Relationship between Quantum Numbers and the Wave-function I recently started learning about quantum mechanics and its applciations in atomic structure in chemistry.
In this inorganic textbook Inorganic Chemistry, it describes
"Each of the wavefunctions obtained by solving the Schrödinger equation for a hydrogenic
atom is uniquely labelled by a set of three integers called quantum numbers."
What is the relationship between the quantum numbers and the wave function? If the above is saying that each wavefunction is uniquely labelled by 3 quantum numbers, then it sounds to me that the 3 quantum numbers can describe the wavefunction on their own. But how can just 3 numbers describe a function?
On this website, it says the following
"A complete solution to the Schrödinger equation, both the three-dimensional wavefunction and energy, includes a set of three quantum numbers ( n,l,ml )."
Is the complete solution to the Schrodinger's Equation the wavefunction in 3D? what is the "energy" mentioned in the last quote to do with solution of the equation?
Thank you.

The hydrogen atom
Indeed, three numbers cannot define a function. There are several more criteria, one of which you mentioned already: it must satisfy the time-independent Schrödinger equation. The Schrödinger equation is a nasty differential equation, which looks something like the following:
$$-\frac{\hbar^2}{2\mu}\left(\frac{\partial^2\psi}{\partial x^2} + \frac{\partial^2\psi}{\partial y^2} + \frac{\partial^2\psi}{\partial z^2}\right) - \frac{e^2}{4\pi\varepsilon_0r}\psi = E\psi$$
On top of this, it must also satisfy the usual criteria for a wavefunction ("boundary conditions"): for example it must be square integrable (i.e. its integral over all space must be finite, meaning that it can be normalised):
$$\int |\psi|^2\,\mathrm{d}\tau < \infty$$
Once you put all of these conditions together, it turns out that not every plain old function $\psi(x,y,z)$ is a permissible atomic orbital. There is a family of functions $\psi$ which do, and they all have a certain form which is completely determined by the three quantum numbers $(n, l, m_l)$.

A simpler example
This is pretty abstract, so why don't we make it more concrete with a simpler example. (The hydrogen atom is not easy to solve explicitly.) Consider the one-dimensional function $f = f(\phi)$, where $\phi$ is an angle. You should recognise at this point that, given no other information, one quantum number is not sufficient to specify what $f$ is, as there is an infinite possible different $f$'s.*
However, assume now that $f$ obeys the differential equation
$$\frac{\mathrm{d}f}{\mathrm{d}\phi} = \mathrm{i}mf$$
The general solution of this differential equation is (after dropping the normalisation factor)
$$f = \exp(\mathrm{i}m\phi)$$
But note also that since $\phi$ is an angle, we have that for all $\phi$
$$f(\phi) = f(\phi + 2\pi)$$
meaning that
$$\begin{align}
\exp(\mathrm{i}m\phi) &= \exp[\mathrm{i}m(\phi + 2\pi)] \\
&= \exp(\mathrm{i}m\phi) \exp(\mathrm{i}m2\pi) \\
1 &= \exp(\mathrm{i}m2\pi)
\end{align}$$
which means that $m$ must be an integer. Consequently, the allowed $f$'s are
$$f = \exp(\mathrm{i}m\phi) \quad\quad m \in \mathbb{Z}$$
and so we can now label the $f$'s by the value of $m$. The quantum numbers $m$ do not represent actual values of $f$ at any point in space, nor do they represent actual values of $\phi$: they instead serve to parameterise the $f$'s. That is to say, the value of $f$ at any value of $\phi$ is completely determined by the value of $m$. We do not write this as $f(m)$, but rather $f_m$, and thus we have:
$$\begin{align}
&\,\,\,\vdots \\
f_{-2} &= \exp(-2\mathrm{i}\phi) \\
f_{-1} &= \exp(-\mathrm{i}\phi) \\
f_{0} &= 1 \\
f_{1} &= \exp(\mathrm{i}\phi) \\
f_{2} &= \exp(2\mathrm{i}\phi) \\
&\,\,\,\vdots
\end{align}$$
Indeed, $m$ can be said to be a quantum number here. Notice how we've used a simplified problem: it has one instead of three dimensions, and it has a much simpler differential equation. However, it illustrates that by enforcing additional conditions on the wavefunction — namely that it obey a differential equation and boundary conditions — we can get a family of solutions which can be labelled with quantum numbers.†

What is the energy?
When you find the allowed $\psi$'s that solve the Schrödinger equation, you can plug it back into the differential equation and evaluate the value of $E$ inside it. That will give you the energy of the atomic orbital.

Footnotes
* Technically, there is an uncountably infinite number of $f$'s. If it were countably infinite, then we could establish a one-to-one mapping between the different $f$'s and the natural numbers, which would allow us to label the $f$'s as $f_1, f_2, f_3, \cdots$ which is exactly what a quantum number does.
† With regards to the previous footnote, this set of permissible $f$'s is still infinite, but it is countably infinite this time. Also, this actually happens to be the exact way in which the quantum number $m_l$ is derived, so my choice of $m$ as the quantum number was not arbitrary.


What is in silico energy optimization? While doing computational chemistry for both organic and inorganic compounds, an energy optimization of that compound is done before starting a study.

For example, if we do drug docking studies using aspirin, first we would make aspirin's chemical structure, and then we would optimize it. So what does "energy optimization" mean in the context of organic and inorganic compounds.

Is energy optimization different for organic and inorganic materials (e.g.,aspirin or gold nanoparticles, respectively)?

Generally, energy optimization is done as iterative process with following steps repeating

calculate forces of nuclei
move nuclei according forces known so final structure were less tense.

Most focus in science is focused on how first step is made, as choice of the second step does not affect final structure except pathological cases.

The main difference is usually tied to computational capabilities of 'silica' available. Complexity of quantum mechanical computations usually grows as $O(n^3)$, $O(n^4)$ or even $O(e^n)$ where n is number of atoms of same type in system. So, the bigger system is, the 'worse' approximation is used. Biological systems are usually so large, that any sort of quantum mechanical computation is not an option at all, so "molecular mechanics/dynamics" approach is used, with empiric formulas of forces based on relative positions of atoms


Propanoic acid reaction with potassium permanganate I was conducting a lab where we have to identify/justify how functional groups affect the oxidation of a molecule.

I used propene, propanone, and propanoic acid for this reaction. From what I have learned in class, ketones and carboxylic acids cannot get oxidized. However, in my experiment the ketone somehow is not getting oxidized while the propanoic acid is. I tried using butanoic acid just to be sure but it turns out that it gets oxidized as well. I have also made sure that there are no contaminants. I used 4 ml of 100% propanoic acid with 0.003 mol of $\ce{KMnO4}$.

Can you detail the concentrations of both the acid as well as $\ce{KMnO4}$ and also the time it took to de-colourise?

I tried an experiment however, I only have Glacial Acetic Acid, which didn't seem to react ( as the smell was still prominent).

I can suggest you a simple trick to analyse whether it is decomposing. Smell Propionic acid before adding $\ce{KMnO4}$. Then smell it after the colour of $\ce{KMnO4}$ is gone. Keep adding $\ce{KMnO4}$ till the pink colour stay(similar to titration). Then smell again. If the smell is gone then the carboxylic acid is oxidised.

There are 2 possible oxidation products of carboxylic acid:

Ketone
Carbon Dioxide

Both can be pretty obvious. Ketone will smell differently than the acid(more like nail polish remover) and Carbon Dioxide should be visible because of effervescence.


How to condense a melted wax in to small aggregate forms like stones of different sizes? I am an Architect trying to do some experiments with wax. Normally when the wax is melted and poured on to a surface and it takes the shape of the surface with some strings formed . But instead I would like to condense the wax before it hits the surface so it becomes a small aggregate of varying sizes in the form of a stone . How can this be achieved ?

I want it to condense quickly before it hits the surface once its relieved from the melting point and gets in contact with the room temperature of 12-14 Degrees celcius.Should any solvent be added to the melted wax to achieve this state I wanted?

Your suggestions are most welcomed.

Part of your problem is that wax is an amorphous solid and a mixture of various compounds.  This causes it to have no specific temperature that can be defined as a melting point, but rather a range of temperatures where it transforms from a hard brittle solid to a soft plyable solid to a liquid.
The best I can recomend is that you first pour very hot wax into a much larger amount of ice cold water to form wax shot.  Then take the shot and a propane torch, torch the wax beads as they fall to partially melt the surface of the shot and that may get you your desired effect.


Why are triamino compounds so rare? I was doing a reaxys substructure search for the compound shown on the left below, with the 3 amines on 1 carbon. However, the closest I found were all trinitro compounds like on the right. I would have thought the nitro compounds would be less stable and likely to explode. But given the lack of triamine compounds, they must be hard to make. All I can figure is that when all three amines are protonated you end up packing a lot of postive charge in a small place.

Now that I look at that nitro structure I'm pretty sure I got the charges wrong. Just ignore that.

You may be familiar with the equilibrium that exists between a gem-diol and the corresponding carbonyl compound, as shown in the figure below.  The carbonyl double bond is very strong, so in most cases the equilibrium lies far to the carbonyl side.

The same type of equilibrium exists in the case of bis- and tris-amino compounds where all of the amino groups are attached to the same carbon atom.  In the case of such a tris-amino compound (see figure) the imino analogue of a carbonyl is generated, but since a third amino group is present, another equilibrium between the imino compound and a nitrile also exists.  Just like in the carbonyl case, the nitrile triple bond is very strong.  So when a tris-amino compound is generated an equilibrium with a very stable nitrile will be set up and the equilibrium will (usually) lie far to the nitrile side.  Most tris-amino compounds prefer to exist as nitriles.  If the amino group is fully alkylated (all of the amino hydrogens are replaced by alkyl groups), then this equilibrium cannot occur and such alkylated tris-amino compounds should be isolable.


Purpose of electrolyte in a galvanic cell I am confused as to why galvanic cells have electrolytes. For example, a zinc-hydrogen galvanic cell has a zinc sulfate anode electrolyte and hydrochloric acid cathode electrolyte. The overall reaction of this may be written as:

$$ \ce{ZnSO4 + 2HCl -> ZnCl2 + H2SO4} $$

But rather, it is written as:

$$ \ce{Zn(s) + 2H+(aq) -> Zn^2+(aq) + H2(g)} $$

Why is this the overall reaction? What is then the purpose of an electrolyte, if only the anodes and cathodes are written in the overall equation?

Unless I'm badly misunderstanding your question, I think your confusion arises from an error in the balanced chemical equation you wrote.  As you have it, a salt is reacting with a strong acid to make another strong acid and a new salt. This reaction may occur spontaneously as written (I haven't figured its Gibb's free energy) , but it's not the oxidation-reduction reaction that drives a galvanic cell.  In fact, it's not an oxidation reduction at all, it's some sort of acid-base reaction, and all of the reactants and products remain in solution anyway.

The actual chemistry that occurs in this system is between the zinc anode and the acid.

$\ce{Zn + 2 HCl -> ZnCl2 + H2}$

From this it is much clearer why the net ionic equation is as you wrote.  The zinc sulfate could be replaced with zinc chloride with almost no effect on the battery's potential.  In fact, the zinc solution exists only to create standard conditions for the cell and to carry electrical current through the solution - if it were replaced by a nonreactive electrolyte solution the cell would still function, albeit at a different (higher) potential.

You might then ask what the purpose of the hydrogen cathode is.  Again, it's to create standard state.  If you did this in reality, you would be using some sort of inert electrode at the hydrogen end to carry the electrical current into the acid solution and likely would just vent the hydrogen to the surroundings.

