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Why Does Activation Energy not change with Temperature change
As The Graph Shows; $E_a$ (Activation Energy) = Energy Of transition State(Threshold Energy) $-$ Energy of Reactants.
So let This be the graph at Temperature $T_1$, Now say We Increase The Temperature to say $T_2$ and so the energy of reactants must increase (as kinetic energy of molecules increases). So at $T_2$, reactants would start with more energy. So Shouldn't the Activation Energy decrease as now less energy is needed to reach Transition State.
Or I can say that since $E_a$ (Activation Energy) = Energy Of transition State(Threshold Energy) $-$ Energy of Reactants, So as Energy of Reactants has changed so should Activation Energy.
I already saw answers on a similar question but couldn't figure out what's wrong with my approach.

Say you have to travel to some place. If you drive faster, you'll get there sooner, but that doesn't mean that the distance got shorter. Same thing here.

The problem is that the activation energy is not derived the way you think. It is obtained by measuring a reaction rate constant at different temperatures. Afterwards, when reporting the logarithm of this rate constant versus the inverse of the absolute temperature, you obtain a line. And the slope of the line is equal to the energy of activation divided by the gas constant R.

You are on the right track. The problem is with the diagram you show, which is just a cartoon, and describes something that is actually not the activation energy, although it is generally interpreted as this and as such is a good 'rule of thumb'.
The activation energy is experimentally derived as $\displaystyle E_A=-R\frac{\ln(k(T)}{d(1/T)}=\frac{RT^2}{k(T)}\frac{d k(T)}{dT}$.
where $k(T)$ is the rate constant at temperature $T$. However, this mathematical method imparts little or no physical information about what the activation energy is.
It is not just the height of the barrier above the bottom of the potential well, but has to reflect the fact that, in any experiment, the molecules are at a given temperature and so have a Boltzmann distribution of energies and conform to a Maxwell–Boltzmann distribution of collision speeds. Tolman interpreted this as shown in the  figure.

Figure 1.  Tolman interpretation of the activation energy. Shown are the rate constant with energy $k(E)\equiv \sigma P(E)$, the Boltzmann factor $\exp(-E / k_BT)$ and the product $f(E)=\sigma P(E)\exp(-E/k_BT)$ which is the probability of having energy $E$ at the transition state. Note that $E_T\ne E_0$ where $E_0$ is the threshold energy for reaction. ($f(E)$ is multiplied by a large factor to make it visible on the graph).

In the Tolman interpretation, the activation energy is the average total energy $\langle E^*\rangle$ (relative translational plus internal) of pairs of species that have reacted, minus the average energy of all $\langle E_R\rangle$  pairs of species whether or not they react, or $\langle E_T\rangle =\langle E^*\rangle -\langle E_R\rangle $.
Direct calculation via the Equipartition theorem gives the average energy of all pairs as $\langle E_R\rangle =3RT/2$ but the average of the reactive pairs is more complicated and depends on knowing how the rate constant $k(E)\equiv \sigma P(E)$ changes with energy and then $\langle E^*\rangle$ can be calculated.
The essential point in this idea is that the activation energy is the difference in the average energy in the reactants and the average energy in the transition state which is the average of $f(E)$ as shown in the figure.

So finally does it depend on temprature or not?...Thanks for the answer.

why yes, both $\langle E^*\rangle$ and $\langle E_R\rangle$ depend on temperature. The former because there is a Boltzmann distribution in the transition state and the latter is 3RT/2 as mentioned in the text.

You have conceptual misunderstanding. In the activation energy context, energy of reactants, products and transition state are all potential energies, independent on the kinetic energy of molecules.
The kinetic energy of molecules is used to reach the peak of the potential energy, similarly as e.g. a acrobatic aircraft uses its kinetic energy to climb to the apex of an acrobatic loop.
Higher temperature means higher kinetic energy, but it does not mean the potential energy barrier is lower. It means there is higher probability to overcome it, leading to a faster reaction.
There is a famour Arrhenius' equation for the reaction rate:
$$k = A \exp{\left(\frac{-E_\mathrm{A}}{kT}\right)}$$
where $A$ is the frequency factor, describing the rate of molecular collisions, potentially able to take part of the reaction and
$\exp{\left(\frac{-E_\mathrm{A}}{kT}\right)}$ is the term from the Boltzmann distribution,
telling us the probability a molecule at temperature $T$ has sufficient energy, if the activation energy is $E_\mathrm{A}$, and $k$ is the Boltzmann constant.

I completely understood your answer and I got why I was wrong, but then I went on to thinking that Due to the vibrational degree of freedom Wouldn't the vibrational energy also increase with with temperature, hence leading to a higher potential energy, so the at higher temperature reactants would start with higher potential energy. And thanks a lot for your help.

Well, I have been simplifying a little. The activation energy is potential energy of another kind, compared to the potential energy of band oscillators. So more properly would be the thermal energy, not just kinetic energy takes part in overcoming the potential wall of the activation energy.

By other words, if a car vibrates, the hill is not smaller .

Sir please take a look at this source especially the last equation https://socratic.org/questions/what-is-activation-energy-what-is-threshold-energy-i-want-to-know-the-difference . This seemingly contradicts your explanation.

I have noticed they incorrently use the total translational kinetic energy, following the Maxwell-Boltzman distribution, while only the KE of 1 degree of freedom ( motion along the line of collision ) is to be considered, following the Boltzmann distribution. the rest may be matter of the terminology confusion.

Sir is it possible that the definitions of Activation Energy and Threshold Energy may differ regionally. What you (and I ) call Activation Energy has another meaning somewhere else ?

For example till now I have seen usage of Ea in the Maxwell curve but here they have used ''Lower Threshold Energy'' .


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Can d(x²-y²) form π bonds with other p- or d-orbitals?
$d_{x^2-y^2}$ doesn't change sign when transformation of $x\rightarrow-x$ or when similar transformation with $y$ is made. However no such thing occurs with any p- or d-orbital. Thus, can we say that $d_{x^2-y^2}$ don't form $π$ bonds with any p/d-orbitals?

You can say that, but it won't be correct.

Likewise, $p_x$ doesn't change sign upon the transformation of $y→−y$.

A p orbital in the x-y plane centered on the $y=x$ line oriented orthogonal to that line has the correct symmetry to interact with your orbital, for example.

source: Stack Exchange

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

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How does a battery create a potential differece (voltage)?
I'm 15 and recently started electronics and I just had a question about batteries.
(I've been told by people on physics stack exchange to come here to have my question answered).
From what I understand and according to the videos I've watched, a battery creates a potential difference (votlage) by accumulating more high concentration the negative plate than the positive plate. Therefore the higher concentration of electrons, the higher potential difference (voltage).
Could someone tell me if this is correct?
Thanks

Development of a potential difference when you connect two dissimilar metals "dipping" in their own solutions is another fundamental phenomenon exhibited by matter. It is a property of matter and batteries utilize this very phenomenon. Perhaps a hardcore physicist can tell you why do you have a mass, in the same way, this is a pure theoretical problem of physics as to why this happens. It just happens and it is related to the structure of matter (in fancier terms ... Fermi levels).
Suppose you have two different metal pieces (= two materials having different electronic energy levels) dipping in their respective solutions of ions. Now you connect them with a piece of wire. Since, by definition, there are two different materials, one of them would have electrons at a higher energy level than the other. I mean there must a be difference in the energies of electrons in material A vs. material B, since these are two different elements. Since the metals are connected, it is one conducting system. The electrons would travel from a higher energy to lower energy in this system. What is the flow of electrons in unit time? It is called current. Your battery is ready.

You identify a system where electron transfer is spontaneous, say, from R to O. This is a redox reaction with a reductant (R) and an oxidant (O).
$$\ce{R + O -> R+ + O-}$$
Now if you are able to physically separate R and O and provide a pathway just for the electrons, then you can harness the flow of electrons in the wire.
You're not really going to get any significant concentration of electrons because any accumulation of charge immediately leads to a high energy system due to charge-charge repulsion. But a higher concentration of either reductant or oxidant should increase the potential difference for thermodynamic reasons.

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Concentration and Half Life of reactants A and B

$3.12$ A reaction $\ce{A + B ->C}$, follows first order kinetics with respect to each reactant, with an overall rate constant $k=\pu{2.0 x 10^{-3} M-1s-1}$, the initial concentrations of A and B are $\pu{0.10 M}$ and $\pu{6.93 M}$. respectively. Given $\ln 2 = 0.693$, choose the correct statement(s).
(A) The concentration of A remaining after $100$ seconds is approximately $\pu{2.5 x 10^{-2} M}$.
(B) The concentration of B remaining after $100$ seconds is approximately $\pu{1.73 M}$.
(C) The half life for A is about $50$ seconds
(D) The half life for B is about $50$ seconds
Since reaction is first order with respect to both reactant A and B, I can write,

$$ k=\frac{1}{t} \ln{\frac{a}{a-x}}
 $$
where a is initial concentration of reactant and a-x is remaining concentration.
On solving the above for A and B after 100 seconds, I get $A_r = \frac{0.1}{e^{0.2}} $ and $B_r = 0.693/e^{0.2}$
The answer does not match with any options for both concentration as well as half life.
Can you please help me with this?

Don't you think a reaction that involves B should somehow depend on B? In your derivation for A, it doesn't.

@IvanNeretin The equation is that of 1st order and depends only on 1 reactant. Even  though the reaction involves A and B both, they both follow 1st order kinetics so the above equation will apply to each A and B seperately

@IvanNeretin Rate of Reaction = k [A][B] that is true

We have one reaction. It is first order in A, so according to you, it should depend only on A. Also, it is first order in B, so it should depend only on B. How do you reconcile that?

In questions like this it always helps to look at all the data you are given including the concentrations.

@porphyrin I am thorough with the topic and yet I am unable to solve this one... Nothing seems to help! :(

have a look at pseudo first order reactions

The whole reaction isn't first order and that can be arrived at using one simple concept. The unit of rate constant $k$ is $\pu{M^{-1} s^{-1}}$ and not $\mathrm s^{-1}$.

