant and solving for $T$ when $\Delta G = 0$. This gives $T = 1042\text{ K}$.

However, I don't understand why is this assumption valid. Isn't it true that $\Delta S_r$ would change at different temperatures? Why is it that the value for $\Delta S$ for this reaction under standard conditions can be used to calculate $\Delta G$ at $1042 \text{ K}$?

The question implicitly assumes that $\Delta H^\circ$ and $\Delta S^\circ$ are independent of temperature.

The temperature dependences depends on the change in heat capcity ($\Delta C_p$) are given by $\Delta H^\circ(T_2) = \Delta H^\circ(T_1) + \Delta C_p(T_2-T_1)$ and $\Delta S^\circ(T_2) = \Delta S^\circ(T_1) + \Delta C_p \ln(T_2/T_1)$.  (These equations assume $\Delta C$ is constant.)

So if $T_2-T_1$ or $\Delta C_p$ is small then $\Delta H^\circ$ and $\Delta S^\circ$ will be fairly constant.


Comparing ionic character of group 1 elements According to Fajan's rule ionic character should increase down the group as the size of cation increase.
So it must be
$$\ce{LiH < NaH < KH < RbH < CsH}$$
However, the following two sources provide two different opinions. Which one is right?
Source 1

Explanation: As the size of the metal cation increases down the group, the lattice energy of the hydrides decreases down the group. Consequently, the reactivity of the alkali towards hydrogen decreases down the group.
iii. The ionic character of the alkali metal hydrides increases from Li to Cs.
Explanation: Since ionisation enthalpy of alkali metals decreases down the group, tendency to form cations as well as the ionic character of the hydrides increases.

Source 2

But it we are asked to compare the ionic character of hydrides (LiH, NaH, KH, RbH, CsH), we can not use the Fajan's rule. This is because, here the anion is very small ($\ce{H-}$) as compared to the cations (alkali metal cations). The ionic character is this case can thus be compared on the basis of Lattice energy.
Since lithium ion and the hydride ion are approximately of same size, the charge density is high between the two, thus the ionic bond so formed will be very strong and hence more will be the lattice energy.
On coming down the group, the difference in the size of cation and anion increases, thereby making the ionic bond weaker, thus reducing the lattice energy.
Also, lesser will be the lattice energy, lesser will be the ionic character.
Thus, down the group, the ionic character of the hydrides decreases.

Well, the second source is certainly more wrong. There is a major difference between ionic character of a bond and the stability (lattice energy) of an ionic lattice. $\ce{LiH}$ is certainly more covalent than $\ce{CsH}$ (meaning $\ce{CsH}$ is more ionic), which can be justified simply by comparing charge densities on $\ce{Li}$, which is very high and draws electron density from the $\ce{H}$ atom, forming a more covalent bond.


Inconsistency in Elementary Bi-Molecular Reaction Rate Expressions Assume the following bi-molecular reaction is elementary as written with rate constant $k_\mathrm{f}$:
$$\ce{A + A -> P}$$
This review, suggests to express the rate in terms of the production of product "P":
\begin{align}
\frac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} &= k_\mathrm{f}[\ce{A}]^{2},&
\text{where }
\frac{\mathrm{d}[\ce{P}]}{\mathrm{d}t} &= \left(-\frac{1}{2}\right)\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t},
\end{align}
which gives:
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -2k_\mathrm{f}[\ce{A}]^{2}.\tag{1}$$
While this other review suggests that the rate be written in terms of the consumption of reactant "$\ce{A}$":
$$\frac{\mathrm{d}[\ce{A}]}{\mathrm{d}t} = -k_\mathrm{f}[\ce{A}]^{2}.\tag{2}$$
Which is correct?