In short, this galvanic cell does not function because of a reaction between its electrolytes, nor because of a reaction between its electrodes.  Instead it is ultimately a reaction between the anode and the cathode's electrolyte.  The anode's electrolyte and cathode are required for electrical continuity and are used in examples of this sort to put the cell in standard state.  Beyond that they don't serve much of a purpose.


Alkylation of alkenes by nucleophilic substitution A double bond/alkene reacts as a nucleophile. I have never seen a reaction where it takes part in an $S_N1$ or $S_N2$ reaction with a halogenoalkane for example. This would presumably be a very useful reaction yet I have never seen it. I assume this is because it doesn't work. Is this the case? If so why?

The pi electrons in an alkene or alkyne can act as a nucleophile.  This class of reactions is often taught a bit later than the standard $\ce{S_{N}1}$ and $\ce{S_{N}2}$ reactions and comes up when neighboring group participation or anchimeric assistance is covered.

The pi electrons in a double or triple bond are not strongly nucleophilic but when you arrange things in their favor, pi electrons can act as a nucleophile.  The reactions involving double bonds that I am familiar with involve an $\ce{S_{N}1}$ mechanism.  Here are a few examples.

(image source)

In example (1) notice how the compound with the double bond solvolyzes $10^{11}$ times faster than the saturated model compound!  Note too that only the product with acetate group anti to the double bond is formed.

Triple bonds and benzene rings can participate as well.  Often in cases involving double or triple bonds new carbon-carbon bonds are formed between the unsaturated linkage and the carbon with the leaving group [examples (2) and (3) above]. In cases involving aromatic rings aromaticity is restored so no new carbon-carbon bonds are formed between the aromatic ring and the carbon with the leaving group.


Bubble formation when aluminium reacts with copper(II) sulfate solution I recently did an experiment putting pieces of $\ce{Al}$ and $\ce{Fe}$ into copper(II) sulfate solution and I noticed when the piece of aluminium was reacting, tiny bubbles were coming out whereas with the piece of iron this didn't happen.

I think it was part of the bare aluminium that was reacting with water producing hydrogen gas but I'm not sure. Can anyone please explain this to me?

As Ivan Neretin noted in a comment, your supposition is absolutely correct.  The aluminium metal is reactive enough with water that it forms hydrogen gas and aluminum hydroxide (or other aluminum oxyhydroxide species) where it's exposed to the solution:

$$
\begin{align}
\ce{6H2O + 6e-} & \ce{->~ 6OH- + 3H2\uparrow} \\
\ce{2Al(s) + 6H2O } & \ce{->~2 Al(OH)3 + 6H+ + 6e-}
\end{align}
$$

Overall reaction:

$$
\ce{2Al(s) + 6H2O -> 2Al(OH)3 + 3H2\uparrow}
$$

If the $\ce{CuSO4}$ solution is acidic enough, the $\ce{Al(OH)3}$ rapidly dissolves, exposing a fresh aluminium surface, ready for further reaction.

On the other hand, the redox reaction of $\ce{Fe(s)}$ and water is not favorable enough for hydrogen evolution to occur, which explains the lack of bubbles you observed.  The single-displacement reaction between $\ce{Cu^2+}$ and $\ce{Fe(s)}$ is quite favorable, however, leading to paired reactions like the following:

$$
\begin{align}
\ce{Fe(s)} & \ce{->~ Fe^2+ + 2e-} \\
\ce{Cu^2+ + 2e-} & \ce{->~ Cu(s)}
\end{align}
$$

Overall reaction:

$$
\ce{Fe(s) + Cu^2+ -> Cu(s) + Fe^2+}
$$

This results in the rapid exchange of metallic Cu for metallic Fe on the iron surface, leading to a film of plated copper on the surface.


Is resonance always stabilizing Does resonance always stabilize a molecule or can resonance effects destabilize molecules?

The only example I can think of in which resonance (more accurately, conjugation) destabilizes molecules is in the cases of anti-aromatic compounds. Is this a valid example, and are there any more such examples?

Resonance only occurs when the result is beneficial, that is when a more-stabilized molecule will result. If resonance (delocalization) occurred in a compound such as cyclobutadiene to produce a square molecule with equivalent bond lengths, an antiaromatic molecule would be the result.  Therefore, resonance does not occur and a square molecule with equivalent bond lengths is not produced.


Why does milk flake? While drinking milk (or better 'while seeing the milk I'm gonna drink') a question came up to my mind:

Why does the milk sometimes flake, even if not in contact with some other substances?

I guess the correct term to use is flocculation, at least this is the word that raises up when talking about cheese production. I just learned the word casein. However, my question is not about cheese production.

My question is also not about flaking milk when poured into a cup of coffee which seems to be dependent on the acidity of the coffee. But casein is again the keyword.

I think the answer to my actual question will also contain the word casein, but here's the scenario:

I open a milk carton for the first time. This package doesn't have any hole and the shelf life isn't reached yet. I pour the milk into a clean cup and then observing some flakes at top of the milk.

What I was told as a child, when curiously asking, flaking is an indicator that the milk is getting acid. And in that case the keyword casein would fit again, but there's still one thing why I'm not persuaded: the milk was quite tasty and my stomach also didn't complain about it.

So, what exactly has happened to my milk when I see some flakes on top of it? Is the milk indeed spoiled and I actually shouldn't have drunk it (and just was lucky) or is there anything else which can lead to flakes?

I think the right word for it is curdle. Have you read, for example, the Wikipedia article on curd? It says:

  “Milk that has been left to sour (raw milk alone or pasteurized milk with added lactic acid bacteria or yeast) will also naturally produce curds, and sour milk cheese is produced this way”

or this excerpt of the “Milk” page:

  When raw milk is left standing for a while, it turns "sour". This is the result of fermentation, where lactic acid bacteria ferment the lactose in the milk into lactic acid.

So, milk naturally contains bacteria which turn it acid, and hence it curdles spontaneously. You see the beginning of this process.

An invaluable source of information on this topic is McGee’s On food and cooking. Great reading if you like chemistry and food.


Why is cooking not potentially lethal? Even the simplest ingredients used in cooking, such as a banana, contain hundreds of chemical compounds. When we cook we mix several of these ingredients and apply heat, which makes a perfect environment for chemical reactions between the various compounds. Actually cooking is the art of initiating the right chemical reactions.

For a typical recipe there are billions of possible reactions between the initial compounds and there are many possible recipes with different initial compounds.

Why are none of the chemical products of this vast space lethal enough to harm a human? In other words why aren't there ingredients that must not be mixed?

Cooking is potentially lethal.
Just like how going out for a walk is potentially lethal (You could get run over by a car, maybe develop a stroke, or get hit by a meteorite).

Cooking or cookery is the art, technology, science and craft of preparing food for consumption.
 From this Wikipedia article on Cooking

The basic idea behind cooking, as a whole, (regardless of the actual physical process) is to either make certain inedible (or indigestible) items  edible (digestible), or otherwise unsavory items more palatable.
Sometimes we have toxic substances that are rendered "safe", through the reactions that the cooking process affords. Other times, the opposite happens.

Now, does cooking (boiling, frying, baking) something generate toxicants?
Almost certainly yes. But the amount and type of toxicant(s) produced is what determines the overall lethality of the item you've cooked.
The potato chips/fries we all hold so dear almost invariably contain acrolein and acrylamides, however in trace amounts. If, you chose to let the potatoes simmer for a little too long, then you'll better be able to taste the wrath of those substances.
It is the dose, that maketh the poison (as I've discussed in a previous answer).

The cooking process does, as you point out, initiate a gamut of reactions: Some simply break down the food into products you can better digest, absorb and assimilate... while others lead to products that aren't as useful.


What is the bond energy $\ce{Cl2}$ bond energy given is as 244.

1/2 $\ce{Cl2}$ will still be 244 or 244/2? My teacher said it is still 244.

The quantity you refer to has units of kJ/mol. Thus, for 1/2 mole of $\ce {\rm Cl}_{2}$, the quantity is halved to 122 kJ/mol. Your teacher might be referring to the quantity on a per-bond basis, but the number you report agrees with references online for a per-mole basis.


Why is enthalpy of atomization of Mn exceptionally low? Transition elements usually have high values, especially so for greater number of unpaired electrons. Why the exception here?

According to "Atomic Structure and Periodicity" by Jack Barrett, the reason is a combination of the $d^5$ electron configuration (half filled subshell) in the gas state and, relative to chromium, "the extra electron in the metal occupying an anti-bonding region of a band, s or d".


Show proposed mechanism is consistent with the experimental rate law The mechanism for the reaction:
\begin{align}
\ce{NO_{(g)} + O_{2(g)}  &<=>[k_{-1}][k_{1}] NO_{3(g)}} &\text{(fast)}\\
\ce{NO_{3(g)} + NO_{(g)} &<=>[K_{2}] 2\,NO_{2(g)}}      &\text{(slow)}\\
\end{align}

The experimental rate law:
$$Rate=-\dfrac{\Delta\ce{[NO]}}{\Delta t}=k\ce{[NO]}^2\ce{[O_{2}]}$$

At equilibrium, the forward and reverse rates are equal to each other. We then have
$$k_{1}\ce{[NO]}\ce{[O_{2}]}=k_{-1}\ce{[NO_{3}]}$$

This can be re-arranged to
$$\ce{[NO_{3}]}=\dfrac{K_{1}}{K_{-1}}\ce{[NO]}\ce{[O_{2}]}$$

Since $2\,\ce{NO}$ disappear in the overall reaction for every NO that reacts in the second step, we multiply the overall reaction rate by two and thus have:

$$2k_{2}\ce{[NO_{3}]}\ce{[NO]}$$

I understand everything up to this point. What I do not understand is how we get to this equation from the previous one.

$$Rate=2k_{2}\dfrac{k_{1}}{k_{-1}}\ce{[NO]}^2\ce{[O_{2}]}$$

The final rate equation is generated from substituting $\ce{[NO3]}$ defined in the third to last equation:

$\ce{[NO3]} = \frac{k_1}{k_{–1}}\ce{[NO][O_2]}$

into the second to last equation:

$2k_2\ce{[NO_3][NO]}$

This will give you the final rate equation defined only in terms of the starting concentrations. The $\ce{[NO_3]}$ term is replaced, $\ce{[NO]}$ squared, and $\ce{[O_2]}$ remains. None of the equilibrium constants cancel out. The intermediate $\ce{[NO_3]}$ is not included in the final equation, as it is consumed in the overall reaction.


What is causing the buffer region in a weak acid - strong base titration? For the buffer region in this, why does it stop the pH from increasing?

At the start:

$$\ce{CH3COOH + NaOH -> CH3COONa + H2O}$$

Which means there will be a high amount of salt in the solution, but why is this stopping the pH from rising? I thought it would actually make it rise further as the disassociated salt it reacts with water to form $\ce{OH-}$?

What's reacting with the $\ce{OH-}$ in solution, that isn't there anyway? Seeing as, with a buffer solution, the acid is needed ($\ce{CH3COOH + OH- -> CH3COO- + H2O}$).

The acid concentration hasn't been increased?

Or is the presence of the dissasocated salt:
$\ce{CH3COONa(aq) -> CH3COO- + Na+}$

Causing the acid to reform:
$\ce{CH3COO- + H2O -> CH3COOH + OH-}$ <<< but there's more $\ce{OH-}$!!

Surely the $\ce{Na+}$ isn't going to react with the $\ce{OH-}$, as it will dissociate in the water.

The linked question's answers were too long for me, so I'll attempt to summarize quickly.

