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Relationship between the symmetry number of a molecule as used in rotational spectroscopy and point group
While doing a problem set I noticed that the symmetry number of a molecule turns out (usually) to be half the number of symmetry elements that the point group the molecule belongs to.
When I say symmetry number I refer to the symmetry number used in rotational spectroscopy — i.e. the number of indistinguishable orientations of the molecule. This accounts for suitable symmetrisation of the rotational wavefunction and the nuclear wavefunction of the molecule.
For example benzene belongs to point group $D_\mathrm{6h}$, which has 24 symmetry elements. It also has symmetry number 12, i.e. 1/2 the number of symmetry elements in it's point group.
Another few examples:

Water. Point group $C_\mathrm{2v}$, with 4 symmetry elements. Has 2 as symmetry number.
Ammonia. Point group $C_\mathrm{3v}$ with 6 symmetry elements. 3 as symmetry number.
Ethene. Point group $D_\mathrm{2h}$ with 8 symmetry elements. 4 as symmetry number.

This seems to be a general trend. It sort of makes sense intuitively that the 2 should have a relationship, but I can't formalise the reason for this.
Obviously for diatomics/linear molecules this doesn't apply. This is because rotational motion around the axis of symmetry is not physical — moment of inertia around this axis is zero, as all mass along axis. Thus there are no rotational levels associated with this axis. But ignoring diatomics/linear molecules, where we have neglected/ignored an axis of rotation, the above relationship seems to hold.
Why is this the case? I'd appreciate both a mathematical derivation and a more intuitive argument.

Table II on this NIST webpage gives the number of symmetry elements for most of the common point groups and a brief explanation of how they can be obtained. It seems to align with the trend you are describing. @SwedishArchitect

This is not in general true
Consider molecules a point group not containing inversion symmetry, e.g. $C_2$ hydrogen peroxide

The $C_2$ group has only two elements, $E$ and $C_2$, and the $C_2$ rotation operation maps between two identical arrangements of atoms.  Both the symmetry number and order of the group are 2.
Matrix representations
All rotations in 3D space can be represented by a orthogonal 3x3 matrix with determinant 1, and the composition of rotations about the origin can be represented by matrix multiplication of the transformation.
More formally: the group SO$(3)$ (3D rotations) is isomorphic to the group of orthogonal real 3x3 matrices with determinant 1 under multiplication.
The number of symmetry elements in a point group formed from rotations (and which hence preserve chirality) will be the same as the number which can be represented by such matrices and the same as the number of equivalent orientations of the molecule (the symmetry number) as they are all representation of the same thing - 3D rotations.
Inversions
The group of all transformations that keep the origin fixed, including reflections as well as rotations is O$(3)$ - or the product of SO$(3)$ with the set $\lbrace I,-I\rbrace$ or the inversion operation.  It maps rotations onto improper rotations (which include reflections - improper rotation by 0 degrees) and hence all symmetries in 3D space.
All point groups with inversion symmetry, such as the 3 listed in the question, contains this $-I$ element.  $-I$ will multiple all the elements of the rotation matrix by $-1$ and create an orthogonal 3x3 matrix with determinant -1.  By composition with each of the rotations elements of the point group it creates another element - an inversion, improper rotation or reflection - doubling the size of the group without increasing the symmetry number.
Conclusion: chirality matters
This rule is true for non-chiral groups.  You can invert the molecule to get another symmetric copy for every rotationally equivalent copy, so the order of the group is twice the symmetry number.
For chiral groups that will create the enantiomer which isn't symmetrically equivalent so the order of the group and symmetry number are the same.

Just to clarify: the molecules in the question do not have inversion symmetry, but they have an improper symmetry operation (a mirror plane, inversion center, or improper rotation). The more direct way of saying this is the symmetry number corresponds to the order of the largest (purely) rotational subgroup of the point group. For $C_{2v}$, that's $C_2$. For $C_{3v}$, that is $C_3$. For $D_{nh}$, it's $D_n$.

I would like to add an intuitive explanation to the trend you have found that is complementary to the previous answer.
It seems to me that the trend you have notice is related to the number of symmetry operations in the point group that lead to different atomic configurations. For example, if you consider $\ce{H2O2}$ molecule, that belongs to $C_2$ point group, you have that $\hat{C}_2^2$ leads to the same configuration that $\hat{E}$, so classicaly you can only distinguish two molecular arragements. The symmetry number and the order of the group are 2.
If we apply this reasoning to $\ce{H2O}$ ($C_{2v}$ point group), we can group the symmetry operations taking into account the final atomic arrangement:
$$
\begin{array}{l}
  \hat{E} \equiv \hat{C}_2^2 \equiv \hat{\sigma}(xz)^2 \equiv \hat{\sigma}(yz)^2 \equiv \hat{\sigma(yz)}^1  \\
  \hat{C}_2^1 \equiv \hat{\sigma}(xz)^1
 \end{array}
$$
Thus, although the point group has 4 classes of operations, we can only obtain 2 different molecular configurations.


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How tartaric acid gives Tollen's test?
I wanted to know how Tollen's test is given by tartaric acid. Google searches say that it belongs to some "Hydroxyl" group. I know that it oxidises aliphatic and aromatic aldehydes along with α-hydroxy ketones and some other compounds such as alkynes and it's mechanism.
Can someone help me out in the mechanism of oxidation or provide some more detail on this reaction?

Related: https://chemistry.stackexchange.com/questions/97592/why-does-3-hydroxy-butan-2-one-give-positive-tollens-test

I know how α - hydroxy ketones are oxidised. But, this is α - hydroxy acid. The answer doesn't explain.

Maxted[1] reported in 1926 that the reaction of tartaric acid with silver produces oxalic acid, formic acid, CO2 and water as products and that the stoichiometry is 6 moles of silver reduced per tartartic acid. He proposed the following possible net reactions:
$\ce{C4H6O6 + 6 Ag+ + 2H2O -> 2(C2H2O4) + 6 Ag + 6 H+}$
$\ce{C4H6O6 + 6 Ag+ + 2H2O -> C2H2O4 + CO2 + HCO2H + 6 Ag + 6 H+}$
$\ce{C4H6O6 + 6 Ag+ + 2H2O -> 2(HCO2H) + 2CO2 + 6 Ag + 6 H+}$
For the mechanism, he concluded that initial decarboxylation to 2-hydroxy-3-oxo-propionate or splitting into two glyoxylic acid molecules were both possibilities. The glyoxylate could be oxidized to oxalic acid or oxidatively decarboxylated to formic acid, while the 2-hydroxy-3-oxo-propionate could split into glyoxylic acid and formic acid.
I'm sure someone must have done more recent mechanistic work on this to pin it down further, but that gives a pretty good idea of the possibilities.
[1] Maxted, D.R. (1926) CCLXXXVIII-The Oxidation of Tartartic Acid by Solutions of Silver Salts; J. Chem Soc. 129:2178.

I outright do not "know" the answer, if there is one. It could just be put down to the old "for every rule there's an exception, except in chemistry where there seems to be hundreds".
I will hazard a guess though.
In solution, aldehydes and alpha-hydroxy ketones can tautomerise through an enediol state. Tartaric acid can also tautomerise into something similar- a "hydroxy-enediol", if you will. Perhaps this is enough outright to allow the reduction reaction to occur, or perhaps enough of the hydroxy-enediol can then tautomerise into aldehyde (however unfavourable this may be) to react with Tollen's and then be driven by equilibrium.
This is speculation, perhaps someone else knows of any literature that confirms the mechanism of Tollen's and tartaric acid.

Well, you surely are not tautomerising the carbonyl of acid group?

Yes I am. The following paper describes the potential for the tartaric acid tautomer described. The tautomer most probably is very unfavourable, especially so when taken out of the context of nano-particle treatments, however if it could be formed a little, this may simply mean a slower reaction to Tollen's driven by equilibrium. https://www.researchgate.net/publication/263508230_Observing_the_in_situ_chiral_modification_of_Ni_nanoparticles_using_scanning_transmission_X-ray_microspectroscopy

Hmm, that is interesting.

I think the relevant reference is Maxted, journal of the chemical society 1926 p 2178, but I don’t have access to the full article right now. Decarboxylation is definitely involved.

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How do I make a solution of spinach to titrate with pottasium permanganate?
My teacher said something about straining it but I didnt quite get it

Also see very similar question by OP -- Redox titration of oxalate from spinach

The best approach is to ask the teacher again after doing some "homework" (i.e. reading and searching). Extracting and then titrating oxalic acid/oxalate is not that easy from spinach. You will need a lot of spinach to extract a titratable amount. Oxalic acid can be extracted by boiling spinach in 0.1 M HCl. However, tons of other redox active substances will also appear in the extract. Direct titration with KMnO4 will give an extra-ordinary large percentage, which is of course erroneous. After extraction, you need to precipitate oxalic acid via some cation. After isolation of the precipitate you would need to bring back oxalic acid into solution, one can do a redox titration. In real-world nobody would use this approach but as an academic project, it could be a good exercise. Again ask your teacher if she/he is willing to go to this extent.

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Kinetic model of proton reduction
One of my colleagues is working on a homogeneous catalyst ($\ce{X}$) for proton reduction. The simplified proposed mechanism for it is the following.
$$\ce{X + 2 e^- ->[$k_1$] X^{2-}}    \tag{1}$$
$$\ce{X^{2-} + H+ <=>[$k_2$,\,slow][$k_{-2}$] XH-}    \tag{2}$$
$$\ce{XH- + H+ <=>[$k_3$][$k_{-3}$] X + H2}   \tag{3}$$
However, this is not a true catalyst and it degrades over time:
$$\ce{X ->[$k_\text{inact}$] X_\text{inact}}$$
We monitor $\ce{H2}$ at different time intervals and we plot the turnover number (TON), defined as the total number of moles of $\ce{H2}$ produced per mole of initial catalyst (so if we produce 20 mmol of $\ce{H2}$ in an experiment where we started with 2 mmol of catalyst, the TON at that time is 10).
Is there a way to derive an equation for the TON vs time plot that only depends on total $\ce{H2}$ and time? (We get something that looks hyperbolic, like Michaelis Menten plots, but we don't know if it's hyperbolic, exponential, or something else).

