text:
Why a TS must have only one negative frequency

The vibrational frequencies are related to second derivatives; a minimum will have only positive frequencies while transition state should have one negative frequency. The vibrational analysis must be performed at the the optimized geometry.
Reproduced from ucsb.com

When finding a minimum (local or global), the vibrational frequencies should all be positive. This makes sense (at least looking at it simplistically) as if we travel in any direction we're always going up (else we're not in a minimum).
When looking for a transition state, the common goal is to end up with one (and exactly one) negative frequency, with all others being positive. This makes less sense to me, as, from a transition state there are (in 2D space) two possible ways to go the wrong way (down in energy) - clearly one is more favourable as it leads to a lower energy intermediate/product, but none the less, the other pathway exists.
Is my analysis too basic (i.e. thinking about a reaction profile curve is too simplistic), or is this okay but i'm missing some fundamental transition state theory?

I'd have to look at the calculus a bit to confirm, but my intuition says that you can probably prove that near a second order saddle point, there is a first order saddle point with a lower energy, given appropriate assumptions about the energy surface.

A second order saddle point has to be higher in energy and will be downhill to a first order saddle point and then to a minimum (think mountain peaks vs a mountain pass).

I don't think I'm understanding. Do you know that each frequency corresponds to one dimension, and each dimension has two directions you can travel in? So there are two downhill paths, but they arise from a single imaginary frequency (note it's the force constant, not the frequency, that's negative).

Related reference: http://www.math.uni-leipzig.de/~quapp/dhTCA86.pdf

Negative frequency/ mode: wrong terminology. Imaginary mode/ frequency: correct terminology. Not too shabby reference: http://computational-chemistry.com/en/blog/transition-state/

AFAIK, one imaginary mode or one negative frequency is an assumption made to apply harmonic approximation to transition state theory (TST). But for generalized TST, you don't need to make this approximation. In that case, your TS will reside on all maxima and the residence time will depend on the barrier height. So, your reactant molecule will cross-over all the available TS that could be calculated using Maxwell-Boltzmann distribution. So, if your TS contains more than one imaginary mode than you are actually overestimating your barrier height and thus underestimating your transition prob.

Well, let's just look at the simple case you mention. If we are at a transition state, then we are at a saddle point on the potential energy surface (PES). As you say, there are two directions where we go down in energy and two where we go up in energy which are along or perpendicular to the gradient vector.
Note that to confirm we are at a saddle point, in mathematics one could do the second-derivative test. Or, in chemistry (and in math) in order to do this, we construct the Hessian matrix and find its determinant. So, let our PES be given by a function $f(x,y)$. Then, the Hessian takes the form,
$$
\begin{bmatrix}
f_{xx} & f_{xy} \\
f_{xy} & f_{yy} \\
\end{bmatrix}
$$
where $f_{xx}$ denotes the second partial derivative of $f$, both with respect to $x$ and likewise for the rest of the terms. Notice that $f_{xy}=f_{yx}$, so this matrix is symmetric and hence all its eigenvalues are real. Also note that sometimes people will write this in terms of the force constants instead of the second-derivatives because a harmonic force-constant is a second derivative.
So, if we take the determinant of this matrix, its sign tells us about whether the function is concave up or concave down at the point we are interested in. This determinant is given by,
$$
f_{xx}f_{yy}-(f_{xy})^2.
$$
The sign of this actually tells us information about the eigenvalues of the matrix because the determinant of a matrix is equal to the product of its eigenvalues.
Notice that in order to find the eigenvalues of this matrix we would solve the following equation:
$$
(f_{xx}-\lambda)(f_{yy}-\lambda)-(f_{xy})^2=0
$$
Or, after rearranging,
$$
\lambda^2-\lambda(f_{xx}+f_{yy})+f_{xx}f_{yy}-f_{xy}^2=0
$$
These eigenvalues can be related to the frequencies of the vibrations if we first mass-weighted each second-derivative. All I want to show here is that for this case there are only two eigenvalues, and these eigenvalues do not have to do with the number of directions to move towards a minimum. One of the vibrations you can imagine as being the mode which in two directions have an increase in energy. So, along this mode, the molecule stays at the transition state and hence there is a physical frequency associated with this motion. If the second derivative test is worth anything, then the overall sign of this determinant will be negative at a transition state and hence there will be one and only one negative eigenvalue. Note that this only works at critical points, so that's why at the transition state there must be only one negative imaginary frequency (or negative eigenvalue).
Because for most every molecule we have more than two modes to worry about, you could imagine splitting a much larger matrix into many 2-d matrices and looking at two modes at a time. From this idea it is clear that a transition state need not be limited to having only one negative frequency. Rather, at second-order saddle points, (i.e. a saddle point where following the eigenvector corresponding to a negative eigenvalue leads you to another saddle point) there will be two negative imaginary eigenvalues.

To summarize:

The number of imaginary frequencies is the number of negative eigenvalues of the Hessian matrix (the matrix of all possible mixed-second-partial derivatives)
There can be more than one negative frequency at a higher-order saddle point, but physically speaking these are much less common than transition states with only one imaginary frequency
One cannot determine the number of imaginary frequencies by thinking about the number of paths to a minimum. You could think about the number of modes which do not kick you off of a saddle point if you prefer this.

Edit:
I mistakenly referred to negative frequencies which is not correct (as alluded to by a commenter). The eigenvalues of this matrix correspond to the force constants (second derivatives with respect to displacement coordinates), so one gets negative force constants, not negative frequencies at a saddle point. Because the force constant for one mode is negative at a transition state, the "frequency" of this mode is evaluated as $\nu=\sqrt{\frac{k}{\mu}}$ as usual, and the frequency is imaginary. So, one is looking for either one negative force constant or one imaginary frequency, not a negative frequency. Sorry I missed that.

I will answer from a practical point of view. First off, let's note that a transition state (TS) is, loosely speaking, in-between a reactant and a product (state). These constitute local minima, possibly with very low barriers. Moving along an appropriately defined reaction coordinate, you go from the reactant through the TS to the product or vice versa (think reaction with equilibrium). So having "two directions", that are actually just one path on which you are going forward or backward, makes sense.
In terms of performing calculations, one finds that it can be difficult to obtain even minima that do not have spurious negative frequencies. The dreaded methyl rotations are the most common example. (One implication is that free enthalpies, which take into account the vibrational component of the partition function, may easily be wrong in cases of methyl rotations.) For a TS, this is even more difficult, because the algorithms used for geometry optimization are less robust than those for minima. However, as noted in the comments to your question, a second-order saddle point (meaning two negative frequencies) is often accompanied by a first-order saddle point close by, which is usually a better TS.
Note however, that this is not a strict requirement. Depending on the chemical species involved, degenerate TS have been observed, essentially existing where two pathways cross. (I do not have a reference on hand.) These are very rare and I assume require a lot of symmetry. So for standard chemistry, the rule of "TS means stationary point and exactly one negative frequency" holds.
Further note that in order to be sure about the TS being connected to the product and reactant you have in mind, one should modify the TS geometry to follow the pseudo-vibrational mode twice (once for each direction) and optimize for a minimum. Some programs even have automation for this. This procedure leads to the reactant and product of that TS.

Instead of negative frequency, could we perhaps use imaginary mode?


text:
Can someone suggest me important references for deriving EOM in MCSCF?
I am currently starting a project dealing with MCSCF method. Can someone please give me references (papers or lecture notes are preferred) which may be considered as the "corner stones" in the history of the development of this method for atoms (for molecule is fine but the for atom will be nice)? I have searched and in fact obtained some papers which look important in this subject but since MCSCF is very popular I am sure there are already loads of papers of various purposes published within the subject, making locating the most important ones difficult. I am also looking for the derivation of the equation of motion (EOM) of the orbitals and coefficients.
Currently I am reading "Computational Atomic Structure - An MCHF Approach" by C. Froesche-Fischer and "The Hartree-Fock Methods for Atoms - A Numerical Approach" also by the same author. But from the terminology the author adopts it looks like these books is not the typical literature a quantum chemistry lecturer would recommend his students to read.

The best way to start learning about MCSCF is to consult a standard text book (a selection can be found in our resources for learning chemistry thread) and follow the citations within. It is possible that the community will regard your question as too broad (or even a duplicate of the linked post). For [tag:reference-request]s, you should be very specific, also outlining what you have read so far, found so far.

Currently I am reading "Computational Atomic Structure - An MCHF Approach" by C. Froesche-Fischer and "The Hartree-Fock Methods for Atoms - A Numerical Approach" also by the same author.

You should [edit] your question and include any and all information you have obtained to make it complete and standalone as comments are not meant to last forever.

source: Stack Exchange

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/

sort: votes


text:
Calculate ozone layer thickness at a given temperature and pressure

The ozone concentration in the upper atmosphere averages $\pu{3.0 \times 10^13 molecules per cm3}$ in a region between $15$ and $\pu{35 km}$ high, at an average pressure of $\pu{0.001 atm}$ and temperature of $\pu{–25 °C}$. How thick would the ozone layer be, if these same molecules were at a pressure of $\pu{1.0 atm}$ and a temperature of $\pu{20 °C}$?

I understand that I need to use the ideal gas law, but I really have no idea how to approach the question, or even how to work with $\pu{molecules per cm3}$. I see that we have been given the values for temperature and pressure, but am unsure how to go about finding volume, and then the thickness.
I am also confused with how to deal with the two separate situations. Would you use $p = \pu{0.001 atm}$ or $p = \pu{1 atm}$. And would you use $T = \pu{248 K}$ or $T = \pu{293 K}$? And where do you find the value for the amount of substance?

Please share with us your thoughts and what exactly you find difficult in this problem. This network isn't about just doing the homework for you, but to help you to understand and proceed on your own when possible.

Start with writing down the equation for ideal gas, then express the unknown entity (volume), unify the units, plug in the known values. Also, think about what formula from the geometry course can you utilize to find the thickness from the obtained volume. You can edit your question at any time and include your progress.

Please be so kind and post the answer that whoever gave you the assignment provides.  @andselisk and I are off by a factor of a thousand, and I believe the reason for that is that we interpreted the question differently. (If you are uncomfortable with the markup you can post a picture and notify me in the comments section of my answer, I'll rewrite it for you.)

@Martin-マーチン The answer provided is 25 mm! Thank you so much for the help.

You are welcome. Would it be possible for you to post the entire solution, it really hinges somewhat on the interpretation and I would like to see the whole explanation of it. If you found that exercise in a textbook, it would be nice if you could provide the source; thank you.

