text:
Mass action law and conservation of mass
I am starting to learn about the use of mass action law in chemistry. I am doing some exercises to practice. In particular, I am considering the reaction
$$2 X \underset{3}{\stackrel{4}{\rightleftharpoons}} Y.$$ By mass action law I have obtained the following system of ODE's
$$ \left\{
 \begin{array}{ l }
    \frac{d}{dt}x(t)=-8x^2+6y\\
    \frac{d}{dt}y(t)=4x^2-3y,
 \end{array}\label{SEDO} \right.$$
where I have used this result.

I am wondering if it is usual that the amount $x(t)+y(t)$ is not conservated.

That should be obvious from the reaction's equation...

I am a Math student, so I don't know too much about chemistry. Do you mean that the equations are wrong?

If you make one Y from two X than why would you expect to still have two?

There is law of mass conservation, there is no law of molar amount conservation. 1 mol of Y can be formed from 2 mol of X with the same mass.

Please MathJax the screenshots

Where does the screenshot come from?

What is $x$ ? What is $y$ ? What is the "system of ODE" ?

The rate laws that you show don't give an identity to $x$ or $y$. I am going to assume they are moles, even though it is uncommon to show rate laws as a function of moles. The time evolution of the amount of substance of both species can be written as
\begin{align}
  \frac{\mathrm{d}x}{\mathrm{d}t} &=
  -8x^2 + 6y = (-2) (4x^2 - 3y) \tag{1} = -2r_1\\
  \frac{\mathrm{d}y}{\mathrm{d}t} &=
  4x^2 - 3y = (+1) (4x^2 - 3y) = r_1 \tag{2}
\end{align}
where we defined $r_1 := 4x^2 - 3y$. Thus:

It is clear that $x(t) + y(t)$ is not a conserved quantity. This is because for every $2$ moles disappeared of $x$, you get only $1$ mol of $y$. But the quantity $x(t)+2y(t)$ is conserved, by using Eqs. (1-2)
\begin{align}
  x + 2y &= \text{constant} \\
  \frac{\mathrm{d}x}{\mathrm{d}t} + 2\frac{\mathrm{d}x}{\mathrm{d}t} &= 0 \\
  (-2r_1) + 2(r_1) &= 0 \rightarrow 0=0 \tag{3}
\end{align}
The mass for a chemical reaction is conserved, but this quantity is not $x(t) + y(t)$ rather it is $M_x x(t) + M_yy(t)$, where $M$'s are the molar masses of the compounds. Again, by Eqs. (1-2)
\begin{align}
  \require{cancel}
  M_xx + M_yy &= \text{constant} \\
  M_x\frac{\mathrm{d}x}{\mathrm{d}t} + M_y\frac{\mathrm{d}x}{\mathrm{d}t} &= 0 \\
  (-2M_x)r_1 + (M_y)r_1 &= 0 \rightarrow -2M_x + M_y = 0 \tag{4} \\
\end{align}
Eq. (4) is always satisfied. To give an example that is similar to your chemical reaction, we can illustrate the decomposition of nitrogen dioxide in the gas phase
$$ \ce{2NO2(g) \rightleftharpoons N2O4(g)} \tag{5} $$
And this chemical equation satisfies
\begin{align}
  -2M_\ce{NO2} + M_\ce{N2O4} &= 0 \\
  (-2) \left(\pu{46.01 \frac{g}{mol}}\right) +
  \left(\pu{92.02 \frac{g}{mol}}\right) &= 0 \\
  (-92.02 + 92.02) \; \pu{\frac{g}{mol}} &= 0 \rightarrow 0=0 \tag{6}
\end{align}

To generalize, for a general chemical reaction that comprises $m$ species, we always have
$$ \boxed{\sum_{j = 1}^m \nu_j M_{j} = 0} \tag{7} $$
So, if you solve a system that has at time $t$ an amount of moles of $x_1(t)$, $x_2(t)$, ..., $x_m(t)$, then $M_1x_1(t) + M_2x_2(t) + \dots + M_m x_m(t)$ is constant. It is a good check to see if the code is satisfying the mass balance by calculating this quantity and viewing that it does not deviate, apart from the absolute/relative errors that you specify.

Thank you! Now is completely clear for me!


text:
Is the compression factor of a gas equal to Vm/Vm-b?
According to the definition, the compressibility factor is the ratio of the molar volume and the molar volume if the gas behaved as an ideal gas
$$ Z = \frac{V_\mathrm{m}(p,T)}{V_\mathrm{m}^\pu{ig}(p,T)} \tag{1} $$
When comparing the ideal gas and the van der Waals equations of state, the ideal volume term is substituted by the real molar volume minus the correction factor $b$ accounting for the size of the molecules
$$ p = \frac{RT}{\underbrace{V_\mathrm{m}^\pu{ig}}} \hspace{1 cm}
   p = \frac{RT}{\underbrace{V_\mathrm{m} - b}} - \frac{a}{V_\mathrm{m}^2}
   \tag{2} $$
So we can say that
$$ V_\mathrm{m}^\pu{ig} = V_\mathrm{m} - b \tag{3} $$
Is this the case only when the constant $a \approx 0 $? Eq. (3) does not account for attraction forces, so it is obviously wrong. What is wrong with my logic?

To address the title question:
No, it is not, as the reality of real gases changes both volume and pressure. Therefore, if we corrected real volume to ideal volume, the real pressure would still differ from ideal pressure and vice versa.
It is approximately true for low pressure and not too low temperature, when repulsion due finite volume is (much) more significant than cohesive forces.

$$a \ne 0 \implies V_\mathrm{m,Ideal} \ne V_\mathrm{m} - b$$
$$(p + \frac{a}{V_\mathrm{m}^2})(V_\mathrm{m}-b) = RT$$
$$(1 + \frac {a}{pV_\mathrm{m}^2})(1 -\frac{b}{V_\mathrm{m}}) = \frac 1Z$$
$$1 -\frac{b}{V_\mathrm{m}} + \frac {a}{pV_\mathrm{m}^2}  - \frac {ab}{pV_\mathrm{m}^3}= \frac 1Z$$
As $1 + x \approx \frac {1}{1 - x}$ if $|x| \ll 1$:
$$Z \approx 1 +\frac{b}{V_\mathrm{m}} - \frac {a}{pV_\mathrm{m}^2}  + \frac {ab}{pV_\mathrm{m}^3}\\=1 +\frac{b}{V_\mathrm{m}} - \frac {a}{pV_\mathrm{m}^2}(1 -  \frac {b}{V_\mathrm{m}})$$

Real Volume
$$
\text{Real volume} \neq V- \mathrm{b}
$$
You need to solve the van der Waals gas equation to obtain the real volume.
Compressibility Factor
According to your definition of compressibility factor $Z$:
$$
Z = \dfrac{V_\text{real}}{V_\text{ideal}}
$$
and not the other way around; $Z \neq \left( \dfrac{V}{V-n\mathrm{b}} \right)$.
The van der Waals (Real) Gas Equation
$$
\left( P_\text{real}+\dfrac{\mathrm{a}n^2}{V_\text{real}^2} \right) \left( V_\text{real} - n\mathrm{b} \right) = n\mathrm{R}T \tag{1}$$
where $P_\text{real}$ and $V_\text{real}$ are the pressure and volume of the real gas, $n$ is the number of moles of gas particles, $\mathrm{a}$ and $\mathrm{b}$ are the van der Waals gas constants (which depend on the gas), and $\mathrm{R}$ is the universal gas constant.
Rearranging Equation (1) gives us:
$$
P_\text{real}\cdot V_\text{real}^3 - (n\mathrm{b}+nRT)\cdot V_\text{real}^2 + n^2\mathrm{a}\cdot V_\text{real}-n^3 \mathrm{ab} = 0 \tag{2}
$$
Notice that this is a cubic equation and, thus, we will obtain three roots.

Example
Here is an example of how $V_\text{real}$ varies with carbon dioxide gas $\ce{CO2}(g)$ for which $\mathrm{a} = \pu {3.658 bar L^2 mol^{-2}}$ and $\mathrm{b} = \pu {0.04286 L mol^{-1}}$. We are going to Assume:
$$
P_\text{real} = \pu{1 bar}\\
T = \pu{300 K}\\
n = \pu{1 mol}
$$
Substituting in Equation (2):
$$
V_\text{real}^3 - 24.985\cdot V_\text{real}^2 + 3.658 \cdot V_\text{real}- 0.157 = 0 \\
\implies V_\text{real} = \pu{24.838 L}
$$
The other two roots are complex and need not be considered. Under the same conditions:
$$
V_\text{ideal} = \dfrac{nRT}{P} = \pu{24.942 L}\\
\implies Z = \dfrac{V_\text{real}}{V_\text{ideal}} = \dfrac{24.838}{24.942} = 0.996
$$

source: Stack Exchange

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

sort: votes


text:
Enthalpy of adsorption and adsorption rate
What I want to know is about the adsorption enthalpy change by the change of surface coverage and the adsorption rate. In general situations, adsorption would be exothermic. So in the Gibbs free energy equation ${\displaystyle \Delta G^{\circ }=\Delta H^{\circ }-T\Delta S^{\circ }}$, adsorption usually fixes molecules on the surface of the adsorbent so $\Delta S^{\circ}$ is negative and therefore, the higher absolute value of negative $\Delta H^{\circ }$ would cause the adsorption to be more spontaneous. Can this change of spontaneity affect the rate of adsorption if the pressure and temperature are same? If not, what causes the adsorption rate to decrease when surface coverage increases in the same temperature, pressure and concentration?

I think you mean $\Delta S^ \circ$ is negative because entropy $S^\circ$ itself is always positive. Are you?

Yes. I think I made a mistake there

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text:
Why can sulfate be oxidized to peroxodisulfate?
I am currently in high school and have encountered the following reaction in my course on electrochemistry:
$$\ce{2 SO4- -> S2O8^2- + 2 e-}$$
Why this reaction happens in the first place? Isn’t the maximum oxidation state of sulfur only +6?

https://chemistry.stackexchange.com/questions/59465/can-we-oxidize-oxygen

The oxidation number of sulfur in peroxydisulfate is still +6.
But the oxidation number of oxygen in peroxy group $\ce{\bond{-}O-O\bond{-}}$ is not −2, but −1.
$$\ce{O^{-}-SO2-O-O-SO2-O^{-}}$$
The reaction $\ce{2 SO4^2- -> S2O8^2- + 2 e-}$ does not oxidize sulfur. It oxidized two oxygen atoms and joins them to the peroxy group.


text:
Do gas particles colliding with each other affect the overall pressure?
Gas pressure is the pressure exerted by gas particles when they collide with the walls of the container. The gas pressure increases with the number of gas particles.
Does the collision between two gas particles also exerts pressure? If so, how does the collision between two gas particles affect the overall gas pressure?

You cannot separate collisions with the walls and collisions between gas particules.

@Ethan If you know how to code, you could solve the equation of motion for $N$ particles subjected to an intermolecular potential (like LJ), i.e. for particle $i$ you have $\sum_j \mathbf{F}{ij} = m_i \mathrm{d}^2\mathbf{r}{ij}/\mathrm{d}t^2$. Then you could calculate $P=(1/A)\mathrm{d}p_n/\mathrm{d}t$, where $p_n$ is the momentum component that is perpendicular to the wall, and apply some averaging. I suggest reading Gould & Tobochnik book if you want quantitative answers.

