text:
Reference request on potassium cyanide production method
I kindly ask for a reference for the following claim about potassium cyanide, found at potassium.atomistry.com:

It is also obtained by the action of a mixture of steam and nitrogen on potassium carbonate.

I'd be very interested to know the temperatures required and yields obtainable, and of course if this works at all.

Related: How to prepare potassium cyanide from potassium ferrocyanide

I did some literature research and I wasn't able to find any methods involving steam (as water in the gas phase) for the synthesis.
This in fact makes sense as alkali metal cyanides are prone to hydrolysis and tend to decompose to formate $\ce{K(HCOO)}$ over time or in hot water:
$$\ce{KCN(aq) + 2 H2O(l) -> K(HCOO)(aq) + NH3(g)}$$
which further decomposes to potash $\ce{K2CO3}$ upon further heating, so presence of water, especially at elevated temperatures, appears to be highly undesirable for the synthesis of cyanides.
I suspect the aforementioned article contains a typo.
Probably it should read "stream of nitrogen" — then it would make more sense as there are several industrial processes utilizing atmospheric nitrogen fixation for making cyanides with coal proposed in late XIX – beginning of XX centuries.
First, there is a German patent by Caro issued around 1900 (I couldn't find the full text or the exact date) mentioning synthesis of the cyanides of group 1 and 2 elements from carbonates using nitrogen and coal [1].
Second, there is Bucher process [2], which, I think, http://potassium.atomistry.com might refer to.
There is a comprehensive review of Bucher process [3]:

The Bucher process, as known, consists in heating a mixture of sodium carbonate, charcoal, and iron powder (catalyst) in a stream of nitrogen at a temperature above 900°. The reaction is written as follows:
$$\ce{Na2CO3 + 4 C + N2 -> 2 NaCN + 3 CO}$$
[…] A stream of nitrogen was passed  through the tube while a certain temperature between 900° and 1000° was maintained for several hours.
  The reaction  product was rapidly cooled by lifting the tube out of the furnace without interrupting the flow of nitrogen and was then examined microscopically.
  In all experiments sodium cyanide, together with some of the ingredients of the original mixture, proved to be present, which confirms the reaction given by Bucher.

There is also a brief overview of industrial methods used for manufacturing alkali metals cyanides throughout the past 170 years in Ullmann’s Encyclopedia of Industrial Chemistry [4]:
1850s–1900s: semi-industrial process (aka "the blood process") of fusion of potassium carbonate with potassium hexacyanoferrate(II) (obtained from animal blood and waste) initially proposed by F. and E. Rodgers [5].
1900s–1960s: Castner–Keller process; primarily employed at Degussa AG [6]:
$$\ce{Na + C + NH3 ->[\pu{600 °C}] NaCN + 3/2 H2}$$
1960s–present sodium and potassium cyanides are produced exclusively by the neutralization process, between a solution of sodium or potassium hydroxide with liquid or gaseous hydrogen cyanide:
$$\ce{MOH + HCN -> MCN + H2O}$$

Gaseous or liquid $\ce{HCN}$ is first introduced continuously into concentrated solutions of $\ce{NaOH}$ or $\ce{KOH}$ at elevated temperature.
  The heat of neutralization is used for the subsequent vaporization of water. The use of special mixing devices and the addition of $0.2–3\%$ excess $\ce{NaOH}$ prevents polymerization.
  Simultaneously or in a second step, water is vaporized under reduced pressure and the alkali metal cyanide is precipitated below $\pu{100 °C}.$
  The shorter the residence time of the cyanide in solution, the cleaner the product and the lower the content of hydrolysis products and iron impurities.

Figure 6. Flow chart of the main production steps for cyanide
  production ($\ce{NaCN}$ as example)

References

Caro, Nikodem. German pat. anmeld. C15812 (8-4-09). Verfahren zur Darstellung von Cyaniden aus Alkali oder Erdalkalioxyden, bezw. Carbonaten, Kohle und Stickstoff.
Bucher, J. E. The Fixation of Nitrogen. J. Ind. Eng. Chem. 1917, 9 (3), 233–253. DOI: 10.1021/ie50087a011.
Posnjak, E.; Merwin, H. E. Note on the Bucher Cyanide Process for the Fixation of Nitrogen. Journal of the Washington Academy of Sciences 1919, 9 (2), 28–30. https://www.jstor.org/stable/24521552
Gail, E.; Gos, S.; Kulzer, R.; Lorösch, J.; Rubo, A.; Sauer, M.; Kellens, R.; Reddy, J.; Steier, N.; Hasenpusch, W. Cyano Compounds, Inorganic. In Ullmann’s Encyclopedia of Industrial Chemistry; Wiley, 2011. DOI: 10.1002/14356007.a08_159.pub3.
F. and E. Rodgers; Ueber Darstellung und Anwendung des Cyankaliums. Ann. Chem. Pharm. 1842, 41 (3), 285–294. DOI: 10.1002/jlac.18420410303.
Degussa, DE 124 977, 1900; DE 126 241, 1900; DE 148 045, 1901; DE 148 046, 1901; DE 149 678, 1901; GB 265 639, 1926 (H. Freudenberg); DE 851
056, 1943 (W. Stephan).

The complete paragraph, of which you took a particular phrase, reads:

The Potassium cyanide, $\ce{KCN}$, is manufactured by methods similar to those employed for the corresponding sodium derivative. Potassium ferrocyanide is heated either in absence of air, or with potassium carbonate and charcoal, or with sodium, the potassium cyanide being extracted by lixiviation with water or dilute alcohol. It is also obtained by the action of a mixture of steam and nitrogen on potassium carbonate or carbide, the manufacture of cyanides by the aid of atmospheric nitrogen being now an important industrial process.

This so-called process, the manufacture of cyanides by the aid of atmospheric nitrogen, is well documented in the book by John Edward Clennell, titled The cyanide handbook (Ref.1). For your convenience, I'd write the sub-chapter titled Production of Cyanides from Atmospheric Nitrogen:

It was observed by Scheele that when nitrogen is passed over a mixture of $\ce{K2CO3}$ and charcoal heated to redness, a cyanide of potassium is formed. Many attempts were made throughout the nineteenth century to utilize this reaction for industrial purposes. One of the earliest was that of Possoz and Boissiere, who used a mixture of charcoal with 30 per cent. of potassium carbonate. This was kept at a red heat in fire-clay cylinders, through which a mixture of $\ce{N}$ and $\ce{CO}$, produced by passing air over red-hot alkalized carbon, was allowed to pass for about $\pu{10 hours}$. The product was then lixiviated with water in presence of ferrous carbonate (spathic iron ore) to give a ferrocyanide.
It was also observed at an early date (by Clark, in 1837), that cyanides are formed as an efflorescence in blast-furnaces, and that the gases of these furnaces contain cyanogen. Bunsen proposed a special blast-furnace for the production of cyanide, in which coke and potash in alternate layers were to be heated by a strong blast, the fused cyanide running off at the bottom of the furnace. It was found, however, that a very high temperature was necessary, as the potassium compound must be reduced to metallic potassium before combination with atmospheric nitrogen takes place.
Better results were obtained by substituting barium carbonate for $\ce{K2CO3}$ as in the process of Margueritte and DeSourdeval (1861). Air (deoxygenated by hot carbon) was passed over a previously ignited mixture of $\ce{BaCO3}$, iron-filings, coal tar, and sawdust, whereby barium cyanide is produced. This is converted into sodium cyanide by fusion with sodium carbonate. The $\ce{BaCO3}$ is first reduced to barium carbide ($\ce{BaC2}$):
$$\ce{BaCO3 + 4C = BaC2 + 3CO}$$
This then combines with nitrogen to form $\ce{Ba(CN)2}$. It has been found, however, that only about 30 per cent. of the barium is converted to cyanide, the remainder forming barium cyanamide by a secondary reaction:
$$\ce{ Ba(CN)2 = C + BaCN2}$$
When calcium is substituted for barium in this process, practically the whole is converted into calcium cyanamide:
$$\ce{ CaC2 + N2 = C + CaCN2}$$
Calcium cyanamide is also formed in the electric resistance furnace by passing nitrogen over a mixture of lime and charcoal:
$$\ce{ CaO + 2C + N2 = CaCN2 + CO}$$
By heating the product at a high temperature with a further quantity of carbon, with the addition of salt to prevent frothing and facilitate the reaction, the cyanamide is converted into cyanide as follows:
$$\ce{ CaCN2 + C = Ca(CN)2}$$
The crude mixture so formed has been used as a substitute for potassium cyanide under the name of “Cyankalium surrogat,” and is equivalent in cyanogen contents to about 30 per cent. $\ce{KCN}$. [See Erlwein and Frank, U. S. patent, 708,333]

This book was written by a Cyanide Chemist to the Creston-Colorada Company, South Africa. Thus, he should have definitely tried out these methods, I guessed. He also have hold following titles at the time book was written: Associate of the Instituti on of Mining and Metallurgy; and Associate of the Chemical, Metallurgical, and Mining Society of South Africa. He is also the author of the book, Chemistry of Cyanide Solutions. Thus, I suppose these method should have worked.
If you interested, the limited version of the book can be downloaded from Forgotten Books website (download › The Cyanide Handbook).
I was unable to find the temperature range of the reactions from either references (Ref.1 and 2). However, I found a another patent, which also use the identical processes to synthesize alkali cyanogen compounds (Ref.3). This patent mentioned the temperature range of $\pu{800-1000 ^\circ C}$. In this process, sodium sulfide has been used in place of sadium carbonate. The authors depicted following equation as the result:

$$\ce{ CaCN2 + Na2S + C = 2NaCN + CaS}$$

References:

J. E. Clennell, The cyanide handbook; First Edn.; McGraw-Hill Book Company: New York, NY, 1910, "Part II, Section IV(b): Production of Cyanides from Atmospheric Nitrogen," pp. 128-130.
Greorg Erlwein, Albert Frank, “Cyanid compound and process of making same,” US Patent 1902, 708,333A (Application filed on: 1902-02-27; Application granted on: 1902-09-02).
John Collins Clancy, “Making of Alkali Cyanogen Compounds,” US Patent 1914, 1,112,893 (Application filed on: June 22, 1912; Application granted on: Oct. 06, 1914).


text:
Ion exchange resins
My problem is explained quite easily: I have no experience with ion exchange resins but I would like to test something out in order to exchange my sulfate ions in solution for nitrate (can also be perchlorate or tetrafluoroborate or something similar). Has anybody worked with such resins? Is there anything available on the market?
EDIT: I have seen that most resins work in water, but my compound is soluble in DCM, MeOH, EtOH and also chloroform, less so in acetonitrile, so I would need something that in the best case works in one of those solvents.

How big and how polar is your compound? You can load an anion exchange resin with hydroxide and then remove all anions in your solution, replacing them with hydroxide. Not sure if you could load it with nitrate instead.