Final conditions ($\pu{1.0 atm}$) give a hint at what altitude the thickness should be estimated – at the Earth's surface. Let's denote the unknown thickness as $h_x$, radius of the Earth as $r_\mathrm{E} = \pu{6371 km}$; also take $h_1 = \pu{15 km}$ and $h_2 = \pu{35 km}$ and put these dimensions on a rough 2D sketch:

Fig. 1. Schematic representation of two ozone layers at given altitudes.
To find $h_x$, we can use the volume of the new hypothetical ozone layer $V_x$ as the difference in volumes of two spheres with corresponding radii:
$$V_x = \frac{4 \pi (r_\mathrm{E} + h_x)^3}{3} - \frac{4 \pi r_\mathrm{E}^3}{3} \tag{1}$$
and express $h_x$ from (1) as follows:
$$h_x = \left(\frac{3V_x}{4 \pi} +r_\mathrm{E}^3\right)^{\frac{1}{3}} - r_\mathrm{E} \tag{2}$$
From ideal gas law for a relatively thin gas layer without temperature and pressure gradients:
$$V_x = \frac{nRT_x}{p_x} \tag{3}$$
where $n$ is the amount of gas in ozone layer, $T_x = \pu{20 ^\circ C}$ and $p_x = \pu{1 atm}$ (new conditions). Let's also denote ozone concentration as $N = \pu{3.0e13 cm-3}$, then total amount of ozone is to be found via initial volume $V_0$ and Avogadro constant $N_\mathrm{A}$ as
$$n = V_0 \times \frac{N}{N_\mathrm{A}} \tag{4}$$
Initial volume can also be found from Fig. 1 using the difference in volumes of the corresponding spheres with radii $r_\mathrm{E} + h_1$ and $r_\mathrm{E} + h_2$:
$$V_0 =  \frac{4 \pi (r_\mathrm{E} + h_2)^3}{3} - \frac{4 \pi (r_\mathrm{E} + h_1)^3}{3} = \frac{4 \pi}{3}((r_\mathrm{E} + h_2)^3 - (r_\mathrm{E} + h_1)^3) \tag{5}$$
Now all we have to do is carefully plug in the values with the corresponding units carefully and find all the unknown variables. From (5):
$$V_0 = \frac{4 \pi}{3}((\pu{6371 km} + \pu{35 km})^3 - (\pu{6371 km} + \pu{15 km})^3) = \pu{1.03e10 km3} = \pu{1.03e25 cm3} \tag{6}$$
Now let's find the amount of ozone using $V_0$ and equation (4):
$$n = \pu{3.0e13 cm-3} \times \frac{\pu{1.03e25 cm3}}{\pu{6.023e23 mol-1}} = \pu{5.12e14 mol} \tag{7}$$
and, subsequently, the unknown volume $V_x$ using (3):
$$V_x = \frac{\pu{5.12e14 mol} \cdot \pu{8.314 J mol-1 K-1} \cdot \pu{293 K}}{\pu{101325 Pa}} = \pu{1.23e13 m3} = \pu{1.23e4 km3} \tag{8}$$
Finally, plugging all the values in (2), we can find the thickness:
$$h_x = \left(\frac{3 \cdot \pu{1.23e4 km3}}{4 \pi} + (\pu{6371 km})^3\right)^{\frac{1}{3}} - \pu{6371 km} = \pu{2.4e-5 km} = \pu{24 mm} \tag{9}$$
The result is somewhat close to the fact posted by NASA:

Around the world, the ozone layer averages about 3 millimeters (1/8 inch) thick, approximately the same as two pennies stacked one on top of the other.

When you forget to hit post and then someone else is faster... now I wonder why I have different results... well, I'm going to have to read your post then.

@Martin-マーチン This site really needs a function that many messengers have: "X is typing..." :) You relied on ideal gas equation to determine the amount, whereas I ditched it and used ozone number concentration instead. For some reason I didn't find an idea of applying the ideal gas law when there is probably a reasonably high gradient of temperature and pressure appealing.

Yeah, I agree. Initially I didn't want to bring the spherical shell into the answer, but then I thought I still should do that (then I was distracted, haha). I found that I am off by a factor of 1000. That is also the discrepancy that I found when I calculated the particle density from the ideal gas. Now that implies that this question is open for interpretation... gosh I hate when that happens. Anyway, you got my vote :D

@Martin-マーチン +1 too, Chemistry Votes Exchange:) Maybe the OP eventually can comment on the situation or submit the work and get an answer (probably, a different one – that'd be fun).

@Martin-マーチン When using ideal gas law for determining $n$, I also arrived at $\pu{23.8 m}$. Either way, I think both approaches are valid depending on the question's interpretation.

The number concentration ($C$, or number density $\rho_N$) is not really relevant for this exercise. (For more on this, see at the bottom.)
You have correctly identified that you will need the ideal gas law (obviously as an approximation):
$$pV = nRT$$
For the ozone layer you have been given the pressure $p_\text{layer}$, temperature $T_\text{layer}$, and the thickness $h_\text{layer}$:
\begin{align}
p_\text{layer} &= \pu{0.001 atm}\\
T_\text{layer} &= \pu{-25 ^\circ C} = \pu{248.15 K}\\
h_\text{layer} &= |\pu{15 km} - \pu{35 km}| = \pu{20000 m}
\end{align}
You are asked to calculate the thickness $h_\text{std}$ at $p_\text{std} = \pu{1 atm}$ and $T_\text{std} = \pu{20 ^\circ C} = \pu{293.15 K}$ ($\text{std}$ is short for standard).
The exercise clearly states "same molecules", therefore $n_\text{layer} = n_\text{std} = n$. You can now assume that you are simply compressing the gas, assuming in both cases the same area $A$, therefore you can express the volume as $$V = A\cdot h.$$
With this you can form the following relations:
\begin{align}
p_\text{layer}Ah_\text{layer} &= nRT_\text{layer}
&\Leftrightarrow&&
\frac{p_\text{layer}h_\text{layer}}{T_\text{layer}} &= \frac{nR}{A}\\
p_\text{std}Ah_\text{std} &= nRT_\text{std}
&\Leftrightarrow&&
\frac{p_\text{std}h_\text{std}}{T_\text{std}} &= \frac{nR}{A}\\
\end{align}
Since the right sides equal each other you arrive at
\begin{align}
\frac{p_\text{layer}h_\text{layer}}{T_\text{layer}}
&=\frac{p_\text{std}h_\text{std}}{T_\text{std}},\\
h_\text{std} &=
\frac{p_\text{layer}}{p_\text{std}}\cdot
\frac{T_\text{std}}{T_\text{layer}}\cdot h_\text{layer}.
\end{align}
Fill in the known values and the thickness is $h_\text{std}\approx\pu{24 m}$.

What should be clear from that exercise is, that it is a very crude approximation. It does not take into account, that the layer is a spherical shell. In that case you would have to calculate the volume differently, and you would have to know $r_\text{earth}=\pu{6371 km}$ (according to a quick Google search):
\begin{align}
V_\text{layer} &= \frac{4}{3}\pi\left[
(r_\text{earth} + \pu{35 km})^3 - (r_\text{earth} + \pu{15 km})^3\right]\\
&\approx 4\pi (r_\text{earth}+\pu{35km})^2\cdot\pu{20 km}\\
&\approx \pu{1E10 km3}\\[2ex]
V_\text{std} &= \frac{4}{3}\pi\left[
(r_\text{earth} + h_\text{std})^3 - r_\text{earth}^3\right]\\
&\approx 4\pi (r_\text{earth} + h_\text{std})^2h_\text{std}&
(r_\text{earth}>>>h_\text{std})\\
&\approx 4\pi r_\text{earth}^2h_\text{std}\\
\end{align}
And with that calculation we would get to the thickness being $h_\text{std}\approx\pu{23 m}$.
Since the ozone layer is in no way a pure substance, the ideal gas also is only a crude approximation as you are treating everything the same. That that is not the case is even given from the values of the exercise.
You can calculate the number concentration $C$ also from the ideal gas:
\begin{align}
&& C &= \frac{N}{V} \\
&& pV &= nRT \\
&& N&= n\cdot N_\mathrm{A}\\[2ex]
\implies&&
C &= \frac{p\cdot N_\mathrm{A}}{RT}\\
&&C_\text{layer}
&\approx \frac{\pu{0.001 atm}\cdot\pu{6.023E23 mol-1}}{
\pu{82.057 cm3 atm K−1 mol−1}\cdot\pu{248.15 K}}\\
&&C_\text{layer} &\approx \pu{3E16 cm-3}
\end{align}
From the given number concentration of ozone $C_\text{ozone} = \pu{3.0E13 cm-3}$ you can see that only every thousandth molecule in that layer is ozone.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/3.0/


text:
Oxy acids of chlorine bond order
In my book it 's stated that the bond order of the oxy acids increases with increasing value of $n$ in $\ce{HClO_n}$ , and also the energy required to break $\ce{Cl-O}$ bond decreases with increasing n leading to the greater strength in acidity . But with increasing bond order bond length should decrease making it tougher for it to break the $\ce{Cl-O}$ bond, so how's that possible?

For starters, your book is wrong; the $\ce{Cl-O}$ bond order for all chlorine oxyacids is very close to 1. Any bond order differences are so minor that they do not play any role. (Draw out the molecular structures observing the octet rule to confirm that.)


text:
Why are indicators used only in the form of dilute solutions?
Why are indicators used only in the form of dilute solutions? I get that indicators are a form of weak acids themselves, and the extent of their dissociation varies with the pH of the solution they are in.

indicators 1) are very costly 2) affect pH of the solution itself.

And visual recognition of their virage interval!

