text:
Can bromo(chloro)fluoro(iodo)methane be synthesized?
Bromo(chloro)fluoro(iodo)methane is supposed to have all of the stable halogens a part of this molecule: fluorine, chlorine, bromine, and iodine. Can it really able to be synthesized and if so could it actually not decompose the moment it goes outside of the lab?

That Wikipedia link says it is a hypothetical compound. So, what's exactly is your question? FWIW, bromochlorofluoromethane has been synthesized and so has been bromochlorodifluoromethane.

It may be possible in theory, but as yet no synthetic route to it has been found.  A compound is known with hydrogen in place of the iodine (bromochlorofluoromethane).  This latter compound has found some use in research on chiral compounds, which both molecules given here are.  Bromochlorofluoromethane proved difficult to separate into its enantioners because it hydrolyzes readily and lacks suitable functional groups, but we now have multiple techniques available to resolve this compound (and which, likely, have found other such uses).  See the following WP articles on Bromochlorofluoroiodomethane and Bromochlorofluoromethane.


text:
Does burning plastic completely disintegrate it? Accidentally melted some plastic in a BBQ, wondering if pyrolytic cleaning is an option
I accidentally barbecued some plastic today, making quite a mess. It's a cushion for a lawn chair, so I imagine probably some sort of plastic or polyurethane.
Rather than cleaning melted plastic manually, I'm wondering if its a good idea to just burn it off? The BBQ is a Kamado oven, so it can reach about $\pu{350 ^\circ C}$ or $\pu{700 ^\circ F}$ for a sustained period of time.
It's a well ventilated area, so smoke isn't a concern. I am somewhat concerned that it may not burn completely and contaminate future cooks.
Can anybody offer some insights on if $\pu{350 ^\circ C}$ is hot enough or if this is a good idea at all?

Am I understanding correctly that you barbecued a lawn chair cushion?

Well, the tags are appropriate. If it was me, I would replace the affected part: even scraping, sanding and heating it for a few hours would not erase the memory of what happened and who last sat on the BBQ’ed cushion.

@EdV is right. As an experienced grillsman living in Texas, I can assure you that you don't want to season your rig improperly, and that precludes using plastic and law furniture components for that task. Nor should you diminish the BBQ experience for your guests by offering charred lawn furniture on which to sit: that's impolite. You wouldn't skip on the marinade, or the sides and the ice-cold beer: do the right thing and respect the entire experience.

I would imagine that burning the plastic would work, but I would probably not recommend this method because it will likely release a ton of toxic smoke. I cannot post this as an answer because I don't feel certain enough on the "burning would work" part.

Reducing everithing to soot than chair than clean as usual seems a good idea, especially if the plastic is in the "furnace" Not really a question for Chemistry SE but worth a comment.

Heh, I did in fact bbq a lawn chair cushion. What can I say, always check the inside before you light a bbq.

@user1543574 Sounds like you might need to contact Kamodo Joe support folks. Good luck and keep on grillin'!

source: Stack Exchange

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sort: votes


text:
Procedure for laboratory synthesis of tertiary amines
I am having difficulty finding a procedure for making tertiary amines in the lab, any is fine but I am specifically interested in tributylamine.

Best way of doing it is by reductive amination between dibutylamine and butanal. Sodium Triacetoxyborohydride is a good reducing agent for this, can be done in dichloromethane. See https://pubs.acs.org/doi/10.1021/jo960057x   Tributylamine is widely commercially available.

Do you have access to database searches eg SciFinder, Reaxys? If you do, you should learn to use these.

Check the chemical inventory of your group / department / shop; a purchase of $\ce{NBu3}$ may be cheaper or / and faster than running a synthesis by your own.  Equally, for learning, consider the relevant entries on organic-chemistry.org for that they set a reaction into context of application / contemporary literature references.  With some names / name reactions retained, you will find your way through the general entry here, too.

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


text:
Stability of α, β, γ allotropes at different temperatures
While reading about elements or compounds I often come across allotropes (or forms) being referred to as α, β, γ forms.
Examples:

Sulfur has two allotropes: α rhombic and β monoclinic. Out of these two α is considered more stable at ordinary temperatures.
$\ce{SO3}$ has three forms out of which again α is considered more stable.
α-Form of iron is considered the most stable at ordinary temperatures.

Based on this small data, it makes me think whether α forms are always in general the more stable ones? Is it true? How does this nomenclature signify stability at different temperatures?

$\alpha$ is simply the first to be found. What color it is or how much does it cost is irrelevant; ditto for the temperature stability.

Sulfur has many more allotropes than two.  Counting for sulfur $\ce{S8}$ alone has already has $\alpha$, $\beta$, and $\gamma$, and if you account for smaller sulfur molecules, the number increases further.

A German speaking mineral atlas states:

«Man unterscheidet die verschiedenen polymorphen Modifikationen durch
Voranstellen der griechischen Buchstaben alpha, beta, gamma usw. Die
Buchstaben entsprechen im Wesentlichen der Erhöhung oder Erniedrigung
der Umwandlungstemperatur.»

which may be translated into

«One discerns the different polymorphs by Greek letters $\alpha$, $\beta$,
$\gamma$, etc.  The letters mainly correspond to an increase or
decrease of the [solid-solid] phase transition.»

To extend the answer by @ĐỨc Lê Hồng (+1):
Eilhard Mitscherlich may be credited to extend Steno's angle law about crystals of one and the same compound to isomorphism (different chemical compounds may yield crystals of at least very similar shape) and eventually allotropes (about elements) / polymorphism (the term equally applicable for compounds) that the same chemical compound may form crystals of distinct different shape and symmetry.
Back around 1825, the experimentally accessible space in terms of temperature and especially pressure was much more constraint, than today.  Thus, the phases were sort in ascending order of the temperature of the corresponding phase transition by Greek letters.
This is the reason why e.g., while the stable form of tin observed at room temperature is labeled by $\beta$, and the one at lower temperatures by $\alpha$.  Staying with elements, cyclic $\ce{S8}$ sulfur is known in the $\alpha$ form stable below $\ce{95.3 ^\circ{}C}$, but by the $\beta$ and $\gamma$ form above this temperature.  Without counting the other allotropes of sulfur. (reference).
Iron is an example suitable to to show early limitations of this nomenclature.  The phase diagram of pure iron still is simple for its $\alpha$, $\gamma$, and $\delta$ form.

(credit)
(There equally is a $\beta$ form of Fe, but this is of much lesser significance, seen in the binary phase diagram Fe-C.)
Today's convention to count polymorphs by roman numbers (if there is more than one) has the advantage that these may reflect the sequence of their discovery, regardless if researchers explore the variation of temperature, pressure, or the combination of the two.  An example for this may be water:

(figure 7 from here, figure 2 from here, both open access)
On occasion, you encounter binary phase diagrams reflecting both approaches, e.g., Fe-C:

(credit)
describing ledeburite I (a micro structure of austenit + $\ce{Fe3C}$) and ledeburite II (a microstructure of Perlit + $\ce{Fe3C}$).

I think most of the time, they sort according to the order which temperature the allotrope/polymorph is more stable. Let's say element Z has two allotropes A and B. A is stable below 100 °C and B is stable above 100 °C. Thus allotrope A would be denoted alpha-Z and allotrope B would be denoted beta-Z.
Gray tin (alpha-tin) and white tin (beta-tin) is an example. Gray tin is stable only below 13.2 °C and white tin is stable above this point. So that's why gray tin is called alpha-tin because it is stable at a lower temperature, and ditto with white tin. As you can see in this case the beta allotrope would be the one stable at room temperature one.
Quartz (a crystalline form of silicon dioxide) is another example. See this from Wikipedia:

Quartz exists in two forms, the normal α-quartz and the high-temperature β->quartz, both of which are chiral. The transformation from α-quartz to β-quartz >takes place abruptly at 573 °C (846 K; 1,063 °F).

I thought the same based on the small data I had however @Ivan Neretin 's comment says otherwise and so if you could cite some source it would be great! One way both of the arguments go hand in hand is if we think that coincidentally the form discovered first (named α) was the one which was found more easier, which would probably be in ordinary conditions. I'm not sure however


text:
How can I weigh the compound that melts at room temperature?
I am conducting an organic synthesis experiment involving 2,3,5-tribromothiophene.
The melting point of 2,3,5-tribromothiophene is about 25-28 ℃, so it melts easily during weighing.
Whenever I weigh this compound, it melts soon and wet the weighing paper. So there arises big uncertainty in weight of compound. How can I weigh the compound that melts at room temperature and move it to the reactor with minimum loss? Is there no other way but to weigh as quickly as possible?

How about using a plastic weighing boat and flush the sample into the reactor with the corresponding solvent?

Weigh it as a liquid, pipette it into a tared vial, dissolve in reaction solvent and add to reactor.

Similar to weighting boats of plastic or glass, a watch glass may do the job, too.  Depending on scale and design of the balance, a beaker (there are already ones for about 20 mL) may do the job, too.

@andselisk Yes, but not with dichloromethane... Learnt that the hard way!

@orthocresol On the positive side, you also discover a nice "glue" for many plastics;)


text:
(E)-2-bromo-1,4-difluoro-3-iodobut-2-ene or (E)-3-bromo-1,4-difluoro-2-iodobut-2-ene
I am struggling with naming this structure either (E)-2-bromo-1,4-difluoro-3-iodobut-2-ene or (E)-3-bromo-1,4-difluoro-2-iodobut-2-ene.

I am aiming at the question whether there is also a prioritization for IUPAC nomenclature given by the order of atomic numbers as we have it for the CIP nomenclature. Or do we just have a prioritization given by the alphabetic order in the case of halides?

Hello, if that answered your question, then you can 'accept' the answer by clicking on the green tick below the voting buttons. Please see: https://chemistry.stackexchange.com/help/someone-answers. (I saw your comment saying thanks, but it got deleted.)

As per the official IUPAC guidelines, namely Nomenclature of Organic Chemistry – IUPAC Recommendations and Preferred Names 2013 (Blue Book):

P-14.4 NUMBERING
When several structural features appear in cyclic and acyclic compounds, low locants are assigned to them in the following decreasing order of seniority:
(…)
(e) saturation/unsaturation:
     (i) low locants are given to hydro/dehydro prefixes (…) and ‘ene’ and ‘yne’ endings;
     (ii) low locants are given first to multiple bonds as a set and then to double bonds (…);
(f) detachable alphabetized prefixes, all considered together in a series of increasing numerical order;
(g) lowest locants for the substituent cited first as a prefix in the name;
(…)

The directly relevant part here is rule (g), which states that the substituent cited  first gets the lowest locant. Since substituents are cited in alphabetical order, this indirectly means that the lowest locant is assigned to the substituent which comes first alphabetically, i.e. 'bromo'.
Note that this is only rule (g), and that in general rules (a)–(f) should be applied first (their full text can likely be found elsewhere on Stack Exchange). In this case, rules (a)–(d) are irrelevant and rules (e)–(f) do not distinguish between the two proposed numbering systems, so (g) is used as the tiebreaker.
In any case, the preferred IUPAC name of the above molecule is (2E)-2-bromo-1,4-difluoro-3-iodobut-2-ene. [Technically, the locant of the double bond must be specified with the stereodescriptor, hence (2E)- instead of just (E)-, although in practice this is often ignored when there is only one double bond.]


text:
Determining priority for a cyclic compound
I have problem in determining the CIP priorities in (S)-ketamine.

What is the order of priorities for esketamine? Does the carbon bonded to the ring with chlorine have higher priority compared to the carbon bonded to oxygen?

N  >  (C=O){O,O,C} > C-arom{C,C,C} > (CH2){C,H,H}. The configuration is S. You don't have to go out as far as the carbon bearing the chlorine to make a decision.