@KarstenTheis pretty big (like 8Å across, but its a bit bent) and quite unpolar due to an extended pi-system

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What is the relation between volume strength and normality of H2O2?
I have been told that:
$$\text{equivalent weight of }\ce{H2O2} =\text{molecular weight of }  \ce{H2O2}$$
as it's $n$-factor is $1$ (from balanced equation).
But in my book it's given that:

Molarity $= \frac {V }{11.2}$
Normality $= \frac {V} {5.6}$

Why are they different if equivalent and molecular weights are equal?

The n- factor must be obviously wrong because two electrons are involved in the reduction of peroxide to oxygen. Write the half cell yourself. Indian chemistry textbooks should have pity on the world and stop teaching these equivalents to young students. This is a thing of the 18th century. Those who invented this concept moved on.

Don't come to such a conclusion of "Indian textbooks" abruptly. How many "Indian" textbooks have you read? And if you have read about equivalents properly, you would come to know that it's a very powerful tool.

@user600016 This might come out harsh, but the concept of equivalence has no use today as we are already able to determine molecular formulas and formula units. An equivalent and its derivative quantities such as gram-equivalent and normality are based on archaic representation. It's only powerful to those who don't know or don't want to know better. I'm not familiar with Indian literature, but judging from the questions posted by other users quoting Indian textbooks, there are way too many erroneous or outdated concepts still being taught (not to mention odd formatting and typos).

user600016, I have seen more Indian textbooks and older British books teaching the concept of equivalents than you can imagine. No modern book ever mentions equivalents. Some analytical chemistry books aimed for technicians use it because equivalents make calculations easier in terms of N1V1=N2V2, which can be easily memorized by those who don't want to think about mole ratios. Yes, it was a powerful tool but in the 18th century.

I did get the n factor wrong.

I just posted an answer in which I am indeed getting the n-factor for the overall decomposition of $\ce{H2O2}$ as $1$, which I believe is the same reaction using which volume strength is defined. The n-factor of $1$ seems to fit the law of chemical equivalence, so am I doing something wrong in my answer or is this answer wrong? Or maybe volume strength is not defined using the reaction $$\ce{H2O2 -> H2O + O2}$$?

@AshishAhuja, I am afraid your answer is wrong. The reason is that the equivalent weight is defined with respect to a chemical reaction. You cannot calculate eq. wt the way you showed it.

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Does reaction of alkali and alkaine earth metals' reaction with water an acid-base reaction?
In my chemistry notes the teacher gave me the note that reaction of alkali, alkaline earh metals' reactivity with water, acids(e.g. H2S) increases down the group and bases(e.g.NH3) decreases down the group,
And also I have some doubts that
-alkali metals such as Li first react with water and produce oxides first and those oxides react with water and produce hydroxides as end product, is this true?
-and normally water has H2O molecules, (H3O+,OH- ions at small concentration). And when alkali metals react with water do they react with H2O molecule or H3O+ molecule?, If someone could give me a reaction mechanism of an alkali metal with water?
-do alkli, alkaline earth metals act as bases when reacting with water?
Any help would be most greatfull

Na and K react with the O-H bond of the water molecule. This removes the H atom that is reduced to an H atom and then to a H2 molecule. It is not an acid/base reaction. It is a redox reaction. Na or K atom is oxidized at the end. No metal oxides are produced as intermediating state.
Alkaline and alkaline earth metals act in the same way.


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Structure of the periodic table
The periodic table I understand is arranged by order of increasing atomic number after Henry Moseley's X-ray experiment.
But I also learned that the periodic table also has its shape due to the types of subshells present in certain sections called blocks of the table.
However, I still do not understand who shaped the modern periodic table so it mimics the filling of lattermost subshells of atoms of each element (in order of increasing number of electrons- from left to right as well as top to bottom for the elements.)
My Question is : was the periodic table Mendeleev and Moseley devised not in a shape that can be used to predict the filling order of each elements, and that the periodic table was later adapted to fit this concept of orbitals in?

What about https://en.wikipedia.org/wiki/Periodic_table#History ?

Mendeleev was the first to realize that properties recur among the elements. To highlight this, he devised the early periodic table based on atomic masses. Moseley came along and stated that the periodic properties of elements was a function of their atomic numbers rather than their atomic masses.

It was Niels Bohr who came up with the idea of shells and subshells. He developed his own take on the periodic table, which looked somewhat like this:

Finally, coming to your question, A lot of derivatives of the periodic table were made by various scientists during the 1920's - 1940's. The version we use today was made by Glenn T. Seaborg and his colleagues.

Seaborg and collaborators had synthetically produced several new elements with atomic numbers beyond uranium, the last naturally occurring element in the table. Seaborg saw that these elements, the transuranics (plus the three elements preceding uranium) demanded a new row in the table, something Mendeleev had not foreseen. Seaborg’s table added the row for those elements beneath a similar row for the rare earth elements, whose proper place had never been quite clear, either.

Further reading:

How the periodic table went from a sketch to an enduring masterpiece
Further development of the Periodic Table - Wikipedia

Mendeleev actually wasn’t the first to realise that but his version of the periodic table was published first.

And Mendeleiev had the merit of having been able to predict the existence of missing elements, like Gallium and Germanium, and to predict their properties.


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Among hydrogen, helium and carbon dioxide, which gas would behave most like ideal gas and why?
I was wondering which gas would behave most ideally out of $\ce{H2},$ $\ce{He},$ $\ce{CO2}.$ All gasses are in the same condition.
I know the answer is either $\ce{H2}$ or $\ce{He}$ because of London dispersion forces. I was wondering which one it was and why.

It may depend on the considered gas property.

@Poutnik which one?

Most helium properties are closer to ideal behaviour, but I am almost sure I have seen one where the opposite is true. But I cannot remember which one.

At least the value of Joule_Thomson effect coefficient in a specific temperature range cca 130-650 K https://en.wikipedia.org/wiki/Joule%E2%80%93Thomson_effect?wprov=sfla1

"Same condition" means you want an answer that is true for all conditions. I don´t think there is one. Also it is not obvious what "most ideally" really means.

@Karl I suppose "most ideally" means having the least deviation from ideal gas behavoir.

@Poutnik You mean density at given pressure/temperature? You also mentioned the Joule Thomson coefficient. Or the heat capacity. Etc. Im pretty sure the answer would not be the same in all cases.

No, I mean whatever real versus ideal gas properties or behaviour. And that is what I say since very beginning, the answer may be case dependent. I would say helium is winning by sum of gathered points.

In the simplest model, a gas is called ideal when its particles are point-like (no volume) and have no interactions. Real gases behave like ideal gases at low pressure (where the particle volume is neglible compared to the total volume) and high temperature (where condensed phases, i.e. interatomic or intermolecular interactions are disfavored).
The size-comparison between helium and dihydrogen is straightforward: Dihydrogen is larger. As for the strength of interparticle interactions, we can compare normal boiling points: Helium's is 4 Kelvin and dihydrogen's is 24 Kelvin.
So this would suggest that helium is "more ideal" as it has the lower boiling point and the smaller size.

With rare exceptions where non zero volume effect and interaction effect by chance partially cancel each other better for hydrogen than for helium, even if  each effect alone is stronger for hydrogen.

A straightforward way to evaluate ideality is to compute the compressibility Z:
$$Z=\frac{PV_m}{RT}$$
Z equals 1 for an ideal gas, so deviations from this condition serve as a measure of non-ideality. If you examine a plot of compressibility for a real gas you will in general notice the existence of two regimes: at low pressure the compressibility is smaller than 1 and shows a minimum before rising again and eventually exceeding 1. The deviation from ideal gas behavior at low P is due to attractive interactions whereas deviations at high P are due to repulsive excluded volume interactions.
One way to infer ideality is to check van der Waals gas parameters. The following are the critical point parameters and vdW parameters for the gases:
$$
\begin{array}{|compound|T_c|P_c|} \hline \textrm{compound}&T_c/\pu{K}&P_c/\pu{bar}&V_c/\pu{L*mol^-1}&a/\pu{L*mol^-1}&b/\pu{bar*L^2*mol^-2} \\ \hline \ce{He} &5.195 &2.275&0.0578 &0.0346&0.0237  \\ \ce{H_2} &32.938&12.838& 0.065& 0.2465&0.0267 \\ \ce{CO_2} &304.14& 73.843&0.094&3.655&0.0428\\ \hline \end{array}
$$
Since helium has the smallest critical volume $V_c$, it also has the smallest van der Waals parameter $b$ (the covolume) and is expected to have the smallest hard-sphere radius. Helium also displays the smallest value of the van der Waals parameter $a$ which reflects the strength of attractive interactions, important at low pressure. This is consistent with helium being the smallest closed shell (inert) monoatomic (spherically symmetric) noble gas, and nearest among the gases to a point particle. If the van der Waals parameters are used to predict the compressibility coefficient (which they might do qualitatively if not quantitatively) using a freely available Matlab function the following figure is obtained at 50 K:

Clearly helium shows the smallest deviations from Z=1 and therefore behaves most ideally.
As hinted in a comment, $\ce{CO2}$ is solid at the temperature and pressure displayed in the above figure. At 250 K however it's behavior is still quite non-ideal:

Interestingly, at 250 K hydrogen is more ideal than helium (if only slightly so), if the vdW prediction is to be trusted.
An interesting aside is that helium has the lowest values of critical temperature $T_c$ and critical pressure $P_c$.

What is happening with the carbon dioxide? Is it a solid, liquid or gas at that temperature and that range of pressures?

@KarstenTheis Hmm, good point. Consider it "supercooled". The phase diagram says that above a few bar at 50 K it's a solid. Originally I had a plot at 200 K but it doesn't reveal the differences in nonideality between the gases as strikingly so for dramatic effect  I chose 50 K. But it's admittedly nonphysical. (https://en.wikipedia.org/wiki/Carbon_dioxide#/media)/File:Carbon_dioxide_pressure-temperature_phase_diagram.svg

I also agree with Karsten Theis that $\ce{He}$ would show more ideal behavior than that of $\ce{H2}$ for longer temperature range, just based on their boiling points. The boiling point of helium (~$\pu{4 K}$ or $\pu{-269 ^\circ C}$ at $\pu{1 atm}$) is more close to absolute zero temperature ($\pu{0 K}$) than that of $\ce{H2}$ (~$\pu{20 K}$ or $\pu{-253 ^\circ C}$ at $\pu{1 atm}$) where ideal gas have zero volume (theoretically; See Figure 1 below):

In general, real gases behave like ideal gases at high temperatures and/or low pressures. However, at lower temperatures or high pressures, the interactions between the molecules and their volumes cannot be ignored (e.g., van der Waals equation). For example, at lower temperatures (suppose at a constant pressure) close to its boiling point, the molecules get very close to each other, and hence, the interactions between molecules such as van der Waals forces start to act on each other. As a result, they tend to condense. This condensation causes a dramatic decrease in volume (see Figure 1 above). Thus, at first sight in comparison of real gas behavior between $\ce{H2}$ and $\ce{He}$, you can speculate that $\ce{He}$ has an extra temperature window ($\pu{16 K}$ more) to behave as a ideal gas than $\ce{H2}$ would.
In comparison to these two gases, $\ce{CO2}$ would not have a chance to have a better ideal gas behavior, based on its boiling point, ~$\pu{195 K}$ or $\pu{-78.5 ^\circ C}$ at standard pressure.
I'd like to end my argument on one important note: According to the literature, both $\ce{CO2}$ and $\ce{H2}$ exhibit a triple point of each, but $\ce{He}$  doesn't. Don't you think it says something? I think $\ce{He}$ is the one most closed to an ideal gas than any other existing gases.