If the reaction is
$$\ce{2A -> P},$$
then the rates are defined to be related as
$$
  -\frac{1}{2}\frac{\mathrm{d}\ce{A}}{\mathrm{d}t}
  = \frac{\mathrm{d}\ce{P}}{\mathrm{d}t}
$$
and this is true throughout the reaction. The rate at which $\ce{P}$ is produced is half that at which $\ce{A}$ is consumed. Normally therefore a factor of 2 is expected with the rate constant. With any other usage an explanation should be added so as not to confuse things.
In the general case
$$\ce{aA + bB -> pP + qQ},$$
then
$$
  -\frac{1}{a}\frac{\mathrm{d}\ce{A}}{\mathrm{d}t}
  = -\frac{1}{b}\frac{\mathrm{d}\ce{B}}{\mathrm{d}t}
  = \frac{1}{p}\frac{\mathrm{d}\ce{P}}{\mathrm{d}t}
  = \frac{1}{q}\frac{\mathrm{d}\ce{Q}}{\mathrm{d}t}
  = \frac{\mathrm{d}x}{\mathrm{d}t},
$$
where at any time $t$, $ax$ moles of $\ce{A}$ and $bx$ of $\ce{B}$  are consumed and $px$ of $\ce{P}$ are produced, where $x$ is the extent of reaction.


How does pressure shift equlibrium Here is a question from my practice problems:

I know that the answer is A, but I don't know why.  Is it because A has the most moles of gas on the left? thanks

I think the most intuitive way to think about this is rates of reaction. A reaction involving 2 molecules in the gas phase only happens when they collide. At higher pressures there will be more collisions between molecules.

This means that the reaction with the most molecules will be sped up more, so if a reaction has 2 molecules reacting to give 1 molecule of product, the forward reaction will be sped up more than the reverse reaction by increasing the pressure.

To do the kinetic calculations properly will require all the mechanistic steps involved. The general principle is that that at equilibrium, both reactions are happening at the same rate, but when you increase the pressure you will speed up the reaction which has more moles of gas, resulting in that reaction happening faster than the other reaction, shifting the equilibrium until the rates of the two reactions are balanced again.


Why don’t all objects melt and liquefy when heated sufficiently? Certain objects such as metals, glass, plastic, sugar etc. melt when heated to a certain temperature but other materials such as a piece of wood, paper, a piece of rock etc. do not. Why is that?

The problem with melting most substances is that you would be doing it in a non-innocent atmosphere and that sometimes even the functional groups inside a compound are non-innocent.

In a perfectly innocent world, everything could be molten given sufficient temperature. Given real world paper, remember that reaction $(1)$ is exothermic and the only reason why it does not spontaneously occur is the lack of activiation energy.

$$\ce{C_nH_mO_l + \frac{$2n + m - l$}{2} O2 -> n CO2 + \frac{$m$}{2} H2O}\tag{1}$$

Once you supply sufficient heat, the activation energy will be overcome and the reaction suddenly becomes spontaneous; for paper, I am told that this happens at around $451~\mathrm{^\circ F}$, which is $233~\mathrm{^\circ C}$ in proper units.

Now you could suggest melting things in an inert argon atmosphere — or maybe even take helium for additional inertness. But then still you have potential reactions that can occur with organic materials. What then will happen might be called smouldering. Technically, this is no more than burning without oxygen.


Total number of octahedral void(s) per atom in a cubic closed pack structure The total number of octahedral void(s) per atom present in a cubic closed pack structure is
(a) 1
(b) 2
(c) 3
(d) 4

I answered with (d), but the key on internet says it is (a). Please clarify the dispute.

Finally got it, it was a misunderstanding. I ignored to read the word ‘per atom’ in a hurry. So clearly the answer is 1.


Why does transition state loses its ability to vibrate? (Transition State Theory) In transition state theory, (according to my textbook) it is assumed that the the transition state loses its structure, and the ability to vibrate and rotate. If the transition state cannot vibrate, then my questions are

how does the previous bonds break and the newer bonds form?
Isn't the instablity and high potential energy of transition state
due to the  extra vibration of the atoms in the molecules?

Please guide me through this.

my book states:

  "The transition state does not represent a real molecule, it is impossible to isolate a transition state. However it is assumed to possess properties common to real molecule, such as molecular weight, intermediate distances, a definite enthalpy, definite composition but loses its structure, and the ability to rotate and vibrate. The transition state may either return to the initial reactant or it proceeds to form products."

