text:
Stereospecificity of the bimolecular electrophilic substitution reaction
We know that $\mathrm{S_N2}$ reactions always go with inversion at the saturated centre (they're stereospecific) and $\mathrm{E2}$ reactions are always anti (also stereospecific).
I recently observed in a paper an example of what was called an $\mathrm{S_E2}$ reaction, but can't understand how the stereochemistry could be predicted in this case? Is it also stereospecific, or is there some other factor involved?

Can you provide a link?

The question I ask was a general one. I can re find my example but it's irrelevant. I'm  asking about the mechanistic / orbitals interpretation of this type of reaction- my examples were just the first time I have seen this named as Se2

Except that SE2 is not a standard description for a mechanism...

@Zhe SE is a standard mechanistic description, theres brief mention on Wikipedia, but IIRC Ansyln goes into all of the various permutations quite nicely (if not in a tonne of detail). That said, I can't immediately think of an example with stereochemistry involved

I totally forgot about this one... This ACR paper is kind of old, but it suggests that there are several ways that the $S_{\mathrm{E}}2$ reaction can take place.

The $\mathrm{S_E2}$ mechanism is most commonly found using organometallic reagents ($\ce{R-Li, R-MgBr}$ etc.) in which the metal is the electropositive element react with electrophiles.
Unlike with $\mathrm{S_N2}$, there are multiple stereochemical outcomes, which can make it difficult to determine whether the reaction is indeed stereospecific. This is described in Modern Physical Organic Chemistry (emphasis mine):

One of the most interesting aspects of these reactions is stereochemistry. Varying results are obtained depending upon the metal, the solvent, and the R-group.
Source: Modern Physical Organic Chemistry, Ansyln and Dougherty [1]

Criterea for a stereospecific reaction
Before going any further, it may be instructive to define what we mean by stereospecific:

stereospecific is used for those reactions where the configuration of the starting material and the configuration of the product are related in a mechanistically constrained way
Source: Molecular Orbitals and Organic Chemical Reactions (Reference Edition), Fleming [2]

What this means in practice is that if one enantiomer of a starting material gives one enantiomer of product, then the opposite enantiomer of starting material must give the opposite enantiomer of product. What stereospecific does not mean is high selectivity, many reactions are highly stereoselective but in no way mechanistically stereospecific (and equally stereospecific reactions are not always $100~\%$ stereoselective).
The empirical evidence
Unlike the $\mathrm{S_E2}$ reaction, where many examples have been studied, it has been considerably more challenging to examine the stereochemical cause of the $\mathrm{S_E2}$ reaction. In general, chiral alkyl-metal species are prone to racemisation (chiral Grignards, as an example, are barely known), and those chiral species that are stable and usually such for steric reasons (such as a constrained ring system).
Many of the experiments that have been conducted are described in March[3], and the overwhelming majority of them go with retention (note that this doesn't necessarily imply a stereospecific process in itself and it has often been hard to study both enantiomers, for various reasons).
The orbital explanation
If we consider the orbitals involved during an $\mathrm{S_E2}$  reaction, it becomes clearer why there are multiple stereochemical outcomes, consider first $\mathrm{S_E2}$  at a saturated centre, such as $\ce{MeLi}$ attacking an electrophile :

Source: Molecular Orbitals and Organic Chemical Reactions (Reference Edition), Fleming [2]
In this reaction, the electrophile is quite able to interact in such a way as to give rise to overall inversion or overall retention, in order to distinguish, these two electrophilic substitutions are labelled $\mathrm{S_E2}_\text{back}$ and $\mathrm{S_E2}_\text{front}$ respectively, though both are formally designated DEAE if the IUPAC system is followed.[4]
This differs from the $\mathrm{S_N2}$ reaction in which inversion is almost always observed due to an anti-bonding interaction in the transition state if the nucleophile approaches in such a way as to give retention:

Source: Molecular Orbitals and Organic Chemical Reactions (Reference Edition), Fleming [2]
Conclusion
The short answer is that it doesn't appear that the reaction is stereospecific. Orbitally speaking, there are two pathways leading to two different products (front or back attack), which may be observed in poor selectivity.
It appears, with the reactions studied so far, that the outcome is far more likely to be down to sterics/electronics as to whether the electrophile approaches from one side or the other

References and Notes:
[1]: Ansyln, E.V.; Dougherty, D.A.  Modern Physical Organic Chemistry; University Science Books:California, 2004
[2]: I. Fleming  Molecular Orbitals and Organic Chemical Reactions (Reference Edition); Wiley:Chichester, 2012
[3]: March, J.; Smith, M.B.  Advanced Organic Chemistry; Wiley:New Jersey, 2013
[4]: IUPAC Recommendations for the Representation of Reaction Mechanisms: Acc. Chem. Res. 1989, 22, 343


text:
Uncertainty Theorem
What is the physical meaning of Heisenberg's Uncertainty Principle? We have derived it mathematically but what is the logical explanation for it?

So...you're saying there isn't any logic in the math? 8D

https://chemistry.stackexchange.com/questions/59822/derivation-of-the-heisenberg-uncertainty-principle

@paracetamol I meant to say is there any logical deduction of the principle(not formula) without involving complex mayhematics.

https://en.wikipedia.org/wiki/Wave%E2%80%93particle_duality#Brief_history_of_wave_and_particle_viewpoints

Possible duplicate - Heisenbergs uncertainty principle. (Digressing, but what a crappy title - bonus points to anybody who edits that question.)

need explanation of principle without mathematics

The physical meaning is that your cannot measure the exact position and exact momentum at the same  time. The principle gives a limit on both quantities if we try and measure both simultaneously. You can accurately measure position and momentum if done separately .

A wave packet is localized in space, but it is a superposition of many waves of different wavelengths.
A simple wave has a distinct wavelength, but it is spread out in space.
The wavelength corresponds to the momentum of a particle: $p = \frac{h}{\lambda}$.
These are examples where you can't know both the position and wavelength (i.e., momentum) of a wave form.

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text:
Group theoretical condition for an integral to be zero
In many aspects of chemistry it is of interest to find out whether a given integral $\int \psi \,\mathrm{d}\tau$ vanishes, i.e. becomes zero. Group theory provides us with a handy condition:

If $\Gamma_\psi$ does not include the totally symmetric irreducible representation (TSIR) of the point group, then $\int \psi \, \mathrm{d}\tau$ must be equal to zero by symmetry.

The utility of this is that it can be easily expanded to different types of integrals, such as matrix elements $\langle i | A | j \rangle$, where the integrand transforms as $\Gamma_i \otimes \Gamma_A \otimes \Gamma_j$. If the direct product does not contain the TSIR, then the matrix element must be zero.
What is the theoretical basis of this statement, i.e. is there a proof of it?
In particular I'd like to see a proof that:

doesn't make assumptions about which symmetry operator(s) the integrand is not symmetric under; and
is able to deal with irreps with dimensionality larger than one.

But if such a proof doesn't exist, I'd be happy to compromise on one or both of these criteria.

I checked FA Cotton's book, but he skirts the issue by saying that "it is a generalisation of the case where $\int y(x)\,\mathrm{d}x = 0$ if $y$ is an odd function of $x$" (p 107, 3rd ed.). That does makes sense, and I have an idea of how it could be mathematically formalised, but it doesn't satisfy either of the two criteria above.

By Maschke's Theorem, every direct product of representations is decomposable into a direct sum of representations, that is, the function you are integrating can be rewritten as a sum of functions with symmetry properties equal to specific irreducible representations.
As you note, integrating an odd function (which has a certain symmetry, namely antisymmetry with respect to $\sigma_y$, the mirror "plane" of the y-axis) over all space is zero, while integrating an even function (which is symmetric with respect to $\sigma_y$) is generally not. Since integration is linear (that is, the integral of a sum is the sum of an integral), we can consider the integral of each irrep separately.
If the function being integrated is antisymmetric with respect to some symmetry, it will be zero when integrated over all space. The only way for the integral not to be zero is if it is symmetric with respect to all possible symmetries, i.e. it is the totally symmetric irrep.
Expressed in equation form:
$$\int{\prod_{i}\Gamma_i\,\mathrm d\tau}=\int{\sum_j\Gamma_j\,\mathrm d\tau}=\sum_j\int{\Gamma_j\,\mathrm d\tau}$$
and
$$\int{\Gamma}\,\mathrm d\tau=\left\{
  \begin{array}{@{}ll@{}}
    0, & \text{if}\ \Gamma \text{ contains any odd (antisymmetry)} \\
    \text{not necessarily } 0, & \text{otherwise}
  \end{array}\right.$$
Then, it is clear that the only time the integral is nonzero is when, amongst the irreducible representations of the direct product (regardless of its dimensionality), the totally symmetric one is present, all other ones integrate out to zero.
I should clarify that when I write the integral of $\Gamma$, I really mean the integral of some awful, complicated looking function that has the symmetry properties of $\Gamma$, which is why I used $\prod$ and $\sum$ instead of $\otimes$ and $\oplus$.
If your question was instead how do you know that the functions with a given antisymmetry really do integrate to zero, let's consider what it means to have a certain symmetry. It means that there is a particular line or plane or point about which the value of the function is redundant. That is, if you specify it on one side, then the value is exactly determined on the other. We are integrating over all space, which means we can choose bounds of integration that utilize this symmetry. This then separates your integral into a non-symmetric piece and a piece using the symmetry. You first integrate over the non-symmetric piece and get something, and then you integrate that over the symmetry-including part. If that symmetry is antisymmetry, as will be the case with at least one symmetry of any irrep other than the totally symmetric one, this will cancel out the non-symmetric part with an equal and opposite non-symmetric part and the whole integral will be zero. Regardless of the dimensionality of the irrep being integrated it has at least one symmetry element over which it can be integrated that is antisymmetric. Hence, only integrals of functions that are totally symmetric are non-zero.
As a basic example, consider your $\int{x\,\mathrm dx}$ integral that you started with. Since this function has a symmetry (namely antisymmetry with respect to $\sigma_y$), I can make the following choice of bounds:
$$\int_{-\infty}^\infty{x \,\mathrm dx} = \int_{-\infty}^0{x \,\mathrm dx}+\int_{0}^\infty{x \,\mathrm dx}  =-\int_{0}^\infty{x \,\mathrm dx}+\int_{0}^\infty{x \,\mathrm dx}=0$$
I can choose to integrate only over the nonsymmetric part and then use the symmetry for the rest of space. The presence of symmetry guarantees that I can do that. Remember, the irrep label is just telling you the symmetry properties of some complicated function. If $x$ in the above equation were some function $f$ that I knew nothing about except that is has this symmetry, I could still make this choice of bounds and use this property to get a 0 integral.
EDIT: To address orthocresol's question below:
How does this work for the $E$ set in $C_{3v}$? Let us first think of a function with $E$ symmetry. The $p_x+p_y$ orbitals are such a candidate. Let's consider a top-down view of a $C_{3v}$ molecule. Now, I will consider one of the mirror planes to lie in the $xz$-plane (see Figure 1). I can do this because I can rotate the $p_x+p_y$ ensemble anywhere in the $xy$-plane. That is, there is some linear combination of $p_x+p_y$ such that there is a new set $p_{x'}+p_{y'}$ where the mirror plane lies on the $x'y'$-plane.

