text:
Why do solutions of electron in a box (and in a ring) predict coefficients for LCAO (linear combination of atomic orbitals) in 1D systems?
The solutions to the 1D particle in a box quantum mechanic system are standing waves (zero at both ends of the box) with 0,1,2... nodes for increasing energy (zero for the ground state).

If I look at the LCAO solutions for a linear system, combining AOs of the same shape and energy (e.g. conjugated double bonds, linear array of sodium atoms), the coefficients follow the pattern of a standing wave with 0,1,2... nodes. For example, when I combine 3 AO's in a row, the solutions are +++, +0-, +-+, i.e. zero, one, and two switches of signs as I go to higher energy states. For the state with one node, the node is in the center of the box, corresponding to the middle AO, which has a coefficient of zero. (More examples, i.e. hexatriene and pentadienyl cation, are on slide 126 and 128 of this document.)

The solutions to the particle in a ring system are standing circular waves (same value at 0 and 360 degrees) with 0,2,4... nodes for increasing energy. The animation shows a standing circular wave with 8 nodes:

If I look at the LCAO solutions for a circular system, again combining AOs of the same shape and energy (e.g. p-orbitals in an aromatic system), the coefficents follow the pattern of a circular wave with 0,2,4... nodes. For benzene, the solutions are ++++++, +++--- and +0--0+, +-++-+ and -0+-0+, +-+-+- (see picture):

What underlies the connection between the unbound electron in a box or corral with the formation of covalent bonds of bound electrons?

You mean why are they qualitatively similar, with more nodes corresponding to higher energy?

@BuckThorn Yes, but it also predicts the positions of the nodes, i. e. which AO will get a negative or positive coefficient or be zero. Maybe it also predicts the relative size of the coefficients, but I’m not sure about that.

When you use Hückel theory to solve for the π MOs you will get coefficients that have some sin/cos dependency, which reproduces the form of the standing wave. It's easy enough to show this for small systems but I think the maths to prove it more generally is non-trivial. Also, I've never heard of it as "particle in a corral" - usually "particle on a ring".

@orthocresol I changed it (if you google it outside of Texas, ring is more popular than corral).

The lack of symmetry of degenerate states such as $\psi_2$ makes me wonder to what extent these drawings are just "cartoons" drawn to illustrate or fit the expected trend of energy being proportional to the number of nodes. Or, more to the point, to what extent the solutions are unique, as opposed to chosen to represent what we would like to emphasize. AFAIK any linear combination of the two benzene $\psi_2$ MOs drawn here will have the same E ie be an eigenfunction of the problem as written.

See eg http://www.chemtube3d.com/BenzeneMOs.html

@BuckThorn- what do you mean by "lack of symmetry " of $\phi_2$? While it is true that other linear combinations can be used, the two degenerate orbitals must be orthogonal (ie net overlap=integral of product = 0) and must be in one of the symmetry groups of the original molecule. So you're always going to end up with a pair that looks fairly similar to the ones shown here.

@Andrew A hunch tells me that you should be able to combine the two $\psi_2$ orbitals shown in the OP into two orthogonal orbitals that are identical in appearance but distinguished by a rotation operation about the benzene highest symmetry axis.

No one has yet mentioned boundary conditions. These determine the nature of the solutions, i.e. ensuring that the wavefunction is zero or has the same amplitude and gradient at the boundary etc.  Without this restriction there are an infinite number of possible solutions. The boundary condition is the reason that there is similarity between the particle on the ring and a circular aromatic such as benzene and the nodes in a particle in a box.

@porphyrin Well, while attempting to address math a little, I wanted to stick to a physchem explanation and not dwell on math. Plus "confining" more or less means that the electron density goes to zero far away from the origin of the confining potential.

@porphyrin Also, try creating a "standing wave" that is not continuous (function + derivatives).

@Buck Thorn I was not particularly suggesting a maths detail because the boundary conditions are imposed by the physics of whatever the problem is that is being studied. To get a  'standing wave' solution implies  boundary conditions have been used.

A short answer is that the functions that satisfy the Schrodinger equation for a system are largely determined by the potential energy part of the function, since that's the part of the Hamiltonian that varies with different scenarios. The particle-in-a-well (= 1D box) is defined by having infinite potential energy outside the walls and a constant finite energy within. That turns out to be a very good approximation of the potential energy function of a pi system or a line of identical metal atoms.
At the end of the line of atoms, an electron faces a steep potential energy curve. Not quite the vertical wall of the 1D box, but pretty close to it. The same is true on the sides, so the system can be treated as 1D. Within the system, the electron moves essentially freely (or exists in a broadly delocalized way) with very little change in potential across the system.
Since the potential functions are nearly identical, the solutions to the Schrodinger equation take the same mathematical form.

"with very little change in potential across the system"... I agree with you up to this point. The position of the nodes in the delocalized-pi ring is determined by the position of the nuclei. The nuclei are what makes a comparison between particle-in-a-box and any real system non-trivial. I would not say that the potential functions are identical!

@BuckThorn - The potential functions are very different, but they are taken out of the equation (the AOs already take account of them). When making MO's from the AO's, all that happens is to delocalize the electrons without changing the electron density, so the Coulomb interactions are the same (= alpha in Hueckel theory).

I want to add a slightly different but related perspective to the excellent answers already posted: Comparing the Hamiltonian of particle in a box and for Hückel theory. First though, I want to compare boundary conditions and potential (Coulomb) energy.
Boundary conditions
For the particle in a box, the boundary conditions are explicit: The particle has one degree of freedom (1D) and has to be inside the box. For LCAO, there are implicit boundary conditions: The electrons can only be where they would be in the atomic orbitals.
Potential energy
For the particle in a box, there are no Coulomb interaction within the box; this is a free electron in that region. For the Hückel theory in its simplest version (all p-orbitals, all carbon atoms) of LCAO, the Coulomb interaction is encoded in the $\alpha$ parameter (pre-computed, or just given as a conceptual parameter). The two scenarios are very similar: there is a region of low energy (inside the box or near the atoms) and a region of high energy (outside the box or away from the atoms). You can either describe this as "confining potential" (as in the answer by Buck Thorn), or you can describe this as absence of Coulomb term in the Hamiltonian because the boundary conditions (and the choice of basis functions for Hückel theory) take care of the "confining".
The Hamiltonian
For the particle in a box, the Hamiltonian is just about the kinetic energy (no electrostatic interactions), so it is simply the second derivative with some constants:
$$ H = -\frac{\hbar^2}{2m} \frac{d^2}{dx^2}$$
For Hückel theory, the Hamiltonian is not explicitly shown, but the result of $⟨ϕ_i|H|ϕ_j⟩$ are given as $\alpha$ for i=j, as $\beta$ if i and j are neighbors, and are neglected (set to zero) otherwise. The $\alpha$ corresponds to the Coulomb interaction, and its contribution is constant. With the atomic orbitals as basis functions, you might say the electrons are free otherwise (this amounts to saying that the Coulomb interaction is the same around each atom). In a way, the Coulomb interaction is taken out of the equation because all the AO's have the same energy, and the MO's are just linear combinations of them (so there will not be more electron density between atoms for the MO's compared to the AO's - still, the electrons are more delocalized and therefore have lower kinetic energy).
The $\beta$ term roughly corresponds to the kinetic energy. If two adjacent AOs are combined with equal signs of coefficients, the resulting wave function is smoother (has a smaller second derivative) than when the signs of coefficients are opposite. Another way of saying that is that if the coefficients "have a node", the kinetic energy is higher. So this models the kinetic energy part of the Hamiltonian.
The solutions
If the Hamiltonian (or the implicit Hamiltonian) are similar because the only relevant term for finding a solution is the kinetic one, then the resulting wave functions are expected to be similar. One difference, though, is that the particle in a box can have arbitrarily high energies (number of nodes) while the LCAO result has a finite number of energies constrained by the available basis functions.

Can we address your last point by considering the limit as the number of atoms in the line or ring goes to infinity rather than just a small system like benzene?

@Andrew I suppose so, for example for the sodium example. As you go from a couple of atoms to the bulk system, the discreet energy levels turn into bands, and the electrons are delocalized over the entire volume (not free, though, but more like hopping from nucleus to nucleus).

Attractive Coulombic interactions impose a confining potential, and single electrons, being wave-like in character, will display standing-wave probability densities within such a potential. In a 1-D system, lower energy corresponds to probability density distributions with less nodes, because highly spatially-varying wavefunctions correspond to a higher electron kinetic energy. All of this is explained here in an excellent answer to another post. In the case of an aromatic ring, the location of the nodes can be inferred from the fact that the approximate MOs are constructed from AOs consisting of p-orbitals, which have density centered  above the nuclei on the plane above the ring.
In summary,

the wave character of electrons accounts for the "standing wave" appearance of the density
the wavefunctions are sensitive to the potential, specifically the location of nuclei, but symmetrical distribution of nuclei (as in a ring or line) will result in nice symmetrical MOs that might make important details less obvious
the 1D character of the potential accounts for the proportionality of number of nodes and energy

Finally, I should point out that the MO's constructed to generate the solution to the aromatic ring problem are obviously finite in number, as the AOs used to construct them are finite in number (#MOs=#p-orbitals). The same is true in a sense of the sodium row problem (#MOs=#AOs=#1s orbitals). This is absolutely not true for the particle in a box. There you don't need more AOs (or atoms) to construct a new wavefunction, and the position of the nodes is determined by the number of nodes and size of the box, not by the position of any nuclei. The analogy breaks down because you are not constrained by any AOs (or nuclei).

I'm sorry, I'm still confused what point you are making about the linear combinations. No matter how the orbitals are recombined or rotated, I believe the result will always resemble a circular standing wave with two nodes (=single nodal plane). It is this resemblance that is the motivation for the question.

@Andrew Upon further inspection it seems my "MO uniqueness" comment was vague at best so I dispensed with it.

To answer this question it is advantageous to treat a molecule as a graph and use the well known adjacency matrix from graph theory.
Here is the wikipedia definition:

For a simple graph with vertex set $V$, the adjacency matrix is a square $|V| × |V|$ matrix $\mathbb{A}$ such that its element $\mathbb{A}_{ij}$ is one when there is an edge from vertex $i$ to vertex $j$, and zero when there is no edge.