Only some of the statements you quote are true, certainly a transition state cannot be isolated  since it lasts for less than a picosecond. In fact there is hardly any direct measurement of transition states, such as a spectrum or time profile or any structure. It is not that they are too fleeting to measure but that they occur very infrequently in any reaction so their population is always vanishingly small. This is a real experimental challenge.

What we do know is that the transition state is just like any other molecule with bonds vibrating/rotating  normally except for the bond(s) that are involved in reaction. They stretch and break then form new ones as the transition state is crossed. This pathway is the lowest point on the trajectory from reactants to products and is usually drawn at the top of the reaction energy profile as a 'saddle point'. (When at the saddle point looking at right angles to the reaction path the energy increases, looking forwards or backwards along the reaction path the energy decreases).  As the new bond has lots of energy it is initially vibrationally hot and rapidly loses energy either to solvent or into other vibrational modes in the product molecule or both.

In a complex molecule you can imagine the reaction starting as the species approach one another. The energy starts to increase as the atoms that are going to react  approach one another, but collisions with solvent can randomly knock them off course, or further on course for a while, or vibrations in the molecule may randomly add or remove some energy and so hinder or accelerate the reaction. One can then imagine the species  behaving somewhat as if they are diffusing along the reaction path, sometimes going forwards but sometimes being hindered. At any point, including the saddle point, they can diffuse around a bit and either return to reactants or proceed to products.

Even in simple atom-diatom reactions in the gas phase, such as $\ce{H + Cl_2}$ vibrations in the diatomic molecule  can inhibit reaction even when there is enough total energy to react; sometimes the energy is just in the wrong place and cannot aid the reaction. This can be the case because the diatomic is both rotating and vibrating and reaction will depend on the angle which it has to the line of approach ( atom to diatom) at the separation at which reaction starts.


Why don't electrons pair in [Fe(NH3)6]2+? Why doesn't electron pairing take place in $\ce{[Fe(NH3)6]^2+}$ though $\ce{NH3}$ is a strong ligand field producer?

(If you know the basics, go to last paragraph and skip the introduction part)

There are two types of ligands. Strong field ligands and weak field ligands. Those which only cause a small extent of crystal field splitting is called weak field ligand and those which cause large splitting is called strong field ligand.

Why does crystal field splitting take place?

Let's take an octahedral complex. The $d_{x^2-y^2}$ and $d_{z^2}$ point along x,y,z axis and the $t_{2g}$ orbitals point between the axis. When the 6 ligands approach along the three axes, the energy of the $e_g$ orbitals increase more than the increase in the energy of $t_{2g}$ orbitals. Hence the d-orbital splits into two. Filling each electron in $t_{2g}$ orbitals decrease energy by $0.4\Delta_0$ and filling electrons in $e_g$ increase energy by $0.6\Delta_0$

Coming back to your question, you know how electrons are filled in the orbitals (according to hund's rule of course). But how does electron pairing take place in your case?

Let the energy required to pair an electron be P.
Let's say there are 4 electrons. The two arrangements are one electron in each sub-orbital (no pairing) and 4 electrons all in $t_{2g}$ itself (instead of 4th electron going to $e_g$, its paired up in $t_{2g}$).
In the first case, the total energy is $-0.6\Delta_0$ and in second case, its $-1.6\Delta_0+P$. So, only strong field ligands can form the second arrangement. $P$ value is constant for a given metal ion.

Finally, ammonia is weak field ligand but in certain cases, it acts as a strong field ligand (example: for Cobalt). For Fe, ammonia is a weak field ligand. Hence no pairing takes place.


Reactions which confirms the presence of cyclic structure of glucose Are there any reactions that can be explained only on the basis of the cyclic structure of glucose rather than the open chain thus confirming its presence in the solution? Like a reaction that can only be given by hemiacetals thus confirming its presence and hence confirming the existence of the cyclic structure.

Any help would be appreciated.

Yes, such a reaction exists. In fact, it can not only confirm the presence of ring forms in solution but also how prevalent they are.

In 1962, Walker Jr, Gee and McCready published Complete Methylation of Reducing Carbohydrates, in which they subject solutions of sugars in dry DMF to silver(I) oxide  and methyl iodide. The reaction proceeds according to scheme 1 below.