Since reaction is first order with respect to both reactant A and B, I can write,
$$ k=\frac{1}{t} \ln{\frac{a}{a-x}} $$

This would have been correct if and only if the whole reaction was a first order reaction. Not a first order reaction with respect to each reactant individually. For example:
$$\ce{A->B}\quad \text{Rate}=k[A]$$
That is not the case here. Here the reaction would be better phrased as:
$$\ce{A + B->C}\quad \text{Rate} = k[A][B]$$
Which means that this is a second order reaction and not a first order reaction.
Now, the other issue is that this second order reaction isn't of the form $\ce{2A-> B}$ where the rate would have been relatively easy to find using a generic law and hence we move to a more basic proof for a reaction $\ce{A + B -> C}$. As given above, the rate law for this expression is "Rate $= k[A][B]$" Now we draw a RICE table for this generic reaction to get:
\begin{array}{|l|c|} \hline
\mathrm{R} & \mathrm A & \mathrm B &\mathrm C \\ \hline
\mathrm I & a & b & 0 \\
\mathrm C & -x & -x & x \\
\mathrm E & a-x & b-x & x \\ \hline
\end{array}
Hence, we can see that for a time t, the rate of reaction can be defined as:
$$ \text{Rate} = -\frac{\mathrm d[A]}{\mathrm dt}=k(a-x)(b-x)$$
Differentiating $[A]$ with respect to $t$, we get:
$$-\frac{\mathrm d[A]}{\mathrm dt} = \frac{\mathrm dx}{\mathrm dt}$$
Substituting this in the previous equation and integrating with respect to x, we get:
$$kt = \frac{1}{b-a}\ln \frac{a(b-x)}{b(a-x)}$$
Now, in the given question, we are asked to find the value of $[A]$ and $[B]$ after $\pu{100 s}$. Therefore, $t=\pu{100 s}$, $a = \pu{0.1 M}\,,b=\pu{6.93 M}$ and $k = \pu{2 \times 10^-3 M^{-1} s^{-1}}$.
Solve for x in this equation and find the value of $a-x$ and $b-x$.
This isn't a complete solution, but the rest is just manipulating the expression using mathematics and has no actual relation to kinetics.
A hint to make calculations simpler would arise from noticing that $a<<b$ and so $b-a$, $b-x$ can both be approximated to $b$.

How are you stating that the first differential of [A] with respect to time is the product of the concentrations of both the reactants after some arbitrary time period?

@AdityaRoychowdhury Isn't that what the rate law states? $-\frac{1}{a} \frac{\mathrm d[A]}{\mathrm dt} = Rate$ for a reaction $\ce{aA +bB -> cC + dD}$ here a=1.

Yes, but am asking why are you multiplying (a-x)(b-x). How do you realise that that is the rate law. Shouldn't it be the product of the initial concetration. Even if you demonstrate it in the form of a rate law by using derivative.

@Aditya This would be the rate at some time t.. after x moles of A and B have reacted.. since rate here would be $k[A_t][B_t]$


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Hydrogen bonding in sugar acids
I recently learnt that the reason for low solubility of mucic acid as compared to glucaric acid was more hydrogen bonding in the crystal state of mucic acid as compared to glucaric acid.

I would like to know why mucic acid has more hydrogen bonding? Does symmetry of the molecule play a role? If so how?

From the abstract of an article titled "The crystal structure of galactaric acid (mucic acid) at −147°: An unusually dense, hydrogen-bonded structure"

[...] D$_m$ = [1.790] $\pu{g cm−3}$ [...] The crystal structure has a system of strong, intermolecular hydrogen-bonds, which accounts for the high crystal density and low solubility in water.

The crystal structure was solved at very low temperature, but the density D$_m$ is quoted for room temperature. For comparison, the density of sucrose or cellulose is around 1.5 $\pu{g cm−3}$.

For glucaric acid, the volume of the asymmetric unit is 421.9 Å$^3$ / 2, while for mucic acid it is 194.0 Å$^3$. As the molar mass of the two stereoisomers is the same, there is a difference in density of about 9%.

A lower density implies more attractive interactions or less repulsive interactions in the crystal state. For a complete analysis, you would also have to compare how well these compounds are solvated in aqueous solution. Typically, it is impossible to predict solubility by just looking at structures because both solvation and crystal structure are hard to predict.


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Formation of Calcite from Calcium carbonate
So, in lab we are forming the different polymorphs of calcium carbonate. It required us to age the sample that was prepared for the formation of calcite for one week. This sample mixture was suspended in water. Whilst the other samples that were prepared for vaterite and aragonite were immediately washed/filtered.
Why do you need to age a calcite sample for a week?

It's my understanding that the production of the calcite takes a decent amount of time as there are 2 stages being undertaken for the crystallization of the amorphous calcium carbonate (ACC). The first of these is relatively fast whilst the second stage is the one that you want to be looking at as it is roughly 10 times slower. This is the stage which looks at a dissolution and precipitation mechanism.


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What exactly are electron states?
From Tro's Chemistry: Structure and Properties [1, p. 93]:

2.5 Quantum Mechanics and the Atom
As we have seen, the position and velocity of the electron are complementary properties—if we know one accurately, the other becomes indeterminate. Since velocity is directly related to energy (recall that kinetic energy equals $\frac 1 2 mv^2),$ position and energy are also complementary properties—the more we
know about one, the less we know about the other. Many of the properties of an element, however, depend on the energies of its electrons. In the following paragraphs, we describe the probability distribution maps for electron states in which the electron has well-defined energy, but not well-defined position. In other words, for each of these states, we can specify the energy of the electron precisely, but not its location at a given instant. Instead, the electron’s position is described in terms of an orbital, a probability distribution map showing where the electron is likely to be found. Since chemical bonding often involves the sharing of electrons between atoms (see Section 4.2), the spatial distribution of atomic electrons is important to bonding.

The textbook says that electrons can exist in two states simultaneously. As I went further, I saw that an electron could exist as both a particle and a wave. So I thought that this must be the states that I was introduced to earlier, until I was learning about quantum mechanics.
In the text above, the writer was talking about electron states with well defined energy. From what I know the electron state that has well defined energy would be one with well defined velocity. And the velocity of an electron is associated with it's wave nature.
So, aren't we supposed to be dealing with electron as waves only? Or what exactly does the writer mean by the electron states?
Reference

Tro, N. J. Chemistry: Structure and Properties, 2nd ed.; Pearson: Hoboken, NJ, 2018. ISBN 978-0-13-429393-6.

Keep in mind people are not very careful in using the terminology. You will find this discussion useful https://chemistry.stackexchange.com/questions/119746/terminology-of-atomic-spectroscopy-difference-among-term-states-and-level

How do you conclude that a defined energy means a defined velocity? That only applies if there is no potential energy term in the Hamiltonian

Kinetic energy=1/2mv2

Although this question can be easily answered, I'm upvoting because it is a key concept for beginners.

An electronic state is defined by the electron configuration of the system and by the quantum numbers of each electron contributing to that configuration. Each electronic state corresponds to one of the energy levels of the molecule (source: chemistry libretexts).

"Energy" includes both kinetic and potential energy terms.

In QM, properties that can be measured (such as energy, spin, position, etc) are each associated with an "operator". Every observation will always return an eigenvalue for the relevant operator. Eigen values are values for $\lambda$ for which the equation $$\hat{O}\Psi = \lambda\Psi$$ is true, where $\hat{O}$ is the operator of interest and $\Psi$ is the wave function of the entity of interest.
For example, let's consider electron "spin", by which we typically mean specifically the component of the spin in the z-direction, given by the operator $\hat{S}_z$. For the electron, this operator has two possible eigenvalues, which we abbreviate as $+1/2$ and $-1/2$. An electron having one of these values of spin is described as being in the eigenstate (or more simply just "state") associated with the given eigenvalue.
So we can say that having $+1/2$ spin and having $-1/2$ spin are two possible states of the electron.
What makes QM different from classical mechanics is that measuring an electron to have $+1/2$ spin does not mean that it was necessarily in the $+1/2$ spin state prior to our taking the measurement. An electron can (and often does) exist in a super-position of states, meaning that it simultaneously has both $+1/2$ and $-1/2$ spin, but our measurement causes it to collapse into one state or the other.
As you described in your question, some pairs of operators have the property that you cannot measure a defined value for both at the same time. Position and momentum are the most famous pair, since they are the example for the Heisenberg uncertainty principle, but many other pairs behave similarly. Energy and position are one such pair. So if we measure energy precisely, we cannot know where the electron with that energy exists. Since energy levels are the basis of the Schrodinger equation and define much of the behavior of atoms, chemists tend to treat electrons as if they are in a state in which the energy has been measured and other properties are unknown.
To address your point of confusion - "wave" and "particle" are descriptions of the behavior of electrons, most useful when understanding the concept of positional uncertainty, since "position" is most associated with particle behavior. When we learn about the Heisenberg uncertainty, we learn about experiments in which electrons behave as waves when their position has not been measured and like particles when their position has been measured, so it's easy to see why you associate wave and particle behavior with velocity and position, respectively, but those are just convenient examples to show the concept of duality. Operators generally should not be thought of as being associated with one description or the other, nor should we think of an electron as switching between particle and wave. It is better just to think of it as having specific position in space or not, and whether or not it has specific position it is neither a wave nor a particle, but rather just a quantum mechanical entity.
The results of any experiments we perform with electrons are predicted very well by the mathematical system that has been developed for quantum mechanics, and some of that math has parallels with classical wave behavior and some with classical particle behavior and some with neither of them. Beyond that, we have very little understanding of what an electron actually is and how its behavior accords with our understanding of the physics of the macroscopic world.
ADDENDUM For completeness, I'll add that in addition to $\hat{S}_z$, two other important operators that measure properties of electrons that can be known at the same time as energy is known are $\hat{L}^2$, the operator for the magnitude of total (quantum) angular momentum and $\hat{L}_z$, the z-component of that angular momentum. Thus, we can theoretically have an electron for which all four properties are fixed at one time, and we define the state described by the four eigenvalues by representing those eigenvalues with the "quantum numbers" (simplified forms of the eigenvalues) with which you are probably familiar:
n is the quantum number which represents the eigenvalues of the energy operator $\hat{H}$ (ie the Hamiltonian, which we associate with a specific shell)
l is the quantum number which represents the eigenvalues of the operator $\hat{L}^2$ (which we equate with s,p, d etc orbitals)
ml is the quantum number which represents the eigenvalues of the operator $\hat{L}_z$ (which we equate with the axial orientation of orbitals)
ms is the quantum number which represents the eigenvalues of the operator $\hat{S}_z$ (which we equate with the above-described spin orientation)


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Is a beta pleated parallel sheet always formed from more than one peptide?
I can conceptually grasp the beta pleated anti-parallel sheet by imagining a curved length of string i.e. one polypeptide, starting with the N- terminus on the left and ending with the C-terminus at the end of the string, opposite the N terminus.
I can't do the same for the beta pleated parallel sheet if it is only formed of one polypeptide. How could a fold in a single strand cause the N and C termini to be parallel with each other?
Or is it that parallel sheets are always composed of more than one polypeptide?

Would single chain parallel sheet look more like this:

No, you just need a long linker, imagine perhaps a long Z instead of U on your diagram.

Thanks - could you elaborate a little?

Yes, that's the schematic diagram I was thinking about. It's very schematic, though, because this linker would circumvent the sheet altogether, perhaps even contain an alpha helix put onto it (for example).