Thank you! Is it possible exchange the CH2 group with the C-aryl so that the CH2 at lowest priority is on the back?

By "back" do you mean draw the structure such that the -CH2- is dotted? Yes you can as long as you don't break any bonds. Make a molecular model and rotate it to see the result. If you dont have a model kit, use a potato and toothpicks arranged as a tetrahedron. Label the toothpicks with the groups.

Silvia95: You can visualize (S)-ketamine here as the last item in the first group. http://ursula.chem.yale.edu/~chem220/chem220js/STUDYAIDS/isomers/RS14272/RSEZ14161.html


text:
Relative humidity and vapor-pressure in equilibrium
I have quite an unclear understanding of relative humidity. I am writing my understanding on the topic and please correct me wherever I'm wrong:
What I understood:
Let's say $\ce{H2O}$ has a (saturated) vapour pressure of $x~\pu{atm}$ at a certain temperature. Now, relative humidity (RH) is the ratio of the partial pressure of water vapour to the equilibrium vapour pressure of water at a given temperature.
So what this means is that if RH = 100% then the maximum amount of water vapours is present in the air. Now if RH = 50% means that the amount of water vapours present in the air is only half of the maximum possible.

So, does this mean the water has not been fully evaporated up to the
limit it can?
Does this mean that the equilibrium has not yet been established, and
the reaction tends to shift in the forward direction? If yes, then
what is stopping the reaction to move forward?
Does 100% RH mean the water and it's vapour are in equilibrium?

Can relative humidity be compared with an example of a sponge? There's a limit of the amount of water a sponge can absorb water. If it has only absorbed 50% means there is still room for more absorption until 100% capacity is reached.

What I understood:
Let's say $\ce{H2O}$ has a (saturated) vapour pressure of $x$ atm at a certain temperature. Now Relative humidity (RH) is the ratio of the partial pressure of water vapour to the equilibrium vapour pressure of water at a given temperature.
So what this means is that if RH = 100% then the maximum amount of water vapours is present in the air. Now if RH = 50% means that the amount of water vapours present in the air is only half of the maximum possible.

So, does this mean the water has not been fully evaporated up to the limit it can?

Less than 100% humidity would mean exactly that. Assuming that there is still liquid water left, then an equilibrium has not been reached.

Does this mean that the equilibrium has not yet been established, and the reaction tends to shift in the forward direction? If yes, then what is stopping the reaction to move forward?

Exactly. Assuming an isothermal system with more liquid water available, then yes, more water would evaporate. Remember when water evaporates it cools the remaining water.

Does 100% RH mean the water and it's vapour are in equilibrium?

Yes, assuming both phases exists, and that the gas and liquid phase are at the same temperature.
Some  extra points:

The saturated vapor pressure of water is a complex function of temperature. So it is $x_1$ at $T_1$ but $x_2$ at $T_2$, and so on.

Ice sublimes and thus will also have an equilibrium water vapor pressure. The equilibrium water vapor pressure again depends on the temperature of the ice.


text:
How does copper oxidise alcohols?
Consider for example ethanol being heated in the presence of copper at 300 °C. How does it give ethanal as a product?
Copper being in its elemental state (oxidation state 0). Since negative oxidation state is not possible for Cu, which means that it should reduce ethanol to  ethane. Certainly, high temperature should have some role here. What could it be?

CuO does the job.

When heated in air, a piece of copper metal gets oxidized into black copper oxide $\ce{CuO}$ according to  $$\ce{2 Cu + O2 -> 2 CuO}$$
When hot copper oxide $\ce{CuO}$ is in contact with some ethanol vapor $\ce{C2H5OH}$, it reacts according to : $$\ce{CuO + C2H5OH -> Cu + CH3CHO + H2O}$$ And the copper metal is regenerated. So if a copper plate is heated in a mixture air + ethanol, it is simultaneously oxidized to $\ce{CuO}$ and reduced to $\ce{Cu}$, so that the only visible effect in the experiment is that the ethanol molecules are transformed into ethanal $\ce{CH3CHO}$ molecules.

Thanks for the answer, just wanted to know if  CuO is strong enough to oxidise aldehydes into acids?

@green_32. At high temperature, CuO is able to oxidize all organic molecules to $\ce{CO2}$ and $\ce{H2O}$. This was even the principle of the traditional analysis of organic compounds in the 19th century. An unknown substance was mixed with a large excess of $\ce{CuO}$. The whole  mixture was heated in a horizontal tube with a very gaz running along it. The produced $\ce{H2O}$ vapor was absorbed in $\ce{CaCl}$2 and $\ce{CO2}$ in $\ce{NaOH}$, for further analysis.

You likely had either or both of the following:

Copper had some oxide on it, and this oxidized the alcohol leaving the metal behind.

If there was air in your system, especially on heating copper could have reacted with it to generate the oxide (a relatively fast reaction with heating) and you're back to (1).  The metal would then, of course, be available to react with more air.  In effect this would be a catalytic oxidation.

Copper is actually known in negative oxidation states, but that does not happen or need to be invoked here.

Copper is a very good dehydrogenating and dehydrating agent.
When a primary alcohol is heated in presence of copper than one hydrogen connected to oxygen and another connected to carbon leave in the form of hydrogen gas converting it into an aldehyde.
Further oxidation of the aldehyde into a carboxylic acid is not favourable.
A secondary alcohol is converted into ketone showing same nature as above.
But tertiary alcohols show the dehydrating nature converting the alcohol into an alkene according Zaitsev's rule.

Here oxidation of compound take place because of escape of hydrogen gas from compound leading to rise of oxidation state of carbon by +1


text:
Titration of weak acid and strong base at low concentrations
Titration continues until one of the reactants (the one in conical flask) turns out to be a limiting agent as far as I know. From the titration of a weak acid and a strong base, I could learn that the end point is above $\mathrm{pH \ 7}$ since weak acids do not ionize completely. Till this point I could understand everything. But I had a confusion at the condition of ions at end point. In this titration, we are basically dealing with $\ce{H+}$ ions dissociated from water (which is very low) and acid. Usually acids remain a bit more concentrated and release more ions than water does. That is at end point we often ignore the auto ionization of water.
But what if we are dealing with such acid whose concentration leads to total amount of dissociated $\ce{H+}$ ions comparable to $\pu{1 \times 10^{-7} M}$. Will titration be an appropriate way to determine the concentration of acid in this case? Or we need to use some other ways to measure the concentration?
Please let me know if there is any point I'm missing out regarding this topic. And this is my first question in this forum so I'm looking for cooperation. Any help will be appreciated.

Your intuition about there being a problem when concentrations are very low is correct: see here for titration curves as a function of concentrations involved.

Thanks a lot @Ed V.  Your answer to that question was amazing.

Yeah it did answer but one thing is missing. That is what will be the alternative procedure to measure the concentration?

Non-aqueous titrations are one possibility: you need a solvent system where there is less auto-ionization of the solvent.


text:
Why does calcined feldspar crumble?
For context, I am a potter and am interested in collecting and processing all materials that I use by myself. One of the processes which I use is calcination of certain minerals like feldspar, so that I can turn them into fine powder.
For reference, most feldspars are mixtures of the three chemicals: $\ce{KAlSi3O8,\ NaAlSi3O8,}$ and $\ce{CaAl2Si2O8}$. I usually collect pieces of granite which have a high feldspar content. I then put the granite into my electric (oxidation) kiln and fire it above $\pu{1000 ^\circ C}$. After it has been calcined, the granite falls apart very easily, with the exception of the pieces of quartz which remain quite hard. I am then able to crush the feldspar up into a fine powder.
I gather that the heat does something to the feldspar which makes it crumble easily but chemically speaking I don't know what this involves. Does the chemical formula of the feldspar change? Does it just break down in some way?

Out of curiosity, what happens to the mica in the granite? Vermiculite (used in gardening and for packing) is mica that has been so strongly heated that it expands (“puffs up”).

@EdV according to wikipedia, mica remains stable up to 900C, but since I calcine the granite above 1000C I guess the mica decomposes in some way. I'm not a chemist, so I don't know what this entails, but I guess it's something similar to what happens to the feldspar.

Most of the mica you'll find is biotite (for the heavy end-member silicates) or muscovite (for the lighter-end silicates). At high-T you are dehydrating the sheet silicates, and reacting the end-member feldspars with water, oxygen, and carbon dioxide at the same time. Essentially you are speeding-up the weathering process (which is well-quantified) of these minerals, which (mostly) become "clays" of various types.

@toddminehardt that smells like an answer to me.

You're replicating the natural weathering process of granite.
Granite is a rock comprising the minerals feldspar, quartz, and mica.
There is a continuous solid-solution ternary diagram here that illustrates the various feldspars as a function of calcium, potassium, and sodium content: the end members are K-spar (potassium variety), albite (sodium variety), and anorthite (calcium variety).
The weathering process is complex and involves so-called incongruent dissolution, where one component in the solid state reacts to form one or more nother solid state components. This is why you see granitic feldspar weathering faster than the quartz or mica components out in nature.
You are reacting various minerals (the exact composition of your feldspars and micas would be determined by XRD or a similar method) with oxygen, carbon dioxide, and water from the atmosphere as well as any waters of hydration that may be in your micas, which are sheet silicates from a general class of silicates known as phyllosilicates.
Back to your query about crumbling: you're dehydrating the micas (muscovite on the light-colored end, biotite on the dark-colored end) if they are hydrated at all, but mostly you're making various clays from the feldspars via the incongruent dissolution pathway. Those new minerals you're making are the crumbly ones, mostly clays and maybe some brittle mica scraps. But it's mostly clay.
Generally speaking, carbon dioxide and water form an aqueous solution of carbonic acid species which reacts in the presence of oxygen and a feldspar to form a clay and aqueous calcium and carbonate ions. There are many, many complex reactions.
Here is an excellent write-up about anorthite hydrolysis, specifically regarding the region around Egersund, Norway. In it, the author does some modeling using PHREEQC and gives specific examples of some of the weathering reactions. Here is one of the reactions that takes a mixed albanitic-anorthitic feldspar to kaolinite:
$$\ce{2NaCaAl3Si5O16 +
      8CO2 +
      9H2O =
      2Ca^2+ +
      2Na^+ +
      6HCO3- +
      4SiO2 +
      3~Al2Si2O5(OH)4 +
      2CO2}$$
High temperature speeds things up, and also will dictate what products you get. Same for water and carbon dioxide, of course.
In summary: you are making mostly crumbly clay from granite by speeding-up the chemical weathering process of (mostly) the feldspar component(s). The particular composition of the clay varies with the entire system composition.
This is a very complex sub-field of study in aqueous geochemistry. An excellent reference (which I used in my grad school days and beyond) is Stumm and Morgan.

Thank you for the great answer, this is just what I needed. You'll probably be able to answer this question as well :)

I'm not an expert but based on Googling around, I suspect what happens is primarily:

Driving off any bound water.

Driving some of the silicate out of the material $(\ce{KAlSi3O8 -> KAlSi2O6 + SiO2}).$

My bet is the second effect is the more important.  You could try firing it at 500 °C and see if there is any appreciable impact. Think all the bound water would leave then.  The second reaction requires temps above 950 °C. Maybe that is the major driver of what you are seeing.
Also, sounds like it's not just feldspar, other junk in there. So a little hard to tell… might by other materials being affected.
If you really wanted to be a chemist about it you could take a chemically pure sample of feldspar (maybe by preparing it, not extracting from nature), fire it at 1000 °C and then look at the morphology (hardness, grain structure). And probably do X-ray powder diffraction or SEM-EDAX to look at what little buggers you made.


text:
I need to find the mechanism of a reaction
As in the title, I have a reaction to witch I need to give the mechanism.

Get out your textbook. Shouldn't be hard to find.

