text:
Pyridinium ion in inositol nicotinate
I have read about inositol nicotinate which is used as a medication for deficiency in vitamin B-3. I think that if acetic acid was applied to this medication it would form pyridinium ion and acetate ion, but I don't know if the acetic acid will react with inositol in a transesterification reaction, and form nicotinic acid and ionsitol acetate, so if the latter happens how could I prevent it and push the reaction towards the first route?

First of all, why you want to make a acetate salt of inositol hexanicotinate? Well, if things happens as you suggested, both nicotinic acid and expected inositol hexanicotinate$\ce{.6CH3CO2H}$ would disolve in aqueous medium while trans-esterated product would dissolve in only organic solvents. So, you may able to remove unwanted byproduct by solvent extraction.

@MathewMahindaratne then I should react acetic acid with inositol hexanicotinate in organic solvent and then collect nicotinic acid and inositol hexapyridinium hexaacetate using water ?But I think  more nicotinic acid would be produced than the desired product. I want to push the reaction towards making inositol hexapyridinium hexaacetate. Note: inositol hexanicotinate is insoluble in water

Sounds like a plan.


text:
How is the pH scale constructed?
How is the pH scale constructed , I mean the strength of an acid is determined by its pKa value and this one can be determined in different solvents if water doesn't suits it but how do you determine the pKa value of a substance in water after you have found the value of its pKa in another solvent ? Also I don't think I quite understand why the pH interval in water is running from 0 to 14 (I know that it has to do with the leveling effect but when you take this scale for other solvents the starting value can be different from 0 , if I am not mistaken for liquid NH3 the starting value is 5 ) can someone explain to me from where does this values come from ?

First of all the scale in water doesn't go from 0 to 14, it can extend beyond those values (values of -2 to 16 are possible without too much effort in a lab). The scale in liquid ammonia is much larger, covering over 30 units. See here and here, as well as several other Chem.SE posts tagged with pH.

The pH is determined experimentally with a glass electrode. A glass electrode is a small glass tube limited by a thin membrane and filled with an aqueous KCl solution with a silver electrode inside. There is also a similar silver electrode out of the tube, in a compartment in contact with the outer solution. When you dip a glass electrode into an aqueous solution, the experiment shows that the tension E read between the two electrodes is proportional to the logarithm of the H+ concentration of the external solution. It is :  E  = Constant + 0.059 pH. The Constant depends of the particular glass membrane.
And the experiment shows that pH is -log[H+] at low acidic concentrations, and is measured as

7 in pure water,
3 in 0.001 M aqueous solutions
2 in 0.01 M aqueous HCl solutions,
11 in 0.001 M aqueous NaOH solutions
12 in 0.01 M aqueous NaOH solutions.

These values are in agreement with the usual expression : pH = -log[H+], or : pH = 14 - log[OH-]. At higher concentration, the measured pH values are not exactly equal to the logarithm of [H+]. The measured pH is said to be the logarithm of the "activity" of H+.
The pH does not go only from zero to 14. It may be smaller than zero in concentrated solutions of HCl and of H2SO4.
In non-conducting solutions, the glass electrode does not give any stable values. The measured tension is fluctuating without any reason and can be ± 3, 5, 10 or 0, and this changes constantly.

Maurice's great experimental depiction clearly notes the measurements are created in DILUTE conditions. Otherwise, the 'activity coefficient' is important which is a function of ion charge, ion size and ionic strength. So, in the presence of high ionic concentrations of salts like FeCl3 or CaCl2 with a large ion charge, even dilute HCl can apparently 'act' as a much stronger acid due to the activity coefficient which is applied to [H+]  to determine the pH (see  http://www.csun.edu/~hcchm003/321/321100313.pdf page 4).

The pH+ scale is a logarithmic scale, based on the number of $\ce{H+}$ ions available (with the caveat that $\ce{H+}$ is theoretical, it may be $\ce{H3O+}$ or other ionic species). I.E., in water at ~300 K, ~10−7 mol dm−3 of the positive and of the negative ions exist, a pH+ of 7 (the negative exponent).
For a strong acid, virtual all, 100%, of the molecules are broken into ions, e.g. $\ce{Cl-}$ and $\ce{H+}$, so the pH+ is 0 [10−0 = 1].
Note that at different temperatures, water may be more or less ionized, so a neutral pH+ is not necessarily 7. At its boiling point, "neutral" is about 6.14.

source: Stack Exchange

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

sort: votes


text:
Cyclisation of acetone on reaction with conc. H2SO4
Mechanism for reaction of acetone with concentrated $\ce{H2SO4}$:
The product is supposed to be mesitylene, and on various sites I found the reason behind it to be aldol condensation And some sites provided me with unusual products( as for a high school student)

Does this reaction have a reasonable reaction mechanism or do we just have to mug this up?

What are these cyclisation kinda mechanisms called?

Do they have a category that is to be understood?

Yeah, it's standard acid catalysed aldol condensation.

Do an acid-catalyzed aldol condensation to get (E)-4,6-dimethylhepta-3,5-dien-2-one. Form the vinylogous enol with a Z-geometry then cyclization by either double bond addition to the carbonyl or hexatriene cyclization. Work on it awhile!

You may have noticed that your question received a downvote. This was probably because you did not show any effort toward a solution of the question. I posted in a Comment some clues to assist you in a solution. In the interim an Answer was posted that showed the compounds formed in the reaction of acetone with H2SO4 but it did not illustrate a mechanism for the formation of mesitylene, which was your question. Here is my contribution to your understanding.
The reaction between the keto form of acetone 1a and its enol 1b forms aldol 2. Acid-catalyzed dehydration of 2 via the enol 3 leads to mesityl oxide 4. The loss of water from 3 may be stepwise but, to save space, I have presented the loss of water in a single operation. The third unit of acetone is incorporated via the vinylogous enol 4b to form the dienone 5a. Enolization of 5a forms 5b' that allows rotation about the red single bond in 5a and 5b' to form conformation 5b". This species may cyclize by an allowed thermal electrocyclization to form cyclohexadiene 8 which is readily dehydrated by acid to mesitylene 9. Alternatively, enol 5b' may reketonize to 6 and undergo cyclization to cation 7 and thence on to mesitylene 9. continued...

Addendum: An old name for acetone was mesite, the source of mesitylene. Read about it here and here.

Thanks it was really helpful.. actually  I am a student and when I heard about acid cataylsed aldol and kinda freaked out cause i was taught that you identify aldol itself with Dil naoh/koh and heat.

Could this be performed with acetaldehyde to get benzene?

I came across this: https://chemistry.stackexchange.com/questions/100340/acetaldehyde-with-few-drops-of-concentrated-sulphuric-acid
Which... makes things confusing.

@McSuperbX1: What exactly is confusing you? Why not pose a new question.

@user55119 after 4a , why doesn't aldol directly happen on the right side methyl group (attached to the ketone group). It rather unexpectedly follows a "sort of" 1-4-michael addition mechanism. why does this happen ?

Certainly the yield of mesitylene is not 100%. Enol 4a is more stable than the enol toward the methyl, for what that is worth. You say "unexpectedly". Apparently, not to Mother Nature.  Base vs. acid-mediated reactions can go in different directions. The bromination of 2-butanone in base leads to a haloform reaction whereas in acid, bromination is autocatalytic  and bromination occurs at C3 via the more stable enol.

Catalytic (acid or base) self-condensation of acetone is a very complex reaction, which has been analyzed several years ago (Ref.1). I don't have an access to the original article, yet, I recommend OP to find this article and read it carefully. Its abstract states that:

Catalytic self-condensation of acetone is a very complex reaction and numerous products are possible via competitive self-condensation and cross-condensation between the same or different ketones that are formed in the reaction. All the major products of the reaction, diacetone alcohol, mesityl oxide, phorone, mesitylene, isophorone, 3,5-xylenol and 2,3,5-trimethylphenol find important industrial applications. The reaction is catalysed by acids as well as bases, and it can be carried out in both liquid and vapour phases. The selectivity of the reaction for the desired product is achieved by proper choice of catalyst and experimental conditions. This paper reviews the recent developments in the process of self-condensation of acetone, evaluating the significance of various parameters for obtaining the desired product.

Nonetheless, I found a sketch of condensation pathways of acetone from that article, which I depicted in following diagram:

As OP recognized, it is based on acid-catalyzed aldol condensation. Condensation of two molecules of acetone gives 4-methylpent-3-en-2-one (the trivial name of which is mesityl oxide) first. Keep in mind that since this is not a base catalyzed reaction, fair amount of 4-hydroxy-4-methylpent-2-one is also exist in the reaction mixture. However, these two compounds and acetone further undergo a plethora of other condensation to give several products, including mesitylene (see the diagram).
Nevertheless, the method is used to synthesize mesitylene (Ref.2) large scale, although the yield is poor (~15-20%).
References:

G. S. Salvapati, K. V. Ramanamurty, Mulpuri Janardanarao, “Selective catalytic self-condensation of acetone,” Journal of Molecular Catalysis 1989, 54(1), 9-30 (https://doi.org/10.1016/0304-5102(89)80134-8).
Submitted by Roger Adams and R. W. Hufferd (Checked by H. T. Clarke and W. W. Hartman), “Mesitylene,” Organic Syntheses, Coll. Vol. 1 1941, p.341 (Org. Synth. 1922, 2, 41) (DOI: 10.15227/orgsyn.002.0041).

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


text:
What is the order of acidity for hydrogens in primary, secondary and tertiary alkanes, alkenes, and cycloalkanes
The bond dissociation energies of C-H bonds in alkanes, alkenes and aromatic compounds decrease in the following order: primary > secondary > tertiary > allylic > benzylic.
The weaker the bond, the lower the bond dissociation energies. Does the acidity of primary, secondary, tertiary, allylic and benzylic hydrogens follow the same order?

No, it correlates rather with how well carboanion is stabilised.

related https://chemistry.stackexchange.com/questions/43468/which-proton-in-cyclohexene-is-the-most-acidic


text:
Is it possible to find the volume of condensed water by temperature change without knowing the initial pressure of the flask?
I found this problem in a practice sheet. The original source doesn't indicate any authorship or whatsoever.
It states the following:

A sealed container of $\pu{6.24 L}$ in capacity has humid air at $\pu{30 ^\circ C}$ with a $RH=90\%$. The container temperature is then lower to $\pu{27 ^\circ C}$. Find the volume of condensed water. The vapor pressure of water is $\pu{32 mmHg}$ at $\pu{30 ^\circ C}$ and $\pu{27 mmHg}$ at $\pu{27 ^\circ C}$. You can consider that the volume of condensed water is insignificant.

