text:
What is the reaction between NaOH and urea?
In this video it is mentioned that urea and sodium hydroxide will react to produce ammonia. However, I can't seem to find a clear explanation of the reaction and the other products generated.
One explanation I found says that the NaOH causes the urea to hydrolyze, yielding sodium carbamate, which will then further hydrolyze to sodium carbonate.
Another says that the urea tautomerizes to ammonium cyanate, which preforms a substitution with the NaOH to yield ammonium hydroxide and sodium cyanate.
Which of the two reactions occurs? Do they both occur? Neither?

Urea can be attacked by nucleophiles like any carbonyl. The immediate intermediate is a tetrahedral orthoacid which will break down to liberate ammonia once. The same process can be repeated a second time to liberate the second molecule of ammonia.

Technically, this is a hydrolysis catalysed by hydroxide. Hydroxide may be the reactive species but it is regenerated in the course of the reaction.
For urea to tautomerise to ammonium cyanate, a significant rearrangement and breaking a $\ce{C-N}$ bond would be required. Neither is likely in any way.

Does the carbamic acid react further and we end up with carbonate ion + more ammonia?

Carbamic acid is inherently unstable and will decompose to ammonia and carbon dioxide rapidly @OscarLanzi

Well, in practice adding sodium hydroxide granules to a concentrated urea solution at room temperature results in the rapid growth of needle crystals, most likely urea being pushed out of solution. Further heating does not result in $\ce{NH4+}$ or $\ce{NH3}$. However the melting of the dry reactants might produce a different result.

There are a lot of possible experiments using urea and sodium hydroxide; you are presenting just one variety. Of course, given concentrated solutions of both, chances are that one of the two becomes over-concentrated and crystallises out but there are a lot more variables.

Is heating compulsory for this reaction to take place, considering dilute solutions of both urea and sodium hydroxide?

When dissolved urea is heated in the presence of around an equal mass of dissolved sodium hydroxide I can say for sure the stench of ammonia is very present and bubbling is nearly instant upon heating, note this was performed in about 5 or 10 parts water. To generate large quantities a equal molar ratio of caustic soda and urea would likely be required as the carbamic acid byproduct can react to convert the sodium hydroxide to sodium carbonate, additional CO2 scribing should also be implemented as CO2 will likely leave the reaction mix and contaminate the ammonia gas

"note this was performed in about 5 or 10 parts water" why in past tense?


text:
Decomposing chemical compounds - Ease of Decomposition
Someone asked me a question, that which of the following can be decomposed chemically:

Ammonia
Iron
Neon
Hydrogen
Fluorine

My Analysis:
At first attempt I answered Ammonia but changed it to 'I can't answer'. My analysis of the question is as follows

If Fluorine (the name implies fluorine gas) is decomposed, but it will require to 242kJ/mol (bond dissociation enthalpy). So, its quite difficult that we will be able to decompose it chemically.
If Hydrogen (the name implies Hydrogen gas) is decomposed, but it will require to 435.88kJ/mol (bond dissociation enthalpy). So, its even more difficult that we will be able to decompose it chemically.
If If Ammonia (the name implies Ammonia gas) is decomposed, but it will require to 389kJ/mol (bond dissociation enthalpy). So, its significantly difficult that we will be able to decompose it chemically.
Its useless to say that we will be decomposing Iron or Neon, because they are already in elemental state.

Why I tempted to answer Ammonia: I thought about Haber's Process, and remembered that the enthalpy change of the main reaction is negative (N2+3H2=NH3), about -46.1kJ/mol; this implies that the enthalpy of the just opposite reaction is positive (Enthalpy change of NH3=N2+3H2). This made me to conclude that the reaction is not feasible, thermodynamically and therefore ammonia cannot be decomposed easily. Therefore I changed my answer to 'I can't Answer'
My Question:
I asked my friend my what is the answer, he said Ammonia. I was surprised but still confused. I have written all of my thoughts above, but neither I am able to justify my answer, nor is my friend able to understand my logic. I want your discussion and a final review of the question understanding
'

It's exactly as "useless" in case of iron as fluorine or hydrogen. You simply don't say than most stable form of element decomposes.

I can't understand what are you trying to convey.

"Its useless to say that we will be decomposing Iron or Neon, because they are already in elemental state." Hydrogen and fluorine also are. You should read your own text.

That means I have wrongly interpreted them as gases, they all are in elemental state, except ammonia, which is a compound and hence can be decomposed chemically.  The answer is therefore ammonia.

It really depends on what the definition of "decompose" is in this case. If you take the definition of "decompose" to be:

... to break down or cause to break down into component elements or simpler constituents

then the only option would be ammonia because the others are elements whereas ammonia is a compound which can be broken down into elemental Hydrogen and Nitrogen by the Haber Process. The energetics is not of concern in this case.

Thank you! It's simply about compounds and elements. Compounds can be broken into elements while there's no such a thing into which element can broken. Am I right?

Well, elements could be broken down into smaller elements by nuclear fission but that's not at the chemical level anymore.

If it's not chemical level, then what name chemists use for it. Is it 'Nuclear level'.

Yep - a nuclear reaction.

One last question from 'IT Tsoi', could you comment on my question language and grammar and way I have asked it. I want to improve the way I ask question because for the last few days I have been criticised for it.

I think this time is all right @ShivanshuGupta

Why can't you decompose, say, $\ce{H2 -> 2H}$... Not trying to pick on your answer. As you said it depends on the definition of decompose. I just think the question is poor.

When I asked my friend, from where did he got the question, he said he got it from an book for SAT. I think at the SAT Exam level, we have to deal the question at the high-school level, in the context of which it's simply about element and Compound.

source: Stack Exchange

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sort: votes


text:
What analytical method(s) to use for finding the contaminations in a fatty base material
In a large batch of a fatty combustible mixture there appear to be two contaminations, since after combustion there are 2 residues clogging up the injectors and exhaust system:
- a calcium salt
- a phosphate
We can easily analyse these residues, but to find a method to remove these contaminations before they are combusted we need to find out what form they have in the base material.
From that information we could try to 'wash them out' and quantify the contamination in both the 'washing sollution' and in the combustion gasses to optimize this process.
But for starters we need to know where the contamination is comming from in the first place, so hence my question: what analytical method(s) could be used to figure out these (probably small) contaminations in this fat ?

I have found that with the ICP-MS method we can at least get a quantitative approach of the elements in these samples.
Unfortunately it is not so easy to figure out the form these contaminations are in, especially the phosphate.

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text:
Why are bonding interactions so much stronger than interactions of other types?
Why are bonding interactions so much stronger than all the other types of interactions like dipole-dipole, London dispersion, hydrogen bond etc. even though they are all of electromagnetic origin?

They aren't.

Bonding interactions effectively share electrons. In a gedankenexperiment, you can think of two isolated species with many unpaired electrons coming together, bonding and forming a species with less unpaired electrons. The exact nature of bonds is slightly different, involves quantum mechanics and orbitals, but the end of the story is that the electrons are in a less energetic state after bonding than before it. The energy gained by this process is rather significant, typically hundreds of kilojoules per mole.
These bonds that are created typically generate a molecule or some type of network solid (ionic, metallic or covalent). If you have generated a molecule, then you can rather easily separate a single molecule from all the surrounding molecules by physical methods such as high vacuum. It is not possible, however, to break the bonds by any simple, low-energy physical methods. (UV irradiation will do, but that is high-energy.)
The other interactions, London dispersion forces, hydrogen bonds, etc. are interactions between molecules. You already have two species that by themselves are in a rather stable, i.e. low energetic, state and you bring these two together. The energy that can still be gained is very minor compared to that obtained by bonding. This is also why this type of interaction is often called secondary interaction.
A second reason is the distance. As you note, most of these interactions can be explained electromagnetically. However, the distances between molecules are typically significantly larger than the distances between atoms of a molecule. Since electromagnetical interactions decline exponentially with rising distance, these forces are a lot weaker at higher distances.

I got your second reason, but I couldn't properly get what you wanted to say as your first reason.

Because the later interactions are just invented to provide a correction to the deficiencies in the description of the complete picture, which is obtained when an approximated treatment of the total interactions is used.
That is, you get most of the description from the former (in an approximate way), and to correct deficiencies you add some corrections. These minor corrections are the London, Debye etc.. interactions.


text:
Acid - Base Equilibrium
Suppose we had a closed system proceeding according to,
$$\ce A +2\ce B \rightleftharpoons 2 \ce C$$
And that three experiments produced the following concentration data.
Experiment 1:
\begin{array}{|c|c|} \hline
{\rm \small Species} & {\rm \small concentration} \\\hline
\ce A & 0.500\rm~M \\
\ce B & 0.700\rm~M \\
\ce C & 0.900\rm~M
\\\hline\end{array}
Experiment 2:
\begin{array}{|c|c|} \hline
{\rm \small Species} & {\rm \small concentration} \\\hline
\ce A & 0.300\rm~M \\
\ce B & 0.420\rm~M \\
\ce C & 0.424\rm~M \\\hline
\end{array}
Experiment 3:
\begin{array}{|c|c|} \hline
{\rm \small Species} & {\rm \small concentration} \\\hline
\ce A & 1.40\rm~M \\
\ce B & {x}\rm~M  \\
\ce C & 1.49\rm~M \\\hline
\end{array}
How would you find the concentration of $\ce B$ from experiment 3 at equilibrium?
I assumed that each experiment was in equilibrium so that I could do the following,
$$K_c = \frac{\ce{[C]}^2}{[\ce{A}][\ce{B}]^2} = \frac{[0.900]^2}{[0.500][0.700]^2}$$
$$\therefore \frac{\ce{[C]}^2}{\ce{[A][B]}^2} = \frac{[0.900]^2}{[0.500][0.700]^2}$$
$$\therefore \ce{[B]} = \sqrt\frac{[1.490]^2[0.500][0.700]^2}{[0.900]^2[0.140]}$$
$$\therefore \ce{[B]} = 2.19\rm~M$$
However, $[\ce{B}] = 2.19\rm~M$ is not the answer. Could someone please provide some insight and direction into this question?

Did you just ignore the experiment 2 altogether?

You have an error somewhere in your math.
Assuming you were supposed to interpolate $K_C$ from both experiment 1 and 2, you would do the following:
Experiment1 yields $K_C=3.31$. Experiement two yields $K_C=3.40$. Interpolation yields $K_C=3.35$.
This would give you $$B=\sqrt{\frac{C^2}{A\cdot K_C}}=\sqrt{\frac{1.49^2}{1.4\cdot 3.35}}=0.69$$

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text:
To heat or not to heat; distillation?
One of the questions in a recent past paper I completed was:

One way to make water from the reservoir suitable for drinking is by distillation.
Describe how water is distilled.

And I answered (perhaps too succinctly) that "The water is vaporised and the condensation is collected."
I would have thought my answer correct, in that it fully addresses the question, however, the mark scheme mentions two points I did not make:

I should have mentioned heat.
I should have mentioned cool (the mark scheme's words, not mine).