While it is true that some antiparallel beta sheets have neighboring strands that are adjacent in primary sequence, other cases do exist. As for parallel beta sheets, the polypeptide chain between strands could be connected via a loop or a helix if the two strands are close in primary sequence.
The mapping of strands on the primary structure has been studied thoroughly, and there are two types of diagrams to quickly show what is going on. For the "TOPS" diagram, we pretend to look down the strands, and up and down triangles represent strands in one or the other direction:

Source: https://onlinelibrary.wiley.com/doi/pdf/10.1002/pro.3285
Circles represent helices. If you look closely at the connecting lines, you can see whether the connection is above or below the plane of the paper.
The other representation has the sheets in the plane of the paper, e.g.:

Source: https://image.slideserve.com/258316/beta-sheet-topology-diagrams-l.jpg
Take-home messages

Sheets are antiparallel, mixed or parallel.
Most neighboring strands are not adjacent in sequence.
When two neighboring parallel strands are adjacent in sequence, there is a loop or a helix between them (or a strand from a different sheet) to connect them in the proper orientation.

The Beta Sheet Topology Diagram helped a lot - thank you :)

@Naj Yes, first comes the hydrophobic collapse, then some shuffling to get secondary structure (main chain interactions) and tertiary stucture (side chain interactions) all sorted out. There is, however, a strong indication which secondary structure a segment of primary sequence will adopt, regardless of the rest of the sequence. For a full answer, ask another question.

@KarstenTheis I actually deleted the comment you responded to above, suddenly thinking it was a stupid question. For the sake of any one turning up here and wondering what you were replying to, I asked if main chain interactions (secondary structure) and side chain interactions (tertiary structure) might happen in a non-temporal sequence i.e. not necessarily secondary then tertiary, maybe tertiary then secondary, maybe both at the same time. 'Shuffling' between the two seems to confirm that I was somewhat correct in my hypothesis. New question coming soon :)

@KarstenTheis Note: New question will be regarding "There is, however, a strong indication which secondary structure a segment of primary sequence will adopt, regardless of the rest of the sequence."

Actually, I came upon a Hussain biology Youtube video which answered that very question. Highly recommended!


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Laboratory and commercial preparations of aniline other than reduction of nitroarenes
Till now I could only find one laboratory preparation of aniline, i.e. reducing nitrobenzene with $\ce{Sn}$ and concentrated $\ce{HCl}$. I'm confused if Hofmann degradation is only commercial.
I'd like to know if there are other processes to prepare aniline in laboratory.

Catalytic hydrogenation of nitrobenzene over Pd/Pt/Ni catalysts

Why would anybody prepare aniline in lab?

Is Hoffmann Bromamide reaction only commercial or a laboratory method? I only want to know, I don't perform these experiments so don't ask me why to prepare in laboratory or whatnot.

Check this 2009 paper which gives loads of palladium catalyzed and copper catalyzed reactions.

@AkshSarma Hoffman Bromamide is pointless from a synthesis perspective; if you have to prepare aniline, it is much easier/cheaper to use nitrobenzene as a raw material than Acetanilide. Chances are if you have acetanilide, you will also have aniline.

@AniruddhaDeb ... also the yields are generally poor in HBR because it requires a lot of distillation, decarboxylation and the usage of already a volatile reactant i.e bromine. This reaction is better studied for academic purpose only.

It is true that until recently, the syntheses of primary anilines mostly rely on the reduction of corresponding nitroarenes. However, recent publications allow the synthesis of primary anilines or their derivatives in laboratory.
One such method is Buchwald–Hartwig amination:

The palladium-catalyzed cross-coupling of amines and aryl (pseudo)halides, now commonly known as the Buchwald–Hartwig amination, was first reported 25 years ago. Since the simultaneous breakthrough reports of Buchwald and Hartwig in 1995, this reaction has transformed the way synthetic chemists think about synthesizing aromatic amines... (Ref.1):

A continuous process for Buchwald–Hartwig amination in different scale synthesis has also been introduced recently (Ref.2). Further, a version of Buchwald–Hartwig amination for the preparation of primary arylamines has been introduced recently as a general method (Ref.3):

The importance of the method is reflected by the development of a synthetic route to manufacture the drug candidate AMG 925 on kilogram scale using Buchwald–Hartwig amination (Ref.4):

Another amine synthetic method not following nitro reduction is Ullmann amine synthesis and its variations (Ref.5 and references therein):

In addition, secondary or tertiary amine can be prepared using Chan-Lam coupling reaction. For example see following reaction (Ref.6):

Other than using aryl halides, there is an another method to convert phenol derivatives to anilines. One such example describes a direct conversion of phenols into primary anilines with hydrazine catalyzed by palladium (Ref.7):

References:

Paola A. Forero-Cortés, Alexander M. Haydl, "The 25th Anniversary of the Buchwald–Hartwig Amination: Development, Applications, and Outlook," Org. Process Res. Dev. 2019, 23(8), 1478–1483 (https://doi.org/10.1021/acs.oprd.9b00161).
Sebastian Falß, Giovanna Tomaiuolo, Antonio Perazzo, Paul Hodgson, Polina Yaseneva, Jacek Zakrzewski, Stefano Guido, Alexei Lapkin, Robert Woodward, Rebecca E. Meadows, "A Continuous Process for Buchwald–Hartwig Amination at Micro-, Lab-, and Mesoscale Using a Novel Reactor Concept," Org. Process Res. Dev. 2016, 20(2), 558–567 (https://doi.org/10.1021/acs.oprd.5b00350).
Giang D. Vo, John F. Hartwig, "Palladium-Catalyzed Coupling of Ammonia with Aryl Chlorides, Bromides, Iodides, and Sulfonates: A General Method for the Preparation of Primary Arylamines," J. Am. Chem. Soc. 2009, 131(31), 11049–11061 (https://doi.org/10.1021/ja903049z).
Caroline Affouard, Richard D. Crockett, Khalid Diker, Robert P. Farrell, Gilles Gorins, John R. Huckins, Seb Caille, "Multi-Kilo Delivery of AMG 925 Featuring a Buchwald–Hartwig Amination and Processing with Insoluble Synthetic Intermediates," Org. Process Res. Dev. 2015, 19(3), 476–485 (https://doi.org/10.1021/op500367p).
Andrea Francesca Quivelli, Paola Vitale, Filippo Maria Perna, Vito Capriati, "Reshaping Ullmann Amine Synthesis in Deep Eutectic Solvents: A Mild Approach for Cu-Catalyzed C–N Coupling Reactions With No Additional Ligands," Frontiers in Chemistry 2019, 7, Article 723 (14 pages) (https://doi.org/10.3389/fchem.2019.00723)(Full Paper).
Carolin Fischer, Burkhard Koenig, "Palladium- and copper-mediated N-aryl bond formation reactions for the synthesis of biological active compounds," Beilstein J. Org. Chem. 2011, 7, 59-74 (https://doi.org/10.3762/bjoc.7.10)(Full Paper).
Zihang Qiu, Leiyang Lv, Jianbin Li, Chen-Chen Li, Chao-Jun Li, "Direct conversion of phenols into primary anilines with hydrazine catalyzed by palladium," Chem. Sci. 2019, 10(18), 4775-4781 (https://doi.org/10.1039/C9SC00595A)(Full Paper).


text:
Could observer effect cause electrons in one orbital to be found in another?
The observer effect states [1] that

when unobserved, absolutely small particles like electrons can simultaneously be in two different states at the same time.

If we look at an atom of any element, all of its electrons are subject to the observer effect. As the electrons are known for wave-particle duality, the Schrödinger equation can be solved to yield a wave function for each electron in the atom. An orbital is a probability distribution map of a wave function squared, and each orbital can accommodate up to two electrons.
For a single electron, it is not hard to imagine that the observer effect operates upon it to the extent of the probability distribution map specified by its orbital. That is, the electron can simultaneously have two positions if unobserved, and these two positions will not go beyond the space occupied by its orbital.
However, I'm wondering if an electron can have two positions when unobserved, where these positions are in different orbitals. For example, an electron falls under both the $\mathrm{2s}$ orbital and the $\mathrm{2p}_x$ orbital, if unobserved, and observation forces it to collapse into either $\mathrm{2s}$ or $\mathrm{2p}_x.$
Reference

Tro, N. J. (2015). Principles of Chemistry: A Molecular Approach, Global Edition. United Kingdom: Pearson Education Limited.

Your intuition is exactly right. Let's consider an atom of neon. It has ten electrons, which we describe as being in 1s, 2s and 2p orbitals. But the electrons are indistinguishable from each other, and accurate wave functions allow all of them to be in all orbitals. Only when we make a measurement (or some other sort of "observation") do the electrons behave as if they are localized to individual orbitals.
For example, if we measure the binding energy of the electrons by photoelectron spectroscopy, the result will always be an eigenvalue of the energy operator (aka Hamiltonian). But that does not mean that any one electron had exactly that energy prior to the measurement. As is always the case in quantum mechanics, systems that have multiple eigenstates exist as a superposition of those states until the measurement, the result of which is always one of the possible eigenvalues rather than, for example, a weighted average of them. So we will measure binding energies consistent with 2p orbitals, 2s orbitals and 1s orbitals with an intensity ratio of 3:1:1.


text:
Comparison of the solubility of lead(II) iodide in water and in lead(II) nitrate solution

Solubility of $\ce{PbI2}$: Is $\ce{PbI2}$ more soluble in $\ce{H2O}$ or in an aqueous solution of Lead(II) nitrate?
$K_\mathrm{sp} (\ce{PbI_{2}})=8.7\times 10^{-9}$

I'm not very good in Chemistry, I study mathematics and now I have this exam (in chemistry), but I have no idea how to proceed in this exercise. Can someone explain the method to solve it?

You have a chemistry exam. This doesn't look like math.

@Karl whay is the problem?

I asked help because I don't understand how to proceed with these exercises. I posted it in the forum of Chemistry so what means "This doesn't look like math" ?

If you are supposed to be able to answer this, then you must know a bit more. You're welcome to ask here if you encounter a problem when revising your lecture notes, but you must show your own effort.

Did you ever heard of common ion effect?

@Mathew Mahindaratne yes, and I know that PbI_{2} is not soluble in water. I wrot the dissociation reaction of PbI2 but I think that there aren't common ions with Lead(II) nitrate. Am I wrong?

Remember, $\ce{PbI2}$ is only sparingly soluble, thus cannot completely dissociate in water.  $\ce{Pb(NO3)2}$, on the other hand very soluble and hence, completely dissociated in water. Therefore, you have $\ce{Pb^2+}$ present in that solution as a common ion.

