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Do all salts of sodium form colorless solutions?
I read all salts from sodium salts form colorless solutions unless the anion is colored. Can someone explain?

There is nothing to explain, really. White paint is white and remains white, unless mixed with some paint colored other than white. Same thing here.

Salts containing a counterion that is coloured will be coloured. the colour won't come from the sodium.

So, you ask what, really? Why they are colorless?

New contributors need to be treated a bit more gently. Their questions generally indicate an introduction to the topic with unconnected facts. If we demonstrate one connection, they are likely to be able to make more connections, on their own.

There are various colors of chemical compounds because color is dependent on all of the interactions of all the components of the mixture, including the concentration. Some materials can appear to be one color in dilute solution, but a significantly different color in concentrated solution, because the level of absorption at a particular wavelength becomes so high that it dominates the color.
In the case of the sodium salts, it turns out that sodium ion does not have an electronic transition in the visible range, so it's invisible! However, all the other materials, ions, compounds, etc., may contribute their own absorptions to the mixture.


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Doubt regarding steam distillation
According to my textbook, in steam distillation the liquid boils when the sum of vapor pressure due to the organic liquid and that due to water becomes equal to the atmospheric pressure. Why? I did not understand.

According to my knowledge, the organic liquid is volatile & is carried by the hot steam with itself. There's something that I'm unsure about in your question. Could you upload a picture of your textbook where it's mentioned this way?

What's to understand? 0.5 is less than 1, and another 0.5 is also less than 1, but 0.5+0.5=1. There is nothing more to it.

Solutions don't boil at constant temperatures just like that unless it's azeotropic? If you're asking why doesn't the organic solvent vaporise if it is volatile etc. then it means you're not clear with the concept of how steam distillation is carried out for which you can read Wikipedia as it has diagrams as well.

Distillation of many organic oils and resins requires temperatures significantly higher than 100$^o$C. If you add water to the oil/resin (say, 50/50), the mixture will boil at 100$^o$C. Well, not quite 100$^o$C, because at the boiling temperature, the oil or resin will have a slight vapor pressure of its own, and the total vapor will therefore have some water molecules and some oil molecules, which together will exert a pressure of one atmosphere (plus a tiny tiny extra pressure that is needed to force this vapor out of the flask).
In practice, steam distillation is essentially using the water to provide nearly one atmosphere of pressure to carry the slightly volatile oil/resin out of the boiling flask. This is not normally done at equilibrium conditions, but at a rapid boil so you can get some product in a reasonable time. However, the theoreticians are correct that the water has a vapor pressure, and the oil/resin has a vapor pressure, and they theoretically add up to one atmosphere (plus whatever added pressure is needed to force the vapor out of the flask at the rate you are using).
Steam distillation is not the best description for a simple separation of water from an organic, nor for an azeotropic distillation, where water is added so a "compound" (azeotrope) of the water plus the organic can be carried over because its composition is more preferred than the original mixture.
Ref: https://www.thoughtco.com/definition-of-steam-distillation-605690

Evaporation happens all the time, at a positive net rate as long as the vapor pressure of that component is larger than the atmospheric partial pressure of the same component.
Boiling is a specific case of evaporation, at the pressure and component specific boiling point, when the total pressure equals the atmospheric pressure. It is fairly trivial to show that at least one of the liquid components must have greater partial pressure than the atmosphere for this case to exist. At this state the temperature of the liquid cannot increase, until the situation is resolved. Either by system pressure rising or by elimination of the liquid.
It would be beneficial to read up on thermodynamics, phases and states, before proceeding with regular chem, it's only going to get worse...

Imagine for illustrative purposes there are two liquids with identical vapour pressure and boiling temperature.
Now consider two scenarios:
Liquids are miscible, in 1:1 molar ratio. Saturated vapor pressure of each is half of the vapour pressure of the pure liquid. But summed, it still makes vapour pressure of either of liquid, so the mixture boiling point hoping stays. For non equal boiling points, the vapor pressure would be the average of pure liquid pressure ( at ideal behaviour ).
Now, liquids are not miscible. Each liquids forms separate phase of the pure liquid and keeps the saturated vapor pressure of pure liquid. But there are 2 liquids, so there is twice as big total vapor pressure. Therefore the boiling occurs at temperature each of saturated pressures is just a half of the external pressure.
For highly non equal boiling points, total vapor pressure is still sum of pressures, with the mixture boiling point somewhat less than if the more volatile liquid.

source: Stack Exchange

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

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Calculating PPM of colloidal material in solution produced by electrolysis
I am running an electrolysis between two silver electrodes.
The “electrolyte” is deionized water.
Ions are still formed as the CO2 of the air generates enough ions for a few µA of current.
During the electrolysis the ions are converted to metallic silver nanoparticles due to strong ultraviolet radiation.
I have the following variables:

Current in Ampere [A]
Time passed in seconds [s]
Coulomb [C or As]
Quantity of deionized water [l (liter)]
I want to know:
On the anode dissolved silver in milligrams [mg]
PPM of silver in the deionized water [PPM]

In practice there will be some silver ions that make it from the anode to the cathode without “colliding” with an photon – so they will be depleted on the cathode.
However for the calculation I assume that all ions will get reduced to metallic nanoparticles and nothing will be depleted on the cathode.
How do I calculate the dissolved silver in mg and the PPM of the solution?
Thanks in advance!

Note that "ppm" is an ambiguous way to describe concentration and its usage is discouraged. See, for example, my answer.

First no electrolysis can be made in deionized water. What voltage ? Under high tension, an ark may be obtained. Is it what you want ? Then you speak of ultraviolet radiation. Why ? Where ?  Why don't you add silver nitrate in the solution ? Finally what are you trying to do ?

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


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What is the double bond equivalence of the following compound?

What is the double bond equivalence of the given compound? The given answer is 14 (8 pi electrons and 6 rings = 14 DBE). But why isn't the lone pair on nitrogen considered? The lone pair on nitrogen is involved in resonance with the other 2 bonds making the ring aromatic, so shouldn't we also consider that lone pair?

If some double bonds are in resonance, that does not make more or fewer of them. It is still the same number of double bonds. Resonance is about as irrelevant as the weather in the street outside. So is the lone pair of N.

As Ivan Neretin pointed out in the comments, $\pi$ bonds will remain same whether or not in resonance.
However, if you have trouble counting the double bond equvalent, you can directly use the formula for it given as $$\ce{DU=\dfrac{2C+2-H+N-X}{2}}$$ so here number of carbons are C=23, H=21, N=1 which also gives 14 as the answer.

6 rings, and 8 double bonds, adds together for 14 DBE. Resonance will not change how many double bonds there are - you can see that the nitrogen has made 3 bonds and so does not contribute to the DBE total

For an alternative method, look here.


text:
How carbon nanotube sensors works?
How carbon nanotube sensors works?
Is the protein (or chemical compound used as part of the sensor) placed inside the tube or where is it placed? and how is the technology used to place the protein there?
In the National Geographic magazine we can read (speaking about a young inventor)

he could interweave antibodies sensitive to mesothelin into a network
of carbon nanotubes. High levels of the protein in a drop of blood
from someone with pancreatic cancer would attach to the antibodies and
cause them to enlarge, thus changing the network's electrical charge
and signaling the cancer's presence.

In the Smithsonian magazine we can read

What if he could lace a nanotube network with mesothelin-specific
antibodies, then introduce a drop of a pancreatic cancer patient’s
blood? The antibodies would bind to the mesothelin and enlarge. These
beefed-up molecules would spread the nanotubes farther apart, changing
the electrical properties of the network: The more mesothelin present,
the more antibodies would bind and grow big, and the weaker the
electrical signal would become.

link

The tubular nature of CNTs allows for internal protein encapsulation,
and also for their external coating by protein cages, affording
bottom-up ordering of molecules in hierarchical structures. To achieve
such complex systems it is imperative to master the intermolecular
forces between CNTs and proteins, including geometry effects (e.g. CNT
diameter and curvature) and how they translate into changes in the
local environment (e.g. water entropy).


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Comparison between bond length of platinum and titanium oxides
This question comes from a problem sheet I recently had.
It concerns the fact that Pt-O bond lengths are typically longer than Ti-O ones, even though the radii of the two metals are similar.
My rationale is that as Ti d-orbitals are more contracted than the ones of Pt, it would form shorter bonds for that reason. I was wondering whether there is something else to it.

hint: TiO2 has Ti-O about 2A, but TiO has Ti-O about 3A . On the other hand, both PtO2 and PtO have close Pt-O about 2A

@permeakra is this to suggest that due to the absence of d-electrons in TiO2, such a compound would be with significantly more ionic character, thus having the atoms closer for electrostatic reasons?


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Uses of Fourier / Laplace transforms in chemistry (apart from spectroscopy)
I'm just now learning about the Fourier transform, which seems like a pretty useful tool, and I know it has uses in spectroscopy (e.g. FTIR) since that is what Google shows when searching for applications of the Fourier transform in chemistry. Interestingly, I found out that the Laplace transform can be used in
chemical kinetics.
I was wondering if there were any other cool uses for the Fourier/Laplace transform in chemistry?
And if so, any links to resources to learn more about that particular use would be appreciated.

(+1) Fourier transform techniques are used extensively in signal processing (check out the signal processing stack exchange for examples) and chemists use various signal processing techniques for things like resolution enhancement of chromatograms. I predict that you will get answers or comments that are informative in this regard.

I don't have the knowledge to go much deeper, but in molecular dynamics simulations Fourier transfroms are used to calculate the electrostatic interaction in periodic systems.  For more details see https://en.wikipedia.org/wiki/Ewald_summation

Generously used in crystallography and for discussing the electronic properties of crystals or other semicrystalline nature, e.g. solids with repeating or symmetrical subunits.

It is essential to NMR,  EPR, and xray diffraction from crystals, as well as  IR /Raman spectroscopy. It is not used in kinetics but various maths transforms may be used in solving these equations and also in solving quantum equations. The Hadamard transform can be used in pump-probe chemical kinetics.

For the principle «what is small, gets large; what is large, gets small» when applying Fourier transformation and inverse Fourier transformation, it is one important tool in crystallography to relate between the regular arrangement of molecules in crystals at Angstrom scale, and the intensity recordings on film (old school) and electronic detector (very often area CCD) where signals are much wider apart from each other.  (This is often is referred to the relationship between direct and reciprocal space, too.)  Thus, scrutinizing the diffraction pattern's symmetry from special directions (figure b) below) allows the trained eye to relate to possible symmetries of the crystalline sample (a) , and the ratio of vector lengths describing the smallest unit of a crystal structure, the unit cell.  Today, assessing position and intensities of these data is typically assisted by a computer.

(credit, open access)
As shown by Tsutaoka et al. (Eur. J. Phys. 35 (2014) 055021, doi 10.1088/0143-0807/35/5/055021), it is possible to simulate such patterns with a laser pointer passing two film gratings slightly rotated to each other; the 2D crystal is then build by the points where the two gratings S1 and S2 intersect each other (note the vector ratio $\vec{a} / \vec{b}$ vs. $\vec{a^*} / \vec{b^*}$, and angles enclosed), e.g.

