text:
Reference for electronegativities of different metal oxidation states
A long time ago I was researching the effect of the self regulatory response in Fe and Co. I found that my results made sense based on the idea of the electronegativity of the ions considered. I found a  webpage that listed the relationship between the different electronegativities for different oxidation states of Fe and Co.
My problem is I didn't save the webpage and I don't know where I found it. I am not a chemist so I don't know in which kind of books I could find a list of values of electronegativities for different oxidized ions.
It basically said that the relationship between the electronegativities was something like
$$\ce{Co^2+} < \ce{Fe^2+} < \ce{Fe^3+} < \ce{Co^3+}$$
Note: I don't remember it well, so what I wrote could be lies. The important thing was that there was a flip of order in electronegativities for Fe and Co when changing oxidation state.
Does anyone know where I could find that kind of information? I tried googling, but I am not finding the webpage and don't remember the terms I used to find it in the first place.
I never understood the why of the order either, and I have seen chemist webpages that tell you which ion is more electronegative by just looking at it. If someone could explain that too I would appreciate it.

Edit:
I forgot to add that it was in octahedral complexes. I was able to find this pdf online (page 32) but it doesn't state a reference.

Pearson conveniently lists cumulative experimental data in the 1988 paper [1], referrring to the earlier work of Moore [2].
Selected values of $I$ (ionization potential), $A$ (electron affinity), $χ$ (absolute electronegativity – probably, that's what you are looking for) and $η$ (absolute hardness) for iron and cobalt cations are:

Table I. Experimental Parameters for Monatomic Cations (eV)
$$
\begin{array}{lcccc}
\hline
\text{ion} & I & A & χ & η \\
\hline
\ce{Fe^2+} & 30.65 & 16.18 & 23.42 & 7.24 \\
\ce{Fe^3+} & 54.8 & 30.65 & 42.73 & 12.08 \\
\ce{Co^2+} & 33.50 & 17.06 & 25.28 & 8.22 \\
\ce{Co^3+} & 51.3 & 33.50 & 42.4 & 8.9 \\
\hline
\end{array}
$$

So it looks like the relation is a bit different:
$$\ce{Fe^2+} < \ce{Co^2+} < \ce{Co^3+} < \ce{Fe^3+}$$
Complete table as a screenshot:

References

Pearson, R. G. Absolute Electronegativity and Hardness: Application to Inorganic Chemistry. Inorganic Chemistry 1988, 27 (4), 734–740. https://doi.org/10.1021/ic00277a030.
Moore, C. E. "Ionization Potentials and Ionization Limits"; Natl. Stand. Ref. Data Ser. (U.S. Natl. Bur. Stand.); 1970, NSRDS-NBS 34. (NIST - PDF)

Thank you so much! The odd thing is that it is contrary to what I remember finding. I think because my case was applied to octahedral complexes but I didn't really know if it made a difference so I didn't mention it in the question. I was just able to find a resource online. I will edit the question to add the image. I will accept your answer later today if no one else answers for taking the time to do so. Sorry for not putting all the information straight away. Hope you can clarify maybe with just the origin of the difference.

@M.O. No prob, I'm glad you found the data you were looking for. AFAIK the values for $χ$ are determined spectroscopically, so, yes, the coordination environment may affect the order since the difference in energies for the same oxidation number is relatively small.


text:
Why does internal rotation of ethane give an energy spectrum asymptotically approaching a rotation based on $\nu=1$ rather than $\nu=0$?
I'm a physicist with very little knowledge of chemistry. Recently a chem colleague left a molecular model in my classroom, so I brought it back to her, and as I was walking down the hall with it, a hydroxy group started spontaneously swinging around in circles. This made me curious about whether such motions could actually be observed. After asking around and doing some web searching, my understanding is that this sort of thing is called an internal rotation, and usually there is a barrier to rotation, which is modeled by a potential $V(\phi)=(1/2)\sum V_n(1-\cos n\phi)$.
If I'm understanding the history correctly, then originally people thought that the spectrum would be that of a free rotor, but that produced incorrect thermodynamic predictions. Then in 1937 Pitzer published a famous paper, "Thermodynamics of Gaseous Hydrocarbons," in which he showed that you could get correct thermodynamic predictions if you put in a barrier to rotation.
The classic textbook example seems to be ethane. I found this explanation  of the physics (di Lauro, "Floppy Molecules," in Rotational Structure in Molecular Infrared Spectra, 2013), which I only partly understand. He gives the following theoretical spectrum of energy states:

So at low energies we have roughly a set of vibrational energy levels with energies $\hbar\omega (\nu+1/2)$, which makes sense to me because I guess we're sitting in a minimum of the potential where the geometry is staggered. He says, "The high levels, on the contrary, correlate and eventually almost coincide with the free internal rotor eigenfunctions, whose energy increases parabolically as $AK_i^2$, where $K_i$ is the eigenvalue of $J_γ$, in ℏ units." This also sort of makes sense to me, because if you put in a lot of energy, you shouldn't see the effect of the little hills and valleys in the potential $V(\phi)$, so it should act more like a free rotor.
Do the units of inverse cm mean $1/\lambda$, or $2\pi/\lambda$? If the former, then $1/\lambda= 200\ \text{cm}^{-1}$ corresponds to have $\hbar\omega_0/k\approx 300$ kelvin, so it makes sense that the barrier to rotation would have significant thermodynamic effects at temperatures on the order of room temperature or below, since the rotational-vibrational modes are not available. Am I getting this right?
But why is it necessary to place this fictitious rotational band on top of the $\nu=1$ state, rather than $\nu=0$? I would think that the fictitious band should be build from the ground state, it wouldn't match the actual energies at low energies, and then at high energies they would match up.

Wavenumber units are cycles-per-cm, which I think corresponds to your $1/\lambda$. So, if you're using $\omega_0$ to represent the wavenumber value, you would need to use $h\omega_0c\over k$ to calculate that representative temperature. By my math $200\mathrm{cyc\over cm} \rightarrow 288\ \mathrm K$.

The rotor potential is cyclic which means that levels in each potential well interact with one another and become split. In addition there is quantised angular momentum both about the C-C axis and perpendicular to this which leads to further sets of energy levels. The unique axis is defined by quantum number $K$.  Look for spectroscopy of symmetric  top molecules. To me it is unclear what the levels in the figure refer to. ($ k_BT \approx 208, \mathrm{cm^{-1}}$ at 300 K)

@porphyrin: To me it is unclear what the levels in the figure refer to. (kBT≈208cm−1 at 300 K) I think that rotational band is supposed to be a fictitious one that he is saying gives the right energy levels in the limit of high spin.

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text:
Nice textbook illustration of microwave spectrum for a nearly free internal rotation?
Is it possible to find a nice textbookish example of experimental data for an internal rotation of a molecule, in which the rotation is nearly free? For example, 1-butene is supposed to have a nearly free rotation about its middle bond.
It would be fun, for example, to be able to show a microwave spectrum for this molecule that demonstrates the evenly spaced photon energies of a rotor.

I did find some data on NIST. Perhaps you could find more there: https://webbook.nist.gov/cgi/cbook.cgi?ID=C106989&Mask=80. Look at "Data at other public NIST sites." I was personally disappointed by the apparent lack of a nice visual spec for it, but perhaps I'm looking in the wrong place.

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text:
400–430 degrees Celsius heated bath
I want to perform a reaction which needs the temperature to be maintained between 400–430 °C. Since it's the only practical solution, I opted for a heated bath.
I'm looking for a hydrocarbon fraction/blend with the boiling point in this range, however, I keep finding conflicting information. If anyone knows such a hydrocarbon product please, let me know.

Those temperatures are well above the flash point of any hydrocarbon I can find, and above the autoignition point of many. Might you be able to get by with a sand bath or tightly-regulated mantle instead?

The point of a heated bath is that the temperature, given sufficient heating, will be exactly the boiling point of the medium. Sand bath would offer no control over the temperature.

Also, to rig such a mantle on ones own would be no small feat. Not to mention being able to afford one.

Keep in mind there is also a variety of molten salt baths; however, you'll need a thermocouple and protect the glassware as the majority of molten salts would etch the glass.

What's wrong with carrying out your reaction in a furnace?

If it doesn't specifically need to be a hydrocarbon, zinc metal melts at 419.5 °C. Could you do an "ice bath" of zinc chunks in molten zinc, maintaining the melt right at its melting point?

I have to say this is an interesting idea. I'll give it a good thought.

In fact, if it works it will be a superior solution, since I won't lose the zinc to the atmosphere the same way I would lose the liquid. I'll let you know how it went. Thank you!

Woods metal can used for this  https://en.wikipedia.org/wiki/Wood%27s_metal

Word of warning -- this idea just came to me out of the blue. I'm not a practitioner. I have no idea if a solid-liquid zinc bath is something that's routinely done; if not, there may be good reasons why not.

You can try $\ce{C30}$ hydrocarbon, triacontane (CAS #: 638-68-6). It is a straight chain hydrocarbon with formula weight of $\pu{422.8 gmol^{-1}}$ ($n$-$\ce{C30H62}$). It is a solid at $\pu{25 ^{\circ}C}$ and NIST Webbook reported its boiling point as $\pu{722.9 K}$ ($\approx\pu{450 ^{\circ}C}$) at $\pu{1 atm}$. More phase changing data can be found in [NIST Webbook] and following reference:

J. S. Chickos, W. Hanshaw, “Vapor Pressures and Vaporization Enthalpies of the n-Alkanes from $\ce{C21}$ to $\ce{C30}$ at $\mathrm{T} = \pu{298.15 K})$ by Correlation Gas Chromatography,” J. Chem. Eng. Data 2004, 49(1), 77–85 (DOI: 10.1021/je0301747).

The vapor pressure would be significant at 400C (a mere 50C below boiling point), even if it didn't start to carbonize. Those vapors probably would ignite.

@Kevin Kostlan: I know the risk. I was thinking about a salt bath to be honest with you. But OP insisted that "he's looking for a hydrocarbon fraction/blend with the boiling point in this range" and then this'd be the answer.


text:
Two cells are connected in series. If x g of silver are deposited in cell 1, what volume of oxygen is given off in cell 2?

I got the answer correct but I'm pretty sure my methodology was wrong. How would you solve such a question?

Silver plates out, i.e. is reduced, at the cell 1 cathode according to
$$\ce{Ag+(aq) + 2 e- → 2 Ag(s)}$$
while oxygen is produced at cell 2's anode via the electrolysis of water
$$\ce{H2O(l) → 0.5 O2(g) + 2 H+ + 2 e-}$$
Since the cells are in series, we know that every electron that flows through cell 1 must flow through cell 2. This allows us to create a relationship between the moles of silver plated and the moles of oxygen evolved. Specifically,
$$\ce{H2O(l) + 2 Ag+(aq) → 2 Ag(s) + 0.5 O2(g) + 2 H+}$$
That is, for every 4 moles of silver that plate out, we evolve 1 mole of oxygen gas. The rest is algebra to give us answer A:
$$V(\ce{O2}) = \frac{\pu{22.7 L/mol}\cdot x}{4\cdot \pu{108 g/mol}}$$

“every electron that flows through cell 1 must flow through cell 2.“ Not really, different electron flow through the middle wire than through the outer wires. It is one to one though.

Trying to keep it simple, but I get your point. However, electrons from the the far right wire could certainly end up in the middle wire.