The image shows the complete mechanism

It would be better if you provide some explanation.


text:
How to write the Nernst Equation for this reaction?
I am required to write the Nernst equation for the following reaction:

$\ce{O_2 + 4H^+ + 4e^- -> 2H_2O}$

Clearly, here $\ce{O_2}$ is getting reduced from an oxidation number of zero to an oxidation number of $\mathrm{-2}$ in the water molecule.
Usually, the examples of chemical reactions provided in the textbooks(at least in high school textbooks) for writing Nernst equations are of the form: $$\ce{M^{n+}(aq) + ne^- -> M(s)}$$ and the Nernst equation for this electrode reaction is given as: $$E_\ce{(M^{n+}/M)} = E^\circ _\ce{(M^{n+}/M)} - \frac{RT}{nF}\mathrm{ln}\frac{1}{[\ce{M^{n+}}]}$$
But here, in this case, the oxidation state for the $\ce{H^+}$ remains the same on both side of the reaction, rather the oxygen gas gets reduced. And this is what is confusing me.
The solution provided states that the equation should be: $$E = E^\circ+\frac{0.059}{4}\mathrm{log}[\ce{H^+}]^4 \cdot P_\ce{O_2}$$
I specifically didn't understand the reasoning behind writing the part that comes after log.
Please explain.

You know that Q, the reaction coefficient, equals 1 divided by x, where x equals the partial pressure of oxygen times the 4th power of the hydrogen ion concentration. These are the only two variables. So the second term on the right is approximately -(0.059 mV/4)log(1/x). But -log(1/x) = +log(x).

That P you wrote after the log is actually included in the log. It's like the expression of Q but reciprocated and the -ve sign adjusted.

@EdV Thanks, this just made it clear for me. I was confused about whether to take the partial pressure of oxygen into account or not.


text:
Synthesis of 1-naphthaleneacetic acid: Methods
I am trying to synthesizing 1-naphthaleneacetic acid from naphthalene and monochloroacetic acid. I found some patents and papers, and I have tried to synthesis but the yield is low. I used iron powder and $\ce{KBr}$ as catalyst. Papers and some patents mentioned about the yield is 70% but I achieved only 35%. I tried different iron compounds such as $\ce{FeCl3, Fe2O3}$ with $\ce{KBr}$ but the maximum yield gained is 35%.
Could you please help me for this synthesis method for high yield?

It'd be helpful if you listed your procedures so we'd see if they have some discrepancies.

If you have read the reference 1 carefully, you would realized that the amounts of $\ce{KBr}$ and $\ce{Fe2O3}$, as well as reaction temperature and time is very important to the reaction yield. That's the reason I asked you to list your procedures so that we can analyzed them and suggest some changes. Well, as reference 1 listed, follow the following procedure exactly and see the results:

Naphthalene with 0.0010% iron $(\pu{57.6 g},\ \pu{0.449 mol})$, chloroacetic acid $(\pu{14.1 g},\ \pu{0.149 mol})$, ferric oxide $(\pu{87.6 mg},\ \pu{0.549 mmol})$, and potassium bromide $(\pu{420 mg},\ \pu{3.53 mmol})$ were placed in a long-necked round-bottom flask fitted with a rubber stopper carrying a thermometer and an air-cooled tube about a $\pu{1 m}$ long, and the mixture was boiled gently on a sand bath for twenty hours. The best yield was obtained when the heating was controlled according to a temperature-time curve in which the temperature of reactants attained $200^\circ$ after ten hours and $218^\circ$ after twenty hours. After the reaction, a greater part of the unreacted naphthalene $(\pu{43 g})$ was recovered by distillation. The residue was extracted with hot sodium hydroxide solution, cooled, and filtered. Acidification of the filtrate with hydrochloric acid yieldeed the brownish precipitate, $\pu{19.4 g}$ (70%), m.p. $108-113^\circ$ ....

It is noted that the melting point can be raised to $132^\circ$ by repeated crystallization. First crystallization the crude product with 250-fold distilled water has yielded $\pu{9.5 g}$ of product with m.p. $124-126^\circ$. This yield is 34% based on  chloroacetic acid used, which is the limiting reagent and has reacted completely, and 45% based on naphthalene reacted.
The condition described above is the optimum conditions regarding to temperature, time, increased amount of $\ce{KBr}$, and the amount ratio of naphthalene to chloroacetic acid. Also note that:

In the absence of $\ce{KBr}$ and $\ce{Fe2O3}$, no naphthaleneacetic acid was formed.
In the absence of $\ce{KBr}$ but the presence of $\ce{Fe2O3}$, only 3% of naphthaleneacetic acid was formed.
In the absence of $\ce{Fe2O3}$ but the presence of $\ce{KBr}$, only 4% of naphthaleneacetic acid was formed.
$\ce{KBr}$ is the most effective additive for the reaction. The promoting action of several other metallic halides such as $\ce{NaCl},$ $\ce{KCl},$ $\ce{KI},$ hydrated $\ce{AlCl3},$ hydrated $\ce{AlBr3},$ hydrated $\ce{FeBr3},$ and $\ce{HgBr2}$ when mixed with $\ce{Fe2O3}$ in place of $\ce{KBr}$ was also observed but they are less effective than $\ce{KBr}$.

References:

Yoshiro Ogata and Jiro Ishiguro, "Preparation of $\alpha$-Naphthaleneacetic Acid by the Condensation of Naphthalene with Chloroacetic Acid," J. Am. Chem. Soc. 1950, 72(9), 4302 (https://doi.org/10.1021/ja01165a533).


text:
Using symmetry and group theory arguments to explain iron(II) in a tetrahedral crystal field
I am trying to figure out how to explain $1s \rightarrow 3d$ spectroscopic transitions for $\ce{Fe^{2+}}$ in $T_\mathrm{d}$ symmetry. These transitions make up the pre-edge region in K edge X-ray absorption spectroscopy (XAS). My goal is to rely on group theory and symmetry arguments as much as possible. Still, I am interested in the problem from a chemist perspective. I will first discuss what I already know and subsequently formulate my question.
$\ce{Fe^{2+}}$ is a $d^6$ ion with the ground state Russell–Saunders term $^5\!D$. In the event of a $1s \rightarrow 3d$ transition we find the final state configuration to be $1s^13d^7$. To find the corresponding terms we couple the core hole $s^1$ and the $d^7$ terms $$^2S \otimes \{{}^4\!F, {}^4\!P, {}^2\!H, {}^2G, {}^2\!F, {}^2\!P, {}^2\!D(2)\} = {{}^{5,3}F, {}^{5,3}\!P, {}^{1,3}\!H, {}^{1,3}G, {}^{1,3}\!F, {}^{1,3}\!P, {}^{1,3}\!D(2)}$$ From the $^5D$ ground state only the quintet $F$ and quintet $P$ terms can be reached following the quadrupole selection rule $\Delta S = 0$. For the isolated ion (without spin orbit coupling) we thus have two transitions: $^5\!D \rightarrow {}^5\!F$ and $^5\!D \rightarrow {}^5\!P$. So far so good. Now we can include the $T_\mathrm{d}$ crystal field.
In a $T_\mathrm{d}$ crystal field the $5D$ terms splits into $^5\!E_u$ and $^5T_{2u}$. Of these, the $^5\!E_u$ represents the lowest energy $(e_2)^3(t_2)^3$ configuration. Our initial state irreducible representations $\Gamma_i$ thus is $\Gamma_i = {}^5\!E_u$.
$$\Gamma_i = {}^5\!E_u$$
The final state irreducible representations $\Gamma_f$ are obtained by branching the $^5\!P$ and $^5\!F$ terms from $O_3$ to $O_h$:
$$\begin{align}
^5\!F &\rightarrow {}^5\!A_{2u} \oplus {}^5T_{2u} \oplus {}^5T_{1u} \\
^5\!P &\rightarrow {}^5T_{1u} \\
\Gamma_f &= {}^5\!A_{2u} \oplus {}^5T_{2u}\oplus {}^5T_{1u}(F) {}\oplus {}^5T_{1u}(P)
\end{align}$$
The quadrupole transition operator in $T_\mathrm{d}$ symmetry is given by $\Gamma_{\hat{T}} = T_{2u} \oplus E_u$. Now I understand that a transition is possible if the matrix element $\langle f|\hat{T}|i \rangle$ is non zero. To see which final states irreducible representations $\Gamma_f$ are accessible through the transition operator $\hat{T}$ we can take the direct product $\Gamma_i \otimes \Gamma_{\hat{T}}$. This will give:
$${}^5\!E_u \otimes (T_{2u} \oplus E_u) = {}^5\!E_u \otimes T_{2u} \oplus {}^5\!E_u \otimes E_u = {}^5\!A_{1u} \oplus {}^5\!A_{2u} \oplus{} ^5\!E_u \oplus {}^5T_{1u} \oplus {}^5T_{2u}$$
Please correct me if I already made some mistakes. From here I am not so sure how to continue. All the final state irreps $\Gamma_f$ are contained in the direct product $\Gamma_i \otimes \Gamma_f$.

Can i conclude that from the $^5\!E_g$ ground state all of the final state irreps $\Gamma_f$ can and will be reached?
If the only possible final states are $(e_2)^4(t_2)^3$ and $(e_2)^3(t_2)^4$ which $\Gamma_f$ irreps correspond to these configurations?
It is well known that the K pre edge gains intensity for non-centrosymmetric ions with respect to centrosymmetric ions. This is because the $4p$ orbitals can mix with the $d_{xy}$, $d_{xz}$ and $d_{yz}$ orbitals. How does this fit the above storyline?

Since 24+ months have gone by with no answers or comments, I think it's fair to let you know about Matter Modeling Stack Exchange which also answers group theory questions, but make sure to delete the question from here if you're cross-posting there (or to at least put a note at the top of your question, saying that it's cross-posted).


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Why does increased conjugation in a molecule cause the wavelength absorbed to decrease (i.e. why does a red-shift occur)?
I was recently looking at a paper about novel organic dyes for efficient dye-sensitized solar cells and one point the authors made was that out of the two dyes mentioned the one with a larger pi-conjugated system had a red-shift compared to the other one.
I understand that a larger pi-conjugated system would increase the stability of the molecule and decrease the gap between the HOMO and LUMO energies, but how would this cause a red-shift?

They were talking about actual color, not wavelength. Larger conjugation gives order like yellow -> orange -> red

In ancient time, when wavelength increases due to any event, it's called a red-shift (wavelength shifting towards red color). Similarly, when wavelength decreases due to any event, it's called a blue-shift (wavelength shifting towards blue color). To date, some of researchers still use this terminology.

Blue/red shift term is used regularly, but rather in physics, e.g. context of the Doppler effect or General relativity. It is not usually used in chemistry, if it is like a  comparison between two different objects.

In ancient times sound a bit to much long time ago. Red or blue shift must be read as "shift to the red" (or to the blue). In general, unless is a single line shifting, there is marginal or no correspondence to the actual colour, it is even somehow complementary. Obviously as red has wl longer than blue, red shift means bathocromic shift and blue means hypsochtomic.shift.

I do assume that now you can make the link to absorption as it is clear to you that generally more extended the effective conjugation is, lower is the energy gap. Your nickname is quite red-shifted :))


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Phenolphthalein + sodium carbonate and phenolphthalein + sodium bicarbonate
What would the word equations be for the following two situations?
Initially $\ce{Na2CO3}$ and $\ce{NaHCO3}$ were present as 2 unknown powders, dissolved in water, and phenolphthalein was added to identify them.