The problem doesn't specifically mentions any other assumptions but I'm assuming that the gas is behaving ideally and since it is referring to water. I will need the density of water, which is $\pu{1 \frac{g}{mL}}$ and the formula weight of water, which is $\pu{18 g mol-1}$.
The strategy which I thought was that the moles of air will remain the same when the vessel is cooled but the total moles will change before and after, as the moles of water initially will be different when the vessel it is cooled.
Then to find the volume what I need to find is that difference in moles and use the formula weight of water and the density to get volume.
However this is a problem because, how can I characterize the vessel if I'm not given an initial pressure?
I'm aware that:
$$RH=\frac{\textrm{partial pressure}}{\textrm{vapor pressure}}\times 100$$
Therefore I can say that the partial pressure for water is:
$$0.9\times \pu{32 mmHg} = \pu{28.8 mmHg}$$
But what I need is the total pressure in the vessel. Is it possible to find it?
If I could find the total moles, what I could do is find the moles of air by difference as follows:
$$n_{air}=n_{t}-n_{water}$$
These moles will not change.
The moles of water could be found by using the ratio of the partial pressure of water to that of the total is equal to the moles of water divided by the total moles.
$$\frac{n_{water}}{n_{total}}=\frac{p_{water}}{p_{total}}$$
But as it can be seen, I'm stuck. I did not included numerical calculations because if I were to use the equation of the ideal gas:
$$PV=nRT$$
What would I had use as the total pressure? Therefore I'm trapped. Is there any way to find what it is being asked?
Edit:
Following the indications by the answer I did the following:
The number of moles of water vapor at $\pu{30 ^\circ C}$ can be found from using the equation of ideal gas:
$\left(0.9\times 32\,mmHg\right)(6.24\,L)=n_{water}(62.4\frac{mmHg\cdot L}{mol\cdot K})(30+273)K$
$n_{water}=0.009505\,mol$
Then the number of moles of water vapor at $\pu{27 ^\circ C}$:
$\left(27\,mmHg\right)(6.24\,L)=n_{water}(62.4\frac{mmHg\cdot L}{mol\cdot K})(27+273)K$
$n_{water}=0.009\,mol$
Then:
$\Delta n_{water}=0.009505-0.009=0.000505\,mol$
From this quantity the grams of water and volume can be obtained using density of water and the formula weight.
$V=0.000505\,mol\times\frac{18\,g}{1\,mol\,\pu{H_{2}O}}\times\frac{1\,mL}{1g}=0.009090\,mL$
Which would be the volume of the condensed water. Judging from the ammount obtained it makes sense that it is a very tiny volume.

Why do you need to mention air at all? Air is quite irrelevant. It could just as well have not been there. You have the pressure of water vapor before and after, you don't need anything else.

Is your answer $\pu{8.99 mg}$?

@MathewMahindaratne I have obtained $0.009,mL$ as the volume since the density of water is $1\frac{g}{mL}$ that would be what you obtained. I'll show it as an edit in my question.

Your calculations are seemingly okay to me. But you must recheck the numerical value of $R$ (gas constant). Also, you must state that your assumption of saturated vapor pressure at $\pu{27 ^\circ C}$, because some water is condensed.

@MathewMahindaratne I rounded the gas constant but the value which is $62.36359 , torr ,L, mol^{-1} K^{-1}$. Were you referring to this?

$R=\pu{8.31446 \times 10^{−2} L bar K−1 mol−1}\times \frac{\pu{760 Torr}}{\pu{1 bar}}=\pu{63.189896 L Torr K−1 mol−1}$.

@MathewMahindaratne Chris's value for R is correct.  760 torr = 1 atm, not 1 bar. The correct conversion is 760 torr = 1.01325 bar. Hence R = 62.3636 L torr/(mol K).

Chris, I get essentially the same answer as you, though you should express it to only two sig figs (based on what was given in the problem): 0.0091 mL. Expressing it to four sig figs, as you did, misrepresents the precision of the answer.

@theorist Okay!. I forgot that fact. I did my calculations in a rush and did not paid attention to that detail.

This problem is sloppily worded and, technically, it is not possible to give a definitive answer based on the information provided.  But the missing information is not what you think -- you don't need to know the air pressure in the flask.  What's missing are two key pieces:  The degree of humidity ("humid" doesn't mean "saturated") and the volume of water in the flask.  If the flask contains no water, and air that, while humid, is sufficiently below 100% $RH$ at $\pu{30 ^\circ C}$, there might be no water condensation when cooling to $\pu{27 ^\circ C}$ (i.e. you'd just be increasing the $RH$ from some number <100% to a value that, while larger, is still <100%). Alternately, if the flask is initially half liquid water, then the volume of water that condenses would be about half that if the flask instead initially contained a very small volume of liquid water. Thus here is what I think the author meant to ask:

A sealed container of $\pu{6.24 L}$ contains an equilibrated mixture of air and liquid water at $\pu{30 ^\circ C}$. The container temperature is then lowered to $\pu{27 ^\circ C}$. Find the volume of water that condenses. Assume the volume of liquid water is insigificant relative to the volume of the container.  The vapor pressure of water is $\pu{32 mmHg}$ at $\pu{30 ^\circ C}$ and $\pu{27 mmHg}$ at $\pu{27 ^\circ C}$.

As to how to approach the question, that's straightforward:  You know the partial pressure of the water vapor at each temperature.  You also know the total volume of the air.  Given this, you can determine the number of moles of water vapor at each temperature (using the ideal gas law).  The difference is the number of moles of water that condenses, from which you can then determine the volume of the water that condenses.

I'm sorry, during my transcription of the problem I forgot to give the relative humidity which was 90 percent. The other piece which I omitted in a rush was that the problem did say to assume that the volume of liquid water is insignificant. But it used the term condensed and because of such I found it misleading because it said that I have to find the volume of condensed water and the it mentions that the volume of condensed water is negligible. I did not understood this part. How does this changes the way to approach this question?

In my question I did mentioned that density of water and the formula weight can be used to find that volume. Am I right with this part?

@ChrisSteinbeckBell (1) Since the RH is 90% (and assuming equil. conditions), there can be no water in the container to start (if there were water, the RH would be 100%).  So the final V of condensed water equals the V that condenses.  What they meant by "negligible" is prob. that the V of condensed water is negligible relative to the V of the container (such that you can use V_container = V_gas even after the water condenses). (2) Yes, you would use the molecular weight and density to convert from moles to volume: molecular weight converts moles to mass, and density converts mass to volume

I did the computation of the volume. If I understood you correctly that would be the volume but I'm still lost at trying to get your idea from your answer about $RH$. Can you explain this part from your answer?. If the flask contains no water, and air that, while humid, is sufficiently below $100% RH$ at $\pu{30 ^\circ C}$, there might be no water condensation when cooling to $\pu{27 ^\circ C}$.

and Alternately, if the flask is initially half liquid water, then the volume of water that condenses would be about half that if the flask instead initially contained a very small volume of liquid water. I don't understand clearly what you meant by those two ideas.

This part I'm still stuck i.e. you'd just be increasing the $RH$ from some number $<100%$ to a value that, while larger, is still $<100%$). Can you please tell me what did you meant by this?

@ChrisSteinbeckBell  (1) Image you're cooling the air from Th to Tc.  Assume at Th and Tc a certain V of air can hold 1 mol and 0.9 mols H2O vapor, respectively.  Further assume the container has no water, so the air is not saturated and that, in addition, the RH at Th is 80%, meaning it is actually holding 0.8 moles H2O vapor. Now we cool it down to Tc.  At Tc the air is capable of holding 0.9 mols vapor, so none condenses.   Further, the RH at Tc is 0.8/0.9 x 100 = 89%.  Hence the RH has changed from some number <100% (80%) to a number that, while larger, is still <100% (89%).

@ChrisSteinbeckBell (2) Suppose instead the flask has water, in which case the air is saturated.  If the flask is half-filled with water, the volume of air is half what it would be if the volume of water were negligible.  Thus half as many moles condense, because the number of moles of water that condense when you cool the flask is proportional to the volume of the air.
Also:  If you found my answer above helpful, you should up-vote it and/or accept it

RH relates pressures, in your example you use moles. Is it equivalent?. RH is the capacity to hold an ammount of water by a gas? or is it vapor? or is it both?. Is this the idea?. In your given example the RH at the lower temperature (Tc) uses the same amount of moles that was on the Th and is divided by 0.9 mol, does it mean that the 80% at the beginning had originaly $0.9\times0.8=0.72,mol$ of water?. How do I interpret HR=100% and values higher than that?. How do these affect condensation?.

Had the problem been posed as in your example (in the sense of being filled to one half with water) could it be approached the same way as if it had no water inside?.

@ChrisSteinbeckBell No. If the air at Th were fully saturated, it would hold 1 mol of H2O. Thus, since its RH at Th is 80%, it actually held 1 mol x 80% RH = 0.8 mols H2O (at Th). It looks like you have some basic confusion about the general relationship between moles, vapor pressure, and relative humidity. I'd recommend posting that as a separate question. As to the flask being half-filled with water as opposed to a tiny amount of water, it's approached the same either way: (1) the air is saturated with water; (b) the volume of air is reduced by half when the flask is half-filled with water.

I'll follow your suggestion but before doing that I'd like to know why am I using the relative humidity times the vapor pressure at $30^{\circ}C$?. Is it because when I'm doing this, I'm finding the moles of water which is in the air (the actual vapor). But in the second step, why the moles of water is found only from the vapor pressure?. With the given information can the $RH$ be found at that cooler temperature ($27^{\circ}C$)?.

@ChrisSteinbeckBell  The first step gives you how much water vapor it actually holds at Th. The second step gives you the max it can hold at Tc.  The difference is what comes out of the air.  Here's an analogy: suppose you have a 1L jar that is 90% full of liquid.  You then pour that liquid into a jar that can only hold a max of 0.8L of liquid.  How much liquid overflows the second jar?  Answer: (1L x 90%) - 0.8L = 0.1L (here the 90% is analogous to RH).  Take some time to carefully think about what I wrote and try to understand it. A more detailed explanation requires a separate question.


text:
What are some other gases that liquify at around the same temperature as ammonia?
I am looking for gases that have a liquefication that falls within or on the edges of the liquefication range of Ammonia. I am mainly looking for gases that are comprised mainly of Carbon, Nitrogen, Oxygen, Hydrogen, Chlorine and Fluorine and liquify in a similar temperature and pressure range. This is for a worldbuilding project, but since the question is real, not theoretical, chemistry, I thought it would be better to ask about it here.

Wikipedia has a large list of gasses in ambient conditions here as well as a list of refrigerants here. You can sort them in order of boiling point by clicking on the small black triangles just underneath the respective column. Then just look for things in the range of ammonia, which boils at -33 °C under 1 bar. Note that a large number of entries in the table for refrigerants are for mixtures of gasses.
I don't think these lists are complete, but there are comparatively few substances with such low boiling points (probably no more than a few thousand that could ever be made), simply because having more than a handful of atoms will already cause the molecules to attract enough to remain as liquids or solids at non-cryogenic temperatures.

I found few gasses under the category of the gases, which have a boiling points (at atmospheric pressure) that falls within the boiling point of Ammonia. since you are mainly looking for gases that are comprised mainly of Carbon, Nitrogen, Oxygen, Hydrogen, Chlorine, and Fluorine, these gases are ideal for your purpose:
$$
\begin{array}{c|c}
\text{Name of the compound (gas)} & \text{Boiling point at } \pu{760 mmHg} \\
\hline
\text{Ammonia, } \ce{NH3} & \pu{-35.5 ^\circ C} \\
\text{Chloromethane (Metyl Chloride), } \ce{CH3Cl} & \pu{-23.9 ^\circ C} \\
\text{1,1-Difluoroethane (Freon refrigerant R-152a), } \ce{CHF2CH3} & \pu{-25 ^\circ C} \\
\text{Methoxymethane (Methyl Ether), } \ce{CH3OCH3} & \pu{-25 ^\circ C} \\
\text{1,1,1,2-Tetrafluoroethane (Freon refrigerant R-134a), } \ce{CF3CH2F} & \pu{-26.3 ^\circ C} \\
\text{Dichlorodifluoromethane (Freon refrigerant R-12), } \ce{CCl2F2} & \pu{-29.8 ^\circ C} \\
\color{Red} {\text{Chlorine, } \ce{Cl2}} & \color{Red} {\pu{-34.4 ^\circ C}}\\
\text{Octafluoropropane (Freon refrigerant R-218), } \ce{CF3CF2CF3} & \pu{-36.7 ^\circ C} \\
\text{Chloropentafluoroethane (Freon refrigerant R-115), } \ce{CClF2CF3} & \pu{-39.1 ^\circ C} \\
\text{Chlorodifluoromethane (Freon refrigerant R-22), } \ce{CHClF2} & \pu{-41.2 ^\circ C} \\
\text{Propane, } \ce{CH3CH2CH3} & \pu{-42 ^\circ C} \\
\text{Propyne, } \ce{CH#CCH3} & \pu{-47.7 ^\circ C} \\
\text{Propene, } \ce{CH2=CHCH3} & \pu{-47.72 ^\circ C} \\
\hline
\end{array}
$$
Among these gases, chlorine has come the most closest to ammonia.
Most data are from this source and most Freon data are from this source.


text:
Oxidation state of Manganese
We know that Mn shows variable oxidation states ranging from +2 to +7 but why is +1 oxidation state of Manganese(Mn) not stable?
The +1 oxidation state of Mn would have a configuration of 4s1 3d5.
What is that I am missing or wrong with?