Electrons from one wire end up in the gas phase (hydrogen) and from the other end up in the solid phase (deposited silver). Of course there could be some exchange reactions that we can’t measure, electrons being indistinguishable.

Ah, true! Forgot about the gaseous species.

@Tunk Note that it's better to provide the final answer (e.g. having "algebra" done) since according to the current policy hints or incomplete answers are more suited for the comment section (see Hint answers revisited and answer to How do I ask homework questions on Chemistry Stack Exchange?).

@andselisk $x$ was not specified.

Thank you so much! I actually did use a similar method but after looking up the oxidation of water, because we aren't given it and I don't know how they'd expect us to know that off the top of our heads.

You can solve by using the Faradays second law: use the following expression:
$$\frac{\text{mass of}~\ce{Ag}}{\text{equivalent mass of}~\ce{Ag}} = \frac{\text{mass of}~\ce{O2}} {\text{equivalent mass of}~\ce{O2}}$$
$$\frac{x}{108} = \frac{\text{mass of}~\ce{O2}} {8 }$$
so :mass of $\ce{O2} =\frac{x}{108}\times{\pu{8 g}}$
and the amount of $\ce{O2} = \frac{x}{108}\times{\frac{\pu{8 g}}{\pu{32g/mol}}}=\frac{x}{108}\times{\frac{\pu{1 mol}}{4}}$
$$V(\ce{O2}) =\frac{x}{108}\times{\frac{1}{4}}\times{\pu{22.7 L}}$$
Another method for solving the problem :by using the balanced chemical equations calculations and the  Faraday first law  as the following :
1-  Write the two half cell reactions ,and make the electrons gained equal to the electrons lost:
$$\ce{4Ag+_\mathrm{(aq)} + 4 e- → 4 Ag_\mathrm{(s)})}$$
$$\ce{2H2O_\mathrm{(l)}  → O2_\mathrm{(g)} +4 H+_\mathrm{(aq)}  + 4 e-}$$
2- Calculate the amount of electricity needed to depsite $\pu{x g}$ of silver($\pu{\frac{x}{108} mol}$ of siver atom) in  the 1st cell,by creating a relationship between the moles of silver plated and the amount of electricity flowing in  the first cell as the following :
\begin{align}
\ce{&4Ag+_\mathrm{(aq)} &+ &4 e- &→ &4 Ag_\mathrm{(s)})}\\
&&& \text{4 electrons}&& \text{4 silver atoms}\\
&&&\pu{4 mol} ~\text{of electrons} &&\pu{4 mol}~\text{of silver atoms}\\
&&&\pu{4 Faraday}  &&\pu{4 mol}~\text{of silver atoms}\\
&&&    ?                &&\pu{\frac{x}{108}mol}
\end{align}
so,the amount of electricity flows through the first cell =$\pu{\frac{x}{108}mol}\times\frac{{\pu{4 Faraday}}}{\pu{4 mol}}=\pu{\frac{x}{108}Faraday} $
Because the cells are in series, the amount of electricity flows through the first cell
equal the amount of electricity flows through the second cell=$\pu{\frac{x}{108}Faraday} $.
3- Calculate the amount of $\ce{O2}$ evolved  in  the second cell
by creating a relationship between the amount of $\ce{O2}$ evolved and the amount of electricity flowed through the second cell as the following :
\begin{align}
\ce{2H2O_\mathrm{(l)}  → &O2_\mathrm{(g)} +4 H+_\mathrm{(aq)}  + &4 e-}\\
&\pu{1 mol}     &\pu{4 faraday}\\
&?                  &\pu{\frac{x}{108}Faraday}
\end{align}
so,the amount of $\ce{O2}$ evolved =$\pu{\frac{x}{108}Faraday}\times\frac{\pu{1 mol }}{\pu{4 Faraday}}=\frac{x}{108}\times\frac{\pu{1 mol}}{4}$
Thus,$$V(\ce{O2}) = \frac{x}{108}\times\frac{1\times{\pu{22.7L}}}{4}$$


text:
How can I find the emf of a cell given the individual reduction and oxidation reactions and their respective potentials?
If I am given the individual oxidation and reduction half-equations, how can I find the emf of the combination? For example:
$$
\begin{align}
E^\circ(\ce{Fe^3+}/\ce{Fe^2+}) &= \pu{0.77 V}\\
E^\circ(\ce{Fe^2+}/\ce{Fe^0}) &= \pu{-0.44V}
\end{align}
$$
How can I find the emf of this redox pair, i.e. $E^\circ(\ce{Fe^3+}/\ce{Fe^0})?$

Determine the Gibbs energy of reaction for each against the standard hydrogen electrode, add them up and determine the standard potential from the Gibbs energy of the summed reactions.

You can’t combine the two in a cell because both are reduction half reactions. You could do a disproportionation or comproportionation, maybe.

This is my first answer in stack exchange if some mistake comment on.
$$E^\circ_\ce{Fe^3+/Fe} \neq E^\circ_\ce{Fe^3+/Fe^2+} + E^\circ_\ce{Fe^2+/Fe}$$
but
$$\Delta_r G^\circ_1 = \Delta_r G^\circ_2 + \Delta_r G^\circ_3$$
for the three reactions:
$$\ce{Fe^3+ + 3/2 H2 -> Fe + 3H+}\tag{1}$$
$$\ce{Fe^3+ + 1/2 H2 -> Fe^2+ + H+}\tag{2}$$
$$\ce{Fe^2+ +  H2 -> Fe + 2H+}\tag{3}$$
With $$\Delta_r G^\circ = - n F E^\circ , $$ we can establish a relationship between the reduction potentials:
$$n_1 F E^\circ_1 = n_2 F E^\circ_2 + n_3 F E^\circ_3$$
Cancelling F and solving for $E^\circ_1$ allows us to calculate the reduction potential of the iron/iron(III) half reaction:
$$n_1 E^\circ_1 = n_2 E^\circ_2 + n_3 E^\circ_3$$
$$E^\circ_1 = \frac{n_2 E^\circ_2 + n_3 E^\circ_3}{n_1}$$
$$ = \frac{1 \cdot \pu{0.77 V} + 2 \cdot (\pu{-0.44 V})}{3}$$
$$ = \pu{-0.367 V}$$

Technically, a half reaction does not have a Gibbs energy of reaction. I think the numerical answer is correct, though,

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Please format this using MathJax for easier readability. This page will get you started. Also, thank you for submitting an answer and, of course, welcome to the site! It's always awesome to see a new contributor answering questions!


text:
Surface chemistry
Why doesn't the process of chemisorption form multimolecular layers whereas he physisorption with lesser force of attraction between molecules form multimolecular layer?

Welcome to Chemistry.SE! What are your thoughts on the question? Please include them as well.

Chemisorption involves chemical attachment (electrostatic or covalent bonding) with the surface. Since there are a finite number of such bonding sites on the surface, the maximum occupancy is one monolayer.
Physisorption involves weaker (dipolar or dispersion) interactions, and additional layers might be able to attach to underlying layers via similar weak interactions.
It should not be assumed that additional layers cannot form following chemisorption, simply that these would not be associated by the same mechanism, rather through weaker interactions (physisorption). Such interactions are often termed "nonspecific" and preparation protocols may call for treating a surface to remove molecules associated that way.


text:
What is the explanation of the structure of H2XeO4?
Using the obsolete VSEPR theory I had a hard time predicting the structure of $\ce{H2XeO4}$. So I looked it up. I found the molecule via chemapps.stolaf.edu:

It looks weird to me. I can't explain the molecular structure. Can anyone explain it using more sophisticated theories like molecular orbital theory (MOT)?
I don't know much MOT, yet.

Interesting, do you know where the structural data (that https://chemapps.stolaf.edu uses) comes from? Atkins' Chemical principles: the quest for insight, p. 661 (ISBN 978-1-4292-1955-6) depicts xenic acid as tetrahedron, and to the best of my knowledge there is still no single crystal diffraction data available elsewhere.

I knew that oxo-acids of Xe are very unstable and don't really exist. This is the reason that upon hydrolysis of XeFₓ we get oxides of Xe as products and not oxo-acids.

https://en.wikipedia.org/wiki/Xenic_acid

@andselisk Even chemtube 3d shows the same. It computed (predicted) the structure I guess. http://www.chemtube3d.com/playground/playground.html

No it doesn't show the same! It shows more or less what I expected the oxide groups are at axial. But chemmaps one oxide and other hydroxide.

Please note that the VSEPR theory is not obsolete. It has it's limitations, and this is one of them, but if you are applying it within reason, then it is a very useful model.

When I think about such structure, I draw all valence electrons around the atoms and connect them somehow reasonably. There are two ways to connect the hydrogens to xenon. Either directly bonded as H-Xe bonds, which for me does not feel right here, or as part of OH groups, which also make more sense in the "acid" part.
By connecting everything that way, you end up with:

This let you end up with a lone pair at xenon which is the reason for the non-tetrahedral structure.
A quick and dirty quantum calculation[1] shows what I mean. By calculating the ELF (Electron Localization Function), you can easily see where the electrons are located. The lone pair is shown by the yellow volume surrounding no atom.

This structure is not clear from the Lewis formula shown on Wikipedia, but can be seen in the ball and stick representation that is shown there (don't care about the "flipped" hydrogens):

[1] ORCA 4.1.2, best functional of the planet B3LYP def2-QZVP // Crest 2.6, Visualization with Chimera, iso value 0.8 ... whatever ... camels? ... what unit has the ELF?

It's a probability distribution, so the unit is 1 (or none), and values are 0 to 1. (And B3LYP.... pffft)


text:
What is the significance of antinode at r=0 in the plot of radial wave functions of s-orbitals?

Why should there be an antinode at r=0?
I can comprehend the presence of node at r=0 for p,d,f orbitals; but I don't understand that of s-orbital.
Is there any intuitive approach?

Welcome to Chemistry.SE! We have a site policy for homework questions. Please edit your question to include your attempt at the problem, where you got stuck, and let us know where you're finding difficulty so we may best help you.

possible duplicate of : 2s orbital wavefunction has non-zero probability at r=0?(duplicate) from another site on the network (physics.SE). Interestingly, that question is also a duplicate.

@ANBENZENE please do research (at least a little!) on google before asking a query, it really helps.

A nice little meta.SE post for other users in doubt

Ask yourself: what do ordinate and abscissa in the plots represent? Then follow the logical implications of a value y=0.


text:
Why is the entropy of solidifying solution higher than that of a pure solvent?
According to a source,

Solutions have a greater entropy than pure liquids as the particles in a solution are more separated and solvent molecules separate the solute particles. If the number of moles of gas increases in a chemical reaction (as shown by the stoichiometry of the reaction) then the entropy also increases.

My questions are as follows:
1) How do we know the particles in a solution are more separated ?
2) How do the solvent molecules actually separate the solute particles?
3) Also, we know that the entropy of solid is less than that of a liquid, so shouldn't the addition of solute particle (which are obviously solid) actually lead to lesser entropy than pure liquid?

1) How do we know the particles in a solution are more separated?

When a solution forms, intermolecular bonds in the solid solute are broken and replaced by interactions with the solvent. The solute is then free to diffuse in the solvent. If the solute molecules still interacted in a highly favorable way they would tend to bond and precipitate out. This is exactly what happens when a solute has a low solubility.