Yes and no.
As a first approximation, a given gas molecule is roughly as likely to gain momentum than to lose momentum when colliding with another molecule. Thus once our molecule hits a wall, the collisions that happened on the way there don’t really matter on average.
This is exactly the point of the ideal gas model, which ignores particle-particle interactions except for exchanging energy (which happens through collisions).
As a second approximation, the finite size of particles and thus collisions matter.
This is one of the main things captured by real gas models such as the van der Waals model.
According to the latter, the pressure of a gas is (in dependence of the molar volume $v:=\frac{V}{N}$):
$$ p(v) = \frac{RT}{v-b} - \frac{a}{v^2}.$$
Here $a$ and $b$ are positive constants where:

$a$ captures the attraction of gas particles to each other (usually via the van der Waals force).
$b$ captures the repulsion of gas particles due to their finite size.

As you can see, $b$ increases the pressure, in particular for small $v$, i.e., high particle densities.
(Both, $a$ and $b$ become zero for an ideal gas.)
However, it is debatable whether this is “collisions increasing the pressure”.
The increase in pressure originates from the particles having less effective volume available; the averaging effect of collisions is still in effect.
From another perspective, it doesn’t matter how many collisions happen in the gas, only that they happen to equilibrate the energies of particles.

how does the collision between two gas particles affect the overall gas pressure?

In some sense, not at all. Pressure captures the total kinetic and potential energy of the gas particles, which is not affected by collisions. The pressure of a gas doesn’t change on its own. (The main exception from this is of course if the gas is not in chemical equilibrium and chemical and kinetic energy can be translated into each other on a macroscopic scale.)

Thats interesting!So can we assume for noble gases that the attraction term(a) is 0?

@Volpina: No. Noble gases are not exempt from the van der Waals force; that’s how they become liquid at low temperatures. To have a gas in the first place, the attraction between molecules needs to be low (with respect to the temperature and pressure); otherwise you get condensation. (Also: Don’t confuse noble and ideal gases.)

Note as the ideal gas model assumes point particles within that model there are no interparticle collisions; you need a more complex model or a real gas for those.

@IanBush: Not really: As I write in my answer, you do need the particle collisions to equilibrate particle energies. This is the same for ideal and real gas models. The difference for real gas models is that they consider the effect of particle interactions on particle energies (taking a snapshot if you so wish), but they do not feature particle collisions explicitly.

The ideal gas is a model of the gas at equilibrium - In the ideal gas model there is no need for collisions to establish equilibrium because it is already at equilibrium.  But nobody ever listens to me on this one so I'll shut up now!

@IanBush: Well, it’s getting a bit semantic, depending on when you consider the ideal gas model to be separate from its derivation, e.g., in equilibrium statistical physics. In the latter, you need some mechanism for microstates to transition into each other. And this requires some exchange of energies between particles – which only collisions really offer. (This is not exactly equilibrating, but it becomes when you, mix gases for example.) Anyway, whatever your take on this, it equally applies to any other model based on equilibrium statistical physics, e.g., the van der Waals model.

Isn't heat the measure of collisions between particles?  (Note: I'm neither a chemist nor physicist nor engineer.)

The mutual collisions of real gas molecules do not have significant direct impact on the gas pressure.
The original thermodynamic notion of an ideal gas considers just wall collisions. Deviation of real gas pressure is at not too high pressure and not too low temperature often negligible. Note that at ambient conditions, typical collision frequency for a molecule in air is roughly $\pu{E10 s-1}$ and the mean free flight distance between collisions is about $\pu{70 nm}$.
But, collisions of molecules have indirect pressure impact. Colliding molecules mean the total volume of molecules is not fully negligible, compared to the volume of free space between them. It causes increased frequency of wall collisions and higher pressure.
When the gas volume becomes comparable to the total molecule volume, molecules become being stuck, not able to freely pass each along other. At that point, the wall collision frequency and pressure raise with the volume decrease much steeper than for ideal gasses.
Another indirect effect, mainly at low temperatures and high pressures, are intermolecular cohesive forces, that manifest the most just before/after collisions. This has the opposite, pressure decreasing effect.
Both effects are addressed by the van der Waals equation, the most known state equation of real gases:
$$\left(P+a \frac{n^2}{V^2}\right)(V-n b)=n R T$$

The pressure in an ideal gas is only due to collisions with the vessel's walls, collisions between molecules do not contribute. This can be shown by considering general forces on a molecule using Newton's second law of motion. If the force on a molecule is $f_x,f_y,f_z$ in directions $x,y,z$ where
$$\displaystyle f_x=m\frac{d^2x}{dt^2}$$
and similarly for the $y$ and $z$ directions. The energy is found by multiplying by $x,y,z$ as appropriate (because energy is force times distance), giving
$$\displaystyle xf_x+yf_y+zf_z=m\left(x\frac{d^2x}{dt^2}+y\frac{d^2y}{dt^2}+z\frac{d^2z}{dt^2} \right)$$
Now we need to manipulate this equation using a 'trick'
$$\displaystyle \frac{d}{dt}\left(x\frac{dx}{dt}\right)=x\frac{d^2x}{dt^2}+\left(\frac{dx}{dt}\right)^2$$
and similarly for $y,z$ and so obtain an expression for the second derivative. Substituting for this gives
$$\displaystyle xf_x+yf_y+zf_z=m\frac{d}{dt}\left(x\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}\right)-m\left(\left(\frac{dx}{dt}\right)^2+\left(\frac{dy}{dt}\right)^2+\left(\frac{dz}{dt}\right)^2\right)$$
the right-hand most term is just the velocity $v$ squared  multiplied by the mass $m$ and is therefore $mv^2$ is twice the energy of the molecule giving
$$\displaystyle xf_x+yf_y+zf_z=m\frac{d}{dt}\left(x\frac{dx}{dt}+y\frac{dy}{dt}+z\frac{dz}{dt}\right)-mv^2$$
The remaining derivate is $\displaystyle x\frac{dx}{dt}=\frac{1}{2}\frac{dx^2}{dt}$  producing
$$\displaystyle xf_x+yf_y+zf_z=\frac{m}{2}\frac{d}{dt}\left(\frac{d}{dt}(x^2+y^2+z^2)\right)-mv^2$$
So far this applies only to a single molecule. Over all molecules while the value of $\displaystyle \frac{d}{dt}(x^2+y^2+z^2)$ must fluctuate in time for any given molecule, overall all these values must average to zero, because if this were not so the gas would be expanding or contractions as time progressed.
The expression becomes, after rearranging and summing over all molecules,
$$\displaystyle \frac{1}{2}\sum mv^2=-\frac{1}{2}\sum (xf_x+yf_y+zf_z)$$
The term of the left is the total kinetic energy which from laws of motion is  $3k_BT/2$ per molecule with $k_B$ being the Boltzmann constant therefore
$$\displaystyle Nk_BT =-\frac{1}{3}\sum (xf_x+yf_y+zf_z)$$
for $N$ molecules.  (The term on the right is sometimes called the 'virial of Clausius').
The right hand side of this equation needs to be calculated. To do this we first find the force on one wall, say wall $xy$, and this force has the value $xyp$ where $p$ is the pressure ($\equiv$ force/area) and $xy$ is the area. The total force the wall exerts is therefore $-xyp$, and similarly for the other walls, $xz$ etc.  The average distance of a molecule from a wall in the $x$ direction is $x/2$ so the energy $xf_x$ has a value $-xyzp/2$, thus over all six walls is $-3pV$ where $V=xyz$. Thus over all walls this is also the value of $\sum (xf_x+yf_y+zf_z)$.
The ideal Gas Law $pV=Nk_BT$ is thereby produced and so in the absence of intermolecular interactions all the pressure is caused by collisions with the wall. In other words, the effect of all the forces the molecules experience, but not those caused by intermolecular interaction, is to produce the ideal gas law.
If intermolecular interactions are added then we must add a similar general term term
$$\displaystyle pV=Nk_BT+\frac{1}{3}\sum (xf_x+yf_y+zf_z)$$
where now $xf_x$ etc. now refer only to intermolecular forces. This term can be recast in a form that measures the pair-wise interaction between molecules $\varphi$ at a radial distance $r$ and has the form $-\sum r\frac{d\varphi}{dr}$. This term is the origin of correction terms such as in the van der Waals equation.

I liked this answer! Just a little detail, in the first paragraph, at the end you would have wanted to write "in directions $x, y, \bf{\color{red}{z}}$ where..."

@Metal Storm, thanks, typo corrected

Think of it this way:
You have a gas filled inside a cubical container closed by all six faces. For sake of simplicity assume you have 1 mole of that gas in the container. Even this much amount of gas means you have approximately $6.022 × 10^{23}$ molecules which is a lot. Now imagine a single molecule to start it's motion from one of the faces of the container to reach to the opposite face. From the third postulate of kinetic theory of gases, it will encounter so many random collisions with so many other randomly moving molecules that before it reaches the opposite face, it has undergone an unimaginable number of collisions.
So when finally it strikes the opposite face to apply pressure on it, it also implies the average pressure from all those collisions it encountered along the way. This averaged value includes the effective as well as non-effective collisions. All those collisions average out because they are completely random. This was just for one molecule and the similar happens to the other $6.022 × 10^{23}$ of them in the container.

So when finally it strikes the opposite face to apply pressure on it, it also implies the pressure from all those collisions it encountered along the way. – This ignores the fact that the collisions can increase as well as decrease the momentum of our molecule and all of this averages out. For example, in an ideal gas, it does not really matter how many of these collisions happened, only that the particles can somehow exchange energy.

I haven't used the word increases or decreases in my answer at all. When I said it implies all those collisions, it itself means that the impact has averaged over the distance travelled.

Then I suggest you clarify that. Because “pressure from all those collisions” suggest that pressure (increase) is accumulated.

@Wrzlprmft I just tried to present a rough image of how the average pressure from collisions is a significant contributor in the total pressure. I think if I missed on something, your and Poutnik's answer would provide that.

The answer reads as though all the molecule-molecule collisions do matter to the pressure: they don't. Only collisions with walls contribute to pressure. The last sentence reads completely wrong.

@matt_black What I mean is that when it collides the wall, it does so with the average pressure it has got from all the previous collisions. The average must include all those that matters and those that do not.

@Proscionexium This confuses what pressure is. Molecules have kinetic energy and create pressure when they collide with the wall of a vessel (you could say that pressure is the average momentum transfer). The only influence gas-gas collisions have is to ensure that the overall kinetic energy in the gas is well distributed across all the molecules (in a boltzmann distribution).


text:
Malleability and Ductility Origins
Malleability and Ductility are both properties of metals. I have learnt that lead is malleable but not ductile. However, don't these properties both arise from metallic bonds? So how can a substance be malleable but not ductile and vice versa?
PDF documents with information to help me learn would be appreciated.

Hmm.. While I admit all solid phase matter can be a subject of chemistry, specialized domains of science and technology have often much superior knowledge of some aspects, like metal processing. Malleability does not involve enduring tensil stress like ductility does.

Those properties arise at a higher level than 'metallic bonds', having more to do with crystal structure and the energetics of point and line defects (vacancy/interstitial and dislocations). Welcome to Materials Science!