But I was wondering, are these two (common) aspects of distillation actually necessary to distill water?

Surely heating and cooling are not really necessary to distill water; couldn't pressure be reduced to evaporate the water, and increased again to condense it?
Or, in the context of the question, is the heating perhaps necessary to sterilise the water (for killing microbes)?

Welcome to chemistry.se! If you ask me, and by extension you are already doing this, your explanation is quite fine. That is the general process of distillation. In the case of water, however, reducing pressure might just not be enough to evaporate it.

@Martin-マーチン According to this table (and my own - perhaps incorrect - presumptions about how vapour pressure works) pressure would need to be reduced to 0.0121 atm for water to boil at 10°C, which should be doable with even just vacuum pumps.

Of course, 10°C being an amount chosen to represent the reservoir water temperature; a higher temperature means water can boil at a higher pressure.

Why, of course you can make water boil and then condense by manipulating pressure alone. It would be slow and cumbersome, but very much possible. The problem is, as some water evaporates, the rest of it cools down, and continues to cool all the way down until it becomes ice, so you'll have to actually heat it to keep it at room temperature, and if you have the heater anyway, why bother with all that pressure stuff.

You can but the instrumentation is far more complicated. I a laboratory I would prefer do not use a noise devise. Reach a pressure low enough is not a problem. In do not know which alternative is cheaper.

You obviously already have a general idea of what is going on in a distillation, so I will spare you the picture and everything.
Your question can be broken into two parts:

Do I need heat for vapourisation?
Do I need cooling for condensation?

I am going to address them one after the other.
Vapourisation
As you probably also know, reducing pressure is one means of evaporating liquids. This is used in rotavaps which use a $40~\mathrm{^\circ C}$ bath and reduce the pressure to make sure the solvent evaporates. You could think that just reducing the pressure should be enough to vapourise any liquid.
But if you vapourise a liquid only by reducing pressure, you will reduce the temperature of the liquid. This is the evaporation cooling that your body uses when sweating: If the water in sweat evaporates it cools the surface it evaporates from. Thus, after a certain time and especially if you attempt to distill large quantities of water, it will freeze and further evaporation will become exponentially harder. Therefore, a practical setup will always require heating.
Condensation
Effectively, this is the same discussion but in reverse. Note that if you boil water in a closed pot, the lid of the pot will get successively hotter due to condensing water releasing heat. After a certain time, the temperature of the surface you want to condense on will be so hot that the liquid will no longer condense.
This is especially a problem if you decided to reduce pressure for evaporation: If you have to reduce the pressure so far that the liquid would evaporate at $10~\mathrm{^\circ C}$, then there is no way that it will condense at a cooling trap of $25~\mathrm{^\circ C}$. Therefore, the condensation area must be cooled which is typically achieved with Liebig coolers or similar.
What about vapourisation followed by condensation in the same vessel with changing pressure?
Well, that is possible. It is not practical, though, since the inside of a single vessel can only have a certain pressure (within slight margins). And vapour spreads out uniformly. Your vessel would have to have an initial ‘bowl’ and a collecting one. You reduce pressure to evaporate out of the initial bowl and then re-raise pressure. The liquid will condense uniformly and you will lose a significant amount of it because it will drip back into the initial bowl. Your yield with a typical common setup is much better.

So you're saying that heating and cooling are necessary processes for distillation?

@theonlygusti Necessary for a practical distillation setup, yeah.

but in terms of the question, is my answer technically fully correct? And so, the mark scheme wrong?

@theonlygusti The question asks ‘describe how water is distilled’. Therefore, it is reasonable to require heat and cooling since the other setup is very impractical and I am not aware of it being used.

Ah, so you're saying I should have read it as "Describe how water is distilled" instead of "Describe how water can be distilled," therefore making my evaluation afterwards incorrect?

Do you have any hard numbers to back up just how impractical my suggested set-up would be?

@theonlygusti Well, let’s put it this way. A single pot which is heated and which has a cooling device can give a constant flow of distillate. Working with pressure will require a stepwise process making it much less efficient per se because it is not constantly flowing.

The question is trying to explore whether you understand how distillation works. Yes, vapour is created and, yes, the condensate is collected. But your answer doesn't describe how that is done.
You are correct that you could use reduced pressure to create vapour, but you will still need some input of heat and some way to cool the vapour. Or you could have described the simpler process of applying heat at normal pressure and condensing the resulting vapour and the examiner would know you understood the basics.
It may be possible to do distillation purely with pressure swings, but the apparatus would be really complicated (how, for example, do you collect the low pressure vapour and then repressurize it to cause condensation?). Besides the reason why distillation used lowered pressure isn't to avoid the need for energy input but to avoid the need for heating the liquid to excessive temperatures to achieve energy input (some things will decompose at high temperatures). In a constant pressure vessel, the only way to condense the vapour is to cool it so you can't avoid the need for either cooling the vapour or heating the liquid.
Show you understand the basics and you will get full marks in the test. Don't overcomplicate.

Obviously, as I sat the test, I didn't think through the question in as much depth as I lay out in my post; if anything, I under-complicated my answer, in failing to mention how distillation can be achieved. I was simply a bit annoyed to find out that I had the "wrong" answer because I had failed to mention two processes which, in contemplation afterwards, don't even seem to be necessary for distillation.

Anyway, I don't feel that this answer answers the question I've asked definitively enough; is heating/cooling necessary for distillation?

@theonlygusti The only way you can possibly avoid heating and cooling is to have some immensely complicated apparatus that allows significant pressure swings. In practice and in real labs you will need heating or coolling or usually both. (BTW the answer does say this.)

@theonlygusti And you have speculated that pressure alone could do the job: it can't (unless you have some very clever design of device, which you have not demonstrated). Low pressure makes the temperature differential required different, not non-existent (in a constant pressure apparatus.)

You will need heat even if you play with pressure: this heat may come from a reservoir which stays at room temperature, yet heat will flow back and forth during evaporation and condensation.
And let us be honest: you gave a half-baked answer, which is half baked even if your teacher didn't write an essay about what is missing

If distilling means vaporization and then condensation of a liquid without applying work to the system, then the energy needed to transform the liquid to the gaseous state has to be supplied as heat.
This is called heat of vaporization, or latent heat of vaporization, since this does not necessarily mean that the temperature has to be raised.

I'm sorry, i don't really understand this answer; heat has to be applied, yet the temperature does not raise?

Consider water at 100°C. You have liquid at 100°C and above it you have vapor at 100°C. If you now apply heat to the liquid water it would not raise the temperature. Instead the heat would be consumed to transform more liquid water to gaseous phase. This is the latent heat of vaporization. Follow the link to get additional information.


text:
How to obtain saturated pressure of water at 200 ºC using fugacities?
The target is to obtain saturated pressure of water at $\pu{200 ^\circ C}$ using PHI-PHI (fugacity coefficients) approach, so it means it's a VLE  of water only, so we write equilibrium condition:
$$f_i(l) = f_i(v),\tag1\label1$$
with $f_i(l)/P=PHI(l)$ and the same for $PHI(v)$ (note that $Y_i=X_i=1$), substituting into \eqref{1} we get:
$$PHI(l)=PHI(v)$$
There are several ways to calculate $PHI$ and here we use the RK equation of state, my problem begins here, if we calculate the $PHI$, then what should we do next? Don't we need $f_i$ to calculate $P$?

Are you talking about $\phi(l)$, maybe $\varphi(l)$? What is a PHI-PHI approach? What does VLE stand for? Where did $X_i$ and $Y_i$ come from? What is $(*)$? What is RK equation of state? What is $P$?

Here's my idea, for whatever it's worth.  For a real material like water, if you plot pressure as a function of decreasing specific volume at any temperature below the critical temperature, the graph will become perfectly flat in the two-phase region (where you have a combination of liquid and vapor).  It will then start to rise again once the specific volume reaches that of the liquid.  In the case of an equation of state like RK, this won't happen.  The graph will be smooth, and will exhibit maxima and minima in the two phase region.  I know that $\mathrm{d}G = V\mathrm{d}P$ at constant $T$, so, if I integrate from a large specific volume downward at constant temperature, I should be able to produce the curve of $G$ vs $V$ (and $p$) for an arbitrary equation of state. At constant $T$, we have:
$$
\left(\frac{\partial p}{\partial V}\right)_T
= -\left(\frac{\partial T}{\partial V}\right)_p
   \left(\frac{\partial p}{\partial T}\right)_V
$$
The first parenthesis on the right is, of course, equal to zero for a real substance in the two-phase region.  If I substitute this into the equation for $\mathrm{d}G$, I get:
$$
\mathrm{d}G
= -V\left(\frac{\partial T}{\partial V}\right)_p
    \left(\frac{\partial p}{\partial T}\right)_V\mathrm{d}V
$$
So, I can integrate this equation downward from high specific volumes (where the ideal gas law applies) through the "two-phase" region of an equation of state.  If I plot a graph of $G$ vs $V$, I can see the points where the values of $G$ are the same and the values of $p$ are also the same.  This should identify the saturated values of the liquid and vapor specific volumes and pressure.  Maybe that's what your equations already do.

Actually setting fugacity coefficients equal to each other is an direct result of no change in G and the equality of values ,We used the same concept by different ways, It was a great approach sir,Thank you for sharing.

Thanks for keeping me involved.  It was a very interesting and challenging problem.

Hi, would you be willing to help edit the question to tidy up a bit of the notation? I will readily admit that I don't know anything about the topic - but ideally readers, even those who don't know much, shouldn't have to try to decipher abbreviations / notations. I've tried asking the user base more generally, but I couldn't find anybody with the necessary knowledge, unfortunately.

We know that fugacity coefficients must be equal, so taking a natural log we have:
$$\ln PHI(l) = \ln PHI(v)$$
As we know in thermodynamics (better to refer Chem. Eng. Thermodynamics by J.Smith):
$$\ln PHI = Z -1 -\ln(Z-\beta)-q \cdot I,$$
Which $\beta=b\cdot \frac{P}{RT}$ and $b$ is an EoS parameter like $q$, about $I$ for $RK$ EoS it is:
$$I=\ln\frac{Z+\beta}{Z}$$
As we see all of them depend on $P$ which is we want, this was the confusing thing for me, note $Z$ and $I$ are different for vapor and liquid,
and we have some correlations for $Z_v$ and $Z_l$ so we can obtain $I_i$ and $I_v$:
\begin{align}
Z_v &= 1+\beta - q\cdot \beta\frac{Z_v-\beta}{Z_v\cdot(Z_v+\beta)}\\
Z_l &= \beta+Z_l \cdot (Z_l+\beta)\frac{1+\beta-Z_l}{q\cdot\beta}
\end{align}
We use top equations for $PHI_v$ and $PHI_l$ in respect,
so we have non-linear multi variate system of equations, for $I_v$, $I_l$, $Z_v$, $Z_l$ and $P$.
Here is the code for solving:
function F = root2d(x)

F(1) = x(1)-5.3726e-04*x(5)-x(1)*(x(1)+5.3726e-04*x(5))*(1+5.3726e-04*x(5)-x(1))/0.0042*x(5);
F(2) = x(2)-1-5.3726e-04*x(5)+7.8914*5.3726e-04*x(5)*((x(2)-5.3726e-04*x(5))/(x(2)*(x(2)+5.3726e-04*x(5))));

F(3) = x(3)-log((x(1)+5.3726e-04*x(5))/x(1));

F(4) = x(4)-log((x(2)+5.3726e-04*x(5))/x(2))

F(5) = x(2)-log(x(2)-5.3726e-04*x(5))-7.8914*x(4)-x(1)+log(x(1)-5.3726e-04*x(5))+7.8914*x(3);

And the solving code:
options = optimset('MaxIter',500,'MaxFunEvals',500);

fun = @root2d;
x0 = [0.1,0.9,0.5,0.5,15];
x = fsolve(fun,x0,options);

$x(1)$ to $x(5)$ are $Z_l$, $Z_v$, $I_l$, $I_v$, $P$.
The answer we get is:
x =

0.0112    0.9348    0.6543    0.0753   15.5033

In which the $P$ is very near to the steam table value (15.54 bar).

