What causes transition from smoldering combustion to flame combustion? What causes the transition from smoldering combustion to flame combustion?

From what I have read about it is that, smoldering combustion gives off a lot of flammable gases, primarily, Carbon Monoxide, however, the lower flammability of carbon monoxide is 12.5% and it takes about 2-3 hours to reach a concentration of 1%, it is lethal to humans in concentration of 1%.

There have been reported experiment studies, where smoldering combustion transitioned into flame combustion, as quickly as 22 minutes. So, there must be some other cause. Carbon Monoxide is, I guess not the reason.

Burning in open air isn't a steady state process, but a very dynamic one. Likely when the fuel is smouldering there isn't enough of a convection current in the air to bring in fresh oxygen. As the smouldering continues something about the geometry of the area smouldering changes and fresh oxygen is brought in by a strong convection current. At that point the fuel bursts into open flame.


Can natural gas appliances run on Hydrogen? Can natural gas appliances run on Hydrogen? If so, does the appliance have to be modified and how significant of a modification would it need?

Thanks

Pure hydrogen, as opposed to hydrocarbons, degrades metals.  $\ce {H_2}$ molecules are tiny, so they wedge themselves into the relatively large metallic crystal structures and form hydrides.  This is the entire reason nickel-metal hydride (NiMH) batteries work - the hydrogen is stored in an iron-hydride matrix. An appliance that might last 20 or 30 years burning natural gas might not make it past ten years burning $\ce{H_2}$, and a failure may take the form of a dangerous leak.


Tools for Molecular Simulation I am a high school student and I don't know a whole lot about molecular simulation. I wish to view different conformers of cyclohexane and the changes that occur when substituents are attached (for example cis-1,4-di-tertbutylcyclohexane attains a twist boat shape) through molecular simulation.

The only easy-to-use tool I could find was MolView, but I can't see the above molecule getting into a twist boat shape when I ran an energy minimization because it swapped the cis to a trans form (which is in a chair conformation). Couldn't find the molecule through PubChem as well, so that I could import it.

I would also like to try out ortho effect on substituted benzene rings. Basically, I want to see a few simple molecules, (not anything like proteins and other macromolecules) and run an energy minimization on them.

So, is there any easy-to-use, free molecular energy minimization simulation program that can help me achieve this?

A very simple way is using molecular mechanics: it only applies to very simple molecules, but it should be enough for your purpose. Avogadro is a free software which allows you to build structures and to minimize them via molecular mechanics, through a variety of "methods" (force fields).
The software outputs an estimate of the energy of your geometry, which you can compare to that if other geometries (for instance, boat vs chair cyclohexane).

Note that these results are very approximate, the methods work properly only in the case of neutral and covalent molecules, but anyway they provide good results for a qualitative analysis.

More accurate methods, capable of working with more complex molecules (in order of accuracy: semiempirical methods, like PM3, PM6, AM1... << Ab initio methods like HF and DFT methods like B3LYP, M06, PBE...) exist in other software packages, but this is a good starting point for learning.

Molecular mechanics works by treating angles, bonds, etc. as subject to forces (Imagine them as springs) which constrain them to an "optimal" conformation, as seen in sets of typical and simple molecules.

The other methods work by solving quantum-mechanical equations, with or without some approximations, and thus can treat "atypical" geometries which don't work with molecular mechanics.

To make a example, molecular mechanics works well for whole sets of typical molecules like our amino acids, but fails to predict the geometry of strange molecules, like tetrahedrane, which shows very atypical angles.

Within these limits, you can use molecular mechanics for free and with not much computational and software hassle


Why H2O2's molecular geometry is what it is...? I came up with this question when I was trying to determine whether H2O2 is polar or not.

H2O2 looks like this when looked parallel to oxygen-oxygen bound:(source: Pubchem)

Each hydrogen is in one of the three positions available around each oxygen(the two nonbonding electron pairs occupy the other two); even the oxygen-oxygen bound can rotate to set hydrogens in any position relative to each other.  Therefore I don't see the problem in following nonpolar structure(pretty neat!)

What concept am I missing while determining molecular geometry?

Is a H2O2 molecule sometimes polar and sometimes nonpolar?

Thanks for your help!

As stated in the link provided by @Shadock, hydrogen peroxide, like many more other molecules, is subject to hindered internal rotation. When you rotate the two OH groups with respect to each other around the O-O bond, this requires a certain energy depending on the relative angle. In the eclipsed configuration (OH groups pointing in the same direction), it's easy to see that there must be a maximum in the potentiall due to the Paulli repulsion. In the anti-eclipsed configuration (both OH groups pointing in opposite directions) there is another maximum as you anti align the two dipoles of the groups. In between these maxima, you'll find a minimum and the potential can be expanded as a Fourier series in the relative bond angle

$$V(\gamma)=\frac{V_2}{2}\cos(2\gamma)+\frac{V_4}{2}\cos(4\gamma)+\ldots$$

where $V_2$ and $V_4$ are related to the heights of the two barriers. If the barriers were infinitely high, the OH groups would act as a quantum harmonic oscillator in one of the potential minima. On the other hand, if the barrier were very small, the OH groups would rotate freely around each other. In the case of hydrogen peroxide, the eclipsed barrier is so high that the OH groups can hardly go over/through it. The lower barrier is a different story. It is high enough to support a few vibrational levels, but not high enough to prevent the OH groups from tunnelling through the barrier and as a consequence, these energy levels are split in two. (In fact, these two levels are again split by the tunnelling through the higher barrier, but since this is so high the splitting is very small). Mathematically, the Schroedinger equation for this periodic potential is equivalent to the Mathieu equation (or more general Hill's equation).

In the case of H$_2$O$_2$ the tunnelling is so fast, that the molecule vibrates around the lower barrier with a period of ~100 fs. As a consequence, to determine the polarity, you have to average over this motion, resulting in a net polarity.

Note that, if the barrier would be so high that tunnelling would be sufficiently slow, H$_2$O$_2$ would be a chiral molecule!


Use of aqueous KOH and alcoholic KOH in dehydrohalogenation reactions What is the difference between aqueous KOH and alcoholic KOH? How do they react differently in dehydrohalogenation?

Aqueous $\ce{KOH}$ is alkaline in nature i.e. it dissociates to produce a hydroxide ion. These hydroxide ions act as a strong nucleophile and replace the halogen atom in an alkyl halide.

$$\ce{RCl + KOH (aq) -> ROH + KCl}$$

This results in the formation of alcohol molecules and the reaction is known as nucleophilic substitution reaction.

Alcoholic, $\ce{KOH}$, specially in ethanol, produces $\ce{C2H5O-}$ ions. The $\ce{C2H5O-}$ ion is a stronger base than the $\ce{OH-}$ ion. Thus,the former abstracts the ß-hydrogen of an alkyl halide to produce alkenes. This reaction is known as elimination reaction.

$$\ce{CH3CH2Br + KOH (alc) -> H2C=CH2 + KBr + H2O}$$


Calculating the EMF of cell Find the EMF of the following cell :
  $$\ce{Pb(s)}, \ce{PbSO_4}|\ce{SO_4^{2-}}(\pu{0.100M})||\ce{Pb^{2+}}(\pu{0.004M})|\ce{Pb(s)}$$
  Given:  $E^0_{\ce{PbSO_4|Pb,SO_4^{2-}}}=\pu{-0.359V}$
  and $E^0_{\ce{Pb^{2+}|Pb}}=\pu{-0.126V}$

I first found the $\ce{Pb^{2+}}$ concentration in the oxidation half cell using the sulphate ion concentration and the solubility product of lead sulpahte ($2.53\times10^{-8}$) and found the concentration to be $(2.53\times10^{-7})$
And accordingly $E^0_{\text{cell}}=0.359-0.126=0.233$
Then using the Nernst equation,
$$E_{\text{cell}}=E^0_{\text{cell}}-\frac{RT}{nF}\ln\frac{[\text{Products}]}{[\text{Reactants}]}$$
And substituting,
$n=2$,
$[\text{Products}]=2.53\times10^{-7}$,
$[\text{Reactants}]=0.004$,
$T=\pu{298K}$,
$R=\pu{8.314JK^{-1}mol^{-1}}$, and $F=\pu{96500C}$. Hence, I got $E_{\text{cell}}=\pu{0.357V}$

Whereas, the answer given in my book is $E_{\text{cell}}=\pu{0.133V}$.

So, is my answer correct or have i  misunderstood something?

Use RT/F = 0.059 or 0.06
It will simplify your calculations.

Note:- I have converted ln into log.
So formula becomes:-
$$E_{\text{cell}}=E^0_{\text{cell}}-\frac{0.059}{n}\log\frac{[\text{Products}]}{[\text{Reactants}]}$$

Just try this in problems. Calculations would become easier.


Suggestion for a Synthesis of Paracetamol — plausible? I have seen a number of different plausible syntheses for paracetamol but they suffer from problems such poor regioselectivity of aromatic substitution (usually when nitrating phenol getting a mixture of undesired ortho and desired para). However, is the following not a plausible synthesis?

Take aniline (I believe it's commercially available but can be synthesised from mononitration then reduction).
React with acetic anhydride to make the amide — still ortho-, para-directing and mildly activating and the directing effects are not acid sensitive as in aniline.
Do a Friedel–Crafts acylation with the an acyl chloride.
React with hydrogen peroxide and sodium hydroxide to get paracetamol via a Dakin reaction.

Of course the acylation will suffer a little with regioselectivity issues but I feel like it will suffer less as the amide group is larger than a hydroxyl so favours para on steric grounds more strongly and is thus a better solution than nitration of phenol and subsequent reduction.

Nitration of phenol then reduction and reaction with acetic anhydride is the most common hit on google so I assume it's the most common method. But what's wrong with what I have suggested?

Your route sounds possible, yes. (Aniline is commercially available.) There would very likely be side reactions of the amide under Baeyer-Villiger conditions but nothing exceptionally out of control—except you lose a little bit of yield.

There seems to have been a little race concerning the shortest and most environmentally favourable route to paracetamol. To the best of my knowledge, and assuming that their NMR spectra are unaltered and directly from the reaction published, Joncour et al. have the shortest route in 1 (one) step starting from (also commercially available) hydroquinone and utilising ammonium acetate in acetic acid.[1]

However, for the industry to decide to use a non-established method, that new method must offer a significant advantage over what is presently used. ‘Significant advantage’ typically means much safer or much cheaper. The industrial paracetamol synthesis as employed by Hoechst–Celanese starts off from phenol (extremely inexpesive), does a Friedel–Crafts type acylation with acetic anhydride and $\ce{HF}$ (cheap materials), transforms the ketone into a ketoxime (cheap $\ce{NH2OH}$) and finally uses catalytic acids ($\ce{CF3COOH, SOCl2}$; the latter being rather cheap) to induce a Beckmann rearrangement.

Viewed zealously, this is technically already a more optimised synthesis than the one you propose—it is a step shorter and uses even cheaper reagents. There is probably not much room left for optimisation along these traditional lines.

Referene:

[1]: R. Joncour, N. Duguet, E. Métay, A. Ferreira, M. Lemaire, Green Chem. 2014, 16, 2997. DOI: 10.1039/C4GC00166D. The openly accessable supplementary information with experimental details and the NMR spectrum is available on the same page.


Presence of precipitates Does a cloudy solution indicate presence of precipitates or not?The solution isn't completely opaque neither it is completely clear.