Hückel
The Hückel-Hamiltonian $H$ can be written as:
$$ H = \alpha \mathbf{1} + \beta \mathbb{A}$$
Where $\alpha$ is the ionization energy, $\beta$ the overlap between adjacent $p_z$ orbitals, $\mathbf{1}$ the unit matrix, and $\mathbb{A}$ the adjacency matrix.
If you look at the eigenvalue equation,
$$H \psi = \lambda \psi $$
it has a solution if and only if $\psi$ is an eigenvector of the adjacency matrix.
The eigenvalue $\lambda$ of $H$ is given by the eigenvalue $\tilde{\lambda}$ of the adjacency matrix using the following equation:
$$(\alpha \mathbf{1} + \beta \mathbb{A}) \psi = \alpha \psi + \beta \mathbb{A} \psi = (\alpha + \beta \tilde{\lambda}) \psi$$
What we see is that the $\alpha$ is just a constant offset for the energy. All the relevant information about level spacing, degeneracy etc. is contained in the $\tilde{\lambda}$ with a scaling factor $\beta$.
To put this in another way: All relevant properties of the Hückel-Hamiltionian are encoded in the adjacency matrix of a graph/molecule.
Particle in a box (Free electron network model)
There is a nice paper by Ruedenberg and Scheer 1953 about that topic here.
The main idea is, that your 1D-particle-in-a-box solutions on each edge/bond have to be constrained at joint vertices/atoms.
These constraints are "derived by intuition" in the cited paper.¹
You want the whole wavefunction, that is piecewise composed of the wavefunctions on each edge, to be continous.
Similar to Kirchhoff's Law for electric circuits you want to assert, that the probability current that flows into an vertex/atom flows out.
For this reason, the constraints are called Kirchhoff conditions.
If $V$ is the set of all vertices/atoms, $\psi^e$ labels the wavefunction at the edge/bond $e$, and $E_v$ contains all edges/bonds on a vertex/atom $v$. Then you can express your constraints as:

Continuity:
$$
\forall v \in V: \forall e_1, e_2 \in E_v: \psi^e_1(v) = \psi^e_2(v)
$$
Flux Preservation:
$$
\forall v \in V: \sum_{e \in E_v} (\psi^e)' (v) = 0
$$

Without going into the details of the derivation: for cyclic graphs/molecules your eigenvalues $\lambda$ are given again by the eigenvalues $\tilde{\lambda}$ of the adjacency matrix $\mathbb{A}$.
The free scaling factor is then $L$ the edge/bond length (Up to some fixed constants like number literals or $h$.)
Conclusion
For cyclic molecules, it can be proven that the essential properties of the spectrum are given just by the adjacency matrix $\mathbb{A}$ of the molecule in both cases.
The parameter $\alpha$ in the Hückel-Formalism introduces a constant offset, which can be usually ignored for chemistry.
This means that both methods have only one free parameter that scales the spectrum of the adjacency matrix. In the case of Hückel it is the overlap between adjacent $p_z$ orbitals, in the case of the free electron network model it is the bond length $L$.

¹ Note that you can rigorously derive general constraints by forcing only self-adjoint/hermitian realisation of the hamiltonian on the graph. This can be found e.g. here. But we are chemists and not mathematicians so let's stick to the Kirchhoff conditions.


text:
What is "dot" sign in •NO?
What is "dot" sign in •NO? I know that it is radical nitric oxide, but I don't know if it is necessary to put the "dot". Is there any difference between •NO and NO?

The dot is the radical (unpaired electron).

The answer is NO.

Usually the dot is put there to emphasize that the nitric oxide is a free radical that includes an unpaired electron.  This is especially notable by comparison with $\ce{NO^+}$, which does not have the unpaired electron.
Note that the nomenclature $\ce{·NO}$ should not be rendered as showing the unpaired electron on nitrogen.  The unpaired electron is actually incorporated into the molecular orbital structure of the nitric oxide molecule, covering both atoms.

It is not necessary, but optional, to express explicitly the radical status.
In other cases, like alkyl radicals, the dot marking is mandatory, not to be confused e.g. with a functional group.
For curiosity, the ground state of the oxygen molecule - triplet oxygen - is a biradical, with 2 unpaired electrons. What we write as $\ce{O=O}$ is singlet oxygen, that is not a radical, but has counter-intuitively higher energy than triplet oxygen.

Nitric oxide ($\ce{NO}$) is a free radical and hence why that dot is for. Explanation for the dot and the reason why its there have been well answered in other responses. Nevertheless, I decided to put the molecular orbital representation of $\ce{NO}$ as depicted below for your convenience (Ref.1):

There are three electrons in antibonding orbitals and eight electrons in bonding orbitals. One single electron occupying the highest energy level is in antibonding $\pi$∗ orbital (HOMO). It is an unpaired electron, and therefore, $\ce{NO}$ is a radical and paramagnetic. This electronic configuration explains the high reactivity of $\ce{NO}$ molecule: $\ce{NO}$ can easily be oxidized to become nitrosonium ion ($\ce{NO+}$) and be reduced to be nitroxide ($\ce{NO-}$). $\ce{NO}$ is unique among the diatomic biomolecules because it can bind to both ferric and ferrous heme iron due to its electronic configuration.
Reference:

Byung-Kuk Yoo, “Investigation of the mechanisms of regulations, activation, and deactivation of Guanylate Cyclase, the endogenous NO-receptor, and NO-sensors,” PhD Thesis, École Polytechnique, Paris, France, 2010 (https://tel.archives-ouvertes.fr/tel-00557106).

The dot represents an unpaired electron.  It's written that way as a reduced Lewis dot diagram.
The reduction works like this:

Start with the typical dot-diagrams for Nitrogen and Oxygen atoms, i.e.$$
\begin{array}{ccc}
   \textbf{Nitrogen}
 & \qquad
 & \textbf{Oxygen}
\\[-10px]
   {\Huge{\begin{array}{rcl} & \cdot \phantom{\cdot} \\[-50px] \raise{0.1ex}{.} &\mathrm{N} & \raise{0.1ex}{:} \\[-500px] &\raise{0.25ex}{\cdot} \phantom{\cdot} \end{array}}}
 &
 &    {\Huge{\begin{array}{rcl} & \cdot \phantom{\cdot} \\[-50px] \raise{0.1ex}{:}&\mathrm{O} & \raise{0.1ex}{:} \\[-500px] &\raise{0.25ex}{\cdot} \phantom{\cdot} \end{array}}}
\end{array}
$$
Combine them to create nitrogen oxide:$$
\textbf{Nitrogen Oxide} \\
{\Huge{
\begin{array}{rcccc}
\raise{0.1ex}{.} &\mathrm{N} & {\rlap{\raise{0.1ex}{: :}}} ~~ &\mathrm{O} & \raise{0.1ex}{:}
\\[-500px]
&\raise{0.25ex}{\cdot \cdot}  & & \raise{0.25ex}{\cdot \cdot}
\end{array}
}}
$$
Condense the representation by drawing the double-bond as in:$$
\textbf{Nitrogen Oxide} \\
{\Huge{
\begin{array}{rcccc}
\raise{0.1ex}{.} &\mathrm{N} & {=} &\mathrm{O} & \raise{0.1ex}{:}
\\[-500px]
&\raise{0.25ex}{\cdot \cdot}  & & \raise{0.25ex}{\cdot \cdot}
\end{array}
}}
$$
Identify the paired electrons,$$
\textbf{Nitrogen Oxide} \\
{\Huge{
\begin{array}{rcccc}
\raise{0.1ex}{.} &\mathrm{N} & {=} &\mathrm{O} & \raise{0.1ex}{\boxed{:}}
\\[-500px]
&\raise{0.25ex}{\boxed{\cdot \cdot}}  & & \raise{0.25ex}{\boxed{\cdot \cdot}}
\end{array}
}}
\rlap{
\qquad
\underbrace{\boxed{\Huge{\cdot \cdot}}}_{\text{electron pair}}
}
$$
then just go ahead and omit them, assuming that the reader is using typical electron-counting rules, such that they'll infer that those electrons are there without us explicitly drawing them, yielding$$
\textbf{Nitrogen Oxide} \\
{\Huge{
\begin{array}{rcc}
\raise{0.1ex}{.} &\mathrm{N} & {=} &\mathrm{O}
\end{array}
}}
$$
Finally, under the same logic for omitting assumed structures, drop the double-bond, yielding$$
\textbf{Nitrogen Oxide} \\
{\Huge{
\raise{0.1ex}{.}\mathrm{NO}
}}
$$

This gives us $`` \, \raise{0.25ex}{.}\mathrm{NO} \, " ,$ which could be further reduced to just $`` \, \mathrm{NO} \, "$ if we further assume that the reader knows about the unpaired electron.

I want to upvote this answer, but I can't condone the misconception stated in the last paragraph. The orbitals can't be treated like a cup with enough volume for two electrons and having only the bottom half full when there's only one; it's less incorrect to treat it like a sealed box that will hold one electron okay, and two under pressure, but blow its safety valve if there are more than that, so that the entire volume is still used but with less pressure when only one electron is present. Hence the dot should be located as if taking up all that space to itself, in the centre.

@Nij I'd be happy to correct it if that's the case, but do you have a reference or something?  I mean, I'd thought that our current models were that the electrons could have one set of quantum numbers identifying a stable orbital, and while resonance might mean that it's not consistently one orbital being filled, the actual electron orbital structures aren't well-described as being the average of the two, but rather their superposition.  But, you're saying that it's less wrong to see the electron's position as the average?

@Nij In other words, I'd say that it's more accurately represented as $\left[\sideset{_{\bullet}}{}{\mathrm{NO}} \rightleftharpoons \sideset{^{\bullet}}{}{\mathrm{NO}}\right] ,$ if we're talking about resonance.  But to say that a resonance structure is its average, e.g. ${\small{\bullet}} \mathrm{NO} ,$ just seems off to me.  I mean, that seems like saying that a benzyl ring doesn't have single- and double- bonds, but rather 1.5-times-bonds all around it, because that's kinda what it looks like if we average it over time.

In a sense, yes, as you don't know (unless by convention which may not be universal and hence either ambiguous or confusing, or by other specific means in a particular case representation) which "side" of the orbital the electron is (currently) occupying. Since the diagram is itself ambiguous between showing an average of the numbers {0,1} and the superposition of an occupied sub-orbital and an unoccupied one, it simply makes more sense to go for the middle.

@Nij That's inconsistent, though.  We don't draw benzyl rings with one-and-a-half bonds all around; we draw them with a single time-dependent snapshot.  Or we explicitly dot the lines to represent time-dependence.  But the halfway-drawn dot is a misconception; it doesn't exist at any time.  Superpositions can't be time-averaged.

Re: benzyl, really? I've always been taught to diagram it in the superposition when structured with the understanding that the three bonds are shared across the six places, hence being all 1.5 bonds each. Re: dot diagrams, additionally, having it centred allows for discussion of the difference between the electron occupying each of the orbitals and providing a representation for the different cases, compare the superposition benzyl with drawing the snapshot for diagramming electron movement in reactions.

Let us continue this discussion in chat.

source: Stack Exchange

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sort: votes


text:
How to reverse copper electroplating?
We did a lab where we built a simple electrolytic cell. In a beaker was a solution of $\ce{CuSO4}.$ A copper strip (anode) and a copper wire (cathode) was place in it. A wire connected the copper strip to the negative DC power source and a wire connected the copper wire to the positive DC power source. The power source was turned on and the copper wire was electroplated.
Considering that both electrodes are the same material, would it be possible to create a galvanic cell by reversing the reaction?
I think it would not be possible. Since galvanic cells use spontaneous reactions to create electrical energy, you would have these two copper electrodes in a solution of $\ce{CuSO4}$ separated by a salt bridge and connected by a wire. Nothing would happen because both coppers are the same and neither one of them can attract electrons more than the other.