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Determining relative changes in a generic reaction
I'm an electrical engineer by trade and I am trying to solve a problem I found in some old course notes.
Given is a reversible, homogeneous reaction:
$\ce{ A + 2B <=> C + 3D }$
A  solution of 0.2 moles of A and a solution of 0.3 moles B are added together. The solution is lengthened until 1.0 liters. At equilibrium, 0.05 moles of C is formed. The question is to determine the equilibrium concentration of D.
How should the 'relative changes' look like? I think it's:

$-x$ for A
$-2x$ for B
$+x$ for C
$+3x$ for D

It is, but it should be rather obvious even for electrical engineers, if A,B,C,D are consider as not chemical compounds, but as whatever spent resources respectively produced items.

Thanks Poutnik. I was doubting that the small amount of A could be a limiting factor. My answer also seemed a bit 'too obvious'.

Yes, A us limiting, but it does not affect ratio of changes. But I see a flow in task  If A has initially 0.02 mol, There cannot be created 0.05 lol of C. Max 0.02, if equilibrium constant K >> 1, as it is 1:1 ratio. Can it be a typo and there is 8nitially 0.2 mol of A ?

Yes, sorry, it's a typo.


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Transition metals and their ability to form coloured ions
I understand that in solution. transition metals form coloured ions, but what about the transition metals in solid compounds that are coloured- is this also due to d-d transition?
can you give any examples of transition metals bonded to ligands in a solid state?

Essentially, ionic compounds are just a different kind of coordination complex in most cases.

Edit: In this answer I've focused on solid solutions of transition metals, which are kind of half-way between pure transition metal solid compounds, and aqueous solutions of transition metal compounds (though more difficult to describe than either extreme...). I see now this is not necessarily what you were asking about, though it does fall under the question scope. Perhaps somebody else can supply the answer for the case of the pure transition metal solids.
Indeed, even in solid solutions, coordinated transition metals can be the source of colour.
The clearest examples are gemstone sapphires. The main component (generally >98% by weight) of sapphires is just aluminium oxide ($\ce{Al2O3}$), also known as alumina. Though extremely widely available (to the tune of over 100 million metric tons produced per year worldwide), it is often used as a fine white powder or white/grey chunks. However if pure aluminium oxide is melted over 2100 °C and cooled slowly, it can form large completely colourless crystals. These can also be found commonly in nature, as the mineral corundum.
(Even though these are colourless crystals, the industry already often refers to them as sapphire. See this YouTube video on the industrial production of "sapphire glass").
However, small amounts of other elements (including, but not limited to, transition metals) can form stable solid solutions with aluminium oxide. Here, the impurity is the solute, and the aluminium oxide is the solvent. These impurities generally bind to the oxygen atoms, so it could be said that these impurities form coordination compounds (oxo complexes) dispersed inside the solid solvent. Note that formation of solid solutions is common in many situations, and not at all limited to aluminium oxide.
The exact characterisation of the coordination geometry for these impurities and the local crystal structure surrounding them is typically more complex relative to coordination compounds in solution. This is because, in most cases, the impurity atoms do not fit perfectly in the crystal lattice of aluminium oxide, and therefore the atomic structure suffers some local distortion. However, these disturbances are commonly limited to only a few tens or hundreds of atoms immediately surrounding the impurity.
The colour generated in these solid solutions can actually come from several sources. The simplest case is where the impurity (e.g. a transition metal) causes virtually no structural or electronic changes to the overall solid solvent, and colour is generated from electronic transitions in internal and largely undisturbed electronic orbitals (e.g., d-d transitions or f-f transitions). I believe this is relatively rare, except when using lanthanide impurities.
In other cases, the small amount of impurity doesn't strongly affect affect the atomic structure of the main component, but it donates electrons to (or pulls electrons from) the main component. This appears to be the case for most coloured sapphires.
Another possibility is that the impurities actually do cause havoc in the overall atomic structure of the crystal. Changing the crystal's atomic structure causes large changes in the electronic band structure. This can trigger absorption of light, even if the original components of the mixture were both colourless and no chemical reaction has taken place. This is generally only the case with addition of a large amount of impurities, though.
There's several other ways to generate colour in solid solutions, though a full list is outside the scope of this answer. For a soft overview of gem colouration, I recommend the site Causes of Color. A more technical analysis can be found at the Gemology Project wiki.


text:
Conversion of barbituric acid to veronal
This is just out of curiosity. I have no intention or authority to do conversions.
I am studying chemistry of tranquilizers, in that an important class is of barbiturates, which are also hypnotics. The structure of Barbituric acid and veronal, are similar, except veronal has two ethyl groups instead of hydrogen at lone $sp^3$ hybridised carbon. How can I convert Barbituric acid to veronal?
I thought that making carbocation or carbanion will help, but I don't think carbon will get negative or positive charge until nitrogen has hydrogen. Probably we'll have to protect nitrogens, am I thinking right?

This paper from Tetrahedron Letters here details synthetic procedures using carbonyl compounds in solution with barbituric acid and Pt or Pd hydrogenation catalysyts to give reductive di-C alklyation. This includes examples of N-unsubstituted barbituric acids R1 = R2 = H. Using barbituric acid plus acetaldehyde should give you veronal in a single step.

What's the mechanism? (I don't have access to the paper)

Neither do I but my guess is aldol-type condensation of the active methylene followed by reduction of the double bond then a second similar condensation/reduction

@user55119  Thank you. I could only see the abstract of the TL paper.

@Waylander: I looked at the TL paper. Reductive monoalkylation with ketones and aliphatic/aromatic aldehydes works. Double alkylation only occurs reductively in the presence of strong acid with aromatic aldehydes. Although no mechanism was given, after monoalkylation, the benzylic alcohol formed from reduction of the aryl aldehyde forms a cation in acid and is alkylated by the enol of the monoalkylated product. This procedure will permit monoethylation but not the introduction of 2 ethyl groups.

@user55119 I wasn't able to get the whole of your comment, I understood firat part but second part of your seemed to be enigmatic. Can you please explain in a more elaborate way...

Symmetrical reductive dialkylation in the Waylander link only works with aromatic aldehydes sometimes in the presence of H2SO4. For barbituric acid itself, condensation occurs followed by hydrogenation. This monoalkylated species can undergo acid catalyzed aldolization followed by hydrogenolysis of the benzylic alcohol leading to the dibenzylated product. After some more thought, the cation I previous suggested is less likely. Aliphatic aldehydes only give monoalkylation because the penultimate aldol cmpd. cannot undergo hydrogenolysis. The formation of veranol this way appears unlikely.


text:
How to recover C18 column after prolonged exposure to phosphate buffer?
A colleague of mine who occasionally does HPLC in a biochemistry lab discovered that a C18 column was not properly flushed after the previous experiment.
It appears that the column was stored for several weeks capped with aqueous phosphate buffer inside.
The main concern is partially crystallized phosphate as well as bacterial growth as there is no clear evidence whether antibiotics were added to the buffer or not.
Is there a way to recover the column, and if so, what procedure is typically recommended (solvents, temperature, flow rate etc., preferably with a reference for the lab report)?

If it’s 100% aqueous, phase collapse is a concern. Regeneration procedure is described in http://files.pharmtech.com/alfresco_images/pharma/2014/08/22/55290841-3e3e-433b-a5d4-adb7f6495a06/article-32973.pdf

The bad news is that HPLC columns often die because this ill-treatment. We have had the same issues and eventually several expensive columns was discarded due to the negligence of a student. Acetonitrile, when used above 90% with a phosphate buffer, leads to precipitation inside the column. I have great deal of experience with column manufacturing technology. You never add antibiotics to mobile phases. So don't worry about that. The solvents themselves, like ACN, and MeOH ensure that bacteria don't survive inside the column. This is the typical storage solvent after washing the column and ensuring all the buffer is out.
The salvaging procedure can be the following:

Run the column at really low flow rates say, 0.1 mL/min with a high aqueous content (90/10 H2O/ACN). If the pressure is extensively high change the flow direction. Columns have a flow direction marked on them. Try the direction which shows a lower pressure.
Record the pressure, if the pressure slowly goes down, this is a very good news.
If the step 2 is going well, wash overnight at 0.1 mL/min. If all went well, pressure is back to normal in the morning, you are good to go.
Never sonicate the column. It will damage the bed permanently.

P.S. Phase collapse has nothing to do with problem of precipitation inside the column. This is a completely different story. Please don't use that suggestion from LC-GC magazine (link in the comment above). Ron Majors is very experienced but here he talks about another issue.
Phase collapse is not a well understood phenomenon, a better word is de-wetting. Since the C18 phase is very hydrophobic, with highly aqueous mobile phases, the pores (where all the retention occurs) exclude water. Chromatographers, especially those who write in commercial magazines for public readership, tend to do a lot of imaginations.
If the column pressure becomes normal, repeat and old separation. If the retention time is similar, no need to worry. If it is reduced, then try the LCGC suggestion. Pressure is the main test to confirm if all ppt is gone or not.


text:
Why is the name 'propylene oxide' preferred for epoxypropane?
This is the molecule I know as epoxypropane:

In my neck of the woods (example), this molecule is used as one of the standard examples of small chiral molecules. The name 'epoxypropane' makes a heck of a lot of sense to me $-$ an epoxy bridge  spanning a single bond of what would otherwise be propane.
It seems, however, that most everyone thinks that a better name for this is propylene oxide, from Wikipedia upwards. Now, I'm an atomic physicist, so bear with me:

I can see how the name 'propylene oxide' fits that structure. You've taken propylene and you've oxidized it.
However, I cannot see why the name 'propylene oxide' doesn't also apply to other ways to oxidize propylene, say, sticking that oxygen over at one end or in the middle,

i.e. what Wikipedia calls propionaldehyde (also propanal) and acetone, or making that bridge over the (1,3) combination in what Wikipedia calls oxetane,

none of which is chiral.

Now, I imagine that IUPAC has all sorts of rules to suss these things out, so:

why is 'propylene oxide' the preferred name for epoxypropane? and
why is epoxypropane the only isomer encompassed by the designation 'propylene oxide'?

Its only a https://en.wikipedia.org/wiki/Retained_name All simple chemicals have such names, and they are often preferred.

why is 'propylene oxide' the preferred name for epoxypropane?

Propylene oxide is not the "preferred" IUPAC name (PIN): Wikipedia lists the PIN as 2-methyloxirane, which is in line with the 2013 IUPAC recommendations.

why is epoxypropane the only isomer encompassed by the designation 'propylene oxide'?