Cloudiness indicates light scattering by small particles, with size on the order of the wavelength of light and larger. If those small particles are denser than the liquid, they may eventually sink and become the settled precipitate.

However, some small particles are so small and so nearly equivalent in density to the liquid that they do not sink, but remain suspended. Brownian motion helps. Some clays can remain suspended indefinitely; in water treatment plants, FeCl3 is used as a flocculant to bring such turbidity down.

Sometimes intermolecular interactions keep particles suspended. For instance, many surfactants form micelles comparable in size to the wavelength of light and may, in pure form, be considerably more, or less, dense than the liquid, and yet remain suspended indefinitely because they are at least partially hydrophillic.

Centrifugation may help bring down suspended particles (or float them if they are lighter than water).


What is the constant of integration in a series reaction? Consider the series reaction

$$\ce{A ->[k_1] B ->[k_2] C}$$

Then

$$\frac{\mathrm d}{\mathrm d t} [\ce{B}] = k_1 [\ce{A}] - k_2 [\ce{B}]$$

We already know that

$$[\ce{A}]=[\ce{A}]_{0} \exp(-k_1 t)$$

Thus far, I have used the method integrating factors, with $u=\exp(k_2t)$, to arrive at the equation:

\begin{equation}
[\ce{B}]= \frac{k_1}{k_2-k_1} \left( \exp(-k_1 t) + C \exp(-k_2 t) \right) [\ce{A}]_{0}
\end{equation}

What would the constant of integration be in this situation?

There might or there might not be a problem with your final equation. Either you have made a mistake while using the integrating factor to solve your differential equation or you made some further substitutions so that your constant $C$ contains something more than just the "standard stuff" I would have expected from the method. Thus it's best I go through the derivation from the beginning so that it'll become clear what I mean.

The method of solving differential equation using an integrating factor is explained well here. You have used the correct integrating factor $u(t)$. For the differential equation at hand, i.e.

\begin{align}
  \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} &= k_1[\ce{A}] - k_2[\ce{B}] \\
  \frac{\mathrm{d}[\ce{B}]}{\mathrm{d}t} + k_2[\ce{B}] &= k_1 [\ce{A}]_{0} \mathrm{e}^{-k_1 t} \ ,
\end{align}

it is indeed

\begin{align}
  u(t) = \exp\biggl( \int_{0}^{t} k_2 \, \mathrm{d} t \biggr) = \mathrm{e}^{k_2 t} \ .
\end{align}

With the integrating factor you can get $[\ce{B}]$ according to

\begin{align}
  u(t) [\ce{B}] &= \int_{0}^{t} u(t) k_1 [\ce{A}]_{0} \mathrm{e}^{-k_1 t} \, \mathrm{d} t + \tilde{C} \\
  [\ce{B}] \mathrm{e}^{k_2 t} &= \int_{0}^{t} k_1 [\ce{A}]_{0} \mathrm{e}^{(k_2 - k_1) t} \, \mathrm{d} t + \tilde{C} \\
  &= k_1 [\ce{A}]_{0} \biggl[\frac{1}{k_2 - k_1} \mathrm{e}^{(k_2 - k_1) t} \biggr]_{0}^{t} + \tilde{C} \\
  &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{(k_2 - k_1) t} - 1 \bigr) + \tilde{C}
\end{align}

Rearranging for $[\ce{B}]$ gives

\begin{align}
  [\ce{B}] &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \underbrace{\frac{ \mathrm{e}^{(k_2 - k_1) t} - 1 }{\mathrm{e}^{k_2 t}}}_{= \, \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t}}  + \frac{\tilde{C}}{\mathrm{e}^{k_2 t}} \\
 &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t} \bigr) + \tilde{C} \mathrm{e}^{-k_2 t} \ .
\end{align}

This equation differs from yours a little if your integration constant $C$ would be the same as my $\tilde{C}$. But if you used some substitutions so that your integration constant is $C = \frac{k_2 - k_1}{k_1 [\ce{A}]_{0}}  \tilde{C} - 1$ then your solution and mine would be equivalent. If you indeed made a mistake then I hope my derivation cleared things up a bit and if you had intended the equation to look that way by introducing the substitution then at least you know what integration constant I mean when I am going to explain the method for determining its value in the following.

The integration constant can now be determined by enforcing boundary conditions. In this particular case this means that at the start of your reaction ($t=0$) you require that $[\ce{B}]$ is equal to the initial concentration of your reactant $\ce{B}$, i.e. $[\ce{B}](t\! = \! 0) = [\ce{B}]_0$. So, plugging $t=0$ and $[\ce{B}](t\! = \! 0) = [\ce{B}]_0$ into the equation for $[\ce{B}]$ you get

\begin{align}
  [\ce{B}]_0 &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{0} - \mathrm{e}^{0} \bigr) + \tilde{C} \mathrm{e}^{0} \\
  [\ce{B}]_0 &= \tilde{C} \ .
\end{align}

In the case of your reaction $\ce{A -> B -> C}$ the initial concentration of $[\ce{B}]$ will usually be $0$ as $\ce{B}$ is only an intermediate product, so $\tilde{C} = [\ce{B}]_0 = 0$ and the final equation is

\begin{align}
  [\ce{B}] &= \frac{k_1}{k_2 - k_1} [\ce{A}]_{0} \bigl( \mathrm{e}^{-k_1 t} - \mathrm{e}^{-k_2 t} \bigr) \ .
\end{align}


Can the "rate coefficient" of a chemical reaction really be called a "rate constant"? Can the rate coefficient of a reaction ever be referred to as a constant? To my limited knowledge, it cannot, because it is dependent on factors such as temperature (I'm getting that info from the Arrhenius Equation).

I'm looking for clarification here.

Yes the rate coefficient can be referred to as a "rate constant".
This is in fact a very, very common name for the rate coefficient.  Encyclopedia Brittanica has an entry for rate constant.  Wikipedia has a reaction rate constant page.  Google's N-gram viewer shows that "rate constant" is used more often in the English written corpus by a factor of more than five, relative to "rate coefficient".
The "constant" in rate constant refers to the fact that it does not vary with concentration of the reactants or products.  However, as you note, it could depend on other variables, such as temperature, pressure, catalyst, light, etc.


If the half-life of an isotope exceeds the age of the Universe, then how is it measured? According to this Wikipedia article, the half-life of Bismuth-209 is 19 billion billion years, which exceeds the age of the Universe by factor on the order of a billion. How is the half-life of an element measured and experimentally verified if the half-life exceeds the age of the Universe?

Well, half-life describes the time after which half of the substance has decayed.

This is all probability and statistics. If you look at a single atom you cannot make any prediction when it will decay. It might be within the next 3 seconds or it might be here even after hundreds of billion years. The probability that it decays during the half-life is 50%, so if we look at a very large number of such atoms we can use statistics to predict how much substance will actually decay in a given time.

Now bismuth-209 has $1.9×10^{19}$ years half-life, So if we have a lot of Bismuth-209 after that time we will only have half of it. But that doesn't mean we don't have any decay right now. 209 grams of Bismuth-209 are 1 mole so $6.022×10^{23}$ particles, a lot of those. Now the decay goes down exponentially but if we just calculate the average decay rate during the first half-life if we start off with 1 mole and we keep in mind that the rate will be much faster in the beginning we see that we have an average of around 16000 decays every year, which means around 44 a day. Again, in the beginning it will be more, approaching the time of half-life it will be less.

Now how do we know the half-life? It's simple in theory, some french researcher put 31 grams bismuth-209 in a box with very sensitive equipment and measured the alpha particles which are send out during decay. This isn't a trivial task to do, but it's possible. They counted 128 particles over 5 days and then did the math to calculate the half-life.

The original paper can be found here.


For a given element do heavier isotopes form stronger bonds than lighter isotopes? I know that $\ce{C-D}$ and $\ce{C-T}$ bonds are stronger than $\ce{C-H}$ bonds, but is this generally true for isotopes of all elements? (By isotopes I mean stable ones with large half lives.)

Isovalent isotopes will have the same force constant. However the different masses of the isotope will affect the position of the vibrational state in its potential well. You can rationalise the difference in well depth briefly using the vibrational frequencies of a classical oscillator as the harmonic approximation to the asymmetric well for low lying states. If the stretching frequency is given by;
\begin{equation}
\nu \propto \frac{1}{\sqrt{\mu}}, \ \ \ \ \ \ \ \ where \ \ \ \ \ \ \ \mu =\frac{m_1\cdot m_2}{m_1+m_2}
\end{equation}
Then as the reduced mass increases the frequency decreases and the vibrational energy decreases $E=\hbar\omega $ with  $\omega\cdot 2\pi=\nu$. We can see that if we were to double the reduced mass then the energy would decrease by root 2.

So we need to examine the reduced masses of some simple systems, lets say the diatomics you suggest. If we let the masses of $H=1$, $D=2$, and $T=3$ then we calculate the reduced masses to be $1/2$, $2/3$ and $3/4$. The reduced mass is increasing as we go along the series and hence we expect the successive molecules to have stationary states lower in the well.

In vibrational spectroscopy this is best seen by the decrease in zero point energies for HCl and DCl. It is often easiest to take the ratios of the frequencies $\omega _{HCl}/\omega _{DCl}$, where using $m_{Cl}=35$ and H and D as above we obtain 1.395. indicating that the deuterated molecule is deeper in its well.

Edit

I have been advised to explain zero point energy. In short it is a irremovable quantity of energy from a quantum oscillator. The classical oscillator has no parallel to this. It is a manifestation of the uncertainty principle in position and momentum. If you want the truth of where it comes from you will need to consult the physics SE, but I will do my best :)

The quantum oscillator has a Hamiltonian of the classical oscillator except just replace the classical position and momentum variables with their quantum operator analogues $p\mapsto \hat p$ and $x\mapsto \hat x$.

The classical Hamiltonian (or total energy) is learnt at school to be kinetic + potential,
\begin{equation}
H=\frac{p^2}{2\mu}+\frac 12 kx^2
\end{equation}
For the reduced two body system. It is a postulate of quantum mechanics that in the position representation (just a way of formulating things) that wherever you see a $p$ and an $x$ you replace them by $\hat p$ and $\hat x$ where,
\begin{equation}
\hat p =-i\hbar \frac{\partial }{\partial x}, \ \ \ \ \ \ \ \ and\ \ \ \ \ \ \ \hat x =\cdot x
\end{equation}
Therefore with these changes in the above classical expression we generate the quantum Hamiltonian of the quantum harmonic oscillator,
\begin{equation}
\hat H=\frac{\hbar^2}{2\mu}\frac{\partial ^2}{\partial x^2}+\frac 12 kx^2
\end{equation}
Where $i=\sqrt -1$ so $i^2=-1$ and $--1=1$.

It is another principle of quantum mechanics that the Schrodinger equation is the equation to solve for quantum systems, just like in classical mechanics you use Newton's second law, $F=ma$, This equation has a bit of a different shape but in essence (decoupled from time) takes the following form.
\begin{equation}
\hat H\psi =E\psi
\end{equation}
Substituting the above expression for $\hat H$ in and doing a bit of maths (wayyyy to advanced for us here) we can solve for the energy $E$ in the Schrodinger equation, when we do this we get,

\begin{equation}
E_n=(n+\frac 12)\hbar \omega
\end{equation}
What this tells us is that the vibrational levels are not continuous like the classical oscillator (above) but can only take discrete values. We see in the maths of the solution for $E$ the need for discrete integers $n$ which we call quantum numbers. These may take in this case values of $0,1,2,...$ etc etc. Each corresponding to the next adjacent energy solution.