Now we set about integrating the upper half plane. Clearly, the half of the negative lobe of the $p_x$ orbital will cancel the half of the positive lobe, leaving only the integral of the positive lobe of $p_y$. Now we consider the lower half plane, the two halves of the $p_x$ lobes cancel again and we have only the integral of the negative lobe of $p_y$. Since the integral of the upper-half plane is the negative of the lower-half plane, the integral over all space is 0.
Now, why was it important to consider $p_x+p_y$ together? This is what allowed us to orient the orbitals with respect to the mirror plane. Consider a $p_x$ orbital alone rotated away from the axis given. It will have no defined symmetries with respect to the plane (reflecting across it will not map the orbital to itself or its negative) and there is no way to rotate it without the $p_y$. This is why they need to transform together in $C_{3v}$. Indeed, the reason why $E$ functions have 0 character under this symmetry is because the $p_x$ got sent to $p_x$ and the $p_y$ to $-p_y$ and so the whole ensemble was sent to $1-1=0$.
Now, hold on you say, can't I use similar arguments to imply that if I integrated a $p_z$ orbital (which is $A_1$ in $C_{3v}$) in all space, I will also get zero. The answer is yes. The $p_z$ orbital is not zero by symmetry in $C_{3v}$. I alluded to this above, the integral of a totally symmetric function in a given point group can still be zero. However, by describing the point group as $C_{3v}$, we are indicating that there is something special about the $z$-axis that breaks the symmetry and therefore precludes coming to this conclusion based on symmetry alone. Consider the deformation of planar $\ce{NH_3}$ to pyramidal $\ce{NH_3}$. In the former ($D_{3h}$), nothing distinguishes the positive $z$ from negative $z$ axis and a $p_z$ orbital is not totally symmetric ($A_2''$) and so we could conclude that it is zero by symmetry. As we pyramidalize to $C_{3v}$ though, the negative $z$ axis becomes different (it's got hydrogens there while the postive $z$ does not) and we can no longer conclude by symmetry that the integral of the $p_z$ orbital will be 0 (indeed, if it mixes even slightly with those hydrogens, it won't be). If it didn't mix though, it's integral would still be 0 (it's still just a $p_z$ orbital), we just couldn't tell that assuming $C_{3v}$ symmetry.

I like your answer. Referring to 2nd eqn, if $\Gamma$ is 'odd' then the integral is always zero, but if not then is it true that it is generally not zero but could be? For example, $\cos(x)$ is even about y axis but the integral is zero over $\pm 2\pi$ but $x^2$ is also even but integral not zero.

Yes, absolutely. An even function need not integrate to a non-zero number. I sort of wrote that in the text, but I will revise the equation.

Thanks! I understand what you're saying; the only part is your statement that "regardless of the dimensionality of the irrep being integrated it has at least one symmetry element over which it can be integrated that is antisymmetric." - how does that apply to, for example, the E irrep in $C_{3v}$?

@orthocresol As an example,  the triple product is needed when trying to determine selection rules, transition i to f e.g. if $\Gamma_i \times \Gamma_\mu \times\Gamma_f $ contains the totally symmetric representation the transition integral is not zero. So if the product is E only then transition not allowed, integral is zero.

@porphyrin yes, I understand that bit. Sorry, I guess should have been clearer - I meant, if the integrand transforms as E, then why would the integral evaluate to zero? It's not obvious (to me, at least) how the integral can be split up into components that are equal in magnitude but opposite in sign (which was the strategy adopted for $\int x ,\mathrm dx$ above).

@orthocresol I added an explanation for this too my answer. Hopefully, it is clear enough. Let me know if you need more elaboration.

I also like this answer, but what I feel like I'm missing is how you describe the oddness of a function directly from the irreducible representation. This is what I'm trying to do in my answer, though I think you are probably more equipped to address this issue than I am.

@Zhe I am calling a function odd if it has the property $f(\hat{O}x)=-f(x)$ for some operator $\hat{O}$. The irrep tells you (clearly for 1D, less so for 2D) whether $f(\hat{O}x)$ is $\pm f(x)$; it is the character under that symmetry. Because we integrate over all space, there is always a coordinate that we can use to take advantage of the antisymmetry (the negative sign). For reflections, the coordinate normal to the reflection, for rotations, it is $\theta$ in cylindrical coordinates, for inversion, it is $\rho$ in spherical coordinates.

@Zhe  In your $C_3$ example below, in cylindrical coords, you could integrate over $r$ and $z$, leaving $\int_0^{2\pi}\phi' d\theta=\int_0^{2\pi/3}\phi' d\theta+\int_{2\pi/3}^{4\pi/3}\phi' d\theta+\int_{4\pi/3}^{2\pi}\phi' d\theta=\int_0^{2\pi/3}\phi' d\theta-1/2\int_0^{2\pi/3}\phi' d\theta-1/2\int_0^{2\pi/3}\phi' d\theta=0$ where the 1/2 come from the relative character under identity and $C_3$ rotation.

@levineds But spanning A (the totally symmetric irrep) isn't a sufficient condition for an integral to be nonzero right?

@adosar Correct, it is necessary, but not sufficient, as described in more detail in the last paragraph.

EDIT: Made this more rigorous. (I think. Feel free to critique).
First, let's decompose $\psi$ into components that correspond to how its representation reduces. That is, if
$$\Gamma_{\psi} = \Gamma_{1}\oplus\Gamma_{2}+\ldots$$
we write
$$\psi = \phi_{1} + \phi_{2} + \ldots $$
where the symmetry of $\phi_{i}$ is captured by the $\Gamma_{i}$.
Suppose $\Gamma_{i}$ is not fully symmetric. There is some symmetry element $x$ whereby $x$ is antisymmetric in this representation. In other words, $$x\phi_{i} = -\phi_{i}$$
Now, consider this operation. You could perform the change of variable $x\tau \rightarrow \tau'$ for the integral.
$$\int \phi_{i}\,\mathrm d\tau = \int x \phi_{i} x^{-1}\,\mathrm d\tau'$$
This step is straightforward because we're basically arguing that if we convert our basis to something else (spcifically, transformed under $x$), the integral will be the same. This must be true since rotating the coordinate system, for example, does not change the value of the integral.
Now, we can substitute our definitions above:
$$\int \phi_{i}\,\mathrm  d\tau = \int x \phi_{i} x^{-1}\,\mathrm d\tau' = \int (-1)\phi_{i} \,\mathrm d\tau = -\int \phi_{i}\,\mathrm  d\tau$$
where I have used $x^{-1}\tau' = \tau$.
The equality means that the integral must be zero.
Now, if the irreducible representation is the totally symmetric representation, then antisymmetric element $x$ does not exist, which invalidates the rationale above, though the other components of $\psi$ will integrate to zero.
This argument works quite well when the irreducible representation $\Gamma_{i}$ is 1-dimensional. What happens when $\Gamma_{i}$ is higher dimensional?
I'm not sure I can rigorously prove this part, but I think I can sketch it out for a simple example.
Consider point group $C_{3}$.
\begin{array}{|c|c|c|}
\hline & E & 2C_{3} \\ \hline
A & 1 & 1 \\
E & 2 & -1 \\ \hline
\end{array}
However, you could also write this character table as:
\begin{array}{|c|c|c|}
\hline & E & C_{3} & C_{3}^{-1} \\ \hline
A & 1 & 1 & 1 \\
E_{1} & 1 & \varepsilon & \varepsilon^{*} \\
E_{2} & 1 & \varepsilon^{*} & \varepsilon \\ \hline
\end{array}
where $\varepsilon = -\frac{1}{2} + \frac{\sqrt{3}}{2}i$. Note that a linear combination of $C_{3}$ and $C_{3}^{-1}$, namely $\frac{\sqrt{2}}{2}C_{3} + \frac{\sqrt{2}}{2}C_{3}^{-1}$, provides a linear combination that is equal to operation $-E$. Wait, you ask, but that is not an element of the $C_{3}$ point group. That's right, but a $p_{x}$ orbital does not have 3-fold symmetry either. But if you rotate it both $120^{\circ}$ and $-120^{\circ}$, you get two orbitals that when mixed, gives the $p_{x}$ orbital rotated $180^{\circ}$.
I think this generalizes out to other 2-dimensional and higher dimensional irreducible representations in that you either have a matrix that inverts (for example character -2 in a 2-dimensional representation) or you are guaranteed that a linear combination of representations gives the inversion (like here). My background here is too weak at present to provide anything more concrete.

This isn't quite right. Let's be specific: say $\phi=x$ and the symmetry $x$ is negation (sending $x$ to $-x$, i.e $\sigma_y$). I agree that (-1)x = -x. The problem comes when you try to change variables. Make the substitution $u=-x$, then $du=-dx$ and $\int{xdx}=\int{-u(-du)}=\int{udu}$ and you've gotten nowhere. I think you conflated change of basis of matrices and change of variable. You applied $x$ and $x^{-1}$ to either side, which is conjugation by some operator and not variable change (but it does resemble basis change in vector spaces).

Consider $\phi_2=x^2$. Then $x\phi x^{-1}$ is the same for both $\phi$ and $\phi_2$ even though one is odd and one is even, so clearly this isn't doing what you think. The way to use symmetry for integrals, as I described in my answer, is to manipulate the bounds of integration and then use change of variables to show that the integral is sum of two integrals (with the same bounds) that are equal and opposite.

But $x^{2}$ is not antisymmetric. Doesn't it belong to the totally symmetric irreducible representation?

But, a change of basis in the vector space is exactly what I want, @levineds. This means that the integral should be the same regardless of the coordinate system... But I see that I didn't take the chain rule correctly to compute $\mathrm{d}\tau'$. There needs to be an invocation of the fact the the volume element is unchanged under the transformation because $x$ needs to be in $O(3)$.