It looks like you were able to solve it correctly.  I am not familiar with the equations you used, so I have trouble relating to what you did.  The method I suggested in my original answer is obviously not the way to go, so I have deleted my answer.  But I have developed an alternative approach that I'm pretty confident will work.

@ChesterMiller I'll appreciate it if you share your way with me,I'll learn a new way...thank you.

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text:
Does the addition of water to N2H4 produce N2H4
I am currently calculating the $\ce{pH}$ of a species.
The reaction is as follows,
$$\ce{N2H4 + H2O <=> N2H_{$A$}+ + OH-}$$
However, I'm not sure what the value of $A$ will be. In my textbook it says $4$, which is stumping me as in every other example with a weak base the subscript increases.
What is the value of $A$ and why?

A little nonsense in a textbook is good for you; it puts your mental immune system on alert. You know that equations must be balanced, right?

Yes, equations must be balanced, thus the only correct way to write this is:
$$\ce{N2H4 + H2O <=> N2H5+ + OH-}$$
Easy to remember: Nitrogen typically has a free electron pair which can act as a Lewis base for a proton.
If your textbook says $4$, it is an error in your book. Good job on spotting it and realising it.


text:
Sign of the electron's wave function
What do the + and - sign of 2 lobes of the p orbital represent? I know that they are the signs of the electron's wave function but what is their significance?

They mean nothing.

There is no physical interpretation of the phase but that doesn't mean it's irrelevant


text:
Ring expansion and carbocation formation
If in a reaction the carbocation is formed outside the ring then why does the ring expansion does not happen as follows:

Is there any reaction that could create a carbocation at that place?

@KennyLau  i have now posted that reaction

To understand which rearrangement pathway is preferred, we must consider the relative stabilities of the intermediates. The rearrangement that puts the carbocation one carbon away from the the oxygen does not have any stabilizing resonance structures:
$\hspace{5.4cm}$
The rearrangement that places the carbocation adjacent to the oxygen delocalizes the positive charge to two atoms, and is much more stable:
$\hspace{3.7cm}$

+1 That's how dihydropyrane can be obtained from tetrahydrofurfuryl alcohol via catalytic dehydration on alumina at elevated temperatures.


text:
Is temperature just the phenomenon of particles colliding against an object?
This is just something that I thought up while doing busy work at my job. Is the sensation of heat simply coming from particles colliding with your, say, hand? Are they transferring energy more to objects because they have higher velocities, and thus heat can hurt? (Sorry, this isn't a more cohesive question...).

Yes, because the proteins in your body have a specific temperature range. Too hot and they get denatured. (That range is quite small)

https://en.wikipedia.org/wiki/Denaturation_%28biochemistry%29

There's also thermal radiation. https://en.wikipedia.org/wiki/Thermal_radiation

Heat transfer occurs when particles collide against an object (which is made of other particles) or when photons impact on the object.
Temperature is a measure of the average amount of kinetic energy of the particles (in a gas this will be simple related to how fast the particles of gas move but in a solid it will be about vibrations of the particles).
The sensation of heat is a product of both thermal radiation (infra-red photons are emitted by the hot object and absorbed by your hand) and contact with something that can transfer some of that energy directly (hot gas particles or a hot solid directly touching your hand). Living things have evolved to sense heat and cold because they only survive if they avoid extremes of both.


text:
Hybridization and excitation of electrons
In case of hybridization of PCl5 why does the electron move to 3d orbital though 4s orbital has a lower energy?

https://www.quora.com/If-4s-orbitals-are-higher-in-energy-than-3d-orbitals-then-why-do-electrons-fill-up-in-4s-before-filling-up-in-3d

In short, 3d>4s only in group I and II.

Bottom of http://www.meta-synthesis.com/webbook/39_diatomics/diatomics.html

In AO level diagrams, it is important to note the large differences of gaps:
Not: 1s < 2s < 2p < 3s < 3p < 4s <3d < 4p <5s <4d with equal distances

But correctly, according to both spectroscopy and quantum chemistry:
1s << 2s < 2p << 3s < 3p << 3d< 4s < 4p << 4d <5s

Note the d < s order given here, which is the correct one for most cases, with the exception of group 1 and 2 atoms.

Because the energy of 3d is less than the energy of 4s.

@kenny but while filling of atomic orbitals 3d>4s... why is that so then

The first website already explained it. "The 4s sublevel is only lower in energy if there are no electrons in the 3d sublevel."

The second website provided extra information: the energy levels of 3d and 4s are very close.

$\ce{PCl5}$ is poorly described by hybridization.

It simply does not work like that - significant contribution of d orbitals in p black compounds was disproved many years ago.

related http://chemistry.stackexchange.com/questions/29142/hypervalency-and-the-octet-rule http://chemistry.stackexchange.com/questions/18427/why-does-f-replace-the-axial-bond-in-pcl5

It doesn’t. It simply doesn’t. The $\mathrm{3d}$ orbital has a similar energy to the $\mathrm{4s}$ one — nobody considers $\mathrm{4s}$ to take part in bonding but historically, invoking $\mathrm{d}$-orbitals was very popular.
The better description is to have an $\mathrm{sp^2}$ hybridised phosphorus atom which has a free p-orbital. The $\mathrm{sp^2}$ orbitals bond with three chlorides in traditional two-electron-two-centre bonds while the remaining $\mathrm{p}$-orbital — populated by two electrons — participates in a four-electron-three-centre bond with the two remaining chlorines. You can think of this as two resonance structures:
$$\ce{Cl-{P+}Cl3\bond{...}Cl- <-> Cl- \bond{...}{P+}Cl3-Cl}$$


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Does an axis of symmetry determine chiralty?
What I've understood about chirality and elements of symmetry:
A molecule that is not superposable on its mirror image is said to be chiral.
Plane of symmetry:
An imaginary plane cutting a compound into two symmetric halves.
Any compound possessing it is achiral, so any compound without it is chiral?
Axis of symmetry:
I have never spent so long trying to understand a concept.
An axis of symmetry is an imaginary axis around which by rotating through a minimum angle of rotation, the original compound is obtained.
1. Any compound, be it chiral, achiral or dissymmetric, can have an AOS.
2. You cannot say a compound is achiral just because it has an AOS.
3. Unless, I just had a thought. If on rotation it gives the same compound, making it superposable with the original, thereby making it achiral.
However, wouldn't all compounds (chiral or achiral) on rotation through a particular degree give back the same compound?
Please tell me if everything I've mentioned is right.

Everything you've mentioned is right, except item 3, which contradicts the rest. Indeed, any molecule with axis of symmetry, chiral or not, would coincide with itself if turned for the appropriate angle around that axis. This has nothing to do with chirality.

Chirality is left hand—right hand. The left hand is a mirror image of your right hand. If you take this mirror image and then rotate or shift it and you are able to superimpose it onto the original image, it is achiral. This is not possible with the hands, the thumb will always be wrong.
A plane of symmetry is sufficient but not mandatory to give an achiral compound. But it should be clear that mirroring an image that contains an intrinsic mirroring plane will give the same.
Another sufficient but not mandatory symmetry operation to generate a chiral compound is inversion. If some compound is point symmetric, it is never chiral.
And there are other elements of symmetry that are rare if not accompanied by planes or centres of symmetry but which are sufficient for generating achiral compounds: Improper rotation. In a nutshell, this is a symmetry element that requires a rotation with a subsequent mirroring perpendicular to the rotation axis. Take a tetrahedral compound such as methane: If you select the correct axis, rotate by $90^\circ$ and then mirror it perpendicular to that axis, you will receive the same product.
Strictly speaking, planes and centres of symmetry are special cases of improper rotation with a rotation of $360^\circ$ and $180^\circ$, respectively. So one could say that improper rotation is sufficient and mandatory to give an achiral compound.

is trivially correct since every object has $C_1$ symmetry, i.e. after rotation around an axis about 360 degrees it is indistinguishable from the starting state.
follows from 1.
is not correct. By definition a molecule is chiral  if it does not have an improper axis of rotation.

Yes. In general an object stays the same if you look at it from a different point of view.


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Parameter effects on kinetics of esterification for bio-diesel production
I want to discuss the effect of parameters on the kinetics of esterification for bio-diesel production. In esterification process, temperature effect and molar ration are two important parameters. During conversion of oliec acid with methanol to bio-diesel, temperature contributes more than molar ratio ( happen before 70 to 80 % conversion to bio-diesel). But after 70 to 80% conversion molar ratio contribute more than temperature.
My question is why this happens. Why the contribution of temperature and molar ratio reverse after 70 to 80% conversion to bio-diesel?


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Why does the solubility of some salts decrease with temperature?
It is well known to us that the Solubility of solute in a solution increases with the increase in the temperature because, when the temperature increases the molecules of the solvent gain more kinetic energy. Thus the molecules move randomly and having greater distance from each other which is responsible for the large voids between them and it gives more space for the solute molecules between them to come in. However, there a few salts  like cerium sulphate , lithium carbonate sodium carbonate monohydrate, etc. whose solubility decreases with the increase in temperature. How it is so ?

Relevant: http://pubs.acs.org/doi/abs/10.1021/ed051p555.1

A saturated solution is at equilibrium (rate of dissolution is equal to rate of crystallization) with some equilibrium constant $K_1$. If you change the temperature of the system at equilibrium, you will observe a different equilibrium constant $K_2$. Whether the equilibrium constant increase or decreases is described by the Van't Hoff equation:
$$\ln \left(\frac{K_2}{K_1}\right) = \frac{-\Delta H}{R}\left(\frac{1}{T_2}-\frac{1}{T_1}\right)$$
So it depends on the sign of the enthalpy of reaction, in this case the enthalpy of dissolution.

It is well known to us that the Solubility of solute in a solution increases with the increase in the temperature because, when the temperature increases the molecules of the solvent gain more kinetic energy.