The most simple and general answer is: because we can! However, it may require explanation.
Indicators will always affect the observable of the solution they are measuring. For example, an acid-base indicator is an acid or a base itself (depending on the pH of the solution). It will take at least one indicator-equivalent of acid/base to protonate/deprotonate the indicator and induce the colour change. (Some indicators may even require two or more equivalents.) Until that full equivalent is supplied, we cannot witness the new colour. Thus, we want the indicator concentration to be as minute as possible because that will give us the lowest indicator error possible.
All other properties and effects of indicators are secondary to the major point above. For example, indicators are typically chosen to be very intensely coloured so that the colour may be observed even at very low concentrations; this allows us to reduce indicator loading and error and thereby increase accuracy.
Synthesising these indicators may be time-consuming and they may be expensive. This further strengthens our wish to use as little as possible.
The colour change of an indicator is ideally abrupt and strong, again meaning we can reduce the loading a lot.
And so on and so forth.
Thus, because we can use indicators at these low concentrations, we do.


text:
Strange green and gold coloured chemical
I have a reagent bottle with an unknown chemical that came with some other lab gear about 30 years ago, I stumbled on it again when clearing out some old stuff.
It is a standard brown glass Merck of about 250ml size.  It has no label and never had any indication of a label and I do not know if it is the original bottle.
The powder is a fine dark free flowing crystalline solid.  It is DARK green with a gold iridescent sheen when in bright light, something like a butterfly wing or some anti-forgery inks on banknotes.
Yet if I wet a very small portion it dissolves completely and creates a vivid dark blue solution that dries blue in a thin film.  If the crystals dry in a clump they form a smutty dark green gold pasty lump that is reminiscent of the original bulk colour.
I had to hand some potassium permanganate crystals and when I mixed a pinch of the two together and added a drop of water a gas evolved and the colour seemed to be somewhere between the permanganate and the mystery chemical blue when spread on paper.  Adding a few drops of acetic acid seemed to have no effect.
Can anyone think of an obvious candidate or some simple experiments that could be used to determine the material?

A picture would really help IMHO...

In general, I don't advise experiments on unknown substances, particularly because it is unknown.

@Zhe But we wouldn't know what we're dealing with until we "experiment" >:-D

That's true, but frankly, from a safety perspective, that's strongly ill-advised.  Will your analysis liberate hydrogen cyanide? Probably not, but for unknown substances, without a lot of context, you can't really know for sure.

The pasty lump seems to indicate an organic molecule. Obviously with quite an extended conjugation. Why not to proceed with analysis?  Obviously with an eye at safety first. I mean really go through solubility test,  Lassaigne. ... It seems you are in a lab....

Does it dissolve in ethanol or acetone? A photo of the substance and its aqueous solution would probably help.

I worked for a time with squaraine dyes that were often greenish in the crystalline state, but ink-blue when in solution.

However, I would suggest that you take an analytical approach to identifying this compound.  Have you done any NMR, IR or even UV/Vis spectroscopy?  Elemental analysis?
These would be a much safer way to go about identifying the mystery compound.
Good luck, and I'd love to know what you discover!

Cool, never heard of these but it is a likely candidate.   The colour shifting seems to be common with them depending on solvents and concentration  -  https://www.chm.uri.edu/levinegroup/documents/pub2.pdf


text:
Which formula to use to find the pH of a solution?
I just had my first exam yesterday, and I'm always tripped with this question that combines neutralization with buffer solution. The task is to find the pH. I did try to understand the general concept behind it. From my understanding, if $\ce{CH3COOH}$ reacts with $\ce{NaOH}$ and the acid has a higher amount of substance ($\pu{0,005 mol}$ vs $\pu{0,004 mol}$), it means that $\pu{0,005 mol}$ of acid is used to neutralize $\pu{0,004 mol}~\ce{NaOH}$, and $\pu{0,001 mol}$ of $\ce{CH3COOH}$ will remain together with $\pu{0,004}~\ce{CH3COONA}$.
To find the pH, one has to use the Henderson-Hasselbach equation and divide the amount of substance with the volume, with $\mathrm{pH} = \mathrm{p}K_\mathrm{a} - (\log c(\ce{CH3COOH})/c(\ce{CH3COONa})$. And then if $\ce{NaOH}$ has a higher amount of substance ($\pu{0,0051 mol}$ vs $\pu{0,005 mol}$ acid), it would create $\pu{0,0001}~\ce{NaOH}$ and $\pu{0,005}~\ce{CH3COONa}$.
So to find the pH, you need to find the pOH of $\ce{NaOH}$ first. So far so good. But in the solution to the problems posted, there's suddenly an acid popping up in the solution and you have to use Henderson/Hasselbach equation. For example $\ce{NH3}$ reacts with $\ce{HCl}$ and creates $\ce{NH4Cl}$, even if you have a bigger amount of substance for $\ce{NH3}$, you still get an acid on the result... I'm so confused. And then there's an example with the coefficient ($\ce{Na2CO3 + 2HCl -> 2NaCL + H2CO3}$). If $\ce{Na2CO3}$ has a higher amount of substance (let's say $\pu{0,224 mol}$) and $\ce{HCl}$ only $\pu{0,0816 mol}$, does it mean that it would result in $0,224 \times 2 - 0,0816 \pu{mol}~\ce{NaCl}$ and $\pu{0,0816 mol}~\ce{H2CO3}$? I feel so stupid now that I can't understand this simple concept and keep making mistakes (and wasted like 8 points during the exam).

I  think there is probably a good question in here. However, it starts by asking about pH calculation, and then the real question seems to be about stoichiometry. I suggest editing this so that the question (could be an exam problem) is clear as well as your approach. I can tell that you're doing some things right (e.g., identifying the limiting reagent) and some things incorrectly. However seems to be obscured by your frustration in the exam result.

From reading your question it is obvious that you are confused. When I studied chemistry in college in the 1970's I don't remember the Henderson-Hasselbach equation being a famous equation. It seems like it is now being used as some magic can opener.
You need to go back to the chemistry and figure out the chemistry first, then do the math.
Ok, let's start by noting that the pKa of acetic acid is 4.76 from Wikipedia. There are two significant figures in the mantissa so a mathematical solution good to 1% will be adequate. Also we are dealing with solutions on the order of $10^{-3}$ molar. This allows us to make two very important assumptions which significantly simply the mathematics.

First we assume that we can ignore the autoionization of water.
Second we assume that since acetic acid is a weak acid the solution ends up being essentially 0.004 molar sodium acetate and 0.001 molar acetic acid.

Because we have excess acetic acid we know that the solution will be acidic. If we wanted a more exact solution we'd ignore the second assumption and use the $K_\mathrm{a}$ expression
$K_\mathrm{a} = \dfrac{\ce{[H^+][Ac^-]}}{\ce{[HAc]}}$
to solve the problem more exactly. Letting $\ce{[H^+]} = x$ then $\ce{[Ac^-]} = 0.004 + x$ and $\ce{[HAc^-]} = 0.001 - x$. This requires solving a quadratic equation which is doable, but mathematically messier.
Now back to using our two assumptions we can go to the Henderson–Hasselbalch equation and solve the problem with a simple formula.
$\mathrm{pH} = \mathrm{pKa} + \log{\dfrac{\ce{[Ac^-]}}{\ce{[HAc]}}} = 4.76 +  \log{\dfrac{0.004}{0.001}} = 5.36$
This means that the $\ce{[H^+]} = 4.4 \times 10^{-6}$.

So our assumption that the final solution ends up being essentially 0.004 molar sodium acetate and 0.001 molar acetic acid is reasonable.
The assumption that the autocatalysis of water can be ignored is a bit dicier.


text:
Is it possible to speed up radioactive decay?
I’m aware that elements like $\ce{^14C}$ have a known half-life, which means that over a span of roughly $5730$ years, half of the $\ce{^14C}$ atoms decay into $\ce{^14N}$.
Are there any substances known to speed up this process, that applying it to $\ce{^14C}$ will catalyze a chemical reaction resulting in $\ce{^14N}$, or sticking them in a particle accelerator and slamming them together will result in $\ce{^14N}$, or otherwise transmute elements?

No sure what you're asking, but it is impossible to transmute an isotope via any chemical reaction. You of course can transmute isotopes in a particle accelerator - that is what a particle accelerator is designed to do.

@MaxW What about in extreme conditions (for instance, abnormally high pressure or heat)?

The sun "burns" hydrogen into helium via high pressure and heat. But that isn't considered to be a chemical reaction.

@MaxW Alright, so transmute it is. Is there a formula to describe how much energy would be required to do something like that?

How is a really opened ended question that would span a number of books. In general you're asking about astrophysics. https://en.wikipedia.org/wiki/Astrophysics There is the big bang, how a sun works, and man's effort to harness energy via fusion reactors instead of conventional nuclear reactors that work via neutron emission.

This is more a nuclear physics question. Nuclear reaction and cross section data can be found at the National Nuclear Data Center, hosted at Brookhaven National Laboratory (nndc.bnl.gov).

It is possible to modify nuclear decay rates using chemistry, though it is rare and the effect is usually very small. Here I summarize the information available in this link. You may want to see the references within.
There is a type of nuclear decay called electron capture, where a nuclide directly captures an electron from the innermost electron shells and transforms a proton into a neutron. Therefore, there is coupling between the nucleus and the wavefunctions of the innermost electrons in this form of radioactive decay.
Usually the core electrons are very weakly disturbed by changes in chemical environment, but they are changed slightly. For especially light atoms, where the core electrons are very close to the valence shell, such as $\ce{^7_4 Be}$, this change results in a measurable difference in decay rate, varying by 0.1-1% compared to the isolated atom.
In a more dramatic case, there is a subset of beta decays called bound-state beta decay, where the electron released by the decaying neutron is immediately captured by the nucleus after decay. Apparently, if the parent atom is stripped completely bare of its electrons, and if the energy involved in the nuclear decay is comparatively low, once again there is a meaningful coupling between nuclear and electronic states. For the case of rhenium-187, the neutral atom $\ce{^187_75 Re}$ has a half-life of 42 billion years, but upon full ionization to $\ce{^187_75 Re^75+}$, the half-life is reduced to 32.9 years! For dysprosium-163, the neutral atom $\ce{^163_66 Dy}$ has half-life so high it has not been measured to decay, but when ionized to $\ce{^163_66 Dy^{66+}}$ its half-life is reduced to 47 days!

I should know better than to say "impossible." Great information!

Perhaps Randall was incorrect after all?

+1, but now we're left wondering what kind of chemical process can completely strip rhenium or dysprosium of electrons!

@ruakh Indeed I am stretching the domain of chemistry by including such ions, but it's interesting enough that I had to do it.

I know this is a very old question, but what is meant by meaningful couplings between electronic and nuclear states for a bare nucleus? I can't see why there should be any electronic states/wavefunction when there are no electrons present. This is a very interesting process so I'm curious to know more.

@jheindel I had a feeling I might get asked this when I saw this question bumped yesterday. I am not using these terms in a strict technical sense. All I mean to say is that there is some connection between what happens inside and outside of the nucleus. Apologies for the confusion. That said, it does appear to be possible to accurately speak of electron wavefunctions in this scenario, if I am reading this article correctly. This is probably because it depends on the wavefunction of the hydrogen-like atom after decay.

@ruakh - Apparently they can get highly stripped in galactic plasmas, so the change in half-life impacts estimates of the age of the universe.