The galvanic cells from the same material are possible. They are called concentration cells and they use two electrolytes of the different concentration of the same electrolyte.
But their voltage is low. In the case of $\ce{CuSO4}$
$$U = 0.0295 \log\frac{c_1}{c_2}\ \mathrm V$$

Note that the electrode naming is based on the direction of the electron flow, and their mutual polarity are switched for electrolysis versus cells (primary cells or by convention discharging mode of secondary cells ).
Anode = electrons flow upward (anabasis), an anode is the place of oxidition. (Xenophon, Anabasis, "The way upwards")
Cathode = electrons flow downwards, a cathode is the place of reduction (katabasis - the way downwards)
During electrolysis, the anode is the positive electrode and the cathode is the negative electrode.
During the cell discharging, the anode is the negative electrode and the cathode is the positive electrode.

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text:
What is the line cutting benzene ring?

There is a line cutting benzene ring what is that called.

The line is just a bond. You draw it like that when you are unsure or don't care where exactly it should connect to the benzene ring.

Just a clarification to @IvanNeretin's comment: It's when it may be of any configuration -- ortho-, meta- or para-substituted.

Related: Structural representation of substituted benzene and For a benzene shown in a skeletal structure, what does a substituent to the center of the ring mean?

The depiction at start is wrong as for there is only one monosubstituted benzene. Avoid that representation.


text:
What kind of algorithm does computational software use to process molecular symmetry?
Background
I'm forced into using GAMESS-UK for an older software package. Using a cs-symmetry molecule, I've noticed that GAMESS will reorient the molecule before running a standard SCF calculation. This is related to the algorithm the software uses to process the symmetry of the molecule. For instance, If I were to input my molecule as having C1 symmetry, no reorientation would occur; the scf calculation would run using the exact same coordinates from the input file.
For some molecules, I've noticed a rotation for the reorientation as large as about 45°. For other molecules, it might only rotate 5° from my input coordinates. This for the same system of molecules—all cs symmetry, same functional groups, only the alkyl backbone has a variable number of carbons. As a result of this, my molecular axis is being defined differently across my various output integral files. It'd be more convenient for me if that were possible to avoid.
I'd like to better understand what conditions determine the reorientation of the molecule. I realize Gamess-UK is old, so any kind of symmetry algorithm from other computational software would also be useful to know about. Maybe I can see if it's similar to the one I'm encountering.
Question
Does anyone know how computational software determines the reorientation of input coordinates prior to a calculation?

Isn't that also a source code distributed package? In that case you're probably more successful trying to read that.

I don't know the algorithm, but I believe the resulting structure is usally called the canonical orientation for the given point group

Have you read the manual? Most packages have detailed references related to the applied algorithms.

This question would fit in on the new Materials Modeling SE: https://materials.stackexchange.com/


text:
Synthases vs Lyases - nomenclature
Lyases are a class of enzymes that break down (lyase: ly- = lysis, -ase = enzyme) bonds like $\ce{C-C, C-S}$ and so on (except hydrolysis), and can eliminate molecules like $\ce{H2O}$ or $\ce{CO2}$. They are referred to as synthases, respectively, some authors categorize synthases among lyases.
My question is, (if word "synthesis" still means joining two substances into one more complex by creation of bonds between them), why are synthases placed among lyases if lyases actually break down bonds and synthases are (by their name) supposed to join, create bonds? Respectively, why was their name derived from 'synthesis' if their role actually is to break down?
Lastly I have even seen that JCBN (Joint Commission on Biochemical Nomenclature) dictated that "synthase" may be used for any enzyme catalysing synthesis, and that even without paying attention to whether the enzyme uses NTP like ATP (ligases = synthetases) or not (synthases).
I'd like to let everybody know that I understand the difference between synthases and synthetases, and that this difference isn't my point of question. I also do not controvert the fact that in the literature synthases are referred to as lyases, but somehow I kind of miss logic of naming enzymes responsible for breakdown after 'synthesis' which is actually opposite reaction than analysis. Until today I thought 'synthesis' stands for joining (syn-) and 'analysis' for breakdown into simpler products.
Here I attach in the images the answers from Wikipedia and forums for you to read.

Wikipedia says: Lyase catalyses breaking, elimination. And synthase catalyses synthesis process.

Enzymes always catalyze reactions in both directions (all catalysts do). When an enzyme is first discovered and named, it is not always clear which direction is physiologically relevant (and sometimes both are).
There is a nice answer on StackExchange Biology talking about the enzyme pyruvate kinase, which acts as a phosphatase (i.e. in the opposite direction than the name suggests) in glycolysis. In the lab, you can study the reaction in both directions, depending on the concentrations of species present.
For the class of enzymes you are talking about, the same idea applies. A lyase can work in the other direction, as a synthase, and vice versa. If you want to classify enzymes in terms of mechanism and structure, there is no reason to separate the lyases from the synthases. They are expected to be highly similar or identical (if we are just using two names for the same enzyme).


text:
Acid and base separation on an anion exchange column
I'm new with this concept and so I would need some advice.
I want to separate an acid and a base in a water solution. The acid has $\mathrm{p}K_\mathrm{a}(\ce{HA}) = 5$ and the corresponding acid of the base has $\mathrm{p}K_\mathrm{a}(\ce{HB}) = 7.$ The column is positively charged. What $\mathrm{pH}$ should I use in order to retain the acid in the stationary phase and the base passing the column?
Well, I'm just thinking that I would want the acid to be negatively charged in order to bind to the stationary phase, and so I would use a $\mathrm{pH} > \mathrm{p}K_\mathrm{a}(\ce{HA})$ because then the solution will gain a proton and leave the acid negatively charged. Right?
What about the base, should it be positively charged in order to escape the stationary phase and be attracted to the polar water?

Is the base neutral before it picks up a proton?. $$\ce{B + H+ <=> BH+}$$ Maybe make a table with pH range (0-5, 5-7, 7-14) and the charge of the two species you want to separate. I'm not sure you can make one negatively charged and the other positively charged at the same pH.

Also, consider whether the charge of the column changes (weak anion exchange column, e.g. primary amines) or not (strong anion exchange column, e.g. quaternary ammonium)

Can you be specific with your acid and base?

I don't have more information than this, only that it is anion exchange.


text:
NMR for an octahedral complexes sources of peaks
I wanted to ask a question about splitting pattern in NMR.
I was given the following information to assign to a Compound C and justify why the splitting patterns are as given:

I was confused about both the peaks.
I recall giving a previous question regarding NMR here.
Using the first bit of information

singlet in the proton-decoupled $\ce{^{31}P}$

I can't seem to understand why, because there is no plane of symmetry going across the plane of the octahedral 4 ligands (from left to right) so they wouldn't see "the same electronic environment."
Also, regarding the $\ce{^1H}$ I tried to justify this by saying that as there were 2 different $\ce{P}$ atoms in axial but they both faced different environments, the splitting would be triplet too, with the same splitting constant, hence a triplet, but I was unsure about this.
Why is the proton-decoupled $\ce{^{31}P}$ only displaying a singlet?

The phosphorus atoms are enantiotopic. Thus they have the same chemical shift.

@Zhe Hi there, I have just googled the definition of "enantiotopic" as being a proton in an enantiomer since I didn't come across this specific term previously. So, does this mean that as we are not in a chiral solvent, they will display the same shift? And one basic question, why is it enantiotopic? I just want to clarify this fully before I can move on!

One P looks up and sees four ligands in the plane, the other one looks down and sees the same arrangement, only mirrored. The two are chemically identical.

@Karl I spent years trying to work out a reason like that and I've just found it! Wow. Many many thanks! Add it as an answer so I can tick it, please!

Feel free to write the answer yourself. But you do see the mirror plane through the metal atom and the four ligands CO,CO,MeCN and H, right?

As a follow up to my first comment, the logic is that enantiotopic groups are in identical but mirrored environments, so their shifts are identical because the NMR environment is achiral. Groups with identical shifts cannot split each other.


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what is mechanism for reaction phenol + dimethylsulphate -> anisole
Dimethyl sulphate breaks the oh bond and forms anisole. How does the reaction proceed?? is it oxygen lone pair attacks on sulphur which has partial positive charge?? the reaction takes place in presence of naoh

Before we can answer your question or provide assistance, the expectation is that you make an effort to solve the problem. And no, dimethyl sulfate does not break the OH bond!

oh- ion from naoh takes away acidic hydrogen from phenol and forms phenoxide ion but in the next step will it attack the electrophilic sulphur atom ??

No, it attacks the carbon of one of the methyl groups

Treat dimethylsulfate as analogous to methyl iodide.

The methylation of Phenol using Dimethyl Sulfate is an Electrophilic Methylation and happens through a SN2 nucleophilic attack. Phenol is first deprotonated by the sodium hydroxide as you already thought. The deprotonation of Phenol is favored because the pKa of Phenol is lower than the pKa of water meaning $\ce{OH-}$ is a stronger conjugated base than $\ce{PhO-}$:

The formed Phenolate is a good nucleophile, and will attack the methyl groups of the Dimethyl Sulfate. Those methyl groups are readily attacked by nucleophiles because of the great induction effect caused by the neighbor atoms which are much more electronegative than the carbon atom, which makes those more positively charged. This makes the methyl groups good electrophiles, hence the reaction name "Electrophilic Methylation": If we focus on the methylating reactant, it's performing an Electrophilic Attack.
Below is the reaction mechanism:

EDIT: Thanks @user55119 for pointing out that only one methyl group of the Dimethyl Sulfate reacts under normal circumstances, requiring 1 equivalent of Dimethyl Sulfate per Hydroxyl Group and not 1:2 as I previously thought

In practice 2+ equivalents of dimethyl sulfate are used per phenolic group owing to the greater reactivity of the first methyl group. See,
R. N. Mirrington and G. I. Feutrill
Org. Synth. 1973, 53, 90, DOI: 10.15227/orgsyn.053.0090

I found that article, is that the one you mentioned? You're right, the procedure describes 0.344mol of Orcinol reacting with 0.750mol of Dimethyl Sulfate, close to 1:1 ratio for each hydroxyl group. So we would have $\ce{NaSO4CH3}$ amongst the products? I'll edit the mechanism image to account for that.

I will correct my comment forgetting that orcinol has 2 phenolic groups. It should read 1+ equivalents of dimethyl sulfate per phenolic group. Also, note in the link that an excess of conc. ammonia is added at the end of the alkylation and the reaction mixture is heated to destroy excess dimethyl sulfate.