I consulted my coworker who has a Ph.D. in algebraic topology. The guidance I received was pretty helpful, so here's the new way I understand the problem. There are a few steps that I might not be able to answer questions about because I'm not a mathematician, but hopefully, you can read up on these if my explanation is not clear enough.
First, let's decide to look at arbitrary point group $G\subseteq O(3)$.
For an arbitrary function $\psi:\mathbb{R}^{3}\rightarrow\mathbb{C}$, we decompose it as follows:
$$\psi = \sum_{i}\phi_{i}$$
where each of the $\phi_{i}$ has the same symmetry as irreducible representation $\Gamma_{i}$ of $G$ and the symmetry of $\psi$ corresponds to representation $\Gamma_{\psi}$ where
$$\Gamma_{\psi} = \bigoplus_{i}\Gamma_{i}$$
We need to show that $\int \phi_{i}$ vanishes if $\Gamma_{i}$ is not the fully symmetric irreducible representation.
So let's consider a function $f : \mathbb{R}^{3}\rightarrow\mathbb{C}$. My coworker told me that the representation of $f$ from a math perspective is $G$ acting on the $\mathbb{C}G$-module generated by $f$:
$$\mathrm{Rep}(f) = G\text{ acting on } (\mathbb{C}G)[f]$$
$\mathbb{C}G$ is the free vector space on $G$. It is composed of vectors $\sum\omega_{g}g$ for $\omega_{g}\in \mathbb{C}$ and $g\in G$.
If $|G|=n$, then $\mathbb{C}G\cong\mathbb{C}^{n}$.
We want to prove the following theorem: Suppose $\mathrm{Rep}(f)$ is irreducible and not trivial, then $\int f = 0$.
If the representation is irreducible, it means that $\mathrm{Rep}(f)\neq V\oplus W$, where $GV \subseteq V, GW \subseteq W$.
Now, we know because of the irreducibility condition that $\mathbb{C}G[f]$ has no fixed points other than 0, or else, we could decompose it
two subspaces that were disjoint under $G$. Therefore, if $v=\sum_{g}\omega_{g}(gf)$ and $gv=v$ for all $g\in G$, then $v = 0$.
Consider the vector $w = \sum_{g\in G} g f$. Note that
$$g'w = g'\sum_{g\in G} g f = w$$
This is because when you construct the multiplication table for the group, each row or column has distinct elements (otherwise, elements in
the group would turn out not to be distinct). There are two conditions under which this statement is true. One is the conclusion above that $w = 0$,
and the other is that $gf = f, \forall g$. The second condition is false by the statement of the problem: if $f$ is fixed $\forall g\in G$, it implies
that $G$ is the totally symmetric irreducible representation. Therefore, the only valid conclusion is that $$w = \sum_{g\in G}gf = 0$$.
Now consider the integral of $f$ under a transformation $g$. Because $g$ is a transformation is $O(3)$, this integral must be just the integral of $f$.
$$\int gf = \int f$$
We now sum the integrals over all elements of $G$:
$$\sum_{g\in G}\int g f = \int \sum_{g\in G}g f = 0 = \sum_{g\in G}\int f = |G|\int f$$
Since $|G|\neq 0$, we conclude that the integral of $f$ is indeed zero.
Circling back, $\int \psi$ reduces to $$\sum_{i\in S}\int \phi_{i}, i\in S \Rightarrow \Gamma_{i}\text{ totally symmetric}$$
This solution works for all point groups, doesn't require a definite basis, and works for arbitrary functions. Just the kind of answer you would expect from an abstract mathematician...

Does this work for groups of infinite order? Forming the G-invariant vector $w$ requires summing over all the elements in the group. Can I really do that for an infinite subgroup of $O_3$?

I love your answer! What does it mean of "G acting on the $\Bbb{C}$ G-module" generated by $f$?

@Zhe and what does it mean of representation of a function in this context?

@KangyuZhang As I hinted at, my colleague, who is an algebraic topologist and much better at mathematics than I am, presented this solution to me. Representations, as I understand it, are just ways to map objects to other kinds of objects with a similar property, the property here being symmetry. I can only provide very limited explanations about details here since I'm not a mathematician. I suggest consulting the internet for actual details.

@KangyuZhang As for your other comment, see here.

I was not completely happy with any answer so far, so here is my take on this:
We consider some integral
$$
\int_{\Omega} f d\Omega
$$
over some volume $\Omega$, ideally over $\mathbb{R}^d$, where $d$ is the dimension in which the function lives.
We next decompose the integrand into functions $f^{\alpha}_q$ spanning row $q$ of irrep $\alpha$ of the relevant (point) group, that is
$$
f = \sum_{\alpha}\sum_{q} f^{\alpha}_q c^{\alpha}_q,\quad(1)
$$
where $c^{\alpha}_q\in\mathbb{C}$ are expansion coefficients.
The behaviour of function $f^{\alpha}_q$ under symmetry operation $\hat{G}$ is given by
$$
\hat{G}f^{\alpha}_q = \sum_{p}f_p^{\alpha} D^{\alpha}_{pq}(\hat{G})
$$
The representation matrices $D^{\alpha}(\hat{G})$ satisfy the great orthogonality theorem (GOT)
$$
\frac{1}{h}\sum_G D^{\alpha}_{pq}(\hat{G}) D^{\beta\ast}_{rs}(\hat{G})=\frac{1}{d_{\beta}}\delta_{\alpha\beta}\delta_{pr}\delta_{qs}
$$
where $h$ is the order of the group and $d_{\beta}$ is the dimension of irrep $\beta$.
Consider now the special case that $\beta$ is the totally symmetric irrep; I will indicate this by $\beta=0$. This is a one-dimensional irrep with representation matrices $D^0_{11}(\hat{G})=1$. Inserted into the GOT we therefore get the relation
$$
\frac{1}{h}\sum_G D^{\alpha}_{pq}(\hat{G}) =\delta_{\alpha 0}\delta_{p1}\delta_{q1}\quad(2)
$$
which is central in the proof.
An integral is a scalar and is thereby invariant under any symmetry operation of the relevant group. Hence
$$
\int_{\Omega} f d\Omega = \frac{1}{h}\sum_G\int_{\Omega} \hat{G}f d\Omega
$$
Consider now a term in our expansion (1)
$$
\begin{array}{lcl}
\int_{\Omega} f^{\alpha}_q d\Omega &=& \frac{1}{h}\sum_G\int_{\Omega} \hat{G}f^{\alpha}_q d\Omega\\
&=&\frac{1}{h}\sum_G\sum_p\int_{\Omega} f_p^{\alpha} D^{\alpha}_{pq}(\hat{G}) d\Omega\\
&=&\sum_p\int_{\Omega} f_p^{\alpha}d\Omega\left[\frac{1}{h}\sum_G D^{\alpha}_{pq}(\hat{G})\right]
\end{array}
$$
Using relation (2) for the term in square brackets, we immediately see that the integral is zero unless $\alpha=0$, that is, the integrand has a totally symmetric component.

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text:
How to determine the dimensions of a bixin molecule?
I need the dimension of a bixin molecule.  I do not know where to look or how to calculate this information.
I understand that this molecule has several rotational axes and conformation, so there is no uniform size to be determined. But I am interested in the standard method for the determination of the molecule's dimensions.
I am doing a series of refractive and scattering experiments in a concentrated bixin medium. I need to determine the most probable cross section area normal to its principal axes.

There are crystal structures of cis- and trans-bixin published (D. R. Kelly, A. A. Edwards et al. J. Chem. Res., 446, 1996, p. 2640), you can roughly estimate the dimensions from CIF-files, I guess. But I have no clue which conformation dominates in solution. If it helps, I can provide the dimensions of both molecules packed in crystal in an answer.

If you need to know the dimensions of a conformationally flexible molecule in solution, the easiest way is to measure it (somehow). Predicting which conformation is likely is very hard to do. The best you could achieve would be to build a 3-D model (either on computer or using modelling kits) and measure the dimensions of the extreme conformations (stretched out or folded up). This would bound possible values but not tell you which is most likely in solution.

There is a paper on bolaamphiphiles (1) where it is stated that measured surface area corresponding to the area of a bixin molecule lying on the water surface with most methyl groups pointing outside is approx. $\pu{1.0 nm^2/molecule}$ at zero pressure. If the methyl groups were parallel to the water surface, a molecular area of approx. $\pu{1.7 nm^2/molecule}$ would be occupied.
Also, there is a linear size of a single molecule stated as the following:

Bixin therefore constitutes a valuable starting material for various bolaamphiphiles which can be tied within lipid membranes. Its molecular length is about $\pu{3.1 nm}$ in the all-trans configuration.

I think it might be worth for OP to have a look at the full-text (1) paper as it also discusses various methods of characterization and altering of bolaamphiphiles.
A few years later a crystal structure has been published (2). There is not much to add, as it seems like there is no contradiction with the data obtained with reflection spectroscopy (1). Here are the notable linear dimensions of cis-bixin and overall geometry in 2 projections:

(1) Fuhrhop, J. H.; Krull, M.; Schulz, A.; Moebius, D. Langmuir 1990, 6 (2), 497–505. DOI: 10.1021/la00092a034
(2) Kelly, D. R.; Edwards, A. A.; Parkinson, J. A.; Olovsson, G.; Trotter, J.; Jones, S.; Malik, K. M. A.; Hursthouse, M. B.; Hibbs D. E. 1996, J. Chem. Res., 446, 2640

What are the units of this dimensions?

@JoseECalderon Ångströms [Å].


text:
why don't carbons use sp3 instead of sp2 orbitals to make double bonds?
I don't have a strong chemistry background, but I was thinking why do carbons have to make different kinds of orbital hybridization for double and triple bonds?
Why doesn't a carbon make two sigma bonds into a double bond?

It costs energy to hybridize and if you can make a stable bond without "paying" the cost the molecule would do that. The p Orbitals can bond through a pi bond without need for hybridization so the atom's orbitals just hybridize to sp2. https://en.m.wikipedia.org/wiki/Orbital_hybridisation the wiki really gives a good idea of how hybridization works. I hope i clarified some of the question.

Should always keep mindful that hybridisation isn't a real thing, just simply a concept to help understand bonding

Only one electron can be in an orbital at a time. There are three p orbitals, one in each of the x, y, and z axes. A pi bond is the overlap of one p oribital from each atom. Like the p orbital, a pi bond also can be rotated from the x axis to the y or z to produce a different unique pi bond. Because a sigma bond has rotational symmetry, rotating it does not give a unique bond. Therefore, you cannot have two of them between the same pair of atoms.
Edit: It appears this question has been asked before and there are much better answers than mine (Why can there not be more than one sigma bond in a set of bonds?)


text:
Is Orca wrongly breaking bonds? Why can it only optimize the molecule with COPT activated?
I would like to ask for help in using Orca (v. 4.0.0) for conducting a simple optimization.
First I have this structure (primarily optimized using MMFF94s force field on Avogadro):

Thus I tried to conduct a simple optimization by using Orca and just as a test I used the following input:
! PAL8 PBE def2-SVP def2/JK RIJONX TightOpt TightSCF CPCM(DMSO) KeepDens NormalPrint Grid4 FinalGrid5
%method
    Z_MaxIter 150
end
%scf
    MaxIter 150
end

*xyz 0 1
C       4.026085     -1.212051      0.000031
C      -3.313454      0.000008     -0.000131
C       5.413748     -1.205853     -0.000365
C      -4.026091      1.212051     -0.000333
C       6.111884      0.000013     -0.000433
C      -5.413752      1.205847     -0.000315
C       5.413733      1.205871     -0.000018
C      -6.111885     -0.000021     -0.000102
C       4.026071      1.212051      0.000379
C      -5.413728     -1.205875      0.000032
C       3.313452     -0.000005      0.000350
C      -4.026066     -1.212050      0.000016
C       1.893686     -0.000008      0.000523
C       0.678508      0.000000      0.000397
C      -0.678520      0.000009      0.000187
C      -1.893672      0.000016     -0.000043
H       3.483003     -2.148319      0.000143
H       5.951276     -2.145874     -0.000768
H      -3.483012      2.148322     -0.000572
H       7.194627      0.000021     -0.001168
H      -5.951286      2.145866     -0.000334
H       5.951249      2.145899     -0.000149
H      -7.194627     -0.000032      0.000311
H       3.482975      2.148312      0.000762
H      -5.951240     -2.145905      0.000285
H      -3.482967     -2.148309      0.000048

Thus after running it, I saw that somehow Orca was "splitting" all bonds and getting this:

However, my job keeps going without detecting any problem and it just ends because hits the maximum allowed iterations.
I also tried to change the basis set, but without success. Also, I have played a lot with functionals, changed between LDA and GGA ones without success, semiempirical also have not shown any good result.
The only thing I could use to solve this problem was by setting COPT instead of OPT only, which changes Orca to cartesian coordinate (but was I already using cartesian coordinates?).
I just wonder why that is happening. Am I missing something? What is wrong with my input to cause Orca to "split" all atoms apart? And why Orca only can optimize this molecule with COPT activated?