Something interesting happens if we choose $n=0$ however. You would think that the lowest energy would be zero right? well no... quantum mechanics says we can never remove a little bit of energy from the system. (try substituting $n=0$ in the above expression and we see that the answer is non zero (1/2 $\hbar \omega$).

This is a zero point energy.

As promised here is the home-made graph of the molecular potential energy. Im sorry its not very good but honestly the best I could do with the available tools! Apologies.

The main points to follow are:

At the base of the well the graph looks like a parabola, i.e. the harmonic oscillator approximation to the molecule works best at low lying states since the anharmonicity term grows with excitation (well increasing quantum number).
Where the harmonic oscillator is valid you can see that the energy levels of are evenly spaced, each one corresponds to the quantum numbers $n$ as given above. You can calculate $\Delta E=E_{n+1}-E_{n}$ to get the difference in energy as $\frac 12\hbar \omega$. In this approximation the equilibrium bond length does not change with excitation. However we can see it does in the anharmonic case.
To answer your question, the energy levels corresponding to the heavier isotope are the blue ones since they sit deeper in the well. So you could imagine the blue ones as C-D while the red ones would be C-H. (normally you would just be looking at one species per graph, so only one set of coloured lines.


Polycarbonate resistant to chemical Why polycarbonate couldn't stand some chemicals i.e. acid and base? What reaction will occur if polycarbonate meets acid or base?

Strong acids or bases will hydrolyze polycarbonate resin--specifically poly(bisphenol A carbonate)--at the ester linkages, first giving shorter polycarbonate chains, and eventually down to bisphenol A and carbonate ion or carbon dioxide.


Strength of hydrogen bonding in phenol or methanol I wanted to know whether hydrogen bonding is stronger in phenol or methanol.
I saw on Wikipedia about the enthalpies of hydrogen bonding in various cases.
but could not find the answer to this question of mine.

I also want to know whether the hydrogen bond strength depends upon the carbon atom to which oxygen is attached.

Let me answer the second question first, because the answer is so much easier: Of course the hydrogen bond strength depends among others upon the carbon oxygen is attached to. Consider phenol and methanol or methanol and formic acid.

The guess as to which hydrogen bond is stronger is — I have to admit it — my speculation. But I think that methanol should be able to undergo stronger hydrogen bonding.

Phenol, of course, is aromatic and capable of delocalising negative (or positive) charges across its entire ring system — the reason why its $\mathrm{p}K_{\mathrm{a}} = 10$, a good five units lower than methanol’s. That means that at the same pH value of the solution, there should be less strong a negative charge on phenol’s oxygen than on methanol’s.

But hydrogen bonds are, to a certain extent, electrostatic interaction. So the system which has a stronger negative charge on the oxygen (methanol) should be able to build up stronger hydrogen bonds by attracting hydrogen more strongly.

One could also picture it differently: Phenol gets deprotonated at a lower pH, so the $\ce{O-H}$-bond must be a lot weaker than methanol’s. But if the $\ce{O-H}$-bond is weak, the oxygen has a larger tendency to simply loose the hydrogen to another electronegative partner (deprotonation/protonation) rather than share it (hydrogen bond).


Cyanide ion non-bonding/lone pair? Where's the lone pair on the cyanide anion in this MO diagram?

In the traditional Lewis picture there are two lone pairs on the cyanide ion. In the MO diagram though there appear to be no non-bonding orbitals that are populated ...

Also how come there appear to be two sigma bonding orbitals populated when in $\ce{NC^-}$ there is only one sigma bond?

As Geoff already pointed out in the comments, the connection between a molecular orbital theory (MO) point of view and a Lewis picture (LT) is tenuous at best.

Maybe the best example for this is the mentioned $\ce{{}^{-}CN}$ and obviously the isoelectronic $\ce{CO}$. But there are many more molecules.

In the MO point of view the electron pair which can be donated to an acceptor orbital is of course the highest occupied molecular orbital (HOMO). Often this comes with the more trivial notation of "lone pair", based on the coefficients of a certain element in that orbital (percentage contribution). In the case of $\ce{CO}$ this becomes fairly obvious, but since this is not the question at hand I will only focus on the isoelectronic counterpart.

Please consider the above MO scheme only as a sketch, it is not to scale and contributions were estimated only be a quick browse of the output file. I also did not compensate the calculation for the negative charge, so even the bonding orbitals are antibonding with respect to fragmentation. But it shall still provide the point.
We can clearly see that the highest coefficient of the HOMO is at the carbon end, this is consistent with the fact, that it will most likely bond with this end to a more electropositive element. But this is, however, not a real lone pair since it is $\sigma$ bonding and also has quite a big coefficient at the nitrogen end. (In $\ce{CO}$ the difference is much bigger.)
Below that we see the two $\pi$ orbitals, followed again by two $\sigma$ orbitals. All five orbitals make up the bonding of the molecules and determine also its reactivity.

Now in the Lewis theory picture this is quite different. We traditionally describe it as a carbon oxygen $\sigma$ bond, two carbon oxygen $\pi$ bonds, a oxygen lone pair and a carbon lone pair. All made up from some kind of hybrid orbitals. In computational chemistry we can analyse the wave function and obtain localised orbitals. One of these methods is natural bond orbital theory (NBO). I have carried that out and sketched the common bonding picture along with it:

While this picture is also consistent with a valence bond theory (VB) approach, it explains bonding in a different way than MO theory. However, the conclusions we will ultimately get are essentially the same.

It shall be noted, that MO theory and VB theory are complementary theories, but the different explanations should not be mixed, even when the conclusions are essentially the same.


What is the mechanism of the uncatalysed epoxide opening with alcohols? The Wikipedia page says this:

  Ethoxylation is an industrial process in which ethylene oxide is added to alcohols and phenols to turn it into a surfactant.

How does the reaction proceed?

The answer to the question "Do epoxide rings react with bases?" regarding epoxide rings reacting with bases does not include alcohols, and does not answer my question.

Because of its special chemical struture Ethyelene oxide (EO) is highly reactive material thus easily participate in chemical reactions which includes polymerization. Mechanism of opening of EO's ring in acidic and alkaline medium is given below

If you see in both cases a molecule with alcoholic group ($\ce{-OH}$) is formed.If the epoxyethane is in excess, the reaction can continue and could lead to polymerization.

A good example is the reaction between EO and water

As explained in wikipedia ethoxylation is an industrial process to make surfactant from EO. Surfactants are chemical molecule which contain both hydrophobic groups (their tails) and hydrophilic groups (their heads).

The end result of reaction with EO and alchol is a long chain molecule with these hydrophobic ($\ce{-R}$) and hydrophilic ($\ce{-OH}$) group


Can sensitive nuclei like 1H be exploited to obtain 13C spectra? Measurement of 13C-NMR is essential for characterisation of compounds in organic synthesis, however, sensitivity can often be an issue, especially as molecules get larger and quantities get smaller, due to the low natural abundance of the NMR active 13C nuclei.

Standard carbon NMR is ran using a single pulsed experiment with heteronuclear decoupling for the 1H nuclei. This decoupling improves sensitivity by focusing the multiplets into a single sharp peak, with added sensitivity gained due to the nOe. Even with this enhancement, it can still often be difficult to obtain satisfactory spectra without resorting to using a stronger magnet (500 MHz/600 MHz is standard, the use of the 800 MHz spectrometer is prohibitively expensive for routine work) or running thousands of scans (there's a limit to how well this works anyway due to natural drift leading to peak broadening over time).

Unlike the standard 13C-NMR experiment, many 2D experiments such as HSQC are able to record data for 13C by observing the 1H nuclei (this is a huge oversimplification, I realise). This does, of course, lose the data for quaternary carbons, but I believe there are ways to get this data back (DEPTQ and various 2D experiments for example).

Why then are we unable to record 1D heteronuclear NMR using a similar technique?

We are able to record 1D heteronuclear experiments using these techniques, and absolutely do make use of this behaviour as a commonplace method for measuring 1D spectra of low sensitivity nuclei, such as $\ce{^13C, ^15N}$ and $\ce{^29Si}$. However, in order to get frequency information for the insensitive nucleus, we still need to use what is called a direct-detect method, where the nucleus of interest is actually the acquisition nucleus.

The most common method for sensitivity enhancement uses the general class of experiments called INEPT experiments; Insensitive nuclei enhanced by polarization transfer. The DEPT series (DEPT45, DEPT90 and DEPT135) are a sub-class of the INEPT experiment.

The signal of some nuclei can be improved by taking advantage of low power 1H decoupling during the relaxation period to allow nOe transfer to occur. This, as you mention, is typically how $\ce{^13C}$ spectra are recorded, and is why the $\ce{CH3}$ peaks ar much bigger than quaternary peaks. This method cannot be used for nuclei that have a negative magnetogyric ratio, such as $\ce{^15N}$ and $\ce{^29Si}$, as the nOe will actually generate a diminished signal. For these nuclei, the INEPT method is the preferred experiment for acquisition.

The INEPT magnetisation transfer step is at the core of most common 2D sequences, such as the HSQC. It provides a much greater signal enhancement than the nOe, although is strongly dependent on correct calibration of pulses.

The theoretical maximum sensitivity gain for nOe is $\displaystyle1+\frac{\gamma_\mathrm{S}}{2\gamma_\mathrm{I}}$
The theoretical maximum sensitivity gain for INEPT is  $\displaystyle\left|\frac{\gamma_\mathrm{S}}{\gamma_\mathrm{I}}\right|$

For $\ce{^13C}$, this equates to $\approx 3\times$ for nOe and $\approx 4\times$ for INEPT. For $\ce{^15N}$, it is $\approx -4\times$ for nOe (yep, negative) and $\approx 25\times$ for INEPT.

Glenn Facey at University of Ottawa has a couple of nice examples of INEPT and DEPT that are well worth looking at.

Both of these classes of experiment (nOE and INEPT) typically rely on interaction with nearby $\ce{^1H}$ signals: nOe is a through space interaction, and INEPT is a through bond interaction. Other nuclear pairs can be used, and many examples such as INEPT using $\ce{^19F-^29Si}$ have been published. The detection of insenstive nucleus environments with no directly attached sensitive nucleus (such as $\ce{^13C}$ quaternary centres) still requires long range coupling, and there are many published long-range INEPT methods.

The 2D HSQC method is an indirect detection method — we get information about the insensitive nucleus by acquiring another nucleus. Equivalent 1D variants of these 2D experiments do exist (so, a 1D HSQC), but they will only give you information about those $\ce{^1H}$ signals that have coupling to $\ce{^13C}$ — they don't provide any information about the $\ce{^13C}$ frequencies, unless you do a selective variant. So, a sel-HSQC will only irradiate a small band of the $\ce{^13C}$ frequency window, and therefore only give $\ce{^1H}$ signals that correlate to a $\ce{^13C}$ nucleus in this band. This is sometimes useful to probe a specfic question of structure, but a very tedious method of stepping through frequency bands of the $\ce{^13C}$ region (which is what a 2D experiment does for you).

For what it is worth, there are a number of cost-effective ways to improve sensitivity without having to fork out the big bucks for an 800. Choice of probe (X-observe, cryoprobe, microprobe) will offer improvements over increase in magnetic field at a fraction of the cost, and Shigemi tubes and micro cell tubes will also provide significant boosts in  sensitivity.