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Why does ordinary water have a triple point?
Triple points and triple lines for pure substances and mixtures
My understanding is that pure substances can have triple points.  But mixtures of substances have triple lines, i.e. there isn't a single point but a one-dimensional path in $PT$ space where three phases can coexist.
This is understanding comes from the Gibbs phase rule that $F = C - P + 2$, where $F$ is the degrees of freedom, $C$ is the number of components, and $P$ is the number of phases.  At a triple point (or triple line) $P = 3$.   For a two-component system like a mixture of ethanol and water, $C = 2$.  If we apply the equation, then we find that in this case, $F = 1$, i.e. there is one degree of freedom.  If we pick $P$ then $T$ is defined for us by the constraints of any two-component, three-phase system.  That's a "triple line".  For a one-component system, the equation says that if $C = 1$ and $P = 3$, then $F = 0$.  A triple point.  We don't get to pick either $P$ or $T$.
Water isn't exactly a pure substance
But, ordinary water is not actually a pure substance.  It is a mixture of different isotopologues, e.g. $\ce{HDO}$ and $\ce{H2^{18}O}$, etc.  It is also a mixture of para- and ortho-isomers that differ only in the relative spin states of the two $\ce{H}$ atoms, as I learned from reading this very site the other day.
Does ordinary water have a triple point?
Pure $\ce{D2O}$ has a triple point distinct from that of $\ce{H2O}$.  For example, one web site says that $\ce{H2O}$ has a triple point of 612 Pa and 0.01 °C, while $\ce{D2O}$ reportedly has a triple point of 661 Pa and 3.82 °C.  But most (all?) water samples in the world are not 100% isotopically pure.  And very little work has been done separating para- and ortho spin isomers of water.
So given that water isn't a pure substance, how does it have a well-defined triple point?  "The" triple point of water was until recently fixed by the definition of SI units, so it seems that for most practical purposes the impurities in water don't matter much.  Why not?  What is the triple point of para-$\ce{H2O}$?  Of ortho-$\ce{H2O}$?  Of para-$\ce{H2^{18}O}$?  Etc. etc. etc.  Why doesn't the phase rule apply to these species?

Interesting question.  The Comité International des Poids et Mesures (CPIM) defines the triple point of water using VSMOW (https://en.wikipedia.org/wiki/Vienna_Standard_Mean_Ocean_Water), and VSMOW has multiple components.  So essentially you're asking, given the Gibbs phase rule, how the international standard for water's triple point can be based on a substance with multiple components.

When a solution of glucose in water boils or freezes, the composition of the solution changes because glucose stays in the liquid phase. I'm not aware that it is possible to enrich isotopologues of water by freezing or boiling (e.g. ice can accommodate the different isotopes and does not crystallize into separate crystal lattices). If that is the case, the composition of "pure" water does not change, so maybe the number of components is one after all.

@theorist From the wikipedia article you quote: "Fresh distilled VSMOW water is also used for making high accuracy measurement of water's physical properties and for defining laboratory standards since it is considered to be representative of average ocean water, in effect representing all water on Earth." So they remove the solutes first before doing any measurements.

@KarstenTheis  The OP's concern about 'multiple components' referred not to solutes, but to the multiple isotopic (and other) components of pure water itself.  And VSMOW's own definition (see, specifically, https://en.wikipedia.org/wiki/Vienna_Standard_Mean_Ocean_Water#Composition) provides a perfect example of this.  So yes, the solutes are removed (as much as possible) from VSMOW but this still leaves, by the VSMOW specification itself, its multiple isotopic  components, each with different physical properties.

Summarizing the relevant point (pun intended) of my answer to a related post: a mixture does not exhibit a "triple line" if its composition is constant. It is only a "triple line" insofar as you get to vary the proportion (mole fractions) of the components. Therefore different samples of water with different isotopic composition will differ in the value of the TP, but for a given sample the TP is just that, a single constant point (not a line).
This question however does point out a very important problem for the implementation of the standards adopted during establishment of an absolute temperature scale based on the triple point of water. A good discussion of this is provided in white papers (1,2) from Isotech (manufaturers of triple point measurement apparatuses):

There are 3 main sources of error in a water triple point cell[6,
7] (excluding errors associated with measurement):

The isotopic composition of the cell’s water may not be that of SMOW.

There may be air trapped inside the cell.

There will be impurities in the water.

Isotopic composition can be measured to great accuracy from a
small sample of the cell’s contents.

SMOW and V-SMOW are water samples of standard composition (this is amply discussed in Isotech white papers, referring also to the documentation behind the International Temperature Scale established in 1990 (ITS-90)):
\begin{array} {|r|r|}
\hline
\text{Component}  &  \text{Mole Fraction} \\
\hline
\ce{^1H} & 0.999 842 \\
\ce{^2H} (\text{deuterium}) & 0.000 158 \\
\ce{^3H} (\text{tritium}) & 0 \\
\ce{^{16}O}  & 0.997 640 \\
\ce{^{17}O} & 0.000 371 \\
\ce{^{18}O} & 0.001 989 \\
\hline
\end{array}
A second white paper provides guidelines on consideration of these effects (see the section "Isotopic Composition"):

We think of water as $\ce{H2O}$, more precisely we should write $\ce{^1H2^{16}O}$. This has been termed ‘light’ water and is unobtainable. If it were obtainable it’s triple point would be approximately +0.008°C. In practice what we drink from the tap, and what is in the ocean is a soup of $\ce{^1H}$, $\ce{^2H}$ with $\ce{^{16}O}$, $\ce{^{17}O}$ and $\ce{^{18}O}$. This means real water does not have a single triple point temperature. Its ‘combination triple point’ will vary as the isotopic composition varies. This  was  understood  more  than  50  years  ago  and  after  much  analysis  of  water  from  different  sources recent temperature scales state that  the water in a triple point cell shall have the isotopic composition  of ocean water.  Although not explicitly stated, there is a Standard Mean Ocean Water (SMOW) and it is assumed that the isotopic content of SMOW fulfils the ITS-90 requirements.

concluding:

Analysing the cell contents for isotopic composition is a useful and valid addition to the usual techniques of intercomparing cells and gives an accurate indication of the cells ideal triple point value.
[...]
Such cells, complete with isotopic analysis provide the closest association to the ITS-90 value of the water triple point and hence the KTTS Scale. It is now possible to produce cells, based on Stimpson’s original design which can be associated to V-SMOW within ±3mK and this is a smaller uncertainty than can be provided by traditional measurements.

References

The Establishment  of  ITS-90 Water Triple Point References  to  ±2μK,  and  the Assessment of 100 Water Triple Point Cells Made Between 2001 and 2006, John P. Tavener and Nick Davies, Isothermal Technology Ltd. (England).

The Water Triple Point - A Reference Cell Close to the ITS-90 Value. John P. Tavener & Anne Blundell. Northern Temperature Primary Laboratory, Pine Grove, Southport, Merseyside, PR9 9AG, England.

Supplementary Information for the International Temperature Scale of 1990, BIPM. ISBN 92-822-2111-3.

Very nice answer, thank you.  For the mixture composition to be constant, doesn't that mean the composition of each phase at the triple point (ice, steam, and liquid water) must be the same as the other two?  At the triple point, would that really be true?  Or am I misinterpreting what you are saying?

@CurtF. Basically all the phases have to remain at constant composition but they don't have to be equivalent, only each component has to have balanced chemical potential across phases according to Raoult's law or equivalent (for practical purposes). In practice you can have problems (see the isotech white papers) with preferential evaporation or freezing of a particular isotopic consituent, so that you essentially deplete that component from your solution (same principle as in distillation). If you operate in a closed chamber apparently there are ways around this, to obtain reproducibility.


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NMR of trigonal bipyramidal molecules
I'm slightly confused about NMR.
I seem to remember at school that if a molecule is symmetrical in $\ce{^1H}$ NMR, that only one peak would be shown e.g. propane's terminal $\ce{-CH3}$.
Consider the two molecules shown below.
My professor remarked that both these isomers would show a doublet and triplet in their proton-decoupled $\ce{^31P}$ NMR and I was struggling to work out why.
Consider the molecule on the left as molecule A and the right as molecule B.
My train of logic was that the axial $\ce{Ph3P}$ group on molecule A would be split by each $\ce{P}$ atom that is equatorial, so using the expression $2nI + 1$ where $I = \frac{1}{2}$ so our singlet would be split into a triplet.
Likewise, the equatorial $\ce{Ph3P}$ groups would both only see one axial $\ce{Ph3P}$ group, hence only a singlet would be observed.
I used the same logic on molecule B.
Is this the correct reasoning for the doublet and triplets observed in both molecules?

“Likewise the equatorial Ph3P groups would both only see one axial Ph3P group, hence only a singlet would be observed” - I think you’ve just made a little mistake. The equatorial groups can see one phosphorus other than themselves, so (2nI + 1) = 2, not 1!


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How do you remove lumps in reagents without contaminating them?
I think desiccating the powdered reagent would help remove lumps (at least in preventing lump neoformation) but would desiccation alone collapse all lumps in a powder (say KCl as it lumps easily), or would the lump formation itself prevent desiccation of the molecules at their core?
What about vacuum?
The challenge is in doing it without adulterating the reagent and I'm not sure how to do that.
Ideally I'd vacuum heat the powder inside of a glass grinder but the budget for this is probably far beyond my means.

This question is btoad and is not clear what the requirements are. Do you assume lumps are due to moisture and dessication is needed? Reaction involves solutions, so lumps are not a big concern. Solid state chemistry requires grinding itself. Etc. Clarify the contest, such as scale, chemicals, purposes. ...

I don't really know what causes lumps, so I'm welcoming any knowledge on the matter (I just assume moisture because I've seen powders lump in moist environments). I probably used the wrong word when writing "reagent": I mean any pure chemical species in solid (powdered) form. How do you grind a powder without contaminating it with whatever matter the grinder is made out of (if not also biological contaminants from it)? Is there any other way than grinding, that would just make the lumps collapse back into fluid powder?

All this depends on what purity is required. The levels of your precautionary concerns is indeed high.  Normally chemicals are supposed to come in contact with quite some containers and tools whatever operation they are submitted to. Assuming that one knows materials and goals, lumps can be ignored, manually crushed by mean of a pestle, milled in machinery, and so on. Yes, most often lumps are caused by moisture, via surface dissolution and merging - and I would guess other interface mechanisms.

@Alchimista: I guess for small quantities a glass pestle would be ok (and they are quite cheap). Do you know if there is machinery that would mimic that (e.g. grinding between rough glass cylinders or something)? What are my contamination risk with that? Si? Bo? I'm wondering if softer or harder materials contaminate the most (the former by a sort of soft erosion and the later by brittleness)?

One cannot really answer without knowing what is the aim. Making basil pesto and pharmaceuticals are totally different things. I guess in your case ceramic or glass pestle might work. It really depends on the level of purity of your chemicals as well as the final purity. At this level of concerns, the vast majority of chemistry labs would be totally unable to do their work.

Finally, comments are normally intended to help improve the question, not to start a discussion.


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Mechanism of DPD (N,N-diethyl-p-phenylenediamine) reaction with free chlorine
NOTE: As I've been finding new informations about the question, I've been editing the question to keep it up to date. If some comment doesn't make sense just check the Edit History.