Does this also happen if you remove CPCM(DMSO)?

About copt: Your input coordinates are transformed to some internal coordinates, which are then used for the optimization.  copt is about controlling these internal coordinates and thus independent of the input.

Hi Feodoran, yes. That also happens if I remove CPCM(DMSO). I also thought that could be the reason, but removing that did cause the same result.

If you inspect the output you can see a lot of "RMS(Cart) increases - taking linear step, building new B-matrix and trying again" messages that ends up with a cleat indication that something went wrong: "Not converged!" Since OP mentioned that switching to optimization in Cartesian coordinates solves the issue, it looks like a bug with optimization in internal coordinates.

@Feodoran, COPT in ORCA means that optimization is done directly in Cartesian coordinates without even constructing internal ones.

It seems to be a bug with optimization in internal coordinates. As I already mentioned in comments, ORCA couldn't converge at geometry relaxation step when optimization is done in internal coordinates. Switching to optimization in Cartesian coordinates (coordsys cartesian) or old internal ones (coordsys redundant_old) in the %geom section of the input, or using the ! COPT simple input as OP has done, solves the issue.
P.S. OP might report it as a bug at ORCA forum so that developers can take a closer look at this case.

In particular, the problem likely comes from the extended linear diyne linkage, possibly in combination with the planar capping benzenes. OP should definitely report this on the ORCA forum; the developers will want to know about the misbehavior.

I'm very pleased to know that. Thank you, Wildcat. I'll definitively report that on Orca forum. Thanks a lot for all of you guys who helped me with that. :D

https://orcaforum.cec.mpg.de/viewtopic.php?f=11&t=3105


text:
In a reaction between Calcium Ammonium Nitrate or Urea with water, does Nitrogen Content and Purity affect how cold the solution gets?
Recently, I bought an ice pack that contained Calcium Ammonium Nitrate, and realized that it really didn't get cold. I came to the conclusion that perhaps Urea is more effective, yet it is still puzzling to me that in a perfectly secure ice pack, it got to refrigerator water temperature.
I was thinking, perhaps purity of compound and nitrogen content (46% in Urea vs ~23% in CAN) could make a difference.

at equilibrium?

Welcome to Chemistry.se! It seems you have accidentally created multiple accounts, please visit this page to find out how to merge them.

Greetings Omar. It looks like you created a second account to ask a slightly different question. If you want to alter your question, you can simply [edit] the current one. Or, if my answer below didn't quite address what you were looking for, you could leave a comment below my answer and I will try to clarify. It looks like you might be particularly interested in whether the nitrogen concentration is responsible for the cooling? In this case, it's really coincidental that the two compounds both contain nitrogen. Again, feel free to leave a comment asking for any clarification you need. Cheers.

The endothermic process is just the dissolution of the calcium ammonium nitrate (CAN) or urea in water.
The most likely explanation for the lack of cooling when you used your pack is that over time, by getting partially squished at some point, or from defective manufacturing, some water leaked out so that the CAN or urea had already been partially/fully dissolved by the time you tried to use it.
These reagents are pretty stable and it is very unlikely that the nitrogen had changed form or that the system had been contaminated.


text:
What is metamerism
According to metamerism in polyvalent functional groups, if both alkyl groups' around functional group are different then they are considered as metamers but in few books it is given that if one of the alkyl group is different then it is a metamer.
In my opinion, for e.g. diethyl ether and methyl propyl ether are metamers.
But methyl propyl ether and methyl isopropyl ether should be position isomers and not metamers.
Which is true?

@Mesentery That link is great, but it has way to many low-voted answers and it's confusing for the OP to know exactly which one is correct. Voting to leave open.

I don't agree with that rationale, but it is consistent with what the SE team thinks, so...

@NilayGhosh I wish to not start another discussion (too tired), but if we can always check the quality of the answer, why is there a voting feature? Have you mouseovered the vote buttons? They say "This answer is/is not useful". Voting is how the community decides the quality of posts. Can you expect the asker, who is just learning isomerism, to decide the quality?

@NilayGhosh see this discussion

Reopened per a flag promising an answer.

There is something wrong in the first paragraph as the two alternatives you propose are in fact the same.

The terms metamerism, or metamer, in a sense of chemical isomerism (there are other meanings in zoology and colorimetry), are very old. Presently, they are not used nor mentioned in authoritative chemistry texts.
Even the old Watts’ Dictionary of Chemistry from 1892 states the related terms do not have conclusive definition:

ISOMERISM.  Even a superficial reader of chemical literature will soon become aware that the term isomeric and the kindred expressions allotropic, metameric, and polymeric are by no means always used in consistent senses, and he will have considerable difficulty in clearly
  realising their exact and relative import; (…)
(…)
METAMERISM. This term is generally applied to those cases if isomerism wherein compounds show identity of elementary composition, but belong to different types of classes (…).
(…)
(…) For example, it has long been a question whether ethylic acetoacetate is to be represented as $\ce{CH3-CO-CH2-CO2Et}$ or by the formula $\ce{CH3-C(OH)=CH-CO2Et}$. A discussion of such causes has been given by Laar (…), who proposes to term such isomerides tautomeric. But, as he practically himself admits, the use of such a term savours of tautology. V. Meyer subsequently brought forward a suggestion of Jacobson's to substitute desmotropic for tautomeric. Hantzsch and Herrmann (…), while using the term tautomerism when speaking of compounds capable of passing from the one type into the other, proposed to limit the term desmotropic to each of the states. As the phenomena in question are the outcome of mobility and not of fixity, the new term also appears to be particularly ill chosen, and bearing in mind the intention of Berzelius in introducing the term metameric (see p. 81), it would appear that this old term
  is a peculiarly appropriate one to use in such cases. (…)

Or S. A. Norton, The elements of chemistry: inorganic and organic (1884)

Isomerism.
Two or more compounds which contain the same elements, and have the same percentage composition, but differ in properties, are said to be isomeric. There are several varieties if Isomerism:

Bodies are physically isomeric when they differ only in certain physical properties, as their odors or their relation to polarized light. Over twenty volatile oils have the composition, $\ce{C10H16}$ (lemons, orange, bergamotte).
Bodies are isomeric in the strict sense of the word when they have the same vapor density, the same percentage composition, and exhibit similar chemical changes under similar circumstances. Thus there are two primary butyl alcohols which are strictly isomeric: $\ce{CH3CH2CH2CH2OH}$, and $\ce{(CH3)2CHCH2OH}$, and which give rise to other compounds, acids, ethers, etc., which are also isomeric.
In general use of the word, metameric bodies are also called isomers. Bodies are metameric when they have the same percentage composition and the same molecular weight, but exhibit dissimilar chemical properties under similar circumstances.
Propionic acid, methyl acetate, and ethyl formate have the same molecular formula, $\ce{C3H6O2}$, but when acted upon by caustic potash yield very dissimar products:
  (…)

Some metamers are so totally different that no resemblances of structure have been imagined to exist. Such are starch and gum arabic, which are pseudo-isomers.
Metamers are found in all terms above $\ce{C3}$, and increase in number very rapidly with each addition of $\ce{CH2}$. There are four butyl alcohols known, two primary, which are strictly isomeric with each other, and two others metameric with these, which do not form corresponding acids and ethers.
Polymeric bodies agree in percentage composition, but do not have the same molecular weight. Their formulæ are multiples of some empirical formulæ common to all, as the $\ce{CH2}$ in the olefines, $\ce{C_nH_{2n}}$.

So, the historical meaning of mesomers is that they are isomers with very diferent properties, different functional groups.
Contrarily, there are several organic chemistry textbooks (mostly Indian provenience), with somewhat contrary definition, e.g. Tewari & Vishnoi: A Textbook of Organic Chemistry, 4th Edition:

Metamerism
Compounds of the same homologous series show this type of isomerism because of unequal distribution of carbon atoms (…)
(…)
(ii) Metamerism. The carbonyl group may occupy different position in the carbon chain to give metamers.
$\ce{CH3CH2COCH2CH3}$      $\ce{CH3CH2CH2COCH3}$

Therefore I think that these terms should not be used.

The question is ill posed. Both alkyl groups are different.... (from each other?) if one alkyl group is different (from what?)... Metamers are isomers which differs by the structural carbon atoms distribution at the sides of a polivalent functional group. As such given a linear chain they are positional isomers too, with respect to the position of the polivalent functional group. All four ethers you posed as example are metamers. Personally I never heard of metamers except in an introductory course, for that matter.

This makes me to think about two ethers differing only by the R or S configuration of one substituent. Are they metamers? LOL! I would say yes as we have (stereo) isomers due to a whatever difference at the sides of O

Your answer may be correct, but please provide a reliable source.

Which source I shall give if I say that isotopes are  baryons with the same atomic number but different mass number? I have no idea, I heard of metamers during my "introduction to organic chemistry course" and then never more but here. If you refer to my doubt in the conment than I just try to draw a conclusion from the sort of definition of metamers I gave

And I honestly don't see the reason to download the answer. Additional info :I remember that a chain of given length has a fixed number of metamers, so obviously metamers are isomers at first. The question is really ill posed even in the statement "if one alkyl group is different". Different from what and in  what? Else every ether would be metamer of another one except itself.


text:
How does partially thawing a frozen solution affect concentration?
I have a mixture of antibiotics at a concentration of 100X and it is in the frozen state. Now some volume of the frozen mixture has been liquefied. Will the concentration of that liquefied part be 100X, or will it be less than that?