Nuclear decay is accelerated (artificially produced) all the time. Per wikipedia,

U-238 is not usable directly as nuclear fuel, though it can produce
  energy via "fast" fission. In this process, a neutron that has a
  kinetic energy in excess of 1 MeV can cause the nucleus of 238U to
  split in two.

So a neutron with 1 MeV kinetic energy can split the nucleus. 1 MeV = 1,000,000 eV.
Under normal conditions, we breathe around an ambient available thermal energy of $k_B T \approx 2.5$ kJ/mol. The neutron above has $96485307$ kJ/mol of energy.
A lot more. It takes a lot of energy to produce artificial radioactive decay; but we can and regularly do it.
I think I recall the US government used $\ce{UF6}$ as the primary energy-producer for nuclear development during/after WWII. Different forms (salts) of Uranium seem to make artificial decay easier (I don't know by how much, though).

+1 .  But are you sure about that last sentence?  What evidence is there that decay rates of U are different in its different salts?

That whole last paragraph is nonsense.

I think it was easier to put energy into said salt -- not fundamentally faster decay. i.e. easier in lab.

Turns out $\ce{UF6}$ is used for enrichment due to its high vapor pressure (processed as a gas at low temperature) -- i.e. isotope separation of U-235 from U-238. No artificial decay in that process; it's just the method by which the desired isotope is obtained in greater quantities


text:
Can aniline autodeprotonate?
I had an argument with an assistant in my lab, as he stated that in a hypothetical reaction two aniline molecules could react in a way, that one takes a proton off of the other one, thus creating a $\ce{R-NH3^+}$ and a $\ce{R-NH^- }$
However, the $\mathrm pK_\mathrm a$ of aniline is 27, which means, that it is, if at all, a very very poor acid, making it in my eyes impossible to deprotonate. Maybe it could be possible with n-BuLi, but that was not part of the reaction. The main topic was a nucleophilic aromatic substitution to 1-bromo-4-nitrobenzene, with aniline.
Is it even hypothetically possible for such a deprotonation in pure aniline to occur?

Looks you're right.

Deprotonation of very weak acids is still very much possible; that's how their pKa's came to be known in the first place. But yes, that would take BuLi or something comparable, which you say you didn't use.

I'm going to side with you @TheChemist.
Your TA might be suggesting this because you need often need a negatively charged nucleophile in order to carry out nucleophilic aromatic substitutions.
Having the nitro group in the 4 position makes the your electrophile much more receptive to attack, since the negatively charged intermediate will be resonance stabilized.
Even if the one aniline does deprotonate another, the concentration of the deprotonated anilide will be so low (on the order of 5 x 10-28 mol/L in pure aniline, if the pKa is 27) that this will have no impact on the reaction.


text:
Explaining why CN- is a soft base
So, I am having difficulty with the following explanation/question.

(Based on your MO diagram and pi bonding theory) explain why $\ce{CN-}$ is a soft base and prefers to bind to soft acids.

Now, I know that the HOMO is the 5 σ MO and that the electron pair in the 5 σ MO is, to a large extent, more concentrated around the carbon.  So, is $\ce{CN-}$ a soft base because it has a low energy HOMO but large magnitude HOMO coefficient?  Does polarizability have anything to do with it?
When $\ce{CN-}$ reacts with a soft acid what physical changes in the cyanide molecule would such a reaction lead to?
Not really sure about this one.  Would it have to do with the donation of the electron pair in the HOMO and therefore change the HOMO?

But since N is more electronegative than C (i.e. it has lower energy AOs than C), wouldn't more electron density be concentrated on N as N contributes more to the bonding MOs of the cyanide ion?

If you look at the electron distribution it surrounds the carbon atom instead of the nitrogen as one would expect. This is caused by the interaction of the atomic orbitals of carbon among each other. They contribute more significantly to this 5σ molecular orbital compared to the influence of the 2p atomic orbital. You should not forget that the energy difference between 2s and 2p in Nitrogen is greater than that in Carbon, so the overlap between them is less significant, and thus the total contribution to the 5σ molecular orbital.

In one of his great answers, Martin gave us the calculated molecular orbitals for carbon monoxide I am copying below. Note that while cyanide has a nitrogen atom in place of the oxygen and features a single negative charge, we can still qualitatively consider the MO schemes identical; thus I will base my discussion on what you see in the image.

In this scheme, the HOMO is referred to as MO number 7; since 5 and 6 are π-type MOs, this corresponds to the 5σ you mentioned.
One thing we immediately note is that the carbon-centred lobe of the HOMO is very diffuse and extends far into space; much farther than any other orbital of the molecule. It is also the orbital that an acid (either proton or Lewis acid) would bond to. In Pearson’s HSAB theory, large and diffuse orbitals — often also called polarisable — are associated with soft acids or bases, as the charge per volume value is considered comparatively low. Therefore, just by looking at the size of this orbital, we can identify it as a ‘soft’ one.
It should also be explained why soft bases tend to create more stable complexes with soft acids. This is a direct consequence of the more diffuse nature of the orbital: the overlap between two orbitals (which will later form a bond) is typically best if the orbitals have similar shapes and sizes. Thus, because the HOMO has a rather diffuse lobe on carbon’s side it will prefer binding to other compounds with rather diffuse orbitals in a Lewis acid-base reaction.
That should be all that is required to explain cyanide’s behaviour under Pearson’s theory.

I wonder if the molecular orbital appears "diffuse" because you set the isocontour value to a particular value.

@EricBrown No matter what the isocontour value is set to, there will always be a good deal more density in a larger lobe on the carbon side.


text:
How do we know that a matter is acidic or basic without knowing their $pH$ value?
There are too many examples for acidic matters as seen below.
$$HCI \tag {1}$$
$$HNO_3 \tag {2}$$
$$H_2SO_4 \tag {3}$$
And the examples for basic matters
$$NH_3 \tag {4}$$
$$NaOH \tag {5}$$
According to this, how do we differenciate between acidic and basic matters without knowing their $pH$ value?
Regards

@Karl Can you please show it? That method seems very very good.

@86BCP2432T What is it?

Might someone show a method? It doesn't seem too hard to explain.

People seem to not understand that this community is not intended only for those with basic/advanced knowledge in Chemistry. There might be users here that are in high school or other majors who need basic chemistry information... @Maxime , you need to know that acids and bases are categorized as such only in comparison to another substance. When this 'comparing' substance is water, it is easy to tell because it will generate H+ if it's an acid and OH- if it's a base. However, if you mix sulphuric acid and nitric acid, the latter will react as a base, because H2SO4 is a stronger acid than HNO3

@Raul Luciano I get what you are saying, and some of the responses could definitely be less terse, but the site arguably isn't meant for questions like this. We aren't trying to replace Google or chemistry textbooks; information that could found from a simple search of either of these isn't well suited to the model of the site.

I will not post this as an answer because I believe the answer to this question is too broad that the one willing to answer will spend at least 2 hours planning it. There are several factors involving acidity and basicity of substances. To use simple terms, in Lewis theory, acidity is the ability of a compound to accept a pair of electrons (the compound AlCl3 will probably react as an acid, for example, because aluminum can accept electrons from a ligand), while basicity is the ability to donate a pair of electrons (NH3 will probably act as a base because it can donate a pair of electrons).

I agree @Tyberius , however, comments like "basic chemistry" add absolutely nothing to the discussion. If someone thinks the answer to the question is too broad that can't be presented in the answer box then the post should be flagged and addressed to a reliable source (textbook or website).

@Maxime , the book Chemical Principles from Peter Atkins have a nice introductory section about acid-base chemistry. I believe it will clarify your doubt about the classification of compounds as acids and bases.

pH is simply a measure of H+/H3O+ concentration in a solution.  I will use the two terms interchangeably below.
The acidity or basicity of a compound has to do with its chemical behaviour.
There are three definitions of acidity.
The simplest definition is the Arrhenius definition:

An acid dissociates in water to form H+.
A base ionizes in
water to form OH-.

This definition is limited to aqueous solutions, and so is usually ignored in favor of the Brønsted-Lowry definition, where:

An acid is a hydrogen ion donor.
A base is a hydrogen ion acceptor.

This definition is more generally, because it can be used to describe acid-base reactions that occur in solvents other than water.
There is a third definition, Lewis:

A Lewis acid is an electron pair acceptor.
A Lewis base is an electron pair donor.

The Lewis definition is the broadest definition of acidity.  However, most entry-level courses stick with the Brønsted-Lowry definition since it does an excellent job of describing what we typically consider to be acid-base chemistry.
To summarize, we recognize acids and bases from their chemical structure.
Recognizing Acids
Acids generally come in one of three forms:
Binary Acids
These have the form HX, like HCl, or HF.
Oxyacids
The have an oxygen-containing polyatomic anion.  Examples include H2SO4 or HNO3.
Organic Acids
You can't tell if an organic (carbon-containing) compound is an acid based on its formula.  However, organic acids are often written RCOOH.  The COOH represents the carboxylic acid functional group, which is what makes the compound an acid.  For example, acetic acid is often written CH3COOH.
Recognizing Bases
Metal Hydroxides
Soluble metal hydroxides are all bases (since when the dissociate in water they generate OH-).
Nitrogen-containing Bases
Nitrogen-containing compounds similar to ammonia (NH3) are also bases.
$$\ce {NH3 + H2O <=> NH4+ + OH-}$$


text:
Why are polysaccharides non-reducing sugars?
My book says that polysaccharides are non-reducing sugars, and they form of condensation of >6 molecules of monosaccharides. The condensation involves the carbonyl groups of the sugars, leaving only one free carbonyl group at the end of a big molecule.
My question is: since there is one free carbonyl group why aren't polysaccharides reducing sugars?!

This question is better suited over at chemistry.

https://chemistry.stackexchange.com/questions/63045/why-is-maltose-a-reducing-sugar-but-not-sucrose-even-though-theyre-both-disacc

For polysaccharides made with only glucose (starch, cellulose, glycogen, etc), only 1 unit can be reduced from hundreds, thousands or tens of thousands of units. With the same mass of dextrose and starch, the amount reduced for starch is 1/100, 1/1000 or 1/10000 the amount for dextrose. This is usually below the detection limit for most ways of detecting the reaction.


text:
Field Keyword and Molecular Orbitals--Gaussian 09
I am using Gaussian 09 to run calculations. I am applying a 0.1 eV field in the positive Z direction to a molecule. I am also using the %chk keyword so that I may analyze the Molecular Orbitals after I apply the field. Only the HOMO and LUMO show absolutely no change. Am I doing something wrong? Should I use a higher value for the field? I am using #Field=z+36 (did the conversion to hartrees)
Theory wise there should be a change in the orbitals, correct?
To add, I aligned the molecule along the Z axis using a script I wrote.