You're underestimating things here https://chemistry.stackexchange.com/questions/118739/what-is-the-oxidation-state-of-mn-in-hmnco5 for example.

@Abhishek There is no need to affront other users, especially if they actually pointed you in the right direction. Note that the question wasn't closed single-handedly; rather, it was a peer-review. In your previous (currently deleted) question, there also was a heated discussion in the comments: you refused to fix factological issues and typos; you ignored clues by other users and demanded an answer. This is not a constructive behavior and is not in line with the guidelines for a respectable discussion.

$\ce{Mn+}$ has configuration $\mathrm{(4s)^0(3d)^6},$ while on the other hand $\ce{Mn^2+}$ $\mathrm{(3d)^5(4s)^0}$ and also, $\ce{Mn}$ in ground state is $\mathrm{(4s)^2(3d)^5}.$
Therefore you can see in the oxidation state +2 orbitals are half-filled, and in +1 they are not half-filled. As half-filled is more stable, therefore +2 and 0 are more stable than +1.
Therefore Mn prefers not to stay in +1 due to higher stability of adjacent states.

Why is the configuration of Mn+ 4s0 3d6 and not 4s1 3d5?

(-1) Writing that Mn(II) is "more stable" than Mn(I) is a very dangerous thing. Note that Fe(II)/(III) are isoelectronic to Mn(I)/(II) respectively. Why, then, is Fe(II) common, and is it "more stable" than Fe(III)? Why can you invoke the half-filled subshell thing to "explain" why Mn(I) is rare, but not why Fe(II) is common? A proper explanation needs to consider many energetic factors, e.g. ionisation energy, lattice energy, ... something similar to https://chemistry.stackexchange.com/questions/45153/crii-and-mniii-their-oxidizing-and-reducing-properties/45168#45168

I think you miss understood my question,I was asking when Mn+ can be in 4s1 3d5( a half filled stable state) , then why does it exist in 4s0 3d6?

Compounds with Mn in formal oxidation state +1 and -1 are well known.  They simply are not stable on air and in water.
In general, air-stable compounds of more active 3d-metals are ionic compounds. Their stability is a result of fine balance of energy of formation of relevant ions and energy of stabilization of said ions due to ion-ion and/or ion-dipole interactions. Naturally, $\ce{Mn^{2+}}$ interacts stronger than $\ce{Mn^{1+}}$, but isn't all that harder to make. However, by giving $\ce{Mn}$ a good partner that isn't hard dipole, but is still capable of interacting with the metal without oxidizing it, low oxidation states can be stabilized. Common partners of such type are $\ce{CO}$ and organic phosphines. If you are interested, go for introductory coordination chemistry book/chapter.


text:
Is there any difference between medical and cleaning ozone?
What are the main differences between medical ozone (auto-hemotherapy, …) and cleaning ozone (for example, the kitchen cleaning devices for vegetables and food). I noticed that they can produce a lower amount of ozone per hour while the medical ones have bigger amount of production.

Welcome to MedicalSciences.SE! Please take the [tour] and read the [help]. For reasons mentioned in this post and in [ask], we require prior research information when asking questions. See this list of helpful resources. Please help us to help you and [edit] your question to provide more information on what you have read on this subject, what made you ask this question, and any problems you are having understanding your research. If you found nothing, what did you Google?

I see as the only difference the ozone concentration and the gas mixture purpose. The ozone itself is the same, there is no medical nor cleaning ozone. Breathing ozone containing gas is harmful. It is a potent gas initiating the SO2 imission damage of spruce trees, and is the key component of the oxidative smog of the Los Angeles type, that can cause defoliation of trees.

“Medical ozone” is a mixture of 0.05 to 5 vol% of ozone produced from medically pure oxygen (usually via silent electrical discharge) and 99.95 to 95 vol% of oxygen.
Using air for medical-grade ozone production is strictly forbidden due to formation of hazardous nitrogen oxides $\ce{NO_x}.$
Note that ozone therapy is still considered an alternative medicine.
Ozone used for cleaning doesn't have such constrains and its concentration and production method vastly depends on application area and regulatory standards therein.
For example, for water treatment Russian federal standards prescribe the ozone concentration in the ozone-air mixture from 15 to 22 mg/l based on the required efficiency and level of pollution of the water source, and average contact time of 5 to 15 min.
Prolonged exposure or increased ozone concentration during water treatment may lead to unwanted effects such as problems with coagulation and extensive phenols formation.


text:
Is hydrogen bonding separate from permanent dipole interactions?
Is hydrogen bonding separate from permanent dipole interactions? What I mean here is that a molecule such as ethanol has hydrogen bonding, but does it also have permanent dipoles other than the ones on the oxygen and hydrogen atoms that cause permanent dipole attraction?

Hydrogen bonds have a small but existing covalent character, dipole interactions in general don't. https://memegen.link/custom/hydrogen_bomb~q/no_--_hydrogen_bond%21.jpg?alt=https://www.extremetech.com/wp-content/uploads/2013/09/Molecules.jpg

Consider the three hydroxyphenols catechol (ortho-hydroxyphenol), resorcinol (meta-), and hydroquinone (para-).  All three engage in hydrogen bonding with their hydroxyl groups.  But they differ in polarity.  Hydroquinone, with its hydroxyl groups symmetrically placed on opposite corners of the benzene ring, has a limited (average) dipole moment between two conformations, one polar and one nonpolar, whereas the other two isomers have an additional dipole from their hydroxyl groups being nonsymmettically located and have larger overall dipole moments.

You mean pyrocatechol.

@poutnik ? ? ? ?

ortho-hydroxyphenol. :-) But I see they are synonyms, and the first time I may missed you mentioned catechol.

Hydroquinone has a dipole moment.

How is a permanent moment rendered?  Will the vector change direction if bonds rotate?

For elaboration on Martin's comment, see: https://chemistry.stackexchange.com/questions/20100/non-zero-dipole-moment-of-hydroquinone


text:
Why I can't use a single Slater determinant to describe the ground state of open-shell atoms and molecules?
If we have a closed-shell atom or molecule (in the Born-Oppenheimer aproximation), and adopt the central field aproximation, we can write the ground state wave function of the system using a Slater determinant, constructed using the spin-orbitals that are the solutions of the equations
$$(h_{i} + V_{i}) \varphi_{i} = \varepsilon_{i} \varphi_{i}$$
with the lowests energies, where $h_{i}$ is the sum of the kinect energy and interaction with the nucleo(s) of the electrons, and $V$ is the potential. However, I read that if the system is open-shell, in general we can't write the ground state wave function using only a single Slater determinant, we have to take linear combinations of Slater determinants. Can someone explaim why this happpen, and indicate a reference? I would like to understand it.

This text goes over Slater determinants for the excited helium atom. Perhaps it helps? https://chem.libretexts.org/Bookshelves/Physical_and_Theoretical_Chemistry_Textbook_Maps/Map%3A_Physical_Chemistry_(McQuarrie_and_Simon)/08%3A_Multielectron_Atoms/8.06%3A_Antisymmetric_Wavefunctions_can_be_Represented_by_Slater_Determinants

You are looking for multi-reference cases.$^1$ Another term usually applied is "strong correlation". See What exactly is meant by 'multi-configurational' and 'multireference'?.
Instead of trying to answer myself on what those cases might be, let me quote Frank Jensen (Introduction to Computational Chemistry, 2nd ed., Wiley, 2007, p. 154):

Consider again the ozone molecule with the two resonance structures shown in Figure 4.9 [which shows a) a zwitterion with a positive charge on the central oxygen b) a triplet configuration with single electrons on terminal oxygens]. Each type of resonance structure essentially translates into a different determinant. If more than one non-equivalent resonance structure is important, this means that the wave function cannot be described even qualitatively correctly at the RHF single-determinant level (benzene, for example, has two equivalent cyclo hexatriene resonance structures, and is adequately described by an RHF wave function). A UHF wave function allows some biradical characters, with the disadvantages discussed [earlier in the text]. Alternatively, a second restricted type [Configurational State Function (CSF)] (consisting of two determinants) with two singly occupied MOs may be included in the wave function. The simplest [multi-configuration SCF] for ozone contains two configurations (often denoted TCSCF), with the optimum MOs and configurational weights determined by the variational principle. The CSFs entering an MCSCF expansion are pure spin states, and MCSCF wave functions therefore do not suffer from the problem of spin contamination.

In some cases, a single determinant may give you an incomplete picture. This is more often the case for triplet or open-shell species, but open-shell singlet cases can also demand a multi-determinant treatment.
Note that multi-reference treatments may also be necessary to obtain accurate quantitative results if the physical ground and excited states are close in energy.

$^1$ The "reference" is due to the use of the Hartree-Fock determinant as the reference state for configuration interaction (CI) post-SCF calculations. There are systems where only bad results can be achieved when using a single reference in a truncated CI. (In full CI, you have every possible reference included.)

I very much agree with your answer. I'd like to add that some open-shell cases can be correctly described by a single determinante (e.g. with ROHF). On the other hand, unrestricted methods use two SD already, but that is quite different from what you have described here.

@Martin-マーチン I often encounter this claim that "unrestricted methods (effectively) use two Slater determinants" but I haven't really found a rigorous proof of this. Do you have any good references?

@eimrek There is no rigorous proof, because rigorously speaking it's not true. However, it is almost true. See for example the Wikipedia Page on UHF, the Pople-Nesbet-Berthier eq: $$\mathbf{F}^\alpha\mathbf{C}^\alpha=\mathbf{S}\mathbf{C}^\alpha\mathbf{\epsilon}^\alpha,\quad\mathbf{F}^\beta\mathbf{C}^\beta=\mathbf{S}\mathbf{C}^\beta\mathbf{\epsilon}^\beta$$ Looks independent enough, but $\mathbf{F}^{\alpha,\beta}$ depend on their own solutions and on each other, so they are coupled. So that's more of a qualitative statement.


text:
Why there is a huge difference between the basicity of urea and guanidine?
The two molecules have very similar structures, but urea has a $\mathrm{p}K_\mathrm{b}$ of $13.9,$ while guanidine have a $\mathrm{p}K_\mathrm{b}$ of $0.4.$ If guanidine cation can be stabilized by resonance, then why can't urea cation do so?

Probably this can be the answer- all three canonical forms of guanidine are equivalent, while of urea aren't.

For the same reason that ammonia is a much stronger base than water, even though their structures are very similar in the same sense.