If this were some simple salt solution you're dealing with, I'd say the liquid portion of the thing also has a concentration of 100x. But since this is some unspecified antibiotic you're dealing with...can't be too sure...

@paracetamol It isn't even the case for simple salts like sodium chloride - see my answer for more details.

In general, no, a frozen solution that partially thaws is not the same concentration as the original solution.
The reason for this is the same reason chemists can purify a compound by crystallizing it, and one of the reasons why north polar ice cap is freshwater, despite floating in salt water.
When something (e.g. water) crystallizes (e.g. forms ice) the crystal lattice tends to reject impurities (e.g. things dissolved in the water). This means that as the solution crystallizes/freezes, the solid portion becomes depleted in the solute, and the remaining liquid portion becomes enriched in the solute. Eventually, though, if things are cold enough, even the concentrated residual liquid will freeze. But this mean that you now no longer have a uniform concentration of solvent in your frozen sample.
The reverse happens when you thaw it. Since you don't have a uniform concentration of solute, not all portions of your solid sample have the same freezing point depression. So what tends to happen is that portions of the solid which are more concentrated in solute will thaw first (as they have a lower melting point), whereas portions which are depleted in solute will thaw later. -- This is all generalization, though. What actually thaws first depends a lot on the details of where these regions are in relation to the heat source, and how thermal energy is flowing through the sample.
The up shot of this is, though, that if you freeze a stock solution, you can't partially thaw it and expect to get the appropriate concentration. You have to completely unthaw the stock solution and mix it thoroughly to make sure you have a homogenous mixture of the correct concentration.
This is why it is typically recommended to make single-use aliquots of stock solutions like antibiotics. Not only do the smaller aliquots thaw faster, but you don't put your entire stock solution into multiple freeze/thaw cycles, which can degrade some of the more delicate compounds.

While the outcome of directed crystallisation you refer to depends on the Roozeboom / phase diagram, if there the remnant liquid will be richer (or poorer) in impurities compared to the solid, it may be worth to pint to the zone melting (which is different from the Czochralski process) as one practical application, too.  Here, like the example of polar ice, the solid is the more pure than the liquid.  The contrary, zone leveling, is known too, which aims to mix e.g. alloys.  Both methods are applicable on organic matter, too; but chromatography obviously is quicker than zone melting.


text:
SN1 reaction in a polar aprotic solvent
I'm seeking a $\ce{S_N1}$ reaction that happens in a polar aprotic solvent.
Do you know any good simple $\ce{S_N1}$ reactions which satisfy the condition?

Try something that forms a tertiary benzyl carbocation in the reaction.

I think something like potassium 2-methylpropan-2-olate and 3-bromopentane in $\ce{SO(CH3)2}$ should do the trick.


text:
Can amines in acidic medium last for long?
I have been thinking of a situation where amines are kept in moderately-to-highly acidic media. In such a case, the amines would assume the protonated forms:
$$\ce{RNH2 + H3O+ <=>> RNH3+ + H2O}$$
However, I don't think that the protonated amine would remain as it is. I feel that thermal agitations may cause the protonated amine to remove an ammonia molecule as a leaving group:
$$\ce{RNH3+ <=>[K] R+ + NH3}$$
The fact that a positively charged nitrogen is highly electron withdrawing, and might tug at the $\ce{C-N}$ $\sigma$-pair, seems to aid in the removal of an ammonia molecule, along with the fact that gaseous ammonia can easily remove itself from the medium, enhancing the reaction's equilibrium towards the forward direction. However the rather unstable carbocation that forms as a result of the deammoniation may prove to hinder this reaction, although the forming carbocation may be immediatly suppressed by the conjugate base of the acid used (like $\ce{Cl-}$, etc.) The overall reaction for an anime placed in conc.$\ce{HCl}$ could be:
$$\ce{RNH2 + HCl <=> RCl + NH3}$$
I have no lab equipments, nor the ability to access sources that confirm or reject my thoughts, so I decided to ask for help here. So, Can amines in an acidic medium remain as such for long periods, or release ammonia? If this doesn't happen at room temperatures, may this happen at a higher temperature?

I suspect it would depend on what the nature of R was (i.e. whether it was able to stabilise a carbocation). In general, quaternary ammonium salts of amines are stable, and indeed isolable (often as nice crystalline solids).

@NotEvans. what about Aminium salts? Are they isolatable as well?

I don't know of that word, aminium, but I'm assuming you mean salts of $\ce{NR3}$, where R could be H or some other group, in that case, often, yes, but it would, as stated above, depend on the exact nature of the groups

Alkyl ammonium ions are generally stable and more stable than the ammonium one due to inductive effects. You get a salt, which is often the way amines are commercially distributed. K is very small in the second reaction you draw. I want be surprised to get some ammonia smell opening whatever contains amines, tough.

Note that according to Wikipedia, "many hydrochlorides of amines have a longer shelf-life than their respective free bases", which, along with making amines more bioavailable, is why so many amine-based medications are delivered as the hydrochloride (or other acid) salt.

@NotEvans. I was told that Ammonium salts are $\ce{R4N+}$ where $\ce{R}$ is alkyl group, and Aminium salts are $\ce{RNH3+}$.

@PrittBalagopal Then you were told wrong.  R4N+ is tetraalkylammonium, RNH3+ is alkylammonium.

Obligatory IUPAC Gold Book reference. The $\ce{NH4+}$ cation is called ammonium and so are the various substitutions of $\ce{R}$ for $\ce{H}$.


text:
How can I calculate the concentration of nanoparticles in water solution?
I have synthesized gold-silver bimetallic nanoparticles in water media by following this article.
How can I calculate the concentration of my nanoparticles solution?

Do you know how much gold and silver you added? And how much water you added?

Obviously you cannot, at least not in terms of number of entities per unit V. This would require to know the size of them, or the "nanoparticular" weight. It is a kind of surprise when sample are real and not always NaCl sugar and so on. It is similar polyeners, with all their mass distribution and aggregations. If you mean mass per volume see comment above.

In the current form, the original question provides too little information to answer the question.

@PrittBalagopal Yes!It is possible to calculate the amount of gold, silver and water from the article.

It's pretty rare to synthesize a nanoparticle solution like that for the purpose of calculating its concentration. What do you want to do with the information about the concentration?

@chipbuster I have synthesized this kind of nanoparticles and also have functionalized them through thiol group. I want to know how many moles of thiol could be attach on each of the nanoparticles.


text:
How to optimize graphene geometry with Gaussian?
I am new to Gaussian and I want to optimize graphene using DFT. After the the geometry optimization it has been severely changed, and some bonds were set as double and triple. Why am I unable to get the structural values similar to the input structure?

Please post your input file.

Possible reasons: wrong input files / unreasonable edge effects

Thank you for your interest.
I didn't know how to post a input file so that I have sent a google drive link.
https://drive.google.com/file/d/0B3OwbLbmgat8YXRRX1huNHhNV2M/view?usp=drive_web

The decision for a bond to be visually represented as a single, double, or triple bond in a molecular visualization program (e.g. GaussView) is based on rather arbitrary distance rules that often do not represent the actual bond order. Unless you are investigating the bond order in other ways (e.g. Wiberg bond index), what you're seeing for the bond order is nothing more than a guess. If you switch visualization programs, you'll likely find the bond orders to change.
If, indeed, the actual geometry of the graphene is completely unphysical after the geometry optimization, here are some questions to ask yourself:

How good is the initial structure you gave the optimization algorithm? Are you using a template graphene structure that should be close to the final geometry? If you made it "from scratch", it would be worthwhile to apply symmetry to your structure prior to running the calculation in Gaussian.
How have you terminated your graphene? You say you're using Gaussian, so I doubt you are using periodic methods. Have you terminated all dangling bonds with hydrogens, for instance?
In line with Point #2, what is the size of your graphene "cluster" model? Perhaps the cluster you chose is too small.
Ensure you don't have various input file errors (e.g charge/multiplicity).

Dear Argon,
Thank you for your comments.
Yes, It may be the reason of input files. I gave it by drawn using chemdraw then save it as gaussian input file. Isn't right? I have deleted all the hydrogen atoms manually.
I really didn't know how to give optimized algorithm without get successful optimization.
There are 54 atoms in my geometry of graphene structure. Isn't enough size?
Thanks in advance.

Nobody is going to be able to help you directly unless you post your input file or at least the atomic coordinates. It sounds like the initial geometry could have been very poor if you made it in ChemDraw. Did you at least visualize it beforehand in a program like Avogadro or GaussView to confirm it looks correct? Also, as I mentioned, you likely need to artificially terminate the undercoordinated C atoms with H atoms. If you just consider a small graphene cluster of 54 atoms, you need to terminate the cluster where it would normally continue periodically, otherwise the coordination is off.

Thank you for your reply. Yes, I have visualized and verified using GAISSVIEW prior to run calculation. It was pretty good to view, all the bonds are single.
Sorry, I am new to this forum also. I didn't know how to upload a input file. The characters also very limited here so i could not copy and paste the input coordinates also. I am unable to know how to overcome this issue.

sgmariappan@gmail.com this is my mail id. will you please mail me? I will send files.


text:
What pulls the electrons from iron in rust formation?
In the first step of formation of rust in iron, there is something that pulls electrons from the iron lattice, resulting in the formation of $\ce{Fe^{2+}}$. The reaction is:
$\ce{Fe  -> Fe^{2+} + 2e-}$
After that positive iron is repelled from the lattice, it is dissolved into water and a lot of reactions take place.
What is the thing that pulls the electrons from iron? (I guess it must be something positive, right?)

This is only "half-reaction" by no means description what really happens.

The iron is being oxidized in this process. You just need something to act as an oxidizing agent. Dioxygen will do quite nicely:
$$\ce{O2 + 4e- -> 2O^{2-}}$$

how $\ce{O2}$ does that? if $\ce{O2}$ is not negative, how can it attract electrons?

Iron isn't negative but it can give up electrons. Shouldn't the opposite also be true? It's not so much that the oxygen attracts electrons. It's just that electrons are more stabilized when they're with oxygen than with iron.


text:
Acidity/basicity of oxides across the periodic table
Recently while self-studying my chemistry book, it dawned on me that metals form basic oxides and nonmetals forma acidic oxides. Why is this?

That is not true.

Manganese heptoxide is an acidic oxide.