Thinking in a classical way, how strong is the force on one electron due the applied external field? And how strong are the coulomb forces between particles?

I guess I'm just confused because there is work being done on the particles, per the field. So why doesn't that reflect in the HOMO and LUMO?

You need a field strong enough

I do not know the field intensity needed in order to get an energy difference large enough, I would expect an energy level splitting

I'm such an airhead. I forgot to divide by the length of the molecule...

So did you get the answer? I am having the same doubt. Please explain.


text:
How to figure out electron wavelength?
This is a sample question in our study guide and it has me perplexed. Any advice on how to solve it?

In a transmission electron microscope, electrons are transmitted through a sample at around 70% the speed of light. What is the wavelength of these electrons? (This wavelength will determine the resolution of the microscope.)
If you wanted to increase the resolution of the microscope, would you want to transmit the electrons faster or slower? Why?

PS i dont want the answer just how to solve it. Thanks

To the best of my knowledge, $\lambda = \frac{v}{f}$, where $\lambda$ is wavelength, $v$ is the speed of the wave, and $f$ is the frequency of the wave. Do you know the frequency of the wave? If you don't I'm not sure how to solve the problem.

Use the de Broglie equation:
$$\lambda = \frac{h}{mv} $$
h is Planck's constant, m is the mass of the electron in kg, and v is the speed of the electron in m/s.


text:
Does possibility of Tunneling influence Ground State Energy of PIB with and without finite energy barriers?
Suppose there are two Particle in a Box models with the same initial energy and same size. Suppose model A has a finite wall or barrier in between the two infinite walls, and the system has sufficient energy for tunneling to occur.
Would the occurrence of Tunneling influence the ground state? If so, how? Which model would have a higher or lower state and why?

You are asking about a double-well potential where the two wells are separated by a finite square wall. This model is very well studied as it is the simplest model which involves tunneling through a barrier, so it provides a conceptual understanding to tunneling along a reaction coordinate.
The potential is visualized below along with the two lowest energy solutions:

Note, however, that in order to make a good comparison to the particle in a box, one must think about two times the particle in a box energy. This is because the double-well potential is really two infinite square-wells with a finite barrier between them. So, this tells you that in the limit that the barrier in the middle becomes infinitely large, there will be two ground state solutions with equal energy. Namely, the particle could be on the left or on the right.
So, as we lower the barrier and the particle is allowed to tunnel between the two wells, this degeneracy will be broken. These are the two wavefunctions which you see in the diagrams above. Namely, the solutions very closely resemble a symmetric and an antisymmetric linear combination of the ground state of a particle in an infinite box.
So, this has the effect of splitting the degeneracy into two states, one of which is higher in energy and one of which is lower in energy than particle in a box.
Why should the ground state of this double well be lower in energy than the ground state of a particle in a box where the particle stays on only one side of the barrier?
One way to think about this is that tunneling allows the particle to "leak" into the other box, which has the effect of making the effective box size of the particle larger.
This is all very qualitative, but the details of the math aren't really all that much more enlightening than simply looking at the qualitative features of the problem. The most important point I think is to recognize that tunneling through this barrier has the effect of splitting an otherwise degenerate state into a lower and higher energy state.

So a few questions first. Are we assuming that both particles have the same surface energy? If so, then I am almost positive that the one with the barrier in it is going to have a lower ground state after tunneling due to the energy required to penetrate the barrier. There is an exponential decay in the wave function when tunneling occurs. Hope this is of some help.


text:
Buffer solution
Prepare an aqueous buffer solution of $\mathrm{pH}=2.85$ that efficiently resists a change in pH yet contains only a small amount of buffering agent. Which one of the following weak acid together with its salt would be best to use?
A) m-chlorobenzoic acid ($\mathrm{p}K_a=3.98$)
B) p-chloro cinnamic acid ($\mathrm{p}K_a=4.41$)
C) 2,5-dihydroxy benzoic acid ($\mathrm{p}K_a=2.97$)
D) Acetoacetic acid ($\mathrm{p}K_a=3.58$)
Looking at the $\mathrm{p}K_a$ values C) is the strongest acid. However I cannot get if the strongest acid is best suited for buffer action which resists change in pH when small amounts of acid and base are added to it.

Possible duplicate of Why is a buffer solution best when pH = pKa i.e. when A-/HA=1

It is not the compound that has the lowest pKa value that is important, but which one has their pKa value closest to the desired buffer's pH. If I wanted a buffer at pH 9, I wouldn't use any of these.
What happens to the distribution of the acid and salt as the pH moves away from the pKa value?
What distribution of acid and salt makes the best buffer?

Answering with counter-questions is basically screaming for an NAA flag.

When $pH$ moves away from $pK_a$ value the concentration of acid and salt vary more.But I couldn't get how this distribution matters for making the best buffer?Thanks.

@Jan The questions guide him to the answer.


text:
Rate of nitration of C6H6, C6D6 and C6T6
Why the reaction rate (nitration) of $\ce{C6H6}$, $\ce{C6D6}$ and $\ce{C6T6}$ are equal? Can anyone explain this using simple words?

I suspect the rates seem to be the same within an error of measurement.

If you measured the rates carefully, you would probably see an inverse kinetic isotope effect (that is, the reaction will very slightly speed up as one goes from $\ce{C6H6}$ to $\ce{C6T6}$) because nitration involves an $sp^2$ to $sp^3$ change in the rate determining step.

In simple words, H, D, and T are isotopes*, and from chemical point of view they are just the same. Only if you measure them with really, really great precision, you may notice some differences.

The rate-determining step is the attack of the benzene aromatic ring onto the electrophile, and the loss of aromaticity is what causes the instability of the transition state, resulting in a higher activation energy and thus a slower rate of reaction. The subsequent deprotonation is after the rate-determining step and thus will not affect the rate of reaction, despite C-T > C-D > C-H bond strength.

In nitration of benzene or $\ce{C6D6}$, the rate determining or slowest step is the formation of the nitronium ion $\ce{NO2+}$ and not the formation of sigma complex as in the other electrophilic aromatic substitutions, therefore the rate is unaffected by the leaving of a hydrogen or deuterium atom.


text:
Where does the additional hydrogen in the Fischer depictions of glucopyranose come from?
My book shows this figure of Fischer formula of D-glucose:

I don't understand this figure, and I wonder why the upper carbon atoms in both the right and left formulas have 5 bonds.
And from where comes this hydrogen atom in the upper left side of the figure (attached to α-D-glucopyranose).
When turning D-glucose to Fischer formula, it seems there's an additional hydrogen atom, isn't it? Why and how does this happen?!!
Could someone explain how to get from the aldehyde form of glucose (the middle structure) to the β-D-Glucopyranose and the α-D-Glucopyranose?

Throw away that book, you have found a very terrible error. Okay the rest of the book may still be good, but at least black out that picture or stick a correct version of the picture over the top.
In the image below, you can see four depictions of α-D-glucose taken from Wikipedia (where a full list of authors is available). From left to right these are: Fischer projection, Haworth projection, chair depiction and ‘flat circle’ (a depiction which follows the guidelines for 2D representations of 3D structures).

As you can see, there is no additional hydrogen on the anomeric carbon and it is not pentavalent.
The mechanism of ring-closure is also mentioned in sufficient detail on the corresponding Wikipedia page to which I hereby point you. Basically, the 5-hydroxy group attacks the $\ce{C=O}$ π system as a nucleophile. Proton transfer will lead to the cyclic product. (In actual aquaeous solution, the carbonyl oxygen will be protonated immediately from a nearby water molecule and one will also take away the hydroxy group’s excess proton, so the two are not the same.)


text:
What do locants with prime symbols (e.g. 4’) mean?
I was reading an article about acetaminophen and it says its chemical name is 4'-hydroxyacetanilide.
What does the 4 signify and most importantly, what does the apostrophe mean? First semester chem student so if you can answer in layman terms that would be great.

4: fourth atom (usually but not necessarily carbon). Prime (not apostrophe): concerning the secondary structure not the primary one. Should be explained in every organic chemistry textbook in the nomenclature section.

@Jan What do you mean by secondary and primary structure? Thanks

It’s a bit hard to explain concisely. In your example, you have an amide (acetanilide). The carbonyl side will be referred to with only the numbers (e.g. 2-hydroxyacetanilide) while the amino side’s carbons are referenced with primed numbers (4'-hydroxy). Another example would be 4,4'-dimethylbisphenyl; here, an arbitrary phenyl group is the primary and the other arbitrary one is the secondary.

See also https://chemistry.stackexchange.com/questions/61155/what-does-the-prime-symbol-mean-in-organic-nomenclature-e-g-2

4 means the fourth atom and the prime symbol specifies the part of the molecule (one ' means the secondary structure), so 4'-hydroxyacetanilide means you have acetanilide with a hydroxide group on the fourth atom of the secondary structure.
Naming organic compounds works like this:

Find the longest continueous chain of carbon atoms. Called the parent chain this chain establishes the base alkane name for the compound.
Number the carbon atoms in the parent chain beginning with the end closest to any substituents.
Using the correct substituent name (methyl, ethyl, chloro,etc.), and the carbon atom number on the parent chain.
If a particular substituent occurs more than once, use a prefix (di-, tri-, tetra-) to indicate the number of those substituents.
List the alkyl substituents in alphabetical order.

Source: http://intro.chem.okstate.edu/1014/lecture/organic.html
Here are the IUPAC guidelines about the nomenclature of organic compounds:
http://old.iupac.org/reports/provisional/abstract04/BB-prs310305/Chapter1.pdf

thank you ! What exactly do you mean by secondary and primary structure? Also, when you say 4th atom, how do you tell what the 4th atom is?

@user54747 The quote should explain how to find the fourth atom. As for the secondary structure...To be honest I don't know it myself as I only added information given in a comment.

Assume you have a biphenyl (Google it) with a bromine in the 3-position (aka, meta) in one ring and a chlorine at the 4-position  (aka, para) in the other, its name is 3-bromo-4'-chloro-1,1'-biphenyl. The primary structure is the one without primes; the secondary with primes. The 1,1' nomenclature is the point of attachment of the two benzene rings.


text:
Is calcium carbide an ionic or covalent compound?
We know that the bond which is formed by exchanging electrons between a metal and non-metal atom is called ionic bond. Calcium is a alkali earth metal and carbon is a non-metal. So, calcium carbide ($\ce{CaC2}$) should be an ionic bond. But it is not an ionic bond, why?

strongly related: https://chemistry.stackexchange.com/questions/5227/is-there-a-word-for-a-compound-that-has-both-ionic-and-covalent-bonds

https://chemistry.stackexchange.com/questions/34099/why-do-different-elements-form-different-types-of-carbides

Because you said so. There is no other reason.