Urea and guanidine are structurally similar except that the carbonyl oxygen of urea has been replaced by an imine $\ce{NH}$ in guanidine. When it come to protonation, carbonyl oxygen of urea is always protonated over its 2 $\ce{NH2}$ groups (Ref.1 and 2). In guanidine, it is the imine $\ce{NH}$ protonated first. This preference in guanidine is not because a lone pair of $\mathrm{sp^2}$ nitrogen is more basic than that of $\mathrm{sp^3}$ nitrogen (actually it is other way around), but the lone pair of $\mathrm{sp^2}$ nitrogen is the most available pair for protonation over those of two $\mathrm{sp^2}$ nitrogens (lone pairs of two $\mathrm{sp^2}$ nitrogen atoms are contributing to resonance stabilization at the time). However, when protonated, guanidinium ion was highly stabilized by resonance as depicted in the diagram:

Thus, basically, it  comes down to protonation ability of oxygen versus nitrogen. Oxygen is more electronegative than that of nitrogen, thus like to retain its lone pare better than nitrogen (cf. basicity of water and ammonia). Therefore, in that sense alone, guanidine is a better base than urea.
References:

Nanping Wen, Murray H. Brooker, “Urea protonation: Raman and theoretical study,” J. Phys. Chem. 1993, 97(33), 8608-8616 (https://doi.org/10.1021/j100135a013).
B. Valentine, T. E. St. Amour, D. Fiat, “A $\ce{^{17}O}$ NMR study of the protonation of urea,” Organic Magnetic Resonance 1984, 22(11), 697-700 (https://doi.org/10.1002/mrc.1270221105).


text:
Why is the Si-F bond one of the strongest single covalent bonds of them all?
I honestly have no clue where to begin explaining why the Si-F bond is one of the strongest single covalent bonds of them all.
I would actually predict the bond to be relatively weak on account of the relatively large size of the Si atom leading to a relatively long bond length and hence poor orbital overlap. Indeed, Si's valence orbitals being large and diffuse further suggest to me that the overlap with F's small concentrated orbitals is poor.

I think you need a source to support that Si–F is indeed the strongest single covalent bond; I was under the impression that B–F was the strongest (although that number is probably unduly influenced by backbonding in BF3 & similar compounds). And http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html agrees with me.

@orthocresol I'm afraid that I have no better source than my lecturer. After looking at this website http://www.wiredchemist.com/chemistry/data/bond_energies_lengths.html, it would appear that the B-F bond is indeed stronger than the Si-F bond. I'm still curious to know why the Si-F bond is so strong.

Before you ask "why", ask "if".

https://chemistry.stackexchange.com/a/21833/9961 https://chemistry.stackexchange.com/questions/24078/how-does-hf-dissolve-glass

A long bond does not necessarily mean that it is weaker. A strong overlap between orbitals will also increase the same-spin (Pauli) repulsion, so more overlap does also not necessarily mean stronger bond. It is usually a delicate balance between many factors.


text:
Limiting fraction of association

Show that the limiting fraction of association of a base in water, as
  the concentration of base approaches 0, is $$\alpha=\frac{10^7 K_b}{(1+10^7\times K_\mathrm{b})}.$$

I have tried this problem for a long time and have made little progress. Here is what I have tried:
We know that:
\begin{align}
K_\mathrm{b} &= \frac{\ce{[HB+]}\ce{[OH-]}}{\ce{[B]}}\tag{definition}\\
\ce{[HB+]} + \ce{[H+]} &= \ce{[OH-]} \tag{charge balance}\\
\ce{[H+][OH-]} &= 10^{-14} \tag{$K_\mathrm{w}$ at $\pu{25^o C}$}\\
\text{initial concentration of base added} &= \ce{[B]} + \ce{[HB+]} \tag{mass balance}
\end{align}
I know that the limiting fraction of association is
$$\alpha = \frac{\ce{[HB+]}}{\text{initial concentration of base added}} = \frac{\ce{[OH-]}}{(\ce{[B]} + \ce{[HB+]})},$$
but I'm confused about how to proceed from here. I tried putting $\alpha$ in terms of $\ce{[OH-]}$ to get a cancellation, but that proved difficult. How do I proceed?

Please use \ce{} only for actual chemical content, please don't wrap whole mathematical statements into it. I have updated your post with chemistry markup. If you want to know more, please have a look here and here. We prefer to not use MathJax in the title field, see here for details.

Start from
$$K_\mathrm{b} = \frac{\ce{[HB+]}\ce{[OH-]}}{\ce{[B]}}$$
Define
$$\alpha = \frac{\ce{[HB+]}}{\ce{[B]} + \ce{[HB+]}} $$
or
$$\frac{\ce{[HB+]}}{\ce{[B]}}=\frac{\alpha}{1-\alpha}$$
so that
$$K_\mathrm{b} = \frac{\alpha\ce{[OH-]}}{1-\alpha}$$
or
$$\alpha = \frac{K_\mathrm{b}}{K_\mathrm{b}+\ce{[OH-]}}$$
But in the limit of infinite dilution the concentration of $\ce{OH^-}$ is solely determined by the autodissociation of water (if the solution is pure unbuffered water) and
$$\ce{[OH-]} = \sqrt{K_\mathrm{w}}$$
so that
$$\alpha = \frac{K_\mathrm{b}}{K_\mathrm{b}+K_\mathrm{w}^{1/2}}$$
But at RTP $K_\mathrm{w}^{1/2}\approx 10^{-7}$ resulting in the final expression
$$\alpha = \frac{10^{7}K_\mathrm{b}}{10^{7}K_\mathrm{b}+1}$$

@andselisk none really, both are (small) oversights resulting during editing of the text.


text:
MO theory used in organic chemistry textbook
I'm currently using Clayden's organic chemistry textbook, and when he shows MO diagrams of functional groups of organic molecules, he sometimes treated it as if only orbitals on two atoms are interacting. For example, the carbonyl chemistry is reduced to looking at the interaction between C and O orbitals. (For conjugated systems the MO diagrams he created make more sense to me)
However, as far as I know, MO theory is used to construct wavefunctions that spread over the whole molecule. Hence, how is it justified that we are combining orbitals centered on a pair of atom rather than all atoms in the molecule? What is that powerful assumption that allows us to apply MO theory locally at functional groups?

It's 1:30 am here and I don't have the desire to run calculations and generate images right now, but the simple answer is that although they are indeed a simplification, they tend to resemble the actual thing closely enough, so are "good enough" for organic chemists. At least, that's true for the carbonyl case. For other situations like C-H σ or σ* orbitals in ethane, it gets a bit messier because of symmetry properties and then it becomes necessary to talk about weird things like https://en.wikipedia.org/wiki/Localized_molecular_orbitals.

@orthocresol Thank you for your quick response and I would love to see your answer when you have time! By the way, can you point to some reference paper or textbook where this simplification is explicitly discussed and justified? That will certainly be the most helpful along with calculation results.

I don't know if it will have what you're looking for, but the go-to book for MO theory in organic chemistry is Ian Fleming's Molecular Orbitals and Organic Chemical Reactions.

Thanks, I will check it out when possible!


text:
If a reactant appears on both sides of an elementary equation, can it be cancelled?
Me and a friend were debating the following. For the elementary reaction
$$\ce{A + B -> A + C}$$
my friend says that it is unimolecular because $\ce{A}$ does not "participate" in the reaction. So, he says that the rate law is
$$\text{rate} = k[\ce{B}].$$
But, I think that that since it is an elementary reaction step, you cannot cancel $\ce{A}$ from both sides. I think that it is a bimolecular reaction with
$$\text{rate} = k[\ce{A}][\ce{B}].$$
Who is correct?

In a real experiment, if one starts with A(aq) + B(aq) and the reaction forms more water, one may, or may not, be actually able to remove the starting water. The issue is does A + B create water, moving the reaction forward, upon heating the dry salts. If so, canceling out the water may be appropriate, otherwise, not.

This is either not an elementary reaction step, or "A" should be excluded from the equation.

@Karl It can be an elementary step. For example in reaction of chiral bromoalkane with bromide.

If the elementary step is indeed written
$$\ce{A + B -> A + C}$$
and assuming that you haven't just written in $\ce{A}$ for fun, i.e. $\ce{A}$ is actually a participant in the step, then yes, this would be considered bimolecular. This sort of step, featuring the same chemical species on both sides of the reaction, does pop up in (for example) the Lindemann mechanism.
With regard to kinetics, however, it is worth noting that steps of this kind can sometimes indicate a reaction that is catalytic with respect to $\ce{A}$. If that is the case, then the concentration of $\ce{A}$ is essentially constant and you would obtain pseudo-first-order kinetics.

You wrote that this "can sometimes [emphasis mine] indicate a reaction that is catalytic with respect to A".  But if A is, as depicted above, both a participant in, and yet unchanged by, the reaction, then wouldn't it always be considered a catalyst?  I.e., are there any reactions in which A is both a participant and unchanged, yet for which it would not be considered a catalyst?

@theorist I use that wording on purpose because I don't want to rule out the possibility that A is further consumed in a separate step of the mechanism. I can't come up with an example right now, but there may well be something out there I don't know.

In that case it seems it would at least be catalytic with respect to that particular elementary reaction.  The only exception I can think of is if A is added to slow down the reaction, in which case I suppose one might say that it must be either catalytic or anticatalytic.  Though until just now I've never thought about anticatalysts in elementary reactions

@theorist You guys are discussing on the basis of a wrong premise in the original question. Catalytic reactions can never be single step! This is not an elementar reaction, if A not in there just for fun.

@orthocresol so suppose I have an overall reaction A + B -> D, with elementary steps (1) A + B -> A + C and (2) A + C -> D. If step 1 was rate determining, then my reaction would be unimolecular, correct?

@DrPepper I’m not sure why you say it would be unimolecular - I wrote bimolecular in my answer.

This is common in biochemistry where A is a protein (an enzyme). Enzymes that alter one substrate only, producing a single product, are called isomerases.
Michaelis-Menten kinetics applies. The reaction speed is proportional to the concentration of the enzyme:
$$v =  k [\ce E] \frac{[\ce S]}{K + [\ce S]} $$
here [E] is the enzyme concentration (would be A) and [S] is the substrate concentration (would be B). k and K are the reaction-specific constants. Hence no, [E] cannot be "abstracted out". The reaction speed would be close to zero regardless of the amount of the substrate.
This obviously assumes that A and B form the complex first. If A is an unrelated substance at the opposite edge of the Universe than B now converting into C, it probably should not be considered as part that conversion.

I didn't use M-M kinetics as an example because the question is asking about an elementary step, not an overall reaction. Even though the overall reaction $$\ce{E + S -> E + P}$$ resembles the original form given in the question, the individual steps $$\ce{E + S <-> ES <-> E + P}$$ aren't quite the same as what OP is asking about, I think.


text:
Why dried, closed coconuts go through carbonation?
Some dried, closed coconuts go through carbonation:

The white meat of the coconut becomes disintegrated and smeary
The liquid becomes carbonated with a bit sour-salty tasted coconut water (which reminds the taste of a water and baking soda solution) and a grey to blue color
If the grey-blue liquid given enough time (inside a dried, closed coconut) it could become yellowish and possibly contaminated (I have no idea if the yellow color is due to further carbonation or bacteria)

Given that a dried, closed coconut is a dry and sealed by hard, fibrous shell,
Why dried, closed coconuts go through carbonation?

Hello to mods; if you think this is a better fit for biology SE, please don't hesitate to migrate. I assumed here it will be better due to biochemical process.

I can recall the bad taste of very old coconut water. It does get fermented. Sugars are converted to carbon dioxide and sugars to organic acid.

Hint 1: what causes champagne to become carbonated? Hint 2: What is the etymology of the English word enzyme?

@DrMoishePippik: Are you 100% sure that yeast causes carbonation of coconut meat and liquid? When I search yeast and coconut, all I find is articles saying that coconut oil can be used as a remedy for yeast infections. There seems to be evidence for destroying yeast, but no evidence for containing yeast. Do you have a source?