Malleability and ductility are properties of the bulk structure of metals not their metallic bonds
It is easy to assume that the key properties of materials are primarily determined by their chemical bonds. But this is often a vast simplification, and especially so for metals.
And the explanation for why metals (or at least some of them) behave the way they do took a long time to work out with some major issues only being clarified in the mid 20th century.
The key to malleability and ductility (to oversimplify greatly) in found in the bulk material structure. Most metals are not single crystals but consists of many small domains of interacting crystals or imperfect crystals. Many of the properties are determined by what happens at the crystal boundaries when the material is under stress.
Again, simplifying a huge amount, in many metals, defects in the boundaries and the crystals can move around in ways that relieve stress and avoid the stress causing cracking in the bulk structure (as it does in brittle materials like glass or diamond). So malleable metals can undergo huge amounts of shape changes (eg you can beat gold into films just nm thick). Chemical bonds may break but in a limited and controlled way limited to specific small regions in the bulk compound.
This also explains why the specific properties of metals can be radically changed by adding small amounts of other compounds to create alloys: despite the small change in the material makeup, the added substances often concentrate at the crystal boundaries, highly influencing the behaviour of defects.
A good read that explains a lot of the science behind this (and many other material properties) and how we discovered them is The New Science of strong Materials by JE Gordon. in addition to being insightful it is also extremely readable and possibly the single best introduction to the subject of Materials Science (which is a kind of mongrel of chemistry, physics and engineering).


text:
Help with Paraffin wax
I am trying so hard to figure out what I am missing or what I need to use to make a lava lamp in a bottle with no heat and is safe?
Ingredients below are specifically for a lava lamp which includes use with heat;
Based on patents, new articles, and user research the original 60's formula is expected to contain distilled water, 38% by mass; chlorinated paraffin, 36%; low molecular weight polyethylene glycol, 13%; kerosene 7%; and microcrystalline wax, 6%.

The lava lamp is based on different thermal dilation of liquids, The organic phase dilates with temperature more and at some temperature switches from more dense to less dense.
To have the threshold temperature lower, without need of heating, you need to make water based phase somewhat denser, e.g. by dissolving the table salt.
But the particular salt concentration has to be determined experimentally.


text:
Classification of mixtures for metals?
From my understanding, Alloys have metallic bonds between different metal elements. How does it make sense to classify alloys as a mixture then? Is there a similarity between an alloy and a mixture such as a salt solution?

If something is not a compound or element, then it's a mixture. Nature of bonding is irrelevant.

Some alloys have wide solubility ranges. Some don’t. That holds for many other systems such as liquid salts and aqueous solutions.

Just to muddy the waters -- there are intermetallic compounds, https://en.wikipedia.org/wiki/Intermetallic , some with elements in stoichiometric ratio, e.g., Ti3Al. Compound or mixture? There is no unalloyed answer.

Sure, but standard entropy of mixing applies to alloys with compositional width in, e.g Calphad models.

If I have an alloy of two materials, will I have a chemical reaction between the two materials; If alloys are homogeneous mixtures, why can't we separate their components?

Yes! Alloys are considered mixtures in a similar way salt solutions are. According to LibreTexts Chemistry:

An alloy is a mixture composed of two or more elements, at least one of which is a metal.

Composition
Consider that composition of a mixture is defined as the amounts of substances present in it:
$$
\text{composition} \equiv n_1, n_2, n_3, \cdots
$$
where $n_i$ is the amount of $i^\text{th}$ substance present in the mixture.
Definitions
Note: Here, we are representing composition in terms of absolute amounts of substances. One can also use relative amounts of substances using concepts such as mole fraction, which reduces the number of variable by 1.
Pure substance: a substance whose composition can be represented by a single number (amount of the substance $n$)
$$
\text{composition}_\text{pure} \equiv n
$$
Mixture: a substance whose composition is represented by a multiple numbers (amounts of the substances present in the mixture $n_1, n_2, n_3, \cdots$)
$$
\text{composition}_\text{mixture} \equiv n_1, n_2, n_3, \cdots
$$
Binary Mixtures
A binary mixture is represented by amounts of two substances:
$$
\text{composition}_\text{binary mixture} \equiv n_1, n_2
$$
Ternary Mixtures
A ternary mixture is represented by amounts of three substances:
$$
\text{composition}_\text{ternary mixture} \equiv n_1, n_2, n_3
$$
and so on...
Examples

Brass is an alloy made by mixing varying amounts of $\ce{Cu}$ and $\ce{Zn}$
Bronze is an multi-component alloy made primarily ($\approx 90\%$) of $\ce{Cu}$ with the remaining percentage composed metals such as $\ce{Al}$, $\ce{Mn}$, $\ce{Ni}$, or $\ce{Zn}$.

In my opinion this doesn't really answer the question. Additionally I'd refer to "official" definitions. Libretexts is pretty hit and miss with their content. https://doi.org/10.1351/goldbook.M03949 is mixture, there's no alloy. And that quote is also verbatim in Wikipedia https://en.m.wikipedia.org/wiki/Alloy

Yes, metal alloys can be mixtures, but it is a lot more complicated than that
There is a big problem in describing alloys as chemical mixtures: material scientists tend to use "alloys" to cover a much broader range of materials than chemists would if they were being strict about what they mean by "mixture".
So there are specific alloys that are, in effect, what chemists would describe as a mixture. But many of the things material scientists would describe as "alloys" are not uniform, homogeneous mixtures at all. The material properties of many alloys (to stick with the material science use of the term) depend on the bulk structure of the material and the way individual crystals in it interact. And the crystals might not all have the same composition.
Stainless steels, for example, can be a mixture of different crystal structures all of which are influenced by the production method and the alloying ingredients. The properties of the material are very highly dependent on the mix of different structures and compositions and these are strongly influenced by the processing history of the material. There can be a mix of relatively pure iron but with different crystal structures. Small amounts of carbon results in some crystals of cementite (which is Fe3C) but also martensite which also has carbon but has a different structure in a matrix of ferrite (what iron is when pure) and austenite (what chemists would call a mixture of Fe and other components). It is the interaction of all these multiple components that determines the key properties of the substance not the individual properties of a single crystal of a simple mixture.
Materials scientists and metallurgists seem to be happy calling such complex structures "alloys". But a chemist might not be as the overall substance is not uniform on a small enough scale.
While a chemist could happily describe austenite as a mixture, this is not helpful in talking to a material scientist who knows that most of the alloys they talk about are not uniform but consist of complicated microstructures containing crystals of different composition.


text:
Is breaking something an irreversible change?
The science textbooks say that, unlike change of state, not all physical changes are reversible. For this, the example provided is grinding wood into sawdust or tearing off paper. They say that we cannot recover paper from its pieces again. However, I think we can. To make it simpler, consider the example of ice. If I break the ice, it is a physical change. In this case, I can melt the pieces and freeze the water again to recover the ice. Hence, this becomes a reversible physical change. I argue that the same goes for the tearing of a paper, too. However, I wonder if, in the process of recovery, we are allowed to use any method, physical or chemical.
My purpose is not to get to the right definitions or examples of physical or chemical change but to understand why we have to call some changes reversible and some irreversible while we can reverse, in my understanding, all physical changes. I wonder if the same is true for chemical changes as well.

If you tear off paper, you may recover paper from its pieces using some chemical solution to dissolve cellulose, and reproduce paper again. But the breaking operation has not been really reversed. Furthermore the obtained recovered paper will not look like original paper. It will not contain the same cellulose molecules at the same place. Also the ink and dyes deposited on the previous paper will not be reproduced.

There is several meanings of reversibility, in different contexts. 1/Being in equilibrium during state changes 2/ Being able to revert the process/change  3/ Being able to revert the process/change by reverting the state of the neighborhood. // For 3/, if you use fuel to melt the ice and then freeze it, the neighborhood changed, so it is not reversible process by 3/ even if the phase change itself can be reversible by 1/ and 2. // Or if freezing/melting is too fast, it may be reversible by 2/ and 3/ in other scenario, but not by 1/.

Personally, I'd avoid thinking about it too much. This sort of definition is not really watertight and the more you look into it, the more problems you're going to find. Reversibility has a formal definition in thermodynamics, but it's not the same as what you've been taught (it's something you see in a more advanced chemistry course), and it also doesn't have any link to the classification of something as a physical or chemical change.

Some physical changes are reversible, if the requirement is simply that materials are substantially similar at beginning and end. Here are two examples.

Crush a salt crystal, dissolve it in water, and let the water evaporate. Substantially the same crystal structure results. However, it would be somewhat ;-) unlikely that each atom of the first crystal returned to the same position in the second. We'd still say the process was reversible.
Boil water, and then condense it. Again, the end-product substantially resembles the starting material.

Some physical changes are not reversible, if the requirement is that long range order in materials are identical at beginning and end. Here are two examples.

If one were to cut Guernica in Museo Reina Sofía into slices in a paper shredder, there is no technique to restore Picasso's work to be the exact original. The reasonable person, looking at the results of an attempted restoration, would say it is not the original.
Look at a piece of wood. Note the whorls and growth rings, the phloem and xylem. If it were burned, another plant could eventually turn the $\ce{CO2}$, $\ce{H2O}$ and minerals back to cellulose and lignin, but it would no longer be the same piece of wood.

So this question is one of philosophy, more than of chemistry. As Plutarch (and others) have asked, if one were to replace every part of a ship, one at a time, would the resulting vessel be the same as the original? And in an ancient Buddhist text, and in modern science fiction, the question arises if we're the same person with parts replaced: blood transfusion? cornea replacement? kidney transplant? brain tissue transplant?

It's good that you start with a functional definition. For the definition of reversible to apply it suffices to say that the initial and final properties of the sample are identical (to within measurable precision). If it's thermodynamic reversibility, then thermodynamic properties such as S, H, etc... then the precise position of the atoms is of little importance provided the macroscopic property is retained (see also orthocresols comment under the other answer).

A good and very sad real example as alternative to destroying Picasso's Guernica: https://www.theguardian.com/artanddesign/2020/jun/22/experts-call-for-regulation-after-latest-botched-art-restoration-in-spain  did the entropy increase or decrease here?


text:
Will alcoholic acid solution with organic acids break down Bakuchiol and reduce it's biological effect?
I wish to use Bakuchiol in a solution of isoproply alcohol and water in the presence of glycolic acid (7%), citric acid (0.5%) and salicylic acid (1.4%) at a pH of 4 to 5, adjusted with TEA.    Wu et al ("Synthesis and Evaluation of Bakuchiol Derivatives as potential Anticancer Agents") claim that Bakuchiol reacts with water or methanol at room temperature in the presence of 'dilute' H2SO4 to form an ether or alcohol from the terminal alkene.   So will my weaker organic acids break down the Bakuchiol terminal alkene by this pathway to either an alcohol or ether (with aq IPA) ?  Or are the organic acids too weak and having too strong a nucleaphilic counter ion to react this way.  Standard theory for ether formation from alkanes requires a strong acid with a poor nucleophilic counter ion .. but Bakuchiol is branched and kind of conjugated, so the terminal alkene may be more reactive.   Can you advise ?  Proceed ?   How do I test for no reaction at home ?  Colour ?

At pH 4-5 I doubt it

Thank you Waylander


text:
Auxochrome groups and red shift
Wikipedia states that "An auxochrome is known as a compound that produces a batochromic shift, also known as a red shift because it increases the absorption wavelength, thus approaching infrared light."
That said, let us consider ethylene (chromophore group) and vinyl chloride (chromophore group to which we have added the auxochrome Cl).  The addition of the auxochrome produces a red shift, as written in the definition above. But what transition does this Red shift refer to?
Does it refer to the π-π* transition, whereby this means that, as a result of the addition of chlorine, there is a reduction in the energy gap between the π orbitals, so the absorption band corresponding to this transition shifts to higher wavelengths?
Or does the Red shift refer to the fact that new non-bonding orbitals are formed, due to the addition of chlorine, so that a new absorption band at higher lambdas is generated, corresponding to the n-π* transition, and thus a reduction in the energy gap between HOMO and LUMO (which are, respectively, n and π*) is observed?