HINT - Start at the beginning. Write out the chemical equations for lead iodide and lead nitrate dissolving in water. Then think about the common ion effect which is a specific case of Le Chatelier's principle.

In water the following equilibrium would establish:
$$\ce{PbI2 (s) <=> Pb^2+ (aq) + 2I- (aq) \tag1}$$
$$\therefore \ K_\mathrm{sp} = [\ce{Pb^2+}][\ce{I-}]^2 = 8.7 \times 10^{−9} \tag2$$
Now, using equations $(1)$ and $(2)$, you can calculate $[\ce{Pb^2+}]$ and $[\ce{I-}]$ in water.
In  $\ce{Pb(NO3)2}$ solution, following ions are present:
$$\ce{Pb(NO3)2 (aq) -> Pb^2+ (aq) + 2NO3- (aq) \tag3}$$
Thus, $\ce{Pb^2+}$ is present abundantly in the solution as the common ion (suppose $[\ce{Pb^2+}]$ is $\pu{0.1 M}$). Still, the equilibrium representing the equation $(1)$ should establish regardless of the presence of common ion. However, the equilibrium concentrations of $[\ce{Pb^2+}]$ and $[\ce{I-}]$ are very different than that of with pure water you calculated before. Yet, $K_\mathrm{sp}$ is still the same. So, you can again calculate $[\ce{I-}]$ using the equation $(2)$ and known $K_\mathrm{sp}$. Here, $[\ce{Pb^2+}] \approx \pu{0.1 M}$. You should think about the reson why we are using $[\ce{Pb^2+}] = \pu{0.1 M}$ here.
Once you find $[\ce{I-}]$ in water and in $\ce{Pb(NO3)2}$ solution, respectively, you would be able to see which solution makes $\ce{PbI2}$ more soluble.

If i didn't do it wrong, I think that PbI_{2} is more soluble in water. Am I right?

Excellent! Keep up with your work. Chemistry is not that bad. :-)


text:
Are those magnetic iron filings in the dirt pure iron?
When I was a kid we used to drag magnets around in the dirt and collect "iron filings". Are they pure iron, or something else?

How in the world would anyone be able to know? There are many different sorts of dirt out there, after all.

I’m voting to close this question because we just don't know.

Metallic iron deposits are exceptionally rare in the natural environment.  The filings were almost certainly not iron, unless they were scraps or powders from human-based industrial operations.

The earth surface is littered* with rust particles, mostly from human sources. $\ce{Fe2O3}$  is magnetic.

More likely they were iron oxides. In moist soil ,iron/steel filings would not last more than a few weeks before turning to oxides. Now the "oxides" are not simply Fe O, Fe 2 O 3, Fe 3 O 4 , they are also hydrates, hydroxides ,etc.

Then again, not all oxides and hydroxides are magnetic.

Agreed, i type slow and did not put in that qualification.


text:
Does the 4s orbital really have a higher energy level compared to 3d?
For writing the electronic configuration of the $\ce{Fe^2+}$ ion, you need to remove two electrons (as it is an ion) which are from the highest energy level.
Apparently, this would be $\mathrm{4s}$ from $n = 4,$ but I see that $\mathrm{4s}$ is actually lower than $\mathrm{3d}$ in the shell $n = 3.$ Is there a flaw in my understanding?

Removal of e takes place from outermost shell, although 4s is in lower energy, it is in the outermist shell. So.

https://chemistry.stackexchange.com/questions/35674/why-do-ironii-ions-and-chromium-have-different-electronic-configurations

https://chemistry.stackexchange.com/questions/8357/why-does-the-3rd-electron-shell-start-filling-up-with-scandium https://chemistry.stackexchange.com/questions/5108/in-which-cases-does-the-3d-subshell-take-precedence-over-the-4s-subshell

Yes, generally, 4s orbital indeed has higher energy than 3d orbital. Based on what I read, that 4s orbital is at lower energy level than 3d orbital is a wrong concept, as that has been proven to be wrong spectroscopically and by calculation.1 But as Poutnik has pointed out, the order of orbital energy is not fixed.
If then, why we normally say that 4s orbital will be filled before 3d orbital?

Although 4s orbital has higher energy than 3d orbital, the most probable distance of 4s electron from the nucleus is longer than than of 3d electrons, so if you put two electrons in 3d orbitals, the two electrons will be closer together and the repulsion will be more significant. Else if you put two electrons in 4s orbitals, yes, you first need to pay higher energy because 4s higher in energy, but the cost is less than the one incurred by strong repulsion when placing two electrons in 3d.1

Why do we remove electron from 4s but not 3d?

Remember that the energy level is not fixed. The 3d repulsion effect mentioned above can be compensated when your effective nuclear charge is big enough (especially when you are discussing about cations).1 Just like you put electron in further 4s, but it got pulled down to 3d which is closer to nucleus (in terms of most probable distance).

1 Peter Atkins, Julio de Paula. Physical Chemistry (8th Edition). OUP. 2006. Page 342.
P.S. I don't know why, based on the reasoning above, K is 3d0 4s1 but not 3d1 4s0

Because for K, 4s has lower energy than 3d. Their energy mutually switches later, for scandium upwards. So does Ca configuration is 3d0 4s2. By other words, for 3d+4s, there is a kind of filling/removing hysteresis.

The order of orbital energies is not fixed, but depends on electron filling, because of different levels of kernel screening and mutual electron repulsion.
When both 3d and 4s are empty, 4s orbitals have lower energy than 3d and get filled first. But being filled , somewhere at scandium 4s orbitals get higher energy than 3d, so they get away first. This get few exceptions when 3d is half or fully filled.

As per $(n+l)$ rule, $\mathrm{4s}$ as a value $n=4$ and $l=0$, therefore $n+l =4$.
In the case of $\mathrm{3d}$, $n=3$ and $l=2$, therefore $n+l=5$.
So, low energy orbital is $\mathrm{4s}$ and not $\mathrm{3d}$. Therefore electrons are easily lost from the $\mathrm{4s}$ level.


text:
Reaction (or lack thereof) between oxygen and halogen gas
I understand that oxygen and halogens are both very electronegative elements and it would makes sense that they don't want to associate with each other at all. However, I know that oxides of halogens ($\ce{Cl2O, ClO2, ClO, Cl2O2}$ etc.) do exist. When reading about halogens, it says that chlorine gas doesn't react with oxygen gas. I tried to find the intuitive reasoning behind this phenomenon online, but to no avail.
I know that chlorine and fluorine gas can react to produce chlorine trifluoride (fluorination of chlorine), proving that two diatomic, covalent and non-polar gases of electronegative elements can react.
$$\ce{Cl2 + 3F2 -> 2ClF3}$$
Also, drawing comparison with the reaction between nitrogen and oxygen gas:
$$\ce{N2 + O2 -> NO}$$
$$\ce{N2 + 2O2 -> NO2}$$
Nitrogen's triple bond is loads stronger than the covalent bonds from oxygen or chlorine, and it can still react with oxygen gas, albeit only under extreme conditions (lightning). Is this the case with chlorine? Even so, it only makes sense that it requires a lower temperature, as the energy needed to atomize them is lower (I haven't done the calculations with enthalpy and entropy yet). I can't find any reference to a reaction such as this online, though.
Most oxides of chlorine can only be made from compounds like chlorites and chlorates or by other novel methods. Is there a reason oxygen and chlorine (or other halogens, but chlorine in particular) can't react with oxygen gas? Is there something stopping them from reacting that I don't know about? Is it due to undesirable entropy changes? Or are the reactions I mentioned special, allowing two relatively nonreactive gases to react?

All chlorine oxides are endothermic, to begin with.

Oxygen and fluorine reacts directly with each other to form dioxygen difluoride, $\ce{O2F2}$.

You cannot get chlorine and oxygen to react directly. Just like oxygen and nitrogen. They simply won't do it, except in an electrical arc (which is terribly inefficient). You need to take detours: chlorine + water + electrolysis, or make ammonia from nitrogen, and burn that afterwards.

Yeah, that goes with what I read. But why do they not want to react, though? What's keeping them from reacting directly? Is it the high activation energy?

@chematwork You already answered that in your first sentence above! See also Ivan's comment.

@Karl It does make sense to just go with that... but chlorine and fluorine are able to directly react. Am I overthinking things? Is it really just because oxygen has a strong double bond and fluorine has a weak single bond? It'll be nice if it really is that simple :)

Fluorine also directly reacts with oxygen. Because firstly it is much more reactive (kinetics!) than chlorine, AND the resulting oxides are more stable (thermodynamics!).

Your on the right track: in order to form a new compound, old bonds must be broken. This requires energy, even though the reaction may, overall , be exothermic. This is called the activation energy.
Think of it as pushing a car a short distance up one side of a hill before it reaches the peak and then rolls a long distance down the other side.
For example, in the dark, at room temperature, $\ce{Cl2}$ and $\ce{H2}$ coexist quietly in a flask... until light activates the chlorine (wait for the bang).
On the other hand, add enough energy, and even endothermic reactions take place, and, if the product is comparatively stable, or metastable, some can be left afterwards. For example $\ce{O2}$ is fairly stable, but given some energy input, e.g. from an electric discharge or from shortwave ultraviolet radiation, $\ce{O3}$, ozone, is produced, though that reaction is not thermodynamically favorable. The ozone is waiting to decompose... once it's activation energy is achieved.

So, my suspicions were right? $\ce{Cl2}$ and $\ce{O2}$ can react, but it requires very high temperatures normally unachievable. Drawing comparison to the reaction of  $\ce{N2}$ and  $\ce{O2}$ in the presence of lightning,  $\ce{N2}$ has a much stronger triple bond. It just doesn't make sense how oxides of nitrogen $\ce{NO2}$ and  $\ce{NO}$ are known to be formed from their elemental gases but not oxides of chloride. Even after radicalization, chlorine is said to react with ozone, not oxygen gas.

@chematwork Not exactly. Chlorine oxides are even more unstable at higher temperatures, just like e.g. mercury sulfide: you heat it, it decomposes into the elements (or other, more stable compounds). Chlorine oxides are usually produced electrochemically.

@Karl So nitrogen oxides are more thermostable? Or is lightning more akin to an electrochemical mechanism?

@chematwork In a lightning, you have radical reactions, and the product cools extremely fast afterwards.Similar thing is done in industry: You burn ammonia (at a vanadium catalyst), and cool down the product NO as fast as possible. NO can then react with more oxygen, then add water and get HNO3.

@Karl Hypothetically speaking, if chlorine and oxygen were subject to the same thing, then will oxides of chlorine form?