2) How do the solvent molecules actually separate the solute particles?

What drives the dissolution process is an overall favorable combination of the change in the energy of intermolecular interactions and motional freedom. This is described by the Gibbs free energy. The solvent provides an environment for the solute to move (diffuse), which increases the solute's entropy, and the solvent forms weak interactions with the solute that make up (perhaps only partially) for the lost interactions when in the solid. In general, there is an enthalpic component (bond-breaking and making) and an entropic component (changes in solute configurational freedom) to the free energy. If the energy of the interactions of a solute with itself in the solid is identical to the interaction energy between solute and solvent in the solution, then the change in free energy is described entirely by the entropic contribution.

3) Also, we know that the entropy of solid is less than that of a liquid, so shouldn't the addition of solute particle (which are obviously solid) actually lead to lesser entropy than pure liquid?

The solid particle is no longer "solid" when dissolved, it is in a new state. The entropy in solution reflects the new state of the solute upon dissolution. The solute does not retain a memory of what it was before. At low concentrations, the solute is diffusing with few other solute molecules nearby. The new entropy is not an average over the entropy of the starting states (solid and solvent).
(That is unless you are referring to a dispersion of solid particles, but that is called a colloidal solution and is another matter.)

I think the first question refers to the separation of the solvent molecules not the solute molecules. The implicit assumption is something close to an ideal solution where each solute molecule occupies the same volume as a solvent molecule.

@Andrew Well, I avoided emphasizing ideality since I don't think it's necessary. I interpret it to mean that the solute is not clustering.

thankyou so much @NightWriter its a great answer indeed !


text:
Why not use all the colours in colorimetry?
Colorimetry (and it's cousin visible-UV spectroscopy) is a form of chemical analysis that works by sending visible light of one wavelength through the sample. It then analyses how much of the light was absorbed in that process so you can determine concentration.
What I don't understand is why only one wavelength is used. Apparently, even for Infrared Spectroscopy, only one wavelength is analysed at one time yet the whole spectrum is needed. Why not send in white light (with all the wavelengths at once)? The output light could be split through dispersion and every wavelength could be analysed for its absorbance. Perhaps some wavelengths of light aren't worth measuring or too many frequencies at once cause interference?

Diode array spectrophotometers work with white light. There are pros and cons to that method.

There is also Fourier-transform (FT) vs scanning instruments.

Why not send in white light (with all the wavelengths at once)?

This is certainly and routinely done today. The main thing is how much price are we willing to pay? In a rigorous sense, colorimetry refers to the fact that we use optical filters to isolate wavelengths rather than monochromators. When the latter are used, you call them spectrophotometers. Basically, the question is referring to scanning instruments versus diode array or multiplexing instruments. Both exist today and both have their advantages and disadvantages.

In scanning spectrophotometers, one wavelength passes through the sample at a time. In diode array instruments, all light passes through the sample at once and it is dispersed later. The dispersed light falls on an array of detectors (thousands of tiny detectors, usually 1024). The detectors do not know what wavelength is falling on them, however, they are calibrated.
Similarly, if you were working in the 1970s-80s you would use a scanning infrared spectrophotometer (one wavelength at a time). One needed a prism made of salt or a very special diffraction grating to disperse the infrared radiation. This made life very difficult. Imagine working with a salt prism!
In Fourier transform infrared spectroscopy, all wavelengths are passed at once, but there is no array of detectors. Unfortunately, there is no detector which can differentiate wavelengths on the basis of their frequency and separate them. We need to modulate them in such a way that an ordinary detector can "follow" the frequency of wavelengths. Here comes in the Michelson interferometer. The raw output does not look like a spectrum at all, but it has all the information "hidden" there. You need Fourier transform to recover a typical infrared spectrum.
Now one may ask why don't we use FT for UV-Vis? The answer is that the required precision of a moving mirror is so high that it is rather difficult to make one.

I'm confused about your last statement. I thought an FT instrument would not require a moving mirror.

No, the interferometer used in FTIR requires a moving mirror. It is a really fancy stuff how it moves on air bearings.

MEMS devices may get into the required positioning precision soon. But the caveats discussed in EdV's and Buck Thorne's answers about the different types of noise stay.

The previous answers by @Buck Thorn and @M. Farooq are very good, but not quite complete. The main advantages of FT UV-Vis spectroscopy would be 1) accuracy of wavelength (or wavenumber) determination and 2) better spectral resolution of bands, relative to conventional UV Vis spectroscopy. But it is very hard to do well, and very expensive.
The disadvantage is that source shot noise, which is white noise, and source 'flicker' noise, which is low frequency noise with power spectral density proportional to $1/f^\alpha$, get modulated by the moving mirror. The modulation frequency for any given wavenumber of light is directly proportional to the light's wavenumber value. But the source noises also get modulated.
The absorbing specimen absorbs the modulated light, as per the absorber's absorption bands, resulting in an interferogram like the one shown in M. Farooq's answer. Then the FT does the demodulation, i.e., 'decoding', resulting in the spectrum of the absorbing specimen. But the modulated source noises also get decoded and cause a reduction in the signal-to-noise ratio (S/N or SNR). The noises on the large spectal bands are the problem: the white noise spreads throughout the spectrum, thereby lowering the SNR and making it harder to resolve the small bands.  The source flicker noise also causes trouble by making 'band shoulders', as it were. Thus, the hoped for advantage of high wavelength accuracy (and improved spectral resolution) gets very significantly compromised by the decreased SNR.
For more details, see E.Voigtman and J.D. Winefordner, “The Multiplex Disadvantage and Excess Low Frequency Noise”, Applied Spectroscopy, 41, 1182-1184 (1987). Also see the comments posted after Buck Thorn's answer.

An explanation is provided in this abstract (1):

Fourier transform spectrometry in the UV-Vis region (FT/UV-Vis),
  because it is source shot-noise limited, has a signal-to-noise ratio
  (S/N) disadvantage in comparison to dispersive spectrometry,
  especially with dense spectra. At the expense of poorer S/N, FT/UV-Vis
  can be satisfactory for high-resolution measurements. However,
  low-resolution spectroscopic studies, such as molecular absorption
  measurements, are not expected to be performed advantageously by
  FT/UV-Vis. The broad, dense spectra with high intensity throughout a
  wide spectral range should show a significant S/N degradation, in
  comparison to results with dispersive spectrometry.

Further explanation of the subject of noise and why FT does not provide an advantage here can be found in course notes made available by U. Delaware Prof. S. L. Neal:

Filters do not always improve the quality of spectral data.
  Measurements that are dominated by noise from the source, e.g.,
  fluorescence, are often distorted by flicker noise.  Whereas white
  noise is frequency independent, flicker noise is larger at low
  frequencies.  In fact, it is sometimes called pink noise (since low
  frequencies are associated with red rather than blue light).
  Filtering techniques are based on the idea that the noise does not, in
  general, consist of the same frequency components as the signal.  In
  the case of flicker noise, this condition is not met.  This is one
  reason that FT-UV-VIS spectrometers are not widely used.  The
  improvement in signal to noise that comes from measuring the entire
  spectrum many times using an interferometer during the time a single
  spectrum can be measured on a dispersive (monochromator based)
  spectrometer is the square root of the number of times the spectrum
  was measured for shot noise limited (FT-IR) measurements.  In other
  words, a ten point spectrum measured by the interferometer has a
  signal to noise ratio that is about three times ($\sqrt{10}$ ) larger than the
  same spectrum measured using a dispersive device because signal
  averaging reduces random noise by the square root of the number of
  acquisitions. (Remember this is called the multiplex or Fellgett
  advantage.)  This improvement is always smaller in the presence of
  flicker noise, and in very crowded spectra, a multiplex disadvantage
  in which the signal-to-noise ratio in the data acquired by the
  interferometer is lower than in the dispersive instrument can be
  observed.

Reference

Mark R. Glick, Bradley T. Jones, Ben W. Smith, and James D. Winefordner
Applied Spectroscopy Vol. 43, Issue 2, pp. 342-344 (1989)
V. Saptari, Fourier-Transform Spectroscopy Instrumentation Engineering, SPIE Press, Bellingham, WA (2003).

Winefordner was a very famous spectroscopist. However, the sentence "source is shot noise limited" is not clear to me. I think he means that the interferogram will be dominated by shot noise. His student, now in his 80s, happened to be a good acquaintance/ coauthor of mine. He is still very active in science and education.

@M.Farooq To be honest I am still trying to understand the argument in the excerpts, which is why I posted them as an answer without further explanation. My understanding is the shot and white (Johnson) noise have similar frequency properties (constant over freq) but one is T dependent. On the other hand, flicker noise has a 1/f dependence and can be limiting in low frequency measurements.

@M.Farooq Pretty cool that I happened accidentally upon one of your colleagues :-) (if in print)

Are you Winefordner's student by chance? It will be a very small world then.

@M.Farooq No, but he sounds exceptional (found a webpage at Florida, https://www.chem.ufl.edu/people/name/james-winefordner/)  :-) And it is kind of a small world... I'll have to look up your work, any recommendations?

His name is Thomas O'Haver (emeritus now). See Google Scholar for recent work https://scholar.google.com/scholar?as_ylo=2018&q=Thomas+O%27Haver&hl=en&as_sdt=1,44&as_vis=1

@EdV Thank you for your comments and for the kind offer to make the article available. As my experience is mainly with high-resolution NMR (mirrorless spectroscopy with white noise) none of this was exactly familiar to me.

@Ed V, it is amazing that Winefordner is still doing well. Tom is also amazing and I feel lucky to have discovered him from a ChemEd list a decade ago. I published three articles with him (writing a fourth one now). I can access online articles if you need one.

@ Ed V,  Thanks. This is a common problem in public forums. It is best to ignore that. You don't know who you are dealing with and the other person does know who he/she is talking to. Yes, I remember that someone was making an argument about LaTeX, however such people think that computer presentation and software is "everything". My response was that far better science was done when we had typewriters (although I never used one for manuscripts).


text:
How can I change the synthetic conditions for powder product into single crystal growth?
I am working on growing metal-organic frameworks crystals based on hexatopic linker for different metals. Upon using some synthetic conditions it gave fast precipitation into powder material after heating for half hour. My question is how to modulate the synthesis to grow single crystals for Zr, Fe, Al in my case? I used metal salt:linker as 10:1 using modulator 100 equivalent to metal salt.

Conditions optimal for seeding crystallization needn't to be the optimum ones for crystal growth.  Attempt to seed slowier (e.g., less concentrated reagents, slower addition) than "crushing out"; warm a little to moderate growth.  If aiming for a crystal structure determination and still less lucky, you still may opt for microfocus instruments (e.g., https://www.diamond.ac.uk/Instruments/Techniques/Diffraction/Microfocus#), electron diffraction (e.g., https://en.wikipedia.org/wiki/Electron_crystallography) or general powder diffraction (https://en.wikipedia.org/wiki/Powder_diffraction).

You, as far synthesis as the common meaning, won't attain a single crystal directly. The fact that you are working with framework complicate things as for it is possible that the standard way of growing crystal won't work. Else normally one prepare diluted solution in clean glassware and wait for solvent evaporation in a vibration free corner of the building. If you are dealing with crystal determination specialist can cut single crystals out of big enough not perfect crystals.


text:
Space group diagram
If you give me the name of a certain space group, for example, the P222 (orthorhombic system). How could I construct the following diagram?