DPD (N,N-diethyl-p-phenylenediamine) is used in free chlorine colorimetric analysis since it reacts with Hypochlorous acid and Hypochlorite ions acquiring a pinkish color that can be quantitatively related to the concentraction of those substances if it's inside the working range.
Even though the analysis method is very well documented, I haven't found much about the reaction itself and its mechanism, and since it involves lots of subjects I was recently studying (like the relationship between π-systems and color in organic molecules) I thought it would be interesting.
First taking a look at the indicator molecule:

There are two amine groups: a primary and a tertiary one. I don't have software here to run a Molecular Orbital analysis, but I'd guess the amines lone pairs plays a big role in the HOMO (probably the tertiary amine a little more since it's more basic). The carbons in the aromatic ring probably also influence the energy and shape of the HOMO since they are very nucleophilic due to the electron density added by both nitrogens lone pairs to the π-system. I'd guess the biggest contributors for the LUMO would be the hydrogens connected to the primary amine and the ethyl groups connected to the tertiary one, since they are more positively charged due to induction.
The reagent is mostly used in its salt form, as N,N-diethyl-p-phenylenediamine sulfate for example.
According to some sources[1] [2], the final products of the DPD oxidation by the free chlorine are the following ones:

The first one (the free radical), being called Würster Dye, is responsable for the pinkish/magenta color that is used to measure the amount of free chlorine in the solution. The second one (Imine) is colourless, and is favored on more strong oxidizing conditions, reason why the DPD-Method can't be used in very high concentrations of free chlorine.
I decided to test this new information and ran a DPD-Analysis on very concentrated sodium hypochlorite and then on wastewater dosed with sodium hypochlorite.
Concentrated Sodium Hypochlorite:
Sample initially yellowish and photometer baseline set before reaction. After adding DPD the color seemed to be a slightly more bright yellow. Measurement of free chlorine was about 0.35mg/L.

Chlorinated Wastewater:
Sample initially colourless and photometer baseline set before reaction. After adding DPD the color changed to a light pink, as expected by the method. Measurement of free chlorine was about 1.50mg/L.

The test confirmed the information on both cited articles that excess free-chlorine would result in the colourless Imine product being favored.
As for the mechanism, I'm still unsure of how it goes. First I thought about an homolytic cleavage of the HO-Cl bond and the N-H bond of the tertiary amine. Later the hydroxyl radical would combine with the hydrogen radical to form water and the chlorine radical would combine with another chlorine radical to form $\ce{Cl2}$. But that doesn't make sense because we would end up replenishing the free-chlorine as the reaction goes. So instead of the hydroxyl radical combining with the hydrogen one, maybe the chlorine radical is the one combining resulting in the release of $\ce{HCl}$. But then we end up with a hydroxyl free radical. It could recombine with another hydroxyl radical forming $\ce{H2O2}$ which would further decompose to $\ce{H2O + O2}$ but it sounds a little odd for some reason. Despite sounding odd it's still the best I could come up with, so I draw the mechanism below:

As for the Imine mechanism, would the following be correct?

I'm not sure about neither, and I'm very confused as how a radicalar reaction takes place instead of the heterolytic one if there's no external influence to favor the formation of a free radical (I always thought we would need UV or a free-radical catalyst to favor a homolytic fission).
As for the HOMO-LUMO gap getting smaller, I guess it makes sense. When we think for example, about the combination of two hydrogen radicals, there's a lowering of the occupied orbital energy. So maybe the formation of a free-radical raises the energy of the HOMO but doesn't raise the LUMO energy of the DPD making the gap between them smaller, reaching visible light spectrum.

So the questions: Are any of those mechanisms correct? If not, what is happening on the DPD/Free Chlorine reaction? Is my reasoning on the HOMO/LUMO gap getting smaller right?
Side-note: The original article that described the DPD method for the first time apparently was this: PALIN, A.T. 1957. The determination of free and combined chlorine inwater by the use of diethyl-p-phenylene diamine.J. Amer. WaterWorks Assoc.49:873. But I'm not sure if there's an open version of it. If there is I couldn't yet find it.
[1] CHLORINE ANALYTICAL TEST ELEMENT AND A STABILIZED N,N-DETHYL-P-PHENYLENEDLAMINE SOLUTION - Patent Application Publication, Pub. No.: US 2015/0050195 A1 - https://patentimages.storage.googleapis.com/ec/5e/50/485b22cea62cdc/US20150050195A1.pdf
[2] 4500-Cl CHLORINE (RESIDUAL)* - http://edgeanalytical.com/wp-content/uploads/Inorganic_SM4500-Cl.pdf

do you have any evidence for the final product being what you drew?  I was wondering about dimerisation via the first chloro compound you drew ... the resulting hydrazine dimer being oxidised up to a N=N compound ... anyone else?

@user1945827 No evidence, I haven't found much information on the reaction, so all the mechanisms and products on the question are mostly deductions. I've just stumbled into an article that I'll examine better, but it indicates a reaction of a protonated DPD salt (which aligns with the fact DPD packages are sulfate salts) with actual $\ce{Cl2}$ to form a radicalar product called "Würster Dye". I'll update the question as soon as I get to read the article enough to have better conclusions.

The mechanism after "As for the Imine ..." is improbable. Loss of hydride from a protonated amine?? Just chlorinate the -NH2 once and then eliminate HCl. As to the structure of the red dye, it apparently remains a mystery.

@user55119 I also thought it would be a bad mechanism, but at first glance I couldn't see how to keep charges without having one of the substituents to leave as a negatively charged species. But you're right, by first chlorinating the primary amine we add a more electronegative atom to the nitrogen making it easier for it to leave negatively charged. I updated the mechanism for the Imine on the question, is it more accurate?

At para0037 they say that the imine is colourless.  That takes you back to the dimer, CCN(CC)C(C=C1)=CC=C1N=NC2=CC=C(N(CC)CC)C=C2 I suggested in my previous comment.  Further this cpd is in the literature.

This reference may be of interest, https://link.springer.com/chapter/10.1007/978-3-662-12512-0_24 I haven't translated it with sufficient rigour to say one way or the other.

@user1945827 I only managed to read the preview of the article you linked, still translating the parts I can access. Do you think this dimerization happens with both products (free radical and Imine)? The free radical seems to be relatively stable, enough to give the method precision. Next time I make the test I'll see if I can leave the final solution sitting long enough to check if the colorful compound is being slowly consumed (maybe through the dimerization process you mentioned).

Hi ... I think you should still explorer the 'Wurrster dye' and as a competing hypothesis the dimerization which I would guess happens with the starting material.  You need to gather more evidence, either through the literature or empirically, for example this paper https://pubs.acs.org/doi/abs/10.1021/ja01249a026 discusses Wurster dyes as does this - both useful.  BTW, Alexandra Elbakyan's work may be of use to you.

@user1945827 I’ll read the work, but I have just ran a small experiment that seems like an indication that you’re right in saying the dimerization happens, at least in some conditions. I added DPD to a chlorinated solution of relatively higher conc., but not enough to make the solution yellow (solution was colourless). After adding the DPD it precipitated as a very pinkish solid, then after mixing the solution the precipitate dissolved but turned the solution yellow. The DPD dimer is probably yellow.

Isolate and characterise the dimer, see if it turns up in your reaction system by LC, try and quantify it against oxidant added.  There seem to be a lot of old chemistry of poorly understood chemistry -  a treasure trove if you're into that sort of thing.

@user1945827 I wish I had the lab equipment to do that. I was able to do the small experiments because I use the DPD reagent frequently at work, but besides the material to run the method itself, I’m very limited :/

Note that hydrogen peroxide is slow to oxidize DPD except at high concentrations but using peroxidase cleaves at the oxygens to form hydroxyl radicals.  As noted in the cited paper [1], this can be used to quantify hydrogen peroxide as it becomes activated to oxidize DPD.
So the hydroxyl radical from hypochlorous acid can form water from the hydrogen on DPD leaving the radical on DPD.  This leaves a chlorine radical which can react with another chlorine radical from a second parallel reaction.  The resulting molecular chlorine in water forms hypochlorous acid and hydrochloric acid.  The net result is the following (the HOCl doesn't normally split to form radicals except by UV photolysis so I show a direct attack instead):
$\ce{2x (HOCl + H-DPD -> H2O + DPD• + Cl•)}$
$\ce{2Cl• -> Cl2}$
$\ce{Cl2 + H2O -> HOCl + HCl}$

$\ce{HOCl + 2H-DPD -> H2O + 2DPD• + HCl}$
Normally hypochlorous acid ($\ce{HOCl}$) oxidizes in a 2-electron transfer but the above has it oxidize via a 1-electron transfer via its hydroxyl radical.  Having two of these reactions occur creates molecular chlorine that then disproportionates into hypochlorous acid and hydrochloric acid but since you started with two hypochlorous acid the net effect is that two reacting DPD behave together like a 2-electron transfer reaction converting hypochlorous acid into hydrochloric acid.
Another way to think about the initial steps in this reaction is as a chlorine substitution reaction that then cleaves into radicals.  That is,
$\ce{HOCl + H-DPD -> H2O + Cl-DPD}$
$\ce{Cl-DPD -> Cl• + DPD•}$
Normally chlorine substitution reactions occur with amine groups, but the nitrogen in this case is cationic so the hydrogen is more easily substituted but the chlorine being electronegative doesn't bond well with the cationic nitrogen.
Reference
[1] : Chang, Q., Deng, K., Zhu, L. et al. Determination of hydrogen peroxide with the aid of peroxidase-like Fe3O4 magnetic nanoparticles as the catalyst. Microchim Acta 165, 299 (2009). Doi: 10.1007/s00604-008-0133-z


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How does 1-bromo-2-fluorobenzene react with lithium amalgam?
How does this organometallic reagent react with 1-bromo-2-fluorobenzene? Does it react like Grignard reagent? If the product of this reaction is then reacted with furan, what will be the final product? I can guess there will be a benzyne intermediate, but I am not sure.

At what temperature is this reaction taking place?

Benzyne does a Diels-Alder with furan

I'm voting to close this question as off-topic because your question lacks context: This question needs revision before it is ready for a great answer. Please edit to include how this question came up and how you tried to answer it. This will help writing an answer that is useful for you and for others

Quoting from Page 1063 of Advanced Organic Chemistry FIFTH EDITION, Part B -
Reactions and Synthesis, FRANCIS A. CAREY, and RICHARD J. SUNDBERG:

Benzyne can also be generated from o-dihaloaromatics. Reaction with lithium amalgam or magnesium results in the formation of transient organometallic compounds that decompose with elimination of lithium halide. o-Fluorobromobenzene is the usual starting material in this procedure.

As @orthocresol said in the comments that Benzyne undergoes Diels-Alder reaction with furan.
The first Diels–Alder reaction of benzyne with furan was performed before giving a 76% yield of the adduct, which was converted to 1-naphthol with acid${^1}$ ${^2}$.

From Diels-Alder adduct to 1-naphthol

In the absence of the furan trapping agent, biphenylene and triphenylene were obtained in yields of 24 and 3% respectively${^3}$.

References
${^1}$:https://www.researchgate.net/publication/273556121_The_Benzyne_Story
${^2}$ :Shashidher et al., Pharm Anal Acta 2011, 2:7
DOI: 10.4172/2153-2435.1000137
${^3}$ :https://www.researchgate.net/publication/273556121_The_Benzyne_Story

It is worth noting that  F is a moderate directing/stabilising group for lithiation so ortho-lithio- fluorobenzene are stable at dry ice/acetone temperatures.