@user1271772  actually yeast can ferment coconut oil see the link in my comment

@JohnDoea see this link : https://www.sciencedirect.com/science/article/pii/0740002086900092

why does it say modified by organicgem and i dont see any modifications?


text:
When Cp = Cv + R
So in a physics test question, there was this question when is $C_p$ not $C_V + R$?
And I wrote: when number of molecules change. Because the derivation of above formula comes from $nC_p\,\mathrm dT - nR\,\mathrm dT = nC_V\,\mathrm dT$, depending on the variable conditions, (for example under constant pressure, volume and temperature might change, and under constant volume, pressure and temperature might change), there might be some chemical reactions that change the $n$ value. So my question is, is there a case where chemicals do NOT react (that is divide or combine) depending where volume is constant or pressure is constant but do when it is not?

Another, trivial answer would be, when the system is not just a gas. Or, imho, when it is not considered as an ideal gas.

Actually That was the correct answer. But then, woould my answer be considered as wrong?

When a system of ideal gases reacts, Cp and Cv are changing, but not their difference.

If chemicals react at constant pressure, they also react at constant volume. The yield will be different. But is is not possible to think that a reaction may happen only at constant pressure and not at constant temperature.

I think.....while deriving this equation we use ideal gas equation PV=nRT which is valid only for ideal gas and not for real gas . So this equation Cp is not equal to Cv+R for real gas.

In an ideal gas mixture, the enthalpy and internal energy of the mixture is equal to the sum of the enthalpies and internal energies of the pure gases at the same temperature and pressure as the mixture.  So this has nothing to do with whether reaction takes place.  As others have indicated, the relationship only applies to ideal gases.


text:
Structure of crystalline boron trioxide
I recently came across the statement

The structure of crystalline $\ce{B2O3}$ consists of $\ce{BO4}$ tetrahedra, two sets of which form two types of interconnected spiral chains, three $\ce{B-O}$ bonds are equivalent, but the fourth one is somewhat longer.

Please explain the meaning of this statement, above is the image of crystalline $\ce{B2O3}$ from Wikipedia.

This statement is relevant for the initially [incorrectly] determined structure of $\ce{B2O3}$ by Berger [1, 2] who used powder x-ray diffraction data.
You can still access the structure and visualize it in 3D (ICSD 24047), but it was shown later by several researchers and summarized in paper by Gurr et al. [3] that this structure has been erroneously determined:

After failing to grow a useful single crystal by this process, Berger (1952, 1953) was forced to use data from powder photographs in an attempt to determine the structure. The basic units of Berger's structure were two different irregular tetrahedra of oxygen atoms each surrounding a boron atom. This structure has been considered unsatisfactory in several respects (e.g. Wells, 1962), particularly as it results in two different $\ce{BO4}$ tetrahedra with six different $\ce{B-O}$ bond lengths ranging from $1.31$ to $\pu{2.14 Å}.$ In this laboratory a least-squares refinement (Hamilton, 1961) on Berger's powder data was attempted from his positional parameters. The attempt was abandoned when the refinement diverged. The tetrahedrally coordinated structure proposed by Berger was also puzzling because in various borate systems there is a regular variation of the fraction of boron atoms in fourfold coordination as a function of increasing alkali content (both in glasses and crystals); yet in each instance there was an obvious trend towards only threefold coordinated boron as the composition approached $\ce{B2O3}.$
[…]
The n.m.r, absorption for $\ce{BO4}$ groups is sharper and stronger than that for $\ce{BO3}$ groups (Bray et al., 1961; Silver & Bray, 1958) and thus the absence of sharp components in the spectrum of $\ce{B2O3}$ I precludes the presence of tetrahedral $\ce{BO4}$ groups in this material.

A new crystal structure was proposed: same space group $P3_1,$ but lacking tetrahedral units (ICSD 16021).
This is the one that is shown on Wikipedia, hence the dissonance.

Figure 1. Comparison between the unit cell content of two $\ce{B2O3}$ structures (boron atoms and boron coordination polyhedra are depicted in yellow color). Left: crystal structure according to Berger (1952). Right: crystal structure according to Gurr et al. (1970).
References

Berger, S. V. The Crystal Structure of $\ce{B2O3}.$ Acta Cryst 1952, 5 (3), 389–389. DOI: 10.1107/S0365110X52001155.
Berger, S. V.; Boeri, E.; Bonnichsen, R. K.; Sörensen, J. S.; Sörensen, N. A. The Crystal Structure of Boron Oxide. Acta Chem. Scand. 1953, 7, 611–622. DOI: 10.3891/acta.chem.scand.07-0611.
Gurr, G. E.; Montgomery, P. W.; Knutson, C. D.; Gorres, B. T. The Crystal Structure of Trigonal Diboron Trioxide. Acta Crystallogr B Struct Crystallogr Cryst Chem 1970, 26 (7), 906–915. DOI: 10.1107/S0567740870003369.


text:
How to increase volatility?
Is it possible to put a substance with low volatility into a highly volatile solvent, thus make the original substance also vaporize quickly?
If not, are there anyway to increase it's volatility?

https://chemistry.stackexchange.com/questions/9130/steam-distillation-how-it-actually-works

The important thing is the solvent does not dissolve the substance. Then their vapour pressures sum up and they boil together.
This technique is massively used for obtaining aromatic oily essences from plants, that are mostly insoluble in water.
The destilate usually forms a stable emulsion, that gets extracted by lipophilic volatile solvent, that is later evaporated to got the substance, or a mixture of them.

Actually any positive azeotrope (with a Bp. minimum) works. E.g. adding ethanol makes it easier to evaporate water.

@Karl Hmm, but azeotropes are hard to find and even them decrease the vapour pressure of the less volatile liquid.

Basically any two substances with different polarity form an azeotrope. What you propose is also one. https://en.wikipedia.org/wiki/Azeotrope#Homogeneous_and_heterogeneous_azeotropes

Methanol does not form an azeotrope with water. For existence of an azeotrope, the deviation from Raoult law is not enough. It must be high enough deviation. Neither nonmiscible liquids form azeotropes. Each maintains its vapour pressure like the other was not present.


text:
Why is phenol more acidic than an alcohol but less reactive as an alcohol?
I understand why phenol is more acidic than ethanol but is less reactive?
Eg: why unlike ethanol it can't react with a carboxylic acid?

It can and does react with carboxylic acids to form phenol esters, but as the phenoxide ion is stable and easily formed it is a good leaving group. Phenol  esters are activated, hydrolyse readily and are sometimes used as acylating agents, example https://en.wikipedia.org/wiki/Pentafluorophenyl_esters

There are no more reactive or less reactive things. Also, somewhat in contradiction with the previous, I feel the need to say that phenol is in fact more reactive.

Quite, phenol is more reactive as a nucleophile than alcohols

I think it phenol is a poorer nucleophile than an alcohol because the unshared electron pair is delocalized into the benzene ring. These reactions are a delicate balance of base strength, nucleophilicity and leaving group. I have wondered if a simple ester could be transesterified with a an excess of phenoxide ion or if phenoxide is just too good a leaving group.


text:
Acidic nature of boric acid when its concentration high
Following statement was given for boric acid $(\ce{H3BO3}):$

At low concentrations $(\leq\pu{0.02 M})$ essentially $\ce{B(OH)3}$ and $\ce{B(OH)4-}$ are present, but at higher concentration the acidity increases and pH studies are consistent with the formation of polymeric species such as $$\ce{B(OH)3 <=> H+ + [B3O3(OH)4]- + H2O}$$

Please explain the meaning of the italicised statement i.e. pH studies are consistent with the formation of polymeric species?

Could you please narrow down as to what exactly needs clarification? The following questions could arise: How the pH studies are conducted?; Why polyoxoborates form and what their structure is?; What does Pourbaix diagram say regarding olation/oxolation?; What are corresponding $\mathrm{p}K_\mathrm{a}$ values? It would be nice if you could point out which one interests you in the first place, otherwise the answer can and should be very lengthy.

@andselisk what is the meaning of the statement pH studies are consistent with the formation of polymeric species?

I suppose predictions of polyanion hypothesis about H+ concentration agrees with observation. Also note borax existence as the solid evidence of boron polyanions

In the equation $\ce{H2O + H3BO3 -> H+ + B(OH)4-}$, one negative charge on the anion is shared among 4 electronegative oxygens. Yes, one positive charge on the proton is also shared with some water molecules, so the orthoborate ion probably shares some negative charge with some waters, but let's keep it simple.
Boric acid is known to complex with sugars to generate higher acidity. If boric acid complexes with itself, as in $$\ce{3 H3BO3 -> H+ + [B3O3(OH)4]- + 2 H2O}$$, the one negative charge on the anion can be considered to be distributed among 7 electronegative oxygens, or maybe still distributed mostly on the OH groups, but since they will be separated more, the stability of the ion will be greater, and the tendency to lose a proton will be greater: greater acidity. The electronegativity of boron is less than that of hydrogen (2.02 vs 2.2, electronegativity, Wikipedia), so when the atoms in the ions are counted, the first ionization equation has one hydrogen and one boron as the metallic atoms getting the positive charge, while the second equation has one hydrogen and 3 borons donating electrons - more sharing of the positive charge here, tho the hydrogen still gets the positive charge.
But this structure may be only a suggestion. Well-known polyborate anions include the triborate(1-), tetraborate(2-) and pentaborate(1-) anions. The condensation reaction for the formation of tetraborate(2-) is as follows Wikipedia
$$\ce{2 B(OH)3 + 2 [B(OH)4]− ⇌ [B4O5(OH)4]2- + 5 H2O}$$
The italicized text mentions an observation of pH that is consistent with the complexation of boric acid with itself. It doesn't prove the existence of the triborate ion only, and perhaps is only a way to keep the text shorter without ignoring complexation completely. Then along comes a student who looks into the sentence and sees that something was omitted or not explained thoroughly. (Keep asking questions!)

A more precise answer to the polymeric structure, at least at neutral pH, is found in Wikipedia on borate, to quote in part:
"At neutral pH boric acid undergoes condensation reactions to form polymeric oxyanions....
The tetraborate anion (tetramer) includes two tetrahedral and two trigonal boron atoms symmetrically assembled in a fused bicyclic structure. The two tetrahedral boron atoms are linked together by a common oxygen atom and each also bears a negative net charge brought by the supplementary OH− groups laterally attached to them. This intricate molecular anion also exhibits three rings: two fused distorted hexagonal (boroxole) rings and one distorted octagonal ring. Each ring is made of a succession of alternate boron and oxygen atoms. Boroxole rings are a very common structural motif in polyborate ions."
Also:
"Simple bonding theories point to the trigonal planar structure. In terms of valence bond theory the bonds are formed by using sp2 hybrid orbitals on boron. Some compounds termed orthoborates do not necessarily contain the trigonal planar ion, for example gadolinium orthoborate..."
My reading of the above is that it is possible, via a condensation reaction, for Boric acid to create polymeric oxyanions, and in particular Boroxole rings at neutral pH.


text:
What happens when you run house electricity through salt water?
What happens when you run house electricity (120VAC) through salt water? (with standard copper wires!)
Well, in https://youtu.be/dcrY59nGxBg , an interesting experiment was conducted to find out!
At the end of the video, however, I noticed the salt water became rather dark and murky, and I'm wondering what happened inside.
Did the copper wires just corrode into some  kind of precipitate?
$$\ce{Cu(s) + H2O(\ell) -> CuO(s) + H2(g)}$$
$$\ce{Cu(s) + 2H2O(\ell) -> Cu(OH)2(s) + H2(g)}$$
Was there chlorine gas even created too?
$$\ce{2H+(aq) + 2Cl^{-}(aq) -> H2(g) + Cl2(g)}$$

Much of $\ce{Cu^2+}$ will initially go to the solution and only gradually form precipitate. Other than that, everything seems right.