Image source: http://img.chem.ucl.ac.uk/sgp/large/sgp.htm
I know it's probably an ambiguous question. I, in the subject of Chemical Crystallography, have given the concepts of sliding plane, helical axis, among others. But, at the time of making this type of diagrams and giving the coordinates of the special positions or those of Wyckoff.

Here is a rough outline how you could proceed:
Symmetry elements along principal, secondary and tertiary direction
The space group symbol summarizes the symmetry operations in the crystal. In your example, there are two-fold axis in three directions (222). Which directions depends on the crystal system; for orthorhombic, they are mutually perpendicular.
For a tetragonal space group, the principal direction is along the $c$-axis, the secondary direction is perpendicular to it along the cell axes, and the tertiary direction is along the diagonal. So for ${P4_3 2_1 2}$, there is a ${4_3}$ axis along $c$, a $\mathrm{2_1}$ axis along $a$ and $b$, and a 2-fold along the diagonals ($a+b$ and $a-b$).
For a cubic space group, the principal direction is along the cell axes ($a$, $b$, and $c$), the secondary along the body diagonal ($a+b+c$) and the tertiary along the face diagonals ($a+b$, $a+c$, etc).
For the complete set of crystal systems, see here or here for example.
Additional symmetry elements by combining rotational and translational symmetry
In your example, there are additional symmetry elements that don't go through the origin, but cut axes at 1/2. These are from combining translation perpendicular to a 2-fold with the 2-fold operation.
Positions
If you start with coordinates $(x,y,z)$ and apply the 2-fold axes running along $a$, $b$, and $c$, you get the four general positions of the space group $P222$. The coordinates are given as fractional coordinates. For example, if you apply a two-fold along a to the coordinates $(x,y,z)$, you get $(x,-y,-z)$, the second general position.
Now, you can show these in the diagram. The coordinates in $a$ and in $b$ are shown by positioning, while the coordinate in $c$ is shown by $(+)$ or $(-)$ in this case. As you can see, there are 4 circles inside the cell, corresponding to four positions.
Centering
Your example space group is primitive (the $P$ of $P222$), so there is no centering. For space groups with symbols starting with $C$, $I$ or other letters, you would have to add the centering to the general positions, and to the circles in the cell. Combining the centering operation with the symmetry elements given in the space group symbol might give you additional symmetry elements in the cell.
International tables of crystallography
There are a finite number of space groups, and they are all known and tabulated in the international tables. So as an exercise, you could go through each one of them, correlating the space groups symbol with the general positions and the arrangement of symmetry elements.
The International Tables are not available for free online, but are available in many chemistry libraries. Online, you can find a list of space groups at http://img.chem.ucl.ac.uk/sgp/mainmenu.htm, and also a guide to the symbols used for the different symmetry elements. Wikipedia also lists the space groups with short names and full names by space group number and crystal system.

First of all, thank you very much. And, secondly, if you know some online resource in which all the possible elements of symmetry that exist appear. Well, for example, I've seen tables that show things like: P6/mmc. And I don't know what the bar or the c means. By the way, the first link doesn't work. Besides, finally, where can I consult all the tables that exist?  @Karsten Theis

I edited my answer to address some of your questions here. As for P6/mcc, first it helps to put some spaces: P 6/m c c. "6/m" means a six-fold rotation with a mirror-plane perpendicular. "c" means a glide plane (1/2 translation in c), in this case perpendicular to the secondary and tertiary direction.

@aprendiendo-a-programar symmetry@otterbein (http://symmetry.otterbein.edu/) is one useful online interactive source to learn about symmetry operators, both at molecular level as well as with crystals.

@Buttonwood - Thanks for formatting, and cool link!

I'm still making out these diagrams. Especially in calculating the possible coordinates. I do understand the elements of symmetry. @KarstenTheis


text:
Does a non-chiral substrate (2-bromo-2-methylpropane) also form two products via an SN1 mechanism?
If I have an alkyl halide, in which its tertiary carbon is non-chiral (for example 2-bromobutane), which reacts via an SN1 mechanism does it also produce 2 products?
I am asking because I know that the carbocation intermediate's plane of symmetry allows the nucleophile to attack equally well from either side and thus the product is a racemic mixture of enantiomers.  Is this also the case for a non-chiral substrate?

@H.Linkhorn We prefer to not use MathJax in the title field, see here for details.

The formation of the product will happen along both pathways, but since the product is as achiral as the reactant, you won't find different products, and you can't tell.

Not sure what you mean here; 2-bromobutane is undoubtedly chiral (we even have a question on it). If you mean something like 2-bromopropane, or 3-bromopentane, then Martin's comment is spot on.

You are asking about a non-chiral substrate, but the example you chose is chiral. Which case are you asking about? Could you edit your question accordingly?

I am sorry, I meant 2-bromo-2-methylpropane. I edited it out.
I must have confused it with another substrate I was looking at.


text:
Reaction of 1 equivalent of Mg in ether with 1,4-dibromobut-2-yne
What will be the product formed the if the reaction of Mg in ether with 1,4-dibromobut-2-yne occurs? At first, I thought the answer would be cyclobutyne, but the correct answer is buta-1,2,3-triene.
Can anyone please explain how?

Cyclobutyne cannot exist.

Cyclobutyne cannot exist.  There is no way to enforce 90° bends at both ends of a triple bond, but the carbanion formed by the Grignard reagent synthesis finds an easy way around this problem.
Let's say the magnesium reacts at position 1.  Then the negative charge would couple with the conjugated triple bond to spread to position 3:
$\ce{Br\overset{+}{Mg}\overset{-}{C}H2-C#C-CH2Br <-> Br\overset{+}{Mg}CH2=C=\overset{-}{C}-CH2Br}$
Then the carbon atom at position 3 can act as a nucleophile, displacing the remaining bromine as a bromide ion to give the product named in the book.

Sorry, I've noticed you've been struggling with formatting for a while and IMO it's visually better to put charges above with \overset{<above>}{<below>} command so that the skeletal structure is preserved – it's then arguably easier to visually compare both structures, like so (you can right-click on the formula and select Show Math As → TeX Commands): $$\ce{Br\overset{\large +}{Mg}-\overset{\large -}{C}H2-C#C-CH2Br <-> Br\overset{\large +}{Mg}-CH2=C=\overset{\large -}{C}-CH2Br}$$ Just a suggestion:)

Does not make sense. The Mg-R bond is part of the anion there is no free electron pair. Too many pentavalent carbons. Experimental details to this synthesis would help out.

I do not understand. Where are the pentavalent carbons? Are you saying neither of the two pi bonds is in plane with the anionic center electron pair to undergo conjugation? Please offer an alternative answer.

Sorry I confused the typing  The possible sigma bond to the Mg does not seem likely to be in resonance with the triple bond the unanswered question, Does this reaction really happen?

I am sure that these type of questions have been found in IIT-JEE entrance examination books, which are mostly written by authors who have not thought well before preparing the question. Before creating  a new question, the author has to look for experimental evidence. Chemistry is not mathematical science, where you can extrapolate and find a solution to question.
Coming to your question, it seems theoretical. If I assert that  buta-1,2,3-triene would form, then it becomes a precedent and  a novice chemistry researcher would try hard to make the same compound.
As I  have not come across any textbook or reference discussing this type of reactions,  answering the question would be inappropriate at this time.
I will update the answer if I come across any reference.

The internet offers no info; what is there is fraught with misstatements. Grignards are reducing agents not oxidizers  so the cumulated  triene is difficult to predict unless bimolecular. Perhaps cyclobutene?


text:
Changes in entropy and Gibbs free energy for sublimation at triple point
The standard enthalpy of sublimation of dry ice (solid carbon dioxide) is $\Delta H=6.03\ \mathrm{kJ/mol}$. The triple point of $\ce{CO2}$ is at $p=5.1\ \mathrm{atm}$, $T=-56.7\ \mathrm{^\circ C}$.
I am trying to find the $\Delta S_\text{sublimation}$ and $\Delta G_\text{sublimation}$ at the triple point. I do not know which formula to use.

Welcome to Chemistry.SE! We have a site policy for homework questions. Please edit your question to include your attempt at the problem, where you got stuck, and let us know where you're finding difficulty so we may best help you.

You can work from the principle that at a phase transition the partial molar Gibbs free energy (chemical potential) of the substance in each phase is equal, so that
$$\Delta_\mathrm{sub} G = 0$$
from which
$$\Delta_\mathrm{sub} S_\mathrm{m} = \frac{\Delta_\mathrm{sub} H_\mathrm{m}}{T_\mathrm{sub}}$$
Note however that the value of $\Delta_\mathrm{sub} H_\mathrm{m}$ that you provide seems a bit off$^\dagger$. NIST gives a value at $\pu{207 K}$ of $\Delta_\mathrm{sub} H_\mathrm{m} = \pu{26.1 kJmol^{-1}}$. Values at other temperatures in the range $\pu{167-195 K}$ are similar. Other internet sources can helpfully confirm this.
$\dagger$ The value you post may be in $\pu{kcal mol-1}$, then it would translate into $\pu{25.3 kJ mol-1}$, in reasonable agreement with the values on the NIST data page.


text:
Help reading a phase diagram?
How does the y axis work? If the pressure I am looking for is 10^‐4 to 10^‐3 nbar? I tried doing log of  the yaxis but its giving me numbers that are not even close to what I am looking for.

We have a policy which states that you should show your thoughts and/or efforts into solving the problem. It'll make us certain that we aren't doing your homework for you.
Basically any question with the wording your question has is considered homework; it needn't be literally one. Self-study questions, puzzles etc. also count as homework.
Don't worry, they're not banned. But, we require a minimal effort. Otherwise, this question may get closed. Please [edit] in your full reasoning or thoughts on this.
See Homework

What base are you using when applying the log? $log_e(10)=2.3$ whereas $log_{10}(10)=1$. Looks like you're using base 10. For sulfur dioxide the triple point is at a pressure in bar of $log_{10}(P)=-1.8$

Here is how the y-axis works :


text:
Is there a quantitative test/analysis technique that can determine if some chloroform has degraded into phosgene?
Before I go to use any chemical that I've had in storage for a while, I try to check if it's degraded into something else (e.g.: checking the $\ce{H2O}$ content of $\ce{H2O2}$, checking for $\ce{H2O2}$ in $\ce{Et2O}$, etc.). This can usually be done using the appropriate test strip.
I have a 3/4 full bottle of chloroform that's been sitting on a shelf for about a year (probably less, actually), in an amber bottle to restrict the amount of light that can get to the chloroform. I'm aware that chloroform can break down into phosgene, and I know there are some ways to test for this, but I was hoping for a quick quantitative test that can be done without any hassle, but I can't seem to find anything out there. The Wikipedia page for chloroform states:

Suspected samples can be tested for phosgene using filter paper (treated with 5% diphenylamine, 5% dimethylaminobenzaldehyde in ethanol, and then dried), which turns yellow in phosgene vapor.

I don't have either of the diphenylamine or the dimethylaminobenzaldehyde, and would prefer not to have to purchase them just for this simple test (I don't have any use for either chemical currently).
Does anyone know if there is a simple quantitative test that accomplishes the same thing? Preferably a test strip.

Phosgene reacts with ethanol. Thus, if your chloroform is ethanol protected then your risk is minimal.