I don't understand.........

What's to understand? It is not ionic because you said so. Before that, it was ionic all right.

I asked what is the reason behind it?

How am I to know the reason behind your words? As to why is it ionic, you just explained it yourself in your question.

@IvanNeretin I don't think your answers are helping here at all. There is an issue with calcium carbide that confuses people and they deserve a proper answer (Nilay Ghosh gives a link to good one).

@matt_black There is an issue with every single established fact in physics and chemistry. Every rule has either exceptions, or some subtle effects that are only visible at 9-digit precision, or some non-obvious aspect that would make it look entirely different, or something like that. Really, open a schoolbook and read a few rules at random; I bet you'll easily point out some exceptions or corrections to each of them. Some time ago I decided against delving into the subtleties too early. Simple things first. It is ionic, period. If someone wants to know more, there's a link by Nilay.

The structure of calcium carbide is actually consistent with both ionic and covalent calcium-carbon bonding.  See https://chemistry.stackexchange.com/questions/55681/why-does-magnesium-form-covalent-bonds-with-carbon/55750#55750.

Myth debunking time!
First, there is no such thing as a purely ionic bond and no such thing as a purely covalent bond. Each bond (yes, even the bond in $\ce{H2}$) has a partially covalent and a partially ionic character. However, it is true that we can classify bonds and interactions as predominantly covalent or predominantly ionic in many cases especially those that occur at a beginner’s level.
While it is a nice explanation for the creation of e.g. sodium chloride to think of a full electron transfer — not an exchange because only one electron per ion pair is transferred and only in one direction — this cannot explain the structures of e.g. sodium sulphate well. Instead, you should think of ionic compounds as those composed of differently charged ions between which the interactions are predominantly electrostatic (non-directional). This is not true for all ionic compounds (ammonium fluoride being an exception) but gets you very far.
Now that I have established that (those that can be classified as predominantly) ionic compounds are composed of charged particles that interact electrostatically, you may be able to understand that not each ionic compound consists of exactly one atom type as the anion and exactly one atom type as the cation. In fact, I want to postulate that the vast majority of ionic compounds do not follow this rule. These compounds contain polyatomic ions meaning that many atoms connected together by covalent bonds form one charged molecule (or molecular ion). One typical example is the sulphate anion $\ce{SO4^2-}$ that I mentioned earlier but also the ammonium cation $\ce{NH4+}$. Thus, a compound can be held together by both covalent and ionic bonds, if for example it is composed of molecular ions.
This is the case in calcium carbide. We do indeed have a calcium cation $\ce{Ca^2+}$. The counterion is a diatomic anion, the ethynediide anion $\ce{C2^2-}$. It consists of two carbon atoms connected by a triple bond with each carbon atom carrying an additional lone pair and thus formal negative charge for an overall charge of $2-$. The bond between the two carbon atoms is covalent. However, the bond between calcium and the $\ce{C2^2-}$ fragment is ionic.
Therefore, calcium carbide can be classified as an ionic compound.

If you dig deeper, you may come across the crystal structure of calcium carbide; see for example the following taken from Wikipedia:

(This structure actually depicts barium peroxide but calcium carbide crystallises in the same structure.)
The interatomic distances given by Atoji and Medrud[1] are:
$$\begin{array}{lccc}\hline
\text{atoms} & \ce{C#C} & \ce{C-Ca}\text{ (lin)} & \ce{C-Ca}\text{ (orth)}\\ \hline
d/\mathrm{pm} & 120 & 259 & 282 \\ \hline\end{array}$$
The linear $\ce{(C#)C-Ca}$ distances are actually short enough to also consider a covalent interaction. Other acetylides can also be characterised with an intermediate or predominantly covalent bond. The picture becomes much more fuzzy once you take a closer look. The interaction of calcium with the $\ce{C2}$ fragment’s π system should also be considered as a potentially bonding interaction; think in the direction of transition metal complexes.
However, the similarity of both $\ce{C-Ca}$ distances (which becomes even more similar if you take the distance from calcium to the centre of the $\ce{C2}$ fragment) can also be seen as support for the argument of non-directional ionic bonding.
Considering all this experimental data, it is clear that the compound should have significant covalent bonding. Significant ionic bonding is not out of the question. The bonds are likely somewhere in the middle between covalent and ionic. Whatever the ‘final result’ e.g. of calculations is, it will not tip the scales strongly.
Reference:
[1]: M. Atoji, C. Medrud, J. Chem. Phys. 1959, 31, 332. DOI: 10.1063/1.1730352.

Well, this looks rather like repeating high-school illusions then "Myth debunking".

Starts like debunking, but "we can classify bonds and interactions as predominantly covalent or predominantly ionic in many cases — practically all cases that occur at a beginner’s level." is well... Most compounds of transition metals and even ones like very CaC2 aren't obvious at all, or are predominantly covalent.

@Mithoron Not sure if I classify coordination compounds as truly beginner’s level, though.

HgS, Au2S - very covalent and rather "beginner".


text:
"Hamiltonian operator has no effect on the spin function" what does it mean?
I have read the Levine Quantum Chemistry book and it says "Hamiltonian operator has no effect on the spin function" in chapter 10 (Electron Spin and the Spin-Statistics Theorem)  and does the Hamiltonian operation like the following. Why the g(ms) state is taken out of the Hamiltonian?? Why is it constant with respect to the Hamiltonian?
$$\hat H[\psi(x,y,z)g(m_s)] = g(m_s)\hat H\psi(x,y,z) = E[\psi(x,y,z)g(m_s)]$$

Because the operator $\mathrm d/\mathrm dx$ only acts on $x$ and not on $y$, we can write
$$\frac{\mathrm d}{\mathrm dx}[f(x)g(y)] = g(y) \left[\frac{\mathrm d}{\mathrm dx}f(x)\right]$$
Likewise, in this context, the Hamiltonian operator $\hat{H}$ only acts on $(x,y,z)$ and not $m_s$. This is explained by Levine1 in the paragraph immediately preceding this

To a very good approximation, the Hamiltonian operator for a system of electrons does not involve the spin variables but is a function only of spatial coordinates and derivatives with respect to spatial coordinates [...]

Is there something about this paragraph which you don't understand? This is pretty much all there is to be said on the topic.

Levine, I. N. Quantum Chemistry (7th ed.), p 268

the spin of an electron is due to the rotational motion about its own axis. Why it should not be there for the Hamiltonian operator?? I mean the electron should also have a rotational kinetic energy due to its motion about its own axis.

Spin doesn't correspond to a physical rotation of the electron! See, e.g. https://chemistry.stackexchange.com/questions/58020/do-electrons-really-spin If you consider both up spin and down spin to be degenerate, there will not be any term depending on $m_s$ that enters the Hamiltonian (depending on where you set your zero of energy, there might be a constant term, but this term doesn't depend on $m_s$).

To clarify, this is not the Hamiltonian, it is a Hamiltonian, specifically the one that acts on a non-relativistic stationary wavefunction (the time-independent Schrödinger equation, particle in a box/on a ring models, ...). Spin appears naturally in other Hamiltonians, such as Dirac's equation and its many reductions, and is "bolted on" for magnetic perturbations in the Schrödinger equation.


text:
Use of basis set in DFT (Density Functional Theory)
Basis sets are used to guess the electronic wave functions for Hartree Fock or similar methods, which are quite legitimate since these methods deal with the wave function of each and every electron.
Density functional theory, on other hand, uses the electron density at every point of space for optimization and the calculation of properties. This has led to two doubts which I want to clarify:

Is the basis set used to estimate the initial electron density of the system? If so, how the basis functions of an STO or GTF basis set change in order to exhibit the electron density? If not, what exactly is the application of basis set in DFT calculation?
In HF SCF, the coefficients of basis functions (for basis sets with GTF) along with geometry changes with change in nuclear geometry while optimising the structure. Does optimisation with DFT follow a similar procedure?

P.S.: Please do not suggest me this question, since this is not the answer I am searching for.

In a rush, but have a look at https://en.wikipedia.org/wiki/Orbital-free_density_functional_theory

You may want to have a look at the paper of Kohn and Sham from 1965. It states: We derive two alternative sets of equations [...] which are analogous, respectively, to the conventional Hartree and Hartree–Fock equations, and, although they also include correlation effects, they are no more difficult to solve. Both HF and KS-DFT are mean-field methods using quite similar equations.

The modern, mostly used DFT is Kohn-Sham DFT, which has more in common with HF than with DFT...

This answer only deals with the most common variety of Density Functional Theory, namely, Kohn-Sham DFT. This is what most people mean by "DFT", but, as noted in the comments, things such as orbital-free DFT exist.
Kohn-Sham DFT was created to solve a historical problem of DFT: The electronic kinetic energy term (as used in the Thomas-Fermi model) was not accurate enough. Kohn and Sham proposed to make use of what is typically called a "fictitious, non-interacting reference", that is, a wave-function whose main purpose would be to yield a density to be used in all other (nucleus-electron attraction, Coulomb, exchange, and correlation energy/potential) and whose secondary function is to provide the kinetic energy, which is accurate from wave-function theory. This wave-function works just like a Slater determinant and is called the Kohn-Sham determinant. See this question for pointers on where to read more: Equivalent of Szabo and Ostlund book for DFT.
As often in DFT, practicality showed a way that was justified later. When implementing this idea, one learns that one can basically use a HF code and replace the exchange term by the exchange-correlation (XC) potential (which is projected back on to the AO basis set). The potential needs to evaluated numerically on a grid, a procedure that yields the XC energy for almost all choices of density functionals. Thus one obtains the energy and the KS operator (the equivalent of the Fock operator) and can perform a SCF procedure in a given basis set. For the mathematical details, see e.g. JA Pople, PMW Gill, BG Johnson Chem Phys Lett 199, 557 (1992).
To answer your questions in this context:
1) The initial density depends on the guess, which is a whole different can of worms, but the same is true for HF calculations. (Procedures based on tabulated atomic densities exist, but the initial guess business is a bit of a dark art, and I don't think this is what you meant by your question.) The basis set and its MO-like coefficients (defining the KS determinant) are used on every iteration to yield the density.
2) Naturally, yes. The MO-coefficients of basis sets change on geometry change for a given system, just like in HF. The procedure and workable algorithms are very similar to HF (because it all is).


text:
Dissociation constant of solid ammonium chloride

When solid ammonium chloride dissociates at a certain temperature in a $\pu{0.500 dm3}$ container, ammonia and hydrogen chloride are formed.
$$\ce{NH4Cl <=> NH3 + HCl}$$
The initial amount of ammonium chloride was $\pu{1.00 mol}$, and when the system had reached equilibrium there was a $\pu{0.300 mol}$ of ammonium chloride.
What is the numerical value of $K_\mathrm{c}$ for this reaction under these conditions?
A 0.490
B 1.63
C 1.96
D 3.27

If my cat asked me that, I would say: "Meteor, why in the world would you want to know about equilibrium constants? You're a cat!"