ImageMagick is one program allowing you to experiment with FT on 2D images but recommend to consult it altogether with the above mentioned open access publication by Aubert and Lecomte (J. Appl. Cryst. (2007) 40, 1153-1165, doi 10.1107/S0021889807043622) for additional examples and background.

I still think it is absolutely magic to remember the TEM transmitted beam visual, (starmap-like) set the aperture to just one particular "star" then flip the magic switch to get a heat map of the structure where just that one phase lights up and the rest are dim. Absolute witchcraft.

The Fourier Transform is important in many applications  beyond spectroscopy, for instance in computational and theoretical chemistry. One example encountered in learning chemistry is in quantum theory. The FT is intimately related to the Heisenberg Uncertainty principle (for instance for position-momentum). You might want to look at a libretext on the subject.
As implied in other answers, a practical utility of the FT lies in revealing or exploiting regularity in data.
When evaluating integrals in some computational QM problems, application of the FT can provide a time-saving device. See for instance Ref 1, a thesis detailing this.
A very important example of its application in computational chemistry is in computing the Ewald summation:

Ewald summation, named after Paul Peter Ewald, is a method for computing long-range interactions (e.g. electrostatic interactions) in periodic systems. It was first developed as the method for calculating electrostatic energies of ionic crystals, and is now commonly used for calculating long-range interactions in computational chemistry. Ewald summation is a special case of the Poisson summation formula, replacing the summation of interaction energies in real space with an equivalent summation in Fourier space. In this method, the long-range interaction is divided into two parts: a short-range contribution, and a long-range contribution which does not have a singularity. The short-range contribution is calculated in real space, whereas the long-range contribution is calculated using a Fourier transform.

If you want to evaluate a Madelung constant you might also employ the FT.
References

Jaime Axel Rosal Sandberg. New efficient integral algorithms for quantum chemistry. Doctoral Thesis in Theoretical Chemistry and Biology, School of Biotechnology, Royal Institute of Technology, Stockholm, Sweden 2014 (open access repository link).

@BuchThorn If this is the reference you intended, keep the link; if not revoke the edit.  For future reference: For various reasons, PhD theses may be more difficult to access, than a publication in a scientific journal.  But if there is a public repository of them, please add one permanent link.

Fourier transform (FT) is simply a bridge which lets us view the data in reciprocal domains. For example, if you collected data with respect to time, FT of this data will allow you to see the same information in the (1/time) domain. In FTIR, you collect an interferogram with respect to distance (length units), FT of this data yields information in (1/length). That is why your FTIR x-axis is in wavenumbers (cm$^{-1}$).

The second advantage of FT is that simplifies many calculations into simple multiplication or division.

You should explore a little bit more about discrete Fourier transform (DFT). That is what is used in spectroscopy and analytical chemistry. FT is for continuous function, DFT is for sampled data. The most important thing to note is that there is no loss or gain of information before or after (D)FT of any data.
FT finds extensive use in denoising analytical chemistry data. All instrumental signals have noise to some extent. In most cases, the signal is present at low frequencies and noise has higher frequencies. You can cut-off high frequency noise and do an inverse transform to get a signal which has higher signal to noise ratio.
FT techniques can resolve overlapping peaks if you know what broadened the signal. Visit Prof. Tom O'Haver website or see his book (free pdf). Window 1 (top) is the broadened signal which could be a chromatogram or a spectrum, Window 2 is a function which caused broadening of the peaks. Window 1 (bottom) is "recovered" signal.
https://terpconnect.umd.edu/~toh/spectrum/Deconvolution.html

FT is also required in determining X-ray crystal structure.

"You can cut-off high frequency noise and do an inverse transform to get a signal which has higher signal to noise ratio." I haven't heard anyone using this technique in practice. I played with this approach a bit and it wasn't satisfactory: either not much smoothing or the signal starts to look wavy. "there is no loss or gain of information before or after (D)FT of any data" - well, except for the noise added due to rounding by computers :) The more calculations the higher the error. Though it's probably nothing relative to the chemical and electronic noise.

Hi Stanislav, I am afraid you are not doing this correctly. I routinely use DFT to denoise analytical data and it is very common in signal processing. It is called "windowing" in the literature. If you want post a separate question with details, and I will try to address it. The loss in info due to computers is negligible for all practical purpose until and unless someone is doing theoretical calculations.

I've found articles on windowing with FFT, thanks. Will dig deeper into the subject someday when I get back to it. Out of curiosity: why do you use FT instead of Savitzky-Golay or Moving Average? They seem to be the default choices for smoothing in analytical chemistry. At least I hear about them most.

People run away from DFT because it is not as simple as moving average or Savitsky-Golay. I have written a long review on DFT applications for chromatography/analytical chemistry with detailed steps. Hopefully it should be out in 2-3 months, if no major problems occur.

Cool. Would you be able to post a link here once it's available?

Yes, but I am assuming it will take at least two more months. Review process is very slow these days. In the meanwhile please explore Tom O'Haver's website and search the relevant topics

Laplace transforms are used all the time in Chemical Process Control (I'm a chemical engineer, not a chemist).  Most (open-loop) chemical processes can modeled as a first order system, possibly with delay. A tank being heated using a heat-exchanger on its input will show the same dynamics as a you see in reaction kinetics.
There are various ways to measure the dynamics of a process.  Once you do that, you can characterize them as first-order + delay and use that Laplace process description to get a dynamic model of the process.
Non open-loop processes (whether because they are under control or because they have a natural loop in the process) can also be modeled using Laplace equations.
Forgive me if I get the terminology wrong - I've been doing mostly software and not process control for the last several decades

Just to add some more information you may find useful.
I have found the use of Laplace transforms for the analytic solution of reaction mechanisms in Chemical Kinetics and in Dynamic Electrochemistry. Basically, the Laplace transforms are used as a method to simplify the solution of a set of coupled differential equations that in this case are assoacited with a reaction mechanism or the solution of the reaction-diffusion equations.
In the case of Chemical Kinetics you can check the introductory article Andraos, J., Streamlined Approach to Solving Simple and Complex Kinetic Systems Analytically, J. Chem. Educ., 76, 1578 (1999)
You can also find some information in advanced Chemical Kinetics textbooks as Jeffrey I. Steinfeld, Joseph S. Francisco, William L. Hase, Chemical Kinetics and Dynamics 2nd ed., Prentice Hall (1999) or Connors, Kenneth A.
K. A. Connors, Chemical kinetics : the study of reaction rates in solution, Wiley-VCH (1990).
Fourier transforms are also used in Quantum Chemical Dynamics for the propagatiion of wave-packets to describe the molecular motion on one or more potential energy surfaces.


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What does "the natural bond angle" mean in the context of Universal Force Field (UFF) potential energy calculation and how are bond angles calculated?
As the title says, what does "the natural bond angle θ0" mean in the context of Universal Force Field (UFF) potential energy calculation? The reference appears in section D. Angular Distortions, 1. Angle Bend). The article also talks about "natural bond length" (Section C. Bond Stretch).
I am curious about how molecular dynamics (MD) simulations estimate/calculate bond angles using the Universal Force Field.
At several points in the original UFF paper they mention a "natural angle" but it is not overwhelmingly clear to me what this means. Is it the empirically observed bond angle(s) in particular molecules? How do MD simulations, then, produce bond angles for molecules that there is no published data for?
I understand that the UFF potential energy calculation contains several terms that account for various properties like hybridization, van der Waals interactions, etc. Which of the terms, which are outlined in the first pages, account(s) for steric interactions between molecules and determine the final bond angles?
I am admittedly not very well versed in the types of maths that are involved, but I would appreciate any explanation possible.

The original article on UFF is from 1992, so it is somewhat dated. The selection of "natural bond" values is explained on page 10028:

a. Standard Bond Angles. The natural angles for the group 15, 17, and 18 main group elements are obtained from standard reference structures of the parent hydrides. Thus 0-3 has Bo = 104.5' from HzO, while S3 has Bo = 92.2' from HzS. Exceptions include 0-32, OR, and N-2. The bond angles for 0-32,O-R, and N-2 are fit to (C13Si)z0, methyl vinyl ether, and dimethyl- diazene angles of 146O, 118.3', and 112.3', respectively. Where structural data are unavailable, the natural angles are extrapolated from the element above it in the periodic table. The remaining elements are all assumed to have regular octahedral, tetrahedral, trigonal, or linear structures. The natural angles are collected in Table I.

So yes, the angles are selected from experimental geometries where available, or where not by extrapolation from geometries for compounds in the same group.
References

A. K. Rappe et al.. UFF, a Full Periodic Table Force Field  for  Molecular Mechanics and  Molecular Dynamics Simulations. J. Am. Chem. Soc. 1992, 114, 10024-10039.


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Comparison of C=O bond strength in metal carbonyls

Find the weakest $\ce{C=O}$ bond among $\ce{[Mn(CO)6]+},$ $\ce{Fe[(CO)5]},$ $\ce{[Cr(CO)6]}$ and $\ce{[V(CO)6]-}.$

I thought the $\ce{C=O}$ bond strength would be lowest in $\ce{[V(CO)6]-}$ since the negative charge meant that vanadium had an excess of electrons, hence the metal-carbon back-bonding would be the strongest. As the result, the $\ce{C=O}$ bond order would be the least.
However, the answer given states that iron in $\ce{Fe[(CO)5]}$ has a $\mathrm{3d^8}$ configuration, which means four electron pairs for back-bonding as compared to $\mathrm{3d^6}$ in $\ce{V-},$ hence it would have a greater degree of back-bonding and thus its carbonyl bonding would be the weakest.
Is this correct or does $\ce{[V(CO)6]-}$ have the strongest metal-carbon synergistic bond? If the former is correct, then does a synergistic bond depend on the number of lone pairs first and a negative charge is only considered if the number of lone pairs is the same?

@LightYagami The first part of your question was easy to understand, I have commented on it there; the second is where my question lies.

@Rishi what does that have to do with the question? Is the given explanation wrong?

@Metaxylene Also see this question, I wonder, why nobody provides a definite answer to this question, maybe due to lack of data of bond lengths? So, it may just become handwaving...

@LightYagami In the isoelectronic case, I see that it depends on the electron density. So maybe in other non-isoelectronic cases, the number of d$\pi$ orbital electrons is deciding. I agree though- it seems like no one knows for sure what the trend/property is.

@Metaxylene Have you tried consulting your teacher regarding this?

Let us continue this discussion in chat.

At first glance I would say this isn't a fair comparison since they don't all contain 6 CO ligands. But your logic of $\ce{[V(CO)6]−}$ having the lowest CO stretching frequency for the octahedral complexes applies.

@Metaxylene Ignore the earlier comment, I was picturing something else in my head, But by the second question do you mean “Whether this works”? Because it does work, you just need to relate your introductory MOT a little bit and read the linked answer a few times to get a grip on what it is fundamentally doing.

@H.Linkhorn so $[V(CO){6}]^{−}$ has a weaker $C=O$ bond than $Fe[(CO){5}]$ ?

@Rishi the linked answer (I'm assuming it's the first comment) shows it as a relation, but I wanted to know about the bond in terms of the strength between the carbon-metal synergistic bond which would decrease the $C=O$ bond strength.

@Metaxylene Bond length of CO $\propto$ 1/(BO of CO)$\propto$ BO of Metal-C, Is this what you are asking about?