Welcome to Chemistry SE. We have the policy which states that you should explicitly show your thoughts, effort and attempts to understand underlying principles and solve the question.  It'll make us certain that we aren't doing your homework for you. ( As homework is considered literal homework, exams, self-study questions, puzzles, worked examples etc.)  Please provide your full reasoning or thoughts on this, otherwise, the  question may get closed. See Homework

Phenolphthalein is colorless for acidic solution, viz. upto $\mathrm{p}H=8$, and for $\mathrm{p}H>8$, it shows pink color. $\ce{NaHCO3}$ is an amphoteric salt with $\mathrm{p}H$ value close to 8, so phenolphthalein will be colorless. But $\ce{Na2CO3}$ being a basic salt, solution will show pink color.


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Confusion regarding Finkelstein Reaction
In Finkelstein reaction we know when a primary alkyl chloride or bromide is treated with NaI in acetone, we get the corresponding alkyl iodide
Now acetone is a polar aprotic solvent and order of nucleophilicity in acetone is F->Cl->Br->I-. I have read some articles on chemistry.stackexchange which explain why the driving force behind Finkelstein reaction is the precipitation of NaCl or NaBr. But one thing I am still not able to get why the Carbon chlorine or Carbon Bromine bond breaks in the first place by a relatively weaker nucleophile ie I-.

Also see: https://chemistry.stackexchange.com/questions/133984/why-is-finkelstein-reaction-not-conducted-using-naf

First, $\ce{NaI}$ is soluble in acetone while $\ce{NaBr}$ and $\ce{NaCl}$ are not. Here, the reaction is $\ce{S_N2}$ so the nucleophile, $\ce{I-}$, will displace the other halide. The other leaving group halide will enter the solution as $\ce{NaBr}$ or $\ce{NaCl}$ which will precipitate. It doesn't matter if $\ce{Cl-}$ or $\ce{I-}$ is a stronger nucleophile in acetone since they are the leaving group and won't be present in the solution once they are precipitated.


text:
How to compare volatility of inorganic acids
Extending This question further

Sulfuric acid because of its low volatility can be used to manufacture more volatile acids from their corresponding salts

To use this statement in practice I need to compare the volatility of the acids. My book states that using this principle above $\ce{HF, HCl, HNO3}$ can be formed as some examples.
I wish to ask how can I compare the volatility of inorganic acid without being provided the data. Is there any reliable method to do so? I see sometimes hydrogen bonding being used for this purpose but are there any more methods which cover a wider range of molecules ?

You will realize that in chemistry, it is usually much, Much, MUCH easier to grab particular experimental data than to make quantitative theoretical predictions from basic general principles. You can later apply some recognized behaviour patterns, but it is very rough estimation with many exceptions and outliers.


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Finding major product after dehydration

here,should I prefer hydride shift or a product with more $\alpha$ hydrogens?
my doubt is...as hydride shift is easier than methyl shift,the answer should be $a$ or $c$.if methyl shift takes place the alkene formed has more $\alpha$ hydrogens than $a$ and $c$.what should I prefer? The question is to find major product.also,irrespective of the answer I would like to know which is more stable $a$ or $c$?

Compound $a$ is stabler than compound $c$ because the alkene in $a$ has $7$ hyperconjugative structures due to $7 \alpha$-Hydrogen, whereas $c$ has only $5$ hyperconjugative structures.
Hyperconjugation or Nathan-Baker Effect is a stabilizing phenomenon due to resonance-like charge delocalisation. Also, the reason why here hydride shift takes place is because it leads to a stabler carbocation.
Hence the product should be $a$.

methyl shift also produces stable carbocation?

True but hydride shift would be preferred, due to more hyperconjugation

how do I know whether to consider stability of intermediate or product

You consider stability of intermediate first. Once you know which carbocation it is, you start considering product. Please be advised though, try to avoid exam-help questions on this site, it is not in accordance with the guidelines.

ok,and this is not exam-help question it is assignment-doubt question :) thanks for answering

Alright fair enough.

Dehydration of alcohols (in case of secondary and tertiary) proceeds through a carbocation intermediate, as you already have mentioned. You can refer this link in Master Organic Chemistry for the full mechanism.
In this case, the formation of carbocation is the slowest step, hence the rate determining step. So the stability of carbocation must be considered in the mechanism. After the formation of the initial carbocation, it can rearrange either by methyl shift or hydride shift. I could spot two reasons why hydride shift will be preferred -

Hydride has more migratory aptitude than methyl
After methyl shift, the new carbocation will have 6 alpha hydrogen for hyperconjugation. By hydride shift it will have 7. The latter is hence more stable and preferred.

After this step, when the new double bond is formed, it can give either A or C as you said. (A) will have more hyperconjugative structures than (C) so (A) is major.


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Oxidizing properties of sulfuric acid: can it oxidize hydrochloric acid to chlorine?
Can $\ce{H2SO4}$ oxidize $\ce{HCl}$ to $\ce{Cl2}$? I know that $\ce{KMnO4}$ can do it, but my question is considering $\ce{H2SO4}$. In sulfuric acid, sulphur is in +6 oxidation state and in $\ce{KMnO4}$, manganese is in +7 oxidation state. But sulphur is more electronegative than manganese, so I guessed that, $\ce{H2SO4}$ should be equally good oxidizing agent as $\ce{KMnO4}$.
However, I got to know that $\ce{KMnO4}$ can oxidising $\ce{HCl}$ to $\ce{Cl2}$ but $\ce{H2SO4}$ can't. Why it is so? I would also like to know which one has more oxidizing power and why?

Oxidation state and central atom electronegativity do not determine oxidation  properties.

See: https://chemistry.stackexchange.com/questions/94128/mixing-hydrochloric-acid-and-sulphuric-acid-both-concentrated

Concentrated sulfuric acid cannot oxidise chloride or fluoride, see here for a detailed explanation here

Concentrated sulfuric acid is not a strong enough oxidizing agent to oxidize fluoride or chloride. In those cases, only the steamy fumes of the hydrogen halide—hydrogen fluoride or hydrogen chloride—are produced. In terms of the halide ions, fluoride and chloride are not strong enough reducing agents to reduce the sulfuric acid. This is not the case for bromides and iodides.


text:
Reason for different chemical shift in chiral solvents for enantiotopic protons
The chemical shift of enantiotopic protons is defined as follows in Spectroscopy/Homotopic, Enantiotopic, Diastereotopic

Enantiotopic protons have the same chemical shift in the vast majority of situations. However, if they are placed in a chiral environment (e.g. a chiral solvent) they will have different chemical shifts.

The second sentence in this question piques my interest. Even after some searching, I only saw this same sentence over and over again.
What makes the chiral solvent cause a change in magnetic environment causing the change in chemical shift for enantiotopic protons?

Generally, it can be assumed that the chiral solvent would interact with enantiotopic protons differently (because they are chiral). This would provide a different chemical environment for the enantiotopic protons. (The solvent need not stay bound to the solute, just the average interaction has to be different). Whether you see this difference in the actual spectrum is another matter entirely.


text:
How does the surface to volume ratio of ions and their distribution in solution compare to that of suspended nanoparticles in nanofluids?
I am interested in understanding the heat transfer capabilities of fluids.
Nano-fluids are suspensions of nanoparticles (1-100nm sized particles) in a base fluid, often water, oil or glycol. These provide anomalously enhanced heat transfer capabilities on account of the surface area to volume ratio of the particles. The percentage of atoms present on the surface of nanoparticles far exceeds that of the percentage of atoms in bulk materials.
Smaller particle sizes (with larger surface areas) have been shown to correlate with higher specific heat capacities (see Ref. 1).
What I would like to understand is how ions are distributed in a solution, and its impact on heat transfer. Are ions in solution smaller than nano-particles? What is the surface area of an ion in solution, and what volume does it occupy?
If I were to dissolve silver nitrate in water, I would get $\ce{Ag+}$ and $\ce{NO3-}$ ions distributed in the solution. Will they be distributed as aggregates of positively charged silver and negatively charged nitrate ions or as uniformly distributed individual ions?
If the latter, would not these ions have very large surface area to volume ratios, more so than that of nanoparticles? I do not know. Intuitively it seems so. If so, specific heat capacities of solutions should reflect this.
However, this does not appear to be the case. Experimental data and theoretical calculations and models for calculating heat capacities for solutions reflect bulk properties. A common and simple method of estimating specific heat capacity of solutions was proposed by Dimoplon as the sum of the component fractions of individual specific heat capacities:
$$           C_p(\textrm{soln}) = w_\textrm{solid} C_p(\textrm{solid}) + w_\textrm{water} C_p(\textrm{water})   $$
where
$C_p(\textrm{soln})$= Heat capacity of the solution,
$C_p(\textrm{solid})$= Heat capacity of the dissolved solid,
$C_p(\textrm{water})$ = Heat capacity of water,
$w_\textrm{solid}$ = Weight percent of dissolved solid,
$w_\textrm{water}$ = Weight percent of water.
This calculation (which agrees closely enough with experimental data) makes use of specific heat capacities of bulk material, so obviously it appears that the surface area to volume ratio of ions is not very high with respect to nanoparticles.
How then are ions distributed in solution?
I would appreciate if anyone could explain the theory/mechanism behind this or point me to a place where I can obtain a better understanding.
References

Madan Singh, Sekhants o'Lara, Spirit Tlali. Effects of size and shape on the specific heat, melting entropy and enthalpy of nanomaterials. Journal of Taibah University for Science. Volume 11, Issue 6, November 2017, Pages 922-929. https://doi.org/10.1016/j.jtusci.2016.09.011

See also https://en.wikipedia.org/wiki/Ion_association and https://en.wikipedia.org/wiki/Intimate_ion_pair

Why would "uniformly distributed individual ions" result in "very large surface area to volume ratios". Specifically, surface and volume of what exactly?

@Buck Thorn: Nano-particles when suspended in a fluid act as excellent heat transfer media because their surface area to volume ratio is extremely high, allowing for heat to be absorbed and circulated by the large number of surface atoms vis à vis volume. When a salt is dissolved in water, it forms ions. Ions individually are smaller than nanoparticles, thus when dissolved in water as non-aggregates they should have larger surface areas. Should they not have better heat transfer capability? Why is this not the case? I would like to understand how salt ions are distributed in a solution.

Remember that molecules and ions do not have temperature, which is statistical parameter defined for (very) large sets of atoms/molecules/ions only.


text:
Oscillator Strength and Absorption Intensity in TDDFT
When I run a linear-response time-dependent DFT calculation in my favourite electronic structure program, for each excitation I'm given an oscillator strength, $f_i$, and transition dipole moment $$\boldsymbol{\mu}_i = \int\Psi_0\hat{\mu}\Psi_i \text{d} \bf{r}$$
What formulae relates the transition dipole moments, the oscillator strength and the observed absorption relative intensity?
And more specifically, how can I calculate the relative experimental intensity of TDDFT transitions from the oscillator strengths and transition dipole moments?

You may consider posting it on matter modelling SE (which is not to say in any way that this is unsuitable for chem SE)

This might help https://www.researchgate.net/deref/http%3A%2F%2Farxiv.org%2Fftp%2Fphysics%2Fpapers%2F0202%2F0202029.pdf then "take me there". Pdf.