Not a quantitative test but a good proof of phosgene. You could use a pH paper strip. Chloroform should show a neutral pH, but if there is phosgene/HCl the paper will show acidity.

I see - That seems simple enough! Thank you

I admire your safety precautions. Like we always check for peroxide content in older ether bottles, we should also test older chloroform bottles, specifically those of alkene-preserved, for the presence of phosgene (Ref.1). The C&E News letter recommended  use of filter paper strips, wetted with 5% diphenylamine, 5% dimethylaminobenzaldehyde and then dried, which turn yellow in phosgene vapor (a qualitative test). Note that as test strips age they become light yellow/brown, they will still show a significant change to bright yellow upon detection of phosgene (Wooley Lab).
There are also Draeger tubes available for phosgene testing, but they are kind of expensive (e.g., check on Shopcross.com). The Draeger tube test uses the general reaction of an aromatic amine with phosgene to obtain the corresponding aromatic isocyanate (I believe), which has long been known. However, their aromatic amine turns red in the presence of phosgene. It has been suggested that such reactions proceed thru the acid salt of the amine such as an amine hydrochloride or an amine-$\ce{CO2}$ adduct either by forming the salt in situ in the process or by starting with the salt, to form the corresponding aromatic carbamyl chloride, which in turn is decomposed to the isocyanate with liberation of hydrogen chloride.
In addition, qualitative (and can develop for quantitative) analysis by GC/MS would work as well. However, Ref.1 has used UV-vis colorimetric quantitation described in Ref.2. The abstract states that:

A reagent consisting of 0.25% 4-p-nitrobenzylpyridine and 0.5% N-benzylaniline in diethylphthalate was tested and proved to be highly sensitive and relatively specific for the spectrophotometric determination of phosgene in air. Concentrations under $\pu{0.1 ppm}$ of phosgene in air may be determined with reliability. The color developed with the reagent was stable for several hours and only slightly affected by normal ambient humidity. None of the following substances interfered with the reaction: hydrogen chloride, chlorine, chlorine dioxide, carbon tetrachloride, chloroform, trichloroethylene, tetrachlorethylene, and dichlorodifluoromethane in concentrations likely to be encountered under industrial conditions.

It seems there is some development in teststrips containing a fluorophore for phosgene detection (See diagram below; Ref.3), which changes the color from blue to green with a hand-held UV light in the presence of phosgene.

References:

Eric Turk, “Phosgene from chloroform,” Chemical & Engineering News 1998, 76(9), p 6 (DOI: 10.1021/cen-v076n009.p006).
M. H. Noweir, E. A. Pfitzer, “An Improved Method for Determination of Phosgene in Air,” American Industrial Hygiene Association Journal 1971, 32(3), 163–169 (https://doi.org/10.1080/0002889718506431).
H. Xie, Y. Wu, F. Zeng, J. Chena, S. Wu, “An AIE-based fluorescent test strip for the portable detection of gaseous phosgene,” Chem. Comm. 2017, 53(70), 9813–9816 (DOI: 10.1039/c7cc05313d).


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Is the phenyl group in benzophenone electron donating/withdrawing?
The ketone carbonyl group is less reactive to nucleophilic attack than aldehyde since the alkyl group in ketone is electron-donating towards the carbon.
Is this the same case for benzophenone, where there is a phenyl group?
Phenyl groups are usually electron-withdrawing, but in the case of benzophenone, would it be electron-donating through resonance?

Carbonyl group are electron withdrawing for a benzene ring. So by converse, phenyl group is electron donation for carbonyl group.

A keto or an aldo group with phenyl ring is less reactive towards nucleophillic addition then their aliphatic counterparts(acetaldehyde or acetone).With phenyl group the positive charge on carbonyl carbon is delocalised via resonance (structures  2 , 3 , 4 , 5 ) .

Resonance with phenyl ring is due to conjugation.

such conjugation is absent in aliphatic aldehydes and ketones.Therefore carbonyl compounds with phenyl rings bonded to carbonyl group are less reactive then their aliphatic counterparts.
With respect to + carbon on carbonyl group pi electrons on phenyl group are polarized towards it , in a way electron donating .


text:
Barron states that 4.18×10⁸ joules equal 1 kcal, is this correct?
Here is a paragraph in Barron's SAT chemistry prep book:

Work itself is measured in Joules, and so is energy. In some problems, however, energy may be expressed in kilocalories. The relationship between these two units is that $4.18\times10^8$ Joules equals 1 kilocalorie (kcal).

Is this correct? Isn't 1 kilocalorie 4180 Joules?

Also note that names of units such as ‘joule’ are spelled with a lower case initial in English. (For SI units, it is only the unit name ‘degree Celsius’ that contains a capital letter.)

It's likely just 3 being mistaken for 8 by typesetter as they are similar in some fonts and handwritings.

This should never have happened in the scientific community, but:

The energy values of food are measured by the Calorie, with a capital C. A Calorie is equal to 1,000 calories or 1 kilocalorie (kcal).

Source: here
It is possible that Barron's got confused between Calorie and calorie. Even so, they still would be off by a factor of 100.
According to Wikipedia:

The SI unit of energy is the joule, with symbol "J"; one small calorie is now defined as exactly 4.184 J, and one large calorie is 4184 J. However, the two units are still used occasionally in technical work, and the large calorie is still widely used in nutrition.

So you are correct in saying that 1 "small" kilocalorie is 4180 joules. 1 "large" kilocalorie is $\pu{4.18e6 J}$, and maybe Barron's is now suggesting an "extra large" calorie.
More likely it is just a typo.


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Need to normalize lattice enthalpy by no. of nearest-neighbor (NN) interactions to equate to NN interaction strength?
I've seen electrostatic interaction energy (from $E\sim \frac{q_1 q_2}{r}$) used to explain differences in lattice enthalpies between different crystalline solids.  For example, I've seen it explained that $\ce{NaF}$'s $\pu{910 kJ/mol}$ lattice enthalpy is less than $\ce{MgCl2}$'s $\pu{2326 kJ/mol}$ because of $\ce{Mg}$'s greater cationic charge (+2 vs. +1).
Question: Isn't the lattice enthalpy defined per stochiometric unit, rather than per nearest-neighbor interaction? And if so, if one is going to use electrostatic models of nearest-neighbor interaction to explain differences in lattice enthalpy, shouldn't the respective lattice enthalpies first be divided by the number of nearest-neighbor interactions per stochiometric unit?

What is "per bond"? There are no individual bonds here anyway.

@IvanNeretin Good point.  I've edited my question.


text:
Why are palladium and platinum carbonyls unstable at room temperature?
By palladium and platinum carbonyls, I mean the mononuclear, homoleptic, charge-neutral, binary carbonyl complexes of the form $\ce{M(CO)_n}$ and not any heteroleptic complexes like $\ce{Pd(CO)(PPh3)3}$ and similar compounds of platinum which are abundant in chemical industry and used as catalysts.

So, I was reviewing carbonyls of transition metals when I encountered this statement:

Group 10 elements with 10 valence electrons form metal carbonyls
  $\ce{Ni(CO)4}$ ($10 + 4 \times 2 = 18$ electrons). Curiously $\mathbf{Pd(CO)_4}$ and $\mathbf{Pt(CO)_4}$ are not stable.

I am proposing that heavy metal carbonyls tend to be unstable(?) and decarbonylate to form polynuclear carbonyls. For example: $\ce{Ru(CO)5}$ and $\ce{Os(CO)5}$ being unstable form $\ce{Ru3(CO)12}$, and $\ce{Os3(CO)12}$. But in the case of palladium and platinum, no carbonyl complex seems to exist in stable condition at room temperature. I reviewed two papers which described the Group 10 carbonyl complexes (links below):

Complexes where the carbonyl group is the only ligand present have
  been reported for all the platinum group metals except
  palladium.[...] No substantiated reports have been published on
  platinum or palladium compounds similar to the nickel carbonyl
  $\ce{Ni(CO)4}$. Thus the polymeric platinum dicarbonyl is the only
  known carbonyl of these two metals and even this compound is unstable
  in air.[...]
The unstable polymeric platinum dicarbonyl $\ce{Pt_n(CO)_{2n}}$ has
  been obtained as a cherry red precipitate by the action of carbon
  monoxide on aqueous or ethanolic solutions of Pt(II) compounds (e.g.
  $\ce{K2PtCl4}$) at temperatures up to 80 °C; or as a purple colloidal
  precipitate by the action of water on a benzene solution of
  $\ce{Pt(CO)2Cl2}$ under an atmosphere of carbon monoxide.

It was not until 1972 that a definite existence of the congeners
  $\ce{Pd(CO)4}$ and $\ce{Pt(CO)4}$ was reported. Using matrix isolation
  and Raman techniques, data for the products of Pd atoms CO and CO/Ar
  cocondensation reaction were able to characterize the series of binary
  complexes $\ce{Pd(CO)_n}$ analogous to $\ce{Ni(CO)4}$, the most stable
  being $\ce{Pd(CO)4}$ but still unstable than $\ce{Ni(CO)4}$ which
  explains the failed attempts in its synthesis. Similar technique were
  used in making $\ce{Pt(CO)4}$ at 4.2-10 K which is stable up to 30
  K.  (slightly abridged)

The question is: why they are unstable/don't exist at room temperature?

Platinum Metals Rev., 1972, 16, (2), 50 "The Carbonyls of the Platinum Group Metals" By C. W. Bradford, Research Laboratories, Johnson Matthey & Co Limited (link)
J. Am. Chem. Soc., 1973, 95 (22), pp 7234–7241 "Binary carbonyls of platinum, Pt(CO)n (where n = 1-4). Comparative study of the chemical and physical properties of M(CO)n (where M = nickel, palladium, or platinum; n = 1-4) DOI: 10.1021/ja00803a009

Tl, dr:  It's not just a matter of size.  A combination of coordination geometry, node distribution in the $d$-orbital wavefunctions, and atomic size all conspire to weaken $\pi$ overlap in palladium and platinum carbonyls, not only relative to the nickel compound but also relative to earlier transition metal carbonyls in their respective periods.
Ziegler et al. [1] studied the relative stabilities of metal-carbonyl bonds for $\text{G6(CO)}_6$, $\text{G8(CO)}_5$ and $\text{G10(CO)}_4$ carbonyl complexes, where $\text{G6}$ in the first formula is a Group 6 metal and similarly for the second and third formulae.  The following general trends were observed:

Within a period, the G6 hexacarbonyl is most stable, with the G10 tetracarbonyl least stable and the G8 pentacarbonyl in-between.

Within each group, the 4th period carbonyl is most stable, the 5th period carbonyl is least stable and the 6th period carbonyl is slightly more stable than the 5th.