My attempt:
Assuming steady-state of the intermediate, $\ce{XH-}$ (and I'm not sure we should!):
$k_2[\ce{X^{2-}}][\ce{H+}] + k_{-3}[\ce{X}][\ce{H2}]= k_{-2}[\ce{XH-}][\ce{H+}] + k_{3}[\ce{XH-}][\ce{H+}] $
$\displaystyle [\ce{XH-}] = \frac{k_2[\ce{X^{2-}}][\ce{H+}] + k_{-3}[\ce{X}][\ce{H2}]}{(k_{-2}+k_3)[\ce{H+}]}$
The rate law is given by:
$\displaystyle \frac{d[\ce{H2}]}{dt} = k_3[\ce{XH-}][\ce{H+}] - k_{-3}[\ce{X}][\ce{H2}]$
So:
$\displaystyle \frac{d[\ce{H2}]}{dt} = k_3\left(\frac{k_2[\ce{X^{2-}}][\ce{H+}] + k_{-3}[\ce{X}][\ce{H2}]}{(k_{-2}+k_3)}\right) - k_{-3}[\ce{X}][\ce{H2}]$
From here I am not sure how to proceed. I think it is safe to assume that $[\ce{X}]$ is constant in the catalytic cycle so we can say that $[\ce{X}] = [\ce{X}]_0 e^{-k_\text{inact}t}$. However, I'm not sure how to treat $[\ce{X^{2-}}]$, and I'm not exactly sure how to solve the integral for $[\ce{H2}]$ from there. Any suggestions are appreciated.

Your equation ($ii$) is neither mass nor charge balanced.  Would you please correct that first.

Oops, I had it all in sequence and later decided to show it in steps for clarity and missed that. It is edited now.

@ralk I would recommend maybe change your "C" for catalyst to something else. Some of the steps could be mistaken for unusual carbon compounds.

Well maybe it is carbon :P

About TON vs time plot, did you try to graph on excel? You may able to check whether it is hyperbolic or exponential easily.

We've actually done the fit for a generic hyperbolic or exponential equation and it fits pretty well for both. That's why we are trying to come up with a model that supports one or the other. (The other problem being we have no idea what the $k$ values are!).

If you're not sure if a species should be a fleeting intermediate for the purposes of steady state analysis, you should draw a reaction coordinate diagram for a single cycle.

Do you have any mechanistic prediction for how the catalyst inactivates? Specifically, which species is involved? For example, it might be more accurate to write your inactivation step as $k_{inact}[X^{2-}]$. That difference could be important in getting a good model for TON.  More generally, I would take the approach of writing the rate laws as diff. eqs and fitting by simulation rather than making SS assumptions (that might not be valid) and trying to solve explicitly. Many computer programs are available to do this.

You definitely make several good points, I think you might be correct by saying it's more likely to be $k_{inact}[\ce{X^{2-}}]$. I guess the best choice is to do as you say since it will probably require many other assumptions to get an explicit solution.


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Why the use of MD simulations is less popular for the study of carbohydrates?
In classical all-atom MD simulations, the force fields for proteins, nucleic acids and lipids seem to be more well developed than carbohydrates. It seems to me that the use of MD simulations is not so popular for the study of carbohydrates. Is there any reason for this?

No reason that I know of. I would guess that there is less funding in that area... unfortunately money drives research interest.

Maybe because there are much more actively studied questions and structural variability with proteins (enzymes), nucleic acids and lipids than carbohydrates? I mean soaking starch in water is interesting and stuff, but carbohydrate macromolecules are generally static structural elements and energy careers, while proteins, nucleic acids, even lipid membranes have much more active, dynamic roles.

Might just be that the force fields for proteins, nucleic acids and lipids are more well developed than carbohydrates - force field generation is not straightforward and out side the experience (and interest) of your average (bio-)chemist.

I am not sure this is true (that it is less popular), and also wonder if either of the following apply: (1) one size fits all (finding a suitable set of ff  parameters for a general carbohydrate is not hard) or (2) just the opposite (describing any individual carbohydrate requires costly parameterizations of dihedrals, slowing down implementation)

Second, what Greg says, but there are glycoproteins though. Maybe it's just that they're aren't all that different. While in lipids that's mostly vdW and hydrophobic effect and DNA/RNA needs charge transfer, carbohydrates use h-bonds like proteins.


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Why can you blow bubbles in milk but not hot cocoa, which is milk based?
My daughter is asking this question and I’m assuming it’s something related to the cocoa added to milk but don’t know what change is happening molecularly and why it would affect the bubbles. Thanks!
Let me know if there is a better place to ask this question!

Milk contains lecitihin, which acts like soap to keep fat globules in milk suspended (although it does slowly separate into cream and skimmed milk, which you could demonstrate with non-homogenized milk).
Lecithin and soap work by having long molecules, one end of which is attracted to water (hydrophilic) and the other is attracted to oil or air more than to water (hydrophobic). This helps the water form a double layer, with air inside and outside and the lecithin holding things together. C. V. Boys Soap-Bubble and the Forces Which Mould Them is worthwhile reading and lists numerous experiments with detergent and water that you might try with your daughter's assistance.

Ah... but you ask why cocoa interferes with bubble blowing! Here are two tentative explanations:

The cocoa particles don't dissolve, so they "poke holes" in the soap/water/soap membrane, bursting the bubbles.
The cocoa adheres more tightly to the hydrophobic end of the milk lecithin than does air, preventing formation of the bubble layers.

You and she could experiment to see if either are correct. Try putting something other than cocoa in the milk (you might substitute dish-washing detergent in water for milk, because it makes longer lasting, cleaner and cheaper bubble stuff). Finely shredded cellulose is hydrophilic, and powdered candle wax is hydrophobic, for example. You could try finer powders, too. If you do experiment, please post your results here - I would be interested in your findings!

Wonderful...this is so great and clear. Will look into that book you shared as well! Thank you!

Thanks, @ScottErickson, and please add any experiments your daughter and you try to the answer. (In fact, she might have fun making a video or book for others to nuse!)


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Dependence of primary alcohol oxidation on presence of water and "strong" or "weak" oxidants
It is known that the oxidation of primary alcohols can result in aldehydes or carboxylic acids. The product obtained from the reaction is described as dependent on reagents and conditions;[1]

"Primary alcohols are oxidized to either aldehydes or carboxylic acids, depending on the reagents chosen and the conditions used."

A distinction is typically made between "weak" (eg. PCC) and "strong" (eg. chromic acid) oxidants[2], the former being capable of giving an aldehyde product while the latter carries out the oxidation to carboxylic acids. However, the accepted mechanism for the reaction shows that the oxidation always results on an aldehyde, which can then be further oxidized to a carboxylic acid through an hemiacetalic aldehyde hydrate forming in the presence of water.[1][3]

"Most other commonly used oxidizing agents [...] oxidize primary alcohols directly to carboxylic acids. An aldehyde is involved as an intermediate in this reaction but can’t usually be isolated because it is further oxidized too rapidly."

This indicates that it is the presence of water in the reaction medium which determines the product of the oxidation. Thus, "weak" oxidants could still yield a carboxylic acid in aqueous solutions (in fact, they are always noted to be used in anhydrous conditions) and "strong" oxidants will result on an aldehyde when used with organic solvants. If so, I fail to understand the classification of "weak" and "strong".
Furthermore, some sources, like Wikipedia, seem to imply the existence of a direct oxidation from primary alcohol to carboxylic acid, but I've been unable to find such a mechanism in bibliography, always ending up with the hydrate intermediary method.
To summarize:

Is the distinction between "strong" and "weak" oxidants meaningful or does the presence of water always determine the reaction product?
Does a method for the direct oxidation of primary alcohols to carboxylic acids without an aldehyde hydrate exist?

McMurry, John. Organic Chemistry. Brooks Cole, 2011. 8th edition.
Alcohol Oxidation: “Strong” and “Weak” Oxidants
Reagent Friday: Chromic Acid, H2CrO4

"Weak" and "strong" in the examples are used because the weak oxidant stops at an aldehyde while the strong oxidant goes on to the carboxylic acid. Water is the critical factor. "Weak" PCC in the presence of water can give carboxylic acids. PS: PCC oxidation of primary alcohols under anhydrous conditions can produce esters via the hemiacetal of the aldehyde formed from unoxidized alcohol.

I'm afraid I still don't understand the difference. If "weak" oxidants can behave as "strong" oxidants, then why use those terms at all? Is it that a single equivalent of "strong" oxidant is enough to obtain a carboxylic acid, while a "weak" one requires two equivalents?

Forget about "weak" and "strong". The oxidation of a primary alcohol with Cr(VI) reagents goes through the aldehyde, then hydration, then it goes on to the carboxylic acid. Without water, the rxn stops at the aldehyde. Look up Collins oxidation and PCC oxidation.

After looking up more information and taking the comments of user55119 into account, I've decided to post an answer myself, hoping to clearly address the main question points.

The determinant factor of the products obtained from an oxidation reaction of primary alcohols is the presence of water in the reaction medium, as it allows the formation of an aldehyde hydrate which can be further oxidized. Thus, the distinction between "weak" and "strong" oxidants is not necessarily informative (and could even be misleading) and emphasis should be placed in the reaction mechanism alongside the reaction medium.
The oxidation of primary alcohols with Cr(VI) reagents takes place through the mechanism depicted in the question or slight variations of it (depending on the reagent chosen). Carboxylic acids are obtained only through an aldehyde intermediate.

Further reading:
McMurry, John. Organic Chemistry. 8th edition.
MasterOrganicChemistry


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Do lithium batteries 'explode'/'burn' because of the electric energy stored within?
I've been watching some videos where people show lithium ion batteries burning/exploding. What usually happens is someone punctures the battery, and a jet of gas comes out the puncture hole, followed by the gas catching fire and the entire thing going up in flames.
I often hear that the energy for this reaction comes from the chemical energy stored in the battery. Thing is; i'm pretty sure that there wouldn't be enough chemical energy in the battery to cause such an intense reaction. I'm thinking that it's probably the lithium inside the battery reacting with the air in an exothermic manner.
So what exactly causes the reaction that we see? Does it have anything to do with the electric energy stored in the battery or is it just a lithium/oxygen reaction?
Edit: It was suggested I actually do the math. So here goes; typical energy density of a lithium ion battery would be around 200 watt-hours per kg of battery mass. Therefore a 1kg battery would have about 720 kJ of chemical energy. Proceeding from here we would need to know at the very least a.) the mass of the substance we see being 'exploded', b.) the specific heat capacity of said substance.
For example; if I assume that from a 1kg battery we'd have 300 grams of material being heated up and expelled from the battery, and that said material has a specific heat capacity of 2 kJ/(kg K) , and that said material starts at 293 kelvin, then I'd know that it would take .6 kJ of energy to heat said material up by 1 degree Kelvin. 720 kJ would be enough to heat it up to 1493 Kelvin, assuming 100% of the energy went into heating up just that 300 grams of material and not the other 700 grams. With a specific heat capacity of 3kj/(kg*kelvin) we'd get a temperature of 1013 Kelvin.
I suppose it'd really come down to what the substance we see shooting out of the battery is, and how much is shooting out of the battery, and what it's heat capacity is (and whether it's approximately constant across the temperature range of interest).