It's more of a tradition than anything else. Note how the two ring carbons in propylene oxide are the same as the two doubly bonded carbons in propylene: this is not a coincidence. A very common way of synthesising an epoxide is to oxidise the corresponding alkene (you can think of it as simply adding an oxygen atom across the double bond), and so "alkene oxide" has become a rather conventional way of referring to an epoxide (see also https://en.wikipedia.org/wiki/Styrene_oxide for another example).
So, the meaning of "propylene oxide" is not really "an oxidation product of propylene". It should pretty much be read as "the epoxide derived from propylene", or "the epoxide formed when you add an oxygen atom across the double bond of propylene".
In the strict sense of oxidation/reduction being addition/loss of oxygen, oxetane is an oxidised form of propylene, but there is no simple chemical oxidant that will perform this transformation. So, calling oxetane "propylene oxide" would never have - and has never - caught on.

For full clarity, can acetone and propanal be considered to be oxidised forms of propylene? Or have you singled out oxetane for a reason, and the double bonds change the extent to which one can speak of oxidation there?

The issue isn’t so much about what is an oxidised form of propylene or not; they all are, including oxetane, which is what I wrote above. But if you’re asking about whether there are chemical transformations to take propylene to acetone or propanal, then yes, there are the Wacker oxidation and hydroboration/oxidation respectively.

...and just to ward off any potential objections: the most common hydroboration/oxidation protocol uses H2O2/NaOH at the end, and this produces an alcohol instead of an aldehyde, but you can use other oxidants (NMO/TPAP) to go straight to the aldehyde.

No, I was just wondering why you had singled oxetane out. Thanks for the detailed response!

The name propylene oxide does not fit the structure. It fits the synthesis, because this substance is made by oxidation of propylene (old name of propene). The oxidant is usually a peracid, or hydrogen peroxide.


text:
Why is the entropy of container A the greatest?
Question

9.35 Container A is filled with 1.0 mol of the atoms of an ideal monatomic gas. Container B has 1.0 mol of atoms bound together as diatomic molecules that are not vibrationally active. Container C has 1.0 mol of atoms bound together as diatomic molecules that are vibrationally active. The containers
  all start at $T_\mathrm{i}$ and the temperature is increased to $T_\mathrm{f}.$ Rank the containers in order of increasing change in entropy. Explain
  your reasoning.

Answer

 B < C < A

My Approach
I thought that you were supposed to assume that container C had the most number of atoms and therefore a greater entropy automatically.

An accurate summary of the relative molar heat capacities of the different species is given in another answer and recapitulated in the following table together with the total heat capacity for each gas:
$$\begin{array}{|case|Cv|} \hline \textrm{molecule} & C_{vm}/R & NC_{vm}/R \\ \hline  \mathrm{A} & 3/2 & 1.5 \\ \mathrm{B} & 5/2 & 1.25  \\\mathrm{C} & 7/2 & 1.75 \\ \hline \end{array}$$
where $N$ is the total amount of gas, which based on the problem described in the OP is either $\pu{1 mol}$ of atoms ($\mathrm{A}$) or $\pu{\frac{1}{2} mol}$ of homonuclear diatomic molecules ($\mathrm{B}$ and $\mathrm{C}$).
Since the entropy change for the isochoric temperature change is $\Delta S = C_v \log\left( \frac{T_2}{T_1} \right)$, it follows that $\Delta S_C > \Delta S_A > \Delta S_B$.

In $\mathrm{A}$, the atoms are only increasing their internal energy by the three degrees of liberty due to translation. There are no rotation $n_i$ vibration. So $C_v = 3/2 R$
In $\mathrm{B}$, there is two degrees of liberty due to rotation. This makes twice $R/2$ to add to the precedent values. $C_v = 5R/2$.
In $\mathrm{C}$, there is two new degrees of liberty due to vibration which increases potential and kinetics energy. This makes a supplement of $2 \cdot R/2 = R$, which has to be added to the precedent values, $C_v = 7R/2$.
For a given change of temperature, more energy $\Delta Q$ has to be given to the gas when changing from $\mathrm{A}$ to $\mathrm{B}$ to $\mathrm{C}$.
As $dS = dQ/T$, the change of entropy follows the same order. So $\mathrm{C}$ has had a greater change of entropy than $\mathrm{B}$ and of course than $\mathrm{A}$.

The important detail is the molar amount of atoms, so B and C contain a half molar amount of substance, compared to A.


text:
Mass Spectra: 1-pentanol vs 3-pentanol

In mass spectrometry, 1-pentanol can be observed to have no peak at $m/z = 59$, whilst 3-pentanol can be observed to have a peak at $m/z = 59$.

Clearly the 3-pentanol peak at $m/z = 59$ is due to the cleavage creating $\ce{CH3CH2CHOH | CH2CH3}$
Why is this the case, as isn't 1-pentanol able to form the fragment $\ce{CH2CH2CH2OH}$ by cleaving at the third carbon?
i.e. fragmenting into $\ce{CH3CH2}$ and $\ce{CH2CH2CH2OH}$ (this should have $m/z = 59$?).

RE: "... Why is this the case, as isn't 1-pentanol able to form the fragment CH2CH2CH2OH by cleaving at the third carbon?..." Of course 1-pentanol can cleave in such a manner. It is just that such cleavage is highly improbable.

In mass spectrometry, organic compounds can be ionized by different methods. The compounds ionized by electron impact ionization display particular fragmentation patterns. When subjected to mass spectroscopy, cleavage of the $\ce{C-C}$ bond next to the oxygen or nitrogen can usually occur, when the compound is ionized by electron impact ionization.
Let’s look at what happens to 3-pentanol:
$$\ce{CH3CH2CH(OH)CH2CH3 ->[e- beam] CH3CH2CH(O^.+H)CH2CH3 }$$
$$\ce{CH3CH2CH(O^.+H)CH2CH3 -> CH3CH2CH=O^+H + ^.CH2CH3}$$
Now, base peak (most probably), would be $\ce{[CH3CH2CH=OH]+}$, which should be at $ 59 \ m/z$. Note that in general, the molecular ion of an alcohol (here it is $ 88 \ m/z$) is either trace or non-existent, probably loosing $\ce{H2O^.}$.
Now, we'll look at what happens to 1-pentanol:
$$\ce{CH3CH2CH2CH2CH2-OH ->[e- beam] CH3CH2CH2CH2CH2-O^.+H }$$
$$\ce{CH3CH2CH2CH2CH2-O^.+H -> CH2=O^+H + ^.CH2CH2CH2CH3}$$
Now, base peak (again, most probably), would be $\ce{[CH2=OH]+}$, which should be at $ 31 \ m/z$ ($88-31=57$ mass loss). Note that here again, the molecular ion is either trace or non-existent, probably loosing $\ce{H2O^.}$, and possibly $[M-18]$ ion should be significant here, since there is equal chance of cleaving $\ce{C_1-O}$ bond (giving peak at $ 70 \ m/z$) or $\ce{C_2-C_3}$ bond (giving peak at $ 31 \ m/z$):

However, according to the mass spectrum given, $\ce{C_5}$-carbon chain has undergone McLafferty type rearrangement (which is common in MS of aldehydes and ketones) to give the base peak at $ 42 \ m/z$ by loosing a water and ethylene simultaneously from parent ion (see the insert in above diagram).


text:
Are all enolised molecules called enolates?
From what I've gathered in my lectures, enolates are enolised cabonyl compounds. However e.g. a nitrile with an H on the alpha C can undergo enolisation. Is the resulting molecule called an enolate or an enolate equivalent, or something else?

There are two questions in here.
First enols/enolates. It is correct that all enolates are enolised carbonyl compounds but not all enolised carbonyl compounds are enolates. Take for example pentan-2,4-dione. While in protic solvents this will indeed have two $\ce{C=O}$ groups, if dissolved in unpolar solvents it will enolise to allow for a cyclic intramolecular hydrogen bond for stabilisation (see scheme).

Obviously, the resulting compound is an enol, as the functional group $\ce{R2C=CR-OH}$ (a hydroxy group bound directly to a vinylic double bound) is present. However, it is not an anion so it is not an enolate.
Most carbonyl compounds prefer the keto-configuration but if there are electronic factors that favour enolisation (e.g. enolisation results in a sufficiently large conjugated system) then an enol will be preferred.
Enols can be easily deprotonated at the oxygen so the generation of an enolate is straightforward. However, since ketone and enols are tautomers of another with the key limiting step being hydrogen transfer ([1,3]-suprafacial hydrogen migration is forbidden), once the proton is removed the difference becomes purely electronic and thus deprotonated ketones are routinely referred to as enolates whether or not the cation seems more attached to the oxygen or the carbon. (Reaction mechanisms differ; sometimes only one attack (often carbon) results in a productive mechanism so it seems as if the other had not reacted or be not present but this is often a hindsight fallacy.)
The term enolate is not limited strictly to ketones. For one of the more synthetically relevant examples, see the Evans enolate used in Evans alkylation or Evans aldol reactions or the Myers enolate used in Myers alkylation reactions; both are amide enolates.

However, these compounds are still characterised by the presence of a $\ce{(C-)C=O}$ bond which upon enolisation becomes a $\ce{(C=)C-OH}$ functionality. In the case of a nitrile or an imine, the oxygen is not present so even though a similar reaction may take place it is not correct to call the resulting product enol. For imines, the compound is known as an enamine and happens to be the standard outcome if a secondary amine is reacted with a carbonyl group.
For nitriles, I am not aware of a similar shorthand and would probably simply call it a nitrile anion (as does Wikipedia). (It is worth pointing out that nitrile anions are somewhat different from enamides/enolates as they are essentially an aza-ketene, not an aza-allyl anion.)


text:
What is more stable, Na+ or Na?
As per noble gas configuration $\ce{Na+}$ should be more stable than $\ce{Na}$. But at the same time $\ce{Na+}$ has a positive charge so it will be easily attracted by negative charge while $\ce{Na}$ will remain unaffected in presence of negative charge. So accordingly, should not $\ce{Na}$ more stable than $\ce{Na+}$? Please help me get out of this confusion.

If you mix NaCl with elemental sodium, nothing will happen. So they have the same stability???

Actually, if you put $\ce{NaCl}$ and metallic $\ce{Na}$ to two separate beakers full of water, which would catch fire?

You can't compare stability of these two things. It simply makes no sense.

In vacuum Na is certainly more stable than its ion and a free electron. This is crystal clear. But see also Ivan Neretin comment (you cannot compare different things, at least one should be aware of the differences) as well as the answer by Poutnik.

context is all. Under what conditions is the question asking about (if it doesn't specify them, it is a bad question).

Relative stability of both depends on environment.
In gaseous atomic phase $\ce{Na}$ is somewhat more stable than $\ce{Na+}$  and  $\ce{e-}$ pair due the needed ionization energy.
In water, Na  is very unstable, as a big  amount of energy is released by reaction of $\ce{e-}$ with water and by hydration of Na+. That leads to very fast up to explosive reaction.
Note that this explosion ( that can be accompanied by the hydrogen-oxygen explosion ) is primarily a physical Coulombic explosion due excessive positive charge of melted sodium.

To Poutnik: what causes melted sodium to get an excessive positive charge? Would an explosion occur in the absence of atmospheric oxygen?