I'll address the last part of your question. I think the first part

What formulae relates the transition dipole moments, the oscillator
strength and the observed absorption relative intensity?

of your question could stand on its own.
The oscillator strength is connected to the transition dipole moment vector via
$$
f_{if} = \frac{2}{3}\omega_{if}|\vec \mu_{if}|^2
$$
in atomic units,i.e. $m_e=\hbar =1$, $\hbar \omega_{if}=(E_f-E_i)$. The equation in SI units can be found on Wikipedia. $i,f$ stand for the initial and final electronic state involved in the transition.
For spontaneous emission, you have to multiply $f_{if}$ by an additional factor of $\omega_{if}^2$ since spontaneous emission is proportional to $\omega_{if}^3|\mu_{if}|^2$.
The oscillator strength is already a dimensionless relative quantity and the ratio of different oscillator strengths should correspond to the ratio's in an absorption spectrum measurement. Just normalize everything by the highest value if you are going for a  plot that is comparable with a relative intensities plot where the peaks are also  normalized .
It is important to note that the calculation only describes a purely electronic transition and yields only a single line at the energy difference of the involved adiabatic electronic states. The result of a real measurement is not a single line and includes many effects that are not described by a simple excited electronic states calculation.
But the calculation gives you an estimate where the center of the peaks should be and how intense they should be. A simple TD-DFT calculation does not give you information about the detailed lineshape. The lineshape is typically assumed to have some form and then added by broadening a stick spectrum based on the oscillator strenghts with a spectral lineshape function. The width of the lineshape function is then adjusted to match the experimental measurement manually.  Common lineshape functions are here.

So the adsorption intensity is directly proportional to the oscillator strength, not with standing in/homogenous broadening, finite temperature and other factors giving finite line width?

Yes, as long as two different electronic stats are involved in the transition. It would not matter for example in a pure infra red spectrum where we do not excite electrons and the electronic state doesn't change. But as long as you also excite electronic states, you get this as a multiplicative factor for your peak intensity.

Temperature effects, Doppler broadening, Life time broadening, etc are all additional effects that have to be incorporated by other means. They are not part of a simple excited state calculation. That is why you need  to broaden the stick spectrum with a suitable lineshape function. This additional broadening with a lineshape function is used to model the effects.  It is non trivial to predict the correct lineshape in detail, and typically its adjusted by comparison with the experimental data.

I'm using an extremely simplified adiabatic approach, instantaneous excitation at the FC point, no relaxation, purely electronic etc.


text:
Enantiomers and diastereomers
I don't understand why on the problem a. B and C are enantiomers, while on the problem b. B is identical to the last molecule.

In the 1,3-disubstituted cyclohexanes, A can jump over the mirror and superimpose on its mirror image. It has a plane of symmetry. B and C cannot superimpose on one another. No symmetry plane in B or C.

Ask yourself this: in problem $a$, are B and C superimposable on each other? You'll find that that is not the case. So those two are enantiomers.
In problem $b$, B and its mirror image are perfectly superimposable on each other, hence identical. This also shows that in problem $a$, B is am optically active molecule, while in problem $b$ it is not.

How I understand that B and C (in problem a) are not superimposable?

If you try to do it, you would see that turning the plane of the mirror image(C) would cause the wedges and dashes to reverse, hence its impossible for it to superimpose with B.


text:
Does Copper(I) Iodide have a dimer?
A qualitative test for the iodide ion involves reacting the salt solution with copper sulphate. This is said to yield the triiodide ion and the white precipitate of $\ce{Cu2I2}$.
I haven't been able to find a proper name for, or any data on this copper iodide... dimer? Wikipedia mentions that it exists in several crystalline forms, so is it something that arises in that context? Or is it just an obscure dimer of copper iodide?

Related. Moreover, when coordinated with monodentate ligands, copper(I) halides can form cyclic dimers, tetramers, zigzag or ladder shaped polymers. See these pdfs: http://polymer.chem.cmu.edu/~kmatweb/2000/October_00/Inorg.Chem/3.pdf ... https://scholarworks.wm.edu/cgi/viewcontent.cgi?article=1578&context=honorstheses

Cu2I2 in this context is just an unnecessarily complicated way to write CuI.

@IvanNeretin or more appropriate, $\ce{2CuI}$ ;)

@NilayGhosh: I saw that exact Quora answer, but he was talking about molecular beams, which (though I know nothing about them) sound way different from anything you'd see with these reaction conditions.


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pH for an HA/NaA solution using a contour map
The following question has troubled me for a while now.

A virtual weak acid $\ce{HA}$ dissociates as follows:
$\ce{HA <=> H^+ + A^-}$
with the acid dissociation constant $K_\mathrm{a} = 10^{-3.00}$.
A solution is prepared using a given concentration of $\ce{HA}$, $C_\ce{HA}$ , and a given concentration of its conjugate base salt $\ce{NaA}$, $C_\ce{NaA}$. The pH of the solution depends on these concentrations. The dependence of the pH on $\mathrm{p}C_\ce{HA} (= −\log C_\ce{HA}$) and $\mathrm{p}C_\ce{NaA} (= −\log C_\ce{NaA}$) is
shown as a contour map below. A doubling in $C_\ce{HA}$ corresponds to a $–0.30$ shift of $\mathrm{p}C_\ce{HA}$ to the left in this contour map.

Figure 1. Contour map of pH for an $\ce{HA/NaA}$ solution.
In this case $C_\ce{NaA}$ is ideally zero and can be neglected. In the above contour map, this situation corresponds to either region $A$, $B$, or $C$, where $\mathrm{p}C_\ce{NaA}$ is large and the following inequalities for $\mathrm{p}C_\ce{HA}$ hold:
Region A: $\mathrm{p}C_\ce{HA} < \mathrm{p}K_\mathrm{a}$
Region B: $\mathrm{p}K_\mathrm{a} < \mathrm{p}C_\ce{HA} < \mathrm{p}K_\mathrm{w}/2$
Region C: $\mathrm{p}K_\mathrm{w}/2 < \mathrm{p}C_\ce{HA}$
Derive analytical solutions for the pH as a function of $\mathrm{p}C_\ce{HA}$ using the charge balance equation:
$[\ce{H^+}] = [\ce{OH^-}] + [\ce{A^-}]$
combined with the above inequalities for each Region. Because doubling $C_\ce{HA}$ corresponds to a $–0.30$ shift in $\mathrm{p}C_\ce{HA}$ as described above, the pH shift can be determined from the derived analytical
solutions. Calculate the pH shift when $C_\ce{HA}$ is doubled for Regions $A$, $B$, and $C$, respectively.

I was able to get the charge balance equation into the following form using $K_\mathrm{a}$ and $K_\mathrm{w}$:
$[\ce{H^+}]^2 = K_\mathrm{w} + K_\mathrm{a}C_\ce{HA}$
$$\mathrm{pH} = \frac{\mathrm{-lg}(K_\mathrm{w} + K_\mathrm{a}C_\ce{HA})}{2}\tag{1}$$
Using this I calculated the pH for region $A$ and got
$$\mathrm{pH} = \frac{\mathrm{-lg}(10^{-14}+10^{-3}\cdot10^{-(1-0.3)})}{2} = 1.85$$
Is this reasoning correct or am I understanding something incorrectly?

Your equation (1) is wrong: the log of a sum is not the sum of the logs.

Thanks for noticing, corrected it by editing the post,

Your second equation  is mathematically still horribly wrong.

Fixed the typo.

The general method described here might help https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566 (Don't be put of by the title)


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Is the effective nuclear charge only defined for valence electrons?
I've just been taught about the effective nuclear charge and screening effect in school. I'd like to ask and confirm if the effective nuclear charge is only defined for valence electrons. If not then why don't we consider the repulsion from the electrons further away from the nucleus than the test electron, which by common sense should add to the attractive force by nucleus?

In chemistry you often worry only about the outermost electrons. In atomic physics they worry about more of them (e.g. levels leading to x-ray generation). As you have noticed, trying to pack all the effects into one parameter (effective nuclear charge) doesn't work so well for all of the electrons.

$Z=Z_{eff}+\sigma$, where $\sigma$ is shielding constant calculated according to Slater's rule. Since Slater's rule can be applied to any electron regardless of whether or not it's a valence electron, $Z_{eff}$ and $Z$ can be defined for any electron too, $Z$ in particular remaining constant. In his paper, which I have attached below, Slater states that the approximations to the wave functions that he gives have been obtained empirically, and do not involve any of the outer electrons.
References:
https://chem.libretexts.org/Courses/Mount_Royal_University/Chem_1201/Unit_2._Periodic_Properties_of_the_Elements/2.06%3A_Slater's_Rules
Original paper by JC Slater ($1930$):
https://web.archive.org/web/20120323031605/http://astrophysics.fic.uni.lodz.pl/100yrs/pdf/04/008.pdf

Can you please elaborate how the Slater's rules explain the effect of outer electrons empirically?

Explain, as in according to slater's rules the calculation of shielding constant only involves inner electrons. That is what I meant.

This still doesn't answer the question. Why don't the outer electrons matter?

The use of the word "think" makes this sound opinion-based. If you remove that and expand on your last sentence plus add some references this would be greatly improved.

@Buck Thorn thanks for the feedback. I have edited.


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Why is ethanol, water and amonium needed for DMG to react with nickel?
I had this question on a test and I'm unsure on why ethanol is required for the reaction.
I understand that you need the ammonia and water to make oxonium ions needed for the reaction to occur.
We had a hint about the solubility of ethanol in water (very high), ammonia (high) and DMG (low)

Have you considered the reverse, solubility DMG and eventually Ni(DMG)2 in ammonia, water and ethanol ?

What do you mean? Sorry, I'm not that good in chemistry...

The DMG - Ni complex is likely reasonably soluble in EtOH/water but not in water. This insolubilty will prevent the reaction going to completion.


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What does fractional notation represent in molecular formulas such as SiO4/2 or H2SiO2/2?
I'm going through some papers about HSQ (hydrogen silsesquioxane) and keep seeing this fractional notation being used:
HSiO3/2 , SiO4/2 , H2SiO2/2 , H3SiO1/2.
Example paper.
I understand that the proper formula for HSQ is [HSiO3/2]2n. Because we can't have fractional atoms, it's always multiplied by an even number and as such the fraction is never a consideration. But then why are the fractions maintained in the other compounds, and not reduced to SiO2 and H2SiO ?
Does keeping the fraction confer some structural information ? I've tried to find out what this notation means, but could not find a proper explanation.
Edit: What do the numerator and denominators mean, formally, in those formulas ? If it's that the oxigen atoms share outside bonds, why not write the compound as SiO4−4 ? If the numerator is the number of atoms, what does the denominator represent ? And is it always /2 ? Can we have /x , where x is a different integer ? I've never seen this type of notation before. Does the denominator mean that the valence electrons are halved ?

Well, yes, it does confer certain structural information. You don't have an entity SiO2, for example. What you have is an entity SiO4 in which all oxygen atoms are shared with others.

So then what does the /2 in $SiO_{4/2}$ mean ? I've never seen the fractional notation before. What is it telling us ? That all of the oxigen atoms are shared ? Why "4/2" then if they all share outside bonds ? Why not write it as $SiO_{4}^{−4}$ ? If the numerator is the number of atoms, what does the denominator represent ? Right until now I've only seen whole numbers in the subscripts, not fractions. Just trying to understand what it represents formally. And is it always /2 ? Can we have /x , where x is a different integer ?

4 stands for four oxygens, 2 means that each oxygen is shared between two SiO4 entities, and we don't write $\ce{SiO4^4-}$ because that's a thing we don't have here either. (We may have it elsewhere, but that's another story.)

"oxygen is shared between two SiO4 entities". Well if you look at the linked paper's figure 1 where they draw out the cage, there's only one $SiO_{4}$ entity. The oxigen atoms are not shared with another $SiO_{4}$ entity, they connect to a $H_{3}SiO_{1/2}$, or a $HSiO_{3/2}$. I am really confused by this notation becaus I've not seen it in any textbook, and it's not formally defined. It just seems like something they came up with and this "number of atoms/shared valence electrons" is a bit hand-wavy.

If the denominator trully means atoms are shared, then isn't it more valid to write $[SiO_{4}]2n$ ? And does the denominator always have to be 2, or can we have $RSiO_{3/X}$ (Taking $HSiO_{3/2}$ as an example) where R = alkyl, halide, alkoxide, etc., and the Si is connected to 3 oxigen atoms but the oxigen is shared in between X entities ? The fraction is really throwing me off.