"Electrochemical" shows the correspondence between chemical and electrical energy. The energy in a cell can be released slowly over time, or rapidly when short-circuited internally or externally, or the ingredients meet under the wrong conditions. No additional chemicals, such as oxygen from the air, are needed.
You state, "there wouldn't be enough electrochemical energy in the battery to cause such an intense reaction." Calculate how much energy is stored, instead of guessing:
The common Li-ion 18650 cell stores about 2.5 ampere-hours at 4 volts (rounded). That's 10 watt-hours, and multiply by 3,600 sec/hr to get 36,000 watt-seconds or joules (J). That is a lot of energy compared to some projectile weapons. For example, the largest cartridge shown in this reference, for the .50 caliber Browning Machine Gun, packs 18,000 J, half that of the inexpensive 18650 cell.
So the question is, does the energy release come from thermal effects due to ohmic heating, or from the direct combination of chemicals. Most likely, once the cell is physically damaged, whether by overheating or mechanical stress, direct chemical combination takes place. Though it can be exacerbated by water or atmospheric oxygen, there is more than enough energy in even small Li-ion cells to cause spectacular fires or explosions.

I can't comment yet, so to add the answer provided by @DrMoishe Pippik.
The initial internal heating of the battery is caused by the reaction of the chemical components within the battery. It is a similar reaction as to what occurs during discharging, but with a few noteworthy aspects:
1) The rate of reaction is limited by the leakage within the battery and not by the current draw of a circuit.
2) There is no external circuit to do any work, so the energy is instead converted to heat within the battery.
As the temperature rises, the barrier separating the components can degrade further, resulting in a run-away reaction. The heat produced during this reaction readily exceeds a few hundred degrees Celsius which is often enough to deteriorate the battery package and ignite the flammable electrolyte. At this point, the energy is derived from reaction with air and is a typical combustion reaction. So, the total energy output in these spectacular explosions is the combined electrochemical potential energy and energy of combustion.


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In a galvanic cell why does using a more reactive metal increase the voltage?
As the title says, why does using magnesium produce a higher voltage compared to using zinc when reacting with iron? Im having trouble figuring out why the voltage changes.
magnesium+copper > zinc+copper in terms of voltage, why?
Any help is appreciated.

What is your understanding so far, did you look up electrode potential tables? It is a good question.

@M.Farooq My current understanding is that the higher the negative or positive value on the electrode table is, the higher the voltage output. I how ever do not understand why this is the case.

It is a good question because textbooks do not discuss it. Scientific explanations are of two types, how and why. The "why" part of the explanation is very difficult to answer. Often we don't know the answer at a fundamental level. A simple example is that of what is universe is made? What is the origin of mass or charge? In the same way the electrode potential can also be understood at various levels. The most advanced discussion of electrode potentials can be done in terms of Fermi levels, but this is not my area of expertise. The macroscopic level discussion is that one can consider the electrode potential as a (thermodynamic) tendency of a substance to gain electrons. If you look up modern electrode reduction tables, all entries are listed as reduction process (all electrochemists agreed to do that after a long academic fight to write it this way). Hydrogen has been given an arbitrary value of 0.0000 V for the sake of comparison. You always need a reference in electrochemistry so electrochemists decided to use H2 electrode as a reference.
a) If you look at the table below, look at lithium entry: It shows that Li+ has very low (or negative electrode potential) tendency to gain electrons. Read it the other way (right to left), that Li metal has a very high tendency to become a lithium ion.
b) Look at the silver entry: It shows that Ag+ has a high tendency to gain electrons (positive electrode potential). Read it the other way (right to left) that Ag has a very low tendency to become Ag+.
If I make a galvanic cell consisting of two beakers connected by a salt bridge, having a lithium electrode in 1 M Li+ solution, and a silver electrode dipped in 1 M Ag+ ions. Connect the leads to an voltmeter.
Now Ag+ has a tendency to accept electrons and become Ag and lithium metal has the tendency to lose electrons and become Li+, what do you think happen? Electrons would feel a "pressure" or better "potential difference" causing them flow from higher potential to a lower potential, which is given by
Ecell=  Ecathode - Eanode= 0.7996 - (-3.040) = + 3.8396 V, this should be your voltmeter readings. Why did we choose Ag+ /Ag as a cathode? Because Ag+ ions have a higher tendency to reduce than lithium ions.
In the same way you can choose Li and Be cell. What would be the cell voltage?
For example, electrode potential tables look like:
Li+(aq) + e−  ⇌  Li(s)      –3.040
Be2+(aq) + 2e−  ⇌  Be(s)    –1.99
Al3+(aq) + 3e−  ⇌  Al(s)    –1.676
2H+(aq) + 2e−  ⇌  H2(g)  0.00
Ag+(aq) + e−  ⇌  Ag(s)       0.7996
Ce4+(aq)+e−⇌Ce3+(aq)        +1.44
PbO2(s) + HSO4−(aq) + 3H+(aq) + 2e− ⇌ PbSO4(s) + 2H2O(l)    +1.690
H2O2(aq) + 2H+(aq) + 2e− ⇌ 2H2O(l) +1.763
F2(g) + 2e−⇌ 2F−(aq)         +2.87


text:
What is the Gibbs free energy difference between conformational states
In statistical mechanics, for a system of $N$ particles $x_1, \ldots, x_N$ in three dimensions, the Gibbs free energy is defined in terms of the Hamiltonian $H$ as
$$ G = -k_\mathrm B T \log \int_{\mathbb{R}^{3N}} \mathrm e^{-H(x_1, \dots, x_N)}\,\mathrm dx_1 \dots \mathrm dx_N.$$
People talk about the "free energy difference between conformations". For instance, for some region $A \subset \mathbb{R}^N$, the particles could be in a certain conformation, and for some other region $B \subset \mathbb{R}^N$, the particles could be in another conformation.
Question

In this paper, equation (3), what is the (implied) definition of $F_A$ or $F_B$ in terms of equation (2)?

Equivalent Question

Is the free energy difference between the two conformations defined as
  $$G_B - G_A = -k_\mathrm B T\log \frac{\int_B \mathrm e^{-H(x)}\,\mathrm dx}{\int_A \mathrm e^{-H(x)}\,\mathrm dx}?$$

Auxiliary details
The above definition made sense to me, until I was told that "it does not make sense to talk about the free energy of a conformation". An alternative I see is to think of two Hamiltonians $H_A$ and $H_B$, and instead define the free energy difference as
$$G_B - G_A = -k_\mathrm B T \log \frac{\int_{\mathbb{R}^{3N}} \mathrm e^{-H_B(x)}\,\mathrm  dx}{\int_{\mathbb{R}^{3N}}\mathrm e^{-H_A(x)}\,\mathrm dx}. $$
Are either/both/neither of these definitions correct?
I apologize if the question is exceedingly obvious. I am unable to find a straightforward answer by Googling. Any references are welcome.

Your question implies that you define the Gibbs free energy as $G=-RT\ln(Z)$ where $Z$ is the partition function, in which case there will be a difference in $G$ only if there is a difference in the partition function for your two configurations, i.e differences in energy levels and/or in the population in these levels. That will depend entirely on your system (and also the temperature).

I'm not sure I understand. In the original post, the boxed-in question assumes a common temperature $T$ and Hamiltonian $H$, which together determine all energy levels across $\mathbb{R}^{3N}$. In particular, $H$ is common to both configuration regions $A$ and $B$. The restatement in terms of a partition function is a mathematical tautology for me, but I understand there may be physical intuition behind it (which I do not have). The question is whether it makes sense or is common to define the "partial partition function" $Z_A = \int_A e^{-H(x)} dx$, and if this is what is implicitly done.

The partition function is normally described as $\sum_i g_i \exp(-E_i/k_BT)$ for energy levels $E_i$ with degeneracy $g_i$. In your case you use the Hamiltonian (energy operator) to give the energy levels, and presumably the sum can be replaced by an integral without error as the energy levels must be closely spaced, as in the translational $Z$; (See Sakur-Tetrode eqn for example). Unless your $H$ and hence $E_i$ are different then there can be no change in $G$. ps thanks for edit.

I am not accepting your answer because it does not address the question of free energy difference between conformational states and you have not addressed whether it makes sense to define $Z_A$. The passage from an integral to a sum is mathematically inconsequential.

I have added a reference in the OP to make the question foolproof. What I want is an expression for $F_A$ or $F_B$.

The direct counting in 3.1 in the paper gives an answer, but it boils down to knowing what the energy levels are and if they have different populations, which I pointed out in my answer. If you don't know that you are stuck no matter how fancy the model is that you use.

I'd say that either way is correct as long as you are using the same potential function for the Hamiltonian in both cases. If you aren't, then you would need to indicate distinct Hamiltonians, but then the free energy difference that is calculated is pretty meaningless. Given that, my preference would be to write it the way you originally did, with the same Hamiltonian integrated over two conformations (or ensembles of conformations).
I don't have a reference on-hand to dispute the assertion that "it does not make sense to talk about the free energy of a conformation", but I'll make an argument against it. First, I don't think it's controversial to say that it is possible to determine an equilibrium constant for interconversion of two conformations and that the interconversion can be treated thermodynamically like any other transition between two states. The distribution between the two states is determined by the difference in free energy.
We can also use the paper you cited as an example. The authors look at an ensemble of states of a given molecule and divide it into two sub-ensembles, one of "bound" molecules and one of "unbound" molecules. They then calculate an ensemble average free energy for each state. There is no reason why the same procedure could not be used to divide a large ensemble into ensembles which each represent a different conformation (even if your definition of conformation is so restrictive that each individual molecule is a unique conformation) and thereby to calculate a difference in free energy between two conformations. That one can calculate a difference in free energy between two conformations suggests quite strongly (to me at least) that there is such a thing as "the free energy of a conformation."


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Reduction using hydrazine and glycol in alkaline medium
Question:

This question was asked in my exam and I have done it properly until the third step. Unfortunately, although the last step looked like Wolff-Kishner reduction to me, I could not do it.
My approach:
At first, via ozonolysis aldehyde groups will come and then aldol reaction would take place and by heating alpha-beta unsaturated aldehyde product will be formed. But then, what if I reduce it in the last step, my answer is not matching with any of the options given.

The Ca(OH)2 step does not look like std conditions for an aldol, more like a decarboxylation.

If you look carefully, the reagent for the first step is $\ce{O3/H2O}$, which is used for Oxidative ozonolysis. If the mechanism of ozonolysis is drawn, you will realise that a molecule of Hydrogen peroxide($\ce{H2O2}$) is released by the oxidation of the alkene to aldehyde and if that $\ce{H2O2}$ is not removed from the reaction mixture, it will oxidize the aldehyde ultimately to acid. That's why during reductive ozonolysis, people use reducing agents like $\ce{Zn}$ or $\ce{Me2S}$ or $\ce{PPh3}$ which can get oxidised themselves by the $\ce{H2O2}$  produced  to $\ce{ZnO}$, $\ce{Me2S=O}$ and $\ce{Ph3P=O}$ respectively and your aldehyde is not oxidized to acid any more. But in this case no such reducing agents is used and thus the product of the first step in the diacid not di-aldehyde. After that heating with $\ce{Ca(OH)2}$ doesn't perform aldol reaction here but eliminates one molecule of $\ce{CaCO3}$ from the compound i.e. essentially decarboxylation and you are left with one carbon less i.e. $\text{Cyclopentanone}$. Then as you recognised correctly is the Wolff-Kisher reduction.
So your final product is $\text{Cyclopentane (B)}$. Here is the schematic of Oxidative ozonolysis releasing $\ce{H2O2}$ responsible for oxidizing the di-aldehyde to di-acid.