Fast electrons leave sodium much faster than slow sodium ions, what charges sodium positively. Just before Coulombic explosion, sodium builds charged spikes of metallic sodium, pinching through isolation layer of water vapour and hydrogen and by sparks ignites the stuff at contact with air.. It was both simulated by quantum calculations and observed by high speed cameras. Search for recent Nature article.

Thanks! I remember now an article about NaK alloy, which is much more reactive.

@James Gaidis. Yes it is. But the main point in using it was it form an eutecticum with low melting point, so they could work with liquid Na-K alloy.

Many years ago, I did an experiment with what I considered a small piece of sodium (about 20 grams). I dropped it into a laundry tub (in our basement)  which contained about 2 inches of COLD water (because I had read that sodium would explode in hot water). I watched it fizz and zip around on the surface for about 15 seconds; then it suddenly melted and became globular. About 5 seconds later, it blew up, throwing little droplets of sodium all over, but they rapidly burned up. I suppose the spikes could cause sparks. Lithium reacts slower and does not seem to cause explosions.


text:
Which gas is a oxidizing agent as well as a reducing agent?
I think it's Hydrogen, because,
$\ce{PbO + H2 -> Pb + H2O}$
Here $\ce{PbO}$ is reduced to $\ce{Pb}$ and $\ce{H2}$ acts as a reducing agent.
But, in $\ce{2K + H2 -> 2KH}$, hydrogen gains a electron and undergoes reduction
Thus, $\ce{K}$ is oxidized to $\ce{KH}$ and hydrogen works as oxidizing agent.
My teacher says it's wrong but doesn't give any explanation.
Can someone explain why my reasoning is wrong.

Considering you made a typo here, and the first reaction is $\ce{PbO + H2 -> Pb + H2O},$ it looks OK to me. Probably you got a picky teacher and they also expected you to add conditions and states of aggregation, but this is something you need to sort out with them.

Like Einstein "said", "Everything is relative..."

@andselisk The question on the test asked for a one word answer. As far as I know everyone else wrote Sulphur Dioxide as the answer. I think I was absent the day she explained redox reactions, she probably mentioned it in class.

By the way, "Sulphur Dioxide" is a two-word answer! :-)

Some teachers are like a tram: they go on rails and won't swerve either way, no matter what. Indeed, SO2 is an oxidizing and a reducing agent. So is H2, and so is HI, and so are many other gases (arguably, more than those that aren't).

@MaxW could you please elaborate :-)

@Padmanava - Look at the position of the half cell reaction in a table of reduction potentials. The relative position of the half-cell reaction tells you for which reactions the half-cell can be an oxidant, and for which reactions can be an reductant.

Another case is N2, forming NH3 and NO with H2 resp. O2. But both need special conditions. Another case is NO2.

I'm guessing your teacher is looking for sulfur dioxide as the answer, but I don't see how or why you're supposed to be able to arrive to this answer logically. Either you'd need to read about it specifically, or maybe you're supposed to stare at a table of standard reduction potentials and notice that $\ce{SO2}$ appears as both a reagent and as a product with comparable voltages:
$$
\begin{array}{}
\ce{SO2(aq) + 4 H+(aq) + 4 e- &<=>& S(s) + 2 H2O(l)} &  E^\circ = \pu{+0.50 V}\\
\ce{SO2(aq) + 2 H2O(l) &<=>& SO4^2-(aq) + 4 H+(aq) + 2 e-} &  E^\circ = \pu{-0.17 V}
\end{array}
$$
(The bottom equation is the inverse of what you would find in a reduction potential table)
Extrapolating from these aqueous potentials, it appears $\ce{SO2}$ is a modest reductant and a weak oxidiser, but both reactions can be accessed in reasonable conditions.
I think you're technically correct (as Ivan mentions), but what makes hydrogen gas a "less correct" answer is that it only rarely acts an oxidiser (in the presence of very strong reducing agents, such as alkali metals), whereas it can act as a reductant much more commonly. Compare these equations with the ones above:
$$
\begin{array}{}
\ce{H2(g) + 2 e- &<=>& 2 H^-(aq, extrapolated)} &  E^\circ = \pu{-2.3 V}\\
\ce{H2(g) &<=>& 2 H+(aq) + 2 e-} &  E^\circ = \pu{+0.00 V}
\end{array}
$$
(Reference for the hydride potential)
Note the much larger gap in potential between the two reactions. In this sense, the reducing nature of $\ce{H2}$ vastly overwhelms its oxidising nature, so most people will think of it as a reducing gas.

My first thought was also hydrogen, and I was once a chemistry teacher myself. If the question was "Name a gas that can both reduce and oxidize reactants," then H2 is a correct answer.

The first candidate that comes to my mind is carbon monoxide. Electropositive metals such as magnesium[1] can use it as a combustion oxidizer, while iron is commonly smelted using carbon monoxide as a reducing agent.
References

E. Ya. SHAFIROVICH & U.I. GOLDSHLEGER (1992) "Combustion of Magnesium Particles in CO2/CO Mixtures". Combustion Science and Technology 84:1-6, 33-43. https://doi.org/10.1080/00102209208951843


text:
Redox reaction: Iron(III) and Iodide ion
I'm a real beginner in chemistry and I'm currently reading about redox reactions. I'm quite confused about this simple redox reaction:

${\text{Iron(III) ion + Iodide ion} \rightleftharpoons \text {Iron(II) ion + Iodine}}$
\begin{array}{|c|c|c|c|c|c|}
\hline
\text{Oxidation numbers}& +3 & -1 &\rightleftharpoons & -1 & 0\\ \hline
 & Fe^{3+}& I^-&\rightleftharpoons & Fe^{2+}&I_2\\ \hline
\end{array}

I understand the principles of transferring electrons. But how would I come from the upper formula to the result that the Iodine exists two times on the right side? After reading the reaction formula I would have written the Iodine just once (not as seen here twice).

You wouldn't. The fact has nothing to do with this reaction (or redox reactions in general, or any other reactions, for that matter). It is just that elementary iodine exists as I2.

Thank you. If you want, you can write that as an answer, I would accept it then.

Google is your friend. Balance atoms first then electrons.

Iodine exists as a diatomic molecule. This reaction indicates how the iodide ion is converted into diiodine. Your reaction just indicates the electron transfer: you still need to balance the reaction.
You need an $\ce{I_2}$ molecule on the right hand side, which would need two iodide ions on the left hand side:
$$\ce{2 I- -> I_2 + 2e-}$$
To accept two electrons, we would need two $\ce{Fe^3+}$ ions:
$$\ce{2 Fe^3+ + 2 e- -> 2 Fe^2+}$$
Now we have a balanced reaction. All the electrons donated by iodine are accepted by iron. Adding up the two reactions and cancelling the electrons gives us:
$$\ce{2 Fe^3+ + 2I- -> 2 Fe^2+ + I_2}$$


text:
Exceptional electromagnetic properties of lanthanide series elements
Why do so many lanthanide series elements have exceptional electromagnetic properties? They can form strong magnets and are also used in superconducting applications. The number of elements in the lanthanide series displaying these properties is far higher than other blocks in the periodic table
One thought is that these properties arise as a result of the magnetic moment of the atoms themselves: filling up the f-orbital contributes to both angular orbital momentum as well as spin orbital momentum. Is this speculation correct?
References:

Another important use of neodymium is as a component in the alloys used to make high-strength neodymium magnets—powerful permanent magnets.

Neodymium - Wikipedia, the Free Encyclopedia

Europium becomes a superconductor when it is cooled below 1.8 K and compressed to above 80 GPa. This occurs because europium is divalent in the metallic state,[8] and is converted into the trivalent state by the applied pressure.

Europium - Wikipedia, the Free Encyclopedia

Samarium doping of iron-based superconductors – the most recent class of high-temperature superconductors – allows enhancing their transition temperature to 56 K, which is the highest value achieved so far in this series.

Samarium - Wikipedia, the Free Encyclopedia

I think it'll be difficult to find a good answer that encompasses superconductivity, because there are still important gaps in our theoretical understanding of high-temperature superconductors.

Also, many of the key properties (for both superconductivity and magnetism) are dependent on the bulk properties of the solids or alloys containing the element and not just the properties of the individual atoms. That makes any explanation much more complex.

A simplistic answer may indeed be the open f-shells. The relativistic effects should not be ignored, either.

@TAR86 Relativistic effects, to my knowledge, are only significant for the 6th period elements and below eg. Gold, Cerium, Mercury and Tungsten.

@AniruddhaDeb I believe that the lanthanides are 6th period elements, and cerium is one of the lanthanides.


text:
Nomenclature - Lanthanoids vs Lanthanides (and Actinides vs Actinoids)
Apologies if this seems like a silly question.
Rare earth elements (the 4d series) are called by the names Lanthanides or Lanthanoids interchangeably. The same applies for the 5d series, which are called Actinides or Actinoids.
Which one of these two terms is more preferred? Is there an IUPAC directive preferring one over the other in scientific documents? If one is to write a paper or book regarding these elements, which term would one use? Or can both of them be used interchangeably?

Briefly, since "-ide" implies an anion, using suffix "-oid" is more preferable. I bet I saw this question before either on Chemistry.SE or on English.SE.

@andselisk I did search for a similar question on Chemistry SE but couldn't find it. Hence I posted the question.

According to p 52, Nomenclature of Inorganic Chemistry: IUPAC Recommendations 2005 (‘Red Book’):

The ending ‘ide’
  normally indicates a negative ion, and therefore lanthanoid and actinoid are preferred to
  lanthanide and actinide.

In practice this is almost never followed, and consequently both terms are (for all intents and purposes) interchangeable, unless the context demands a differentiation.

Let me just add that lanthanide/lanthanoid anions do not exist (at least under any reasonable conditions) so there is no concern of confusion.


text:
How are the Van der Waals constants a and b related to each other?
I know that the constant $a$ represents the attractive forces between the molecules while the constant $b$ represents the size/repulsion between them. Does that mean that $a$ and $b$ are "inversely proportional" or are they independent of each other? If we have two gases and we can compare their sizes and attractive natures, can we compare $a$ and $b$ for the two gases? If $a$ is greater for gas 1 will $b$ always be lesser for gas 1 as compared to gas 2?

They are independent.

I love how Ivan Neretin says so much with just few words.

Yes, they are certainly correlated as shown in the following plot based on data from the Wikipedia Van der Waals constants data page

The correlation should not be entirely surprising: the covolume b is a measure of the size of the gas molecules, whereas a is a measure of the strength of intermolecular interactions which is expected to depend through polarizability and dispersion interactions on the size of the molecules (molecular volume).
For a given gas parameters a and b tend to increase together. The shown correlation is empirical but since both a and b depend on molecular size and apparently in a very similar functional way, the result is a linear correlation.
Following the hint from Karl, here is a version of the plot with log(a) along the abscissa, and with superimposed labels.

Thank you! I expected them to be decreasing functions of each other but it instead seems like a line with positive slope. I still have two questions: 1) Why are they related through an increasing function and 2) Does this mean that if a is greater for gas1 b will also be greater for gas1?