The statement about solubility is not always true, and the explanation leaves out something. A lot of things change when you increase the temperature. The solvent gains kinetic energy, the solute gains kinetic energy, and the solid gains kinetic energy. How this influences the solubility depends on the specific system, and is hard to predict.

However, there a few salts like cerium sulphate , lithium carbonate sodium carbonate monohydrate, etc. whose solubility decreases with the increase in temperature. How it is so ?

Now that we introduced the Van't Hoff equation, we know it must have to do with the enthalpy of dissolution. Usually, you expect the interactions in the solid that need to be broken to be stronger than the gains from solvating the solute. In these cases, it is apparently the opposite.

Note that you can use "stretchy parentheses" with \left( and \right) pairs to enclose an entire fraction, e.g. \left(\frac{K_1}{K_2}\right) gives $$\left(\frac{K_1}{K_2}\right)$$ Also, you can combine different brackets (e.g. [ ) or ( }) and use a placeholder (invisible bracket) such as \left. to leave a bracket on one side, for example \left.\frac{K_1}{K_2}\right\}: $$\left.\frac{K_1}{K_2}\right}$$

@KarstenTheis enthalpy of dissolution for both, lithium carbonate and sodium hydroxide, is negative. But solubility of former decreases and that for later increases upon increasing of temperature.

@Apurvium Apparently, for a saturated solution of sodium hydroxide, dissolution is endothermic even though iti is exothermic at low concentrations (see Andrew's comment for Airhuff's answer). That would explain the distinct temperature dependency of solubilities of lithium carbonate vs sodium hydroxide.

see also https://chemistry.stackexchange.com/q/4440/72973

enthalpy of dissolution of infinitely dilute solution: https://diverdi.colostate.edu/C433/miscellanea/CRC%20reference%20data/enthalpies%20of%20solution%20of%20electrolytes.pdf

What you refer to, the decreased in a solid's solubility in a liquid with increased temperature, is frequently called retrograde or inverse solubility, and occurs when the dissolution of the solute is exothermic.
The explanation for this can be viewed as a manifestation of Le Chatelier's principle.  Essentially, since evolved heat can be viewed as a product of an exothermic reaction, the addition of more heat (e.g. a higher temperature) is equivalent to adding a product to the product side of the chemical equation for dissolution, driving the equilibrium back toward the reactants, in this case toward the undissolved compound.
Take the exothermic dissolution of calcium sulfate in water for example:
$$\ce{CaSO4_{(s)} <--> Ca^{2+}_{(aq)} + SO4^{2-}_{(aq)} + heat}$$
In this case, the solubility decreases with increasing temperature because by increasing the temperature you are adding heat to the product side.  Inversely, if you pull heat from this system, e.g. cool it, you drive the equilibrium toward the products side and solubility is increased.

I've heard this argument many times before, but I don't know if it's very good. For example, sodium hydroxide dissolves very exothermically in water, but nevertheless its solubility increases very steeply with temperature (from 418 g/L at 0 °C to 3370 g/L at 100 °C according to Wikipedia). Rather than enthalpy of dissolution, in most cases it is presumably the entropy of dissolution that dictates whether a compound has increased or decreased solubility with increasing temperature.

@NicolauSakerNeto, I certainly didn't expect the Le Chatelier's principle argument to go without opposition ;) And it's not a perfect model, as your exception points out. But I have a problem with the entropy argument also. You would intuitively think that NaOH to Na+ and OH- would have a smaller entropy change than $\ce{CaSO4}$ to $\ce{Ca^{2+}}$ and $\ce{SO4^{2-}}$, based on the lower degree of symmetry of d $\ce{SO4^{2-}}$ as compared to OH-. So shouldn't $\ce{CaSO4}$ solubility increase with temperature if entropy change controls the temperature dependence of solubility?

The solvation of species with high charge density forces water molecules in their vicinity to adopt a more strict subset of conformations, which is entropically less favourable. Both the anion and cation in $\ce{CaSO4}$ are divalent, and thus may fall in this category. Meanwhile, $\ce{NaOH}$ contains only monovalent species and thus restricts the conformation of solvating water molecules less (though I am sure that the hydroxide ion in water is special due its Grotthuss-type mobility, which likely results in a particularly diffuse solvation shell and even less restricted conformations).

More generally, the solubility is governed by Gibbs free energy, which  has both enthalpic and entropic components. In the example in this answer, the enthalpic component is dominant. The entropy continues to dominate in the NaOH counter-example though.

@NicolauSakerNeto I realize this is an old post, but I thought it useful to point out that NaOH dissolution is only exothermic at very low concentrations. As soon as you add enough that the pH goes up, the exothermic reaction $\ce{HO- + H+ -> H2O}$ essentially stops happening, and the dissolution switches to endothermic. Try making a saturated solution of NaOH in water. It will get quite cold.

As others noted, the effect of temperature is governed completely by $\Delta H$ of dissolution. The reason that dissolution can have either positive or negative $\Delta H$ is because the favorability is determined by the Gibbs free energy $\Delta G = \Delta H - T \Delta S$. If the $\Delta S$ term is negative (before multiplication by -T) and $\Delta H$ is also negative, dissolution is favorable only if $|\Delta H|>|T\Delta S|$.
It is important to note also that the magnitude of $\Delta G^\circ$ does NOT correlate with the solubility, since the change in concentration per change in $\Delta G^\circ$ is not constant. That is why $\Delta S^\circ$ does not determine the effect of temperature.
This can be seen by noting that the important value is $K$. Since $\Delta G^\circ=-RT\ln K=\Delta H^\circ - T\Delta S^\circ$, we can rearrange and find that $\ln K = -\frac{\Delta H^\circ}{R}\left(\frac{1}{T}\right)+\frac{\Delta S^\circ}{R}$, so a plot of $\ln K$ vs $\frac{1}{T}$ (which is the temperature dependence of solubility) has a slope whose sign is determined only by $\Delta H^\circ$.

Can we say that, when we dissolve something, $ΔG_{sys}<0$ till saturation point and then $ΔG_{sys}>0$?


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Preparation of alcohols from Monosubstituted cyclic ethers
Can methyl oxirane(epoxide) (a mono substituted 3 membered cyclic ether) form alcohols by reacting it with Grignard reagent? If yes, can you please explain the reaction with mechanism? If no, why?

``Ethers react with Grignard reagent to form alcohols```- where did you get this from? Grignards are usually prepared in ether solvents. They do not react with them.

Grignard Reagents only react with three-member cyclic ethers (epoxides).  As bon says, they don't react with any other kind of ether.

Technically, they react with oxetanes as well: http://chemistry.stackexchange.com/questions/21921/grignard-reagents-react-with-epoxides-but-dissolve-in-thf-what-happens-in-oxeta but that's hardly the point :)

Sorry. I have edited the question. Please forgive me

Yes, Girgnard reagents can act as nucleophiles in the ring opening of epoxides. Nucleophilic attack occurs much faster at the less hindered carbon, so the product of this reaction will predominate. The reaction results in the formation of an alkoxide, and does not form the alcohol until treated with a protic acid. The mechanism for Grignard addition to epoxypropane (methyl oxirane) is:
$\hspace{4cm}$
For 2-methyl-2,3-epoxybutane, a more hindered epoxide, the addition still occurs at the less hindered carbon:
$\hspace{3.8cm}$


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Mixing oxalic acid and glycerin
I am doing an experiment for a miticide for honey bees. I will be dissolving 600 g of 99.6 % oxalic acid in one liter vegetable glycerin at 150 °F (65 °C). I have been told it will give off formic acid fumes while mixing and heating. I don't think it will until it get to the melting temperature of oxalic acid. Is there any safety concern? Also what would the shelf life of the mixture be?

You might want to raise the temperature a little bit as the reaction proceeds at 100-110°C to form formic acid. If the temperature is further raised to 260-280°C, then the reaction will proceed to form allyl alcohol.

Formic acid has almost unlimited shelf life if unopened and is protected from heat and light sources. Allyl alcohol is stable in normal storing conditions but it may polymerize into a thick syrup if left untouched for several years.
Discussing about safety concerns is pretty much off-topic here but you might want to check the safety data-sheets: formic acid and allyl alcohol.
Here is a research paper for further reading.
Reaction sources: Google Images

"Formic acid has almost unlimited shelf life if unopened and is protected from heat and light sources."  I have read it needs to be stored with a vent because of  slow decomposition $\ce{HCOOH -> H2O + CO}$


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Why is the boiling point of rectified spirit lower than alcohol absolute?
The boiling point of rectified spirit is ($78.1~\mathrm{^\circ C}$) while that of absolute alcohol is ($78.3~\mathrm{^\circ C}$). I know its too small, but shouldn't the addition of impurities increase the boiling point?
What qualifies as an impurity? Is it that solids like salt only can be called an impurity? — which increase the boiling point of water?

I'd be surprised you could even measure the difference between those boiling points in normal conditions. Is the difference really significant?

Ethanol and water have weaker intermolecular interactions than water with water and ethanol with ethanol. For this reason ethanol and water molecules can more easily escape from the liquid mixture, called azeotrope, to the gaseous phase than it is the case for the pure alcohol. Consequently the azeotrope has a lower boiling point.
What is considered as an impurity depends on your use case. E.g. 96% pure ethanol may qualify as extra fine spirit.

Thanks! so there is no rule as such on the impurity thing?

Added a remark on impurity.


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Have "molecular clusters" for azeotropes been identified?
A different question about azeotropes got me thinking about this point again. Azeotropes have a very specific composition so it seems that the azeotrope ought to have some sort of physical structure. It seems to be a "molecular cluster" of some sort.
The azetrope for water and ethanol is about 95.5% ethanol by weight. A little fiddling and it seems that the ratio is 8 ethanol molecules to 1 water molecule. Is there some particular physical configuration of molecules to which this would correspond?

For  the cluster of 8 my perception would be something like a cube where the water is in the center and the 8 ethanol molecules are on the corners with the OH groups pointing in. Such a structure would be a cube, but something in between a more like a ball. I can't visualize how the OH groups would pack though.

No, it doesn't work like that.

In positive azeotropes for which the boiling point is less than the boiling points of any of the constituents the intermolecular interaction of the different molecules is less than in the pure liquid phase. Therefore it is not very likely that clusters involving the different molecules will be formed.
In negative azeotropes that have a higher boiling point than the constituents cluster structures in the liquid phase are more likely. E.g. for concentrated hydrochloric acid the existence of clusters has been suggested1.

1 Agmon, N., Structure of Concentrated HCl Solutions, J. Phys. Chem. A 1998, 102, 192-199

Azeotropes are not about molecular clusters at all. In the relatively rare cases where certain intermolecular compounds with specific compositions indeed do exist (examples include $\ce{H2O + N2H4}$ and $\ce{H2O + HClO4}$), they manifest themselves on the phase diagram in a quite different manner. First and foremost, they should be expected to appear as peritectics on the melting diagram, which is not the case for the water+alcohol system.