@Metaxylene that’s not what I’m saying.

@Rishi Yes. I want to compare it in those terms between $[V(CO){6}]^{−}$ and $Fe[(CO){5}$

@H.Linkhorn Okay, so what are you saying?

That comparing $\ce{[Fe(CO)5]}$ and the octahedral complexes isn't a fair comparison. So your explanation that you deduced in your question only applies to the octahedral complexes and you should then treat $\ce{[Fe(CO)5]}$ separately. The drop is coordination number also means the metals $d$ electrons are "shared" among less CO ligands and so the degree of backbonding can be expected to be greater.

A common measure of $\ce{C-O}$ bond strength is $\ce{C-O}$ distance. For $\ce{[Fe(CO)5]}$ the data can be found, for example, in paper 1, where values of 1.12-1.14 Å are reported. For $\ce{[V(CO)6]-}$ in paper 2 values around 1.17-1.28 Å are reported.
As for rationalization, there are several factors in play here with opposite effect. It is not possible to guess which one wins beforehand. Here are two factors

The iron complex shares 8 electrons between 5 carbonyl ligands while the vanadium complex shares 6 electrons between 6 ligands, so iron provides more electrons per ligand
The vanadium complex has the central atom withlower electronegativity and higher negative charge, so donation from vanadium is easier.

First of all your answer, :
The order of CO bond order is : $\ce{[Mn(CO)6]+}>\ce{[Fe(CO)5]}>\ce{[Cr(CO)6]}>\ce{[V(CO)6]-}$
Surprising position of $\ce{[Fe(CO)5]}$ !!

Explanation:
CO bond length is compared by studying it's stretching frequency .
I won't go into the details, but to give an idea,
CO bond order is directly proportional to it's stretching frequency.
[Note: Although it's frequency, but typical units of wave number are used in IR Spectra]
For details on stretching frequency:

The Wikipedia link.
Refer Organic Chemistry by Jonathan Clayden, Nick Greeves & Stuart Warren; 2nd ed , Ch 3 (Determining organic structures) ad Ch 18 (Review of Spectroscopic Methods)

Stretching frequencies of some complexes are:

$\ce{[V(CO)6]-} = 1859 cm^{-1}$
$\ce{[Cr(CO)6]} = 2000 cm^{-1}$
$\ce{[Mn(CO)6]+}= 2100 cm^{-1}$

$\ce{[Fe(CO)5]}$ complex is characterized by two stretching frequencies of 2034 and 2014 $cm^{-1}$ due to axial and equatorial CO ligands. But both the stretching frequencies are greater than that of $\ce{[Cr(CO)6]}$.
Hence the order of CO bond strength is $\ce{[Mn(CO)6]+}>\ce{[Fe(CO)5]}>\ce{[Cr(CO)6]}>\ce{[V(CO)6]-}$.
So weakest bond of CO is in $\ce{[V(CO)6]-}.$

Theoretically, the order $\ce{[Fe(CO)5]}>\ce{[Cr(CO)6]}>\ce{[V(CO)6]-}$ can be understood as:

$\ce{[V(CO)6]-}$ has negative charge as compared to other to so it donates most electron density in $\pi^{*}$ orbital of CO ligand and hence CO has least bond order in $\ce{[V(CO)6]-}$
Other two complexes are neutral. The order of effective nuclear charge is $\ce{Fe} > \ce{Cr} $ due to more $d e^{-}s $ in in $\ce{Fe}$. Hence $\ce{Fe}$ has less donation tendency then $\ce{Cr}$ and so it donates less electron density in $\pi^{*}$ orbital of CO and therefore CO has higher bond order in $\ce{[Fe(CO)5]}$.

Hope helpful :)

Source for $\ce{[Fe(CO)5]}$ stretching frequency: Adams, R. D.; Barnard, T. S.; Cortopassi, J. E.; Wu, W.; Li, Z. "Platinum-ruthenium carbonyl cluster complexes" Inorganic Syntheses 1998, volume 32, pp. 280-284.

The given answer is incorrect. From the article:

As we move left to right, the d orbital energies decrease and the energies of the $\textrm{d}\pi$ and $\pi^{*}$ orbitals separate. As a result, the backbonding orbital interaction becomes worse (remember that strong orbital interactions require well-matched orbital energies) as we move toward the more electronegative late transition metals!

You should explain it fully ,,and not just give a vague reference


text:
Can atomic orbitals combine in a manner intermediate between bonding and antibonding?
When atomic orbitals combine in phase (0°) we get bonding molecular orbitals, and when they combine with opposite phases (180°) we get antibonding molecular orbitals.
Why can't atomic orbitals combine in phases between 0 and 180?

You may combine orbitals whatever you like, but not all combinations make sense.

Take a look at https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Quantum_Tutorials_(Rioux)/Chemical_Bonding/183:_Semi-empirical_Molecular_Orbital_Calculation_on_XeF2 where you get a bonding, a non-bonding and an anti-bonding orbital.


text:
How do acids prevent ferrous sulfate from getting oxidised?
If $\ce{FeSO4.7H2O}$ is dissolved in an acidic environment, for example in 0.05 M aqueous solutions of methanoic or ethanoic acid, would that prevent it from getting oxidized if an oxidizing agent like bromine water was added to it? If so, how? Since this is not examining oxidation with respect to air or oxygen.

Diluted solutions of $\ce{FeSO4}$ are partially hydrolyzed, producing $\ce{[Fe(OH)(H2O)5]+}$ ion. In the presence of an acid in aqueous solution, the main ion containing iron is $\ce{[Fe(H2O)6]^2+}.$ This ion must be more resistant to oxidation than the partially hydrolyzed ion.

@Maurice can you please provide the equations for the formation of the respective ions?

@user109975 You should be able to do it yourself, shouldn't you ?  Ferric ions have much greater hydrolysis tendency than ferrous ions, what leads to decreasing ratio of ferric/ferrous ions and decreasing redox potential, making it some what less vulnerable to oxidation in acidic solutions. But is is not sufficient not to be oxidized by bromine. Carboxylic acids presence brings another factor, forming complexes with iron. See https://en.wikipedia.org/wiki/Iron%28III%29_acetate  and https://en.wikipedia.org/wiki/Iron%28II%29_acetate

No, adding acid to a ferrous sulfate solution will not prevent its being oxidized by bromine water.
Fresh FeSO$_4$ solutions are light blue-green, almost colorless, but pick up O$_2$ readily:
Fe$^3$$^+$ + e$^-$ --> Fe$^2$$^+$       +0.770 V and
O$_2$ + 2 H$_2$O + 4e$^-$ --> 4 OH$^-$    +0.401 V
(The half cell voltages are only indicative of what's going on, because we are not at standard concentrations of 1 M.)
The light-colored solution quickly becomes yellowish and consumes all the oxygen in the vessel, and then sucks in more air and consumes that oxygen, and then things slow down a little, but diffusion still lets a bit more oxygen get in, and the "ferrous" becomes somewhat ferric, and the liquid becomes a muddy color and you wish you could preserve the nice blue-green color somehow.
You could preserve the ferrous if you could somehow complex it and keep it from oxidizing to ferric. But nature prefers to complex the ferric ion: smaller ion, bigger charge. Typical ferric stability constants are larger than ferrous by ten orders of magnitude.
It is instinctive to suppose that metals will oxidize in the presence of O$_2$, and if higher valences are possible, well, then, the metal will go as far as possible. But the oxidation of ferrous to ferric is unfavored (+0.770 V) and Fe$^2$$^+$ needs help from O$_2$. The reduction of water is also unfavored, but is driven forward by the consumption of OH$^-$ ions by precipitation by Fe$^+$$^3$, which also keeps the Fe$^+$$^3$ low.
LeChatelier predicted as much: by increasing the acidity, you reduce the availability of electrons from OH$^-$, and O$_2$ is more able to take electrons from Fe$^+$$^2$.
So you could say that adding acid stabilizes the ferrous ion, but what it really does is inhibit the oxygen molecule. Bromine is totally unaffected.

How about just adding metallic iron to the ferrous sulfate solution and, of course, keeping the solution in a sealed bottle to reduce oxygen issues? Ferric ions will certainly oxidize iron, with ferrous ions being the result.

@Ed V: I like that approach to keeping the ferrous as Fe+2. Why have I never heard of it before? We put desiccants in solvents to keep them dry. Sodium in benzene. Iron in ferrous sulfate solution sounds great!

We used to do it in labs I was in. One little problem: iron powder is readily available, and I even used to have two 5 pound bottles of it. But high purity iron is a bit harder to find. We used small pieces of high purity iron wire: just drop a clean piece in the solution.


text:
How can platinum function as an "inert electrode" while having a standard reduction potential of 1.2 from its 2+ oxidation state?
In lecture I was taught that, in a galvanic cell at standard conditions, platinum is often used as an inert electrode when the species being oxidized or reduced exists only in solution, because platinum is "inert," which means it is very stable at and resists reduction or oxidation. This concept confuses me for two reasons:

Reduction means to lose charge, a.k.a. gain an electron(s). When the electrode has an electron passing through it in order to be transferred to another material, isn't it being reduced, albeit for a very miniscule amount of time? So, if we were to pause the cell during electron transfer and examine the iron wire and platinum electrode, we'd see a charge of -1 on both of them, correct?

I Googled the SRP of platinum, and this link states that $\ce{Pt^2+ -> Pt}$ has an SRP of $\pu{1.2 V}$. This translates to an "oxidation" potential of $\pu{-1.2 V}$. This one confuses me because, if we were to use a platinum electrode during the oxidation of $\ce{Mn^3+  +e^--> Mn^2+}$, which has an SRP of $\pu{1.54 V}$ (An "oxidation" potential of $\pu{-1.54 V}$), wouldn't the platinum electrode be the one to oxidize, as it has the higher value?

One concept that I'd like to imagine is that inertness is more about a lack of change in properties of the element as it reduces or oxidizes rather than it is about the "inability" to oxidize or reduce. This still doesn't explain how platinum can function as an electrode for species with a lesser standard reduction potential, though.

There is no way to oxidize fluoride in water solution. Water gets oxidized to O2 instead.

Hint: what is the electrochemical window in your reaction medium, i.e. what electrochemical potentials are experimentally accessible without destroying your electrolyte?

To comment 1: Thanks for the tip, I added oxidation of fluoride for the sake of example in asking the question; if there is an ion with a higher SRP than platinum with both states existing only in solution, let me know, I'll edit to use that as an example.
To comment 2: I think this comment is meant to hint that I should be using a different electrode if its SRP is lower than the oxidizing electrolyte's?

Keep in mind that a redox potential is thermodynamic quantity. It does not say if related reactions go fast, slow or without noticeable change at all.

Upon all, the  inertness  is a relative term here.

$\ce{Mn^3+ + e- -> Mn^2+}$, $E^\circ = \pu{1.54 V}$.