The equation that tells you everything you need to know is the Henderson-Hasselbalch equation:

$$\mathrm{pH} = \mathrm{p}K_{\mathrm a} - \log\frac{\ce{[A-]}}{\ce{[HA]}}$$

Here $\ce{HA}$ is a weak acid, and it's conjugate base is also weak. But we can imagine also a weak base with a weak conjugate acid. The magic happens when you've half neutralized a weak acid or base, creating a mixture of a weak acid and a weak base. The fraction on the right is fairly close to 1 (within an order of magnitude, let's say).

As you add more acid, the weak base can neutralize it. As you add more base, the weak acid can neutralize it. Assuming these amounts are reasonably small, the logarithm of the fraction on the right doesn't change by much, so the overall pH of the solution does not change by much. For example, changing the fraction on the right by a factor of 10 only affects the pH value by $\pm 1$.

Of course, if you add enough acid or base to neutralize (or close to neutralize) one of the buffer components, then the buffer stops working. This is also when the fraction gets very large or goes to zero and even the logarithm can't change that.


What are the differences between the Hilbert space and real space representations in chemistry? I would like to understand the difference between Hilbert space and real space in a molecular code.

My understanding is that the Hilbert space of a system (such as a molecule) can be represented by a set of atomic orbitals (AOs) while the real space represents a system through a grid of points in the specific space of a system. Is this true?

For a more concrete example, consider electronic excitations. Using the entire (orbital) space of a reference system, one can project out (excite) an electron to the virtual space of a reference ground state. However, I'm not sure how this is handled in a "real space" program, or what that even means.

A Hilbert space is a kind of linear vector space.

In chemistry which encounter it when quantum mechanics, when we can represent wavefunctions by their contributions from different orthonormal single particle states.  It is these single particle states which we build up out our atomic orbitals (AOs).

Example:

Consider a two level systems - minimal basis $\ce{H2}$:

Each atom has a $1s$ AO on its hydrogen: $\phi_1(\textbf{r})$, $\phi_2(\textbf{r})$.  These AOs aren't orthogonal though:
$$
\iiint d^3\textbf{r}\ \phi_1^*(\textbf{r}) \phi_2(\textbf{r}) \ne 0
$$but we can construct orthonormal states:
$$
\psi_\pm(\textbf{r}) = \frac{1}{\sqrt{2}}\left(\phi_1(\textbf{r})\pm\phi_2(\textbf{r})\right)
$$
such that like functions integrate to $1$ and functions of unlike functions vanish:
$$
\iiint d^3\textbf{r}\ \psi_\pm^*(\textbf{r}) \psi_\mp(\textbf{r}) = 0 \\
\iiint d^3\textbf{r}\ \psi_\pm^*(\textbf{r}) \psi_\pm(\textbf{r}) = 1
$$.
These two orthogonal function can be represented as state vectors:
$$
\psi_+(\textbf{r}) = \pmatrix{1\\0},\; \psi_-(\textbf{r}) = \pmatrix{0\\1}
$$
This defines a Hilbert or vector space, where our single particle states $\psi_\pm$ are our vectors and the inner product of two vectors, $\psi_i$ and $\psi_j$, is defined by the integral over all space:
$$
\langle\psi_i\vert\psi_j\rangle = \iiint d^3\textbf{r}\ \psi_i^*(\textbf{r}) \psi_j(\textbf{r})
$$
and is equivalent to the dot product of two state vectors.

In a electronic structure (specifically DFT) code, a "real space code" means using atom orbitals as basis functions rather than plane waves.

The Hilbert space described above is complete - it describes all possible wavefunctions which can be made from the AOs included.  If you included the infinite number of hydrogenic wavefunctions on each atom you would be able to have an exact description of all possible wavefunctions for $\ce{H_2}$ - as the hydrogenic wavefunctions form a complete set, instead you include as many as practical.

An alternative complete set of functions is the infinite set of all plane waves of all frequencies.  In cases where you are performing calculations on a periodic system, such as a solid, plane waves can be a more natural choice as they themselves are also periodic.  In a box of width $L$, the set of waves with wavelengths:
$$\lambda = \frac{2L}{n}$$
for all integral n, form a complete set.  They are more commonly described by their wavevectors, $\textbf{k}$, a vector in what is called reciprocal space, which describes the momentum of the wave.

In DFT, when calculating integrals over density, in a plane wave code you manipulate the density in terms of wavevectors, $\textbf{k}$.  In a real space code, you compose your density from Kohn-Sham orbitals in real space and calculate integrals of the density over quadrature grids in real (Euclidean) space.

Hence "real space" is used as a descriptor, for DFT codes that work with AOs and basis functions defined on atoms in Euclidean space, in contrast to "plane wave" codes which used basis functions defined in position space.


Adding Standard Reduction Potentials I am working the following USNCO problem (#41 from 2002). Based on this question and answer Deriving a reduction potential from two other reduction potentials, it seems that $\Delta G$ must be calculated and then added. However, when I do this, I get an answer that is not one of the given choices.

The problem is:

  Use the given standard reduction potentials to determine the reduction potential for this half-reaction:
  $$\ce{MnO4- + 3e- +4H+ -> MnO2 + 2H2O}$$

  The given reactions are:

  $$\begin{align}
\ce{MnO4- + e-} &\rightarrow \ce{MnO4^2-} & E &= +0.564~\mathrm{V} \\
\ce{MnO4^2- + 2e- + 4H+} &\rightarrow \ce{MnO2 + 2H2O} & E &= +2.261~\mathrm{V}
\end{align}$$

  The possible answers are: $1.695~\mathrm{V}$, $2.825~\mathrm{V}$, $3.389~\mathrm{V}$, and $5.086~\mathrm{V}$.

The correct answer is A. Using $\Delta G$, I got $E = 1.928$, but this is not a choice. Am I doing the math wrong, or is my method incorrect? Thanks!

It's somewhat unclear what you mean by "using $\Delta G$", but here's the full working. The information you are given is:

  $$\begin{align}
\ce{MnO4- + e-} &\rightarrow \ce{MnO4^2-} & E &= +0.564~\mathrm{V} \tag{1}  \\
\ce{MnO4^2- + 2e- + 4H+} &\rightarrow \ce{MnO2 + 2H2O} & E &= +2.261~\mathrm{V} \tag{2}
\end{align}$$

Converting these to Gibbs free energy changes is fundamentally the correct step. We have $\Delta G = -nFE$. (Standard state symbols omitted for clarity, they are implied throughout.) For half-equation (1), $n = 1$, and for half-equation (2), $n = 2$. So:

$$\begin{align}
\Delta G_1 &= -(1)(96485~\mathrm{C~mol^{-1}})(0.564~\mathrm{V}) = -54417.7~\mathrm{J~mol^{-1}} \\
\Delta G_2 &= -(2)(96485~\mathrm{C~mol^{-1}})(2.261~\mathrm{V}) = -436306.7~\mathrm{J~mol^{-1}} \\
\end{align}$$

You need to now sum up these two equations, which also means summing up the changes in Gibbs free energy. Let's define a new reaction $(3) = (1) + (2)$:

  $$\begin{align}
\ce{MnO4- + 3e- + 4H+} &\rightarrow \ce{MnO2 + 2H2O} & E &= E_3 \tag{3}
\end{align}$$

As you can see, all I have done is to add the two half-equations together and cancel out any terms that appear on both sides. We have $n = 3$ for this half-reaction.

$$\begin{align}
\Delta G_3 &= \Delta G_1 + \Delta G_2 \\
&= -490724.4~\mathrm{J~mol^{-1}} \\
E_3 &= -\frac{\Delta G_3}{nF} \\
&= 1.695~\mathrm{V}
\end{align}$$

Last thing, please always quote your units in your answer! You cannot have an electrode potential that is unitless.


What is the correctly diagrammed electron probability distribution for p orbital? This is what my book shows (the darker the more probable):

The obvious places where the electron probability is zero is the midsection of the dumbbell.

My question is how the probability distributed inside the dumbbell.

The probability distribution of an electron in the p sub-shell is determined from its wavefunction. The wavefunction of a system contains all the information about the quantum states of the system. Once of the most important properties of the wavefunction is that the squared modulus of the wave function, $|ψ|^2$, is a real number, interpreted as the probability density of the particle. Here is the wavefunction for the 2p orbital:

So if you square this function you will get the probabily distribution which looks like this:

As you can see, this corresponds with the p orbital drawn in your book, as there is a node in the middle, on either side is high probability density and at minus and positive infinity probability density is actually very, very small despite being shown as 0 on the wavefunction. The reason for this is that you need to remember that the orbital shows where an electron will be 90% of the time, not where it will be at all times. This means that the electron can be found outside the orbital.


Why Birch reduction of alkyne is E-selective? From Organic Chemistry(Clayden, Greeves and Warren, 2nd edition), pp. 682:

  The sodium donates an electron to the LUMO of the triple bond (one of the two orthogonal $\pi^\ast$ orbitals). The resulting radical anion can pick up a proton from the ammonia solution to give a vinyl radical. A second electron, supplied again by the sodium, gives an anion that can adopt the more stable trans geometry. A final proton quench by a second molecule of ammonia or by an added proton source (t-butanol is often used, as in the Birch reduction) forms the E alkene.

I have some questions on this reaction:

(1) Double bond on the vinyl anion blocks rotation, but how can anion adopts trans geometry between cis and trans without rotation?

(2) Why anion with trans geometry is more stable? If it is because of steric repulsion between $R^1$ and $R^2$, why electronic repulsion between lone pair and $R^1$ can be ignored?

It's not rotation but inversion (remember amine lone pair inversion) from a linear $sp$ form to trigonal planar $sp^2$ that occurs. Lone pair and radical electron are held close to atom and repulse each other a lot so stable to keep them in anti dihedral positions. If the lone pair next to a radical electron was less repulsive, then it would have a lower energy transition state and be more readily formed according to Hammond's Postulate. But experimental data indicates that that is not the case.

Picture Citation: Kurti (2019)


Why and when is lysine charged I'm reading about Lysine and doing some calculations on it's charge under various $\ce{pH}$.

Depending on the source, Lysine is usually stated as a charged amino acid with a $\ce{NH3+}$ group. However the video I'm watching is calculating the charge at 5 pH and it writes Lysine out with $\ce{NH2}$ and goes on to say that because the solution is pH 5, then $\ce{NH2 -> NH3+}$ and arguments that THIS is why Lysine is charged.

Does the literature state Lysine as a charged acid, assuming it is at pH 7? Or is the video I am watching wrong and it should not gain a charge as $\ce{NH3}$ will not accept a hydrogen?

Bonus: Why are only SOME groups given a Pka value?

http://www.geneinfinity.org/images/aapks.jpg

http://www.cem.msu.edu/~cem252/sp97/ch24/ch24aa.html

How would I calculate the charge for Isoleucin for example?

At pH 7 lysine has a net charge of very close to +1.

The carboxylic acid group is fully deprotonated (-1 charge).

The alpha amino group is about 99% protonated (+1 charge).

The side chain amino group is fully protonated (+1 charge).

You need to look at the pK of each ionizable group, consider what fraction of the group is in each ionization state (for example using the Henderson Hasselbach equation), and add the charges of each group to get the net charge.

For isoleucin, at pH 7, the net charge will be almost neutral, because 7 is well above the carboxylic acid pK of 2.32 and significantly below the amino pK of 9.76.  There will be a small fractional negative charge because 7 is closer to the pK of 9.76 than to the pK of 2.32.  Note in the first linked table in the question, there is a column for "pI".  The pI will be the only pH where the amino acid has a net neutral charge.