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Degree of unsaturation of naphthalene

Double Bond Equivalent/Degree of Unsaturation
25. Degree of unsaturation in

  is
(a) 6; (b) 7; (c) 8.

There are two rings and six double bonds in structure so degree of unsaturation should be 8, but it is not. Why?

Note that it is generally not acceptable to use curves in two adjacent fused rings, since such diagrams are at best ambiguous in terms of the character of the shared fusion bond between the two rings.

Could you edit your question to tell us why you think there are six double bonds (I don't see any) and how you know that the answer is not 8?

You determined the number of double bonds incorrectly, probably due to structure representation with aromatic rings, which, as Loong pointed out, is not entirely correct.
You can redraw the structure of naphthalene as such:

and make sure that there are five double bonds and two rings giving degree of unsaturation
$$\text{DU} = n_\text{rings} + n_\text{π-bonds} = 2 + 5 = 7$$
You can also compare formula of naphthalene $\ce{C10H8}$ with fully saturated alternative $\ce{C10H22}$ and proceed from there:
$$\text{DU} = (H_\mathrm{sat} - H)/2 = (22 - 8)/2 = 7$$
You again arrive at $\text{DU} = 7$ using a simplified formula:
$$\text{DU} = C - H/2 + 1 = 10 - 8/2 + 1 = 7$$


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Does enthalpy of dissolution change with temperature?
Dissolution can be either endothermic or exothermic. At higher temperature dissolution is much faster. However, supplying heat also leads to an increase in the internal energy of the system. Ultimately, this affects the enthalphy of product and reactant. I want to know whether an increase in temperature affects the enthalphy of dissolution?

I want to know whether an increase in temperature affects the enthalphy of dissolution?

It certainly can. You can treat the temperature dependence of dissolution with a formalism analogous to that applied to chemical reactions. The key parameter is the difference in heat capacity of the solid and dissolved solute, since the enthalpy of solubilization can be written as
$$\Delta H(T) = \Delta H(T_\mathrm{ref}) + \int_{T_\mathrm{ref}}^{T} \Delta C_p\,\mathrm{d}T$$
where $\Delta C_p = C_{p,\mathrm{solute}} - C_{p,\mathrm{s}}$. The $\Delta C_p$ term accounts for the different temperature sensitivities of the enthalpy (which for condensed phases approximates the energy) of solid and solute. If the heat capacity of the solute is greater than that of the solid, the enthalpy of dissolution will increase (become more positive) with an increase in $T.$
Note that if we associate a solubility constant $K_\mathrm{sp}$ with the dissolution process
$$\ce{A(solid) <=> A(solution)}$$
then we can write
$$\frac{\mathrm{d}\log K_\mathrm{sp}}{\mathrm{d}T}=\frac{\Delta H^\circ}{RT^2}$$
as suggested in a comment. Here both $K_\mathrm{sp}$ and $\Delta H^\circ$ are functions of $T.$

This is a great answer, but I think it's worth mentioning that from a practical standpoint, the change in enthalpy is typically very small unless $\Delta T$ is quite large (greater than 100 K is a good ballpark), so you can usually ignore it. This is why, for example, you can treat the standard enthalpy change as a constant in the van't Hoff equation even though it's technically only accurate for 298 K. Similarly, you can do the correction described here under the assumption that $\Delta C_P$ is constant over T or you can be even more precise and include the variation of $\Delta C_P$ with T.

@Andrew Thanks for the comment. I thought of what you mention, but decided to provide a general answer.


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Determine Which has the higher density the pure liquids or solutions?
What i know that it immensely depends on their individual density, but am not really sure since doing research they was no solid answer. i just want a conformation if what am suggesting is correct.

For pure liquids and corresponding solutions there is a distinct answer, but I don't see any argumentation from your side. Can you re-formulate your thoughts more clearly?

well i think, it depends on the density of either one of them. Pure liquids such as water has 1g/ml of density whereas solution (methanol etc..) vary with temperature and pressure that determine their density. Therefore either one could be with higher and lower density. does that make sense?,

Not really, the question is not about density dependency on conditions, rather on composition. Density is essentially mass per volume. Liquid and solution would have approx. same volume, but what about the mass?

It can be even tricky with similar density of pure liquids and important volumetric effects, at least in some ranges. Water and alcohol might be one  example.  As for determination one needs a cylinder and a scale. "On paper" one need tables or complex simulation. The general meaning of the Q is not so evident to me, if we all agree what density is.

Also d of water shows T dependency.  But as @andselisk pointed out conditions are not mentioned in your question, that, let apart what I have commented above, is self standing as it is.

okay thank you, it means the question is therefore no use unless given the conditions right? then we can  predict which has a higher or lower density.


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Metallic Versus Covalent Bonding
Given that a metal's low ionisation energy is often used to explain the metallic bond (as are other factors such as temperature and pressure and possibly a number of other factors), is there any known mathematical and/or empirically proven theory that predicts that under the given set of conditions, a transition will occur from covalent to metallic or vice versa? A metal's proximity to a stable electron shell structure (be it noble gas or complete/semi-complete d- or f- shell structure) is a generic explanation, but (without going into high level mathematical explanations) is there a firmer basis that can make the prediction of that transition, given a set of conditions? If not, is there any research into this issue?

There surely is a firm basis, but it goes all the way into pretty high-level math, and those who tried to walk around it were never heard of again.

That's a pretty all-encompassing "dismissal" (for want of a better word) of the issue. Do you have any references that would back that up? Unfortunately I don't have the mathematical knowledge or ability to understand detailed mathematical explanations, but perhaps there is a narrative somewhere which describes, in more simple terms, the efforts taken to address this. To me, it's an area which lies at the core of Chemistry as a science and I find it difficult to believe that it's been placed in the "too hard" basket.


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What kind of plastic is (really) safe to store vegetable oil?
Storing colza or olive oil in a plastic bottle might be risky since the oil can leach out phthalates and BPA (and others molecules?).
Some manufacturers sell "phthalates and BPA free bottles", but are they safe to store oil?
What kind of plastic would you use to store oil?

I would use amber glass.

@andselisk I need something light and unbreakable (that would fit in a backpack)

Glass would be better than any plastic, but if you need a lightweight container for a short period of time, I suspect that any food-grade plastic is going to be acceptable. Although I don't see the problem with glass: my colleague used to carry a couple of 1 L lass bottles with some lemonade each day to the university in his backpack without any issues.

@andselisk try to walk few thousand kilometers with a glass bottle on your backpack, you'll start to love plastic, very much. (it's robust and above all, super light)

How long do you need to store the oil?  For long-term storage, all plastics will leach, in which case you might need to sacrifice some weight and consider a stainless steel water bottle.

@theorist, "all plastics will leach". Oh that's bad new. But how long is "long-term storage"?

A stainless steel water bottle is not much heavier than plastic, especially when full

I just use the plastic bottle the olive oil came in from the store. For backpacking you buy the tiny little bottle and get a good easy-pour dripless spout good for pouring olive oil with the bottle. Remember to store in a ziploc bag since you really don't want olive oil all over everything should the lid fail.

It would be helpful if you could edit your question to tell us exactly how you will be using this. E.g. : "I want to carry 4 oz. of olive oil, for cooking purposes, on weekend backpacking trips." Or: "I'm hauling supplies, on foot, to my cabin that's a 2-day hike into the backcountry. On my next trip I'd like to leave a quart of olive oil there."  See also this report from the UC Davis Olive Oil Center: https://1.oliveoiltimes.com/library/olive-oil-packaging.pdf

If you are concerned, I would suggest that high-density poly ethylene (HDPE) or poly propylene (PP) to be better options. HDPE is a heavy duty plastic resistant to most chemicals and physical deformation- it is often used for household products including bleach, detergent etc, but is also commonly used for milk bottles and the like.

"There is evidence indicating that phthalates can leach from PET and HDPE bottles into their contents" https://hrcak.srce.hr/file/283768

But, some manufacturer sell oil (or milk, which contains lipids) in plastic bottles which could mean either that they are completely safe, or they are safe for some time, or they leach out chemicals but they (most manufacturers/consumers) don't care.

I appreciate the reference. Most oil products are triglycerides and are non-acidic, so I would question the relevance, however, I will point to the conclusion in that paper which still supports the idea that HDPE is safe: "... PET- and HDPE- bottled liquids do not represent a relevant ingestion source of phthalate... and therefore, the concentrations of phthalates observed in them are not a matter of concern from the standpoint of adverse health effects".

So I am confused as to why would you reference an unreliable paper that you agree isn't relevant- but I suppose this thread is moving away from the question, so I will leave it at that.


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Direct reaction of fluorine with benzene will give what product?
My book says fluorine will react with benzene vigorously and involve C–C bond breakage. What does that mean and what product will be formed? It also said that the reaction is of no preparative use.

OK, you surely must have heard about the reaction known as burning, right? Like C6H6+O2$\to$... Well, with fluorine it is pretty much the same, only more violent.


text:
When Tildens Reagent is added to an alkene does the product change to an oxime always?
I was reading about Tildens Reagent when I came across this..

Does this happen always?
Or is it when there is only a single hydrogen attached to the carbon?

No, this will happen when the starting material is a terminal alkene giving two hydrogens on the carbon, an aldoxime will be formed.

So the inter conversion happens only when there is 2 hydrogens r8?

And why is it unstable when hydrogen is present?

The interconversion will take place whenever there is at least one hydrogen present

But you told in the beginning that it will happen when the starting material is a terminal alkene giving 2 hydrogens?

You misunderstand. The reaction will occur will any sibstrate where there is a hydrogen on the carbon, only tertiary compounds will not give this reaction


text:
how can I predict the rate of water ions precipitation on a surface?
Consider a titanium pot that contains 1kg water with the following ions:
$[Ca^{2+}]=22000$ ppm, $[Mg^{2+}]=1500$ ppm,$[Sr^{2+}]=791$ ppm,$[Na^{1+}]=48000$ ppm,$[Cl^{1-}]=120000$ ppm,$[SO_4^{2-}]=22000$ ppm
And predict the rate of component precipitation like CaSO4, CaCl2 and ... on the inside wall of the titanium surface at the 20 C, 30C, 50C.
Calculate the rate of component precipitation in water not only on the titanium wall.
After that, calculate the final steady-state ion contents that exists in the water.

I have trouble understanding your question. Could you describe the system you are looking at? Which surface, what do you mean when you say "water ions", what do you mean by "precipitation on a surface", and what do you mean by "saturation state of that compound"? It would be great if you could edit your question to clarify it.