With high concentration of chlorides, Pt and Au can be anodically dissolved like Pt + 6 Cl- -> [PtCl6]^2-, + 4e-    or Au + 4 Cl- -> [AuCl4]- + 3e-. Otherwise, water oxidation is preferred.


text:
"HBr is donating a proton to LiOH": Why doesn't it produce LiOH2?
In the section Brønsted-Lowry acid and base theory from Khan Academy:

In the reaction $\ce{LiOH(aq) + HBr(aq) -> H2O(l) + LiBr(aq)}$, $\ce{HBr}$ is donating a proton to $\ce{LiOH}$, thus acting as a Brønsted-Lowry acid. Since $\ce{LiOH}$ accepts the proton, it is acting as a Brønsted-Lowry base.
Image source: https://i.stack.imgur.com/fjV8R.png

I don't understand how $\ce{HBr}$ is donating a proton to $\ce{LiOH}$, because I think it would produce something like "$\ce{LiOH2}$", but the reaction gives $\ce{LiBr}$.
So where is my misunderstanding?

If you see something that needs improvement, please don't just down vote it, and kindly leave a comment so that I can pay attention to the insufficiency. Thank you

You need to understand there is no LiOH nor HBr in water solution as molecules. LiOH is ionic compound , HBr is strongly polar compound, both fully dissociation to respective ions. So the reaction is just a reaction of the B-L acid H3O+(aq) and B-L base + OH-(aq).

The quoted text describes chemistry in a non-traditional language. It could be totally understandable if you kept to certain limits, but then, there's one effect that needs different terminology, and the whole language falls apart. It's like using poetry to describe mathematical operations.

I think it would produce something like "LiOH2,"

You are missing the charge. It should be something like $\ce{LiOH2+}$, which is an unconventional way of writing $\ce{Li+(aq) + H2O(l)}$.

So where is my misunderstanding?

You have to get used to interpreting the (aq) designation after ionic compounds. To help you out, here is the equation written with the ions shown as individual species (more typing, but closer to the actual situation):
$$\ce{Li+(aq) + OH-(aq) + H+(aq) + Br-(aq) -> Li+(aq) + H2O(l) + Br-(aq)}$$


text:
Reactivity of unsaturated organolithium compounds
I'm particularly looking in the context of alkyl lithiums compared to vinyl-lithiums and alkynyl-lithiums.
Why is it that shifting from sp3 to sp2, and even further to sp carbanions, reduces the basicity and allows for nucleophilic attack to occur preferably?
I understand that a greater s-character means the negative charge is held closer to the nucleus and therefore further stabilised, decreasing its drive to obtain a proton, but wouldn't this also decrease its drive to attack an electrophile?

The C-Li bond is so reactive in alkyl lithium compounds, that they tend to form aggregates except in solvents with donor atoms (like THF). This aggregate only exposes the alkyl groups to the outside, and buries the nucleophilic Li-C bonds inside. My guess is that this type of aggregation does not happen in alkynes, so they are more anionic in character, and their nucleophilic part is exposed.

Basicity decreases but acidity increases when going from $\mathrm{sp^3}$-$\ce{C-H}$ to  $\mathrm{sp^2}$-$\ce{C-H}$ to $\mathrm{sp}$-$\ce{C-H}$ regardless of it is an organo-lithium compound or not.

@ShoubhikRMaiti aggregation does occur in $sp^2$ and $sp$ substrates, for example allyl lithium is highly aggregated ($n>=10$). My guess would be that there is a slight drop in nucleophilicity as the charge is better stabilised since it is held in a lower energy orbital, but the change is very small the change in nucleophilicity isn't really noticeable. All organolithium compounds are very reactive so a small drop/increase in reactivity isn't easily noticeable.

@H.Linkhorn I don't know if allyl lithium can be really considered sp2, because if you consider the C-Li bond covalent, then it is sp3. Also I found this paper: https://www2.chem.wisc.edu/areas/reich/papers/Reich-2013-CR-113,7130-Lithium-aggregates-Review.pdf . It seems that sp2 (vinyl lithium) and sp (lithium acetylides) do aggregate, but the number is low, usually dimer or tetramer. So, I guess my original point still stands. There's more aggregation in alkyl substrates, usually n=4,6,8 etc.


text:
Origin of the D1 line of sodium
According to the fine structure formula for exact energies (Sommerfield formula)
$$E_{jn} = -m_\mathrm{e}c^2\left[1 - \left(1 + \left[\frac{\alpha}{n - j - \frac 1 2 + \sqrt{\left(j + \frac 1 2\right)^2 - \alpha^2}}\right]^2\right)^{-1/2}\right],$$
the energy depends only on $n$ and the absolute value of the total angular momentum $j.$ In the absence of magnetic field, there are two spectral lines for the sodium atom $\mathrm{D1}$ and $\mathrm{D2}.$ I am confused about the origin of the $\mathrm{D1}$ line since it corresponds to a transition $3P_{1/2}\rightarrow 3S_{1/2},$ both states have total angular momentum $j=1/2.$ However, these should be degenerate according to Dirac theory.
What am I missing here? Is the $\mathrm{D1}$ line purely due to electron-electron repulsion inside the sodium atom?

This link says spin-orbit coupling is the source of the splitting.

The equation you give is for a one-electron atom, if you have more electrons, such as in sodium, the potential is no longer purely Coulombic and the levels split. Note that in hydrogen this also happens due to QED effects and this gives rise to the famous Lamb shift.

I have seen that, but since the spin-orbit coupling, along with other effects, arises from the non-relativistic limit of Dirac's equation, it looks wrong to me to say that spin-orbit coupling is the origin of the D1 line. The exact solution to Dirac equation, as seen above, predicts that the 3S1/2 and the 3P1/2 are degenerate. That's why I am confused

Thanks, Paul. I think you're right, but I just wanted to check, since I did not find this written in the quantum mechanics textbooks I know. I believe the Lamb's shift should be of the same order as the hyperfine structure, right? If so, the origin of the D1 line is due to electron-electrion repulsion.

More or less. More precisely, it is caused by a phase shift in the electron wave function as the electron scatters from the ionic core. The magnitude of this phase shift is related to the quantum defect in Rydberg's formula.

OK, thank you. I didn't know about that. Is this related to the fact that the nucleus is an extended object rather than a point-like positive charge?


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How to learn about hydrogen fuel cells
Problem: I am a physics and engineering major going to work for an engineering company this summer. My boss wants me to do a comprehensive review of hydrogen fuel cells and report back on the cutting-edge advancements and potential future advances. I tried checking out some papers on hydrogen fuel cells on the web of science, but there are more words I don't understand than ones I do.
Question: What chemistry should I learn to understand hydrogen fuel cells better? (I have taken one college chemistry class) Obviously, I can't become a chemist in a few months, but I'm wondering if it is possible to learn enough to understand what others are doing from an abstract level.
Any advice is welcome including books, papers, or telling me to give up!

I'd suggest to open relevant Wikipedia page and look at the References and Categories sections at the very bottom.

You don't have a sufficient background in chemistry and material science to do a worthwhile job. I suggest talking with your boss upfront before starting this too ambitious project and be sure that appropriate expectations are set.

I would familiarize myself with and master the following:

The actual concepts of oxidation and reduction (OIL RIG - Oxidation Is Loss, Reduction Is Gain of electrons)

The electrochemical series
Which metals are more and less prone to oxidation/reduction than hydrogen

Half equations using electrons

The Galvanic cell and electrolytes

Electrolysis of water (without sodium chloride)

Hydrogen fuel cell concept

More advanced details like membranes, electrolytes, materials for electrodes etc.

Dedication to the topic should see you up and running in less than a month.
I'm no expert, but it sounds like you need enough electrochemistry to get going. A good introductory book (although not right up to date) is Understanding Batteries by the Royal Society of Chemistry. Not a page tuner by any means, but has the basics.
Other sources are any search for electrochemical hydrogen fuel cell.
Good luck.


text:
Rusting of iron pipes
I have a basketball support pole that is 5" in diameter welded from iron pipes that supports the hoop. There is a vertical component welded to a horizontal one that supports the backboard and hoop. The pole developed two rust holes approximately 1 1/2" in diameter in the horizontal section. Based on responses I have seen, rust cannot form without the presence of moisture and oxygen. Therefore, I sprayed an expansive foam to occupy the air voids within the pipe (only in the horizontal portion where the rust holes are). I sprayed a permatex rust neutralizer on the exterior location of the two holes. I plan on closing the holes with J B Weld (a very strong 2 part resin). Lastly, I will paint the poles with Rustoleum protective enamel paint. The thought is to seal the pipe from air and slow or discontinue the rusting process. I even considered spraying the pipe before painting with the rubberized Flex Seal to completely encapsulate the pipe. This may not be necessary but your opinion would be helpful. Please comment on the statements I have made in terms of accuracy and procedure. Thank you for your thoughts in advance.

Some DIY pages might be more useful than this site. It seems you are planning carefully and I guess your pole will last

You might try https://diy.stackexchange.com/ , since the question is not simply about chemistry per se. For example, can water/air enter under those coatings? What has been the actual experience in such repairs?


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What would be the product (B) in the following sequence of reaction?

Here is how I approached this:
As we know, alc. KOH results in dehydrohalogenation of the compound which takes place as an anti elimination. And then, we're simply conducting halogenation of the compound which would result in a dibromo substituded product.

Where I'm facing problems is deciding the final orientation of the product. How will we represent the product in the fischer representation as the initial reactant is given in its fischer representation as well.

For reference, here's the product (B) given in my book:

A quick way to solve this particular question would be to notice that the first reaction is anti elimination, and the second is anti addition, so you can simply replace the atoms being eliminated (H, Br) with those being added (Br, Br).

@AshishAhuja Oh. So in a way we can say that the first anti elimination then again an anti addition sort of balance out and it appears as if it was a syn addition?

Yes, something like that. Also, there are tricks for figuring out the final product (since there are only four cases, cis,trans + syn,anti) without actually drawing the newman projections which you can use in objective examinations, which would make solving such a question a very fast process, unlike the newman approach.

This is a fun one, and a common question in this course.  First, as you say, H and Br are lost by anti elimination, since they helpfully tell you base and E2 are involved.  That means the H and Br are lost from opposite faces of the future double bond, leaving us with a trans intermediate that has a mirror plane (the plane of the molecule)

Now we need to add Br+ to make a bromonium intermediate.  Notice that both carbons end up as (R) or else both end up as (S), so two enantiomers are possible here.

But, whichever carbon is attacked by Br- ends up with an inversion of symmetry, so you now have two carbons of opposite chirality.  The result, as shown in your answer, is a molecule with a mirror plane.  In other words, adding bromine to opposite sides of a trans double bond makes both sides come out as mirror images, once you turn the bromines around to line up.  This compound is a meso compound.  But ... what about the other isomer if the Br+ originally goes to the other side of the double bond?  Well, just turn your Fisher structure upside down. :)


text:
Should Birge–Sponer plots always have a negative gradient?
I have a set of data I would like to use the Birge–Sponer extrapolation with, but the data is the wavenumbers of the bands in the vibrational-electronic spectrum of a diatomic molecule for $v' = 0, 1, 2, 3, 4, 5, 6.$
I thought the $y$ axis should have the wavenumber values given as they are (in $\pu{cm^-1}),$ but the values given increase with $v'$ giving a positive gradient of the graph.
Every plot I have seen has had a negative gradient. Have I interpreted what should go on the $y$-axis wrongly?

Yes, they should have a negative gradient. The energy gap between each level decreases as the vibrational quantum number increases. Maybe your data is for the energy difference between $v=0$ and $v$ as opposed to the difference between the levels?