Predicting bonds from properties of compounds You are given compounds $x, y$ and $z$. $x$ conducts electricity in aqueous solution as well as in molten state. $y$ conducts electricity in aqueous form only. $z$ conducts electricity in neither the aqueous nor the molten state. Predict the intramolecular bonds in each compound.

Compound $x$ is clearly ionic. In water, the ions disassociate and form free charges for flow of electricity. In the molten state, the ionic bonds break, and again free charges can flow.

I'm not sure about the other two compounds.

A is an ionic salt like NaCl.  Z is covalent and does not solvate to ions when dissolved in water, e.g., sugar, poly(ethylene glycol).  Y is polar covalent and requires solvation to ionize, e.g., HCl.


What is Molar Conductivity of an electrolytic solution Please explain the physical meaning it carries ? And how is it that it can increase with dilution ?

And why is it called Molar ?

And what is equivalent conductivity ?

I will answer your question about the definition of molar conductivity and why it increases with decreasing concentration.

As Jerry pointed out, you can probably learn what you need about equivalent conductivity using a Google search.

The molar conductivity of an electrolyte solution is the conductivity of the solution divided by the molar concentration of the dissolved electrolyte.

Conductivity ($\kappa$, in $\text{S/cm}$) is proportional to the square root of the concentration of ions ($\sqrt{[\text{ions}]}$, withy $\ce{[ions]}$ in $\text{mol/L}$). In the following equation, A is an empirical proportionality constant with units needed to convert from $\text{M}^{1/2}$ to $\text{S/cm}$, namely $\frac{\text{S/cm}}{\text{M}^{1/2}}$

$$\kappa = A\sqrt{\ce{[ions]}}$$

The ion concentration is related to the electrolyte concentration ($C$) by the van't Hoff factor ($i$).

$$[\ce{ions}]=iC$$

$$\therefore \ \kappa = A\sqrt{iC}$$

The molar conductivity $\kappa_M$ is the conductivity divided by the molar concentration of electrolyte ($C$).

$$\kappa_M=\frac{\kappa}{C}=\frac{A\sqrt{iC}}{C}=A\frac{\sqrt{i}}{\sqrt{C}}$$

As $C$ decreases, so does $\sqrt{C}$, and thus $\frac{1}{\sqrt{C}}$ increases, while $A$ remains constant. The van't Hoff factor changes a little with concentration, but not dramatically. ($A$ probably also varies a little with concentration. It certainly varies with temperature.)

For example, if you have a series of solutions of sodium chloride, concentration decreases faster than conductivity, so conductivity divided by concentration increases. Note that the conductivity data are fabricated with an arbitrary value of $A$. I could not find the specific value of $A$ for sodium chloride. The van't Hoff Factors come from the "Colligative Properties" Chapter of Chemistry: Atoms First by Burdge and Overby, McGraw Hill, 2012.


Why doesn't chemical equilibrium lead to uncertainties? Let's consider a reaction

$$\ce{a A + b B <=> c C + d D}$$

We will also assume that the concentrations are low and the activity coefficient is one. Now let's say initially one has $p_0$, $q_0$, $r_0$, $s_0$ as the concentrations of $\ce{A}$, $\ce{B}$, $\ce{C}$, $\ce{D}$, respectively. Let the equilibrium constant be $K$. Let's investigate about the equilibrium composition.

If $x$ equivalents of reactants reacted, then the equation for $x$ would be:

$$K = \frac{(r_0 + cx)(s_0 + dx)}{(p_0 - ax)(q_0 - bx)}.$$

Now, my question is how do we know all the time such equations have proper roots for $x$? I mean all the conditions $x$ needs to satisfy, like $ax < p_0$ and first of all, how do we know the equation has roots in first place? It's a quadratic and it need not have real roots necessarily. This is just a simple case of 2 reactants and 2 products.

I would like to see some rigorous proof that such equations always contain proper (roots which satisfy) all the conditions or a counter example of such a chemical equation. If such counter examples exist, how does the chemical reaction proceed?

The reaction quotient Q approaches zero when at least one of the product concentration approaches zero (as the reaction goes to completion in the reverse direction and the extent of reaction x is at the lowest possible value), and it approaches infinity when at least one of the reactant concentrations approaches zero (as the reaction goes to completion in the forward direction and the extent of reaction x is at the highest possible value). Also, the reaction quotient increases monotonically with x.

No matter what K is, there is always a Q that matches K because Q assumes all values between zero and infinity.

The only real case where this argument does not work (real as in concentrations have to be non-negative real numbers) is when at least one reactant and one product concentration is zero. In that case, there will be no reaction at all, and Q is not defined.


Determining the formula of zirconium oxide I am doing the following question (from the 1988 HSC Chemistry paper):

  In the 19th Century the atomic weight of zirconium (Zr) was known but there was considerable doubt about the formula of zirconium oxide. Experiments determined the relative weight of the volatile chloride to be about 233.

  Assuming that zirconium in the oxide has the same valency, this evidence suggests that the formula for the oxide is:

  (a) $\ce{ZrO}$

  (b) $\ce{Zr2O3}$

  (c) $\ce{ZrO2}$

  (d) $\ce{ZrO4}$

I am not sure what is meant by the sentence 'Experiments determined the relative weight of the volatile chloride to be about 233'. By 'volatile chloride', are they referring to zirconium oxide or something else? It seems that they are referring to something else since a chloride is a compound of chlorine with another element or group.

Also, by the statement 'assuming that zirconium in the oxide has the same valency' are they saying that zirconium has the same valency as chlorine (i.e. -1)?

Any insights are appreciated.

First, let's solve the problem.
Both relative atomic weight $A_\mathrm{r}$ and relative molecular weight $M_\mathrm{r}$ are a historical dimensionless terms denoting relative atomic mass and molecular mass, respectively:

$$A_\mathrm{r} = \frac{m_\mathrm{a}}{m_\mathrm{u}}$$

$$M_\mathrm{r} = \frac{M_\mathrm{B}}{M_\mathrm{u}}$$

where $m$ corresponds to atomic mass, $M$ – to molecular mass, and index $\mathrm{u}$ refers to unified atomic mass unit ($= m_\mathrm{a}(\ce{^{12}C})/12$).
Since $\ce{Cl}$ in chlorides is monovalent, then formula of unknown zirconium chloride is $\ce{ZrCl_x}$, where $x$ coincides with the valency of $\ce{Zr}$.
From this the relative weight of $\ce{ZrCl_x}$ is

$$M_\mathrm{r}(\ce{\ce{ZrCl_x}}) = A_\mathrm{r}(\ce{Zr}) + xA_\mathrm{r}(\ce{Cl})$$

relative atomic masses of both elements were known ($A_\mathrm{r}(\ce{Zr})\approx 91$, $A_\mathrm{r}(\ce{Cl})\approx 36$), so that the valency of zirconium can be found via $x$:

$$91 + 36x = 233 \implies x \approx 4$$

Since oxygen in oxides has valency of $2$, then the formula of unknown zirconium oxide is $\ce{ZrO2}$.

  By 'volatile chloride', are they referring to zirconium oxide or something else?

No, they refer to zirconium chloride $\ce{ZrCl4}$.
They mention volatility probably to underline how zirconium and hafnium are separated — by differences in sublimation of their volatile compounds.

  Also, by the statement 'assuming that zirconium in the oxide has the same valency' are they saying that zirconium has the same valency as chlorine (i.e. -1)?

No, this means that zirconium in both oxide $\ce{ZrO2}$ and chloride $\ce{ZrCl4}$ has the same valency.


Proper use of vacuum pump during vacuum distillation I am preparing to do a vacuum distillation and am unsure exactly how to manage the vacuum. The vacuum pump I have is a two-stage rotary vane pump which is capable of 250 microns or lower. It has a triple inlet, one 1/4", 3/8" and 1/2".

My glassware is typical distillation glassware including a Graham condenser and a vacuum takeoff joint that connects between the condenser and the receiver.

My concern is that if I apply too much vacuum that it will cause a glass implosion. Is this not a concern and I can just operate the vacuum pump freely, or do I need to take special steps to limit the degree of vacuum?

I watched a YouTube video of a guy doing a vacuum distillation and I noticed that he did not use a capillary bubbler. Under what circumstances can the bubbler be dispensed with?

Another concern is the need for a manometer. I have read that I need to use a closed-end manometer (very expensive) to measure the pressure in a vacuum distillation. Why can't I use a normal vacuum gauge? Also, if I am doing an azeotropic separation such as a water-ethanol separation, is it critical I monitor the pressure? I am assuming it is.

This pump is way overkill for a vacuum distillation. The problem is not glass implosion necessarily (round bottom jointed glassware is designed to be put under vacuum, less clear on the condenser), but rather that you may not be able to condense your product given the low vacuum. Vacuum distillation is typically done with an aspirator that gets down to 25 Torr (25,000 microns) or so. At too low a pressure, you may need a very cold receiving vessel and condenser, and depending upon the solvent dry ice slush may not even be enough. You need to determine the boiling point of your solvent at low pressures to figure this out. Your pump is for Schlenk lines and similar applications.

I don't see a reason not to use a vacuum gauge (and you are not supposed to use mercury anymore in any application), though again, depending upon the setup, it is possible to harm the gauge. You should try to monitor pressure as part of the experimental process.

Not using the capillary bubbler seems foolish, as bumping is a major problem with vacuum distillations. A capillary bubbler is a very thin tube (like when you draw glass tubing after melting it) that goes into one joint of the stillpot (drawn end in the reaction mixture). When vacuum is applied, air is forced through the tube, but because the drawn end is so narrow it only allows small bubbles into the mixture. These take the place of boiling chips which will not form bubbles as all gases are extracted from the chips via the vacuum.


Rust removal ended with black oxide on the metal? I have tried to remove common red iron rust from a small piece of metal that looks like steel.

I submerged it into hot water with a bit of citric acid, and let it sit overnight. In the morning I have no rust, but the thing is now covered in dark-black oxide.

Was that possibly steel that I was cleaning or something else? How do I remove that black oxide?

How should remove rust in situations like this in the future?!

left - piece from the same metal that has no rust, and was not cleaned, right - piece that was cleaned

In the absence of oxygen iron reacts with water forming $\ce{Fe(OH)_2}$, which can react with rust forming $\ce{Fe_3O_4}$ via the Schikorr reaction.  See https://en.m.wikipedia.org/wiki/Schikorr_reaction.

You should use a stronger acid, which will dissolve the rust and bring the iron ions into solution.  You will, of course, lose the original metal that rusted instead of recovering it, but you can't make such a recovery by "home remedies" anyway.  With such an aggressive approach rethink soaking overnight.


Role of NaOH in synthesis of Dithiocarbamate? What is the role of NaOH in a given reaction. whether It act as a base to removal  proton from amine or just to provide Na ion for the synthesis of sod. dithiocarbamate?

Proposed mechanism

$\ce{NaOH}$ is used as a base to remove the amine proton, which improves the nucleophilic power of the molecule. This way the addition to the double bond in $\ce{CS2}$ is favoured.


How does an irreversible process maintain internal energy while dissipating excess heat? From my understanding, if you were to look at a pressure-volume graph of a reversible, isothermal expansion, the area under it would represent reversible work. Similarly, if you were to graph an irreversible, isothermal expansion on top of this graph, you’d see a smaller rectangular area representing the irreversible work.

What I have been told is that the area between these areas, irreversible and reversible, is wasted energy dissipated as heat. This confuses me, however. If the work the irreversible system does is less, then it should draw less heat from the surroundings (w=-q). So, if it still releases the same quantity of energy, just more of it as heat and less as work, then how would it maintain a constant internal energy? In other words, if the system dissipates the unused work as heat, which is not replenished, then how does it stay at constant T?