Your thinking sounds in line with an old theory abandoned centuries ago.  This theory is still evident in outdated conventions. https://chemistry.stackexchange.com/questions/16267/why-do-we-write-nh3/43412#43412

It is one of the characteristics of the metals as for electrical conductivity, heat conductivity, mechanical properties. All these basically originate from the outer electrons to be "loosely" bound to the nucleus, when compared to non metals. Bonds in metal oxides are ionic, with positive charge on the metal. The same is for their hydroxides, i.e. their compounds containing an OH group. This leads to their basic dissociation in water  giving the stable hydroxide ion.
Contrary non metals have tightly bound electrons and are prone to fill their outer shell (Octet rule). On reaction with a non metal as oxygen the situation is therefore a bit more complicated depending on what specific element the non metal is. Non metals oxides are essentially molecular, although the covalent bond is polar to various extents. However, their compounds containing an OH group will retain electrons on the fragment consisting of the two non metals (of which, one is oxygen of course) leading to acid reaction, i.e. H+
(Note that HClO would be better wrote as ClOH).
It should be clear that things are going as ionisation potential - electron affinity - atom size (distance of the outer electrons from the nucleus). For a better explanation I would need more time, it is basically a big chapter of introductory general chemistry, which goes under the name of periodic law and properties of the elements, or something like that.

Why HCIO would be better wrote as CIOH?

It does not mean you have to abandon writing HClO. The same is true for H2SO4, etc. They are perfect as minimal formula, indicating the exact number  and nature of the atoms forming the molecule. I wanted to point out that in the frame of my answer would better to think of the structural formula or even the Lewis structure of the non metal acid. H+ in the acid reaction is coming from the dissociation of a H-O bond rather than a not existing (in water) H-Cl one. Look for HClO, or H2SO4 etc Lewis formulae. Hope this help.

Moreover what you concluded is right. Just you will find out that acid base reactions do not always directly involve protons exchange or water. There are reaction based on electron pair sharing between Lewis acid and bases. If a metal is involved in that, it is in an oxidation states that actually it is not a metal anymore. You will soon notice (or be teached) amphoterism and finer details.

@SaraAsfar see this: https://chemistry.stackexchange.com/questions/53568/what-is-the-structural-formula-of-alkali-hypohalite-mox-or-mxo

Well nice. As you see one should not confuse notation with "reality". Just I made clear I wasn't speaking about notation. At the and everything is Ok saying that benzene has (minimal) formula C6H6 but no one will conclude that it consists of 6 carbon followed by 6 hydrogens in lane :)

Neglect down voting. It comes from those convinced that what we are talking about was dismissed centurieS ago. Let us assume they are right in the given contest (which is not) it would be exactly 101 years, as we are in 2017.

Yup.I was also thinking same why  they downvote my questions.But anyhow let them be.

@Sara Asfar. As I assume that you are a motivated student, do not hurry up and just go through your book with calm. If  comments as "it is not true" generated doubts , look at the Wikipedia page Acid Base reactions. Is compact and it will provide you a quick fix.


text:
What is the mechanism that creates a potential between Zn and Cu when they are connected with a wire?
When people explain a galvanic reaction, they usually stop the explanation in the dogma that one is more reactive/electronegative/nobler then the other. But that doesn't really explain the mechanism. I mean, why there would be a electrostatic force between elements with different reactivity?  What mechanism creates a potential there? Why electrons get repelled from $\ce{Zn}$ and get attracted to $\ce{Cu}$?

You have a net transfer of electrons from zinc to copper(II) ions. That's not the same thing as what you've written.

There is a layering of ions and oriented water dipoles at the interfaces between water and metal. That creates a voltage difference between the electrodes, depending on the ion concentrations in the electrolyte and the polarisability of the metal atoms at the interface. Wherever the local electric field is high enough,  metal ions can be formed out of the metal or coalece into it, creating a current. (My personal understanding of the process.)

There's also https://chemistry.stackexchange.com/questions/20084/why-do-electrons-leave-the-zinc-in-a-galvanic-cell


text:
Should I add more of a trihydrate chemical to a solution if recipe calls for hemihydrate?
I am running phosphate by the ascorbic acid method (SM 4500-PE, PDF), and making the antimony potassium tartrate reagent. Standard methods uses $\pu{1.3715 g}$ of $\ce{K(SbO)C4H4O6 * 1/2 H2O}$ per $\pu{500 mL}$, while the chemical I have is $\ce{C8H4K2O12Sb2 * 3 H2O}$.
Should I add proportionately more of the trihydrate chemical to make up for the weight of water? I'm not sure what the exact active ingredient here is, and I realize that there are different chemical formulas besides the water content. But, I am hoping that if I account for the difference in weight of water, then I will get at least closer to the appropriate amount of necessary active reagent.

I changed the volume from $\pu{500 mL}$ to $\pu{400 mL}$ in your question according to the datasheet info I also provided a link for (I guess you used this one). $\pu{500 mL}$ is a suggested flask volume, not the volume of antimony potassium tartrate solution.

The formatting was a nice change, but the 500mL final volume is correct (they're just suggesting using 400mL as a starting point before diluting to volume). Regardless, still unsure about what proper weight to use. Seems to be working well so far by just using the same 1.3715g as listed in SM, but using the chemical that we bought (trihydrate). I'm not entirely sure what this chemicals purpose is in the color-forming step, but it's not causing any major problems. I'm probably thinking too far into this.

Your two products are very different, à part from the crystallization water (1/2 or 3). The first product is $\ce{K(SbO)C4H4O6}$⋅1/2 H2O, which can be doubled in $\ce{K2(SbO)2C8H8O12}$⋅H2O. And the second product is $\ce{C8H4K2O12Sb2}$ ⋅3H2O. The number of $\ce{H}$ atoms are not the same in the main tartrate formula : $\ce{C4H4O6}$ in the first formula, and twice, that is $\ce{C8H4O12}$ in the second formula: it is all right, except the number of $\ce{H}$ atoms

The mass percent of Sb in both the hemihydrate and trihydrate is exactly the same. You can verify this by using https://www.webqc.org. Further, upon dissolving in water the chemical species with both these compounds will be identical. Hence, no change in chemistry between these two crystal forms. Hence, weighing the same amount of either hemihydrate or trihydrate will not make any difference in your chemistry. Also, in any analytical analyses, your analyte has to be the limiting reactant (by necessity) and everything else must in excess. In your case you want to determine the concentration of P as o-phosphate. So, P content is the limiting reagent. Acidic molybdenum, ascorbic acid, and antimony potassium tartrate (all under acidic conditions) must be in excess. Hope this clarifies.

The second formula is factually wrong, as it formally moves structural hydrogen (C-H, C-OH) and oxygen ( SbO+ ) among the crystal water.

https://i.stack.imgur.com/tdNN4.jpg according to Aldrich. @Poutnik

@Karsten Yeah, the picture is right, I revoke my prior comment. :-)

The two are synonyms for the same compound. Potassium antimonyl tartrate sesquihydrate is another synonym.
Tartrate occurs in a fully deprotonated state (ie -4) complexing the antimony(III) with four oxygen atoms each. In other words, tartrate acts as tetradentate tartrato(4-) ligand.

For a 3D structure, see https://en.m.wikipedia.org/wiki/Antimony_potassium_tartrate
Source: https://www.chemwhat.com/potassium-antimonyl-tartrate-sesquihydrate-cas-28300-74-5/


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What effect does hydrogen bonding have on H NMR spectra?
My thinking is that an increase in electron density caused by hydrogen bonding should cause the chemical shift of participating atoms to be moved downfield. Am I right or wrong?
Can someone provide an example showing this phenomenon?

Related: Is there intramolecular hydrogen bonding in salicylaldehyde? (It doesn't quite sound related, but there's an NMR spectrum there.)

There is also scalar J-coupling between the hydrogen and the acceptor atom, source

Hydrogen bonding is well known to cause large changes in chemical shift values– this is evident if you look up the chemical shift of an alcohol in a data-book (for an aliphatic alcohol this is anywhere between 0.5 and ca. 6.0 ppm depending on solvent/concentration), where a large range of values are given.[1]
In general,  formation of hydrogen bonds causes shifts to move to a higher frequency (higher ppm) due to deshielding.[2] In itself this is counter-intuitive, as in a hydrogen bonding interaction the H is actually receiving electron density from the donor atom (and therefore should be shielded). In the case of NMR however, the actual electron density isn’t whats causing this, but rather the field set up from the presence of the interaction.[#]
Intramolecular vs intermolecular hydrogen bonding
Although hydrogen bonding does cause a change in chemical shift, it is often only evident (/causes an issue) with intermolecular hydrogen bonding (between two molecules). With intramolecular bonding, the system is sufficiently stable that the chemical shift of the proton is consistent (and doesn't change with concentration/solvent, much).
To give an example,[##] consider the 1H-NMR of methanol and hexafluoroacetylacetone:[3]

On the top is the spectra of the neat compounds, where hydrogen bonding is present. On the bottom, the spectra of the compounds with added tetrachloromethane (which breaks down hydrogen bonding). Whilst the intramolecular hydrogen bonding in hexafluoroacetylacetone is unchanged, we can see immediately that the chemical shift of the hydroxyl proton to move.
Applications
Hydrogen bonding in NMR may seem rather inconvenient (and indeed it can be in routine structure elucidation work where the shift keeps on moving depending on how dry the NMR solvent is on any given day), however it has played a vital role in biochemistry and supramolecular chemistry, where the change in proton shifts can be used to determine whether or not multiple components are interacting.
A classic example is the interaction between cytosine and guanidine (C pairs with G in DNA via a hydrogen bonding interaction), which can be observed by NMR when the two compounds are mixed in an equimolar ratio:

[1]: Presch, E.; Buhlmann, P.; Badertscher, M. Structure Determination of Organic Compounds- Tables of Spectral Data; Springer: Berlin, 2009
[2]: Günther, H. NMR Spectroscopy- Basic Principles, Concepts, and Applications in Chemistry; Wiley: Weinheim, 2013
[#]: This is covered in Spin Dynamics by Levitt if you have any interest in the theoretical reasoning behind this
[##]: In this other answer, i’d previously given the example of salicaldehyde, which shows a consistent intramolecular H-bond free from solvent/concentration effects
[3]: Figure reproduced from Breitmaier, E. Structure Elucidation by NMR in Organic Chemistry; Wiley: Chichester, 2002


text:
Help identifying glassware
I am helping clean out my lab and there is some glassware that I can't identify. Attached are the photos:

Thank you!

Ok. Finally found some time to look up references.
The first piece of glassware is a bubbler. See:
https://www.chemglass.com/product_view.asp?pnr=AF-0513
You use this to help keep positive pressure in your inert atmosphere system. You need to fill it with mineral oil or mercury and you can vent your sealed system though this. This allows you to keep piping in say nitrogen or argon and keep positive pressure without blowing anything up.
The second piece of glassware is a fritted filter used for inert atmosphere chemistry. See:
https://www.chemglass.com/product_view.asp?pnr=AF-0540
The side arm allows you to equilize the pressure between the two sides (otherwise the liquid may not flow from top to bottom). There are ground glass joints on both ends so that you can create a sealed system free from water or oxygen.

Is filling the bubbler with mercury up to date?

As of 5 years ago, yes. It's anecdotal, but I think mineral oil is fine most most applications, though I've also known some inorganic chemists that say it's not enough for what they're doing.

Could the first one be used as a trap, as one that is used before a vacuum pump?