@chematwork You cannot do "the same thing" with chlorine and oxygen. Have a look on e.g. wikipedia how chlorates etc. are produced, don't reason by drawing questionable analogies. ;-)


text:
Reaction of copper with nitric acid
When copper reacts with dilute nitric acid,
$$\ce{3Cu + 8HNO3 -> 3Cu(NO3)2 + 2NO +4H2O}$$
while with concentrated nitric acid,
$$\ce{Cu + 4HNO3 -> Cu(NO3)2 + 2NO2 + 2H2O}$$
As we see that with dilute acid oxidation state of nitrogen changes from $+5$ to $+2$ and with concentrated acid it changes from $+5$ to $+4$ but shouldn't it be vice- versa because since concentrated nitric acid is a good oxidizing agent and should show a huge change in oxidation number ?
Where am I going wrong?

Though not related, hot, alkaline potassium permanganate changes the O.N. of manganese from 7+ to 6+ to 4+ in MnO4. So, its O.N. change is not as drastic as in acidic medium i.e. from 7+ to 2+ to form Mn2+ ions. But it is a much stronger oxidising agent in basic medium than in acidic medium.

Do you know the reason

You may reconcile both equations by stating that your second reaction occurs in all nitric acid solutions (concentrated or dilute). And in dilute solutions, you may admit that there is enough water to destroy $\ce{NO2}$ according to the following equation :
$$\ce{3NO2 + H2O⟶ NO + 2HNO3}$$ And multiplying this new equation by $2$, and adding to your second equation (after multiplication by $3$) gives your first equation

Because nitric acid is such a good oxidizing agent, when sufficiently concentrated it could oxidize any nitric oxide that tries to form.
There are various ways to view this effect.  See answers to this question for some ideas.


text:
Can a reaction have more than one rate determining step?
There are many organic reactions in which two steps have comparable rates. Can we say they are both RDS?
For example, the rate of iodination of benzene and deuterated benzene is different even though the step is not considered RDS.


text:
Is there a standard scholarly reference for lattice constants of crystals of the elements?
I need to discuss the lattice constants of bulk crystals of several metal and semiconductor elements. I can find plenty of tables and numbers that are probably "close enough" but for a paper I'd like to cite a standard source.
From  https://periodictable.com/Properties/A/LatticeConstants.html I have the following numbers, but I don't want to use a "dot com" as a scholarly source, and I can not figure out how to use Wolfram Alpha (the source for this website) or understand where WA gets its numbers. They list several references here https://reference.wolfram.com/language/note/ElementDataSourceInformation.html , but it will be a challenge to track them all down one by one. I'm hoping someone will recognize one of them or simply be able to mention it.
I only need (at a minimum) three decimal places for the lattice constants in Angstroms (though four is better if they are known), but what I can't find is a citable, schollarly source that covers all of these and is likely to cover other common elemental crystals when I need them in the future.
element.   lattice constant
             (a, b and c)
              Angstroms

   Au          4.0782
   Ag          4.0853
   Pb          4.9508
   Ge          5.6575
   Si          5.4309

Wikipedia cites a lot of sources. Elements I'm pretty sure you will find in the CDC handbook. http://www.crystallography.net/cod/ https://it.iucr.org/

@Karl I'm having trouble understanding that link. I won't be purchasing an on-line copy, is this a book I can find in some library or is it strictly a paid access online database? I'd like to be able to cite something that one could actually check and confirm in a reasonably well-sized and libraried university. I can't even figure out which of the eight volumes (A through H) has an actual table of lattice constants with numbers in it.

@Karl I'm one of those people who won't cite something unless I've actually check it myself.

It's two links, and any uni library should give you access. The CRC (sorry, typo) handbook used to be printed, now it's online too, and your library should have access, too. http://hbcponline.com/faces/contents/ContentsSearch.xhtml;jsessionid=3068098853C27DB9ADAD2661032F9515

@Karl okay got it! Let me try a device that's connected to the library, this one isn't. :-)

The CRC Handbook of Chemistry and Physics contains a dedicated compilation by H. W. King, titled «Crystal Structures and Lattice Parameters of Allotropes of the Elements».  In case your research library is closed, you may access some of its editions freely or borrow them with the library card of archive.org.
In case of the 97th edition (by 2016), the section starts by page 12-16.  (Elements liquid or gaseous at ambient conditions are included.)

source

Thanks, this may be what I end up using. It's a real, recognized and trusted physical reference I can hold in my hand and read, and it lists its source from where these numbers come from, which I can also put my hands on and confirm. Thanks!

The lattice constants of all five elements have been published in a single reference in 1925 (Ref.1):

The lattice constant $a$ has been determined within 0.1 percent (.03 percent for $\ce{W}$) for aluminum, iron, nickel, copper, molybdenum, palladium, silver, tungsten, platinum, gold, lead, and bismuth, by direct comparison with $\ce{NaCl}$,  $a(\ce{NaCl}) = \pu{2.814 \mathring A}$. As pure samples as could be obtained were used, from 99.55 percent for $\ce{Ni}$ to 99.9995 percent for tungsten, and in many cases commercially pure samples were also measured for comparison. The results for the purest samples are summarized in Table XIII. The density from the x-ray data is in each case (except $\ce{Al}$ and $\ce{Ag}$) greater than the density of the bulk metal as given in the literature, the difference being rather large for $\ce{Mo}$ (10.21 vs 9.1), for $\ce{Pd}$ (12.25 vs 11.9) and $\ce{W}$ (19.32 vs 18.77). For pure $\ce{W}$ remarkably sharp lines were obtained. For $\ce{Bi}$ a piece of a single artificial crystal was used.

The relevant $a$ values are listed in a Wikipedia article as follows:
$$
\begin{array}{c|ccc}
\hline
\text{Metal}  & \text{Lattice constant}^a & \text{Crystal structure} & \text{Lattice constant given} \\
\hline
\ce{Au} & \pu{4.065 \mathring A} & \text{FCC} & \pu{4.0782 \mathring A} \\
\ce{Ag} & \pu{4.079 \mathring A} & \text{FCC} & \pu{4.0853 \mathring A} \\
\ce{Pb} & \pu{4.920 \mathring A} & \text{FCC} & \pu{4.9508 \mathring A} \\
\ce{Ge} & \pu{5.658 \mathring A} & \text{Diamond (FCC)} & \pu{5.6575 \mathring A} \\
\ce{Si} & \pu{5.4310205 \mathring A} & \text{Diamond (FCC)} & \pu{5.4309 \mathring A} \\
\ce{Cu} & \pu{3.597 \mathring A} & \text{FCC} & \pu{3.6149 \mathring A} \\
\ce{Pt} & \pu{3.912 \mathring A} & \text{FCC} & \pu{3.9242 \mathring A} \\
\hline
\end{array}\\
^a \ \text{Values from: Phys. Rev. 1925, 25(6), 753-761 (Ref.1; as listed in Wikipedia)}
$$
For comparison purposes, I have included $\ce{Cu}$ and $\ce{Pt}$ as well. Lattice Constants of Crystals of three elements of the 5 listed have also been discussed in a relatively new article (Ref.2).
References:

Wheeler P. Davey, “Precision Measurements of the Lattice Constants of Twelve Common Metals,” Phys. Rev. 1925, 25(6), 753-761 (https://doi.org/10.1103/PhysRev.25.753).
D. N. Batchelder, R. O. Simmons, “X‐Ray Lattice Constants of Crystals by a Rotating‐Camera Method: $\ce{Al, Ar, Au, CaF2, Cu, Ge, Ne, Si,}$” Journal of Applied Physics 1965, 36(9), 2864-2868 (https://doi.org/10.1063/1.1714595).

To resolve the discrepancy between Wolfram Alpha and (Davey 1925) that are tens to a a hundred times larger than three decimal places in Angstroms specified in the question, I'm going to need a more modern source which reasonably represents some generally agreed-upon values by the crystallographic community. I don't think it's reasonable to cite a single measurement in a single author paper from 1925 in a paper to be published in 2021.

-1 Checking both columns (1925 numbers and the ones from the question) against a CRC from 2012 (93rd Ed.) show that some of the numbers from 1925 are pretty poor; for the three metals mine are very close to CRC, the 1925 numbers differ by 0.3 to 0.6%! It is certainly up to you, but if you would consider deleting this answer I'd like to ask in a different SE site instead. Thanks!

@uhoh: You are completely wrong on this down voting. I have found a reliable literature satisfactory for your request "I only need (at a minimum) three decimal places for the lattice constants in Angstroms." What not reliable is your sited dot com page, which even not site any references for their claim. You didn't even appreciate my effort. Shame on you on that.

I'm confident that my down vote accurately represents my view, which is that this answer is "not useful." When writing a paper no reasonable author chooses a 100 year old measurement to cite when much newer, more careful and more carefully reviewed data is available. Please recheck the complete text of my question post and my previous comment.

Considering how badly those number match modern values, how can you claim this 1925 paper to be a "standard, reliable" source?

Without those papers you won't be here arguing this. Don't underestimate their dedication. The sited are peer-reviewed papers. According to your argument, you cannot trust Diels-Alder or Claisen rearrangement or Friedel-Crafts. The original work of those are also more than a century older. Have some respect.

The comparison of this early experimental result to that body of work does not make sense. A New X-Ray Diffraction Apparatus (Davey 1921) describes the apparatus in question and here's what it looked like XRD lattice constant determinations have improved tremendously over the last century. Those numbers are inaccurate by amounts many times larger than their stated repeatabilities.

Measurement equipment gets better over time, we don't cite 100 year old measurements when far better values using better equipment is available. This is just common sense. They did great and pioneering research at GE Schenectady in the 1920's. We can honor that but that doesn't mean we should use those measurements and ignore a century of improvements!


text:
Can we use PV^y = constant equation for irreversible adiabatic processes?
Example: One mole of monoatomic gas expands at initial temperature $T$ at constant external pressure of $\pu{1atm}$ from $\pu{1L}$ to $\pu{2L}$.
To find final temperature, why can't we use $PV^y$ or $TV^{y-1} = \text{constant}$?
I am able to get final Temperature using $W+U=0$.

Imagine reversible expansion against a piston, versus irreversible expansion to another contained with vacuum. Is there any difference in W for the former and for the latter ? Se also Joule-Thompson efect

It is extremely difficult to carry out any process in an irreversible way.

@Maurice Why ? It is extremely easy. It is any process where we cannot consider the system being in equilibrium in every moment. What is extremely difficult to manage is a reversible process.

@ Poutnik. To transform a gas sample irreversibly, you have to adapt the inner pressure to the outer pressure during the whole transformation. You also have to equalize the temperature of the inner gas and the temperature of the surroundings. That is not easy to carry out.

Sure, I fully agree, but I want to point out you switched the meanings of the terms reversible and irreversible, so where you say irreversibly should be reversibly.

https://en.wikipedia.org/wiki/Reversible_process_(thermodynamics):  "Throughout the entire reversible process, the system is in thermodynamic equilibrium with its surroundings."..."Since it would take an infinite amount of time for the reversible process to finish, perfectly reversible processes are impossible."