Nice graph! If you label the dots and plot log(a) vs b, you notice a few notorious weirdos who stick out: H2, He, Ne, mercury and water.

H2 and He are actually nicely on the line, while water and mercury stick out, which is somewhat tbe. I wonder whats wrong with neon: https://i.stack.imgur.com/B9s6W.png

Neon is decidedly odd. It just wants to be left alone :-)

@Karl: The dataset from Wikipedia gives neon a lower $b$ than helium, which seems a priori surprising (and apparent I'm not the only one surprised by it). I haven't checked the literature, but I wouldn't be too surprised if the value for neon was simply inaccurate (and/or poorly constrained by available data, or a result of some behavior that does not actually fit the van der Waals equation well).

Just for curiosity, I also plotted a vs b using the same date set used by Buck Thorn. What I want to know is what is the correlation shown in Buck Thorn's plot and what are the two outliers clearly in the plot:

As I marked in my plot, they are mercury and idobenzene. As Karl pointed out in one of his comments, mercury is the biggest outlier. Other than these two, other compounds have shown fairly good linear relationship with $R^2 = 0.9308$.
Again, based on few comments, I picked just the noble gasses and hydrogen to see how they would be doing since they are closed to ideal gas behavior. First, again following Karl's suggestion, I checked with $\log a$ versus $b$. The best fit ($R^2 = 0.999$) is the $2^\circ$-degree polynomial curve, but best fit come with the lost of hydrogen and helium (see the left top plot). Both values for helium and hydrogen are way off (not shown here). However, when I plot all $a$ and $b$ data of noble gasses and hydrogen (without log values), it gives a nice linear relationship ($R^2 = 0.9571$). Still, there is one outlier: Neon. As Buck Thorn nicely put it, neon just wants to be left alone!
So I eliminated the data for neon and re-plot the remaining data and it came out even a better fit linearly (compatible with the whole data set):

Should not them be rather noble gases (like nobleman ) than Nobel gases ? :-) It makes an impression they may be Nobel prize laureates. :-)

I always seems to make that mistake! :-) Will make the corrections.

You make a quadratic fit in the half logarithmised data in your upper left plot. Im afraid that is highly unphysical. Also: 5 datapoints, four of which correlate very well, three parameters. Dont need much luck to get a good fit there.

@Karl: Well, tha's what happens when a non-physical chemist try to be a one.:-) Nonetheless, I can't resist it's such a good relationship ($R^2=0.999$).

As Ivan writes $a$ and $b$ are independent, but it does not mean that they are not correlated; however, so are a child's shoe size and reading ability which clearly does not mean that buying a child larger shoes will increase its reading ability. A correlation does not imply cause and effect.
Plots of the compression factor with reduced pressure show that many gasses follow a general trend when using the van der Waals equation. The Critical temperature is defined as $T_\mathrm c=8a/(27R b)$ ($R$ is gas constant) which shows the connection between $a$ and $b$ for any gas but the values of $T_\mathrm c$ vary between gases. A plot of $a$ vs $b$ is thus effectively a plot of (scaled) $T_\mathrm c$ values. The other critical constants are $V_\mathrm{cm}=3b$ (molar volume) and $P_\mathrm c=a/(27b^2)$. Look for the Law of Corresponding States in a good phys. chem. texbook, such as Engel & Reid for more detail.

So the relation between A and B was a false alarm?

@aditya_stack No. The critical pressure and temperature both contain a and b. That is a strong hint that these two parameters are not independent at all.

Of course you can make a hen and egg problem out of this: Are Pc and Tc independent? And they are of course not. After all, we don´t have statistical data here, but physical parameters of basic matter.

@Karl, if $a$ and $b$ are fundamentally related then what is the physical basis for this, in other words, what is the derivation of $a$ and $b$ from first principles that unveils their relationship?

Is Engel & Reid for Undergrad level courses or can it be digested by a highschooler also?


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How to make elemental silicon?
I want to extract elemental silicon from pure silicon dioxide powder.
From what I've read, one way to do it is by adding carbon from graphite at very high temperatures, which is difficult.
A better way to do it is to mix with magnesium powder and heat it to about $800 ^\circ \rm C$.
Some places mention the use of a hydrogen atmosphere, is that necessary?
What is the best way to do it, the best temperature? If I use a torch, wouldn't the exhaust gases purge the oxygen and serve as an inert atmosphere?
How do I know the reaction is over?

The exhaust gases, mainly CO2 will react with Mg at 800C. Nitrogen reacts with Mg at 800C to form Magnesium Nitride https://en.wikipedia.org/wiki/Magnesium_nitride. This is why an unreactive and reducing atmosphere of hydrogen is used. Argon might would probably be OK. This is not something to be attempted without the proper equipment

Making elemental silicon is easy, but not for amateurs. Since it is cheap (~$2/kg) there is little reason to do this yourself.
Industrial processes do indeed mix silica with a source of carbon like coal. These react when very hot to give silicon. But the process usually uses an electric arc furnace and controlled conditions with carefully controlled use of gasses (including extra oxygen late in the process to enhance the amount of impurities that gather in the surface slag). It is unlikely that an amateur can easily reproduce this control.
You might get a result at lower temperatures using magnesium. But you will struggle to do this safely without some control of the atmosphere above the mix as magnesium will burn violently in air (or even in carbon dioxide). You might even need to use argon to blanket the mix as even nitrogen will react with magnesium.
So, basically, this is not something to contemplate as an amateur outside a well-equipped lab.

Assuming you do not burn the magnesium, it can still react with the silicon forming $\ce{Mg2Si}$.  Carbon is less prone to a similar side reaction.

Silicon dioxide is a very thermodynamically stable compound. In most labs I have been in, $\ce{SiO2}$ is equated to sand whenever we aren’t explicitly talking about silica gel (or sometimes celite, but that tends to get referred to as celite). Considering how hard it is to make anything out of sand that isn’t essentially still a $\ce{SiO2}$-derived compound should point to how difficult it is to break the $\ce{Si-O}$ bonds.
To make elemental silicon from the dioxide, you need to reduce it, so in theory all your standard reduction agents should work. Carbon is considered to be one of the cheapest, easiest to use and most readily available reducing agents (except maybe the cation which requires liquid phases). That, the equally high stability of $\ce{CO2}$ and the fact that the transformation is entropically favourable as a gas is generated from two solids make carbon an obvious choice in industry. (Note: I did not go through the trouble of looking at the actual processes, I am just deriving the obvious constraints from my chemical background knowledge.)
However you are going to synthesise elemental silicon, you need to ensure that it doesn’t get reoxidised rapidly. The higher the temperature, the more activation energy is supplied and the more likely it is for the back reaction to occur. A hydrogen atmosphere serves a double purpose: not only does it displace oxygen preventing reoxidation but it is also a reductant in itself and may aid (given appropriate conditions) the reduction of $\ce{SiO2}$. In case of a highly reactive reductant such as magnesium, it also prevents oxidative side reactions of magnesium to a certain extent.
All that said, industry (especially semiconductor industry) has a huge demand for silicon, meaning a lot is synthesised industrially; the raw materials (sand and coal) are almost literally dirt cheap and thus so is the price of industrial silicon. As very high purity silicon is needed in certain applications, you can likely acquire it from crude unpurified to extremely high purity, depending on what your situation requires. There is really little point in trying to make it oneself.

Another way of getting Silicon, is first to prepare SiCl4 by having a current of Chlorine gas crossing a horizontal tube filled with a mixture of silica SiO2 + charcoal C in a flame, but not at a as high temperature as the electric furnace mentioned by Matt black. The following reactions happens : $$\ce{SiO2  +  2 C  + 2 Cl2 ->SiCl4 + 2 CO}$$  SiCl4 is a gas above 57°C. It is condensed by cooling the gas mixture getting out of the reaction tube, than purified from impurities like AlCl3.
Later on, SiCl4 reacts with magnesium at high temperature, according to the reaction : $$\ce{SiCl4  +  2 Mg ->  Si  +  2MgCl2}$$ This produce pure Silicon, and MgCl2 which liquid à 112°C.

Magnesium chloride is not liquid at 112°C.  Wikipedia gives a melting point of 117°C for the hexahydrate if heated rapidly, but this system is clearly anhydrous.  The relevant melting point, again from WP, is then 714°C.


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Why don’t compounds with high boiling points not necessarily have high attraction constants (a) in the Van der Waals equation?
You can find a list of constants $a$ and $b$ for the van der Waals equation in a Wikipedia website. Accordingly, hexane has a higher attraction constant ($a$) than that for water (24.71 vs 5.536), but water has a higher boiling point than hexane. So, intuitively water would have stronger attraction between gas molecules. In general, the alkanes demonstrate high $a$ values. Why is this?
My guess is that since the alkanes are symmetrical, it does not matter how the molecules near the edge of a container are oriented, they will be attracted toward the middle of the container no matter what. On the other hand, water’s asymmetry means water molecules have to be oriented a certain way in relation to the other molecules in order to be attracted to them. While in liquid form, water is more often oriented in this attractive way, but in gas form water’s orientation is more random and therefore there is less attraction while water is in gas form.

Note that a hexane molecule is quite a bit larger than a water molecule. Why is this information significant?

Any effects due to size I would expect to come into play for boiling point too

vdW parameter a is not just a measure of attractive forces, but 8s a measure of overall "being busy with other molecules, Much bigger and planar hexane molecule is much more busy.It Is a kind of a dynamic aspect of molecule size and shape.

Could you elaborate on this “busy-ness” and how it contributes to a lower measured pressure when the compressibility factor (Z) is below 1?

Comparing vdW parameters of very different molecules is imo an excercise in futility. It is a qualitatively rather bad model, and the tabulated values are calculated from the critical parameters, i.e. they belong to different temperatures.

I took all the data on the Wikipedia datapage and normalised the constant a by the molecular weight of the compounds as a very rough measure of molecular "size". While compounds with inherently stronger intermolecular forces tend to have higher values of a, the agreement is only qualitative, and there are considerable sections of outliers (for example, water has the second highest MW-normalised value of a, but hexane is the fifth highest). There is surely a more accurate form of normalising the data, but there could also be something missing.

Closely related to "How are the Van der Waals constants a and b related to each other?" https://chemistry.stackexchange.com/q/125331/35806

@Ryan By "busyness" I mean general interaction,decreasing frequency of wall hitting. It means either dynamic creation of real dimers, either creation a kind of "mechanical dimers" due being obstacles for each other. E.g 2 n-haxane molecules are like 2 fisherman nets catching each other.