Also, "You don't have an entity SiO2, for example". I was under the impression that the structure of $SiO_{2}$ was a Si-O cage. That means a Si connected to 4 Oxigen atoms, exactly the same way our $SiO_{4/2}$ is structured. The reason it reduces to $SiO_{2}$ is because those structures repeat and are connected through Si-O-Si bonds to others, making the overall balance of atoms in the compound to equal $SiO_{2}$. Link to the Si-O cage in $SiO_{2}$ showing the same structure as the $SiO_{4/2}$ entity

Would it be correct then to say that silicon dioxide $SiO_{2}$ is made up of $SiO_{4/2}$ entities "in which all oxygen atoms are shared with others." ? Where "4 stands for four oxygens, 2 means that each oxygen is shared between two SiO4 entities," ?

Indeed, you'd better accept it in a more hand-wavy and less formal way. Even if there is a formal definition out there, this notation is far from being used universally. OK, SiO4/2 means that each oxygen atom is shared between two entities which do not have to be similar. Say, between SiO4/2 and HSiO3/2.

I see, thank you. Would it be fair then to say that the /2 is used to show that the O atom is a bridging ligand ? If so, can these be re-written as $HSi(μ_{2}−O){3}$, $Si(μ{2}−O){4}$, $H{2}Si(μ_{2}−O){2}$, and $H{3}Si(μ_{2}−O)$? Or is that wrong because these formulae refer to particular types of Si environment, and are not empirical formulae for the substance as a whole ?

The terminology of metal clusters and ligands... well, I guess it can technically be used here, but then it doesn't sit well with writing "clusters" by parts. When you call your "ligand" $\mu_2$, you imply that it is $\mu_2$ with respect to the whole "cluster".


text:
Can any one list the set of fundamental unsolved questions in coordination chemistry?
I am interested in unsolved problems in coordination chemistry, from particular cases to the most fundamental phenomena.
Thanks in advance

If there even are any "fundamental" unsolved questions in chemistry at all is rather debatable. 2) How you define "coordination" chemistry? It's also debatable what is or isn't a coordination compound.

Hello, Stack Exchange isn't really the right place for open-ended discussion-style questions such as these. There aren't really objective criteria by which one can measure the fundamentalness of a question. There are obviously a lot—possibly even an infinite number!—of unsolved research questions, and whether each of them deserve to be included in a "compilation" of questions cannot be answered easily. In conclusion, there's no "one true right answer" to your question, which is why I've closed it.


text:
Finding the pOH of a buffer from the scratch without the Henderson Hasselbalch equation
I found this question on the internet that said:

Find the pOH of the solution obtained by mixing $\pu{0.1 mol}$ of $\mathrm{NH_4 OH}$ $(K_\mathrm b = 10^{-5})$ and $\pu{0.1 mol}$ of $\ce{(NH4)2SO4}$ in a $\pu{500 mL}$ solution.

What my approach was to do it without applying the Henderson Hasselbalch equation and doing it from the scratch using the "common ion" effect. Here is how I did it:
$\ce{NH4OH}$ is a weak acid and $\ce{(NH4)2SO4}$ is a strong electrolyte.
$\alpha$= The dissociation constant per mole
$$\ce{NH4OH <<=> NH4+ + OH-}$$
\begin{array}{l|c|c|c} \hline
n_\mathrm{initial} & 0.1& 0 &0\\
n_\mathrm{equilibrium} &0.1(1-\alpha) &0.1\alpha& 0.1\alpha \\
\hline
\end{array}
$$\ce{(NH4)2SO4 -> 2 NH4+ + SO4^2-}$$
\begin{array}{l|c|c|c} \hline
n_\mathrm{initial} & 0.1& 0 &0\\
n_\mathrm{dissociation} &0 &0.2 & 0.1\\
\hline
\end{array}
Now, we can say that
$$K_\mathrm b = \frac{[\ce{NH4+}] [\ce{OH-}]}{[\ce{NH_4OH}]}$$
$$10^{-5}= \frac{\dfrac{0.2+0.1 \alpha}{0.5} \times \dfrac{0.1 \alpha}{0.5}}{\dfrac{0.1}{0.5}}$$
As, $\alpha$ is very small for a weak acid, $0.1 \alpha \approx 0$.
Doing the further calculations, we get:
$$\boxed{\alpha = \frac{5}{2} \times 10^{-5}}$$
$$\implies \boxed{\ce{[OH-]} = 0.1 \alpha = \frac{5}{2} \times 10^{-6} \frac{\pu{mol}}{\pu{500 mL}}}$$
Now, $$\mathrm p\ce{OH}=\log[\ce{OH-}] = -[\log(5)-\log(2)-6] = \boxed{5.6}$$
But if we directly apply the Henderson Hasselbalch equation for the buffer, we get:
$$\mathrm p\ce{OH} = \mathrm pK_\mathrm b + \log\left(\frac{[\mathrm{salt}]}{[\mathrm{base}]}\right)$$
$$\mathrm p\ce{OH} = 5 + \log\left(\frac{\dfrac{0.2}{0.5}}{\dfrac{0.1}{0.5}} \right) = 5+ \log(2)=\boxed{5.3}$$
[.] = molarity
Where did I go wrong in the calculation or what mistake did I do while solving this question?

Using photos/screenshots of text ( especially if handwritten ) instead of typing text itself is highly discouraged.  The image text content cannot be indexed nor searched for, cannot be reused nor referred in answers. Additionally, it  can be challenge to decipher.  Consider  copy/pasting or rewriting of the essential parts and using of MathJax for eventual formatting of mathematical/chemical formulas or equations.

Thank you for your reply @Poutnik. So shall I delete the question or edit it and try to write it down in the form of LaTeX?

Rather edit it, using MathJax with mhchem extension.  Notation basics and How can I format math/chemistry expressions on Chemistry SE.
See also upright vs italic and Math SE MathJax tutorial.

@Poutnik I do not know how to write the moles initial and at the equilibrium, there seem to be nothing I can do about it.

Note that $K_\mathrm{b, \ce{NH3}}=\pu{e-5}$ is wrong and $\ce{NH4OH}$ does not exist. $\mathrm{p}K_\mathrm{b}=4.75$, so $K_\mathrm{b}=\pu{10^{-4.75}}$

It is $\ce{NH_3 + H_2O -> NH_4^{+} + OH^{-}}$ and hence we can define K_b for it right?

Let us continue this discussion in chat.

There is a general method described here and in the comments ( don't be put off by the title) https://chemistry.stackexchange.com/questions/60068/how-to-set-up-equation-for-buffer-reaction/95566

@porphyrin it still doesn't answer why do I have different answers with different methods opted.

The method I showed does work with base 0.4 M. and acid 0.2 M, (answer = 5.301) so you will get the correct answer and can then look for errors in your calculation

I found out where the mistake was in my method while solving the question. Here it goes.
• $\ce{NH4OH}$ is a weak acid and $\ce{(NH4)2SO4}$ is a strong electrolyte.
• $\alpha$= The dissociation constant per mole.
$$\ce{NH4OH <<=> NH4+ + OH-}$$
\begin{array}{l|c|c|c} \hline
n_\mathrm{initial} & 0.1& 0 &0\\
n_\mathrm{equilibrium} &0.1(1-\alpha) &0.1\alpha& 0.1\alpha \\
\hline
\end{array}
$$\ce{(NH_4)_2SO_4 -> 2 NH_4^{+} + SO4^{2-}}$$
\begin{array}{l|c|c|c} \hline
n_\mathrm{initial} & 0.1& 0 &0\\
n_\mathrm{dissociation} &0 &0.2 & 0.1\\
\hline
\end{array}
Now, we can say that:
$$K_\mathrm b = \frac{[\ce{NH4+}] [\ce{OH-}]}{[\ce{NH_4OH}]}$$
Plugging in the values mentioned and derived,
$$10^{-5}= \left(\frac{\dfrac{0.2+0.1 \alpha}{0.5} \times \dfrac{0.1 \alpha}{0.5}}{\dfrac{0.1(1-\alpha)}{0.5}}\right)$$
As $\alpha$ is very small for a weak acid, $0.1 \alpha \approx 0$ and $(1- \alpha) \approx 1$. Doing the calculations above, we get that:
$$\boxed{\alpha = \frac{5}{2} \times 10^{-5}}$$
Which further implies that, there are $0.1 \alpha$ moles of $\ce{OH^{-}}$ ions in the 0.5L solution.
$$ \therefore \ \boxed{\ce{[OH-]} = \frac{0.1 \alpha \ \text{mol}}{0.5 \ \text{L}} = \frac{0.1}{0.5} \cdot \frac{5}{2} 10^{-5} = \frac{10^{-5}}{2} \frac{\text{mol}}{\text{L}}}$$
Now, $$\text{pOH} = -\log{[\ce{OH^{-}}]}= -\log{\left(\frac{10^{-5}}{2}\right)}= -[-\log(2)-5]=\boxed{5.3}$$

Find the $\mathrm{pOH}$ of the solution obtained by mixing $\pu{0.1 mol}$ of $\ce{NH4OH}$ $(K_\mathrm{b} = 10^{-5})$ and $\pu{0.1 mol}$ of $\ce{(NH4)2SO4}$ in a $\pu{500 mL}$ solution.

OP wants to find $\mathrm{pOH}$ without applying Henderson-Hasselbalch equation, and to do it from the scratch. Here is how I would did it:
Aqueous ammonia solution, still erroneously called it as $\ce{NH4OH}$, is a weak base, and $\ce{(NH4)2SO4}$ is a strong electrolyte of conjugate acid of aqueous ammonia. In aqueous solution, $\ce{(NH4)2SO4}$ is completely dissociate:
$$\ce{(NH4)2SO4(aq) -> 2 NH4+(aq) + SO4^2-(aq)}$$
The molarity of each of $\ce{(NH4)2SO4}$ and aqueous ammonia is $\dfrac{0.1}{0.5} = \pu{0.2 M}$. Accordingly, initial concentration of $\ce{NH4+}$ due to $\ce{(NH4)2SO4}$ dissociation is $\pu{0.4 M}$. Hence:
$$\begin{array}{l|c|c|c}
 & [\ce{(NH4)2SO4}] & [\ce{NH4+}] & [\ce{SO4^2-}]\\
\hline
n_\mathrm{ini.} & 0.2 & 0 & 0\\
n_\mathrm{diss.} & 0 & 0.4 & 0.2\\
\hline
\end{array}$$
The ionization of ammonia in water can be written as:
$$\ce{NH3 (aq) + H2O (l) <=> NH4+ (aq) + OH- (aq)}$$
If $\alpha$ is the amount of ammonia ionized to give $\ce{NH4+}$ ion in $\pu{M}$:
$$\begin{array}{l|c|c|c}
& [\ce{NH3}] & [\ce{NH4+}] & [\ce{OH-}]\\
\hline
n_\mathrm{ini.} & 0.2 & 0.4 & 0\\
n_\mathrm{equi.} & (0.2 - \alpha) & (0.4 + \alpha) & \alpha \\
\hline
\end{array} $$
By definition,
$$K_\mathrm b = \frac{[\ce{NH4+}] [\ce{OH-}]}{[\ce{NH3}]}$$
$$10^{-5} = \frac{(0.4 + \alpha) \times \alpha}{0.2 - \alpha}$$
Since $\alpha \lt \lt 0.2$ for a weak base, $(0.2 - \alpha) \approx 0.2$ and $(0.4 + \alpha) \approx 0.4$.
$$\therefore \ 10^{-5} = \frac{0.4 \times \alpha}{0.2} \ \Rightarrow \ \alpha = 10^{-5} \times \frac{0.2}{0.4} \tag1$$
Since $\alpha = [\ce{OH-}]$, we can rewrite the equation $(1)$ as:
$$ [\ce{OH-}] = 10^{-5} \times \frac{0.2}{0.4} \tag2$$
Take $-\log$ on both sides of the equation $(2)$:
$$ -\log [\ce{OH-}] = -\log (10^{-5}) - \log \left( \frac{0.2}{0.4}\right) = -\log (10^{-5}) + \log \left( \frac{0.4}{0.2}\right) \tag3$$
The equation $(3)$ is technically the Henderson-Hasselbalch equation for the basic buffer
since $ -\log [\ce{OH-}] = \mathrm{pOH}$, $ -\log (10^{-5}) = \mathrm{p}K_\mathrm{b}$, $0.2 = [\ce{NH3}]$, and $0.4 = [\ce{NH4+}]$. If rewrite the equation $(3)$ with these notation, you got:
$$ \mathrm{pOH} = \mathrm{p}K_\mathrm{b} + \log \left( \frac{[\ce{NH4+}]}{[\ce{NH3}]}\right) \tag4$$
So, you have end up with The Henderson-Hasselbalch equation after all. That's because, it is the way the Henderson-Hasselbalch equation derived. If it's not given, you can derived it and use it. Remember the major assumption, $\alpha \lt \lt [\ce{NH3}]$ and $\alpha \lt \lt [\ce{NH4+}]$ in this case, has to be valid.