Your original equation depends on the ideal gas law being satisfied, and the ideal gas law is not satisfied in an irreversible process (exxcept in the initial and final thermodynamic equilibrium states).  The ideal gas law is satisfied only at thermodynamic equilibrium (or in a reversible process, which is comprised of a continuous sequence of thermodynamic equilibrium states).


text:
Ideal gas equation derivation proportionality
While deriving the Ideal gas Equation using Boyle's, Charles', and Avogadro Laws,
how are each of the laws, which have different proportionality constants and different quantities as constant combined?
(eg: Boyle's  law which assumes the temperature and moles of gas to be constant, but in Charles' law temperature changes. How is it possible?)

If you consider the ideal gas state equation $pV=nRT$, and if you keep either of the 3 state variables p,V,T constant, involving its value into the equation constant, you get each of the respective original gas laws.

Related question: https://chemistry.stackexchange.com/questions/58330/how-can-we-combine-all-the-three-gas-laws-into-a-single-ideal-gas-equation.

I understand your concern because this is the way it is taught in South Asian schools and colleges and even general chemistry texts taught elsewhere are silent about it. They state all the individual gas laws using a proportionality sign "$\propto$" and magically all gas laws are combined and that becomes $R$- the universal gas constant. This is done in a single paragraph.
This is not the proper way how the ideal gas law was derived. Out of curiosity, I checked the original paper " Mémoire on the Motive Power of Heat" where the ideal gas law appeared in 1834!!. It is in French, but one can see the equations. Not once he used any proportionality sign but there is a lot of calculus there.
https://gallica.bnf.fr/ark:/12148/bpt6k3414331n/f34.image.r=M%C3%A9moire%20sur%20la%20puissance%20motrice%20de%20la%20chaleur
You should scroll the original paper and feel the length of it to appreciate that the ideal gas law was not derived as shown in your school / college textbooks.
You would see a formal derivation of the ideal gas law in higher classes using the kinetic theory of gases.

Thank you for your help everyone. I knew there were other derivations, but I wanted to clarify the validity of the method.


text:
Challenges Associated with Earth's field NMR
The ACS recently hosted a virtual conference and there was an interesting talk on NMR experiments using the Earth's magnetic field. What surprised me was the number of molecules studied with Earth's field NMR so far is around 22. Another interesting aspect was that NMR at low magnetic field strength is richer in the number of lines as compared to the extremely strong magnets used today. I think spectroscopists might be happier if they get a rich spectrum (good for finger printing).
This site has many NMR researchers, my question is why Earth's field NMR never caught up? The number of molecules studied so far is very low. Is it theoretically very challenging or the analytical sensitivity is hopelessly low?

Hmm, I have never been an NMR expert, but AFAIK, typical frequences for x T magnetic fields are tens of MHz, what may look like(low?) hundreds of Hz for Earth field. It may be technologically challenging to work quantitatively with such low frequencies.

Poutnik, I do not think technology is the issue. An Earth's field NMR for advanced physics teaching lab costs just < $ 10,000.

Earth’s field is <1 gauss, so  >10,000x lower than even benchtop NMR, which today is always >1 T. At such low fields there is essentially no spectral dispersion from chemical shift, so we give up the most important analytical information provided by NMR.
At such low fields sensitivity is extremely low, requiring a large sample volume of order 1 L, at least for inductive detection.
It is difficult to preserve the high homogeneity of the Earth’s field in normal laboratory and industrial environments.
Rare-earth benchtop magnets at 1-3 T do not cost much more than an earth’s field instrument, and they provide much more analytical power.

[disclosure: I’m in the benchtop NMR biz qmagnetics.com]

Interesting to see your systems. Is it an HPLC system or a flow system? What is the magnetic field strength in your NMRs?

125 MHz is almost 3 T, the highest of any benchtop. It is a proton NMR spectrometer with the sample fluid delivered to the coil by flow. You can inject a sample with a syringe for “walk-up” use or integrate it with other systems. The software interface will be open to allow anyone to develop hyphenated applications.

One question I'd ask in return is, do you expect ultralow-field NMR to replace high-field magnets? You'd likely agree that this is impossible, given how dependent other fields of research are on high-field NMR. So it's not plausible for universities or other institutions to get rid of their high-field setups and replace them with ultralow-field NMR.
Once we accept that, then the question becomes: what is the benefit of adding ultralow-field setups to an existing high-field NMR facility? Given that the low-field spectra are difficult to interpret (after all we're talking about magnetic fields several orders of magnitude smaller than what we're used to, and possibly with multiple nuclei overlapping with each other), I don't see why people would be in a rush to get their hands on low-field setups, when they can get their 500 MHz spectra and go off happy. On top of that, there's an obvious barrier to entry in terms of technical knowledge required to build and operate such a spectrometer.
I certainly don't mean to discredit the existing research that has been done on it, of course. It is not bad, or useless, to perform research on this. However, realistically speaking, and as of the time of writing, I don't see the incentive for this to be "the next big thing"... yet.
The current trend in "lower"-field NMR is probably benchtop NMR spectroscopy, and (apart from arguments like portability) that's almost certainly only because it still gives you pretty decent spectra (e.g.., it's suitable for undergraduate chemists to look at and apply the $n+1$ rule, etc.). If benchtop spectrometers were 1 kHz I doubt anybody would be buying them.

Certainly, High field magnets are not going anywhere! I forgot, what is the current record now in terms of magnetic strength for NMRs? I was told there are only two or three such NMRs in the world. This Earth field NMR, if it costs less  $10,000, should be present in all universities for teacing purposes. Spectral intrepretation is another story.

@M.Farooq I think the record is 1.2 GHz now.

This is amazing! 28 Tesla magnets. I wonder that just like very magnetic field NMR is strange, if we keep on increasing the magnetic field, newer phenomena may start to appear.


text:
Why aren't solid pure transition metals colourful?
I understand that the compounds of transition metal are colourful because of the appropriate energy gap between different d orbitals. But why aren't they also colourful in their solid pure state? For example, why isn't Fe colourful whereas, its compounds are?

The colour of TM compounds is typically because of electronic transitions. The electronic structure of bulk metal is not the same as the electronic structure of an isolated metal ion surrounded by ligands. Look up "band theory" for example.

I don't understand. In a piece of metal, the metal atoms form metallic bonds, leaving the valence shell electrons. This is also, often, the case in some TM compounds (like in CuNO3). Hence, the electronic configuration of the metal atom is the same. And since we generally exclude the anions or ligands in determining colour, why the discrepancy?

We dont exclude the ligands, they cause crystal field splitting to different extents. So we are including them. Individual differences are also observed like in pi acid ligands.

If there are no ligands, there is no splitting typical to crystal field. So the d orbitals are degenerate. And no absorption occurs in the visible region.

Don't you consider copper (red in color) as a transition metal?

Yes, nevertheless, its compounds are blue. I was just looking for why there is a difference after all.

@ Mathew. That is a good question. Do you know why copper is red ? And why Copper is an exception (apart from gold, which is due to relativist contraction) ?

You said it already: it’s the metallic bonds that make things different. Do you expect a nitrogen atom and N2 to absorb light of the same wavelength? No, because the covalent bonds lead to a different set of orbitals with a different set of energies. Same with the metallic bonds. Note also that in a metal there are lots of metallic bonds, and each neighbour is further bonded to lots more atoms. In Cu(I) compounds there’s only one Cu-Cu bond and you can treat it as an isolated Cu2(2+) unit.

@orthocresol how is copper(I) dimeric when each ion has a closed subshell configuration?  Zinc(I) is dimeric, but the ions bond to each other with electrons whose counterparts do not exist in copper/Group 11.

@OscarLanzi I am mortified. Was obviously too sleepy, probably thinking of Hg(I)....... Thanks for pointing out...

Colour is a property of electronic transitions: the transitions in metals are different to those in their compounds
In many discrete compounds of transition metals the colour arises because there are accessible electronic transitions between molecular orbitals in the molecules (usually involving metal d-orbitals) matching the energy of wavelengths of visible light. These transitions are often fairly narrow so giving a wide range of possible colours.
Pure metals are not like that. To simplify a lot, the definition of a metal is having a continuous band of electronic states (not a simple molecular orbital) for the electrons involved in conduction of electricity (the conduction band). This band involves the whole bulk substance, not just isolated atoms of the metal. So the transitions that might involve colour (d-d transitions in a molecule) are not present in the same way: instead transitions occur inside the conduction band or between orbitals and the conduction band. Either way, the narrow transitions giving colour in compounds are now much broader because  a wide range of continuous electron energies exist in the conduction band.
This is a gross simplification, but it is good enough to explain the key observation if you don't want to get into the theory behind the electronic structure of metals.


text:
Which chiral carbon do L and D in L- or D-glucose refer to?
I see and D- and L- glucose are mirror images of each other. and I understand D and L are for chirality. But it is not clear which carbon do D and L refer to. Could anybody let me know? Thanks.
What are the differences between D and L glucopyranose?

For some more context, see e.g. https://chem.libretexts.org/Courses/Purdue/Purdue%3A_Chem_26200%3A_Organic_Chemistry_II_(Wenthold)/Chapter_22._Carbohydrates/22.03%3A_The_D_and_L_Notation

Tbe discussions in both places are still confusing to me. I don't think they directly answer my question. So D and L do not refer to any specific carbon in glucose?

They always refer to the chiral carbon furthest from the carbonyl, so carbon 5 in this case.

I think your confusion arises from trying to locate a particular carbon atom, which is D or L. This is not the case.
Points to remember:
The reference for D and L notation is glyceraldehye molecule not a particular carbon atom.
D and L are not equivalent to R and S which are individually applicable to each chiral center in a molecule.
Similarly, D and L cannot tell you the direction of rotation of light before hand.
Capital “D” and “L” notations have no connection with small d and small l, which tell you the direction of rotation of light.

So D and L can only be determined by experiment (or complicated computational program). This is no way that I can easily determine whether a molecule is D or L manually? So the answer by @KarstenTheis "carbon 5 in this case" is wrong? Thanks.

No, Karsten is also right. Yes, in order to assign D or L, all you need to know is the correct Fisher projection and compare it with glyceraldehyde.


text:
Can mixing too much citric acid and bicarbonate of soda create dangerous levels of carbon dioxide in the home?
I am not a chemist. In order to clean hard limescale in my toilet, I usually put down half a cup of bicarbonate of soda and half a cup of citric acid.
Someone told me this is dangerous because it creates too much $\ce{CO2}$. Is this true?

No, it's not.

And it doesn't sound like a good way to clear limescale either.