What's the definition for mole before Avogadro's constant? I have been searching for about 3 days how Avogadro's constant is found, and as all of us know the value is determined by division of charge of one mole of electrons (Faraday's constant) by the charge of one electron.
What confuses me is that, Faraday's constant is 96500 (it's not accurate value)
and its unit is C/mole. If the mole is defined by Avogadro's Constant
and Avogadro used Faraday's constant to determine his constant,
what was the definition of mole for Faraday?

These constants have been named in honour of these people. They did not even know the existence of mole concept.

In order to search how early precursors to Avogadro's number's were found, read about Loschmidt constant and then move to Avogadro's story. There were many $independent$ methods to determine Avogadro's constant. Loschmidt was trying to estimate size of the "air molecule" in 1865.  Number of particles in air in a given volume.

The determination of Faraday's constant does not require the knowledge of Avogadro' ideas or his laws. This is a modern misconception. However, it does require the concept of gram equivalents (precursors to atomic mass or formula weights). In Faraday's time, measuring current/voltage and masses of substances was not a big deal. By electrolyzing a solution of silver salt, you can deposit a known mass of silver on the electrode. I am showing a page how Faraday's constant was found. As you can see we never needed to invoke Avogadro's law or numbers and constant. Faraday did such neat experiments! Clever and simple.

Electro-chemistry: General theory, Volume 1, pg 4, By Robert Alfred Lehfeldt, T. S. Moore. Longmans, Green and Company, 1904.


Crystallization of Glycerin Is it possible to crystallize vegetable glycerin?

I saw a myth about it online and was wondering if it is possible.

yes it is possible, for example by adding a seed crystal under the right conditions.

See the reference for further information: http://www.google.com/patents/US2233606


Boron enolates double bond geometry The usage of dialkylboron triflates produces Z enolates from ketones predominately, while the usage of dialkylboron chlorides changes the selectivity favoring the E enolate:

How is this trend explained?

The image below (Taken from the Evans' lectures, which you might like to consult as there is a whole lot more detail in them) sums up the situation.
Essentially you're interested in how good a leaving group is attached to the boron. The triflate is excellent and as such fully dissociates, whilst the chloride remains (at least partially) in association.
Depending on whether the group on boron dissociates, the 'boron' is significantly more/less bulky, changing the conformation from which enolisation is preferred, which in turn leads to E/Z enolates.
The choice of base is also important. With boron triflates, DIPEA is often used as it is just that bit bulkier and so biases even more towards the Z enolate. In the boron chloride aldols, the bulky DIPEA is a bit too much, and so Et3N is commonly used.
If you study the TS's below, it might be apparent why the boron chloride aldols are typically a little less selective than the boron trifoliate aldols.

Source: Dave Evans' Harvard CHEM207 Lecture notes.


Geometry optimization of a small molecule inside of a larger one I'm very new to this and trying to optimize the structure of a larger molecule that "wraps" around a smaller molecule. I'm using ORCA. I'm having two issues with the task:

a) I've attempted to use the following:
! RI BP86 def2-SVP def2-SVP/J D3BJ TIGHTSCF Opt Grid3 FinalGrid5
but have not had any success.

b) How can I reliably export a two-molecule starting configuration in .xyz format starting from ChemDraw? I've tried exporting in .cml, loading in JMOL/Avogadro/Molekel and printing in xyz, but it invariably destroys the bonds.

I don't think you've provided nearly enough detail about your problems, particular "a".

But let me comment about "exporting from ChemDraw." Keep in mind that ChemDraw is a 2D tool. Consequently, the lengths between atoms are designed for depiction on paper, not necessarily for accurate representation of the 3D geometry necessary for calculations.

Avogadro can import 2D geometries from ChemDraw and attempt to assign reasonable 3D geometries. In your case, this seems especially troublesome since you want to have one molecule inside another.

I'd either:

Sketch the two molecules separately in ChemDraw and then use Avogadro to "insert" the smaller one into the larger after geometry optimization.
Use Avogadro to create the full 3D geometries of each molecule and then, select the smaller molecule and use the manipulate tool to move it into position.

As several people have commented, if you want to preserve the appearance of bonds, XYZ is not the right format. It only stores coordinates, not bonding information. Avogadro and other tools will attempt to re-perceive bonds and bond orders, but this is not a perfect task, particularly in the presence of multiple nested molecules.

Instead, I'd keep a file in CML format with the bond orders intact. This is more-or-less the "native" file format for Avogadro.


What is the definition of organic compounds? According to the definition of organic compounds on Wikipedia:

  An organic compound is any member of a large class of gaseous, liquid, or solid chemical compounds whose molecules contain carbon.

Carbonates, simple oxides of carbon, and cyanide molecules do have carbon, but they are not regarded as organic.  The same Wikipedia article on organic compounds reads:

  The modern meaning of "organic compound" is any one of them that contains a significant amount of carbon.

The current main definition for organic compounds doesn't fit for carbonates, simple oxides of carbon, steel, etc. So, what should be the redefined definition for organic compounds?

So, what should be the redefined definition for organic compounds?

We should not define organic compounds. As F'x pointed out in the comments, the distinction between organic and inorganic is largely arbitrary and mostly based on a distortion of historical precedent.

Rather, we should classify compounds by the type(s) of bonding interaction present in the substance. By this criterion, we have several classes of compounds:

Molecular (or covalent) compounds - the particles that make up these compounds are molecules held together by covalent bonds. This class of compound includes traditionally inorganic $\ce{CO2}$ and organic $\ce{CH4}$ carbon compounds. Covalent compounds also the majority of organometallic compounds, so we do not need a separate category for these. Other traditionally "inorganic" compounds, like glass, also belong in this category.
Ionic compounds - the particles that make up these compounds are largely ions held together by Coulombic forces. For the sake of simplicity most people would include in this class of compounds those ionic compounds where one or both ions is a small molecular ion held together by covalent bonds $($e.g. the $\ce{NH4^+}$ ion in $\ce{NH4Cl}$ or the  $\ce{CO3^{2-}}$ ion in $\ce{Na2CO3})$.
Mixed covalent/ionic compounds - those in which there are some ionic functional groups, but mostly covalent bonds, like methyl orange. Many folks might list these substances in Category 1. This category would include most coordination compounds including such things as metal-organic frameworks and probably many composite and/or ceramic materials.
Metallic compounds - those substances with extremely delocalized bonding, typical of metals. Alloys, including steel, belong in this category.

Under this system, what we traditionally study as organic chemistry becomes redefined as the study of the structure, properties, and reactions of covalent compounds (Classes 1 and 3), especially the nature of the covalent bond and how it can be formed and broken selectively. What you might learn about the structure, properties, and reactions of  organic compounds can easily be extended to the structure, properties, and reactions of molecular inorganic compounds. For example, we do not need two different branches of chemistry to study benzene and borazine.


Solvents adequate for dissolving overheated polyethylene terephthalate glycol-modified (PETG) residue? I work in 3D printing. One of the materials we use is polyethylene terephthalate glycol-modified, or PETG. It is a pretty good material, both in strength and appearance. It is melted and pushed through a "hot end" nozzle, typically either brass or hardened steel, at about 230 degrees C.

Unfortunately, it is rather sticky, and tends to accumulate on the hot metal surfaces, where it slowly degrades due to the heat. The resulting residue is black, and somewhat hard to remove. I have tried blowtorching it, which works with polylactic acid (PLA), but that just results in the same black substance.

I have seen recommendations to use either acetone, dichloromethane, or phenol. Other substances like toluene are recommended - some say they work, some say they don't. However, all of these seem to be recommended for dissolving/welding the plastic in its original form. There doesn't seem to be any advice for dissolving the residue after it has degraded from heat, at which point its chemical composition has changed.

From my research, the residue may be composed primarily of acetaldehyde. I was not able to find information (that I can understand as a lay person) about how to dissolve that.

So far, I've tried putting a brass nozzle caked with PETG residue into a 50C acetone bath for an hour. A little of it came off, but not much. The ideal solvent would attack the residue so completely that it would come off without too much force. The inside bore of the nozzles (the melt zone) is typically on the order of 1-2 millimeters, so the best option is to use an abrasive cleaning filament. Something tougher, like a wire brush, is not possible in that confined space, as far as I know.

If anyone has a good answer, I'll spread it around the 3D printing community, so you won't just be helping me. Thanks in advance!

It is difficult to tell when you consider PETG overheated (how long? what temperature?). Thermal degradation likely produces several reaction products each of which having different solubility in different solvents. So I don't know whether I ever encountered "overheated" PETG during my own 3D printing experiments.

But I have had very good experience with Tetrahydrofuran (THF), which I happened to have in my basement for a long time (beware of peroxides!) and I have finally found an application for. It dissolves freshly printed PETG (and glue parts together) as well as the residues on the print nozzle.

Side notes: Acetone is wellknown for being absolutely useless for dissolving PETG and I can confirm this from my own experience. All you can expect is to roughen the surface a little bit.

You can read several reports that say dichloromethane is not working as a solvent for PETG. I have found a report that claims Cyclohexanone is working. But I haven't got personal experience with these both. Toluene neither.

But in general, the best method for cleaning nozzles is brute force, i.e. pyrolysis by bringing the (brass) nozzle to red heat with a gas torch. Fast (takes only a few minutes) and reliable (absolutely no residues, expecially at the nozzle hole). One thing I have noticed is that it is a good idea to polish the nozzle with finest steel wool after pyrolysis, especially the inside. On the inside this polish reduces friction for the filament (the brass gets "hazy" after heating it with the gas torch). On the outside, having a shiny surface may reduce emissivity/absorbtivity a little bit, which helps keeping set temperature during printing.


Organic chemistry polymerisation I have a doubt regarding the polymerisation of ethene, in the case of polymerisation, we basically are arranging the molecules in a systematic order and thus decreasing its entropy so shouldn't we be providing a low temperature in order to keep the process spontaneous...
Whereas, in reality the conditions for the reaction are high temperature and pressure?
Please help...

Rate of a reaction depends on the temperature conditions. So to make a reaction proceed significantly, we have to take the reactants up to the level of activation energy. One more thing to note here is that low temperature is not a fixed temperature, for example - consider 2 reactions A and B, now if low temperature for reaction A is 400 Kelvin, then it is certainly possible that for reaction B low temperature is say 800 Kelvin.


What is a stereochemically active or inactive s orbital? In Concise Inorganic Chemistry by JD Lee (4th edition; adapted by Sudarshan Guha), page 73 under the section 3.6 (VSEPR Theory chapter Chemical Bonding):

...according to this theory, the position of the lone pair in $\ce{XeF6}$ molecule should be in axial position and the shape should be pentagonal pyramidal. But this is not observed and the actual shape of molecule is octahedral. This implies that the lone pair is placed at equatorial position, and this explains the structure of $\ce{XeF6}$ shown in figures 3.7 a,b,c.
According to Figure 3.7, $\mu_\text{expected}$ for $\ce{XeF6}$ is not equal to zero. But $\mu_\text{real}(\ce{XeF6})\sim0$ . To explain the above facts, it is considered that the actual molecule is in dynamic equilibrium of all the three structures shown in Figure 3.7. Here it is considered that the lone pair is present in the stereochemically active $s$ orbital and similar structure is observed as $\ce{IF6-}$. But anions like $[\ce{SbX6}]^{3-}$, $[\ce{TeX6}]^{2-}$ ($\ce{X}=\ce{Cl, Br, I}$), $[\ce{BrF6}]^-$,$[\ce{ICl6}]^-$ have been assigned perfectly octahedral structure on the basis of X-ray crystallography and the lone pair is present in stereochemically inactive $s$ orbital.