Why did you do the calculations considering everything happening for 1 litre of the stuff and not considering 500 mL itself?

Didn't you see I have calculated the initial concentration in first hand so I don't have to deal with volumes?

Thank you for your reply. But why did you bring the "0.2 M" is what I am not able to understand. As in you wrote 0.2 mol of $\ce{NH_4OH}$ where as it was given that it was 0.1 mol in the question. I thought what you might have done would be "if 0.1 mol in 500mL then, 0.2 mol in 1L". Am I missing something? What I wanted to discuss  was the "extra factor of 0.5" I had when I solved it the way I explained.

$\pu{1 M} = \pu{1 mol L-1}$. Your solutions are $\pu{0.1 mol}$ in $\pu{0.5 L}$ for both compounds. So, $\pu{M} = \frac{\pu{mol}}{\pu{L}}$, and do your calculations.

That makes sense to me now, thank you. But I still don't know the "extra factor" of 0.5 I had in the way where I did not convert into standard concentration. Is there an explanation to that?

To me it seems like you have not counted each molecule of $\ce{(NH4)2SO4}$ gives 2 ions of $\ce{NH4+}$. Therefore, $[\ce{NH4+}]$ of the solution is $\pu{0.4 M}$ (not $\pu{0.2 M}$; see my calculations).

I did calculate that, you can see it below the reaction I wrote up, I wrote that 0.1 mol of $\ce{(NH_4)_2 SO_4}$ gives 0.2 mol of $\ce{NH_4^{+}}$.


text:
Diastereomers - cyclic compounds
Why in this problem there isn't a diastereomer with the $\ce{-CH3}$ in the back (dash) and the $\ce{-CH2CH3}$ in front (wedge)?

The structure that you are asking about is essentially the first structure (the left-most one) given.

Try assigning (R) and (S) to the two stereocenters.

To answer this question I would first assign the R-S configuration to the original molecule and the molecule proposed by you and show you that both are identical. At the end I will also show you a simpler way to tell that both are identical.
Original molecule:

Proposed molecule:

As you can see, both the molecules have the same R-S configuration and thus are identical.
Now to see this easily and intuitively just assume you are looking at the original molecule from inside the screen. What do you see? You will observe that the original molecule will look just as same as the proposed molecule which means that both are identical. This method works because our frame of reference doesn't change the optical properties of a molecule until and unless we change the configuration of the molecule by doing some reflections.

Actually for this question if we draw the isomer with methyl group in the back and ethyl group in the front as in:

we would end up getting the enantiomer of the given compound in the question (you could visualise it by flipping the above compound (plane of compound) by 180 degree and then rotate the compound through an angle of 180 degree in the same plane)
And if we draw the isomer like this:

we would end up getting the same initial compound!!( you visualise it in the same way as mentioned before)
So in either case the compound with methyl in front and ethyl at back would not yield a diastereoisomer.
Hope it helps!!!


text:
Why are electrons filled in fourth shell before the third shell is filled?
I am a beginner and am learning about atomic structure and I am getting confusions regarding the aufbau principle.
It is said that the electrons will be filled first in lowest available energy level. But according to aufbau principle electrons are filled in $\mathrm{4s}$ orbital before filling the $\mathrm{3d}$ orbital. Similarly,electrons are filled in $\mathrm{5s}$ before $\mathrm{4f}$.
But I am also taught that if $n$(principle quantum number) increases,the size of the orbital also increases. By size I mean the maximum distance from the nucleus.
So accordingly, $\mathrm{4s}$ orbital should have greater size than $\mathrm{3d}$. Similarly $\mathrm{5s}$ should have greater size than $\mathrm{4f}$. Hence $\mathrm{4s}$ as well as $\mathrm{5s}$ are farther from nucleus as compared to $\mathrm{3d}$ and $\mathrm{4f}$ respectively. Thus they have more total energy and hence electrons should be filled in $\mathrm{3d}$ and $\mathrm{4f}$ before $\mathrm{4s}$ and $\mathrm{5s}$ respectively.
But the above stated doesn't happen. My confusion is, "Why isn't this line of thinking correct?"
Here is a diagram depicting that $\mathrm{4s}$ is bigger than $\mathrm{3d_{xy}}$.

It seems like you are confused about why the principal quantum number is sometimes secondary to the angular number with respect to energy ordering.  If rephrased, it's a reasonable question, why is 4s lower in energy than 3d, for instance.  It's answered entirely in: https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Supplemental_Modules_(Physical_and_Theoretical_Chemistry)/Electronic_Structure_of_Atoms_and_Molecules/Electronic_Configurations/The_Order_of_Filling_3d_and_4s_Orbitals .

@Bertram,thanks for the link you shared,but it still does not clear my confusion about why is this so,the website you mentioned shows the aufbau rule and its exception,but why does it happens so,its not yet clear to me!though I thank for your contribution

The maximum distance of an orbital doesn't mean that the electron is always going to be on the border. You are thinking in terms of Bohr's model, which is wrong.

@Nisarg Bhavsar,could you please elaborate where am I understanding it wrong?I am trying to understand and had searched all way long on internet but didn't find anything!  Please explain more about the 'border' sentence too.

The quantum orbitals are not orbits which an electron takes while "revolving" around the nucleus. It's just an area where the probability of finding an electron is high. In fact the electrons do not "revolve" around the nucleus. An electron is just everywhere in the atom until we measure it's position. So, talking about the "border radius" and relating it with the electrostatic binding energy with the nucleus doesn't hold water in a system with more than one electron.

Note that, as you keep filling electrons, the tendency of an electron to enter the same shell keeps decreasing to reduce the inter-electronic repulsion.

@Nisarg Bhavsar,could you please explain why is n+l rule used to find the energy of the orbital,.  And as you said,orbital is probability where electron may be found at its highest,so what is relationship between energy of electron and it's orbitals,I mean,If energy is more,does that electron probability area increase?


text:
My EDTA titration isn't giving the right titre
I am, as a fun off-syllabus investigation, looking at hard water and how effective the EDTA titration method is, for permanent and temporary hard water.
I ran a titration today and had horrific results. I actually knew the concentration of my water sample: $\pu{0.005 M}$ made up with $\ce{Ca(NO3)2}.$ My EDTA was $\pu{0.01 M}.$
I'm clearly looking at a $1:1$ ratio between $\ce{Ca^2+}$ ions and EDTA molecules. I had $\pu{50 cm^3}$ in my conical flask, so at this concentration, I'd expect my titre to be about $\pu{25 cm^3}.$ I added a little Eriochrome Black T, 8 drops of $\mathrm{pH}~10$ buffer, and my colour change was more or less instant.
If I'd actually bothered to work out that I should aim for a $\pu{25 cm^3}$ titre, I wouldn't have wasted my time but instead, I diluted my EDTA to $\pu{0.001 M},$ and my next few titres were $\pu{4.8 cm^3}.$
Now, I know very little about the buffer. Am I putting far too little in? Or is my EDTA actually, $\pu{0.5 M}$ instead of $\pu{0.01}?$

We cannot predict what went wrong until and unless you add exact details. How much mass did you weigh out? How was the calcium standard made? Calcium nitrate is not a primary standard. What were your calculations? Did you use di-sodium EDTA or EDTA? If you add those details, I will tell you how to make a calcium standard.

8 buffer drops is suspicious. Buffers are not added in drops, but in substantial relative volume. Combining 8 drops with 25 mL 0.01 M EDTA is like combining it formally with about 8 drops of 1 M EDTA, so resulting pH woild be very different to expected pH=10.

0.05 M Na2H2-EDTA is about the top practical concentration, regarding solubility. Solubility of H4-EDTA is significantly less than of the disodium salt.


text:
How can I set up and process my excitation-emission matrices properly?
I'm running EEMS on some water samples on a Horiba Fluoromax 4 spectrofluorometer. Method below.

$\pu{240-580nm}$ emission, $\pu{5nm}$ interval, $\pu{1nm}$ width.

$\pu{240-480nm}$ excitation, $\pu{5nm}$ interval, $\pu{1nm}$ width.

$\pu{0.01 mg/L}$ quinine sulfate standard in $\pu{0.5M}~\ce{H2SO4}$ + water.

Raman EEMS to use as references for my unknowns.

I use factory correction files for both detectors and then use the 3D profile button on my output charts. My data is only letting me export it as ASCII, and my quinine sulfate EEMS looks like the water Raman EEMS.
My questions are:

Is it possible my quinine sulfate is expired? I don't know how old it is (at least 2+ years, but it's been in the fridge in amber plastic). There's also a chance I set it up wrong which is why the signal just looks like water showing some Rayleigh scatter.

How can I export the images of my EEMs intensity profiles? I get an error saying I need to convert my matrix to a worksheet first, but I don't know how to do that.

Finally, how do I get an intensity map of my EEMs (heat map)? Does that happen in Matlab/R? There's a button in FluorEscence to get a 2D intensity map, but it doesn't seem to do anything. Maybe I'm only supposed to export ASCII and the Matlab/R package makes an intensity map. I'm not sure, but the manuals and help for the Horiba software aren't helpful in the least.

You have three questions and need rather detailed and specific information. But you can easily check the quinine fluorescence issue since quinine water fluoresces very well: see here. An inexpensive UV flashlight, such as is used for scorpion hunting, counterfeit specie examination or pet accident detection, works fine if you just want to see if your old quinine solution has gone south.

@EdV Thank you; I will test the flashlight tomorrow - I believe we have one in the lab. Do you know anything about EEMs processing or Horiba software?

I know nothing about Horiba software and probably very little about EEM processing. I know a little about what we used to call synchronous luminescence spectroscopy, where you scan both the excitation and emission and get fluorescence intensity as a function of the excitation and emission wavelengths. But this is not useful: you have a fancy instrument, presumably, and the adventure of mastering it! Best of success!


text:
Acidic character and anion stability across periods and groups of the periodic table
I understand that to compare relative acidity one must consider the stability of the conjugate bases.
Across a period the electronegativity of an element increases. And that is for example why $\ce{HF}$ is more acidic than $\ce{H2O}$, as the negative charge of the fluoride anion is drawn closer to the positively charged nucleus and thus more stabilized than that of an oxide anion.
Down the group the atom radius increases. And that is for example why $\ce{HI}$ is more acidic than $\ce{HF}$, as the negative charge of the iodide anion is distributed over a greater volume and thus more stabilized than that of a fluoride anion.
To me, these two trends are contradictory.
How can the stability of an anion increase as the negative charge gets closer to the nucleus(observed trend across periods) but at the same time decreases as the negative charge gets further away from the nucleus(observed trend down groups)?
Is there a concept I have misunderstood?