Adding Cl is essentially no different to the H atom that it replaces, except that the orbital energy may be lowered and so affecting the $\pi-\pi^$ energy gap. You can observe $n-\pi^$ transitions in molecule such as acetone or formaldehyde where there are lone pairs not involved in bonding.

@porphyrin but also Cl has lone pairs


text:
Obtaining transition density matrix
I'm trying to assemble a transition density matrix on quantum computer using one-body excitation $\widehat{E}$ operator on 2 active spatial orbitals, i.e. on 4 spin-orbitals. The problem is, I need to rewrite all the four parts of $\widehat{E}$ only via expectation values
I have a one-body excitation operator defined as
$$
\begin{align}
\widehat{E}_{00} &= \widehat{a}_0^\dagger \widehat{a}_0 + \widehat{a}_2^\dagger \widehat{a}_2\\
\widehat{E}_{01} &= \widehat{a}_0^\dagger \widehat{a}_1 + \widehat{a}_2^\dagger \widehat{a}_3^\dagger\\
\widehat{E}_{10} &= \widehat{a}_1^\dagger \widehat{a}_0 + \widehat{a}_3^\dagger \widehat{a}_2\\
\widehat{E}_{11} &= \widehat{a}_1^\dagger \widehat{a}_1 + \widehat{a}_3^\dagger \widehat{a}_3
\end{align}
$$
and I need to evaluate a transition density matrix
$$
\begin{bmatrix}
\langle \psi_0|\widehat{E}_{00} |\psi_1 \rangle & \langle \psi_0|\widehat{E}_{01} |\psi_1 \rangle\\
\langle \psi_0|\widehat{E}_{10} |\psi_1 \rangle &  \langle \psi_0|\widehat{E}_{11} |\psi_1 \rangle
\end{bmatrix}
$$
without computing $\langle \psi_i|\widehat{E}_{ij} |\psi_j \rangle$ explicitly - as in real-world setting I have no statevectors.
For $\langle \psi_0|\widehat{E}_{00} |\psi_1 \rangle$ and $\langle \psi_0|\widehat{E}_{11} |\psi_1 \rangle$ I'm using the formula from (https://journals.aps.org/prresearch/pdf/10.1103/PhysRevResearch.1.033062)
$$\begin{align}
Re\left(\langle \psi_i | \widehat{U}^\dagger(\theta^*) \widehat{A} \widehat{U}(\theta^*)| \psi_j \rangle \right) &= \langle +^x_{ij} | \widehat{U}^\dagger(\theta^*) \widehat{A} \widehat{U}(\theta^*)| +^x_{ij} \rangle - \frac{1}{2} \langle \psi_i | \widehat{U}^\dagger(\theta^*) \widehat{A} \widehat{U}(\theta^*)| \psi_i \rangle - \frac{1}{2} \langle \psi_j | \widehat{U}^\dagger(\theta^*) \widehat{A} \widehat{U}(\theta^*)| \psi_j \rangle\\
|+^x_{ij}\rangle &= \frac{|\psi_i\rangle + | \psi_j \rangle}{\sqrt{2}},
\end{align}
$$
allowing me to evaluate expectation values on a quantum computer and obtaining the transition amplitude.
But, while  the formula works for $\widehat{E}_{00}$, $\widehat{E}_{11}$, it doesn't work for $\widehat{E}_{01}$, $\widehat{E}_{10}$. Do you know what's wrong and how could I fix it?

My guess
I'm suspecting, that the problem lies in the fact, that $\widehat{E}_{00}$ and $\widehat{E}_{11}$ are Hermitian, while $\widehat{E}_{01}$ and $\widehat{E}_{10}$ are not. So, is there any way around this?

My guess 2
Well, it seems, that the problematic part (at least from the "proof" what I've seen) is, that it works only with symmetric operators. But, according to the paper it shouldn't be the case.


text:
Is an AuCl4 / Zn battery possible? (similar to CuSo4 / Zn Daniell Cell) - expense is no consideration
I have been approached by an interesting fellow who collects anything made from Gold.
He has asked me to make him the Gold/Zinc equivalent of the Copper/Zinc Daniell Gravity Cell.
Would this be possible using gold chloride instead of copper sulphate?
The standard electrode potential for the
Zn2+ + 2 e− ⇌   Zn  (-0.76 V)
will be the same but the electrode potential for the gold is approx 0.6 volts higher than the Copper electrode in a regular Daniell cell.
[AuCl4]− + 3 e− ⇌   Au(s) + 4Cl− (0.93 V)
Any tips, insights, and pitfalls would be appreciated before I accidently do something dumb with this fellow's Gold.

At some current density or too high charging voltage, it may develop chlorine at charging, when diffusion or charge transfer are not fast enough.

Thanks for the tip.  Has anyone ever made one of these batteries that you know of?  I cannot find any reference articles...probably because no one in their right mind would want such a kooky battery.

Making a functional galvanic cell is easy. Making a good one is very difficult, regarding capacity, power, longevity, energy and power density. The critical factor is making as big active electrode surface as possible. For rechargable ones, the electrolyte and electrode material must endure as many cycles as possible without mechanical or chemical deterioration.

From what I gather, this fellow is interested in having this battery form gold metal crystals in the same manner as the Daniell Cell he saw in my lab grows copper metal crystals.  That should make the battery simpler since no recharging will be needed.  If I can get it to blink an LED through a step up circuit that would definitely amuse him so ideally... 1 V @ 20 mA.


text:
Distance of Orbitals from Nucleus relation with Energy level and penetration power
I have studied that according to Aufbau rule the energy of subshells is dependent on the sum of $ n $ and $l$ values. This would imply that the energy of subshells in a shell varies as
$$ ns \lt np \lt nd \lt nf  $$
and this makes sense (intuitively from Bohr model) too as the closer the electron is to the nucleus the lesser energy it would have.
And this idea corresponds to penetration power as well where the trend is
$$ nf \lt nd \lt np \lt ns  $$
as farther the subshell is from nucleus lesser it experiences its force and hence lesser penetration power.
Everything made sense until I saw the graphs of Radial Probability Distribution.

and also the expression for average radius of subshell
$$ 1/2(3n^2 - l(l+1)) $$
According to these both my conclusions about penetration power and energy levels should be wrong as for greater $l$ value average radius and even most probable radius both are smaller than corresponding values of smaller $l$ values. What is wrong with my reasoning?

Just a point to note - When you say closer to the nucleus, the lesser the energy, it means the potential energy is more and more negative.

@Proscionexium yes exactly my point, not talking about magnitude.

Note the energy E is proportional to -1/r, so it will help if the radial probability was weighted by multiplying by 1/r.  Parts near a nucleus would become more prominent. Similarly, if the weighted average radius is calculated, it would be closed to nucleus, (to be formally equal to equivalent radius of circular orbit with the same el. energy). It would be more closer for orbitals with more significant near nucleus maxima.

@Poutnik could you explain that in simpler terms as I have only studied this topic at just a little more than high school level and don't have that much knowledge on the same.

The mean or most probable radius and the energy are directly comparable for circular orbits, what is not our case. Probabilities for being closer to a nucleus are more significant for the average energy.

@Poutnik so is it not reasonable to compare radii and energies for orbitals of different shapes?

I have studied that according to Aufbau rule the energy of subshells
is dependent on the sum of $n$ and $l$ values. This would imply that the energy of subshells in a shell varies as

$$ n_s<n_p<n_d<n_f $$
** Let us try a few examples, from the picture diagram. **
$ 1 S : n=1, l=0; n+l = 1$
$ 2 S : n=2, l=0; n+l = 2$
$ 2 P : n=2, l=1; n+l = 3$
$ 3 S : n=3, l=0; n+l = 3$
Here is a surprise, different from your understanding.  For the $ 2 P $ and $ 3 S $ examples, the sum of $n+l$ is equal instead of the increasing of order that you had expected.
$ (n+1)_{2P} = 3 = (n+l)_{3S} $

And this makes sense (intuitively from Bohr model) too as the closer
the electron is to the nucleus the lesser energy it would have. And
this idea corresponds to penetration power as well where the trend is
$ n_f < n_d <n_p < n_s $

Well that quoted statement needs to be checked carefully.  Unfortunately a little bit of background in Quantum Mechanics and the Schrödinger equation are needed for a more intuitive understanding of what is going on.
There are a few different forms of the Schrödinger equation. To fundamentally understand electron-ionization of atoms, the time-independent form is most useful 2 :
The simplest way to write the time--independent Schrödinger equation is
$H\psi = E\psi$, however, with the radially-symmetric Hamiltonian operator expanded it becomes:
$\sum_i{(-\frac{\hbar^2}{2m_e} \frac{d^2 \psi_i}{dr_i^2})} + \sum_i{V_i\psi_i} = E_i\psi_i\\ $
The left-hand-most side of the equation is calculated from momentum contributions (of the wave-functions themselves) versus radius.  In Quantum Mechanics, the electron can be found anywhere when it is ionized with some probability, but until it is ionized is is found no-where.  There is just some probability that it it exists at any particular place to enable wave-function expected-energy calculations. The wave-function, in short, corresponds to a probability distribution for each $i_{th}$ electron.
On the left-side is also a sum of the wave-function multiplied by a potential function that accounts for the energy that the electron has because of the protons at the nucleus or because of other electrons.
To get to understand this potential (labeled as $V$ in the equation), there are a couple of interactions that are important, for the moment I will describe for Helium, for you to generalize to other atoms:

The potential of the inner-electron due to the two protons in the
center of the nucleus and its probability density.  (This well is
so deep that the wave-function of the inner electron is not much influenced by the outer-electron.  For a good first calculation of ionization energy, this inner-electron wave-function can be assumed to not change at all.)
The potential of the outer-electron including the shielding of the nucleus by going to a value of $e^{2+}$ to $e^{1+}$ because the inner electron has a shielding value of $e^{1-}$.  Ironically, for a basic good calculation, the outer-shielded-electron wave-function does not change very much either, and for a first start, this change can also be entirely neglected.
The interaction of the inner electron
and outer electron as a coulomb correction to the Schroedinger
model for Hydrogen, which accounts only for a point-source nucleus
versus a spread-out expectation of the electric field and the
expectation of the outer electron charge density integrated
(on the entire volume).
For this Coulomb interaction of the inner electron and outer-electron, the inner-electron wave-function is that from $e_{2+} hydrogen$  and the outer-electron wave-function is that from $e_{1+} hydrogen$.

Thus regarding:

as farther the sub-shell is from nucleus lesser it experiences its
force and hence lesser penetration power. Everything made sense until
I saw the graphs of Radial Probability Distribution.

The component of the Schrödinger equation that deals with the nuclear attraction is the fully-shielded atom, because the outer-electron that is ejected for the first-ionization is always out-most. However, the near-field Coulomb energy contribution changes quite a bit as the electrons are pushed further and further from each-other, as is shown in the below diagram.
Outer-most Ionization Energies:
.
Like you stated, we can expect that for higher element-numbers in the Periodic-Table, as a general trend, the outer electron ionization energy is lower because the inner-electron-shielding is more complete.
I have not yet plotted out your diagrams, but instead I think this reference about Hydrogen Wave Functions might be more-complete.
One of the illustrations is that reference 6 is very much easier to understand:

Also, I found for you this reference websites.umich.edu/~chem461/QMChap7.pdf.  It also has the linear plots in it with a more detailed explanation about how to calculate the probability distributions which should be more insightful than the previous diagrams you referenced.
From the reference Hydrogen Electron Density Versus Radius R, there is a plot on page 19:

It does seem that the expected radius for the $3 S$ - wave-function is greater than the expected-radius for the $3 P$ - wave-function, which is also apparently greater than the expected-radius for the $3 D$ - wave-function.  So the expected trend from your plots seem somewhat confirmed, in terms of expected radius.  The problem with your understanding is that expected radius for Hydrogen is not the only thing determining the first ionization potential.  Indeed for Hydrogen, only the quantum number $n$ determines potential ionization energy, independent of $l$ and $s$ quantum numbers.  The plots you have seem to be similar (in trends) to those for pure hydrogen wave-functions, all with expected energy just from $n=3$ and nothing else, making the ionization energies for the three first ionization energies exactly the same because of other effects, such as distributions versus spherical coordinates (aside from the radial coordinate $r$).

thanks again for your answer but I think you misunderstood my question (or maybe I haven't explained it well), I'm not bothered with the size comparison of orbitals of different shells (K,L,M...) , my confusion arises from the fact that according the average radius formula mentioned in my question and also the plots (even in the reference you shared) depict that the $s$ subshells is larger than $d$ for the same $n$ value, despite the fact that it has less energy and less penetration power. Is there a reason for that or the information provided to me is false?