From these trends it is seen that palladium and platinum carbonyls are literally in a bad place.
The authors attribute the stability order primarily to $\pi$ backbonding.  We may expect such backbonding to be strongest for the G6 hexacarbonyls where the octahedral geometry about the metal atom causes the metal-atom donor orbitals to have pure $\pi$ symmetry.  The five- and four-coordinate geometries (trigonal bipyramidal and tetrahedral, respectively) are progressively less favorable with regard to this symmetry.  This accounts for effect (1).
Effect (2) is related to relative atomic sizes within each respective group, but also to a more subtle effect.  In the $3d$ subshell, which is used as the valence subshell in the 4th period, the lobes have no radial nodes, a condition that allows maximum overlap with the molecular orbitals coming from the carbon monoxide.  The $4d$ and $5d$ orbitals used by the heavier metals have radial nodes so poorer overlap.  With the 6th period metals, relativistic effects intervene, but this effect (which I cannot tell from the abstract, and the full paper is behind a paywall) is less potent than having more compact orbitals with no radial nodes at all in the $3d$ series.
Reference:
1.
Tom Ziegler, Vincenzo Tschinke, and Charles Ursenbach, "Thermal stability and kinetic lability of the metal carbonyl bond. A theoretical study on M(CO)6 (M = chromium, molybdenum, tungsten), M(CO)5 (M = iron, ruthenium, osmium), and M(CO)4 (M = nickel, palladium, platinum)", J. Am. Chem. Soc. 1987, 109, 16, 4825–4837. Link

I don't know definitively but we know that carbon monoxide acts as a strong sigma donor and  pi acceptor of electrons when acting as a ligand. Palladium being huge, has dispersed valence electrons and perhaps due to this, sigma donation to the metal is repelled by the orbitals and/or the dispersed valence electrons do not back bond as strongly as lighter metals such as nickel (whose carbonyl complex is quite stable). Metals such as nickel may be able to donate more e density to those empty P orbitals on carbons monoxides carbon due to more compact and dense valence orbitals and thus form stronger bonds.

You'll get more upvotes with a reference to back up this argument.

Palladium is smaller than Molybdenum, and the same size as Chromium, Iron and Manganese (https://en.wikipedia.org/wiki/Atomic_radius) all of which form stable carbonyls. Platinum is smaller than palladium


text:
Why is porphyrin aromatic?
I was asked which of the following were aromatic:

My understanding of aromatic nature is, that the molecule must be cyclic, planar, every atom in the ring must be $sp^2$-hybridized and the molecule must have [4n+2] π electrons, where n∈N (Hückel’s rule). The molecular orbitals also have to have the correct orientation and phase for aromaticity.

A and B are aromatic since they follow conditions as above (including empty orbital of boron in conjugation)
azulene's aromaticity is due to the “fusion” of
two aromatic ions, both with 6 π electrons, cycloheptatrienyl (tropylium) cation and cyclopentadienyl anion.

Based on the above My answer was a, b, and c.
Given answer: a, b, c, d.
I checked for d. It was aromatic as described in this link. 1. The diagram indicates only π electrons along the blue need to be considered amounting to 18 π electrons in conjugation.

My question is why only consider π electrons in blue and why not others? How do we explain the aromaticity of porphyrin?
References
https://en.wikipedia.org/wiki/Porphyrin

Simply that a circuit satisfying those rules is attained. Though one can image system in which fused moiety can pin electrons and reduce the aromaticity of the whole ensemble.

@Alchimista why cant pi electrons in black be used instead?

Simply because that would lead to a higher energy of the system. By the way, as usual and even more with a complex ring, first one assess the aromaticity, then moves to explain. For instance in this case coordination properties and reactivity at the various position point to aromaticity. Then you identify the blue system. Your exercise requires a previous knowledge or a good feeling for aromaticity. Is the same with Hueckel rules.

Counting electrons in a ring is a rather crude way to determine whether a structure is aromatic. There are far more direct and reliable ways to prove aromaticity, such as unusual NMR chemical shifts due to the presence of aromatic ring currents. Typically, once a compound is shown to be aromatic by these more stringent measures, then a posteriori arguments regarding electron count tend to be made.

The aromatic properties are evident from the heats of combustion, the fact that all bonds have the same length, the ability of porphyrins and metalloporphyrins to undergo electrophilic substitution reactions, the stability of the porphyrin nucleus in electron impact mass spectrometry and the characteristic chemical shifts observed in proton NMR spectra1.
The 18π electron substructure (shown in blue in your figure) was thought to be responsible for the aromatic characteristics. In support of this view it was noted that the outermost bonds in the pyrrolenine rings(the not blue double bonds in your figure) take part in cycloaddition reactions(Fig.1, taken from ref.2) and therefore have a significant double bond character suggesting that their electrons are more localized.
This 18π model is also called "the diaza[18]annulene model" because it was speculated that the aromatic pathway found in porphyrins resembles [18]annulene(fig.2, ref.1). You can also note the similarities in the spectra of [18]annulene(fig.3, ref.3) and porphyrins: both have a major peak in the UV region and a less intense absorption at longer wavelengths.
Other supportive data to the the diaza[18]annulene model are proton NMR studies all mentioned in ref.1

https://www.worldscientific.com/doi/abs/10.1142/S1088424611004063
http://www.scielo.br/pdf/jbchs/v15n6/22664.pdf
Chem. Commun., 2016, 52, 4710


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Which is more metallic: boron or silicon?
I read somewhere that boron is more metallic. Is it correct? If so, can you please elaborate?
The reason why I'm confused is because 2 factors come into play here:

When you move towards right, metallic character decreases.
When you move downwards, metallic character increases.

So, which one would win?
Are there any exception cases I should look out for?

Welcome to Chemistry.SE! We have a site policy for homework questions which can also apply to self-study questions. Please edit your question to include your thoughts, where you got stuck, and let us know where you're finding difficulty so we may best help you.

It's either metallic or it isn't. "More metallic" makes no sense. Both are non-metals.

@IanBush, not nonmetals but semimetal, semiconductor or metalloid! See https://en.wikipedia.org/wiki/Metalloid

Both Boron and Silicon in their standard forms at STP have a band gap.  As such. they are non-metals. As the link provided says "No widely accepted definition of a metalloid exists", as such I don't think it is a useful term. Semimetal, with a positive direct band gap but negative indirect one, is of some use but not relevant here.

Since both are metaloids, or semiconductors, look for the criteria that define them. In particular, electronegativity or ionization energy are important.


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Why do glow-in-the-dark substances dim gradually?
Related: How do things glow in the dark?
Growing up, I had glow-in-the-dark stars on my ceiling and after being "charged" by having my lights on, they would phosphoresce for quite some time.
However, throughout the night, a star would gradually dim. I imagine that the cause of the stars becoming dimmer is that some electrons are relaxing down before others.
I thought the thickness of the star could be part of this as the inner parts likely receive less light, so perhaps this bit would completely dim out before the surface of the star. However, my thought is that the star should either be completely "charged" or cast no light at all, like a switch.
The dimming effect is gradual. Why?

If you drew the curve of light emission over time you would find something like a first-order exponential decay. This should be true for all the stars and–assuming they are all made of the same stuff–the rate of the decay (the time to decay to half the intensity) would be the same. But some stars contain more phosphor than others. They will be brighter at the start and take longer to decay to the visibility threshold even though the half-life of the decay is the same.

Source of light

I imagine that the cause of the stars becoming dimmer is that some electrons are relaxing down before others.

This statement could be interpreted in two different ways, one incorrect and one correct. It could mean that electrons have an "on" state where they continuously emit light. That is not the case. Instead, they emit light once while going from an excited to the ground state. It could also mean that different electrons relax at different times. As electrons relax, the total number of excited electrons decreases. As a consequence, there are fewer excited electrons after some time, and as such fewer electrons that can emit at that later time. This is the correct interpretation.
Kinetics of decay

my thought is that the star should either be completely "charged" or cast no light at all, like a switch.

This is a conceptual model that goes against the OP's observations, so we already know it does not work this way, but the question is why.
To address this, here is a setup that does have the properties of a switch. I will use this alternate setup to compare and contrast. Take a star lit up by multiple light-emitting diodes (LEDs) powered by a battery. In LEDs, the electron flow continuously gets electrons into excited states, which then relax to ground states, giving off light. As a consequence, the LED will emit light until the battery is discharged to a degree that it no longer provides the necessary voltage. As a consequence, all the LED's will turn off at the same time.
Now let's compare and contrast with the glow-in-the-dark stars. They contain electrons that get excited in the first phase (shining light on them) and then they go back to the ground state once in a slow process that takes hours. You can measure the light emitted, as in this experiment:

Source: https://www.cei.washington.edu/education/lessons/phosphorescent-decay/
Rather than all emitting at the same time (which would result in one flash of light) or emitting in a zero-order fashion (which would result in constant intensity that abruptly turns off when all electrons are at ground state), the intensity diminishes gradually, with the largest change at the outset.
This is not what we perceive when looking at glow-in-the-dark stars in a darkened room (probably because our eyes adjust continuously because the is no background or reference of constant brightness). Also, at one point we will probably fall asleep.
Mechanism at the atomic level
The Journal of Chemical Education has a nice article on this (Lisensky et al 1996, DOI: 10.1021/ed073p1048). Here is a quote from the abstract that connects kinetics with mechanism:

[J Chem Ed] The phosphorescence decay of ZnS:Cu fits a second-order equation, consistent with recombination of relatively equal populations of holes and electrons (n ~ p).

And as an aside, here is a diagram from Wikipedia (article: Light-emitting diode physics) showing how LEDs are continuously providing excited electrons and holes they can fall into:

@virolino - I tried to address your criticism of my previous answer, and wrote a more complete new answer here. One comment on you comment: "they return to their original state and emit light when it is dark". They actually emit light independent of whether the room is dark or light. It's just that we can see it more clearly when it's dark.

There are two potential questions here: one is why is dimming a slow drawn-out process (incoherent, unlike a "switch"). The second question is why different stars may appear to lose their intensity at a different rate.
The light-emission process behind your stars involves phosphorescence. The process involves relaxation of an electronically excited system during which it undergoes a "forbidden" triplet$\rightarrow$ singlet transition. For an individual system, light emission is controlled by the probability of the quantum-mechanical interconversion (or equivalently by the lifetime of the triplet state). It is a stochastic event described by Poisson statistics or "shot" noise.
However, the luminosity of one star represents the collective behavior of a large number of emitters (systems). One star, therefore, appears uniform in luminosity and its time-dependent luminosity decays exponentially.
Variability might be expected in the luminosity of different stars. This is not likely to be QM statistics since, as already noted, you are observing a very large ensemble of relaxing molecules, and one fairly uniform decay constant is expected for all stars, assuming the phosphorescent material is uniform. It seems more likely that luminosity differences among or within stars may occur due to size effects or differences in the concentration of phosphorescent material, or both$^\daleth$. Stickers that are thicker or containing a greater concentration of excited phosphorescent material, due for instance to variability in the manufacturing process, should appear to glow longer simply because they contain more emitters$^\dagger$.
$\dagger$ As helpfully pointed out in the commentaries, more emitters lead simply to the detection limit being reached later.
$\daleth$ As helpfully emphasized in yet another comment, and alluded to in the OP, it is the concentration of excited phosphorescent material that determines the luminosity, and excitation may also depend on the dimensions of the star.


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Why are glow-in-the-dark things usually green?
I've noticed that many glow-in-the-dark objects you can purchase in the store are green. Occasionally I would come across something that was orange or blue, but mostly it seems that green is the ubiquitous glow-in-the-dark color.
Does this have anything to do with common wavelengths of light emitted from the relaxing of electrons in some of the more common phosphorescent compounds? Are some phosphorescent compounds just more readily available or "safe" for use in toys?

Maybe most glow-in-the-dark things are based on the same active molecule?

Google for  glow stick for some of them and zinc sulphate , strontium aluninate for the others.