Thanks for your comment and it cleans it off just fine.

Since it is about chemistry here: fine it cleans but the mixing does not make sense / It does create CO2 but  unless you do that on tub scale see Mithoron above.

Using both acid and bicarbonate does not make sense. Limestone and limescale is destroyed by citric acid or any other acid. Using citric acid is a good way of getting rid of this deposit. But adding bicarbonate is counter-productive. Because bicarbonate destroys citric acid (as any other acid). So adding bicarbonate to citric acid is similar to removing citric acid, and preventing it to dissolve limescale or limestone. It produces huge amounts of foam, but it does not clean your toilet.

I have used citric acid in toilets lots of times, but never with sodium bicarbonate. At work, the incoming water (rusty pipes?) was so discolored that I would add a spoonful (not a half cup!) of citric acid and stir with a brush - the bowl sparkled.
At home, a spoonful of citric acid would clean off any grunge that had accumulated, but I admit I had no hard deposits.
If you assume a half cup of sodium bicarbonate is about $\pu{200 grams}$, and add enough citric acid to react completely with it, and it all comes out of solution, you could get $\pu{53 liters}$ of $\ce{CO2}$ gas, or about 14 gallons, which is indeed, quite a lot. In fact, unless you have some special technique, I would expect your toilet bowl to bubble over. Perhaps your cup is somewhat smaller than I imagine.
A normal concentration of $\ce{CO2}$ in air is about $\pu{400 parts per million (ppm)}$, but can get up to $\pu{1000 ppm}$ indoors, safely. The workplace limit (8 hour exposure) is $\pu{5000 ppm}$. Possible headaches at this level. Over $\pu{40,000 ppm}$, coma, death, or even worse, brain damage (See What are safe levels of $\ce{CO}$ and $\ce{CO2}$ in rooms?). That's $4$% percent. The $\pu{14 gallons}$ of gas would occupy $\pu{1.87 cubic feet}$, so if your bathroom were larger than $\pu{46.8 cubic feet}$, even with the door closed, you could escape death - and you would probably feel a strong desire to get more air. Now even a small bathroom $(\pu{ 8 \times \, 8 \times \, 7 feet})$ would have almost 10 times the volume, so you would be below the workplace limit, and you could get away with only headaches or dizziness. With any kind of ventilation (open a window or turn on the vent), there should be no significant health issue. (I am still imagining $\pu{14 gallons}$ of foam rising in a toilet bowl.)
Nevertheless, I suggest you use the citric acid alone - well, not completely alone - help it with a brush and let the acidic solution remain in the bowl for some time ($\pu{10 - 15 minutes}$ if needed). The bubbling action of the bicarbonate is probably not much help (opportunity for an experiment!). Although $\ce{CO2}$ does dissolve limestone (stalagtites and stalagmites), the citric acid does better.

I think citric acid is limiting, and there is only about 1 mole of it, enough to produce about 1 mole of carbon dioxide if you consider the speciation diagram. So I think your estimate is too high by a factor of about two.


text:
Supporting electrolytes and increased conductivity?
Hi a question about supporting electrolytes (SE) I hope you can help with.  I found some related questions here through searching but looking for a simple answer (if there is one!):
If SE are electroinactive (in the potential range of interest) and so do not pass or receive electrons at the electrodes to complete the circuit then how to they "increase conductivity"?  I am referring to typical DC experiments, e.g. chronoamperometry.  It's not like electrons hop through the solution from one electrode to the other using SE ions as a shuttle.
I understand that one role is that they migrate to the opposite electrodes to neutralize ion charge imbalances that would build up due to reagent consumption or product generation from the electroactive species you are measuring (e.g oxidizing an enzyme product), thereby facilitating these redox reactions that do occur to continue. Is this the answer?  So in just water + SE, for example, it is more conductive because they facilitate the electrolysis of water (H2 and O2 generation (e.g for SE without Cl-) by this method?
I also understand that movement of ions is considered current, but with DC wouldn't this migration slow down and exhaust as time went on? Is it this transient conductivity due to ion movement that they refer to?
Thanks!

Thanks for your answer Maurice.  What you describe sounds like this (maybe not clearly written) part of my own possible answer to my question: "I understand that one role is that they migrate to the opposite electrodes to neutralize ion charge imbalances that would build up due to reagent consumption or product generation from the electroactive species you are measuring,  thereby facilitating these redox reactions that do occur to continue".
I think my confusion arises because the IUPAC definition is:  "A supporting electrolyte, in electrochemistry is an electrolyte containing chemical species that are not electroactive (within the range of potentials used) and which has an ionic strength and conductivity much larger than those due to the electroactive species added to the electrolyte", which makes it sound like the the SE ions themselves are conductive.  Plus the most common explanation offered is along the lines of "because they are ions and ions conduct".
If I understand correctly now as discussed, they actually facilitate the reaction of the true electroactive species (which produce faradaic currents through redox - the current being a measure of conductivity), by maintaining neutrality around the electrode, rather than being "conductive" themselves (along with other roles such as making results more purely diffusion controlled, etc).
Put simply, the SE ions do not increase conductivity directly by contributing to the current (like an electroactive ion would), but increase conductivity by increasing the rate of faradaic processes of the actual electroactive species that are present (by using the SE ions to maintain electrical neutrality around the electrode).
Is this it?
Again, thanks for your time!

You are mixing many things. Start from here https://www.youtube.com/watch?v=QiCSTlgFwCY and perhaps refine your query.

Thank you for you attention to my question.  That video discusses the role of SE in making the results being controlled by diffusion, rather than migration and diffusion, which i understand OK.  My question is how electroinactive ions of the SE "increase the conductivity" of the solution, as is commonly stated, (when if they are electroinactive they do not produce any (Faradaic) current themselves).

@ Bob Tomas. Your supplementary information is correct. One might add that the negative pole (cathode) attracts positive ions (cations) and that the positive pole (anode) attracts negative ions. The movements of these ions create the electric currant in solution.

Thanks for the followup and informative answer!  I would accept your answer too if I knew how to accept more than one!

Remember that in electrochemistry whenever a direct current is passing through the solution there must be electrolysis occurring at the electrode. Now, all you want to see in electrochemical experiment, is current due to the oxidation or the reduction of the desired solute which is diffusing to the electrode in an unstirred cell.
The gist of your excellent and thought provoking question is:

My question is how electroinactive ions of the SE "increase the conductivity" of the solution, as is commonly stated, (when if they are electroinactive they do not produce any (Faradaic) current themselves).

After giving it a little thought and some literature search, it is clear that supporting electrolytes increase conductivity is an erroneous conclusion which is rampant everywhere. In the electrochemical experiment, the flux of the supporting electrolyte must be zero, it should not be carrying any current, because by definition, it is non-reactive in that potential window. The mass transport of the electro-inactive species must be zero.
*Yes, separately, if we were to measure the conductivity of a solution to which an supporting electrolyte, the conductivity would be higher than the pure solvent..but that is a separate measurement. Remember that conductivity is measured by very high frequency AC current.
The book, Electrochemical Engineering by Thomas F. Fuller, John N. Harb, explains it better, pg 69:

It is frequently reported or implied that the role of the supporting electrolyte is to reduce the electric field by carrying the current. This conclusion is in error since, at steady state, the flux of any nonreacting species must be zero, and the supporting electrolyte is by definition non- reacting. Therefore, only the fluxes of the reacting species contribute to the current (see Equation 4.4). That said, the presence of the supporting electrolyte does diminish the magnitude of the potential gradient. To explore this further, consider a system at steady state with no bulk flow where any transport of the supporting electrolyte due to migration must be offset by diffusion. Concentration gradients of the nonreacting species must be connected to those of the minor reacting species in order to preserve electroneutrality. Therefore, concentration gradients of the supporting species cannot be large. Furthermore, the conductivity of the supporting electrolyte is high. These two facts together make it impossible to support a large potential gradient in solution. As a result, migration can be neglected for the reacting species when its concentration is significantly less than that of the supporting electrolyte.

Many, many thanks!  This has been bugging me for ages and many hours of searching.  Finally, you say "the conductivity would be higher [with SE] than the pure solvent..but that is a separate measurement". Is there a simple explanation of what process is being measured in this other AC conductivity measurement (that is increased by SE)?

In those experiments, one is measuring the resistance of the solution by using AC current. The purpose of the AC current is to "undo" the electrolysis- remember whenever direct current is flowing in a solution, there must be electrolysis. AC trick is to avoid it the by very quickly changing the electrode polarity. You can accept the answer if you want.

Great! again thanks for taking the time to explain this.

It is not easy to understand what you are looking for. Maybe an example could help. Suppose you are using a solution of sodium sulfate $\ce{Na2SO4}$. The corresponding ions $\ce{Na+}$ and $\ce{SO4^{2-}}$are not discharged at the electrodes. On the contrary $\ce{H2O}$ is reacting at both electrodes. Let's start with the cathode.
At the cathode, $\ce{H2O}$ reacts with electrons coming from the outer cell, and produces $\ce{H2}$ and $\ce{OH-}$ ions. But this reaction can only happen if there is enough positive ions around the cathode to realize the electric neutrality of the solution. That is where $\ce{Na+}$ intervenes. Forget about $\ce{H2}$. If there is enough $\ce{Na+}$ at the cathode, $\ce{H2O}$ can react and produce $\ce{OH-}$ ions in solution. The rate of this reaction is proportional to the concentration of $\ce{Na+}$ ions in solution.  This is why the presence of the electrolyte increases the reaction rate. Simultaneously, the negative ions $\ce{SO4^{2-}}$ are repelled by the cathode and attracted by the anode.
At the anode, $\ce{H2O}$ reacts in a similar manner to produce electrons. To do it, $\ce{H2O}$ is decomposed and produces $\ce{H+}$ ions in solution (or $\ce{H3O+}$)  and $\ce{O2}$ gas. But this production of $\ce{H+}$ ions can only happen if there are enough negative ions in solution near the anode. If there are no negative ions, the first $\ce{H+}$ ions repell the next ones, and the reaction stops. This is where the sulfate ions intervene. The $\ce{H+}$ ions attract the $\ce{SO4^{2-}}$, and so the neutrality of the solution is maintained. The rate of the reaction is proportional to the concentration of the sulfate ions.
As a consequence, there is a general movement of positive ions towards the anode, and of negative ions towards the cathode. The amount of gases produced at the electrode is proportional to the concentration of electrolyte in solution.

Thanks Maurice, my question has been updated to respond to you (please see above).


text:
Reaction of a cyclic tertiary alcohol with chromic acid
We know that $\ce{H2CrO4}$ cannot oxidize tertiary alcohols. The mechanism is:

Now the ${\alpha}$ carbon does not have any hydrogen to proceed to next step. Therefore, this reaction would not occur.
Is there any possibility of dehydration due to acidic medium?