To explain the dipole moment, it is considered that the molecule is in dynamic equilibrium between three structures, where it is considered that the lone pair is present in the stereochemically active s orbital. However, I don't get what the author means by stereochemically active (or inactive) s orbital.

Generally, these terms are used to refer to lone pairs rather than s-orbitals.

In the context of VSEPR theory, lone pairs usually affect molecular geometry. For example $\ce{NH3}$ adopts a trigonal pyramidal structure, because of the presence of the lone pair on nitrogen; if there was no lone pair, then it would be trigonal planar, cf. $\ce{BF3}$. These are so-called stereochemically active lone pairs, since they affect the stereochemistry at nitrogen.

In a few rare cases lone pairs do not affect the molecular geometry, and these are termed stereochemically inactive. The only example I can think of, off the top of my head, is in $\ce{[XeF8]^2-}$, which should have 8 bond pairs and 1 lone pair around xenon. Despite the presence of the lone pair, the dianion adopts a square antiprismatic structure, without any obvious position in space for the lone pair to occupy. Looking in Shriver and Atkins' Inorganic Chemistry (6th ed.) there are a couple more example of $\ce{[SeF6]^2-}$ and $\ce{[TeCl6]^2-}$, which both have octahedral geometries despite having six bond pairs and one lone pair around the central atom.

In VSEPR theory, these stereochemically inactive lone pairs are rationalised by placing them in an s-orbital, which is non-directional. An electron pair in an s-orbital has a spherically symmetric electron density, and hence cannot influence the shape of a molecule. However, it is not exactly the orbital that is stereochemically (in)active, it is the electrons, and hence the more frequently seen term is "stereochemically (in)active lone pair".

There is one more context which should be mentioned, because it occurs frequently, and that is in tin(II) and lead(II) compounds. The electronic configurations of these two ions end with $n\mathrm{s^2}$, and this s-orbital lone pair tends to generate some distortions in stereochemistry.

A great example is that of the oxides. Solid $\ce{MgO}$ has a structure in which $\ce{Mg^2+}$ is octahedrally coordinated by six $\ce{O^2-}$ ions (rock salt/halite structure), but $\ce{SnO}$ and $\ce{PbO}$ have an unevenly distributed, square pyramidal coordination of $\ce{M^2+}$ by four $\ce{O^2-}$ ions. Traditionally this is attributed to the presence of a stereochemically active lone pair on Sn(II) or Pb(II).

This description cannot actually be complete, though, because if the geometry is solely determined by the lone pair on the metal, then one would expect PbS to have the same structure - and it doesn't. In fact, PbS has the same structure as MgO, with regular octahedral coordination, so clearly the anion has a role to play. A full explanation is beyond the scope of this answer, but one way of rationalising this is by invoking a second-order Jahn–Teller distortion. The interested reader may consult Chem. Soc. Rev. 2011, 40 (9), 4455–4463.

However, the "stereochemically active lone pair" is a recurrent theme in Sn(II) and Pb(II) structural chemistry (sometimes Tl(I) and Bi(III) also). For example, the metallocenes $\ce{SnCp2}$ and $\ce{PbCp2}$ adopt a bent structure, rather than the linear structure of $\ce{MgCp2}$ (and the more famous $\ce{FeCp2}$). Again, though, VSEPR theory is not the best way of explaining any of this; a proper explanation would involve MO theory.


Hydrogen bonds - why not in HCl? In a textbook1, I found the following clear definition of hydrogen bonds:

  The strongest secondary bonding type, the hydrogen bond, is a special case of polar molecule bonding. It occurs between molecules in which hydrogen is covalently bonded to fluorine (as in $\ce{HF}$), oxygen (as in $\ce{H_2O}$), and nitrogen (as in $\ce{NH_3}$). For each $\ce{H-F}$, $\ce{H-O}$, or $\ce{H-N}$ bond, the single hydrogen electron is shared with the other atom. Thus, the hydrogen end of the bond is essentially a positively charged bare proton that is unscreened by any electrons.

This is not an "etc." definition. It is made clear that we only call it a hydrogen bond when we have either of the three bonds: $\ce{H-F}$, $\ce{H-O}$ or $\ce{H-N}$.

My question is why this definition excludes compounds such as $\ce{HCl}$? $\ce{HCl}$ and $\ce{HF}$ are very similar with one covalent bond that pulls the hydrogen electron towards their "shared" region, exposing the hydrogen atom as a positive end of this permanent dipole. Both can create secondary dipole bonds to adjacent molecules of the same type.

Why is $\ce{HF}$ called a hydrogen bond while $\ce{HCl}$ is not?

1 Materials Science and Engineering by W. D. Callister & D. G. Rethwisch, Wiley, 8th edition

Hydrogen bonds to chlorine atoms are known. In the gas phase HCl dimers are hydrogen bonded but the bond has only about half the dissociation energy of $\ce{(HF)2}$ which has a larger electrostatic attraction energy. The adduct $\ce{HCl\cdots OH2}$ is also known and has a similar hydrogen bond energy to that of  $\ce{HOH\cdots OH2}$.

Using neutron diffraction hydrogen bonds have been studied in many crystals. From a study of many types of amino acid crystals (and derivatives thereof) the normal $\ce{HOH\cdots O}$ hydrogen bond is found to have  a range of lengths, from about 0.175 to 0.2 nm with most lengths at around 0.185 nm. The $\ce{OH\cdots Cl}$ hydrogen bond peaks about 0.2 nm with an range of $\approx \pm$ 0.05 nm. The $\ce{NH^+\cdots Cl}$ bond is longer peaking at 0.21 nm.

Calculations (Singh & Kollman, J. Chem. Phys. v83, p4033, 1985) show that the linear hydrogen bond, X-H-X, has an energy profile that is not that dissimilar to any chemical bond when energy is plotted against X to X separation. There are electrostatic attraction terms and exchange repulsion terms. These form a potential well with repulsion dominating (positive energy) at close separation of the two X atoms, a minimum energy (negative) at about 0.18 nm for water dimers, and a gradual increase in energy as the distance X - X increases reaching zero at large separation. However, there are also polarisation terms and those due to dispersion and charge transfer to take into account. These terms are all attractive and become important only at small separation of the X atoms. However, they are not insignificant and  contribute  a few tens of percentage to the total energy. The result of these terms is to move the minimum of the potential energy to smaller values that that due to exchange and electrostatic terms alone. Thus one can imagine the H atom residing in the potential energy between the two X atoms and being close to the minimum energy. Having described this ideal case,  IR spectroscopy does  indicate that in some cases the potential may have two minima rather than just one as described so the nature of these bonds is still a subject of debate.


Exothermic: enthalpy or heat? I am used to thinking of exothermic reactions as being defined as having negative $ΔH.$  However, I also am used to hearing/thinking of exothermic reactions as "releasing heat".

Since change in enthalpy $∆H = Q + VΔP,$ it seems that enthalpy and heat flow $(Q)$ are not strictly the same thing. Is it not therefore possible to have a reaction which absorbs heat $(+Q),$ but has a decrease in pressure, such that the product $-VΔP > Q,$ thus both absorbing heat and being exothermic (or if exo-/endothermicity is determined by $Q,$ then an exothermic reaction with positive $∆H)?$

As a sort of corollary, is there an example of a thermodynamic process that changes only the pressure (thus $V∆P)$ with no heat flow or $PV$ work? $(Q = 0 = PΔV)$

"Since change in enthalpy ∆H = Q + VΔP, it seems that enthalpy and heat flow (Q) are not strictly the same thing."

First, note that, when you write "∆H = Q + VΔP," you are assuming the only type of work is PV-work.  That's a reasonable assumption for these discussions.  It's just important to make that assumption explicit.

Given this, the quoted statement is correct.  Note, however, that many chemical reactions are carried out in an open container, such that the system stays at constant pressure (∆P = $0$) throughout the reaction. Thus, under this condition,  ∆H = Q + VΔP = Q.  I.e., the reason ∆H is a particularly useful thermodynamic function is that, under the common condition of ∆P = $0$, it corresponds directly to an important quantity: the amount of heat that flows into or out of the system.

  "Is it not therefore possible to have a reaction which absorbs heat (+Q), but has a decrease in pressure, such that the product -VΔP > Q, thus both absorbing heat and being exothermic (or if exo-/endothermicity is determined by Q, then an exothermic reaction with positive ∆H)?"

Here it comes down to a definition of terms.  The IUPAC Gold Book defines an "exothermic reaction" as "A reaction for which the overall standard enthalpy change ΔH⚬ is negative" ( https://goldbook.iupac.org/html/E/E02269.html ), where "standard" means P = 1 bar. Thus it doesn't cover how one would define exothermic or endothermic if the process is not at constant pressure.*   Having said that you could, in principle, have a reaction that takes place, say, at constant volume, and that releases thermal energy (Q < $0$) but, because of a sufficiently large and positive VΔP term, has ΔH > $0$. But, in practice, $\mid{Q}\mid$ would have to be very small in order for the VΔP term to dominate.

  "As a sort of corollary, is there an example of a thermodynamic process that changes only the pressure (thus $V∆P)$ with no heat flow or $PV$ work? (Q = $0$ = PΔV)"

Yes.  Consider, for instance, an isolated system, which is a system in which no flow of heat or work is allowed, and that is closed (no flow of matter).  The pressure could increase in an isolated system if, for instance, a chemical reaction is taking place in that system that evolves gas, and/or for which (were we to remove the adiabatic constraint) Q would be negative.

*Indeed, the IUPAC definition doesn't even seem to cover how one would define exothermic or endothermic if the reaction took place under constant P $\neq$ 1 bar, which seems quite limiting!  Further, it only defines the term for a "reaction", which IUPAC seems to take to mean a "chemical reaction" (typing "reaction" into the search field redirects you to "chemical reaction"), thus excluding important processes like phase changes.


How does acetone remove pencil marks? In another question it was noted that acetone is a useful solvent for removing pencil marks from wood surfaces.

(Source: https://woodworking.stackexchange.com/a/1870/763)

I'm used to acetone being used as a solvent for things like spray foam, nail polish, etc. but was surprised it could 'dissolve' pencil lead. I take it pencil lead is a mixture of graphite & clay substances. What enables the acetone to remove pencil marks?

Perhaps another way to think about this is, why does pencil lead adhere to surfaces like wood or paper in the first place? This is discussed in How are pencil marks "adhered" to the paper? but without mention of using solvents to remove it. How does the acetone reverse or interfere with the adhesion?

Pencil lead writes on rough surfaces by abrading into tiny particles and getting pushed into surface irregularities. If the surface isn't rough, you would get no visible residue of graphite. A pencil might sometimes even leave a scratch on dense wood that is so smooth that it has almost no blackness from graphite.

Wood is hydrophilic and even contains some water (unless bone dry). Graphite is not so hydrophilic or polar. Acetone wets the wood surface and either swells it by diffusing into watery regions or simply gets between a graphite particle and the wood surface, loosening any van Der Waals forces holding the graphite to the wood. While the surface is wet, the graphite can be wiped off (or smeared out so much that the original mark is not visible any more). Then the acetone evaporates quickly and the wood looks clean(er). Ethanol or isopropanol might do as well - even water (especially with a surfactant to coat and disperse the graphite) might work.

A grease pencil would be much more difficult to remove because so much material is pushed into crevices that a solvent can't easily loosen it without spreading it.