@EashaanGodbole That seems highly unlikely. First, there is no banana for scale, but that appears to be quite small, smaller than any trap I've seen. Second, a trap would have a collection vessel that was detachable, so that the trapped contents could be easily cleaned out. I suppose you could rig it to serve as a trap if you needed to, but I would be concerned about pulling vacuum on a piece that was not intended for that purpose.


text:
Harmonic frequencies (IR) and basis set size
Usually vibrations (harmonic), derived from harmonic approximations are bigger than for experimental values. The question is, how does this depend on the size of the basis set. Many papers give that as a statement in their introduction without any explanation.
My idea would be that, if I use variational methods like DFT, then my energy will always be bigger than the real energy, as long as I use a finite basis set. And also increasing the set gives better results in DFT. The harmonic frequencies I determine computationally are derived from the Hessian matrix (or better the force constants for the harmonic oscillator) and this matrix is the second derivative of the energy, which is much too big for small basis sets.
So can I say that, because a smaller basis set will give me a bigger energy, the derived frequency will have to be bigger as well?

I'm curious: Why do you think that there's a correlation between the second derivative and the absolute value of a minimum?

Thinking about it like that...well yeah you are right. But then I still don't understand why there is a dependency on the basis set size.

No, you cannot say that larger energies will necessarily give larger frequencies.
Consider, for instance, a hypothetical (and purely fictional) method of determining the electronic energy of a system for a given geometry. Call this set of nuclear coordinates (geometry) $\mathbf{G}$, and our hypothetical method is a function which acts on these coordinates to produce an energy, i.e. $$f(\mathbf{G})=E$$
This method, however, is a very special kind of method. It is one which gives an absolutely unreliable energy. It gives you a number that has absolutely no basis in reality and can't be trusted at all. But, one thing we know about this method is that the error, which is truly horrendous and unpredictable, is constant! From this fact, it follows that this method predicts the correct frequencies, the experimental ones. That is, at the minimum of the potential predicted by our hypothetical method and the actual minimum, the difference between our energy, $E_0$, and the true energy, $\mathcal{E_0}$, is,
$$E_0-\mathcal{E_0}=\epsilon$$
or,
$$f(\mathbf{G})-\mathcal{E}_{\mathbf{G}}=\epsilon$$
For all real methods, however, we would need to write the above equation as,
$$h(\mathbf{G})-\mathcal{E}_{\mathbf{G}}=\epsilon({\mathbf{G}})$$
The point of all this being that the hypothetical method I described was not variational or anything else you might desire out of a quantum chemical method. It only predicted energies for which the error term was constant. Whereas the real method, $h(\mathbf{G})$, predicts energies for which the error term is a function of the geometry $\mathbf{G}$.
The concise way of saying this is that frequencies depend only on the curvature of the potential well and not on the depth of the potential well.

With that established, we have answered the question that smaller basis sets, with their larger energies, do not necessarily give higher frequencies.
For instance, if you look at Sinha and Wilson's paper,[1] you will learn of something very empirical which computational chemists do sometimes that bothers me a lot but is illustrative for our purposes. In that paper, a big set of 42 organic molecules is taken, and the harmonic frequencies are calculated at a bunch of different methods and basis sets, then these frequencies are compared with the experimental values. A least squares fit is then performed to come up with scaling values to map the harmonic frequency onto the experimental frequency. As you say, these constants are numbers less than one, say 0.95. The reason these constants are less than one is what bothers me about doing this. Experimental values are anharmonic frequencies. We don't live in a harmonic world. Anharmonicity shifts the frequency down, by lowering one of the potential walls to accommodate bond-breaking, hence we need to scale our harmonic frequencies downwards. So we're basically comparing apples with oranges and there's no way to be sure that this will work in all cases.
More to your question, in this paper, the correction factors do not increase monotonically with basis set size as you suggested (this is the same as frequencies decreasing with basis set size). I won't reproduce their numbers here, because I don't know if that's frowned upon, but for both MP2 and B3LYP, the cc-pVnZ and aug-cc-pVnZ (n=D,T,Q) basis sets have T < D < Q for the scaling factor. So, the triple-zeta basis set invariably predicts a lower energy, but predicts a higher frequency.
Even more interestingly, if one only considers low-frequency modes, which are particularly poorly described by normal modes (these are torsions usually), the ordering becomes Q < T < D for the scaling factors. That's really a failure of the harmonic model more than anything, but it illustrates the point well that lowering the energy really doesn't guarantee you anything in terms of frequencies.
To address your comment, the reason there is any dependence on basis set size is because the error away from the equilibrium structure will change with basis set size, and hence the curvature of the potential changes which affects the frequency.
[1] Pankaj Sinha, Angela K. Wilson, J. Phys. Chem. A, 2004, 108 (42), 9213-9217. DOI: 10.1021/jp048233q

As a comment on your anharmonicity concerns: One quite often observes that anharmonic frequency corrections (eg from VSCF) lower the high and increase the low frequencies. I am not entirely sure how that happens exactly but as a matter of fact, that type of mismatch is also what you observe when you compare scaled computed harmonic frequencies with exptl. ones.

@R_Berger this makes sense though because if the low frequency is a torsion or an umbrella mode, the largest correction will come from the coefficient on the x^4 term of a Taylor expansion. Whereas the high frequencies are Morse-like which has a first-order anharmonicity that decreases the frequency. Very good point though. I hope I didn't imply anything to the contrary.

Not as far as I can see.

smaller basis set would give a bigger energy that is sum of BO electron energy,but the frequency is derived from this energy

This has the makings of a decent answer, but could you expand on it a bit, please?


text:
What can be used to thionate a monosubstituted urea?
I'm looking to make a thiourea from a monosubstituted urea. I haven't found any references that use Lawesson's Reagent, which I have been using for the thionation of di and trisubstituted ureas. Is there something that would prevent Lawesson's Reagent from working on a monosubstituted urea?

It's described at least in 1 patent: WO2004031161 and one article: Korkusuz, Elif; Yildirim, Ismail, Journal of Heterocyclic Chemistry, 2010, vol. 47, # 2, p. 472 - 476. But I'm not sure about the correctness of the data in the article.

I believe Lawesson's Reagent could do this. It is useful to thionate a large number of carbonyl compounds including amides, so I think it can do for urea as well. I don't see any effect on a monosubstituted urea that would negatively affect the reaction.


text:
Order of basicity of halides
The order of basicity of halides is $\ce{F-  >Cl- > Br- > I-}$
My teacher explained this on the basis that the smaller is the ion, the greater is the charge/size ratio, hence it more unstable and more likely to gain a proton.
However if this is the case, why is the magnitude of electron gain enthalpy in the order $\ce{Cl- > F- > Br- > I-}$ and not $\ce{F- < Cl- < Br- < I-}$ since we are calling $\ce{I-}$ the most stable and so on?
(The reaction is going on in an aqueous solution.)

There are so many competing effects going on here. Also, how these measurements are made. Are they gas phase? In aqueous solution?

The electron gain enthalpy you've mentioned is measured in the gas phase, not in aqueous solution like the order of basicity.
When you go down on a group, two important properties are increased: nuclear charge and atomic radius. However, the increase in atomic radius is much more pronounced than of the nuclear charge. Considering this, as you go down on the halogens group the electrons added feel less attracted to the nucleus and, therefore, the electron gain enthalpy become less negative.
The order is only apparently changed for $\ce{Cl-}$ and $\ce{F-}$ due to fluorine small size. The additional electron would be added to a "small" $\ce{2p}$ orbital in fluorine (and there would be more electron-electron repulsion) while in the chlorine atom it would be added to a "big" $\ce{3p}$ orbital that can better accommodate the additional charge, so the enthalpy is less negative for the electron gain in $\ce{F}$ then for $\ce{Cl}$.

So is it possible to measure electron gain enthalpy in aqueous solutions, and what is (are) the factors that change the order?

@InternetGuy That's fine, but in solution, entropic effects would be very much relevant. Solvation is the big factor that makes this a complicated problem.

Maybe there is a way of measuring it in aqueous solution that I'm not aware. But I can guarantee it won't be as easy as in the gas phase. In solution there are other interactions that would affect the enthalpy measured, not to mention the difficulty to produce elementar X atoms in aqueous solution. Just the water-halogen interaction in aqueous solution would make your measurement much harder.

So basically there is no exact theoretical reason for the observed order of basicity?

@User , Ofcourse there is. In an aqueous solution,  although halide ions are formed , There is no actual gain or loss in electron.  Therefore HF dissociating into H+ and I- cannot be considered that Iodine was physically forced to accept one electron in its orbitals.


text:
Specific conductance of 0.1 M acetic acid

Find specific conductance of $\pu{0.1 M}$ acetic acid given that its molar conductivity is $\pu{4.6 S cm^2 mol^-1}$.

I used the formula $\Lambda_\mathrm m = \kappa/c$, where $\Lambda_\mathrm m$ is the molar conductivity, $\kappa$ is the specific conductance and $c$ is the concentration of the electrolyte in $\pu{mol/L}$.
What's confusing me is the value of $c$. At first I thought it should be the given molarity, but according to my text book $c$ is the concentration of the electrolyte. Now the acetic acid molecule does not conduct electricity, rather its constituent ions do. So shouldn't the value of $c$ be the concentration of its ions which turns out to be $\pu{\sqrt{1.8} \times 10^{-3} mol/L}$ (using the equation of the acidic dissociation of acetic acid)?
Also the answer given was $\pu{4.6E-4 S cm^-1}$.

With the data provided with your problem I suspect you are only supposed to carefully use provided formula with the corresponding units of measurements. You need to convert the concentration from $\pu{mol L^-1}$ to $\pu{mol  cm^-3}$ considering $\pu{1 L} = \pu{10^3 cm^3}$:
$$\kappa = \Lambda_\mathrm m c = (\pu{4.6 S cm^2 mol^-1})(\pu{0.1E-3 mol cm^-3}) = \pu{4.6E-4 S cm^-1}$$

So is my reasoning correct, that is, should 'c' actually be substituted as (sqrt 1.8) x 10^-3?

@Azulene No, I have no idea where this  (sqrt 1.8) x 10^-3 value comes from. I used $\pu{0.1 M}$. You are given neither the dissociation constant of acetic acid, nor the temperature. And please get to know how to use LaTeX for proper formatting of math and chemical equations.

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


text:
How to re-suspend silver from the dried out silver paste?
Kindly suggest any method or solvent to re suspend silver from old dried out silver paint/paste. It's a 25 years old "Dotite©" D-550 bottle. Strangely there is no expiry date mentioned on the bottle.

Generally speaking, with difficulty. After 25 years perhaps it is time to get a new bottle?

A current advert (https://www.2spi.com/item/05003-ab/) displays the jar labelled with expiry date.

According to SPI Supplies:

If the Dotite Silver Paint dries out, unlike some other silver paints, it can be re-suspended with the addition of SPI #05004 Silver Paint Thinner followed by an ultrasonic bath treatment.