Does this answer your question? Electrolysis puzzle

https://chemistry.stackexchange.com/questions/49578/copper-chloride-color-in-water-electrolysis

@Mithoron Not quite, it seems those experiments were conducted under much milder/less extreme conditions, thus allowing other more gradual coloration effects to be demonstrated. Ivan Neretin's answer is spot on, pointing to the eventual result of running this setup to completion and likely accelerated by the large 120V supply.

On a side note, would it be a bad idea to just pour the final solution down the drain, as copper ions may be an environmental hazard? (The video did not demonstrate how it was disposed)


text:
NMR in pyridine no meta splitting
Consider the NMR of Brominated Pyridine shown below:

When I looked at the NMR data, I was confused because I thought that H2 would also be split in the meta position by H6 but the data says that this is not present in the overall NMR Spectra, although from previous NMR lessons I thought that this would be observed.
Why does H2 not couple to H6 in the NMR shown above?

Can you speculate why that coupling constant is quite different from the 4J coupling in the other direction?

@Karl i'm not too sure what you mean. can you explain your question because I'm unsure

Of course there is a 4J coupling from H2 to H6, but it seems to be much smaller than the 2.5 Hz from H2 to H4. Agreed?

@Karl yes. I was wondering though if that was due to the presence of the Heteroatom present (N) on the pyridine ring?

Yes, of course. Hint: J coupling works through electrons. ;) N is more .... than C.

@Karl ok so because the electrons in the bonds around the heteroatom N in the ring experience a dipole moment towards the N, this means that the J coupling would be low because the electrons are being pulled towards the N atom and away from the H2 and the H6 respectively? I think that's the reasoning I needed, unless I am mistaken?

This article would be helpful: J. Chem. Phys. 1967, 46, 327; https://doi.org/10.1063/1.1840389.


text:
Does a bulky nuclephile react faster by Michael or direct addition?
Between Michael or direct addition on a $\alpha,\beta$-unsaturated ketone, I think that Michael addition is faster than the direct addition as the nucleophile can attack orthrogonally to the $\ce{C=C}$ which is a less hindered approach than the Bürgi-Dunitz angle required to attack the $\ce{C=O}$.

First, you got wrong thinking Michael addition is faster than the direct addition. Actually, Michael addition (thermodinamic control) is comparatively slower than the direct addition (kinetic control) here.
Second, to my understanding, Bürgi-Dunitz angle (Ref.1) has nothing to do here (may be it affect both additions, you never knows). However, there are several factors control the outcome.  The way that nucleophiles react depends on three factors:

The conditions of the reaction;
The nature of the $\alpha,\beta$-unsaturated carbonyl compound; and
The type of the nucleophile.

An example for the conditions of the reaction: The reaction of but-3-en-2-one and $\ce{KCN}$  in the presence of an acid catalyst gives 1,1-cyanohydrin (2-cyanobut-3-en-2-ol) at $\pu{10-20 ^\circ C}$ by 1,2-addition (kinetic product), but would give 4-cyanobuta-2-one at $\pu{80 ^\circ C}$ by 1,4-addition (thermodynamic product). Thus, it is clear that the Direct addition to the carbonyl group (which forms at low temperature) is faster than conjugate addition (forms at higher temperature), giving cyanohydrin. However, that reaction is reversible and, as a result, converted to more stable cyanobutanone at high temperatute, which is essentially an irreversible reaction.
Why is direct addition (1,2-addition) faster than conjugate addition (1,4-addition)?: Simple molecular orbital calculation shows both, the carbon atom of the $\ce{C=O}$ group and the carbon atom $\beta$ to it, carry some positive charges. However, the carbon atom of the $\ce{C=O}$ group carries comparatively more charge, so electrostatic attraction for the charged nucleophiles will encourage it to attack the carbonyl group directly rather than undergo conjugate addition.
On the other hand, with direct 1,2-addition, we gain a new $\ce{C–C}$ $\sigma$-bond, but lose the $\ce{C=O}$ $\pi$-bond and keep the $\ce{C=C}$ $\pi$-bond. Meantime, with conjugate 1,4-addition, we still gain a new $\ce{C–C}$ $\sigma$-bond, but lose the $\ce{C=C}$ $\pi$-bond instead of $\ce{C=O}$ $\pi$-bond, and keep the $\ce{C=O}$ $\pi$-bond instead of $\ce{C=C}$ $\pi$-bond. Overall, the $\ce{C=O}$ $\pi$-bonds ($\pu{369 kJmol-1}$) are stronger than $\ce{C=C}$ $\pi$-bonds ($\pu{280 kJmol-1}$), hence the conjugate addition gives comparatively more stable product.
An example for the nature of the $\alpha,\beta$-unsaturated carbonyl compound: The reaction of 2-butenal (an $\alpha,\beta$-unsaturated aldehyde) and butyllithium in THF at $\pu{-70 ^\circ C}$ gives oct-2-en-4-ol by 1,2-addition exclusively. Yet, under the identical conditions, N,N-dimethyl-2-butenamide (an $\alpha,\beta$-unsaturated amide) would give N,N-dimethyl-2-butylbutanamide by 1,4-addition exclusively.
An example for the type of the nucleophile (Hard or Soft): The reaction of 4-methylpent-3-en-2-one (an $\alpha,\beta$-unsaturated ketone) and thiophenol at $\pu{25 ^\circ C}$ gives 4-methyl-4-thiophenoxypent-2-one exclusively by 1,4-addition (Remarkably, no acid or base catalyst is needed for this reaction,which would be needed for the alcohol additions, instead).

Here, it has been explained that the attraction between nucleophiles and electrophiles is governed by two related interactions: (1) Electrostatic attraction between positive and negative charges, and (2) Orbital overlap between the HOMO of the nucleophile and the LUMO of the electrophile. Successful reactions usually result from a combination of both, but sometimes reactivity can be dominated by one or the other. The dominant factor, either it be electrostatic or orbital control, depends on the nucleophile and electrophile involved. Nucleophiles containing small, electonegative atoms (such as $\ce{O}$ or $\ce{Cl}$) tend to react under predominantly electrostatic control, while nuclophiles containing larger atoms (including the sulfur of thiols) are predominantly subject to control by orbital overlap. The terms hard and soft have been coined to describe these two types of reagents. Hard nucleophiles are typically from the early rows of the periodic table and have higher charge density, while soft nucleophiles are from the later rows of the periodic table—they are either uncharged or have larger atoms with higher-energy, more diffuse orbitals (Ref.2).

Few examples:
$$
\begin{array}{c|c}
\hline
\text{Hard nucleophiles} & \text{Borderline nucleophiles} & \text{Soft nucleophiles} \\
\hline
 \ce{F-, \mathbf{HO^-, RO^-}, SO4^2-, Cl-} & \ce{N3-, \mathbf{CN^-}} & \ce{\mathbf{I^-, RS^-}, RSe-, S^2-}\\
\ce{\mathbf{H_2O, ROH}, ROR', RCOR'} & \ce{\mathbf{RNH_2, RR'NH}} & \ce{\mathbf{RSH, RSR'}, R3P}\\
\ce{\mathbf{NH_3, RMgBr, RLi}} & \ce{\mathbf{Br^-}} & \text{Alkenes, aromatic rings}\\
\hline
\end{array}
$$
Electrophiles can also classified in this manner. For example, $\ce{H+}$ is a very hard electrophile because its small size and charged. Bromine $\ce{Br2}$ is, on the other hand, a soft electrophile due to its large size and having no charge.
In general, hard nucleophiles prefer to react with hard electrophiles, and soft nucleophiles with soft electrophiles.
Overall, Hard/Soft reactivity:

Reactions of hard species are dominated by charges and electrostatic effects
Reactions of soft species are dominated by orbital effects
Hard nucleophiles tend to react well with hard electrophiles
Soft nucleophiles tend to react well with soft electrophiles

References:

H. Bürgi, J. Dunitz, J. Lehn, G. Wipff, "Stereochemistry of reaction paths at carbonyl centres", Tetrahedron 1974, 30(12), 1563-1572 (http://dx.doi.org/10.1016/S0040-4020(01)90678-7).
Jonathan Clayden, Nick Greeves, Stuart Warren, Organic Chemistry; 2nd Edn; Oxford University Press: Oxford, United Kingdom, 2012.


text:
Problem with synthesis of calcium formate
I just tried preparing calcium formate by adding $\ce{Ca(OH)2}$ to ammonium formate, but I observed no bubble formation and no smell of ammonia. Why doesn't calcium hydroxide, being a strong base, displace ammonia?

There is a multitude of possibilities. For one thing, Ca(OH)2 isn't that strong a base. For another, maybe it does displace ammonia, but the solution is too dilute to detect the smell, let alone the bubbles. NH3 never forms bubbles.

Did you check the solubilities of calcium formate,16.1 g/100 g (0 °C), and calcium hydroxide, 0.173 g/ 100 ml (20 °C)?

You probably did get your anticipated reaction, but unlike carbon dioxide or hydrogen ammonia is highly soluble in water.  Unless your reactants are more highly concentrated than is possible with the limited solubility of calcium hydroxide, the ammonia remains in solution instead of giving off bubbles or a smell.
There are various ways to render this reaction visible:

Heat the solution, driving out any dissolved ammonia.  If you generate a smell of ammonia, or better if the solution emits a gas that turns moist red litmus paper blue, you found the ammonia. You should compare with a blank where you just have the unreacted ammonium formate.

If you used solid calcium hydroxide, double-check how much you
dissolved.  Calcium hydroxide has only a solubility of 1-2 grams
per liter, so if
you dissolved more than that it must have reacted.  If you were
using magnesium hydroxide the solubility difference is even more
evident, for magnesium hydroxide is also strongly basic enough to
displace ammonia but its solubility without the reaction is very
low.

Check the pH.  Ammonia is a weaker base than the calcium hydroxide that displaces it, so you get a lower pH than you would have gotten just adding the same amount of calcium hydroxide to plain water (which in this case you would do for comparison).

Chemistry is an experimental science.  Therefore experimental design is essential for chemistry.


text:
Is there a standardized way to assign a label to the symbol of a chemical element (in a published work)?
When referring to a particular atom, both in text and when using a graphical representation, several conventions are being applied.
In the field of crystallography an arabic numeral either follows the element symbol without any space, e.g. $\mathrm{C}42$ for carbon atom number $42$ (usually assigned during solve/refine routine by SHELX), or is being placed in braces, e.g. $\mathrm{C}(42).$
Sometimes hydrogens attached to the same carbon atom inherit its number and receive an additional letter, e.g. $\mathrm{H42c}$ for the third hydrogen in  $\mathrm{C42}$ methyl group.
In NMR spectroscopy an element is normally labeled with a script-sized symbol (arabic number or lowercase letter) in the bottom right corner, e.g. $\ce{H_3}$ or $\ce{H_c}.$
Mestrelab Mnova places numerical labels underneath the element symbol, e.g. $\underset{3}{\ce{H}}$, or directly in place of the element if the symbol for the latter is hidden (carbon, in most cases): ${\scriptsize 3}.$
These labeling schemes used in both instrumental methods (which I cherry-picked solely for the illustrative purposes) seem to work OK in a sense that they are compatible with expanded AZE-notation required to depict any element and its attributes, such as chemical symbol $\ce{El},$ mass number $A,$ atomic number $Z,$ oxidation number O.N., charge $q$ and number of atoms of given kind $x$:
$$\Large\ce{\overset{\text{O.N.}}{^{$A$}_{$Z$}El^{$q$}_x}}$$
and won't cause any confusion (except for $\ce{H_3},$ maybe).
However, it would be nice to know whether there is a standardized way to attribute a label to an element that can be uniformly displayed and recognized (main text, titles, structures, illustrations etc.).
Does such robust labeling scheme exist?
P.S. If you are a software developer and read this, also feel free to comment or add an answer as to which labeling scheme you've came up with for calling out a particular atom.