I have edited my question


text:
Order of reactivity of hydrogen halides
Solomon and Fryhle $11^{th}$ edition says :
Order of reactivity of HX in the acid catalyzed conversion of alcohols to alkyl halides follows the order : HI>HBr>HCl
We can account for this by saying that the acidity order is also HI>HBr>HCl
and hence the $-OH$ is easily protonated by the more acidic halide but the reaction is already acid catalyzed by a strong acid and the conversion of $-OH$ to  $-H_2O^+$ is fast so the acidity of $HX$ should make a negligible difference.
The slow step of the reaction is the formation of the carbocation and that does not seem to involve the halide so how can we explain this order ?

See this

The halide ion is involved, indirectly.  The hydrogen halide has to protonate the alcohol first, and this requires dissociation of the proton from  the halide ion.  Chloride ion is more strongly basic than iodide ion, an effect we do not see in water (leveling effect) but can see it in the reaction with the alcohol.  So we can protonate more of the alcohol at any time, with hydrogen iodide, making more opportunities to then form the carbocation, than with hydrogen chloride.
Hydrogen fluoride would be way off the scale here, in the direction of less reactivity.  It is only a weak acid in water, let alone in this reaction mechanism.  It's one manifestation of fluorination being more difficult than other halogenations.


text:
Is it possible to obtain n-butane by treating halopropane grignard reagent with methanol?

The answer is a,b.

My mechanism however doesn't give n-butane when R is propyl. Is this a case of textbook errata?

Himanshu Pandey, Advanced Problems In Organic Chemistry, 9th edition 2017

This is clearly an error. b & c will give n-butane

@Waylander Won't sec-butyl give isobutane?

No. s-bu grignard has the Mg on the second carbon of a n-butyl chain (it is made from 2-halo-butane) so replacing that with H gives n-butane.

@AvnishKabaj Not surprisingly, isobutyl (not listed here) (and also tert-butyl) would give isobutane.

@Zhe How? Some fancy mechanism?

It's just quenching the Grignard. Take the anion equivalent and add a proton. @AvnishKabaj


text:
Aufbau's principle violation at the time of complex formation

My doubt is why don't we write 4s orbital before 3d as it is lower in energy according to Aufbau's principle? (I'm getting this info from other sites that 3d comes before 4s in transition metals but I'm not pretty sure about this thing)
Further, why are 4p and 4d come directly after 4s? Should have 5s come before 4d according to Aufbau's Principle?
Please help me I'm very much confused

Because hybridisation in transition metal complexes is a load of bogus rubbish.

This might help: http://chemguide.co.uk/atoms/properties/3d4sproblem.html

If this hybridisation thing is rubbish in complexes how am I gonna understand complex formation? Can you explain this particular complex formation from some other theory?

Where did you get the image? Is that from a textbook? Even if you do use the concept of hybridization, it would be 2 of the 3d orbitals, not the 4d that are involved in bonding. If you look at the orbtial splitting diagram for an octahedral complex that only considers sigma interactions, you'll see three nonbonding orbitals that are from the metal 3d and two higher energy antibonding orbitals that are also mostly metal 3d. These two come about because two of the metal 3d orbitals contribute to bonding orbitals. The 4d's are not involved at all.


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Why was silver considered valuable in history?
Silver has been used for several millenia and (according to the german wikipedia page) was at some point even considered more valuable than gold. Today we know that it has the highest thermal conductivity and so on but these things were not known to the people back then.
What property made ancient civilizations agree that silver is something valuable compared to other metals like tin, gold or copper which were also known in the early days of humanity? Was is just rarity? Ductility? The ability to be polished? I assume that it was not just corrosion resistance as polished silver becomes dark rather quick.

It was fairly easy to separate, easy to work, and held up reasonably well. But this isn't really a chemistry question.

Well, I am asking about properties of silver that would have been known to ancient civilizations that justify this valuation.

Well, it was fairly easy to separate from the ores, was easy to work as a metal, and held up reasonable well in use. Only the first is chemistry, but the ancient civilizations had no idea of the actual chemistries of the ore and its separate.

Why people valued properties is not a chemistry question, but what those properties are clearly is. This should not be closed.

I'm voting to close this question as off-topic because it was just about rarity.


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Lattice type from image
What type of 2D lattice is depicted in the image shown below? Is it possible to determine without knowing the length of the unit vectors and the angles in between them? I'm confused because it seemingly doesn't fit with either monoclinic or orthorhombic description as $\mathrm{a \ne b}$ and $\mathrm{c\ne d}$

The picture is not a lattice, it is a periodic pattern. In a lattice, all lattice points are related by the translational symmetry given by the unit cell vectors. I can take any two lattice points and take the vector between them. Then, when I add that vector to any other lattice points, I should land on another lattice point.
This is not the case here. If I take the vector between the pair of closest points, and add the vector to the upper right point of the pair, I land on empty space.
So let's paint half of the points a different color. The points of different color are related by two-fold axis (located for instance at the center of two close pairs). The two-fold symmetry applies not only to the two closest points, but to the entire pattern, so it is called a crystallographic symmetry (as opposed to non-crystallographic symmetry).
This crystallographic symmetry imposes no restraints on the dimensions of the lattice. Other symmetry elements would impose restraints, for example in the presence of a four-fold crystallographic symmetry, the lattice would have to be tetragonal.
For the example at hand, the convention is to put the origin on one of the two-fold axes. It looks like you could have one unit cell vector along x and one along y, at approximately right angle (this is not imposed by point group symmetry, so it is "by accident" and you would have to measure carefully to see whether the angle is exactly 90 or maybe 90.0001 degrees).
To summarize, it is monoclinic with an angle that happens to be close to 90 degrees.
For specifics on the lattice types, see e.g. Wikipedia.
Thanks, @Ivan_Neretin, for the helpful comment.

Everything you say is true, except one thing. The points are equivalent.

Oh, yes I see it now. There is a two-fold in the middle of the close pairs, right?

I updated my answer based on the comment by @Ivan_Neretin

Now it is all right. True, they are equivalent by two-fold axis, but not by translations.


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Mechanism of the Gatterman Reaction
Aryl diazonium salt on treatment with copper and the corresponding hydrohalic acid, gives the aryl halide. This reaction is called the Gatterman reaction.
What is the mechanism of this reaction? Is it similar to Sandmeyer reaction?

That's not the Gatterman reaction. https://en.wikipedia.org/wiki/Gattermann_reaction

Have you Googled this? I think you have described the Sandmeyer rxn.

@Zhe Indeed that is the more famous Gattermann reaction. The Wikipedia link that you have provided also contains a link to the Gatterman reaction I am referring to right at the top of the page.

@user55119 No, the reagent used in Sandmeyer reaction is cuprous halide. March's Advanced Organic Chemistry specifically mentions Gatterman reaction as an alternative to Sandmeyer reaction.

March notwithstanding, the Sandmeyer rxn is an improvement on the Gattermann conditions that uses mineral acid (HCl, HBr) and copper salts in the second step. The Sandmeyer rxn uses only CuX. This version of the "Gattermann rxn" is lost to history. The Gattermann and Gattermann-Koch reactions are methods for formylating aromatic rings in the absence of diazonium salt.

@user55119 Thank you, I wasn't aware of this. Does this mean that the 'obsolete' Gattermann reaction doesn't have any accepted mechanism? Would you please suggest a probable mechanism anyway?

It's probably $\mathrm{S}_{N}\mathrm{Ar}$.

Listen to @Zhe!

@Zhe But, what will be the nucleophile? Also, can we rule out the possibility of an aryl radical intermediate which is formed in the Sandmeyer reaction?

@Zhe Hey, I just found this - https://onlinelibrary.wiley.com/doi/pdf/10.1002/9780470638859.conrr265 while I was googling the reaction. Unfortunately, I can't access these papers. If you can, would you check if it talks about any mechanism? Thanks

I can't access it, but a radical pathway is probably the only other reasonable thing to consider. And frankly, if the paper can substantiate it, it is radical, at least under their conditions.

@Zhe Thanks a lot for your help. Hopefully, someone who can access it will see this question :)

The reference for the Kochi paper is DOI: 10.1021/ja01568a066

The Gattermann reaction you mentioned in the question uses Copper powder and Hydrohalous acid to substitute diazonium group with a halide in an aromatic diazonium salt.
It uses $ArSN_1$  pathway, similar to Sandmeyer Reaction which uses Copper Halide and corresponding Hydrohalous acid instead of copper powder.
As diazonium group is a very good leaving group, it departs leading to a phenyl carbocation which is very unstable, this is in fact the slow step of the reaction.
The Nucleophile (Halide in this case) attacks the phenyl carbocation leading to the formation of Halobenzene(fast step).
Nitrogen gas and Hydrogen chloride are also formed.

Can you tell me the source of this information?

It's what our teacher taught us in the class. @AnubhabDas


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Periodic Table and Atomic Radius
Well Wikipedia says that the atomic radius is lowered for elements in a group closer to noble gases (same row). It also says that those elements want to keep their valence electrons more than metals, for example. But the atomic shielding of those elements is bigger than the metals one because there are more electrons to repel. How is that possible?

Some insights: http://www.crystalmaker.com/support/tutorials/atomic-radii/

More tutorials: https://chem.libretexts.org/Bookshelves/Inorganic_Chemistry/Supplemental_Modules_(Inorganic_Chemistry)/Descriptive_Chemistry/Periodic_Trends_of_Elemental_Properties/Periodic_Trends

Unfortunately, that's not how shielding works.
Shielding is the ability of electrons to shield each other from the nuclear charge.
As you move right across the periodic table, the number of electrons increases but the new electrons are all in the same energy level and thus, roughly at the same distance from the nucleus. Therefore, they cannot shield each other from nucleus as well as, say, an inner (core) electron.
You've now increased the nuclear charge by a full unit of charge, but the increase in shielding is not commensurate. That means that the effective nuclear charge ($Z_{\mathrm{eff}}$) has increased, pulling all of the valence electrons closer to the nucleus.


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Why benzene has more heat of hydrogenation than cyclohexene?
According to me, benzene is more stable than cyclohexene due to its aromatic nature. Therefore the heat of hydrogenation of cyclohexene should be more than benzene (because I read that heat of hydrogenation is inversely proportional to stability).
But then also the heat of hydrogenation of benzene is more than that of cyclohexene.
Does it indicate the idea that non-aromatic compounds may be more stable than aromatic ones?

Please include relevant data (and sources where appropriate) for these types of questions.

I think this is relevant to your question as it seems you neglect the different degree of unsaturation. https://chemistry.stackexchange.com/questions/108726/why-total-heat-of-hydrogenation-of-1-3-cyclohexadiene-is-more-than-that-of-benze


text:
Does a tertiary carbocation rearrange to another tertiary carbocation?
I strongly believe that a carbocation should not rearrange to another if there are no immediate benefits(like a Greedy Algorithm).