Generally, ketones are used as diluents, also for conductive paints (acetone, methyl ethyl ketone (MEK), methyl isobutyl ketone (MIBK)).


text:
Can the reaction intermediate be lower in energy than the reactants or products?
It makes sense to me that reaction intermediates exist transiently if they are of higher energy than the reactants and the products, my organic chemistry textbook seems to give me information that doesn’t seem to match up.
My knowledge is that the overall activation energy of the reaction is the energy change between reactants and the highest-energy transition state and the overall energy change is he energy difference between the reactants and the final products. If intermediates could be of lower energy levels than either the reactants or products, would this definition change?
I am confused because the solutions manual to one of the problems gives me an energy diagram with an intermediate that is of lower energy than both the reactants (but not the products), and labels the overall activation energy as the energy change between the intermediate state and the highest-energy transition state. The textbook also gives me an energy diagram of a catalyzed reaction that has several intermediates of lower energy than the reactants, but not the products.
(The textbook I’m referring to is Organic Chemistry. 8th edition by John McMurry)
Would it make sense for intermediates to only be lower in energy than the reactants but not the products since the products would still be more favorable and stable than any of the intermediates? Would this only happen if there is a catalyst?

The intermediate is a proper chemical species but a reactive one under 'normal' conditions and is not a transition state, i.e. it has a short, but measurable, lifetime. The intermediate can change only by crossing one of two transition states, one back to reactants and the other to another intermediate or to product.
Any intermediate must not be lowest energy species, including reactant and product, but can be lower or higher in energy than the reactant (but less than any transition state) and higher than product. The limiting activation energy will be the largest one in the whole scheme, so may be that between reactant & intermediate or between intermediate & product. Which it is depends on the exact species involved.


text:
Why is oxygen oxidized with lead instead of being reduced?
Oxygen is more electro negative yet in the redox reaction
$$\ce{2PbO(s) -> 2Pb(s) + O2(g)}$$
my text book says lead is reduced and oxygen is oxidized. How can this be? It doesn't make sense to me. Oxygen is usually $2-$ while lead is usually $2+$.

Oxidation is by definition a loss of electrone(s). $\ce{2O^{2-} - 4e -> O2^0}$ is an oxidation of oxygen. I honestly don't see what confuses you so much.

I agree with the OP that oxidation is a confusing word, especially if applied to the element oxygen. It would be better to say, in the example given, that the lead(+2) cation is reduced to elemental lead (0) and the oxide (-2) anion is oxidized to elemental oxygen (0). By analogy: converting chloride ion to chlorine gas is clearly oxidation, so the same is true for oxygen.

Your confusion stems from that fact that you are trying to look at what reactions are favorable instead of the reaction as written.
Oxygen is readily reduced because it is a strong oxidant. You can check any reduction half reaction with oxygen on the left side. Chances are that the reduction potential will be fairly positive.
In the reaction you have written, oxide is being oxidized to oxygen. Nowhere (including your textbook) do we claim that this reaction is favorable or spontaneous in the forward direction. In fact, if you consult a table of free energies of formation (example), you would find an entry like:
$$\ce{PbO (s, red)}, \Delta G^{\varnothing}_{f}=-189.3\ \mathrm{kJ}\ \mathrm{mol}^{-1}$$
or
$$\ce{PbO (s, yellow)}, \Delta G^{\varnothing}_{f}=-188.5\ \mathrm{kJ}\ \mathrm{mol}^{-1}$$
These values indicate that indeed, the reverse reaction is the one that's favorable, the one where oxygen serves as an oxidizing agent and is reduced.
However, that is not the reaction as written. The notation of chemistry allows us to describe reactions that are hypothetical or unfavorable.


text:
How to identify lewis acids and bases
What is the general methodology to identify Lewis acids and bases in an acid base reaction? Do we need to draw structures to identify which species is the electron donor and which one the acceptor, or is there any other way of doing this?
If you consider $\ce{HClO_4}$, oxygen $\ce{O}$ has lone pair(s), so there is a confusion whether it can donate the pair to other species or not. So in that case, although it is an acid, one may say it is a base. Similarly, in $\ce{H2S}$, sulfur $\ce{S}$ has got lone pairs and I am confused whether it will act as base or acid.

You need to draw the structure and identify the most acid and most basic site(s).
Notice how I didn't say "the most acid molecule".
Boron tribromide ($ \ce{BBr3} $) contains both a super-strong acidic site (the trivalent boron atom) and potentially basic sites (the lone pairs of the three bromine atoms).
Whether a molecule (e.g. $ \ce{H2S} $) will act as an acid or as a base depends on the reagents.

@above:

You need to draw the structure and identify the most acid and most basic site(s). Notice how I didn't say "the most acid molecule". Boron tribromide (BBr3) contains both a super-strong acidic site (the trivalent boron atom) and potentially basic sites (the lone pairs of the three bromine atoms). How do I identify the most acidic and basic site(s)?

@DhrubajyotiGhosh Welcome to Chemistry.SE! Take the [tour] to get familiar with this site. Mathematical expressions and equations can be formatted using $\LaTeX$ syntax. I have converted your answer to a comment here. With a bit more rep, you will be able to post comments yourself. Please don't add answers that do not actually answer the question, they will be deleted and no-one will be any wiser.


text:
How are these two organic resonance structures different?
I've been working through some questions online (self study, not homework) predicting the various forms for molecules. One of the questions involved the following molecule (see image below). It had four structures listed in total, but the two below have me baffled. They appear to be identical if rotated by 180 degrees along the horizontal axis.
Please explain the difference between these two structures.

They are the same as they are superimposable by rotation and reflection and the energy of the molecule cannot change by simply doing this.

@porphyrin Do you mean that resonance structures must have different energies?

@andselisk no ! exactly the opposite, because they are related by simply rotating etc they must have the same energy

@porphyrin Ahh! Now we are on the same page, sorry, I misunderstood your comment.

@porphyrin In resonance structures, the atoms are regarded as fixed, so the two structure are not the same. Born-Oppenheimer appoximation. Of course the two have the same energy, and if you regarded them as actual molecules, they'd be identical.

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Resonance structures do not describe the conformation (=energy state) of a molecule, but are didactic. They show how you can just flip schemactic $\pi$ electrons around the structure, without breaking any connection. The bond lenghts and angles are constant in the actual molecule, at values somewhere between those indicated by the resonance structures.
Between resonance structures, you don't look at a possible symmetry of the location of atoms, but regard them as unique and fixed (see also https://en.wikipedia.org/wiki/Born-Oppenheimer_approximation). Resonance structures can have different energy or be symmetric (=conformationally identical) to each other. If one resonance structure has higher energy than the others, that is just a hint that the actual structure might by more closely described by the other resonance structures.

The B-O approximation is not relevant here, and the structures are more than didactic. In a variational approach we would consider various covalent and ionic structures, the true energy will lie below any of these and the superposition corresponds to a 'species of resonance' (Coulson 'Valence') as a result of which energy is lowered. The amount of lowering below the lowest component energy is the resonance energy.

@porphyrin You regard the atoms as fixed and wonder what electronic structure(s) could fit. How is that not B-O?

While drawing resonance structures, we keep the orientation of the molecule same and assume that the bonds have changed their positions. This is done to pictorially show the movement of electrons. The two structures given are 'equivalent' since they can be superimposed but not the same since the double bond is now between two different atoms.

Oh, ok. Let me see if I've understood correctly. So with the benzene ring, even though both structures are the 'same' under rotation, they are different if you fix the orientation. Fixing this makes it possible to write two sensible forms of the ring that, when considered together, show all the C-C bonds are equivalent. Is that right?

So the two 'different' resonance forms in my question show that all the bonds in the ring are 'equivalent' in the same sense the C-C benzene bonds are 'equivalent'? And this would be difficult to communicate without fixing the orientation?

This answer is ambigous, on the border to incomplete or just plain wrong.

@Guest You are right in saying that the orientation is kept fixed to show the movement of electrons/bonds. However, unlike benzene, not $ all $ bonds in the compound you have given are equivalent, only those opposite to each other about an axis passing through the B and N atoms are. Overall the two structures are 'equivalent' but not the same.


text:
Does a positive electron affinity actually correspond to the release of energy?
I read about the periodic properties — ionisation energy and electron affinity — in my book. It was quite intuitive that energy is required to remove an electron from the atom but I wonder why energy is released when an electron is added to an atom? Is it actually released or is it because of our convention that energy of electron at infinite distance from nucleus is 0, that we consider it to be released?

It may help to take a look at the two processes you are comparing in terms of reactions. Ionisation follows the following reaction:
$$\ce{A ->[][$\Delta E_\mathrm{i}$] A+ + e-}\tag{1}$$
This process involves tearing charges apart, separating a positive and a negative one. Even if the original atom was somewhat unstable, it will be unfavourable to remove that electron simply due to electrostatic interaction. You write in the question that you were able to deduce this already.
The electron affinity on the other hand describes the process in this following reaction:
$$\ce{A + e- ->[][EA] A-}\tag{2}$$
Here, a charged particle is approaching a neutral particle and we have the same number of charged particles (and the same charges) on both sides of the equation. This means that we don’t have a strong, obvious physical effect that will make all electron affinities sign-equal for very simple reasons. Instead, we need to consider the details more.
The process often involves a release of energy because most atomic configurations have space for extra electrons at little or no extra cost. Except for the noble gases and the second group, atoms don’t have completely populated sub-shells so there should always be a position available in a subshell at an acceptable exothermic energy level. While adding an electron will increase electron-electron repulsion and increase all energy levels by a certain amount, this increase is typically less than the first ionisation energy which corresponds to the energy level of the highest occupied atomic orbital. Thus in electron affinity, we in principle gain the entire ionisation enthalpy minus the destabilising contribution of the new electron. (Caution! This is a very simplified picture and not suitable for generalisation or even strictly for explanation. It gets the basic point across though.)
Therefore, most electron affinities are exothermic with the notable exceptions of group 2, group 18 and nitrogen.

Energy has to be released when electrons are added to atoms in most cases.
Otherwise, why would they ever form covalent bonds?
The whole point of a covalent bond is that electrons are stabilized by the attraction to multiple nuclei, which helps to hold the system together. Therefore, the system must be more stable when you add an electron to an atom.
Note that in the case of something like a neon atom, where adding an electron is not favorable is not the case, we don't form covalent bonds.
From the atomic perspective, even for a neutral atom, the nuclear charge is not completely shielded by all of the electrons. Some electrons are going to be on the other side of the nucleus from the new electron you're adding.

This is because potential energy is more in this case when electron is further from nucleus as the forces between them are attractive. As the distance is decreased between nucleus and electron, potential energy decreases. Since energy is always conserved, the net decrease in potential energy is released as any form of energy. However the energy is not released in all the cases. It depends on the net effect of the interactions of the electron with the nucleus and the electrons that are already present in the atom.