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How do I interpret GAMESS thermochemistry output?
In GAMESS's thermochemistry output, which numbers mean what?
The gaussian docs have a nice description of how to use thermochemistry data, but for gaussian's output. This appears to be different, eg, it mentions thermal corrections to energy, enthalpy, and Gibbs free energy, but GAMESS outputs only one, vibrational, thermal correction.
Example output:
 USING IDEAL GAS, RIGID ROTOR, HARMONIC NORMAL MODE APPROXIMATIONS.
 P=  1.01325E+05 PASCAL.
 ALL FREQUENCIES ARE SCALED BY   1.00000
 THE MOMENTS OF INERTIA ARE (IN AMU*BOHR**2)
              0.00000            35.04363            35.04363
 THE ROTATIONAL SYMMETRY NUMBER IS  2.0
 THE ROTATIONAL CONSTANTS ARE (IN GHZ)
      0.00000    51.45266    51.45266
      6 -      6 VIBRATIONAL MODES ARE USED IN THERMOCHEMISTRY.
 THE HARMONIC ZERO POINT ENERGY IS (SCALED BY   1.000)
        0.000110 HARTREE/MOLECULE       24.044128 CM**-1/MOLECULE
        0.068746 KCAL/MOL                0.287632 KJ/MOL

               Q               LN Q
 ELEC.     1.00000E+00       0.000000
 TRANS.    1.12480E+05      11.630533
 ROT.      6.03147E+01       4.099576
 VIB.      4.82855E+00       1.574546
 TOT.      3.27579E+07      17.304656

              E         H         G         CV        CP        S
           KJ/MOL    KJ/MOL    KJ/MOL   J/MOL-K   J/MOL-K   J/MOL-K
 ELEC.      0.000     0.000     0.000     0.000     0.000     0.000
 TRANS.     3.718     6.197   -28.831    12.472    20.786   117.487
 ROT.       2.479     2.479   -10.163     8.314     8.314    42.400
 VIB.       2.490     2.490    -3.616     8.277     8.277    20.478
 TOTAL      8.687    11.166   -42.610    29.063    37.378   180.365
 VIB. THERMAL CORRECTION E(T)-E(0) = H(T)-H(0) =      2202.424 J/MOL

              E         H         G         CV        CP        S
         KCAL/MOL  KCAL/MOL  KCAL/MOL CAL/MOL-K CAL/MOL-K CAL/MOL-K
 ELEC.      0.000     0.000     0.000     0.000     0.000     0.000
 TRANS.     0.889     1.481    -6.891     2.981     4.968    28.080
 ROT.       0.592     0.592    -2.429     1.987     1.987    10.134
 VIB.       0.595     0.595    -0.864     1.978     1.978     4.894
 TOTAL      2.076     2.669   -10.184     6.946     8.933    43.108
 VIB. THERMAL CORRECTION E(T)-E(0) = H(T)-H(0) =       526.392 CAL/MOL

Is this your GAMMES output? What exactly is the problem? I see "E", "H", "G", S", which are energy, enthalpy, Gibbs free energy, and entropy. I can show many examples here at the Chemistry SE, where questions have been closed due to being about a specific program (like GAMMES). Your question might be suited for a dedicated Stack Exchange just for computational material science. You might like to help support the motion to make this stack exchange: https://area51.stackexchange.com/proposals/122958/materials-modeling

@user1271772 So G - TOTAL is the Gibbs free energy 298K? Or 0K?

From the GAMESS manual supplied by the Gordon Group: "TEMP = an array of up to ten temperatures at which the thermochemistry should be printed out. The default is a single temperature, 298.15 K. To use absolute zero, input 0.001 degrees." So unless you have supplied a temperature, it would be 298.15 K.

So does it include the electronic energy, or do I have to add that? To be more precise, are these numbers that I can subtract between two runs to get the ∆G?

@Dan The electronic energy, as determined by the SCF (or what have you) is not included because 1) The $G$ etc. are called "corrections" 2) The total energy of the SCF is much higher than 0.5 kcal/mol. Note that it is most likely given in units of hartree.


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Why is Pyridinium Chlorochromate soluble in organic solvents?
I would presume that Pyridinium Chlorochromate is not organic soluble due to the relatively strong ionic bonding between the anion and cation on account of their similarly large size. Although the charge on the pyridine is not stabilised by resonance, the charge on the chromium molecule is. Does this weaken the ionic bond enough so that the weak London forces between organic solvent molecules and the ions is enough to pull them apart? Are there other factors at play here?

Weak London forces? It's soluble in polar solvents, not hexane.

Apart from the fact noted by Mithoron that it's soluble in polar organic solvents, we may consider what lattice binding energy the solvation forces must overcome.  In a compound like sodium chloride with compact, monatomic ions the lattice forces are too strong for all but the most aggressive solvents (but lithium salts have a way around this problem).  With pyridinium chlorochromate (PCC), however, the ions are multiatomic, bulky, and differently shaped as well as having only one charge, so the crystal lattice is only weakly bound compared with sodium chloride.  PCC is far from unique; to cite another example, the triangular cyclopropenyl cation was identified through its nmr specta when salts were dissolved in polar solvents.

The fact that PCC is a protonated salt (of pyridine and the theoretical chlorochromic acid) may also mean that the ionicity is only "partial", with formation of tightly bound neutral pairs in solution in low polarity solvents. A somewhat related case is ethylammonium nitrate (EAN), which is even a room-temperature ionic liquid. and probably has substantial solubility in some organic solvents. Hydrocarbons also have comparatively high solubility in EAN.

If you look at my references answer you see that lithium salts are rendered more soluble than heavier alkali metal salts due to ion pairing/association.

I see your point for the lithium salts, but there's a slight subtlety in that simple lithium salts are highly dependent on solvents with good donor atoms to become soluble. For example, lithium iodide has poor solubility in DCM, even though it is relatively polar, and even though LiI is extremely soluble in things like diethyl ether, etc. Lithium favours covalency, as you say, but it needs to bind to a donor solvent to form a tight solvate, which essentially acts as a bulky organic cation. A proton salt like PCC or EAN favours covalency more, and would in principle be less solvent-dependent.

Protic cations could also hydrogen bond.  That could mean a huge solvent dependency, actually.

I suppose that's true, protic salts just shift the dissolution enhancement (from donor solvents for lithium salts) to solvents capable of supporting hydrogen bonding even if they do not hydrogen bond themselves (which I suppose covers most donor solvents...), and due to the higher covalency of a hydrogen bond (higher polarising power of a proton compared to a lithium cation), dissolution can still be supported in non-donor/weakly polar solvents by forming a strong internal partially covalent hydrogen bond between anion and cation.


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Confusion regarding the positively charged copper electrode in a Daniell cell
Suppose I bought a newly manufactured Daniell cell then in this situation, can someone explain what all will be going on at the zinc and copper electrode? (Since it is newly manufactured and not used in external circuit the question of charges being transferred from one electrode to another does not arise) So would there be negative charge accumulation at zinc and positive charge at copper electrode? (Actually I am confused over situation at cathode a. I have read it to be positively charged but I could not understand how as the electrons $\ce{cu^{2+}}$ gain to get reduced actually comes from zinc and not from copper electrode)
Can someone please clarify my both doubts?

This web page has nice demonstration to show how it works.

If it may help you to understand how the Daniell cell works, you may use a language full of imagery and say that the zinc metal absolutely wants to loose its electrons, so as to go back to its initial state on Earth, that is $\ce{Zn^2+}$ like in the mineral $\ce{ZnS}$ called blende. To be allowed to carry out this transformation, $\ce{Zn}$ needs somebody or something that would accept its electrons. Fortunately, copper is not like $\ce{Zn}$. It has not the same tendency. Copper does not want so eagerly to loose its electrons. In case of necessity like here, $\ce{Cu}$ even accepts foreign electrons to discharge its own ions  $\ce{Cu^{2+}}$ and make some new neutral atoms $\ce{Cu}$.
That is how the Daniell cell works. On the left-hand side, a plate of metal $\ce{Zn}$ is dipped into water. It looses its electrons, so that the ion $\ce{Zn^{2+}}$ could pass into solution. The electrons are sent through the outer circuit to the next plate, which can be in copper or in platinum or in silver. This is not important, The only important thing is that the copper or platinum plate dips in a $\ce{Cu^{2+}}$ solution, for example a solution made by dissolving $\ce{CuSO4}$ in water. When the electrons have reached the copper or silver plate, they attract $\ce{Cu^{2+}}$ ions, discharge them, and some Copper metal is deposited on the plate.
One should notice that there are two different solutions, one in contact with the $\ce{Zn}$ plate, one in contact with the copper or silver plate. These solutions should not be mixed. If the two solutions are mixed, the Daniell cell will not produce any current, because the zinc atoms would find next to them ions that accept their electrons. No need to send them through a long outer circuit.
The only thing happening then is the fate of the negative ion $\ce{SO4^{2-}}$ remaining in the solution near the copper plate. It cannot stay here. It is attracted by the  $\ce{Zn^{2+}}$ ion newly appeared near the $\ce{Zn}$ place. There must be a wet junction or a bridge to let the negative ion move to the other compartment.

That all I know. What I am trying to understand is how positive charge buildup occurs on cathode?

It does not matter if the cell was previously already used in circuit or not.( unless the cell is exhausted )
For both electrodes, there is ongoing bidirectional exchange of the positive metallic cations, what changes the electrode potential.
At equilibrium, both processes go by the equal rate with the equilibrium potential, given by the Nernst equation.
If the electrode potential is more positive then the equilibrium one, the net effect is metal dissolution. If it is more negative, the net effect is metal deposition.
That may happen by applying the external voltage ( electrolysis or charging ) or by galvanic electrode connection ( power provisioning ).
The conventional electrode positive/negative potential is not absolute, but relative wrt the standard hydrogen electrode(SME).  The absolute SME potential is $\pu{+4.44 \pm 0.02  V}$. So both electrodes have the same charge sign.
Note the terms cathode/anode:

A cathode is the electrode from which a conventional current leaves a polarized electrical device.
  An anode is an electrode through which the conventional current enters into a polarized electrical device.

Be aware that conventional current has for historical reasons the opposite direction than the electron flow.
It means electrons leaving an anode by wire during the primarily intended process(electrolysis or power provision), respectively coming by a wire to the cathode.
A help to remember is the Greek word anabasis=way upwards(for electrons), catabasis=way downwards.
That also means, cathode is the more negative electrode at electrolysis, but the more positive in cell providing power, as for both electrons "go downwards from a wire". And vice versa for an anode.
Naming for secondary cells stays the same as for primary ones.


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Why is the Sulphur Dioxide molecule polar?
I am in 12th standard and was revising the concept of determining which polyatomic molecule can be polar and which can be non polar. So when this example was given we were told that vertical dipole moment will not be cancelled out. But isn't it possible that it can be cancelled out? is there any other molecule which has a shape like $\ce{SO_2}$ i.e. the central atom is $sp^2$ hybridised and has one lone pair, and is non polar? Also where is the net dipole moment in $\ce{SO_2}$, upwards or downwards and why?

You are confusing two different (though related) concepts. Being polar is not the same as having dipole moment. CO2 has none, but still is pretty polar.