Yes! I apologise if this seems a basic question. I am finding this field quite tricky- how should I think about converting this to the energy gap between the levels?

Don't apologies, we're all here to learn!!! To get the energy gaps between each of the levels you will need to subtract the values to leave you the difference. I would recommended drawing a diagram. If you need more help I can write a longer answer explaining how to get each value.

Let me know when you've got your full plot all looking good :)

Yay! I think I have it. It certainly looks nice and gives sensible values. Thank you again for your help, I wouldn't have understood this properly at all without you.

Let us continue this discussion in chat.


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Why is the hydrolysis of ATP to AMP more energetic than the hydrolysis of ATP to ADP?
I am confused as to why the hydrolysis of ATP to AMP and pyrophosphate is more energetic than hydrolysis to ADP and inorganic phosphate, especially since pyrophosphates are more unstable than inorganic phosphates.
My only guess was that the pyrophosphate undergoes further hydrolysis to inorganic phosphates due to its instability and this drives the reaction forward via Le Chatelier's principle, but I think my logic is flawed. I would like to know the factors which cause this difference of -2.6 kcal/mol between the two reactions.

Dr. Larry Moran (Professor Emeritus in the Department of Biochemistry at the University of Toronto) has posted an nice article about the Free Energy of ATP Hydrolysis. Accordingly, some important concepts in biochemistry may be widely misunderstood and/or not well described in most textbooks. One of them is the free energy of ATP hydrolysis:

The traditional standard Gibbs free energy changes do not apply in biochemical system because there are $\ce{Mg^2+}$ ions, ionic strength, and $\mathrm{pH}$ involve in the transformation. The traditional standard Gibbs free energy changes are at $\pu{25 ^\circ C}$ $(\pu{298 K})$ and $\mathrm{pH} \ 7$. For reactions like ATP hydrolysis, we want a new "standard" that includes free $\ce{Mg^2+}$ ion concentration of $\pu{10^{-3} M}$ and an ionic strength $(I)$ of $\pu{0.25 M}$ (Ref.1).
The transformed Gibbs energy (denoted as $G'$) is defined by the following Legendre transform of the Gibbs energy $G$:
$$G' = G - n'(\ce{H+})\mu(\ce{H+}) - n'(\ce{Mg^2+})\mu(\ce{Mg^2+}) \tag1$$
where $\mu(\ce{H+})$ and $\mu(\ce{Mg^2+})$ are the specified chemical potential of $\ce{H+}$ and  $\ce{Mg^2+}$ at equilibrium, respectively, and $n'(\ce{H+})$ and $n'(\ce{Mg^2+})$ are the total amount of these ions (free and bound) in the system. Accordingly, following are the calculated new values for specific reactions:
$$
\begin{array}{l|c}
\text{Biochemical reaction} & \Delta G'_\text{hydrolysis} \ (\pu{kJ mol-1})  \\
\hline
\ce{ATP + H2O -> ADP + P_$i$ + H+} & -32  \\
\ce{ATP + H2O -> AMP + PP_$i$ + H+} & -45  \\
\ce{AMP + H2O -> Adinosine + P_$i$ + H+} & -13  \\
\ce{PP_$i$ + H2O -> 2P_$i$} & -29  \\
\hline
\end{array}
$$

References:

Robert A. Alberty, Robert N. Goldberg, “Standard thermodynamic formation properties for the adenosine 5'-triphosphate series,” Biochemistry 1992, 31(43), 10610–10615 (https://doi.org/10.1021/bi00158a025).


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Struvite crystals: how do they form?
I'm in general chemistry, and I was wondering how ionic species crystallize and form the shapes they do, specifically in regard to the struvite precipitate $\ce{MgNH_4PO_4. 6H_2O}$.
Deconstructing the ionic species, we get $\ce{NH_4^+}$, which does not have a dipole moment (I think), $\ce{PO_4^{3-}}$, which also does not have a dipole moment, $\ce{H_2O}$, which has its dipole moment pointing towards the oxygen, and $\ce{Mg^{2+}}$, which is just an ion floating in solution.
Drawing the Lewis structures of these molecules and writing the Lewis structures, I'm not really sure how the crystal would form. But I do have a guess.
I feel as if the $+1$ formal charge on the nitrogen in $\ce{NH_4^+}$ will be attracted to one of the $-1$ formal charges on the oxygens in $\ce{PO_4^{3-}}$, while the magnesium ion will take in some of the free electrons on the other two oxygens. The $\ce{H_2O}$ will then form hydrogen bonds with the net dipole moment generated?
But this all sounds so hand-wavy to me, and I know it's probably wrong. Furthermore, I can't really visualize what this would look like. I honestly don't have a good understanding of how things work with crystallization. This is, after all, my first semester of chemistry.
If someone could give me tips and hints on how a struvite unit arranges its ions via electrostatic interactions to make something stable, I would appreciate it. Resources and pages in textbooks would be appreciated too!

Have you searched for the crystal structure of struvite?

I looked at this article briefly: https://www.sciencedirect.com/science/article/pii/S0045653517319653 but I didn't understand the diagram given.

The abstract notes that "conditions that result in struvite formation are highly dependent on the ionic compositions, temperature, pH, and ion speciation characteristics." Unfortunately, that is way beyond any first semester general chemistry course. Maybe you will be able to find, with lots of searching, a crystal structure figure showing the unit cell. But you cannot get it by just thinking about it and using electrostatic interactions.

To get an idea of how apparently simple compounds can have unexpectedly complicated structures, check out copper sulfate pentahydrate: https://chemistry.stackexchange.com/a/133955/79678. These kinds of things are what make chemistry to interesting: there are surprises everywhere and we get to study them!

An old Acta Cryst. publication about struvite is here, its animated motiv here.  There are multiple entries on COD like this about struvite as such and other minerals now known as struvite type (COD entries 2021185, 2106462, 2218903, 2220514, 4000911, 9007573, 9007674, 9010847, 9011168, 9011199; just adjust the address accordingly) if you know to display .cif files (e.g., Jmol).

As Ed V points out, the conditions of formation of struvite have many variables and the struvite crystals produced are variable. On the one hand, this could be a pleasant exercise of playing with atomic models, but on the other, it may be unproductive unless you were to use computer graphing to model various configurations.
Perhaps you could start with a simpler, but similar compound and develop a crystallization mechanism from that. Ammonium phosphate seems simpler, but isn't, because the third proton of H$_3$PO$_4$ is so weak that pH is still a significant variable, and you are also dealing with diammonium phosphate and double salts (Ref 1).
Another possibility is plain magnesium phosphate, but it appears in several forms and with several degrees of hydration, so it is not a simple crystal formation. (Ref 2) The anhydrous form is too unrelated to struvite. (Ref 3)
Stretching your imagination further, replacing the ammonium with potassium gives an interesting material that has applications in cementing. One article gives a simple picture of the "components" of KMgPO$_4$.6H$_2$O:

The octahedron of Mg(H$_2$O)$_6$$^+$$^2$, the tetrahedron of PO$_4$$^3$$^-$, and the sphere of K$^+$ are objects that could be played with to make a model crystal. The sphere could also suggest how the NH$_4$$^+$ ion (just assume it is almost spherical) fits into struvite. (Ref 4)
Struvite was the basis for a rapid repair patch for roadways some years ago (Ref 5), and has current applications in immobilizing radioactive waste. (Ref 6)
Ref 1. https://en.wikipedia.org/wiki/Ammonium_phosphate
Ref 2. https://en.wikipedia.org/wiki/Magnesium_phosphate
Ref 3. https://byjus.com/chemistry/magnesium-phosphate/
Ref 4. https://pubs.acs.org/doi/10.1021/acs.jpcc.7b01396#
Ref 5. US Patent 3,960,580 J.M. Gaidis et al.
Ref 6. https://www.sciencedirect.com/topics/engineering/magnesium-potassium-phosphate


text:
In a pair of diastereomers, which of the following is true?
While going through my book, I came across this question and got really confused. Here's what the problem was:

In a pair of diastereomers,

Both can be laevo-rotatory or both can be dextro-rotatory
Both can be optically inactive
One can be optically active other can be optically inactive
Presence of two chiral carbons is must in both of the isomer

As you can see, this is a multiple correct problem.

I tried going with the definition, Diastereomers are stereoisomers whose molecules are not mirror images of each other
Also, to satisfy this, I believe there should at least be two chiral carbons. So option (4) seems right. But I'm really not able to figure out the others. How can we determine this?

You can have chirality around heteroatoms e.g. Esomeprazole

This might be easiest to think about if you picture a polyol, like glycerol or xylitol, but even longer.  Every carbon except the ones on the end has a chiral center.  But can you make more than one stereoisomer with a mirror plane that is optically inactive?  Can they rotate light the same direction?  It should be easy to think about from there.

I think all the options are correct. The definition directly leads to all of them.

They’re all true except for 4. You don’t need chiral carbons for optical activity. Other heteroatoms will work. And even molecules with restricted rotation around a signal bond can show optical activity.


text:
Coordination number of 3D close packing
Can anyone explain me that how the coordination number of corner sphere of this 3-D packing is 6 ? it should be 3

The 3D packing is considered infinite, so there  is no corner sphere.

As @Poutnik said, the packing is infinite, so there are no corner spheres.
Consider it this way-

Each sphere is in contact with four other spheres in it's own layer, as shown-

Also note that each sphere is in contact with one sphere on the layer just above it and one sphere on the layer just below it. A total of two additional spheres thus.

Hence the co-ordination number is 4+2=6.

As @Poutnik mentioned and @Meta xylene confirmed, the packing is infinite, so there are no corners. In case you have any difficulty in understanding this, consider it this way: the coordination number will be the maximum number of adjacent spheres...So now if you take the apparent corner sphere, the number of adjacent spheres is 3, taking one on the edge, the answer will be 4 and for a central sphere, it will be 6. The maximum out of 3,4 and 6 is obviously 6, so that is the coordination number. (Just remember that this is just a simple hack, not the correct theory or explanation!)

Here, the first figure is Square Close Packing and the second one is Hexagonal Close Packing of spheres in 2D. Now, stacking two or more of the square close packing (AA type) one on top of the other, we get ccp, which stands for cubic close packing (AAA type)

Consider a sphere in ccp, it will have 6 adjacent spheres, i.e, one on the top, one on  the bottom, and 4 in the same plane (see it as North, South, East and West of the sphere you are considering maybe) [the spheres are marked in the square close packing figure]
So, the coordination number is 6. I hope my answer helps you, the "hack" I mentioned worked for me and is really simple! :)

Please note that spheres diagonal to each other are NOT considered adjacent. Also, we have only taken into account Cubic close packing here...


text:
Ring Expansion in a bicylic compound
The following compound undergoes ring expansion and contraction simultaneously.
I am not able to figure out a clear mechanism for the same, and stuck for a lot of time to get this clear. Kindly help me out with the same by using appropriate mechanism or models.

My Approach: I could think of the following. A carbocation would be formed after the protonation of -OH group and after it leaves. Now carbocation would not be able to directly rearrange at the bridging carbon. So probably the bond between the bridgehead carbon and the adjacent carbon breaks and a new bond is formed (ring contraction) leaving behind a "3 degree" carbocation at the original bridgehead.