If you are talking about normal dehydration conditions, then yes.. however under mild conditions no reaction takes place.

@Apurvium Take a look at the phenomenon of Carbonyl Transposition. That will be the case here.

@Safdar I am unable to find any example for dehydration through chromic acid in any condition.

It is normally given as $\ce{H+} $and not $\ce{H2CrO4}$

See Reference 2 here on carbonyl transposition: https://chemistry.stackexchange.com/questions/134863/why-cant-we-oxidise-allylic-alcohol-to-aldehyde-using-pyridinium-chlorochromate/134868#134868


text:
Number of chiral centers in penicillin V
I have often seen that in many drugs chirality plays a role. So I was trying to analyze the structure of Penicillin V.

What is number of chiral centers in the structure of penicillin V? Is the nitrogen which is  a part of the four membered ring to be counted here because I have read that the nitrogen atom without pyramidal inversion is treated as chiral. So can the nitrogen at bridgehead position be treated as chiral? Because that makes the stereocenter count 4 while on pubchem website the count is mentioned  to be 3.

Where did you read that?

Well, the beta-lactam nitrogen is pyramidal: https://pubs.acs.org/doi/pdf/10.1021/ja00721a032

For this structure in question, count only the sites where carbon is the stereogenic centre.

Nitrogen atoms in amides (peptide bond) are usually considered planar.

Yes, here Nitrogen has to be considered as a stereocenter. Also, you can assign R and S to it. The rule is that the lone pair has the least priority and then you can assign R and S as you do for other molecules.

In terms of CIP-rules, I agree.  For the triple substituted nitrogen, however, I assumed nitrogen inversion still could occur, thus discounting this site.  Maybe N's simultaneous participation in a five and a four membered ring does not permit the large «umbrella flip» otherwise seen / taught with amines thrice substituted by alkyl chains so ChemSE users with experience in quantum chemistry can provide an estimate for the activation barrier to interconvert the two states into each other to support either perspective about this substrate.

You cannot assign R and S to the nitrogen; the geometry of the amide nitrogen is closer to trigonal planar (sp2).

You cannot consider the bridge-head nitrogen as it's lone pair is in resonance with the adjacent oxygen, thus the nitrogen becomes sp2 rather than being sp3, which is a necessary condition to consider any atom to be a chiral center.
The same thing applies to the other nitrogen in pencilin.

Please expand your answer and cite references. Unsupported answer may get deleted.


text:
Actual color of colloids
I have read that the color of colloid depends upon the wavelength of light scattered by the dispersed phase and also in the sense in which the receiver receives the light.
If all of this depends on wavelength scattered and in the sense we see, then what is the actual color of a colloid?

Gold can produce colloids which are blue or red. So there is no simple relationship between the color of the bulk substance and the colloid made by dispersing this substance in tiny particules.

Colloidal solutions scatter light of different wavelenghts in different directions, and that's also dependent on the size(!) of the colloidal particles.
To see any scattering at all, there only has to be a difference in the refractive index of pure solution and dissolved particles.
The individual colour (i.e. absorption spectrum) of the particles is irrelevant to the scattering, but of course it additionally changes the colour that you see. Absorbed light does not get scattered, obviously.


text:
Normal (Regular) or Inverted Terms?
I have a question on the order of the spin-orbit molecular states. I understand how to find the terms that split from a parent molecular state in a diatomic molecule. But my question is on how to find whether the multiplet states are normal (regular) or inverted. For example, if we take $^3\Pi$, this state will split into $^3\Pi_{0^+}$, $^3\Pi_{0^-}$, $^3\Pi_1$, and $^3\Pi_2$. How can we know the order of these states, that is to say normal (regular) or inverted? If they are in this order: $^3\Pi_{0^+}$, $^3\Pi_{0^-}$, $^3\Pi_1$, and $^3\Pi_2$; or in that order: $^3\Pi_2$, $^3\Pi_1$, $^3\Pi_{0^-}$, and $^3\Pi_{0^+}$.
In Spectra of Diatomic Molecules book, by Herzberg, he pointed out these terms, and gave the following equation:
$$T_e=T_0+A\Lambda\Sigma$$
where $A$ is a constant for a given multiplet term.
He also stated that if $A$ is positive, this will give rise to normal (regular) terms. However, if $A$ is negative, we get inverted terms.
So is there another way to find out other than this way? And by the way, how can we determine $A$?
Thanks a lot...

What page is this on? I looked but could not find it from the index on spin orbit.

Pages 215 and 216

In case that you can get your hands on a copy of Brown and Carrington's "Rotational Spectroscopy of Diatomic Molecules" you should check out section 7.8.3 (page 357) where they discuss a simple orbital model to estimate the value of the spin-orbit parameter.


text:
How do we get g and E values for various levels to calculate electronic partition function
When we calculate electronic partition function using the formula:
$$q_\mathrm{elec}=\sum^\infty_{n=1}g_ne^{-E_n/k_BT}$$
How can I get the $g_n$ value and $E_n$ values for $n=0,1,2,3...$ levels?
I need to calculate the $q_\mathrm{elec}$ for various species like $\ce{H, H2, OH, H2O, O2, O, O+, H+}$.

First you must understand what the $n$ represents then you will need to look up the values for $E_n$ and $g_n$ at least for all but a couple of of the species you quote.

$E_n$ represents the energy of the $n$th electronic state relative to the ground state (i.e. the ground-state electronic energy $E_0 = 0$). Often, electronic states are very high in energy and the excited-state energies $E_1, E_2, \cdots$ are far greater than $k_\mathrm{B}T$, such that $\mathrm{e}^{-E_n/k_\mathrm{B}T} \approx 0$.
Under these conditions, the sum can be collapsed to just a ground-state term
$$q_\mathrm{elec} = \sum_n g_n \mathrm{e}^{-E_n/k_\mathrm{B}T} \approx g_0 \mathrm{e}^{-E_0/k_\mathrm{B}T} = g_0$$
where the latter equality is because $E_0 = 0$. $g_0$ is the degeneracy of the electronic ground state: to find this out you will need to draw some MO diagrams and/or construct the term symbols.
If the ground state is nondegenerate (e.g. $\ce{H2O}$) the whole thing is just equal to 1. Otherwise, for an atom with ground-state term symbol $^{2S+1}L_J$, the degeneracy is $2J + 1$. Similar considerations can be applied to molecules.
However, you will have to watch out for cases where there are low-lying energy states where $E_n$ is on the order of $k_\mathrm{B}T$. I am not aware of an easy way to predict this, though – you will have to know some typical examples, failing which you will just have to look it up, as porphyrin suggests. The NIST WebBook is a good starting place for this.


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How to determine rate law for a reaction?
I am having trouble understanding this problem.

A proposed mechanism for the decomposition of $\ce{N2O5}$ is as follows
\begin{align}
\ce{N2O5 &->[$k_1$]NO2 + NO3}          &&\text{(slow step)} \tag1\\
\ce{NO2 + NO3 &->[$k_2$]NO2 + O2 + NO} &&\text{(fast step)} \tag2\\
\ce{NO + N2O5 &->[$k_3$]3NO2}          &&\text{(fast step)} \tag3\\
\end{align}
What is the rate law predicted by this mechanism?
A. $\quad \text{Rate} = k[\ce{N2O5}]$
B. $\quad \text{Rate} = k[\ce{NO2}][\ce{NO3}]$
C. $\quad \text{Rate} = k[\ce{NO}][\ce{N2O5}]$
D. $\quad \text{Rate} = k[\ce{N2O5}][\ce{NO2}][\ce{NO3}]$
E. $\quad \text{Rate} = k[\ce{N2O5}]^2$

I can determine the rate law for each individual reaction:
\begin{align}
\text{Rate} &= k_1[\ce{N2O5}] \tag{1'}\\
\text{Rate} &= k_2[\ce{NO2}][\ce{NO3}]\tag{2'}\\
\text{Rate} &= k_3[\ce{NO}][\ce{N2O5}]\tag{3'}
\end{align}
But I am having trouble understanding how to incorporate all of this reaction rates into one complete reaction rate.
Do I multiply all the rates?

As the system is described, we can suppose $k_1 \ll k_2$ ans $k1 \ll k_3$, as Martin correctly noted and I have omitted to explicitly mention.
For such cases, we can consider for intermediate products to be in a steady state, i.e. $\frac {\mathrm{d}[A]}{\mathrm{d}t} \simeq 0$. So the rate of their creation is about equal to the rate of there destruction.
E.g. the rate of $\ce{NO3}$ production in reaction (1) is the same as the rate of its consumption in the reaction (2). Similarly, the rate of $\ce{NO}$ production in the reaction (2) is the same as the rate of its consumption in the reaction (3)
$$\frac {\mathrm{d}[\ce{NO3}]}{\mathrm{d}t} = k_1 [\ce{N2O5}] - k_2 [\ce{NO2}][\ce{NO3}]=0 \tag{1}$$
$$\frac {\mathrm{d}[\ce{NO}]}{\mathrm{d}t} = k_2 [\ce{NO2}][\ce{NO3}]- k_3 [\ce{NO}][\ce{N2O5}]=0 \tag{2}$$
Then try to express concenrations of intermediate products as function of  concentration of reagents and final products.
$$[\ce{NO3}] = \frac{k_1 [\ce{N2O5}]}{k_2 [\ce{NO2}]} \tag{3}$$
$$ \ce{[NO}] = \frac{ k_2 [\ce{NO2}][\ce{NO3}] }{k_3 [\ce{N2O5}]} \tag{4}$$
$$ \ce{[NO}] = \frac{ k_2 [\ce{NO2}]\left( \frac{k_1 [\ce{N2O5}]}{k_2 [\ce{NO2}]} \right) }{k_3 [\ce{N2O5}]}=\frac{  k_1 [\ce{N2O5}]  }{k_3 [\ce{N2O5}]}=\frac{k_1}{k_3} \tag{5}$$
From (5) and  (2):
$$ k_2 [\ce{NO2}][\ce{NO3}]= k_1[\ce{N2O5}] \tag{6}$$
$$ [\ce{NO3}]= \frac{k_1[\ce{N2O5}]}{k_2 [\ce{NO2}]} \tag{7}$$
$$\frac {\mathrm{d}[\ce{NO2}]}{\mathrm{d}t} = k_1 [\ce{N2O5}] + 3 \cdot k_3 [\ce{NO}][\ce{N2O5}] = \\
k_1 [\ce{N2O5}] + 3 \cdot k_1 [\ce{N2O5}] = k [\ce{N2O5}] \tag{8}$$

So the answer is A.

How does eq(4) come?, from eq (5), you say that concentration of NO is fixed!?

Directly from (2)