Sulfur dioxide has a dipole moment because of its shape
Bonds between dissimilar atoms in molecules inevitably have dipole moments (though they are sometimes small). The only way the whole molecule can be free of a net dipole is if the individual bond dipoles are exactly cancelled out. This means the one criterion for determining whether a molecule has a net dipole is geometric ie it depends on the shape of the molecule.
If the symmetry of the molecule means the individual bond dipoles cancel exactly, then there will be no net dipole. Compare carbon dioxide with sulfur dioxide. Carbon dioxide is a linear molecule so the two (relatively strong) C-O bond dipoles exactly cancel; the formula for sulfur dioxide looks superficially similar but the molecule is bent. Because the molecule is bent the bond dipoles cannot cancel and the molecule has a net dipole moment (with with a dipole resulting from net positive charge on the sulfur and negative on the oxygen).
It is possible, in principle, that a bunch of dipoles in a complex molecule could cancel out, but this is incredibly unlikely. Geometry essentially always predicts whether a molecule has a dipole, though calculating how big it is can sometimes be a challenge. None of this applies in simple cases like sulfur or carbon dioxide where the net molecular dipole is intuitively obvious once you know the shape of the molecule.


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Perturbational Molecular Orbital Theory - How to derive the results?
I am reading Albright's Orbital Interactions in chemistry. The qualitatively perturbational treatment of molecular orbitals seem to be a central tool used throughout the book. I've referred to the Appendix for a thorough derivation of the theory. However, I got stuck at equation 22(a). I don't understand how it follows from everything that is spelled out before it.
Hopefully, somebody who has read the same book or is knowledgeable about this topic can help me out. Thank you!
Remark: I know how to derive the nondegenerate perturbational theory in QM but it seems that with MO theory the derivation is more complex, since we are combining AOs to form MOs rather than working with totally independent, orthonormal eigenstates as in QM.

Why don't you at least put the relevant equation(s) here?

@Antimon Because there is a lot of equations... anyway, I have recently figured out the derivation process. I’ll probably come back to answer my question later

For those interested, eq 22(a) is $$[(e^\circ+\tilde{H})-e_i(1+\tilde{S})]T_i=0$$


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Difference in transition lifetimes for near-resonant v/s non-resonant Raman effect
I have read that a virtual state is created when Raman scattering occurs. Since this state is forbidden, the molecule upon absorbing a photon (not enough to take it to the next electronically excited level) immediately gives off the stokes or anti-stokes shifted photon with delay in the order of femto-seconds. If the photon were to be energetic enough to take it close to the excited state(near-resonant), would there be a change in the lifetime of the transition?

The dephasing of the excited molecule+light field is important here, but this is complicated; see eqns 9.15-9.19 chapter 9 in S. Mukamel, 'Nonlinear Optical Spectroscopy', (Publ OUP), not an easy read but he connects theory with experiment very clearly.


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Formyl chloride naming
How is the naming of formyl chloride $\ce{CHOCl}$ justified? What is the functional group in it?
Why can't I name it as methanoyl chloride as acid halide is given more preference than aldehyde?

Welcome to parallel worlds of trivial and systematic names. We also use formic acid, acetic acid, acetyl chloride.
You can name them all systematically as methanoyl chloride, methanoic acid,  ethanoic acid and  ethanoyl chloride. But in the real lab life, nobody call them like that.
Formyl chloride has  its name from
formyl-  as acyl derived from formic acid, the same as acetyl- from acetic acid.
Technically, formyl chloride is an aldehyde as well, as formyl is de facto an aldehyde group, being formally a member of the serie :
$$\ce{Cl-CH2OH <> Cl-CHO <> Cl-COOH}$$
But by naming conventions, both trivial and systematic, being a derivate of an acid replacing $\ce{OH}$ by $\ce{Cl}$ takes precedence to naming it as an aldehyde.
As a sidenote, it is rather funny aldehyde names are derived from the respective acids and not alcohols.  Because the aldehyde etymology says aldehyde="alcohol dehydrogenatus", so it could be as well methylaldehyde and ethylaldehyde, instead of formaldehyde and acetaldehyde.

Why cant I take CHO in formyl chloride as aldehyde?

Technically, it is an aldehyde as well, as formyl is de facto an aldehyde group. But by naming conventions, both trivial and systematic, being a derivate of an acid replacing OH by Cl takes precedence.

Only thanks to your answer did I notice that analysing formyl chloride as $\ce{CHO}+\ce{Cl}$ makes sense, too =D


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Preparation of Barium Hexaaluminate using co-precipitation reaction mechanism

I am preparing Barium Hexaaluminate (BHA) using $\ce{Ba(NO3)2, Al(NO3)3}$, and $\ce{(NH4)2CO3}$. For $\pu{1 mol}$ of each $\ce{Ba(NO3)2}$ and $\ce{Al(NO3)4}$, I am using $\pu{1.5 mol}$ of $\ce{(NH4)2CO3}$.

Following are my questions:

What temperature to select for reaction, according to some research papers we should use $\pu{60 ^\circ C}$. But the problem with keeping reaction temperature at $\pu{60 ^\circ C}$ will cause $\ce{(NH4)2CO3}$ to decompose into $\ce{NH3}$ and $\ce{CO2}$ and it will not take place in double displacement reaction.
Can I replace $\ce{(NH4)2CO3}$ with some other carbonates?

@Maurice, I always thought that barium nitrate and aluminium nitrate will react with ammonium carbonate and barium carbonate and aluminium hydroxide will precipitate down.
$$
\begin{align}
\ce{2 Al(NO3)3 (aq) + 3 (NH4)2CO3 (aq) &-> Al2(CO3)3 (s) + 6 NH4NO3 (aq)}\tag{R1}\\
\ce{Ba(NO3)2 (aq) + (NH4)2CO3 (aq) &-> BaCO3 (s) + 2 NH4NO3 (aq)}\tag{R2}
\end{align}
$$
(As aluminium carbonate is not stable and readily decomposes to form aluminium hydroxide.)
After that, if we keep the ratio of barium and aluminium $1:12$ and raise the temperature above $\pu{1200 °C}$ by calcinating the dried precipitate, barium hexaaluminate forms.
Also, can you clarify what $\ce{BaAl6(OH)20}$ is, because according to my understanding the general formula of barium hexaluminate is $\ce{BaO.6(Al2O3)}.$
Reference

Li, J. Q.; Wang, R. K.; Chen, C. Y. Preparation of Barium Aluminate with $\ce{BaCO3}$ and $\ce{Al(OH)3}$. AMR 2015, 1096, 156–160. DOI: 10.4028/www.scientific.net/AMR.1096.156.

Heating a mixture of Barium and Aluminum salts with $\ce{(NH4)2CO3}$ is done on purpose, because $\ce{(NH4)2CO3}$ is decomposed at 58°C into $\ce{NH3}$ and $\ce{CO2}$ and $\ce{NH3}$ is the reagent wanted to react with Barium and Aluminium to make them precipitate together. This is due to the two successive reactions: $$\ce{NH3  +  H2O <=> NH4+  +  OH-}$$
$$\ce{20 OH^- + Ba^{2+} + 6Al^{3+} -> BaAl6(OH)20}$$
If you would have used $\ce{(NH4)2CO3}$ at ordinary temperature, it would have made a mixed precipitate of $\ce{BaCO3}$ and $\ce{Al(OH)3}$.
The oddity of this reaction is the fact that you use 1 mol Barium, 1 mole Aluminium, and 1.5 mole $\ce{(NH4)2CO3}$. Barium is in great excess, but it will not precipitate, as $\ce{Ba(OH)2}$ is quite soluble in water. I wonder why you use such a great excess of barium.


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why does a pot of boiling water quiet down after it reaches boiling point but is still roiling?
I've noticed that when water boils, it makes a lot of noise. After the water reaches boiling point, even though the water is still ""rolling", the water quiets down. What is all the noise about and why does it go away?

It makes several different kinds of noise which peak at different moments. The whole thing is complicated and not necessarily well understood.

One noise might be due to all the gases that were dissolved in the water (mostly carbon dioxide, some oxygen) that are driven out before it boils.

See also https://physics.stackexchange.com/questions/375243/what-is-the-reason-for-this-pattern-of-noise-in-electric-kettles and https://physics.stackexchange.com/questions/28069/why-is-boiling-water-loud-then-quiet

Oh, you've come to the right place.  Many years ago I asked this question as a kid on a TV write-in programme called Don't Ask Me.  The TV people invited me on the show and built a series of demonstrations to explain.
The first thing that happens when you warm up a cold pot of water is that the pan bottom gets to boiling temperature.  The water starts to boil.  Liquids boil when their vapour pressure equals atmospheric pressure.  Otherwise bubbles could not form in the the liquid.  So, the pan bottom gets hot, bubbles form, but then they move up into cooler water and implode as the vapour pressure drops.  This makes a sharp snapping sound.  Lots and lots of these imploding bubbles make a roaring sound.
As convection currents then start to mix up the water, it becomes more uniform in temperature, and the bubbles actually rise to the top of the liquid and burst.  The implosions stop and you get the softer bubbling sound.
Hope this answers your question!


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Why glucose and other similar carbohydrates are oxidised fully by HIO4?
$\ce{HIO4_{(aq)}}$ requires the vicinal diols to have syn configuration, as it forms a cyclic periodate ester.

But glucose does not have syn configuration at C3, and if you consider cyclic forms of glucose it shouldn't oxidise fully at all because alternate hydroxy groups with anti configuration will be unreactive. But it still gets oxidised by $\ce{HIO4}$ to give 5 molecules of methanoic acid, $\ce{HCOOH}$, and one molecule of formaldehyde, $\ce{HCHO}$.

I googled "syn anti configuration" and it said "did you mean syn anti conformation". So the search engine has a good point: In the linear form, you can get a conformation where the hydroxyls point the same way.

"Anti" looks impossible in a Haworth projection. Use a chair conformation of glucose. Anti is slower than syn both with an ~60 deg. dihedral angle. HIO4 can catalyze beta to alpha. Once syn C1-C2 is cleaved in alpha-D-glucose, the ring is gone and there is no anti/syn issue. Or it may all happen in the open form. Look at my answer here: https://chemistry.stackexchange.com/questions/270/effect-of-periodic-acid-on-cyclohexane-derivatives

Glucose can react with $\ce{HIO4}$, which can be understood easily by looking at chair conformation of glucose.
Here is example of a diol, for more information looks at this answer https://chemistry.stackexchange.com/a/114836/67254


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Optical activity of cis-trans isomers of a co-ordination compound
Why do geometrical (cis and trans) isomers of [Pt(NH3)(Br)(Cl)(py)] not show optical activity ? Thanks In Advance...

Because each of them is a superimposable mirror image of itself.

Your statement of isomers of $\ce{[Pt(NH3)Br Cl py)]}$ giving non-superimposable mirror images is incorrect. Bottom line is, if they are not  they are non-superimposable mirror images, then they should be opticlly active. Basically all three possible geometrical isomers of said compound are suire planner and have superimposable mirror image of each. I have graphically illustrate one of them it for your benefit:

Suppose three isomers, I, II, and III, respective mirror images of I', II', and III'. Consider only mirror image I'. If you rotate it $90^\circ$ anti-clockwise around the axis perpendicular to the paper, you get the structure I". Now flip structure I" around the dioganal axis indicated ($100^\circ$). Now you got the structure I back, meaning the isomer I and its mirror image are identical.