Scheme 1: Methylation of glucose with $\ce{Ag2O}$ and $\ce{MeI}$ as published by Walker Jr, Gee and McCready.

Before reporting on the products, they note:

  It was somewhat unexpected that methylation of
  aldoses would proceed so smoothly in the presence
  of silver oxide. Glucose, shaken with silver oxide in
  DMF at room temperature, is partially decomposed
  in several hours.

Usually, silver(I) salts in solution are used in the Tollens test in which reducing sugars such as glucose reduce the silver ions to give elemental silver which forms a silver film on the inside of the glass. The presence of methyl iodide completely suppresses this reaction as they state in the next sentence:

  When methyl iodide is present
  in the reaction mixture, however, no silver mirror
  (indicating oxidation-reduction) is formed

As for the two step process that I have indicated, the authors state:

  In this reaction methylation
  of hydroxyl of the hemi-acetal group of glucose is
  essentially complete in one hour (as shown
  by loss of reducing function), but complete
  etherification of the remaining hydroxyls requires
  considerably longer.

With the experimental section detailing that the reactions were run ‘at least $\pu{16h}$’ which permits drawing the mechanism as a two-stage process as 1 can be isolated if the reaction is cut short.

Mechanistically, this reaction process is different from an acid-catalysed acetalisation with methanol. In the latter, an entire $\ce{OMe}$ group is attached to the sugar and the reactive species is an aldehyde form. This could be the open-chain aldehyde form (requires an additional reaction step) but most likely occurs via a cyclic aldehyde as depicted below. When using $\ce{Ag2O/MeI}$, the acetal oxygen is supplied by glucose as methyl iodide does not have any available oxygen. The two competing mechanisms are shown in scheme 2.

Scheme 2: Mechanisms of the acetalisation of glucose. (a) $\ce{H+/MeOH}$ and cyclic aldehyde intermediate; (b) $\ce{H+/MeOH}$ and chain-form aldehyde intermediate; (c) $\ce{Ag2O/MeI}$ single-step mechanism.

As these mechanisms show, acid-catalysed acetalisation need not start from the cyclic form to get acetal 1; it is possible to draw the reaction mechanism as originating from the chain form of glucose. But methylation with $\ce{MeI}$ requires the oxygen to be in place as an $\ce{OH}$ function that is deprotonated and attacks $\ce{MeI}$ which is not possible if the chain form were present in solution.

As to why the $\ce{C{1}}$ oxygen is methylated first, and the explanation is the higher acidity of an acetal hydroxy function when compared to an alcoholic one as the authors state:

  This difference in rate is expected
  from the difference in acidity (and ease of
  anion formation) of the hydrogens on the two different
  types of hydroxyl groups.

Furthermore, it bears stating that this reaction not only proves cyclic forms are present but also that they are prevalent. Methylation of the $\ce{C{1}}$ oxygen is rapid compared to all the other hydroxy groups meaning that they should be present in at least appreaciable ratios. Furthermore as mentioned above, under these conditions the reduction of silver(I) to a silver mirror – which requires the open-chain aldehyde form mechanistically – is practically entirely suppressed. Given that the Tollens test is a standardised method to detect glucose, the complete absence of all traces of silver film serves as a good indicator of the low concentration of open-chain aldehyde in solution.

The original paper is: H. G. Walker Jr, M. Gee, R. M. McCready, J. Org. Chem. 1962, 27, 1000–1002. DOI: 10.1021/jo01053a047.


What is the CNC bond angle in methyl isothiocyanate? According to VSEPR theory, for $\mathrm{sp^2}$ hybridisation the bond angle is approximately $120^\circ$. In $\ce{H3CNCS}$, the nitrogen has one lone pair. As lone-pair-bond-pair repulsion is greater than bond-pair-bond-pair repulsion, the bond angle of $\ce{CNC}$ should be less than $120^\circ$,  but it is actually greater. How can this be rationalised?