@IvanNeretin So why in the world is your cat called Meteor?

@Jan Well, he has a shape resembling the letter M on his forehead, and he tends to run around wildly and wreak havoc. So the name kinda fits.

You must note that $[\ce{NH4Cl}]$ does not change as the density of solids almost remains constant throughout reaction.
At equilibrium, number of moles of $\ce{NH3} $ and $\ce{HCl}$ will be same and equal to $0.7$ from stoichiometry of reaction.
So the expression of $\ce{K_c}$ will be
$$\ce{K_c = [NH3][HCl] = \left(\frac{0.7 mol}{0.5 L}\right)^2 } $$
which is $1.96 \, \ce{\frac{mol^2}{L^2}}$


text:
What do you chemists need from a periodic table?
I am developing a website and later on an app with the periodic table. The reason I started was that I thought almost all the other periodic tables was awfully ugly.
I do not only want to make it a well-looking periodic table, but I want to make it easy and smart to use combined with the design.
My question to you:
Which features do you require and want from an online periodic table?
I am also asking for the smallest features you can come up with. The more the better!

Features I already have got:

The periodic table view with number, symbol, name, and type of element (like noble gas)
Detailed information on an element on a specific page for each element including:

Number, name, and symbol
Discovery, etymology, and meaning
Electron configuration
Electronegativity
Density
Melting and boiling point
Phase at room temperature

I hope you chemists can help me. By the way, the service will be free of charge to use when it is developed.
Right now it looks like this:

and when you click on one of the elements:

Comments are not for extended discussion; this conversation has been moved to chat.

Look on wikipedia. Every information their infobox on elements give is supposedly relevant.

@Karl Do you think all the information should be included?

Why make your app unuseful for half the people? If I can't find everything, I would'nt use it at all, but stick to wikipedia.

I’m going to vote to close as primarily opinion based. Reason: I don’t need the periodic table for anything; I know where each element is, I know the atomic masses of the most common ones and if I am missing a mass I look it up on Wikipedia. However, others will most certainly disagree; therefore every chemist will use the periodic table for something different.

Well. Before you do that. Could you do me the favor of telling me what people like you who for the most doesn't need a periodic table require when you are using one?

@Karl thank you. I will include all the information I can find!


text:
Comparision of the Ionisation Energy Change of the alkali metals and noble gases?
Currently I compare the ionisation energys of the alkali metals:
Li: $\pu{0.52 eV}$
Na: $\pu{0.50 eV}$
K: $\pu{0.42 eV}$
Where only a slight drop (appr. remain constant) occurs.
This ionisation energys occur in comparision at the noble gases:
Ne: $\pu{2.08 eV}$
Ar: $\pu{1.52 eV}$
Kr: $\pu{1.35 eV}$
Here we can observe a faster drop in the ionisation energy.
This diagram visualizes the numbers:

Source: https://www.tf.uni-kiel.de/matwis/amat/mw1_ge/kap_2/illustr/b2_1_1.html
How can we explain that? My first attempt was for the alkali metals: Due to the rising effective nuclear charge (values of Clementi: Li 1.2, Na 2.5, K: 3.5) the orbital energys of the "highest" s-Orbital raughly remain constant (despite of rising quantum number n) and so the ionisation energy does (Koopmanns-Rule).
But the same explanation can be taken for the noble gases ($Z_\mathrm{eff}$ for highest s orbital: Ne 5.7, Ar 7,7, Kr 11.3) Here we have even a faster rise of the effective nuclear charge … but an immense drop of the ionisation energy!
So something is missing in my explanation. Does the energy of the s orbitals influenced by another important factor? Does somebody have a better explaination?
Data from Huheey: "Inorganic chemistry"

https://chemistry.stackexchange.com/questions/66492/why-is-the-first-ionisation-energy-of-a-sodium-cation-bigger-than-that-of-a-neon

@Nilay Ghosh As far as I can see relates this linked question Not much with mine. My question deals with the enormous change of the IE of the noble gases (relative to the Change of the alkali metals).


text:
Benzilic rearangment
Why does benzilic rearrangment occurs? Is there any explanation?

That's the wrong question to ask. As with any reaction, it takes place because the kinetics and thermodynamics are favorable.

We can perform a very crude and very basic analysis of reactant and product to determine what is different and whether this corresponds to something more stable or not.
The difference is that one $\ce{C=O}$ double bond has been sacrificed to create two $\ce{C-O}$ single bonds. In terms of standardised bond enthalpies, two $\ce{C-O}$ single bonds are more stable than one $\ce{C=O}$ double bond.

Deprotonation of carboxylic acid should pull equilibrium to the product side too, imo.


text:
Number of intermediates and transition states in hydration of given alkene

What is the number of transition states and intermediates formed in the following reaction?

My thought:

First the alkene extracts $\ce {H^+}$ from the acid to form a carbocation. This step involves a transition state and the formation of an intermediate which is a secondary carbocation.
Next, we have an alkyl shift to this carbon, resulting in a ring expansion. The carbocation is now tertiary. Again, involving an intermediate and a transition state.
Finally $\ce{H2O}$ attacks on the tertiary carbocation. We again get a transition state and intermediate here.
A proton now leaves, which involves a transition state. The product now forms.

So I think there will be 4 transition states and 3 intermediates, but I am not sure.

I think that’s a good brain-and-paper theoretical analysis. A computational chemist would probably use this as a starting point.

Haha I had no idea that it was even slightly correct. I got it all by fluke, I think.


text:
Recharge battery using heat
A galvanic cell works by having two electrochemical equilibriums which have a different potential.
If you (theoretically) had a galvanic cell with two electrodes which have almost identical potentials and are sensitive to heat (at the anode the oxidation is exothermic and at the cathode the reduction is), connected the two half cells with a diode contrary to the normal current and heated the whole thing up, shouldn't it be possible to reverse the current and recharge the cell?
Please excuse language mistakes I might have made; I am not a native English speaker and therefore don't know all technical terms in English.

An electrochemical cell works on the principle that the Gibbs free energy gets converted into electrical energy by making the electron transfer indirect. This is done by itself to make the 'heat energy' that would normally be released in a redox reaction into 'electrical energy' so ideally heat shouldn't be lost.
Also you say you want to charge the cell by heating it so as to make the reaction go in the opposite direction. But for a spontaneous cell reaction, increasing the temperature makes the Gibbs free energy more negative and $\Delta G^\circ = -nFE^\circ$ ($E^\circ$ is standard cell potential).
So the $E^\circ$ value increases, and so does the concentration (because of backwards reaction), so the cell doesn't actually get charged up.
Ideally just charge it by passing current in the other direction.

Thanks for your answer. Of course I know how to recharge a battery but I was just interested if it would theoretically possible. Also could you please link some sources where I can look this up?

I don't know of any sources, although I wrote this answer with just high school knowledge so probably some good physical chemistry textbook will do.


text:
Should I provide a citation for atomic masses?
I am writing my first lab report ever, and the handbook I was given did not say anything in particular about this issue. I will use the atomic masses for my calculations and I am not sure if I should cite the source from which I obtained the data for those figures.
Should I cite periodic tables? If yes, how?

Usually citing molecular weights is not necessary, especially for a lab report (check the guidelines with your TA though). If you really want to cite anyways, I personally would use CRC handbook of chemistry and physics.  There is data for standard atomic weights with ESDs, also tons of other useful information on various constants and properties.

How would you put the citation markings? For example, right after every calculation using atomic masses?

I would put a single note in the "Methods and Defnitions" section. But what I'd do is irrelevant. I insist you check the guidelines first as your university might have completely different requirements.

You can’t cite ‘a’ or ‘the periodic table’ because neither are published books. What you can do is state that you took the atomic masses as given by <book> or <source>.
I personally would not do that and I have not yet seen a research article that includes citations for what is considered a well-determined atomic property such as atomic mass or even electronegativity — even though the values for the latter vary greatly depending on source and method used. (Papers that discuss the difference between different definitions being the obvious exception.) Whether you are required to do that depends on what your TA or professor thinks. So go and ask them.


text:
What do you call the number of atoms in a molecule?
I just bought a molecular modeling kit for my toddler ("Snatoms").
To use the kit to play with her, I want to start out by memorizing a list of all molecules (isomers) fitting these criteria:

only contains C, H, and/or O
contains a maximum of 6 atoms

I started out by noodling around with Lewis diagrams and wikipedia... but then I ran into the various isomers of C2H2O2:

I wanted to do this search on a chemical database (like chemspider.com) but I couldn't find a criterion that would filter the search by number of atoms in the molecule (isomer).
Is there a name for this "number"?

For molecules of an element, it's called the “atomicity,” and from a bit of web searching, it looks like that word (also terms like “hexatomic”) is sometimes applied to non-elemental compounds as well.

Are you only interested in structures that are real chemicals, or do you want to include all possible ways of connecting the atoms? That is, do you also want very chemically unlikely compounds that will not be in a database?

@SteveKass thanks much! Certainly that should be the answer. As helpful as SteffX's answer was, I would "accept" your "answer" if you added one for the sake of site conformity.

@gilleain thanks for your interest! For the didactic purpose at hand, I was thinking that I would want to list and memorize any chemical compound with enough relevance and existence to have a reasonably canonically accepted name in English.
I wanted to make the list as best I could myself, before asking the question you are describing on the stackexchange, as a separate question. But feel free to ask it yourself in the meantime if you're curious now!

I would like an answer to this question, because I'd like to make it easier for me to write the group theory paper on which I'm currently working. I don't want to keep saying things like "the number of atoms in the molecule". I'd like to be able to say something like "the cardinality" or the "atomic number" (although the latter clashes with an existing meaning for the proposed term). Instead of "For each axial point group, we determine the range of numbers of atoms allowed", I'd like to say something like "For each axial point group, we determine the allowed cardinalities".