If the S-shell radius is larger than the D-shell radius, that would put the S-shell-energy conceptually lower.  It is the nucleus that binds the electron;  so if the distance is expected to be further from the nucleus, then the nucleus's effect is weaker.  The energy you are talking about is the binding energy to the nucleus.  Far away from the nucleus, this energy is Hydrogen-like, corresponding to $e^{+1}$.  Closer to the nucleus this binding energy is greater because of the lower shielding of the other electrons.  Please keep on asking be questions if this is not clear.

How would the energy of S-shell be lower if its radius is greater than that of D-shell, wouldn't it be greater as further from the nucleus the magnitude of Columbic attractive force would decrease and hence the energy would decrease, as closer to the nucleus more negative the energy is and hence smaller. I'm not sure the energy we are talking about is the same but I think they have the same magnitude, as the energy the electron has in an orbital should be the same as its binding/ionization energy.

@Banaj Mahajan :  please see my updated answer to your question.  The ionization-energy picture is not clear from just the radial-coordinate plot.  Plots versus the 3 spherical coordinates are necessary to get a more complete picture about contributions to the energy aside from the radial-distribution contribution to the ionization-energy. I suspect that a good part of the periodic table can be calculated straight-forwardly, but I have not had the time yet to take on this expanded task versus just the Helium calculations that were very successful.

So considering all the factors including the spherical coordinates what would be the size comparison for the subshells of the 3rd shell?

The Aufbau Principle reference is summarized with the following picture:

Keep in mind that the Aufbau principle is founded on 1) the energy levels of Hydrogen, which are only determined by the quantum number $n$.  But that is not the entire picture.  Because of the shielding effect as well as near-nucleus penetration, there is an additional perturbation that comes in to play.
I have calculated this for Helium, and I discovered that the binding energy taking into account this addition perturbation could be calculated with around $3$ $Percent$ $accuracy$ which I considered a great success. There are two electrons in this model, an inner electron and an outer electron.  For Helium, the inner electron is so closer to the nucleus, and even as the radius approaches infinity, it is still inner, meaning that it bears the $2e^+$ charge attraction. And the outer electron, even as the radius approaches infinity is still outer, meaning that it bears $1e^+$, being shielded by the inner electron.
The two spins are opposite for Helium for the inner and outer electron, so there is no major problem with spin interaction.  The perturbation comes almost entirely from the effect of the outer electron penetrating into the shielding of the inner at the near field, close to the nucleus.
So the Aufbau Principle also considers this shielding perturbation of energies to arrive at an accurate ordering of shells, which makes it different than the pure Hydrogen calculations that just depend on the quantum number $n$.
There are a couple of other considerations.  First of all, when the Aufbau Principle references the $2$ $P$ $Orbital$, it is really a hybridized orbital, which is some combination of $2$ $S$ and $2$ $P$ orbitals (which for idealized Hydrogen have instead equal energy levels).  This hybridization pushes the $S-P$ orbital outward, with more focus. This same hybridization effect increases with $S-P-D$ hybridized orbitals and so forth.
Taking account with the VESPR model, which includes this Hybridization-focusing effect of electron bonds, one can see that as that from the order of $1S$ which is spherically symmetrical, going on to $2S, 2P$ which has 4-fold symmetry, that the electrons of that level are able to distance themselves out further from one another in the near field to the nucleus.  The Aufbau Principle is a kind of short-hand notation, labeling $2P$ as what is really a $2S-2P$ hybridized orbital that allows for this additional separation.  The additional separation in the near-field to the nucleus of the electrons allow for lower interaction energy there.
This is my understanding of how all of this works.  I am welcome to additional insight and comments, as I have not yet gone through the calculations for Lithium and so forth, which I estimate are not hard calculations, just tedious ones.

Hello. The figures in the OP just refer to the radial distribution function in the H-atom (multiplied strangely by $4\pi$), no need to analyze $\ce{Be}$, an so on. I saw the link to "calculated". You wrote a book about the helium atom?

@Metal Storm :  Yes (pdf free), I went through the symbolic math for Helium using this inner-electron, outer-electron model.  Because all the integrals are powers of radius and an exponential, the symbolic solution is straight-forward.  The inner electron is just like Hydrogen in this model, except it faces a nucleus charge of $ 2e^+ $. The outer electron's wave-function is a polynomial times an exponential that in the far-field is bound by a shielded $ 1 e^+ $ field.  One can calculate the interaction energy using the expectation for the Electric field (it is after all spherically symmetric).

Never heard of the inner-electron, outer-electron model. So you were able to obtain an analytical (symbolic in you terms) expression for the $2$-electron wavefunction $\Psi(\mathbf{r}_1,\mathbf{r}_2)$?

@StephenElliott thanks for your answer, but it didn't really clear my doubts, additionally I haven't studied such in depth didn't even know aside from penetration and shielding there was something called perturbation.

@Metal Storm :  Introductory College Chemistry courses talk about the inner electron and the outer electron.  But it is hard to believe the texts or the teachers! There is actually an inner electron for Helium.  When the outer electron is ionized, the inner electron does not change its distribution very much.  There is not really $ \Psi(r_1,r_2) $ but rather $ \Psi_1(r_1) $ and $ \Psi_2(r_2) $ which are somewhat independent wave functions for electron one and electron two.  Except the Energy of the system with these two electrons is different than with Hydrogen.

@Metal Storm :  To clarify, there is a $\Psi(r_1,r_2)$ contribution that can be calculated.  But this is not the only contribution to the first ionization energy of the atom.  Without the other energy components that I mentioned in the more detailed answer, the energy calculation is incomplete and I estimate more inaccurate.  I would welcome an updated answer or comment demonstrating otherwise.

@StephenElliott Is it possible to download your book and check it out?

@Metal Storm :  please reference the link underlined in blue called calculated to get a download of the book.  It seems that the pdf version requires an account setup, but is otherwise still free from Lulu at https://www.lulu.com/shop/stephen-elliott/magnetics-and-radially-symmetric-helium-bh/ebook/product-18npv8vz.html?page=1&pageSize=4.  The integrals are all taken from standard integration tables and are symbolically expanded.  I had to make a couple of attempts before I got the right answer, but at the end I was happy that the accuracy seemed very good.

Where did you get this diagrams from?There is something definitely suspicious with your diagrams.The more  the wavefunction of a particle  changes over r the more energy the particle has.From your diagrams obviously the "3s" wavefunction has more energy than the other two because it has more valleys and troughs which according to the Aufbau principle cannot be true.
Also if you look at some of your diagrams at some points the wavefunction $\Psi(r)$ takes a value $>1$ which isnt possible as well.

I got this from JD LEE but I have checked several other sources and they too have a similar curve

the figures appear to be correct they show $4\pi r^2 R_{n\ell}(r)^2$

@porphyrin even if the x axis is $r^2$ diagram A still has more valleys and troughs which indicates a bigger energy of the wavefunction.Remember the more the wavefunction changes the bigger the energy of the particle which is described by that wavefunction

@Volpina the x axis is $r$ in the diagram and I checked again the diagrams are similar from other sources as well

@BanajMahajan can you site all  your sources of the diagrams?Thanks.I could be wrong and if I am wrong I want to learn.

@Volpina The radial probability diagrams are correct. but are misleading due non-homogeneous potential gradient.

@Volpina, the energy depends only on the principle quantum number,  so is the same for the H atom wavefunctions described in the question.

@Volpina I have taken the diagrams from Concise Inorganic Chemistry by JD LEE ( 4th Edition) you can once google only radial probability distribution curves for 3s,3p and 3d and it would yield similar results


text:
How does enzyme velocity decrease as reaction quotient approaches equilibrium constant?
I created a simulation which approximates molecule concentrations in a cell. One part of it are enzymes that can catalyze pre-defined reactions, e.g. $\ce{A <=> B}$.
I based the kinetics for these reactions on Michaelis-Menten kinetics, so every enzyme has a $V_\text{max}$ and $K_\mathrm M$, and has velocity $v=V_\text{max}\frac x{K_\mathrm M+x}$ at a given time point, where $x$ is the concentration of substrates.
I use the Reaction Quotient $Q$ at any point in time to determine in which direction a reaction would proceed. E.g. would (net/overall) A be converted to B or vice versa. Each reaction is defined with a reaction energy which is used to calculate Equilibrium constant $K_\mathrm e$.
Here is my question
My understanding is that Miachelis-Menten kinetics only describe the case in which $Q\ll K_\mathrm e$. What happens if $Q\to K_\mathrm e$?
Intuitively I would assume that the overall reaction velocity slows down as $Q$ approaches $K_\mathrm e$.
Is there a term that I can add to the Michaelis-Menten equation which would describe this decreasing $v$?

Here's a good description: http://www.bio-physics.at/wiki/index.php?title=Reversible_Michaelis_Menten_Kinetics Also, I don't think you mean the Nernst equation. That is only applicable to redox reactions.

Then, I have a follow up question to this. In the link's equation, for simplicity let's assume $V_{max}^f = V_{max}^b = V_{max}$ and $K_{M,1} = K_{M,2} = K_M$, we get $v = V_{max} \frac{S - P}{K_M + S + P}$. So that means even if $S \rightleftharpoons P | -100kJ$, the enzyme would stop ($v = 0$) once $P = S$? My understanding was always that the reaction would continue until $K_e = Q$

Or in other words: Say $K_{M,2} >> K_{M,1}$. So, even though a lot of $P$ might exist, it will not have a big effect and $v \approx V_{max}^f$. Then, would $v \approx V_{max}^f$ until the last molecule $S$ was converted to $P$ that sets $Q = K_e$, then suddenly $v = 0$? I would have imagined that there is a kind of slow-down in $v$ as $Q \rightarrow K_e$

If P is present, you are no longer under Vmax conditions, since P and S are competing for the enzyme. Also, if the Vmax and Km values are the same for forward and reverse reaction, then Ke necessarily equals 1, so your first hypothetical is not possible.

I see your point about the enzyme being occupied by P and this is something I should also add to my simulation. However, I still don't see the connection to $K_e$. Aren't the $K_M$s properties of the enzyme, while $K_e$ is a thermodynamic property? Maybe I should formulate the question without any enzymes.

The kinetics and thermodynamics are related by the fact that equilibrium means that the conditions are such that the forward and reverse reaction are occurring at identical rates, thus no net reaction in either direction is observed. If the Km's and Vmax's are the same, then the rates will be the same when S=P

I see, so the relations of both $K_M$s and $V_{max}$s would have to result in $K_e$ for a consistent simulation.