IUPAC's green book "Quantities Units and Symbols in Physical Chemistry" might be worth a look. It is a 165-paged book.

@M.Farooq Already did. According to the Green Book, labels (in general) must be typed in upright typeface, that's all I found.

I can imagine that this can become a headache for a reader if the numbering scheme is not obvious, but that does not appear to be a consideration in this question. To draw a comparison, protein sequences are numbered N->C ie Ala42 is the C-terminal residue in the A$\beta$(1-42) peptide. This rarely is a problem although numbering conventions may refer to the original gene rather than a specific construct such as the A$\beta$ in question. But since the sequences are linear the numbering convention is otherwise clear. Now consider branching and rings in compounds and the story changes.

In any case your question is about convention, not about the best way to do this for the sake of clarity. I would say stick to the  crystallography convention but if the molecule is complex provide a structure file or figure with numbering.

@BuckThorn Yeah, I have the same feeling that crystallographic notation would be easier to maintain. And I actually do care about reader's ergonomics, but I'd put standards/recommendations/guidelines first, if there are any, to avoid confusion and develop a coherent labeling scheme. Ideally, the one that is interdisciplinary understood:)

Well, if IUPAC has not dictated a final opinion on the subject (or doesn't seem to care enough to put clear guidelines where people can readily find them) then maybe clarity should be your guideline. I don't know who beside IUPAC can be expected to provide official guidance (or should be payed close attention to) on the matter. If your molecule is complex, then aggregating opinion from others might be a valuable alternative.


text:
How to understand the (non-)degeneracy of states of singlet Σ and Δ terms for a ππ configuration of two nonequivalent electrons?
As instructed in Levine's Quantum Chemistry 7th edition, Chapter 13.8, page 377, for a $\pi \pi$ configuration of two nonequivalent electrons, four functions $\pi_{+1}(1) \pi_{-1}^{\prime}(2)$, $\pi_{+1}(2) \pi_{-1}^{\prime}(1)$, $\pi_{-1}(1) \pi^{\prime}_{+1}(2)$, $\pi_{-1}(2) \pi^{\prime}_{+1}(1)$
 are used to construct the wave functions of the singlet $\Sigma$ terms
$$
\begin{array}{ll}{^{1} \Sigma^{+}:} & {\pi_{+1}(1) \pi_{-1}^{\prime}(2)+\pi_{+1}(2) \pi_{-1}^{\prime}(1)+\pi_{-1}(1) \pi_{+1}^{\prime}(2)+\pi_{-1}(2) \pi_{+1}^{\prime}(1)} \\ {^{1} \Sigma^{-}:} & {\pi_{+1}(1) \pi_{-1}^{\prime}(2)+\pi_{+1}(2) \pi_{-1}^{\prime}(1)-\pi_{-1}(1) \pi_{+1}^{\prime}(2)-\pi_{-1}(2) \pi_{+1}^{\prime}(1)}\end{array}
$$
The text following these equations says "Clearly, these two spatial functions have different energies." Could you suggest how to understand this point ?
The discussion further below suggests that when four functions $\pi_{+1}(1) \pi_{+1}^{\prime}(2)$, $\pi_{+1}(2) \pi_{+1}^{\prime}(1)$, $\pi_{-1}(1) \pi^{\prime}_{-1}(2)$, $\pi_{-1}(2) \pi^{\prime}_{-1}(1)$
 are used to construct the ${}^1 \Delta^{+}$ and ${}^1 \Delta^{-}$ wave functions of the singlet $\Delta$ terms, "Since they have the same energy, there is no
point in using the $+$ and $-$ superscripts". Could you suggest how to understand this point and the reason of the difference from the above point ?
Many thanks for your suggestions !

Brief note: we prefer to not use MathJax in the title field due to issues it gives rise to; see here for details.

I believe the sigma singlet terms describe states with electron density on different atoms in the diatomic molecule, so if they are different elements, the energies are different. In the delta singlet states, the difference between them is only the orientation relative to the x and y axes, so the energies are the same regardless of whether you have hetero or homodiatomic. I need to draw it out to be sure.

The setup here is a diatomic molecule with two $\pi$ electrons in different subshells. A typical scenario would be one in a $\pi$ orbital and the other in a higher energy $\pi^*$ orbital.
In a diatomic molecule aligned with its molecular axis on the z-axis, the $\pi$ subshell consists of two orbitals, one with a node in the x-z plane and one with a node in the y-z plane. The $\pi^*$ subshell likewise consists of two orbitals, which are the antibonding counterparts to the two $\pi$ orbitals.
In the cited text, "+1" and "-1" indicate the two orientations, so it might be simpler to replace them with x and y. Likewise, the "$\pi'$" can be changed to $\pi^*$ for clarity.
Now consider the first $^1\Sigma$ term:
$$^1\Sigma^+: \pi_{+1}(1)\pi'_{-1}(2) + \pi_{+1}(2)\pi'_{-1}(1)+\pi_{-1}(1)\pi'_{+1}(2)+\pi_{-1}(2)\pi'_{+1}(1)$$
We can rewrite this in the friendlier notation and use only every other term, since the others are just exchange within the same orbitals. (What follows here is conceptual, not mathematically rigorous.):
$$^1\Sigma^+: \pi_x\pi^*_y + \pi_y\pi^*_x$$
The second $^1\Sigma$ term likewise simplifies to:
$$^1\Sigma^-: \pi_x\pi^*_y - \pi_y\pi^*_x$$
Now we further separate the contributions from each axis:
$$^1\Sigma^+: \pi_x+ \pi^*_x+\pi_y + \pi^*_y $$
$$^1\Sigma^-: \pi_x- \pi^*_x+\pi^*_y - \pi_y $$
We see that what we have done is to "undo" the formation of the $\pi$ and $\pi^*$ orbitals and essentially restored the original atomic $p$ orbitals:
$$^1\Sigma^+: \pi_x+ \pi^*_x+\pi_y + \pi^*_y=p_{Ax} + p_{Ay} $$
$$^1\Sigma^-: \pi_x- \pi^*_x+\pi^*_y - \pi_y=p_{Bx} + p_{By}$$
where A and B are the two atoms in the molecule. If A and B are different elements, these terms will have different energy.
If you do the same exercise with the $^1\Delta$ terms, you'll get something like:
$$^1\Delta^+: p_{Ax} + p_{By}$$
$$^1\Delta^-: p_{Bx} + p_{Ay}$$
Since both terms have equal contributions from A and B, they will have the same energies whether or not A and B are the same element. And since the orbitals on the two axes are orthogonal to each other, the change in sign does not affect the energy.
Conceptually, the sigma terms are diradicals where either both electrons are on A or both electrons are on B in orthogonal p orbitals. The delta terms are diradicals in which one electron is on each atom.


text:
Why does lithium aluminium hydride reduce a ketone over enolising the ketone?
By lithium aluminium hydride enolising the ketone, hydrogen gas is produced.
This is very entropically favourable, so why does this not happen and rather we get addition of the hydride ion to the carbonyl?

LAH is better as reductant then base. Precise answer needs mechanism of reduction.

If the things were solved by thermodynamics alone, then most part of organic chemistry would not exist.

First you get Aluminium binding to the ketone - this is very important


text:
Sum of spins of electrons in a wavefunction
I wanted to ask a question about spin in an electron.
We were presented earlier today with the following information:

Using my synoptic knowledge, I recall that spin up $\alpha$ had a value of $+\frac{1}{2}$ and spin down $\beta$ had a value of $-\frac{1}{2}$.
Also, using the information above, $\alpha(1)\alpha(2)$ would have a total Spin quantum number $S$ of $\frac{1}{2} + \frac{1}{2} = 1$, $\beta(1)\beta(2)$ however I thought would result to being $S$ of $-\frac{1}{2} - \frac{1}{2} = -1$ and even so, one spin up and one spin down $\alpha(1)\beta(2) +  \alpha(2)\beta(1)$ would total $\frac{1}{2} - \frac{1}{2} + \frac{1}{2} - \frac{1}{2}= 0$. These calculations contradict the data given above for the triplets and singlets states.
I'm confused. How are the values of $S$ for $\beta(1)\beta(2)$ and $\alpha(1)\beta(2) +  \alpha(2)\beta(1)$ exactly derived?

There are some pictures of singlet and triplets  in this answer that might help https://chemistry.stackexchange.com/questions/50277/how-does-spin-flipping-of-triplet-carbenes-occur/54259#54259

related: https://chemistry.stackexchange.com/questions/111760/what-is-the-difference-between-a-configuration-state-function-and-a-slater-deter/111947#111947

I think the reason you're getting confused is because when we talk about "spin", we actually are referring to two separate observable values, the z-component of the spin (properly designated $\hat{S}_z$) and the square of the magnitude of the three-component spin vector (properly designated as $\hat{S}^2$).
The commonly used values of $+\frac12$ and $-\frac12$ are shorthand for $+\frac12 \hbar$ and $-\frac12\hbar$, the only values possible to measure for $\hat{S}_z$.
Because of quantum uncertainty requirements, we can know the values of both the z-component and the square of the magnitude of spin at the same time, but we cannot know either the x- or y-component while we know the z-component. As a result, we can only say that the spin vector lies somewhere on a the surface of a cone centered on the  z-axis that expands upwards from the origin ($+\frac12$) or downwards from the origin ($-\frac12$).
For two uncorrelated spins (like in a triplet state), the z-components either add (2 positive) or (2 negatives) or cancel (1 positive and 1 negative), but that only means that the net z-component is 1, 0, or -1. Even if $\hat{S}_z=0$, there is still a net spin from the x- and y-components. The value of the magnitude (or technically its square) comes from $\hat{S}^2$, which is conserved as 1 for all three possibilities. That's the S value given in your notes. For the $\hat{S}_z=0$ case, that means the spin vector lies in the x-y plane as one of the possible radial vectors of a circle with radius 1 centered on the origin in that plane.
For the correlated spins (singlet state), the magnitude squared is 0, which occurs because the two spin vectors exactly oppose each other, and all three components cancel out, not just $\hat{S}_z$, though we have no measure of the x- and y- components that are cancelling. We only know the resulting magnitude is 0.