The doubt hit me while solving this question. The answer mentioned in the book answers (b) but my answer comes out to be (a).
Here is the mechanism:

Ring expansion step gives you secondary carbocation, which is in higher energy state than original tertiary carbocation.

@MathewMahindaratne I am talking about the first shift.

Ring expansion and release of some strain presumably outweighs going from a tertiary to secondary carbocation.

Option B is the correct one. The process of dehydration of alcohols follow E1 elimination mechanism which involves two steps out of which the first one is slow ionisation of C-X bond( where X is any hetro atom) and the second step is  fast removal of $H^+$ ion .Since the first step is slow it provides enough time for the carbocation to rearrange and form a stable carbocation . The mechanism itself is self explanatory. Firstly as Oxygen possess lone pairs of electrons so the $H^+$ ion attacks the OH group as a result the $H_2O $ molecular gets eliminated resulting in a positive charge there.  Now hydride shift takes place as the so formed $ \textbf{Carbocation is more stable as it has 7 alpha hydrogens compared to the 5 alpha hydrogens of the previous carbocation}$. Now the C-C bond undergoes ionisation to form a three degree Carbocation and a carbanion . Now this carbanion forms a bond with the Carbocation formed earlier to form a six member ring . Now the resulting carbocation undergoes methyl shift which produces more stable three degree Carbocation. Now a $H^+$ ion from the neighbouring carbon atom is ejected as a result the pie bond is formed.

Yes b is the correct answer.
The carbocation will do a hydride shift to the tertiary carbon. It will then have a ring expansion. Then it will have a methyl shift. It will then lose its H+ ion to give option b.
Actually its forming a much stable product on hydride shift. Ergo it does shift to another tertiary carbon


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SN2 reactions with primary alcohols
A problem I was recently working on poses the following scenario:

A student realizes that protonation of an alcohol makes the -OH into a better leaving group. Therefore, he suggested the following reaction between 1-propanol, sulfuric acid, and trimethylamine. However, this process does not yield the product predicted here. Why not?

The only reason I can think of deals with the leaving group, which I usually only see in tertiary alcohols/SN1 reactions. Therefore, I supppose that since the reaction proceeds via an SN1 mechanism, the carbocation formed in this case (which is primary) would not form at an appreciable rate --> reaction would not proceed to form product predicted.
Is this line of reasoning correct or is there a better reason why the predicted reaction does not occur? Any ideas will be highly appreciated.

@AdvilSell The diagrams were provided in the question. I am not sure what the question makers used to make the diagrams.

Here the reason is not $S_N1$. In fact the $S_N1$ mechanism is quite difficult to observe with $1$-propanol. The main problem is the sulfuric acid, which not only protonates the alcohol but also protonates the trimethylamine, which is a strong base also. Therefore, being protonated, its nucleophilicity is greatly diminished and the reaction can't occur. The protonation and consequent loss of nucleophilicity of trimethylamine is the main reason.


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Linear Combination of Atomic Orbitals
In Atkins Physical Chemistry 10th edition in chapter number 10B it is given that within the Born-Oppenheimer approximation, the Schrödinger equation can be solved for the molecular species $\ce{H2+}$ but to simplify things and to obtain a method more readily generalizable to more complex species, we use the linear combination of atomic orbitals concept.
But my question is: how do we know that the wave function obtained from the linear combination obtained (say of two 1s orbitals of each atom) satisfies the Hamiltonian of this new system involving two nuclei? (The modified Hamiltonian is also given on the same page in the book).
Or do we need to make some iterative calculation as in the Hartree-Fock self consistent field method.
PS. I am a beginner so please try to keep the answer technically simple if possible.

Using a Linear Combination of Atomic Orbitals (LCAO) will not exactly satisfy the Schrödinger Equation / be an eigenfunction of the Hamiltonian
We can write down the Hamiltonian for the $\ce{H2+}$ cation with bond length $2R$, nuclei at $\vec{R_1}$ and $\vec{R_2}$, using the Born-Oppenheimer approximation:
$$
H\Psi(\vec{r}) = \left[-\frac{\nabla^2 }{2} - \frac{1}{\vert\vec{r}-\vec{R_1}\vert} - \frac{1}{\vert\vec{r}-\vec{R_2}\vert} \right]\Psi(\vec{r}) = E \Psi(\vec{r})
$$
If you feed in a linear combination of 2 $1s$ orbitals into $H\Psi$ you get a mess that is not a multiple of the linear combination you put in - i.e. a linear combination of 1s orbitals is not the exact solution.
In this case, because we only have one electron, the Schrödinger Equation can be solved to arbitrary accuracy using a variety of numerical techniques for second order partial differential equations in 3D; such as solution in an basis of functions or on a grid in space.
Solving in Practice
A suitable infinite basis of functions can describe all possible functions in 3D.  Two different examples are spherical harmonics with suitable radial components (Slater type orbitals) such as the $1s$ orbital, or Gaussian functions with suitable angular components (Gaussian type orbitals more commonly used in computational chemistry today).
With a infinite number of functions we would have an exact solution, but with only the first few terms in a suitable series we can be remarkably accurate.
In practical electronic structure theory, this means 3-30 functions per atom with suitable chosen parameters, centering the functions on atomic nuclei, as that's where the electrons will be mostly due to attraction to nuclei.
Variational Principle
Seeing as using a finite number of orbitals will not give an eigenfunction of the Hamiltonian, and certainly not the lowest energy ground state, we need another way to confirm a solution.
Instead we solve the Schrödinger equation "projectively".  Hamiltonians are hermitian operators.  One of the properties of hermitian operators is that their eigenfunctions are  orthogonal.  What this means is that for a Hamiltonian $H$ and two exact solutions $\Psi_i$ and $\Psi_j$ the integral:
$$
<\Psi_i\vert H\vert\Psi_j> = \iiint_{\text{all space}} \Psi_i^* H \Psi_j \;\text{d}V = \cases{E_i \text{ if } \Psi_i = \Psi_j\\0 \text{ otherwise}}
$$
Taking the integral of two wavefunctions, one multiplied by the complex conjugate of the other, is often referred to as projecting one wavefunction onto another.
Another property of the Hamiltonian operator is that the value of the integral with some generalised wavefunction $\Psi$:
$<\Psi\vert H\vert\Psi>$, the "expectation value" of the energy of $\Psi$, is minimised by the exact ground state wavefunction.  This property is know as the variational principle.  Hence in self consistent field/Hartree Fock calculations, we solve equations to minimise that integral of our LCAO wavefunctions by varying the coefficents to get the best LCAO approximation to the true wavefunction.

As you requested a technically simple answer, I'll describe every step thoroughly to be safe. I'm also assuming we're dealing with the time-independent Schrödinger equation, and will refer to it below just as the Schrödinger equation (I'm happy to explain why in the comments if you're interested).
By solving the Schrödinger equation for your system, you find the eigenstates/eigenfunctions of your hamiltonian, as well as the energy values of each of those eigenstates. Because your state function (the specific wave function you're interested in -- the one that describes your system in its current state) obeys the Hamiltonian of the system, then you are guaranteed that the state function can be exactly written (with no loss of accuracy or approximation) as a linear combination of the eigenfunctions you found by solving the Schrödinger equation.
For any system containing two or more electrons, the electron-electron interaction terms render it impossible to separate the Schrödinger equation into separable variables, meaning you'll need to use further approximations to solve the problem. However, within the Born-Oppenheimer approximation, the Schrödinger equation can be solved exactly when there's a single electron. That being said, the dihydrogen cation is a great example of where that doesn't mean it can be done easily.
A linear combination of atomic orbitals is a workaround to having to solve the Schrödinger equation for your system. Using them is similar in mindset to using the eigenfunctions to compose your state function. However, this time, there is no guarantee that your wave function of interest will exactly be a mixture of the eigenfunctions. This is because here, your eigenfunctions are for the hydrogen atoms, which is not exactly the system we are looking at! As such, you often need to use numerous basis functions to get closer to agreement with the exact solution. The usage of this substitution will certainly come with an optimization through a self consistent field or other theoretical method, in order to find the best possible combination of the eigenfunctions to reflect your state function (you wouldn't expect a 2p and a 1s atomic basis function to equally contribute to the dihydrogen ground state wave function).
TLDR: The atomic orbitals are not eigenfunctions of the dihydrogen cation, and so substituting the actual state function with a linear combination of atomic orbitals will be approximate. However, if you use  the complete basis set (CBS) limit and use a sufficient level of theory, you can recover excellent agreement with the exact solution.

Why are we dealing with the time-independent Schrodinger equation? Is it just to simplify or is there a deeper reason?

Sorry for the late reply. Assuming the Hamiltonian were time independent, then my answer can be trivially extrapolated to the TDSE (so yes, just for simplicity, UNLESS the Hamiltonian is time-dependent, then the form of the eigenfunctions change and it gets messy).

I am paraphrasing from an article in Wikipedia and applying it to your specific case.

To simplify things and to obtain a method more readily generalizable to more complex species, we use the linear combination of atomic orbitals concept.

The special thing about the ground state solution of the Schrödinger equation is that it has the lowest energy. The variational method uses the idea that any other wave function will have a higher energy.
If you have a bunch of wave functions and want to know which one is closest to the real one for the ground state, you calculate the energy each one of these would have (using the Hamiltonian for $\ce{H2+}$ in your case). Instead of taking unrelated functions, you can also build trial wave functions from linear combinations of basis functions, in your case just the 1s functions of the two hydrogen atoms, each multiplied by a coefficient. This is called a linear combination. Then, you optimize those two coefficients until you get the lowest energy.

How do we know that the wave function obtained from the linear combination obtained (say of two 1s orbitals of each atom) satisfies the Hamiltonian of this new system involving two nuclei?

They don't satisfy the Hamiltonian exactly (i.e. they are not eigenfunctions of the Hamiltonian) but they come pretty close. We know that by comparing the calculated energy to experimental data or to more sophisticated computational results. In the case of the $\ce{H2+}$ ion, you can also compare it to the exact solution.
With the 1s function centered on each atom as basis functions, you get a pretty good solution (with two positive coefficients) and a pretty awful solution (with coefficients of opposite signs), the latter of which is called the anti-bonding case (i.e. an excited state that does not hold the atoms together).

Do we need to make some iterative calculation as in the Hartree-Fock self consistent field method.

No, you don't need to iterate because there is only one electron, so there is no need for a potential energy coming from the other electrons (i.e. the "self-consistent field"). Also, with only two coefficients to vary, you will be able to find the global minimum in a straightforward manner.
If you want better solutions, you add more basis functions to your trial function (like 2s, 2p etc, or something completely different). The calculation will get more complicated, but the solutions will be closer to the actual ground state wave function. Now it will be more difficult to find the set of coefficients that give you the wave function with the lowest energy, and there might be some iteration to search for the global minimum, but that iteration is different from the ever-changing Hamiltonian in the Hartree-Fock method.
There are some tricks to cut down on computational time (or to do some of this qualitatively in your head or on paper). One central idea is to sort basis functions by symmetry, and keep different symmetries separate from each other (certain terms will be zero when the two basis functions appearing in the are zero). Another trick is to pre-compute parts of some terms, and then just vary the coefficients.
In your case no tricks are necessary because you are just combining two basis functions.