Yes. $K_e$ must be equal to the ratio of the forward $k_{cat}/K_m$ to the backward $k_{cat}/K_m$.

The Michaelis-Menten mechanism is the following:
\begin{align}
  \ce{E + S &<-->[$k_1$][$k_{-1}$] ES}  \tag{R1} \\
  \ce{ES    &->[$k_2$] P + E} \tag{R2}
\end{align}
If you apply: (1) steady-state approximation for the enzyme-substrate complex $\ce{ES}$, and (2) enzyme concentration balance $[E_\mathrm{t}] = [E] + [ES]$, then you arrive at the rate law or the Michaelis-Menten equation
\begin{equation}
  r_\mathrm{P} = \frac{\mathrm{d}[P]}{\mathrm{d}t} =
  \frac{k_2[E_\mathrm{t}] [S]}{[S] + K_\mathrm{M}} =
  \frac{r_\mathrm{P,max} [S]}{[S] + K_\mathrm{M}} \tag{1}
\end{equation}
where $K_\mathrm{M} = (k_2 + k_{-1})/k_1$ and $r_\mathrm{P,max}:=k_2[E_\mathrm{t}]$, which is a constant.

[OP] I based the kinetics for these reactions on Michaelis-Menten kinetics, so every enzyme has a $V_{max}$ and $K_M$, and has velocity $v=V_{max}x /(K_M+x)$

The product $P$ has a maximum rate of $r_\mathrm{P,max}$, not the enzyme. This value is obtained when the substrate is so abundant that $[S] \gg K_\mathrm{M}$.

[OP] I use the Reaction Quotient $Q$ at any point in time to determine in which direction a reaction would proceed. E.g. would (net/overall) $A$ be converted to $B$ or vice versa. Each reaction is defined with a reaction energy which is used to calculate Equilibrium constant $K_e$.

You have as many reaction quotients $Q$ as reversible chemical reactions. You can do this for $\text{R1}$ because this step is reversible. However, you cannot do this for $\text{R2}$, because for the M-M mechanism this step is irreversible. Thus, once any quantity of $\ce{ES}$ complex has been formed, $\text{R2}$ will proceed, independently if you had a lot of $\pu{P}$ at $t=0$.

My understanding is that Miachelis-Menten kinetics only describe the case in which $Q<<K_e$. What happens if $Q→K_e$?

A reaction quotient is not defined for irreversible reactions.

Intuitively I would assume that the overall reaction velocity slows down as $Q$ approaches $K_e$. Is there a term that I can add to the Michaelis-Menten equation which would describe this decreasing $v$?

I propose to modify the reaction mechanism in the second step
\begin{align}
  \ce{ES    &<-->[$k_2$][$k_{-2}$]P + E} \tag{new R2}
\end{align}
However, the math has to be carried again, because the old $\text{R2}$ had a rate of $k_2[ES]$, and now we have
\begin{align}
  r_\mathrm{P} = \frac{\mathrm{d}[P]}{\mathrm{d}t} =
  k_2[ES] - k_{-2} [P][E] \tag{2}
\end{align}
This adds an additional rate constant $k_{-2}$ to your simulation. Another option is to consider the Briggs-Haldane rate law, where every step is reversible. If you need help with the math just tell in the comments.

Is this the Briggs-Haldane rate law? I find it mentioned a lot on google but no explicit formulas.

@mRcSchwering No. That web has in fact the mechanism I proposed to you. The one I am referring to is $\ce{E + S <--> ES <--> PE <--> P + E}$, where a product-enzyme complex is added.

Ah, yes now I see it. I think I will take your mechanism. It was mentioned in comments earlier as well. Does it have a name?

@mRcSchwering Yes, my name. Joking! Just say that is the classic M-M mechanism, but you allow the possibility of reversibility on the second step. This means that the presence of $P$ and $E$ will hinder the rate of reaction for the formation of your desired product $P$.

You can use the formula you have for the rate of the forward reaction. Once you have product, you need a second set of kinetic parameters for the reverse reaction. As you reach equilibrium, one rate increases and the other decreases until they are equal. So you can’t say the reaction slows down. Instead, the net reaction tends to zero while the forward and reverse rate remain non-zero and equal.


text:
What equation can be used to determine the mass of water evaporation over time, when cooking?
When baking, cooking or writing recipes, it would be useful to have a ballpark estimate of evaporation rates. This question asks about water, although the answer would ideally include an interchangeable variable to account for other substances, such as cooking oils.
Essentially, I would like to build a calculator program that I could just type in these variables, and be able to answer the following questions:

How much water will evaporate after 10 minutes of boiling?
How long will it take for my 500 grams of water to boil down to 300 grams?

There are some other questions on StackExchange which kind of ask similar questions, but the scope is too broad. For example, there is no wind in an indoor kitchen, so we would be able to use an established value for wind activity. Another example: we know there's negligible solar radiation in a kitchen that doesn't meaningfully change the equation. There's also multiple citations referencing one particular webpage, The Engineering ToolBox. However, that website and its javascript calculator seems to account for large bodies of water (such as swimming pools) and it's unclear how to scale it down into something useful for a kitchen.
Basically, in a kitchen, there are known variables:

The substance (water)
The initial mass of the substance (500 grams)
The initial temperature of the substance (23°C)
The evaporative surface area in centimetres ($\frac{1}{4}\pi d^2$, where  is the diameter of your saucepan, 16 cm → 201 cm²)
The air temperature (23°C)
The air humidity percentage (65%)

Some more complicated, yet still accessible variables are:

The kitchen's position above sea level (therefore, the air pressure)
The cooking medium's heat capacity (stainless steel, copper, cast iron)
The heat output of your stove, in BTUs or joules (induction, electric, gas)

Perhaps, though this is speculative, the theoretical equation would be simpler if it assumed the saucepan and water begins at boiling temperature (100°C, perhaps?).
Surely, using these variables, there must be an equation that can be used to determine the mass of water evaporation over time, when cooking. Does anybody know how to do this?

As a rule of thumb, the rate of water evaporation = heating power / water specific heat of evaporation. Otherwise, there is many corrective factors that are easier to describe than evaluate.


text:
How does the canonization method of RDKit operate when valence is exceeded?
RDKit's canonization method1 allows nitrogen to bond with 3 hydrogen atoms and one oxygen atom
>>> mol = Chem.MolFromSmiles('[NH3]=[O]')
>>> Chem.rdmolfiles.MolToSmiles(mol, allHsExplicit=True, allBondsExplicit=True)
[NH3+]-[O-]

but not with two hydrogen atoms and one oxygen
>>> mol = Chem.MolFromSmiles('[NH2]=[O]')
>>> Chem.rdmolfiles.MolToSmiles(mol, allHsExplicit=True, allBondsExplicit=True)
Explicit valence for atom # 0 N, 4, is greater than permitted

Is it taking into account the morphology of the atom and bonds, not just its valence? Does the first one represent a stable compound, but not the second?

It's not exceeding, more like the opposite. Your second thing is a hydroxylamine radical.

This is probably a better question for either the rdkit e-mail list or the GitHub Discussions

@Mithoron but, how can this be predicted for a given compound? for example, [C]=[N]#[N] > [C]=[N+]=[N-], is it a property that only affects nitrogen?


text:
Crown ether complex formation with cesium carbonate
I am searching for some papers which form complex between crown ether and caesium carbonate. But Most papers only use caesium carbonate as a base in the synthesis of crown ether, not ionic compound for chelation. Can't caesium carbonate form complex with crown ether? Does the fact that caesium carbonate is a base have to do with it? More generally, does the anion of ionic compound also affect the formation of complex between crown ether and metal cation?

Size matters in crown ethers. You have to find the right crown ether where Cs ion can fit. The are scores of articles on this topic. Did you try Google Scholar?

@AChem I know there are appropriate sizes of crown ether according to the target cation. But I wonder that despite of right size of cavity whether cesium carbonate cannot form complex. Most papers uses cesium chloride or cesium nitrate. Does the anion 'carbonate' or basicity of cesium carbonate matter?

Krang Lee, Yes, the anions can significantly affect extraction / crown ether's selecitivity. Anion effect on selectivity in crown ether extraction of alkali metal cations
Uriel Olsher, Matthew G. Hankins, Youn Doo Kim, and Richard A. Bartsch
Cite this: J. Am. Chem. Soc. 1993, 115, 8, 3370–3371

If nobody has specifically studied CsCO3 and crown ether complexation it could be a small and useful publication for you just like one cited in J. Am. Chem. Soc. This is why countless Chem Dept. in the world are still alive and thriving. Not all is known or published.


text:
Why do different orbitals have different shapes?
P and S orbitals have different shapes defining what kind of bond they have like pi bond, sigma bond etc. So I was wondering what actually determines the shape and each region for different types of orbitals?

Have you reviewed a targeted search like site:stackexchange.com OR site:libretexts.org  orbital shapes  ?

Why do the solutions of certain differential equation have such-and-such properties? This is a deep question which can't be answered easily.

 A solution to the Schrödinger equation maybe written as a product of radial and spherical functions. The size of different orbitals varies according to the radial function and the shapes of different orbitals are a result of different spherical functions.

The Schrödinger equation
You might be familiar with the following (magic) equation:
$$
\hat{H}\mathrm{\Psi} = E\mathrm{\Psi}
$$
where $\hat{H}$ is the Hamiltonian operator, and $\mathrm{\Psi}$ is the eigenfunction corresponding to its eigenvalue, which is, energy $E$ of the system. $\mathrm{\Psi}$s, often called wavefunctions, are obtained by solving the Schrödinger equation.
The Born Interpretation
The different orbitals are actually isosurfaces of the square-magnitudes $|\mathrm{\Psi}|^2$ of these $\mathrm{\Psi}$s:
$$
|\mathrm{\Psi}|^2 = \mathrm{\Psi}^*\mathrm{\Psi}
$$
which, for an electron, represent the probability of finding the electron on that surface. Mostly, the maximum-probability isosurface is shown.
Separation of Variables
$\mathrm{\Psi}$ is a function of space-coordinates and time. We may represent space-coordinates in radial form $(r\text{, }\theta\text{, and }\phi)$ and apply separation of variables:
$$
\mathrm{\Psi}(r, \theta, \phi, t) = R(r)Y(\theta, \phi)f(t) \tag{1}
$$
where, the radial component, $R(r)$, of $\mathrm{\Psi}$, roughly speaking, represents size of the orbital, the angular component, $Y(\theta, \phi)$, (again) roughly speaking, represents shape of the orbital, and we also have a time component $f(t)$ to represent how this wavefunction varies with time. From Equations (1) and the Schrödinger equation:
$$
\hat{H}(R(r)Y(\theta, \phi)f(t)) = E(R(r)Y(\theta, \phi)f(t)) \tag{2}
$$
Spherical Harmonics
We need to focus on the shape of orbitals; therefore, our discussion will be on $Y(\theta, \phi)$. When we solve Equation (2) for $Y$, we obtain spherical harmonics which are dependent on two variables called $l$ and $m$. Yes! these correspond to the two principal quantum numbers of the four $n$, $l$, $m$, and $s$, where $l\in [0,n-1]$ and $m\in [-l,l]$:
$$
Y(\theta, \phi) = Y_l^m(\theta, \phi)
$$
The solutions, if visualized, look (something) like this:

Notice how the shape of orbitals emerges naturally from the sign and values of spherical harmonics.
Reference
Levine, I. N. (2014). Quantum Chemistry, 7$^\textit{th}$ edition. Pearson Education Inc.