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What value should I use for P in the ideal gas equation for an unknown molar quantity?
I did a chemistry lab, in which baking soda and vinegar were reacted in a ziplock bag, and I was to calculate the proper quantity of reactants to fill the bag with CO2. The molar quantity of gas was found using the ideal gas law, where volume, temperature, and pressure were all known. Pressure, however, was set to the rooms air pressure, and I was wondering why that would be the case. Wouldn't a plastic bag full of gas have a fairly high pressure? If its pressure were in equilibrium with the air, wouldn't none of the gas flow out when the bag is opened?

Strictly speaking, the pressure inside must be certainly higher inside due to the restoring forces generated by the stretching of the plastic bag. It is this force that causes the gases to gush out when punctured or opened.

Bottom line: the assumption of the pressure inside being equal to atmospheric pressure can be considered correct with a considerably large error margin, provided we are talking about a ziploc bag and not the tyre of a tractor.


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What is the cause for the difference in pH between NaCl and Ringer Lactate
Reading this article, I came across a passage which got me to think of this question:
"Normal saline contains approximately $\pu{154mM}$ of each $\ce{Na+}$ and $\ce{Cl-}$ with average pH of 5.0 and osmolarity of $\pu{308mOsm/L}$. LR solution has electrolytes $\pu{130mM}$ of $\ce{Na+}$, $\pu{109mM}$ of $\ce{Cl-}$, and $\pu{28mM}$ of lactate. The Average pH of LR is 6.5 and is slightly hypo-osmolar with $\pu{272mOsm/L}$. 6,7"
Regular $\ce{NaCl}$ solutions should have 7 as a pH, yet this solution is 5.
A lactate ringer which has a similar composition to regular $\ce{NaCl}$ is 6.5.
I thought that the pH of 5 doesn't have anything to do with the $\ce{NaCl}$ but with the reaction of blood $\ce{CO2}$ with the water in the $\ce{NaCl}$ solution lowering the pH, but clearly that was wrong.

Note that pH of water in equilibrium with air is expected to be about 5.6. NaCl itself does not cause pH lowering, but CO2 and possible impurities do.

Note that presence of lactate as salt of weak acid breaks the similarity from acid-base equilibrium point of view.


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Why are yields of some chemical reactions so dependent on the stoichiometry of the reagents?
Back in chemistry class in high school I was working with a partner doing a chemical reaction. We measured the amounts of reagents well, but not perfect, and our yield suffered greatly. On the other hand my friend and his partner came up with quite a high yield. He told me it was because they measured each amount of reagent exactly.
So I wonder... Why are these reactions so dependent on perfect measuring of reagents? Why can't you just use more than enough of one reagent to completely react with the other and remove the excess reagent?
Also does this apply to all chemical reactions, or just a select few?

It depends.....

And I'll say your friend and his partner got lucky. Or rather you did some small but crucial mistake they didn't. A lot of reasons come to mind, but exact stochiometry of the reagents  is not among the first.

The proper stoichiometry may not be the only reason for the friend's success, but also the overall preciseness of their preparation procedure.
Chemical reactions are not generally that much critical, but there are lot of  cases, that are.
There are few aspects to consider:

Excessive reagents may affect the effectiveness of the product isolation.
Excessive reagents may alter chemical environment in final mixture, affecting formation of products.
Excessive reagent may not remain, but may form another product, instead of the desired one

Typical case for 3. is e.g reaction of $\ce{NaOH}$ with $\ce{H3PO4}$. Phosphoric acid forms 3 salts.
$$\ce{NaH2PO4 + Na2HPO4 + Na3PO4}$$
Let suppose you want to produce
$\ce{NaH2PO4 }$
$$\ce{NaOH + H3PO4 -> NaH2PO4 + H2O}$$
Extra  $\ce{NaOH}$ will create from the wanted salt  other salt...
$$\ce{NaOH + NaH2PO4 -> Na2HPO4 + H2O}$$
... and the amount of the final product decreases.
More detailed answer depends on providing your particular scenario.


text:
Addition v/s elimination with double bond and OH together

If a compound has both an hydroxyl group and a double bond and we add $\ce{HBr}$ to it,
(a) substitution reaction to the hydroxyl group can take place, or
(b) an addition reaction of a double bond can happen.

How to decide which one will occur?
What kind of conditions favour which type? Eg : How are the products affected by different conditions like solvent medium , Temperature, Structure , etc

Which is more likely to protonate first, the hydroxyl group or the double bond?

@Waylander I don't know..., I read organic from a book , and these were in 2 different sections...

It is important to specific the context. Is it dry HBr gas or aqueous HBr? Is it carried out at high temperature or is the reaction done at room temperature?

@TanYongBoon Actually , I have read about these 2 reactions in different sections of a book. There was no combined form given. But in my tests , these questions will surely be asked. Can you please tell me how would the different conditions like Temp/Solvent affect the reaction. I would be really grateful. Thank You

In almost all conditions the hydroxy group will protonate first. You then need to consdier whether an external or internal nucleophile is more likely to react with it.


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Bond lengths in P₄O₁₀
According to JD Lee's Concise Inorganic Chemistry:

The double bond $\ce{P=O}$ in $\ce{P4O10}$ contains p(π)–d(π) backbonding, where a full p-orbital on oxygen overlaps with an empty d-orbital on $\ce{P}$.

The four oxygens around each $\ce{P}$ form a tetrahedral shape. Then one lone pair from  $\ce{P}$ in it's $\mathrm{sp^3}$ orbital will donate to an empty p-orbital on $\ce{O}$ (created by pairing of electrons in it's remaining two p-orbitals), only then can the "doubly-bonded" oxygen attain it's octet. There is no "single electron from each" bond here. Is my understanding correct?
In earlier similar questions (Hybridisation of POCl3) and (Bonding in the phosphate ion) the d-orbital theory has been said to be disproven, but then how do you explain the different bond lengths in the double (1.43 Å) and single (1.65 Å) $\ce{P-O}$ bonds?


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How can we find out the type of bonding present in a compound or a molecule?
For example, if the bonding in diamond, ice, MgO or CO2 is to determine, how can I find it out? How to determine van der Waals forces?

By learning chemistry, you'll start to recognize that some compounds are rather covalent, and others are more on the ionic side. As for van der Waals forces, they are always present.

The basic types of chemical bonds are learned already at school: atomic bond (single bond, double bond, triple bond), ionic bonding, atomic bond with partial ionic character, metallic bonding. The first three mentioned types depend on the difference of electronegativity difference. Single bond, double bond and triple bond depend on the outer electrons (valence electrons) of the atoms involved in the chemical bond.
Look up these keywords in Wikipedia.
At high school, you learn the octet rule and that there are also double bonds and triple bonds with partial ionic character.
At studying chemistry at the college, you learn the chemical bonds in the main chemical compounds. You learn that for some chemical bonds, more complex models are required that depend on the orbital structure. You also learn that the chemical bonds are only models for describing some facets of the chemical compounds and that for some chemical bonds there is no adequate model.
See Wikipedia: Chemical bond.
Van der Waals forces are only one kind of intermolecular forces. There are models for calculating them. In principle, intermolecular forces as a whole can today be measured by atomic force microscopy.


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Decrease in molar conductivity with increase in concentration
Why does molar conductivity $Λ$ for strong electrolytes decrease with increase in concentration? I mean, more concentration would mean higher number of solute particles present in the solution, and as it is a strong electrolyte all of them would  disassociate and can help in conduction of electricity, so the $Λ$ should ultimately increase, but $Λ$ is given by
$$Λ_\mathrm{m} = Λ^°_\mathrm{m} - Ac^{1/2}$$
What am I missing here?

If I understand the question correctly, first, it's getting too crowded in the solution as the concentration rises (so that solvated ions interfere with each other), and second, note that this is an expression for molar (hence subscript "m") conductivity.

The equation you quote is the empirical relationship discovered by Kohlrausch (1907) for the conductivity of strong electrolytes, and is valid for concentrations up to $\sqrt{c} \approx 0.3$. $\Lambda_m$ is the molar conductivity usually given with units $\mathrm{\Omega^{-1}\,cm^2\,mol^{-1}}$ ($\Omega$ = ohm).  The $\Lambda_m^0$ is the molar conductivity at infinite dilution which effectively means that of pure water.
The reason that the conductivity falls with concentration is due to inter-ionic forces on the ion mobilities. There are two related effects, the Electrophoretic and Relaxation Field effects.
Each ion is surrounded by a 'cloud' of oppositely charged ions and when a field is applied there is simultaneous movement of an ion in one direction and of the opposite movement of the ionic atmosphere; the electrophoretic effect. Both the central ion and its atmosphere drag solvent molecules along with them  and this results in a retardation of the movement of the central ion. The effect can be evaluated using Stoke's Law $v=f/6\pi\eta L$, where velocity is $v$ and force $f=zeE$ for an ion of charge $z$ in field $E$. The solution viscosity is $\eta$ and $L$ the Debye length which is the effective radius of the ionic atmosphere, typically a few nm in millimolar solution. The velocity is identified with the deceleration of the ion caused by the ionic atmosphere moving in the opposite direction.
The second effect is a relaxation effect. Suppose the central ion is randomly moved, the overall spherical symmetry that existed before any movement occurred will be broken, i.e. the central ion is asymmetrically located with respect to its atmosphere, and because of this a 'relaxation field' is set up. The symmetry is restored by the effect of the relaxation field which takes only ten to hundreds of nanoseconds.
When the ion is under the influence of the applied field the time average of the forces on the ion are therefore no longer zero. The external field driving the ions through solution is reduced by the continual effect of the relaxation field. This seems strange since this field relaxes so very rapidly compared to motion of the central ion through its atmosphere, $\approx 10^{-3}$ sec, however, its magnitude is huge $10^5$ V/cm compared to the applied field and even small changes in symmetry do have a measurable effect in acting against the external field.


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What is the mechanism of the reaction of ethyl prop-2-ynoate to ethyl (2Z)-2,3-dibromoprop-2-enoate?

It's really weird for the product is 99％ Z-comformation?
Please tell me why?
This reaction can be found in scifinder.

Welcome to chem SE, try to add some attempt at the question and don't just ask for an answer. let us know your thoughts on the problem

In regards to the question, you are adding one molecules of $\ce{Br2}$ across the triple bond to leave a double bond. Hence it is an electrophilic addition reaction

But why the product is z-comformation is the problem as in usual the product is e-confirmation

I would assume steric hindrance will have a large factor to play in the reaction as the bromine is a large-ish thing to have on the side of the molecule when next to the carbonyl

A search of SciFinder will also produce the formation of the (E)-dibromide 3. The conditions described for the formation of the (Z)-dibromide 2 are taken from a paper by Myers1 as reported earlier by Tippett.2 Rossi3 has reported the formation of the (E)-dibromide 3 using pyridinium hydrobromide perbromide.
Given the more vigorous conditions for the formation of the (Z)-isomer 2 compared to the milder conditions for the formation of the (E)-isomer 3 suggests that the former reaction is under thermodynamic control while the latter is kinetically controlled.  There is the distinct possibility that bromine radicals may effect the isomerization via species 4 as may surreptitious acid.