Now this is what I'm facing problem understanding:
i) Mechanism of ring expansion using the newly formed carbocation.
ii) What provides energy for the bond to break and the ring to contract based on the mechanism I suggested earlier (does carbocation stability provide enough energy?). Also is my proposed mechanism even correct ?

This may help.(Ending part of the first sub topic)


text:
What are the limitations of the D, L system and the (+), (-) system for naming organic compounds
In organic chemistry, both the D, L system and the (+), (-) system are widely used, in conjunction with the modern (R, S) system.
The D, L system names an organic molecule based on the direction that its bottom -OH group points at, when the molecule is drawn with the most oxidized carbon on top in the Fischer projection. Is it true that organic molecules that cannot be drawn and analyzed in this way cannot be named according to this system? Also, is this system incapable of specifying diastereomers and epimers among stereoisomers, as it was first invented to address enantiomers?
As for the (+), (-) system, I think as long as an organic molecule does not have optical activity, it cannot be named in accordance with this system. Does this sound right?

Well many rules and terms are specific for a class. Actually is the aim, in a way. You won't use ciclo for a linear molecule. The +/- signs refers, on the other hand, to the actual optical activity.

+/− is only designed to work for chiral molecules. Saying that it can't work with achiral molecules is true, but pointless, because it was never meant to do so. A bigger problem is that the sign of optical rotation tells you nothing about the chemical structure.


text:
Photochemical Bromination of Cyclohexenes
I am still very confused with the bromination of alkenes. Can I ask help in finding the structure of the main product when each of these 3 are subjected to photochemical bromination? I know for a fact that the Br will replace an H atom, but I don't know what to do anymore if there's another functional group present at one/both ends of the alkenes.

This is a radical driven process, so you need to identify the most stable radical intermediate. Note that allylic radicals are pretty stable.

All 3 of the compounds are unsaturated so it is also worth noting what the source of bromine is so we don't see electrophilic addition.


text:
Entropy between elements and compounds
I have encountered this reaction when I am learning entropy:

I don't understand why there is an increase in entropy (said to be slightly larger than 0).
What I am thinking is:

both reactants and products are in the same states, so there isn't any change in entropy in terms of states.
both the reactant and products have the same number of moles i.e. the same number of molecules, which means in a constant volume, means the 'disorderliness is the same throughout the reaction, hence the entropy.

According to my lecturer and the internet, it is said that "Pure substances tend to have lower entropy than compounds" and "the pure substances are arranged 'tidier' than its compounds", which I don't understand why, especially the meaning of 'tidier'.
Therefore, I would like to ask what makes this reaction has an increase in entropy, and how can I determine that?
Thank you.

Have you considered change of entropy due exchange of heat ? dS=dQrev/T. Entropy change due structural changes adds to this one.


text:
Color of Flame Test for Alkali Metals
Lithium is known to have the highest ionization energy among Group 1 elements. Also, characteristic colours in the flame test arise due to the excitation and de-excitation of electrons. Then why Lithium gives Crimson Red colour in the Flame test, why not a colour corresponding to a higher frequency like Purple or Blue colour even though the highest amount of energy is involved in the excitation and de-excitation of electrons of lithium when compared to the energy involved in the same process for other Alkali metals?

Nobody pretends that the color of the flame is due to the transition from the first excited state to the ground state. It may be due to a transition between higher excited states of the Lithium atom.

@Maurice Does that mean in the case of metals like Caesium the electron can excite to higher orbits and in the case of Lithium the electrons get excited to comparatively less energetic orbits and hence the observation?

Look up “Grotian diagram” and “energy level diagram”. Sodium is a good place to start. The lithium one is also easy to find, so happy searching!

Instead of «de-excitation», you may use «relaxation», too.

The point is that the emission has nothing to do with ionization. Simple as this. Ions are involved in test flames, but the colour never involves ion-atom transition. It is a transition X* —> X that does not involve a change in charge/oxidation state.

See: https://chemistry.stackexchange.com/questions/50467/why-lithium-gives-flame-coloration


text:
Can the enrichment of Uranium be done via manipulating the boiling/melting point difference between differing isotopes of the same element?
I have a Physics question for which I could not find the answer on the Internet, so I would like to see if y'all know the answer to it.
I have found that heavy water will have a different boiling point compared to light water, and this trend is present between various isotopes of an element. Based on this research, I have hypothesized that Uranium can be enriched using the procedure below.

Make UF6 (s)

Sublime it in a glass tube

Lower temperature with such precision, that one isotope of Uranium still remains in the gaseous form while another isotope has turned back into a solid.

Separate the gas and solid without disrupting the fragile temperature balance.

Although this link is for Germanium isotopes, it states that there is a 0.15 Celcius change of melting points between isotopes. https://www.sciencedirect.com/science/article/abs/pii/S0921452618300863
So, here are my questions.
Would this apply similarly to Uranium, and how would the use of UF6 affect this boiling point difference?
How exactly does the boiling/melting point change between differing isotopes? And is there a particular formula for finding the melting point for a particular isotope? (I assume it would take in the number of neutrons and some other info)
In addition, is it possible to adjust the temperature of the entire system with decimal grade precision?
Furthermore, would the above setup work at all?
Also, if UF6 is too sensitive, would melting Uranium metal in a kiln with decimal precision temperature adjustability work?
Finally, is there a good, reliable source for melting point differences between isotopes?
I would like to first say that this is all purely theoretical and that I will take no action upon the information obtained; it is all merely a thought experiment to enhance my understanding of Chemistry and Physics. I mean, it's not like I'm gonna make a nuclear bomb; I'm just a normal HS student in the US. If you still think that this information might be dangerous out here, please PM me so we might work something out, like meeting via Zoom/Skype/Phone or something; I am very flexible about that.
I hope y'all have a great day.

Note that there is a separation method that uses an IR laser to do the separation, but obviously, the details are classified.

Since your question is basically about distillation, you might get more practical advice from [chemistry.se] rather than here at [physics.se]. We can migrate the question for you if you'd prefer to have it only in one place. If you end up posting your question on both sites, it's polite to link each post to its partner and to explain why you might hope for different advice from the different audiences.

@rob Yes, that would be wonderful, thank you very much!

@DavidWhite While that is true, enrichment via lasers(AVLIS, or SILEX for uranium) requires extreme precision. That is why I was seeking an alternative method, and it seemed like boiling/melting point differences seemed slightly more manageable than the 0.01nm difference between the excitement energy differences between U-235 and U-238. Thank you for your contribution, and I hope you have a nice day! I think you will also find this link entertaining.

@DavidWhite https://en.wikipedia.org/wiki/Atomic_vapor_laser_isotope_separation

During WWII, there was used gaseous diffusion seperation technique for UF6, with the diffused part of the gas flow enriched by 235UF6.

@Poutnik This is a very good point. However, I am very curious if this mass difference(and consequently, phase change temperature difference) can be manipulated not just over a series of diffusion filters, but also via careful manipulation of the system's temperature within one or two glass tubes. If this were possible, it would have the effect of making isotope enrichment much more accessible. Thank you very much, and have a great day!

It is possible, but the separation would be less effective, that is why it was not used. It would not be question of careful temperature manipulation, but of very many repeated evaporation and condensation steps. The gaseous phase would be just very slightly enriched by 235UF6.

@Poutnik So would the process be as follows?

@Poutnik 1. Create UF6 and sublime in glass tube.    2. Oscillate the temperature rapidly between a very small interval somewhere around 2 degrees Celsius between the phase changing temperature of UF6.     3. Create some sort of process using a microprocessor that will let out a much higher percentage of solid U-238 than solid U-235, by letting out initial solid UF6 while trapping later solid UF6.     4. Repeat until needed.

Of course isotopes may be separated by distillation, but it's useful, AFAIR, only for hydrogen isotopes.
For uranium, gas centrifuge is proven much cheaper, to some orders of magnitude: https://ru.wikipedia.org/wiki/Газовая_центрифуга
(translate this page, it contains more information than English Wikipedia article).
The essence of this technology is known for decades, and included in basic chemistry textbooks. Only engineering/technological know-hows are sort of classified.
But in the politics, mass media, and perhaps on stackexchange, the whole topic is still considered taboo and is surrounded by fear.

Comments are not for extended discussion; this conversation has been moved to chat.


text:
Can secondary alkyl halides be used in the Corey–House synthesis?
I've been told by my teacher that the Corey–House synthesis cannot proceed with secondary alkyl halides; they will prefer to undergo elimination over $\mathrm{S_N2}$ substitution. [used as the pseudo halide, not the diorganyl cuprate.]. My book, however says that secondary alkyl halides do undergo the reaction. A quick google search on Wikipedia says:

[...]RX, R = methyl, benzylic, allylic, primary, or cyclic secondary alkyl, aryl, or alkenyl

These seemingly varied responses have really left me rather confused. Is the applicability different for different secondary halides? If yes, why? And if no, then which is the correct answer?
Secondly, my teacher also tells me that in R-X, using Br, Cl, or F as the halogen will not produce a satisfactory yield. Again, Wikipedia goes on to say:

[...] X = Br, I, OTs, or OTf; X = Cl is marginal.

Again,  could anyone help me out in resolving these differences? I've never really had the opportunity to see these reactions in the lab, so I feel that someone with real experience would be able to help me out.


text:
Why do fluoride molecules seem to have abnormally low boiling points?
As far as I know, to non-polar molecules like CF4 and C2H6,
boiling point is mainly affected by the London dispersion force.
As a result, molecules with more protons and electrons normally have higher boiling points.
However, CF4 (34 electrons) has a boilng point of -128°C, while C2H6 (18 electrons) has -89°C.
Also, I find it puzzling that fluorides tend to have abnormally low boiling points.
 For example (the below are all boiling points of  compounds with a benzene ring)
C6H6 80.1 °C
C6F6 80.2 °C
C6Cl6 322 °C
C6Br6 418 °C
C6I6 531 °C
 It seems to me that fluorides are the abnormal ones. Despite having a lot more electrons than benzene, the boiling point is almost the same.
Is there something I'm missing?

See this question and the various links therein.


text:
How do I write a balanced redox reaction for the oxidation of chalcocite [(Cu2S) (s)] by O2 (g) to form Cu2+ (aq) and SO42- (aq)
I'm not exactly sure how to get the oxidation and reduction half reactions. Any help will be much appreciated!

Do you know what is the oxidation states of Cu and S in the reagents?