The simplest answer is, of course, that this is what solving the equations produces, which is not that helpful.
When the Schroedinger eqn. for the H atom is solved instead of using $x,y,z$ coordinates we use $r,\theta,\phi$ as mathematically the solution is easier this way, where $r$ is radial distance and $\theta, \phi$ angles. As well as radial solutions, depending on distance from nucleus $r$ (and $n,\ell$ quantum numbers), there are also mathematically well known angular functions called spherical harmonics which can be complex numbers and have various shapes depending on the quantum numbers. To be able to draw these, and to make chemical sense, we have to combine them together so that they are no longer complex numbers, and this is what is done to form the p orbitals etc. Mathematically this 'trick' is ok as the sum of solutions can be shown also to be a solution of the Schroedinger eqn.
So its not simple, but once we have the orbital shapes we can then begin to join them together to form bonds or even combine them again to form hybrids such as tetrahedral $sp^3$.


text:
I stained my lab workstation with Coomassie blue. How do I clean it?
I stained my work station with Coomassie blue solution during my SDS-PAGE protocol. I have been cleaning but the solution seems to can't be removed. What should I use to remove all these stains?

What is the surface of the work-station made of? How long have the stains been there?  What are you using to clean the stains?

What is a SDS-PAGE protocol ?

Coomassie Blue is colorless at pH 1 - 3. So by adding a $\ce{HCl}$ solution diluted to about 0.01 M, it should make the blue dye colorless.

The color will come back once HCl evaporates. A very dilute solution of bleach (used as per manufacturer's and lab instructions) should remove dye stains.

Use organic solvents, like ethanol, DMSO, etc.

Why bother at least you will know its yours :)


text:
Brittle Properties of Ionic Compounds?
From my understanding, ionic compounds are brittle due to the alternating cations and anions within their lattice. Hence when stress is applied, similarly charged ions will repel causing the structure to shatter. However, by this logic, shouldn't ionic compounds also be able to come back together this way? Consider grains of table salt, why can't they join together via electrostatic forces?
Edit: I don't think the links answer my question.

Why are so many ionic compounds brittle?

I think these textbook stories of ionic repulsion and the answers in the linked question are partial stories (looking at one side of a coin). Perhaps a material scientist here can give a better picture rather than handwaving answers. Why is ordinary sulfur brittle? Why are certain plastics brittle?

https://chemistry.stackexchange.com/questions/174002/malleability-and-ductility-origins

You logic is thoroughly wrong.

Does this answer your question? Malleability and Ductility Origins


text:
Why does this show E1CB mechanism?

The following is a solution to a "match the following" question from IITJEE 2006.
I'm confused why does it show E1CB mechanism? Isn't the mechanism exclusively for highly acidic Hydrogen atom, which eventually creates a carbanion intermediate? I don't know why the following setup creates such a carbanion intermediate.

As well you should be confused. A Chem. Absts. search found no hits, even with the para-nitro analog. The question maybe made up to illustrate E1CB and a deuterium isotope effect.

I agree the info is confusing. If the major product is deuterated it implies a rapid exchange at the benzyl position forming a carbanion that did NOT eliminate. The kinetics would be of interest. Extensive studies of these compounds generally support an E2 mechanism. In any case were the mechanism E1CB formation of the carbanion should give more immediate elimination since the deuterated compound should react much more slowly. A reference should have been cited. Unfortunately I have no access to chem abstracts to research in depth but one reference that pops up states that deuterium exchange does not take place.
If this is a fabricated question it is more than unfortunate it is professional malfeasance.  One far out possibility is a paper was misread and a 1-phenylethyl bromide was reacted. This compound might have an active benzyl H that exchanges faster than the E2 reaction.  It might be an interesting reaction to study.

Conditions for E1CB mechanism are as follows: -

The leaving group must be bad (Fluorine, etc.)
The carbanion should be resonance stabilized.

Since the leaving group in the question is Bromine (which is a good leaving group), this will not show E1CB. Apparently, the question is wrong.

The substrate having a good leaving group definitely doesn't prevent E1cB from happening, it just means E2 and E1 can compete.

Your statement is self contradictory or at best all inclusive, a simultaneous E1cB, E2 and E1 mechanism. Maybe that is what is going on.


text:
Determine the pKa of a diprotic acid by titration (if pKa1 is similar to pKa2)
Determining the acidity constants of a diprotic acid (in my case, tartaric acid) by titration with NaOH and pH-meter is not possible if pKa2 - pKa1 < 2.
This is because there are no "pH jumps" at the equivalent points, so you cannot exploit the method that uses the "midpoints":

In fact, at pH=pKa1 the concentration of H2A is not equal to the concentration of HA- (the simplification performed in the equilibrium constant equation is not valid), as can be seen from the speciation diagram:

This is how the titration curve of 1.5 g of anhydrous tartaric acid VS 1 M NaOH looks like.

How could this be done?

If the experimental numerical data for the curves are available, try the numerical analysis of the 1st and second derivative of the curve before the endpoint, i.e. with higher resolution. As derivatives from numerical data are usually noisy, you may try kernel filtering by e.g. Savitzky-Golay filter available for MATLAB as sgolayfilt ( in compatible Octave too)


text:
Help with drawing an Molecular Orbital (MO) diagram for organometallic complex
I do not understand how to draw the MO diagram for this question, i appreciate any help you can give on my 3rd yr Chem.
Have i understood this correctly?
QUESTION: Using an MO bonding picture and resonance structures explain the reactivity of the following complexes:
(i)  (CO)5W=C(OMe)Ph and Et3N
(ii)   (Np)3Ta=CH(tBu) and (CH3)2C=O (Np = neopentyl)
So for (i), would (CO)5 be written on one side and W on the other of the MO diagram?
Or would (CO)5W be written on one side and C(OMe)Ph on the other of the MO diagram?
Steps needed for ANSWER:
1.
Identify the atomic orbitals: Here, we're dealing with a transition metal (Tungsten, W) bonded to a carbonyl group (CO) and a carbene (=C). W has 5d, 6s, and 6p orbitals, C has 2s and 2p orbitals, and O has 2s and 2p orbitals. The atomic orbitals involved in bonding would be a subset of these.
2.
Identify the ligands: Here, the ligands are CO and =C(OMe)Ph. The CO ligands are known as pi-acceptor ligands because they can accept electron density from the metal into their pi* antibonding orbitals. This back-bonding interaction would result in the formation of metal-carbon pi bonds, in addition to the metal-carbon sigma bonds. The carbene =C(OMe)Ph would also form a sigma bond and potentially a pi bond with the metal.
3.
Combine atomic orbitals of the same symmetry: Next, atomic orbitals of the same symmetry can overlap to form molecular orbitals. For instance, the s orbital of W can overlap with the s orbitals of C and O to form sigma bonds. Similarly, the p orbitals of C and O can overlap with the d orbitals of W to form pi bonds.
4.
Fill in the electrons: Electrons are then added to the molecular orbitals, starting with the lowest energy orbital. In this case, the number of electrons would be determined by the sum of the valence electrons from W, C, and O.
5.
Identify the HOMO and LUMO: These are the highest occupied molecular orbital and lowest unoccupied molecular orbital, respectively. To determine the reactivity of the molecule.

*Side note: More general information if this is all new to you A transition metal ion has five d orbitals of equal energy, which split
into different energy levels when ligands bond to the metal ion. Some
ligands cause a large splitting of the d orbitals (known as
strong-field ligands, like CN- and CO), while others cause a small
splitting (known as weak-field ligands, like I- and Br-).
In a complex ion, the metal atom or ion is at the center of the
molecule, while the ligands are atoms, ions, or molecules that are
bonded to it. The d-orbitals of transition metals are the last set of
orbitals to be filled in accordance with the Aufbau Principle. These
orbitals have specific shapes and orientations and are part of the
outermost (valence) electron shell, which participates in chemical
bonding.
The type of bonding in transition metal complexes involves
coordination bonds, which are a type of covalent bond where the metal
ion (acting as a Lewis acid) accepts a pair of electrons from a ligand
(acting as a Lewis base). This is different from typical covalent
bonding, where each atom contributes one electron to the bond. In the
case of transition metal complexes, the ligand donates both electrons.
Now, let's discuss the orbitals: s orbitals: These are spherical in
shape. Every energy level has one s orbital, and it can hold up to 2
electrons.
p orbitals: These have a dumbbell shape. Each energy level from n=2
onwards has three p orbitals, each of which can hold up to 2
electrons. d orbitals: These have more complex shapes. Each energy
level from n=3 onwards has five d orbitals, each of which can hold up
to 2 electrons. These orbitals are critical for transition metal
chemistry, because they can participate in various types of bonding.
The bonding between the d-orbitals of the transition metal and the
ligands can take several forms. As I mentioned before, sigma and pi
bonding can occur. Sigma bonding involves the head-on overlap of
orbitals and is generally stronger. Pi bonding involves the side-on
overlap of p or d orbitals and is generally weaker. In many cases,
transition metal complexes exhibit sigma bonding with their ligands
and also have pi back-bonding, particularly with pi-acceptor ligands
like CO.*

Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer.

It looks like the question has already been solved online. Here is a screenshot of how it looks

If these don't show, please feel free to follow this reference link


text:
Why are the derivatives of the concentrations divided by stoichiometric coefficients?
Consider this simple dioxygen to ozone chemical equation
$$\ce{3 O2 -> 2 O3}$$
with the average rate of the reaction
$$r = -\frac{1}{3}\frac{\Delta[\ce{O2}]}{\Delta t} = \frac{1}{2}\frac{\Delta[\ce{O3}]}{\Delta t}.$$
So, the rate of change of concentration of the reactant $\ce{O2}$ must be divided by its stoichiometric coefficient, i.e., by $3$ and so does that of $\ce O_3$, i.e., by $2$.
But what I think after doing this is that by doing this division, we are no longer referring to this particular reaction as it requires $3$ molecules of $\ce{O2}$ and to form $2$ molecules of $\ce{O3}$. Why should the reaction rate, if in terms of reactants here, be divided by the stoichiometric coefficient $3$ if there are $3$ molecules taking part in the reaction and thus $3$ molecules contributing to the rate?

I've taken a liberty to correct technicalities, but the title needs to be rewritten to reflect the question, which I'm afraid I don't quite understand. You appear to apply the rate expression by its definition, whereas the mechanism is a bit more complex (see also Chapman mechanism). This might possible the source of confusion.

Rate is the property of the reaction, and not of O2 or O3. There is no point in treating them separately.

For $$\ce{3 O2 -> 2 O3}$$ and
$$r = -\frac{1}{3}\frac{\mathrm{d}[\ce{O2}]}{\mathrm{d} t} = \frac{1}{2}\frac{\mathrm{d}[\ce{O3}]}{\mathrm{d} t},$$
you can consider the reaction rate as the rate of formal elementary operations of taking 3 molecules of $\ce{O2}$ to form 2 molecules of $\ce{O3}$. The reaction rate $\pu{1 mol/L/s}$ is then formally $\pu{1 mol/s}$  of such operations in $\ce{1 L}$. You should consider that whatever component concentration rate you take, it has to lead to the same reaction rate, as the reaction rate does not depend on what component you choose.
1 operation takes $\ce{3 O2}$ molecules to make $\ce{2 O3}$ molecules. $\pu{1 mol}$ of operations takes $\pu{3 mol}$ of $\ce{O2}$ to make $\ce{2 mol}$ of $\ce{O3}$.
Now all what remains is making rates from them, as above.
It is closely related to $\ce{3 O2 -> 2 O3}$ reaction enthalpy $\Delta H_\mathrm{r} [\pu{kJ/mol}]$,
which is enthalpy change for "$\pu{1 mol} \text{ of reaction operations}"$, leading from $\ce{\pu{3 mol} O2}$ to $\ce{\pu{2 mol} O3}$.