No, in the absence of extra data, there is no reason to suppose that there is any vapor-phase cluster formation.
Cluster formation in the gas phase would demand very, very strong departures from ideal-gas behavior.  To the contrary, the ideal gas law is an excellent descriptor of gas phase mixtures of ethanol and water.
Check out a Wolfram Demonstration for the ethanol-water system.  It says:

You can vary the pressure $P$ to any value between 50 kPa and 200 kPa  (i.e., low to moderate pressure so that the ideal gas-phase assumption holds).

If the ideal-gas assumption holds, then there is no significant structure formation in the vapor phase.  The "ideal" gas law describes negligibly small particles that have no attraction or repulsion to each other.  Structure formation means that molecules must be strongly attracted to each other in order for arrangement into a persistent structure to occur.
An "extended" form of Raoult's law that is valid for non-ideal vapor as well as non-ideal liquids, and thus is applicable to azeotropes, is
$y_i \phi_iP = x_i \gamma_i p_{i, \mathrm{sat}^{\star}}$
Here, $\phi_i$ is the fugacity coefficient and takes into account vapor-phase non-idealities (i.e. deviations from the ideal gas law), and $\gamma_i$ is an activity coefficient and  takes into account liquid-phase non-idealities.
For many, many systems of interest, $\gamma_i$ is the driver of non-ideality, including azeotropic behavior.  Fugacity coefficients $\phi_i$ are negligible (except at enormous pressures) a much higher percentage of the time than activity coefficients $\gamma_i$.  This is because liquid phases are often far more dense than vapor phases, meaning that intermolecular forces govern behavior to a much stronger degree than in vapors.


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Is the concentration of negative ions always equal to that of positive ions?
Is there any way in which a solution can have more negative charge than positive charge, or vice versa? For example, in the context of acid-base chemistry, is there any scenario in which $[\ce{H+}]\ne[\ce{A-}]$? I have never heard of a solution with a non-neutral overall charge, but is it possible in some extreme scenario?

Well, a trivial and unhelpful answer would be to consider a solution of $\ce{MgCl2}$: there are twice as many negative ions as there are positive ions. A solution that is electrically neutral does not necessarily have the same number of negative ions and positive ions.

You're right, of course. Question edited to say charge, not ions.

The only point that $[\ce{H+}]=[\ce{A-}]$ is when the $\mathrm{pH}=\mathrm{p}K_\mathrm{a}$ of a given acid, so indeed in most instances $[\ce{H+}] \neq [\ce{A-}]$. In addition to $\ce{A-}$, the solution also contains negatively charged $\ce{OH-}$ that balances the positive charge of the $\ce{H+}$ ions. If the acid is titrated with a strong base (e.g. $\ce{NaOH}$), the positively charged $\ce{Na+}$ ions balance the negative charge of $\ce{OH-}$ and $\ce{A-}$.
To answer the question in your title though, yes, charges are always balanced. To quote another good post on this topic:

While it is true that solutions are not always locally electroneutral, that non-neutrality is small, and does not extend to larger scale. Thus, it does not impact chemical concentrations: you cannot separate positive and negative charges in a solution at a macroscopic level.


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If I can't draw resonance structures for a pi system, does that mean it doesn't exist?
Whilst trying to answer this question I came across an interesting situation regarding the relationship between resonance structures and molecular orbitals. There are two places that you can deprotonate this compound. The two different deprotonated forms are shown below along with their resonance structures.
Both compounds have a continuous system of p orbitals which can align in a plane and overlap to form a $\pi$ system. However, in the first example it is not possible to draw resonance structures that delocalise the negative charge over all five atoms that have the available p orbitals. Does this mean that the negative charge is not delocalised over all five atoms? @Martin said in his comment on this answer that valence bond theory and molecular orbital theory ultimately give the same result so this should also apply here. In the molecular orbital description, what is the reason why the negative charge is not delocalised over all five atoms? What do the molecular orbitals for these anions look like?

Perhaps it would be simpler to start with the basic allyl anion system. In that system you can only delocalise the negative charge onto C-1 and C-3; this is consistent with the HOMO having no coefficient on C-2. That means that if you were to react the allyl anion with an electrophile you would get no substitution on C-2. http://www.chemtube3d.com/orbitalsallylanion.htm

The fact that you cannot delocalise negative charge onto every single carbon is significant and I wouldn't call it a "limitation". Another example is the o,p-directing effect of the phenoxide anion, which can be somewhat explained by showing that the negative charge can only be delocalised onto the ortho and para carbons (a more rigorous explanation would talk about the intermediate or transition state instead of the starting material, but the argument is the same).

@orthocresol I think I phrased that badly. What I meant was that, is it actually true that the charge is not fully delocalised or does resonance give the wrong answer (i.e. it has a limitation)?

If I deprotonated in the kinetic direction (first set of resonance structures, i.e. BuLi or LDA), I would expect only alkylation on that carbon or the oxygen possible. If I deprotonated in the thermodynamic direction (a Lewis acid and triethylamine), I expect either central alkylation or vinylogous alkylation (or oxygen, but that is harder with a Lewis acid sitting ther). So yes, I think there is a strong physical implication there.

If there are adjacent, parallel, overlapping p-orbitals in a molecule, the electrons will be delocalized over all such atoms. While the negative charge isn't delocalized over every atom in the conjugated system, it's more than you have shown. What you are failing to consider is:
$\hspace{2.4cm}$
What's important to note about these resonances structures is that the net charge remains constant. These structures are consistent with both the reactivities of the enolate and $\alpha$,$\beta$-unsaturated carbonyl compounds. There negative charge is delocalized between the carbonyl oxygen and its $\alpha$-carbon, and the positive charge is delocalized between the carbonyl carbon (giving rise to 1,2-addition) and its $\beta$-carbon (giving rise to 1,4-addition).


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Why are protic acids not used for the formation of thioacetals?
The standard procedure for protecting a carbonyl group as an acetal uses dry acid, such as $\ce{TsOH}$, but when forming a thioacetal, the Lewis acid $\ce{BF3}$ seems to be the catalyst of choice. For example, see Corey and Seebach's original demonstration of dithiane acidity: Org. Synth. 1970, 50, 72.
Is there any particular theoretical or practical reason why this is the case? I am guessing that the boron trifluoride coordinates exclusively to the harder oxygen atom, making it a better leaving group and hence favouring the formation of products (since the reaction is under thermodynamic control). However, I don't see any reason why acid wouldn't be able to do the job equally well.

The pKa/leaving group ability between alcohols and thiols may have something to do with it.

Alcohol pKa around 17, thiol around 13.

As you pointed out the whole reaction is in equilibrium. But in the case of the thioacetal, after the initial addition theres a strong driving force to kick back out the sulfur to reform the carbonyl rather than committing to the thioacetyl formation since the sulfur is a relatively stable leaving group.
Protic acids are almost always wet, and as such the water would prevent the reaction, whereas Lewis acids are generally either neat, or in ethereal solutions.


text:
Which class of polymerization reactions reaches high conversion under one second?
Are there polymerization reactions that can reach high conversion very fast, say under 1 s? i.e. to polymerize "immediately"?

The closest thing that comes to mind is the explosive polymerization of 4-nitroaniline, though it takes more than a second, and the product has a poorly defined structure.

Nylon also polymerizes pretty quickly

@Benzene, how quick?

https://youtu.be/T0ndxrYD1WM


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Nucleophile in a reaction
How can we determine which atom is a nucleophile in a compound in a reaction?
I know we see resonance, but I am little confused between electronegativity, formal charge, solvent, and size.

Here are some rules i remember from organic chemistry in Collage.
When trying to distinguish between nucleophile and base,

Look for resonance
If atom is involved in resonsance, this makes ato/group more nucleophilic,
if electrons are more localized its more basic.
Eg. A carboxylate is a nucleophile, the -ve charge is delocalized (i.e - charge is  spread out across carbonyl and oxyanion. Compared to nitrogen on an amine,  electron are restricted.

Sterics,
more sterics less nucleophilic.
Smaller radius atoms, valence electrons are more localized in space, thus more basic. Big radius atoms, poor localization relative to small, thus more nucleophilic.
More electronegastive, less basic
Electronegativity is a measure of the nuclear positive charge effect on near -ve charges.
The more electronegative the less basic. Because the atom doesnt want to give up its electrons. Why? Because it has high electronegativity
Conduction effects
If there are electronegative atoms within 3 bonds (taking into account electronegativities, and proximity) more nucleophilic less basic.

Nitrogens are usually bases, oxygens nucleophiles. I swear it was always like that. Halogens are weak nucleophiles.
Will edit for legibility when i get to computer

Among halogens which is more nucleophic and which is more basic


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Does grape juice change to vinegar?
If I add cooked grape juice to vinegar, does the juice undergo fermentation and also change to vinegar? Are there other conditions like exposure to air? How long would such a process roughly take?

Fermentation is what creates alcohols and vinegar from the sugars in grape juice. Just adding grape juice to vinegar won't cause fermentation. You need to look up the correct conditions for fermentation (which requires the right concentration of sugar and yeast). Try a wine-making manual.

The process that transforms sugars (e.g. the ones in grape juice) into ethyl alcohol and then this latter into acetic acid is a biological one, that is, performed by micro-organisms. The first step is usually performed by yeasts such as Saccharomyces Cerevisiae, whereas the second step is performed by bacteria of the Acetobacter genus. Thus, your grape juice won't change to vinegar if you mix both liquids. All you will get is a sweeter vinegar, or a sour grape juice.


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Grapefruit juice to extract herbs for tincture
I tried searching my question on google and did not find anything. I am looking to extract herbs with pure 100% grape juice http://www.lakewoodjuices.com/product_detail/id-32/  ( it is pasteurized ) instead of using the old school method of extracting with a 50% alcohol solution.
My theory as of now would be that it would be more beneficial and it would  ferment or extract the same plus the extra beneficial minerals of the juice. I am wonder what your opinion on this is. Do you think it would extract the same and be just as beneficial as a tincture? Would it be more beneficial? How long should I let it sit for? should I let it sit at room temp or in cooler temp say, $45 ^\circ \mathrm{F}$?

Juice is no better than water in this regard.

Use the old school method and extract with aqueous ethanol. It is more efficient and more healthy:

The solubility of most compounds to be extracted from herbs is likely to be much higher in ethanol than in water (or juice). You get more out of the plants.
The juice was initially pasteurized, but once opened, it will not necessarily stay free from microbes and/or might get mouldy. The risk of infections is definitively lower when staying with aqueous ethanol for extraction.


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Does halogenation by phosphorus tribromide lead to an inverted product?

I think it follows the SNi mechanism and would not lead to Walden inversion.
Hence, the product should be (S)‐2‐bromobutane:

But the answer says otherwise (the problem didn't indicate the presence of pyridine in the reaction mixture though).

Yes, the haloalkane product will have the inverse stereochemistry of the alcohol that it was produced form. $\ce{PBr3}$ does this by converting the $\ce{-OH}$ into a good leaving group and providing a nucleophile all in one. This is can be seen in the two step mechanism:
$\hspace{2.4cm}$
When $\ce{Br-}$ ejects $\ce{PBr2OH}$ in an $\mathrm{S_N2}$ reaction, the stereochemistry is inverted:
$\hspace{3.1cm}$

This means it does not follow the SNi mechanism ?