Apparently I'm supposed to arrive at this answer: Cu2S +5/2O2 +(2H+) gives 2Cu2+ + SO4^2- +H2O

I'm just not sure how to write the reduction and oxidation half reactions to arrive at the above given answer. Its a question from the book ''Soil and water chemistry'' by Michael Essington

I think that the half-reaction method for this reaction would look something like this:

$\ce{Cu2S + O2 + H2SO4 -> CuSO4 + H2O}$

$\ce{2Cu^+ -2e^- -> 2Cu^2+}$ (oxidation)
$\ce{S^2- + 4H2O - 8e^- -> SO4^- + 8H^+}$ (oxidation)
$\ce{O2 + 4H^+ +4e^- -> 2H2O}$ (reduction)

$\ce{2Cu2S + 5O2 + 2H2SO4 -> 4CuSO4 + 2H2O}$

I think  that the following method would look parallel to the method  mentioned in the above post:

$\ce{4H2O + Cu2S + H2SO4 -> 2CuSO4 + 10H+ + 10e^- }$ (oxidation)
$5(\ce{Cu2S + 3O2 + H2SO4 + 2H+ + 2e^- ->2CuSO4 +  2H2O})$ (reduction)

$\ce{6Cu2S + 15O2 + 6H2SO4 -> 12CuSO4 + 6H2O}$

$\ce{2Cu2S + 5O2 + 2H2SO4 -> 4CuSO4 + 2H2O}$


text:
Getting an “index” of 7 for a set of powder diffraction data
I tried to index powder diffractogram from the following problem:

A metal (of density $\pu{0.862 g cm^-3})$ gives the following powder X-ray diffraction data $(\lambda = \pu{1.5418 Å}).$
$$2\theta \qquad 23.6^\circ \quad 33.6^\circ \quad 41.5^\circ \quad 48.3^\circ \quad 54.4^\circ \quad 60.1^\circ \quad 65.5^\circ $$

I calculated $\displaystyle d = \frac{\lambda}{2\sin\theta}$ and compiled a table:
$$
\begin{array}{cclccc}
\hline
2\theta/^\circ & d/\pu{Å} & \frac{1}{d^2}/\pu{Å^-2} & \frac{1}{d^2}/Z & h^2 + k^2 + l^2 & hkl \\
\hline
23.6 & 3.769754932 & 0.07036776606~(= Z) & 1 & 1 & 100 \\
33.6 & 2.667180594 & 0.1405708123 & 1.997659159 & 2 & 110 \\
41.5 & 2.175894723 & 0.2112147493 & 3.001583838 & 3 & 111 \\
48.3 & 1.884256436 & 0.2816566342 & 4.002637145 & 4 & 200 \\
54.4 & 1.686509508 & 0.3515785836 & 4.996301621 & 5 & 210 \\
60.1 & 1.539473673 & 0.4219446333 & 5.996277229 & 6 & 211 \\
65.5 & 1.425021033 & 0.4924446812 & 6.998157094 & 7 & \color{red}{?} \\
\hline
\end{array}
$$
I know a value of $h^2 + k^2 + l^2 = 7$ is not allowed, but I cannot understand where I am going wrong with the indexing here. Any insight would be greatly appreciated as I am wondering if this is made up data that has a typo?

Assuming metal = element (which may be wrong), and this table here, did you give potassium a trial (COD, a = 5.247 A), bcc?  And, what is the source of this table you cite?

I'm not sure there is any problem as the unit cell parameters were assumed to be identical, such as in the case of a cubic space group. How about in the case of a tetragonal space group?

@z1273 Three nitpicks: #1 a space group (typically one of the 230 standard settings, like $P2_1$, or $P2_1/c$) may belong to a crystal class (like for the two mentioned, both monoclinic). #2 The tetragonal class is of lower symmetry than the cubic one. It is possible to describe anything in monoclinic $P1$, the lowest symmetry.  But even then, one strives to identify evidence for higher symmetry in the data (some computer programs proceed this way early / lately, e.g. Platon's ADDSYM).

@z1273 #3:  It is possible the experimental data leading to the above mentioned COD .cif entry differ to the (maybe constructed) diffractogram of the above question differ.  If however the above mentioned .cif is feed, e.g. into CCDC's Mercury, after adjusting for this approx Cu K$\alpha$ radiation, then the position of the predicted diffraction peaks along $2\theta$ are very close to the ones predicted.  (Only, that the diffraction indices don't match the ones in the OP ...)  By this & the proximity of the density stated in the OP and in the .cif (calculated Rx density of 0.859), K is likely.

@Buttonwood These are many interesting bits of information, but I'm not sure of their relevancy to the question at hand. I am also leery of coincidence, such as the stated density matching K's figure. It seems to me a≠c is a simpler explanation.

@z1273 True, #1 and #2 are not required here, address a meta level.  And on occasion, with the information given, there may be more than one solution.

TL;DR: The data from the assignment is correct.
You confused yourself by assigning $h_i^2 + k_i^2 + l_i^2$ to the ratio of $\sin^2\theta_i$ values, and keep in mind $(100)$ might be absent.

First, let's get bold and assume that we are indexing a cubic system.
This is usually the first step in analytical indexing, plus it's an extremely common system for metals.
Combining Wulff–Bragg's condition
$$2d_i\sin\theta_i = n\lambda\tag{1}$$
with the relationship between the quadratic form of the relation between interplanar spacing $d_{hkl}$ and the cell parameter $a$
$$\frac{1}{d_i^2} = \frac{h_i^2 + k_i^2 + l_i^2}{a^2},\tag{2}$$
the following relation between the scattering angles and Miller indices is evident:
$$\sin^2\theta_i = \frac{\lambda^2}{4a^2}\left(h_i^2 + k_i^2 + l_i^2\right)\tag{3}$$
Since $h_i, k_i, l_i \in \mathbb{Z}$ and $\lambda^2/(4a^2)$ is a constant, normalized series of $\sin^2\theta_i$ ratios corresponds to the $h_i^2 + k_i^2 + l_i^2$ sums.
Your mistake was that you took normalized $d_i^2/Z$ values (by the way, $Z$ is the number of formula units in crystallography) as $h_i^2 + k_i^2 + l_i^2.$
$$
\begin{array}{cccccccc}
\hline
i & 2\theta/^\circ & d_i/Å & \sin^2\theta_i & \displaystyle\frac{\sin^2\theta_i}{\sin^2\theta_1} & h_i^2 + k_i^2 + l_i^2 & hkl & a_i/Å \\
\hline
1 & 23.6 & 3.77 & 0.042 & 1 & 2 & 110 & 5.33 \\
2 & 33.6 & 2.67 & 0.084 & 2 & 4 & 200 & 5.33 \\
3 & 41.5 & 2.18 & 0.126 & 3 & 6 & 211 & 5.33 \\
4 & 48.3 & 1.88 & 0.167 & 4 & 8 & 220 & 5.33 \\
5 & 54.4 & 1.69 & 0.209 & 5 & 10 & 310 & 5.33 \\
6 & 60.1 & 1.54 & 0.251 & 6 & 12 & 222 & 5.33 \\
7 & 65.5 & 1.43 & 0.293 & \color{red}{7} & 14 & 321 & 5.33 \\
\hline
\end{array}
$$
As you can see, we have a number seven again which is … good news!
Not only our initial suggestion to go with the cubic system was right (integer relations), but we also stumbled upon $\sin^2\theta_7/\sin^2\theta_1 = 7$ thanks to the author of the assignment who was kind enough to give us at least seven peaks.
As you have rightfully suggested, it is impossible to find the sum of the squares of three integers for the number seven, but we don't have to — these are ratios, and can be multiplied by two.
At this point I pre-calculated cell parameter
$$a_i = d_i\sqrt{h_i^2 + k_i^2 + l_i^2}\tag{4}$$
to find the average from each signal, but thanks to the rounding error that wasn't necessary, and $a = \pu{5.33 Å}.$
Multiplying by two is equivalent to assuming that the first reflection in the diffractogram is obtained as a result of x-ray scattering from the atomic plane with indices $(110).$
At this point any number can be represented as the sum of the squares of three integers and we populate the table to the end with Miller indices.
Now thanks to that seven we also know what out of three cubic systems we have exactly: body-centered cubic (bcc) with $Z = 1 + 8 \cdot 1/8 = 2:$

This means we can check ourselves and deduce what metal has been analyzed by finding its molar mass $M$:
$$
\begin{align}
M &= \frac{\rho a^3N_\mathrm{A}}{Z} \\
  &= \frac{\left(\pu{0.862 g cm^-3}\right)\left(\pu{5.33E-8 cm}\right)^3\left(\pu{6.02E23 mol^-1}\right)}{2} \\
  &\approx \pu{39.3 g mol^-1},\tag{5}
\end{align}
$$
which leads us to believe the metal we analyzed was potassium with the density $\pu{0.89 g cm^-3}$ as reported by Wikipedia, which in my opinion is close enough to the value of $\pu{0.862 g cm^-3}$ from the assignment.
Final note: keep track of significant figures.
You are not allowed to present values with ten sig figs calculated from the scattering angles given with three sig figs accuracy.
“Indexing is more an art than a science” —A. Le Bail [1], aka the quote that crystallographers love to insert into presentations and emails for no reason:)
Reference

Bail, A. L. Monte Carlo Indexing with McMaille. Powder Diffraction 2004, 19 (3), 249–254. DOI: 10.1154/1.1763152.


text:
D-orbitals of point group C2v
I'm trying to construct an MO diagram for cisplatin, which has C2v symmetry. However, there is no Mulliken label for the d(x^2-y^2) orbital in the character table. Does this orbital not form molecular orbitals in C2v symmetry?

X^2 and Y^2 lie in A1, so linear combinations of those do, too.


text:
Why can you sanitize paper money notes with a Clothes Iron?
I have some Canadian and British bank notes. The first issue is that I don't know if they're paper or plastic.
Second, I don't understand this answer. Why's it safe  to iron paper bills? Won't the heat or pressure ruin the paper bill?

You can iron paper bills, like the American ones, on rather high heat. Just the same iron you use for your clothing, at the hottest setting, but no steam. It will kill all viruses.
Plastic bills can be hand washed, with the same soap you would use on your hand, you can dry them the same way you dry your hands, with a clean cloth or even the hot air dryer you use for your hands. Or you can let them air dry.
These bills can withstand some heat but I have heard not enough to iron them clean.

Ever hear of a book titled “Fahrenheit 451”? Paper money in the USA has to meet stringent requirements, by design, so a quick ironing, without steam, will not damage it. Of course, this is a testable matter: iron a one dollar bill. As for polymer currencies, I would not do that: too risky. But, again, this is testable.

@EdV Even this got an entry on wikipedia...

For the same reason that a laser printer doesn't ruin its printouts - paper can withstand the heat and pressure. Viruses can't

ps; the corona situation has people doing some crazy things.. but i suppose if one is sticking rolled up paper notes up one's nose for some reason, one would want them to be virus free, I'm sure!

In short: This is possible because «paper money» actually is not referring to the paper like paper from the office supply for (your) desktop printer or xerox, but to cotton paper.
Because of the cotton, if folded and forgotten in a pocket of your jeans close to $\pu{40 ^\circ{}C}$, they may physically survive a pass in the washing machine with a mild non-bleaching detergent (including the spinning).  They obviously suffer from the dye if e.g., the jeans blead, or mechanically if you try to unfold them after if passing a dryer.  Of course different to the ones made of polymers as (but not limited to) in Australia.

Among other things, USA paper money has to withstand submersion in 5% NaOH solution for 24 hours.

While I don't let my cloths this long in the tide (TM), it is good to know.