1) A. G. Myers, M. M. Alauddin, M. A. M. Fuhry, P. S. Dragovich, N. S. Finney, and P. M. Harrington, Tetrahedron Letters, 1989, 30, 6997.
2) R. G. Hall and S. Trippett, Tetrahedron Letters, 1982, 23, 2603.
3) F. Bellina, A. Carpita, M. De Santis and R. Rossi, Tetrahedron Letters, 1994, 35, 6913.

Although the following reaction is not syn dibromination, the following reaction sequence is a example of syn dichlorination from https://www.semanticscholar.org/paper/Catalytic%2C-stereospecific-syn-dichlorination-of-Cresswell-Eey/a53c83d35d5acf0a7ac1311bd9462dedd2dc666f .
The possible mechanism is follows.


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HOMO orbitals of B-N from an MO diagram
I want to ask a question about the MO diagram of BN.
I was provided with the following question and answer and describe the situation occurring:

After performing the s-p mixing for BN, I found this diagram worked as the $\ce{1\pi}$ MO was the HOMO, but I thought using atomic structure knowledge that as B and N are in the same group, and N has a greater nuclear charge (and hence greater effective nuclear charge) its orbital size would be smaller, not bigger as the diagram shows above.
Am I right in saying that the label above for these orbitals are wrong, so it should be showing $\ce{N-B}$ from left to right?
The whole question is given here below

If we knew the other answer options, we would be better equipped for making guesses as to what was the author's intention.

@IvanNeretin done!

Great. Now what do they really mean by orbitals' size? You think it is the physical size of the corresponding orbitals. Maybe so, but there is another possibility. In many situations, people draw their AOs bigger or smaller to reflect the contribution of the corresponding AO to the MO.

@IvanNeretin So,how should I decide which AO contributes more? Please give me a few more hints to solve this problem

@yusuf the lowest energy orbital with a given symmetry favors the atom whose own contributing orbital is itself lower (typically the more electronegative atom), in this case nitrogen as opposed to boron.  Here we have that with the lowest energy pi orbital.


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why lithium is less reactive than sodium?
Lithium lies above sodium in a group and is also smaller in size. According to periodic trend reactivity decreases from left to right in period and down the group.

Reactivity decreases down the group for halogens and increases down the group for alkali metals.
As you have said, Lithium is smaller than Sodium, the valence electron on sodium is further away from the positive nucleus and therefore it has a weaker electrostatic attraction and is more easily lost in a chemical reaction, making it more reactive. This is due to the shielding effect of the second energy level (electron shell) which makes the valence electron on Sodium further away from the nucleus and hence the larger atomic radii.


text:
Does Suncream have any affect or help protect against radiation
So I was just watching the HBO series Chenoybl and saw that people suffered various degreesvof burns from radiation form Uranium 235 and some were very minor and looked like a normal sun burn.
My question is, does Suncream protect against radiation burns?

There are quite a few different kinds of ionizing radiation, to begin with.

I suspect the general answer would be no as the suncreams are designed to protect from UV-light only (and most of them only cover this range partially).

Your best bet would be to make yourself a nice lead coat (insulated by a coating on the outside not to expose yourself to lead either), though that wouldn't protect you from inhaling and assimilating radioactive compounds, and many of them would not be excreted immediately, accumulating in your thyroid (in the case of radioactive iodine), or for a very long term in your bones.

The sunscreen would have tiny effect against Alfa radiation, as any other matter, because Alfa gets absorbed in fraction of mm. The bad luck is that the most of the radiation from uranium fission is beta and gamma radiation that reach much deeper, especially the latter.

The only radiation sunscreen protects you against is near UV light not radioactive emission
Sunscreens are designed to protect from the near UV radiation present in sunlight. Some block both UVA (315-400nm) and UVB (280-315nm) light but some don't work well for UVA. All UV light at ground level is harmful to skin to some extent and too much of it will cause skin damage that looks a little like a burn (though small amounts are good for encouraging vitamin D production).
Radioactivity can cause similar looking burns. But the cause of the burns is not near UV light but gamma rays (hard X-rays of much shorter wavelengths and much higher energy than UV light) and alpha and beta particles. All of these are much more dangerous than UV light (as the amount of energy deposited in tissues is far greater per photon or particle).
And sunscreen will offer no protection at all to radioactive emissions. Clothing can protect against alpha particles. But you will need several millimetres of aluminium to stop beta particles (and even then the secondary gamma rays from the interaction of those particles with the aluminium can be harmful). And gamma rays might need a thick chunk of lead to keep them out.


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Colour of metal complexes with the same type and number of ligands
Crystal Field Theory says that because of d-orbital splitting caused by different ligands we have 2 different energy levels for d-orbitals.
It also says that transition metals are coloured because photons excite electrons, and when they have the crystal field splitting energy through a specific wavelength of light, the electrons to go to the higher energy d-orbital.
When electrons come back down they emit the same wavelength of light and that's the colour we see.
But then why is $\ce{[Ni(H2O)6]^2+}$ green and $\ce{[Cu(H2O)_6]^2+}
$ blue? They have the same type and number of ligands.

The Cu2+ and Ni2+ complexes, while having the same ligands, have different amounts of d-electrons (8 for Ni and 9 for Cu), meaning their ∆o value is going to be different.

For both complexes the t2g is completely filled meaning the ∆o only increases based on the electrons in the eg. This is why Cu2+ will absorb higher energy light (orange light) in order to excite an electron to the eg orbital where as Ni will absorb lower energy light (red light) for its promotion. I believe this change in ∆o is rather minor though in comparison to ligand effects and charge and this may be one of the few cases where the change really is due to the number of d electrons. Also keep in mind the ∆o increases as you go down any group of the periodic table which is why you often see more low spin complexes.
Someone please correct me if I'm wrong here, but this is at least my understanding of it.

When electrons come back down they emit the same wavelength of light and that's the colour we see.

I think this is a major misconception held by students. The color of these transition metal complexes is due to the absorption of light in the visible region. Very crudely, imagine you have six colors in the rainbow, if you remove a certain color portion from that rainbow, the transmitted light is no longer white to our eyes. For example, copper (II) absorbs red, that is why it appears blue to us. Yes, the electron has to relax back to its ground state, it does so non-radiatively.
The spacing (∆o) in the energy level determines the wavelength which will be absorbed.

but if 2 complexes absorb the same wavelength of light wouldn't they show the same colour?

@muhammadhaider yes, but the colour depends on the exact energy difference which will, in general, not be the same even for different metal complexes with a similar structure.


text:
Bond angle and bond lengths in Arginine

Based on the Lewis structure of arginine, the terminal primary amine attached to the guanidine group seems to be $\mathrm{sp^3}$ hybridized and trigonal pyramidal. Also, in quantum mechanical computations, it is shown  that all carbon-nitrogen bonds in the guanidine group are different lengths.
However, in the 3d structure below, the entire guanidine group is planar, indicating an sp2 hybridization. Why is $\ce{N}$ $\mathrm{sp^2}$ and not $\mathrm{sp^3}$ hybridized? Why is the guanidine group planar? Why are all the bond lengths different if all the $\ce{N}$ atoms are $\mathrm{sp^2}$?

Have you considered resonance?

The paper you linked with the reference to quantum mechanical calculations appears to be instead a description of a crystal structure doped with arginine. In that structure the C-N bonds are identical within error, and the guanidino group is described as planar, so all sp2 hybridized.

@Andrew, what do you mean "All C-N bonds are identical within error"?

In the paper you linked, the bond lengths are given as 1.351, 1.340 and 1.322. That’s identical as far as I’m concerned. Compare to the C-N bond of the primary amine at 1.48. Clearly a difference in bond order.

Atoms cannot be hybridised, only orbitals can. Therefore it is incorrect to state N is sp² hybridised, even though it is used in that way (too) often. Consider it lab jargon.

L-arginine is the most basic amino acid since it carries guanidyl group in addition to the $\alpha$-amino group. The arginine molecule is a zwitterion with the guanidyl group, rather than the primary $\alpha$-amino group, accepting extra proton from carboxylic acid group (Ref.1). Protonated or not, when consider guanidine as a molecule, there shouldn't be any $\mathrm{sp^3}$ nitrogen in the molecule because it is resonance-stabilized molecule. As a guanidyl group, this principle stays the same and it stays flat. There are few excellent answers to similar question you could find in here. In one answer it states that:

Guanidine is isolobal to urea, where the carbonyl oxygen has been replaced by an imine $\ce{NH}$. However, in principle it is still the same flat, resonance-stabilised molecule. The main difference is that there is no ‘preferred’ site for the double bond — it could point towards any of the three nitrogens in theory; you could say the diamide resonance is enhanced (Author: Jan).

And, accepted answer showed following diagram to depict the concept of resonance stabilization (Author: Ron):

Further, I'll attached recent crystal structure of L-arginine, which shows clear equivalent three nitrogen atoms with hydrohen bonding network (Ref.2):

Two planes characterize the arginine molecule: one through the acid group and the other through the extended side chain, which contains the guanidyl group. The dihedral angle between these two planes is $74^\circ$ (Ref.1).

References:

I. L. Karle, J. Karle, “An application of the symbolic addition method to the structure of L-arginine dehydrate,” Acta Cryst. 1964, 17, 835–841 (https://doi.org/10.1107/S0365110X64002250).
Emilie Courvoisier, P. Andrew Williams, Gin K. Lim, Colan E. Hughes,  Kenneth D. M. Harris, “The crystal structure of L-arginine,” Chem. Commun. 2012, 48, 2761–2763 (DOI: 10.1039/C2CC17203H).

So the nitrogens in the guanidine group are all sp2 and planar, correct?

Also, are all the nitrogen bond lengths the same in the guanidine group, or no?

@DrPepper: See Andrew's comments to your question.

That answer by Ron, and especially the image, is terrible. Apart from some hand waving and a much too literal use of hybridisation, there is not much more to it. I doubt there is a large enough barrier (if at all) for the protonation step to be able to identify kinetics versus thermodynamics.

@Mithoron: To your information, I didn't copy and paste and pretend it's my answer.  Since you didn't know me, search google my name and find who I am. I'm a genuine researcher and do not disrespect me. You have no rights to order me.

Wow, seems I need to write who know how much for anyone to understand... You cited Jan's answer - this is not a good practice. One could "cite" whole answer this way...  and  our site isn't a source for itself (Wikipedia learned it ages ago). One should write as much as possible with own words, etc. etc.

@Mithoron: If you know so much, why you don't give your own answer and mind your own business. There are moderators to do these kind of critical work.


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What makes a compound basic?
Aluminum chlorohydrate is often referred to as basic aluminum chloride.
The pH value of ACH is about 4. How is basicity in this context defined?

https://en.wikipedia.org/wiki/Aluminium_chlorohydrate It should be rather called chloride hydroxide.