@AryanJakhar No, it does not. $\ce{SOCl2}$ does, however, if you are looking for something that does retain stereochemistry.

Yes, halogenation by $\ce{PBr3}$ will lead to an inverted product, as it follows $\mathrm{S_N2}$ mechanism. In an $\mathrm{S_N2}$ mechanism, the product is formed through inversion as the incoming nucleophile attacks from the side opposite of the leaving group.


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Heating of mixture of gases with one gas barely above its boiling point
What happens when mixture of two gases is heated in a closed container?
If one gas expands faster than the other and increases pressure of the container enough so that the boiling point of the other gas is decreased below the gas temperature, will second gas start turning into liquid?
This phase change must also give out heat, causing the overall temperature and pressure to further increase.
Is this valid? or mixture of two gases work lot different than two of them separately?

You could certainly have such a situation; that is, heating two compounds in their gaseous state at constant volume such that that the total pressure of the system exceeds the vapor pressure of the lower boiling compound causing it to condense.
As the gas condenses, it will of course give off heat as you said, but it would require continued heat input to sustain the increasing pressure and condensation. I believe that is what you meant in your second question anyway.

It would be interesting to see if the increased volume due to the liquid phase taking up much less volume than it did in the gaseous phase would decrease the pressure too quickly to sustain the condensation.

Yea, a couple big and important unknowns in this scenario are the relative volumes of the gases and their enthalpies of vaporization. Do we have 98% of the original vapor as the non-condensing gas and only 2% is the condensing gas so that the decrease in total pressure isn't prohibitive? But then of course the partial pressure would be correspondingly small, so you need two gases with very different enthalpies of vaporization, with the condensing gas having a much higher vapor pressure. Would be fun to play with if I had the equipment and some ideas for the right gases...I have neither ;)

oops, cc @J.Ari


text:
Optimization and DFT-D3 energy calculation in GamessFireFly version 8.1.1
My input was
$CONTRL SCFTYP=RHF DFTTYP=B2PLYP DFTD=.T. d5=.T. RUNTYP=OPTIMIZE $END
$DFTD VERSN=3 $END
$SYSTEM TIMLIM=3000 MEMORY=30000000 $END
$BASIS EXTFIL=.T.  GBASIS=acc-pvdz $END
$STATPT HSSEND=.T. NPRT=-2 $END
$FORCE NVIB=2 VIBSIZ=0.005 $END
$DATA
 H2O
C1
O  8.0                   -3.46278   2.54045   0.
H  1.0                  -2.50278   2.54045   0.
H  1.0                  -3.78324   3.44539   0.
 $END

I am getting the following error:
FATAL ERROR: DFTTYP = B2PLYP   DOES NOT YET HAVE GRADIENTS PROGRAMMED.
So is it not possible to perform optimization with this functional?

In original GAMESS-US code there are no analytic gradients for "double hybrids" and it looks like it is true for Firefly as well. You can use numerical gradients by adding NUMDER=.T. in $CONTRL section. See this for details.


text:
How should I think about potential energy surfaces for reactions of complex systems?
I've recently started doing some computational chemistry, studying the reaction dynamics of some moderately complex organic and inorganic reactions. I'm doing things like finding maxima, minima, TS's, saddle points, geom. optimizations, and finding conformational isomers of all of the above. I'm really enjoying myself.
One thing I would like to better understand, though, is how to think of the potential energy surface (PES) of these sorts of systems. The way I see it, a PES is just a potential energy profile (PEP) with one or more dimensions beyond energy and reaction coordinate. I understand that you cannot really compute a complete PES for a complex system. Most of the PES's I've seen are for simple systems like this solvent study of an SN2 reaction:

(R. Otto et al. 2011, Nature Chemistry 4, 534-538)
This is simple enough to understand. They have added the dihedral angle as another dimension of their picture of the reaction dynamics. But what about a large system, especially a non-rigid one? Since there are so many more possible ways the system can move, does that mean its 'true' PES is polydimensional? I like to think of these mechanistic studies of big reactions as us 'exploring the PES', even though we cannot see the whole thing. How should I think about this surface? Am I understanding this concept properly?

Yes, in general the PES has many dimensions and only some of them can be displayed in any one diagram.

Is the dimensionality of the PES related to the number of motions, i.e.; 3N-6 for a given molecule? How should I think of the PES in the context of my computations? For example, sometimes I will look for a saddle point. Obviously this saddle point must exist on a surface of some sort. How should I visualize a saddle point on a polydimensional surface?

In order to understand what the PES really is, it is useful to see where this concept comes from.
A system of $M$ nuclei and $N$ electrons should be described by a multidimensional wavefunction $\Psi(\vec{r}_1,\dots,\vec{r}_N,\vec{R}_1,\dots,\vec{R}_M)$, where $\vec{r}$ denotes electronic coordinates and $\vec{R}$ denotes nuclear coordinates (we neglect spin for simplicity). Within the Born-Oppenheimer approximation, the total wavefunction is split in an electronic part $\phi$ and a nuclear part $\chi$:
$$
\Psi(\vec{r}_1,\dots,\vec{r}_N,\vec{R}_1,\dots,\vec{R}_M) = \phi(\vec{r}_1,\dots,\vec{r}_N;\vec{R}_1,\dots,\vec{R}_M)\chi(\vec{R}_1,\dots,\vec{R}_M)
$$
where the electronic wavefunction depends parametrically on nuclear positions (as it will be clear in a moment).
If you inject the Born-Oppenheimer ansatz in Schrödinger equation for a system of interacting nuclei and electrons and you integrate out electronic degrees of freedom you get an equation for the nuclear wavefunction only
$$
\left[ -\sum_I \frac{1}{2M_I} \nabla^2 + \sum_{I<J}     \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|} + E(\vec{R}_1,\dots,\vec{R}_M) \right]\chi = E_\text{tot} \chi
$$
where $E(\vec{R}_1,\dots,\vec{R}_M)$ is the solution of the electronic problem
$$
\left[ -\sum_i \frac{1}{2}\nabla_{i}^2 +   \sum_{i<j} \frac{1}{|\vec{r}_i-\vec{r}_j|} -\sum_{i,I} \frac{Z_I}{|\vec{r}_i-\vec{R}_I|} \right]\phi = E(\vec{R}_1,\dots,\vec{R}_M)\phi
$$
As you can easily see the solution of the electronic problem (both the energy and the wavefunctions) depend parametrically on nuclear coordinates. The potential energy surface is the potential energy appearing in the equation of the nuclear wavefunction, i.e.
$$
U(\vec{R}_1,\dots,\vec{R}_M)=E(\vec{R}_1,\dots,\vec{R}_M) + \sum_{I<J}  \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|}.
$$
It is clear that finding the PES is a formidable task, since you have to solve the electronic problem for each possible set of nuclear coordinates. So, yes the PES is a multi-dimensional function of nuclear coordinates. If you have $M$ nuclei, your PES will be a function of at least $3M-6$ coordinates (once you remove translation and rotation of the whole system).
Since a multi-dimensional function is impossible to visualize, usually we study the PES by considering few interesting degrees of freedom (as the O-C distance and O-C-I angle in your example). Therefore, as you say, you can think as if you are exploring the PES locally and in a reduced space.

ADDITIONAL INFORMATIONS
Since the nuclear wavefunction is sharply peaked, (heavy) nuclei can usually be considered as classical particles. Thus instead of the nuclear quantum Hamiltonian
$$
H_n = -\sum_I \frac{1}{2M_I} \nabla^2 + \sum_{I<J}  \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|} + E(\vec{R}_1,\dots,\vec{R}_M)
$$
you can consider its classical counterpart
$$
H_n^\text{cl} = \sum_I \frac{\vec{P}_I^2}{2M_I} + \sum_{I<J}    \frac{Z_IZ_J}{|\vec{R}_I-\vec{R}_J|} + E(\vec{R}_1,\dots,\vec{R}_M) = \sum_I \frac{\vec{P}_I^2}{2M_I} + U(\vec{R}_1,\dots,\vec{R}_M)
$$
where $\vec{P}_I$ is the momentum of the nucleus $I$. From the laws of classical mechanics you can easily verify that the force acting on nucleus I is simply given by
$$
\vec{F}_I = -\nabla_{\vec{R}_I} U(\vec{R}_1,\dots,\vec{R}_M).
$$
So even if you consider the nuclei as classical particles, the PES has the same meaning, i.e. is the potential felt by the nuclei. Here you can easily apply a classical picture. A transition state is simply a state along your reaction coordinate where you have a local maxima of the PES for the reduced coordinates you are considering(as the C-O bond in you example).
I never looked to this into details but I think computational programs climb the PES uphill from the initial and final state of your system (i.e. they follows -$\vec{F}_I$) in order to find the TS.

This is the sort of answer I was looking for. Based on your explanation, I think my conceptual understanding was alright, but seeing some of the math and seeing how Born-Oppenheimer fits into things is very helpful! One thing, though; if I'm exploring a reaction mechanism by modelling species and TS's using DFT, am I really exploring the PES at all, since I am just looking at the system as a sort of 'fluid of electrons', rather than solving nuclear wavefunctions?

@gannex Maybe the fact that I considered the nuclei as quantum particles was a bit confusing... I added a section to my answer to (hopefully) clear this confusion.

@gennax Basically you use DFT to solve the electronic problem and find E and then you add to E the classical nucleus-nucleus repulsion in order to obtain the total PES (which I called U).

It can also be noted that often the coordinates used for a system are determined by using principle component analysis and making these reaction coordinates simply the coordinate transformations which are most likely to occur.

Just to add that for TS search usually some eigenvector following based algorithm is used.

thx R.M. I looked at some of your other posts about q. mech. and comp. chem. and I like the explanations.

I think QST2 in g09 tries to average out the set of redundant coordinates and uses that as a guess for a TS search. (It fails a lot.) I am pretty sure they also use Newton Raphson for a TS search.