There are two factors determining acidity of a compound, the electronegativity factor and the size factor. Down the period, the size factor dominates, and the reason $\ce{HI}$ is more acidic than $\ce{HF}$ is because of this factor, as in the former molecule, the orbitals of iodide participating in bonding are larger and more diffuse, hence bond is weaker.
Across the period, however, size does not vary as significantly as down the group. Hence, we compare using the electronegativity factor. If electronegativity difference is higher, there is larger ionic character, and hence it is easier to deprotonate.
Also, basicity of an ion is proportional to its charge density. So you misunderstood the first concept. You should compare fluoride with hydroxide ion, and not oxide.

There is still something that I still can´t get my head around. You argued that HI is more acidic as its bond is weaker than that of HF. But this method of argument wouldn't support the observation that alkynes (which have stronger C-H bonds) are more acidic that alkanes (see source). So I believed that the only way of explaining the trend down a group is by talking about the bigger atom radius contributing to the stabilization of the negative charge.

mentioned source:  https://www.masterorganicchemistry.com/2018/01/19/hybridization-and-bond-strengths/#six

In case of HI and HF(diatomic molecules) hybridization doesn't apply, so so concept of 'higher s character' cannot be applied here.

There is one more thing I would like to ask you. You argued that across a period size is not a significant contributing factor anymore. So that is the reason why we have to rely on electronegativity. But I believed that the increasing electronegativity across a period is based on the decreasing atomic radius and on the valence shell being more closer to the nucleus and thus keener on accepting an extra electron.

Sure. Hybridization cannot indeed be applied here. But why can´t the concept of stronger bonds, in general, attributing to higher acidity be applied in the example of HI and HF as HF being the one having a stronger bond?

Electronegativity (by the Allred-Rochow scale) is dependent upon both $Z_{eff}$ as well as atomic size. The first factor is more dominating across a period, but be advised, these are all general arguments, exceptions would always exist.

The main reason alkynes are more acidic is because of higher s-character. There is, however, no such consideration in the case of hydrogen halides. The overlap of (pure) orbitals is weaker for $\ce{HI}$, hence it is easier to deprotonate.

I am sorry but I think I might need to be clarified on sth more. So the weaker the overlap of two orbitals is, the easier the deprotonation. In alkanes the sp3-orbital of carbon would overlap with the s-orbital of hydrogen. This overlap is weaker than the one in an alkyne. And furthermore, the negative charge in an sp3-orbital would be found among a larger volume, wouldn't it? Somehow I am always encountering the same contradiction.

The fallacy is that you are applying the same principle to two different cases, one where group trend is being analyzed, and another where we are looking at different hybridizations of the same atom. The contradiction is avoided, because there is no universal principle on deciding acidity. There are general principles and dominating factors, which are different for groups and periods.


text:
What product is created in a zinc - nitrate reaction?
Question:

Place 2 mL of Nitric acid in a test tube and add a small piece of Zinc turning

I thought that the overall net ionic equation would be:
$$\ce{Zn + 2H+ -> Zn^2+ + H2}$$
Is this correct? Is the zinc oxidized and the hydrogen reduced? Or have I got this completely wrong?

You have to consider oxidative properties of nitric acid, being reduced to NO2, NO, or even NH4+.

You are correct.
The net ionic reaction between zinc metal and nitric acid can be written:
$$ \ce{Zn + 2 H+(aq) -> Zn^2+(aq) + H2(g)} $$
The oxidation state of zinc changes as it is oxidized from 0 to +2.
Similarly, hydrogen's state changes as it reduces from +1 to 0.


text:
Does lead carry the negative charge in plumbane?
It says in Wikipedia that lead tetrahydride or hydrogen plumbide or plumbane ($\ce{PbH4}$) is supposed to have lead in its -4 state and the hydrogen is in its +1 state because supposedly lead has a higher electronegativity than hydrogen. Is this possible to be synthesized and actually be stable for long enough to at least view it?

As an oxidation number is just a formal state, it is rather a moot point. Pb in PbH4 definitely has neither +4 neither -4 charge.

$\ce{Pb^{4+}}$ is very unstable due to the inert pair effect, wherein for p-block elements lower oxidation state becomes stabler as we move down the group. Also, the main factor is the electronegativity of lead being more than hydrogen, which you stated in your question (although they are very close). Hence the oxidation states would not be flipped.

For something that's unstable, we seem to see a lot of Pb(IV) in car batteries.

As for experimental synthesis of plumbane, it most certainly has been done by Krivstun et al.[1], preceded by Jin and Taga[2]. The plumbane so obtained was stable enough to allow IR spectroscopy in [1] and chemical analysis in [2]. From Ref. [1]:

For the hydride generation of $\ce{PbH4}$ , we used the
scheme similar to that described in [2]. Figure 1 shows
a scheme of the laboratory setup for synthesis of plumbane. The basic reagents were a 5% solution of $\ce{NaBH4}$ in 0.1 g-equiv./l $\ce{Ba(OH)2}$ and a solution of lead nitrate in 0.7 M $\ce{HNO3}$ . To increase the yield of $\ce{PbH4}$ , a $\ce{K2S2O8}$ oxidizer was added to the nitrous acid solution to saturate the solution at 10°C. Lead nitrate was added to the solution directly before the performance of the reaction. The reagents were mixed in a reactor R, which represented five coils of a glass tube with inner diameter of 5 mm. The mixing zone of reagents and a spiral reactor were cooled by means of a water jacket at T = 5–10°C. The reaction products were directed to a gas–liquid separator V3 from which $\ce{PbH4}$ was carried out by the $\ce{H2}$ flow, which was produced in excess during the reaction. Then, gaseous products of the reaction were directed to a flow gas cell through a trap V4 cooled to T = –100°C. The trap served to decrease the amount of water vapors filling a measuring cell.
Special attention was paid to the uniform supply of
reagents to the reaction zone and the absence of bubbles and solid particles, because in the case of deviation
in the ratio of components (the excess of $\ce{NaBH4}$),
reduced elementary lead appears, which causes fast
decomposition of $\ce{PbH4}$. A contact of reaction products with metals and grease also results in the fast decomposition of $\ce{PbH4}$. We determined the amount of prepared $\ce{PbH4}$ by measuring the mass of Pb carried out by the $\ce{H2}$ flow from the reactor. For this purpose, we used traps V5 and V6 cooled with liquid nitrogen. The traps were evacuated after termination of the reaction and then were heated to room temperature, resulting in a complete decomposition of collected $\ce{PbH4}$. The instant of the decomposition could be observed visually by the sharp formation of an opaque black film of amorphous
lead on the trap walls, which were transparent at low temperature. The amount of the lead formed was measured be the chemical method and was 3.2 mg per 1 g of $\ce{NaBH4}$ under optimum conditions of the synthesis.

References
1.
Krivtsun, V. M.; Kuritsyn, Y. A.; Snegirev, E. P. (1999). "Observation of IR absorption spectra of the unstable PbH4 molecule" Opt. Spectrosc. 86 (5): 686–691.
2.
Jin, K. and Taga, M. (1982). "Determination of lead by continuous-flow hydride generation and atomic absorption spectrometry". Anal. Chim. Acta 143: 229.

Another reference (although behind paywall): https://pubs.rsc.org/en/content/articlelanding/1925/ct/ct9252702845


text:
Which salts give the sodium carbonate bead test?
Apart from the borax and the phosphate bead tests, there's also this one.
It's  done by heating the white bead of sodium carbonate and the salt with potassium nitrate; green and yellow colours indicate the presence of manganese and chromium respectively.
I've only seen these two examples for this test. Are there any more?


text:
Coordination complex - nitrito-N & nitrito-O
I am currently learning the coordination compounds and encountered a new isomerism called linkage isomerism.
I am thinking that if I want to get some $\ce{[Co(NO2)6]^3-}$, I should have to dissolve $\ce{CoCl3}$ in water to get $\ce{[Co(H2O)6]^3+}$, then add some concentrated $\ce{HNO2}$ to displace the six water molecules away (although I don't know why $\ce{NO2-}$ is a stronger ligand than $\ce{H2O}$).
Questions:

Will I get a 50-50 mixture of nitrito-N & nitrito-O co-coordination complexes? Or will there be a chance of getting random combinations of nitrito-N & nitrito-O coordinate bonds to a $\ce{Co^3+}$ (e.g. 4 nitrito-O + 2 nitrito-N)?
If I just want to get nitrito-N complex, are there ways to prevent the formation of nitrito-O complex? Or the mixture must form and then we need to do some other reactions to get rid of nitrito-O complexes?

See here.

Linkage isomerism is of major interest in the field of inorganic chemistry. Linkage isomerism occurs when an ambidentate ligand such as $\ce{NO2, SCN,}$ etc. binds to the metal center. Nitro-nitrito metal complexes can be regarded as good examples of linkage isomerism. It must be noted that the $\ce{NO2}$ group can be coordinated to the metal atom in different ways such as nitro (via the nitrogen), nitrito (via oxygen), chelating via both oxygens (nitrito-O,O′), and bridging nitro, usually via nitrogen and oxygen atoms (Ref.1):

Source of diagram: Modified from Ref.1
Amongst various coordination modes, above diagram shows not only $\ce{NO2}$ is capable of having the wide range of binding modes in mononuclear metal complexes, but also its potential tendency to act as a bidentate ligand in some cases.

Will I get a 50-50 mixture of nitrito-N & nitrito-O co-coordination complexes? Or will there be a chance of getting random combinations of nitrito-N & nitrito-O coordinate bonds to a Co3+ (e.g. 4 nitrito-O + 2 nitrito-N)?

Not necessarily! It is not easy (at least not as easy as OP described it) to synthesize specific linkage isomer. It has been shown that the coordination mode of nitrite ion depends on its nature and the stereochemical environment which surrounds the metal ion (Ref.1 & references there in).

If I just want to get nitrito-N complex, are there ways to prevent the formation of nitrito-O complex? Or the mixture must form and then we need to do some other reactions to get rid of nitrito-O complexes?

These specific compound have been synthesized. Thus, there should be ways to such synthesis. However, I'm not qualified to give you a definitive answer for that question. Inorganic coordination chemistry expert would be the most qualified one. Yet, references in here including Ref.1-3, and references there in would give you better view. For instance, Ref.3 has described a ways to synthesize $\ce{Cu(NO2)2}$ and $\ce{Mg(NO2)2}$ complexes. Both metal showed nitrito-O,O′ (as bidentate) coordination with different stabilities.

References:

Hoda Pasdar, Bahare Hedayati Saghavaz, Naser Foroughifar, and Mehran Davallo, "Synthesis, Characterization and Antibacterial Activity of Novel 1,3-Diethyl-1,3-bis(4-nitrophenyl)urea and Its Metal(II) Complexes," Molecules 2017, 22(12), 2125 (12 pages) (DOI: https://doi.org/10.3390/molecules22122125).
Andrew J. Timmons and Mark D. Symes, "Converting between the oxides of nitrogen using metal–ligand coordination complexes," Chem. Soc. Rev. 2015, 44, 6708-6722 (DOI: 10.1039/C5CS00269A).
Andrew J. Timmons and Mark D. Symes, "Coordination of $\ce{NO2}$ to $\ce{Cu}$ and $\ce{Mg}$ in $\ce{M(NO2)2}$ Complexes. A Theoretical Study," Chem. Soc. Rev. 2015, 44, 6708-6722 (DOI: 10.1039/C5CS00269A).