How many vibrations are expected for butatriene and cyclobutadiene? I have butatriene and cyclobutadiene, the formula for finding the number of vibrations expected are

Linear: $3N-5$
Nonlinear: $3N-6$

where $N$ is the number of atoms.

Both of these molecules are $\ce{C4H4}$ so they both have 8 atoms (because I think you include the $\ce{H}$ atoms).
Butatriene is nonlinear so the vibrations expected would be
$$3N-5 = 3(8)-6 = 18,$$
and Cyclobutadiene is nonlinear so
$$3N-6 = 3(8)-6 = 18.$$

Is this correct? Or do you only count $C$ atoms when you do this calculation?
Would all nonlinear $\ce{C4H4}$ isomers have the same number of vibrations expected?

Perhaps an explanation of where these mysterious equations come from will help you understand how to apply them properly.

I only need you to make one, single (correct) assumption: all atoms in a molecule are capable of moving in each of the three Cartesian coordinate directions.

That is, each of those little hydrogens can wiggle around in the x, y, and z coordinates as much as it pleases.

That means that we have a total $3N$ ways that the whole molecule can move because each of $N$ atoms can move in three directions.

Now, let's consider how molecules move. If every one of those atoms happened to be moving in the same direction, what would that look like?

Well, that would be translation of the molecule (i.e. moving). Any molecule can do this in three directions.

So, excluding translation we have $3N-3$ ways of moving around.

Now, another thing we know molecules do is rotate. By the same reasoning, we can say that a molecule can rotate in all three coordinate directions.

This leaves us with $3N-6$ ways to move around.

But, what if that molecule was linear? Well, then that molecule can only rotate in two unique directions because spinning around like a cylinder is known to be a zero energy process. Another way of saying that is that it doesn't happen.

So for a linear molecule there $3N-5$ ways of moving around still because we only subtracted out two rotational modes.

Of course, now we see that the only other way our molecules can move is by vibrating.

Thus, for a linear molecule, $$Number\ of\ normal\ modes=3N-5$$and for a nonlinear polyatomic molecule,$$Number\ of\ normal\ modes=3N-6$$

So, using these equations which now hopefully make some more sense, it's quite obvious that we do indeed include the hydrogens in our $N$ and find there are 18 normal modes of vibration for both molecules.

And, if there were an isomer of the formula $\ce{C4H4}$ which was linear then there would be 19 normal modes, but there is no such molecule so, yes all molecules with the molecular formula $\ce{C4H4}$ have 18 vibrational modes.


Why can't an electrochemical cell be a single cell? If I put two electrodes of two different metals in 1 solution containing ions for both of the metals and connect these electrodes with a wire, do I still have an electrochemical cell ? Or does it have to be made of two half cells in 2 different solutions joined by a salt bridge? If so,why ?

Yes, if you dip two different metals in an aqueous solution containing ions, you will have an electrochemical cell. This is how all lemon batteries work, which consist of zinc or copper inserted into lemons (approximately an aqueous citric acid solution). You can see a potential difference between zinc and copper. The key point is that one cannot determine Nernstian potential in such cases.
Half cell is just a mental construct as it cannot exist alone. If there is a reduction, something must be oxidized. Electrochemists usually separate the "half-cells" in order to conveniently calculate Nernst potential.

Imagine a hypothetical cell, which is partitioned into half cells with a plantinum sheet. Each compartment contains an electrode Cu and Zn, dipped in their respective salt solutions. What will happen in that case?


Comparing Electron affinity and electron gain enthalpy at 0 K In my book it is given that first electron gain enthalpy is greater than second for elements. Should we compare the magnitudes in such cases or the actual numbers with signs? Does the same comparision hold with Electron Affinity at 0K?
For example Be has positive Electron Affinity and F has negative Electron Affinity. So should I consider signs and say Be has more Electron Affinity than F or consider magnitude and say F has more Electron Affinity than Be?
How to compare their Electron gain enthalpies also at 0K?

In my book it is given that first electron gain enthalpy is greater than second for elements. Should we compare the magnitudes in such cases or the actual numbers with signs?

Successive Electron Gain Enthalpies
After the addition of one electron atom becomes negatively charged and second electron is to be added to a negatively charged ion. Thus, due to electrostatic repulsion energy is always required for addition of second electron. First electron gain enthalpy is always more negative than second electron gain enthalpy (which is essentially positive). Author made an ambiguous statement, as we can't just compare the magnitudes. For more clarity see:
$$\ce{S_{(g)} + e- -> S- _{(g)}}             \Delta_{eg} H^- _1 = \pu{-200kJ mol^{-1}}$$
$$\ce{S- _{(g)} + e- -> S^{2-}_{(g)}}             \Delta_{eg} H^- _2 = \pu{600kJ mol^{-1}}$$.

Does the same comparision hold with Electron Affinity at 0K?

Generally, if energy is released when an electron is added to an atom, electron affinity is taken as positive, contrary to thermodynamic convention. That is, if $\Delta_{eg} H^-$ is negative EA is positive (since temperature mentioned is absolute zero).
At absolute zero, $\Delta_{eg} H^- = \pu{-EA}$. Thus if $\Delta_{eg} H^- _1$ is negative, EA is positive and successive EAs will be increasingly negative. Here, you can say first electron affinity is greater than second (actual numbers with sign).

For example, Be has positive Electron Affinity and F has negative Electron Affinity. So should I consider signs and say Be has more Electron Affinity than F or consider magnitude and say F has more Electron Affinity than Be?

First, $\ce{Be}$ has negative electron affinity ($\pu{-0.5 eV atom^{-1}}$) and $\ce{F}$ has positive electron affinity ($\pu{+3.4 eV atom^{-1}}$). So, fluorine has more electron affinity than beryllium, consider the actual numbers with signs).

How to compare their Electron gain enthalpies also at 0K?

Since, at 0K, $\Delta_{eg} H^- = \pu{-EA}$. Thus, first electron enthalpy of fluorine is more negative than that of beryllium. Strictly speaking, it's value is less than that of beryllium.


What do (R) and (Z) mean in chemical names? So, I'm confused on the notation my professor uses in a certain question; could someone explain.
He says:

Draw the structure: (R)-(Z)-chloro-4,6,-dimethyl-3-octene

I get everything after '(z)'

I'm just unsure what he means by '(R)-(Z)'...

Any input is appreciated. Thanks.

The name (R,Z)-chloro-4,6-dimethyloct-3-ene is ambiguous. This could be (R,Z)-1-chloro-4,6-dimethyloct-3-ene, (R,Z)-3-chloro-4,6-dimethyloct-3-ene, (R,Z)-6-chloro-4,6-dimethyloct-3-ene, or (R,Z)-8-chloro-4,6-dimethyloct-3-ene. (2-chloro-4,6-dimethyloct-3-ene, 5-chloro-4,6-dimethyloct-3-ene, and 7-chloro-4,6-dimethyloct-3-ene have one additional stereocenter.) For example:


Molpro - Simple explanation of OCC, CORE, FROZEN and CLOSED orbitals I'm beginning with quantum chemistry and I'm supposed to do some ab initio calculations with Molpro. The problem is, I don't understand the orbital specification very well.

Let's say, I have the following computation:

 {mrci;
  occ,4,2,2,0,4,2,2,0;
  core,1,0,0,0,1,0,0,0;
  wf,13,5,1;
  state,3;
  option,nstati=5;
  option,maxit=200;
  option,maxiti=200;
 }

I don't understand much the flags OCC, CORE, CLOSED and FROZEN. I've read about them in the documentation, but I have troubles understanding it. Moreover, they possess different meaning in different methods, i.e. MCSCF and MRCI.

Could you, please, explain it in a basic way and graphically, if possible? I'm not looking for the deep understanding of methods now (I'll study them later), but I need to be able to use Molpro for some simple (di/tri-atomic) systems.

In MCSCF we have:

occ are occupied orbitals, it refers to all orbitals that may be occupied in any configuration. For a FCI calculation this equals the total number of orbitals (per IRREP). If occ is smaller than that, the difference will be the number of virtual (unoccupied) orbitals. Virtual orbitals are never occupied in all the configurations.
closed orbitals are doubly occupied in all configurations. This means occ - closed equals the number of active orbitals. closed orbitals are optimized in an MCSCF calculation.
frozen is similar to closed, but they are not optimized in MCSCF, i.e. their MO coefficients will be kept fixed.

In MRCI there are:

occ and closed being similar to MCSCF. They are used to define the reference space of configurations.
Multi-reference means additional external configurations are created by excitations from the reference space. Here excitations from closed orbitals are considered, but not from core orbitals. In other words: closed orbitals are only doubly occupied in the reference space but may have different occupations in the external configurations, while core orbitals are doubly occupied in both, the reference space and external configurations.
frozen is not used here, as orbitals are not optimized anyway.

The "lower" set of orbitals is always included in the "higher" set, e.g. occ includes all closed. Thus we always have core/frozen <= closed <= occ <= total number of orbitals.

Here is a graphic for the MRCI case. The curved arrows to the left indicate considered (black) and neglected (red) external excitations into the virtual orbital space. For MCSCF replace core with frozen and ignore the curved arrows.

One more note: The picture shows only excitations into the virtual space. But in principle you could also have excitations from closed to active orbitals.


Empirical formula, simplifying problems Need some help with second part.....can't get it to be balanced and therefore  can't find the formula and name! Thanks in advance!

What is the empirical formula of each of the following compounds?

a) A liquid containing 2.0g of hydrogen, 32.1g of sulfur and 64.0g of oxygen.

•   Hydrogen = 2.0g / 1.0g = 2 / 1 = 2
•   Sulfur = 32.1g / 32.1g = 1 / 1 = 1
•   Oxygen = 64.0g / 16.0g = 4 / 1 = 4

Empirical formula of this compound is: $\ce{SH_2O_4}$ and the name of the compound is Sulfuric Acid.

b) A white solid containing 4.0g of calcium, 3.2g of oxygen and 0.2g of hydrogen.

****•   Calcium = 4.0g / 40.1g = 0.09975062344 =
•   Oxygen = 3.2g / 16g = 0.2 =
•   Hydrogen = 0.2g / 1g = 0.2 =****

Empirical formula of this compound is:

c) A white solid containing 0.243g of magnesium and 0.710g of chlorine

•   Magnesium = 0.243g / 24.3g = 0.01 / 0.01 = 1
•   Chlorine = 0.710g / 35.5g = 0.02 / 0.01 = 2

Empirical formula of this compound is: $\ce{MgCl_2}$ and the name of the compound is Magnesium Chloride.

You find #2 difficult because #1 and #3 are ridiculously easy!

In these two questions the mass of every element present just happens to match exactly with the atomic weight of that element.  So when you divide each mass by the corresponding atomic weight, the resulting ratios are very close to integers, or integers divided by 100.  So the (almost) final step of dividing each ratio by the smallest ratio goes by almost unnoticed.

But consider for the moment what would happen if you used three or four decimal place atomic weights.  All those nice neat exact answers would disappear in a haze of "sort of" whole numbers, and you would have to use some judgement as to whether a number is close enough to a whole number.

In #2, the questioner made the "drastic" change of using 4.0 and 40.1 as the two numbers for an element.  If you had carried on mechanically, dividing all ratios by the smallest ratio, you would have gotten:

Ca ... $1.0$
O .....$2.005$
H......$2.005$

Is $2.005$ close enough to a whole number?  Or should you multiply everything by $1000$ to give $1000, 2005, 2005$ as the empirical formula?