The change in entropy of the system in both cases is the same, but, in the irreversible case, the entropy of the surroundings decreased less.  This is because less heat was transferred from the surroundings to the system.  In this sense, all the entropy generated within the system in the irreversible process is transferred to the surroundings.  So the interpretation would be that the "heat generated" within the system causes less heat to be transferred from the surrounding to the system and less work to be done.  In my judgment, this is a desperate attempt to provide some physical interpretation to what is happening.


Why does bleach feel slippery? My high-school chemistry teacher taught us the mantra "bases are slippery-soapy-slimy!"

This question and this answer in Quora say this is due to saponification - conversion of lipids from the top layer dead skin cells into a soap-like substance.

During some household cleaning I ended up with a small amount of a fairly strong solution of bleach in water between my thumb and forefinger and noticed it felt quite slippery, and it took several seconds of rubbing under water for it to stop.

In this case I would guess that saponification is not the explanation for the very slippery feeling. Is there another explanation?

Actually it is saponification.  Bleach has alkali added to it, to stabilize it against decomposion to chlorine gas.

To wash your hands after contact with bleach was a wise move.


Steric Effects in Friedel-Crafts Reactions I recently came across a Friedel-Crafts reaction happening between benzoyl chloride and mesitylene with $\ce{AlCl3}$ as a catalyst. Apparently this reaction doesn't occur due to steric effects even though the mesitylene ring is highly activated by the three methyl groups. This got me wondering whether the reaction between benzoyl chloride with two methyl groups at ortho positions and benzene with $\ce{AlCl3}$ catalyst would occur or not. Most importantly, my question is: where all do steric effects become the major factor in the reaction, and how do we identify that? Why is it so?

Based on literature, I'd say the unsuccessful Friedel-Crafts reaction between benzoyl chloride and mesitylene is not due to steric reasoning. My argument is based on at least one report of Friedel-Crafts acylation on following polymethylated aromatic compounds (Ref.1):

The mono- and di-acylation have been taken place between 1,3-substitutions in each occasion as indicated in the abstract:

Diacylation of mesitylene, durene and isodurene by the Friedel–Crafts method, using the acylating agent in excess (as both solvent and reagent), gives derivatives which are of potential use as monomers. The bis(p-fluorobenzoyl) derivatives, for example, can function as the electrophilic components in poly(ether ether ketone)(PEEK)-type Polymerisations, and the bisnitrobenzoyl compounds are potential sources of novel diamines. Tribenzoylation [although not tris(p-fluorobenzoylation)] of mesitylene has been observed, but dibenzoylation of m-xylene is a much less efficient process than previously reported.

The reasoning for the resistance to dibenzoylation on m-xylene must be rationalized due to the deactivation of in-plane carbonyl $\pi$-electrons of benzoyl group on aromatic ring after first benzoylation (highly possible 1-benzoyl-2,4-dimethylbenzene). All other instances (2-benzoyl-1,3-dimethyl substitutions), the benzoyl carbonyl group is in out of plane with the corresponding aromatic nucleus (Ref.2), probably due to steric hindrance:
$$
\begin{array}{c|ccc}
\text{Compound} & \nu_\ce{C=O} \ (\ce{CH3C=O}) & \nu_\ce{C=O} \ (\ce{C6H5C=O}) & \text{ECA} \ (c.a.) \\
\hline
\text{Acetophenone}  & \pu{1688 cm-1} & - & -\\
\text{Benzophenone}  & - & \pu{1664 cm-1} & 42^\circ\\
\text{2,4,6-trimethylacetylbenzene}  & \pu{1703 cm-1} & - & 73^\circ\\
\text{2,4,6-trimethylbenzylbenzene}  & -& \pu{1672 cm-1}  & 90^\circ, 40^\circ\\
\text{2,4,6-trimethyl-1,3-diacetylbenzene}  & \pu{1705 cm-1} & - & -\\
\hline
\end{array}
$$
$^a$ ECA: Effective Conformational Angle (whenever two angles are given, first refer to the angle with mesitylene ring).
A comprehensive study of mono- and di-acylation of mesitylene have been done and in all cases, at least mono-acylation was achieved in $\gt 90\%$ whenever anticipated Ref:2). The abstract states:

The Friedel–Crafts acylation of mesitylene has been studied in detail. In the acetylation, propionylation or benzoylation reactions mono- or di-ketones may be formed. Mesitoylation leads only to the monoketone. A two-stage acylation procedure is described for the synthesis of acetylbenzoylmesitylene and acetylpropionylmesitylene. Reversibility has been shown to be a factor in these acylations, especially where an aliphatic acyl group is involved. The relationship between the stereochemistry of mesityl ketones and their u.v., i.r., and $\ce{^1H}$ n.m.r. spectra is discussed.

Overall, these results suggest that Friedel-Crafts acylation reaction would not effected by steric hindrence, at least created by 1,3-dimethyl substitutions. For example, 2,4,6-trimethylacetylbenzene (acetylmesitylene) was achieved in 92% yield; 2,4,6-trimethylbenzoylbenzene (benzoylmesitylene) in 83% yield; 2,4,6-trimethyl-1,3-diacetylylbenzene (diacetylmesitylene) in 95% yield; and 2,4,6-trimethyl-1,3-dibenzoylbenzene (dibenzoylmesitylene) in 89% yield (Ref.2).
It is also noteworthy that the nitration of mesitylene is not effected by steric effects as well. You can find more details on this aspect in here.

References:

Brian N. Hendy,   Kevin H. Patterson,   David M. Smith,   Susan E. Gardner, Neil J. Nicolson, “Polyalkylated aromatic monomers and polymers. Part 1.—Friedel–Crafts diacylation of polymethylbenzenes as a means of monomer synthesis,” J. Mater. Chem. 1995, 5(2), 199-204 (https://doi.org/10.1039/JM9950500199).
P. H. Gore, J. A. Hoskins, “Friedel–Crafts acylations of aromatic hydrocarbons. Part VIII. Mono- and di-acylation of mesitylene,” J. Chem. Soc. C 1970, (4), 517-522 (https://doi.org/10.1039/J39700000517).


What is the chemical name of this derivative of aniline? What is the chemical name of the following compound?

$$\ce{(C6H5)-N=CH2}$$

I doubt that it might be N-phenyl methanal oxime . But  , it can't be oxime of OH ion is absent.  What would it be?

The preferred name is N-phenylmethanimine, but often it is named N-methyleneaniline (or N-methylenaniline, more correct one would be N-methylideneaniline though).

UPDATE a typo (excessive "e" in bogus "phenylmethaneimine") fixed. (thanks, Loong)


Adjusting the ph of salicylic acid using sodium hydroxide Question
  How many milliliters of $0.202\ \mathrm M$ NaOH should be added to $25.0\ \mathrm{mL}$ of $0.0233\ \mathrm M$ salicylic acid to adjust the $\mathrm{pH}$ to $3.50$?
  The answer from the text book is $2.2\ \mathrm{mL}$ NaOH
  My attempt
  I believe the correct equation to use for this is the Henderson–Hasselbalch equation

$$\mathrm{pH}=\mathrm pK_\mathrm a + \log\left(\frac{[\ce B]}{[\ce{BH}]}\right)$$

$\mathrm pK_\mathrm a = 2.972$

So I set it up as:

$$3.50=2.972+\log\left(\frac{x}{0.0233}\right)$$

Solving for $x$ I get $0.0785$

I don't know where to go from here
since $x$ is the concentration I figure that if I multiply by $0.025$ (the volume) and then divide by the molarity ($0.202$) I would get the answer but that's not right.

The Henderson-Hasselbalch equation is the correct equation, but ask yourself what you are putting in there:
$$pH=pK_a + \log\left(\frac{c(A^-)}{c(AH)}\right)$$

Since $A^-$ and $HA$ occupy the same volume, you can rewrite:
$$pH=pK_a + \log\left(\frac{n(A^-)}{n(AH)}\right)$$

When adding Sodium Hydroxide consider the following equation
$$\ce{AH + NaOH <=> A- + H2O + Na+}$$

Given that you will find
$$ n(A^-) = n(\ce{NaOH})\\
   n(HA) = n_0(HA)-n(\ce{NaOH})$$

Then you can substitute and solve for $n(\ce{NaOH})$:
$$pH=pK_a + \log\left(\frac{n(\ce{NaOH})}{n_0(HA)-n(\ce{NaOH})}\right)$$

You then divide by the concentration, yielding your volume.

Intermediate results:

Initial amount of $HA$:

       $n_0(HA) = 0.5825\ \text{mmol}$

Amount of $\ce{NaOH}$:

   $n(\ce{NaOH}) \approx 0.4993\ \text{mmol}$

Final Volume:

   $V(\ce{NaOH}) = 3.22\ \text{mL}\ \text{(3sf)}$


Proof of buffers inability to maintain exact same pH I was wondering if the following is a legitimate proof of the statement above. Assume we have

HA -> A- + H+

where [HA] + [A-] is our buffer.

If we add a small amount of H+ one of two things will happen:

1) The H+ will not react with A- ...thereby raising the pH so we are done.

2) The H+ will react with A- ...converting to [HA]

looking at the HH equation we know that

pH = pKa + log(ratio)

since ratio_before != ratio_after

pH_after = pKa + log(ratio_after) and pH_before = pKa + log(ratio_before)

so therefore

pH_after - log(ratio_after) = pH_before -log(ratio_before)

Now assume pH_after == pH_before.... we get -log(ratio_after) = -log(ratio_before) which is a contradiction of above, so pH_after != pH_before.

What are your thoughts?

A buffer is a solution that can resist pH change upon the addition of
  an acidic or basic components. It is able to neutralize small amounts
  of added acid or base, thus maintaining the pH of the solution
  relatively stable.

Buffers do not maintain exact same pH, but keep it relatively stable i.e. close to the original value.
log(ratio) << pKa, therefore, pH changes will be minimal, unless substantial amount of acid has been added to the solution.

For instance, when the ratio between the conjugate base/acid is equal to 1, the pH = pKa.

If the ratio between the two is 0.10, the pH drops by 1 unit from pKa since log (0.10) = -1.

If a ratio increases to a value of 10, then the pH increases by 1 unit since log (10) = 1.

As you can see, changing ratio by 10 times, affects pH value only by 1 unit.


Why do unripe papayas taste less bitter after being boiled? An unripe papaya tastes very bitter if you try to eat it. Actually, the juice that comes from it is so corrosive that it can burn your skin, causing severe pain. I've experienced this first hand when cutting raw papayas.

However, after boiling, these papayas taste quite bland. Very little of the original bitterness remains.

What chemical substances cause the original bitterness, and what happens to them when boiled?

Unripe papaya contains a high concentration of Papain enzyme, which is an enzyme that breaks down peptide bonds and therefore "digests" proteins, including potentially human tissues.  Enzymes themselves are proteins, and proteins tend to have a bitter taste.

Papain can self digest into amino acids over time, and this process could be accelerated by heating, as chemical digestion reactions will approximately double in rate for every 10K increase in temperature.  So one mechanism is that Papain may break itself down quickly when heated.  The other is that at a high enough temperature, Papain may denature.


Do indigo and leuco-indigo have fully conjugated pi systems? I've just come across a figure indicating the positions of conjugated bonds in indigo and leucoindigo, and I'm quite confused about why both molecules aren't fully conjugated.

My reasoning is as follows:

Both carbons above and below the bond are sp2 hybridized and have π electrons.
The carbon to the left is sp2 hybridized and has π electrons, and the nitrogen to the right has a lone pair.
Same as 2.
Same as 1.

I also want to ask about why leuco-indigo isn't planar. According to Wikipedia:

  In indigo white, the conjugation is interrupted because the molecule is nonplanar.