I have just answered What is the CNC bond angle in methyl isocyanate? Pretty much everything written there also applies to methyl isothiocyanate, just a bit more extreme.
Here is the structure calculated on the RI-BP86/def2-SVP level of theory:

The bond angle is $\angle(\ce{CNC}) = 156^\circ$, so we expect a much larger contribution of the right side in the following resonance description:
$$\ce{H3C-N=C=S <-> H3C-N^+\bond{3}C-S^-}$$
Reasons for this are not trivial, but that would exceed the scope of this answer.
TL;DR From VSEPR we would expect a trigonal coordination and a $120^\circ$ angle. Refining with resonance, we'll expect a widening of the bond angle. Everything beyond that  needs much more involved models.


Electronic configuration of chromium(II) Which is the electron configuration for $\ce{Cr^2+}$

  (A) $\mathrm{[Ar]4s^23d^4}$
  (B) $\mathrm{[Ar]4s^23d^2}$
  (C) $\mathrm{[Ar]4s^03d^4}$
  (D) $\mathrm{[Ar]4s^43d^2}$

I referred to the periodic table and got the orbital notation for $\ce{Cr}$ as $$\mathrm{[Ar]4s^23d^4}\tag1$$
So $\ce{Cr^2+}$ would be $\mathrm{[Ar]4s^23d^2}$, or (B). But the teacher counted that wrong and said it was actually (C).

  Question: Shouldn't the answer be (B)? Why does $\ce{Cr}$ lose the $\mathrm{s}$ shell first?

Chromium’s standard electron configuration as given by the aufbau principle is actually more of an exception than you may think initially. See for example this question to catch a small glimpse of the truth behind the simplifications.

The aufbau principle tells you to fill in the order $\ce{1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, }\dots$ That order, however, is only valid for neutral atoms (and not always as the linked question implies). As soon as you add charge to an atom, the relative orbital energies shift. The $\ce{3d}$ orbitals, belonging to the M shell, are stabilised more than the $\ce{4s}$ orbital which belongs to the N shell. This is a general rule: charges typically stabilise orbitals belonging to lower shells more than those in higher shells.

Thus, while in neutral chromium the $\ce{4s}$ and the $\ce{3d}$ shell are very close in energy, this balance completely tips towards $\ce{3d}$ (which then has a lower energy) upon oxidation. Therefore, one should always assume an empty $\ce{4s}$ orbital when discussing the electronic configuration of charged d-block elements.


How does PEDOT:PSS work (mechanism) in Organic Electrochemical Transistors? I have a question about the fairly fundamental theory behind the application of PEDOT:PSS in an organic electrochemical transistor (OECT) device.

As at least I understand it, the polyethylenedioxythiophene (PEDOT) chain is rich with holes and then is doped with polystyrene sulfonate (PSS) which has sulfonate anions that compensate these holes. Now according to the literature I have read, PEDOT:PSS is conductive (p-type) in its initial state (i.e. no gate voltage is applied). This confuses me a bit because I would have thought compensating these holes by adding PSS would have made the PEDOT less conductive.

Then of course when you apply a gate voltage, cations compensate the PSS anions instead which according to literature makes the PEDOT:PSS less conductive, but I would have thought this would make it more conductive as it liberates the holes in the PEDOT backbone.

What do I seem to be missing here?

This is about the use of well known words in semiconductor physics in the relatively new context of organic electronics. This is not always strictly justified, and at least confusing in many occurrences.

I will try not to slip in any deep debate about communities. Just let me say that obviously general physics applies both in classical semiconductors and organic semiconductors*, in particular when the output of a device - rather than its microscopic structure - is considered. It is fine to me to use whatever approach and treat these systems up bottom (semiconductors people do that) or vice versa, but keeping in mind that the constituents are covalent molecules.

Reality is in the middle, and organic semiconductors are likely to be compared to amorphous inorganic semiconductors even in the case of the firsts having rather crystal like structures. Molecules will be always therein, and a kind of hopping processes will be more or less crucial to conduction anyway.

PEDOT:PSS is in fact an ionic compound, a "poly(salt)", let me say. I want to say more than general ionomer, I do NOT want to propose a new name. It will be clear as you read. Be careful of italic.

For those not familiar with the material here is the link to it in Wikipedia: https://en.m.wikipedia.org/wiki/PEDOT:PSS

I would have placed a posive charge already in the structure, sketching a bit more than one EDOT ring.