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Which will have more Lewis acid character out of BCl3 and AlCl?

Which will have more Lewis acid character out of  $\ce{BCl3}$ and $\ce{AlCl3}$ ?

My attempt: In $\ce{BCl3}$ backbonding takes place and there is $\mathrm{3pπ-2pπ}$ overlap, and in $\ce{AlCl3}$ there is $\mathrm{3pπ-3dπ}$ overlap which is less effective than the first. So $\ce{Al}$ is more electron deficient as compared to $\ce{B}$ hence $\ce{AlCl3}$ must have higher Lewis acid character.
But the answer is the opposite.
Please tell me where I'm wrong.

Eventually, the same proposition applies. Boron, despite being more electron enriched by back bonding, can also accept other lone pairs more easily.
Most lone pair donors are from the 2nd or 3rd periods in the Periodic Table, as such the $\pi$ overlap with boron will be stronger than that in aluminium.
You should also notice that boron trichloride is not as substantially backbonded as boron trifluoride, hence the stability gained through back bonding is not strong enough. Because of this, the boron in the trichloride will still look out for more sources of electrons.

Okay.. thanks for the help! But I was also thinking that AlCl3 dimerizes to Al2Cl6 and this does not happen in BCl3 that's why Boron is more electron deficient?

That may account too. These data are often results of a multitude of factors, because of which it becomes difficult to predict which factors dominate and which don't.  Also, thank you for reminding this.

(In $\ce{AlCl3}$ the possible backbonding is between $\ce{3p_z-3p_z}$ orbitals but due to large size of orbitals, it is not efficient)
Applying Pearson's HSAB concept-
Toward hard Lewis bases such as ethyl acetate, the Lewis acidities of the halides of group 13 elements would decrease as the softness of the acceptor element increases:
$\ce{BX3 > AlX3 > GaX3 > InX3}$
Toward soft Lewis bases such as dimethyl sulphide, the Lewis acidities of the halides of group 13 elements would increase as the softness of the acceptor element increases:
$\ce{BX3 < AlX3 < GaX3 < InX3}$


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What are the effects of salt additives and impurities on the CMC?
I understand what a critical micelle concentration (CMC) is, however I do not know why or how impurities or salt additives affect it when measured in specific ways. Any help is much appreciated.
So what determines whether they are stabilizing or destabilizing? If it interacts with the head group, how does this change the CMC and why? The head group already has a counter ion present?

This is a chemistry question. I have asked the question on a different forum and gotten plenty of helpful responses. There was absolutely no valid reason for this question to be put on hold. I am taking physical chemistry and we were assigned this question. Your moderators are rude and simply block questions with absolutely no constructive feedback or advice. Just because you don't know the answer, doesn't mean others don't. Let them answer for heavens sake.

Every chemical present can interact with the free lipid and/or with the micelles. Organic chemicals will tend to interact with the lipophilic portion of the lipids, while ionic ones will interact with the polar head.
Depending on their structure, all those species can have a stabilizing or destabilizing effect on the formation of micelles and so the CMC will change accordingly.
Of particular importance is the "ionic strength" of the solution (you should research that term).


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Flame Test Spectrograms Not Lining Up With Reality
In my AS Level chemistry class we had tried some DIY spectroscopy of flame tests with the aim of identifying anions in certain salts. We got as far as building the spectrometer inspired by Paper Craft Spectrometer (though we built a much bigger version such that a DSLR could be used) and performing the experiment and collecting the data. The problem was that I couldn't manage to get many of the spectrograms to line up with those sourced from the web. Many of our spectrograms, while plenty bright and with lines covering the whole visible range, had too few lines and those that were there didn't align with any real data we could find.
For instance with the following recorded spectrogram:

Some processing was done on the above image to produce the spectrum below it. The scale is calibrated from an LED which was also recorded and then confirmed my matching it against the emission of sodium which also roughly lined up to this scale +-10nm. From the colour of the flame we assumed it was calcium, given that all salts tested were chlorine salts, and the metal is one of: Ca, Cu, Cs, K, Li, Na, Sr. I don't know if there was an experimental error or if we were not looking at the correct reference spectrograms but nothing I could find would line up with that,does anyone have any insight into this?

Your resolution is low so you'll see bands not lines. // The green band about 555 and the red band about 620 are calcium. The yellow band at 590 is sodium contamination. Sodium has a relatively strong emission and it doesn't take much to contaminate the spectra. // The way to test this is to use relatively pure standards.

Welcome to Chemistry.SE! While we appreciate the cordiality, it is okay and in fact preferred that questions cut to the point and not contain superfluous content such as "Thanks!" and signatures. Cheers!

The first step you can take to make the spectral bands narrow is to reduce the slit width. It is slightly wide at this moment. We are not sure where you were looking for wavelengths? In the compiled tables in handbooks or schematic diagrams on the web, the emission wavelengths are listed for the excited atoms using very high temperature flames using air acetylene, or nitrous oxide acetylene flames. The ordinary flame temperatures in a laboratory are too low to electronically excite the atoms of any other element other than alkali metals. For Li, Na, K, Cs, you would see a line spectrum for each element (one or two lines). For potassium is extremely difficult to see the far red wavelengths (the resonance wavelengths listed in tables) until and unless your eyes are sensitive to red. The flame colors Ca, Ba, Sr, Cu are usually due to molecular compounds of these metals in the ordinary flames.
The key point is that the flame temperature is too low to see the line spectra you were expecting. However, if you look at discharge tubes, then you can do a good comparison.

I agree completely with M. Farroq.
The quality of your data might improve greatly if you reduce the slit width of your spectrometer, you will see less background light and obtain higher resolution.
Lookig at your image, it seems to me that your design may have a problem with the angle of incidence of the light at the difraction grating. It should be at an angle such that   the bands of the spectrum appear at the centre of the image - guess around 45º-.
You can find an image of the emission spectrum of a sodium lamp in this link Low Pressure Sodium spectrum with labels. Not the spectrum of a flame but maybe helpful to claibrate your experiment.


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Are all lift-off resist negative photoresists?
Are all lift-off resist negative photoresists? I'm a bit confused as I can't find anything in my textbook/lecture slides about the type of resist used for lift-off.

In the old days there were many different positive and negative resists. That selection has dropped of late since the requirements on resists in the deep-sub-micron production world are much more stringent. But, no, I remember using positive lift-off processes in the dim dark past...

There are definitely positive photo resists, but there tends to be little incentive to use them. Usually you use a negative mask with a negative resist to create a positive part. The pattern to be made tends to be a relatively small area of the photo resist on the mask and thus it takes less time for the lithography machine to produce a negative mask which can be a bottle neck in an R&D setting. Also Positive resists tend to be perminate as they have chemically inter-reacted to stay in place not be dissolved later (like during lift-off), thus another reason positive resists are not used frequently.


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Birch reduction- 1,4-cyclohexadiene over 1,3-cyclohexadiene

While reading of birch reduction I was confused as of why  1,4-cyclohexadiene is prefered over 1,3-cyclohexadiene,  the latter which ought to have been the major product due to conjugation.
One of the reason I felt could be of this outcome was "symmetry".
The symmetry produced by the product could cancel the dipole forces acting on the cis- bonds, making the molecule stable. But is this sole reason sufficient to overpower the conjugation? Please explain.
Edit: another possible reason I feel could be is the stereochemistry of the product. The product could facilitate more efficient structure in space about it's double bond (again symmetry), where as 1,3-cyclohexadiene formation could face issues of rotational barriers and might increase torsional strain on it.

Symmetry does not inherently lead to any stability. The product is not thermodynamically favoured, but is kinetically favoured: https://en.wikipedia.org/wiki/Birch_reduction#Reaction_regioselectivity see the last paragraph in this section.

Orthocresol thank you. I got it. So as it is kinetically favoured, the reaction was too quick to occur. Instead of stabilising it spontaneously formed the kinetically favoured product, 1,4-cyclohexadiene. I hope I got the right track.

So would there be a notable amount of 1,3-cyclohexadiene at low and controlled temperature and pressure

The temperature isn't the key thing here. The only requirement for a kinetically controlled reaction is that the reaction is not reversible. The last step in the Birch reduction, protonation of a highly basic anion, is for all intents and purposes completely irreversible.

Thank you. Yeah I didn't consider the protonation step here. I got it


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Is the water absorbed by sodium polyacrylate available to react?
I'm curious whether the water absorbed by sodium polyacrylate would react with alkali metals, or be trapped by the sodium polyacrylate.

"React" in what way? Absorbing water is also a reaction, it just depends on who wants it more.

This is too broad. In hydrate of CuSO4 water is covalently bound, in super-absorbent it's just hydrogen bonds.


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Why would CuSO4-EDTA drops be used in Pb titrations?
We use CuSO4-EDTA drops (4) when preparing our lead titration samples for final titration by EDTA.  What mechanism would there be for adding this to a titration?  What benefit does it produce on Pb titration?  We digest galena with nitric and sulfuric, clean up samples with a 5% sulfate wash, and remaining sample is washed into original beaker with 50% ammonium acetate solution (ammonium acetate/acetic acid buffer).  Samples are boiled then cooled before final steps.  The samples have ~1.5g ascorbic acid, 4 copper-EDTA drops, and 4 xylenol orange drops added to each sample prior to titration with EDTA.

Galena is the mineral form of lead sulfide. I'd guess that the copper is added to remove any residual sulfide in solution since the Ksp for copper sulfide is $6\times10^{-37}$ which is more insoluble than lead sulfide which has a Ksp of $3\times10^{-28}$

@MaxW and both and extremely insoluble. I suspect non-equilibrium and kinetic effects might dominate here.