Let's just start with stating the very obvious: Figure 2.23 has more errors than there should be in an entire chapter.

Indigo

With the exception of bonds to hydrogen, all other bonds are nicely conjugated in indigo. The π molecular orbital show this very clearly.[1] The molecule has a fantastic C2h symmetry, which also means it is in one plane. Intramolecular hydrogen bonds certainly further stabilise this conformation. They can be visualised with the help of the Quantum Theory of Atoms in Molecules (QTAIM).[2]

 (click for full image)

Interestingly eleven out of twenty π orbitals are occupied, which leaves the HOMO being anti-bonding with respect to the central carbon-carbon bond, which is 138 pm long.

Going further I will first add electrons, then add protons according to the following scheme:

Indigo2-

I am happy to report, that the molecule stays flat. The central carbon-carbon bond elongates a little to 141 pm. It basically still has comparable orbitals, although the location of the nodal planes already accommodate the divide of the central bond. I am guessing that the intramolecular hydrogen bonds are accountable for this since the molecule is highly charged.

Now the basic question is how do the protons behave, where do they go, will they break the symmetry.

Leucoindigo

A bit to my surprise, the flat version of this compound is also most stable.

Preliminary conclusions

There are several points, where this project could have derailed. Level of theory is only one of them. The evidence presented here is not conclusive enough. In due time I might be able to do some more, but for now, at least the textbook picture is debunked.

Calculated at the DF-BP86/def2-SVP level of theory. Occupied orbitals are blue and orange, virtual orbitals (unoccupied) are red and yellow.
Analysis with MultiWFN 3.3.8, see http://sobereva.com/multiwfn
Nuclei are brown dots, bond critical points are blue dots, ring critical bonds are orange dots. Blue lines are zero-flux surfaces that seoarate the molecule into atomic basins.


Why is acetic acid stronger than p-nitrophenol? I know that acetic acid is a stronger acid than phenol because in acetic acid, the charge is delocalized on two oxygen atoms which is better than 1 oxygen and 3 carbon.

But in case of para-nitrophenol, $\ce{NO2}$ reduces the negative charge in the ring, thereby stabilizing the intermediate formed.

So does this mean that the stabilization caused by the presence of $\ce{NO2}$ is less than the delocalization caused by 2 oxygen atoms?

Kindly correct me if I am going in the wrong direction.

The difference in acidity between acetic acid and phenol can be attributed to the electron-withdrawing properties of a carbonyl function versus a phenyl ring. In both cases, the deprotonated oxygen can be depicted in a resonance structure as forming a double bond to the carbon in question and the charge being delocalised within the ring.

It should immediately strike anybody that a phenyl ring primarily provides a resonance body (i.e. a greater volume into which the charge can be delocalised) while a carbonyl group is also actively withdrawing (both inductively and mesomericly). This is reflected when comparing the $\mathrm{p}K_\mathrm{a}$ values of $\ce{CH3CO-CH3}$ (acetone) and $\ce{Ph-CH3}$ (toluene) — $16.5$ logarithmic units in $\mathrm{DMSO}$ according to the Evans table[1] or approximately $24$ logarithmic units in water ($\mathrm{p}K_\mathrm{a}$ of acetone in water taken from the Brodwell data).[2]

Adding a nitro group to the phenyl ring reduces this difference drastically to $-6.1$ logarithmic units (i.e. para-nitrotoluene is more acidic than acetic acid) — in $\ce{DMSO}$. One should always remember that acidity differences in one solvent can be drastically different in another solvent. For example, the difference between $\ce{MsOH}$ and $\ce{TfOH}$ is $11.4$ logarithmic units in water but only $1.3$ logarthmic units in $\ce{DMSO}$ (that is not a typo). One should also note the difference between acetic acid in water ($\mathrm{p}K_\mathrm{a}\approx4.76$) and $\ce{DMSO}$ ($\mathrm{p}K_\mathrm{a}\approx12.3$).

$$\textbf{Table 1: }\text{Acidity constants of various compounds in various media}\\\text{All data taken from the Evans data unless noted otherwise in the text.}$$
\begin{array}{lrc}\hline
\text{compound} & \mathrm{p}K_\mathrm{a}(\ce{H2O}) & \mathrm{p}K_\mathrm{a}(\ce{DMSO})\\ \hline
\ce{CH3CO-CH3} & 19.3\phantom{000} & 26.5\\
\ce{Ph-CH3} & 31.5\phantom{000} & 30.6\\
\ce{$p${-}NO2-C6H4-CH3} & & 20.4\\
\ce{MsOH} & -2.6\phantom{000} & \phantom{0}1.6 \\
\ce{TsOH} & -14\phantom{.0000} & \phantom{0}0.3 \\
\ce{AcOH} & 4.76\phantom{00} & 12.3\\ \hline \end{array}

Given that a $\mathrm{p}K_\mathrm{a}$ value of para-nitrotoluene has not been reported in water (although it’s $\mathrm{p}K_\mathrm{b}$ value has been),[3] we are not in any position to draw definite conclusions. However, water being the solvent it is, I would still assume the electron-withdrawing ability of a nitro group which relies solely on mesomeric effects to be weaker than the effect of a directly neighbouring carbonyl group. Remember that water is excellent at stabilising charges; $\ce{DMSO}$ not so much.

References:

[1]: The Evans $\mathrm{p}K_\mathrm{a}$ table

[2]: The Brodwell $\mathrm{p}K_\mathrm{a}$ data

[3]: J. C. D. Brand, J. Chem. Soc. 1950, 997. DOI: 10.1039/JR9500000997.


Polymers which are miscible in oil Can someone suggest me a list of polymers or some paper where I may find polymers that are soluble in oils like silicon oil, mustard oil, coconut oil, bean oil or petroleum ether, etc.?

Plastics Design Library series include tabulated parameters for the variety of polymers.
One of such parameters is a PDL number rating from 0 to 9:

0: solvent dissolved disintegrated
1: decomposition
2: severe distortion; oxidizer and plasticizer deteriorated
…
9: highest resistance, no change

In the table below I assembled data on thermoplastics chemically unstable towards various oils with PDL numbers below 3:

$$
\begin{array}{lrcll}
\hline
\textbf{Polymer} & \boldsymbol{T/\pu{°C}} & \textbf{PDL} & \textbf{Oil} & \textbf{Resistance Note} \\
\hline
\text{CAB} & 20 & 0 & \text{spearmint oil} & \text{Limited Resistance} \\
\hline
\text{LDPE} & 20 & 1 & \text{aniseed oil} & \text{Not resistant; tensile strength at yield and elongation at break greatly reduced} \\
\hline
\text{HDPE} & 20 & 2 & \text{camphor oil} & \text{Not resistant} \\
\hline
\text{PP} & 100 & 2 & \text{two-stroke engine oils} & \text{Unsatisfactory/severe effect} \\
  & 20 & 2 & \text{turpentine oil} & \text{Not resistant} \\
  & 100 & 2 & \text{vaseline oil} & \text{Unsatisfactory/severe effect} \\
\hline
\text{LLDPE} & 20 & 1 & \text{aniseed oil} & \text{Not resistant; tensile strength at yield and elongation at break greatly reduced} \\
\hline
\text{PS} & 22 & 2 & \text{clove oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 22 & 2 & \text{lemon peel and oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 22 & 2 & \text{orange peel and oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 22 & 2 & \text{pine needle oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 22 & 2 & \text{spearmint oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
\hline
\text{SAN} & 23 & 2 & \text{citronella oil} & \text{Severe attack; softened in few hrs} \\
  & 23 & 2 & \text{clove oil} & \text{Severe attack; softened in few hrs} \\
  & 22 & 2 & \text{lemon peel and oil} & \text{Not resistant; plastic severely crazed; softened or dissolved} \\
  & 52 & 2 & \text{pine needle oil} & \text{Severe attack; softened in few hrs} \\
  & 23 & 2 & \text{spearmint oil} & \text{Severe attack; softened in few hrs} \\
\hline
\text{ABS} & 52 & 2 & \text{pine needle oil} & \text{Severe attack; softened in few hrs}  \\
  & 23 & 2 & \text{spearmint oil} & \text{Severe attack; softened in few hrs} \\
\hline
\end{array}
$$

Acronyms

$$
\begin{array}{ll}
\text{CAB} & \text{Cellulose Acetate Butyrate} \\
\text{LDPE} & \text{Low Density Polyethylene} \\
\text{HDPE} & \text{Polyethylene} \\
\text{PP} & \text{Polypropylene} \\
\text{LLDPE} & \text{Linear Low Density Polyethylene} \\
\text{PS} & \text{Polystyrene} \\
\text{SAN} & \text{Styrene Acrylonitrile Copolymer} \\
\text{ABS} & \text{Acrylonitrile Butadiene Styrene}
\end{array}
$$

References

Chemical Resistance of Thermoplastics; Woishnis, W. A., Ebnesajjad, S., Eds.; Plastics Design Library; William Andrew: Norwich, N.Y, 2012.


Equation for heat loss over distance? Is there an equation to figure out what the temperature of an the area will be a certain distance from the heat source?

For example, if I know there is a heat source at $0~\mathrm{m}$, and the temperature is $100~\mathrm{^\circ C}$, is there a way to find out what the temperature will be at $50~\mathrm{m}$ away from the heat source? Is there some sort of equation for this? All of this is happening in oxygen.

This is a heat transfer related problem. You can certainly do that. For solid medium heat transfer you need to use Fourier heat transfer equation
\[q=-k\cdot\nabla T\]
Here, k= material's conductivity and $\nabla T$ = 1st derivative of the Temp. along the driving force or heat
For Liquid and gas phase you should use law for thermal convection. And for vacuum medium you should use radiation law. There are lots of ways to do this but from your problem statement i think it's a numerical grid type problem. For your better understanding, i would suggest 'Engineering heat transfer' by william s janna (I may be biased here because i like this book) but you can certainly use any other standard heat transfer book or internet.


Alternative explanation of ethene's pi bond This tutorial about ethene's bonds has the following paragraph that I hope someone can paraphrase in a better way:

The pi bond dominates the chemistry of ethene. It is very vulnerable to attack - a very negative region of space above and below the plane of the molecule. It is also somewhat distant from the control of the nuclei and so is a weaker bond than the sigma bond joining the two carbons.

What does it mean by:

"vulnerable to attack"?
"negative region of space"?

Here is the meaning of the words ""vulnerable attack" and "negative region of space".
The nucleus of the six atoms (2 C and 4 H) are situated in the same plane, which is usually defined as horizontal. The electrons of the 4 H atoms and the three first electrons from each Carbon atom are all together included in sigma bonds, and their maximum probability of presence is situated in this plan. These bonds are made of negative charges, that should be able to attract positive charges. But these bonds can hardly attract any arriving positively charged ion, because this attraction would be entirely compensated by the repulsion due to the neighboring nuclei.
The pi electrons are very different. Their maximum probability of presence is not situated in the plan. They make a sort of double long cloud high above and much under the plan of the molecule. They produce a sort of negative cloud (or a negative region of space) surrounding the molecule above and under the plan.
Electrons are negatively charged. So they may attract positive charges. If a positive ion like $\ce{H+}$ or $\ce{Br+}$ approaches an ethene molecule, it will be attracted by this negative $\pi$ cloud. If this approach leads to the formation of a new bond between the ethene molecule and the coming ion, it makes sense to state that the coming ion attacks the ethene. One may also state that the $\pi$ cloud attacks the coming ion. In other words, it also makes sense to say that the $\pi$  bond of ethene is vulnerable to attack, at least more vulnerable than the sigma bonds in the plane.