Aluminum chlorohydrate (ACH) is a group of specific aluminum salts having the general formula $\ce{Al_nCl_{(3n-m)}(OH)_m}$ (Wikipedia). It is used in cosmetics as an antiperspirant and as a coagulant in water purification. Specifically in wastewater treatment processes, ACH is used as a coagulant to remove dissolved organic matter and colloidal particles present in suspension. In his paper on Water Treatment Coagulants, Peter Gebbie describes what coagulants do (Ref.1):

With aluminium-based coagulants, the metal ion is hydrolyzed to form aluminium hydroxide floc as well as hydrogen ions. The hydrogen ions will react with the alkalinity of the water and in the process, decrease the pH of the water as can be seen from Equation [...].  Similarly for ACH, which is described as being a pre-hydrolyzed coagulant, the following reaction takes place:
  $$\ce{Al2(OH)5Cl -> Al2(OH)5+ + Cl- + H2O -> 2Al(OH)3 + H+ + Cl- }$$
  Less hydrogen ions are produced with ACH, reflecting the hydroxylated nature of this compound.

There it says it all: ACH is more basic than other aluminum salts because it is pre-hydroxylated.
Aluminum chlorohydrate is one of the most common active ingredients in commercial antiperspirants. The variation most commonly used in deodorants and antiperspirants is $\ce{Al2Cl(OH)5}$, similar to the one used in water purification.
Note: Your statement of "The pH value of ACH is about 4" is evidently not correct. However, W. R. Grace & Company (producer of solution form of ACH) listed solubility of ACH as $\pu{1000 g/L}$ at $\pu{20 ^\circ C}$ and $\mathrm{pH} = 3.3$, the optimum condition.
Reference:

Peter Gebbie, “An Operator’s Guide to Water Treatment Coagulants,” 31st Annual Water Industry Workshop – Operations Skills; University Central Queensland Campus: Rockhampton, Australia, 4 to 6 July, 2006, pp. 14–20 (http://wioa.org.au/conference_papers/06_qld/documents/PeterGebbie.pdf).

Thank you very much!
So is it correct that aluminum chlorohydrate is still acidic, but less than other aluminum chloride compounds due to hydroxylation? Did I understand you correctly that acidity is not defined by the pH value?
The statement that the pH value of ACH is about 4 can be found here:
https://books.google.de/books?id=FAYNAwAAQBAJ&pg=PA505&lpg=PA505&dq=sweat+gland+aluminum&source=bl&ots=stwNnvq18L&sig=ACfU3U1YC_uYq9wWi4E28sBusKITFevOIA&hl=de&sa=X&ved=2ahUKEwiAhLDlpLXiAhVP-aQKHdhRByI4HhDoATADegQICRAB#v=onepage&q=sweat%20gland%20aluminum&f=false (page 506)

It says "buffered" ACH has pH around 4. Not ACH itself. $\ce{Al2(OH)5+}$ ion should be basic since $\ce{Cl-}$ is neutral.

@MathewMahindaratne Commonly we say that pKa = optimal pH of buffer, so that would make ACH acidic not basic. This is consistent with the equation in your answer, no?

@Buck Thorn: You are probably correct on that argument. Yet, I think it must be fairly close to neutral to be able to use as  an ingredient in antiperspirants because original $\ce{AlCl3}$ based antiperspirants had side effects due to acidity.


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Why isn't trinitramide planar?

The lewis structure of trinitramide has four major resonance structures, whereby the lone pairs of the central nitrogen can resonate with the nitro groups.
This seems to indicate that each of the nitrogen atoms would be sp2 hybridized, and the entire molecule would be planar. In fact, the Wikipedia ball-and-stick image for trinitramide seems to indicate it so.

However, based on quantum mechanical calculations, trinitramide is in fact non-planar. Why is this so? What drives its non-planarity?

Steric hindrance, maybe?

I would assume it’s a combination of steric and charge repulsion between the oxygen atoms. The experimentally determined structure in https://doi.org/10.1002/anie.201007047 shows the nitro groups essentially at right angles to each other.

I have extensively written about the isoelectronic trinitromethanide anion: https://chemistry.stackexchange.com/q/32354/4945

Doesn't the paper you have linked tell more than what we can presumably say? Interestingly, they not only say isn't planar, but pyramidal. There should be a comment atop the calculations. Otherwise take it at the level of their work. Perhaps one could calculate a different situati

Are there 4 electron domains around the central nitrogen (3 bonds to NO2 groups and one lone pair)? That would suggest trigonal pyramidal geometry.

The presence of nitro groups attached to the central nitrogen would lead to resonance between each nitro group and the central notrogen, which would have in fact made us expect trigonal planar geometry infact. The reason then, for the actual non-planarity, is best explained by the reasoning put forward by Andrew's comment above. Essentially, in the 3D confirmer as shown here the nitro groups are at right angles at each other to minimise steric as well as charge repulsion between the oxygens

A similar rationale regarding steric repulsion follows for the propeller-like arrangement of the three phenyl groups in triphenylphosphine

Also a similar repulsion between ortho hydrogens is the usual explanation for non planarity of biphenyl systems


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Where do negative ions migrate from the salt bridge to which electrode in the voltaic cell?
In the voltaic cell with $\ce{Mg/Mg^2+}$ and $\ce{Ag+/Ag}$, I am not sure whether the ions move from the salt bride to the $\ce{Mg}$ cathode or the $\ce{Mg^2+}$ cathode. I distantly recall my teacher saying the cathode (in this case) is a metal, and so it can't be $\ce{Mg^2+}$? I am confused, and please help ASAP. Final is tomorrow.
Somewhat related question:
At the cathode during electrolysis, what is its charge and what type of reaction occurs?
Is the correct answer: positive charge, and reduction?

Cathode or anode, $\ce{Mg^2+}$ is not a electrode.

So is Mg2+ the cathode? Isn't the cathode the site of reduction, where Mg2+ would reduce?

Mg is the cathode, you were already told Mg2+ is not. But Mg2+ is NOT reduced, at least not in water environment. There reduction 2 H+ + 2 e- -> H2. See also cathode=catabasis of electrons = reduction by incoming electrons

"what is its charge and what type of reaction occurs?"the negative charge, and reduction

Oxidation half reaction (happens at the anode)
$$\ce{Mg -> Mg^2+ + 2e-}$$
The electrons get removed via the wire, and the magnesium cations enter the aqueous solution, giving it a net positive charge.
Reduction half reaction (happens at the cathode)
$$\ce{Ag+ + e- -> Ag}$$
The electrons are supplied via the wire, and the silver cations leave the aqueous solution to be deposited on the cathode, giving the solution a net negative charge.
Salt bridge
The salt bridge provides anions (negative charged particles) to the solution of the oxidation half cell, and cations (positively charged particles) to the solution of the reduction half reaction.
Silly mnemonics
Oil rig: oxidation is loss, reduction is gain (of electrons)
Red cat: reduction occurs at the cathode.

An ox: oxidation occurs at the anode.

Where do negative ions migrate from the salt bridge?

They don't migrate to any electrode (neither cathode nor anode). Instead, they migrate to the solution of the solution of the oxidation half cell.

A voltaic cell is an electrochemical cell that uses a chemical reaction to produce electrical energy. An electrochemical cell consists of two electrodes (anode and cathode), two electlolyte solutions, and a salt bridge:

The anode is an electrode where oxidation occurs and the cathode is an electrode where reduction occurs. The oxidation and reduction reactions are separated into compartments, which are called half-cells (each half-cell consists of electrode and the appropriate electrolyte solution). The salt bridge connecting half-cells is a chamber of electrolytes, which are needed to maintain the neutrality of the solutions in both half-cells. The external circuit is used to conduct the flow of electrons between electrodes, which usually go through a load (e.g., a light bulb or voltmeter) to close the circuit (see the diagram).
Now, let's see what chemical reactions are happening in your voltaic cell. Your two half cells are $\ce{Mg/Mg^2+}$ (suppose your $\ce{Mg^2+}$ solution is an aqueous $\ce{Mg(NO3)2}$ solution) and  $\ce{Ag/Ag+}$ (suppose your $\ce{Ag+}$ solution is an aqueous $\ce{AgNO3}$ solution). Suppose your salt bridge is $\ce{NaNO3}$. To find where oxidation and reduction happening, you may need to look table at the end of your textbook, which is called electrochemical series, giving reduction potentials of many elements such as following:
$$\ce{Ag+ + e- <=> Ag} \qquad \mathrm{E}^\circ = \pu{0.7996 V} \tag{1}$$
$$\ce{Mg^2+ + 2e- <=> Mg} \qquad \mathrm{E}^\circ = \pu{-2.372 V} \tag{2}$$
To have a positive electron flow (or spontaneous redox reaction), $\mathrm{E}_\mathrm{cell}^\circ$ should be positive.  So, you can rearrange equation (2) such that complete redox reaction gives positive $\mathrm{E}_\mathrm{cell}^\circ$:
$$\ce{Mg  <=> Mg^2+ + 2e-} \qquad \mathrm{E}^\circ = \pu{+2.372 V} \tag{3}$$
Now see, in equation (1), $\ce{Ag+}$ gains $\ce{e-}$s, so that it reduces. Therefore, $\ce{Ag/Ag+}$ half-cell is the cathode where reduction happens.
In equation (3), $\ce{Mg^2+}$ gives away $\ce{e-}$s, so that it oxidizes. Therefore, $\ce{Mg/Mg^2+}$ half-cell is the anode where oxidation happens.
Solving (1) and (3) in order to cancel $\ce{e-}$s, you'll get total redox reaction:
$$\ce{Mg + 2Ag+ -> Mg^2+ + 2Ag} \qquad \mathrm{E}_\mathrm{cell}^\circ = \pu{+3.172 V} \tag{3}$$
Overall, at anode, the $\ce{Mg}$ electrode slowly dissolve, and the magnesium cations enter the aqueous solution, giving it a net positive charge. The electrons left at the anode will travel to the cathode through the external circuit (connected to the load).
At the same time, $\ce{Ag+}$ travel to the cathode, and gains $\ce{e-}$s comes through the wire to become $\ce{Ag}$. $\ce{Ag}$ deposits on cathode, and thus, the aqueous solution loses the silver cations, giving it a net negative charge. The salt bridge, then does his duty by neutralizing these two aqueous solutions (see the diagram).

Note: The galvanic cell is courtesy of ELI JONES. The original one was modified in order to suit to the current question in hand.

@ Ed V: Made the correction. So, you can enjoy your coffee! :-)

@ Ed V: I also included the link to the original source of my modified galvanic cell.

https://chem.libretexts.org/Bookshelves/General_Chemistry/Book%3A_Chemistry_(OpenSTAX)/17%3A_Electrochemistry/17.2%3A_Galvanic_Cells Figure 17.2.2 from OpenStax chemistry

Both answers are very good, but Mathew Mahindaratne's answer uses a galvanic cell figure that needs some work. I respectfully request that his galvanic cell figure be replaced and I hereby offer up my handmade figure (see below), either as a temporary filler or for whatever period Mathew may decide. It is entirely up to him. I do not claim my handmade figure is wonderful, but I can truthfully say it is my work and anyone can use it.

Nice cartoon! I like your version. But, I'd make changes to my version in few min. Please don't erase yours. Thanks for your remarks.