Why does delocalization of π electrons bring stability? My book mentions that

  Delocalization of $\pi$ electrons makes a compound stable as the process of delocalization involves loss in internal energy. This is the synopsis of resonance.

But how does shifting of $\pi$ electrons let the body lose internal energy? What is the cause?

While studying resonance, I observed that one structure possesses charge while in other no charge. But in the final structure, there is, according to the book, partial charge present.

Eg.
In phenol, three resonating structures contain '+' charge on $\ce{O}$ . Thus, in the final structure, there is a slightly positive charge on oxygen and also, small negative charge on the main body.

Now my question is, why there is partial negative charge on the body? And why does oxygen carry partial positive charge? What is the cause?

In simple terms electrons repel each other, so distributing them over a geater number of bonds will reduce the repulsive force between them so resulting in a lower energy state. In the case of phenol the oxygen p orbital lone pair at right angles to the ring can overlap with the delocalised p orbital above and below the ring enabling a degree of delocalisation into the ring system so leaving a slight + charge on the oxygen. This in turn will reduce the electron density in the OH bond so weakening it and accounting for the weak acidity of phenol.


Determining k value of a Reaction You and a co-worker have developed a molecule that has shown potential as cobra anti-venom ($\ce{AV}$). This anti-venom works by binding to the venom ($\ce{V}$), thereby rendering it non-toxic. This reaction can be described by the rate law, $\text{rate} = k\left[\ce{AV}\right]^1\left[\ce{V}\right]^1$.

You have been given the following data from your co-worker: $\left[\ce{V}\right]_0=0.20\ \mathrm M$, $\left[\ce{AV}\right]_0=1.0\times10^{-4}\ \mathrm M$. A plot of $\ln\left[\ce{AV}\right]$ versus $t$ (s) gives a straight line with a slope of $-0.32\ \mathrm s^{-1}$. What is the value of the rate constant ($k$) for this reaction?

My logic tends to fail me on this question. So to start off, the overall  reaction order is 2, thus the equation is: $${A}^{-1}=kt+{a}^{-1}$$ However, we're only given a $k$ value for the law: $\text{rate}=k\left[\ce{AV}\right]$. I'm lost as to how to proceed next. Given the k we can calculate the initial rate of $\left[\ce{AV}\right]$ but how is this going to help me find the $k$ for $rate=k\left[\ce{AV}\right]^1\left[\ce{V}\right]^1$? I suspect something tiny that I missed in this problem, any directions or hints would be appreciated!

You can regard the reaction to be pseudo-first order, as the concentration of $\left[\ce{V}\right]$ is much higher than the concentration of $\left[\ce{AV}\right]$. Therefore, $k\cdot\left[\ce{V}\right]_0=\text{const.}=k_{2}$. The rate law of this reaction would change to $r=k_{2}\cdot\left[\ce{AV}\right]$. Integrating this would lead to the expression: $\ln\left[\ce{AV}\right]=\ln\left[\ce{AV}\right]_0-k_{2}\cdot t$. The dependence of $\ln\left[\ce{AV}\right]$ on time has the slope $-k_{2}=-0.32\ \mathrm s^{-1}$, so you can easily calculate $k$ from $k_{2}$ and $\left[\ce{V}\right]_0$.


Which rate is faster: one step with Ea or two steps each with Ea/2? I have been struggling with the following problem posed at a prior lab meeting. The answer seems trivial to me, so can you tell me if I'm doing this wrong? Thanks!

Consider the following two scenarios. They represent a small portion of the same catalytic cycle.

There is a reaction step of that cycle wherein $A\rightarrow P$ with an activation energy $E_{a}$.
Upon further inspection, the above reaction is actually $A\rightarrow I\rightarrow P$, where each of the two steps has an activation barrier of $E_{a}/2$ such that their sum is equal to the barrier height in the first scenario. Refer to the schematic here for a depiction if needed.

The question is: which scenario is faster? My thought process is as follows.

First, we need to define which rate we are talking about. I am going to assume the rate of production of product, $dP/dt$, is compared. I need information about pre-factors, but I'll have them all equal an arbitrary value of 1 for the time being. I have set $[A]_{t=0}=1$, $[P]_{t=0}=0$, and $[I]_{t=0}=0$. I now write out my ODEs and solve them. I've assumed $E_{a}=RT$ for demonstration purposes.

Case 1:

Case 2:

Comparison of Rates:

My Issue:

Can the question not be definitively answered? It seems like the answer (whether case 1 is faster than case 2) is that it depends on what conversion you're running to since the answer changes at a certain point. However, this was asked in the context of catalysis. I've posed this problem to multiple people and they all brought up the point that they don't usually think about time or conversion in this regard. They just want to know which one of these two cases will be faster when looking at two potential energy profiles. Isn't that too naive, or am I thinking about this all wrong? At full disclosure, this is mildly related to a prior question I posed in the comments located here.

Edit: To emphasize, if I change the pre-factors or the value of $E_{a}$, the order of which case is faster will also change. So, the answer to the question in the title is that "it depends".

Instead of looking at $\mathrm{d}[\ce{P}]/\mathrm{d}t$ against time $t$, a better idea might be to plot $[\ce{P}]_t$ against $t$, which you already did. By eyeballing the graphs I'm fairly sure that $[\ce{P}]_t$ in case 2 is always below $[\ce{P}]_t$ in case 1.

In org chem, people probably aren't too interested in what the absolute rate of the reaction at time $t$ is, they are probably more interested in how long the reaction will take to run. Because of the maths, the reaction will never hit 100%, so you can't use that as a benchmark, but you can use e.g. 95% complete (i.e. $[\ce{P}]_t = 0.95$) as a benchmark.


How to identify the number of pi electrons in a conjugated system to calculate the HOMO-LUMO gap with the particle in the box approach? It's my understanding that for a particle in a 1-D box, the change in energy from the HOMO to LUMO can be approximated with the following equation:
$$\Delta E=\frac{h^2 ( N + 1 )}{8 m_\mathrm{e} L^2}$$
I am told that $N$ represents the number of π electrons in the molecule.

If you're studying a molecule like the one shown below and you want to predict the absorption wavelength from UVVIS, would you count all of the π electrons, or just the conjugated ones? It doesn't make much sense to me why you'd want to include the π electrons from the phenyl group next to the neutrally charged Nitrogen because it's not conjugated with the rest of the molecule to the same extent the other phenyl group is, and thus seems like it would be grossly inaccurate to assume that those π electrons contribute the same as the others.

In a very basic first order approximation you can treat dyes with an extended conjugated π-system with the one-dimensional particle in the box approach. I simplified your system and calculated it at the DF-BP86/def2-SVP level of theory. The length of the box is about $L=1.20~\mathrm{nm}$, the following graphic shows it in Ångström.

With that there are a number of problems associated. Counting the number of electrons in the π-system should not be among them. Just assume that all non-hydrogen atoms are sp² hybridised. Then count the electrons that occupy π/p-type orbitals. Here is an image of the lowest occupied π-type orbital of that molecule; you can see that the box stretches over the whole molecule. The thumbnail next to it gives you all occupied π-type orbitals and the lowest unoccupied molecular orbital.

   In this example each phenyl ring contributes six π electrons (12), two sulfurs contribute two each (4), the nitrogen double bond (2), the carbon-carbon double bond (2) and the last nitrogen (2). That makes a total of 22 π electrons.


Latimer diagram for a metal in basic solution Write the four half cell reactions from the Latimer diagram for a metal $(\ce{M})$ in basic solution.
$$\ce{\overset{+6}{M}O4^2-(aq) ->[\pu{0.25 V}] \overset{+5}{M}O4^3-(aq) ->[\pu{0.55 V}] \overset{+3}{M}O(OH)(s) ->[\pu{0.42 V}] \overset{+2}{M}(OH)2(s) ->[\pu{-0.89 V}] \overset{0}{M}(s)}$$

I can only come up with the first half reaction, which I believe is:
$$\ce{MO4^2- + e- -> MO4^3-}$$
But for the others I'm having trouble with knowing whether the $\ce{MO4^3-}$ will react with water or hydroxide ion to create, for example, $\ce{MO(OH)}?$

Let's look at how you would add in the water and hydroxide ions for the $\ce{MO4^{3-}->MO(OH)}$ reaction.

Step 1:  Start with the given metal species, with the metal oxidation states included:

$\ce{M^VO4^{3-}->M^{III}O(OH)}$

Step 2:  Add the electrons according to the oxidation states.  Here the oxidation state on the metal drops by two, so there must be two electrons on the left.  Ignore the fact that the overall charges may be unbalanced, that comes next.

$\ce{M^VO4^{3-} + 2e^- ->M^{III}O(OH)}$

Step 3:  Now we are ready to attack the charge balance.  Add the hydroxide ions, which do not couple with the electron exchange, to balance the charges.  Counting out the charges we have going into this step we need five hydroxide ions on the right.

$\ce{M^VO4^{3-} + 2e^- ->M^{III}O(OH) + 5OH^-}$

Step 4:  Finally add water to balance hydrogen; if you rendered the oxidation states and did the math correctly, the oxygen balance falls into place simultaneously.  With six hydrogen atoms to balance we use three molecules of water to get the final answer for this reduction:

$\ce{M^VO4^{3-} + 3H2O + 2e^- ->M^{III}O(OH) + 5OH^-}$

Can you now do the remaining reductions?


Why N2O is a "laughing gas"? It is (mostly) used as an "oxidizer" in motors and engines, I understand that much. The oxygen "oxidizes" with other chemicals, while the nitrogen returns to the air.

But $\ce{N2O}$ also produces "euphoria" in human beings, works as a mild anesthetic, and causes people to laugh. Why does this happen?

In contrast to $\ce{O2}$ (or $\ce{CO}$), $\ce{N2O}$ does not bind to hemoglobin and it is not metabolized in the human body. It is taken up through the lungs, gets in the blood and leaves the body the same way being exhaled.

According to a 2004 study by P. Nagele in PNAS (DOI), $\ce{N2O}$ acts as an antagonist for the  $N$-methyl-$D$-aspartate (NMDA) receptor. Other infamous antagonists are drugs like ketamine or phencyclidine.

Further information on NDMA receptor antagonists can be found here.


How to stop goggles from fogging? I wear corrective eyeglasses, and because of this, I frequently experience problems with fogging when I wear safety goggles. How can I prevent or reduce fogging?

The safety goggles I use are the standard-issue goggles supplied by the campus store, but I can use other goggles rated at least Z87+D3 ("+"=impact, "D3"=splash) if I want.

One method of reducing the fogging I have tried is wedging a folded paper towel under the top edge of the goggles. It works, but I don't feel comfortable with it because it compromises the splash protection of the goggles.

I have also tried placing desiccant packets (of the sort found in various packaging) in the goggles. This has helped somewhat but doesn't solve the problem completely.

One possible solution would be to get safety goggle with prescription corrective lenses built-in (making them Z87-2D3 instead of Z87D3), but I have not been able to find a supplier online for such eyewear.

What can I do to reduce or eliminate fogging?

Be careful with desiccant: silica may create tiny dust particles that could scratch the cornea, and hygroscopic salts may liquefy, possibly irritating.

A traditional anti-fog treatment is to rub the interior of the lens with bar soap (or a drop of mild detergent) and polish it. The soap decreases the surface tension of water, causing droplets to make a smoother, more transparent film. Wikipedia suggests some other anti-fog treatments.