Why do cyclic hydrocarbons have higher boiling points than their acyclic isomers? As pointed out in the comments to this question, cyclic hydrocarbons have higher boiling points than their acyclic isomers. The major attractive force for hydrocarbons should be the London forces, which scale with surface area. My intuition has me thinking the cyclic molecules have less surface area, but they might enable more productive interactions over larger parts of their surface area due to their fixed conformations. I would love to see a study or some data that supports my intuition.

Boiling point data

  Three carbons

Cyclopropane - BP $=-33\ ^\circ\text{C}$
Propene - BP $=-47.6\ ^\circ\text{C}$
Propane - BP $=-42\ ^\circ\text{C}$

  Four carbons

Cyclobutane - BP $=12.5\ ^\circ\text{C}$
Methylcyclopropane - BP $=4\ ^\circ\text{C}$
1-butene - BP $=-6.7\ ^\circ\text{C}$
Butane - BP $=-1\ ^\circ\text{C}$

  Five carbons

Cyclopentane - BP $=49.2\ ^\circ\text{C}$
1-Pentene - BP $=30\ ^\circ\text{C}$
Pentane - BP $=35.9\ ^\circ\text{C}$

  Six carbons

Cyclohexane - BP $=80.7\ ^\circ\text{C}$
1-Hexene - BP $=63\ ^\circ\text{C}$
Hexane - BP $=68.5\ ^\circ\text{C}$

Your intuition is indeed correct! Several sources provide the same answer (1, 2, 3). Perhaps the simplest and most direct evidence comes from comparing the densities of the liquid unbranched alkanes and cycloalkanes:

(Source)

The cycloalkanes have slightly lower molecular mass than their parent unbranched alkanes with the same number of carbon atoms, yet the cycloalkanes are consistently more dense in the liquid phase by a factor of ~20%. The only way this can happen is if each individual cycloalkane molecule effectively occupies less volume in the liquid phase than the related alkane. This means cycloalkane molecules are closer on average in the liquid, and as van der Waals forces are attractive (at intermolecular separations) and their strength varies inversely with distance, the intermolecular attractions are stronger in cycloalkanes, and so the boiling point of the cycloalkanes should be higher.

This can be explained by considering that the unbranched linear chains in alkanes have many possible conformations, and thus are in more geometrically irregular contact in the liquid, leaving more space on average between the molecules. Cycloalkanes do not have as many conformations available, and so their molecules will approach each other more orderly and effectively, leaving less intermolecular space.

Presumably this effect can be extended to all substances (including the ethers mentioned in the question you linked), but it need not always be the dominant factor in determining boiling points.


Populations of a two level system Take the two level system in atomic iodine where the ground state is four fold degenerate and the excited state at $7603 \,\mathrm{cm}^{-1}$ above the ground state (which is taken to have zero energy) and is two fold degenerate from the $^2P_\frac32$ and $^2P_\frac12$ terms arising from the configuration of atomic iodine.

I want to find the temperature at which the ratio of the population of the upper state to the lower state is 1:50.

Using statistical mechanics is it correct to say that this is the case when:

$$
\frac{n_u}{n_l} = \frac{g_{u}\mathrm{exp}(-E_u/k_{\mathrm{B}}T)}{g_{l}\mathrm{exp}(-E_l/k_{\mathrm{B}}T)} = \frac{2\mathrm{exp}(-7603.hc/k_{\mathrm{B}}T)}{4} = \frac{1}{50}
$$

Rearranging for T and substituting in the values of the constants gives me $3401.9 \,\mathrm{K}$ which seems alarmingly high. Where have I gone wrong?

I'm getting basically the same answer when I work out the math. I got $T=3398.45\ K$, but we probably just rounded the constants differently.

This is a good lesson in how much thermal energy is carried around at any given time by molecules in the air, as well as the importance of distinguishing between thermal processes and photo-processes.

That is, if we wish to sustain an ensemble of excited particles, there must be sufficient energy such that when we spread the energy out evenly, which is one of the things that this equation implies given that it claims all possible microstates are equally likely, some of the energy must be stored in the form of electronic excitations.

The reason this probably seems like a very high temperature is because we use equipment all the time that relies on electronic excitations and we never have problems with the equipment melting due to it being at thousands of degrees Celsius. That is simply because in these photo-process, meaning the energy is carried in the form of photons, the temperature never is that hot.

That is, if I shoot a photon in, or even a stream of photons at a sample with the transition you are describing, each photon carries a very small amount of energy, and even though there may be quite a lot of them, the energy is quickly dispersed because the excited state is very short-lived, so the photon gets re-emitted, and likely just makes it way out into an infinite reservoir known as the surroundings.

Thus, it is very difficult to maintain a population of excited states thermally, because the system needs to basically be a closed system or else it will eventually just radiate all the energy away.

Another way of thinking about this is that the value of $k_bT$ at room-ish temperature of $298\ K$ (in a familiar unit) is $207\ cm^{-1}$. This makes it clear then why rotationally excited states are frequently occupied at room temperature, because these transitions are on the order of tens of wavenumbers. On the other hand, we often get off free assuming everything is in the vibrational ground state because these transitions are on the order of thousands of wavenumbers.

This is also where the idea of rotational temperature and vibrational temperature come from. These can be understood physically as,

  an estimate of the temperature at which thermal energy (of the order of $k_bT$) is comparable to the spacing between rotational [or vibrational] energy levels.


Relation between various bond angles of Nitrogen compounds I was thinking what could be order of bond angles of NH3, NF3, N(CH3)3 and N(C2H5)3.

Considering  NF3 and NH3:

Since there is backbonding between 2p-2p orbitals of N and F, there will be a partial double bond character between N and F and therefore, since repulsion between lone pair and double bond is more than that between lone pair and single bond and thus bond angle in  NF3 will be less than NH3

But, the size of F is very small, and F has lone pairs while H does not. If the N-F bond acquires partial double bond character, the lone pair of other F atoms should repel the bond making the F-N-F angle larger? Would the two effects cancel each other making the bond angle of  NF3 equal to  NH3

It is known that bond angle of  NF3 is less than NH3. Why? How do I decide which factor will dominate over other?

Now since N(CH3)3 and N(C2H5)3 have bulky groups, the groups will repel each other making the bond angle greater than NH3.

But between N(CH3)3 and N(C2H5)3 , which will have greater bond angle? I suppose C2H5 will repel other C2H5 groups more than CH3 groups.

Again there is another factor to be considered here. There are more numbers of bonds in C2H5 and shape is tetrahedral which might (there is a possibility) bring the CH3 connected to the C (which is connected to N) closer to N hence increasing repulsion between the group and N and thus decreasing C2H5-N-C2H5 bond angle.

Again, why does this not happen?

Can anyone explain the correct order between these molecules?

$\ce{NH3}$ and $\ce{NF3}$ are easier to analyse and predict since there is only one set of atoms around the nitrogen — not three groups of four or three groups of seven as for trimethyl- and triethylamine, respectively.

Note that you did not (but should!) include bond lengths in your analysis. It is important to remember them. For ease, here they are as taken from Wikipedia (along with the bonding angles):

$$\textbf{Table 1: Bond lengths and angles}\\\textbf{of nitrogen compounds}\\
\begin{array}{lcc}
\hline
\text{compound} & d(\ce{N-X}) & \angle (\ce{X-N-X})\\
\hline
\ce{NH3} & 102~\mathrm{pm} & 107.8^\circ\\
\ce{NF3} & 137~\mathrm{pm} & 102.5^\circ\\
\ce{NMe3} & 146~\mathrm{pm} & 110.9^\circ\\
\hline
\end{array}$$

Starting with the difference between $\ce{NH3}$ and $\ce{NF3}$ — oh wait, I already answered that once. The gist is both compounds would love a $90^\circ$ bond angle but cannot due to too small bond lengths — the outer atoms would clash into each other, figuratively speaking. Therefore, the bond angles are expanded (at a loss of energy) to reduce this steric strain (gain of energy) up to an optimal sweet spot. Since the $\ce{N-H}$ bond is shorter than $\ce{N-F}$, the fluorines are comfortable with a slightly smaller bond angle as they are further away from each other (even though they are larger).

The difference when going from fluorine atoms to methyl groups (or ethyl groups) — as highlighted above — is the additional set of atoms. These cause further crowding, even though the slightly larger carbon atom (with respect to fluorine) allows for an even longer bond length. The steric strain of the additional hydrogens means that the bond angle must be extended even further to accomodate for all these.

Not much changes when going from trimethyl to triethylamine — except that the steric strain is larger still, and that therefore the bond angle is probably increased slightly again. One source I found including a crystal structure including triethylamine did not, however, give notably different bond angles when compared to trimethylamine, so this point may be moot. Note, however, that the angles reported are already larger than the perfect tetrahedral angle, indicating that the lone pair has an even lower s-proportion which would destabilise the overall molecule. Only by reducing steric strain can this low s-proportion be explained in any way.

As I already commented on the question: Back-bonding will not happen since there are no accessable empty orbitals on nitrogen that could accept a back-bond. Try drawing a back-bond in a Lewis formula without exceeding nitrogen’s octet to see that it is impossible.


Deviation of real gases from ideal behaviour From the Amagat's curve we can see two types of result for real gases.

If we increase pressure,the value of $PV$ decreases initially and then the value starts increasing for the gases like- carbon dioxide, nitrogen, oxygen, methane, etc..
If we increase pressure,the value of $PV$ starts increasing initially for the gases like - hydrogen, helium, etc.

Now please explain what're the reasons behind this kinda behaviour of real gases?

The ideal gas law can be derived using kinetic molecular theory by making two very important assumptions that are not true for real gases:

There is no attractive force between gas molecules
The gas molecules have no volume

Think about how the PV curve would change when you remove one or both of these conditions. Then think about which types of molecules the assumption of "no attractive force" would be more important for.


On biological detergent breaking down organic material This is again related to the crime story I'm working on.

I would like to know how much biological detergent (in kilograms) and water (in liters) and how long it would take to dissolve $60 \,\rm{Kg}$ of organic material at $50^\circ$ Celsius with standard store-bought bio-detergent. Just an estimate of the time required.

Bonus questions:

Would it dissolve hair?
Do I need a special container as to not commit the same mistake Jesse had in Breaking Bad?

In the household, hair and fat residues in the pipes are typically dissolved using strongly alkaine drain cleaners.

In addition to the hints and directions given in the commentaries, namely this and that you might want to consider, at least in part and for the dramatic effect, a biological solution:

In the preparation of skeletons for museum displays, adult insects and larvae of the bacon beetle (Dermestes lardarius) are frequently used. After gutting the carcass (I'll leave the gory details to the novellist), plan a couple of weeks for them.


Compressibility factor of hydrogen at low pressure and constant temperature As we all know, the compressibility factor $Z$ of hydrogen and helium is always greater than $1$ at a constant moderate temperature. But, if we look at the virial equation of compressibility factor

$$Z = 1 + \frac{1}{V_\mathrm{m}}\left(a - \frac{b}{RT}\right) + \frac{b^2}{V_\mathrm{m}^2} + \cdots$$

At extremely low pressure $V_\mathrm{m}$ would be extremely large. So, the equation simplifies to

$$Z = 1 + \frac{1}{V_\mathrm{m}}\left(a - \frac{b}{RT}\right)$$

Now, the values of $a$ for hydrogen and helium are extremely small. So, we can take them as zero. The equation further simplifies to

$$Z = 1 - \frac{1}{V_\mathrm{m}}\left(\frac{b}{RT}\right)$$

Now, the value of $b$, $R$, $T$ and $V_\mathrm{m}$ are all positive. Though the value of $b$ for hydrogen is extremely small, still it is positive. So, shouldn't the value of $Z$ at some small pressure be less than $1$?