In this context, the PES is not proper of a reaction but the nuclei involved.
You should take in mind that, in absence of external fields (and taking apart spin), it is just a functional of the nuclei positions, and as there are many ways to describe nuclei positions there are many ways of visualize the PES.
For a "large" number of nuclei, the dimension of the PES will be very large. Ideally, 3$N$-6 with the considerations above.
How it looks like? It is hard to imagine and represent in an 2D screen. By the way, I never found it useful, no more than to imagine that the reaction is a ball rolling down a hill.
To turn it into something that can be useful, the PES is projected into a subspace spanned by the vectors linked to the coordinates of interest for the case (for example, two bond distances).
Does it answer your question?
Edit:
- Why make a distinction between a 'reaction' and the movement of the system's nuclei
It could be done, but with "the PES is not proper of a reaction but the nuclei involved" I meant that the whole PES is independent of the trajectory followed by the nuclei during the reaction.
- Can you show me an example of a projection of a PES? (Or do you just mean the PEP, as it is shown in most computational journals?)
I mean the projections of the PES usually plotted. With PES is projected into a subspace spanned by the vectors linked to the coordinates of interest for the case I meant the following. The whole functional cannot be plotted in high dimensional cases. So, the trick is chose a set of variables that allow the description of the nuclei positions BUT also allow just use a few to describe the important part (which depends on your needs).
A simple example would be the PES corresponding to the model of a benzene molecule (built for teaching purposes) in a laboratory when you are interested in teaching parabolic movement. Of course you can describe it with Cartesian coordinates of each nuclei in R^3 (12*3=36 coordinates). But they are not very useful, you need to make appear the $x^2$ somehow for your young students. So you make a change of basis and get a new set of coordinates. Three translational of the geometrical centre and internal coordinates.
Then, you can make a matrix representation of this vector:
[int1, int2, ..., int33, z, x, y]$^T$.
It is horrible to write here (you'll see why soon) so fortunately found a little ball. And you reduce it to
[z, x, y]$^T$
But your student have to plot the famous parabolic case in a paper, and you only need $x$ and $y$, so you can just discard $z$ or make a projection by applying the matrix representation of an operator $P= P^2$.
$$ \begin{bmatrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\  0 & 0 & 1 \end{bmatrix} \times \left[ \begin{array}{c} z \\ x \\ y \end{array} \right] = \left[ \begin{array}{c} 0 \\ x \\ y \end{array} \right]$$
now is simpler. The change is subtle (maybe pedantic), one case is an approximated plot of something exact, the second one is an exact plot of something approximated. By the way, the plot is the same...
My advise is: Just look at the PES as a simple function $R^N  \rightarrow R$ with $N$ at least as 3$n$-6. There is nothing special about it that makes it different from what you studied in your algebra/calculus classes. The only one particularity is what the variables represents (nuclear positions), it was well described by the @R.M. answer.
I really never find it too useful more than a fast inspection of simple energy curve. I found the mathematical machinery of the optimisation algorithms much more useful. But of course, I looks very nice in papers, I would recommend a good colour scheme.

It has sort of helped. Why make a distinction between a 'reaction' and the movement of the system's nuclei. That seems a bit pedantic to me, but I get what you mean.
Can you show me an example of a projection of a PES? (Or do you just mean the PEP, as it is shown in most computational journals?)
What I really want to do is get a better understanding of how I am interacting with the PES of a system when I am looking for certain minima/maxima in my calculations. I'm just looking for a better qualitative understanding of what I am doing when I apply this process.

Your welcome @gannex

Why down-voted ?

I didn't downvote, but why you say that the PES is an $R^N\mapsto R^M$ function?

@R.M. Because I was still thinking in the the PES simplification. I should put $R^N \rightarrow R$. Thank you for notice the mistake.


text:
Formation of Bands in Semi-Conductors
At school, we are learning about semi-conductors and their applications in modern electronics. One of the features of semi-conductors is that there is a small energy gap between the valence and conduction band. My teacher explains that he bands occur because in the compound, since there are millions of atoms close together, due to the Pauli Exclusion Principle the energy levels of the atoms slightly change in value, forming band. When electrons are excite form the valence to conduction band, the substance is able to conduct an electric current.
However I fairly certain this is a superficial explanation as it doesn't really explain what the conduction and valence band is and doesn't explain why electrons in the conduction band are able to conduct electricity while electrons in the valence band can't.
After some researching on the web this is my rough understanding of semi-conductors now:
The s and p orbitals of the metals combine together to form MOs. As the AOs of millions of atoms combine together, these MOs become a continuum, forming two bands. One band is the valence band which consists of 2s sigma bonding and anti-bonding orbitals while the other band is the conduction band which consists of 2p pi bonding and anti-bonding orbitals. This explains why electrons in the conduction band are able to conduct an electric current since they are conjugated (not sure if this is the correct term) across the entire substance and hence explains why electrons in metals are described as a 'sea of de-localised electrons'.
This is just my understanding which isn't that solid. Could someone please offer a more in-depth explanation into how these bands form and correct any wrong things that I said. Also I still have some questions.
One website stated that 'it is only when these bands become filled with 2p electrons that the elements lose their metallic character'. Which I interpret as that when the conduction band is completely filled with electrons, the substance isn't able to conduct electricity any more. However why is this so? Aren't there still electrons in the pi MOs which are 'de-localised' along the substance?
Also the above explanation using MOs seems to explain why these bands form. Does this mean that the explanation that my teacher gave me about the Pauli Exclusion Principle being responsible for the formation of bands is wrong and not needed?

doesn't explain why electrons in the conduction band are able to conduct electricity while electrons in the valence band can't

Electrons in the valence band can conduct electricity, as long as the valence band is not completely filled. This occurs in semiconductors, when you thermally promote electrons from the valence to the conduction band, you are leaving behind some holes in the valence band. The electrons in the valence band can then conduct electricity as well by "jumping" through the holes. The conductivity of semiconductors arises therefore not only from electrons in the conduction band, but also holes in the valence band.
The origin of conductivity lies in a partially-filled band, and it doesn't matter which band it is. For a more rigorous treatment google for "free electron model". The idea is that each crystal orbital (i.e. MO of the crystal) has an associated value of electron momentum. For each orbital with a positive electron momentum, there is an orbital with an equal but negative momentum. When the band is fully filled, all of the crystal orbitals are filled, and there is no net electron momentum because the positives and negatives cancel out exactly. When the band is incompletely filled, the application of a potential difference in a certain direction across the metal causes the electrons to preferentially populate orbitals with positive momentum in that direction. This leads to a net positive electron momentum in that direction, meaning that electrons "flow".

The s and p orbitals of the metals combine together to form MOs. [...]

Yes. Bands are simply a collection of crystal orbitals that have such small energy gaps between them such that they are almost continuous. I don't see how the Pauli exclusion principle comes into play except for when you are discussing the population of the orbitals (obviously any one orbital can have a maximum of two electrons). The two ways that I know of to obtain the bands (nearly free electron model, and the LCAO/tight-binding approach) both do not invoke the PEP. However I will add a disclaimer that physicists approach solid state chemistry from a very different direction and there could very well be some explanation that I do not know of.
However, yes, the electrons populate orbitals that are "delocalised" across the entire solid and this is the origin of the "sea of delocalised electrons" description of metals.

'it is only when these bands become filled with 2p electrons that the elements lose their metallic character'

I honestly don't see how that is relevant or even correct at all. Let's use the example of sodium versus chlorine. The band structure of Na metal is vastly different from that of chlorine since the molecular structures are different to begin with. Sodium is a metal and chlorine is a molecular gas, you wouldn't use a band structure to describe chlorine. You cannot just take a diagram for $\ce{Na}$ and say that it applies to $\ce{Cl2}$ and try to make statements such as "filling of the 3p bands leads to decrease in metallic character" because those bands don't even exist in chlorine. Band structures are just large MO diagrams, it is almost like trying to compare the physical properties of sodium and chlorine using MO diagrams for $\ce{Na2}$ and $\ce{Cl2}$, which is plain absurd.
However, if chlorine were to adopt a structure that has close-packed atoms like in sodium, then you would be right that it would still be metallic due to the partially-filled 3p bands.
If you want to read further I would recommend finding a solid state chemistry textbook, all of them would cover basic band theory. In particular I find this book by Moore and Smart clear and accessible.

It is perhaps also worth noting that the atomic orbitals may not simply merge in to the solid state bands - odd things can happen in the process. It has been some time since I looked at a system in detail, but more than one had fairly deep AOs end up above the Fermi level while a nominally unoccupied AO dropped into the valence band.

Thanks for the great answer! This might be completely irrelevant but is the combination of AOs to form bands of  MOs also responsible for high pressure/density gases having a continuous spectrum? I asked this question on Physics Stack Exchange and they said the energy levels become bands due to the PEP, allowing for different electron transitions (my school teacher said the same thing as well).

@Nano Sorry but you'll have to ask someone else, I don't know about that. Seems my hunch was right, it is a physicist's explanation..


text:
Transition Metals colours and electron absorbtion
I understand that when d-orbitals split some electrons can occupy the higher orbitals by absorbing frequencies of light, but do these electrons then fall back to their original ground states? If so, why? Why don't they remain in a permanently excited state?

@KennyLau ΔE does not represent the difference between the excited state and the ground state, it represents the uncertainty in the energy of the excited state. Furthermore, it only gives a minimum bound on the lifetime, not a maximum. The energy-time uncertainty relationship is not incompatible with a persistent excited state.

Oh, sorry for the misinformation. I've deleted them.

Related: Color imparted to flame and color of coordination compounds

I've heard someone mention that the electrons do indeed fall back to ground state but, you don't see that part of the emission spectra since the energy is dispersed everywhere and not just towards your eyes. This is why the colour comes from the absorbance spectra.
The reason is does not stay in the excited state is simply because it is unstable. It will gain energy and then release it as mentioned above, then repeat.


text:
Typography of Orbitals
For mhchem, I need to decide on a typographical representation of orbitals.
Wikipedia tells me, there are atomic orbitals with letters s, p, d, f and g. In a later table, there is even h.
Then, there are molecular orbitals with greek letters.
Orbital hybridisation uses the letters s, p and sp, sp², sp³.
The IUPAC green book
on (printed) page 32 uses upright characters for 1s, 2s, 2p etc.
However, in the footnotes on page 19, ‘orbitals r and s’ are in italic. (r? Wikipedia did not tell me about this.)
That is the only occurrence in an IUPAC document I could find via search. Are there more?
There are other occurences in the Blue Book in italic, but they are stereodescriptors. (1s,3s)-cyclobutane-1,3-diol.
A quick search at Google Books revealed the convention of italic font for orbitals,
e.g. Organic Chemistry, 2008
or Essentials of Organic Chemistry, 2006,
or Orbital Interactions in Chemistry, 2013,
all published by Wiley.
However, there are also
Molecular Orbitals and Organic Chemical Reactions, 2011
and
Discovering Chemistry With Natural Bond Orbitals, 2012
by the same publisher that use an upright font.
Anyhow, I need to decide for one way or the other. Is there any IUPAC document, I could base my decision on? One, that defines the meaning of the letters and not just uses them.
Second question: I need to distinuish orbitals form other notations. Are my following assumptions correct to reliably detect orbital notations?

Orbitals are always written as number + lower-case letter
Letters can be s, p, d, f, g, h.
There is no element entity in chemical notations with these (lower-case) letters. But there is p for proton.
The letters for abbreviations for organic structure fragments such as ‘c’, ‘m’, ‘p’, ‘r’, ‘s’, and ‘t’ overlap, but they are never written with a number in front of them.

Disclaimer:
I am not a chemist and am not familiar with the meaning of all of this.
I am interested in the typography, though.

Judging from the context of the footnotes on P.19, I would say that the r and s are general orbitals?

Nope. Notation like $sp^3d^2$ is pretty common.
Nope. There are Cs, Gd, Hf, Ag.