While not exactly number of atoms; IHD (also known as 'Degree of Unsaturation'), is a concept which measures number of H2 pairs needed to saturate a compound. You can check it out.

The "number of atoms"'s name is... the number of atoms. I am not aware of any special word for that. Databases may not let you search/filter that kind of criterion, i.e. the total number of atoms, but usually let you search by "chemical formula" which is, in your case, C2H2O2.
I have just tried with Chemspider by typing "C2H2O2" in the search box and it disappointingly gave me just 3 results, i.e. glyoxal (OHC-CHO, not in your list), 42879-41-4 and 16005-17-7.
Did you know? Just to expand the scope, did you know that the last digit in CAS# is actually a check digit? It helps any software checking that you did not mistyped the number.

Interesting idea -- thanks! I guess I can try some different databases and exhaustively search all the different combinatorically possible molecular formulas.
I suppose that might work, but as you noted, with that method, ChemSpider doesn't even report all the isomers that wikipedia's image seems to identify.
I guess I'll let the question stand for a little while in case someone comes up with a jargon term before I accept the "negative result". :)
Three cheers for negative results in academic literature though!!!

Awesome!
So PubChem seems to do pretty well. I typed in C2H2O2 and it gave me 978 results... seems that asks for any molecule including at least 2 carbons, 2 hydrogens, 2 oxygens. But after filtering those by the molecular weight of C2H2O2, got down to 17 results, of which 13 aren't ions.
It's a little more effort than I imagined it'd be, but no pain, no gain I guess!

Glyoxal is in the asker's picture, see top middle, has number 107-22-2.

So this addresses the part about going from (number of atoms, element symbols) -> set of structure spaces. For this case, the input is (6, "CHO").
The steps I would go through are:

List all the (weak?) compositions of 6 - e.g. [3, 2, 1], [2, 3, 1], [2, 2, 2], ...
Turn each of these into a formula like C3H2O, C2H3O, C2H2O2, ...
List all the structures with that formula

Of course, step 3 might involve looking them up in a database to see if they are real structures or not.
For step 1, a 'composition' is a breakdown of a number into parts (like a partition) but where the order matters. So [3, 2, 1] is not the same as [2, 3, 1]. A weak composition allows zeros, so you can have [3, 3, 0] which means C3H3 - but that depends on whether you want to include all atoms or not.

all_licenses: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Why doesn't D block contraction affect the other P block groups 14 15 16 etc
Usually we come across anomalies in the trends of periodic properties for example the size [Atomic radius value from Raymond Chang Chemistry 9th Edition] of Gallium (135 pm) is smaller than Aluminium (143 pm) as opposed to the basic trend that atomic size increases down the group. This is usually attributed to the fact that the poor shielding of nuclear charge by intervening D orbitals leads to a greater $Z_\mathrm{eff}$ hence the radius is less as opposed to the trend.
If that's the reason why don't we observe similar effects for other groups like group 15 having Phosphorus (110 pm) and Arsenic (120 pm) [Here the size doesn't increase much but is still significant as compared to changes taking place in group 13] , where the size actually increases even when there are D orbitals to intervene. Or in short why are the effects so profound for group 13 and not the other members?
There's another example of In (166 pm) and Tl (171 pm). Both belong to group 13. As compared to other groups these two members have the lowest increase in atomic radius.

As a property showing d-block contraction, the question cited atomic radius. However, d-block contraction can manifest itself in other properties as well:

$\ce{GeO}$ is described as much more stable than $\ce{SiO}$. $\ce{Sn(II)}$ compounds are described as reducing. [Riedel]
The pentachlorides of $\ce{P, As, Sb}$ show a kink in their stability: $\ce{PCl5}$ melts at 167°C, $\ce{AsCl5}$ decomposes at $-50\mathrm{°C}$, $\ce{SbCl5}$ melts at 4°C (and decomposes at 140°C) [HoWi]
The redox potentials in acidic solution for the (0)/(IV) oxyanions and oxides of chalkogens show a peak at $\ce{Se}$ [HoWi]:

\begin{align}
\ce{SO2} + \ce{4e-} + \ce{4H+} &\leftrightharpoons \ce{S} + \ce{2H2O} \quad &E^0 = +0.50 \ \mathrm{V} \\
\ce{SeO_3^{2-}} + \ce{4e-} + \ce{6H+} &\leftrightharpoons \ce{Se} + \ce{3H2O} \quad &E^0 = +0.74 \ \mathrm{V}\\
\ce{TeO2} + \ce{4e-} + \ce{4H+} &\leftrightharpoons \ce{S} + \ce{2H2O} \quad &E^0 = +0.57 \ \mathrm{V}
\end{align}

the oxidative strength (at both low and high pH) of halogen oxyacids follows the pattern $\ce{BrO4-} > \ce{H_5IO6} > \ce{ClO4-}$. [HoWi]

I think that stability of oxidation states, stability of comparable compounds, and redox potentials of similar compounds are valid examples of d-block contraction. Of course, other chemists may arrive at different conclusions.

Literature:
[Riedel] E Riedel, Anorganische Chemie, Vol. 6, de Gruyter (2004).
[HoWi] AF Holleman, E Wiberg, Lehrbuch der Anorganischen Chemie, Vol. 101, de Gruyter (1995).


text:
How exactly do sulfites prevent oxidation of wine?
I know that sulfites are used in wines to prevent oxidation, but how exactly does this happen, which reactions take place and where do the products go?

Sulphites work as anti-oxidants in wine because they are very easy to oxidise to sulphates. Schematically:
$$\ce{SO_3^{2-} + \text{oxidant}\to SO_4^{2-}}$$
This ease with which the sulphite ion is oxidised to the sulphate ion means that sulphite added to wine tends to 'mop up' any oxidisers (like air oxygen) that might get into the wine, rather than the alcohol (and maybe other substances in the wine) becoming oxidised.
The sulphite method has the added advantage that it, as well as its sulphate reaction product, is quite harmless to humans.

https://www.ncbi.nlm.nih.gov/pmc/articles/PMC4017440/

In addition to @Gert 's answer, Sulphites also denature (damage so they are ineffective) enzymes (released when the fruit is crushed) which accelerate oxidation.

We spray metabisulfite on the equipment that is used to crush and press and transfer and ferment the grape juice. It is used to kill any naturally present yeast or enzymes that may make distasteful wine.


text:
why Nitrocellulose does not explode
my question is simple why Nitrocellulose does not explode since it contains Nitrogen and oxygen in the structure of the cotton why it does burn ?

Why would you expect a compound to be explosive just because it contains both nitrogen and oxygen?

But does explode... A particular form is known as guncotton. https://en.wikipedia.org/wiki/Nitrocellulose#Munitions

@MaxW Guncotton as well as all kinds of gunpowder are not meant to detonate. They just burn rapidly. Detonation of gunpowder will just destroy the gun.

@permeakra Would the word be "deflagrate" (= burn rapidly)? O:)

It does detonate in certain scenarios.
Detonation is a case of rapid decomposition where reaction front travels at supersonic speed. Obviously, for such decomposition to occur, the compounds needs a way to decompose with big enough release of energy. Also, in this case decomposition front travels not by means of heat transfer, but by means of pressure wave. This means, that the detonating solid must be, well, solid and not fluffy. Another requirements is generation of the initial pressure wave. Some compounds can create it on their own, some require a primer and some require a pretty damn big primer.

so the main reason is that the Detonation  wave not occurring instead it's a heat spreading

@iMyth Close enough. Transition from deflagration to detonation is a thing for nitrocellulose when it's used as a solid propellant, but commonly used nitrocellulose is in a form that does not help travel of detonation wave.


text:
What does "atoms make no compromises" mean?
I was watching The Ring of Truth - 5 Atoms and in the last part of the movie (the link starts at this part), Dr. Morrison said

"When it comes to energy, atoms make no compromises. It is all or nothing for them." - Dr. Philip Morrison (MIT)

What does this quote mean? Why can't atoms make compromises in Energy? When he says it is all or nothing for them, does that just mean that atoms can only absorb certain energy levels?
Please help me as I'm a very confused Chemistry student who is even more confused after watching a video my teacher told me would help me understand atoms better.

You have it right, atoms can only absorb certain energies and not others. That is the quantum in quantum mechanics.

@Tyberius Thanks, I understand now, after watching the whole movie. If you or someone else can add an answer stating how atoms can absorb only certain energies, I will accept it.

The subject we're looking at here is the quantization of energy absorbed and released by atoms, which is a fancy word describing the fact that atoms can only absorb or emit very specific amounts of energy (in the form of light) and nothing in between.  This curious fact has everything to do with how electrons orbit the nucleus.
The electrons whirling around the nucleus do so only at specific distances and in specific patterns.  Some orbit in spheres, others in dumbell-shapes, and still others in even odder patterns (a search for "orbital shapes" turns up a few relevant images).  On top of that, smaller spheres (and other shapes) are nested within larger ones, in a manner Niels Bohr compared to the orbits of planets in the solar system (https://en.wikipedia.org/wiki/Bohr_model).
Now here's where the energy comes in.  Electrons can "jump" between orbitals (unlike the planets, thank goodness!) when they absorb energy and move farther from the nucleus.  When electrons jump orbitals, they gain potential energy, just like when we move farther from the Earth, we gain potential energy.  The farther up we go, the more energy we have to expend on our way down: think jumping out of a 1st story window or a 2nd story window, or even a 3rd story window.  Note that the electrons can only jump between the first, second, third, etc. "floors" (orbitals) of an atom.  No mezzanines or mid-levels.
Obviously, the more electrons an atom has, the harder it is to map all of the specific energies it absorbs and emits.  Hydrogen atoms are easy, however, as this equation depicts.  Just plug in the starting shell #, the final shell #, and you've got your specific wavelength (energy) of light.
http://chemed.chem.purdue.edu/genchem/topicreview/bp/ch6/graphics/6_2.gif
The logical question to ask now is "why do electrons orbit only in those specific levels?"
And that's a tough one, involving a crazy (interesting) amount of mathematics.  Equations we've worked out from quantum mechanical theory describe "wavefunctions" for electrons that predict the probability of them being in a certain area (orbital) around the nucleus.  They're convoluted, but brilliant.  Take a gander here if you're curious: https://en.wikipedia.org/wiki/Wave_function.

Electrons don’t orbit.

@Jan I used the term "orbit" as a mere simplification.  I figured saying that they exist in a quantum mechanical probability field with indeterminate velocity and location might have been overkill.  For the purposes of this answer, an imagined planetary orbit suffices.