There are only a small number of really good, cheap phosphorescent materials for long lasting phosphorescent behaviour. Zinc sulphide is cheap and dominates novelty products, strontium aluminate lasts longer and dominates professional signage. Both are green. Most other colours are achievable with other materials but why use them if you don't need a different colour. So the basic explanation for the preponderance of green is economics not chemistry.

The human eye is most sensitive in green wavelengths. Therefore, if one were to put the same amount of light energy into different wavelengths, the green portion of the spectrum would appear brightest, though it contained no more energy than the rest.
If a phosphor or glow-stick manufacturer is seeking the "greatest bang for the buck", i.e. the most light for the least amount of chemical, green is a good choice.
There is a variety of phosphor colors, such as these listed in Wikipedia:

Red: Yttrium oxide-sulfide activated with europium is a red phosphor used for CRT's.
Yellow: Zinc cadmium sulfide (Zn,Cd)S:Ag, emits yellow light.
Green: Zinc sulfide with copper, ZnS:Cu, emits green light, ~530 nm. It was widely used in CRT oscilloscopes and radar displays because of its long persistence and simple chemistry. It is known as phosphor P31.
Blue: Zinc sulfide with few ppm of silver, ZnS:Ag,emits blue light with peaking at 450 nm. It is known as phosphor P22B and, unlike ZnS:Cu, has a very short persistence, so is suitable for slit-photography CRT's.

For a more complete list of phosphors and their colors, see this useful PDF or Phosphor Technology.
Many stores carry a variety of inexpensive chemiluminescent products, such as glow-in-the-dark necklaces and bracelets, in colors ranging from red through violet.


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Can caustic soda (sodium hydroxide) be used to dissolve urine odour?
I have an area of artificial grass at home in the back garden and during the summer months, the area emits a smell which I can only think is from the dog's urine when he pees all year long. The smell isn't a problem in the winter or during wet days.
I read a lot of posts online about using white vinegar and baking/bicarbonate of soda
I don't have any of that at home, but I do have caustic soda from unblocking drains/pipes that I use once in a while. I wonder if caustic soda can be used in a Pump Sprayer with Water or some other cleaning agent to help break down the smell from urine?
I'm a complete noob with chemistry but I thought this is the right forum to ask the question.

Caustic soda, if anything, will make the stench worse.

And possibly dissolve your artifical grass

Is the odor ammoniacal or a different stench? I think NH3 is not so objectionable at low concentrations, but bacterial stench would be.

I wouldn't know. It doesn't stink of the usual piss odour you'd find in an underground passage or a NYC subway. It stinks somewhat different.

Maybe you could go to a nearby lab and have a whiff of ammonia, and compare the smells.

Well, it would actually make things a little awkward.
Why do Vinegar and baking soda work?
The reason for the usage of vinegar or sodium bicarbonate must first be understood.

Vinegar ($\ce{CH3COOH}$) is an acid, and hence protonates the carbonyl oxygen present in urea[1].

$\ce{NaHCO3}$ (baking soda) is amphoteric, meaning it can react with both acids and bases. Urea is basic, and hence forms a salt. This answer on our site talks exclusively about the deodorizing property of baking soda.

Both actions are quite much the same, and prevent the hydrolysis of urea (which happens to release ammonia ($\ce{NH3}$) which is the "foul" smelling substance)

Why wouldn't sodium hydroxide work?
Not that $\ce{NaOH}$ wouldn't react, but because it would accelerate the hydrolysis of urea[2].

So if you're looking to stop the hydrolysis of urea (and the inevitable generation of ammonia) $\ce{NaOH}$ might not be the right guy.
What I wouldn't advise
But, hey, if you're looking to remove all the urea once and for all, you could pour a few buckets of $\ce{NaOH}$ and let the ammonia create a riot in your unsuspecting neighborhood.
What I would advise
Urea hydrolysis is catalysed by the enzyme urease (so it is dead slow without it)
From reference [3] (emphasis mine)

Urea is a stable compound with a decomposition half-life in aqueous media of 3.6 years. However, hydrolysis of urea is 104 times faster when the urease enzyme is present.

If we could inhibhit its action, we'd have got what we needed.
If you're wondering where urease comes from and why the dog urine stench is greatly enhanced in summer months, you're looking at the same answer: microorganisms.
It has been published that Metals and fluorides could be used to inhibhit urease. I quote further from source [3].

2.3. Inhibiting chemicals
Seven different chemicals, silver nitrate, zinc nitrate, sodium fluoride, glacial acetic acid, vinegar, citric acid, and sulfuric acid, were used to inhibit urea hydrolysis.

The list is by no means exhaustive.
From source 4, different classes of compounds which might be used:

The classes of urease inhibitors include: amides and esters of phosphoric acid, thiols, hydroxamic acids, phosphinic and thiophosphinic acids, boric and boronic acids, phosphate, heavy metal ions, bismuth compounds, quinones, and fluoride.

If you're not keen on purchasing baking soda, I'm sure you'll find any one of the above an interesting buy.

References:

Interaction of urea with weak acids and water, Prabhat K. Das Gupta and S. P. Moulik, J. Phys. Chem. 1987, 91, 5826-5832 DOI:10.1021/j100306a061

What is the reaction between NaOH and urea?

Characterization of urea hydrolysis in fresh human urine and inhibition by chemical addition, Hannah Ray, Daniella Saetta, Treavor H. Boyer, Environ. Sci.: Water Res. Technol., 2018, 4, 87-98 DOI: 10.1039/C7EW00271H

A combined temperature-pH study of urease kinetics. Assigning pKa values to ionizable groups of the active site involved in the catalytic reaction, Barbara Krajewska, J. Mol. Cat. B: Enzymatic Volume 124, Feb. 2016, 70-76 DOI: 10.1016/j.molcatb.2015.11.021

EDIT
There has been some discussion on if the stench is truly ammoniacal or is just bacterial odor. In that case, I would suggest sprinkling of significantly higher concentration of acid. Extreme pH is known to disrupt bacterial life forms and the acid will also prevent to a great extent the possibility of urea hydrolysis, thus preventing the odor anyway.

Hi - thanks so much for the detailed analysis. So baking soda + vinegar in a pump sprayer is what we're saying? and how effective is this? how often should it be repeated ? Also what sort of quantities do we need? e.g for 5 L pump spray - how many grams of bicarbonate soda, how much white vinegar and how much water? Excuse my naiveness, but I know nothing about chemistry.

Either baking soda or vinegar. not both(that would give you a volcano of bubbles!). As for quantities, its difficult to speculate. How much area do you have to cover?

The area is about 25-30 square meters.I had read online somewhere that vinegar with baking soda does wonders - but maybe not for removing urea?

It seems all this is missing the point as urea is not the source of urine's stench. Fresh urine isn't hardly stinky, it's the decomposition, by bacteria that's creates truly foul smell and it's not from ammonia. High pH can kill bacteria - that's the point.

@Mithoron I haven't tried it really, but I wrote it under the assumption that the smell was from ammonia being released by bacteria hydrolysing the urea. In that case we could ask the OP to spray caustic soda and see if it would work? I would suggest acid instead of base, as we'd still be inhibhiting urea hydrolysis and might also kill the bacteria causing the stench.

So just to clarify - we're either spraying either with vinegar or with baking soda. Can I confirm how much baking soda and how much water? my pump sprayr is 5L. or if spraying just with vinegar, am I just adding 5L of distilled white vinegar?

@kkudi Yes, I think that should do, no water. Hope it goes well. If it doesn't I can help you nuke it xD. Do get back to us and tell us if it worked.

So if using Vinegar, I just spray Vinegar onto the grass.  If using baking soda/bicarbonate soda... do I just sprinkle it on top of the surface or do I mix that with water? If the latter, how many parts water to baking soda?

@kkudi Well I meant vinegar without water. soda will definitely need water. And I found another way: pine oil should be a good disinfectant too.

@WilliamR.Ebenezer any chance you can advise how much water compared to how much baking soda? I think anything other than vinegar or bicarbonate soda will be expensive.

For soda, use ~95 grams in 1 L, this is the highest concentration you'll get. So 95*5 grams for 5 L. Use 100 grams, nothing will go wrong

thanks so much! will be spraying next weekend. Will probably have to wait a week or two and a hot day to report on results, but will definitely come back and report my findings! thanks once more!

This has made such a huge difference. I sprayed twice about 2 weeks ago. So basically almost a kg of bicarbonate soda in 10 litres of water. We had two sunny weekends and I haven’t been able to smell that ugly smell. So bicarbonate soda has worked wonders!

@kkudi I'm glad to hear that.

Hydrogen peroxide (3%) is recommended for cleaning up animal pee inside a house. It could work outside as well, perhaps by killing bacteria, or maybe oxidizing the stuff that bacteria feed on. It should be safe on plastic.
Chlorox Multi-Surface cleaner is very useful. It contains quaternary ammonium surfactants, not bleach. These are bacterial killers too. It comes in a spray container
And the last resort would be bleach, but you might want to test it on a less visible area. It could be diluted quite a bit. And you might need to rinse the area (after a an hour?) to remove the solids that would otherwise crystallize out when the water evaporates. The solids might be noticeable as a white spot, and if they still contain hypochlorite, there might be continued action which weakens the plastic.

From a practical standpoint for this fellows particular problem, i think instead of inhibiting the reaction, you may consider actually accelerating the reaction to once and for all get rid of the smell for good.  sure you will increase the ammonia smell for a day or so, but this is outside, and once the urea is consumed, the problem is solved.
I vote for YES for the use of Sodium Hydroxide for this practical everyday problem.  as a matter of fact, i’m gonna try it right now !


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If change in free energy (G) is positive, how do those reactions still occur?
I was doing a couple of problems for homework:
Calculate $K_\mathrm{sp}$ of $\ce{AgI}$ at $55.0\ \mathrm{^\circ C}$
Calculate $K_\mathrm{b}$ of $\ce{NH3}$ at $36.0\ \mathrm{^\circ C}$
I have to use $\Delta G^\circ= -RT\ln K$ and $\Delta G= \Delta H-T\,\Delta S$
When I did this $\Delta G^\circ$ is positive ($89.59\ \mathrm{kJ/mol}$ and $28.037831\ \mathrm{kJ/mol}$ respectively), yet $K_\mathrm{sp}$ for $\ce{AgI}$ is $5.5\times10^{-15}$ and $K_\mathrm{b}$ for $\ce{NH3}$ is $1.8\times10^{-5}$, indicating that there are some products and the reactions do occur. Plus, $1.0\ \mathrm M$ $\ce{NH3}$ in solution has a $\mathrm{pH}$ of $11.6$ so it must react a little.
According to the second law of thermodynamics, if $\Delta G$ is positive, the reaction is not spontaneous, right? But clearly, they, in fact, are to a certain extent. What is going on?

For a Q vs K argument (instead of $\Delta_r G$ vs $\Delta_r G^\circ$) see https://chemistry.stackexchange.com/questions/26251/can-a-reaction-with-positive-gibbs-free-energy-change-ever-occur?rq=1

Individual molecules know nothing of Gibbs, or temperatures, or entropy. They just collide and perhaps react, by exchanging energy. As long as molecules do what the conservation of energy and impulse dictates them (within the realm of structures that are possible by QM), the laws of thermodynamics emerge naturally for a macroscopic sample.