Let's start with a neutral PEDOT chain itself. It is poly(ethylene-dioxythiophene), a conjugated polymer with quite low energy gap and low oxidation potential.

No drastic oxidative process, and even photoinduced electron transfer process (transfer to whatever available e-acceptor within the sample, purposefully added or not) lead to relatively stable radical-cations delocalised within the pi system along the polymer chain (there is an adapted or adopted name for these too, polarons).

If the anionic species formed are relatively stable too, then long living charges are formed.

In presence of very stable ionic species, either added or formed by the same process, a polimeric salt is obtained. In this polimeric salt charges of one sign are in chain, not in pending functional groups which is a much trivial case for which coincidentally polystyrene sulphonate PSS is an example.

PEDOT:PSS is right  one example of the polimeric salts that I have just introduced. The double dot here means a certain ratio depending on formulation, as the PPS providing the counterions (neutrality) can be in excess. The degree of oxidation can also varies depending on formulation.

You said " As at least I understand it, the PEDOT chain is rich with holes and then is doped with PSS which has sulfonate anions that compensate these holes."

Is the use of "doped", or alternatively "then", that confuses you.

The oxidized PEDOT chain is rich with holes in the sense that is rich of positive radical-cations and they behave as the carriers. However the chain is far to be neutral, is effectively positively charged, as you know.

Actually there are positive charges (positive carriers) by virtue of the presence of the external compensating PSS counterions. The latters are strictly necessary for the so called holes to exist within the PEDOT chains.
This is not the case for inorganic semiconductors, e.g. doped silicon is inherently neutral.

PSS stabilize the charges that in its absence will be either transient or prone to further chemical reactions as for radicals are generally very reactive.

I think this description should help you in getting the subtleties of this field and those of its dictionaries.

I could have simply written that

in conjugated pi system the so called doping is a redox process, transferring electron between the system and a second material

(generally from - to respectively, with notable exceptions of which fullerenes are probably the best known examples).

The carrier in oxidized PEDOT are positive and their motion is the opposite motion of electrons residing on the highest singly occupied MO, that ideally spread all along a perfect chain. In reality along a chain their will be a distribution of segments with different effective conjugated length, each with its corresponding  singly occupied orbital.
Positive charges move i.e. electrons move between these in chain or inter-chain SOMOs, which can be regarded as a band. In such a sense these carriers are called holes.  They move by a deficit of electrons in the normally occupied levels.
As for the convenience of calling  these carriers holes or not is up to you.
The point is that PEDIT:PSS can be regarded as a whole neutral material with mobile positive charges.

Then you said " Then of course when you apply a gate voltage, cations compensate the PSS anions instead which according to literature makes the PEDOT:PSS less conductive, but I would have thought this would make it more conductive as it liberates the holes in the PEDOT backbone. What do I seem to be missing here?".

I do not know what device structure you refer to. The conductivity of PEDOT:PSS is high enough to shorts two metals or doped Si - metal contacts, and to work as antistatic layer.

I believe that the behaviour of your diode will be dictated mostly by a second material with conductivity/mobility compatible to a transistor function.

PEDOT:PSS is a conductor tout court (between metals) or is in general applied*** as a selective contact when its SOMO levels mentioned above match (or mismatch) the relevant conduction levels of other materials.

In a sense the term semiconductor should be reserved to neutral PEDOT (although its bulk conductivity will be rather that of an insulator) while PEDOT:PSS has metal like bulk conductivity but a hole** type conduction mechanism as described above.

*it is clear what one mean saying "organic semiconductors", but I would prefer "extended pi conjugated systems". The same for "doped", I would say "oxidised (or reduced)". And so on. And not, there are not dampling bonds in molecules, even when it happens they are organic semiconductors.

**this should clarify that sometimes an adopted nomenclature is the first solution at hand ;)

***PEDOT:PSS is also reported as smoothing, adhesion promoter, etc all uses not pertinent to this Q.

EDIT: I realised you refer specifically to electrochemical transistors. Well, the above is still answering. Indeed, upon negative bias electrolyte cations infiltrate the material thereby neutralising the PSS anions. For the above discussion this reduces back the PEDOT, diminishing the available amount of carriers and as such the drain current.