@permeakra sp, sp², sp³ can easily be recognized, as can Cs, Gd, ..., of course. I was talking about the one-letter orbitals. Where can I find an (authorative) example for $d^2$?

@KennyLau Are "General orbitals" some kind of (mathematical) variable? $r \neq s$ looks like that.

I would deduce so (I am not sure at all)

mhchem, strictly speaking orbitals are just functions of the coordinates of only one electron (spatial and eventually spin). They are used/combined in different context to get a more important function, and can be chosen with some arbitrariness. In the case of Hydrogen atom, they also can represent stationary states, and in under a particular chose they depends parametrically of 3 numbers. One of them is called s, p, d, f, g, h i, ... when its value is 0, 1, 2, 3, 4,...  respectively.

They hardly can be confused with elements names, I can not imagine a context that generate this ambiguity .

Not at all. Specially in more advanced treatments.
If you are refering to hydrogen like orbitals, yes, they are normally used.
Every symbol of an element start with upper-case. Proton (in chemistry) is normally denoted by $H^+$. The characters in lower case are present in the symbols of the elements.
I would say yes, but I am not sure.

I never experienced confusion with this while reading books, papers etc. Also, I do not remember to have detected any confusion of this kind while teaching chemistry.

@user1420303 You completely misunderstood the intention of my question. It is not about human confusion and understanding, but about a computer program detecting certain elements (not chemical elements, but semantic entities) and creating correct typography.

Orbitals r and s in the book are generic names. In more theoretical treatments there are a lot of kinds of orbitals. Many times they are denoted by a greek character with a letter as subscript, like $\phi_i$. Both varies according to author choice. Subscripts are normally loosely divided with phrases like "i,j,k subscripts represent unoccupied orbitals while letter like a,b,c represent occupied ... " Sometimes it is needed to write too much symbols and the greek letter is completely avoided and the latin letters are used directly.

@mhchem, I apologize by the misunderstanding.

@mhchem I know your LaTeX package, Incidentally I was trying to get some results with it right now. What exactly are you trying to do? Or, what problem are you trying to solve?

@user1420303 Don't worry! Should we move to the Chat? http://chat.stackexchange.com/rooms/3229/the-periodic-table

See also: On formatting electron configurations on meta.chem.SE.

In guidelines published by the International Union of Pure and Applied Physics or IUPAP$^{[1]}$, upright letters are recommended. Whether these letters are capitalised depends on their use. Lowercase is for quantum numbers of a single particle, and uppercase for describing a system.

A letter symbol indicating a quantum number of a single particle should be printed in lower case upright type.  A letter symbol indicating a quantum number of a system should be printed in capital upright type.$^{[1]}$

Almost identical text is found in the IUPAC Green Book.$^{[2]}$
The letters themselves ignoring capitalisation are $\mathrm{s}$ – sharp, $\mathrm{p}$ – principal, $\mathrm{d}$ – diffuse, $\mathrm{f}$ – fundamental, $\mathrm{g}$, $\mathrm{h}\ \ldots$
Continue according to the English alphabet skipping letter j.$^{[1]}$
In summary, for orbital angular momentum quantum numbers$^{[1]}$

\begin{align}\begin{split} l\ =\ &0\ \ &1\ \ &2 \ \ &3\ \ &4\ \ &5 \ \ &6\ \ &7\ \ &8 \ \ &9\ \ &10\ \ &11\ \ldots\\ \text{symbol}\ \ \ \  &\mathrm{s}\ \ &\mathrm{p}\ \ &\mathrm{d}\ \ &\mathrm{f}\ \ &\mathrm{g}\ \ &\mathrm{h}\ \ &\mathrm{i}\ \ &\mathrm{k}\ \ &\mathrm{l}\ \ &\mathrm{m}\ \ &\mathrm{n}\ \ &\mathrm{o}\end{split}\tag{for a single particle}\end{align}
\begin{align}\begin{split} L\ =\ &0\ \ &1\ \ &2 \ \ &3\ \ &4\ \ &5 \ \ &6\ \ &7\ \ &8 \ \ &9\ \ &10\ \ &11\ \ldots\\ \text{symbol}\ \ \ \  &\mathrm{S}\ \ &\mathrm{P}\ \ &\mathrm{D}\ \ &\mathrm{F}\ \ &\mathrm{G}\ \ &\mathrm{H}\ \ &\mathrm{I}\ \ &\mathrm{K}\ \ &\mathrm{L}\ \ &\mathrm{M}\ \ &\mathrm{N}\ \ &\mathrm{O}\end{split}\tag{for a system}\end{align}

As an example, here are two ways$^{[1]\ [2]\ [3]}$ of writing the predicted ground state electronic structure$^{[4]}$ of oganesson, element $118$:
$$\ce{Og: [Rn] {5f^{14}}{6d^{10}}{7s^2}{7p^6}\\ Og: [Rn] (5f)^14(6d)^10(7s)^2(7p)^6}$$
Typography regarding spacing is not explicitly stated, but is clear from the examples provided. Nevertheless, if you prefer to use
$$\ce{Og: [Rn] 5f^14 6d^10 7s^2 7p^6},$$
it should be fine as well.

$[1]$ E. Richard Cohen, Pierre Giacomo. $(1987)$. Symbols, Units, Nomenclature and Fundamental Constants in Physics. ('IUPAP Red Book'). International Union of Pure and Applied Physics. Commission C2 – Sunamco. ($2010$ reprint) (pp 12$-$13)
$[2]$ E. R. Cohen, T. Cvitas, J. G. Frey, B. Holmström, K. Kuchitsu,
R. Marquardt, I. Mills, F. Pavese, M. Quack, J. Stohner, H. L. Strauss, M. Takami, A. J. Thor. ($2007$). Quantities, Units, and Symbols in Physical Chemistry. ('IUPAC Green Book'). Third Edition. International Union of Pure and Applied Chemistry. ($2008$ IUPAC & RSC reprint). ISBN: 978-0-85404-433-7. (p 32)
$[3]$ Anne M. Coghill, Lorrin R. Garson. ($2006$). The ACS Style Guide. Effective Communication of Scientific Information. American Chemical Society. DOI: 10.1021/bk-2006-STYG, ISBN: 9780841239999 (print), 9780841228306 (online). (p 256)
$[4]$ 'The Element Oganesson'. $(2016)$. Steve Gagnon. Thomas Jefferson National Accelerator Facility – Office of Science Education. (February 1, 2017)

Here's hoping that IUPAP qualifies as close enough ;-)

Are there any guidelines on that spacing between (for example) $\mathrm{5f^{14}}$ and $\mathrm{6d^{10}}$?

@orthocresol Based on the documents I have glanced through I would say no, at least not explicitly. There's an example from IUPAP uses almost no spacing whatsoever; IUPAC in at least one place has used parentheses instead (similarly to the notation for molecular orbital configuration). I will look over a few more  sources, more to follow tomorrow (technically later today).

@orthocresol In the app, my option is very much like the IUPAP example whereas on desktop there are rather big spaces. Perhaps I will have to add a picture instead to emphasise?

Hmm, I didn't mean to be skeptical or question the validity of what you wrote; I don't know anything about it myself. The only thing I remember is this meta post: http://meta.chemistry.stackexchange.com/questions/2735/on-formatting-electron-configurations

@orthocresol I didn't mean to make it seem as if I am offended in any way; to the contrary, you pointed out something that I admittedly didn't give much thought. To be sure, here are upvotes to your comments :)


text:
Identifying base in a chemical equation

According to the Brønsted Lowry principle, "base" is a proton acceptor. I've tried to apply this idea to every choice as you can see these red circles in each equation below, but they are all loses protons, aren't they? How could it be a base?

However, if I choose water in a blue circle, I will get the underlined substance as a base for both choice B and C instead. Also, if you have a look at other substances involve in equation B and C, they do support the assumption that they act as acids that involved in donating protons. Hence, both B and C are correct. But the solution said the correct answer is C.

What is the change from $\ce{HCN}$ to $\ce{CN-}$? Is $\ce{HCN}$ acting as an acid or a base?

I would write down the half equations for each reaction:
A

$\ce{ HNO3 + H+ -> H2NO3+ }$ (Base)
$\ce{H2SO4 -> HSO4- + H+}$  (Acid)

B

$\ce{  HSiO3- -> SiO2 + OH- }$  (Base)
$\ce{HCN  -> CN- + H+ }$ (Acid)

C

$\ce{HNO2 -> NO2- + H+ }$ (Acid)
$\ce{HCO3- -> CO2 + OH- }$(Base)

D

$\ce{C6H5O- + H+ -> C6H5OH }$ (Base)
$\ce{CH2ClCO2H -> CH2ClCO2- + H+ }$(Acid)

There are multiple definitions for bases and acids.
The definition that most people thing of is the Brønsted-Lowry definition. An acid is a proton donor and a base is a proton acceptor. Which for bases implies a reaction of sort $$\ce{H2O + B- -> OH- + HB}$$
However there is also the Lewis definition of acids and bases. In this case acids act as electron acceptors and bases act as electron donors. (A way to remember this is that it is the inverse of Brønsted-Lowry! A base gives electrons instead of taking protons) An example of this would be $$\ce{H- + BH3 -> BH4-}$$ Here, hydride is acting as a lewis base and donating its electrons to borane, a very common lewis acid. Hydride is among the strongest, naturally observed lewis bases.
There are also Arrhenius acids and bases. Arrhenius acid's lose a proton whereas Arrhenius base's lose an $\ce{OH-}$. For example $$\ce{LiOH -> Li+ + OH-}$$
In the case of your problems. Equation C is an Arrhenius base. A quick way to check what is acting as a base is to write down the half reactions. Here, bicarbonate $\ce{HCO3-}$ is giving up a hydroxide ion to and nitrous acid is giving up a proton.


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Intermolecular forces - Covalent bonding
Im confused. How to relate permanent dipole dipole moment, temporary dipole induced dipole and Van der waals forces of attractrion?

You can setup a perturbative expansion to include all these dipoles, except covalent bonding.

Induced dipoles are much like credit money: you can buy goods and use them all right, but still they are not quite yours.

permanent dipole dipole is like a water molecule. temporary dipole is a transient feature that appears to relax the systems energy, Van der waals are a consequence of induced dipoles in larger molecules with more freedom.

Permanent dipoles are the result of charge separation within a bond, and normally results from bonds where the two elements have a large difference in electronegativity. Dipole moment is a measurement of the polarity of a compound. For example, the carbon-oxygen bond in acetone gives the compound a permanent dipole, explaining its large dipole moment. A temporary dipole occurs when electrons randomly group together in one part of a bond. These dipoles result in London Dispersion Forces, the weakest king of Van der Waals force. Van der Waals forces are a collective term for all residual attractive and repulsive forces between compounds that are not the result of ionic or covalent interactions. Induced dipoles are dipoles that result from a non polar compound becoming polarized as a result of interactions with an ion or polar compound.

So  both of them have VDW between molecules?