I'm not an expert on coordination chemistry, and in particularly, metal-indicator complexes in EDTA titrations for heavy metals. But, I assume, the addition of $\ce{[Cu(EDTA)]^2-}$ solution to your unknown $\ce{Pb}$ sample is to get a sharp end-point (example of Replacement or Substitution Titration). Since, the extra metal ion added ($\ce{Cu^2+}$) has already complexed with EDTA, it won't interfere with the calculations. Meantime, both $K_{\mathrm{f}}$s of $\ce{[Cu(EDTA)]^2-}$ and $\ce{[Pb(EDTA)]^2-}$ are so closed ($5\times 10^{18}$ and $2\times 10^{18}$, respectively) that following equilibrium exists:
$$\ce{[Cu(EDTA)]^2- + Pb^2+ <=> [Pb(EDTA)]^2- + Cu^2+}$$
Where, $$K_{\mathrm{eq}} = \frac {[\ce{(Pb(EDTA))^2-}][\ce{Cu^2+]}}{[\ce{(Cu(EDTA))^2-}][\ce{Pb^2+]}} = \frac{K_{\mathrm{f}(\ce{(Pb(EDTA))^2-})}}{K_{\mathrm{f}(\ce{(Cu(EDTA))^2-})}} = \frac{2}{5} = 0.4$$
Thus, released $\ce{Cu^2+}$ cpmplexed with Xylenol Orange to give stable red color at operating pH (e.g., Figure 1A, Ref.2). My assumption is based on EDTA titration of $\ce{Pb^2+}$ and $\ce{Bi^3+}$  ion mixtures in the presence of Xylenol Orange as an indicator. In that titration, pH of a test sample solution containing $\ce{Pb^2+}$ and $\ce{Bi^3+}$ is adjusted to $3$  and add the indicator (Xylenol Orange). The test solution becomes red due to the complex formation of $\ce{Bi^3+}$ and Xylenol Orange (Figure 1B). When titrate with EDTA, at the first end point, the solution becomes yellow, the color of the free indicator at that pH. To titrate for $\ce{Pb^2+}$ at the same solution, it is needed to adjust pH to 5 - 5.5 until the appearance of red color again (indicating the complexation of $\ce{Pb^2+}$ with Xylenol Orange). Now, the slution can be titrated with EDTA until the turning of red to yellow. It is important to note that solution of $\ce{Pb^2+}$ ions does not become red with Xylenol Orange until pH is above 5. That means, the [$\ce{Pb^2+}$ + Xylenol Orange] complex is unstable below pH 5.

Also note that use of ammonium acetate ($\ce{NH4OAc}$) as a neutral buffer is misleading. $\ce{NH4+}$  and $\ce{AcO-}$  are not a conjugate acid/base pair, which means that they do not constitute a buffer at pH 7. However, dissolution of ammonium acetate salt in water results in pH 7, but this pH is highly labile (Ref.1). Ammonium acetate does provide buffering around pH 4.75 (the $\mathrm{p}K_{\mathrm{a}}$ of acetic acid) and around pH 9.25 (the $\mathrm{p}K_{\mathrm{a}}$ of ammonium). Thus, advantage for using $\ce{NH4OAc}$ is a drop from pH 7 to around pH 4.75, which is considerably less dramatic (e.g., addition of $\pu{1 mM} \: \ce{H+}$ to a $\pu{10 mM}$ neutral pH solution will decrease the pH to 5.8) than doing the same to pure water. Remember, during the titration, with out "real" buffer, pH of the solution changed due to release of $\ce{H+}$ ions by complexation.
References:

L. Konermann, “Addressing a Common Misconception: Ammonium Acetate as Neutral pH “Buffer” for Native Electrospray Mass Spectrometry,” J. Am. Soc. Mass Spectrom. 2017, 28(9), 1827–1835 (https://doi.org/10.1007/s13361-017-1739-3).
Interesting reading: O. J. V. Belleza, A. J. L. Villaraza, “Ion charge density governs selectivity in the formation of metal–Xylenol Orange (M–XO) complexes,” Inorganic Chemistry Communications 2014, 47(1), 87–92 (https://doi.org/10.1016/j.inoche.2014.07.024).
Relevant: S. Murakami, “Semi-Xylenol Orange Complexes with Bivalent Metal Ions,” J. inorg, Nucl, Chem. 1981, 43(2), 335–343 (https://doi.org/10.1016/0022-1902(81)90020-8).

Your chemical equation for the equilibrium depicts the charges on the metal ions wrongly. They should be "2+", not "2-".

This makes more sense than my guess about sulfide ppts.

@Tan Yong Boon: Thanks for the finding. It was a typo and changed accordingly. I appreciate very careful reading.

@MaxW: I glad somebody like it. Sulfide precipitation is a believable suggestion as well.

Thank you for the detailed response.  I did some experimentation with Ascorbic and the copper drops on lead nitrate after my post.  I did not find any discernable difference between the endpoint characteristics with or without either of these reagents.  I need to do further testing with our digested mine concentrate samples to evaluate any variance (e.g. sulfide precipitation).


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Optical activity
I was studying about the optical activity of compounds. And I found , it is written that the chirality of a molecule is the necessary and sufficient condition for a molecule to be optically active. I mean if  a molecule is chiral then it ia optically active. But to check chirality of a compound we have assure that the molecule is unsymmetrical and to check whether it is super imposable on ita mirror image or not. I think this process of checking the optical activity of a molecule is very cumbersome. I mean when there will be more than one chiral centres in the compound . Is there any other aid to check the optical activity easily?

Meh, I wonder what you'd tell about chiral compounds without chiral center...

If you find imagining and superimposing mirror images cumbersome, buy a molecular set. This will help. There is no short-cut in science and you are right sometimes very experienced people have difficulties in imagining these things especially if the molecules are complex. Models help them. There are various types of chiralities-having a chiral center is not a necessary condition. You will study a subject called group theory which will provide you a rigorous test for predicting optical activity. There are many molecules which do not have a single chiral center yet they are optical active. A simple example is that of binapthol (check Google images to see its structure). They are called atropoisomers and they are optically active.
Measuring optical activity by an experiment hardly takes 5-10 minutes in a polarimeter.


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Is the analyte included in the ionic strength equation?
I don't quite understand the calculation for ionic strength of a solution. The formula includes the concentration and charges of the ions in solution, however, is the analyte (whatever we're trying to analyze) also supposed to be incorperated into the ionic strength equation?
If so, how would we include it if the analyte has an unknown concentration (which we have to figure out, using equilibrium and ion activity)?

If an ionic analyte has an unknown concentration then you have more unknowns and need more equations to solve the problem. You might be able to get around the problem if the analyte concentration is low compared to the other salts in solution.

Dilute the analyte and/or work at high ionic strength so you can neglect the contribution by the analyte. Once you have an approximate concentration of the analyte from the analysis, check whether neglecting its contribution made sense.


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Characterisation of gold-silver alloy nanoparticle
I have various compositions of gold-silver alloy nanoparticles.
I would like to characterize my nanoparticles and confirm that they are alloys.
Aside from the qualitative analysis of the colour and using UV/vis to identify the number of plasmon bands, what other methods can I use to confirm the presence of alloys?

If sedimentation is an option, then you can try XRF or powder XRD.

I would suggest NMR

@andselisk XRF would not tell if the nanoparticle was an alloy or not, XRD might, but I think XANES/XAFS would give the needed quality for a confident result.

@A.K. XRF is going to show presence of other elements, which is IMO good enough. XANES/EXAFS would be indeed better, but from my experience those instruments are less abundant. In return I don't see how NMR is helpful here:)

@andselisk just because a nanopartiple is of a certain composition, does not mean it alloyed. You can get shell-core structures because of the surface energies. NMR allows you to see this "directly".

Try TEM or SEM. For example, read: https://warwick.ac.uk/fac/sci/physics/current/postgraduate/regs/mpagswarwick/ex5/techniques/structural/tem/

Well, my guess is something that provides a measure of the density of individual nanoparticles would work. Separation of the particles by sedimentation, for instance.

I Think you are in right track using UV-vis spectroscopy as a tool to identify your alloys. Other than UV-vis spectroscopy, researchers around the world have used many other analytical techniques to evaluate the synthesized nanomaterials, including X-ray diffractometry (XRD), Fourier transform infrared spectroscopy (FTIR), X-ray photoelectron spectroscopy (XPS), dynamic light scattering (DLS), scanning electron microscopy (SEM), transmission electron microscopy (TEM), atomic force microscopy (AFM), and etc. (For review of these techniques: Read Ref.1).
According to your notes, it seems like your main concern is how to confirm that your samples are alloys. UV-vis spectroscopy is a very useful and powerful tool in that regards, and reliable technique for the primary characterization of synthesized nanoparticles as well. Silver absorbs around $\pu{400 nm}$ and gold around $\pu{550 nm}$, yet these values vary depends on particle sizes of the dispersion (See Figure 1A & 1B, respectively). For example, absorption spectra calculated using the quasi-static approximation for silver and gold nanospheres with diameters $\mathrm{D} = 26$ and $\pu{20 nm}$, respectively, have shown absorption maxima (surface plasmon resonance wavelength $\lambda_R$) at ~$\pu{375 nm}$  for $\ce{Ag}$ and ~$\pu{525 nm}$  for $\ce{Au}$ in water (Ref.2). This reference also discussed the behavior of $\lambda_R$ as a function of the dielectric constant of the surrounding matrix ($\epsilon_m$), which is 1.77 for water.

This phenomenon was used to the best in identifying $\ce{Ag-Au}$ alloys in various compositions in an article (Ref.3), abstract of which states:

Gold−silver alloy nanoparticles with varying mole fractions are prepared in aqueous solution by the co-reduction of chlorauric acid $\ce{HAuCl4}$ and silver nitrate $\ce{AgNO3}$ with sodium citrate. As the optical absorption spectra of their solutions show only one plasmon absorption it is concluded that mixing of gold and silver leads to a homogeneous formation of alloy nanoparticles. The maximum of the plasmon band blue-shifts linearly with increasing silver content. This fact cannot be explained by a simple linear combination of the dielectric constants of gold and silver within the Mie theory. On the other hand, the extinction coefficient is found to decrease exponentially rather than linearly with increasing gold mole fraction $x_{\ce{Au}}$. Furthermore, the size distribution of the alloy nanoparticles is examined using transmission electron microscopy (TEM). High-resolution TEM (HRTEM) also confirms the formation of homogeneous gold−silver alloy nanocrystals.

According to this reference, only one plasmon band for particular alloy (Two bands would be expected for the case of a mixture of gold and silver nanoparticles). The band shows a blue shift with increasing $\ce{Ag}$ amount in the alloy (See Figure 2) and a linear relationship with increasing gold mole fraction, $x_{\ce{Au}}$ (See the insert in Figure 2):

References:

X.-F. Zhang, Z.-G. Liu, W. Shen, S. Gurunathan, “Silver Nanoparticles: Synthesis, Characterization, Properties, Applications, and Therapeutic Approaches,” Intl. J. Molecular Sci. 2016, 17, 1534 (34 pages) (doi:10.3390/ijms17091534).
F. Vallée, “Chapter 7: Optical Properties of Metallic Nanoparticles,” In Nanomaterials and Nanochemistry; C. Bréchignac, P. Houdy, M. Lahmani, Eds.; Springer-Verlag: Berlin, Germany, pp. 197-228, 2007.
S. Link, Z. L. Wang, M. A. El-Sayed, “Alloy Formation of Gold−Silver Nanoparticles and the Dependence of the Plasmon Absorption on Their Composition,” J. Phys. Chem. B 1999, 103(18), 3529–3533 (DOI: 10.1021/jp990387w).