This is a popular piece of confusion, because of a terseness in notation. Basically, we are secretly talking about two quantum numbers here: The total magnitude of the electron spin, $S$, and the relative orientation of the electron spins, $M_S$.
As you correctly identify, for the $\alpha\beta + \beta\alpha$ configuration you get some quantum number that is $0$. This quantum number is, in fact, $M_S$; commonly interpreted as the orientation of the overall electron spin along the z-axis. Note, however, that $M_S$ for the negative linear combination is also $0$. So something else must be going on here - why do we call one of those a triplet and the other one a singlet state?
At this point, it makes sense to recall another case where you have already learned the difference between the total magnitude of an angular momentum, and its orientation in space; and that is atomic orbitals. For an s orbital, the angular momentum quantum number is $l=0$, and the magnetic quantum number $m_l$ is also $0$. For a p orbital, $l=1$, and $m_l$ can take values of $-1,0,+1$. In other terms, for a state with a given angular momentum quantum number $l$, we have $2l+1$ different ways to orient that momentum in space, which is described by the magnetic quantum number $m_l$.
Now, let's go back to our electron spins. First, it is important to realize that we are not talking about single electron quantum numbers any more, but instead the overall multi-electron system as a whole. For this, we add the magnetic quantum numbers $m_s$ of all electrons to arrive at the total quantum number $M_S$ (note the uppercase letters). Since each electron can have an individual $m_s$ of $+\frac{1}{2}$ or $-\frac{1}{2}$ (i.e., $\alpha$ or $\beta$ spin), we can have different "configurations" with different values of $M_S$.
In your two-electron system, the simplest case are the configurations with $M_S=+1$ ($\alpha\alpha$) and $-1$ ($\beta\beta$), and with our knowledge about the orientation of angular momenta above in hand, we see that these must belong to an overall $S=1$ state. But to such a state must also belong a $M_S=0$ configuration. Luckily, we have two of those at hand with out linear combinations, and since the $\alpha\beta+\beta\alpha$ one obeys the same exchange symmetry as the $\alpha\alpha$ and $\beta\beta$ configurations, we group the three of them together and thus have completed the $S=1$ triplet state.
We are left with the $\alpha\beta - \beta\alpha$ configuration. Its $M_S$ value is $0$, and there are no other configurations available to go along with it, so we have identified an $S=0$ singlet state.

looks like we were typing at the same time. . .

Crazy, isn't it! :)


text:
What am I assuming when I make this approximation?
Suppose you dissolve $\ce{AgBr}$ in a $1$ litre solution of M molar $\ce{Na2S2O3}$. What is the solubility of $\ce{AgBr}$?
I am given that the $\beta_2$ of $\ce{Ag(S2O3)2^3-}$ is $2.8*10^{13}$, and the $\ce{Ksp}$ of $\ce{AgBr}$ is $5.4*10^{-13}$.
So, I have the equations:
$\ce{\frac{[Ag(S2O3)2^3-]}{[S2O3^2-]^2*[Ag+]}} = 2.8*10^{13}$
$\ce{[Ag+][Br-]} = 5.4*10^{-13}$
$\ce{[Br-] = [Ag+] + [Ag(S2O3)2^3-]}$ (mass balance for $\ce{AgBr}$)
$\ce{M = [S2O3^2-] + 2[Ag(S2O3)2^3-]}$ (mass balance for $\ce{S2O3^2-}$)
How would I solve this system of equations for any $M$? What approximations could I make?

Some information is missing, for example about the concentration of bromide ion.

@Maurice Concentration of bromide is the reformulated primary question.If OP knew it, the question would not be asked.

The approximations you can do are the following :
1 - Practically all silver atoms are included in the complex ion. So that the concentration of that ion is equal to [Br-]
2 - The amount of AgBr dissolved is negligible with respect to the amount of Na2S2O3 available. So that, at the end, $\ce{[S2O3^{2-}]}$ is equal to the concentration M of the thiosulfate.
With these approximations, you may rewrite $\beta{_2}$, so that [Ag+] is :
$$\ce{[Ag+]= {[Ag{_2}(S_2O_3^{3-}]}/{(M·{\beta_2})}} = [Br^-]/(M·{\beta_2})$$
With the solubility product of AgBr , you have a system of two equations with two unknowns. The first solution [Ag+] will be very small. But the second solution, namely [Br-] gives you the solubility of AgBr you are looking for.

Thank you! This was very informative and helpful :))


text:
What is the mechanism of this reaction?

I've been stuck with this for some time. I just don't see it. As the reactant is Pd(0), I think oxidative addition will happen to the bromine. Then I would say an insertion reaction happens. However this would form a 7-membered ring which isn't as stable as a 5-or 6-membered ring and isn't even in the final product.
Then I tried complexating it with the double bond next to the ether. However I don't see what use the Pd could have from there. Maybe deprotonate the alpha-carbon which would give a hapto-3 complexation? But what next, I wouldn't know. Any help would be appreciated!

Nicco: Of course you can't "see it". If you were given the correct structure, the solution to your post would be straightforward. There should be a double bond in the lactam ring as shown in compound 1. The reaction is conducted in acetonitrile (bp. 82oC) at reflux. These conditions allow for a Claisen rearrangement to occur to form  ketolactam 2. Subsequent oxidative addition of palladium(0) [ligands ignored] to the C-Br bond creates the Pd(II) species 3 which adds to the double bond to afford cyclized species 4. Reductive elimination of Pd(0) regenerates the catalyst as alkene 5 is formed. The role of $\ce{Et3N}$ is to neutralize HBr that is formed. Addendum: Just found the original paper: https://doi.org/10.1016/0040-4039(94)88380-7


text:
Confusion with E/Z configuration
Could someone please illustrate how to determine the E/Z configuration of this molecule?

What I understand so far is that if the two highest priority substituent on each side are on the same side, then it's Z and vice versa.
Now, how can I handle such a structure?
I've considered it and I thought it might be $\ce{CH-CH3}$ on one side (up), and $\ce{CH2CH2}$ on the other. However, I don't think it's right.
Bottom line is that I don't understand how to fractionate the ring into two substituents.

This molecule has not isomers about the double bond. Simple as this.

The CH3 and C2H5 groups are both smaller than the rest of the cycle. So these two groups can be forgotten and this molecule is Z.

While the configuration is "Z", the IUPAC 2013 Blue Book recommends no designation.: "P-91.2.2 Omission of stereodescriptors
The  omission  of stereodescriptors specifying  double  bonds  is  recommended  in the case  of  three-  through  seven-membered  unsaturated  alicyclic  compounds  where  any double bond has a fixed configuration, 'Z'  in the case of hydrocarbons...."

Thank you. So, just to understand this properly, we consider the cycle as a shared branch for the two pi-bond carbons? (each of them is considered to bear the same ring).
In other terms, I want to know if I could break down this structure to something like this   (a,b)>C=C<(e,d)
if (a) is the methyl branch (CH3) and e is the C2H5, would it be correct to say b and d are the same?

I have added to your post your structure numbered. At C1, C6 ranks higher than the CH2 of the ethyl group because it is attached to C5 (CHH) while the ethyl group terminates with CH3 (HHH). AT C2, C3 (CCH) ranks higher than the methyl group (HHH). Therefore, the ring is unofficially, as I commented above, of the Z-configuration. Read up on Cahn-Ingold-Prelog (CIP) rules.

Thank you for both the clarification, I do have a better understanding now, and the recommendation on the CIP rules; that's exactly what I need.


text:
How to find the final pressure excerted by a mixture of gasses which are then released from a vessel after succesive fillings?
The problem is as follows:

$5\ \mathrm l$ of oxygen at $2\ \mathrm{atm}$ of pressure and $10\ \mathrm l$ of nitrogen at $4\ \mathrm{atm}$ are collected in a vessel of $2\ \mathrm l$ in capacity. Then $25\ \mathrm l$ of the mixture is released from the vessel at the pressure of $760\ \mathrm{mmHg}$. Find the final pressure of the mixture if the temperature doesn't change.

What I've attempted to do to solve this problem was to use this equation from gasses. However, it seems tricky as there's no specific information of the temperature only it mentions that it doesn't change.
$$pV=nRT$$
The way how I thought to approach this problem was that if I could obtain the moles of each gas and add them both I could find the final pressure in the container.
Then as the temperature is held constant I could use this equation using Boyle's law to get the new pressure.
$$\frac{p_{1}V_{1}}{T_{1}}=\frac{p_{2}V_{2}}{T_{2}}$$
(Assuming $T_1=T_2$). However, I'm confused at exactly why was I given the pressure of $760\ \mathrm{mmHg}$. Is it to stress the fact that gas can escape from the vessel if the pressure outside is lower?
Well here's what I did:
For oxygen:
$$(2\ \mathrm{atm})(5\ \mathrm l)=\left(n_{\ce{O2}}\right)(0.082\ \mathrm{atm\ L \ mol^{-1}\ K^{-1}})(T_1)$$
$$n_{\ce{O2}}= \frac{121.9512}{T_1}$$
For nitrogen:
$$(4\ \mathrm{atm})(10\ \mathrm l)=\left(n_{\ce{N2}}\right)(0.082\ \mathrm{atm\ L\  mol^{-1}\ K^{-1}})(T_1)$$
$$n_{\ce{N2}}= \frac{487.8048}{T_1}$$
Then adding those moles together results into:
$$n_\text{total}= \frac{121.9512}{T_1} + \frac{487.8048}{T_1}$$
$$n_\text{total}= \frac{609.7560}{T_1}$$
Using this amount of moles, I am given the volume of the vessel hence I can know the pressure.
$$(p)(2\ \mathrm l)=\frac{609.7560}{T_1} (0.082\ \mathrm{atm\ L\ mol^{-1}\ K^{-1}})(T_1)$$
$$p= 25\ \mathrm{atm}$$
Finally this pressure can be used to get the new one using the equation shown above:
$$\frac{25\ \mathrm{atm} \cdot 2\ \mathrm l}{T_{1}}=\frac{p_2\cdot 25\ \mathrm l}{T_1}$$
Therefore I'm obtaining $2\ \mathrm{atm}$.
But this option doesn't appear in any of the alternatives for this given problem. What could I be overlooking, can someone help me?

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


text:
Indigo and the Environment: is it that bad?
TL;DR: In what aspects are the products of the indigo vat dyeing process (sodium sulphate, sulphite and thiosulphate ions) harmful to the environment or to living beings?

I was recently looking into the wastewater created in the process of vat dying using indigo. I found out that since indigo is quite insoluble ($\pu{990\mu g}$ at $\pu{25^{\circ}C}$) in water for reasons of intra-molecular hydrogen bonding, a reducing agent like sodium dithionite ($\ce{Na2S2O4}$) is required, which gives leucoindigo by reducing indigo in an alkaline medium.

A related passage from a distantly related paper:

For this process, sodium dithionite ($\ce{Na2S2O4}$) has been considered as a best reducing agent in traditional and industrial dying but the reducing agent releases sodium sulphate, sulphite and thiosulphate ions into the media. Those ions not only cause environmental concerns as major pollutants of the modern textile industry waste waters but they present also no negligible risks of toxicity against living organisms.

A few papers suggest that the "wastewater" generated by the process is not suitable for disposal as such. But I still find no clarity, despite prolonged efforts at scouring journals, in why these exact ions cause the suggested adverse effects to the environment. I would be grateful if someone could please share any relevant knowledge.
These ions appear pretty harmless to me..

References:

The role of natural indigo dye in alleviation of genotoxicity of sodium dithionite as a reducing agent, Bektas I, Karaman S, Diraz E, Celik M, Cytotechnology, 2016 Dec; 68(6): 2245–2255, DOI: 10.1007/s10616-016-0018-7

BONUS: Are there any other harmful substances or environmental hazards in this nice indigo broth?

For this process, sodium dithionite (Na2S2O4) has been considered as a best reducing agent in traditional and industrial dying but the reducing agent releases sodium sulphate, sulphite and thiosulphate ions into the media. Those ions not only cause environmental concerns as major pollutants of the modern textile industry waste waters but they present also no negligible risks of toxicity against living organisms.

It is a very convoluted sentence with so many negations. My simple definition of environmentally dangerous stuff is that anything which is not found in nature is bad for the environment in excessive amounts. Sodium thiosulfate and sulfite are strong reducing agents, in excess amounts, they will slowly deplete the oxygen levels in water. Read the concept of "Chemical Oxygen Demand" in natural waters. These ions will heavily affect the COD, which is a measure of water quality. Sodium sulfate should be fine. You have to check the half-life of these ions (thiosulfate and sulfite) in natural waters.
Untreated textile waste is worse in the sense because it is colored. If large amounts of untreated textile waste, such as those of indigo, are discarded into natural streams they might block the light reaching the depths.