Honestly, "may physically survive a pass in the washing machine" seems like an understatement.  Dollar bills go through the wash just fine.


text:
Finding a Mistake Calculating the pH of Ammonium Acetate
I'm trying to calculate the $\mathrm{pH}$ of a $\pu{0.10 M}$ solution of ammonium acetate, and I am struggling to find what mistake I am making within the following calculations:
We know that $\dfrac{[\ce{NH3}][\ce{H^{+}}]}{[\ce{NH4^{+}}]} = K_\mathrm{a} = 5.6 \times10^{-10}$, and that $\dfrac{[\ce{CH_3COOH}]}{[\ce{CH3COO^{-}}][\ce{H+}]} = K_\mathrm{b} = 5.56\times10^{-10}$, and so with an ICE chart, we can find that
$$\dfrac{x(x-y)}{0.1 - x}\approx\dfrac{x(x-y)}{0.1} = K_\mathrm{a}\text{, and }\dfrac{y}{(0.1-y)(x-y)}\approx\dfrac{y}{0.1(x-y)} = K_\mathrm{b}.$$
Solving these equations, we find that $x = \sqrt{0.1\times K_\mathrm{a}\times (0.1\times K_\mathrm{b} + 1)}\approx 4.48\times 10^{-6}$ and $y = x - \dfrac{0.1}{x}\times K_\mathrm{a} \approx 4.16 \times 10^{-16}$, so that $\mathrm{pH} = -\log{(x-y)} = 5.13$.
It should be that $\mathrm{pH} = 7.00$, and so where am I making a mistake?

Perhaps look here: https://chemistry.stackexchange.com/q/75047/79678.

What you have marked as the acetic acid Kb has about the correct value, but does not match the concentration expression on the left. That expression is equal to 1/Ka, while Kb=Kw/Ka.

As @Ed V pointed out in his comment, answer for this question can be found here.

This was answered in the comments, but these second equilibrium expression should not actually be a $K_b$ value. Rather, it should be $1/K_a = 55600$. Using this value in the same calculations, we find that $x=5.5805\times 10^{-4}$ and $y = 5.5795\times 10^{-4}$, so that $\ce{pH} = -\log(x-y) = 7.00$.


text:
Why is the reported melting point of formaldehyde so different from the triple point?
All of the (believable) sources at PubChem and most of the sources at NIST give 181 K for the melting point.
The one source I can find for the triple point is NIST, which gives 155 K.
There are two possibilities I can think of, but I can't confirm or deny either:

The triple point value is wrong.
The density of the solid is very different from the density of the liquid, as suggested by the Clapeyron equation.

The densities of two of the solid forms, oxidane and paraformaldehyde, are quite different from the liquid density, but it doesn't look to me like they form spontaneously by simple cooling. Maybe I'm wrong about this.
What is the true reason for the difference?
Edit: To clarify, I understand that the triple point has two values, one of pressure and one of temperature. I'm not so interested in talking about the pressure of the triple point of formaldehyde unless it helps answer my question, and that's why I didn't mention it at first.
I think my question is interesting because in the data I've seen so far, the difference between the temperature value of formaldehyde's triple point, and the temperature of its melting point, is much greater than that of any other compound.
Maybe I've been extremely unlucky and formaldehyde's difference actually isn't unusual, and the compounds I've reviewed are the unusual ones. This would also be interesting to learn.

The triple point is one point in P-T space. The melting point is given at P = 1 atm. So the better question would be at what pressure and temperature is the triple point.

The 1 atm melting point is usually quite close to the triple point temperature regardless of the triple point pressure. Are you suggesting a third possibility, that the triple point pressure is extremely low?

“Usually”? Helium’s triple point is about 0.03 atm if I recall properly.

The 1 atm melting point is usually quite close to the triple point temperature. That means the melting point temperature at 1 atmosphere of pressure is usually close to the triple point temperature. For example, helium's 1 atm melting point and triple point temperature are 3 degrees apart. The differences for water, formic acid, acetic acid, ethanediol, formamide, dimethyl sulfoxide, methane, ethane, all chloromethanes, ammonia, dioxygen, carbon monoxide, carbon dioxide, and carbon disulfide are all about 1 degree. Formaldehyde and ethanol are the only counterexamples I've found so far.

So yes, "usually".

One of the sources at your NIST link says the melting point is 156.15 K. One of the sources for 188 is "anonymous", and another is from 1901.  Is there a chance this data is for, um, more than one kind of formaldehyde?


text:
How does the relationship between carbonate, pH, and dissolved carbon dioxide work in water?
Fish keepers use a table like the one below to determine dissolved $\ce{CO2}$ by observing the color of a pH indicator in water with a known concentration carbonate solution (typically 4 degrees KH ~= 70ppm carbonate).
I'm curious where the equation in the top of the picture comes from though.  Below is my reasoning for wondering where that equation comes from and knowing would (I think) answer all my questions below too.
It appears to me that the equation doesn't work with pure water (water lacking carbonate).
I can think of a few reasons for this:

The equation implies that either pure water cannot hold any $\ce{CO2}$
$\ce{CO2}$ concentration and carbonate concentration are not truly independent
Depending on how the algebra was done, it could be that pH isn't well defined without the presence of carbonate.
It's an approximation that doesn't work everywhere.

My gut says its #4 or #2 but then the table would be kind of worthless unless $\ce{CO2}$'s contribution to carbonate is low relative to the amount added to solution.  If this is the case, I wonder where does the table become inaccurate.

Hello, welcome to Chemistry! You can use the $\ce{...} command to typeset chemical formulae correctly: see https://chemistry.meta.stackexchange.com/q/86/16683. We also try to avoid MathJax in the title, unless absolutely necessary.

What is a degree KH ?

@Maurice See KH and dGH. It is unit of German origin from old good times of "real" chemistry, equivalent to 10 mg CaO / 1 L, as bicarbonate ( the former ), or totally ( the latter). It was widely used in 20th century, until replaced by mmol/L.  1 mmol/L Ca/Mg is equivalent 5.6 degrees of KH ( bicarbonates only) or  dGH, It is still used by many aquarists or old school chemists in water treatment.

$^{\circ}KH$ or $dKH$ is called (degree of )carbonate hardness, but it is usually about bicarbonates $\ce{HCO3-}$, not carbonates $\ce{CO3^2-}$ . Carbonates are present in water just in traces, unless it is e.g. highly alkaline water from soda lakes of East Africa rift (like where cichlidae from lake Malawi live).
There is the equilibrium for carbon dioxide solution acidity, keeping the expression below constant:
$$K_\mathrm{a1,H2CO3^{*}} = \  \frac{\ce{[H+][HCO3-]}}{[\ce{CO2(aq,total}]}=10^{-\mathrm{pH}}\cdot \frac{\ce{[HCO3-]}}{[\ce{CO2(aq,total}]}=const \cdot 10^{(7-\mathrm{pH})}\cdot \frac{KH}{\mathrm{ppm}  \ce{CO2}},$$
where [] denotes the molar concentration.
Therefore $$\mathrm{ppm}  \ce{CO2} = \frac{\mathrm{const}}{K_\mathrm{a1,H2CO3^{*}}} \cdot 10^{(7-\mathrm{pH})}\cdot KH \approx 3 \cdot 10^{(7-\mathrm{pH})}\cdot KH$$

Pure water in equilibrium with air has theoretical $\mathrm{pH=5.6}$, given by dissolved $\ce{CO2}$ from air. There is established equilibrium
$\ce{CO2(g) <=>[H2O] CO2(aq) <=>[H2O] H2CO3(aq) <=> H+(aq) + HCO3-(aq)}$, but all components are present just in traces.

Addressing feedback:
Carbonate hardness implies there are dissolved bicarbonates $\ce{Ca(HCO3)2}$ resp. $\ce{Mg(HCO3)2}$. These are determined indirectly by titration, that determines bicarbonate content.
This of course fails, if there is sodium bicarbonate or carbonate presence, or if all bicarbonate comes from $\ce{CO2}$. Both cases have carbonate hardness lower or zero, compared to what would say its measurement.
In fact, measuring carbonate hardness ( dKH ) is measuring alkality:
$$\ce{HCO3-(aq) + H3O+(aq) -> 2 H2O + CO2(aq,g)},$$
assuming it is equivalent to content of $\ce{Ca(HCO3)2}$ or $\ce{Mg(HCO3)2}$. But it can be aslo sodium carbonate/bicarbonate, if added to water to increase alkality.

pH of water is determined by this equation:
$$\mathrm{pH} = 6.35 + \log {\frac{[\ce{HCO3-}]}{[\ce{CO2}]}}$$
both in molar concentration.
Calcium carbonate has solubility about 15 mg/L, what makes roughly 1 dKH.
$$\ce{CaCO3(s) <=> Ca^2+(aq) + CO3^2-}$$
$$\ce{CO3^2-(aq) + H2O <=>HCO3-(aq) + OH-(aq)}$$
This solubility increases in presence of dissolved carbon dioxide,which decreases concentration of carbonate:
$$\ce{CO3^2-(aq) + CO2(aq)  + H2O <=> 2 HCO3-(aq)}$$
There is direct relation between pH and relative ratio carbon dioxide : bicarbonate : carbonate.
pH ( in usual aquarium range ) is directly determined carbon dioxide / bicarbonate ratio. If pH change is forced  by other chemicals, the new ratio is established.
BTW, there are just 2 degrees of freedom. You choose 2 of 3 parameters. The third one is determined by the former 2.

So, the $K_H$ for $\ce{CO2}$ is 29.4 atm/M.  Assuming there's 400ppm in the atmosphere $\ce{CO2}$ that's 400ppm/29.4 M or 0.6ppm at equilibrium.  so 0.008 dKH of (bi)carbonate came from the dissolved $\ce{CO2}$.  If I add enough carbonate to make pure water 1 dKH and leave it out in the atmosphere does that mean i'll get 1.008 dKH when it reaches equilibrium?  or does the added carbonate change the the rates that aqueous $\ce{CO2}$ becomes $\ce{H2CO3}$?

After writing the previous comment/question.  I think I figured out what disturbs me.  Perhaps the "correct" question to ask is: the equation has 3 degrees of freedom (pH, [carbonate], and [$\ce{CO2}$]).  Aquarists add a set amount of carbonate, and measure pH and assume the carbonate doesn't change.  This clearly isn't exactly right and that can be seen in the case of pure water exposed to $\ce{CO2}$.  If we know the amount of carbonate prior to exposure to $\ce{CO2}$ can we compute the new carbonate and dissolved $\ce{CO2}$. ie are they truly dependent on pH ?

See the answer update.

The equation is CO2 (ppm) = 3 x ∘KH x 10^(7.0 - pH)

1 mol/L CO2 = 44 g/L = 44000 ppm by weight
dKH is 0.17832 mmol/L MCO3
10^-pH is approximately [H+]

So the equation above can be rewritten as 44000 [CO2] = (3 x 1E7 x 0.17832 x 1E-3) [HCO3-][H+], or
[HCO3-][H+]/[CO2] = 8.2 mol/L
That's just the equilibrium constant for H2O + CO2 = H+ + HCO3-
If you compare a professional-level calculation, you'll see that this is fairly close to K1 on page 147, because ''most'' of the CO2 is converted to carbonic acid. (K0)  But we don't really need to split that into two reactions - the thermodynamics is the same whether we know all the intermediates or not.  Nor do we need to worry about whether the solution was originally made with carbonate vs. bicarbonate, because we're simply measuring the pH, though of course if that pH is high things could get more complicated (see the part about K2), especially for the fish.