Any help would be appreciated.

In my textbook the opposite meaning is given to $a$ (an attractive parameter) and $b$ (a size parameter), but this may be a matter of differences in labelling, not in meaning. I will for the sake of consistency stick to your convention. For $\ce{He}$ the van der Waals parameters have the following values:

$b = \pu{3.4598 J mol-1 M-1}$
$a = \pu{0.023733 M-1}$

As you can see from the following figure at $RT$ ($\pu{298K}$), while your first approximation is quite accurate for $\ce{He}$ already at $V_\mathrm m \approx \pu{1 L/mol}$, your second approximation fails until higher  $V_\mathrm m$. It is negative, but negliglibly so, and at that point is not particularly useful as an approximation as already $Z \approx 1$.

The approximation fails (at RT) because the condition $b/RT \gt a$ is never observed:
$$\frac{b}{RT} = \pu{0.0014 M-1}$$
whereas
$$a = \pu{0.0237 M-1}$$

You can solve for the temperature at which these two become equal:

$$T_b = \frac{b}{aR}$$

For $\ce{He}$, $T_b = 17.53\ \mathrm K$. Below $T_b$ you may observe $Z<1$.

Sure enough, below that $T$ your second approximation starts to hold, as shown here for $T = 10\ \mathrm K$:

or in terms of $P$:

Note: it is not clear that the virial equation and van der Waals parameters should capture the behavior of the gases over a wide range of $P$ or $V_m$ at such low $T$.


Heating CuO and Urea My little brother had to do this experiment at school, but couldn't finish it because of a lack of time. My chemistry knowledge isn't as well as it used to be, so I couldn't be much of an help.

There is a test tube (Tube A) filled with $\ce{CuO}$ and urea. With a glass straw, this is connected to a second test tube (Tube B), which is filled with lime water (which is if I recall correctly $\ce{Ca(OH)2}$?)

A quick sketch:

I really wanted to recreate this at home but I sadly don't have the tools for it.

The question here is: what happens when tube A is heated? (What happens to the material in tube A? What happens to tube B? What is left behind in A? etc).

My guess would be that $\ce{CO2}$ would flow from tube A to B, but I am not sure even about that.

Heating a hydrocarbon with $\ce{CuO}$ generally evolves $\ce{CO2}$ and $\ce{H2O}$. Urea is no exception for this rule.
The $\ce{CO2}$ evolved will turn lime water milky(this is a test for $\ce{CO2}$ gas).


Do dendrites effect lithium-ion polymer batteries? Do dendrites effect lithium-ion polymer batteries?

I recently came across this video which shows dendrites at a microscopic level durring the charge and discharge cycles of a standard Li-Ion battery.

I have come to understand that the primary difference between Li-Ion and Li-Poly batteries is the electrolyte used between electrodes - Li-Ion using a liquid electrode, and Li-Poly using some type of polymer chain.

  Do dendrites form across the polymer the same as the do through a liquid electrolyte?

According to this Berkley Labs publication, "Roots of the Lithium Battery Problem: Berkeley Lab Researchers Find Dendrites Start Below the Surface":

  Using X-ray microtomography at Berkeley Lab’s Advanced Light Source (ALS), a team led by Nitash Balsara, a faculty scientist with Berkeley Lab’s Materials Sciences Division, observed the seeds of dendrites forming in lithium anodes and growing out into a polymer electrolyte during cycling. It was not until the advanced stages of development that the bulk of dendrite material was in the electrolyte. Balsara and his colleagues suspect that non-conductive contaminants in the lithium anode trigger dendrite nucleation.
  “Contrary to conventional wisdom, it seems that preventing dendrite formation in polymer electrolytes depends on inhibiting the formation of subsurface dendritic structures in the lithium electrode,” Balsara says.

So, it would seem that dendrite formation through polymer electrodes is indeed a problem, though it may require a "non-conductive contaminant" to initiate nucleation if these authors are correct.


How to ionize a liquid and apply a force using electrohydrodynamics I would like to create a contactless force on the surface of a liquid by ionizing it and placing the container between a cathode and anode, however I'm not sure how I would go about actually ionizing it.

The liquid is a photopolymer resin, although the specifics of it can be tailored to whatever we need so it makes the question rather broad. So I'm wondering if someone could give me some advice on how to create a surface force on a liquid using electricity.

Though it is possible to ionize the liquid, it is doubtful you want to do that: the effect of ionization would likely be equivalent to exposing the photopolymer resin to bright light.

A System for Cooling Electronic Elements with an EHD Coolant Flow from iopscience.iop.org describes setting up and testing a pump using isopropanol as the working fluid.

If you do decide to try electrohydrodynamic pumping, either of an atomized spray or EHD pumping of dielectric liquids, take precautions against ignition of the liquid and test whether ionization ruins the photoresist.


How to store methanol We will be buying methanol by a about 200 liters at a time, I was wondering what I should consider when storing it.

So far I know that it should be

in the shade
earthed with a copper stake
in a secondary plastic container
wash empty containers and drain them

But I am wondering what else should be considered? How far should it be away from a house?

One of the first places to look at for advice is the Materials Safety Data Sheet (MSDS, or just SDS). Make sure these are kept in a handy place and are up to date. These should be provided by the supplier (but are easy enough to get online).

A pertinent and thorough guide is produced by the aptly named Methanol Institute, the document is called Methanol Safe Handling Manual, the chapter Health and Safety, especially the Safety Precautions subchapter outlines the following:

  Smoking must be prohibited.
  Vehicle access should be strictly controlled.
  Ventilation must be sufficient to cope with the maximum expected vapor levels in buildings.
  Positive pressure may be required for methanol-free areas, such as control, switch, and smoking rooms.
  Storage tank vents to atmosphere should be sized for fire-heated emergency vapor release.
  Electrical equipment must be explosion-proof to meet national electrical code requirements.
  Alcohol-resistant Aqueous Film-forming Foam (AR-AFFF) with 6% foam proportioning (with water) equipment is advised for use on methanol
  fires.
  Dry chemical extinguishers should be accessible for small fires. An adequate supply of handheld and wheeled types should be available.
  Hydrants should be strategically placed with adequate hoses.
  Small spills should be remediated with sand, earth, or other non-combustible absorbent material, and the area then flushed with
  water. Larger spills should be diluted with water and diked for later
  disposal.
  Lighting should be grounded. Tall vessels and structures should be fitted with lightning conductors that are securely grounded.

These kind of directives should be on signs that are easily read alongside hazard signage.

I would suggest having a readily accessed first aid kit and make a poster with first aid instructions, also included in the handbook - First Aid Measures

In terms of storage, the subchapter Methanol Storage, the handbook states that

  In general, fire protection for gasoline tanks is sufficient for
  methanol tanks, provided extra precaution is made for leak detection, toxic hazard, and
  availability of alcohol-compatible fire suppression foam.

In terms of proximity to the house, I would consider storing these tanks as far as feasible from the house.


How does binding energy "manifest" itself? The concept of binding energy is, that it is equal to the energy that is needed to separate atoms in a covalent bond. I think I have understood that. The electrostatic powers between the atoms create a bond which needs to be "broken" to free the atoms.

My question is: If for example an exothermic covalent bond is created how does this bond energy get transferred to the surroundings? Atoms witch want to fulfill the octet rule create an electrostatic bond. HOW/WHERE is energy released and where did it come from?

If you want to break a bond you first need to excited the right degrees of freedom so the molecule can move along the reaction coordinate of interest. In other words, the correct vibrational modes need to be excited. At equilibrium distance this will be reflected in the kinetic energy of the atoms. As you move towards the dissociation limit this kinetic energy gets transformed to potential energy, much like you give a ball a push to move up a hill. Any excess energy (if you excite more than binding energy is required) will result in kinetic energy after the dissociation.

Forming bonds happens in reverse: Two fragment come closer and convert their potential energy into kinetic energy.

After such an reaction (bond forming or breaking) several options can happen:

the excited states go into lower states by emitting photons (e.g. fire, chemoluminescence)
the energy dissipates into other degrees of freedom (e.g. by momentum transfer in collisions with other molecules)
the reaction may happen backwards and you get your initial educts again
(there are probably further options ...)


Bohr Radius Confusion I saw online that the radius for each level is defined by the formula:

$$r_n=n^2 r_1$$

Where $r_1$ is approximately $53\ \mathrm{pm}$.

However, on the Barron's AP chemistry book 7th edition, it says "The radii of the other orbits are whole-number multiples of the Bohr radius" (Page 89). It even gives an example of when $n=2$ that the radius is $106\ \mathrm{pm}$.

But that's different from what the formula above says, which when $n=2$ radius should be $212\ \mathrm{pm}$. Is this an error in the book? I'm quite confused by the discrepancy.

Bohr's theory of the atom precedes quantum theory but does describe the Balmer and other lines in atomic hydrogen.  I've outlined the calculation so that you can see where the Bohr radius comes from.

He  assumed that the electron orbits the nucleus in a circle with a radius r and is determined by the balance of centripetal acceleration and Coulomb attraction towards the nucleus. For an electron of mass $m_e$ and speed $v$ this is
$$\frac{m_ev^2}{r}=\frac{e^2/(4\pi\epsilon _0)}{r^2}$$
 where e is the charge on the electron and $4\pi\epsilon_0$ puts us into SI units, ( $\epsilon_0$ is the permittivity of free space).

The total energy is the sum of kinetic and potential parts $$E=\frac{m_ev^2}{2}-\frac{(e^2/4\pi\epsilon _0)}{r}$$
and using the first equation gives
$$E=\frac{e^2/(4\pi\epsilon _0)}{2r}$$

Bohr had to make the assumption that electrons circling in an orbit do not radiate energy, which is contrary to classical electrodynamics, but he then needed to determine which orbits were allowed. He assumed that the angular momentum is in quantised units of Plancks constant h and a (principal) quantum number n. This gives
$$ m_evr = n\frac{h}{2\pi}$$

which when combined with the first equation gives

$$ r=a_0n^2$$

which is the equation you ask about and $a_0$ is conventionally the symbol for the Bohr radius and is 0.0529 nm.
It is calculated as
$$a_0 = \frac{\hbar^2}{e^2/(4\pi\epsilon_0)m_e}$$

where $\hbar = h/2\pi$.
Re-expressing the  energy using $a_0$ gives
$$E=-\frac{e^2/(4\pi\epsilon_0)}{2a_0}\frac{1}{n^2}$$
which is the famous Bohr equation and explains the energy levels in hydrogen atoms and so the transitions between them.


Is dextrose an isomer of glucose? Is dextrose an optical isomer of glucose?

If a microorganism metabolises glucose, would it also be capable of metabolising dextrose? Do both 'fit' into the same enzyme that binds to glucose?

There are two enantiomeric forms of glucose, D- and L, with D-glucose being the naturally occurring form. Dextrose is just another common name for D-glucose, so if a microorganism can metabolise D-glucose then it can by definition metabolise dextrose as they are just two words for the same thing.

As for the question of whether the enzyme would recognise both optical isomers (enantiomers), this would probably not be the case. Enzymes are 'chiral', as are the two enantiomers of glucose. The active sites of enzymes/proteins are often specific for a given isomer, hence why many drugs are only biologically active in one enantiomeric form, with the other enantiomer being either inactive, less active, or in certain cases toxic.

In particular, the first step in glucose metabolism is phosphorylation of the C-6 hydroxyl group by the enzyme hexokinase. However, hexokinase is incapable of phosphorylating L-glucose.