There are also vented anti-fog goggles (the vents should have foam or other filter to make them splash resistant.

My personal choice for lab work or mechanical work is a full-face splash shield, which may offer more protection and does not fog.


Would silica have the same effect as activated charcoal? Silica found in "do-not-eat" packets is a desiccant (I'm not sure about this); and so is activated charcoal. Would consuming it have the same effect if one were to have a stomach ache?

The silica gel granules are granular silicon dioxide (SiO2), the same matrial from what is the normal sand.

The pure SiO2 has the same effect to health as when you eat sand.

But because the granules are larger, may cause appendix problems (im not an doc). Anyway, this is not chemical related problem.

It can cause dehydratation or other problems, what is caused by it absorbtion capabality.

The bigger problem is, than the standard silica gel granules aren't pure SiO2.  The common additivas are what causes color change when it is moistured,  like cobalt chloride or others and thise compounds can be toxic!

So, don't eat it!


Is this reaction of 4-chloro-1,1-dimethylcyclohexane SN1, SN2, E1, or E2? In the following question, which I tried to solve, I became confused by which way I should proceed.
Here is my thought process:

SN1: If Cl detaches and forms a secondary carbocation, then there's no nearby groups to stabilize it or make a shift to turn it into a tertiary cation.
SN2: I can't make out how exactly SN2 would occur
E1: I can't make out how exactly E1 would occur
E2: There's no stereospecific data given in the question whether the methyl group and the hydrogen lie in one plane or not, so I couldn't make out a way to do anti-elimination.

The Answer is given : E1

so @benzene , This is the E1 product as per the mechanism , Am i right ?

E1 is the correct answer. There are a couple of factors that lead me to this conclusion. First of all, heat favors elimination over substitution, meaning that this reaction will either be E1 or E2. Now that we know that this is an elimination reaction, it is just a matter of deciding if it is E1 or E2. E2 requires a strong base to occur and since $\ce{H2O}$ is a week base, we know that this reaction must follow an E1 mechanism. While E1 is favored by a tertiary carbocation intermediate, both E1 and E2 can occur with a secondary carbocation.


How to calculate the pH of a 0.325 M solution of pyridine hydrofloride? How to calculate the $\mathrm{pH}$ of a $0.325~\mathrm{M}$ solution of $\ce{C5H5NHF}$? The $K_\mathrm{b}$ for $\ce{C5H5N}$ and $\ce{C5H5NH+}$ is $1.7\times 10^{-9}$.

Here's my work, but it's wrong (at least I tried):

$$\ce{C5H5NHF -> C5H5NH+ + F- }$$
$$K_\mathrm{a}(\ce{HF})= 4.0\cdot10^{-4}$$
$$K=\frac{K_\mathrm{w}}{K_\mathrm{a}}$$
$$K=\frac{1\cdot10^{-14}}{4.0\cdot10^{-4}}=2.5\cdot10^{-11}$$

Construct an ice table:

$$
\begin{array}{|c|c|c|c|}\hline
&\ce{F-}&\ce{+H2O(l)->}&\ce{HF}&+\ce{OH-}\\\hline
\text{I}&\pu{0.325M}&&0&0 \\\hline
\text{C}&-x&&x&x\\\hline
\text{E}&\pu{0.325M}-x&&x&x\\\hline
\end{array}
$$

Set up the equation (small x-approximation):
$$2.5\cdot10^{-11} = \frac{x^2}{0.325-x}$$
Now, cross multiply:
$$x^2 = 5.87\cdot10^{-12}$$
Take the square root of both sides:
$$[\ce{OH-}] = x = 2.42\cdot10^{-6}$$
You want to get from $[\ce{OH-}]$ to $\mathrm{pOH}$:
$$\mathrm{pOH} = -\log\left(2.42\cdot10^{-6}\right) = 5.616$$
Then, from $\mathrm{pOH}$ to $\mathrm{pH}$:
$$\mathrm{pH} = 14 - 5.616 = 8.384$$
Which is the wrong answer.

What am I doing wrong? How do I even use the $K_\mathrm{b}$ value in my calculation?

I think part of the issue is that you are looking at the wrong reaction. I believe the reaction should be $$\ce{C5H5NHF_{(s)} ->C5H5NH+_{(aq)} + F-_{(aq)}->C5H5N_{(aq)} + HF_{(aq)}}$$ When looking at this, we can start by noting that $\ce{HF}$ is a stronger acid than $\ce{C5H5NH+}$, so the reaction will favor the reactants. Then, we can compare the reactants and note that $\ce{C5H5NH+}$ has a greater acid strength than the base strength of $\ce{F-}$  (ie $\mathrm{K_a} $ of $\ce{C5H5NH+} >\mathrm{K_b} \text{ of } \ce{F-} $) so the solution will be acidic. Its always useful to have this sort of sanity to check to make sure your answer makes sense. (For more details on this portion, Cengage has a chapter discussing acid/base mixtures)

A good approximation for the pH can be found in this paper, as shown here: $$[\ce{H3O+}]=\mathrm{K_a}*\sqrt{\frac{\mathrm{K_w}}{\mathrm{K_a * K_b}}}$$ where $\mathrm{K_a}$ is for $\ce{C5H5NH+}$ and $\mathrm{K_b}$ is for
$\ce{F-}$. This leads to $[\ce{H3O+}]=4.85*10^{-5}\to\mathrm{pK_a}=4.314$

As a comparison, the $\text{p}\ce{H}$ of $.325$M pyridinium is $2.86$, which it makes sense that the $\text{p}\ce{H}$ is higher when it is mixed with a base.


How does aqua regia dissolve gold? I recently learned that aqua regia can dissolve less reactive or noble metals like gold. Aqua regia is a mixture of $\ce{HCl}$ and $\ce{HNO3}$, but separately, neither of these acids dissolve gold.

So why don't concentrated $\ce{HCl}$ or $\ce{HNO3}$ dissolve gold, though combined in aqua regia they can?

The wikipedia article of aqua regia has beautifully mentioned the mechanism of gold dissolving in aqua regia:

Aqua regia dissolves gold, though neither constituent acid will do so alone, because, in combination, each acid performs a different task. Nitric acid is a powerful oxidizer, which will actually dissolve a virtually undetectable amount of gold, forming gold ions ($\ce{Au^3+}$). The hydrochloric acid provides a ready supply of chloride ions ($\ce{Cl−}$), which react with the gold ions to produce tetrachloroaurate(III) anions, also in solution. The reaction with hydrochloric acid is an equilibrium reaction which favors formation of chloroaurate anions ($\ce{AuCl4−}$). This results in a removal of gold ions from solution and allows further oxidation of gold to take place. The gold dissolves to become chloroauric acid. In addition, gold may be dissolved by the free chlorine present in aqua regia. Appropriate equations are:
$$\ce{Au + 3HNO3 + 4HCl <=>> AuCl4- +  3NO2 + H3O+ + 2H2O}$$
$$\ce{Au + HNO3 + 4HCl <=>> AuCl4- + NO + H3O+ + H2O}$$
If the aqua regia solution only contains gold, solid tetrachloroauric
acid may be prepared by boiling off excess aqua regia, and removing
residual nitric acid by repeatedly heating with hydrochloric acid.


Stability of MnO4- in basic vs acidic conditions Why is $\ce{MnO4-}$ more stable in basic than acidic conditions? I know that electrode potential values suggest this, but what is the underlying chemical explanation for the phenomenon?

The obviously feasible and spontaneous disproportionation reaction can be explained by considering the standard electrode potentials (standard reduction potential) involved (quoted as half–cell reductions, as is the convention).

$\displaystyle \ce{MnO4- (aq) + e- -> MnO4^2- (aq)} \quad\quad (E^0 = +0.56~\mathrm{V})$
$\displaystyle \ce{MnO4^2- (aq) + 4H+(aq) + 2e-  -> MnO2(s) + 2H2O(l)} \quad\quad  (E^0 = +1.70~\mathrm{V}, \text{in acid solution})$

2. has the more positive potential, so this will be the reduction half–cell reaction.

1. has the less positive potential, so this will be (reversed) the oxidation half–cell reaction.

$$E^0_\text{reaction} = E^0_\text{reduction} – E^0_\text{oxidation} = (+1.70~\mathrm{V}) - (+0.56~\mathrm{V}) = +1.14~\mathrm{V} \gg 0~\mathrm{V}$$

Therefore, the reaction is very feasible.

Incidentally: Given the two half–cell reactions, you get the complete balanced equation by adding 2. plus 2 x 1. reversed.

The greater stability of the manganate(VI) ion in alkali can also be explained by considering the electrode potential for 2. in an alkaline media.

$\displaystyle \ce{MnO4^2- (aq) + 2H2O(l) + 2e-  -> MnO2(s) + 4OH- (aq)}\quad \quad  (E^0 = +0.59~\mathrm{V}, \text{in alkaline solution})$

so, re–calculating gives

$$E^0_\text{reaction} = E^0_\text{reduction} - E^0_\text{oxidation} = (+0.59~\mathrm{V}) - (+0.56~\mathrm{V}) = +0.03~\mathrm{V} > 0~\mathrm{V}$$

Therefore just feasible! but on the basis of an equilibrium argument, here, the far lower $E^0_\text{reaction}$, suggests the $\ce{MnO4^2-}$ ion is far more likely to exist, i.e. more stable, in a very high pH solution and in practice it is stable for a few hours in alkali.


Transition metal order of filling electron There is a specific order for filling electron for transition metals, but I don't know if it applies to other elements too?

Some thing to take note of in this picture is the order of 3d and 4s. My belief is: 3d has a higher energy level than 4s, and but 4s is still the outermost shell.

The reason that transition metal would form colorful ions is that there are empty or half filled d orbitals, so when excited electrons drop back to the d orbitals, they emit energy in form of light. I am a bit confused about this.( it is from a book) what electron is it referring to? Is it the d electron or electron of lower energy level orbitals, as for instance 3p?

This is the order of electrons filling for all elements but typically this diagram is only used for transition elements because when writing electron configurations for any other element its simple and intuitive but transition metals are somewhat irregular so a helpful diagram like that is useful.

Its transition metal compounds or complexes that are coloured not the metal on its own. Its coloured because when it forms compounds or complexes the orbitals in the 3d subshell aren't the same energy as they would be if it wasn't part of a compound or complex. The electrons make transitions between these different energy 3d orbitals when they absorb photons of visible light. The colour of the compound will be the complementary colour of the colour absorbed.

The 3d orbital is typically seen as the higher energy orbital but when a transition metal is ionized the 4s electrons are taken first.


Looking for a chemical to calibrate a temperature sensor The best method for calibration would be a bath filled with a chemical at its melting point with a solid/liquid composition.
The constraints being that it must be hydrophobic and relatively non-corrosive and non-toxic with a melting point between (say) 20 °C and 60 °C. Needless to say, but I’ll say it anyway, the chemical must have a very precise melting point. Any ideas?

The standard point recommended in that range by the CIPM's ITS-90 calibration standard is the melting point of gallium at 29.7646 °C. Gallium is not known to be particularly toxic, but it can embrittle metals so is ideally used/stored in glass or plastic containers. It's used as a standard because it is relatively easy to make very pure and is stable in storage. Also, by picking an ITS-90 standard, you can ensure your thermometer is calibrated to the same standard as many other high precision thermometers. N.B. ITS-90 specifies specifically the melting point of gallium, not the freezing point, unlike some of the other calibration standards.