Be aware it is not an elementary reaction and the rates for O2 and O3 do not exactly fit to each other with 3:2 ratio.


text:
Derivation of the conversion of Gaussian type orbitals (in cartesian coordinates) to Gaussian type orbitals (in polar coordinates)
While studying basis sets, I came across Gaussian type orbitals (GTOs), which are used as atomic basis sets. I reviewed the polar form of these orbitals, which are very similar to hydrogen atom orbitals. I then looked at the Cartesian form of the same orbitals and tried to derive the polar form from the Cartesian form using simple polar substitutions.
\begin{align}
x &= r\cos(\theta)\sin(\phi);\\
y &= r\sin(\theta)\sin(\phi);\\
z &= r\cos(\theta)
\end{align}
To my surprise, I was unable to get the same polar expression. Both the polar and Cartesian forms of the GTO are attached for reference. Would someone who has worked on this derivation be willing to share it? It would be very helpful.
\begin{align}
\text{Cartesian form: }&&
G_{\ell,m,\alpha}(\mathbf{r}) &= N_{\ell,m} x^{\ell_x} y^{\ell_y} z^{\ell_z} e^{-\alpha r^2} \\
\text{polar form: }&&
 G_{\ell,m,\alpha}(\mathbf{r}) &= N_{\ell,m} r^{\ell} e^{-\alpha r^2} Y_{\ell,m}(\theta,\phi)
\end{align}

These expressions are not exactly the same. They do, however, form different bases for the same space.

https://mattermodeling.stackexchange.com/questions/9133/how-to-build-cartesian-representations-of-spherical-gaussian-basis-functions/9135#9135 may be of use

Could you please provide the reference from where you are studying basis sets?


text:
Neighbouring Group Participation of Aldehyde group
Does the aldehyde group -CHO give neighbouring group participation? For example in the following compound:

Why or why not will the formyl group show neighboring group participation? Although I have never seen an example of a -CHO group showing SN-NGP, it does seem feasible due to the lone pair of O, but I am not sure if it occurs or not.

If NGP occurs, the resultant ring would be highly unstable and strained due to Oxygen also doubly bonded to a Carbon.


text:
Iodide ion is good leaving group and a good nucleophile. How can both be correct?
Leaving Group Tendency of Iodide

$\ce{I-}$ size is big.

$\ce{I-}$ has stable negative charge.

$\implies$ $\ce{I-}$ is good leaving group.
Nucleophilicity of Iodide

$\ce{I-}$ size is big.

$\ce{I-}$ has high polarizability because it has many electrons that are distant from the nucleus.

$\implies$ $\ce{I-}$ is good nucleophile.

$\implies$ $\ce{I-}$ is good leaving group and also good nucleophile because of big size. How can both be correct and valid; 'stable' and 'can be highly polarized'?
Am I thinking wrong?

Well, $\ce{F2}$ is perfectly stable. And highly reactive.

Nucleophilicity
Because smaller anions are very well solvated in polar protic solvents, such as water and methanol, iodide is a very good nucleophile:
$$
\textbf{Polar Protic Solvents}\\
\begin{align}
\text{Solvation: }& \ce{I- < Cl- < Br- < F-}\\
\text{Nucleophilicity: }& \ce{I- > Cl- > Br- > F-}
\end{align}
$$
For polar aprotic solvents such as DMSO, DMF, and acetonitrile that do not solvate anions nearly as well and the cations, still, to an appreciable extent, the order of nucleophilicity reverses:
$$
\textbf{Polar Aprotic Solvents}\\
\begin{align}
\text{Solvation: }& \ce{I- > Cl- > Br- > F-}\\
\text{Nucleophilicity: }& \ce{I- < Cl- < Br- < F-}
\end{align}
$$
Leaving Group Tendency
Leaving group tendency is largely decided on the basicity of the group. Less basic groups make better leaving groups. Thus, order of leaving group tendency follows the same trend in any solvent.
$$
\begin{align}
\text{Basicity: }& \ce{I- < Cl- < Br- < F-}\\
\text{Leaving Group Tendency: }& \ce{I- > Cl- > Br- > F-}
\end{align}
$$
References

Solomons T. W. G., Fryhle C. B. (2011). Organic Chemistry, 10th ed. John Wiley & Sons.
LibreTexts Chemistry: Nucleophilicity and Solvent Effects

Iodide can be both a good leaving group and also a good nucleophile. By heterolytic cleavage, the iodide ion would leave off the molecule with a net –1 charge.
Take the example of $\ce{H-I}$ which may explain both these effects. Since iodine has a larger size, the longer bond length of $\ce{H-I}$ causes an easy dissociation of it and iodide can easily leave. After leaving, it acquires an extra negative charge. Iodide has larger size to carry this extra charge that provides some stability. Moreover, the extra electron resides in 5p orbital which is outermost shell for iodine atom. This causes a significantly lesser $\ce Z_{eff}$ on it and readily available for an electrophile to attack. You may say that the combination of larger size and low $\ce Z_{eff}$ (consequently lesser negative $\ce \Delta_{eg} =-296$ kJ/mol for iodine) in iodide makes it a better nucleophile.

'stable' and 'can be highly polarized'?

It can be stable and highly polarised if the conditions are just right. Aqueous medium is way to stabilize the polar molecules such as hydronium ion that goes too far to stabilize by making very large associations.

Re. your first sentence, that's not correct — iodide is a good leaving group. Where did you get this from?

I was referring to molecules where iodine has either no charge or partial charge such as HI. That is also my example. Maybe I would edit that to include iodide too.

Yes, but it leaves with a negative charge, so iodide is the leaving group. This is standard usage in every textbook and paper I've seen.


text:
Mechanism of S2Cl2 and NaHSO3
I learned that NaHSO3 reacts with S2Cl2 to give Na2S4O6 as product. After looking up the structure of Na2S4O6, it confuses me that four sulfur atoms connect together. I couldn't figure out the exact mechanism for this reaction.

Consider $\ce{SO3^2- + S -> S2O3^2-}$ (sulfur scavenging in B/W photography fixers) and  $\ce{I2 + 2 S2O3^2-  -> 2 I- + S4O6^2-}$ (nature of iodometric titrations)

Looks straightforward, just consider that S atom in bisulfate is nucleophilic.

Nothing odd happens. The $S-S$ covalent bond is relatively strong and sulfur atoms don't mind at all linking up in chains of four or longer. In that respect they're quite differengt from oxygen atoms, which have smaller, highly localized lone pairs that repel each other strongly.
Mechanistically, the reaction is a simple nucleophilic substitution, where both the nucleophile (in $HSO_3^-$) and the electrophilic center (in $S_2Cl_2$) are sulfur atoms. The bisulfite ion is electron rich due to its charge, and attacks one of the electron poor (due to the electron withdrawing chlorine atom) sulfurs in $S_2Cl_2$, expelling the chlorine as $Cl^-$ and forming a $S-S$ bond in addition to the one already present in $S_2Cl_2$.
$$\ce{S_2Cl_2 + HSO_3^- -> HS_3O_3Cl + Cl^-}$$
This reaction will proceed smoothly as chloride is a weak base and good leaving group, and sulfur has diffuse electron pairs and is a good nucleophile, especially when charged. The resulting intermediate is much more acidic and easily gives off a proton; if there is no other base present sulfurous acid will form, which will partially decompose into $SO_2$ and $H_2O$:
$$\ce{HS_3O_3Cl + HSO_3^- <=>> S_3O_3Cl^- + H_2SO_3}$$
$$\ce{H_2SO_3 <=> H_2O + SO_2}$$
Now repeat the exact same events on the second sulfur atom in $S_2Cl_2$ with a second $HSO_3^-$, forming the fourth $S-S$ bond, and you get tetrathionate:
$$\ce{S_3O_3Cl^- + HSO_3^- -> S4O6^2- + H+ +Cl-}$$
$$\ce{H+ + HSO_3^- <=>>H_2O + SO2}$$
Combining both steps we get an overall stochiometry:
$$\ce{S2Cl2 + 4 HSO3- ->S4O6^2- + 2 Cl- + 2SO2 + 2H2O}$$

I just want to make sure but in Greenwood Inorganic Chemistry book, the reaction didn't include SO2 as byproduct

The simplest explanation is that the central sulfur atoms in $\ce{S2Cl2}$ are acting as substrates for a nucleophilic substitution reaction. Additional bisulfite ions, or an added base, removes the protic hydrogens. A similar mechanism with alkyl halides as the substrate is one way to introduce sulfonate functions into organic compounds.


text:
Chemical Equilibrium and Sponteneity
Consider the reaction $$\ce{N2O4 <=> 2 NO2}.$$ The forward direction of this reaction is non-spontaneous and under standard conditions $\Delta G^\circ = \pu{4.76 kJ/mol}$. Suppose we begin with $\pu{1.00 bar}$ of $\ce{N2O4}$ and no products. Then, we would expect the reaction to proceed to equilibrium in which some small concentration of products is formed. I am struggling to understand why it is not a contradiction that the reaction is non-spontaneous, but still proceeds in the forward direction (which indicates that the forward direction is spontaneous).

Because the criterion of a negative value of standard Gibbs free energy change to signify a spontaneous reaction is only a rule of thumb.

@Johnny Smith, the spontaneity criterium is $\Delta G > 0$. Just recall that  $ \Delta G = \Delta G^0 + \ln Q$$

@PAEP do you mean $\Delta G <0$? Also, doesn't assuming that the reactant begins with a partial pressure of 1 bar mean that $\Delta G = \Delta G^\circ$ or am I missing something. I appolgize in advance if this is a very stupid question.

"standard conditions" mean that there is 1 bar of products as well as reactants.  $\Delta G^\circ >0$ simply means that the net reaction is from products to reactants when all start at 1 bar.

Yes, I meant $\Delta G^0 <0$. As @Andrew pointed, your initial conditions do not correspond to standard conditions.

Chem+Math Expression formatting reference:
MathJax Basics / Chem+Math expressions/formulas/equations / Upright vs italic / Math SE Mathjax tutorial
// MathJax is preferred not to be used in CH SE Q titles.

@PAEP You likely meant $\Delta G^\circ < 0$ because $\Delta G^0 \equiv 1.$

See this answer also   https://chemistry.stackexchange.com/questions/173788/can-free-energy-of-a-spontaneous-reaction-be-positive/173805#173805

A reaction will proceed to achieve an equilibrium state, this only happens when some amount of both reactants and products are present.

Consider a general reaction:
$$
\sum_ir_i\ce{R}_i \ce{->} \sum_jp_j\ce{P}_j
$$
The equilibrium constant is given by:
$$
K = \dfrac{\prod_{j}[\ce{P}_j]_\text{eq}^{p_j}}{\prod_i[{\ce{R}_{i}}]_\text{eq}^{r_i}}
$$
If no product is formed then:
$$
\prod_{j}[\ce{P}_j]_\text{eq}^{p_j} = 0 \implies K = 0
$$
Calculating the free energy for such a reaction:
$$
\Delta G^\circ = -\mathrm{R}T\ln{K}
\implies \lim_{K \rightarrow 0} \Delta G^\circ = \infty
$$
Thus, for a process in which no products are formed, the standard free energy is infinity, which isn't logical, and some products are always formed, regardless of whether the reaction is spontaneous, $\Delta G^\circ <0$, or non-spontaneous, $\Delta G^\circ >0$.

Is this the thinking that caused the alchemists to fail in turning Pb into Au? There never was any product! Now the reason is clear!  DeltaG = +infinity