$\Delta G^\circ_m$ is the difference in molar Gibbs free energy between the reagents and products in their standard states (in the case of $\ce{AgI(s)}$, the standard state for the reagent is the pure solid and that for the product is the solute in a $\pu{1 molal}$ ideal solution, all at $p^\circ =\pu{ 1 bar}$).
If $\Delta G^\circ_m>0$ you cannot convert the reactants completely into products without doing some accompanying work (for instance, applying an electrical voltage, in the case of an electrochemical cell).
If you want to see where the reaction is headed, for instance, if you start with only reactants (say $\ce{AgI(s)}$), then you should compute the following:
$$\Delta G_m=\Delta G^\circ_m+RT\log(Q)$$
For a reaction $\ce{A<->B}$ this is illustrated in the following plot (generated assuming an equilibrium constant $K_{eq}=10$, and $\Delta G$ in arbitrary units):

Here $\zeta$ represents a progress coordinate ($\zeta=Q/(1+Q)= \chi _B$).
Note that some authors place a "tilde" following $\Delta G_m$ to emphasize that it is a partial molar quantity, that is, it measures the change in free energy with reaction progress.
At the start of the reaction, when there is no solute dissolved, Q=0 and $\Delta G_m=- \infty$. In other words, the driving force for the reaction to occur is theoretically unlimited. At equilibrium that force goes to zero. Past the equilibrium point, the reverse reaction becomes more favorable.
See "What is the difference between ∆G and ∆G°?" for an elaborate answer addressing a similar question.

The reaction Gibbs energy says what is $\Delta G$ for the conversion of reagents to products.
But it does not (directly) say what is $G$ value between these 2 states.
The $G$ for the transition from reagents to products is not a linear function, but it has the minimal value at the reaction equilibrium.
If there is reaction $\ce{A <=> B}$ with $\Delta G^\circ_\mathrm{r}=0$ and $K=1$, then the $G_\mathrm{min}$ will be at $x_\mathrm{a}=x_\mathrm{b}=0.5$
If the above reaction has $\Delta G^\circ_\mathrm{r}$ slightly positive, then $K\lt 1$.
The reaction will not magically stop happening, but the equilibrium with $G_\mathrm{min}$ will be slightly shifted toward reactants.

It is only a "rule of thumb" that a reaction is spontaneous only if the standard change in free energy is negative.  In reality, a reaction can be considered spontaneous if, starting with stoichiometric proportions of reactants, the amounts of reactants remaining at final equilibrium is small (i.e., the equilibrium constant of the reaction is high, $\Delta G^0$ is negative, and the reaction goes nearly to completion).
Conversely, a reaction is considered non-spontaneous if, starting with stoichiometric proportions of reactants, the amounts of products produced when final equilibrium is achieved is small (i.e., the equilibrium constant of the reaction is low, $\Delta G^0$ is positive, and the reaction does not approach completion).
It was a stupid idea in the first place for teachers to introduce the concept that the sign of $\Delta G^0$ can be used as an on-off switch for determining whether a reaction is spontaneous or not; it has caused an endless amount of confusion to students.  In reality, all reactions are spontaneous to some extent.


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Do all zero order reactions occur as combination of multiple elementary reactions?
In a  reaction, the rate determining step (RDS) gives the order of reaction. There might be 2 or 3 molecules colliding to give a product (2nd or 3rd order) or a molecule breaks for order 1. But for zero order reactions, how do they occur? Are all zero order reactions complex (i.e. are there no simple 1 step zero order reactions )

Sorry, but I am not able to get the content of the first line. Could you possibly rewrite it?

All reaction orders are one smaller if the rate-determining step involves a catalyst with constant, small concentration, so it becomes the limiting factor.


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Why are tertiary amines less basic than secondary and primary amines in water than when vapourised?
In gaseous state:
$$\ce{(CH3)3N} > \ce{(CH3)2NH} > \ce{CH3NH2} > \ce{NH3}$$
However, when dissolved in water:
$$\ce{(CH3)2NH} > \ce{CH3NH2} > \ce{(CH3)3N} > \ce{NH3}$$
My notes mentioned the reason for this is because of the increased steric hindrance from the greater number of alkyl group which makes the lone pair of electrons on the N atom less available. But, if this is the case, why doesn't steric hindrance affect the basicity of tertiary amines when in a gaseous state?
After some research, I found that the conjugate acid of an amine can be stabilised by solvation. But, shouldn't the stabilisation be the same, because of the attraction of electronegative O in a water molecule to the positive charge on $\ce{N}$ in an ammonium salt? (can ammonium salts still be stabilised by hydrogen bonding? I thought ions don't form hydrogen bonds with water, they form ion-dipole attractions)
Unless the effect of steric hindrance comes in here, where the alkyl groups reduce the strength of the ion-dipole attraction between the ammonium ion and thus the stability?
I'm really confused how everything links up; appreciate it if someone can kind of clear this up.

A possible answer from what I've searched on the internet so far..
The strength of basicity of amines depends on two factors: 1) the availability of the lone pair for donation to a proton, 2) the stability of the conjugate acid of the amine.
In methylamine, the conjugate acid, $CH_3NH_3^+$ will attract water molecules and ion-dipole interactions will form between the cation (the conjugate acid of methylamine) and the partial negative charge on O in the water molecule. This ion-dipole interaction or attractive force helps to stabilize the positive charge on the cation.
So, the more water molecules you have solvating your cation, the more stable the conjugate base. However, the more alkyl groups you have, the less stable the conjugate base, because these alkyl groups are hindering water molecules from coming and stabilizing the conjugate base.
[The N atom is "shielded" by the bulky alkyl groups which take up a lot of space and prevent any water molecules from stabilizing the conjugate acid. Water molecules can only come in from where the H atom is where the shape of the conjugate acid is that of a pyramid. Note that there is a positive charge on N.]

As methylamine has the least number of alkyl groups as compared to dimethylamine and trimethylamine, at this point, it would be the strongest base. However, basicity also depends on the availability of the lone pair of electrons on the N atom of the amine itself. As methylamine has the least number of alkyl groups, the inductive effect by the alkyl group is the least out of methylamine, dimethylamine, and trimethylamine.
So, it really is a measure of which factor is more dominant or which factor outweighs the other. In this case, although trimethylamine has the most number of alkyl groups, the three bulky alkyl groups prevent a lot of water molecules from stabilizing the conjugate acid. And, for dimethylamine, although dimethylamine has a fewer number of alkyl groups than trimethylamine, the inductive effect of the alkyl groups outweighs the steric hindrance from the 2 alkyl groups as compared to the three in trimethylamine.
Hence, in an aqueous solution, $(CH_3)_2NH>CH_3NH_2>(CH_3)_3N>NH_3$
However, in the gaseous state, there are no water molecules to stabilize the conjugate acid and basicity depends on the inductive effect of alkyl groups; there are no water molecules to hinder. As trimethylamine is trigonal bipyramidal in shape, the factor of steric hindrance does not apply here. Hence, as trimethylamine has the most number of alkyl groups, the lone pair of electrons on trimethylamine is most available for donation.

In the gaseous state the basicity can simply be compared on the basis of availability of lone pair.
In tertiary amine the nitrogen atom receives +I effect from all the three methyl groups which increases the electron density on the nitrogen atom ,thus its lone pair is readily available. Whereas in secondary and primary amines the nitrogen atom receives+I effect from two and one methyl group respectively. So there lone pair are not as readily available as of tertiary amines.
When in aqueous phase  the amines interact with water. This interaction also includes hydrogen bonding. This interaction causes steric repulsion. Thus it affects the basicity and causes the order to be so.
Note : the shape of amines is Tetrahedral with three surrounding atoms and one lone pair. So there is minimal steric repulsion in gaseous state. The main reason for steric repulsion in aqueous phase is the interaction with water.

How does this interaction with water cause steric hindrance? Does the water molecules kind of form like a shield around the conjugate acid of the amine(cation)? But, if that's the case, would there also be steric hindrance for the other 2 molecules?


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Can carbon dioxide be used as oxidant for iron?
Determine whether it is possible for $\ce{CO2}$ to be an oxidant for $\ce{Fe}$. Justify your choice by comparing the corresponding half reactions.
I found the following half-reactions, both on reduction form:
\begin{align}
\tag{I}\label{I}
 \ce{Fe^2+ + 2e^- &<=> Fe(s)} & E^\circ &= \pu{-0.44 V}\\
\tag{II}\label{II}
 \ce{CO2(g) + 2H^+ + 2e^- &<<=> CO(g) + H2O}& E^\circ &= \pu{-0.1 V}
\end{align}
As iron should be oxidized, we turn \eqref{I} the other way and change the sign:
\begin{align}
\ce{Fe(s) &<=> Fe^2+ + 2e^-}& E^\circ &= \pu{0.44V}\\
&&E^\circ(\text{total}) &= \pu{0.44 V} - \pu{0.1 V} = \pu{0.34V}
\end{align}
$E^\circ(\text{total}) > 0$, and the reaction is spontaneous. But according to the solution manual $E^\circ(\text{total})$ should be less than $0$. Have I misinterpreted the task?

@HarshWasnik We prefer to not use MathJax in the title field, see here for details.

Anyone able to help me ?


text:
Why is Ni[(PPh₃)₂Cl₂] tetrahedral?
Since PPh₃ is strong field ligand and, the famous Wilkinson's catalyst, which also possess this ligand is square planar, then what makes the above complex tetrahedral?

We sometimes call this type of complex 'pseudotetrahedral' since there is an isomerism from a tetrahedral to a square planar complex possible. I was unable to find the original work here but this link gives some information. As you already mentioned there are two strong and two weak ligands so it's hard to tell how strong the ligand field splitting will be. For your particular complex it seems to be right on the spot where it would change from one to the other so depending on what you do you can influence the equilibrium. From what I read this may depend on the ability of the solvent to coordinate to the complex as well, the temperature, etc.
This is also mentioned in Earnshaw's Chemistry of the elements

Planar-tetrahedral equilibria. Compounds
  such as $\ce{[NiBr2(PEtPh2)2]}$ mentioned above as
  well as a number of sec-alkylsalicylaldiminato
  derivatives (i.e. Me in Fig. 27.6b replaced by
  a sec-alkyl group) dissolve in non-coordinating
  solvents such as chloroform or toluene to give
  solutions whose spectra and magnetic properties
  are temperature-dependent and indicate the presence
  of an equilibrium mixture of diamagnetic
  planar and paramagnetic tetrahedral molecules.

Can it be justified by saying that due to steric repulsions of the bulky PPh3, to minimise repulsions, the equilibrium is on the side of tetrahedral form?

Dichlorobis(triphenylphosphine)nickel(II), or $\ce{NiCl2[P(C6H5)3]2}$ in square planar form is red and diamagnetic. The blue form is paramagnetic and features tetrahedral Ni(II) centers. Both tetrahedral and square planar isomers coexist in solutions. Weak field ligands, favor tetrahedral geometry and strong field ligands favor the square planar isomer. Both weak field ($\ce{Cl−}$) and strong field ($\ce{PPh3}$) ligands comprise $\ce{NiCl2(PPh3)2}$, hence this compound is borderline between the two geometries.

Steric effects also affect the equilibrium; larger ligands favoring the less crowded tetrahedral geometry.[1]
Reference

Greenwood, Norman N.; Earnshaw, Alan (1997). Chemistry of the Elements (2nd ed.).